PROBLEM 2.1

Bianca Silva

https://www.ijert.org/sustainable-solid-waste-management-case-study-of-nagpur-india https://www.ijert.org/research/sustainable-solid-waste-management-case-study-of-nagpur-india-IJERTV9IS110278.pdf Nagpur is the third largest city and the winter capital of the Indian state of Maharashtra..It is the 13th largest city in India by population and according to an Oxford Economics report, Nagpur is projected to be the fifth fastest growing city in the world from 2019 to 2035 with an average growth of 8.41%.It has been proposed as one of the Smart Cities in Maharashtra and is one of the top ten cities in India in Smart City Project execution. Rapid economic growth, increasing population and change in living standards of city causes a high generation rate of different forms of wastes. Waste management and treatment is a major problem faced by municipal bodies all over the world especially in India. There are no specific centres for the disposal of e-waste and medical waste. Waste management techniques are not up to the mark as far as the increasing population of the city is concerned. This case study has discussed the sustainable solid waste management in Nagpur City,Maharashtra,India.

PROBLEM 2.1 Two forces P and Q are applied as shown at Point A of a hook support. Knowing that P = 75 N and Q = 125 N, determine graphically the magnitude and direction of their resultant using (a) the parallelogram law, (b) the triangle rule. SOLUTION (a) Parallelogram law: (b) Triangle rule: We measure: R = 179 N, α = 75.1° R = 179 N 75.1° ! PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 3 PROBLEM 2.2 Two forces P and Q are applied as shown at Point A of a hook support. Knowing that P = 60 lb and Q = 25 lb, determine graphically the magnitude and direction of their resultant using (a) the parallelogram law, (b) the triangle rule. SOLUTION (a) Parallelogram law: (b) Triangle rule: We measure: R = 77.1 lb, α = 85.4° R = 77.1 lb 85.4° ! PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 4 PROBLEM 2.3 The cable stays AB and AD help support pole AC. Knowing that the tension is 120 lb in AB and 40 lb in AD, determine graphically the magnitude and direction of the resultant of the forces exerted by the stays at A using (a) the parallelogram law, (b) the triangle rule. SOLUTION We measure: (a) Parallelogram law: (b) Triangle rule: We measure: α = 51.3° β = 59.0° γ = 67.0° R = 139.1 lb, R = 139.1 lb 67.0° ! PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 5 PROBLEM 2.4 Two forces are applied at Point B of beam AB. Determine graphically the magnitude and direction of their resultant using (a) the parallelogram law, (b) the triangle rule. SOLUTION (a) Parallelogram law: (b) Triangle rule: We measure: R = 3.30 kN, α = 66.6° R = 3.30 kN 66.6° ! PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 6 PROBLEM 2.5 The 300-lb force is to be resolved into components along lines a-a′ and b-b′. (a) Determine the angle α by trigonometry knowing that the component along line a-a′ is to be 240 lb. (b) What is the corresponding value of the component along b-b′? SOLUTION (a) Using the triangle rule and law of sines: sin β sin 60° = 240 lb 300 lb sin β = 0.69282 β = 43.854° α + β + 60° = 180° α = 180° − 60° − 43.854° = 76.146° (b) Law of sines: Fbb′ 300 lb = sin 76.146° sin 60° α = 76.1° Fbb′ = 336 lb PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 7 PROBLEM 2.6 The 300-lb force is to be resolved into components along lines a-a′ and b-b′. (a) Determine the angle α by trigonometry knowing that the component along line b-b′ is to be 120 lb. (b) What is the corresponding value of the component along a-a′? SOLUTION Using the triangle rule and law of sines: (a) sin α sin 60° = 120 lb 300 lb sin α = 0.34641 α = 20.268° (b) α = 20.3° α + β + 60° = 180° β = 180° − 60° − 20.268° = 99.732° Faa′ 300 lb = sin 99.732° sin 60° Faa′ = 341 lb PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 8 PROBLEM 2.7 Two forces are applied as shown to a hook support. Knowing that the magnitude of P is 35 N, determine by trigonometry (a) the required angle α if the resultant R of the two forces applied to the support is to be horizontal, (b) the corresponding magnitude of R. SOLUTION Using the triangle rule and law of sines: (a) sin α sin 25° = 50 N 35 N sin α = 0.60374 α = 37.138° (b) α = 37.1° α + β + 25° = 180° β = 180° − 25° − 37.138° = 117.86° R 35 N = sin117.86 sin 25° R = 73.2 N PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 9 PROBLEM 2.8 For the hook support of Problem 2.1, knowing that the magnitude of P is 75 N, determine by trigonometry (a) the required magnitude of the force Q if the resultant R of the two forces applied at A is to be vertical, (b) the corresponding magnitude of R. PROBLEM 2.1 Two forces P and Q are applied as shown at Point A of a hook support. Knowing that P = 75 N and Q = 125 N, determine graphically the magnitude and direction of their resultant using (a) the parallelogram law, (b) the triangle rule. SOLUTION Using the triangle rule and law of sines: (a) (b) Q 75 N = sin 20° sin 35° Q = 44.7 N α + 20° + 35° = 180° α = 180° − 20° − 35° = 125° R 75 N = sin125° sin 35° R = 107.1 N PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 10 ! PROBLEM 2.9 A trolley that moves along a horizontal beam is acted upon by two forces as shown. (a) Knowing that α = 25°, determine by trigonometry the magnitude of the force P so that the resultant force exerted on the trolley is vertical. (b) What is the corresponding magnitude of the resultant? SOLUTION Using the triangle rule and the law of sines: (a) (b) 1600 N P = sin 25° sin 75° P = 3660 N 25° + β + 75° = 180° β = 180° − 25° − 75° = 80° 1600 N R = sin 25° sin 80° R = 3730 N ! PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 11 PROBLEM 2.10 A trolley that moves along a horizontal beam is acted upon by two forces as shown. Determine by trigonometry the magnitude and direction of the force P so that the resultant is a vertical force of 2500 N. SOLUTION Using the law of cosines: Using the law of sines: P 2 = (1600 N)2 + (2500 N)2 − 2(1600 N)(2500 N) cos 75° P = 2596 N sin α sin 75° = 1600 N 2596 N α = 36.5° P is directed 90° − 36.5° or 53.5° below the horizontal. P = 2600 N 53.5° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 12 PROBLEM 2.11 A steel tank is to be positioned in an excavation. Knowing that α = 20°, determine by trigonometry (a) the required magnitude of the force P if the resultant R of the two forces applied at A is to be vertical, (b) the corresponding magnitude of R. SOLUTION Using the triangle rule and the law of sines: (a) β + 50° + 60° = 180° β = 180° − 50° − 60° = 70° (b) 425 lb P = sin 70° sin 60° P = 392 lb 425 lb R = sin 70° sin 50° R = 346 lb PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 13 ! PROBLEM 2.12 A steel tank is to be positioned in an excavation. Knowing that the magnitude of P is 500 lb, determine by trigonometry (a) the required angle α if the resultant R of the two forces applied at A is to be vertical, (b) the corresponding magnitude of R. SOLUTION Using the triangle rule and the law of sines: (a) (α + 30°) + 60° + β = 180° β = 180° − (α + 30°) − 60° β = 90° − α sin (90° − α ) sin 60° 425 lb (b) = 500 lb 90° − α = 47.40° α = 42.6° R 500 lb = sin (42.6° + 30°) sin 60° R = 551 lb PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 14 ! PROBLEM 2.13 For the hook support of Problem 2.7, determine by trigonometry (a) the magnitude and direction of the smallest force P for which the resultant R of the two forces applied to the support is horizontal, (b) the corresponding magnitude of R. SOLUTION The smallest force P will be perpendicular to R. (a) P = (50 N)sin 25° (b) R = (50 N) cos 25° P = 21.1 N R = 45.3 N PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 15 PROBLEM 2.14 For the steel tank of Problem 2.11, determine by trigonometry (a) the magnitude and direction of the smallest force P for which the resultant R of the two forces applied at A is vertical, (b) the corresponding magnitude of R. PROBLEM 2.11 A steel tank is to be positioned in an excavation. Knowing that α = 20°, determine by trigonometry (a) the required magnitude of the force P if the resultant R of the two forces applied at A is to be vertical, (b) the corresponding magnitude of R. SOLUTION The smallest force P will be perpendicular to R. (a) P = (425 lb) cos 30° (b) R = (425 lb)sin 30° P = 368 lb R = 213 lb PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 16 PROBLEM 2.15 Solve Problem 2.2 by trigonometry. PROBLEM 2.2 Two forces P and Q are applied as shown at Point A of a hook support. Knowing that P = 60 lb and Q = 25 lb, determine graphically the magnitude and direction of their resultant using (a) the parallelogram law, (b) the triangle rule. SOLUTION Using the triangle rule and the law of cosines: 20° + 35° + α = 180° α = 125° R 2 = P 2 + Q 2 − 2 PQ cos α R 2 = (60 lb)2 + (25 lb) 2 − 2(60 lb)(25 lb) cos125° R 2 = 3600 + 625 + 3000(0.5736) R = 77.108 lb Using the law of sines: sin β sin125° = 25 lb 77.108 lb β = 15.402° 70° + β = 85.402° R = 77.1 lb 85.4° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 17 PROBLEM 2.16 Solve Problem 2.3 by trigonometry. PROBLEM 2.3 The cable stays AB and AD help support pole AC. Knowing that the tension is 120 lb in AB and 40 lb in AD, determine graphically the magnitude and direction of the resultant of the forces exerted by the stays at A using (a) the parallelogram law, (b) the triangle rule. SOLUTION 8 10 α = 38.66° 6 tan β = 10 β = 30.96° tan α = Using the triangle rule: Using the law of cosines: Using the law of sines: α + β + ψ = 180° 38.66° + 30.96° + ψ = 180° ψ = 110.38° R 2 = (120 lb)2 + (40 lb) 2 − 2(120 lb)(40 lb) cos110.38° R = 139.08 lb sin γ sin110.38° = 40 lb 139.08 lb γ = 15.64° φ = (90° − α ) + γ φ = (90° − 38.66°) + 15.64° φ = 66.98° R = 139.1 lb 67.0° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 18 ! PROBLEM 2.17 Solve Problem 2.4 by trigonometry. PROBLEM 2.4 Two forces are applied at Point B of beam AB. Determine graphically the magnitude and direction of their resultant using (a) the parallelogram law, (b) the triangle rule. SOLUTION Using the law of cosines: R 2 = (2 kN)2 + (3 kN)2 − 2(2 kN)(3 kN) cos80° R = 3.304 kN Using the law of sines: sin γ sin 80° = 2 kN 3.304 kN γ = 36.59° β + γ + 80° = 180° γ = 180° − 80° − 36.59° γ = 63.41° φ = 180° − β + 50° φ = 66.59° R = 3.30 kN 66.6° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 19 PROBLEM 2.18 Two structural members A and B are bolted to a bracket as shown. Knowing that both members are in compression and that the force is 15 kN in member A and 10 kN in member B, determine by trigonometry the magnitude and direction of the resultant of the forces applied to the bracket by members A and B. SOLUTION Using the force triangle and the laws of cosines and sines: We have γ = 180° − (40° + 20°) = 120° Then R 2 = (15 kN) 2 + (10 kN)2 − 2(15 kN)(10 kN) cos120° = 475 kN 2 R = 21.794 kN and Hence: 10 kN 21.794 kN = sin α sin120° 10 kN ! sin α = " # sin120° $ 21.794 kN % = 0.39737 α = 23.414 φ = α + 50° = 73.414 R = 21.8 kN 73.4° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 20 PROBLEM 2.19 Two structural members A and B are bolted to a bracket as shown. Knowing that both members are in compression and that the force is 10 kN in member A and 15 kN in member B, determine by trigonometry the magnitude and direction of the resultant of the forces applied to the bracket by members A and B. SOLUTION Using the force triangle and the laws of cosines and sines We have γ = 180° − (40° + 20°) = 120° Then R 2 = (10 kN) 2 + (15 kN)2 − 2(10 kN)(15 kN) cos120° = 475 kN 2 R = 21.794 kN and Hence: 15 kN 21.794 kN = sin α sin120° 15 kN ! sin α = " # sin120° $ 21.794 kN % = 0.59605 α = 36.588° φ = α + 50° = 86.588° R = 21.8 kN 86.6° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 21 PROBLEM 2.20 For the hook support of Problem 2.7, knowing that P = 75 N and α = 50°, determine by trigonometry the magnitude and direction of the resultant of the two forces applied to the support. PROBLEM 2.7 Two forces are applied as shown to a hook support. Knowing that the magnitude of P is 35 N, determine by trigonometry (a) the required angle α if the resultant R of the two forces applied to the support is to be horizontal, (b) the corresponding magnitude of R. SOLUTION Using the force triangle and the laws of cosines and sines: We have β = 180° − (50° + 25°) = 105° Then R 2 = (75 N) 2 + (50 N) 2 − 2(75 N)(50 N) cos 105° 2 R = 10066.1 N 2 R = 100.330 N and Hence: sin γ sin105° = 75 N 100.330 N sin γ = 0.72206 γ = 46.225° γ − 25° = 46.225° − 25° = 21.225° R = 100.3 N 21.2° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 22 PROBLEM 2.21 Determine the x and y components of each of the forces shown. SOLUTION Compute the following distances: OA = (600) 2 + (800) 2 = 1000 mm OB = (560)2 + (900) 2 = 1060 mm OC = (480) 2 + (900)2 = 1020 mm 800-N Force: 424-N Force: 408-N Force: Fx = + (800 N) 800 1000 Fx = +640 N ! Fy = +(800 N) 600 1000 Fy = +480 N ! Fx = −(424 N) 560 1060 Fx = −224 N Fy = −(424 N) 900 1060 Fy = −360 N Fx = + (408 N) 480 1020 Fx = +192.0 N Fy = −(408 N) 900 1020 Fy = −360 N PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 23 ! PROBLEM 2.22 Determine the x and y components of each of the forces shown. SOLUTION Compute the following distances: OA = (84) 2 + (80) 2 = 116 in. OB = (28)2 + (96)2 = 100 in. OC = (48)2 + (90)2 = 102 in. 29-lb Force: 50-lb Force: 51-lb Force: Fx = + (29 lb) 84 116 Fx = +21.0 lb Fy = +(29 lb) 80 116 Fy = +20.0 lb Fx = −(50 lb) 28 100 Fx = −14.00 lb Fy = +(50 lb) 96 100 Fy = + 48.0 lb Fx = + (51 lb) 48 102 Fx = +24.0 lb Fy = −(51 lb) 90 102 Fy = −45.0 lb ! PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 24 PROBLEM 2.23 Determine the x and y components of each of the forces shown. SOLUTION 40-lb Force: 50-lb Force: 60-lb Force: Fx = + (40 lb) cos 60° Fx = 20.0 lb Fy = −(40 lb)sin 60° Fy = −34.6 lb Fx = −(50 lb)sin 50° Fx = −38.3 lb Fy = −(50 lb) cos 50° Fy = −32.1 lb Fx = + (60 lb) cos 25° Fx = 54.4 lb Fy = +(60 lb)sin 25° Fy = 25.4 lb PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 25 PROBLEM 2.24 Determine the x and y components of each of the forces shown. SOLUTION 80-N Force: 120-N Force: 150-N Force: Fx = + (80 N) cos 40° Fx = 61.3 N Fy = + (80 N) sin 40° Fy = 51.4 N Fx = + (120 N) cos 70° Fx = 41.0 N Fy = +(120 N) sin 70° Fy = 112.8 N Fx = −(150 N) cos 35° Fx = −122. 9 N Fy = +(150 N) sin 35° Fy = 86.0 N PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 26 PROBLEM 2.25 Member BD exerts on member ABC a force P directed along line BD. Knowing that P must have a 300-lb horizontal component, determine (a) the magnitude of the force P, (b) its vertical component. SOLUTION P sin 35° = 300 lb (a) P= (b) Vertical component 300 lb sin 35° P = 523 lb Pv = P cos 35° = (523 lb) cos 35° Pv = 428 lb PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 27 ! PROBLEM 2.26 The hydraulic cylinder BD exerts on member ABC a force P directed along line BD. Knowing that P must have a 750-N component perpendicular to member ABC, determine (a) the magnitude of the force P, (b) its component parallel to ABC. SOLUTION (a) 750 N = P sin 20° P = 2193 N (b) P = 2190 N PABC = P cos 20° = (2193 N) cos 20° PABC = 2060 N PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 28 PROBLEM 2.27 The guy wire BD exerts on the telephone pole AC a force P directed along BD. Knowing that P must have a 120-N component perpendicular to the pole AC, determine (a) the magnitude of the force P, (b) its component along line AC. SOLUTION (a) Px sin 38° 120 N = sin 38° P= = 194.91 N (b) or P = 194.9 N or Py = 153.6 N Px tan 38° 120 N = tan 38° Py = = 153.59 N PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 29 PROBLEM 2.28 The guy wire BD exerts on the telephone pole AC a force P directed along BD. Knowing that P has a 180-N component along line AC, determine (a) the magnitude of the force P, (b) its component in a direction perpendicular to AC. SOLUTION (a) P= Py cos 38° 180 N = cos 38° = 228.4 N (b) P = 228 N Px = Py tan 38° = (180 N) tan 38° = 140.63 N Px = 140.6 N PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 30 ! PROBLEM 2.29 Member CB of the vise shown exerts on block B a force P directed along line CB. Knowing that P must have a 1200-N horizontal component, determine (a) the magnitude of the force P, (b) its vertical component. SOLUTION We note: CB exerts force P on B along CB, and the horizontal component of P is Px = 1200 N: Then (a) Px = P sin 55° Px sin 55° 1200 N = sin 55° = 1464.9 N P= (b) P = 1465 N Px = Py tan 55° Px tan 55° 1200 N = tan 55° = 840.2 N Py = Py = 840 N PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 31 PROBLEM 2.30 Cable AC exerts on beam AB a force P directed along line AC. Knowing that P must have a 350-lb vertical component, determine (a) the magnitude of the force P, (b) its horizontal component. SOLUTION (a) P= Py cos 55° 350 lb cos 55° = 610.2 lb = (b) P = 610 lb Px = P sin 55° = (610.2 lb) sin 55° = 499.8 lb Px = 500 lb PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 32 PROBLEM 2.31 Determine the resultant of the three forces of Problem 2.22. PROBLEM 2.22 Determine the x and y components of each of the forces shown. SOLUTION Components of the forces were determined in Problem 2.22: Force x Comp. (lb) y Comp. (lb) 29 lb +21.0 +20.0 50 lb –14.00 +48.0 51 lb +24.0 –45.0 Rx = +31.0 Ry = + 23.0 R = Rx i + R y j = (31.0 lb) i + (23.0 lb) j Ry tan α = Rx 23.0 31.0 α = 36.573° 23.0 lb R= sin (36.573°) = = 38.601 lb R = 38.6 lb 36.6° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 33 PROBLEM 2.32 Determine the resultant of the three forces of Problem 2.24. PROBLEM 2.24 Determine the x and y components of each of the forces shown. SOLUTION Components of the forces were determined in Problem 2.24: Force x Comp. (N) y Comp. (N) 80 N +61.3 +51.4 120 N +41.0 +112.8 150 N –122.9 +86.0 Rx = −20.6 Ry = + 250.2 R = Rx i + Ry j = ( −20.6 N)i + (250.2 N) j Ry tan α = Rx 250.2 N 20.6 N tan α = 12.1456 α = 85.293° tan α = R= 250.2 N sin 85.293° R = 251 N 85.3° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 34 PROBLEM 2.33 Determine the resultant of the three forces of Problem 2.23. PROBLEM 2.23 Determine the x and y components of each of the forces shown. SOLUTION Force x Comp. (lb) y Comp. (lb) 40 lb +20.00 –34.64 50 lb –38.30 –32.14 60 lb +54.38 +25.36 Rx = +36.08 Ry = −41.42 R = Rx i + Ry j = ( +36.08 lb)i + (−41.42 lb) j Ry tan α = Rx 41.42 lb 36.08 lb tan α = 1.14800 α = 48.942° tan α = R= 41.42 lb sin 48.942° R = 54.9 lb 48.9° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 35 PROBLEM 2.34 Determine the resultant of the three forces of Problem 2.21. PROBLEM 2.21 Determine the x and y components of each of the forces shown. SOLUTION Components of the forces were determined in Problem 2.21: Force x Comp. (N) y Comp. (N) 800 lb +640 +480 424 lb –224 –360 408 lb +192 –360 Rx = +608 Ry = −240 R = Rx i + Ry j = (608 lb)i + (−240 lb) j tan α = Ry Rx 240 608 α = 21.541° = 240 N sin(21.541°) = 653.65 N R= R = 654 N 21.5° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 36 PROBLEM 2.35 Knowing that α = 35°, determine the resultant of the three forces shown. SOLUTION Fx = +(100 N) cos 35° = +81.915 N 100-N Force: Fy = −(100 N)sin 35° = −57.358 N Fx = +(150 N) cos 65° = +63.393 N 150-N Force: Fy = −(150 N) sin 65° = −135.946 N Fx = −(200 N) cos 35° = −163.830 N 200-N Force: Fy = −(200 N)sin 35° = −114.715 N Force x Comp. (N) y Comp. (N) 100 N +81.915 −57.358 150 N +63.393 −135.946 200 N −163.830 −114.715 Rx = −18.522 Ry = −308.02 R = Rx i + Ry j = (−18.522 N)i + (−308.02 N) j tan α = Ry Rx 308.02 18.522 α = 86.559° = R= 308.02 N sin 86.559 R = 309 N 86.6° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 37 ! PROBLEM 2.36 Knowing that the tension in cable BC is 725 N, determine the resultant of the three forces exerted at Point B of beam AB. SOLUTION Cable BC Force: 500-N Force: 780-N Force: and 840 = −525 N 1160 840 Fy = (725 N) = 500 N 1160 Fx = −(725 N) 3 Fx = −(500 N) = −300 N 5 4 Fy = −(500 N) = −400 N 5 12 = 720 N 13 5 Fy = −(780 N) = −300 N 13 Fx = (780 N) Rx = ΣFx = −105 N R y = ΣFy = −200 N R = (−105 N)2 + (−200 N) 2 = 225.89 N Further: tan α = 200 105 α = tan −1 200 105 = 62.3° R = 226 N Thus: 62.3° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 38 PROBLEM 2.37 Knowing that α = 40°, determine the resultant of the three forces shown. SOLUTION 60-lb Force: Fx = (60 lb) cos 20° = 56.38 lb Fy = (60 lb)sin 20° = 20.52 lb 80-lb Force: Fx = (80 lb) cos 60° = 40.00 lb Fy = (80 lb)sin 60° = 69.28 lb 120-lb Force: Fx = (120 lb) cos 30° = 103.92 lb Fy = −(120 lb)sin 30° = −60.00 lb and Rx = ΣFx = 200.30 lb R y = ΣFy = 29.80 lb R = (200.30 lb) 2 + (29.80 lb) 2 = 202.50 lb Further: tan α = 29.80 200.30 α = tan −1 29.80 200.30 R = 203 lb = 8.46° 8.46° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 39 ! PROBLEM 2.38 Knowing that α = 75°, determine the resultant of the three forces shown. SOLUTION 60-lb Force: Fx = (60 lb) cos 20° = 56.38 lb Fy = (60 lb) sin 20° = 20.52 lb 80-lb Force: Fx = (80 lb) cos 95 ° = −6.97 lb Fy = (80 lb)sin 95° = 79.70 lb 120-lb Force: Fx = (120 lb) cos 5 ° = 119.54 lb Fy = (120 lb) sin 5° = 10.46 lb Then Rx = ΣFx = 168.95 lb R y = ΣFy = 110.68 lb and R = (168.95 lb) 2 + (110.68 lb) 2 = 201.98 lb 110.68 168.95 tan α = 0.655 α = 33.23° tan α = R = 202 lb 33.2° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 40 PROBLEM 2.39 For the collar of Problem 2.35, determine (a) the required value of α if the resultant of the three forces shown is to be vertical, (b) the corresponding magnitude of the resultant. SOLUTION Rx = ΣFx = (100 N) cos α + (150 N) cos (α + 30°) − (200 N) cos α Rx = −(100 N) cos α + (150 N) cos (α + 30°) (1) Ry = ΣFy = −(100 N) sin α − (150 N)sin (α + 30°) − (200 N)sin α Ry = −(300 N) sin α − (150 N)sin (α + 30°) (a) (2) For R to be vertical, we must have Rx = 0. We make Rx = 0 in Eq. (1): −100 cos α + 150cos (α + 30°) = 0 −100cos α + 150 (cos α cos 30° − sin α sin 30°) = 0 29.904cos α = 75sin α 29.904 75 = 0.3988 α = 21.74° tan α = (b) α = 21.7° Substituting for α in Eq. (2): Ry = −300sin 21.74° − 150sin 51.74° = −228.9 N R = | Ry | = 228.9 N R = 229 N PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 41 PROBLEM 2.40 For the beam of Problem 2.36, determine (a) the required tension in cable BC if the resultant of the three forces exerted at Point B is to be vertical, (b) the corresponding magnitude of the resultant. SOLUTION Rx = ΣFx = − Rx = − 21 TBC + 420 N 29 Ry = ΣFy = Ry = (a) (1) 800 5 4 TBC − (780 N) − (500 N) 1160 13 5 20 TBC − 700 N 29 (2) For R to be vertical, we must have Rx = 0 Set Rx = 0 in Eq. (1) (b) 840 12 3 TBC + (780 N) − (500 N) 1160 13 5 − 21 TBC + 420 N = 0 29 TBC = 580 N Substituting for TBC in Eq. (2): 20 (580 N) − 700 N 29 Ry = −300 N Ry = R = | Ry | = 300 N R = 300 N PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 42 PROBLEM 2.41 Determine (a) the required tension in cable AC, knowing that the resultant of the three forces exerted at Point C of boom BC must be directed along BC, (b) the corresponding magnitude of the resultant. SOLUTION Using the x and y axes shown: Rx = ΣFx = TAC sin10° + (50 lb) cos 35° + (75 lb) cos 60° = TAC sin10° + 78.46 lb (1) Ry = ΣFy = (50 lb)sin 35° + (75 lb)sin 60° − TAC cos10° Ry = 93.63 lb − TAC cos10° (a) (2) Set Ry = 0 in Eq. (2): 93.63 lb − TAC cos10° = 0 TAC = 95.07 lb (b) TAC = 95.1 lb Substituting for TAC in Eq. (1): Rx = (95.07 lb) sin10° + 78.46 lb = 94.97 lb R = Rx R = 95.0 lb PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 43 ! PROBLEM 2.42 For the block of Problems 2.37 and 2.38, determine (a) the required value of α if the resultant of the three forces shown is to be parallel to the incline, (b) the corresponding magnitude of the resultant. SOLUTION Select the x axis to be along a a′. Then Rx = ΣFx = (60 lb) + (80 lb) cos α + (120 lb)sin α (1) Ry = ΣFy = (80 lb)sin α − (120 lb) cos α (2) and (a) Set Ry = 0 in Eq. (2). (80 lb) sin α − (120 lb) cos α = 0 Dividing each term by cos α gives: (80 lb) tan α = 120 lb 120 lb 80 lb α = 56.310° tanα = (b) α = 56.3° Substituting for α in Eq. (1) gives: Rx = 60 lb + (80 lb) cos 56.31° + (120 lb)sin 56.31° = 204.22 lb Rx = 204 lb PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 44 PROBLEM 2.43 Two cables are tied together at C and are loaded as shown. Knowing that α = 20°, determine the tension (a) in cable AC, (b) in cable BC. SOLUTION Free-Body Diagram Law of sines: Force Triangle TAC T 1962 N = BC = sin 70° sin 50° sin 60° (a) TAC = 1962 N sin 70° = 2128.9 N sin 60° TAC = 2.13 kN ! (b) TBC = 1962 N sin 50° = 1735.49 N sin 60° TBC = 1.735 kN ! PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 45 PROBLEM 2.44 Two cables are tied together at C and are loaded as shown. Determine the tension (a) in cable AC, (b) in cable BC. SOLUTION Free-Body Diagram Law of sines: Force Triangle TAC T 500 N = BC = sin 60° sin 40° sin 80° (a) TAC = 500 N sin 60° = 439.69 N sin 80° TAC = 440 N (b) TBC = 500 N sin 40° = 326.35 N sin 80° TBC = 326 N PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 46 PROBLEM 2.45 Two cables are tied together at C and are loaded as shown. Knowing that P = 500 N and α = 60°, determine the tension in (a) in cable AC, (b) in cable BC. SOLUTION Free-Body Diagram Law of sines: Force Triangle TAC T 500 N = BC = sin 35° sin 75° sin 70° (a) TAC = 500 N sin 35° sin 70° TAC = 305 N (b) TBC = 500 N sin 75° sin 70° TBC = 514 N PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 47 PROBLEM 2.46 Two cables are tied together at C and are loaded as shown. Determine the tension (a) in cable AC, (b) in cable BC. SOLUTION Free-Body Diagram Force Triangle W = mg = (200 kg)(9.81 m/s 2 ) = 1962 N Law of sines: TAC TBC 1962 N = = sin 15° sin 105° sin 60° (a) TAC = (1962 N) sin 15° sin 60° TAC = 586 N (b) TBC = (1962 N) sin 105° sin 60° TBC = 2190 N ! PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 48 PROBLEM 2.47 Knowing that α = 20°, determine the tension (a) in cable AC, (b) in rope BC. SOLUTION Free-Body Diagram Law of sines: Force Triangle TAC T 1200 lb = BC = sin 110° sin 5° sin 65° (a) TAC = 1200 lb sin 110° sin 65° TAC = 1244 lb (b) TBC = 1200 lb sin 5° sin 65° TBC = 115.4 lb PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 49 PROBLEM 2.48 Knowing that α = 55° and that boom AC exerts on pin C a force directed along line AC, determine (a) the magnitude of that force, (b) the tension in cable BC. SOLUTION Free-Body Diagram Law of sines: Force Triangle FAC T 300 lb = BC = sin 35° sin 50° sin 95° (a) FAC = 300 lb sin 35° sin 95° FAC = 172.7 lb (b) TBC = 300 lb sin 50° sin 95° TBC = 231 lb PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 50 PROBLEM 2.49 Two forces P and Q are applied as shown to an aircraft connection. Knowing that the connection is in equilibrium and that P = 500 lb and Q = 650 lb, determine the magnitudes of the forces exerted on the rods A and B. SOLUTION Free-Body Diagram Resolving the forces into x- and y-directions: R = P + Q + FA + FB = 0 Substituting components: R = −(500 lb) j + [(650 lb) cos 50°]i − [(650 lb) sin 50°] j + FB i − ( FA cos 50°)i + ( FA sin 50°) j = 0 In the y-direction (one unknown force) −500 lb − (650 lb)sin 50° + FA sin 50° = 0 Thus, FA = 500 lb + (650 lb) sin 50° sin 50° = 1302.70 lb In the x-direction: Thus, FA = 1303 lb (650 lb) cos 50° + FB − FA cos 50° = 0 FB = FA cos 50° − (650 lb) cos50° = (1302.70 lb) cos 50° − (650 lb) cos 50° = 419.55 lb FB = 420 lb PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 51 PROBLEM 2.50 Two forces P and Q are applied as shown to an aircraft connection. Knowing that the connection is in equilibrium and that the magnitudes of the forces exerted on rods A and B are FA = 750 lb and FB = 400 lb, determine the magnitudes of P and Q. SOLUTION Free-Body Diagram Resolving the forces into x- and y-directions: R = P + Q + FA + FB = 0 Substituting components: R = − Pj + Q cos 50°i − Q sin 50° j − [(750 lb) cos 50°]i + [(750 lb)sin 50°] j + (400 lb)i In the x-direction (one unknown force) Q cos 50° − [(750 lb) cos 50°] + 400 lb = 0 Q= (750 lb) cos 50° − 400 lb cos 50° = 127.710 lb In the y-direction: − P − Q sin 50° + (750 lb) sin 50° = 0 P = −Q sin 50° + (750 lb) sin 50° = −(127.710 lb)sin 50° + (750 lb) sin 50° = 476.70 lb P = 477 lb; Q = 127.7 lb PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 52 ! PROBLEM 2.51 A welded connection is in equilibrium under the action of the four forces shown. Knowing that FA = 8 kN and FB = 16 kN, determine the magnitudes of the other two forces. SOLUTION Free-Body Diagram of Connection ΣFx = 0: With 3 3 FB − FC − FA = 0 5 5 FA = 8 kN FB = 16 kN FC = 4 4 (16 kN) − (8 kN) 5 5 Σ Fy = 0: − FD + With FA and FB as above: FC = 6.40 kN 3 3 FB − FA = 0 ! 5 5 3 3 FD = (16 kN) − (8 kN) ! 5 5 FD = 4.80 kN ! PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 53 PROBLEM 2.52 A welded connection is in equilibrium under the action of the four forces shown. Knowing that FA = 5 kN and FD = 6 kN, determine the magnitudes of the other two forces. SOLUTION Free-Body Diagram of Connection 3 3 ΣFy = 0: − FD − FA + FB = 0 5 5 3 FA 5 or FB = FD + With FA = 5 kN, FD = 8 kN FB = 5' 3 ( 6 kN + (5 kN) * 3 )+ 5 , ΣFx = 0: − FC + FB = 15.00 kN 4 4 FB − FA = 0 5 5 4 ( FB − FA ) 5 4 = (15 kN − 5 kN) 5 FC = FC = 8.00 kN PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 54 PROBLEM 2.53 Two cables tied together at C are loaded as shown. Knowing that Q = 60 lb, determine the tension (a) in cable AC, (b) in cable BC. SOLUTION ΣFy = 0: TCA − Q cos 30° = 0 With Q = 60 lb (a) TCA = (60 lb)(0.866) (b) ΣFx = 0: P − TCB − Q sin 30° = 0 With TCA = 52.0 lb P = 75 lb TCB = 75 lb − (60 lb)(0.50) or TCB = 45.0 lb PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 55 PROBLEM 2.54 Two cables tied together at C are loaded as shown. Determine the range of values of Q for which the tension will not exceed 60 lb in either cable. SOLUTION Free-Body Diagram ΣFx = 0: −TBC − Q cos 60° + 75 lb = 0 TBC = 75 lb − Q cos 60° (1) ΣFy = 0: TAC − Q sin 60° = 0 TAC = Q sin 60° (2) TAC # 60 lb: Requirement Q sin 60° # 60 lb From Eq. (2): Q # 69.3 lb TBC # 60 lb: Requirement From Eq. (1): 75 lb − Q sin 60° # 60 lb Q $ 30.0 lb 30.0 lb # Q # 69.3 lb PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 56 PROBLEM 2.55 A sailor is being rescued using a boatswain’s chair that is suspended from a pulley that can roll freely on the support cable ACB and is pulled at a constant speed by cable CD. Knowing that α = 30° and β = 10° and that the combined weight of the boatswain’s chair and the sailor is 900 N, determine the tension (a) in the support cable ACB, (b) in the traction cable CD. SOLUTION Free-Body Diagram ΣFx = 0: TACB cos 10° − TACB cos 30° − TCD cos 30° = 0 TCD = 0.137158TACB (1) ΣFy = 0: TACB sin 10° + TACB sin 30° + TCD sin 30° − 900 = 0 0.67365TACB + 0.5TCD = 900 (a) Substitute (1) into (2): 0.67365 TACB + 0.5(0.137158 TACB ) = 900 TACB = 1212.56 N (b) From (1): (2) TCD = 0.137158(1212.56 N) TACB = 1213 N TCD = 166.3 N PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 57 PROBLEM 2.56 A sailor is being rescued using a boatswain’s chair that is suspended from a pulley that can roll freely on the support cable ACB and is pulled at a constant speed by cable CD. Knowing that α = 25° and β = 15° and that the tension in cable CD is 80 N, determine (a) the combined weight of the boatswain’s chair and the sailor, (b) in tension in the support cable ACB. SOLUTION Free-Body Diagram ΣFx = 0: TACB cos 15° − TACB cos 25° − (80 N) cos 25° = 0 TACB = 1216.15 N ΣFy = 0: (1216.15 N) sin 15° + (1216.15 N) sin 25° + (80 N) sin 25° − W = 0 W = 862.54 N (a) W = 863 N (b) TACB = 1216 N PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 58 PROBLEM 2.57 For the cables of Problem 2.45, it is known that the maximum allowable tension is 600 N in cable AC and 750 N in cable BC. Determine (a) the maximum force P that can be applied at C, (b) the corresponding value of α. SOLUTION Free-Body Diagram (a) Law of cosines Force Triangle P 2 = (600) 2 + (750)2 − 2(600)(750) cos (25° + 45°) P = 784.02 N (b) Law of sines P = 784 N sin β sin (25° + 45°) = 600 N 784.02 N β = 46.0° α = 46.0° + 25° = 71.0° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 59 PROBLEM 2.58 For the situation described in Figure P2.47, determine (a) the value of α for which the tension in rope BC is as small as possible, (b) the corresponding value of the tension. PROBLEM 2.47 Knowing that α = 20°, determine the tension (a) in cable AC, (b) in rope BC. SOLUTION Free-Body Diagram Force Triangle To be smallest, TBC must be perpendicular to the direction of TAC . (a) (b) Thus, α = 5° = 5.00° TBC = (1200 lb) sin 5° TBC = 104.6 lb PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 60 PROBLEM 2.59 For the structure and loading of Problem 2.48, determine (a) the value of α for which the tension in cable BC is as small as possible, (b) the corresponding value of the tension. SOLUTION TBC must be perpendicular to FAC to be as small as possible. Free-Body Diagram: C Force Triangle is a right triangle To be a minimum, TBC must be perpendicular to FAC . (a) We observe: α = 90° − 30° α = 60.0° TBC = (300 lb)sin 50° (b) or TBC = 229.81 lb TBC = 230 lb PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 61 PROBLEM 2.60 Knowing that portions AC and BC of cable ACB must be equal, determine the shortest length of cable that can be used to support the load shown if the tension in the cable is not to exceed 870 N. SOLUTION Free-Body Diagram: C (For T = 725 N) ΣFy = 0: 2Ty − 1200 N = 0 Ty = 600 N Tx2 + Ty2 = T 2 Tx2 + (600 N)2 = (870 N) 2 Tx = 630 N By similar triangles: 2.1 m BC = 870 N 630 N BC = 2.90 m L = 2( BC ) = 5.80 m L = 5.80 m PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 62 PROBLEM 2.61 Two cables tied together at C are loaded as shown. Knowing that the maximum allowable tension in each cable is 800 N, determine (a) the magnitude of the largest force P that can be applied at C, (b) the corresponding value of α. SOLUTION Free-Body Diagram: C Force Triangle Force triangle is isosceles with 2 β = 180° − 85° β = 47.5° P = 2(800 N)cos 47.5° = 1081 N (a) P = 1081 N Since P . 0, the solution is correct. (b) α = 180° − 50° − 47.5° = 82.5° α = 82.5° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 63 ! PROBLEM 2.62 Two cables tied together at C are loaded as shown. Knowing that the maximum allowable tension is 1200 N in cable AC and 600 N in cable BC, determine (a) the magnitude of the largest force P that can be applied at C, (b) the corresponding value of α. SOLUTION Free-Body Diagram (a) Law of cosines: Force Triangle P 2 = (1200 N) 2 + (600 N) 2 − 2(1200 N)(600 N) cos 85° P = 1294 N Since P . 1200 N, the solution is correct. P = 1294 N (b) Law of sines: sin β sin 85° = 1200 N 1294 N β = 67.5° α = 180° − 50° − 67.5° α = 62.5° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 64 ! PROBLEM 2.63 Collar A is connected as shown to a 50-lb load and can slide on a frictionless horizontal rod. Determine the magnitude of the force P required to maintain the equilibrium of the collar when (a) x = 4.5 in., (b) x = 15 in. SOLUTION (a) Free Body: Collar A Force Triangle P 50 lb = 4.5 20.5 (b) Free Body: Collar A P = 10.98 lb Force Triangle P 50 lb = 15 25 P = 30.0 lb ! PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 65 PROBLEM 2.64 Collar A is connected as shown to a 50-lb load and can slide on a frictionless horizontal rod. Determine the distance x for which the collar is in equilibrium when P = 48 lb. SOLUTION Free Body: Collar A Force Triangle N 2 = (50) 2 − (48) 2 = 196 N = 14.00 lb Similar Triangles x 48 lb = 20 in. 14 lb x = 68.6 in. PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 66 ! PROBLEM 2.65 A 160-kg load is supported by the rope-and-pulley arrangement shown. Knowing that β = 20°, determine the magnitude and direction of the force P that must be exerted on the free end of the rope to maintain equilibrium. (Hint: The tension in the rope is the same on each side of a simple pulley. This can be proved by the methods of Chapter 4.) SOLUTION Free-Body Diagram: Pulley A ΣFx = 0: 2P sin 20° − P cos α = 0 and cos α = 0.8452 or α = ± 46.840° α = + 46.840 For ΣFy = 0: 2P cos 20° + P sin 46.840° − 1569.60 N = 0 or P = 602 N 46.8° α = −46.840 For ΣFy = 0: 2P cos 20° + P sin( −46.840°) − 1569.60 N = 0 or P = 1365 N 46.8° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 67 PROBLEM 2.66 A 160-kg load is supported by the rope-and-pulley arrangement shown. Knowing that α = 40°, determine (a) the angle β, (b) the magnitude of the force P that must be exerted on the free end of the rope to maintain equilibrium. (See the hint for Problem 2.65.) SOLUTION Free-Body Diagram: Pulley A (a) ΣFx = 0: 2 P sin sin β − P cos 40° = 0 1 cos 40° 2 β = 22.52° sin β = β = 22.5° (b) ΣFy = 0: P sin 40° + 2 P cos 22.52° − 1569.60 N = 0 P = 630 N PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 68 PROBLEM 2.67 A 600-lb crate is supported by several rope-andpulley arrangements as shown. Determine for each arrangement the tension in the rope. (See the hint for Problem 2.65.) SOLUTION Free-Body Diagram of Pulley (a) ΣFy = 0: 2T − (600 lb) = 0 T= 1 (600 lb) 2 T = 300 lb (b) ΣFy = 0: 2T − (600 lb) = 0 T= 1 (600 lb) 2 T = 300 lb (c) ΣFy = 0: 3T − (600 lb) = 0 1 T = (600 lb) 3 T = 200 lb (d) ΣFy = 0: 3T − (600 lb) = 0 1 T = (600 lb) 3 T = 200 lb (e) ΣFy = 0: 4T − (600 lb) = 0 T= 1 (600 lb) 4 T = 150.0 lb ! PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 69 ! PROBLEM 2.68 Solve Parts b and d of Problem 2.67, assuming that the free end of the rope is attached to the crate. PROBLEM 2.67 A 600-lb crate is supported by several rope-and-pulley arrangements as shown. Determine for each arrangement the tension in the rope. (See the hint for Problem 2.65.) SOLUTION Free-Body Diagram of Pulley and Crate (b) ΣFy = 0: 3T − (600 lb) = 0 1 T = (600 lb) 3 T = 200 lb (d) ΣFy = 0: 4T − (600 lb) = 0 T= 1 (600 lb) 4 T = 150.0 lb ! PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 70 ! PROBLEM 2.69 A load Q is applied to the pulley C, which can roll on the cable ACB. The pulley is held in the position shown by a second cable CAD, which passes over the pulley A and supports a load P. Knowing that P = 750 N, determine (a) the tension in cable ACB, (b) the magnitude of load Q. SOLUTION Free-Body Diagram: Pulley C ΣFx = 0: TACB (cos 25° − cos 55°) − (750 N) cos 55° = 0 (a) TACB = 1292.88 N Hence: TACB = 1293 N ΣFy = 0: TACB (sin 25° + sin 55°) + (750 N) sin 55° − Q = 0 (b) ! (1292.88 N)(sin 25° + sin 55°) + (750 N) sin 55° − Q = 0 Q = 2219.8 N or Q = 2220 N ! PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 71 PROBLEM 2.70 An 1800-N load Q is applied to the pulley C, which can roll on the cable ACB. The pulley is held in the position shown by a second cable CAD, which passes over the pulley A and supports a load P. Determine (a) the tension in cable ACB, (b) the magnitude of load P. SOLUTION Free-Body Diagram: Pulley C ΣFx = 0: TACB (cos 25° − cos 55°) − P cos 55° = 0 P = 0.58010TACB or (1) ΣFy = 0: TACB (sin 25° + sin 55°) + P sin 55° − 1800 N = 0 1.24177TACB + 0.81915 P = 1800 N or (a) (2) Substitute Equation (1) into Equation (2): 1.24177TACB + 0.81915(0.58010TACB ) = 1800 N TACB = 1048.37 N Hence: TACB = 1048 N (b) P = 0.58010(1048.37 N) = 608.16 N Using (1), P = 608 N PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 72 PROBLEM 2.71 Determine (a) the x, y, and z components of the 750-N force, (b) the angles θx, θy, and θz that the force forms with the coordinate axes. SOLUTION Fh = F sin 35° = (750 N)sin 35° Fh = 430.2 N (a) Fx = Fh cos 25° = (430.2 N) cos 25° Fx = +390 N, (b) Fy = F cos 35° = (750 N) cos 35° Fy = +614 N, cos θ x = cos θ y = cos θ z = Fx +390 N = 750 N F Fy F = +614 N 750 N Fz +181.8 N = 750 N F Fz = Fh sin 25° = (430.2 N) sin 25° Fz = +181.8 N θ x = 58.7° θ y = 35.0° θ z = 76.0° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 73 ! PROBLEM 2.72 Determine (a) the x, y, and z components of the 900-N force, (b) the angles θx, θy, and θz that the force forms with the coordinate axes. SOLUTION Fh = F cos 65° = (900 N) cos 65° Fh = 380.4 N (a) Fx = Fh sin 20° = (380.4 N)sin 20° Fx = −130.1 N, (b) Fy = F sin 65° = (900 N)sin 65° Fy = +816 N, cos θ x = cos θ y = cos θ z = Fx −130.1 N = 900 N F Fy Fz = Fh cos 20° = (380.4 N) cos 20° Fz = +357 N θ x = 98.3° +816 N 900 N θ y = 25.0° Fz +357 N = 900 N F θ z = 66.6° F = PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 74 ! PROBLEM 2.73 A horizontal circular plate is suspended as shown from three wires that are attached to a support at D and form 30° angles with the vertical. Knowing that the x component of the force exerted by wire AD on the plate is 110.3 N, determine (a) the tension in wire AD, (b) the angles θ x, θ y, and θ z that the force exerted at A forms with the coordinate axes. SOLUTION (a) Fx = F sin 30° sin 50° = 110.3 N (Given) F= (b) cos θ x = 110.3 N = 287.97 N sin 30° sin 50° F = 288 N Fx 110.3 N = = 0.38303 F 287.97 N θ x = 67.5° Fy = F cos 30° = 249.39 cos θ y = Fy F = 249.39 N = 0.86603 287.97 N θ y = 30.0° Fz = − F sin 30° cos 50° = −(287.97 N)sin 30°cos 50° = −92.552 N cos θ z = Fz −92.552 N = = −0.32139 F 287.97 N θ z = 108.7° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 75 PROBLEM 2.74 A horizontal circular plate is suspended as shown from three wires that are attached to a support at D and form 30° angles with the vertical. Knowing that the z component of the force exerted by wire BD on the plate is –32.14 N, determine (a) the tension in wire BD, (b) the angles θx, θy, and θz that the force exerted at B forms with the coordinate axes. SOLUTION (a) Fz = − F sin 30° sin 40° = 32.14 N (Given) F= (b) 32.14 = 100.0 N sin 30° sin 40° F = 100.0 N Fx = − F sin 30° cos 40° = −(100.0 N)sin 30°cos 40° = −38.302 N cos θ x = Fx 38.302 N = = −0.38302 100.0 N F θ x = 112.5° Fy = F cos 30° = 86.603 N cos θ y = Fy F = 86.603 N = 0.86603 100 N θ y = 30.0° Fz = −32.14 N cos θ z = Fz −32.14 N = = −0.32140 F 100 N θ z = 108.7° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 76 PROBLEM 2.75 A horizontal circular plate is suspended as shown from three wires that are attached to a support at D and form 30° angles with the vertical. Knowing that the tension in wire CD is 60 lb, determine (a) the components of the force exerted by this wire on the plate, (b) the angles θ x , θ y , and θ z that the force forms with the coordinate axes. SOLUTION (a) (b) Fx = −(60 lb) sin 30°cos 60° = −15 lb Fx = −15.00 lb Fy = (60 lb) cos 30° = 51.96 lb Fy = +52.0 lb Fz = (60 lb) sin 30° sin 60° = 25.98 lb Fz = +26.0 lb cos θ x = cos θ y = cos θ z = Fx −15.0 lb = = −0.25 60 lb F Fy θ x = 104.5° 51.96 lb = 0.866 60 lb θ y = 30.0° Fz 25.98 lb = = 0.433 F 60 lb θ z = 64.3° F = PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 77 PROBLEM 2.76 A horizontal circular plate is suspended as shown from three wires that are attached to a support at D and form 30° angles with the vertical. Knowing that the x component of the force exerted by wire CD on the plate is –20.0 lb, determine (a) the tension in wire CD, (b) the angles θx, θy, and θz that the force exerted at C forms with the coordinate axes. SOLUTION (a) Fx = − F sin 30° cos 60° = −20 lb (Given) F= (b) cos θ x = 20 lb = 80 lb sin 30°cos 60° F = 80.0 lb Fx −20 lb = = − 0.25 80 lb F θ x = 104.5° Fy = (80 lb) cos 30° = 69.282 lb cos θ y = Fy F = 69.282 lb = 0.86615 80 lb θ y = 30.0° Fz = (80 lb)sin 30° sin 60° = 34.641 lb cos θ z = Fz 34.641 = = 0.43301 80 F θ z = 64.3° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 78 PROBLEM 2.77 The end of the coaxial cable AE is attached to the pole AB, which is strengthened by the guy wires AC and AD. Knowing that the tension in wire AC is 120 lb, determine (a) the components of the force exerted by this wire on the pole, (b) the angles θx, θy, and θz that the force forms with the coordinate axes. SOLUTION (a) Fx = (120 lb) cos 60° cos 20° Fx = 56.382 lb Fx = +56.4 lb Fy = −(120 lb)sin 60° Fy = −103.923 lb Fy = −103.9 lb Fz = −(120 lb) cos 60° sin 20° Fz = −20.521 lb (b) cos θ x = cos θ y = cos θ z = Fz = −20.5 lb Fx 56.382 lb = F 120 lb Fy F = −103.923 lb 120 lb Fz −20.52 lb = F 120 lb θ x = 62.0° θ y = 150.0° θ z = 99.8° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 79 PROBLEM 2.78 The end of the coaxial cable AE is attached to the pole AB, which is strengthened by the guy wires AC and AD. Knowing that the tension in wire AD is 85 lb, determine (a) the components of the force exerted by this wire on the pole, (b) the angles θx, θy, and θz that the force forms with the coordinate axes. SOLUTION (a) Fx = (85 lb)sin 36° sin 48° = 37.129 lb Fx = 37.1 lb Fy = −(85 lb) cos 36° = −68.766 lb Fy = −68.8 lb Fz = (85 lb)sin 36° cos 48° Fz = 33.4 lb = 33.431 lb (b) cos θ x = cos θ y = cos θ z = Fx 37.129 lb = F 85 lb Fy F = −68.766 lb 85 lb Fz 33.431 lb = F 85 lb θ x = 64.1° θ y = 144.0° θ z = 66.8° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 80 PROBLEM 2.79 Determine the magnitude and direction of the force F = (320 N)i + (400 N)j − (250 N)k. SOLUTION F = Fx2 + Fy2 + Fz2 F = (320 N)2 + (400 N) 2 + (−250 N) 2 cos θ x = cos θ y = cos θ y = Fx 320 N = F 570 N Fy F = F = 570 N θ x = 55.8° 400 N 570 N θ y = 45.4° Fz −250 N = F 570 N θ z = 116.0° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 81 PROBLEM 2.80 Determine the magnitude and direction of the force F = (240 N)i − (270 N)j + (680 N)k. SOLUTION F = Fx2 + Fy2 + Fz2 F = (240 N) 2 + (−270 N) 2 + (680 N) cos θ x = cos θ y = cos θ z = Fx 240 N = F 770 N Fy F = F = 770 N θ x = 71.8° −270 N 770 N Fz 680 N = F 770 N θ y = 110.5° θ z = 28.0° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 82 PROBLEM 2.81 A force acts at the origin of a coordinate system in a direction defined by the angles θx = 70.9° and θy = 144.9°. Knowing that the z component of the force is –52.0 lb, determine (a) the angle θz, (b) the other components and the magnitude of the force. SOLUTION (a) We have (cos θ x ) 2 + (cos θ y )2 + (cos θ z ) 2 = 1 (cos θ y ) 2 = 1 − (cos θ y ) 2 − (cos θ z ) 2 Since Fz , 0 we must have cos θ z , 0 Thus, taking the negative square root, from above, we have: cos θ z = − 1 − (cos 70.9°) 2 − (cos144.9°) 2 = 0.47282 (b) θ z = 118.2° Then: F= and Fz 52.0 lb = = 109.978 lb cos θ z 0.47282 Fx = F cos θ x = (109.978 lb) cos 70.9° Fx = 36.0 lb Fy = F cos θ y = (109.978 lb) cos144.9° Fy = −90.0 lb F = 110.0 lb PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 83 ! PROBLEM 2.82 A force acts at the origin of a coordinate system in a direction defined by the angles θy = 55° and θz = 45°. Knowing that the x component of the force is − 500 lb, determine (a) the angle θx, (b) the other components and the magnitude of the force. SOLUTION (a) We have (cos θ x ) 2 + (cos θ y )2 + (cos θ z ) 2 = 1 (cos θ y ) 2 = 1 − (cos θ y ) 2 − (cos θ z ) 2 Since Fx , 0 we must have cos θ x , 0 Thus, taking the negative square root, from above, we have: cos θ x = − 1 − (cos 55) 2 − (cos 45) 2 = 0.41353 (b) θ x = 114.4° Then: Fx 500 lb = = 1209.10 lb cos θ x 0.41353 F = 1209 lb Fy = F cos θ y = (1209.10 lb) cos 55° Fy = 694 lb Fz = F cos θ z = (1209.10 lb) cos 45° Fz = 855 lb F= and PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 84 PROBLEM 2.83 A force F of magnitude 210 N acts at the origin of a coordinate system. Knowing that Fx = 80 N, θz = 151.2°, and Fy < 0, determine (a) the components Fy and Fz, (b) the angles θx and θy. SOLUTION Fz = F cos θ z = (210 N) cos151.2° (a) = −184.024 N Then: So: Hence: F 2 = Fx2 + Fy2 + Fz2 (210 N) 2 = (80 N) 2 + ( Fy ) 2 + (184.024 N)2 Fy = − (210 N) 2 − (80 N) 2 − (184.024 N) 2 = −61.929 N (b) Fz = −184.0 N cos θ x = cos θ y = Fy = −62.0 lb Fx 80 N = = 0.38095 F 210 N Fy F = 61.929 N = −0.29490 210 N θ x = 67.6° θ y = 107.2 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 85 PROBLEM 2.84 A force F of magnitude 230 N acts at the origin of a coordinate system. Knowing that θx = 32.5°, Fy = − 60 N, and Fz > 0, determine (a) the components Fx and Fz, (b) the angles θy and θz. SOLUTION (a) We have Fx = F cos θ x = (230 N) cos 32.5° Then: Fx = −194.0 N Fx = 193.980 N F 2 = Fx2 + Fy2 + Fz2 So: Hence: (b) (230 N) 2 = (193.980 N) 2 + (−60 N) 2 + Fz2 Fz = + (230 N) 2 − (193.980 N) 2 − (−60 N) 2 Fz = 108.0 N Fz = 108.036 N Fy −60 N = − 0.26087 F 230 N F 108.036 N cos θ z = z = = 0.46972 F 230 N cos θ y = = θ y = 105.1° θ z = 62.0° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 86 PROBLEM 2.85 A transmission tower is held by three guy wires anchored by bolts at B, C, and D. If the tension in wire AB is 525 lb, determine the components of the force exerted by the wire on the bolt at B. SOLUTION ! BA = (20 ft)i + (100 ft) j − (25 ft)k BA = (20 ft) 2 + (100 ft) 2 + (−25 ft)2 = 105 ft F=F BA ! BA =F BA 525 lb = [(20 ft)i + (100 ft) j − (25 ft)k ] 105 ft F = (100.0 lb)i + (500 lb) j − (125.0 lb)k Fx = +100.0 lb, Fy = +500 lb, Fz = −125.0 lb PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 87 PROBLEM 2.86 A transmission tower is held by three guy wires anchored by bolts at B, C, and D. If the tension in wire AD is 315 lb, determine the components of the force exerted by the wire on the bolt at D. SOLUTION ! DA = (20 ft)i + (100 ft) j + (70 ft)k DA = (20 ft) 2 + (100 ft) 2 + ( +70 ft) 2 = 126 ft F=F DA ! DA =F DA 315 lb = [(20 ft)i + (100 ft) j + (74 ft)k ] 126 ft F = (50 lb)i + (250 lb) j + (185 lb)k Fx = +50 lb, Fy = +250 lb, Fz = +185.0 lb PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 88 PROBLEM 2.87 A frame ABC is supported in part by cable DBE that passes through a frictionless ring at B. Knowing that the tension in the cable is 385 N, determine the components of the force exerted by the cable on the support at D. SOLUTION ! DB = (480 mm)i − (510 mm) j + (320 mm)k DB = (480 mm) 2 + (510 mm 2 ) + (320 mm) 2 = 770 mm F=F DB ! DB =F DB 385 N = [(480 mm)i − (510 mm)j + (320 mm)k ] 770 mm = (240 N)i − (255 N) j + (160 N)k Fx = +240 N, Fy = −255 N, Fz = +160.0 N PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 89 PROBLEM 2.88 For the frame and cable of Problem 2.87, determine the components of the force exerted by the cable on the support at E. PROBLEM 2.87 A frame ABC is supported in part by cable DBE that passes through a frictionless ring at B. Knowing that the tension in the cable is 385 N, determine the components of the force exerted by the cable on the support at D. SOLUTION ! EB = (270 mm)i − (400 mm) j + (600 mm)k EB = (270 mm) 2 + (400 mm) 2 + (600 mm) 2 = 770 mm F=F EB ! EB =F EB 385 N = [(270 mm)i − (400 mm)j + (600 mm)k ] 770 mm F = (135 N)i − (200 N) j + (300 N)k Fx = +135.0 N, Fy = −200 N, Fz = +300 N PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 90 PROBLEM 2.89 Knowing that the tension in cable AB is 1425 N, determine the components of the force exerted on the plate at B. SOLUTION ! BA = −(900 mm)i + (600 mm) j + (360 mm)k BA = (900 mm) 2 + (600 mm) 2 + (360 mm) 2 = 1140 mm TBA = TBA BA ! BA BA 1425 N = [ −(900 mm)i + (600 mm) j + (360 mm)k ] 1140 mm = −(1125 N)i + (750 N) j + (450 N)k = TBA TBA (TBA ) x = −1125 N, (TBA ) y = 750 N, (TBA ) z = 450 N PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 91 PROBLEM 2.90 Knowing that the tension in cable AC is 2130 N, determine the components of the force exerted on the plate at C. SOLUTION ! CA = −(900 mm)i + (600 mm) j − (920 mm)k CA = (900 mm)2 + (600 mm)2 + (920 mm) 2 = 1420 mm TCA = TCA λCA ! CA = TCA CA 2130 N TCA = [−(900 mm)i + (600 mm) j − (920 mm)k ] 1420 mm = −(1350 N)i + (900 N) j − (1380 N)k (TCA ) x = −1350 N, (TCA ) y = 900 N, (TCA ) z = −1380 N PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 92 PROBLEM 2.91 Find the magnitude and direction of the resultant of the two forces shown knowing that P = 300 N and Q = 400 N. SOLUTION P = (300 N)[− cos 30° sin15°i + sin 30° j + cos 30° cos15°k ] = − (67.243 N)i + (150 N) j + (250.95 N)k Q = (400 N)[cos 50° cos 20°i + sin 50° j − cos 50° sin 20°k ] = (400 N)[0.60402i + 0.76604 j − 0.21985] = (241.61 N)i + (306.42 N) j − (87.939 N)k R = P+Q = (174.367 N)i + (456.42 N) j + (163.011 N)k R = (174.367 N)2 + (456.42 N)2 + (163.011 N) 2 = 515.07 N cos θ x = cos θ y = cos θ z = R = 515 N Rx 174.367 N = = 0.33853 515.07 N R Ry θ x = 70.2° 456.42 N = 0.88613 515.07 N θ y = 27.6° Rz 163.011 N = = 0.31648 R 515.07 N θ z = 71.5° R = PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 93 ! PROBLEM 2.92 Find the magnitude and direction of the resultant of the two forces shown knowing that P = 400 N and Q = 300 N. SOLUTION P = (400 N)[ − cos 30° sin15°i + sin 30° j + cos 30° cos15°k ] = − (89.678 N)i + (200 N) j + (334.61 N)k Q = (300 N)[cos 50° cos 20°i + sin 50° j − cos 50° sin 20°k ] = (181.21 N)i + (229.81 N)j − (65.954 N)k R = P+Q = (91.532 N)i + (429.81 N) j + (268.66 N)k R = (91.532 N)2 + (429.81 N) 2 + (268.66 N) 2 = 515.07 N cos θ x = cos θ y = cos θ z = R = 515 N Rx 91.532 N = = 0.177708 R 515.07 N Ry θ x = 79.8° 429.81 N = 0.83447 515.07 N θ y = 33.4° Rz 268.66 N = = 0.52160 R 515.07 N θ z = 58.6° R = PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 94 ! PROBLEM 2.93 Knowing that the tension is 425 lb in cable AB and 510 lb in cable AC, determine the magnitude and direction of the resultant of the forces exerted at A by the two cables. SOLUTION ! AB = (40 in.)i − (45 in.) j + (60 in.)k AB = (40 in.) 2 + (45 in.) 2 + (60 in.) 2 = 85 in. ! AC = (100 in.)i − (45 in.) j + (60 in.)k AC = (100 in.) 2 + (45 in.) 2 + (60 in.) 2 = 125 in. ! (40 in.)i − (45 in.) j + (60 in.)k ! AB = (425 lb) " TAB = TAB AB = TAB # AB 85 in. $ % TAB = (200 lb)i − (225 lb) j + (300 lb)k ! AC (100 in.)i − (45 in.) j + (60 in.)k ! = (510 lb) " TAC = TAC AC = TAC # 125 in. AC $ % TAC = (408 lb)i − (183.6 lb) j + (244.8 lb)k R = TAB + TAC = (608)i − (408.6 lb) j + (544.8 lb)k Then: and R = 912.92 lb R = 913 lb cos θ x = 608 lb = 0.66599 912.92 lb cos θ y = 408.6 lb = −0.44757 912.92 lb θ y = 116.6° cos θ z = 544.8 lb = 0.59677 912.92 lb θ z = 53.4° θ x = 48.2° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 95 ! PROBLEM 2.94 Knowing that the tension is 510 lb in cable AB and 425 lb in cable AC, determine the magnitude and direction of the resultant of the forces exerted at A by the two cables. SOLUTION ! AB = (40 in.)i − (45 in.) j + (60 in.)k AB = (40 in.) 2 + (45 in.) 2 + (60 in.) 2 = 85 in. ! AC = (100 in.)i − (45 in.) j + (60 in.)k AC = (100 in.) 2 + (45 in.) 2 + (60 in.) 2 = 125 in. ! (40 in.)i − (45 in.) j + (60 in.)k ! AB = (510 lb) " TAB = TAB AB = TAB # AB 85 in. $ % TAB = (240 lb)i − (270 lb) j + (360 lb)k ! AC (100 in.)i − (45 in.) j + (60 in.)k ! = (425 lb) " TAC = TAC AC = TAC # 125 in. AC $ % TAC = (340 lb)i − (153 lb) j + (204 lb)k R = TAB + TAC = (580 lb)i − (423 lb) j + (564 lb)k Then: and R = 912.92 lb R = 913 lb cos θ x = 580 lb = 0.63532 912.92 lb cos θ y = −423 lb = −0.46335 912.92 lb cos θ z = 564 lb = 0.61780 912.92 lb θ x = 50.6° θ y = 117.6° θ z = 51.8° ! PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 96 PROBLEM 2.95 For the frame of Problem 2.87, determine the magnitude and direction of the resultant of the forces exerted by the cable at B knowing that the tension in the cable is 385 N. PROBLEM 2.87 A frame ABC is supported in part by cable DBE that passes through a frictionless ring at B. Knowing that the tension in the cable is 385 N, determine the components of the force exerted by the cable on the support at D. SOLUTION ! BD = −(480 mm)i + (510 mm) j − (320 mm)k BD = (480 mm) 2 + (510 mm) 2 + (320 mm) 2 = 770 mm ! BD FBD = TBD BD = TBD BD (385 N) = [−(480 mm)i + (510 mm) j − (320 mm)k ] (770 mm) = −(240 N)i + (255 N) j − (160 N)k ! BE = −(270 mm)i + (400 mm) j − (600 mm)k BE = (270 mm) 2 + (400 mm) 2 + (600 mm) 2 = 770 mm ! BE FBE = TBE BE = TBE BE (385 N) = [−(270 mm)i + (400 mm) j − (600 mm)k ] (770 mm) = −(135 N)i + (200 N) j − (300 N)k R = FBD + FBE = −(375 N)i + (455 N) j − (460 N)k R = (375 N)2 + (455 N) 2 + (460 N) 2 = 747.83 N R = 748 N cos θ x = −375 N 747.83 N θ x = 120.1° cos θ y = 455 N 747.83 N θ y = 52.5° cos θ z = −460 N 747.83 N θ z = 128.0° ! PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 97 PROBLEM 2.96 For the cables of Problem 2.89, knowing that the tension is 1425 N in cable AB and 2130 N in cable AC, determine the magnitude and direction of the resultant of the forces exerted at A by the two cables. SOLUTION TAB = −TBA (use results of Problem 2.89) (TAB ) x = +1125 N (TAB ) y = −750 N (TAB ) z = − 450 N TAC = −TCA (use results of Problem 2.90) (TAC ) x = +1350 N (TAC ) y = −900 N (TAC ) z = +1380 N Resultant: Rx = ΣFx = +1125 + 1350 = +2475 N Ry = ΣFy = −750 − 900 = −1650 N Rz = ΣFz = −450 + 1380 = + 930 N R = Rx2 + Ry2 + Rz2 = (+2475) 2 + (−1650) 2 + ( +930)2 = 3116.6 N cos θ x = cos θ y = cos θ z = R = 3120 N Rx +2475 = R 3116.6 Ry θ x = 37.4° −1650 3116.6 θ y = 122.0° Rz + 930 = R 3116.6 θ z = 72.6° R = PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 98 ! PROBLEM 2.97 The end of the coaxial cable AE is attached to the pole AB, which is strengthened by the guy wires AC and AD. Knowing that the tension in AC is 150 lb and that the resultant of the forces exerted at A by wires AC and AD must be contained in the xy plane, determine (a) the tension in AD, (b) the magnitude and direction of the resultant of the two forces. SOLUTION R = TAC + TAD = (150 lb)(cos 60° cos 20°i − sin 60° j − cos 60° sin 20°k ) + TAD (sin 36° sin 48°i − cos 36° j + sin 36° cos 48°k ) (a) (1) Since Rz = 0, The coefficient of k must be zero. (150 lb)( − cos 60° sin 20°) + TAD (sin 36° cos 48°) = 0 TAD = 65.220 lb (b) TAD = 65.2 lb Substituting for TAD into Eq. (1) gives: R = [(150 lb) cos 60° cos 20° + (65.220 lb) sin 36° sin 48°)]i − [(150 lb) sin 60° + (65.220 lb) cos 36°]j + 0 R = (98.966 lb)i − (182.668 lb) j R = (98.966 lb)2 + (182.668 lb) 2 = 207.76 lb R = 208 lb cos θ x = 98.966 lb 207.76 lb θ x = 61.6° cos θ y = 182.668 lb 207.76 lb θ y = 151.6° cos θ z = 0 θ z = 90.0° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 99 ! PROBLEM 2.98 The end of the coaxial cable AE is attached to the pole AB, which is strengthened by the guy wires AC and AD. Knowing that the tension in AD is 125 lb and that the resultant of the forces exerted at A by wires AC and AD must be contained in the xy plane, determine (a) the tension in AC, (b) the magnitude and direction of the resultant of the two forces. SOLUTION R = TAC + TAD = TAC (cos 60° cos 20°i − sin 60° j − cos 60° sin 20°k ) + (125 lb)(sin 36° sin 48°i − cos 36° j + sin 36° cos 48°k ) (a) (1) Since Rz = 0, The coefficient of k must be zero. TAC (− cos 60° sin 20°) + (125 lb)(sin 36° cos 48°) = 0 TAC = 287.49 lb (b) TAC = 287 lb Substituting for TAC into Eq. (1) gives: R = [(287.49 lb) cos 60° cos 20° + (125 lb) sin 36° sin 48°]i − [(287.49 lb) sin 60° + (125 lb) cos 36°]j + 0 R = (189.677 lb)i − (350.10 lb) j R = (189.677 lb)2 + (350.10 lb) 2 R = 398 lb = 398.18 lb cos θ x = 189.677 lb 398.18 lb θ x = 61.6° cos θ y = 350.10 lb 398.18 lb θ y = 151.6° cos θ z = 0 θ z = 90.0° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 100 ! PROBLEM 2.99 Three cables are used to tether a balloon as shown. Determine the vertical force P exerted by the balloon at A knowing that the tension in cable AB is 259 N. SOLUTION The forces applied at A are: TAB , TAC , TAD , and P where P = Pj. To express the other forces in terms of the unit vectors i, j, k, we write ! AB = − (4.20 m)i − (5.60 m) j AB = 7.00 m ! AC = (2.40 m)i − (5.60 m) j + (4.20 m)k AC = 7.40 m ! AD = − (5.60 m) j − (3.30 m)k AD = 6.50 m ! AB and TAB = TAB AB = TAB = (− 0.6i − 0.8 j)TAB AB! AC TAC = TAC AC = TAC = (0.32432 − 0.75676 j + 0.56757k )TAC AC! AD TAD = TAD AD = TAD = ( − 0.86154 j − 0.50769k )TAD AD PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 101 PROBLEM 2.99 (Continued) Equilibrium condition ΣF = 0: TAB + TAC + TAD + Pj = 0 Substituting the expressions obtained for TAB , TAC , and TAD and factoring i, j, and k: (− 0.6TAB + 0.32432TAC )i + (−0.8TAB − 0.75676TAC − 0.86154TAD + P ) j + (0.56757TAC − 0.50769TAD )k = 0 Equating to zero the coefficients of i, j, k: − 0.6TAB + 0.32432TAC = 0 (1) − 0.8TAB − 0.75676TAC − 0.86154TAD + P = 0 (2) 0.56757TAC − 0.50769TAD = 0 (3) Setting TAB = 259 N in (1) and (2), and solving the resulting set of equations gives TAC = 479.15 N TAD = 535.66 N P = 1031 N PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 102 PROBLEM 2.100 Three cables are used to tether a balloon as shown. Determine the vertical force P exerted by the balloon at A knowing that the tension in cable AC is 444 N. SOLUTION See Problem 2.99 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3) below: − 0.6TAB + 0.32432TAC = 0 (1) − 0.8TAB − 0.75676TAC − 0.86154TAD + P = 0 (2) 0.56757TAC − 0.50769TAD = 0 (3) Substituting TAC = 444 N in Equations (1), (2), and (3) above, and solving the resulting set of equations using conventional algorithms gives TAB = 240 N TAD = 496.36 N P = 956 N PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 103 PROBLEM 2.101 Three cables are used to tether a balloon as shown. Determine the vertical force P exerted by the balloon at A knowing that the tension in cable AD is 481 N. SOLUTION See Problem 2.99 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3). − 0.6TAB + 0.32432TAC = 0 (1) − 0.8TAB − 0.75676TAC − 0.86154TAD + P = 0 (2) 0.56757TAC − 0.50769TAD = 0 (3) Substituting TAD = 481 N in Equations (1), (2), and (3) above, and solving the resulting set of equations using conventional algorithms gives TAC = 430.26 N TAB = 232.57 N P = 926 N PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 104 PROBLEM 2.102 Three cables are used to tether a balloon as shown. Knowing that the balloon exerts an 800-N vertical force at A, determine the tension in each cable. SOLUTION See Problem 2.99 for the figure and analysis leading to the linear algebraic Equations (1), (2), and (3). − 0.6TAB + 0.32432TAC = 0 (1) − 0.8TAB − 0.75676TAC − 0.86154TAD + P = 0 (2) 0.56757TAC − 0.50769TAD = 0 (3) From Eq. (1) TAB = 0.54053TAC From Eq. (3) TAD = 1.11795TAC Substituting for TAB and TAD in terms of TAC into Eq. (2) gives: − 0.8(0.54053TAC ) − 0.75676TAC − 0.86154(1.11795TAC ) + P = 0 2.1523TAC = P ; P = 800 N 800 N 2.1523 = 371.69 N TAC = Substituting into expressions for TAB and TAD gives: TAB = 0.54053(371.69 N) TAD = 1.11795(371.69 N) TAB = 201 N, TAC = 372 N, TAD = 416 N PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 105 PROBLEM 2.103 A crate is supported by three cables as shown. Determine the weight of the crate knowing that the tension in cable AB is 750 lb. SOLUTION The forces applied at A are: TAB , TAC , TAD and W where P = Pj. To express the other forces in terms of the unit vectors i, j, k, we write ! AB = − (36 in.)i + (60 in.) j − (27 in.)k AB = 75 in. ! AC = (60 in.) j + (32 in.)k AC = 68 in. ! AD = (40 in.)i + (60 in.) j − (27 in.)k AD = 77 in. ! AB TAB = TAB AB = TAB and AB = (− 0.48i + 0.8 j − 0.36k )TAB ! AC TAC = TAC AC = TAC AC = (0.88235 j + 0.47059k )TAC ! AD TAD = TAD AD = TAD AD = (0.51948i + 0.77922 j − 0.35065k )TAD Equilibrium Condition with W = − Wj ΣF = 0: TAB + TAC + TAD − Wj = 0 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 106 PROBLEM 2.103 (Continued) Substituting the expressions obtained for TAB , TAC , and TAD and factoring i, j, and k: (− 0.48TAB + 0.51948TAD )i + (0.8TAB + 0.88235TAC + 0.77922TAD − W ) j + (− 0.36TAB + 0.47059TAC − 0.35065TAD )k = 0 Equating to zero the coefficients of i, j, k: −0.48TAB + 0.51948TAD = 0 0.8TAB + 0.88235TAC + 0.77922TAD − W = 0 −0.36TAB + 0.47059TAC − 0.35065TAD = 0 Substituting TAB = 750 lb in Equations (1), (2), and (3) and solving the resulting set of equations, using conventional algorithms for solving linear algebraic equations, gives: TAC = 1090.1 lb TAD = 693 lb W = 2100 lb PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 107 ! PROBLEM 2.104 A crate is supported by three cables as shown. Determine the weight of the crate knowing that the tension in cable AD is 616 lb. SOLUTION See Problem 2.103 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3) below: − 0.48TAB + 0.51948TAD = 0 0.8TAB + 0.88235TAC + 0.77922TAD − W = 0 −0.36TAB + 0.47059TAC − 0.35065TAD = 0 Substituting TAD = 616 lb in Equations (1), (2), and (3) above, and solving the resulting set of equations using conventional algorithms, gives: TAB = 667.67 lb TAC = 969.00 lb W = 1868 lb PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 108 PROBLEM 2.105 A crate is supported by three cables as shown. Determine the weight of the crate knowing that the tension in cable AC is 544 lb. SOLUTION See Problem 2.103 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3) below: − 0.48TAB + 0.51948TAD = 0 0.8TAB + 0.88235TAC + 0.77922TAD − W = 0 − 0.36TAB + 0.47059TAC − 0.35065TAD = 0 Substituting TAC = 544 lb in Equations (1), (2), and (3) above, and solving the resulting set of equations using conventional algorithms, gives: TAB = 374.27 lb TAD = 345.82 lb W = 1049 lb PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 109 PROBLEM 2.106 A 1600-lb crate is supported by three cables as shown. Determine the tension in each cable. SOLUTION See Problem 2.103 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3) below: −0.48TAB + 0.51948TAD = 0 0.8TAB + 0.88235TAC + 0.77922TAD − W = 0 −0.36TAB + 0.47059TAC − 0.35065TAD = 0 Substituting W = 1600 lb in Equations (1), (2), and (3) above, and solving the resulting set of equations using conventional algorithms, gives: TAB = 571 lb TAC = 830 lb TAD = 528 lb PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 110 ! PROBLEM 2.107 Three cables are connected at A, where the forces P and Q are applied as shown. Knowing that Q = 0, find the value of P for which the tension in cable AD is 305 N. SOLUTION ! ΣFA = 0: TAB + TAC + TAD + P = 0 where P = Pi ! AB = −(960 mm)i − (240 mm)j + (380 mm)k AB = 1060 mm ! AC = −(960 mm)i − (240 mm) j − (320 mm)k AC = 1040 mm ! ! AD = −(960 mm)i + (720 mm) j − (220 mm)k AD = 1220 mm ! 48 12 19 ! AB = TAB " − i − j + k # TAB = TAB AB = TAB AB 53 53 53 % $ ! AC 12 3 4 ! TAC = TAC AC = TAC = TAC " − i − j − k # 13 13 13 AC $ % 305 N [( −960 mm)i + (720 mm) j − (220 mm)k ] TAD = TAD AD = 1220 mm = −(240 N)i + (180 N) j − (55 N)k Substituting into ΣFA = 0, factoring i, j, k , and setting each coefficient equal to φ gives: i: P = 48 12 TAB + TAC + 240 N 53 13 (1) j: 12 3 TAB + TAC = 180 N 53 13 (2) k: 19 4 TAB − TAC = 55 N 53 13 (3) Solving the system of linear equations using conventional algorithms gives: TAB = 446.71 N TAC = 341.71 N P = 960 N PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 111 ! PROBLEM 2.108 Three cables are connected at A, where the forces P and Q are applied as shown. Knowing that P = 1200 N, determine the values of Q for which cable AD is taut. SOLUTION We assume that TAD = 0 and write ΣFA = 0: TAB + TAC + Qj + (1200 N)i = 0 ! AB = −(960 mm)i − (240 mm)j + (380 mm)k AB = 1060 mm ! AC = −(960 mm)i − (240 mm) j − (320 mm)k AC = 1040 mm ! AB 48 12 19 ! TAB = TAB AB = TAB = " − i − j + k # TAB AB $ 53 53 53 % ! AC 12 3 4 ! TAC = TAC AC = TAC = " − i − j − k # TAC AC $ 13 13 13 % Substituting into ΣFA = 0, factoring i, j, k , and setting each coefficient equal to φ gives: i: − 48 12 TAB − TAC + 1200 N = 0 53 13 (1) j: − 12 3 TAB − TAC + Q = 0 53 13 (2) k: 19 4 TAB − TAC = 0 53 13 (3) Solving the resulting system of linear equations using conventional algorithms gives: TAB = 605.71 N TAC = 705.71 N Q = 300.00 N 0 # Q , 300 N Note: This solution assumes that Q is directed upward as shown (Q $ 0), if negative values of Q are considered, cable AD remains taut, but AC becomes slack for Q = −460 N. ! PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 112 PROBLEM 2.109 A transmission tower is held by three guy wires attached to a pin at A and anchored by bolts at B, C, and D. If the tension in wire AB is 630 lb, determine the vertical force P exerted by the tower on the pin at A. SOLUTION Free Body A: We write ΣF = 0: TAB + TAC + TAD + Pj = 0 ! AB = −45i − 90 j + 30k AB = 105 ft ! AC = 30i − 90 j + 65k AC = 115 ft ! AD = 20i − 90 j − 60k AD = 110 ft ! AB TAB = TAB AB = TAB AB 3 6 2 ! = " − i − j + k # TAB 7 7 % $ 7 ! AC TAC = TAC AC = TAC AC 6 18 13 ! j + k # TAC =" i− 23 23 % $ 23 ! AD TAD = TAD AD = TAD AD 2 9 6 ! = " i − j − k # TAD 11 11 11 $ % Substituting into the Eq. ΣF = 0 and factoring i, j, k : 3 6 2 ! " − TAB + TAC + TAD # i 7 23 11 $ % 6 18 9 ! + " − TAB − TAC − TAD + P # j 23 11 $ 7 % 2 13 6 ! + " TAB + TAC − TAD # k = 0 23 11 $7 % PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 113 PROBLEM 2.109 (Continued) Setting the coefficients of i, j, k , equal to zero: i: 3 6 2 − TAB + TAC + TAD = 0 7 23 11 (1) j: 6 18 9 − TAB − TAC − TAD + P = 0 7 23 11 (2) k: 2 13 6 TAB + TAC − TAD = 0 7 23 11 (3) Set TAB = 630 lb in Eqs. (1) – (3): 6 2 TAC + TAD = 0 23 11 (1′) 18 9 TAC − TAD + P = 0 23 11 (2′) 13 6 TAC − TAD = 0 23 11 (3′) −270 lb + −540 lb − 180 lb + Solving, TAC = 467.42 lb TAD = 814.35 lb P = 1572.10 lb P = 1572 lb PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 114 ! PROBLEM 2.110 A transmission tower is held by three guy wires attached to a pin at A and anchored by bolts at B, C, and D. If the tension in wire AC is 920 lb, determine the vertical force P exerted by the tower on the pin at A. SOLUTION See Problem 2.109 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3) below: 3 6 2 − TAB + TAC + TAD = 0 7 23 11 (1) 6 18 9 − TAB − TAC − TAD + P = 0 7 23 11 (2) 2 13 6 TAB + TAC − TAD = 0 7 23 11 (3) Substituting for TAC = 920 lb in Equations (1), (2), and (3) above and solving the resulting set of equations using conventional algorithms gives: Solving, 3 2 − TAB + 240 lb + TAD = 0 7 11 (1′) 6 9 − TAB − 720 lb − TAD + P = 0 7 11 (2′) 2 6 TAB + 520 lb − TAD = 0 7 11 (3′) TAB = 1240.00 lb TAD = 1602.86 lb P = 3094.3 lb P = 3090 lb PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 115 PROBLEM 2.111 A rectangular plate is supported by three cables as shown. Knowing that the tension in cable AC is 60 N, determine the weight of the plate. SOLUTION We note that the weight of the plate is equal in magnitude to the force P exerted by the support on Point A. Free Body A: ΣF = 0: TAB + TAC + TAD + Pj = 0 We have: ! AB = −(320 mm)i − (480 mm)j + (360 mm)k AB = 680 mm ! AC = (450 mm)i − (480 mm) j + (360 mm)k AC = 750 mm ! AD = (250 mm)i − (480 mm) j − ( 360 mm ) k AD = 650 mm Thus: TAB = TAB AB TAC = TAC AC TAD = TAD AD ! AB 8 12 9 ! = − i − j + k TAB AB "$ 17 17 17 #% ! AC = TAC = ( 0.6i − 0.64 j + 0.48k ) TAC AC ! 5 9.6 7.2 ! AD = TAD =" i− j− k TAD 13 13 #% AD $ 13 = TAB Dimensions in mm Substituting into the Eq. ΣF = 0 and factoring i, j, k : 8 5 ! " − TAB + 0.6TAC + TAD # i 13 $ 17 % 12 9.6 ! TAD + P # j + " − TAB − 0.64TAC − 17 13 $ % 9 7.2 ! TAD # k = 0 + " TAB + 0.48TAC − 13 $ 17 % PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 116 PROBLEM 2.111 (Continued) Setting the coefficient of i, j, k equal to zero: i: − 8 5 TAB + 0.6TAC + TAD = 0 17 13 (1) j: − 12 9.6 TAB − 0.64TAC − TAD + P = 0 7 13 (2) 9 7.2 TAB + 0.48TAC − TAD = 0 17 13 (3) 8 5 TAB + 36 N + TAD = 0 17 13 (1′) 9 7.2 TAB + 28.8 N − TAD = 0 17 13 (3′) k: Making TAC = 60 N in (1) and (3): − Multiply (1′) by 9, (3′) by 8, and add: 554.4 N − 12.6 TAD = 0 TAD = 572.0 N 13 Substitute into (1′) and solve for TAB : TAB = 17 5 ! 36 + × 572 # " 8 $ 13 % TAB = 544.0 N Substitute for the tensions in Eq. (2) and solve for P: 12 9.6 (544 N) + 0.64(60 N) + (572 N) 17 13 = 844.8 N P= Weight of plate = P = 845 N PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 117 PROBLEM 2.112 A rectangular plate is supported by three cables as shown. Knowing that the tension in cable AD is 520 N, determine the weight of the plate. SOLUTION See Problem 2.111 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3) below: 8 5 TAB + 0.6TAC + TAD = 0 17 13 (1) 12 9.6 TAB + 0.64 TAC − TAD + P = 0 17 13 (2) 9 7.2 TAB + 0.48TAC − TAD = 0 17 13 (3) − − Making TAD = 520 N in Eqs. (1) and (3): 8 TAB + 0.6TAC + 200 N = 0 17 (1′) 9 TAB + 0.48TAC − 288 N = 0 17 (3′) − Multiply (1′) by 9, (3′) by 8, and add: 9.24TAC − 504 N = 0 TAC = 54.5455 N Substitute into (1′) and solve for TAB : TAB = 17 (0.6 × 54.5455 + 200) TAB = 494.545 N 8 Substitute for the tensions in Eq. (2) and solve for P: 12 9.6 (494.545 N) + 0.64(54.5455 N) + (520 N) 17 13 Weight of plate = P = 768 N = 768.00 N P= PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 118 PROBLEM 2.113 For the transmission tower of Problems 2.109 and 2.110, determine the tension in each guy wire knowing that the tower exerts on the pin at A an upward vertical force of 2100 lb. SOLUTION See Problem 2.109 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3) below: 3 6 2 − TAB + TAC + TAD = 0 7 23 11 (1) 6 18 9 − TAB − TAC − TAD + P = 0 7 23 11 (2) 2 13 6 TAB + TAC − TAD = 0 7 23 11 (3) Substituting for P = 2100 lb in Equations (1), (2), and (3) above and solving the resulting set of equations using conventional algorithms gives: 3 6 2 − TAB + TAC + TAD = 0 7 23 11 (1′) 6 18 9 − TAB − TAC − TAD + 2100 lb = 0 7 23 11 (2′) 2 13 6 TAB + TAC − TAD = 0 7 23 11 (3′) TAB = 841.55 lb TAC = 624.38 lb TAD = 1087.81 lb TAB = 842 lb TAC = 624 lb TAD = 1088 lb PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 119 ! PROBLEM 2.114 A horizontal circular plate weighing 60 lb is suspended as shown from three wires that are attached to a support at D and form 30° angles with the vertical. Determine the tension in each wire. SOLUTION ΣFx = 0: −TAD (sin 30°)(sin 50°) + TBD (sin 30°)(cos 40°) + TCD (sin 30°)(cos 60°) = 0 Dividing through by sin 30° and evaluating: −0.76604TAD + 0.76604TBD + 0.5TCD = 0 (1) ΣFy = 0: −TAD (cos 30°) − TBD (cos 30°) − TCD (cos 30°) + 60 lb = 0 TAD + TBD + TCD = 69.282 lb or (2) ΣFz = 0: TAD sin 30° cos 50° + TBD sin 30° sin 40° − TCD sin 30° sin 60° = 0 ! or 0.64279TAD + 0.64279TBD − 0.86603TCD = 0 (3) Solving Equations (1), (2), and (3) simultaneously: TAD = 29.5 lb TBD = 10.25 lb ! ! ! TCD = 29.5 lb ! PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 120 PROBLEM 2.115 For the rectangular plate of Problems 2.111 and 2.112, determine the tension in each of the three cables knowing that the weight of the plate is 792 N. SOLUTION See Problem 2.111 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3) below. Setting P = 792 N gives: 8 5 TAB + 0.6TAC + TAD = 0 17 13 (1) 12 9.6 TAB − 0.64TAC − TAD + 792 N = 0 17 13 (2) 9 7.2 TAB + 0.48TAC − TAD = 0 17 13 (3) − − Solving Equations (1), (2), and (3) by conventional algorithms gives TAB = 510.00 N TAB = 510 N TAC = 56.250 N TAC = 56.2 N TAD = 536.25 N TAD = 536 N PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 121 PROBLEM 2.116 For the cable system of Problems 2.107 and 2.108, determine the tension in each cable knowing that P = 2880 N and Q = 0. SOLUTION ΣFA = 0: TAB + TAC + TAD + P + Q = 0 Where P = Pi and Q = Qj ! AB = −(960 mm)i − (240 mm) j + (380 mm)k AB = 1060 mm ! AC = −(960 mm)i − (240 mm) j − (320 mm)k AC = 1040 mm ! AD = −(960 mm)i + (720 mm) j − (220 mm)k AD = 1220 mm ! 48 12 19 ! AB TAB = TAB AB = TAB = TAB " − i − j + k # 53 53 53 % AB $ ! 12 3 4 ! AC TAC = TAC AC = TAC = TAC " − i − j − k # 13 13 13 AC $ % ! 48 36 11 ! AD = TAD " − i + TAD = TAD AD = TAD j− k# 61 61 61 % AD $ Substituting into ΣFA = 0, setting P = (2880 N)i and Q = 0, and setting the coefficients of i, j, k equal to 0, we obtain the following three equilibrium equations: i: − 48 12 48 TAB − TAC − TAD + 2880 N = 0 53 13 61 (1) j: − 12 3 36 TAB − TAC + TAD = 0 53 13 61 (2) k: 19 4 11 TAB − TAC − TAD = 0 53 13 61 (3) PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 122 PROBLEM 2.116 (Continued) Solving the system of linear equations using conventional algorithms gives: TAB = 1340.14 N TAC = 1025.12 N TAD = 915.03 N TAB = 1340 N TAC = 1025 N TAD = 915 N PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 123 ! PROBLEM 2.117 For the cable system of Problems 2.107 and 2.108, determine the tension in each cable knowing that P = 2880 N and Q = 576 N. SOLUTION See Problem 2.116 for the analysis leading to the linear algebraic Equations (1), (2), and (3) below: − 48 12 48 TAB − TAC − TAD + P = 0 53 13 61 (1) − 12 3 36 TAB − TAC + TAD + Q = 0 53 13 61 (2) 19 4 11 TAB − TAC − TAD = 0 53 13 61 (3) Setting P = 2880 N and Q = 576 N gives: − 48 12 48 TAB − TAC − TAD + 2880 N = 0 53 13 61 (1′) 12 3 36 TAB − TAC + TAD + 576 N = 0 53 13 61 (2′) 19 4 11 TAB − TAC − TAD = 0 53 13 61 (3′) − Solving the resulting set of equations using conventional algorithms gives: TAB = 1431.00 N TAC = 1560.00 N TAD = 183.010 N TAB = 1431 N ! TAC = 1560 N TAD = 183.0 N PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 124 PROBLEM 2.118 For the cable system of Problems 2.107 and 2.108, determine the tension in each cable knowing that P = 2880 N and Q = −576 N. (Q is directed downward). SOLUTION See Problem 2.116 for the analysis leading to the linear algebraic Equations (1), (2), and (3) below:! ! − 48 12 48 TAB − TAC − TAD + P = 0 53 13 61 (1) − 12 3 36 TAB − TAC + TAD + Q = 0 53 13 61 (2) 19 4 11 TAB − TAC − TAD = 0 53 13 61 (3) Setting P = 2880 N and Q = −576 N gives: − 48 12 48 TAB − TAC − TAD + 2880 N = 0 53 13 61 (1′) 12 3 36 TAB − TAC + TAD − 576 N = 0 53 13 61 (2′) 19 4 11 TAB − TAC − TAD = 0 53 13 61 (3′) − Solving the resulting set of equations using conventional algorithms gives: TAB = 1249.29 N TAC = 490.31 N TAD = 1646.97 N TAB = 1249 N TAC = 490 N TAD = 1647 N PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 125 PROBLEM 2.119 Using two ropes and a roller chute, two workers are unloading a 200-lb cast-iron counterweight from a truck. Knowing that at the instant shown the counterweight is kept from moving and that the positions of Points A, B, and C are, respectively, A(0, –20 in., 40 in.), B(–40 in., 50 in., 0), and C(45 in., 40 in., 0), and assuming that no friction exists between the counterweight and the chute, determine the tension in each rope. (Hint: Since there is no friction, the force exerted by the chute on the counterweight must be perpendicular to the chute.) SOLUTION From the geometry of the chute: N= N (2 j + k ) 5 = N (0.8944 j + 0.4472k ) The force in each rope can be written as the product of the magnitude of the force and the unit vector along the cable. Thus, with ! AB = (40 in.)i + (70 in.)j − (40 in.)k and AB = (40 in.) 2 + (70 in.) 2 + (40 in.)2 = 90 in. ! AB TAB = T AB = TAB AB TAB [(−40 in.)i + (70 in.)j − (40 in.)k ] = 90 in. 4 7 4 ! TAB = TAB " − i + j − k # 9 9 % $ 9 ! AC = (45 in.)i + (60 in.)j − (40 in.)k AC = (45 in.) 2 + (60 in.)2 + (40 in.)2 = 85 in. ! AC TAC = T AC = TAC AC TAC [(45 in.)i + (60 in.)j − (40 in.)k ] = 85 in. 9 12 8 ! TAC = TAC " i + j − k # $ 17 17 17 % Then: ΣF = 0: N + TAB + TAC + W = 0 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 126 PROBLEM 2.119 (Continued) With W = 200 lb, and equating the factors of i, j, and k to zero, we obtain the linear algebraic equations: i: j: k: 4 9 − TAB + TAC = 0 9 17 (1) 7 12 2 TAB + TAC + − 200 lb = 0 9 17 5 (2) 4 8 1 − TAB − TAC + N =0 9 17 5 (3) Using conventional methods for solving linear algebraic equations we obtain: TAB = 65.6 lb TAC = 55.1 lb PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 127 PROBLEM 2.120 Solve Problem 2.119 assuming that a third worker is exerting a force P = −(40 lb)i on the counterweight. PROBLEM 2.119 Using two ropes and a roller chute, two workers are unloading a 200-lb cast-iron counterweight from a truck. Knowing that at the instant shown the counterweight is kept from moving and that the positions of Points A, B, and C are, respectively, A(0, –20 in., 40 in.), B(–40 in., 50 in., 0), and C(45 in., 40 in., 0), and assuming that no friction exists between the counterweight and the chute, determine the tension in each rope. (Hint: Since there is no friction, the force exerted by the chute on the counterweight must be perpendicular to the chute.) SOLUTION See Problem 2.119 for the analysis leading to the vectors describing the tension in each rope. 4 7 4 ! TAB = TAB " − i + j − k # 9 9 % $ 9 9 12 8 ! TAC = TAC " i + j − k # 17 17 17 $ % Then: Where and ΣFA = 0: N + TAB + TAC + P + W = 0 P = −(40 lb)i W = (200 lb)j Equating the factors of i, j, and k to zero, we obtain the linear equations: 4 9 i : − TAB + TAC − 40 lb = 0 9 17 j: k: 2 7 12 N + TAB + TAC − 200 lb = 0 9 17 5 1 5 4 8 N − TAB − TAC = 0 9 17 Using conventional methods for solving linear algebraic equations we obtain TAB = 24.8 lb TAC = 96.4 lb PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 128 PROBLEM 2.121 A container of weight W is suspended from ring A. Cable BAC passes through the ring and is attached to fixed supports at B and C. Two forces P = Pi and Q = Qk are applied to the ring to maintain the container in the position shown. Knowing that W = 376 N, determine P and Q. (Hint: The tension is the same in both portions of cable BAC.) SOLUTION Free Body A: TAB = T AB " AB =T AB (−130 mm)i + (400 mm) j + (160 mm)k =T 450 mm 13 40 16 ! =T "− i + j+ k# 45 45 % $ 45 TAC = T AC " AC =T AC ( −150 mm)i + (400 mm) j + (−240 mm)k =T 490 mm 15 40 24 ! = T "− i + j− k# 49 49 49 % $ ΣF = 0: TAB + TAC + Q + P + W = 0 Setting coefficients of i, j, k equal to zero: i: − 13 15 T − T +P=0 45 49 0.59501T = P (1) j: + 40 40 T + T −W = 0 45 49 1.70521T = W (2) k: + 16 24 T − T +Q =0 45 49 0.134240 T = Q (3) PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 129 PROBLEM 2.121 (Continued) Data: W = 376 N 1.70521T = 376 N T = 220.50 N 0.59501(220.50 N) = P P = 131.2 N 0.134240(220.50 N) = Q Q = 29.6 N PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 130 ! PROBLEM 2.122 For the system of Problem 2.121, determine W and Q knowing that P = 164 N. PROBLEM 2.121 A container of weight W is suspended from ring A. Cable BAC passes through the ring and is attached to fixed supports at B and C. Two forces P = Pi and Q = Qk are applied to the ring to maintain the container in the position shown. Knowing that W = 376 N, determine P and Q. (Hint: The tension is the same in both portions of cable BAC.) SOLUTION Refer to Problem 2.121 for the figure and analysis resulting in Equations (1), (2), and (3) for P, W, and Q in terms of T below. Setting P = 164 N we have: Eq. (1): 0.59501T = 164 N T = 275.63 N Eq. (2): 1.70521(275.63 N) = W W = 470 N Eq. (3): 0.134240(275.63 N) = Q Q = 37.0 N PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 131 ! PROBLEM 2.123 A container of weight W is suspended from ring A, to which cables AC and AE are attached. A force P is applied to the end F of a third cable that passes over a pulley at B and through ring A and that is attached to a support at D. Knowing that W = 1000 N, determine the magnitude of P. (Hint: The tension is the same in all portions of cable FBAD.) SOLUTION The (vector) force in each cable can be written as the product of the (scalar) force and the unit vector along the cable. That is, with ! AB = −(0.78 m)i + (1.6 m) j + (0 m)k AB = (−0.78 m) 2 + (1.6 m) 2 + (0)2 = 1.78 m TAB = T AB = TAB ! AB AB TAB [−(0.78 m)i + (1.6 m) j + (0 m)k ] 1.78 m = TAB (−0.4382i + 0.8989 j + 0k ) = and TAB ! AC = (0)i + (1.6 m) j + (1.2 m)k and AC = (0 m) 2 + (1.6 m) 2 + (1.2 m) 2 = 2 m ! AC TAC [(0)i + (1.6 m) j + (1.2 m)k ] TAC = T AC = TAC = AC 2 m TAC = TAC (0.8 j + 0.6k ) ! AD = (1.3 m)i + (1.6 m) j + (0.4 m)k AD = (1.3 m)2 + (1.6 m)2 + (0.4 m) 2 = 2.1 m ! T AD TAD = T AD = TAD = AD [(1.3 m)i + (1.6 m) j + (0.4 m)k ] AD 2.1 m TAD = TAD (0.6190i + 0.7619 j + 0.1905k ) PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 132 PROBLEM 2.123 (Continued) ! AE = −(0.4 m)i + (1.6 m) j − (0.86 m)k Finally, AE = (−0.4 m) 2 + (1.6 m) 2 + (−0.86 m) 2 = 1.86 m ! AE TAE = T AE = TAE AE T = AE [−(0.4 m)i + (1.6 m) j − (0.86 m)k ] 1.86 m TAE = TAE (−0.2151i + 0.8602 j − 0.4624k ) With the weight of the container W = −W j, at A we have: ΣF = 0: TAB + TAC + TAD − Wj = 0 Equating the factors of i, j, and k to zero, we obtain the following linear algebraic equations: −0.4382TAB + 0.6190TAD − 0.2151TAE = 0 (1) 0.8989TAB + 0.8TAC + 0.7619TAD + 0.8602TAE − W = 0 (2) 0.6TAC + 0.1905TAD − 0.4624TAE = 0 (3) Knowing that W = 1000 N and that because of the pulley system at B TAB = TAD = P, where P is the externally applied (unknown) force, we can solve the system of linear Equations (1), (2) and (3) uniquely for P. P = 378 N PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 133 ! PROBLEM 2.124 Knowing that the tension in cable AC of the system described in Problem 2.123 is 150 N, determine (a) the magnitude of the force P, (b) the weight W of the container. PROBLEM 2.123 A container of weight W is suspended from ring A, to which cables AC and AE are attached. A force P is applied to the end F of a third cable that passes over a pulley at B and through ring A and that is attached to a support at D. Knowing that W = 1000 N, determine the magnitude of P. (Hint: The tension is the same in all portions of cable FBAD.) SOLUTION Here, as in Problem 2.123, the support of the container consists of the four cables AE, AC, AD, and AB, with the condition that the force in cables AB and AD is equal to the externally applied force P. Thus, with the condition TAB = TAD = P and using the linear algebraic equations of Problem 2.131 with TAC = 150 N, we obtain (a) P = 454 N (b) W = 1202 N PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 134 ! PROBLEM 2.125 Collars A and B are connected by a 25-in.-long wire and can slide freely on frictionless rods. If a 60-lb force Q is applied to collar B as shown, determine (a) the tension in the wire when x = 9 in., (b) the corresponding magnitude of the force P required to maintain the equilibrium of the system. SOLUTION Free Body Diagrams of Collars: A: B: AB ΣF = 0: Pi + N y j + N z k + TAB λ AB = 0 Collar A: Substitute for ! AB − xi − (20 in.) j + zk = = 25 in. AB AB and set coefficient of i equal to zero: P− AB and set coefficient of k equal to zero: 60 lb − x = 9 in. (a) From Eq. (2): (b) (1) ΣF = 0: (60 lb)k + N x′ i + N y′ j − TAB λ AB = 0 Collar B: Substitute for TAB x =0 25 in. From Eq. (1): TAB z =0 25 in. (2) (9 in.)2 + (20 in.) 2 + z 2 = (25 in.) 2 z = 12 in. 60 lb − TAB (12 in.) 25 in. P= (125.0 lb)(9 in.) ! 25 in. TAB = 125.0 lb P = 45.0 lb PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 135 ! PROBLEM 2.126 Collars A and B are connected by a 25-in.-long wire and can slide freely on frictionless rods. Determine the distances x and z for which the equilibrium of the system is maintained when P = 120 lb and Q = 60 lb. SOLUTION See Problem 2.125 for the diagrams and analysis leading to Equations (1) and (2) below: P= TAB x =0 25 in. (1) 60 lb − TAB z =0 25 in. (2) For P = 120 lb, Eq. (1) yields TAB x = (25 in.)(20 lb) (1′) From Eq. (2) TAB z = (25 in.) (60 lb) (2′) x =2 z Dividing Eq. (1′) by (2′): Now write (3) x 2 + z 2 + (20 in.) 2 = (25 in.) 2 (4) Solving (3) and (4) simultaneously 4 z 2 + z 2 + 400 = 625 z 2 = 45 z = 6.708 in. From Eq. (3) x = 2 z = 2(6.708 in.) = 13.416 in. x = 13.42 in., z = 6.71 in. PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 136 ! PROBLEM 2.127 The direction of the 75-lb forces may vary, but the angle between the forces is always 50°. Determine the value of for which the resultant of the forces acting at A is directed horizontally to the left. SOLUTION We must first replace the two 75-lb forces by their resultant R1 using the triangle rule. R1 = 2(75 lb) cos 25° = 135.946 lb R1 = 135.946 lb α + 25° Next we consider the resultant R 2 of R1 and the 240-lb force where R 2 must be horizontal and directed to the left. Using the triangle rule and law of sines, sin (α + 25°) sin (30°) = 240 lb 135.946 sin (α + 25°) = 0.88270 α + 25° = 61.970° α = 36.970° α = 37.0° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 137 ! PROBLEM 2.128 A stake is being pulled out of the ground by means of two ropes as shown. Knowing the magnitude and direction of the force exerted on one rope, determine the magnitude and direction of the force P that should be exerted on the other rope if the resultant of these two forces is to be a 40-lb vertical force. SOLUTION Triangle rule: Law of cosines: Law of sines: P 2 = (30) 2 + (40) 2 − 2(30)(40) cos 25° P = 18.0239 lb sin α sin 25° = 30 lb 18.0239 lb α = 44.703° 90° − α = 45.297° P = 18.02 lb 45.3° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 138 ! PROBLEM 2.129 Member BD exerts on member ABC a force P directed along line BD. Knowing that P must have a 240-lb vertical component, determine (a) the magnitude of the force P, (b) its horizontal component. SOLUTION (a) P= (b) Px = Py sin 35° Py tan 40° = 240 lb sin 40° or P = 373 lb = 240 lb tan 40° or Px = 286 lb PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 139 ! PROBLEM 2.130 Two cables are tied together at C and loaded as shown. Determine the tension (a) in cable AC, (b) in cable BC. SOLUTION Free Body Diagram at C: ΣFx = 0: − 12 ft 7.5 ft TAC + TBC = 0 12.5 ft 8.5 ft TBC = 1.08800TAC ΣFy = 0: (a) 3.5 ft 4 ft TAC + TBC − 396 lb = 0 12 ft 8.5 ft 3.5 ft 4 ft (1.08800TAC ) − 396 lb = 0 TAC + 12.5 ft 8.5 ft (0.28000 + 0.51200)TAC = 396 lb TAC = 500.0 lb (b) TBC = (1.08800)(500.0 lb) TAC = 500 lb TBC = 544 lb PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 140 ! PROBLEM 2.131 Two cables are tied together at C and loaded as shown. Knowing that P = 360 N, determine the tension (a) in cable AC, (b) in cable BC. SOLUTION Free Body: C (a) ΣFx = 0: − (b) ΣFy = 0: 12 4 TAC + (360 N) = 0 13 5 TAC = 312 N 5 3 (312 N) + TBC + (360 N) − 480 N = 0 13 5 TBC = 480 N − 120 N − 216 N TBC = 144 N PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 141 ! PROBLEM 2.132 Two cables are tied together at C and loaded as shown. Determine the range of values of P for which both cables remain taut. SOLUTION Free Body: C ΣFx = 0: − 12 4 TAC + P = 0 13 5 TAC = ΣFy = 0: Substitute for TAC from (1): 13 P 15 (1) 5 3 TAC + TBC + P − 480 N = 0 13 5 3 ! 5 "! 13 " # 13 $# 15 $ P + TBC + 5 P − 480 N = 0 % &% & TBC = 480 N − 14 P 15 (2) From (1), TAC . 0 requires P . 0. From (2), TBC . 0 requires 14 P , 480 N, P , 514.29 N 15 0 , P , 514 N Allowable range: PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 142 ! PROBLEM 2.133 A force acts at the origin of a coordinate system in a direction defined by the angles θ x = 69.3° and θ z = 57.9°. Knowing that the y component of the force is –174.0 lb, determine (a) the angle θ y , (b) the other components and the magnitude of the force. SOLUTION (a) To determine θ y , use the relation cos 2 θ y + cos 2 θ y + cos 2 θ z = 1 or cos 2 θ y = 1 − cos 2 θ x − cos 2 θ y Since Fy , 0, we must have cos θ y , 0 cos θ y = − 1 − cos 2 69.3° − cos 2 57.9° θ y = 140.3° = − 0.76985 (b) Fy −174.0 lb = 226.02 lb −0.76985 F = 226 lb Fx = F cos θ x = (226.02 lb) cos 69.3° Fx = 79.9 lb Fz = F cos θ z = (226.02 lb) cos 57.9° Fz = 120.1 lb F= cos θ y = PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 143 ! PROBLEM 2.134 Cable AB is 65 ft long, and the tension in that cable is 3900 lb. Determine (a) the x, y, and z components of the force exerted by the cable on the anchor B, (b) the angles θ x , θ y , and θ z defining the direction of that force. SOLUTION 56 ft 65 ft = 0.86154 cos θ y = From triangle AOB: θ y = 30.51° Fx = − F sin θ y cos 20° (a) = −(3900 lb)sin 30.51° cos 20° Fx = −1861 lb Fy = + F cos θ y = (3900 lb)(0.86154) ! Fz = + (3900 lb)sin 30.51° sin 20° (b) cos θ x = From above: Fx 1861 lb =− = − 0.4771 3900 lb F θ y = 30.51° cos θ z = Fz 677 lb =+ = + 0.1736 3900 lb F Fy = +3360 lb Fz = +677 lb θ x = 118.5° θ y = 30.5° θ z = 80.0° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 144 ! PROBLEM 2.135 In order to move a wrecked truck, two cables are attached at A and pulled by winches B and C as shown. Knowing that the tension is 10 kN in cable AB and 7.5 kN in cable AC, determine the magnitude and direction of the resultant of the forces exerted at A by the two cables. SOLUTION !!!" AB = −15.588i + 15 j + 12k AB = 24.739 m !!!" AC = −15.588i + 18.60 j − 15k AC = 28.530 m !!!" AB TAB = TAB !AB = TAB AB −15.588i + 15 j + 12k TAB = (10 kN) 24.739 TAB = (6.301 kN)i + (6.063 kN) j + (4.851 kN)k !!!" −15.588i + 18.60 j − 15k AC TAC = TAC !AC = TAC (7.5 kN) 28.530 AC TAC = −(4.098 kN)i + (4.890 kN) j − (3.943 kN)k R = TAB + TAC = −(10.399 kN)i + (10.953 kN) j + (0.908 kN)k R = (10.399)2 + (10.953)2 + (0.908) 2 R = 15.13 kN = 15.130 kN cos θ x = cos θ y = cos θ z = Rx −10.399 kN = = − 0.6873 15.130 kN R Ry θ z = 133.4° 10.953 kN = 0.7239 15.130 kN θ y = 43.6° Rz 0.908 kN = = 0.0600 R 15.130 kN θ z = 86.6° R = PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 145 ! PROBLEM 2.136 A container of weight W = 1165 N is supported by three cables as shown. Determine the tension in each cable. SOLUTION Free Body: A We have: ΣF = 0: TAB + TAC + TAD + W = 0 !!!" AB = (450 mm)i + (600 mm) j AB = 750 mm !!!" AC = (600 mm) j − (320 mm)k AC = 680 mm !!!" AD = (500 mm)i + (600 mm) j + (360 mm)k AD = 860 mm !!!" 600 " AB ! 450 =# TAB = TAB λ AB = TAB i+ j TAB AB % 750 750 $& = (0.6i + 0.8 j)TAB !!!" 320 " 8 " AC ! 600 ! 15 TAC = TAC λ AC = TAC =# j− k $ TAC = # j − k $ TAC 680 & AC % 680 % 17 17 & !!!" 600 360 " AD ! 500 = #− TAD = TAD λ AD = TAD i+ j+ k TAD AD % 860 860 860 $& 30 18 " ! 25 TAD = # − i + j + k $ TAD 43 43 & % 43 Substitute into ΣF = 0, factor i, j, k, and set their coefficient equal to zero: 0.6TAB − 0.8TAB + − 25 TAD = 0 43 TAB = 0.96899TAD 15 30 TAC + TAD − 1165 N = 0 17 43 8 18 TAC + TAD = 0 17 43 TAC = 0.88953TAD (1) (2) (3) PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 146 PROBLEM 2.136 (Continued) Substitute for TAB and TAC from (1) and (3) into (2): 15 30 " ! # 0.8 × 0.96899 + 17 × 0.88953 + 53 $ TAD − 1165 N = 0 % & 2.2578TAD − 1165 N = 0 TAD = 516 N From (1): TAB = 0.96899(516 N) TAB = 500 N From (3): TAC = 0.88953(516 N) TAC = 459 N PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 147 ! PROBLEM 2.137 Collars A and B are connected by a 525-mm-long wire and can slide freely on frictionless rods. If a force P = (341 N)j is applied to collar A, determine (a) the tension in the wire when y = 155 mm, (b) the magnitude of the force Q required to maintain the equilibrium of the system. SOLUTION For both Problems 2.137 and 2.138: Free Body Diagrams of Collars: ( AB) 2 = x 2 + y 2 + z 2 Here (0.525 m) 2 = (.20 m) 2 + y 2 + z 2 y 2 + z 2 = 0.23563 m 2 or Thus, why y given, z is determined, Now !AB !!!" AB = AB 1 (0.20i − yj + zk )m 0.525 m = 0.38095i − 1.90476 yj + 1.90476 zk = Where y and z are in units of meters, m. From the F.B. Diagram of collar A: ΣF = 0: N x i + N z k + Pj + TAB λ AB = 0 Setting the j coefficient to zero gives: P − (1.90476 y )TAB = 0 P = 341 N With TAB = 341 N 1.90476 y Now, from the free body diagram of collar B: ΣF = 0: N x i + N y j + Qk − TAB !AB = 0 Setting the k coefficient to zero gives: Q − TAB (1.90476 z ) = 0 And using the above result for TAB we have Q = TAB z = 341 N (341 N)( z ) (1.90476 z ) = y (1.90476) y PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 148 PROBLEM 2.137 (Continued) Then, from the specifications of the problem, y = 155 mm = 0.155 m z 2 = 0.23563 m 2 − (0.155 m) 2 z = 0.46 m and 341 N 0.155(1.90476) = 1155.00 N TAB = (a) TAB = 1.155 kN or and 341 N(0.46 m)(0.866) (0.155 m) = (1012.00 N) Q= (b) Q = 1.012 kN or PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 149 ! PROBLEM 2.138 Solve Problem 2.137 assuming that y = 275 mm. PROBLEM 2.137 Collars A and B are connected by a 525-mm-long wire and can slide freely on frictionless rods. If a force P = (341 N)j is applied to collar A, determine (a) the tension in the wire when y = 155 mm, (b) the magnitude of the force Q required to maintain the equilibrium of the system. SOLUTION From the analysis of Problem 2.137, particularly the results: y 2 + z 2 = 0.23563 m 2 341 N TAB = 1.90476 y 341 N Q= z y With y = 275 mm = 0.275 m, we obtain: z 2 = 0.23563 m 2 − (0.275 m) 2 z = 0.40 m and TAB = (a) 341 N = 651.00 (1.90476)(0.275 m) TAB = 651 N or and Q= (b) 341 N(0.40 m) (0.275 m) Q = 496 N or PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 150 ! CHAPTER 3 PROBLEM 3.1 A foot valve for a pneumatic system is hinged at B. Knowing that a = 28°, determine the moment of the 16-N force about Point B by resolving the force into horizontal and vertical components. SOLUTION Note that and θ = α − 20° = 28° − 20° = 8° Fx = (16 N) cos8° = 15.8443 N Fy = (16 N) sin 8° = 2.2268 N Also x = (0.17 m) cos 20° = 0.159748 m y = (0.17 m) sin 20° = 0.058143 m Noting that the direction of the moment of each force component about B is counterclockwise, MB = xFy + yFx = (0.159748 m)(2.2268 N) + (0.058143 m)(15.8443 N) = 1.277 N ⋅ m or M B = 1.277 N ⋅ m PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 153 PROBLEM 3.2 A foot valve for a pneumatic system is hinged at B. Knowing that a = 28°, determine the moment of the 16-N force about Point B by resolving the force into components along ABC and in a direction perpendicular to ABC. SOLUTION First resolve the 4-lb force into components P and Q, where Q = (16 N) sin 28° = 7.5115 N Then M B = rA/B Q = (0.17 m)(7.5115 N) or M B = 1.277 N ⋅ m = 1.277 N ⋅ m PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 154 PROBLEM 3.3 A 300-N force is applied at A as shown. Determine (a) the moment of the 300-N force about D, (b) the smallest force applied at B that creates the same moment about D. SOLUTION (a) Fx = (300 N) cos 25° = 271.89 N Fy = (300 N) sin 25° = 126.785 N F = (271.89 N)i + (126.785 N) j ! r = DA = −(0.1 m)i − (0.2 m) j MD = r × F M D = [−(0.1 m)i − (0.2 m) j] × [(271.89 N)i + (126.785 N) j] = −(12.6785 N ⋅ m)k + (54.378 N ⋅ m)k = (41.700 N ⋅ m)k M D = 41.7 N ⋅ m (b) The smallest force Q at B must be perpendicular to ! DB at 45° ! M D = Q ( DB ) 41.700 N ⋅ m = Q (0.28284 m) Q = 147.4 N 45° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 155 PROBLEM 3.4 A 300-N force is applied at A as shown. Determine (a) the moment of the 300-N force about D, (b) the magnitude and sense of the horizontal force applied at C that creates the same moment about D, (c) the smallest force applied at C that creates the same moment about D. SOLUTION (a) See Problem 3.3 for the figure and analysis leading to the determination of MD M D = 41.7 N ⋅ m (b) Since C is horizontal C = C i ! r = DC = (0.2 m)i − (0.125 m) j M D = r × C i = C (0.125 m)k 41.7 N ⋅ m = (0.125 m)(C ) C = 333.60 N (c) C = 334 N The smallest force C must be perpendicular to DC; thus, it forms α with the vertical 0.125 m 0.2 m α = 32.0° tan α = M D = C ( DC ); DC = (0.2 m) 2 + (0.125 m) 2 = 0.23585 m 41.70 N ⋅ m = C (0.23585 m) C = 176.8 N 58.0° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 156 PROBLEM 3.5 An 8-lb force P is applied to a shift lever. Determine the moment of P about B when a is equal to 25°. SOLUTION First note Px = (8 lb) cos 25° = 7.2505 lb Py = (8 lb)sin 25° = 3.3809 lb Noting that the direction of the moment of each force component about B is clockwise, have M B = − xPy − yPx = −(8 in.)(3.3809 lb) − (22 in.)(7.2505 lb) = −186.6 lb ⋅ in. or M B = 186.6 lb ⋅ in. PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 157 PROBLEM 3.6 For the shift lever shown, determine the magnitude and the direction of the smallest force P that has a 210-lb ⋅ in. clockwise moment about B. SOLUTION For P to be minimum, it must be perpendicular to the line joining Points A and B. Thus, α =θ 8 in. 22 in. = 19.98° = tan −1 and Where M B = dPmin d = rA/B = (8 in.) 2 + (22 in.)2 = 23.409 in. Then 210 lb ⋅ in. 23.409 in. = 8.97 lb Pmin = Pmin = 8.97 lb 19.98° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 158 PROBLEM 3.7 An 11-lb force P is applied to a shift lever. The moment of P about B is clockwise and has a magnitude of 250 lb ⋅ in. Determine the value of a. SOLUTION By definition M B = rA/B P sin θ where θ = α + (90° − φ ) and φ = tan −1 also 8 in. = 19.9831° 22 in. rA/B = (8 in.) 2 + (22 in.) 2 = 23.409 in. Then or 250 lb ⋅ in = (23.409 in.)(11 lb) x sin(α + 90° − 19.9831°) sin (α + 70.0169°) = 0.97088 or α + 70.0169° = 76.1391° and α + 70.0169° = 103.861° α = 6.12° 33.8° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 159 PROBLEM 3.8 It is known that a vertical force of 200 lb is required to remove the nail at C from the board. As the nail first starts moving, determine (a) the moment about B of the force exerted on the nail, (b) the magnitude of the force P that creates the same moment about B if a = 10°, (c) the smallest force P that creates the same moment about B. SOLUTION (a) M B = rC/B FN We have = (4 in.)(200 lb) = 800 lb ⋅ in. or MB = 800 lb ⋅ in. (b) By definition M B = rA/B P sin θ θ = 10° + (180° − 70°) = 120° Then 800 lb ⋅ in. = (18 in.) × P sin120° or P = 51.3 lb (c) For P to be minimum, it must be perpendicular to the line joining Points A and B. Thus, P must be directed as shown. Thus M B = dPmin d = rA/B or or 800 lb ⋅ in. = (18 in.)Pmin Pmin = 44.4 lb Pmin = 44.4 lb 20° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 160 PROBLEM 3.9 A winch puller AB is used to straighten a fence post. Knowing that the tension in cable BC is 1040 N and length d is 1.90 m, determine the moment about D of the force exerted by the cable at C by resolving that force into horizontal and vertical components applied (a) at Point C, (b) at Point E. SOLUTION (a) Slope of line EC = 12 (TAB ) 13 12 = (1040 N) 13 = 960 N Then TABx = and TABy = Then 0.875 m 5 = 1.90 m + 0.2 m 12 5 (1040 N) 13 = 400 N M D = TABx (0.875 m) − TABy (0.2 m) = (960 N)(0.875 m) − (400 N)(0.2 m) = 760 N ⋅ m (b) We have or M D = 760 N ⋅ m M D = TABx ( y ) + TABx ( x) = (960 N)(0) + (400 N)(1.90 m) = 760 N ⋅ m or M D = 760 N ⋅ m PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 161 ! PROBLEM 3.10 It is known that a force with a moment of 960 N ⋅ m about D is required to straighten the fence post CD. If d = 2.80 m, determine the tension that must be developed in the cable of winch puller AB to create the required moment about Point D. SOLUTION Slope of line EC = 0.875 m 7 = 2.80 m + 0.2 m 24 Then TABx = 24 TAB 25 and TABy = 7 TAB 25 We have M D = TABx ( y ) + TABy ( x) 24 7 TAB (0) + TAB (2.80 m) 25 25 = 1224 N 960 N ⋅ m = TAB or TAB = 1224 N PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 162 PROBLEM 3.11 It is known that a force with a moment of 960 N ⋅ m about D is required to straighten the fence post CD. If the capacity of winch puller AB is 2400 N, determine the minimum value of distance d to create the specified moment about Point D. SOLUTION The minimum value of d can be found based on the equation relating the moment of the force TAB about D: M D = (TAB max ) y (d ) where M D = 960 N ⋅ m (TAB max ) y = TAB max sin θ = (2400 N) sin θ Now sin θ = 0.875m (d + 0.20) 2 + (0.875) 2 m 0.875 960 N ⋅ m = 2400 N " " (d + 0.20)2 + (0.875) 2 $ or (d + 0.20) 2 + (0.875) 2 = 2.1875d or (d + 0.20) 2 + (0.875) 2 = 4.7852d 2 or ! # (d ) # % 3.7852d 2 − 0.40d − .8056 = 0 Using the quadratic equation, the minimum values of d are 0.51719 m and − .41151 m. Since only the positive value applies here, d = 0.51719 m or d = 517 mm PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 163 PROBLEM 3.12 The tailgate of a car is supported by the hydraulic lift BC. If the lift exerts a 125-lb force directed along its centerline on the ball and socket at B, determine the moment of the force about A. SOLUTION First note dCB = (12.0 in.) 2 + (2.33 in.) 2 = 12.2241 in. Then and 12.0 in. 12.2241 in. 2.33 in. sin θ = 12.2241 in. cos θ = FCB = FCB cos θ i − FCB sin θ j = 125 lb [(12.0 in.) i − (2.33 in.) j] 12.2241 in. Now M A = rB/A × FCB where rB/A = (15.3 in.) i − (12.0 in. + 2.33 in.) j = (15.3 in.) i − (14.33 in.) j Then M A = [(15.3 in.)i − (14.33 in.) j] × 125 lb (12.0i − 2.33j) 12.2241 in. = (1393.87 lb ⋅ in.)k = (116.156 lb ⋅ ft)k or M A = 116.2 lb ⋅ ft PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 164 PROBLEM 3.13 The tailgate of a car is supported by the hydraulic lift BC. If the lift exerts a 125-lb force directed along its centerline on the ball and socket at B, determine the moment of the force about A. SOLUTION First note dCB = (17.2 in.) 2 + (7.62 in.) 2 = 18.8123 in. Then and 17.2 in. 18.8123 in. 7.62 in. sin θ = 18.8123 in. cos θ = FCB = ( FCB cos θ )i − ( FCB sin θ ) j = 125 lb [(17.2 in.)i + (7.62 in.) j] 18.8123 in. Now M A = rB/A × FCB where rB/A = (20.5 in.)i − (4.38 in.) j Then MA = [(20.5 in.)i − (4.38 in.) j] × = (1538.53 lb ⋅ in.)k = (128.2 lb ⋅ ft)k 125 lb (17.2i − 7.62 j) 18.8123 in. or M A = 128.2 lb ⋅ ft PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 165 PROBLEM 3.14 A mechanic uses a piece of pipe AB as a lever when tightening an alternator belt. When he pushes down at A, a force of 485 N is exerted on the alternator at B. Determine the moment of that force about bolt C if its line of action passes through O. SOLUTION We have M C = rB/C × FB Noting the direction of the moment of each force component about C is clockwise. M C = xFBy + yFBx Where and x = 120 mm − 65 mm = 55 mm y = 72 mm + 90 mm = 162 mm FBx = FBy = 65 2 (65) + (72) 2 72 2 (65) + (72) 2 (485 N) = 325 N (485 N) = 360 N M C = (55 mm)(360 N) + (162)(325 N) = 72450 N ⋅ mm = 72.450 N ⋅ m or M C = 72.5 N ⋅ m PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 166 PROBLEM 3.15 Form the vector products B × C and B′ × C, where B = B′, and use the results obtained to prove the identity sin a cos β = 1 1 sin ( a + β ) + sin ( a − β ). 2 2 SOLUTION Note: B = B(cos β i + sin β j) B′ = B(cos β i − sin β j) C = C (cos α i + sin α j) By definition: Now | B × C | = BC sin (α − β ) (1) | B′ × C | = BC sin (α + β ) (2) B × C = B(cos β i + sin β j) × C (cos α i + sin α j) = BC (cos β sin α − sin β cos α )k and (3) B′ × C = B(cos β i − sin β j) × C (cos α i + sin α j) = BC (cos β sin α + sin β cos α ) k (4) Equating the magnitudes of B × C from Equations (1) and (3) yields: BC sin(α − β ) = BC (cos β sin α − sin β cos α ) (5) Similarly, equating the magnitudes of B′ × C from Equations (2) and (4) yields: BC sin(α + β ) = BC (cos β sin α + sin β cos α ) (6) Adding Equations (5) and (6) gives: sin(α − β ) + sin(α + β ) = 2cos β sin α or sin α cos β = 1 1 sin(α + β ) + sin(α − β ) 2 2 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 167 PROBLEM 3.16 A line passes through the Points (20 m, 16 m) and (−1 m, −4 m). Determine the perpendicular distance d from the line to the origin O of the system of coordinates. SOLUTION d AB = [20 m − ( −1 m)]2 + [16 m − ( −4 m)]2 = 29 m Assume that a force F, or magnitude F(N), acts at Point A and is directed from A to B. Then, Where F = F λ AB λ AB = = By definition Where Then rB − rA d AB 1 (21i + 20 j) 29 M O = | rA × F | = dF rA = −(1 m)i − (4 m) j M O = [ −(−1 m)i − (4 m) j] × F [(21 m)i + (20 m) j] 29 m F [−(20)k + (84)k ] 29 & 64 ' = ( F )k N ⋅ m * 29 + = Finally & 64 ' ( 29 F ) = d ( F ) * + 64 d= m 29 d = 2.21 m PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 168 PROBLEM 3.17 The vectors P and Q are two adjacent sides of a parallelogram. Determine the area of the parallelogram when (a) P = −7i + 3j − 3k and Q = 2i + 2j + 5k, (b) P = 6i − 5j − 2k and Q = −2i + 5j − k. SOLUTION (a) We have A = |P × Q| where P = −7i + 3j − 3k Q = 2i + 2 j + 5k Then i j k P × Q = −7 3 −3 2 2 5 = [(15 + 6)i + ( −6 + 35) j + ( −14 − 6)k ] = (21)i + (29) j(−20)k A = (20) 2 + (29) 2 + (−20) 2 (b) We have A = |P × Q| where P = 6i − 5 j − 2k or A = 41.0 Q = −2i + 5 in. j − 1k Then i j k P × Q = 6 −5 −2 −2 5 −1 = [(5 + 10)i + (4 + 6) j + (30 − 10)k ] = (15)i + (10) j + (20)k A = (15) 2 + (10) 2 + (20) 2 or A = 26.9 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 169 PROBLEM 3.18 A plane contains the vectors A and B. Determine the unit vector normal to the plane when A and B are equal to, respectively, (a) i + 2j − 5k and 4i − 7j − 5k, (b) 3i − 3j + 2k and −2i + 6j − 4k. SOLUTION (a) We have where = A×B |A × B| A = 1i + 2 j − 5k B = 4i − 7 j − 5k Then i j k A × B = 1 +2 −5 4 −7 −5 = (−10 − 35)i + (20 + 5) j + (−7 − 8)k = 15(3i − 1j − 1k ) and |A × B | = 15 (−3)2 + (−1) 2 + (−1)2 = 15 11 = (b) We have where = 15(−3i − 1j − 1k ) 15 11 = or 1 11 (−3i − j − k ) A×B |A × B| A = 3i − 3 j + 2k B = −2i + 6 j − 4k Then i j k A × B = 3 −3 2 −2 6 −4 = (12 − 12)i + (−4 + 12) j + (18 − 6)k = (8 j + 12k ) and |A × B| = 4 (2) 2 + (3) 2 = 4 13 = 4(2 j + 3k ) 4 13 or = 1 13 (2 j + 3k ) PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 170 PROBLEM 3.19 Determine the moment about the origin O of the force F = 4i + 5j − 3k that acts at a Point A. Assume that the position vector of A is (a) r = 2i − 3j + 4k, (b) r = 2i + 2.5j − 1.5k, (c) r = 2i + 5j + 6k. SOLUTION (a) i j k M O = 2 −3 4 4 5 −3 = (9 − 20)i + (16 + 6) j + (10 + 12)k (b) i j k M O = 2 2.5 −1.5 4 5 −3 = (−7.5 + 7.5)i + (−6 + 6) j + (10 − 10)k (c) M O = −11i + 22 j + 22k MO = 0 i j k MO = 2 5 6 4 5 −3 = (−15 − 30)i + (24 + 6) j + (10 − 20)k M O = −45i + 30 j − 10k Note: The answer to Part b could have been anticipated since the elements of the last two rows of the determinant are proportional. PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 171 PROBLEM 3.20 Determine the moment about the origin O of the force F = −2i + 3j + 5k that acts at a Point A. Assume that the position vector of A is (a) r = i + j + k, (b) r = 2i + 3j − 5k, (c) r = −4i + 6j + 10k. SOLUTION (a) i j k MO = 1 1 1 −2 3 5 = (5 − 3)i + (−2 − 5) j + (3 + 2)k (b) i j k MO = 2 3 − 5 −2 3 5 = (15 + 15)i + (10 − 10) j + (6 + 6)k (c) M O = 2i − 7 j + 5k M O = 30i + 12k i j k M O = −4 6 10 −2 3 5 = (30 − 30)i + ( −20 + 20) j + (−12 + 12)k MO = 0 Note: The answer to Part c could have been anticipated since the elements of the last two rows of the determinant are proportional. PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 172 PROBLEM 3.21 A 200-N force is applied as shown to the bracket ABC. Determine the moment of the force about A. SOLUTION We have M A = rC/A × FC where rC/A = (0.06 m)i + (0.075 m) j FC = −(200 N) cos 30° j + (200 N)sin 30°k Then i j k M A = 200 0.06 0.075 0 0 − cos 30° sin 30° = 200[(0.075sin 30°)i − (0.06sin 30°) j − (0.06 cos 30°)k ] or M A = (7.50 N ⋅ m)i − (6.00 N ⋅ m) j − (10.39 N ⋅ m)k PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 173 PROBLEM 3.22 Before the trunk of a large tree is felled, cables AB and BC are attached as shown. Knowing that the tensions in cables AB and BC are 555 N and 660 N, respectively, determine the moment about O of the resultant force exerted on the tree by the cables at B. SOLUTION We have M O = rB/O × FB where rB/O = (7 m) j FB = TAB + TBC TAB = = TBC = = BATAB −(0.75 m)i − (7 m) j + (6 m)k (.75) 2 + (7) 2 + (6) 2 m (555 N) BC TBC (4.25 m)i − (7 m) j + (1 m)k (4.25) 2 + (7) 2 + (1) 2 m (660 N) FB = [−(45.00 N)i − (420.0 N) j + (360.0 N)k ] + [(340.0 N)i − (560.0 N) j + (80.00 N)k ] = (295.0 N)i − (980.0 N) j + (440.0 N)k and i j k MO = 0 7 0 N⋅m 295 980 440 = (3080 N ⋅ m)i − (2070 N ⋅ m)k or M O = (3080 N ⋅ m)i − (2070 N ⋅ m)k PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 174 PROBLEM 3.23 The 6-m boom AB has a fixed end A. A steel cable is stretched from the free end B of the boom to a Point C located on the vertical wall. If the tension in the cable is 2.5 kN, determine the moment about A of the force exerted by the cable at B. SOLUTION First note d BC = (−6)2 + (2.4) 2 + (−4) 2 = 7.6 m 2.5 kN (−6i + 2.4 j − 4k ) 7.6 Then TBC = We have M A = rB/A × TBC where rB/A = (6 m)i Then M A = (6 m)i × 2.5 kN (−6i + 2.4 j − 4k ) 7.6 or M A = (7.89 kN ⋅ m) j + (4.74 kN ⋅ m)k PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 175 PROBLEM 3.24 A wooden board AB, which is used as a temporary prop to support a small roof, exerts at Point A of the roof a 57-lb force directed along BA. Determine the moment about C of that force. SOLUTION We have M C = rA/C × FBA where rA/C = (48 in.)i − (6 in.) j + (36 in.)k and FBA = BA FBA −(5 in.)i + (90 in.) j − (30 in.)k ! # (57 lb) =" 2 2 2 " # (5) (90) (30) in. + + $ % = −(3 lb)i + (54 lb) j − (18 lb)k i j k M C = 48 6 36 lb ⋅ in. 3 54 18 = −(1836 lb ⋅ in.)i + (756 lb ⋅ in.) j + (2574 lb ⋅ in.) or M C = −(153.0 lb ⋅ ft)i + (63.0 lb ⋅ ft) j + (215 lb ⋅ ft)k PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 176 PROBLEM 3.25 The ramp ABCD is supported by cables at corners C and D. The tension in each of the cables is 810 N. Determine the moment about A of the force exerted by (a) the cable at D, (b) the cable at C. SOLUTION (a) We have M A = rE/A × TDE where rE/A = (2.3 m) j TDE = = DE TDE (0.6 m)i + (3.3 m) j − (3 m)k (0.6) 2 + (3.3)2 + (3)2 m (810 N) = (108 N)i + (594 N) j − (540 N)k i j k MA = 0 2.3 0 N⋅m 108 594 −540 = −(1242 N ⋅ m)i − (248.4 N ⋅ m)k or M A = −(1242 N ⋅ m)i − (248 N ⋅ m)k (b) We have M A = rG/A × TCG where rG/A = (2.7 m)i + (2.3 m) j TCG = = CG TCG −(.6 m)i + (3.3 m) j − (3 m)k (.6) 2 + (3.3) 2 + (3) 2 m (810 N) = −(108 N)i + (594 N) j − (540 N)k i j k M A = 2.7 2.3 0 N⋅m −108 594 −540 = −(1242 N ⋅ m)i + (1458 N ⋅ m) j + (1852 N ⋅ m)k or M A = −(1242 N ⋅ m)i + (1458 N ⋅ m) j + (1852 N ⋅ m)k PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 177 PROBLEM 3.26 A small boat hangs from two davits, one of which is shown in the figure. The tension in line ABAD is 82 lb. Determine the moment about C of the resultant force RA exerted on the davit at A. SOLUTION We have R A = 2FAB + FAD where and FAB = −(82 lb) j ! AD 6i − 7.75 j − 3k FAD = FAD = (82 lb) AD 10.25 FAD = (48 lb)i − (62 lb) j − (24 lb)k Thus R A = 2FAB + FAD = (48 lb)i − (226 lb) j − (24 lb)k Also rA/C = (7.75 ft) j + (3 ft)k Using Eq. (3.21): i j k M C = 0 7.75 3 48 − 226 −24 = (492 lb ⋅ ft)i + (144 lb ⋅ ft) j − (372 lb ⋅ ft)k M C = (492 lb ⋅ ft)i + (144.0 lb ⋅ ft) j − (372 lb ⋅ ft)k PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 178 PROBLEM 3.27 In Problem 3.22, determine the perpendicular distance from Point O to cable AB. PROBLEM 3.22 Before the trunk of a large tree is felled, cables AB and BC are attached as shown. Knowing that the tensions in cables AB and BC are 555 N and 660 N, respectively, determine the moment about O of the resultant force exerted on the tree by the cables at B. SOLUTION We have where | M O | = TBA d d = perpendicular distance from O to line AB. Now M O = rB/O × TBA and rB/O = (7 m) j TBA = = BATAB −(0.75 m)i − (7 m) j + (6 m)k (0.75) 2 + (7) 2 + (6) 2 m (555 N) = −(45.0 N)i − (420 N) j + (360 N)k i j k MO = 0 7 0 N⋅m −45 −420 360 = (2520.0 N ⋅ m)i + (315.00 N ⋅ m)k and | M O | = (2520.0) 2 + (315.00) 2 = 2539.6 N ⋅ m 2539.6 N ⋅ m = (555 N)d or d = 4.5759 m or d = 4.58 m PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 179 PROBLEM 3.28 In Problem 3.22, determine the perpendicular distance from Point O to cable BC. PROBLEM 3.22 Before the trunk of a large tree is felled, cables AB and BC are attached as shown. Knowing that the tensions in cables AB and BC are 555 N and 660 N, respectively, determine the moment about O of the resultant force exerted on the tree by the cables at B. SOLUTION We have where | M O | = TBC d d = perpendicular distance from O to line BC. M O = rB/O × TBC rB/O = 7 mj TBC = = BC TBC (4.25 m)i − (7 m) j + (1 m)k (4.25)2 + (7) 2 + (1) 2 m (660 N) = (340 N)i − (560 N) j + (80 N)k i j k MO = 0 7 0 340 −560 80 = (560 N ⋅ m)i − (2380 N ⋅ m)k and | M O | = (560)2 + (2380) 2 = 2445.0 N ⋅ m 2445.0 N ⋅ m = (660 N)d d = 3.7045 m or d = 3.70 m PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 180 PROBLEM 3.29 In Problem 3.24, determine the perpendicular distance from Point D to a line drawn through Points A and B. PROBLEM 3.24 A wooden board AB, which is used as a temporary prop to support a small roof, exerts at Point A of the roof a 57-lb force directed along BA. Determine the moment about C of that force. SOLUTION We have where | M D | = FBA d d = perpendicular distance from D to line AB. M D = rA/D × FBA rA/D = −(6 in.) j + (36 in.)k FBA = = BA FBA (−(5 in.)i + (90 in.) j − (30 in.)k ) (5)2 + (90) 2 + (30) 2 in. (57 lb) = −(3 lb)i + (54 lb) j − (18 lb)k i j k M D = 0 −6 36 lb ⋅ in. −3 54 −18 = −(1836.00 lb ⋅ in.)i − (108.000 lb ⋅ in.) j − (18.0000 lb ⋅ in.)k and | M D | = (1836.00) 2 + (108.000)2 + (18.0000)2 = 1839.26 lb ⋅ in. 1839.26 lb ⋅ in = (57 lb)d d = 32.268 in. or d = 32.3 in. PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 181 PROBLEM 3.30 In Problem 3.24, determine the perpendicular distance from Point C to a line drawn through Points A and B. PROBLEM 3.24 A wooden board AB, which is used as a temporary prop to support a small roof, exerts at Point A of the roof a 57-lb force directed along BA. Determine the moment about C of that force. SOLUTION We have where | M C | = FBA d d = perpendicular distance from C to line AB. M C = rA/C × FBA rA/C = (48 in.)i − (6 in.) j + (36 in.)k FBA = = BA FBA ( −(5 in.)i + (90 in.) j − (30 in.)k ) (5) 2 + (90) 2 + (30) 2 in. (57 lb) = −(3 lb)i + (54 lb) j − (18 lb)k i j k M C = 48 −6 36 lb ⋅ in. −3 54 −18 = −(1836.00lb ⋅ in.)i − (756.00 lb ⋅ in.) j + (2574.0 lb ⋅ in.)k and | M C | = (1836.00) 2 + (756.00) 2 + (2574.0)2 = 3250.8 lb ⋅ in. 3250.8 lb ⋅ in. = 57 lb d = 57.032 in. or d = 57.0 in. PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 182 PROBLEM 3.31 In Problem 3.25, determine the perpendicular distance from Point A to portion DE of cable DEF. PROBLEM 3.25 The ramp ABCD is supported by cables at corners C and D. The tension in each of the cables is 810 N. Determine the moment about A of the force exerted by (a) the cable at D, (b) the cable at C. SOLUTION We have where | M A | = TDE d d = perpendicular distance from A to line DE. M A = rE/A × TDE rE/A = (2.3 m) j TDE = = DE TDE (0.6 m)i + (3.3 m) j − (3 m)k (0.6) 2 + (3.3) 2 + (3)2 m (810 N) = (108 N)i + (594 N) j − (540 N)k i j k MA = 0 2.3 0 N⋅m 108 594 540 = − (1242.00 N ⋅ m)i − (248.00 N ⋅ m)k and |M A | = (1242.00)2 + (248.00) 2 = 1266.52 N ⋅ m 1266.52 N ⋅ m = (810 N)d d = 1.56360 m or d = 1.564 m PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 183 PROBLEM 3.32 In Problem 3.25, determine the perpendicular distance from Point A to a line drawn through Points C and G. PROBLEM 3.25 The ramp ABCD is supported by cables at corners C and D. The tension in each of the cables is 810 N. Determine the moment about A of the force exerted by (a) the cable at D, (b) the cable at C. SOLUTION We have where |M A | = TCG d d = perpendicular distance from A to line CG. M A = rG/A × TCG rG/A = (2.7 m)i + (2.3 m) j TCG = = CG TCG −(0.6 m) i + (3.3 m) j − (3 m) k (0.6) 2 + (3.3) 2 + (3) 2 m (810 N) = −(108 N) i + (594 N) j − (540 N) k i j k M A = 2.7 2.3 0 N⋅m −108 594 −540 = −(1242.00 N ⋅ m)i + (1458.00 N ⋅ m) j + (1852.00 N ⋅ m)k and |M A | = (1242.00) 2 + (1458.00) 2 + (1852.00)2 = 2664.3 N ⋅ m 2664.3 N ⋅ m = (810 N)d d = 3.2893 m or d = 3.29 m PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 184 PROBLEM 3.33 In Problem 3.26, determine the perpendicular distance from Point C to portion AD of the line ABAD. PROBLEM 3.26 A small boat hangs from two davits, one of which is shown in the figure. The tension in line ABAD is 82 lb. Determine the moment about C of the resultant force RA exerted on the davit at A. SOLUTION First compute the moment about C of the force FDA exerted by the line on D: From Problem 3.26: FDA = − FAD = −(48 lb) i + (62 lb) j + (24 lb)k M C = rD/C × FDA = + (6 ft)i × [−(48 lb)i + (62 lb) j + (24 lb)k ] = −(144 lb ⋅ ft) j + (372 lb ⋅ ft)k M C = (144) 2 + (372)2 = 398.90 lb ⋅ ft Then M C = FDA d Since FDA = 82 lb d= = MC FDA 398.90 lb ⋅ ft 82 lb d = 4.86 ft PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 185 PROBLEM 3.34 Determine the value of a that minimizes the perpendicular distance from Point C to a section of pipeline that passes through Points A and B. SOLUTION Assuming a force F acts along AB, |M C | = |rA / C × F| = F ( d ) d = perpendicular distance from C to line AB Where F= = AB F (24 ft) i + (24 ft) j − (28) k (24) 2 + (24)2 + (18) 2 ft F F (6) i + (6) j − (7) k 11 = (3 ft)i − (10 ft) j − (a − 10 ft)k = rA/C i j k F M C = 3 −10 10a 11 6 6 −7 = [(10 + 6a )i + (81 − 6a) j + 78 k ] |M C | = |rA/C × F 2 | Since or F 11 |rA/C × F 2 | = ( dF ) 2 1 (10 + 6a) 2 + (81 − 6a) 2 + (78)2 = d 2 121 Setting d da (d 2 ) = 0 to find a to minimize d 1 [2(6)(10 + 6a) + 2(−6)(81 − 6a)] = 0 121 Solving a = 5.92 ft or a = 5.92 ft PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 186 PROBLEM 3.35 Given the vectors P = 3i − j + 2k , Q = 4i + 5 j − 3k , and S = −2i + 3j − k , compute the scalar products P ⋅ Q, P ⋅ S, and Q ⋅ S. SOLUTION P ⋅ Q = (3i − 1j + 2k ) ⋅ (4i − 5 j − 3k ) = (3)(4) + (−1)(−5) + (2)(−3) =1 or P ⋅Q =1 or P ⋅ S = −11 or Q ⋅ S = 10 ! P ⋅ S = (3i − 1j + 2k ) ⋅ (−2i + 3j − 1k ) = (3)(−2) + (−1)(3) + (2)( −1) = −11 Q ⋅ S = (4i − 5 j − 3k ) ⋅ ( −2i + 3j − 1k ) = (4)(−2) + (5)(3) + ( −3)(−1) = 10 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 187 PROBLEM 3.36 Form the scalar products B ⋅ C and B′ ⋅ C, where B = B ′, and use the results obtained to prove the identity cos a cos β = 1 1 cos (a + β ) + cos (a − β ). 2 2 SOLUTION By definition B ⋅ C = BC cos(α − β ) B = B [(cos β )i + (sin β ) j] C = C [(cos α )i + (sin α ) j] where (B cos β )( C cos α ) + ( B sin β )(C sin α ) = BC cos (α − β ) cos β cos α + sin β sin α = cos(α − β ) or (1) B′ ⋅ C = BC cos (α + β ) By definition B′ = [(cos β )i − (sin β ) j] where (B cos β ) (C cos α ) + (− B sin β )(C sin α ) = BC cos (α + β ) or cos β cos α − sin β sin α = cos (α + β ) (2) Adding Equations (1) and (2), 2 cos β cos α = cos (α − β ) + cos (α + β ) or cos α cos β = 1 1 cos (α + β ) + cos (α − β ) 2 2 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 188 PROBLEM 3.37 Section AB of a pipeline lies in the yz plane and forms an angle of 37° with the z axis. Branch lines CD and EF join AB as shown. Determine the angle formed by pipes AB and CD. SOLUTION First note ! AB = AB (sin 37° j − cos 37°k ) CD = CD(− cos 40° cos 55° j + sin 40° j − cos 40° sin 55°k ) Now ! ! AB ⋅ CD = ( AB)(CD ) cos θ or AB (sin 37° j − cos 37°k ) ⋅ CD (− cos 40° cos 55°i + sin 40° j − cos 40° sin 55°k ) = (AB)(CD) cos θ or cos θ = (sin 37°)(sin 40°) + (− cos 37°)(− cos 40° sin 55°) = 0.88799 or θ = 27.4° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 189 PROBLEM 3.38 Section AB of a pipeline lies in the yz plane and forms an angle of 37° with the z axis. Branch lines CD and EF join AB as shown. Determine the angle formed by pipes AB and EF. SOLUTION First note Now or ! AB = AB (sin 37° j − cos 37°k ) ! EF = EF (cos 32° cos 45°i + sin 32° j − cos 32° sin 45°k ) ! ! AB ⋅ EF = ( AB )( EF ) cos θ AB (sin 37° j − cos 37°k ) ⋅ EF (cos 32° cos 45° j + sin 32° j − cos 32° sin 45°k ) = ( AB )( EF ) cos θ or cos θ = (sin 37°)(sin 32°) + ( − cos 37°)( − cos 32° sin 45°) = 0.79782 or θ = 37.1° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 190 PROBLEM 3.39 Consider the volleyball net shown. Determine the angle formed by guy wires AB and AC. SOLUTION First note AB = (−6.5)2 + (−8)2 + (2) 2 = 10.5 ft AC = (0) 2 + (−8) 2 + (6)2 = 10 ft and By definition or or ! AB = −(6.5 ft)i − (8 ft) j + (2 ft)k ! AC = −(8 ft) j + (6 ft)k ! ! AB ⋅ AC = ( AB )( AC ) cos θ (−6.5i − 8 j + 2k ) ⋅ (−8 j + 6k ) = (10.5)(10) cos θ (−6.5)(0) + ( −8)( −8) + (2)(6) = 105cos θ cos θ = 0.72381 or θ = 43.6° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 191 PROBLEM 3.40 Consider the volleyball net shown. Determine the angle formed by guy wires AC and AD. SOLUTION First note AC = (0)2 + (−8) 2 + (6) 2 = 10 ft AD = (4) 2 + ( −8) 2 + (1) 2 and By definition or = 9 ft ! AC = −(8 ft)j + (6 ft)k ! AD = (4 ft) j − (8 ft) j + (1 ft)k ! ! AC ⋅ AD = ( AC )( AD ) cos θ (−8 j + 6k ) ⋅ (4i − 8 j + k ) = (10)(9) cos θ (0)(4) + ( −8)( −8) + (6)(1) = 90 cos θ or cos θ = 0.777 78 or θ = 38.9° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 192 PROBLEM 3.41 Knowing that the tension in cable AC is 1260 N, determine (a) the angle between cable AC and the boom AB, (b) the projection on AB of the force exerted by cable AC at Point A. SOLUTION (a) First note AC = (−2.4) 2 + (0.8)2 + (1.2) 2 = 2.8 m AB = (−2.4) 2 + (−1.8)2 + (0) 2 and By definition or (b) = 3.0 m ! AC = −(2.4 m)i + (0.8 m) j + (1.2 m)k ! AB = −(2.4 m)i − (1.8 m) j ! ! AC ⋅ AB = ( AC )( AB) cos θ (−2.4i + 0.8 j + 1.2k ) ⋅ (−2.4i − 1.8 j) = (2.8)(30) × cos θ or (−2.4)(−2.4) + (0.8)(−1.8) + (1.2)(0) = 8.4cos θ or cos θ = 0.514 29 We have or θ = 59.0° (TAC ) AB = TAC ⋅ λ AB = TAC cos θ = (1260 N)(0.51429) or (TAC ) AB = 648 N PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 193 PROBLEM 3.42 Knowing that the tension in cable AD is 405 N, determine (a) the angle between cable AD and the boom AB, (b) the projection on AB of the force exerted by cable AD at Point A. SOLUTION (a) First note AD = (−2.4) 2 + (1.2) 2 + (−2.4) 2 = 3.6 m AB = (−2.4) 2 + (−1.8) 2 + (0) 2 = 3.0 m and # = −(2.4 m)i + (1.2 m) j − (2.4 m)k AD AB = −(2.4 m)i − (1.8 m) j # ⋅ AB = ( AD )( AB) cos θ AD By definition, (−2.4i + 1.2 j − 2.4k ) ⋅ (−2.4i − 1.8 j) = (3.6)(3.0) cos θ (−2.4)(−2.4) + (1.2)(−1.8) + (−2.4)(0) = 10.8cos θ cos θ = (b) 1 3 θ = 70.5° (TAD ) AB = TAD ⋅ λ AB = TAD cos θ &1' = (405 N) ( ) * 3+ (TAD ) AB = 135.0 N PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 194 PROBLEM 3.43 Slider P can move along rod OA. An elastic cord PC is attached to the slider and to the vertical member BC. Knowing that the distance from O to P is 6 in. and that the tension in the cord is 3 lb, determine (a) the angle between the elastic cord and the rod OA, (b) the projection on OA of the force exerted by cord PC at Point P. SOLUTION First note OA = (12)2 + (12)2 + (−6) 2 = 18 in. OA 1 = (12i + 12 j − 6k ) OA 18 1 = (2i + 2 j − k ) 3 Then λ OA = Now 1 OP = 6 in. - OP = (OA) 3 The coordinates of Point P are (4 in., 4 in., −2 in.) ! PC = (5 in.)i + (11 in.) j + (14 in.)k so that PC = (5) 2 + (11) 2 + (14)2 = 342 in. and (a) We have or or ! PC ⋅ λ OA = ( PC ) cos θ 1 (5i + 11j + 14k ) ⋅ (2i + 2 j − k ) = 342 cos θ 3 1 [(5)(2) + (11)(2) + (14)( −1)] 3 342 = 0.324 44 cos θ = or θ = 71.1° (b) We have (TPC ) OA = TPC ⋅ λ OA = (TPC λ PC ) ⋅ λ OA PC ⋅ λ OA PC = TPC cos θ = TPC = (3 lb)(0.324 44) or (TPC )OA = 0.973 lb PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 195 PROBLEM 3.44 Slider P can move along rod OA. An elastic cord PC is attached to the slider and to the vertical member BC. Determine the distance from O to P for which cord PC and rod OA are perpendicular. SOLUTION First note OA = (12)2 + (12)2 + (−6)2 = 18 in. Then λ OA = OA 1 = (12i + 12 j − 6k ) OA 18 1 = (2i + 2 j − k ) 3 Let the coordinates of Point P be (x in., y in., z in.). Then ! PC = [(9 − x)in.]i + (15 − y )in.] j + [(12 − z )in.]k Also, and ! d OP = dOP λ OA = OP (2i + 2 j − k ) 3 ! OP = ( x in.)i + ( y in.) j + ( z in.)k 2 2 1 x = dOP y = dOP z = dOP 3 3 3 The requirement that OA and PC be perpendicular implies that ! λ OA ⋅ PC = 0 or 1 (2 j + 2 j − k ) ⋅ [(9 − x)i + (15 − y ) j + (12 − z )k ] = 0 3 or 2 2 & ' & ' & 1 '! (2) ( 9 − dOP ) + (2) (15 − dOP ) + (−1) "12 − ( − dOP ) # = 0 3 3 * + * + * 3 +% $ or dOP = 12.00 in. PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 196 PROBLEM 3.45 Determine the volume of the parallelepiped of Fig. 3.25 when (a) P = 4i − 3j + 2k, Q = −2i − 5j + k, and S = 7i + j − k, (b) P = 5i − j + 6k, Q = 2i + 3j + k, and S = −3i − 2j + 4k. SOLUTION Volume of a parallelepiped is found using the mixed triple product. (a) Vol = P ⋅ (Q × S) 4 −3 2 = −2 −5 1 in.3 7 1 −1 = (20 − 21 − 4 + 70 + 6 − 4) = 67 or Volume = 67.0 (b) Vol = P ⋅ (Q × S) 5 −1 6 = 2 3 1 in.3 −3 −2 4 = (60 + 3 − 24 + 54 + 8 + 10) = 111 or Volume = 111.0 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 197 PROBLEM 3.46 Given the vectors P = 4i − 2 j + 3k , Q = 2i + 4 j − 5k , and S = S x i − j + 2k , determine the value of S x for which the three vectors are coplanar. SOLUTION If P, Q, and S are coplanar, then P must be perpendicular to (Q × S). P ⋅ (Q × S) = 0 (or, the volume of a parallelepiped defined by P, Q, and S is zero). −2 3 4 −5 = 0 −1 2 Then 4 2 Sx or 32 + 10S x − 6 − 20 + 8 − 12S x = 0 Sx = 7 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 198 PROBLEM 3.47 The 0.61 × 1.00-m lid ABCD of a storage bin is hinged along side AB and is held open by looping cord DEC over a frictionless hook at E. If the tension in the cord is 66 N, determine the moment about each of the coordinate axes of the force exerted by the cord at D. SOLUTION First note z = (0.61) 2 − (0.11)2 = 0.60 m Then d DE = (0.3) 2 + (0.6) 2 + (−0.6) 2 = 0.9 m 66 N (0.3i + 0.6 j − 0.6k ) 0.9 = 22[(1 N)i + (2 N) j − (2 N)k ] and TDE = Now M A = rD/A × TDE where rD/A = (0.11 m) j + (0.60 m)k Then i j k M A = 22 0 0.11 0.60 −2 1 2 = 22[(−0.22 − 1.20)i + 0.60 j − 0.11k ] = − (31.24 N ⋅ m)i + (13.20 N ⋅ m) j − (2.42 N ⋅ m)k M x = −31.2 N ⋅ m, M y = 13.20 N ⋅ m, M z = −2.42 N ⋅ m PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 199 PROBLEM 3.48 The 0.61 × 1.00-m lid ABCD of a storage bin is hinged along side AB and is held open by looping cord DEC over a frictionless hook at E. If the tension in the cord is 66 N, determine the moment about each of the coordinate axes of the force exerted by the cord at C. SOLUTION First note z = (0.61) 2 − (0.11)2 = 0.60 m Then dCE = (−0.7) 2 + (0.6) 2 + (−0.6) 2 = 1.1 m 66 N (−0.7i + 0.6 j − 0.6k ) 1.1 = 6[−(7 N)i + (6 N) j − (6 N)k ] and TCE = Now M A = rE/A × TCE where rE/A = (0.3 m)i + (0.71 m) j Then i j k M A = 6 0.3 0.71 0 −7 −6 6 = 6[ −4.26i + 1.8 j + (1.8 + 4.97)k ] = − (25.56 N ⋅ m)i + (10.80 N ⋅ m) j + (40.62 N ⋅ m)k M x = −25.6 N ⋅ m, M y = 10.80 N ⋅ m, M z = 40.6 N ⋅ m PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 200 PROBLEM 3.49 To lift a heavy crate, a man uses a block and tackle attached to the bottom of an I-beam at hook B. Knowing that the moments about the y and the z axes of the force exerted at B by portion AB of the rope are, respectively, 120 N ⋅ m and −460 N ⋅ m, determine the distance a. SOLUTION First note ! BA = (2.2 m)i − (3.2 m) j − ( a m)k Now M D = rA/D × TBA where rA/D = (2.2 m)i + (1.6 m) j Then TBA = TBA (2.2i − 3.2 j − ak ) (N) d BA MD = i j k TBA 2.2 1.6 0 d BA 2.2 −3.2 − a = Thus TBA {−1.6a i + 2.2a j + [(2.2)(−3.2) − (1.6)(2.2)]k} d BA M y = 2.2 TBA a d BA M z = −10.56 Then forming the ratio TBA d BA (N ⋅ m) (N ⋅ m) My Mz T 2.2 dBA (N ⋅ m) 120 N ⋅ m BA = −460 N ⋅ m −10.56 TdBA (N ⋅ m) or a = 1.252 m BA PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 201 PROBLEM 3.50 To lift a heavy crate, a man uses a block and tackle attached to the bottom of an I-beam at hook B. Knowing that the man applies a 195-N force to end A of the rope and that the moment of that force about the y axis is 132 N ⋅ m, determine the distance a. SOLUTION d BA = (2.2) 2 + (−3.2) 2 + (−a ) 2 First note = 15.08 + a 2 m 195 N (2.2i − 3.2 j − a k ) d BA and TBA = Now M y = j ⋅ (rA/D × TBA ) where rA/0 = (2.2 m)i + (1.6 m) j Then 0 1 0 195 My = 2.2 1.6 0 d BA 2.2 −3.2 − a = 195 (2.2a ) (N ⋅ m) d BA Substituting for My and dBA 132 N ⋅ m = or 195 15.08 + a 2 (2.2a ) 0.30769 15.08 + a 2 = a Squaring both sides of the equation 0.094675(15.08 + a 2 ) = a 2 or a = 1.256 m PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 202 PROBLEM 3.51 A small boat hangs from two davits, one of which is shown in the figure. It is known that the moment about the z axis of the resultant force RA exerted on the davit at A must not exceed 279 lb ⋅ ft in absolute value. Determine the largest allowable tension in line ABAD when x = 6 ft. SOLUTION First note R A = 2TAB + TAD Also note that only TAD will contribute to the moment about the z axis. Now AD = (6) 2 + (−7.75) 2 + (−3) 2 = 10.25 ft ! AD =T AD T (6i − 7.75 j − 3k ) = 10.25 Then, TAD Now M z = k ⋅ (rA/C × TAD ) where rA/C = (7.75 ft) j + (3 ft)k Then for Tmax , 0 0 1 Tmax 279 = 0 7.75 3 10.25 6 −7.75 −3 = Tmax | − (1)(7.75)(6)| 10.25 or Tmax = 61.5 lb PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 203 PROBLEM 3.52 For the davit of Problem 3.51, determine the largest allowable distance x when the tension in line ABAD is 60 lb. SOLUTION From the solution of Problem 3.51, TAD is now TAD = T = AD AD 60 lb x 2 + ( −7.75) 2 + (−3)2 ( xi − 7.75 j − 3k ) Then M z = k ⋅ (rA / C × TAD ) becomes 279 = 279 = 60 x 2 + (−7.75) 2 + ( −3) 2 60 0 0 1 0 7.75 3 x −7.75 −3 | − (1)(7.75)( x) | 2 x + 69.0625 279 x 2 + 69.0625 = 465 x 0.6 x 2 + 69.0625 = x Squaring both sides: 0.36 x 2 + 24.8625 = x 2 x 2 = 38.848 x = 6.23 ft PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 204 PROBLEM 3.53 To loosen a frozen valve, a force F of magnitude 70 lb is applied to the handle of the valve. Knowing that θ = 25°, Mx = −61 lb ⋅ ft, and M z = −43 lb ⋅ ft, determine φ and d. SOLUTION We have ΣM O : rA/O × F = M O where rA/O = −(4 in.)i + (11 in.) j − (d )k F = F (cos θ cos φ i − sin θ j + cos θ sin φ k ) For F = 70 lb, θ = 25° F = (70 lb)[(0.90631cos φ )i − 0.42262 j + (0.90631sin φ )k ] i M O = (70 lb) −4 −0.90631cos φ j k 11 −d in. −0.42262 0.90631sin φ = (70 lb)[(9.9694sin φ − 0.42262d ) i + (−0.90631d cos φ + 3.6252sin φ ) j + (1.69048 − 9.9694cos φ )k ] in. and M x = (70 lb)(9.9694sin φ − 0.42262d )in. = −(61 lb ⋅ ft)(12 in./ft) (1) M y = (70 lb)(−0.90631d cos φ + 3.6252sin φ ) in. (2) M z = (70 lb)(1.69048 − 9.9694cos φ ) in. = −43 lb ⋅ ft(12 in./ft) (3) From Equation (3) & 634.33 ' φ = cos −1 ( ) = 24.636° * 697.86 + or From Equation (1) & 1022.90 ' d =( ) = 34.577 in. * 29.583 + or d = 34.6 in. φ = 24.6° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 205 PROBLEM 3.54 When a force F is applied to the handle of the valve shown, its moments about the x and z axes are, respectively, M x = −77 lb ⋅ ft and M z = −81 lb ⋅ ft. For d = 27 in., determine the moment My of F about the y axis. SOLUTION We have ΣM O : rA/O × F = M O Where rA/O = −(4 in.)i + (11 in.) j − (27 in.)k F = F (cos θ cos φ i − sin θ j + cos θ sin φ k ) MO = F i −4 cos θ cos φ j 11 − sin θ k −27 lb ⋅ in. cos θ sin φ = F [(11cos θ sin φ − 27 sin θ )i + (−27 cos θ cos φ + 4 cos θ sin φ ) j + (4sin θ − 11cos θ cos φ )k ](lb ⋅ in.) and M x = F (11cos θ sin φ − 27sin θ )(lb ⋅ in.) (1) M y = F (−27 cos θ cos φ + 4cos θ sin φ ) (lb ⋅ in.) (2) M z = F (4sin θ − 11cos θ cos φ ) (lb ⋅ in.) (3) Now, Equation (1) cos θ sin φ = 1 & Mx ' + 27sin θ ) 11 (* F + (4) and Equation (3) cos θ cos φ = M ' 1& 4sin θ − z ) ( 11 * F + (5) Substituting Equations (4) and (5) into Equation (2), .0 M '! 1& 1 &M ' ! /0 M y = F 1−27 " ( 4sin θ − z ) # + 4 " ( x + 27 sin θ ) # 2 F +% + % 04 03 $11 * $11 * F or My = 1 (27 M z + 4 M x ) 11 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 206 PROBLEM 3.54 (Continued) Noting that the ratios 27 and 114 are the ratios of lengths, have 11 27 4 (−81 lb ⋅ ft) + (−77 lb ⋅ ft) 11 11 = 226.82 lb ⋅ ft My = or M y = −227 lb ⋅ ft PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 207 PROBLEM 3.55 The frame ACD is hinged at A and D and is supported by a cable that passes through a ring at B and is attached to hooks at G and H. Knowing that the tension in the cable is 450 N, determine the moment about the diagonal AD of the force exerted on the frame by portion BH of the cable. SOLUTION M AD = λ AD ⋅ (rB/A × TBH ) Where and 1 λ AD = (4i − 3k ) 5 rB/A = (0.5 m)i d BH = (0.375)2 + (0.75) 2 + (−0.75) 2 = 1.125 m Then Finally 450 N (0.375i + 0.75 j − 0.75k ) 1.125 = (150 N)i + (300 N) j − (300 N)k TBH = MAD 4 0 −3 1 = 0.5 0 0 5 150 300 −300 1 = [(−3)(0.5)(300)] 5 or M AD = − 90.0 N ⋅ m PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 208 PROBLEM 3.56 In Problem 3.55, determine the moment about the diagonal AD of the force exerted on the frame by portion BG of the cable. SOLUTION M AD = λ AD ⋅ (rB/A × TBG ) Where and 1 λ AD = (4i − 3k ) 5 rB/A = (0.5 m) j BG = (−0.5) 2 + (0.925)2 + (−0.4)2 = 1.125 m Then Finally $ 450 N ( −0.5i + 0.925 j − 0.4k ) T BG = 1.125 = −(200 N)i + (370 N) j − (160 N)k MAD 4 0 −3 1 0.5 0 0 = 5 −200 370 −160 1 = [(−3)(0.5)(370)] 5 M AD = −111.0 N ⋅ m PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 209 PROBLEM 3.57 The triangular plate ABC is supported by ball-and-socket joints at B and D and is held in the position shown by cables AE and CF. If the force exerted by cable AE at A is 55 N, determine the moment of that force about the line joining Points D and B. SOLUTION First note d AE = (0.9) 2 + (−0.6) 2 + (0.2) 2 = 1.1 m Then TAE = Also Then Now where Then 55 N (0.9i − 0.6 j + 0.2k ) 1.1 = 5[(9 N)i − (6 N) j + (2 N)k ] DB = (1.2) 2 + ( −0.35) 2 + (0)2 λ DB = 1.25 m ! DB = DB 1 (1.2i − 0.35 j) = 1.25 1 = (24i − 7 j) 25 M DB = λ DB ⋅ (rA/D × TAE ) TDA = −(0.1 m) j + (0.2 m)k M DB 24 −7 0 1 = (5) 0 −0.1 0.2 25 −6 9 2 1 = (−4.8 − 12.6 + 28.8) 5 or M DB = 2.28 N ⋅ m PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 210 PROBLEM 3.58 The triangular plate ABC is supported by ball-and-socket joints at B and D and is held in the position shown by cables AE and CF. If the force exerted by cable CF at C is 33 N, determine the moment of that force about the line joining Points D and B. SOLUTION First note dCF = (0.6) 2 + (−0.9) 2 + (−0.2) 2 = 1.1 m Then TCF = Also DB = (1.2) 2 + ( −0.35) 2 + (0)2 33 N (0.6i − 0.9 j + 0.2k ) 1.1 = 3[(6 N)i − (9 N) j − (2 N)k ] = 1.25 m ! DB = DB 1 (1.2i − 0.35 j) = 1.25 1 = (24i − 7 j) 25 Then λ DB Now M DB = λ DB ⋅ (rC/D × TCF ) where rC/D = (0.2 m) j − (0.4 m)k Then M DB 24 −7 0 1 = (3) 0 0.2 −0.4 25 6 −9 −2 = 3 (−9.6 + 16.8 − 86.4) 25 or M DB = −9.50 N ⋅ m PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 211 PROBLEM 3.59 A regular tetrahedron has six edges of length a. A force P is directed as shown along edge BC. Determine the moment of P about edge OA. SOLUTION We have M OA = λOA ⋅ (rC/O × P) where From triangle OBC (OA) x = a 2 (OA) z = (OA) x tan 30° = Since a& 1 ' a ( )= 2* 3 + 2 3 (OA) 2 = (OA) 2x + (OA) 2y + (OAz )2 2 or & a ' &a' a 2 = ( ) + (OA) 2y + ( ) *2+ *2 3+ (OA) y = a 2 − Then and rA/O = OA = 2 a2 a2 2 − =a 4 12 3 a a 2 i+a j+ k 2 3 2 3 1 2 1 i+ j+ k 2 3 2 3 P = λBC P = P (a sin 30°)i − (a cos 30°)k ( P) = (i − 3k ) a 2 rC/O = ai M OA 1 2 = 1 2 3 0 2 3 &P' (a) ( ) 0 *2+ 1 0 − 3 = 1 aP & 2 ' aP (( − )) (1)(− 3) = 2 * 3+ 2 M OA = aP 2 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 212 PROBLEM 3.60 A regular tetrahedron has six edges of length a. (a) Show that two opposite edges, such as OA and BC, are perpendicular to each other. (b) Use this property and the result obtained in Problem 3.59 to determine the perpendicular distance between edges OA and BC. SOLUTION (a) For edge OA to be perpendicular to edge BC, ! ! OA ⋅ BC = 0 where From triangle OBC (OA) x = a 2 a& 1 ' a ( )= 2* 3 + 2 3 ! &a' & a ' OA = ( ) i + (OA) y j + ( )k *2+ *2 3+ ! BC = ( a sin 30°) i − (a cos 30°) k (OA) z = (OA) x tan 30° = and = Then or so that (b) Have M OA a a 3 a i− k = (i − 3 k ) 2 2 2 & a ' ! a a " i + (OA) y j + ( ) k # ⋅ (i − 3k ) = 0 2 *2 3+ % $2 a2 a2 + (OA) y (0) − =0 4 4 ! ! OA ⋅ BC = 0 ! ! OA is perpendicular to BC . ! ! = Pd , with P acting along BC and d the perpendicular distance from OA to BC . From the results of Problem 3.57 M OA = Pa 2 Pa 2 or d = = Pd a 2 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 213 ! PROBLEM 3.61 A sign erected on uneven ground is guyed by cables EF and EG. If the force exerted by cable EF at E is 46 lb, determine the moment of that force about the line joining Points A and D. SOLUTION First note that BC = (48) 2 + (36) 2 = 60 in. and that BE BC = 45 60 = 34 . The coordinates of Point E are then ( 34 × 48, 96, 34 × 36 ) or (36 in., 96 in., 27 in.). Then d EF = (−15) 2 + (−110) 2 + (30)2 = 115 in. 46 lb ( −15i − 110 j + 30k ) 115 = 2[−(3 lb)i − (22 lb) j + (6 lb)k ] Then TEF = Also AD = (48)2 + ( −12)2 + (36)2 Then Now λ AD = 12 26 in. ! AD = AD 1 (48i − 12 j + 36k ) = 12 26 1 = (4i − j + 3k ) 26 M AD = λ AD ⋅ (rE/A × TEF ) PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 214 PROBLEM 3.61 (Continued) where Then rE/A = (36 in.)i + (96 in.) j + (27 in.)k M AD = = −1 3 4 (2) 36 96 27 26 −3 −22 6 1 2 26 (2304 + 81 − 2376 + 864 + 216 + 2376) or M AD = 1359 lb ⋅ in. PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 215 ! PROBLEM 3.62 A sign erected on uneven ground is guyed by cables EF and EG. If the force exerted by cable EG at E is 54 lb, determine the moment of that force about the line joining Points A and D. SOLUTION First note that BC = (48) 2 + (36) 2 = 60 in. and that BE BC = 45 60 = 34 . The coordinates of Point E are then ( 34 × 48, 96, 34 × 36 ) or (36 in., 96 in., 27 in.). Then d EG = (11) 2 + (−88) 2 + (−44) 2 = 99 in. Then Also 54 lb (11i − 88 j − 44k ) 99 = 6[(1 lb)i − (8 lb) j − (4 lb)k ] TEG = AD = (48)2 + ( −12)2 + (36)2 = 12 26 in. ! AD = AD 1 (48i − 12 j + 36k ) = 12 26 1 (4i − j + 3k ) = 26 Then λ AD Now M AD = λ AD ⋅ (rE/A × TEG ) PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 216 PROBLEM 3.62 (Continued) where Then rE/A = (36 in.)i + (96 in.) j + (27 in.)k M AD = = 4 −1 3 (6) 36 96 27 26 1 − 8 −4 1 6 26 (−1536 − 27 − 864 − 288 − 144 + 864) or M AD = −2350 lb ⋅ in. ! PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 217 ! PROBLEM 3.63 Two forces F1 and F2 in space have the same magnitude F. Prove that the moment of F1 about the line of action of F2 is equal to the moment of F2 about the line of action of F1 . SOLUTION First note that F1 = F1λ1 and F2 = F2 λ 2 Let M1 = moment of F2 about the line of action of M1 and M 2 = moment of F1 about the line of action of M 2 Now, by definition M1 = λ1 ⋅ (rB/A × F2 ) = λ1 ⋅ (rB/A × λ 2 ) F2 M 2 = λ2 ⋅ (rA/B × F1 ) = λ2 ⋅ (rA/B × λ1 ) F1 Since F1 = F2 = F and rA/B = −rB/A M1 = λ1 ⋅ (rB/A × λ 2 ) F M 2 = λ 2 ⋅ (−rB/A × λ1 ) F Using Equation (3.39) so that λ1 ⋅ (rB/A × λ 2 ) = λ 2 ⋅ ( −rB/A × λ1 ) M 2 = λ 1⋅ (rB/A × λ 2 ) F ! M12 = M 21 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 218 ! PROBLEM 3.64 In Problem 3.55, determine the perpendicular distance between portion BH of the cable and the diagonal AD. PROBLEM 3.55 The frame ACD is hinged at A and D and is supported by a cable that passes through a ring at B and is attached to hooks at G and H. Knowing that the tension in the cable is 450 N, determine the moment about the diagonal AD of the force exerted on the frame by portion BH of the cable. SOLUTION From the solution to Problem 3.55: TBH = 450 N TBH = (150 N)i + (300 N) j − (300 N)k | M AD | = 90.0 N ⋅ m 1 λ AD = (4i − 3k ) 5 Based on the discussion of Section 3.11, it! follows that only the perpendicular component of TBH will contribute to the moment of TBH about line AD. Now (TBH )parallel = TBH ⋅ λ AD 1 = (150i + 300 j − 300k ) ⋅ (4i − 3k ) 5 1 = [(150)(4) + (−300)(−3)] 5 = 300 N Also so that TBH = (TBH ) parallel + (TBH )perpendicular (TBH )perpendicular = (450) 2 − (300) 2 = 335.41 N Since λ AD and (TBH )perpendicular are perpendicular, it follows that M AD = d (TBH ) perpendicular or 90.0 N ⋅ m = d (335.41 N) d = 0.26833 m d = 0.268 m PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 219 ! PROBLEM 3.65 In Problem 3.56, determine the perpendicular distance between portion BG of the cable and the diagonal AD. PROBLEM 3.56 In Problem 3.55, determine the moment about the diagonal AD of the force exerted on the frame by portion BG of the cable. SOLUTION From the solution to Problem 3.56: ΤBG = 450 N TBG = −(200 N)i + (370 N) j − (160 N)k | M AD | = 111 N ⋅ m 1 λ AD = (4i − 3k ) 5 Based on the discussion of Section 3.11, it" follows that only the perpendicular component of TBG will contribute to the moment of TBG about line AD. Now (TBG ) parallel = TBG ⋅ λ AD 1 = ( −200i + 370 j − 160k ) ⋅ (4i − 3k ) 5 1 = [(−200)(4) + (−160)(−3)] 5 = −64 N Also so that TBG = (TBG ) parallel + (TBG ) perpendicular (TBG ) perpendicular = (450) 2 − ( −64) 2 = 445.43 N Since λ AD and (TBG ) perpendicular are perpendicular, it follows that M AD = d (TBG ) perpendicular or 111 N ⋅ m = d (445.43 N) d = 0.24920 m d = 0.249 m PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 220 PROBLEM 3.66 In Problem 3.57, determine the perpendicular distance between cable AE and the line joining Points D and B. PROBLEM 3.57 The triangular plate ABC is supported by ball-and-socket joints at B and D and is held in the position shown by cables AE and CF. If the force exerted by cable AE at A is 55 N, determine the moment of that force about the line joining Points D and B. SOLUTION From the solution to Problem 3.57 ΤAE = 55 N TAE = 5[(9 N)i − (6 N) j + (2 N)k ] | M DB | = 2.28 N ⋅ m λ DB = 1 (24i − 7 j) 25 Based on the discussion of Section 3.11, it! follows that only the perpendicular component of TAE will contribute to the moment of TAE about line DB. Now (TAE )parallel = TAE ⋅ λ DB = 5(9i − 6 j + 2k ) ⋅ 1 (24i − 7 j) 25 1 = [(9)(24) + (−6)(−7)] 5 = 51.6 N Also so that TAE = (TAE ) parallel + (TAE ) perpendicular (TAE )perpendicular = (55) 2 + (51.6) 2 = 19.0379 N Since λ DB and (TAE )perpendicular are perpendicular, it follows that M DB = d (TAE ) perpendicular or 2.28 N ⋅ m = d (19.0379 N) d = 0.119761 d = 0.1198 m PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 221 PROBLEM 3.67 In Problem 3.58, determine the perpendicular distance between cable CF and the line joining Points D and B. PROBLEM 3.58 The triangular plate ABC is supported by ball-and-socket joints at B and D and is held in the position shown by cables AE and CF. If the force exerted by cable CF at C is 33 N, determine the moment of that force about the line joining Points D and B. SOLUTION From the solution to Problem 3.58 ΤCF = 33 N TCF = 3[(6 N)i − (9 N) j − (2 N)k ] | M DB | = 9.50 N ⋅ m λ DB = 1 (24i − 7 j) 25 Based on the discussion of Section 3.11, it! follows that only the perpendicular component of TCF will contribute to the moment of TCF about line DB. Now (TCF )parallel = TCF ⋅ λ DB = 3(6i − 9 j − 2k ) ⋅ 1 (24i − 7 j) 25 3 [(6)(24) + (−9)( −7)] 25 = 24.84 N = Also so that TCF = (TCF ) parallel + (TCF ) perpendicular (TCF )perpendicular = (33) 2 − (24.84) 2 = 21.725 N Since λ DB and (TCF )perpendicular are perpendicular, it follows that | M DB | = d (TCF ) perpendicular or 9.50 N ⋅ m = d × 21.725 N or d = 0.437 m PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 222 PROBLEM 3.68 In Problem 3.61, determine the perpendicular distance between cable EF and the line joining Points A and D. PROBLEM 3.61 A sign erected on uneven ground is guyed by cables EF and EG. If the force exerted by cable EF at E is 46 lb, determine the moment of that force about the line joining Points A and D. SOLUTION From the solution to Problem 3.61 TEF = 46 lb TEF = 2[−(3 lb)i − (22 lb) j + (6 lb)k ] | M AD | = 1359 lb ⋅ in. λ AD = 1 26 (4i − j + 3k ) Based on the discussion of Section 3.11, it! follows that only the perpendicular component of TEF will contribute to the moment of TEF about line AD. Now (TEF ) parallel = TEF ⋅ λ AD = 2(−3i − 22 j + 6k ) ⋅ = 1 26 (4i − j + 3k ) 2 [(−3)(4) + (−22)(−1) + (6)(3)] 26 = 10.9825 lb Also so that TEF = (TEF ) parallel + (TEF )perpendicular (TEF ) perpendicular = (46) 2 − (10.9825) 2 = 44.670 lb Since λ AD and (TEF ) perpendicular are perpendicular, it follows that M AD = d (TEF ) perpendicular or 1359 lb ⋅ in. = d × 44.670 lb or d = 30.4 in. PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 223 ! PROBLEM 3.69 In Problem 3.62, determine the perpendicular distance between cable EG and the line joining Points A and D. PROBLEM 3.62 A sign erected on uneven ground is guyed by cables EF and EG. If the force exerted by cable EG at E is 54 lb, determine the moment of that force about the line joining Points A and D. SOLUTION From the solution to Problem 3.62 TEG = 54 lb TEG = 6[(1 lb)i − (8 lb) j − (4 lb)k ] | M AD | = 2350 lb ⋅ in. 1 λ AD = 26 (4i − j + 3k ) Based on the discussion of Section 3.11, it! follows that only the perpendicular component of TEG will contribute to the moment of TEG about line AD. Now (TEG ) parallel = TEG ⋅ λ AD = 6(i − 8 j − 4k ) ⋅ = Thus, 6 26 1 26 (4i − j + 3k ) [(1)(4) + ( −8)( −1) + (−4)(3)] = 0 (TEG ) perpendicular = TEG = 54 lb Since λ AD and (TEG ) perpendicular are perpendicular, it follows that | M AD | = d (TEG ) perpendicular or 2350 lb ⋅ in. = d × 54 lb or d = 43.5 in. PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 224 ! PROBLEM 3.70 Two parallel 60-N forces are applied to a lever as shown. Determine the moment of the couple formed by the two forces (a) by resolving each force into horizontal and vertical components and adding the moments of the two resulting couples, (b) by using the perpendicular distance between the two forces, (c) by summing the moments of the two forces about Point A. SOLUTION (a) We have where ΣM B : − d1C x + d 2 C y = M d1 = (0.360 m) sin 55° = 0.29489 m d 2 = (0.360 m) sin 55° = 0.20649 m C x = (60 N) cos 20° = 56.382 N C y = (60 N)sin 20° = 20.521 N M = −(0.29489 m)(56.382 N)k + (0.20649 m)(20.521 N)k = −(12.3893 N ⋅ m)k (b) We have M = Fd (−k ) = 60 N[(0.360 m) sin(55° − 20°)]( −k ) = −(12.3893 N ⋅ m)k (c) We have or M = 12.39 N ⋅ m or M = 12.39 N ⋅ m ΣM A : Σ(rA × F) = rB/A × FB + rC/A × FC = M i j k sin 55° 0 M = (0.520 m)(60 N) cos 55° − cos 20° − sin 20° 0 i j k + (0.800 m)(60 N) cos 55° sin 55° 0 cos 20° sin 20° 0 = (17.8956 N ⋅ m − 30.285 N ⋅ m)k = −(12.3892 N ⋅ m)k or M = 12.39 N ⋅ m PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 225 PROBLEM 3.71 A plate in the shape of a parallelogram is acted upon by two couples. Determine (a) the moment of the couple formed by the two 21-lb forces, (b) the perpendicular distance between the 12-lb forces if the resultant of the two couples is zero, (c) the value of α if the resultant couple is 72 lb ⋅ in. clockwise and d is 42 in. SOLUTION (a) M1 = d1 F1 We have d1 = 16 in. where F1 = 21 lb M1 = (16 in.)(21 lb) = 336 lb ⋅ in. (b) (c) M1 + M 2 = 0 We have or 336 lb ⋅ in. − d 2 (12 lb) = 0 d 2 = 28.0 in. M total = M1 + M 2 We have or or M1 = 336 lb ⋅ in. −72 lb ⋅ in. = 336 lb ⋅ in. − (42 in.)(sin α )(12 lb) sin α = 0.80952 α = 54.049° and or α = 54.0° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 226 PROBLEM 3.72 A couple M of magnitude 18 N ⋅ m is applied to the handle of a screwdriver to tighten a screw into a block of wood. Determine the magnitudes of the two smallest horizontal forces that are equivalent to M if they are applied (a) at corners A and D, (b) at corners B and C, (c) anywhere on the block. SOLUTION (a) M = Pd We have or 18 N ⋅ m = P(.24 m) P = 75.0 N or Pmin = 75.0 N d BC = ( BE ) 2 + ( EC ) 2 (b) = (.24 m) 2 + (.08 m) 2 = 0.25298 m M = Pd We have 18 N ⋅ m = P(0.25298 m) P = 71.152 N or P = 71.2 N d AC = ( AD) 2 + ( DC ) 2 (c) = (0.24 m) 2 + (0.32 m) 2 = 0.4 m M = Pd AC We have 18 N ⋅ m = P(0.4 m) P = 45.0 N or P = 45.0 N PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 227 PROBLEM 3.73 Four 1-in.-diameter pegs are attached to a board as shown. Two strings are passed around the pegs and pulled with the forces indicated. (a) Determine the resultant couple acting on the board. (b) If only one string is used, around which pegs should it pass and in what directions should it be pulled to create the same couple with the minimum tension in the string? (c) What is the value of that minimum tension? SOLUTION M = (35 lb)(7 in.) + (25 lb)(9 in.) = 245 lb ⋅ in. + 225 lb ⋅ in. (a) M = 470 lb ⋅ in. (b) With only one string, pegs A and D, or B and C should be used. We have 6 8 tan θ = θ = 36.9° 90° − θ = 53.1° Direction of forces: (c) With pegs A and D: θ = 53.1° With pegs B and C: θ = 53.1° The distance between the centers of the two pegs is 82 + 62 = 10 in. Therefore, the perpendicular distance d between the forces is 1 ! d = 10 in. + 2 " in. # $2 % = 11 in. M = Fd We must have 470 lb ⋅ in. = F (11 in.) F = 42.7 lb PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 228 PROBLEM 3.74 Four pegs of the same diameter are attached to a board as shown. Two strings are passed around the pegs and pulled with the forces indicated. Determine the diameter of the pegs knowing that the resultant couple applied to the board is 485 lb·in. counterclockwise. SOLUTION M = d AD FAD + d BC FBC 485 lb ⋅ in. = [(6 + d )in.](35 lb) + [(8 + d )in.](25 lb) d = 1.250 in. PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 229 PROBLEM 3.75 The shafts of an angle drive are acted upon by the two couples shown. Replace the two couples with a single equivalent couple, specifying its magnitude and the direction of its axis. SOLUTION Based on where M = M1 + M 2 M1 = −(8 lb ⋅ ft)j M 2 = −(6 lb ⋅ ft)k M = −(8 lb ⋅ ft)j − (6 lb ⋅ ft)k and |M| = (8) 2 + (6) 2 = 10 lb ⋅ ft or M = 10.00 lb ⋅ ft M |M| −(8 lb ⋅ ft)j − (6 lb ⋅ ft)k = 10 lb ⋅ ft = −0.8 j − 0.6k = or M = |M| = (10 lb ⋅ ft)(−0.8 j − 0.6k ) cos θ x = 0 cos θ y = −0.8 θ x = 90° θ y = 143.130° cos θ z = −0.6 θ z = 126.870° or θ x = 90.0° θ y = 143.1° θ z = 126.9° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 230 PROBLEM 3.76 If P = 0, replace the two remaining couples with a single equivalent couple, specifying its magnitude and the direction of its axis. SOLUTION We have where M = M1 + M 2 M1 = rG/C × F1 rG/C = −(0.3 m)i F1 = (18 N)k M1 = −(0.3 m)i × (18 N)k = (5.4 N ⋅ m) j Also, M 2 = rD/F × F2 rD/F = −(.15 m)i + (.08 m) j F2 = λED F2 = (.15 m)i + (.08 m) j + (.17 m)k (.15)2 + (.08) 2 + (.17) 2 m (34 N) = 141.421 N ⋅ m(.15i + .08j + .17k ) i j k M 2 = 141.421 N ⋅ m −.15 .08 0 −.15 .08 .17 = 141.421(.0136i + 0.0255 j)N ⋅ m PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 231 PROBLEM 3.76 (Continued) and M = [(5.4 N ⋅ m)j] + [141.421(.0136i + .0255 j) N ⋅ m] = (1.92333 N ⋅ m)i + (9.0062 N ⋅ m)j | M | = (M x )2 + (M y )2 = (1.92333)2 + (9.0062) 2 = 9.2093 N ⋅ m λ= or M = 9.21 N ⋅ m M (1.92333 N ⋅ m)i + (9.0062 N ⋅ m) j = |M| 9.2093 N ⋅ m = 0.20885 + 0.97795 cos θ x = 0.20885 θ x = 77.945° or θ x = 77.9° or θ y = 12.05° or θ z = 90.0° cos θ y = 0.97795 θ y = 12.054° cos θ z = 0.0 θ z = 90° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 232 PROBLEM 3.77 If P = 0, replace the two remaining couples with a single equivalent couple, specifying its magnitude and the direction of its axis. SOLUTION M = M1 + M 2 ; F1 = 16 lb, F2 = 40 lb M1 = rC × F1 = (30 in.)i × [−(16 lb) j] = −(480 lb ⋅ in.)k M 2 = rE/B × F2 ; rE/B = (15 in.)i − (5 in.) j d DE = (0) 2 + (5) 2 + (10) 2 = 5 5 in. F2 = 40 lb 5 5 (5 j − 10k ) = 8 5[(1 lb) j − (2 lb)k ] i j k M 2 = 8 5 15 −5 0 0 1 −2 = 8 5[(10 lb ⋅ in.)i + (30 lb ⋅ in.) j + (15 lb ⋅ in.)k ] M = −(480 lb ⋅ in.)k + 8 5[(10 lb ⋅ in.)i + (30 lb ⋅ in.) j + (15 lb ⋅ in.)k ] = (178.885 lb ⋅ in.)i + (536.66 lb ⋅ in.) j − (211.67 lb ⋅ in.)k M = (178.885) 2 + (536.66) 2 + (−211.67) 2 = 603.99 lb ⋅ in M = 604 lb ⋅ in. M = 0.29617i + 0.88852 j − 0.35045k M cos θ x = 0.29617 λ axis = cos θ y = 0.88852 θ x = 72.8° θ y = 27.3° θ z = 110.5° cos θ z = −0.35045 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 233 PROBLEM 3.78 If P = 20 lb, replace the three couples with a single equivalent couple, specifying its magnitude and the direction of its axis. SOLUTION From the solution to Problem. 3.77 16 lb force: M1 = −(480 lb ⋅ in.)k 40 lb force: M 2 = 8 5[(10 lb ⋅ in.)i + (30 lb ⋅ in.) j + (15 lb ⋅ in.)k ] P = 20 lb M 3 = rC × P = (30 in.)i × (20 lb)k = (600 lb ⋅ in.) j M = M1 + M 2 + M 3 = −(480)k + 8 5 (10i + 30 j + 15k ) + 600 j = (178.885 lb ⋅ in)i + (1136.66 lb ⋅ in.) j − (211.67 lb ⋅ in.)k M = (178.885) 2 + (113.66) 2 + (211.67) 2 = 1169.96 lb ⋅ in M = 1170 lb ⋅ in. M = 0.152898i + 0.97154 j − 0.180921k M cos θ x = 0.152898 λ axis = cos θ y = 0.97154 θ x = 81.2° θ y = 13.70° θ z = 100.4° cos θ z = −0.180921 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 234 PROBLEM 3.79 If P = 20 N, replace the three couples with a single equivalent couple, specifying its magnitude and the direction of its axis. SOLUTION We have M = M1 + M 2 + M 3 where i j k M1 = rG/C × F1 = −0.3 0 0 N ⋅ m = (5.4 N ⋅ m)j 0 0 18 M 2 = rD/F i j k × F2 = −.15 .08 0 141.421 N ⋅ m −.15 .08 .17 = 141.421(.0136i + .0255 j)N ⋅ m (See Solution to Problem 3.76.) i j k M 3 = rC/A × F3 = 0.3 0 0.17 N ⋅ m 0 20 0 = −(3.4 N ⋅ m)i + (6 N ⋅ m)k M = [(1.92333 − 3.4)i + (5.4 + 3.6062) j + (6)k ] N ⋅ m = −(1.47667 N ⋅ m)i + (9.0062 N ⋅ m) j + (6 N ⋅ m) | M | = M x2 + M y2 + M z2 = (1.47667) + (9.0062) + (6)2 or M = 10.92 N ⋅ m = 10.9221 N ⋅ m PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 235 PROBLEM 3.79 (Continued) M −1.47667 + 9.0062 + 6 = |M| 10.9221 = −0.135200i + 0.82459 j + 0.54934k = cos θ x = −0.135200 θ x = 97.770 or θ x = 97.8° cos θ y = 0.82459 θ y = 34.453 or θ y = 34.5° cos θ z = 0.54934 θ z = 56.678 or θ z = 56.7° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 236 PROBLEM 3.80 Shafts A and B connect the gear box to the wheel assemblies of a tractor, and shaft C connects it to the engine. Shafts A and B lie in the vertical yz plane, while shaft C is directed along the x axis. Replace the couples applied to the shafts with a single equivalent couple, specifying its magnitude and the direction of its axis. SOLUTION Represent the given couples by the following couple vectors: M A = −1600sin 20° j + 1600cos 20°k = −(547.232 N ⋅ m) j + (1503.51 N ⋅ m)k M B = 1200sin 20° j + 1200 cos 20°k = (410.424 N ⋅ m) j + (1127.63 N ⋅ m)k M C = −(1120 N ⋅ m)i The single equivalent couple is M = M A + M B + MC = −(1120 N ⋅ m)i − (136.808 N ⋅ m) j + (2631.1 N ⋅ m)k M = (1120) 2 + (136.808)2 + (2631.1)2 = 2862.8 N ⋅ m −1120 cos θ x = 2862.8 −136.808 cos θ y = 2862.8 2631.1 cos θ z = 2862.8 M = 2860 N ⋅ m θ x = 113.0° θ y = 92.7° θ z = 23.2° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 237 PROBLEM 3.81 The tension in the cable attached to the end C of an adjustable boom ABC is 560 lb. Replace the force exerted by the cable at C with an equivalent forcecouple system (a) at A, (b) at B. SOLUTION (a) Based on ΣF : FA = T = 560 lb FA = 560 lb or 20° ΣM A : M A = (T sin 50°)(d A ) = (560 lb)sin 50°(18 ft) = 7721.7 lb ⋅ ft M A = 7720 lb ⋅ ft or (b) Based on ΣF : FB = T = 560 lb FB = 560 lb or 20° ΣM B : M B = (T sin 50°)(d B ) = (560 lb) sin 50°(10 ft) = 4289.8 lb ⋅ ft M B = 4290 lb ⋅ ft or PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 238 PROBLEM 3.82 A 160-lb force P is applied at Point A of a structural member. Replace P with (a) an equivalent force-couple system at C, (b) an equivalent system consisting of a vertical force at B and a second force at D. SOLUTION (a) Based on ΣF : PC = P = 160 lb or PC = 160 lb 60° ΣM C : M C = − Px d cy + Py dCx where Px = (160 lb) cos 60° = 80 lb Py = (160 lb)sin 60° dCx = 138.564 lb = 4 ft dCy = 2.75 ft M C = (80 lb)(2.75 ft) + (138.564 lb)(4 ft) = 220 lb ⋅ ft + 554.26 lb ⋅ ft = 334.26 lb ⋅ ft (b) Based on or M C = 334 lb ⋅ ft ΣFx : PDx = P cos 60° = (160 lb) cos 60° = 80 lb ΣM D : ( P cos 60°)( d DA ) = PB (d DB ) [(160 lb) cos 60°](1.5 ft) = PB (6 ft) PB = 20.0 lb or PB = 20.0 lb PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 239 PROBLEM 3.82 (Continued) ΣFy : P sin 60° = PB + PDy (160 lb)sin 60° = 20.0 lb + PDy PDy = 118.564 lb PD = ( PDx ) 2 + ( PDy ) 2 = (80) 2 + (118.564) 2 = 143.029 lb PDy ! # $ PDx % 118.564 ! = tan −1 " # $ 80 % = 55.991° θ = tan −1 " or PD = 143.0 lb 56.0° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 240 PROBLEM 3.83 The 80-N horizontal force P acts on a bell crank as shown. (a) Replace P with an equivalent force-couple system at B. (b) Find the two vertical forces at C and D that are equivalent to the couple found in Part a. SOLUTION (a) ΣF : FB = F = 80 N Based on or FB = 80.0 N ΣM : M B = Fd B = 80 N (.05 m) = 4.0000 N ⋅ m M B = 4.00 N ⋅ m or (b) If the two vertical forces are to be equivalent to MB, they must be a couple. Further, the sense of the moment of this couple must be counterclockwise. Then, with FC and FD acting as shown, ΣM : M D = FC d 4.0000 N ⋅ m = FC (.04 m) FC = 100.000 N or FC = 100.0 N ΣFy : 0 = FD − FC FD = 100.000 N or FD = 100.0 N PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 241 PROBLEM 3.84 A dirigible is tethered by a cable attached to its cabin at B. If the tension in the cable is 1040 N, replace the force exerted by the cable at B with an equivalent system formed by two parallel forces applied at A and C. SOLUTION Require the equivalent forces acting at A and C be parallel and at an angle of α with the vertical. Then for equivalence, ΣFx : (1040 N)sin 30° = FA sin α + FB sin α (1) ΣFy : −(1040 N) cos 30° = − FA cos α − FB cos α (2) Dividing Equation (1) by Equation (2), ( FA + FB ) sin α (1040 N) sin 30° = −(1040 N) cos 30° −( FA + FB ) cos α Simplifying yields α = 30° Based on ΣM C : [(1040 N) cos 30°](4 m) = ( FA cos 30°)(10.7 m) FA = 388.79 N FA = 389 N or 60° Based on ΣM A : − [(1040 N) cos 30°](6.7 m) = ( FC cos 30°)(10.7 m) FC = 651.21 N FC = 651 N or 60° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 242 PROBLEM 3.85 The force P has a magnitude of 250 N and is applied at the end C of a 500-mm rod AC attached to a bracket at A and B. Assuming α = 30° and β = 60°, replace P with (a) an equivalent force-couple system at B, (b) an equivalent system formed by two parallel forces applied at A and B. SOLUTION (a) ΣF : F = P or F = 250 N Equivalence requires 60° ΣM B : M = −(0.3 m)(250 N) = −75 N ⋅ m The equivalent force-couple system at B is F = 250 N (b) M = 75.0 N ⋅ m 60° Require Equivalence then requires ΣFx : 0 = FA cos φ + FB cos φ FA = − FB or cos φ = 0 ΣFy : − 250 = − FA sin φ − FB sin φ Now if FA = − FB & −250 = 0 reject cos φ = 0 or and Also φ = 90° FA + FB = 250 ΣM B : − (0.3 m)(250 N) = (0.2m) FA or FA = −375 N and FB = 625 N FA = 375 N 60° FB = 625 N 60° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 243 PROBLEM 3.86 Solve Problem 3.85, assuming α = β = 25°. SOLUTION (a) Equivalence requires ΣF : FB = P or FB = 250 N 25.0° ΣM B : M B = −(0.3 m)[(250 N)sin 50°] = −57.453 N ⋅ m The equivalent force-couple system at B is FB = 250 N (b) 25.0° M B = 57.5 N ⋅ m Require Equivalence requires M B = d AE Q (0.3 m)[(250 N) sin 50°] = [(0.2 m) sin 50°]Q Q = 375 N Adding the forces at B: FA = 375 N 25.0° FB = 625 N 25.0° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 244 PROBLEM 3.87 A force and a couple are applied as shown to the end of a cantilever beam. (a) Replace this system with a single force F applied at Point C, and determine the distance d from C to a line drawn through Points D and E. (b) Solve Part a if the directions of the two 360-N forces are reversed. SOLUTION (a) We have ΣF : F = (360 N) j − (360 N) j − (600 N)k or F = −(600 N)k ΣM D : (360 N)(0.15 m) = (600 N)(d ) and d = 0.09 m or d = 90.0 mm below ED (b) We have from Part a F = −(600 N)k ΣM D : −(360 N)(0.15 m) = −(600 N)( d ) and d = 0.09 m or d = 90.0 mm above ED PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 245 PROBLEM 3.88 The shearing forces exerted on the cross section of a steel channel can be represented by a 900-N vertical force and two 250-N horizontal forces as shown. Replace this force and couple with a single force F applied at Point C, and determine the distance x from C to line BD. (Point C is defined as the shear center of the section.) SOLUTION Replace the 250-N forces with a couple and move the 900-N force to Point C such that its moment about H is equal to the moment of the couple M H = (0.18)(250 N) = 45 N ⋅ m Then or M H = x(900 N) 45 N ⋅ m = x(900 N) x = 0.05 m F = 900 N x = 50.0 mm PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 246 PROBLEM 3.89 While tapping a hole, a machinist applies the horizontal forces shown to the handle of the tap wrench. Show that these forces are equivalent to a single force, and specify, if possible, the point of application of the single force on the handle. SOLUTION Since the forces at A and B are parallel, the force at B can be replaced with the sum of two forces with one of the forces equal in magnitude to the force at A except with an opposite sense, resulting in a force-couple. Have FB = 2.9 lb − 2.65 lb = 0.25 lb, where the 2.65 lb force be part of the couple. Combining the two parallel forces, M couple = (2.65 lb)[(3.2 in. + 2.8 in.) cos 25°] = 14.4103 lb ⋅ in. and M couple = 14.4103 lb ⋅ in. A single equivalent force will be located in the negative z-direction Based on ΣM B : −14.4103 lb ⋅ in. = [(.25 lb) cos 25°](a ) a = 63.600 in. F′ = (.25 lb)(cos 25°i + sin 25°k ) F′ = (0.227 lb)i + (0.1057 lb)k and is applied on an extension of handle BD at a distance of 63.6 in. to the right of B PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 247 PROBLEM 3.90 Three control rods attached to a lever ABC exert on it the forces shown. (a) Replace the three forces with an equivalent force-couple system at B. (b) Determine the single force that is equivalent to the force-couple system obtained in Part a, and specify its point of application on the lever. SOLUTION (a) First note that the two 20-lb forces form A couple. Then F = 48 lb θ where θ = 180° − (60° + 55°) = 65° and M = ΣM B = (30 in.)(48 lb) cos55° − (70 in.)(20 lb) cos 20° = −489.62 lb ⋅ in The equivalent force-couple system at B is F = 48.0 lb (b) 65° M = 490 lb ⋅ in. The single equivalent force F ′ is equal to F. Further, since the sense of M is clockwise, F ′ must be applied between A and B. For equivalence. ΣM B : M = − aF ′ cos 55° where a is the distance from B to the point of application of F′. Then −489.62 lb ⋅ in. = −a (48.0 lb) cos 55° or a = 17.78 in. F ′ = 48.0 lb 65.0° and is applied to the lever 17.78 in. To the left of pin B PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 248 PROBLEM 3.91 A hexagonal plate is acted upon by the force P and the couple shown. Determine the magnitude and the direction of the smallest force P for which this system can be replaced with a single force at E. SOLUTION From the statement of the problem, it follows that ΣM E = 0 for the given force-couple system. Further, for Pmin, must require that P be perpendicular to rB/E . Then ΣM E : (0.2 sin 30° + 0.2)m × 300 N + (0.2 m)sin 30° × 300 N − (0.4 m) Pmin = 0 or Pmin = 300 N Pmin = 300 N 30.0° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 249 PROBLEM 3.92 A rectangular plate is acted upon by the force and couple shown. This system is to be replaced with a single equivalent force. (a) For α = 40°, specify the magnitude and the line of action of the equivalent force. (b) Specify the value of α if the line of action of the equivalent force is to intersect line CD 300 mm to the right of D. SOLUTION (a) The given force-couple system (F, M) at B is F = 48 N and M = ΣM B = (0.4 m)(15 N) cos 40° + (0.24 m)(15 N)sin 40° or M = 6.9103 N ⋅ m The single equivalent force F′ is equal to F. Further for equivalence ΣM B : M = dF ′ 6.9103 N ⋅ m = d × 48 N or d = 0.14396 m or F ′ = 48 N and the line of action of F′ intersects line AB 144 mm to the right of A. (b) Following the solution to Part a but with d = 0.1 m and α unknown, have ΣM B : (0.4 m)(15 N) cos α + (0.24 m)(15 N) sin α = (0.1 m)(48 N) or 5cos α + 3sin α = 4 Rearranging and squaring 25 cos 2 α = (4 − 3 sin α )2 Using cos 2 α = 1 − sin 2 α and expanding 25(1 − sin 2 α ) = 16 − 24 sin α + 9 sin 2 α or Then 34 sin 2 α − 24 sin α − 9 = 0 24 ± (−24) 2 − 4(34)(−9) 2(34) sin α = 0.97686 or sin α = −0.27098 sin α = α = 77.7° or α = −15.72° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 250 PROBLEM 3.93 An eccentric, compressive 1220-N force P is applied to the end of a cantilever beam. Replace P with an equivalent force-couple system at G. SOLUTION We have ΣF : − (1220 N)i = F F = − (1220 N)i Also, we have ΣM G : rA/G × P = M i j k 1220 0 −.1 −.06 N ⋅ m = M −1 0 0 M = (1220 N ⋅ m)[(−0.06)(−1) j − ( −0.1)( −1)k ] or M = (73.2 N ⋅ m) j − (122 N ⋅ m)k PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 251 PROBLEM 3.94 To keep a door closed, a wooden stick is wedged between the floor and the doorknob. The stick exerts at B a 175-N force directed along line AB. Replace that force with an equivalent force-couple system at C. SOLUTION We have ΣF : PAB = FC where PAB = = AB PAB (33 mm)i + (990 mm) j − (594 mm)k (175 N) 1155.00 mm or FC = (5.00 N)i + (150 N) j − (90.0 N)k We have ΣM C : rB/C × PAB = M C i j k M C = 5 0.683 −0.860 0 N ⋅ m −18 1 30 = (5){(− 0.860)(−18)i − (0.683)(−18) j + [(0.683)(30) − (0.860)(1)]k} or M C = (77.4 N ⋅ m)i + (61.5 N ⋅ m) j + (106.8 N ⋅ m)k PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 252 PROBLEM 3.95 An antenna is guyed by three cables as shown. Knowing that the tension in cable AB is 288 lb, replace the force exerted at A by cable AB with an equivalent force-couple system at the center O of the base of the antenna. SOLUTION We have d AB = (−64)2 + (−128) 2 + (16) 2 = 144 ft Then TAB = Now 288 lb ( −64i − 128 j + 16k ) 144 = (32 lb)( −4i − 8 j + k ) M = M O = rA / O × TAB = 128 j × 32(−4i − 8 j + k ) = (4096 lb ⋅ ft)i + (16,384 lb ⋅ ft)k The equivalent force-couple system at O is F = −(128.0 lb)i − (256 lb) j + (32.0 lb)k M = (4.10 kip ⋅ ft)i + (16.38 kip ⋅ ft)k PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 253 PROBLEM 3.96 An antenna is guyed by three cables as shown. Knowing that the tension in cable AD is 270 lb, replace the force exerted at A by cable AD with an equivalent force-couple system at the center O of the base of the antenna. SOLUTION We have d AD = ( −64) 2 + (−128)2 + (−128) 2 = 192 ft Then Now 270 lb (−64i − 128 j + 128k ) 192 = (90 lb)(−i − 2 j − 2k ) TAD = M = M O = rA/O × TAD = 128 j × 90(−i − 2 j − 2k ) = −(23, 040 lb ⋅ ft)i + (11,520 lb ⋅ ft)k The equivalent force-couple system at O is F = −(90.0 lb)i − (180.0 lb) j − (180.0 lb)k M = −(23.0 kip ⋅ ft)i + (11.52 kip ⋅ ft)k PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 254 PROBLEM 3.97 Replace the 150-N force with an equivalent force-couple system at A. SOLUTION Equivalence requires ΣF : F = (150 N)(− cos 35° j − sin 35°k ) = −(122.873 N) j − (86.036 N)k ΣMA : M = rD/A × F where Then rD/A = (0.18 m)i − (0.12 m) j + (0.1 m)k i j k −0.12 0.1 N⋅m M = 0.18 −122.873 −86.036 0 = [( −0.12)(−86.036) − (0.1)(−122.873)]i + [−(0.18)(−86.036)]j + [(0.18)(−122.873)]k = (22.6 N ⋅ m)i + (15.49 N ⋅ m) j − (22.1 N ⋅ m)k The equivalent force-couple system at A is F = −(122.9 N) j − (86.0 N)k M = (22.6 N ⋅ m)i + (15.49 N ⋅ m) j − (22.1 N ⋅ m)k PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 255 ! PROBLEM 3.98 A 77-N force F1 and a 31-N ⋅ m couple M1 are applied to corner E of the bent plate shown. If F1 and M1 are to be replaced with an equivalent force-couple system (F2, M2) at corner B and if (M2)z = 0, determine (a) the distance d, (b) F2 and M2. SOLUTION (a) ΣM Bz : M 2 z = 0 We have k ⋅ (rH /B × F1 ) + M 1z = 0 where (1) rH /B = (0.31 m)i − (0.0233) j F1 = EH F1 (0.06 m)i + (0.06 m) j − (0.07 m)k (77 N) 0.11 m = (42 N)i + (42 N) j − (49 N)k = M1z = k ⋅ M1 M1 = = EJ M 1 − di + (0.03 m) j − (0.07 m)k d 2 + 0.0058 m (31 N ⋅ m) Then from Equation (1), 0 0 1 ( −0.07 m)(31 N ⋅ m) 0.31 −0.0233 0 + =0 2 0.0058 + d 42 42 −49 Solving for d, Equation (1) reduces to (13.0200 + 0.9786) − From which 2.17 N ⋅ m d 2 + 0.0058 d = 0.1350 m =0 or d = 135.0 mm PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 256 PROBLEM 3.98 (Continued) (b) F2 = F1 = (42i + 42 j − 49k )N or F2 = (42 N)i + (42 N) j − (49 N)k M 2 = rH /B × F1 + M1 i j k (0.1350)i + 0.03j − 0.07k (31 N ⋅ m) = 0.31 −0.0233 0 + 0.155000 −49 42 42 = (1.14170i + 15.1900 j + 13.9986k ) N ⋅ m + (−27.000i + 6.0000 j − 14.0000k ) N ⋅ m M 2 = − (25.858 N ⋅ m)i + (21.190 N ⋅ m) j or M 2 = − (25.9 N ⋅ m)i + (21.2 N ⋅ m) j PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 257 PROBLEM 3.99 A 46-lb force F and a 2120-lb ⋅ in. couple M are applied to corner A of the block shown. Replace the given force-couple system with an equivalent force-couple system at corner H. SOLUTION We have Then Also d AJ = (18) 2 + (−14) 2 + (−3) 2 = 23 in. 46 lb (18i − 14 j − 3k ) 23 = (36 lb)i − (28 lb) j − (6 lb)k F= d AC = (−45) 2 + (0)2 + (−28)2 = 53 in. 2120 lb ⋅ in. ( −45i − 28k ) 53 = −(1800 lb ⋅ in.)i − (1120 lb ⋅ in.)k Then M= Now M ′ = M + rA/H × F where Then rA/H = (45 in.)i + (14 in.) j i j k 0 M ′ = (−1800i − 1120k ) + 45 14 36 −28 −6 = (−1800i − 1120k ) + {[(14)(−6)]i + [−(45)( −6)]j + [(45)(−28) − (14)(36)]k} = (−1800 − 84)i + (270) j + (−1120 − 1764)k = −(1884 lb ⋅ in.)i + (270 lb ⋅ in.)j − (2884 lb ⋅ in.)k = −(157 lb ⋅ ft)i + (22.5 lb ⋅ ft) j − (240 lb ⋅ ft)k F′ = (36.0 lb)i − (28.0 lb) j − (6.00 lb)k The equivalent force-couple system at H is M ′ = −(157 lb ⋅ ft)i + (22.5 lb ⋅ ft) j − (240 lb ⋅ ft)k PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 258 PROBLEM 3.100 The handpiece for a miniature industrial grinder weighs 0.6 lb, and its center of gravity is located on the y axis. The head of the handpiece is offset in the xz plane in such a way that line BC forms an angle of 25° with the x direction. Show that the weight of the handpiece and the two couples M1 and M2 can be replaced with a single equivalent force. Further, assuming that M1 = 0.68 lb ⋅ in. and M2 = 0.65 lb ⋅ in., determine (a) the magnitude and the direction of the equivalent force, (b) the point where its line of action intersects the xz plane. SOLUTION First assume that the given force W and couples M1 and M2 act at the origin. Now W = − Wj and M = M1 + M 2 = − ( M 2 cos 25°)i + ( M 1 − M 2 sin 25°)k Note that since W and M are perpendicular, it follows that they can be replaced with a single equivalent force. F = W or F = − Wj = − (0.6 lb) j or F = − (0.600 lb)j (a) We have (b) Assume that the line of action of F passes through Point P(x, 0, z). Then for equivalence M = rP/0 × F where rP/0 = xi + zk − ( M 2 cos 25°)i + ( M1 − M 2 sin 25°)k i j = x 0 0 −W Equating the i and k coefficients, (b) For z= k z = (Wz )i − (Wx)k 0 − M z cos 25° W and x = −" $ M 1 − M 2 sin 25° ! # W % W = 0.6 lb M1 = 0.68 lb ⋅ in. M 2 = 0.65 lb ⋅ in. x= 0.68 − 0.65sin 25° = 0.67550 in. − 0.6 or z= − 0.65cos 25° = − 0.98183 in. 0.6 or z = − 0.982 in. x = 0.675 in. PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 259 PROBLEM 3.101 A 4-m-long beam is subjected to a variety of loadings. (a) Replace each loading with an equivalent forcecouple system at end A of the beam. (b) Which of the loadings are equivalent? SOLUTION (a) (a) We have ΣFy : −400 N − 200 N = Ra or and R a = 600 N ΣM A : 1800 N ⋅ m − (200 N)(4 m) = M a or M a = 1000 N ⋅ m (b) We have ΣFy : − 600 N = Rb or and ΣM A : − 900 N ⋅ m = M b or (c) We have M b = 900 N ⋅ m ΣFy : 300 N − 900 N = Rc or and R b = 600 N R c = 600 N ΣM A : 4500 N ⋅ m − (900 N)(4 m) = M c or M c = 900 N ⋅ m PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 260 PROBLEM 3.101 (Continued) (d) We have and (e) We have ΣFy : − 400 N + 800 N = Rd or R d = 400 N or M d = 900 N ⋅ m ΣM A : (800 N)(4 m) − 2300 N ⋅ m = M d ΣFy : − 400 N − 200 N = Re or and ΣM A : 200 N ⋅ m + 400 N ⋅ m − (200 N)(4 m) = M e or ( f ) We have M e = 200 N ⋅ m ΣFy : − 800 N + 200 N = R f or and R e = 600 N R f = 600 N ΣM A : − 300 N ⋅ m + 300 N ⋅ m + (200 N)(4 m) = M f or M f = 800 N ⋅ m (g) We have ΣFy : − 200 N − 800 N = Rg or and R g = 1000 N ΣM A : 200 N ⋅ m + 4000 N ⋅ m − (800 N)(4 m) = M g or M g = 1000 N ⋅ m (h) We have ΣFy : − 300 N − 300 N = Rh or and ΣM A : 2400 N ⋅ m − 300 N ⋅ m − (300 N)(4 m) = M h or (b) R h = 600 N M h = 900 N ⋅ m ! Therefore, loadings (c) and (h) are equivalent. PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 261 PROBLEM 3.102 A 4-m-long beam is loaded as shown. Determine the loading of Problem 3.101 which is equivalent to this loading. SOLUTION We have and ΣFy : − 200 N − 400 N = R or R = 600 N ΣM A : −400 N ⋅ m + 2800 N ⋅ m − (400 N)(4 m) = M M = 800 N ⋅ m or Equivalent to case ( f ) of Problem 3.101 Problem 3.101 Equivalent force-couples at A Case R (a) 600 N 1000 N ⋅ m (b) 600 N 900 N ⋅ m (c) 600 N 900 N ⋅ m (d) 400 N 900 N ⋅ m (e) 600 N 200 N ⋅ m (f ) 600 N 800 N ⋅ m (g) 1000 N 1000 N ⋅ m (h) 600 N 900 N ⋅ m M PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 262 ! PROBLEM 3.103 Determine the single equivalent force and the distance from Point A to its line of action for the beam and loading of (a) Problem 3.101b, (b) Problem 3.101d, (c) Problem 3.101e. PROBLEM 3.101 A 4-m-long beam is subjected to a variety of loadings. (a) Replace each loading with an equivalent force-couple system at end A of the beam. (b) Which of the loadings are equivalent? SOLUTION (a) For equivalent single force at distance d from A ΣFy : −600 N = R We have or R = 600 N and ΣM C : (600 N)(d ) − 900 N ⋅ m = 0 or d = 1.500 m (b) ΣFy : − 400 N + 800 N = R We have or R = 400 N and ΣM C : (400 N)( d ) + (800 N)(4 − d ) − 2300 N ⋅ m = 0 (c) d = 2.25 m or R = 600 N ΣFy : − 400 N − 200 N = R We have and or ΣM C : 200 N ⋅ m + (400 N)(d ) − (200 N)(4 − d ) + 400 N ⋅ m = 0 or d = 0.333 m PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 263 PROBLEM 3.104 Five separate force-couple systems act at the corners of a piece of sheet metal, which has been bent into the shape shown. Determine which of these systems is equivalent to a force F = (10 lb)i and a couple of moment M = (15 lb ⋅ ft)j + (15 lb ⋅ ft)k located at the origin. SOLUTION First note that the force-couple system at F cannot be equivalent because of the direction of the force [The force of the other four systems is (10 lb)i]. Next move each of the systems to the origin O; the forces remain unchanged. A: M A = ΣM O = (5 lb ⋅ ft) j + (15 lb ⋅ ft)k + (2 ft)k × (10 lb)i = (25 lb ⋅ ft) j + (15 lb ⋅ ft)k D : M D = ΣM O = −(5 lb ⋅ ft) j + (25 lb ⋅ ft)k + [(4.5 ft) j + (1 ft) j + (2 ft)k ] × 10 lb)i = (15 lb ⋅ ft)i + (15 lb ⋅ ft)k G : M G = ΣM O = (15 lb ⋅ ft)i + (15 lb ⋅ ft) j I : M I = ΣM I = (15 lb ⋅ ft) j − (5 lb ⋅ ft)k + [(4.5 ft)i + (1 ft) j] × (10 lb) j = (15 lb ⋅ ft) j − (15 lb ⋅ ft)k The equivalent force-couple system is the system at corner D. PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 264 ! PROBLEM 3.105 The weights of two children sitting at ends A and B of a seesaw are 84 lb and 64 lb, respectively. Where should a third child sit so that the resultant of the weights of the three children will pass through C if she weighs (a) 60 lb, (b) 52 lb. SOLUTION (a) For the resultant weight to act at C, Then ΣM C = 0 WC = 60 lb (84 lb)(6 ft) − 60 lb(d ) − 64 lb(6 ft) = 0 d = 2.00 ft to the right of C (b) For the resultant weight to act at C, Then ΣM C = 0 WC = 52 lb (84 lb)(6 ft) − 52 lb(d ) − 64 lb(6 ft) = 0 d = 2.31 ft to the right of C PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 265 ! PROBLEM 3.106 Three stage lights are mounted on a pipe as shown. The lights at A and B each weigh 4.1 lb, while the one at C weighs 3.5 lb. (a) If d = 25 in., determine the distance from D to the line of action of the resultant of the weights of the three lights. (b) Determine the value of d so that the resultant of the weights passes through the midpoint of the pipe. SOLUTION For equivalence ΣFy : − 4.1 − 4.1 − 3.5 = − R or R = 11.7 lb ΣFD : − (10 in.)(4.1 lb) − (44 in.)(4.1 lb) −[(4.4 + d )in.](3.5 lb) = −( L in.)(11.7 lb) 375.4 + 3.5d = 11.7 L (d , L in in.) or d = 25 in. (a) We have 375.4 + 3.5(25) = 11.7 L or L = 39.6 in. The resultant passes through a Point 39.6 in. to the right of D. L = 42 in. (b) We have 375.4 + 3.5d = 11.7(42) or d = 33.1 in. PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 266 PROBLEM 3.107 A beam supports three loads of given magnitude and a fourth load whose magnitude is a function of position. If b = 1.5 m and the loads are to be replaced with a single equivalent force, determine (a) the value of a so that the distance from support A to the line of action of the equivalent force is maximum, (b) the magnitude of the equivalent force and its point of application on the beam. SOLUTION For equivalence ΣFy : −1300 + 400 a − 400 − 600 = − R b a! R = " 2300 − 400 # N b $ % or ΣM A : a a! 400 # − a (400) − (a + b)(600) = − LR " b% 2$ 1000a + 600b − 200 L= or (1) 2300 − 400 a2 b a b 10a + 9 − Then with b = 1.5 m L = 4 2 a 3 (2) 8 23 − a 3 Where a, L are in m (a) Find value of a to maximize L dL "$ = da 8 ! 8 ! 4 ! 8! 10 − a #" 23 − a # − "10a + 9 − a 2 #" − # 3 %$ 3 % $ 3 %$ 3 % 8 ! " 23 − 3 a # $ % 2 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 267 PROBLEM 3.107 (Continued) 230 − or 16a 2 − 276a + 1143 = 0 276 ± (−276) 2 − 4(16)(1143) 2(16) Then a= or a = 10.3435 m and a = 6.9065 m Since (b) 184 80 64 2 80 32 a− a+ a + a + 24 − a 2 = 0 3 3 9 3 9 or AB = 9 m, a must be less than 9 m a = 6.91 m 6.9065 1.5 Using Eq. (1) R = 2300 − 400 and using Eq. (2) 4 10(6.9065) + 9 − (6.9065)2 3 = 3.16 m L= 8 23 − (6.9065) 3 or R = 458 N R is applied 3.16 m to the right of A. PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 268 ! ! PROBLEM 3.108 Gear C is rigidly attached to arm AB. If the forces and couple shown can be reduced to a single equivalent force at A, determine the equivalent force and the magnitude of the couple M. SOLUTION We have For equivalence ΣFx : −18sin 30° + 25cos 40° = Rx or Rx = 10.1511 lb ΣFy : −18cos 30° − 40 − 25sin 40° = Ry or Then Ry = −71.658 lb R = (10.1511) 2 + (71.658) 2 = 72.416 tanθ = or Also 71.658 10.1511 θ = 81.9° R = 72.4 lb 81.9° ΣM A : M − (22 in.)(18 lb)sin 35° − (32 in.)(40 lb) cos 25° − (48 in.)(25 lb) sin 65° = 0 M = 2474.8 lb ⋅ in. or M = 206 lb ⋅ ft PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 269 PROBLEM 3.109 A couple of magnitude M = 54 lb ⋅ in. and the three forces shown are applied to an angle bracket. (a) Find the resultant of this system of forces. (b) Locate the points where the line of action of the resultant intersects line AB and line BC. SOLUTION (a) We have ΣF : R = (−10 j) + (30 cos 60°)i + 30 sin 60° j + (−45i ) = −(30 lb)i + (15.9808 lb) j or R = 34.0 lb (b) 28.0° First reduce the given forces and couple to an equivalent force-couple system (R , M B ) at B. We have ΣM B : M B = (54 lb ⋅ in) + (12 in.)(10 lb) − (8 in.)(45 lb) = −186 lb ⋅ in. Then with R at D or and with R at E or ΣM B : −186 lb ⋅ in = a(15.9808 lb) a = 11.64 in. ΣM B : −186 lb ⋅ in = C (30 lb) C = 6.2 in. The line of action of R intersects line AB 11.64 in. to the left of B and intersects line BC 6.20 in. below B. PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 270 PROBLEM 3.110 A couple M and the three forces shown are applied to an angle bracket. Find the moment of the couple if the line of action of the resultant of the force system is to pass through (a) Point A, (b) Point B, (c) Point C. SOLUTION In each case, must have M iR = 0 (a) M AB = ΣM A = M + (12 in.)[(30 lb) sin 60°] − (8 in.)(45 lb) = 0 M = +48.231 lb ⋅ in. (b) M = 48.2 lb ⋅ in. M BR = ΣM B = M + (12 in.)(10 lb) − (8 in.)(45 lb) = 0 M = +240 lb ⋅ in. (c) M = 240 lb ⋅ in. M CR = ΣM C = M + (12 in.)(10 lb) − (8 in.)[(30 lb) cos 60°] = 0 M =0 M=0 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 271 PROBLEM 3.111 Four forces act on a 700 × 375-mm plate as shown. (a) Find the resultant of these forces. (b) Locate the two points where the line of action of the resultant intersects the edge of the plate. SOLUTION (a) R = ΣF = (−400 N + 160 N − 760 N)i + (600 N + 300 N + 300 N) j = −(1000 N)i + (1200 N) j R = (1000 N) 2 + (1200 N)2 = 1562.09 N 1200 N ! tan θ = " − # $ 1000 N % = −1.20000 θ = −50.194° (b) R = 1562 N 50.2° M CR = Σr × F = (0.5 m)i × (300 N + 300 N) j = (300 N ⋅ m)k (300 N ⋅ m)k = xi × (1200 N) j x = 0.25000 m x = 250 mm (300 N ⋅ m) = yj × ( −1000 N)i y = 0.30000 m y = 300 mm Intersection 250 mm to right of C and 300 mm above C PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 272 PROBLEM 3.112 Solve Problem 3.111, assuming that the 760-N force is directed to the right. PROBLEM 3.111 Four forces act on a 700 × 375-mm plate as shown. (a) Find the resultant of these forces. (b) Locate the two points where the line of action of the resultant intersects the edge of the plate. SOLUTION R = ΣF (a) = ( −400N + 160 N + 760 N)i + (600 N + 300 N + 300 N) j = (520 N)i + (1200 N) j R = (520 N) 2 + (1200 N) 2 = 1307.82 N 1200 N ! tan θ = " # = 2.3077 $ 520 N % θ = 66.5714° R = 1308 N 66.6° M CR = Σr × F (b) = (0.5 m)i × (300 N + 300 N) j = (300 N ⋅ m)k (300 N ⋅ m)k = xi × (1200 N) j x = 0.25000 m or x = 0.250 mm (300 N ⋅ m)k = [ x′i + (0.375 m) j] × [(520 N)i + (1200 N) j] = (1200 x′ − 195)k x′ = 0.41250 m or x′ = 412.5 mm Intersection 412 mm to the right of A and 250 mm to the right of C PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 273 PROBLEM 3.113 A truss supports the loading shown. Determine the equivalent force acting on the truss and the point of intersection of its line of action with a line drawn through Points A and G. SOLUTION We have R = ΣF R = (240 lb)(cos 70°i − sin 70° j) − (160 lb) j + (300 lb)(− cos 40°i − sin 40° j) − (180 lb) j R = −(147.728 lb)i − (758.36 lb) j R = Rx2 + Ry2 = (147.728) 2 + (758.36) 2 = 772.62 lb Ry ! # $ Rx % −758.36 ! = tan −1 " # $ −147.728 % = 78.977° θ = tan −1 " or R = 773 lb We have ΣM A = dRy where ΣM A = −[240 lb cos 70°](6 ft) − [240 lbsin 70°](4 ft) 79.0° − (160 lb)(12 ft) + [300 lb cos 40°](6 ft) − [300 lb sin 40°](20 ft) − (180 lb)(8 ft) = −7232.5 lb ⋅ ft −7232.5 lb ⋅ ft −758.36 lb = 9.5370 ft d= or d = 9.54 ft to the right of A PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 274 PROBLEM 3.114 Pulleys A and B are mounted on bracket CDEF. The tension on each side of the two belts is as shown. Replace the four forces with a single equivalent force, and determine where its line of action intersects the bottom edge of the bracket. SOLUTION Equivalent force-couple at A due to belts on pulley A We have ΣF : − 120 lb − 160 lb = RA R A = 280 lb We have ΣM A : −40 lb(2 in.) = M A M A = 80 lb ⋅ in. Equivalent force-couple at B due to belts on pulley B We have ΣF: (210 lb + 150 lb) 25° = R B R B = 360 lb We have 25° ΣM B : − 60 lb(1.5 in.) = M B M B = 90 lb ⋅ in. Equivalent force-couple at F We have ΣF: R F = ( − 280 lb) j + (360 lb)(cos 25°i + sin 25° j) = (326.27 lb)i − (127.857 lb) j R = RF 2 2 = RFx + RFy = (326.27)2 + (127.857)2 = 350.43 lb RFy ! # $ RFx % −127.857 ! = tan −1 " # $ 326.27 % = −21.399° θ = tan −1 " or R F = R = 350 lb 21.4° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 275 PROBLEM 3.114 (Continued) We have ΣM F : M F = − (280 lb)(6 in.) − 80 lb ⋅ in. − [(360 lb) cos 25°](1.0 in.) + [(360 lb) sin 25°](12 in.) − 90 lb ⋅ in. M F = − (350.56 lb ⋅ in.)k To determine where a single resultant force will intersect line FE, M F = dRy d= MF Ry −350.56 lb ⋅ in. −127 ⋅ 857 lb = 2.7418 in. = or d = 2.74 in. ! PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 276 PROBLEM 3.115 A machine component is subjected to the forces and couples shown. The component is to be held in place by a single rivet that can resist a force but not a couple. For P = 0, determine the location of the rivet hole if it is to be located (a) on line FG, (b) on line GH. SOLUTION We have First replace the applied forces and couples with an equivalent force-couple system at G. ΣFx : 200cos 15° − 120 cos 70° + P = Rx Thus Rx = (152.142 + P) N or ΣFy : − 200sin 15° − 120sin 70° − 80 = Ry Ry = −244.53 N or ΣM G : − (0.47 m)(200 N) cos15° + (0.05 m)(200 N)sin15° + (0.47 m)(120 N) cos 70° − (0.19 m)(120 N)sin 70° − (0.13 m)( P N) − (0.59 m)(80 N) + 42 N ⋅ m + 40 N ⋅ m = M G M G = −(55.544 + 0.13P) N ⋅ m or (1) Setting P = 0 in Eq. (1): Now with R at I ΣM G : − 55.544 N ⋅ m = − a(244.53 N) a = 0.227 m or and with R at J ΣM G : − 55.544 N ⋅ m = −b(152.142 N) b = 0.365 m or (a) The rivet hole is 0.365 m above G. (b) The rivet hole is 0.227 m to the right of G. PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 277 PROBLEM 3.116 Solve Problem 3.115, assuming that P = 60 N. PROBLEM 3.115 A machine component is subjected to the forces and couples shown. The component is to be held in place by a single rivet that can resist a force but not a couple. For P = 0, determine the location of the rivet hole if it is to be located (a) on line FG, (b) on line GH. SOLUTION See the solution to Problem 3.115 leading to the development of Equation (1) M G = −(55.544 + 0.13P) N ⋅ m Rx = (152.142 + P) N and P = 60 N For We have Rx = (152.142 + 60) = 212.14 N M G = −[55.544 + 0.13(60)] = −63.344 N ⋅ m Then with R at I ΣM G : −63.344 N ⋅ m = −a(244.53 N) a = 0.259 m or and with R at J ΣM G : −63.344 N ⋅ m = −b(212.14 N) b = 0.299 m or (a) The rivet hole is 0.299 m above G. (b) The rivet hole is 0.259 m to the right of G. PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 278 PROBLEM 3.117 A 32-lb motor is mounted on the floor. Find the resultant of the weight and the forces exerted on the belt, and determine where the line of action of the resultant intersects the floor. SOLUTION We have ΣF : (60 lb)i − (32 lb) j + (140 lb)(cos 30°i + sin 30° j) = R R = (181.244 lb)i + (38.0 lb) j or R = 185.2 lb We have 11.84° ΣM O : ΣM O = xRy − [(140 lb) cos 30°][(4 + 2 cos 30°)in.] − [(140 lb) sin 30°][(2 in.)sin 30°] − (60 lb)(2 in.) = x(38.0 lb) x= and 1 (− 694.97 − 70.0 − 120) in. 38.0 x = −23.289 in. Or, resultant intersects the base (x axis) 23.3 in. to the left of the vertical centerline (y axis) of the motor. PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 279 PROBLEM 3.118 As follower AB rolls along the surface of member C, it exerts a constant force F perpendicular to the surface. (a) Replace F with an equivalent force-couple system at the Point D obtained by drawing the perpendicular from the point of contact to the x axis. (b) For a = 1 m and b = 2 m, determine the value of x for which the moment of the equivalent forcecouple system at D is maximum. SOLUTION (a) The slope of any tangent to the surface of member C is dy d ( x2 = *b ""1 − 2 dx dx *, $ a ! ) −2b ## + = 2 x % -+ a Since the force F is perpendicular to the surface, dy ! tan α = − " # $ dx % −1 = a2 1 ! 2b "$ x #% For equivalence ΣF : F = R ΣM D : ( F cos α )( y A ) = M D where cos α = 2bx (a 2 ) 2 + (2bx)2 x2 ! y A = b ""1 − 2 ## $ a % x3 ! 2 Fb 2 " x − 2 # a % $ MD = 4 2 2 a + 4b x Therefore, the equivalent force-couple system at D is R=F a2 ! tan −1 "" ## $ 2bx % x3 ! 2 Fb2 " x − 2 # a % $ M= a 4 + 4b 2 x 2 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 280 PROBLEM 3.118 (Continued) (b) To maximize M, the value of x must satisfy dM =0 dx a = 1 m, b = 2 m where, for M= 8F ( x − x3 ) 1 + 16 x 2 (1 ) 1 + 16 x 2 (1 − 3x 2 ) − ( x − x3 ) * (32 x)(1 + 16 x 2 ) −1/ 2 + dM 2 , - =0 = 8F 2 dx (1 + 16 x ) (1 + 16 x 2 )(1 − 3x 2 ) − 16 x( x − x3 ) = 0 32 x 4 + 3x 2 − 1 = 0 or x2 = −3 ± 9 − 4(32)(−1) = 0.136011 m 2 2(32) Using the positive value of x2 x = 0.36880 m and − 0.22976 m 2 or x = 369 mm PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 281 ! PROBLEM 3.119 Four forces are applied to the machine component ABDE as shown. Replace these forces by an equivalent force-couple system at A. SOLUTION R = −(50 N) j − (300 N)i − (120 N)i − (250 N)k R = −(420 N)i − (50 N)j − (250 N)k rB = (0.2 m)i rD = (0.2 m)i + (0.16 m)k rE = (0.2 m)i − (0.1 m) j + (0.16 m)k M RA = rB × [−(300 N)i − (50 N) j] + rD × (−250 N)k + r × ( − 120 N)i i j k i j k = 0.2 m 0 0 + 0.2 m 0 0.16 m −300 N −50 N 0 0 0 −250 N i j k + 0.2 m −0.1 m 0.16 m −120 N 0 0 = −(10 N ⋅ m)k + (50 N ⋅ m) j − (19.2 N ⋅ m) j − (12 N ⋅ m)k Force-couple system at A is R = −(420 N)i − (50 N) j − (250 N)k M RA = (30.8 N ⋅ m) j − (220 N ⋅ m)k PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 282 PROBLEM 3.120 Two 150-mm-diameter pulleys are mounted on line shaft AD. The belts at B and C lie in vertical planes parallel to the yz plane. Replace the belt forces shown with an equivalent forcecouple system at A. SOLUTION Equivalent force-couple at each pulley Pulley B R B = (145 N)(− cos 20° j + sin 20°k ) − 215 Nj = − (351.26 N) j + (49.593 N)k M B = − (215 N − 145 N)(0.075 m)i = − (5.25 N ⋅ m)i Pulley C R C = (155 N + 240 N)(− sin10° j − cos10°k ) = − (68.591 N) j − (389.00 N)k M C = (240 N − 155 N)(0.075 m)i = (6.3750 N ⋅ m)i Then R = R B + R C = − (419.85 N) j − (339.41)k or R = (420 N) j − (339 N)k M A = M B + M C + rB/ A × R B + rC/ A × R C i j k 0 0 N⋅m = − (5.25 N ⋅ m)i + (6.3750 N ⋅ m)i + 0.225 0 −351.26 49.593 i j k + 0.45 0 0 N⋅m 0 −68.591 −389.00 = (1.12500 N ⋅ m)i + (163.892 N ⋅ m) j − (109.899 N ⋅ m)k or M A = (1.125 N ⋅ m)i + (163.9 N ⋅ m) j − (109.9 N ⋅ m)k PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 283 ! PROBLEM 3.121 While using a pencil sharpener, a student applies the forces and couple shown. (a) Determine the forces exerted at B and C knowing that these forces and the couple are equivalent to a force-couple system at A consisting of the force R = (2.6 lb)i + Ry j − (0.7 lb)k and the couple M RA = M x i + (1.0 lb · ft)j − (0.72 lb · ft)k. (b) Find the corresponding values of Ry and M x . SOLUTION (a) From the statement of the problem, equivalence requires ΣF : B + C = R or ΣFx : Bx + C x = 2.6 lb (1) ΣFy : − C y = R y (2) ΣFz : − C z = −0.7 lb or C z = 0.7 lb and ΣM A : (rB/A × B + M B ) + rC/A × C = M AR or 1.75 ! ΣM x : (1 lb ⋅ ft) + " ft # (C y ) = M x $ 12 % (3) 3.75 ! 1.75 ! 3.5 ! ΣM y : " ft # ( Bx ) + " ft # (C x ) + " ft # (0.7 lb) = 1 lb ⋅ ft $ 12 % $ 12 % $ 12 % or Using Eq. (1) 3.75Bx + 1.75C x = 9.55 3.75Bx + 1.75(2.6 Bx ) = 9.55 or Bx = 2.5 lb and C x = 0.1 lb 3.5 ! ΣM z : − " ft # (C y ) = −0.72 lb ⋅ ft $ 12 % C y = 2.4686 lb or B = (2.5 lb)i C = (0.1000 lb)i − (2.47 lb) j − (0.700 lb)k (b) Eq. (2) & Using Eq. (3) Ry = −2.47 lb 1.75 ! 1+ " # (2.4686) = M x $ 12 % or M x = 1.360 lb ⋅ ft PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 284 PROBLEM 3.122 A mechanic uses a crowfoot wrench to loosen a bolt at C. The mechanic holds the socket wrench handle at Points A and B and applies forces at these points. Knowing that these forces are equivalent to a force-couple system at C consisting of the force C = (8 lb)i + (4 lb)k and the couple M C = (360 lb · in.)i, determine the forces applied at A and at B when Az = 2 lb. SOLUTION We have ΣF : or Fx : Ax + Bx = 8 lb A+B=C Bx = −( Ax + 8 lb) (1) ΣFy : Ay + By = 0 Ay = − By or (2) ΣFz : 2 lb + Bz = 4 lb Bz = 2 lb or (3) ΣM C : rB/C × B + rA/C × A = M C We have i j 8 Bx 0 By k i j 2 + 8 2 Ax 0 Ay k 8 lb ⋅ in. = (360 lb ⋅ in.)i 2 (2 By − 8 Ay )i + (2 Bx − 16 + 8 Ax − 16) j or + (8By + 8 Ay )k = (360 lb ⋅ in.)i From i-coefficient j-coefficient k-coefficient 2 By − 8 Ay = 360 lb ⋅ in. −2 Bx + 8 Ax = 32 lb ⋅ in. 8 By + 8 Ay = 0 (4) (5) (6) PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 285 PROBLEM 3.122 (Continued) From Equations (2) and (4): 2 By − 8(− By ) = 360 By = 36 lb From Equations (1) and (5): Ay = 36 lb 2(− Ax − 8) + 8 Ax = 32 Ax = 1.6 lb From Equation (1): Bx = −(1.6 + 8) = −9.6 lb A = (1.600 lb)i − (36.0 lb) j + (2.00 lb)k B = −(9.60 lb)i + (36.0 lb) j + (2.00 lb)k PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 286 PROBLEM 3.123 As an adjustable brace BC is used to bring a wall into plumb, the force-couple system shown is exerted on the wall. Replace this force-couple system with an equivalent force-couple system at A if R = 21.2 lb and M = 13.25 lb · ft. SOLUTION We have where ΣF : R = R A = RλBC BC = RA = or We have where (42 in.)i − (96 in.) j − (16 in.)k 106 in. 21.2 lb (42i − 96 j − 16k ) 106 R A = (8.40 lb)i − (19.20 lb) j − (3.20 lb)k ΣM A : rC/A × R + M = M A rC/A = (42 in.)i + (48 in.)k = 1 (42i + 48k )ft 12 = (3.5 ft)i + (4.0 ft)k R = (8.40 lb)i − (19.50 lb) j − (3.20 lb)k M = −λBC M −42i + 96 j + 16k (13.25 lb ⋅ ft) 106 = −(5.25 lb ⋅ ft)i + (12 lb ⋅ ft) j + (2lb ⋅ ft)k = Then i j k 3.5 0 4.0 lb ⋅ ft + (−5.25i + 12 j + 2k )lb ⋅ ft = M A 8.40 −19.20 −3.20 M A = (71.55 lb ⋅ ft)i + (56.80 lb ⋅ ft)j − (65.20 lb ⋅ ft)k or M A = (71.6 lb ⋅ ft)i + (56.8 lb ⋅ ft)j − (65.2 lb ⋅ ft)k PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 287 PROBLEM 3.124 A mechanic replaces a car’s exhaust system by firmly clamping the catalytic converter FG to its mounting brackets H and I and then loosely assembling the mufflers and the exhaust pipes. To position the tailpipe AB, he pushes in and up at A while pulling down at B. (a) Replace the given force system with an equivalent force-couple system at D. (b) Determine whether pipe CD tends to rotate clockwise or counterclockwise relative to muffler DE, as viewed by the mechanic. SOLUTION (a) Equivalence requires ΣF : R = A + B = (100 N)(cos 30° j − sin 30° k ) − (115 N) j = −(28.4 N) j − (50 N)k and where ΣM D : M D = rA/D × FA + rB/D × FB rA/D = −(0.48 m)i − (0.225 m) j + (1.12 m)k rB/D = −(0.38 m)i + (0.82 m)k Then i j k i j k M D = 100 −0.48 −0.225 1.12 + 115 −0.38 0 0.82 0 cos 30° − sin 30° 0 −1 0 = 100[(0.225sin 30° − 1.12cos 30°)i + (−0.48sin 30°) j + (−0.48cos 30°)k ] + 115[(0.82)i + (0.38)k ] = 8.56i − 24.0 j + 2.13k PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 288 PROBLEM 3.124 (Continued) The equivalent force-couple system at D is R = −(28.4 N) j − (50.0 N)k M D = (8.56 N ⋅ m)i − (24.0 N ⋅ m) j + (2.13 N ⋅ m)k (b) Since ( M D ) z is positive, pipe CD will tend to rotate counterclockwise relative to muffler DE. PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 289 PROBLEM 3.125 For the exhaust system of Problem 3.124, (a) replace the given force system with an equivalent force-couple system at F, where the exhaust pipe is connected to the catalytic converter, (b) determine whether pipe EF tends to rotate clockwise or counterclockwise, as viewed by the mechanic. SOLUTION (a) Equivalence requires ΣF : R = A + B = (100 N)(cos 30° j − sin 30° k ) − (115 N) j = −(28.4 N) j − (50 N)k and where M F : M F = rA/F × A + rB/F × B rA/F = −(0.48 m)i − (0.345 m) j + (2.10 m)k rB/F = −(0.38 m)i − (0.12 m) j + (1.80 m)k Then i j k i j k M F = 100 −0.48 −0.345 + 115 −0.38 0.12 1.80 2.10 0 cos 30° − sin 30° 0 −1 0 M F = 100[(0.345 sin 30° − 2.10 cos 30°)i + (−0.48 sin 30°) j + (−0.48 cos 30°)k ] + 115[(1.80)i + (0.38)k ] = 42.4i − 24.0 j + 2.13k PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 290 PROBLEM 3.125 (Continued) The equivalent force-couple system at F is R = −(28.4 N) j − (50 N)k M F = (42.4 N ⋅ m)i − (24.0 N ⋅ m) j + (2.13 N ⋅ m)k (b) Since ( M F ) z is positive, pipe EF will tend to rotate counterclockwise relative to the mechanic. PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 291 PROBLEM 3.126 The head-and-motor assembly of a radial drill press was originally positioned with arm AB parallel to the z axis and the axis of the chuck and bit parallel to the y axis. The assembly was then rotated 25° about the y axis and 20° about the centerline of the horizontal arm AB, bringing it into the position shown. The drilling process was started by switching on the motor and rotating the handle to bring the bit into contact with the workpiece. Replace the force and couple exerted by the drill press with an equivalent forcecouple system at the center O of the base of the vertical column. SOLUTION We have R = F = (11 lb)[( sin 20° cos 25°)]i − (cos 20°) j − (sin 20° sin 25°)k ] = (3.4097 lb)i − (10.3366 lb)j − (1.58998 lb)k or R = (3.41 lb)i − (10.34 lb) j − (1.590 lb)k We have M O = rB/O × F × M C where rB/O = [(14 in.) sin 25°]i + (15 in.) j + [(14 in.) cos 25°]k = (5.9167 in.)i + (15 in.) j + (12.6883 in.)k M C = (90 lb ⋅ in.)[(sin 20° cos 25°)i − (cos 20°) j − (sin 20° sin 25°)k ] = (27.898 lb ⋅ in.)i − (84.572 lb ⋅ in.) j − (13.0090 lb ⋅ in.)k i j k 15 12.6883 lb ⋅ in. M O = 5.9167 3.4097 −10.3366 1.58998 + (27.898 − 84.572 − 13.0090) lb ⋅ in. = (135.202 lb ⋅ in.)i − (31.901 lb ⋅ in.) j − (125.313 lb ⋅ in.)k or M O = (135.2 lb ⋅ in.)i − (31.9 lb ⋅ in.) j − (125.3 lb ⋅ in.)k PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 292 PROBLEM 3.127 Three children are standing on a 5 × 5-m raft. If the weights of the children at Points A, B, and C are 375 N, 260 N, and 400 N, respectively, determine the magnitude and the point of application of the resultant of the three weights. SOLUTION We have ΣF : FA + FB + FC = R −(375 N) j − (260 N) j − (400 N) j = R −(1035 N) j = R or We have R = 1035 N ΣM x : FA ( z A ) + FB ( z B ) + FC ( zC ) = R ( z D ) (375 N)(3 m) + (260 N)(0.5 m) + (400 N)(4.75 m) = (1035 N)(z D ) z D = 3.0483 m We have or z D = 3.05 m ΣM z : FA ( x A ) + FB ( xB ) + FC ( xC ) = R ( xD ) 375 N(1 m) + (260 N)(1.5 m) + (400 N)(4.75 m) = (1035 N)( xD ) xD = 2.5749 m or xD = 2.57 m PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 293 PROBLEM 3.128 Three children are standing on a 5 × 5-m raft. The weights of the children at Points A, B, and C are 375 N, 260 N, and 400 N, respectively. If a fourth child of weight 425 N climbs onto the raft, determine where she should stand if the other children remain in the positions shown and the line of action of the resultant of the four weights is to pass through the center of the raft. SOLUTION We have ΣF : FA + FB + FC = R −(375 N) j − (260 N) j − (400 N) j − (425 N) j = R R = −(1460 N) j We have ΣM x : FA ( z A ) + FB ( z B ) + FC ( zC ) + FD ( z D ) = R( z H ) (375 N)(3 m) + (260 N)(0.5 m) + (400 N)(4.75 m) + (425 N)(z D ) = (1460 N)(2.5 m) z D = 1.16471 m We have or z D = 1.165 m ΣM z : FA ( x A ) + FB ( xB ) + FC ( xC ) + FD ( xD ) = R ( xH ) (375 N)(1 m) + (260 N)(1.5 m) + (400 N)(4.75 m) + (425 N)(xD ) = (1460 N)(2.5 m) xD = 2.3235 m or xD = 2.32 m PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 294 PROBLEM 3.129 Four signs are mounted on a frame spanning a highway, and the magnitudes of the horizontal wind forces acting on the signs are as shown. Determine the magnitude and the point of application of the resultant of the four wind forces when a = 1 ft and b = 12 ft. SOLUTION We have Assume that the resultant R is applied at Point P whose coordinates are (x, y, 0). Equivalence then requires ΣFz : − 105 − 90 − 160 − 50 = − R or R = 405 lb ΣM x : (5 ft)(105 lb) − (1 ft)(90 lb) + (3 ft)(160 lb) + (5.5 ft)(50 lb) = − y (405 lb) or y = −2.94 ft ΣM y : (5.5 ft)(105 lb) + (12 ft)(90 lb) + (14.5 ft)(160 lb) + (22.5 ft)(50 lb) = − x(405 lb) x = 12.60 ft or R acts 12.60 ft to the right of member AB and 2.94 ft below member BC. PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 295 PROBLEM 3.130 Four signs are mounted on a frame spanning a highway, and the magnitudes of the horizontal wind forces acting on the signs are as shown. Determine a and b so that the point of application of the resultant of the four forces is at G. SOLUTION Since R acts at G, equivalence then requires that ΣM G of the applied system of forces also be zero. Then at G : ΣM x : − (a + 3) ft × (90 lb) + (2 ft)(105 lb) + (2.5 ft)(50 lb) = 0 or a = 0.722 ft ! ΣM y : − (9 ft)(105 ft) − (14.5 − b) ft × (90 lb) ! + (8 ft)(50 lb) = 0 or b = 20.6 ft PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 296 ! PROBLEM 3.131* A group of students loads a 2 × 3.3-m flatbed trailer with two 0.66 × 0.66 × 0.66-m boxes and one 0.66 × 0.66 × 1.2-m box. Each of the boxes at the rear of the trailer is positioned so that it is aligned with both the back and a side of the trailer. Determine the smallest load the students should place in a second 0.66 × 0.66 × 1.2-m box and where on the trailer they should secure it, without any part of the box overhanging the sides of the trailer, if each box is uniformly loaded and the line of action of the resultant of the weights of the four boxes is to pass through the point of intersection of the centerlines of the trailer and the axle. (Hint: Keep in mind that the box may be placed either on its side or on its end.) SOLUTION For the smallest weight on the trailer so that the resultant force of the four weights acts over the axle at the intersection with the center line of the trailer, the added 0.66 × 0.66 × 1.2-m box should be placed adjacent to one of the edges of the trailer with the 0.66 × 0.66-m side on the bottom. The edges to be considered are based on the location of the resultant for the three given weights. We have ΣF : − (224 N) j − (392 N) j − (176 N) j = R R = −(792 N) j We have ΣM z : − (224 N)(0.33 m) − (392 N)(1.67 m) − (176 N)(1.67 m) = ( −792 N)( x) xR = 1.29101 m We have ΣM x : (224 N)(0.33 m) + (392 N)(0.6 m) + (176 N)(2.0 m) = (792 N)( z ) z R = 0.83475 m From the statement of the problem, it is known that the resultant of R from the original loading and the lightest load W passes through G, the point of intersection of the two center lines. Thus, ΣM G = 0. Further, since the lightest load W is to be as small as possible, the fourth box should be placed as far from G as possible without the box overhanging the trailer. These two requirements imply (0.33 m # x # 1 m)(1.5 m # z # 2.97 m) PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 297 PROBLEM 3.131* (Continued) xL = 0.33 m With at G : ΣM Z : (1 − 0.33) m × WL − (1.29101 − 1) m × (792 N) = 0 WL = 344.00 N or Now must check if this is physically possible, at G : ΣM x : ( Z L − 1.5)m × 344 N) − (1.5 − 0.83475)m × (792 N) = 0 Z L = 3.032 m or which is not acceptable. Z L = 2.97 m: With at G : ΣM x : (2.97 − 1.5)m × WL − (1.5 − 0.83475)m × (792 N) = 0 WL = 358.42 N or Now check if this is physically possible at G : ΣM z : (1 − X L )m × (358.42 N) − (1.29101 − 1)m × (792 N) = 0 or X L = 0.357 m ok! WL = 358 N The minimum weight of the fourth box is And it is placed on end (A 0.66 × 0.66-m side down) along side AB with the center of the box 0.357 m from side AD. PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 298 PROBLEM 3.132* Solve Problem 3.131 if the students want to place as much weight as possible in the fourth box and at least one side of the box must coincide with a side of the trailer. PROBLEM 3.131* A group of students loads a 2 × 3.3-m flatbed trailer with two 0.66 × 0.66 × 0.66-m boxes and one 0.66 × 0.66 × 1.2-m box. Each of the boxes at the rear of the trailer is positioned so that it is aligned with both the back and a side of the trailer. Determine the smallest load the students should place in a second 0.66 × 0.66 × 1.2-m box and where on the trailer they should secure it, without any part of the box overhanging the sides of the trailer, if each box is uniformly loaded and the line of action of the resultant of the weights of the four boxes is to pass through the point of intersection of the centerlines of the trailer and the axle. (Hint: Keep in mind that the box may be placed either on its side or on its end.) SOLUTION First replace the three known loads with a single equivalent force R applied at coordinate ( X R , 0, Z R ) Equivalence requires ΣFy : − 224 − 392 − 176 = − R or R = 792 N ΣM x : (0.33 m)(224 N) + (0.6 m)(392 N) + (2 m)(176 N) = z R (792 N) or z R = 0.83475 m ΣM z : − (0.33 m)(224 N) − (1.67 m)(392 N) − (1.67 m)(176 N) = xR (792 N) or xR = 1.29101 m From the statement of the problem, it is known that the resultant of R and the heaviest loads WH passes through G, the point of intersection of the two center lines. Thus, ΣM G = 0 Further, since WH is to be as large as possible, the fourth box should be placed as close to G as possible while keeping one of the sides of the box coincident with a side of the trailer. Thus, the two limiting cases are xH = 0.6 m or z H = 2.7 m PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 299 PROBLEM 3.132* (Continued) Now consider these two possibilities With xH = 0.6 m: at G : ΣM z : (1 − 0.6)m × WH − (1.29101 − 1)m × (792 N) = 0 WH = 576.20 N or Checking if this is physically possible at or G : ΣM x : ( z H − 1.5)m × (576.20 N) − (1.5 − 0.83475)m × (792 N) = 0 z H = 2.414 m which is acceptable. With z H = 2.7 m at or G : ΣM x : (2.7 − 1.5) WH − (1.5 − 0.83475)m × (792 N) = 0 WH = 439 N Since this is less than the first case, the maximum weight of the fourth box is WH = 576 N and it is placed with a 0.66 × 1.2-m side down, a 0.66-m edge along side AD, and the center 2.41 m from side DC. PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 300 PROBLEM 3.133 Three forces of the same magnitude P act on a cube of side a as shown. Replace the three forces by an equivalent wrench and determine (a) the magnitude and direction of the resultant force R, (b) the pitch of the wrench, (c) the axis of the wrench. SOLUTION Force-couple system at O: R = Pi + Pj + Pk = P (i + j + k ) M OR = aj × Pi + ak × Pj + ai × Pk = − Pak − Pai − Paj M OR = − Pa (i + j + k ) Since R and M OR have the same direction, they form a wrench with M1 = M OR . Thus, the axis of the wrench is the diagonal OA. We note that cos θ x = cos θ y = cos θ z = a a 3 = 1 3 R = P 3 θ x = θ y = θ z = 54.7° M1 = M OR = − Pa 3 Pitch = p = (a) (b) (c) M 1 − Pa 3 = = −a R P 3 R = P 3 θ x = θ y = θ z = 54.7° –a Axis of the wrench is diagonal OA PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 301 PROBLEM 3.134* A piece of sheet metal is bent into the shape shown and is acted upon by three forces. If the forces have the same magnitude P, replace them with an equivalent wrench and determine (a) the magnitude and the direction of the resultant force R, (b) the pitch of the wrench, (c) the axis of the wrench. SOLUTION ( ) First reduce the given forces to an equivalent force-couple system R, M OR at the origin. We have ΣF : − Pj + Pj + Pk = R R = Pk or "5 # ! ΣM O : − (aP) j + $ −( aP)i + & aP ' k % = M OR (2 ) + * 5 # " M OR = aP & −i − j + k ' 2 ) ( or (a) Then for the wrench R=P and axis = R =k R cos θ x = 0 cos θ y = 0 cos θ z = 1 or (b) θ x = 90° θ y = 90° θ z = 0° Now M1 = λ axis ⋅ M OR 5 # " = k ⋅ aP & −i − j + k ' 2 ) ( 5 = aP 2 Then P= M1 25 aP = R P or P = 5 a 2 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 302 PROBLEM 3.134* (Continued) (c) The components of the wrench are (R , M1 ), where M1 = M1 λ axis , and the axis of the wrench is assumed to intersect the xy plane at Point Q whose coordinates are (x, y, 0). Thus require M z = rQ × R R Where M z = M O × M1 Then 5 # 5 " aP & −i − j + k ' − aPk = ( xi + yj) + Pk 2 ) 2 ( Equating coefficients i : − aP = yP or j: − aP = − xP or y = −a x=a The axis of the wrench is parallel to the z axis and intersects the xy plane at x = a, y = −a. PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 303 PROBLEM 3.135* The forces and couples shown are applied to two screws as a piece of sheet metal is fastened to a block of wood. Reduce the forces and the couples to an equivalent wrench and determine (a) the resultant force R, (b) the pitch of the wrench, (c) the point where the axis of the wrench intersects the xz plane. SOLUTION First, reduce the given force system to a force-couple system. We have ΣF : − (20 N)i − (15 N) j = R We have ΣM O : Σ(rO × F ) + ΣM C = M OR R = 25 N M OR = −20 N(0.1 m)j − (4 N ⋅ m)i − (1 N ⋅ m)j = −(4 N ⋅ m)i − (3 N ⋅ m) j R = −(20.0 N)i − (15.0 N)j (a) (b) We have R R = (−0.8i − 0.6 j) ⋅ [−(4 N ⋅ m)]i − (3 N ⋅ m)j] M1 = λR ⋅ M OR λ= = 5 N⋅m Pitch p= M1 5 N ⋅ m = = 0.200 m R 25 N or p = 0.200 m (c) From above note that M1 = M OR Therefore, the axis of the wrench goes through the origin. The line of action of the wrench lies in the xy plane with a slope of y= 3 x 4 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 304 PROBLEM 3.136* The forces and couples shown are applied to two screws as a piece of sheet metal is fastened to a block of wood. Reduce the forces and the couples to an equivalent wrench and determine (a) the resultant force R, (b) the pitch of the wrench, (c) the point where the axis of the wrench intersects the xz plane. SOLUTION First, reduce the given force system to a force-couple at the origin. We have ΣF : − (10 lb) j − (11 lb) j = R R = − (21 lb) j We have ΣM O : Σ(rO × F ) + ΣM C = M OR M OR i j k i j k = 0 0 20 lb ⋅ in. + 0 0 −15 lb ⋅ in. − (12 lb ⋅ in) j 0 −10 0 0 −11 0 = (35 lb ⋅ in.)i − (12 lb ⋅ in.) j R = − (21 lb) j (a) (b) We have or R = − (21.0 lb) j R R = (− j) ⋅ [(35 lb ⋅ in.)i − (12 lb ⋅ in.) j] M1 = R ⋅ M OR R = = 12 lb ⋅ in. and M1 = −(12 lb ⋅ in.) j and pitch p= M 1 12 lb ⋅ in. = = 0.57143 in. R 21 lb or p = 0.571 in. PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 305 PROBLEM 3.136* (Continued) (c) We have M OR = M1 + M 2 M 2 = M OR − M1 = (35 lb ⋅ in.)i Require M 2 = rQ/O × R (35 lb ⋅ in.)i = ( xi + zk ) × [ −(21 lb) j] 35i = −(21x)k + (21z )i From i: From k: 35 = 21z z = 1.66667 in. 0 = − 21x z=0 The axis of the wrench is parallel to the y axis and intersects the xz plane at x = 0, z = 1.667 in. PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 306 ! PROBLEM 3.137* Two bolts at A and B are tightened by applying the forces and couples shown. Replace the two wrenches with a single equivalent wrench and determine (a) the resultant R, (b) the pitch of the single equivalent wrench, (c) the point where the axis of the wrench intersects the xz plane. SOLUTION First, reduce the given force system to a force-couple at the origin. We have ΣF : − (84 N) j − (80 N)k = R R = 116 N and ΣM O : Σ(rO × F) + ΣM C = M OR i j k i j k 0.6 0 .1 + 0.4 0.3 0 + ( −30 j − 32k )N ⋅ m = M OR 0 84 0 0 0 80 M OR = − (15.6 N ⋅ m)i + (2 N ⋅ m) j − (82.4 N ⋅ m)k R = − (84.0 N) j − (80.0 N)k (a) (b) We have M1 = R ⋅ M OR R = R R −84 j − 80k ⋅ [− (15.6 N ⋅ m)i + (2 N ⋅ m) j − (82.4 N ⋅ m)k ] 116 = 55.379 N ⋅ m =− and Then pitch M1 = M1λR = − (40.102 N ⋅ m) j − (38.192 N ⋅ m)k p= M 1 55.379 N ⋅ m = = 0.47741 m R 116 N or p = 0.477 m PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 307 PROBLEM 3.137* (Continued) (c) We have M OR = M1 + M 2 M 2 = M OR − M1 = [(−15.6i + 2 j − 82.4k ) − (40.102 j − 38.192k )] N ⋅ m = − (15.6 N ⋅ m)i + (42.102 N ⋅ m) j − (44.208 N ⋅ m)k Require M 2 = rQ/O × R (−15.6i + 42.102 j − 44.208k ) = ( xi + zk ) × (84 j − 80k ) = (84 z )i + (80 x) j − (84 x)k From i: or From k: or −15.6 = 84 z z = − 0.185714 m z = − 0.1857 m −44.208 = −84 x x = 0.52629 m x = 0.526 m The axis of the wrench intersects the xz plane at x = 0.526 m y = 0 z = − 0.1857 m PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 308 ! PROBLEM 3.138* Two bolts at A and B are tightened by applying the forces and couples shown. Replace the two wrenches with a single equivalent wrench and determine (a) the resultant R, (b) the pitch of the single equivalent wrench, (c) the point where the axis of the wrench intersects the xz plane. SOLUTION First, reduce the given force system to a force-couple at the origin at B. (a) We have 15 # " 8 ΣF : − (26.4 lb)k − (17 lb) & i + j ' = R 17 17 ( ) R = − (8.00 lb)i − (15.00 lb) j − (26.4 lb)k and We have R = 31.4 lb ΣM B : rA/B × FA + M A + M B = M RB M RB i j k 15 # " 8 0 − 220k − 238 & i + j ' = 264i − 220k − 14(8i + 15 j) = 0 −10 ( 17 17 ) 0 0 − 26.4 M RB = (152 lb ⋅ in.)i − (210 lb ⋅ in.)j − (220 lb ⋅ in.)k (b) We have R R −8.00i − 15.00 j − 26.4k = ⋅ [(152 lb ⋅ in.)i − (210 lb ⋅ in.) j − (220 lb ⋅ in.)k ] 31.4 = 246.56 lb ⋅ in. M1 = R ⋅ M OR R = PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 309 PROBLEM 3.138* (Continued) and Then pitch (c) We have M1 = M1λR = − (62.818 lb ⋅ in.)i − (117.783 lb ⋅ in.) j − (207.30 lb ⋅ in.)k p= M 1 246.56 lb ⋅ in. = = 7.8522 in. 31.4 lb R or p = 7.85 in. M RB = M1 + M 2 M 2 = M RB − M1 = (152i − 210 j − 220k ) − ( − 62.818i − 117.783j − 207.30k ) = (214.82 lb ⋅ in.)i − (92.217 lb ⋅ in.) j − (12.7000 lb ⋅ in.)k Require M 2 = rQ/B × R i j k 214.82i − 92.217 j − 12.7000k = x 0 z −8 −15 −26.4 = (15 z )i − (8 z ) j + (26.4 x) j − (15 x)k From i: 214.82 = 15 z From k: −12.7000 = −15 x The axis of the wrench intersects the xz plane at z = 14.3213 in. x = 0.84667 in. x = 0.847 in. y = 0 z = 14.32 in. PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 310 ! PROBLEM 3.139* Two ropes attached at A and B are used to move the trunk of a fallen tree. Replace the forces exerted by the ropes with an equivalent wrench and determine (a) the resultant force R, (b) the pitch of the wrench, (c) the point where the axis of the wrench intersects the yz plane. SOLUTION (a) ( ) First replace the given forces with an equivalent force-couple system R, M OR at the origin. We have d AC = (6) 2 + (2) 2 + (9) 2 = 11 m d BD = (14) 2 + (2)2 + (5) 2 = 15 m Then 1650 N = (6i + 2 j + 9k ) 11 = (900 N)i + (300 N) j + (1350 N)k TAC = and 1500 N = (14i + 2 j + 5k ) 15 = (1400 N)i + (200 N) j + (500 N)k TBD = Equivalence then requires ΣF : R = TAC + TBD = (900i + 300 j + 1350k ) +(1400i + 200 j + 500k ) = (2300 N)i + (500 N) j + (1850 N)k ΣM O : M OR = rA × TAC + rB × TBD = (12 m)k × [(900 N)i + (300 N)j + (1350 N)k ] + (9 m)i × [(1400 N)i + (200 N)j + (500 N)k ] = −(3600)i + (10800 − 4500) j + (1800)k = −(3600 N ⋅ m)i + (6300 N ⋅ m)j + (1800 N ⋅ m)k The components of the wrench are (R , M1 ), where R = (2300 N)i + (500 N) j + (1850 N)k PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 311 PROBLEM 3.139* (Continued) ! (b) We have R = 100 (23)2 + (5) 2 + (18.5) 2 = 2993.7 N Let λ axis = Then 1 R (23i + 5 j + 18.5k ) = R 29.937 M1 = λ axis ⋅ M OR 1 (23i + 5 j + 18.5k ) ⋅ (−3600i + 6300 j + 1800k ) 29.937 1 [(23)( −36) + (5)(63) + (18.5)(18)] = 0.29937 = −601.26 N ⋅ m = Finally P= M1 −601.26 N ⋅ m = 2993.7 N R or P = − 0.201 m (c) We have M1 = M 1 axis = (−601.26 N ⋅ m) × 1 (23i + 5 j + 18.5k ) 29.937 or M1 = −(461.93 N ⋅ m)i − (100.421 N ⋅ m) j − (371.56 N ⋅ m)k Now M 2 = M OR − M1 = (−3600i + 6300 j + 1800k ) − ( −461.93i − 100.421j − 371.56k ) = − (3138.1 N ⋅ m)i + (6400.4 N ⋅ m)j + (2171.6 N ⋅ m)k For equivalence PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 312 PROBLEM 3.139* (Continued) Thus require M 2 = rP × R r = ( yj + zk ) Substituting i j k −3138.1i + 6400.4 j + 2171.6k = 0 y z 2300 500 1850 Equating coefficients j : 6400.4 = 2300 z or k : 2171.6 = −2300 y or The axis of the wrench intersects the yz plane at z = 2.78 m y = − 0.944 m y = −0.944 m z = 2.78 m PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 313 ! PROBLEM 3.140* A flagpole is guyed by three cables. If the tensions in the cables have the same magnitude P, replace the forces exerted on the pole with an equivalent wrench and determine (a) the resultant force R, (b) the pitch of the wrench, (c) the point where the axis of the wrench intersects the xz plane. SOLUTION (a) First reduce the given force system to a force-couple at the origin. We have ΣF : P BA +P DC +P DE =R 3 # "3 4 # " −9 4 12 # ! "4 R = P $& j − k ' + & i − j ' + & i − j + k ' % 5 ) (5 5 ) ( 25 5 25 ) + *( 5 R= R= We have 3P (2i − 20 j − k ) 25 3P 27 5 (2) 2 + (20) 2 + (1)2 = P 25 25 ΣM : Σ(rO × P) = M OR 3P # 4P # 4P 12 P # " −4 P " 3P " −9 P j− k ' + (20a) j × & i− j ' + (20a) j × & i− j+ k ' = M OR (24a) j × & 5 5 5 5 25 5 25 ( ) ( ) ( ) M OR = 24 Pa ( −i − k ) 5 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 314 PROBLEM 3.140* (Continued) (b) We have M1 = λR ⋅ M OR where λR = Then M1 = and pitch R 3P 25 1 = = (2i − 20 j − k ) (2i − 20 j − k ) R 25 27 5 P 9 5 p= 1 9 5 (2i − 20 j − k ) ⋅ M 1 −8Pa " 25 # −8a = & '= R 15 5 ( 27 5 P ) 81 M1 = M1λR = (c) Then M 2 = M OR − M1 = 24 Pa −8 Pa (−i − k ) = 5 15 5 or p = − 0.0988a −8 Pa " 1 # 8Pa (−2i + 20 j + k ) & ' (2i − 20 j − k ) = 675 15 5 ( 9 5 ) 24 Pa 8Pa 8 Pa ( −i − k ) − (−2i + 20 j + k ) = (−430i − 20 j − 406k ) 5 675 675 M 2 = rQ/O × R Require " 8Pa # " 3P # & 675 ' (−403i − 20 j − 406k ) = ( xi + zk ) × & 25 ' (2i − 20 j − k ) ( ) ( ) " 3P # =& ' [20 zi + ( x + 2 z ) j − 20 xk ] ( 25 ) From i: 8(− 403) Pa " 3P # = 20 z & ' z = −1.99012a 675 ( 25 ) From k: 8(−406) Pa " 3P # = −20 x & ' x = 2.0049a 675 ( 25 ) The axis of the wrench intersects the xz plane at x = 2.00a, z = −1.990a PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 315 ! PROBLEM 3.141* Determine whether the force-and-couple system shown can be reduced to a single equivalent force R. If it can, determine R and the point where the line of action of R intersects the yz plane. If it cannot be so reduced, replace the given system with an equivalent wrench and determine its resultant, its pitch, and the point where its axis intersects the yz plane. SOLUTION First, reduce the given force system to a force-couple at the origin. We have ΣF : FA + FG = R (40 mm)i + (60 mm) j − (120 mm)k ! R = (50 N)k + 70 N $ % 140 mm * + = (20 N)i + (30 N) j − (10 N)k and We have R = 37.417 N ΣM O : Σ(rO × F) + ΣM C = M OR M OR = [(0.12 m) j × (50 N)k ] + {(0.16 m)i × [(20 N)i + (30 N) j − (60 N)k ]} + (10 N ⋅ m) $ * (160 mm)i − (120 mm) j ! % 200 mm + (40 mm)i − (120 mm) j + (60 mm)k ! + (14 N ⋅ m) $ % 140 mm * + R M 0 = (18 N ⋅ m)i − (8.4 N ⋅ m) j + (10.8 N ⋅ m)k To be able to reduce the original forces and couples to a single equivalent force, R and M must be perpendicular. Thus, R ⋅ M = 0. Substituting ? (20i + 30 j − 10k ) ⋅ (18i − 8.4 j + 10.8k ) = 0 ? or (20)(18) + (30)(−8.4) + (−10)(10.8) = 0 or 0=0 R and M are perpendicular so that the given system can be reduced to the single equivalent force R = (20.0 N)i + (30.0 N) j − (10.00 N)k PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 316 PROBLEM 3.141* (Continued) ! Then for equivalence Thus require M OR = rp × R rp = yj + zk Substituting i j k 18i − 8.4 j + 10.8k = 0 y z 20 30 −10 Equating coefficients j: − 8.4 = 20 z k: or z = −0.42 m 10.8 = −20 y or y = −0.54 m The line of action of R intersects the yz plane at x=0 y = −0.540 m z = −0.420 m PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 317 ! PROBLEM 3.142* Determine whether the force-and-couple system shown can be reduced to a single equivalent force R. If it can, determine R and the point where the line of action of R intersects the yz plane. If it cannot be so reduced, replace the given system with an equivalent wrench and determine its resultant, its pitch, and the point where its axis intersects the yz plane. SOLUTION First determine the resultant of the forces at D. We have d DA = (−12) 2 + (9) 2 + (8) 2 = 17 in. d ED = (−6) 2 + (0)2 + (−8)2 = 10 in. Then 34 lb = (−12i + 9 j + 8k ) 17 = −(24 lb)i + (18 lb) j + (16 lb)k FDA = and 30 lb = (−6i − 8k ) 10 = −(18 lb)i − (24 lb)k FED = Then ΣF : R = FDA + FED = (−24i + 18 j + 16k + ( −18i − 24k ) = −(42 lb)i + (18 lb)j − (8 lb)k For the applied couple d AK = ( −6) 2 + (−6) 2 + (18) 2 = 6 11 in. Then M= 160 lb ⋅ in. ( −6i − 6 j + 18k ) 6 11 160 [−(1 lb ⋅ in.)i − (1 lb ⋅ in.)j + (3 lb ⋅ in.)k ] = 11 To be able to reduce the original forces and couple to a single equivalent force, R and M must be perpendicular. Thus ? R ⋅ M =0 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 318 PROBLEM 3.142* (Continued) Substituting (−42i + 18 j − 8k ) ⋅ 160 or 11 160 11 ? (−i − j + 3k ) = 0 ? [(−42)(−1) + (18)(−1) + (−8)(3)] = 0 0 =0 or R and M are perpendicular so that the given system can be reduced to the single equivalent force R = −(42.0 lb)i + (18.00 lb) j − (8.00 lb)k Then for equivalence M = rP/D × R Thus require where rP/D = −(12 in.)i + [( y − 3)in.] j + ( z in.)k Substituting i j k ( −i − j + 3k ) = −12 ( y − 3) z 11 −42 −8 18 = [( y − 3)( −8) − ( z )(18)]i 160 + [( z )(−42) − (−12)(−8)]j + [( −12)(18) − ( y − 3)(−42)]k Equating coefficients j: − k: 160 or = − 42 z − 96 11 480 = −216 + 42( y − 3) or 11 The line of action of R intersects the yz plane at x=0 z = −1.137 in. y = 11.59 in. y = 11.59 in. z = −1.137 in. PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 319 PROBLEM 3.143* Replace the wrench shown with an equivalent system consisting of two forces perpendicular to the y axis and applied respectively at A and B. SOLUTION Express the forces at A and B as A = Ax i + Az k B = Bx i + Bz k Then, for equivalence to the given force system ΣFx : Ax + Bx = 0 (1) ΣFz : Az + Bz = R (2) ΣM x : Az ( a) + Bz ( a + b) = 0 (3) ΣM z : − Ax (a) − Bx (a + b) = M (4) Bx = − Ax From Equation (1), Substitute into Equation (4) − Ax ( a) + Ax ( a + b) = M M M and Bx = − Ax = b b From Equation (2), Bz = R − Az and Equation (3), Az a + ( R − Az )(a + b) = 0 " a# Az = R &1 + ' ( b) and a# " Bz = R − R &1 + ' ( b) a Bz = − R b "M A=& ( b Then a# # " ' i + R &1 + b ' k ) ( ) "M B = −& ( b # "a # 'i − & b R'k ) ( ) PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 320 ! PROBLEM 3.144* Show that, in general, a wrench can be replaced with two forces chosen in such a way that one force passes through a given point while the other force lies in a given plane. SOLUTION First, choose a coordinate system so that the xy plane coincides with the given plane. Also, position the coordinate system so that the line of action of the wrench passes through the origin as shown in Figure a. Since the orientation of the plane and the components (R, M) of the wrench are known, it follows that the scalar components of R and M are known relative to the shown coordinate system. A force system to be shown as equivalent is illustrated in Figure b. Let A be the force passing through the given Point P and B be the force that lies in the given plane. Let b be the x-axis intercept of B. The known components of the wrench can be expressed as R = Rx i + Ry j + Rz k and M = M x i + M y j + M z k while the unknown forces A and B can be expressed as A = Ax i + Ay j + Az k and B = Bx i + Bz k Since the position vector of Point P is given, it follows that the scalar components (x, y, z) of the position vector rP are also known. Then, for equivalence of the two systems ΣFx : Rx = Ax + Bx (1) ΣFy : Ry = Ay (2) ΣFz : Rz = Az + Bz (3) ΣM x : M x = yAz − zAy (4) ΣM y : M y = zAx − xAz − bBz (5) ΣM z : M z = xAy − yAx (6) Based on the above six independent equations for the six unknowns ( Ax , Ay , Az , Bx , By , Bz , b), there exists a unique solution for A and B. Ay = Ry From Equation (2) PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 321 PROBLEM 3.144* (Continued) Equation (6) "1# Ax = & ' ( xRy − M z ) ( y) Equation (1) "1# Bx = Rx − & ' ( xRy − M z ) ( y) Equation (4) "1# Az = & ' ( M x + zRy ) ( y) Equation (3) "1# Bz = Rz − & ' ( M x + zRy ) ( y) b= Equation (5) ( xM x + yM y + zM z ) ( M x − yRz + zRy ) PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 322 ! ! PROBLEM 3.145* Show that a wrench can be replaced with two perpendicular forces, one of which is applied at a given point. SOLUTION First, observe that it is always possible to construct a line perpendicular to a given line so that the constructed line also passes through a given point. Thus, it is possible to align one of the coordinate axes of a rectangular coordinate system with the axis of the wrench while one of the other axes passes through the given point. See Figures a and b. We have R = Rj and M = Mj and are known. The unknown forces A and B can be expressed as A = Ax i + Ay j + Az k and B = Bx i + By j + Bz k The distance a is known. It is assumed that force B intersects the xz plane at (x, 0, z). The for equivalence ΣFx : 0 = Ax + Bx (1) ΣFy : R = Ay + By (2) ΣFz : 0 = Az + Bz (3) 0 = − zBy (4) ΣM x : ΣM y : M = − aAz − xBz + zBx ΣM z : (5) 0 = aAy + xBy (6) Since A and B are made perpendicular, A ⋅ B = 0 or There are eight unknowns: Ax Bx + Ay B y + Az Bz = 0 (7) Ax , Ay , Az , Bx , By , Bz , x, z But only seven independent equations. Therefore, there exists an infinite number of solutions. 0 = − zBy Next consider Equation (4): If By = 0, Equation (7) becomes Ax Bx + Az Bz = 0 Ax2 + Az2 = 0 Using Equations (1) and (3) this equation becomes PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 323 PROBLEM 3.145* (Continued) Since the components of A must be real, a nontrivial solution is not possible. Thus, it is required that By ≠ 0, so that from Equation (4), z = 0. To obtain one possible solution, arbitrarily let Ax = 0. (Note: Setting Ay , Az , or Bz equal to zero results in unacceptable solutions.) The defining equations then become 0 = Bx (1)′ R = Ay + By (2) 0 = Az + Bz (3) M = − aAz − xBz (5)′ 0 = aAy + xBy (6) Ay By + Az Bz = 0 (7)′ Then Equation (2) can be written Ay = R − By Equation (3) can be written Bz = − Az Equation (6) can be written x=− aAy By Substituting into Equation (5)′, " R − By M = − aAz − & − a & By ( M Az = − By aR or # ' ( − Az ) ' ) (8) Substituting into Equation (7)′, " M #" M # ( R − By ) By + & − By '& By ' = 0 ( aR )( aR ) or By = a 2 R3 a2 R2 + M 2 Then from Equations (2), (8), and (3) a2 R2 RM 2 = a2 R2 + M 2 a2 R2 + M 2 # M " a 2 R3 aR 2 M = − Az = − && 2 2 ' aR ( a R + M 2 ') a2 R2 + M 2 Ay = R − Bz = aR 2 M a2 R2 + M 2 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 324 PROBLEM 3.145* (Continued) In summary A= RM ( Mj − aRk ) a R2 + M 2 B= aR 2 (aRj + Mk ) a2 R2 + M 2 2 Which shows that it is possible to replace a wrench with two perpendicular forces, one of which is applied at a given point. Lastly, if R . 0 and M . 0, it follows from the equations found for A and B that Ay . 0 and By . 0. From Equation (6), x , 0 (assuming a . 0). Then, as a consequence of letting Ax = 0, force A lies in a plane parallel to the yz plane and to the right of the origin, while force B lies in a plane parallel to the yz plane but to the left to the origin, as shown in the figure below. ! PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 325 PROBLEM 3.146* Show that a wrench can be replaced with two forces, one of which has a prescribed line of action. SOLUTION First, choose a rectangular coordinate system where one axis coincides with the axis of the wrench and another axis intersects the prescribed line of action (AA′). Note that it has been assumed that the line of action of force B intersects the xz plane at Point P(x, 0, z). Denoting the known direction of line AA′ by λ A = λx i + λ y j + λz k it follows that force A can be expressed as A = Aλ A = A(λx i + λ y j + λz k ) Force B can be expressed as B = Bx i + By j + Bz k Next, observe that since the axis of the wrench and the prescribed line of action AA′ are known, it follows that the distance a can be determined. In the following solution, it is assumed that a is known. Then, for equivalence ΣFx : 0 = Aλx + Bx (1) ΣFy : R = Aλ y + By (2) ΣFz : 0 = Aλz + Bz (3) ΣM x : 0 = − zBy (4) ΣM y : M = − aAλz + zBx − xBz (5) ΣM x : 0 = − aAλ y + xBy (6) Since there are six unknowns (A, Bx, By, Bz, x, z) and six independent equations, it will be possible to obtain a solution. PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 326 PROBLEM 3.146* (Continued) Case 1: Let z = 0 to satisfy Equation (4) Now Equation (2) Equation (3) Equation (6) Aλ y = R − By Bz = − Aλz x=− aAλ y By " a = −& & By ( # ' ( R − By ) ' ) Substitution into Equation (5) " a M = − aAλz − $ − & $* &( By A=− ! # ' ( R − By )(− Aλz ) % ' %+ ) 1 "M # B λz &( aR ') y Substitution into Equation (2) R=− By = Then 1 "M # B λ + By λz &( aR ') y y λz aR 2 λz aR − λ y M MR R = λz aR − λ y M λ − aR λ y z M λx MR Bx = − Aλx = λz aR − λ y M A=− Bz = − Aλz = λz MR λz aR − λ y M A= In summary B= and R λz aR − λ y M P λA aR λy − λz M (λx Mi + λz aRj + λz M k ) " R # x = a &1 − ' & By ') ( " λz aR − λ y M = a $1 − R & & λ aR 2 $* z ( #! '' % ) %+ or x = λy M λz R Note that for this case, the lines of action of both A and B intersect the x axis. PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 327 PROBLEM 3.146* (Continued) Case 2: Let By = 0 to satisfy Equation (4) Now Equation (2) A= R λy Equation (1) "λ Bx = − R & x & λy ( # ' ' ) Equation (3) "λ Bz = − R & z & λy ( # ' ' ) Equation (6) aAλ y = 0 which requires a = 0 Substitution into Equation (5) "λ M = z $− R & x $* &( λ y #! "λ '% − x $− R & z '% $* &( λ y )+ #! "M # ' % or λz x − λx z = & ' λ y '% ( R) )+ This last expression is the equation for the line of action of force B. In summary " R A=& & λy ( # ' λA ' ) " R B=& & λy ( # ' ( −λ x i − λx k ) ' ) Assuming that λx , λ y , λz . 0, the equivalent force system is as shown below. Note that the component of A in the xz plane is parallel to B.! PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 328 PROBLEM 3.147 A crate of mass 80 kg is held in the position shown. Determine (a) the moment produced by the weight W of the crate about E, (b) the smallest force applied at B that creates a moment of equal magnitude and opposite sense about E. SOLUTION (a) By definition We have W = mg = 80 kg(9.81 m/s 2 ) = 784.8 N ΣM E : M E = (784.8 N)(0.25 m) M E = 196.2 N ⋅ m (b) For the force at B to be the smallest, resulting in a moment (ME) about E, the line of action of force FB must be perpendicular to the line connecting E to B. The sense of FB must be such that the force produces a counterclockwise moment about E. Note: We have d = (0.85 m) 2 + (0.5 m) 2 = 0.98615 m ΣM E : 196.2 N ⋅ m = FB (0.98615 m) FB = 198.954 N and " 0.85 m # ' = 59.534° ( 0.5 m ) θ = tan −1 & FB = 199.0 N or 59.5° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 329 ! PROBLEM 3.148 It is known that the connecting rod AB exerts on the crank BC a 1.5-kN force directed down and to the left along the centerline of AB. Determine the moment of the force about C. SOLUTION Using (a) M C = y1 ( FAB ) x + x1 ( FAB ) y " 7 # " 24 # = (0.028 m) & × 1500 N ' + (0.021 m) & × 1500 N ' ( 25 ) ( 25 ) = 42 N ⋅ m or M C = 42.0 N ⋅ m Using (b) M C = y2 ( FAB ) x " 7 # = (0.1 m) & × 1500 N ' = 42 N ⋅ m ( 25 ) or M C = 42.0 N ⋅ m PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 330 ! PROBLEM 3.149 A 6-ft-long fishing rod AB is securely anchored in the sand of a beach. After a fish takes the bait, the resulting force in the line is 6 lb. Determine the moment about A of the force exerted by the line at B. SOLUTION We have Then Txz = (6 lb) cos 8° = 5.9416 lb Tx = Txz sin 30° = 2.9708 lb Ty = TBC sin 8° = − 0.83504 lb Tz = Txz cos 30° = −5.1456 lb Now M A = rB/A × TBC where rB/A = (6sin 45°) j − (6cos 45°)k = 6 ft 2 (j − k) i j k −1 0 1 2 2.9708 −0.83504 −5.1456 6 6 6 = (−5.1456 − 0.83504)i − (2.9708) j − (2.9708)k 2 2 2 6 Then MA = or M A = −(25.4 lb ⋅ ft)i − (12.60 lb ⋅ ft) j − (12.60 lb ⋅ ft)k PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 331 ! PROBLEM 3.150 Ropes AB and BC are two of the ropes used to support a tent. The two ropes are attached to a stake at B. If the tension in rope AB is 540 N, determine (a) the angle between rope AB and the stake, (b) the projection on the stake of the force exerted by rope AB at Point B. SOLUTION First note BA = (−3) 2 + (3)2 + (−1.5) 2 = 4.5 m BD = (−0.08) 2 + (0.38)2 + (0.16) 2 = 0.42 m λ BD (a) TBA (−3i + 3j − 1.5k ) 4.5 T = BA (−2i + 2 j − k ) 3 BD 1 = = ( −0.08i + 0.38 j + 0.16k ) BD 0.42 1 = (−4i + 19 j + 8k ) 21 TBA = Then We have or or TBA ⋅ λ BD = TBA cos θ TBA 1 ( −2i + 2 j − k ) ⋅ (−4i + 19 j + 8k ) = TBA cos θ 3 21 1 [(−2)( −4) + (2)(19) + (−1)(8)] 63 = 0.60317 cos θ = or θ = 52.9° (b) We have (TBA ) BD = TBA ⋅ λ BD = TBA cos θ = (540 N)(0.60317) or (TBA ) BD = 326 N PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 332 ! PROBLEM 3.151 A farmer uses cables and winch pullers B and E to plumb one side of a small barn. If it is known that the sum of the moments about the x axis of the forces exerted by the cables on the barn at Points A and D is equal to 4728 lb ⋅ ft, determine the magnitude of TDE when TAB = 255 lb. SOLUTION The moment about the x axis due to the two cable forces can be found using the z components of each force acting at their intersection with the xy-plane (A and D). The x components of the forces are parallel to the x axis, and the y components of the forces intersect the x axis. Therefore, neither the x or y components produce a moment about the x axis. We have ΣM x : (TAB ) z ( y A ) + (TDE ) z ( yD ) = M x where (TAB ) z = k ⋅ TAB = k ⋅ (TAB λ AB ) " − i − 12 j + 12k # ! = k ⋅ $ 255 lb & '% 17 ( )+ * = 180 lb (TDE ) z = k ⋅ TDE = k ⋅ (TDE λDE ) " 1.5i − 14 j + 12k # ! = k ⋅ $TDE & '% 18.5 ( )+ * = 0.64865TDE y A = 12 ft yD = 14 ft M x = 4728 lb ⋅ ft (180 lb)(12 ft) + (0.64865TDE )(14 ft) = 4728 lb ⋅ ft and TDE = 282.79 lb or TDE = 283 lb PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 333 ! PROBLEM 3.152 Solve Problem 3.151 when the tension in cable AB is 306 lb. PROBLEM 3.151 A farmer uses cables and winch pullers B and E to plumb one side of a small barn. If it is known that the sum of the moments about the x axis of the forces exerted by the cables on the barn at Points A and D is equal to 4728 lb ⋅ ft, determine the magnitude of TDE when TAB = 255 lb. SOLUTION The moment about the x axis due to the two cable forces can be found using the z components of each force acting at the intersection with the xy plane (A and D). The x components of the forces are parallel to the x axis, and the y components of the forces intersect the x axis. Therefore, neither the x or y components produce a moment about the x axis. We have ΣM x : (TAB ) z ( y A ) + (TDE ) z ( yD ) = M x Where (TAB ) z = k ⋅ TAB = k ⋅ (TAB AB ) " − i − 12 j + 12k # ! = k ⋅ $306 lb & '% 17 ( )+ * = 216 lb (TDE ) z = k ⋅ TDE = k ⋅ (TDE DE ) " 1.5i − 14 j + 12k # ! = k ⋅ $TDE & '% 18.5 ( )+ * = 0.64865TDE y A = 12 ft yD = 14 ft M x = 4728 lb ⋅ ft (216 lb)(12 ft) + (0.64865TDE )(14 ft) = 4728 lb ⋅ ft and TDE = 235.21 lb or TDE = 235 lb PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 334 ! PROBLEM 3.153 A wiring harness is made by routing either two or three wires around 2-in.-diameter pegs mounted on a sheet of plywood. If the force in each wire is 3 lb, determine the resultant couple acting on the plywood when a = 18 in. and (a) only wires AB and CD are in place, (b) all three wires are in place. SOLUTION In general, M = ΣdF, where d is the perpendicular distance between the lines of action of the two forces acting on a given wire. (a) We have M = d AB FAB + dCD FCD 4 " # = (2 + 24) in. × 3 lb + & 2 + × 28 ' in. × 3 lb 5 ( ) or M = 151.2 lb ⋅ in. (b) We have M = [d AB FAB + dCD FCD ] + d EF FEF = 151.2 lb ⋅ in. − 28 in. × 3 lb or M = 67.2 lb ⋅ in. PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 335 ! PROBLEM 3.154 A worker tries to move a rock by applying a 360-N force to a steel bar as shown. (a) Replace that force with an equivalent force-couple system at D. (b) Two workers attempt to move the same rock by applying a vertical force at A and another force at D. Determine these two forces if they are to be equivalent to the single force of Part a. SOLUTION (a) (a) We have (b) ΣF : 360 N(− sin 40°i − cos 40° j) = −(231.40 N)i − (275.78 N) j = F or F = 360 N We have where 50° ΣM D : rB/D × R = M rB/D = −[(0.65 m) cos 30°]i + [(0.65 m)sin 30°]j = −(0.56292 m)i + (0.32500 m) j i j k M = −0.56292 0.32500 0 N ⋅ m −231.40 −275.78 0 = [155.240 + 75.206)N ⋅ m]k = (230.45 N ⋅ m)k (b) We have where or M = 230 N ⋅ m ΣM D : M = rA/D × FA rB/D = −[(1.05 m) cos 30°]i + [(1.05 m)sin 30°]j = −(0.90933 m)i + (0.52500 m) j i j k FA = −0.90933 0.52500 0 N ⋅ m 0 −1 = [230.45 N ⋅ m]k 0 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 336 PROBLEM 3.154 (Continued) (0.90933FA )k = 230.45k or FA = 253.42 N We have or FA = 253 N ΣF : F = FA + FD −(231.40 N)i − (275.78 N) j = −(253.42 N) j + FD (− cos θ i − sin θ j) From i : 231.40 N = FD cos θ (1) j: 22.36 N = FD sin θ (2) Equation (2) divided by Equation (1) tan θ = 0.096629 θ = 5.5193° or θ = 5.52° Substitution into Equation (1) FD = 231.40 = 232.48 N cos 5.5193° or FD = 232 N 5.52° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 337 PROBLEM 3.155 A 110-N force acting in a vertical plane parallel to the yz plane is applied to the 220-mm-long horizontal handle AB of a socket wrench. Replace the force with an equivalent forcecouple system at the origin O of the coordinate system. SOLUTION We have ΣF : PB = F where PB = 110 N[− (sin15°) j + (cos15°)k ] = −(28.470 N) j + (106.252 N)k or F = −(28.5 N) j + (106.3 N)k We have where ΣM O : rB/O × PB = M O rB/O = [(0.22 cos 35°)i + (0.15) j − (0.22sin 35°)k ]m = (0.180213 m)i + (0.15 m) j − (0.126187 m)k i j k 0.180213 0.15 0.126187 N ⋅ m = M O 0 106.3 −28.5 M O = [(12.3487)i − (19.1566) j − (5.1361)k ]N ⋅ m or M O = (12.35 N ⋅ m)i − (19.16 N ⋅ m)j − (5.13 N ⋅ m)k PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 338 PROBLEM 3.156 Four ropes are attached to a crate and exert the forces shown. If the forces are to be replaced with a single equivalent force applied at a point on line AB, determine (a) the equivalent force and the distance from A to the point of application of the force when α = 30°, (b) the value of α so that the single equivalent force is applied at Point B. SOLUTION We have (a) For equivalence ΣFx : −100 cos 30° + 400 cos 65° + 90 cos 65° = Rx or Rx = 120.480 lb ΣFy : 100 sin α + 160 + 400 sin 65° + 90 sin 65° = R y or R y = (604.09 + 100sin α ) lb With α = 30° R y = 654.09 lb Then R = (120.480) 2 + (654.09) 2 (1) = 665 lb Also 654.09 120.480 or θ = 79.6° tan θ = ΣM A : (46 in.)(160 lb) + (66 in.)(400 lb) sin 65° + (26 in.)(400 lb) cos 65° + (66 in.)(90 lb) sin 65° + (36 in.)(90 lb) cos 65° = d (654.09 lb) or ΣM A = 42, 435 lb ⋅ in. and d = 64.9 in. R = 665 lb 79.6° and R is applied 64.9 in. To the right of A. (b) We have d = 66 in. Then or Using Eq. (1) ΣM A : 42, 435 lb ⋅ in = (66 in.) Ry Ry = 642.95 lb 642.95 = 604.09 + 100sin α or α = 22.9° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 339 PROBLEM 3.157 A blade held in a brace is used to tighten a screw at A. (a) Determine the forces exerted at B and C, knowing that these forces are equivalent to a force-couple system at A consisting of R = −(30 N)i + Ry j + Rz k and M RA = − (12 N · m)i. (b) Find the corresponding values of Ry and Rz . (c) What is the orientation of the slot in the head of the screw for which the blade is least likely to slip when the brace is in the position shown? SOLUTION (a) Equivalence requires or Equating the i coefficients Also or Equating coefficients ΣF : R = B + C −(30 N)i + Ry j + Rz k = − Bk + (−Cx i + C y j + C z k ) i : − 30 N = −Cx or Cx = 30 N ΣM A : M RA = rB/A × B + rC/A × C −(12 N ⋅ m)i = [(0.2 m)i + (0.15 m)j] × (− B)k +(0.4 m)i × [−(30 N)i + C y j + Cz k ] i : − 12 N ⋅ m = −(0.15 m) B k : 0 = (0.4 m)C y or B = 80 N or C y = 0 j: 0 = (0.2 m)(80 N) − (0.4 m)C z or Cz = 40 N B = −(80.0 N)k C = −(30.0 N)i + (40.0 N)k (b) Now we have for the equivalence of forces −(30 N)i + Ry j + Rz k = −(80 N)k + [(−30 N)i + (40 N)k ] Equating coefficients j: R y = 0 Ry = 0 k : Rz = −80 + 40 (c) or ! Rz = −40.0 N First note that R = −(30 N)i − (40 N)k. Thus, the screw is best able to resist the lateral force Rz when the slot in the head of the screw is vertical. PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 340 PROBLEM 3.158 A concrete foundation mat in the shape of a regular hexagon of side 12 ft supports four column loads as shown. Determine the magnitudes of the additional loads that must be applied at B and F if the resultant of all six loads is to pass through the center of the mat. SOLUTION From the statement of the problem it can be concluded that the six applied loads are equivalent to the resultant R at O. It then follows that ΣM O = 0 or ΣM x = 0 ΣM z = 0 For the applied loads. Then ΣM x = 0: (6 3 ft) FB + (6 3 ft)(10 kips) − (6 3 ft)(20 kips) − (6 3 ft) FF = 0 FB − FF = 10 or (1) ΣM z = 0: (12 ft)(15 kips) + (6 ft) FB − (6 ft)(10 kips) − (12 ft)(30 kips) − (6 ft)(20 kips) + (6 ft) FF = 0 FB + FF = 60 or (2) Then (1) + (2) - FB = 35.0 kips and FF = 25.0 kips PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 341 CHAPTER 4 PROBLEM 4.1 A 2100-lb tractor is used to lift 900 lb of gravel. Determine the reaction at each of the two (a) rear wheels A, (b) front wheels B. SOLUTION (a) Rear wheels ΣM B = 0: + (2100 lb)(40 in.) − (900 lb)(50 in.) + 2 A(60 in.) = 0 A = +325 lb (b) Front wheels ΣM A : − (2100 lb)(20 in.) − (900 lb)(110 in.) − 2 B(60 in.) = 0 B = +1175 lb Check: A = 325 lb B = 1175 lb +ΣFy = 0: 2 A + 2 B − 2100 lb − 900 lb = 0 2(325 lb) + 2(1175 lb) − 2100 lb − 900 = 0 0 = 0 (Checks) ! PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 345 PROBLEM 4.2 A gardener uses a 60-N wheelbarrow to transport a 250-N bag of fertilizer. What force must she exert on each handle? SOLUTION Free-Body Diagram: ! ΣM A = 0: (2 F )(1 m) − (60 N)(0.15 m) − (250 N)(0.3 m) = 0 F = 42.0 N ! ! ! PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 346 PROBLEM 4.3 The gardener of Problem 4.2 wishes to transport a second 250-N bag of fertilizer at the same time as the first one. Determine the maximum allowable horizontal distance from the axle A of the wheelbarrow to the center of gravity of the second bag if she can hold only 75 N with each arm. PROBLEM 4.2 A gardener uses a 60-N wheelbarrow to transport a 250-N bag of fertilizer. What force must she exert on each handle? SOLUTION Free-Body Diagram: ! ΣMA = 0: 2(75 N)(1 m) − (60 N)(0.15 m) − (250 N)(0.3 m) − (250 N) x = 0 x = 0.264 m ! ! ! PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 347 PROBLEM 4.4 For the beam and loading shown, determine (a) the reaction at A, (b) the tension in cable BC. SOLUTION Free-Body Diagram: (a) Reaction at A: ΣFx = 0: Ax = 0 ΣMB = 0: (15 lb)(28 in.) + (20 lb)(22 in.) + (35 lb)(14 in.) + (20 lb)(6 in.) − Ay (6 in.) = 0 Ay = +245 lb (b) Tension in BC A = 245 lb ΣM A = 0: (15 lb)(22 in.) + (20 lb)(16 in.) + (35 lb)(8 in.) − (15 lb)(6 in.) − FBC (6 in.) = 0 FBC = +140.0 lb Check: FBC = 140.0 lb ΣFy = 0: − 15 lb − 20 lb = 35 lb − 20 lb + A − FBC = 0 −105 lb + 245 lb − 140.0 = 0 0 = 0 (Checks) ! PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 348 PROBLEM 4.5 Two crates, each of mass 350 kg, are placed as shown in the bed of a 1400-kg pickup truck. Determine the reactions at each of the two (a) rear wheels A, (b) front wheels B. SOLUTION Free-Body Diagram: W = (350 kg)(9.81 m/s2 ) = 3.434 kN Wt = (1400 kg)(9.81 m/s 2 ) = 13.734 kN (a) Rear wheels ΣM B = 0: W (1.7 m + 2.05 m) + W (2.05 m) + Wt (1.2 m) − 2 A(3 m) = 0 (3.434 kN)(3.75 m) + (3.434 kN)(2.05 m) + (13.734 kN)(1.2 m) − 2 A(3 m) = 0 A = +6.0663 kN (b) Front wheels A = 6.07 kN ΣFy = 0: − W − W − Wt + 2 A + 2 B = 0 −3.434 kN − 3.434 kN − 13.734 kN + 2(6.0663 kN) + 2B = 0 B = + 4.2347 kN B = 4.23 kN PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 349 PROBLEM 4.6 Solve Problem 4.5, assuming that crate D is removed and that the position of crate C is unchanged. PROBLEM 4.5 Two crates, each of mass 350 kg, are placed as shown in the bed of a 1400-kg pickup truck. Determine the reactions at each of the two (a) rear wheels A, (b) front wheels B. SOLUTION Free-Body Diagram: W = (350 kg)(9.81 m/s2 ) = 3.434 kN Wt = (1400 kg)(9.81 m/s 2 ) = 13.734 kN (a) Rear wheels ΣM B = 0: W (1.7 m + 2.05 m) + Wt (1.2 m) − 2 A(3 m) = 0 (3.434 kN)(3.75 m) + (13.734 kN)(1.2 m) − 2 A(3 m) = 0 A = + 4.893 kN (b) Front wheels A = 4.89 kN ΣM y = 0: − W − Wt + 2 A + 2 B = 0 −3.434 kN − 13.734 kN + 2(4.893 kN) + 2B = 0 B = +3.691 kN B = 3.69 kN PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 350 PROBLEM 4.7 A T-shaped bracket supports the four loads shown. Determine the reactions at A and B (a) if a = 10 in., (b) if a = 7 in. SOLUTION Free-Body Diagram: ΣFx = 0: Bx = 0 ΣM B = 0: (40 lb)(6 in.) − (30 lb)a − (10 lb)(a + 8 in.) + (12 in.) A = 0 (40a − 160) 12 A= (1) ΣM A = 0: − (40 lb)(6 in.) − (50 lb)(12 in.) − (30 lb)(a + 12 in.) − (10 lb)(a + 20 in.) + (12 in.) B y = 0 By = Bx = 0 B = Since (a) (b) (1400 + 40a) 12 (1400 + 40a) 12 (2) For a = 10 in. Eq. (1): A= (40 × 10 − 160) = +20.0 lb 12 Eq. (2): B= (1400 + 40 × 10) = +150.0 lb 12 B = 150.0 lb Eq. (1): A= (40 × 7 − 160) = +10.00 lb 12 A = 10.00 lb Eq. (2): B= (1400 + 40 × 7) = +140.0 lb 12 B = 140.0 lb A = 20.0 lb For a = 7 in. PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 351 PROBLEM 4.8 For the bracket and loading of Problem 4.7, determine the smallest distance a if the bracket is not to move. PROBLEM 4.7 A T-shaped bracket supports the four loads shown. Determine the reactions at A and B (a) if a = 10 in., (b) if a = 7 in. SOLUTION Free-Body Diagram: For no motion, reaction at A must be downward or zero; smallest distance a for no motion corresponds to A = 0. ΣM B = 0: (40 lb)(6 in.) − (30 lb)a − (10 lb)(a + 8 in.) + (12 in.) A = 0 A= (40a − 160) 12 A = 0: (40a − 160) = 0 a = 4.00 in. PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 352 ! PROBLEM 4.9 The maximum allowable value of each of the reactions is 180 N. Neglecting the weight of the beam, determine the range of the distance d for which the beam is safe. SOLUTION ΣFx = 0: Bx = 0 B = By ΣM A = 0: (50 N) d − (100 N)(0.45 m − d ) − (150 N)(0.9 m − d ) + B (0.9 m − d ) = 0 50d − 45 + 100d − 135 + 150d + 0.9 B − Bd d= 180 N ⋅ m − (0.9 m) B 300 A − B (1) ΣM B = 0: (50 N)(0.9 m) − A(0.9 m − d ) + (100 N)(0.45 m) = 0 45 − 0.9 A + Ad + 45 = 0 (0.9 m) A − 90 N ⋅ m A (2) d$ 180 − (0.9)180 18 = = 0.15 m 300 − 180 120 d $ 150.0 mm " d# (0.9)180 − 90 72 = = 0.40 m 180 180 d= Since B # 180 N, Eq. (1) yields. Since A # 180 N, Eq. (2) yields. Range: d # 400 mm " ! 150.0 mm # d # 400 mm PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 353 PROBLEM 4.10 Solve Problem 4.9 if the 50-N load is replaced by an 80-N load. PROBLEM 4.9 The maximum allowable value of each of the reactions is 180 N. Neglecting the weight of the beam, determine the range of the distance d for which the beam is safe. SOLUTION ΣFx = 0: Bx = 0 B = By ΣM A = 0: (80 N)d − (100 N)(0.45 m − d ) − (150 N)(0.9 m − d ) + B(0.9 m − d ) = 0 80d − 45 + 100d − 135 + 150d + 0.9 B − Bd = 0 d= 180 N ⋅ m − 0.9 B 330N − B (1) ΣM B = 0: (80 N)(0.9 m) − A(0.9 m − d ) + (100 N)(0.45 m) = 0 d= 0.9 A − 117 A (2) Since B # 180 N, Eq. (1) yields. d $ (180 − 0.9 × 180)/(330 − 180) = 18 = 0.12 m 150 d = 120.0 mm " Since A # 180 N, Eq. (2) yields. d # (0.9 × 180 − 112)/180 = Range: 45 = 0.25 m 180 d = 250 mm " 120.0 mm # d # 250 mm PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 354 PROBLEM 4.11 For the beam of Sample Problem 4.2, determine the range of values of P for which the beam will be safe, knowing that the maximum allowable value of each of the reactions is 30 kips and that the reaction at A must be directed upward. SOLUTION ΣFx = 0: Bx = 0 B = By ΣM A = 0: − P(3 ft) + B (9 ft) − (6 kips)(11 ft) − (6 kips)(13 ft) = 0 P = 3B − 48 kips (1) ΣM B = 0: − A(9 ft) + P (6 ft) − (6 kips)(2 ft) − (6 kips)(4 ft) = 0 P = 1.5 A + 6 kips (2) Since B # 30 kips, Eq. (1) yields. P # (3)(30 kips) − 48 kips P # 42.0 kips " Since 0 # A # 30 kips. Eq. (2) yields. 0 + 6 kips # P # (1.5)(30 kips)1.6 kips 6.00 kips # P # 51.0 kips " Range of values of P for which beam will be safe: 6.00 kips # P # 42.0 kips PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 355 PROBLEM 4.12 The 10-m beam AB rests upon, but is not attached to, supports at C and D. Neglecting the weight of the beam, determine the range of values of P for which the beam will remain in equilibrium. SOLUTION Free-Body Diagram: ΣM C = 0: P(2 m) − (4 kN)(3 m) − (20 kN)(8 m) + D(6 m) = 0 P = 86 kN − 3D (1) ΣM D = 0: P(8 m) + (4 kN)(3 m) − (20 kN)(2 m) − C (6 m) = 0 P = 3.5 kN + 0.75C (2) For no motion C $ 0 and D $ 0 For C $ 0 from (2) P # 3.50 kN For D $ 0 from (1) P # 86.0 kN Range of P for no motion: 3.50 kN # P # 86.0 kN PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 356 PROBLEM 4.13 The maximum allowable value of each of the reactions is 50 kN, and each reaction must be directed upward. Neglecting the weight of the beam, determine the range of values of P for which the beam is safe. SOLUTION Free-Body Diagram: ΣM C = 0: P(2 m) − (4 kN)(3 m) − (20 kN)(8 m) + D(6 m) = 0 P = 86 kN − 3D (1) ΣM D = 0: P(8 m) + (4 kN)(3 m) − (20 kN)(2 m) − C (6 m) = 0 P = 3.5 kN + 0.75C (2) For C $ 0, from (2): P $ 3.50 kN " For D $ 0, from (1): P # 86.0 kN " P # 3.5 kN + 0.75(50 kN) P # 41.0 kN " P $ 86 kN − 3(50 kN) P $ −64.0 kN " For C # 50 kN, from (2): For D # 50 kN, from (1): Comparing the four criteria, we find 3.50 kN # P # 41.0 kN PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 357 PROBLEM 4.14 For the beam and loading shown, determine the range of the distance a for which the reaction at B does not exceed 100 lb downward or 200 lb upward. SOLUTION Assume B is positive when directed Sketch showing distance from D to forces. ΣM D = 0: (300 lb)(8 in. − a ) − (300 lb)(a − 2 in.) − (50 lb)(4 in.) + 16 B = 0 −600a + 2800 + 16B = 0 (2800 + 16B) 600 (1) [2800 + 16(−100)] 1200 = = 2 in. 600 600 a $ 2.00 in. " a= For B = 100 lb = −100 lb, Eq. (1) yields: a$ For B = 200 = +200 lb, Eq. (1) yields: a# Required range: [2800 + 16(200)] 6000 = = 10 in. 600 600 a # 10.00 in. " 2.00 in. # a # 10.00 in. PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 358 PROBLEM 4.15 Two links AB and DE are connected by a bell crank as shown. Knowing that the tension in link AB is 720 N, determine (a) the tension in link DE, (b) the reaction at C. SOLUTION Free-Body Diagram: ΣM C = 0: FAB (100 mm) − FDE (120 mm) = 0 FDE = (a) For (1) FAB = 720 N FDE = (b) 5 FAB 6 5 (720 N) 6 FDE = 600 N 3 ΣFx = 0: − (720 N) + C x = 0 5 C x = +432 N 4 ΣFy = 0: − (720 N) + C y − 600 N = 0 5 C y = +1176 N C = 1252.84 N α = 69.829° C = 1253 N 69.8° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 359 PROBLEM 4.16 Two links AB and DE are connected by a bell crank as shown. Determine the maximum force that may be safely exerted by link AB on the bell crank if the maximum allowable value for the reaction at C is 1600 N. SOLUTION See solution to Problem 4.15 for F. B. D. and derivation of Eq. (1) FDE = 5 FAB 6 (1) 3 3 ΣFx = 0: − FAB + C x = 0 C x = FAB 5 5 ΣFy = 0: − 4 FAB + C y − FDE = 0 5 4 5 − FAB + C y − FAB = 0 5 6 49 Cy = FAB 30 C = C x2 + C y2 1 (49) 2 + (18) 2 FAB 30 C = 1.74005FAB = For C = 1600 N, 1600 N = 1.74005FAB FAB = 920 N PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 360 PROBLEM 4.17 The required tension in cable AB is 200 lb. Determine (a) the vertical force P that must be applied to the pedal, (b) the corresponding reaction at C. SOLUTION Free-Body Diagram: BC = 7 in. (a) ΣM C = 0: P(15 in.) − (200 lb)(6.062 in.) = 0 P = 80.83 lb (b) P = 80.8 lb ΣFy = 0: C x − 200 lb = 0 C x = 200 lb ΣFy = 0: C y − P = 0 C y − 80.83 lb = 0 C y = 80.83 lb α = 22.0° C = 215.7 lb C = 216 lb 22.0° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 361 PROBLEM 4.18 Determine the maximum tension that can be developed in cable AB if the maximum allowable value of the reaction at C is 250 lb. SOLUTION Free-Body Diagram: BC = 7 in. ΣM C = 0: P (15 in.) − T (6.062 in.) = 0 P = 0.40415T ΣFy = 0: − P + C y = 0 − 0.40415 P + C y = 0 C y = 0.40415T ΣFx = 0: −T + C x = 0 Cx = T C = C x2 + C y2 = T 2 + (0.40415T ) 2 C = 1.0786T For C = 250 lb 250 lb = 1.0786T T = 232 lb PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 362 PROBLEM 4.19 The bracket BCD is hinged at C and attached to a control cable at B. For the loading shown, determine (a) the tension in the cable, (b) the reaction at C. SOLUTION At B: Ty Tx = 0.18 m 0.24 m 3 Ty = Tx 4 (a) (1) ΣM C = 0: Tx (0.18 m) − (240 N)(0.4 m) − (240 N)(0.8 m) = 0 Tx = +1600 N Eq. (1) Ty = 3 (1600 N) = 1200 N 4 T = Tx2 + Ty2 = 16002 + 12002 = 2000 N (b) T = 2.00 kN ΣFx = 0: Cx − Tx = 0 Cx − 1600 N = 0 C x = +1600 N C x = 1600 N ΣFy = 0: C y − Ty − 240 N − 240 N = 0 C y − 1200 N − 480 N = 0 C y = +1680 N C y = 1680 N α = 46.4° C = 2320 N C = 2.32 kN 46.4° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 363 PROBLEM 4.20 Solve Problem 4.19, assuming that a = 0.32 m. PROBLEM 4.19 The bracket BCD is hinged at C and attached to a control cable at B. For the loading shown, determine (a) the tension in the cable, (b) the reaction at C. SOLUTION At B: Ty Tx = 0.32 m 0.24 m 4 Ty = Tx 3 ΣM C = 0: Tx (0.32 m) − (240 N)(0.4 m) − (240 N)(0.8 m) = 0 Tx = 900 N Eq. (1) Ty = 4 (900 N) = 1200 N 3 T = Tx2 + Ty2 = 9002 + 12002 = 1500 N T = 1.500 kN ΣFx = 0: C x − Tx = 0 C x − 900 N = 0 C x = +900 N C x = 900 N ΣFy = 0: C y − Ty − 240 N − 240 N = 0 C y − 1200 N − 480 N = 0 C y = +1680 N C y = 1680 N α = 61.8° C = 1906 N C = 1.906 kN 61.8° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 364 PROBLEM 4.21 Determine the reactions at A and B when (a) h = 0, (b) h = 200 mm. SOLUTION Free-Body Diagram: ΣM A = 0: ( B cos 60°)(0.5 m) − ( B sin 60°)h − (150 N)(0.25 m) = 0 37.5 B= 0.25 − 0.866h (a) (1) When h = 0: B= Eq. (1): 37.5 = 150 N 0.25 B = 150.0 N 30.0° ΣFy = 0: Ax − B sin 60° = 0 Ax = (150)sin 60° = 129.9 N A x = 129.9 N ΣFy = 0: Ay − 150 + B cos 60° = 0 Ay = 150 − (150) cos 60° = 75 N A y = 75 N α = 30° A = 150.0 N (b) A = 150.0 N 30.0° When h = 200 mm = 0.2 m Eq. (1): B= 37.5 = 488.3 N 0.25 − 0.866(0.2) B = 488 N 30.0° ΣFx = 0: Ax − B sin 60° = 0 Ax = (488.3) sin 60° = 422.88 N A x = 422.88 N ΣFy = 0: Ay − 150 + B cos 60° = 0 Ay = 150 − (488.3) cos 60° = −94.15 N A y = 94.15 N α = 12.55° A = 433.2 N A = 433 N 12.6° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 365 PROBLEM 4.22 For the frame and loading shown, determine the reactions at A and E when (a) α = 30°, (b) α = 45°. SOLUTION Free-Body Diagram: ΣM A = 0: ( E sin α )(8 in.) + ( E cos α )(5 in.) − (20 lb)(10 in.) − (20 lb)(3 in.) = 0 (a) When α = 30°: E= 260 8sin α + 5cos α E= 260 = 31.212 lb 8sin 30° + 5cos 30° E = 31.2 lb 60.0° ΣFx = 0: Ax − 20 lb + (31.212 lb) sin 30° = 0 Ax = +4.394 lb A x = 4.394 lb ΣFy = 0: Ay − 20° + (31.212 lb) cos 30° = 0 Ay = −7.03 lb A y = 7.03 lb A = 8.29 lb 58.0° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 366 PROBLEM 4.22 (Continued) (b) When α = 45° : E= 260 = 28.28 lb 8sin 45° + 5cos α E = 28.3 lb 45.0° ΣFx = 0: Ax − 20 lb + (28.28 lb)sin 45° = 0 Ax = 0 Ax = 0 ΣFy = 0: Ay − 20 lb + (28.28 lb) cos 45° = 0 Ay = 0 Ay = 0 A=0 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 367 PROBLEM 4.23 For each of the plates and loadings shown, determine the reactions at A and B. SOLUTION (a) Free-Body Diagram: ΣM A = 0: B(20 in.) − (50 lb)(4 in.) − (40 lb)(10 in.) = 0 B = +30 lb B = 30.0 lb ΣFx = 0: Ax + 40 lb = 0 Ax = −40 lb A x = 40.0 lb ΣFy = 0: Ay + B − 50 lb = 0 Ay + 30 lb − 50 lb = 0 Ay = +20 lb A y = 20.0 lb α = 26.56° A = 44.72 lb A = 44.7 lb 26.6° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 368 PROBLEM 4.23 (Continued) (b) Free-Body Diagram: ΣM A = 0: ( B cos 30°)(20 in.) − (40 lb)(10 in.) − (50 lb)(4 in.) = 0 B = 34.64 lb B = 34.6 lb 60.0° ΣFx = 0: Ax − B sin 30° + 40 lb Ax − (34.64 lb) sin 30° + 40 lb = 0 Ax = −22.68 lb A x = 22.68 lb ΣFy = 0: Ay + B cos 30° − 50 lb = 0 Ay + (34.64 lb) cos 30° − 50 lb = 0 Ay = +20 lb A y = 20.0 lb α = 41.4° A = 30.24 lb A = 30.2 lb 41.4° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 369 PROBLEM 4.24 For each of the plates and loadings shown, determine the reactions at A and B. SOLUTION (a) Free-Body Diagram: ΣM B = 0: A(20 in.) + (50 lb)(16 in.) − (40 lb)(10 in.) = 0 A = +20 lb A = 20.0 lb ΣFx = 0: 40 lb + Bx = 0 Bx = −40 lb B x = 40 lb ΣFy = 0: A + By − 50 lb = 0 20 lb + By − 50 lb = 0 By = +30 lb α = 36.87° B = 50 lb B y = 30 lb B = 50.0 lb 36.9° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 370 PROBLEM 4.24 (Continued) (b) ΣM A = 0: − ( A cos 30°)(20 in.) − (40 lb)(10 in.) + (50 lb)(16 in.) = 0 A = 23.09 lb A = 23.1 lb 60.0° ΣFx = 0: A sin 30° + 40 lb + Bx = 0 (23.09 lb) sin 30° + 40 lb + 8 x = 0 Bx = −51.55 lb B x = 51.55 lb ΣFy = 0: A cos 30° + By − 50 lb = 0 (23.09 lb) cos 30° + By − 50 lb = 0 By = +30 lb B y = 30 lb α = 30.2° B = 59.64 lb B = 59.6 lb 30.2° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 371 PROBLEM 4.25 Determine the reactions at A and B when (a) α = 0, (b) α = 90°, (c) α = 30°. SOLUTION (a) α =0 ΣM A = 0: B(20 in.) − 750 lb ⋅ in. = 0 B = 37.5 lb ΣFx = 0: Ax = 0 ΣFy = 0: Ay + 37.5 lb = 0 Ay = −37.5 lb B = 37.5 lb A = 37.5 lb (b) α = 90° ΣM A = 0: B (12 in.) − 750 lb ⋅ in. = 0 B = 62.5 lb ΣFA = 0: Ax − 62.5 lb = 0, Ax = 62.5 lb ΣFy = 0: Ay = 0 A = 62.5 lb (c) B = 62.5 lb α = 30° ΣM A = 0: ( B cos 30°)(20 in.) + ( B sin 30°)(12 in.) − 750 lb ⋅ in. = 0 B = 32.16 lb ΣFx = 0: Ax − (32.16 lb) sin 30° = 0 Ax = 16.08 lb ΣFy = 0: Ay + (32.16 lb) cos 30° = 0 Ay = −27.85 lb A = 32.16 lb α = 60.0° A = 32.2 lb 60.0° B = 32.2 lb 60.0° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 372 PROBLEM 4.26 A rod AB hinged at A and attached at B to cable BD supports the loads shown. Knowing that d = 200 mm, determine (a) the tension in cable BD, (b) the reaction at A. SOLUTION Free-Body Diagram: (a) Move T along BD until it acts at Point D. ΣM A = 0: (T sin 45°)(0.2 m) + (90 N)(0.1 m) + (90 N)(0.2 m) = 0 T = 190.92 N (b) T = 190.9 N ΣFx = 0: Ax − (190.92 N) cos 45° = 0 Ax = +135 N A x = 135.0 N ΣFy = 0: Ay − 90 N − 90 N + (190.92 N) sin 45° = 0 Ay = +45 N A y = 45.0 N A = 142.3 N 18.43° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 373 PROBLEM 4.27 A rod AB hinged at A and attached at B to cable BD supports the loads shown. Knowing that d = 150 mm, determine (a) the tension in cable BD, (b) the reaction at A. SOLUTION Free-Body Diagram: tan α = (a) 10 ; α = 33.69° 15 Move T along BD until it acts at Point D. ΣM A = 0: (T sin 33.69°)(0.15 m) − (90 N)(0.1 m) − (90 N)(0.2 m) = 0 T = 324.5 N (b) T = 324 N ΣFx = 0: Ax − (324.99 N) cos 33.69° = 0 Ax = +270 N A x = 270 N ΣFy = 0: Ay − 90 N − 90 N + (324.5 N)sin 33.69° = 0 Ay = 0 Ay = 0 A = 270 N PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 374 PROBLEM 4.28 A lever AB is hinged at C and attached to a control cable at A. If the lever is subjected to a 500-N horizontal force at B, determine (a) the tension in the cable, (b) the reaction at C. SOLUTION Triangle ACD is isosceles with C = 90° + 30° = 120° A= D= 1 (180° − 120°) = 30° 2 Thus DA forms angle of 60° with horizontal. (a) We resolve FAD into components along AB and perpendicular to AB. ΣM C = 0: ( FAD sin 30°)(250 mm) − (500 N)(100 mm) = 0 (b) FAD = 400 N ΣFx = 0: − (400 N) cos 60° + C x − 500 N = 0 C x = +300 N ΣFy = 0: − (400 N) sin 60° + C y = 0 C y = +346.4 N C = 458 N 49.1° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 375 PROBLEM 4.29 A force P of magnitude 280 lb is applied to member ABCD, which is supported by a frictionless pin at A and by the cable CED. Since the cable passes over a small pulley at E, the tension may be assumed to be the same in portions CE and ED of the cable. For the case when a = 3 in., determine (a) the tension in the cable, (b) the reaction at A. SOLUTION Free-Body Diagram: (a) ΣM A = 0: − (280 lb)(8 in.) 7 T (12 in.) 25 24 − T (8 in.) = 0 25 T (12 in.) − (12 − 11.04)T = 840 (b) ΣFx = 0: T = 875 lb 7 (875 lb) + 875 lb + Ax = 0 25 Ax = −1120 ΣFy = 0: Ay − 280 lb − Ay = +1120 A x = 1120 lb 24 (875 lb) = 0 25 A y = 1120 lb A = 1584 lb 45.0° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 376 PROBLEM 4.30 Neglecting friction, determine the tension in cable ABD and the reaction at support C. SOLUTION Free-Body Diagram: ΣM C = 0: T (0.25 m) − T (0.1 m) − (120 N)(0.1 m) = 0 T = 80.0 N ΣFx = 0: C x − 80 N = 0 C x = +80 N C x = 80.0 N ΣFy = 0: C y − 120 N + 80 N = 0 C y = +40 N C y = 40.0 N C = 89.4 N 26.6° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 377 PROBLEM 4.31 Rod ABC is bent in the shape of an arc of circle of radius R. Knowing the θ = 30°, determine the reaction (a) at B, (b) at C. SOLUTION Free-Body Diagram: ΣM D = 0: C x ( R) − P( R) = 0 Cx = + P ΣFx = 0: C x − B sin θ = 0 P − B sin θ = 0 B = P/sin θ B= P sin θ θ ΣFy = 0: C y + B cos θ − P = 0 C y + ( P/sin θ ) cos θ − P = 0 1 ! Cy = P 1 − ! tan θ # " For θ = 30°: (a) (b) B = P/sin 30° = 2 P B = 2P 60.0° Cx = + P Cx = P C y = P(1 − 1/tan 30°) = − 0.732/P C y = 0.7321P C = 1.239P 36.2° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 378 ! PROBLEM 4.32 Rod ABC is bent in the shape of an arc of circle of radius R. Knowing the θ = 60°, determine the reaction (a) at B, (b) at C. SOLUTION See the solution to Problem 4.31 for the free-body diagram and analysis leading to the following expressions: Cx = + P 1 % $ Cy = P 1 − tan θ !# " P B= sin θ For θ = 60°: (a) (b) B = P/sin 60° = 1.1547 P B = 1.155P 30.0° Cx = + P Cx = P C y = P(1 − 1/tan 60°) = + 0.4226 P C y = 0.4226 P C = 1.086P 22.9° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 379 ! PROBLEM 4.33 Neglecting friction, determine the tension in cable ABD and the reaction at C when θ = 60°. SOLUTION ΣM C = 0: T (2a + a cos θ ) − Ta + Pa = 0 T= P 1 + cos θ (1) ΣFx = 0: C x − T sin θ = 0 C x = T sin θ = P sin θ 1 + cos θ ΣFy = 0: C y + T + T cos θ − P = 0 C y = P − T (1 + cos θ ) = P − P 1 + cos θ 1 + cos θ Cy = 0 C y = 0, C = C x Since C=P sin θ 1 + cos θ (2) For θ = 60°: P P = 1 + cos 60° 1 + 12 Eq. (1): T= Eq. (2): C=P sin 60° 0.866 =P 1 + cos 60° 1 + 12 T= 2 P 3 C = 0.577P PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 380 ! PROBLEM 4.34 Neglecting friction, determine the tension in cable ABD and the reaction at C when θ = 45°. SOLUTION Free-Body Diagram: Equilibrium for bracket: ΣM C = 0: − T (a ) − P(a ) + (T sin 45°)(2a sin 45°) + (T cos 45°)(a + 2a cos 45°) = 0 T = 0.58579 or T = 0.586 P ΣFx = 0: C x + (0.58579 P) sin 45° = 0 C x = 0.41422 P ΣFy = 0: C y + 0.58579 P − P + (0.58579 P) cos 45° = 0 Cy = 0 or C = 0.414P PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 381 ! PROBLEM 4.35 A light rod AD is supported by frictionless pegs at B and C and rests against a frictionless wall at A. A vertical 120-lb force is applied at D. Determine the reactions at A, B, and C. SOLUTION Free-Body Diagram: ΣFx = 0: A cos 30° − (120 lb) cos 60° = 0 A = 69.28 lb A = 69.3 lb ΣM B = 0: C (8 in.) − (120 lb)(16 in.) cos 30° + (69.28 lb)(8 in.)sin 30° = 0 C = 173.2 lb C = 173.2 lb 60.0° B = 34.6 lb 60.0° ΣM C = 0: B(8 in.) − (120 lb)(8 in.) cos 30° + (69.28 lb)(16 in.) sin 30° = 0 B = 34.6 lb Check: ΣFy = 0: 173.2 − 34.6 − (69.28)sin 30° − (120)sin 60° = 0 0 = 0(check) ! PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 382 PROBLEM 4.36 A light bar AD is suspended from a cable BE and supports a 50-lb block at C. The ends A and D of the bar are in contact with frictionless vertical walls. Determine the tension in cable BE and the reactions at A and D. SOLUTION Free-Body Diagram: ΣFx = 0: A= D ΣFy = 0: TBE = 50.0 lb We note that the forces shown form two couples. ΣM = 0: A(8 in.) − (50 lb)(3 in.) = 0 A = 18.75 lb A = 18.75 lb D = 18.75 lb PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 383 ! PROBLEM 4.37 Bar AC supports two 400-N loads as shown. Rollers at A and C rest against frictionless surfaces and a cable BD is attached at B. Determine (a) the tension in cable BD, (b) the reaction at A, (c) the reaction at C. SOLUTION Similar triangles: ABE and ACD AE BE ; = AD CD (a) ΣM A = 0: Tx (0.25 m) − 0.15 m BE ; BE = 0.075 m = 0.5 m 0.25 m $ 0.075 % Tx ! (0.5 m) − (400 N)(0.1 m) − (400 N)(0.4 m) = 0 " 0.35 # Tx = 1400 N 0.075 (1400 N) 0.35 = 300 N Ty = T = 1432 lb PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 384 PROBLEM 4.37 (Continued) (b) ΣFy = 0: A − 300 N − 400 N − 400 N = 0 A = + 1100 N (c) A = 1100 N ΣFx = 0: −C + 1400 N = 0 C = + 1400 N C = 1400 N PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 385 ! PROBLEM 4.38 Determine the tension in each cable and the reaction at D. SOLUTION tan α = 0.08 m 0.2 m α = 21.80° tan β = 0.08 m 0.1 m β = 38.66° ΣM B = 0: (600 N)(0.1 m) − (TCF sin 38.66°)(0.1 m) = 0 TCF = 960.47 N TCF = 96.0 N ΣM C = 0: (600 N)(0.2 m) − (TBE sin 21.80°)(0.1 m) = 0 TBE = 3231.1 N TBE = 3230 N ΣFy = 0: TBE cos α + TCF cos β − D = 0 (3231.1 N) cos 21.80° + (960.47 N) cos 38.66° − D = 0 D = 3750.03 N D = 3750 N PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 386 ! PROBLEM 4.39 Two slots have been cut in plate DEF, and the plate has been placed so that the slots fit two fixed, frictionless pins A and B. Knowing that P = 15 lb, determine (a) the force each pin exerts on the plate, (b) the reaction at F. SOLUTION Free-Body Diagram: ΣFx = 0: 15 lb − B sin 30° = 0 B = 30.0 lb 60.0° ΣM A = 0: − (30 lb)(4 in.) + B sin 30°(3 in.) + B cos 30°(11 in.) − F (13 in.) = 0 −120 lb ⋅ in. + (30 lb) sin 30°(3 in.) + (30 lb) cos 30°(11 in.) − F (13 in.) = 0 F = + 16.2145 lb F = 16.21 lb ΣFy = 0: A − 30 lb + B cos 30° − F = 0 A − 30 lb + (30 lb) cos 30° − 16.2145 lb = 0 A = + 20.23 lb A = 20.2 lb PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 387 ! PROBLEM 4.40 For the plate of Problem 4.39 the reaction at F must be directed downward, and its maximum allowable value is 20 lb. Neglecting friction at the pins, determine the required range of values of P. PROBLEM 4.39 Two slots have been cut in plate DEF, and the plate has been placed so that the slots fit two fixed, frictionless pins A and B. Knowing that P = 15 lb, determine (a) the force each pin exerts on the plate, (b) the reaction at F. SOLUTION Free-Body Diagram: ΣFx = 0: P − B sin 30° = 0 B = 2P 60° ΣM A = 0: − (30 lb)(4 in.) + B sin 30°(3 in.) + B cos 30°(11 in.) − F (13 in.) = 0 −120 lb ⋅ in.+ 2 P sin 30°(3 in.) + 2 P cos 30°(11 in.) − F (13 in.) = 0 −120 + 3P + 19.0525P − 13F = 0 P= 13E + 120 22.0525 For F = 0: P= 13(0) + 120 = 5.442 lb 22.0525 For P = 20 lb: P= 13(20) + 120 = 17.232 lb 22.0525 For 0 # F # 20 lb: (1) 5.44 lb # P # 17.231 lb PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 388 ! PROBLEM 4.41 Bar AD is attached at A and C to collars that can move freely on the rods shown. If the cord BE is vertical (α = 0), determine the tension in the cord and the reactions at A and C. SOLUTION Free-Body Diagram: ΣFy = 0: − T cos 30° + (80 N) cos 30° = 0 T = 80 N T = 80.0 N ΣM C = 0: ( A sin 30°)(0.4 m) − (80 N)(0.2 m) − (80 N)(0.2 m) = 0 A = + 160 N A = 160.0 N 30.0° C = 160.0 N 30.0° ΣM A = 0: (80 N)(0.2 m) − (80 N)(0.6 m) + (C sin 30°)(0.4 m) = 0 C = + 160 N PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 389 ! PROBLEM 4.42 Solve Problem 4.41 if the cord BE is parallel to the rods (α = 30°). PROBLEM 4.41 Bar AD is attached at A and C to collars that can move freely on the rods shown. If the cord BE is vertical (α = 0), determine the tension in the cord and the reactions at A and C. SOLUTION Free-Body Diagram: ΣFy = 0: − T + (80 N) cos 30° = 0 T = 69.282 N T = 69.3 N ΣM C = 0: − (69.282 N) cos 30°(0.2 m) − (80 N)(0.2 m) + ( A sin 30°)(0.4 m) = 0 A = + 140.000 N A = 140.0 N 30.0° C = 180.0 N 30.0° ΣM A = 0: + (69.282 N) cos 30°(0.2 m) − (80 N)(0.6 m) + (C sin 30°)(0.4 m) = 0 C = + 180.000 N PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 390 ! PROBLEM 4.43 An 8-kg mass can be supported in the three different ways shown. Knowing that the pulleys have a 100-mm radius, determine the reaction at A in each case. SOLUTION ! W = mg = (8 kg)(9.81m/s 2 ) = 78.48 N ! (a) ΣFx = 0: Ax = 0 ΣFy = 0: Ay − W = 0 ! A y = 78.48 N ΣM A = 0: M A − W (1.6 m) = 0 ! M A = + (78.48 N)(1.6 m) ! M A = 125.56 N ⋅ m A = 78.5 N ! (b) ! M A = 125.6 N ⋅ m ΣFx = 0: Ax − W = 0 A x = 78.48 ΣFy = 0: Ay − W = 0 A y = 78.48 A = (78.48 N) 2 = 110.99 N ! 45° ΣM A = 0: M A − W (1.6 m) = 0 ! M A = + (78.48 N)(1.6 m) ! ! A = 111.0 N ! (c) 45° M A = 125.56 N ⋅ m M A = 125.6 N ⋅ m ΣFx = 0: Ax = 0 ΣFy = 0: Ay − 2W = 0 ! ! Ay = 2W = 2(78.48 N) = 156.96 N ΣM A = 0: M A − 2W (1.6 m) = 0 M A = + 2(78.48 N)(1.6 m) A = 157.0 N M A = 125.1 N ⋅ m M A = 125 N ⋅ m ! PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 391 PROBLEM 4.44 A tension of 5 lb is maintained in a tape as it passes through the support system shown. Knowing that the radius of each pulley is 0.4 in., determine the reaction at C. SOLUTION From f.b.d. of system ΣFx = 0: C x + (5 lb) = 0 Cx = −5 lb ΣFy = 0: C y − (5 lb) = 0 C y = 5 lb Then C = ( Cx ) 2 + (C y ) 2 = (5) 2 + (5) 2 = 7.0711 lb and θ = tan −1 $ +5 % ! = −45° " −5 # or C = 7.07 lb 45.0° ΣM C = 0: M C + (5 lb)(6.4 in.) + (5 lb)(2.2 in.) = 0 M C = − 43.0 lb ⋅ in or M C = 43.0 lb ⋅ in. PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 392 ! PROBLEM 4.45 Solve Problem 4.44, assuming that 0.6-in.-radius pulleys are used. PROBLEM 4.44 A tension of 5 lb is maintained in a tape as it passes through the support system shown. Knowing that the radius of each pulley is 0.4 in., determine the reaction at C. SOLUTION From f.b.d. of system ΣFx = 0: C x + (5 lb) = 0 Cx = −5 lb ΣFy = 0: C y − (5 lb) = 0 C y = 5 lb Then C = (Cx ) 2 + (C y ) 2 = (5) 2 + (5) 2 = 7.0711 lb and θ = tan −1 $ 5 % ! = −45.0° " −5 # or C = 7.07 lb 45.0° ΣM C = 0: M C + (5 lb)(6.6 in.) + (5 lb)(2.4 in.) = 0 M C = − 45.0 lb ⋅ in. or M C = 45.0 lb ⋅ in. PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 393 ! PROBLEM 4.46 A 6-m telephone pole weighing 1600 N is used to support the ends of two wires. The wires form the angles shown with the horizontal and the tensions in the wires are, respectively, T1 = 600 N and T2 = 375 N. Determine the reaction at the fixed end A. SOLUTION Free-Body Diagram: ΣFx = 0: Ax + (375 N) cos 20° − (600 N) cos10° = 0 Ax = +238.50 N ΣFy = 0: Ay − 1600 N − (600 N)sin10° − (375 N) sin 20° = 0 Ay = +1832.45 N A = 238.502 + 1832.452 1832.45 θ = tan −1 238.50 A = 1848 N 82.6° ΣM A = 0: M A + (600 N) cos10°(6 m) − (375 N) cos 20°(6 m) = 0 M A = −1431.00 N ⋅ m M A = 1431 N ⋅ m PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 394 PROBLEM 4.47 Beam AD carries the two 40-lb loads shown. The beam is held by a fixed support at D and by the cable BE that is attached to the counterweight W. Determine the reaction at D when (a) W = 100 lb, (b) W = 90 lb. SOLUTION W = 100 lb (a) From f.b.d. of beam AD ΣFx = 0: Dx = 0 ΣFy = 0: D y − 40 lb − 40 lb + 100 lb = 0 Dy = −20.0 lb or D = 20.0 lb ΣM D = 0: M D − (100 lb)(5 ft) + (40 lb)(8 ft) + (40 lb)(4 ft) = 0 M D = 20.0 lb ⋅ ft or M D = 20.0 lb ⋅ ft W = 90 lb (b) From f.b.d. of beam AD ΣFx = 0: Dx = 0 ΣFy = 0: D y + 90 lb − 40 lb − 40 lb = 0 Dy = −10.00 lb or D = 10.00 lb ΣM D = 0: M D − (90 lb)(5 ft) + (40 lb)(8 ft) + (40 lb)(4 ft) = 0 M D = −30.0 lb ⋅ ft or M D = −30.0 lb ⋅ ft PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 395 PROBLEM 4.48 For the beam and loading shown, determine the range of values of W for which the magnitude of the couple at D does not exceed 40 lb ⋅ ft. SOLUTION For Wmin , From f.b.d. of beam AD M D = − 40 lb ⋅ ft ΣM D = 0: (40 lb)(8 ft) − Wmin (5 ft) + (40 lb)(4 ft) − 40 lb ⋅ ft = 0 Wmin = 88.0 lb For Wmax , From f.b.d. of beam AD M D = 40 lb ⋅ ft ΣM D = 0: (40 lb)(8 ft) − Wmax (5 ft) + (40 lb)(4 ft) + 40 lb ⋅ ft = 0 Wmax = 104.0 lb or 88.0 lb # W # 104.0 lb PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 396 PROBLEM 4.49 Knowing that the tension in wire BD is 1300 N, determine the reaction at the fixed support C of the frame shown. SOLUTION T = 1300 N 5 Tx = T 13 = 500 N 12 Ty = T 13 = 1200 N ΣM x = 0: C x − 450 N + 500 N = 0 C x = −50 N ΣFy = 0: C y − 750 N − 1200 N = 0 C y = +1950 N C x = 50 N C y = 1950 N C = 1951 N 88.5° ΣM C = 0: M C + (750 N)(0.5 m) + (4.50 N)(0.4 m) − (1200 N)(0.4 m) = 0 M C = −75.0 N ⋅ m M C = 75.0 N ⋅ m PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 397 PROBLEM 4.50 Determine the range of allowable values of the tension in wire BD if the magnitude of the couple at the fixed support C is not to exceed 100 N · m. SOLUTION ΣM C = 0: (750 N)(0.5 m) + (450 N)(0.4 m) − − $ 5 % T ! (0.6 m) " 13 # $ 12 % T ! (0.15 m) + M C = 0 " 13 # 375 N ⋅ m + 180 N ⋅ m − T= $ 4.8 % m !T + M C = 0 " 13 # 13 (555 + M C ) 4.8 For M C = −100 N ⋅ m: T = 13 (555 − 100) = 1232 N 4.8 For M C = +100 N ⋅ m: T = 13 (555 + 100) = 1774 N 4.8 For |M C | # 100 N ⋅ m: 1.232 kN # T # 1.774 kN PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 398 PROBLEM 4.51 A vertical load P is applied at end B of rod BC. (a) Neglecting the weight of the rod, express the angle θ corresponding to the equilibrium position in terms of P, l, and the counterweight W. (b) Determine the value of θ corresponding to equilibrium if P = 2W . SOLUTION (a) Triangle ABC is isosceles. We have CD = ( BC ) cos θ 2 = l cos θ 2 θ% $ ΣM C = 0: W l cos ! − P(l sin θ ) = 0 2# " Setting sin θ = 2sin θ θ θ W − 2 P sin (b) For P = 2W : sin θ 2 θ 2 or θ θ cos : Wl cos − 2 Pl sin cos = 0 2 2 2 2 2 θ 2 = θ 2 =0 θ = 2sin −1 $ W % ! 2 " P# W W = = 0.25 2 P 4W θ = 29.0° = 14.5° = 165.5° θ = 331°(discard) PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 399 PROBLEM 4.52 A slender rod AB, of weight W, is attached to blocks A and B, which move freely in the guides shown. The blocks are connected by an elastic cord that passes over a pulley at C. (a) Express the tension in the cord in terms of W and θ. (b) Determine the value of θ for which the tension in the cord is equal to 3W. SOLUTION (a) From f.b.d. of rod AB &$ 1 % ' ΣM C = 0: T (l sin θ ) + W ( ! cos θ ) − T (l cos θ ) = 0 2 *" # + T= W cos θ 2(cos θ − sin θ ) Dividing both numerator and denominator by cos θ, T= (b) For T = 3W , or 3W = W$ 1 % 2 " 1 − tan θ !# or T = ( W2 ) (1 − tan θ ) ( W2 ) (1 − tan θ ) 1 1 − tan θ = 6 θ = tan −1 $5% ! = 39.806° "6# or θ = 39.8° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 400 PROBLEM 4.53 Rod AB is acted upon by a couple M and two forces, each of magnitude P. (a) Derive an equation in θ, P, M, and l that must be satisfied when the rod is in equilibrium. (b) Determine the value of θ corresponding to equilibrium when M = 150 N · m, P = 200 N, and l = 600 mm. SOLUTION Free-Body Diagram: (a) From free-body diagram of rod AB ΣM C = 0: P(l cos θ ) + P(l sin θ ) − M = 0 or sinθ + cosθ = (b) For M Pl M = 150 lb ⋅ in., P = 20 lb, and l = 6 in. 150 lb ⋅ in. 5 sin θ + cos θ = = = 1.25 (20 lb)(6 in.) 4 sin 2 θ + cos 2 θ = 1 Using identity sin θ + (1 − sin 2 θ )1/2 = 1.25 (1 − sin 2 θ )1/2 = 1.25 − sin θ 1 − sin 2 θ = 1.5625 − 2.5sin θ + sin 2 θ 2sin 2 θ − 2.5sin θ + 0.5625 = 0 Using quadratic formula sin θ = = or −( −2.5) ± (625) − 4(2)(0.5625) 2(2) 2.5 ± 1.75 4 sin θ = 0.95572 and sin θ = 0.29428 θ = 72.886° and θ = 17.1144° or θ = 17.11° and θ = 72.9° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 401 PROBLEM 4.54 Rod AB is attached to a collar at A and rests against a small roller at C. (a) Neglecting the weight of rod AB, derive an equation in P, Q, a, l, and θ that must be satisfied when the rod is in equilibrium. (b) Determine the value of θ corresponding to equilibrium when P = 16 lb, Q = 12 lb, l = 20 in., and a = 5 in. SOLUTION Free-Body Diagram: ΣFy = 0: C cos θ − P − Q = 0 C= ΣM A = 0: C (a) P+Q cos θ a − Pl cos θ = 0 cos θ P+Q a ⋅ − Pl cos θ = 0 cos θ cos θ (b) For cos3 θ = a( P + Q) Pl P = 16 lb, Q = 12 lb, l = 20 in., and a = 5 in. (5 in.)(16 lb + 12 lb) (16 lb)(20 in.) = 0.4375 cos3 θ = cos θ = 0.75915 θ = 40.6° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 402 PROBLEM 4.55 A collar B of weight W can move freely along the vertical rod shown. The constant of the spring is k, and the spring is unstretched when θ = 0. (a) Derive an equation in θ, W, k, and l that must be satisfied when the collar is in equilibrium. (b) Knowing that W = 300 N, l = 500 mm, and k = 800 N/m, determine the value of θ corresponding to equilibrium. SOLUTION First note T = ks Where k = spring constant s = elongation of spring l = −l cos θ l (1 − cos θ ) = cos θ kl T= (1 − cos θ ) cos θ (a) From f.b.d. of collar B or (b) For ΣFy = 0: T sin θ − W = 0 kl (1 − cos θ )sin θ − W = 0 cos θ or tan θ − sin θ = W kl W = 3 lb l = 6 in. k = 8 lb/ft 6 in. l= = 0.5 ft 12 in./ft tan θ − sin θ = Solving numerically, 3 lb = 0.75 (8 lb/ft)(0.5 ft) θ = 57.957° or θ = 58.0° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 403 PROBLEM 4.56 A vertical load P is applied at end B of rod BC. The constant of the spring is k, and the spring is unstretched when θ = 90°. (a) Neglecting the weight of the rod, express the angle θ corresponding to equilibrium in terms of P, k, and l. (b) Determine the value of θ corresponding to equilibrium when P = 14 kl. SOLUTION First note T = tension in spring = ks where s = elongation of spring = ( AB)θ − ( AB)θ = 90° $θ % $ 90° % ! − 2l sin 2 ! 2 " # " # & $ θ % $ 1 %' = 2l (sin ! − !) * " 2 # " 2 #+ = 2l sin & $ θ % $ 1 %' T = 2kl (sin !) !− * " 2 # " 2 #+ (a) From f.b.d. of rod BC (1) & $ θ %' ΣM C = 0: T (l cos ! ) − P(l sin θ ) = 0 " 2 #+ * Substituting T from Equation (1) & $ θ % $ 1 %' & $ θ %' 2kl (sin ! − ! ) (l cos ! ) − P ( l sin θ ) = 0 " 2 #+ * " 2 # " 2 #+ * & $ θ % $ 1 %' & $θ % $θ % $ θ %' 2kl 2 (sin ! − ! ) cos ! − Pl ( 2sin ! cos ! ) = 0 2 2 2 " # " # " 2 #+ * * " # " 2 #+ Factoring out 2l cos $θ % ! , leaves "2# & $ θ % $ 1 %' $θ % kl (sin ! ) − P sin ! = 0 !− 2 2 "2# #+ * " # " or sin $ θ % 1 $ kl % != ! 2 " kl − P # "2# & ' ) * 2(kl − P) + θ = 2sin −1 ( kl PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 404 PROBLEM 4.56 (Continued) (b) P= kl 4 & kl ' kl ) kl − 2 (* ( 4 )) + & ' kl $ 4 % = 2sin −1 ( !) * 2 " 3kl # + θ = 2sin −1 ( = 2sin −1 $ 4 % ! "3 2 # = 2sin −1 (0.94281) = 141.058° or θ = 141.1° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 405 PROBLEM 4.57 Solve Sample Problem 4.56, assuming that the spring is unstretched when θ = 90°. SOLUTION First note T = tension in spring = ks where s = deformation of spring = rβ F = kr β From f.b.d. of assembly or ΣM 0 = 0: W (l cos β ) − F (r ) = 0 Wl cos β − kr 2 β = 0 cos β = For kr 2 β Wl k = 250 lb/in. r = 3 in. l = 8 in. W = 400 lb cos β = or (250 lb/in.)(3 in.)2 β (400 lb)(8 in.) cos β = 0.703125β Solving numerically, β = 0.89245 rad or β = 51.134° Then θ = 90° + 51.134° = 141.134° or θ = 141.1° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 406 PROBLEM 4.58 A slender rod AB, of weight W, is attached to blocks A and B that move freely in the guides shown. The constant of the spring is k, and the spring is unstretched when θ = 0. (a) Neglecting the weight of the blocks, derive an equation in W, k, l, and θ that must be satisfied when the rod is in equilibrium. (b) Determine the value of θ when W = 75 lb, l = 30 in., and k = 3 lb/in. SOLUTION Free-Body Diagram: Spring force: Fs = ks = k (l − l cos θ ) = kl (1 − cos θ ) ΣM D = 0: Fs (l sin θ ) − W (a) $l % cos θ ! = 0 "2 # kl (1 − cos θ )l sin θ − kl (1 − cos θ ) tan θ − (b) For given values of W l cos θ = 0 2 W =0 2 or (1 − cos θ ) tan θ = W 2kl W = 75 lb l = 30 in. k = 3 lb/in. (1 − cos θ ) tan θ = tan θ − sin θ 75 lb = 2(3 lb/in.)(30 in.) = 0.41667 Solving numerically: θ = 49.710° or θ = 49.7° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 407 PROBLEM 4.59 Eight identical 500 × 750-mm rectangular plates, each of mass m = 40 kg, are held in a vertical plane as shown. All connections consist of frictionless pins, rollers, or short links. In each case, determine whether (a) the plate is completely, partially, or improperly constrained, (b) the reactions are statically determinate or indeterminate, (c) the equilibrium of the plate is maintained in the position shown. Also, wherever possible, compute the reactions. SOLUTION 1. Three non-concurrent, non-parallel reactions (a) Plate: completely constrained (b) Reactions: determinate (c) Equilibrium maintained A = C = 196.2 N 2. Three non-concurrent, non-parallel reactions (a) Plate: completely constrained (b) Reactions: determinate (c) Equilibrium maintained B = 0, C = D = 196.2 N 3. Four non-concurrent, non-parallel reactions (a) Plate: completely constrained (b) Reactions: indeterminate (c) Equilibrium maintained A x = 294 N , D x = 294 N ( A y + D y = 392 N ) 4. Three concurrent reactions (through D) (a) Plate: improperly constrained (b) Reactions: indeterminate (c) No equilibrium (ΣM D ≠ 0) PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 408 PROBLEM 4.59 (Continued) 5. Two reactions (a) Plate: partial constraint (b) Reactions: determinate (c) Equilibrium maintained C = R = 196.2 N 6. Three non-concurrent, non-parallel reactions (a) Plate: completely constrained (b) Reactions: determinate (c) Equilibrium maintained B = 294 N 7. 8. , D = 491 N 53.1° Two reactions (a) Plate: improperly constrained (b) Reactions determined by dynamics (c) No equilibrium (ΣFy ≠ 0) For non-concurrent, non-parallel reactions (a) Plate: completely constrained (b) Reactions: indeterminate (c) Equilibrium maintained B = D y = 196.2 N (C + D x = 0) PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 409 PROBLEM 4.60 The bracket ABC can be supported in the eight different ways shown. All connections consist of smooth pins, rollers, or short links. For each case, answer the questions listed in Problem 4.59, and, wherever possible, compute the reactions, assuming that the magnitude of the force P is 100 lb. SOLUTION 1. Three non-concurrent, non-parallel reactions (a) Bracket: complete constraint (b) Reactions: determinate (c) Equilibrium maintained A = 120.2 lb 2. 3. 56.3°, B = 66.7 lb Four concurrent, reactions (through A) (a) Bracket: improper constraint (b) Reactions: indeterminate (c) No equilibrium (ΣM A ≠ 0) Two reactions (a) Bracket: partial constraint (b) Reactions: determinate (c) Equilibrium maintained A = 50 lb , C = 50 lb 4. Three non-concurrent, non-parallel reactions (a) Bracket: complete constraint (b) Reactions: determinate (c) Equilibrium maintained A = 50 lb , B = 83.3 lb 36.9°, C = 66.7 lb PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 410 PROBLEM 4.60 (Continued) 5. 6. Four non-concurrent, non-parallel reactions (a) Bracket: complete constraint (b) Reactions: indeterminate (c) Equilibrium maintained (ΣM C = 0) A y = 50 lb Four non-concurrent, non-parallel reactions (a) Bracket: complete constraint (b) Reactions: indeterminate (c) Equilibrium maintained A x = 66.7 lb , B = 66.7 lb ( A y + B y = 100 lb ) 7. Three non-concurrent, non-parallel reactions (a) Bracket: complete constraint (b) Reactions: determinate (c) Equilibrium maintained A = C = 50 lb 8. Three concurrent, reactions (through A) (a) Bracket: improper constraint (b) Reactions: indeterminate (c) No equilibrium (ΣM A ≠ 0) PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 411 PROBLEM 4.61 Determine the reactions at A and B when a = 180 mm. SOLUTION Reaction at B must pass through D where A and 300-N load intersect. ∆BCD : Free-Body Diagram: (Three-force member) 240 180 β = 53.13° tan β = Force triangle A = (300 N) tan 53.13° = 400 N 300 N cos 53.13° = 500 N ! A = 400 N B= B = 500 N 53.1° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 412 ! PROBLEM 4.62 For the bracket and loading shown, determine the range of values of the distance a for which the magnitude of the reaction at B does not exceed 600 N. SOLUTION Reaction at B must pass through D where A and 300-N load intersect. Free-Body Diagram: (Three-force member) a= 240 mm tan β (1) Force Triangle (with B = 600 N) 300 N = 0.5 600 N β = 60.0° cos β = Eq. (1) 240 mm tan 60.0° = 138.56 mm a= For B # 600 N a $ 138.6 mm PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 413 ! PROBLEM 4.63 Using the method of Section 4.7, solve Problem 4.17. PROBLEM 4.17 The required tension in cable AB is 200 lb. Determine (a) the vertical force P that must be applied to the pedal, (b) the corresponding reaction at C. SOLUTION Free-Body Diagram: (Three-Force body) Reaction at C must pass through E, where D and 200-lb force intersect. 6.062 in. 15 in. β = 22.005° tan β = Force triangle (a) P = (200 lb)tan 22.005° P = 80.83 lb (b) C= P = 80.8 lb 200 lb = 215.7 lb cos 22.005° C = 216 lb 22.0° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 414 ! PROBLEM 4.64 Using the method of Section 4.7, solve Problem 4.18. PROBLEM 4.18 Determine the maximum tension that can be developed in cable AB if the maximum allowable value of the reaction at C is 250 lb. SOLUTION Free-Body Diagram: (Three -Force body) Reaction at C must pass through E, where D and the force T intersect. 6.062 in. 15 in. β = 22.005° tan β = Force triangle T = (250 lb) cos 22.005° T = 231.8 lb ! T = 232 lb PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 415 ! PROBLEM 4.65 The spanner shown is used to rotate a shaft. A pin fits in a hole at A, while a flat, frictionless surface rests against the shaft at B. If a 60-lb force P is exerted on the spanner at D, find the reactions at A and B. SOLUTION Free-Body Diagram: (Three-Force body) The line of action of A must pass through D, where B and P intersect. 3sin 50° 3cos 50° + 15 = 0.135756 α = 7.7310° tan α = Force triangle 60 lb sin 7.7310° = 446.02 lb 60 lb B= tan 7.7310° = 441.97 lb A= A = 446 lb 7.73° B = 442 lb ! PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 416 PROBLEM 4.66 Determine the reactions at B and D when b = 60 mm. SOLUTION Since CD is a two-force member, the line of action of reaction at D must pass through Points C and D. Free-Body Diagram: (Three-Force body) Reaction at B must pass through E, where the reaction at D and 80-N force intersect. 220 mm 250 mm β = 41.348° tan β = Force triangle Law of sines 80 N B D = = sin 3.652° sin 45° sin131.348° B = 888.0 N D = 942.8 N ! B = 888 N 41.3° D = 943 N 45.0° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 417 PROBLEM 4.67 Determine the reactions at B and D when b = 120 mm. SOLUTION Since CD is a two-force member, line of action of reaction at D must pass through C and D . Free-Body Diagram: (Three-Force body) Reaction at B must pass through E, where the reaction at D and 80-N force intersect. 280 mm 250 mm β = 48.24° tan β = Force triangle Law of sines 80 N B D = = sin 3.24° sin135° sin 41.76° B = 1000.9 N D = 942.8 N ! B = 1001 N 48.2° D = 943 N 45.0° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 418 PROBLEM 4.68 Determine the reactions at B and C when a = 1.5 in. SOLUTION Since CD is a two-force member, the force it exerts on member ABD is directed along DC. Free-Body Diagram of ABD: (Three-Force member) The reaction at B must pass through E, where D and the 50-lb load intersect. Triangle CFD: Triangle EAD: Triangle EGB: Force triangle 3 = 0.6 5 α = 30.964° tan α = AE = 10 tan α = 6 in. GE = AE − AG = 6 − 1.5 = 4.5 in. GB 3 = GE 4.5 β = 33.690° tan β = B D 50 lb = = sin120.964° sin 33.690° sin 25.346° B = 100.155 lb D = 64.789 lb ! B = 100.2 lb 56.3° C = D = 64.8 lb 31.0° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 419 ! PROBLEM 4.69 A 50-kg crate is attached to the trolley-beam system shown. Knowing that a = 1.5 m, determine (a) the tension in cable CD, (b) the reaction at B. SOLUTION Free-Body Diagram: Three-Force body: W and TCD intersect at E. 0.7497 m 1.5 m β = 26.56° tan β = Force triangle 3 forces intersect at E. W = (50 kg) 9.81 m/s 2 = 490.5 N Law of sines TCD 490.5 N B = = sin 61.56° sin 63.44° sin 55° TCD = 498.9 N B = 456.9 N TCD = 499 N (a) B = 457 N (b) 26.6° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 420 ! PROBLEM 4.70 Solve Problem 4.69, assuming that a = 3 m. PROBLEM 4.69 A 50-kg crate is attached to the trolley-beam system shown. Knowing that a = 1.5 m, determine (a) the tension in cable CD, (b) the reaction at B. SOLUTION W and TCD intersect at E Free-Body Diagram: Three-Force body: AE 0.301 m = AB 3m β = 5.722° tan β = Force Triangle (Three forces intersect at E.) W = (50 kg) 9.81 m/s 2 = 490.5 N Law of sines TCD 490.5 N B = = sin 29.278° sin 95.722° sin 55° TCD = 997.99 N B = 821.59 N TCD = 998 N (a) B = 822 N (b) 5.72° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 421 PROBLEM 4.71 One end of rod AB rests in the corner A and the other end is attached to cord BD. If the rod supports a 40-lb load at its midpoint C, find the reaction at A and the tension in the cord. SOLUTION Free-Body Diagram: (Three-Force body) The line of action of reaction at A must pass through E, where T and the 40-lb load intersect. EF 23 = AF 12 α = 62.447° tan α = 5 EH = DH 12 β = 22.620° tan β = Force triangle A T 40 lb = = sin 67.380° sin 27.553° sin 85.067° A = 37.1 lb 62.4° T = 18.57 lb PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 422 PROBLEM 4.72 Determine the reactions at A and D when β = 30°. SOLUTION From f.b.d. of frame ABCD ΣM D = 0: − A(0.18 m) + [(150 N) sin 30°](0.10 m) + [(150 N) cos 30°](0.28 m) = 0 A = 243.74 N or A = 244 N ! ΣFx = 0: (243.74 N) + (150 N) sin 30° + Dx = 0 Dx = −318.74 N ΣFy = 0: D y − (150 N) cos 30° = 0 Dy = 129.904 N Then D = ( Dx ) 2 + Dx2 = (318.74) 2 + (129.904) 2 = 344.19 N and Dy ! # $ Dx % 129.904 ! = tan −1 " # $ −318.74 % θ = tan −1 " = −22.174° or D = 344 N 22.2° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 423 PROBLEM 4.73 Determine the reactions at A and D when β = 60°. SOLUTION From f.b.d. of frame ABCD ΣM D = 0: − A(0.18 m) + [(150 N) sin 60°](0.10 m) + [(150 N) cos 60°](0.28 m) = 0 A = 188.835 N or A = 188.8 N ! ΣFx = 0: (188.835 N) + (150 N) sin 60° + Dx = 0 Dx = −318.74 N ΣFy = 0: D y − (150 N) cos 60° = 0 Dy = 75.0 N Then D = ( Dx ) 2 + ( Dy ) 2 = (318.74) 2 + (75.0) 2 = 327.44 N and Dy ! # $ Dx % θ = tan −1 " 75.0 ! = tan −1 " # $ −318.74 % = −13.2409° or D = 327 N 13.24° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 424 PROBLEM 4.74 A 40-lb roller, of diameter 8 in., which is to be used on a tile floor, is resting directly on the subflooring as shown. Knowing that the thickness of each tile is 0.3 in., determine the force P required to move the roller onto the tiles if the roller is (a) pushed to the left, (b) pulled to the right. SOLUTION See solution to Problem 4.73 for free-body diagram and analysis leading to the following equations: T= P 1 + cos θ C=P (1) sin θ 1 + cos θ (2) For θ = 45° P P = 1 + cos 45° 1.7071 Eq. (1): T= Eq. (2): C=P sin 45° 0.7071 =P 1 + cos 45° 1.7071 T = 0.586 P C = 0.444P PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 425 PROBLEM 4.75 Member ABC is supported by a pin and bracket at B and by an inextensible cord attached at A and C and passing over a frictionless pulley at D. The tension may be assumed to be the same in portions AD and CD of the cord. For the loading shown and neglecting the size of the pulley, determine the tension in the cord and the reaction at B. SOLUTION Free-Body Diagram: Reaction at B must pass through D. 7 in. 12 in. α = 30.256° 7 in. tan β = 24 in. β = 16.26° tan α = Force triangle Law of sines T T − 72 lb B = = sin 59.744° sin13.996° sin106.26 T (sin13.996°) = (T − 72 lb)(sin 59.744°) T (0.24185) = (T − 72)(0.86378) T = 100.00 lb sin 106.26° sin 59.744 = 111.14 lb T = 100.0 lb B = (100 lb) B = 111.1 lb 30.3° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 426 PROBLEM 4.76 Member ABC is supported by a pin and bracket at B and by an inextensible cord attached at A and C and passing over a frictionless pulley at D. The tension may be assumed to be the same in portions AD and CD of the cord. For the loading shown and neglecting the size of the pulley, determine the tension in the cord and the reaction at B. SOLUTION Free-Body Diagram: Force triangle Reaction at B must pass through D. tan α = 120 ; α = 36.9° 160 T T − 75 lb B = = 4 3 5 3T = 4T − 300; T = 300 lb 5 5 B = T = (300 lb) = 375 lb 4 4 B = 375 lb 36.9°! PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 427 PROBLEM 4.77 Rod AB is supported by a pin and bracket at A and rests against a frictionless peg at C. Determine the reactions at A and C when a 170-N vertical force is applied at B. SOLUTION Free-Body Diagram: (Three-Force body) The reaction at A must pass through D where C and 170-N force intersect. 160 mm 300 mm α = 28.07° tan α = We note that triangle ABD is isosceles (since AC = BC) and, therefore CAD = α = 28.07° Also, since CD ⊥ CB, reaction C forms angle α = 28.07° with horizontal. Force triangle We note that A forms angle 2α with vertical. Thus A and C form angle 180° − (90° − α ) − 2α = 90° − α Force triangle is isosceles and we have A = 170 N C = 2(170 N)sin α = 160.0 N A = 170.0 N 33.9° C = 160.0 N 28.1° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 428 ! PROBLEM 4.78 Solve Problem 4.77, assuming that the 170-N force applied at B is horizontal and directed to the left. PROBLEM 4.77 Rod AB is supported by a pin and bracket at A and rests against a frictionless peg at C. Determine the reactions at A and C when a 170-N vertical force is applied at B. SOLUTION Free-Body Diagram: (Three-Force body) The reaction at A must pass through D, where C and the 170-N force intersect. 160 mm 300 mm α = 28.07° tan α = We note that triangle ADB is isosceles (since AC = BC). Therefore Also A= B = 90° − α . ADB = 2α Force triangle The angle between A and C must be 2α − α = α α = 28.07° Thus, force triangle is isosceles and A = 170.0 N C = 2(170 N) cos α = 300 N A = 170.0 N 56.1° C = 300 N 28.1° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 429 ! PROBLEM 4.79 Using the method of Section 4.7, solve Problem 4.21. PROBLEM 4.21 Determine the reactions at A and B when (a) h = 0, (b) h = 200 mm. SOLUTION ! ! Free-Body Diagram: (a) h=0 Reaction A must pass through C where 150-N weight and B interect. Force triangle is equilateral (b) A = 150.0 N 30.0° B = 150.0 N 30.0° h = 200 mm 55.662 250 β = 12.552° ! tan β = ! Force triangle Law of sines A B 150 N = = sin17.448° sin 60° sin102.552° A = 433.247 N B = 488.31 N ! ! A = 433 N ! B = 488 N 12.55° 30.0° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 430 ! PROBLEM 4.80 Using the method of Section 4.7, solve Problem 4.28. PROBLEM 4.28 A lever AB is hinged at C and attached to a control cable at A. If the lever is subjected to a 500-N horizontal force at B, determine (a) the tension in the cable, (b) the reaction at C. SOLUTION Free-Body Diagram: (Three-Force body) Reaction at C must pass through E, where FAD and 500-N force intersect. Since AC = CD = 250 mm, triangle ACD is isosceles. We have C = 90° + 30° = 120° and A= D= 1 (180° − 120°) = 30° 2 Dimensions in mm On the other hand, from triangle BCF: CF = ( BC )sin 30° = 200 sin 30° = 100 mm FD = CD − CF = 250 − 100 = 150 mm From triangle EFD, and since D = 30° : EF = ( FD ) tan 30° = 150 tan 30° = 86.60 mm From triangle EFC: 100 mm CF = EF 86.60 mm α = 49.11° tan α = Force triangle Law of sines FAD C 500 N = = sin 49.11° sin 60° sin 70.89° FAD = 400 N, C = 458 N FAD = 400 N (a) C = 458 N (b) 49.1° ! PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 431 PROBLEM 4.81 Knowing that θ = 30°, determine the reaction (a) at B, (b) at C. SOLUTION Reaction at C must pass through D where force P and reaction at B intersect. In ∆ CDE: Free-Body Diagram: (Three-Force body) ( 3 − 1) R R = 3 −1 β = 36.2° tan β = Force triangle Law of sines P B C = = sin 23.8° sin126.2° sin 30° B = 2.00 P C = 1.239 P (a) B = 2P 60.0° (b) C = 1.239P 36.2° ! PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 432 PROBLEM 4.82 Knowing that θ = 60°, determine the reaction (a) at B, (b) at C. SOLUTION Reaction at C must pass through D where force P and reaction at B intersect. In ∆CDE: tan β = R− =1− R 3 Free-Body Diagram: (Three-Force body) R 1 3 β = 22.9° Force triangle Law of sines P B C = = sin 52.9° sin 67.1° sin 60° B = 1.155P C = 1.086 P (a) B = 1.155P 30.0° (b) C = 1.086P 22.9° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 433 ! PROBLEM 4.83 Rod AB is bent into the shape of an arc of circle and is lodged between two pegs D and E. It supports a load P at end B. Neglecting friction and the weight of the rod, determine the distance c corresponding to equilibrium when a = 20 mm and R = 100 mm. SOLUTION Free-Body Diagram: yED = xED = a, Since Slope of ED is slope of HC is 45° 45° Also DE = 2 a and a 1! DH = HE = " # DE = 2 $2% For triangles DHC and EHC sin β = a 2 R = a 2R Now c = R sin(45° − β ) For a = 20 mm and R = 100 mm sin β = 20 mm 2(100 mm) = 0.141421 β = 8.1301° and c = (100 mm) sin(45° − 8.1301°) = 60.00 mm or c = 60.0 mm PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 434 ! PROBLEM 4.84 A slender rod of length L is attached to collars that can slide freely along the guides shown. Knowing that the rod is in equilibrium, derive an expression for the angle θ in terms of the angle β. SOLUTION As shown in the free-body diagram of the slender rod AB, the three forces intersect at C. From the force geometry Free-Body Diagram: tan β = xGB y AB where y AB = L cos θ and xGB = tan β = 1 L sin θ 2 1 2 L sin θ L cos θ 1 = tan θ 2 or tan θ = 2 tan β PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 435 ! PROBLEM 4.85 An 8-kg slender rod of length L is attached to collars that can slide freely along the guides shown. Knowing that the rod is in equilibrium and that β = 30°, determine (a) the angle θ that the rod forms with the vertical, (b) the reactions at A and B. SOLUTION (a) As shown in the free-body diagram of the slender rod AB, the three forces intersect at C. From the geometry of the forces Free-Body Diagram: tan β = xCB yBC where xCB = 1 L sin θ 2 and yBC = L cos θ tan β = 1 tan θ 2 or tan θ = 2 tan β For β = 30° tan θ = 2 tan 30° = 1.15470 θ = 49.107° or θ = 49.1° ! W = mg = (8 kg)(9.81 m/s2 ) = 78.480 N (b) From force triangle A = W tan β = (78.480 N) tan 30° = 45.310 N and B= or W 78.480 N = = 90.621 N cos β cos 30° A = 45.3 N or B = 90.6 N ! 60.0° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 436 ! PROBLEM 4.86 A slender uniform rod of length L is held in equilibrium as shown, with one end against a frictionless wall and the other end attached to a cord of length S. Derive an expression for the distance h in terms of L and S. Show that this position of equilibrium does not exist if S . 2L. SOLUTION ! From the f.b.d. of the three-force member AB, forces must intersect at D. Since the force T intersects Point D, directly above G, yBE = h For triangle ACE: S 2 = ( AE ) 2 + (2h) 2 (1) For triangle ABE: L2 = ( AE ) 2 + (h) 2 (2) Subtracting Equation (2) from Equation (1) S 2 − L2 = 3h 2 ! (3) ! h= S 2 − L2 3 ! or ! As length S increases relative to length L, angle θ increases until rod AB is vertical. At this vertical position: ! h+L=S or h = S − L Therefore, for all positions of AB h$S − L or S 2 − L2 $S −L 3 or S 2 − L2 $ 3( S − L) 2 (4) = 3( S 2 − 2 SL + L2 ) ! = 3S 2 − 6 SL + 3L2 or 0 $ 2S 2 − 6SL + 4 L2 and 0 $ S 2 − 3SL + 2 L2 = ( S − L)( S − 2 L) ! PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 437 PROBLEM 4.86 (Continued) For S−L=0 S =L Minimum value of S is L For S − 2L = 0 S = 2L Maximum value of S is 2L Therefore, equilibrium does not exist if S . 2 L PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 438 ! PROBLEM 4.87 A slender uniform rod of length L = 20 in. is held in equilibrium as shown, with one end against a frictionless wall and the other end attached to a cord of length S = 30 in. Knowing that the weight of the rod is 10 lb, determine (a) the distance h, (b) the tension in the cord, (c) the reaction at B. SOLUTION From the f.b.d. of the three-force member AB, forces must intersect at D. Since the force T intersects Point D, directly above G, yBE = h For triangle ACE: S 2 = ( AE ) 2 + (2h) 2 (1) For triangle ABE: L2 = ( AE ) 2 + (h) 2 (2) Subtracting Equation (2) from Equation (1) S 2 − L2 = 3h 2 ! ! h= or ! (a) S 2 − L2 3 L = 20 in. and S = 30 in. For (30) 2 − (20) 2 3 = 12.9099 in. h= (b) ! W = 10 lb We have 2h ! # $ s % θ = sin −1 " and ! or h = 12.91 in. & 2(12.9099) ' = sin −1 ( ) 30 * + θ = 59.391° ! ! ! ! ! PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 439 PROBLEM 4.87 (Continued) From the force triangle W sin θ 10 lb = sin 59.391° = 11.6190 lb T= (c) or T = 11.62 lb W tan θ 10 lb = tan 59.391° B= = 5.9161 lb or B = 5.92 lb PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 440 ! PROBLEM 4.88 A uniform rod AB of length 2R rests inside a hemispherical bowl of radius R as shown. Neglecting friction, determine the angle θ corresponding to equilibrium. SOLUTION Based on the f.b.d., the uniform rod AB is a three-force body. Point E is the point of intersection of the three forces. Since force A passes through O, the center of the circle, and since force C is perpendicular to the rod, triangle ACE is a right triangle inscribed in the circle. Thus, E is a point on the circle. Note that the angle α of triangle DOA is the central angle corresponding to the inscribed angle θ of triangle DCA. α = 2θ The horizontal projections of AE , ( x AE ), and AG , ( x AG ), are equal. x AE = x AG = x A or ( AE ) cos 2θ = ( AG ) cos θ and (2 R) cos 2θ = R cos θ cos 2θ = 2 cos 2 θ − 1 Now then or 4 cos 2 θ − 2 = cos θ 4 cos 2 θ − cos θ − 2 = 0 Applying the quadratic equation cos θ = 0.84307 and cos θ = − 0.59307 θ = 32.534° and θ = 126.375°(Discard) or θ = 32.5° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 441 ! PROBLEM 4.89 A slender rod of length L and weight W is attached to a collar at A and is fitted with a small wheel at B. Knowing that the wheel rolls freely along a cylindrical surface of radius R, and neglecting friction, derive an equation in θ, L, and R that must be satisfied when the rod is in equilibrium. SOLUTION Free-Body Diagram (Three-Force body) Reaction B must pass through D where B and W intersect. Note that ∆ABC and ∆BGD are similar. AC = AE = L cos θ In ∆ABC: (CE ) 2 + ( BE )2 = ( BC )2 (2 L cos θ ) 2 + ( L sin θ )2 = R 2 2 R! 2 2 " # = 4cos θ + sin θ $L% 2 R! 2 2 " # = 4cos θ + 1 − cos θ $L% 2 R! 2 " # = 3cos θ + 1 $L% 2 ' 1& cos 2 θ = (" R !# − 1) 3 *$ L % + PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 442 ! PROBLEM 4.90 Knowing that for the rod of Problem 4.89, L = 15 in., R = 20 in., and W = 10 lb, determine (a) the angle θ corresponding to equilibrium, (b) the reactions at A and B. SOLUTION See the solution to Problem 4.89 for free-body diagram and analysis leading to the following equation 2 ' 1& cos 2 θ = (" R !# − 1) 3 *$ L % + For L = 15 in., R = 20 in., and W = 10 lb. 2 ' 1 & 20 in. ! cos 2 θ = (" # − 1) ; θ = 59.39° 3 ($ 15 in. % ) * + (a) In ∆ABC: θ = 59.4° 1 BE L sin θ = = tan θ CE 2 L cos θ 2 1 tan α = tan 59.39° = 0.8452 2 α = 40.2° tan α = Force triangle A = W tan α = (10 lb) tan 40.2° = 8.45 lb (10 lb) W = = 13.09 lb B= cos α cos 40.2° A = 8.45 lb (b) B = 13.09 lb (c) 49.8° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 443 ! PROBLEM 4.91 A 4 × 8-ft sheet of plywood weighing 34 lb has been temporarily placed among three pipe supports. The lower edge of the sheet rests on small collars at A and B and its upper edge leans against pipe C. Neglecting friction at all surfaces, determine the reactions at A, B, and C. SOLUTION rG/B = 3.75 1.3919 i+ j+k 2 2 We have 5 unknowns and 6 Eqs. of equilibrium. Plywood sheet is free to move in z direction, but equilibrium is maintained (ΣFz = 0). ΣM B = 0: rA/B × ( Ax i + Ay j) + rC/B × ( −Ci ) + rG/B × ( − wj) = 0 i 0 Ax j 0 Ay k i j k i j k 4 + 3.75 1.3919 2 + 1.875 0.696 1 = 0 0 0 0 0 −C −34 0 −4 Ay i + 4 Ax j − 2Cj + 1.3919Ck + 34i − 63.75k = 0 Equating coefficients of unit vectors to zero: i: − 4 Ay + 34 = 0 j: − 2C + 4 Ax = 0 k : 1.3919C − 63.75 = 0 Ay = 8.5 lb Ax = 1 1 C = (45.80) = 22.9 lb 2 2 C = 45.80 lb C = 45.8 lb ΣFx = 0: Ax + Bx − C = 0: Bx = 45.8 − 22.9 = 22.9 lb ΣFy = 0: Ay + By − W = 0: By = 34 − 8.5 = 25.5 lb A = (22.9 lb)i + (8.5 lb) j B = (22.9 lb)i + (25.5 lb) j C = −(45.8 lb)i PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 444 PROBLEM 4.92 Two tape spools are attached to an axle supported by bearings at A and D. The radius of spool B is 30 mm and the radius of spool C is 40 mm. Knowing that TB = 80 N and that the system rotates at a constant rate, determine the reactions at A and D. Assume that the bearing at A does not exert any axial thrust and neglect the weights of the spools and axle. SOLUTION Dimensions in mm We have six unknowns and six Eqs. of equilibrium. ΣM A = 0: (90i + 30k ) × (−80 j) + (210i + 40 j) × (−TC k ) + (300i ) × ( Dx i + Dy j + Dz k ) = 0 −7200k + 2400i + 210TC j − 40TC i + 300 D y k − 300 Dz j = 0 Equate coefficients of unit vectors to zero: i: j: 2400 − 40TC = 0 TC = 60 N 210TC − 300 Dz = 0 (210)(60) − 300 Dz = 0 Dz = 42 N k : −7200 + 300 Dy = 0 Dy = 24 N ΣFx = 0: Dx = 0 ΣFy = 0: Ay + D y − 80 N = 0 Ay = 80 − 24 = 56 N ΣFz = 0: Az + Dz − 60 N = 0 Az = 60 − 42 = 18 N A = (56.0 N) j + (18.00 N)k D = (24.0 N) j + (42.0 N)k PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 445 PROBLEM 4.93 Solve Problem 4.92, assuming that the spool C is replaced by a spool of radius 50 mm. PROBLEM 4.92 Two tape spools are attached to an axle supported by bearings at A and D. The radius of spool B is 30 mm and the radius of spool C is 40 mm. Knowing that TB = 80 N and that the system rotates at a constant rate, determine the reactions at A and D. Assume that the bearing at A does not exert any axial thrust and neglect the weights of the spools and axle. SOLUTION Dimensions in mm We have six unknowns and six Eqs. of equilibrium. ΣM A = 0: (90i + 30k ) × (−80 j) + (210i + 50 j) × ( −TC k ) + (300i) × ( Dx i + D y j + Dz k ) = 0 −7200k + 2400i + 210TC j − 50TC i + 300 Dy k − 300 Dz j = 0 Equate coefficients of unit vectors to zero: i: j: 2400 − 50TC = 0 TC = 48 N 210TC − 300 Dz = 0 (210)(48) − 300 Dz = 0 Dz = 33.6 N k : − 7200 + 300 Dy = 0 ΣFx = 0: Dy = 24 N Dx = 0 ΣFy = 0: Ay + Dy − 80 N = 0 Ay = 80 − 24 = 56 N ΣFz = 0: Az = 48 − 33.6 = 14.4 N Az + Dz − 48 = 0 A = (56.0 N) j + (14.40 N)k D = (24.0 N) j + (33.6 N)k PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 446 PROBLEM 4.94 Two transmission belts pass over sheaves welded to an axle supported by bearings at B and D. The sheave at A has a radius of 2.5 in., and the sheave at C has a radius of 2 in. Knowing that the system rotates at a constant rate, determine (a) the tension T, (b) the reactions at B and D. Assume that the bearing at D does not exert any axial thrust and neglect the weights of the sheaves and axle. SOLUTION Assume moment reactions at the bearing supports are zero. From f.b.d. of shaft (a) (b) ΣM x -axis = 0: (24 lb − 18 lb)(5 in.) + (30 lb − T )(4 in.) = 0 T = 37.5 lb ΣFx = 0: Bx = 0 ΣM D ( z -axis) = 0: (30 lb + 37.5 lb)(6 in.) − By (12 in.) = 0 By = 33.75 lb ΣM D ( y -axis) = 0: (24 lb + 18 lb)(20 in.) + Bz (12 in.) = 0 Bz = −70.0 lb or B = (33.8 lb) j − (70.0 lb)k ΣM B ( z -axis) = 0: − (30 lb + 37.5 lb)(6 in.) + D y (12 in.) = 0 Dy = 33.75 lb ΣM B ( y -axis) = 0: (24 lb + 18 lb)(8 in.) + Dz (12 in.) = 0 Dz = −28.0 lb ! or D = (33.8 lb) j − (28.0 lb)k ! PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 447 PROBLEM 4.95 A 200-mm lever and a 240-mm-diameter pulley are welded to the axle BE that is supported by bearings at C and D. If a 720-N vertical load is applied at A when the lever is horizontal, determine (a) the tension in the cord, (b) the reactions at C and D. Assume that the bearing at D does not exert any axial thrust. SOLUTION Dimensions in mm We have six unknowns and six Eqs. of equilibrium–OK ΣM C = 0: (−120k ) × ( Dx i + Dy j) + (120 j − 160k ) × Ti + (80k − 200i ) × (−720 j) = 0 −120 Dx j + 120 D y i − 120Tk − 160Tj + 57.6 × 103 i + 144 × 103 k = 0 Equating to zero the coefficients of the unit vectors: k: −120T + 144 × 103 = 0 i: 120 Dy + 57.6 × 103 = 0 j: − 120 Dx − 160(1200 N) = 0 (b) ΣFx = 0: Cx + Dx + T = 0 ΣFy = 0: C y + Dy − 720 = 0 ΣFz = 0: Cz = 0 (a) T = 1200 N Dy = −480 N Dx = −1600 N Cx = 1600 − 1200 = 400 N C y = 480 + 720 = 1200 N C = (400 N)i + (1200 N) j D = −(1600 N)i − (480 N) j PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 448 PROBLEM 4.96 Solve Problem 4.95, assuming that the axle has been rotated clockwise in its bearings by 30° and that the 720-N load remains vertical. PROBLEM 4.95 A 200-mm lever and a 240-mmdiameter pulley are welded to the axle BE that is supported by bearings at C and D. If a 720-N vertical load is applied at A when the lever is horizontal, determine (a) the tension in the cord, (b) the reactions at C and D. Assume that the bearing at D does not exert any axial thrust. SOLUTION Dimensions in mm We have six unknowns and six Eqs. of equilibrium. ΣM C = 0: (−120k ) × ( Dx i + Dy j) + (120 j − 160k ) × T i + (80k − 173.21i ) × (−720 j) = 0 −120 Dx j + 120 D y i −120T k −160T j + 57.6 × 103 i + 124.71 × 103 k = 0 Equating to zero the coefficients of the unit vectors: k : − 120T + 124.71 × 103 = 0 i: T = 1039 N 120 Dy + 57.6 × 103 = 0 Dy = −480 N j: − 120 Dx − 160(1039.2) (b) T = 1039.2 N ΣFx = 0: Cx + Dx + T = 0 ΣFy = 0: C y + Dy − 720 = 0 ΣFz = 0: Cz = 0 Dx = −1385.6 N Cx = 1385.6 − 1039.2 = 346.4 C y = 480 + 720 = 1200 N C = (346 N)i + (1200 N) j D = −(1386 N)i − (480 N) j PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 449 PROBLEM 4.97 An opening in a floor is covered by a 1 × 1.2-m sheet of plywood of mass 18 kg. The sheet is hinged at A and B and is maintained in a position slightly above the floor by a small block C. Determine the vertical component of the reaction (a) at A, (b) at B, (c) at C. SOLUTION rB/A = 0.6i rC/A = 0.8i + 1.05k rG/A = 0.3i + 0.6k W = mg = (18 kg)9.81 W = 176.58 N ΣM A = 0: rB/A × Bj + rC/A × Cj + rG/A × (−Wj) = 0 (0.6i ) × Bj + (0.8i + 1.05k ) × Cj + (0.3i + 0.6k ) × ( −Wj) = 0 0.6 Bk + 0.8Ck − 1.05Ci − 0.3Wk + 0.6Wi = 0 Equate coefficients of unit vectors of zero: 0.6 ! i : 1.05C + 0.6W = 0 C = " #176.58 N = 100.90 N 1.05 $ % k : 0.6 B + 0.8C − 0.3W = 0 0.6 B + 0.8(100.90 N) − 0.3(176.58 N) = 0 B = −46.24 N ΣFy = 0: A + B + C − W = 0 A − 46.24 N + 100.90 N + 176.58 N = 0 A = 121.92 N (a ) A = 121.9 N (b) B = −46.2 N (c) C = 100.9 N PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 450 PROBLEM 4.98 Solve Problem 4.97, assuming that the small block C is moved and placed under edge DE at a point 0.15 m from corner E. PROBLEM 4.97 An opening in a floor is covered by a 1 × 1.2-m sheet of plywood of mass 18 kg. The sheet is hinged at A and B and is maintained in a position slightly above the floor by a small block C. Determine the vertical component of the reaction (a) at A, (b) at B, (c) at C. SOLUTION rB/A = 0.6i rC/A = 0.65i + 1.2k rG/A = 0.3i + 0.6k W = mg = (18 kg) 9.81 m/s 2 W = 176.58 N ΣM A = 0: rB/A × Bj + rC/A × Cj + rG/A × (−Wj) = 0 0.6i × Bj + (0.65i + 1.2k ) × Cj + (0.3i + 0.6k ) × (−Wj) = 0 0.6 Bk + 0.65Ck − 1.2Ci − 0.3Wk + 0.6Wi = 0 Equate coefficients of unit vectors to zero: 0.6 ! i : −1.2C + 0.6W = 0 C = " #176.58 N = 88.29 N $ 1.2 % k : 0.6 B + 0.65C − 0.3W = 0 0.6 B + 0.65(88.29 N) − 0.3(176.58 N) = 0 B = −7.36 N ΣFy = 0: A + B + C − W = 0 A − 7.36 N + 88.29 N − 176.58 N = 0 A = 95.648 N (a ) A = 95.6 N (b) − 7.36 N (c) 88.3 N PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 451 PROBLEM 4.99 The rectangular plate shown weighs 80 lb and is supported by three vertical wires. Determine the tension in each wire. SOLUTION Free-Body Diagram: ΣM B = 0: rA/B × TA j + rC/B × TC j + rG/B × (−80 lb) j = 0 (60 in.)k × TA j + [(60 in.)i + (15 in.)k ] × TC j + [(30 in.)i + (30 in.)k ] × ( −80 lb) j = 0 −60TAi + 60TC k − 15TC i − 2400k + 2400i = 0 Equating to zero the coefficients of the unit vectors: i: 60TA − 15(40) + 2400 = 0 TA = 30.0 lb k: 60TC − 2400 = 0 TC = 40.0 lb ΣFy = 0: TA + TB + TC − 80 lb = 0 30 lb + TB + 40 lb − 80 lb = 0 TB = 10.00 lb PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 452 PROBLEM 4.100 The rectangular plate shown weighs 80 lb and is supported by three vertical wires. Determine the weight and location of the lightest block that should be placed on the plate if the tensions in the three wires are to be equal. SOLUTION Free-Body Diagram: Let −Wb j be the weight of the block and x and z the block’s coordinates. Since tensions in wires are equal, let TA = TB = TC = T ΣM 0 = 0: (rA × Tj) + (rB × Tj) + (rC × Tj) + rG × (−Wj) + ( xi + zk ) × ( −Wb j) = 0 or, (75 k ) × Tj + (15 k ) × Tj + (60i + 30k ) × Tj + (30i + 45k ) × (−Wj) + ( xi + zk ) × (−Wb j) = 0 or, −75Ti − 15T i + 60T k − 30T i − 30W k + 45W i − Wb × k + Wb zi = 0 Equate coefficients of unit vectors to zero: i: −120T + 45W + Wb z = 0 (1) k: 60T − 30W − Wb x = 0 (2) ΣFy = 0: 3T − W − Wb = 0 (3) Eq. (1) + 40 Eq. (3): 5W + ( z − 40)Wb = 0 (4) Eq. (2) – 20 Eq. (3): −10W − ( x − 20)Wb = 0 (5) Also, PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 453 PROBLEM 4.100 (Continued) Solving (4) and (5) for Wb /W and recalling of 0 # x # 60 in., 0 # z # 90 in., (4): Wb 5 5 = $ = 0.125 W 40 − z 40 − 0 (5): Wb 10 10 = $ = 0.5 W 20 − x 20 − 0 Thus, (Wb ) min = 0.5W = 0.5(80) = 40 lb (Wb ) min = 40.0 lb Making Wb = 0.5W in (4) and (5): 5W + ( z − 40)(0.5W ) = 0 z = 30.0 in. −10W − ( x − 20)(0.5W ) = 0 x = 0 in. PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 454 PROBLEM 4.101 Two steel pipes AB and BC, each having a mass per unit length of 8 kg/m, are welded together at B and supported by three wires. Knowing that a = 0.4 m, determine the tension in each wire. SOLUTION W1 = 0.6m′g W2 = 1.2m′g ΣM D = 0: rA/D × TA j + rE/D × (−W1 j) + rF/D × (−W2 j) + rC/D × TC j = 0 (−0.4i + 0.6k ) × TA j + (−0.4i + 0.3k ) × (−W1 j) + 0.2i × (−W2 j) + 0.8i × TC j = 0 −0.4TAk − 0.6TA i + 0.4W1k + 0.3W1i − 0.2W2 k + 0.8TC k = 0 Equate coefficients of unit vectors to zero: 1 1 i : − 0.6TA + 0.3W1 = 0; TA = W1 = 0.6m′g = 0.3m′g 2 2 k : − 0.4TA + 0.4W1 − 0.2W2 + 0.8TC = 0 −0.4(0.3m′g ) + 0.4(0.6m′g ) − 0.2(1.2m′g ) + 0.8TC = 0 TC = (0.12 − 0.24 − 0.24)m′g = 0.15m′g 0.8 ΣFy = 0: TA + TC + TD − W1 − W2 = 0 0.3m′g + 0.15m′g + TD − 0.6m′g − 1.2m′g = 0 TD = 1.35m′g 2 m′g = (8 kg/m)(9.81m/s ) = 78.48 N/m TA = 0.3m′g = 0.3 × 78.45 TA = 23.5 N TB = 0.15m′g = 0.15 × 78.45 TB = 11.77 N TC = 1.35m′g = 1.35 × 78.45 TC = 105.9 N PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 455 PROBLEM 4.102 For the pipe assembly of Problem 4.101, determine (a) the largest permissible value of a if the assembly is not to tip, (b) the corresponding tension in each wire. SOLUTION W1 = 0.6m′g W2 = 1.2m′g ΣM D = 0: rA/D × TA j + rE/D × (−W1 j) + rF/D × (−W2 j) + rC/D × TC j = 0 (− ai + 0.6k ) × TA j + (− ai + 0.3k ) × (−W1 j) + (0.6 − a)i × (−W2 j) + (1.2 − a)i × TC j = 0 −TA ak − 0.6TA i + W1ak + 0.3W1i − W2 (0.6 − a)k + TC (1.2 − a )k = 0 Equate coefficients of unit vectors to zero: 1 1 i : − 0.6TA + 0.3W1 = 0; TA = W1 = 0.6m′g = 0.3m′g 2 2 k : − TA a + W1a − W2 (0.6 − a ) + TC (1.2 − a) = 0 −0.3m′ga + 0.6m′ga − 1.2m′g (0.6 − a ) + TC (1.2 − a) = 0 TC = (a) 0.3a − 0.6a + 1.2(0.6 − a) 1.2 − a For Max a and no tipping, TC = 0 −0.3a + 1.2(0.6 − a) = 0 −0.3a + 0.72 − 1.2a = 0 1.5a = 0.72 a = 0.480 m PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 456 PROBLEM 4.102 (Continued) (b) Reactions: m′g = (8 kg/m) 9.81 m/s 2 = 78.48 N/m TA = 0.3m′g = 0.3 × 78.48 = 23.544 N TA = 23.5 N ΣFy = 0: TA + TC + TD − W1 − W2 = 0 TA + 0 + TD − 0.6m′g − 1.2m′g = 0 TD = 1.8m′g − TA = 1.8 × 78.48 − 23.544 = 117.72 TD = 117.7 N PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 457 PROBLEM 4.103 The 24-lb square plate shown is supported by three vertical wires. Determine (a) the tension in each wire when a = 10 in., (b) the value of a for which the tension in each wire is 8 lb. SOLUTION rB/A = ai + 30k rC/A = 30i + ak rG/A = 15i + 15k By symmetry: B = C ΣM A = 0: rB/A × Bj + rC × Cj + rG/A × (−Wj) = 0 (ai + 30k ) × Bj + (30i + ak ) × Bj + (15i + 15k ) × (−Wj) = 0 Bak − 30 Bi + 30 Bk − Bai − 15Wk + 15Wi = 0 Equate coefficient of unit vector i to zero: i : − 30 B − Ba + 15W = 0 B= 15W 30 + a C=B= 15W 30 + a (1) ΣFy = 0: A + B + C − W = 0 & 15W ' A + 2( ) − W = 0; * 30 + a + (a) For a = 10 in. Eq. (1) C=B= Eq. (2) A= A= aW 30 + a (2) 15(24 lb) = 9.00 lb 30 + 10 10(24 lb) = 6.00 lb 30 + 10 A = 6.00 lb B = C = 9.00 lb PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 458 ! PROBLEM 4.103 (Continued) (b) For tension in each wire = 8 lb Eq. (1) 8 lb = 15(24 lb) 30 + a 30 in. + a = 45 a = 15.00 in. PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 459 PROBLEM 4.104 The table shown weighs 30 lb and has a diameter of 4 ft. It is supported by three legs equally spaced around the edge. A vertical load P of magnitude 100 lb is applied to the top of the table at D. Determine the maximum value of a if the table is not to tip over. Show, on a sketch, the area of the table over which P can act without tipping the table. SOLUTION r = 2 ft b = r sin 30° = 1 ft We shall sum moments about AB. (b + r )C + (a − b) P − bW = 0 (1 + 2)C + (a − 1)100 − (1)30 = 0 1 C = [30 − (a − 1)100] 3 If table is not to tip, C $ 0 [30 − ( a − 1)100] $ 0 30 $ (a − 1)100 a − 1 # 0.3 a # 1.3 ft a = 1.300 ft Only ⊥ distance from P to AB matters. Same condition must be satisfied for each leg. P must be located in shaded area for no tipping PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 460 PROBLEM 4.105 A 10-ft boom is acted upon by the 840-lb force shown. Determine the tension in each cable and the reaction at the ball-and-socket joint at A. SOLUTION We have five unknowns and six Eqs. of equilibrium but equilibrium is maintained (ΣM x = 0). Free-Body Diagram: ! BD = (−6 ft)i + (7 ft) j + (6 ft)k BD = 11 ft BE = (−6 ft)i + (7 ft) j − (6 ft)k BE = 11 ft ! BD TBD TBD = TBD (−6i + 7 j + 6k ) = BD 11 ! BE TBE (−6i + 7 j − 6k ) TBE = TBE = BE 11 ΣM A = 0: rB × TBD + rB × TBE + rC × ( −840 j) = 0 6i × TBD T (−6i + 7 j + 6k ) + 6i × BE (−6i + 7 j − 6k ) + 10i × (−840 j) = 0 11 11 42 36 42 36 TBD k − TBD j + TBE k + TBE j − 8400k 11 11 11 11 Equate coefficients of unit vectors to zero. i: − k: 36 36 TBD + TBE = 0 TBE = TBD 11 11 42 42 TBD + TBE − 8400 = 0 11 11 42 ! 2 " TBD # = 8400 $ 11 % TBD = 1100 lb TBE = 1100 lb ! ! PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 461 ! PROBLEM 4.105 (Continued) ΣFx = 0: Ax − 6 6 (1100 lb) − (1100 lb) = 0 11 11 Ax = 1200 lb ΣFy = 0: Ay + 7 7 (1100 lb) + (1100 lb) − 840 lb = 0 11 11 Ay = −560 lb ΣFz = 0: Az + 6 6 (1100 lb) − (1100 lb) = 0 11 11 Az = 0 A = (1200 lb)i − (560 lb) j PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 462 PROBLEM 4.106 A 2.4-m boom is held by a ball-and-socket joint at C and by two cables AD and AE. Determine the tension in each cable and the reaction at C. SOLUTION Free-Body Diagram: Five Unknowns and six Eqs. of equilibrium. Equilibrium is maintained (ΣMAC = 0). rB = 1.2k TAD TAE rA = 2.4k ! AD = −0.8i + 0.6 j − 2.4k AD = 2.6 m ! AE = 0.8i + 1.2 j − 2.4k AE = 2.8 m ! AD TAD = = (−0.8i + 0.6 j − 2.4k ) AD 2.6 ! AE TAE = = (0.8i + 1.2 j − 2.4k ) AE 2.8 ΣM C = 0: rA × TAD + rA × TAE + rB × (−3 kN) j = 0 i j k i j k TAD T 0 0 2.4 0 2.4 AE + 1.2k × (−3.6 kN) j = 0 + 0 2.6 2.8 0.8 1.2 −2.4 −0.8 0.6 −2.4 Equate coefficients of unit vectors to zero. i : − 0.55385 TAD − 1.02857 TAE + 4.32 = 0 (1) j : − 0.73846 TAD + 0.68671 TAE = 0 TAD = 0.92857 TAE Eq. (1): (2) −0.55385(0.92857) TAE − 1.02857 TAE + 4.32 = 0 1.54286 TAE = 4.32 TAE = 2.800 kN TAE = 2.80 kN ! ! PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 463 PROBLEM 4.106 (Continued) Eq. (2): TAD = 0.92857(2.80) = 2.600 kN TAD = 2.60 kN 0.8 0.8 (2.6 kN) + (2.8 kN) = 0 2.6 2.8 0.6 1.2 (2.6 kN) + (2.8 kN) − (3.6 kN) = 0 ΣFy = 0: C y + 2.6 2.8 2.4 2.4 (2.6 kN) − (2.8 kN) = 0 ΣFz = 0: C z − 2.6 2.8 ΣFx = 0: C x − ! Cx = 0 C y = 1.800 kN C z = 4.80 kN C = (1.800 kN) j + (4.80 kN)k PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 464 PROBLEM 4.107 Solve Problem 4.106, assuming that the 3.6-kN load is applied at Point A. PROBLEM 4.106 A 2.4-m boom is held by a ball-and-socket joint at C and by two cables AD and AE. Determine the tension in each cable and the reaction at C. SOLUTION Free-Body Diagram: Five unknowns and six Eqs. of equilibrium. Equilibrium is maintained (ΣMAC = 0). ! AD = −0.8i + 0.6 j − 2.4k AD = 2.6 m ! AE = 0.8i + 1.2 j − 2.4k AE = 2.8 m ! AD TAD (−0.8i + 0.6 j − 2.4k ) TAD = = AD 2.6 ! AE TAE (0.8i + 1.2 j − 2.4k ) TAE = = AE 2.8 ΣM C = 0: rA × TAD + rA × TAE + rA × (−3.6 kN) j Factor rA : or: Coefficient of i: rA × (TAD + TAE − (3.6 kN) j) TAD + TAE − (3 kN) j = 0 − (Forces concurrent at A) TAD T (0.8) + AE (0.8) = 0 2.6 2.8 TAD = Coefficient of j: 2.6 TAE 2.8 (1) TAD T (0.6) + AE (1.2) − 3.6 kN = 0 2.6 2.8 2.6 0.6 ! 1.2 TAE " TAE − 3.6 kN = 0 #+ 2.8 $ 2.6 % 2.8 TAE " $ 0.6 + 1.2 ! # = 3.6 kN 2.8 % TAE = 5.600 kN TAE = 5.60 kN PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 465 PROBLEM 4.107 (Continued) Eq. (1): TAD = 2.6 (5.6) = 5.200 kN 2.8 0.8 0.8 (5.2 kN) + (5.6 kN) = 0; 2.6 2.8 0.6 1.2 (5.2 kN) + (5.6 kN) − 3.6 kN = 0 ΣFy = 0: C y + 2.6 2.8 2.4 2.4 (5.2 kN) − (5.6 kN) = 0 ΣFz = 0: Cz − 2.6 2.8 ΣFx = 0: C x − TAD = 5.20 kN ! C = (9.60 kN)k ! Cx = 0 Cy = 0 Cz = 9.60 kN Note: Since forces and reaction are concurrent at A, we could have used the methods of Chapter 2. PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 466 PROBLEM 4.108 A 600-lb crate hangs from a cable that passes over a pulley B and is attached to a support at H. The 200-lb boom AB is supported by a ball-and-socket joint at A and by two cables DE and DF. The center of gravity of the boom is located at G. Determine (a) the tension in cables DE and DF, (b) the reaction at A. SOLUTION Free-Body Diagram: WC = 600 lb WG = 200 lb We have five unknowns (TDE , TDF , Ax , Ay , Az ) and five equilibrium equations. The boom is free to spin about the AB axis, but equilibrium is maintained, since ΣM AB = 0. " We have BH = (30 ft)i − (22.5 ft) j BH = 37.5 ft " 8.8 DE = (13.8 ft)i − (22.5 ft) j + (6.6 ft)k 12 = (13.8 ft)i − (16.5 ft) j + (6.6 ft)k DE = 22.5 ft " DF = (13.8 ft)i − (16.5 ft) j − (6.6 ft)k DF = 22.5 ft PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 467 PROBLEM 4.108 (Continued) TBH = TBH Thus: TDE = TDE TDF = TDF (a) " BH 30i − 22.5 j = (600 lb) = (480 lb)i − (360 lb) j 37.5 BH " DE TDE (13.8i − 16.5 j + 6.6k ) = DE 22.5 " DF TDE (13.8i − 16.5 j − 6.6k ) = DF 22.5 ΣM A = 0: (rJ × WC ) + (rK × WG ) + (rH × TBH ) + (rE × TDE ) + (rF × TDF ) = 0 − (12i ) × (−600 j) − (6i ) × (−200 j) + (18i) × (480i − 360 j) i j k i j k TDE TDF 5 0 6.6 + 5 0 + −6.6 = 0 22.5 22.5 13.8 −16.5 6.6 13.8 −16.5 −6.6 7200k + 1200k − 6480k + 4.84(TDE − TDF )i or, + 58.08 82.5 (TDE − TDF ) j − (TDE + TDF )k = 0 22.5 22.5 Equating to zero the coefficients of the unit vectors: i or j: TDE − TDF = 0 k : 7200 + 1200 − 6480 − TDE = TDF * 82.5 (2TDE ) = 0 22.5 TDE = 261.82 lb TDE = TDF = 262 lb (b) 13.8 ! ΣFx = 0: Ax + 480 + 2 " # (261.82) = 0 $ 22.5 % 16.5 ! ΣFy = 0: Ay − 600 − 200 − 360 − 2 " # (261.82) = 0 $ 22.5 % ΣFz = 0: Az = 0 Ax = −801.17 lb Ay = 1544.00 lb A = −(801 lb)i + (1544 lb) j *Remark: The fact that TDE = TDF could have been noted at the outset from the symmetry of structure with respect to xy plane. PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 468 PROBLEM 4.109 A 3-m pole is supported by a ball-and-socket joint at A and by the cables CD and CE. Knowing that the 5-kN force acts vertically downward (φ = 0), determine (a) the tension in cables CD and CE, (b) the reaction at A. SOLUTION Free-Body Diagram: By symmetry with xy plane TCD = TCE = T ! CD = −3i + 1.5 j + 1.2k CD = 3.562 m TCD TCE rB/A = 2i " −3i + 1.5 j + 1.2k CD =T =T CD 3.562 −3i + 1.5 j − 1.2k =T 3.562 rC/A = 3i ΣMA = 0: rC/A × TCD + rC/A × TCE + rB/A × (−5 kN) j = 0 i j k i j k i j k T T + 3 0 + 2 0 0 =0 3 0 0 0 3.562 3.562 −3 1.5 1.2 −3 1.5 −1.2 0 −5 0 Coefficient of k: T ' & 2 (3 × 1.5 × ) − 10 = 0 T = 3.958 kN 3.562 * + ΣF = 0: A + TCD + TCE − 5 j = 0 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 469 PROBLEM 4.109 (Continued) Coefficient of k: Az = 0 Coefficient of i: Ax − 2[3.958 × 3/3.562] = 0 Ax = 6.67 kN Coefficient of j: Ay + 2[3.958 × 1.5/3.562] − 5 = 0 Ay = 1.667 kN (a) TCD = TCE = 3.96 kN A = (6.67 kN)i + (1.667 kN) j (b) PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 470 PROBLEM 4.110 A 3-m pole is supported by a ball-and-socket joint at A and by the cables CD and CE. Knowing that the line of action of the 5-kN force forms an angle φ = 30° with the vertical xy plane, determine (a) the tension in cables CD and CE, (b) the reaction at A. SOLUTION Free-Body Diagram: Five unknowns and six Eqs. of equilibrium but equilibrium is maintained (ΣM AC = 0) rB/A = 2i rC/A = 3i Load at B. = −(5cos 30) j + (5sin 30)k = −4.33j + 2.5k ! CD = −3i + 1.5 j + 1.2k CD = 3.562 m ! CD T ( −3i + 1.5 j + 1.2k ) TCD = TCD = CD 3.562 Similarly, TCE = T (−3i + 1.5 j − 1.2k ) 3.562 ΣMA = 0: rC/A × TCD + rC/A × TCE + rB/A × ( −4.33j + 2.5k ) = 0 i j k i j k i j k TCD TCE + 3 +2 3 0 0 0 0 0 0 =0 3.562 3.562 −3 1.5 1.2 −3 1.5 −1.2 0 −4.33 2.5 Equate coefficients of unit vectors to zero. j: − 3.6 TCD T + 3.6 CE − 5 = 0 3.562 3.562 −3.6TCD + 3.6TCE − 17.810 = 0 (1) PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 471 PROBLEM 4.110 (Continued) k : 4.5 TCD T + 4.5 CE − 8.66 = 0 3.562 3.562 4.5TCD + 4.5TCE = 30.846 (2) + 1.25(1): Eq. (1): (2) 9TCE − 53.11 = 0 TCE = 5.901 kN −3.6TCD + 3.6(5.901) − 17.810 = 0 TCD = 0.954 kN ΣF = 0: A + TCD + TCE − 4.33j + 2.5k = 0 i : Ax + 0.954 5.901 (−3) + (−3) = 0 3.562 3.562 Ax = 5.77 kN j: Ay + 0.954 5.901 (1.5) + (1.5) − 4.33 = 0 3.562 3.562 Ay = 1.443 kN k : Az + 0.954 5.901 (1.2) + ( −1.2) + 2.5 = 0 3.562 3.562 Az = −0.833 kN Answers: (a) TCD = 0.954 kN TCE = 5.90 kN A = (5.77 kN)i + (1.443 kN) j − (0.833 kN)k (b) PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 472 PROBLEM 4.111 A 48-in. boom is held by a ball-and-socket joint at C and by two cables BF and DAE; cable DAE passes around a frictionless pulley at A. For the loading shown, determine the tension in each cable and the reaction at C. SOLUTION Free-Body Diagram: Five unknowns and six Eqs. of equilibrium but equilibrium is maintained (ΣMAC = 0). T = Tension in both parts of cable DAE. rB = 30k rA = 48k ! AD = −20i − 48k AD = 52 in. ! AE = 20 j − 48k AE = 52 in. ! BF = 16i − 30k BF = 34 in. ! AD T T TAD = T = (−20i − 48k ) = ( −5i − 12k ) AD 52 13 ! AE T T TAE = T = (20 j − 48k ) = (5 j − 12k ) AE 52 13 ! T BF TBF TBF = TBF = (16i − 30k ) = BF (8i − 15k ) BF 34 17 ΣM C = 0: rA × TAD + rA × TAE + rB × TBF + rB × (−320 lb) j = 0 i j k i j k i j k T T T + 0 0 48 + 0 0 30 BF + (30k ) × (−320 j) = 0 0 0 48 13 13 17 −5 0 −12 0 5 −12 8 0 −15 Coefficient of i: − 240 T + 9600 = 0 13 T = 520 lb PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 473 PROBLEM 4.111 (Continued) Coefficient of j: − 240 240 T+ TBD = 0 13 17 TBD = 17 17 T = (520) TBD = 680 lb 13 13 ΣF = 0: TAD + TAE + TBF − 320 j + C = 0 Coefficient of i: − 20 8 (520) + (680) + Cx = 0 52 17 −200 + 320 + Cx = 0 Coefficient of j: 20 (520) − 320 + C y = 0 52 200 − 320 + C y = 0 Coefficient of k: Cx = −120 lb − C y = 120 lb 48 48 30 (520) − (520) − (680) + Cz = 0 52 52 34 −480 − 480 − 600 + Cz = 0 Cz = 1560 lb Answers: TDAE = T TDAE = 520 lb ! TBD = 680 lb ! C = −(120.0 lb)i + (120.0 lb) j + (1560 lb)k PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 474 PROBLEM 4.112 Solve Problem 4.111, assuming that the 320-lb load is applied at A. PROBLEM 4.111 A 48-in. boom is held by a ball-andsocket joint at C and by two cables BF and DAE; cable DAE passes around a frictionless pulley at A. For the loading shown, determine the tension in each cable and the reaction at C. SOLUTION Free-Body Diagram: Five unknowns and six Eqs. of equilibrium but equilibrium is maintained (ΣMAC = 0). T = tension in both parts of cable DAE. rB = 30k rA = 48k ! AD = −20i − 48k AD = 52 in. ! AE = 20 j − 48k AE = 52 in. ! BF = 16i − 30k BF = 34 in. ! AD T T TAD = T = (−20i − 48k ) = ( −5i − 12k ) AD 52 13 ! AE T T TAE = T = (20 j − 48k ) = (5 j − 12k ) AE 52 13 ! T BF TBF TBF = TBF = (16i − 30k ) = BF (8i − 15k ) BF 34 17 ΣM C = 0: rA × TAD + rA × TAE + rB × TBF + rA × ( −320 lb) j = 0 i j k i j k i j k T T T + 0 0 48 + 0 0 30 BF + 48k × (−320 j) = 0 0 0 48 13 13 17 −5 0 −12 0 5 −12 8 0 −15 Coefficient of i: − 240 T + 15360 = 0 13 T = 832 lb PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 475 PROBLEM 4.112 (Continued) Coefficient of j: − 240 240 T+ TBD = 0 13 17 TBD = 17 17 T = (832) 13 13 TBD = 1088 lb ΣF = 0: TAD + TAE + TBF − 320 j + C = 0 − Coefficient of i: 20 8 (832) + (1088) + C x = 0 52 17 −320 + 512 + Cx = 0 20 (832) − 320 + C y = 0 52 Coefficient of j: 320 − 320 + C y = 0 Coefficient of k: − Cy = 0 48 48 30 (832) − (852) − (1088) + Cz = 0 52 52 34 −768 − 768 − 960 + Cz = 0 Answers: Cx = −192 lb TDAE = T Cz = 2496 lb TDAE = 832 lb ! TBD = 1088 lb ! C = −(192.0 lb)i + (2496 lb)k PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 476 PROBLEM 4.113 A 20-kg cover for a roof opening is hinged at corners A and B. The roof forms an angle of 30° with the horizontal, and the cover is maintained in a horizontal position by the brace CE. Determine (a) the magnitude of the force exerted by the brace, (b) the reactions at the hinges. Assume that the hinge at A does not exert any axial thrust. SOLUTION Force exerted by CD F = F (sin 75°)i + F (cos 75°) j F = F (0.2588i + 0.9659 j) W = mg = 20 kg(9.81 m/s 2 ) = 196.2 N rA/B = 0.6k rC/B = 0.9i + 0.6k rG/B = 0.45i + 0.3k F = F (0.2588i + 0.9659 j) ΣM B = 0: rG/B × (−196.2 j) + rC/B × F + rA/B × A = 0 i j k i j k i 0.45 0 0.3 + 0.9 0 0.6 F + 0 0 0.2588 +0.9659 0 −196.2 0 Ax j 0 Ay k 0.6 = 0 0 Coefficient of i : +58.86 − 0.5796 F − 0.6 Ay = 0 (1) Coefficient of j: +0.1553F + 0.6 Ax = 0 (2) Coefficient of k: −88.29 + 0.8693F = 0: F = 101.56 N Eq. (2): +58.86 − 0.5796(101.56) − 0.6 Ay = 0 Eq. (3): +0.1553(101.56) + 0.6 Ax = 0 Ay = 0 Ax = −26.29 N F = 101.6 N A = −(26.3 N)i ΣF : A + B + F − Wj = 0 Coefficient of i: 26.29 + Bx + 0.2588(101.56) = 0 Bx = 0 Coefficient of j: By + 0.9659(101.56) − 196.2 = 0 By = 98.1 N Bz = 0 Coefficient of k: B = (98.1 N) j PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 477 PROBLEM 4.114 The bent rod ABEF is supported by bearings at C and D and by wire AH. Knowing that portion AB of the rod is 250 mm long, determine (a) the tension in wire AH, (b) the reactions at C and D. Assume that the bearing at D does not exert any axial thrust. SOLUTION ∆ABH is equilateral Free-Body Diagram: Dimensions in mm rH/C = −50i + 250 j rD/C = 300i rF/C = 350i + 250k T = T (sin 30°) j − T (cos 30°)k = T (0.5 j − 0.866k ) ΣM C = 0: rH/C × T + rD × D + rF/C × (−400 j) = 0 i j k i j 0 T + 300 0 −50 250 0 0.5 −0.866 0 Dy Coefficient i: k i j k 0 + 350 0 250 = 0 0 −400 0 Dz −216.5T + 100 × 103 = 0 T = 461.9 N Coefficient of j: T = 462 N −43.3T − 300 Dz = 0 −43.3(461.9) − 300 Dz = 0 Dz = −66.67 N PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 478 PROBLEM 4.114 (Continued) Coefficient of k: −25T + 300 D y − 140 × 103 = 0 −25(461.9) + 300 Dy − 140 × 103 = 0 D y = 505.1 N D = (505 N) j − (66.7 N)k ΣF = 0: C + D + T − 400 j = 0 Coefficient i: Cx = 0 Cx = 0 Coefficient j: C y + (461.9)0.5 + 505.1 − 400 = 0 C y = −336 N Coefficient k: C z − (461.9)0.866 − 66.67 = 0 C z = 467 N C = −(336 N) j + (467 N)k PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 479 PROBLEM 4.115 A 100-kg uniform rectangular plate is supported in the position shown by hinges A and B and by cable DCE that passes over a frictionless hook at C. Assuming that the tension is the same in both parts of the cable, determine (a) the tension in the cable, (b) the reactions at A and B. Assume that the hinge at B does not exert any axial thrust. SOLUTION rB/A (960 − 180)i = 780i Dimensions in mm 960 ! 450 rG/A = " − 90 # i + k 2 $ 2 % = 390i + 225k rC/A = 600i + 450k T = Tension in cable DCE ! CD = −690i + 675 j − 450k ! CE = 270i + 675 j − 450k CD = 1065 mm CE = 855 mm T (−690i + 675 j − 450k ) 1065 T TCE = (270i + 675 j − 450k ) 855 W = −mgi = −(100 kg)(9.81 m/s 2 ) j = −(981 N) j TCD = ΣM A = 0: rC/A × TCD + rC/A × TCE + rG/A × (−Wj) + rB/A × B = 0 i j k i j k T T 600 0 450 450 + 600 0 + 1065 855 270 675 −450 −690 675 −450 i + 390 0 j 0 −981 k i 225 + 780 0 0 j 0 k 0 =0 By Bz PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 480 PROBLEM 4.115 (Continued) Coefficient of i: −(450)(675) T T − (450)(675) + 220.725 × 103 = 0 1065 855 T = 344.6 N Coefficient of j: (−690 × 450 + 600 × 450) T = 345 N 344.6 344.6 + (270 × 450 + 600 × 450) − 780 Bz = 0 1065 855 Bz = 185.49 N Coefficient of k: (600)(675) 344.6 344.6 + (600)(675) − 382.59 × 103 + 780 By 1065 855 By = 113.2N B = (113.2 N) j + (185.5 N)k ΣF = 0: A + B + TCD + TCE + W = 0 690 270 (344.6) + (344.6) = 0 1065 855 Coefficient of i: Ax − Ax = 114.4 N Coefficient of j: Ay + 113.2 + 675 675 (344.6) + (344.6) − 981 = 0 1065 855 Ay = 377 N Coefficient of k: Az + 185.5 − 450 450 (344.6) − (344.6) = 0 1065 855 Az = 141.5 N A = (114.4 N)i + (377 N) j + (144.5 N)k PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 481 ! PROBLEM 4.116 Solve Problem 4.115, assuming that cable DCE is replaced by a cable attached to Point E and hook C. PROBLEM 4.115 A 100-kg uniform rectangular plate is supported in the position shown by hinges A and B and by cable DCE that passes over a frictionless hook at C. Assuming that the tension is the same in both parts of the cable, determine (a) the tension in the cable, (b) the reactions at A and B. Assume that the hinge at B does not exert any axial thrust. SOLUTION See solution to Problem 4.115 for free-body diagram and analysis leading to the following: CD = 1065 mm CE = 855 mm T (−690i + 675 j − 450k ) 1065 T (270i + 675 j − 450k ) TCE = 855 W = −mgi = −(100 kg)(9.81 m/s 2 ) j = −(981 N)j TCD = Now: ΣM A = 0: rC/A × TCE + rG/A × ( −Wj) + rB/A × B = 0 i j k i j k i j T + 390 600 0 450 0 225 + 780 0 855 270 675 −450 0 −981 0 0 By Coefficient of i: −(450)(675) k 0 =0 Bz T + 220.725 × 103 = 0 855 T = 621.3 N Coefficient of j: Coefficient of k: (270 × 450 + 600 × 450) (600)(675) T = 621 N 621.3 − 980 Bz = 0 Bz = 364.7 N 855 621.3 − 382.59 × 103 + 780 B y = 0 B y = 113.2 N 855 B = (113.2 N)j + (365 N)k PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 482 PROBLEM 4.116 (Continued) ΣF = 0: A + B + TCE + W = 0 270 (621.3) = 0 855 Coefficient of i: Ax + Coefficient of j: Ay + 113.2 + 675 (621.3) − 981 = 0 855 Ay = 377.3 N Coefficient of k: Az + 364.7 − 450 (621.3) = 0 855 Az = −37.7 N ! ! Ax = −196.2 N A = −(196.2 N)i + (377 N)j − (37.7 N)k ! PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 483 ! PROBLEM 4.117 The rectangular plate shown weighs 75 lb and is held in the position shown by hinges at A and B and by cable EF. Assuming that the hinge at B does not exert any axial thrust, determine (a) the tension in the cable, (b) the reactions at A and B. SOLUTION rB/A = (38 − 8)i = 30i rE/A = (30 − 4)i + 20k rG/A = 26i + 20k 38 = i + 10k 2 = 19i + 10k ! EF = 8i + 25 j − 20k EF = 33 in. ! AE T T =T = (8i + 25 j − 20k ) AE 33 ΣM A = 0: rE/A × T + rG/A × (−75 j) + rB/A × B = 0 i j k i j k i j T + 19 0 10 + 30 0 26 0 20 33 8 25 −20 0 −75 0 0 By −(25)(20) Coefficient of i: Coefficient of j: Coefficient of k: (160 + 520) (26)(25) T + 750 = 0: 33 k 0 =0 Bz T = 49.5 lb 49.5 − 30 Bz = 0: Bz = 34 lb 33 49.5 − 1425 + 30 By = 0: By = 15 lb 33 B = (15 lb)j + (34 lb)k ! ! ! ! PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 484 ! PROBLEM 4.117 (Continued) ! ΣF = 0: A + B + T − (75 lb)j = 0 ! Ax + Coefficient of i: Coefficient of j: Coefficient of k: Ay + 15 + 8 (49.5) = 0 33 25 (49.5) − 75 = 0 33 Az + 34 − 20 (49.5) = 0 33 Ax = −12.00 lb Ay = 22.5 lb Az = −4.00 lb ! A = −(12.00 lb)i + (22.5 lb)j − (4.00 lb)k ! PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 485 ! PROBLEM 4.118 Solve Problem 4.117, assuming that cable EF is replaced by a cable attached at points E and H. PROBLEM 4.117 The rectangular plate shown weighs 75 lb and is held in the position shown by hinges at A and B and by cable EF. Assuming that the hinge at B does not exert any axial thrust, determine (a) the tension in the cable, (b) the reactions at A and B. SOLUTION rB/A = (38 − 8)i = 30i rE/A = (30 − 4)i + 20k = 26i + 20k 38 i + 10k 2 = 19i + 10k rG/A = ! EH = −30i + 12 j − 20k EH = 38 in. ! EH T T=T = (−30i + 12 j − 20k ) EH 38 ΣM A = 0: rE/A × T + rG/A × (−75 j) + rB/A × B = 0 i j k i j k i j T 26 0 20 + 19 0 10 + 30 0 38 0 −75 0 0 By −30 12 −20 −(12)(20) Coefficient of i: Coefficient of j: Coefficient of k: (−600 + 520) (26)(12) T + 750 = 0 38 T = 118.75 k 0 =0 Bz T = 118.8lb 118.75 − 30 Bz = 0 Bz = −8.33lb 38 118.75 − 1425 + 30 By = 0 By = 15.00 lb 38 B = (15.00lb)j − (8.33 lb)k PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 486 PROBLEM 4.118 (Continued) ΣF = 0: Coefficient of j: Coefficient of k: 30 (118.75) = 0 38 Ax = 93.75 lb 12 (118.75) − 75 = 0 38 Ay = 22.5 lb Ax − Coefficient of i: Ay + 15 + A + B + T − (75 lb)j = 0 Az − 8.33 − 20 (118.75) = 0 38 Az = 70.83 lb A = (93.8 lb)i + (22.5 lb)j + (70.8 lb)k ! PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 487 ! PROBLEM 4.119 Solve Problem 4.114, assuming that the bearing at D is removed and that the bearing at C can exert couples about axes parallel to the y and z axes. PROBLEM 4.114 The bent rod ABEF is supported by bearings at C and D and by wire AH. Knowing that portion AB of the rod is 250 mm long, determine (a) the tension in wire AH, (b) the reactions at C and D. Assume that the bearing at D does not exert any axial thrust. SOLUTION Free-Body Diagram: ∆ABH is Equilateral Dimensions in mm rH/C = −50i + 250 j rF/C = 350i + 250k T = T (sin 30°) j − T (cos 30°)k = T (0.5 j − 0.866k ) ΣM C = 0: rF/C × (−400 j) + rH/C × T + ( M C ) y j + ( M C ) z k = 0 i j k i j k 350 0 250 + −50 250 0 T + (M C ) y j + (M C ) z k = 0 0 −400 0 0 0.5 −0.866 Coefficient of i: Coefficient of j: +100 × 103 − 216.5T = 0 T = 461.9 N T = 462 N −43.3(461.9) + ( M C ) y = 0 ( M C ) y = 20 × 103 N ⋅ mm (M C ) y = 20.0 N ⋅ m PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 488 PROBLEM 4.119 (Continued) Coefficient of k: −140 × 103 − 25(461.9) + ( M C ) z = 0 ( M C ) z = 151.54 × 103 N ⋅ mm (M C ) z = 151.5 N ⋅ m ΣF = 0: C + T − 400 j = 0 M C = (20.0 N ⋅ m)j + (151.5 N ⋅ m)k Coefficient of i: Coefficient of j: Coefficient of k: Cx = 0 C y + 0.5(461.9) − 400 = 0 C y = 169.1 N C z − 0.866(461.9) = 0 C z = 400 N C = (169.1 N)j + (400 N)k ! PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 489 ! PROBLEM 4.120 Solve Problem 4.117, assuming that the hinge at B is removed and that the hinge at A can exert couples about axes parallel to the y and z axes. PROBLEM 4.117 The rectangular plate shown weighs 75 lb and is held in the position shown by hinges at A and B and by cable EF. Assuming that the hinge at B does not exert any axial thrust, determine (a) the tension in the cable, (b) the reactions at A and B. SOLUTION rE/A = (30 − 4)i + 20k = 26i + 20k rG/A = (0.5 × 38)i + 10k = 19i + 10k ! AE = 8i + 25 j − 20k AE = 33 in. ! AE T = (8i + 25 j − 20k ) T =T AE 33 ΣM A = 0: rE/A × T + rG/A × (−75 j) + ( M A ) y j + ( M A ) z k = 0 i j k i j k T 26 0 20 + 19 0 10 + ( M A ) y j + ( M A ) z k = 0 33 8 25 −20 0 −75 0 −(20)(25) Coefficient of i: Coefficient of j: Coefficient of k: (160 + 520) (26)(25) T + 750 = 0 33 T = 49.5 lb 49.5 + ( M A ) y = 0 ( M A ) y = −1020 lb ⋅ in. 33 49.5 − 1425 + ( M A ) z = 0 33 ( M A ) z = 450 lb ⋅ in. ΣF = 0: A + T − 75 j = 0 Ax + Coefficient of i: 8 (49.5) = 0 33 M A = −(1020 lb ⋅ in)j + (450 lb ⋅ in.)k Ax = 12.00 lb Coefficient of j: Ay + 25 (49.5) − 75 = 0 33 Ay = 37.5 lb Coefficient of k: Az − 20 (49.5) 33 Az = 30.0 lb A = −(12.00 lb)i + (37.5 lb)j + (30.0 lb)k PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 490 PROBLEM 4.121 The assembly shown is used to control the tension T in a tape that passes around a frictionless spool at E. Collar C is welded to rods ABC and CDE. It can rotate about shaft FG but its motion along the shaft is prevented by a washer S. For the loading shown, determine (a) the tension T in the tape, (b) the reaction at C. SOLUTION Free-Body Diagram: rA/C = 4.2 j + 2k rE/C = 1.6i − 2.4 j ΣM C = 0: rA/C × (−6 j) + rE/C × T (i + k ) + ( M C ) y j + ( M C ) z k = 0 (4.2 j + 2k ) × (−6 j) + (1.6i − 2.4 j) × T (i + k ) + ( M C ) y j + ( M C ) z k = 0 Coefficient of i: Coefficient of j: Coefficient of k: 12 − 2.4T = 0 T = 5 lb −1.6(5 lb) + (M C ) y = 0 ( M C ) y = 8 lb ⋅ in. 2.4(5 lb) + ( M C ) z = 0 ( M C ) z = −12 lb ⋅ in. M C = (8 lb ⋅ in.)j − (12 lb ⋅ in.)k ΣF = 0: C x i + C y j + C z k − (6 lb)j + (5 lb)i + (5 lb)k = 0 Equate coefficients of unit vectors to zero. C x = −5 lb C y = 6 lb C z = −5 lb C = −(5.00 lb)i + (6.00 lb)j − (5.00 lb)k ! PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 491 ! PROBLEM 4.122 The assembly shown is welded to collar A that fits on the vertical pin shown. The pin can exert couples about the x and z axes but does not prevent motion about or along the y axis. For the loading shown, determine the tension in each cable and the reaction at A. SOLUTION Free-Body Diagram: First note TCF = CF TCF −(0.08 m)i + (0.06 m) j = (0.08) 2 + (0.06)2 m TCF = TCF (−0.8i + 0.6 j) TDE = DE TDE = (0.12 m)j − (0.09 m)k (0.12) 2 + (0.09)2 m TDE = TDE (0.8 j − 0.6k ) (a) From f.b.d. of assembly ΣFy = 0: 0.6TCF + 0.8TDE − 480 N = 0 0.6TCF + 0.8TDE = 480 N or (1) ΣM y = 0: − (0.8TCF )(0.135 m) + (0.6TDE )(0.08 m) = 0 TDE = 2.25TCF or (2) Substituting Equation (2) into Equation (1) 0.6TCF + 0.8[(2.25)TCF ] = 480 N TCF = 200.00 N TCF = 200 N or ! and from Equation (2) TDE = 2.25(200.00 N) = 450.00 TDE = 450 N or PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 492 ! PROBLEM 4.122 (Continued) (b) From f.b.d. of assembly ΣFz = 0: Az − (0.6)(450.00 N) = 0 Az = 270.00 N ΣFx = 0: Ax − (0.8)(200.00 N) = 0 Ax = 160.000 N or A = (160.0 N)i + (270 N)k ΣM x = 0: MAx + (480 N)(0.135 m) − [(200.00 N)(0.6)](0.135 m) − [(450 N)(0.8)](0.09 m) = 0 M Ax = −16.2000 N ⋅ m ΣM z = 0: MAz − (480 N)(0.08 m) + [(200.00 N)(0.6)](0.08 m) + [(450 N)(0.8)](0.08 m) = 0 M Az = 0 or M A = −(16.20 N ⋅ m)i PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 493 PROBLEM 4.123 The rigid L-shaped member ABC is supported by a ball-and-socket joint at A and by three cables. If a 450-lb load is applied at F, determine the tension in each cable. SOLUTION Free-Body Diagram: In this problem: We have Thus a = 21 in. ! CD = (24 in.)j − (32 in.)k CD = 40 in. ! BD = −(42 in.)i + (24 in.)j − (32 in.)k BD = 58 in. ! BE = (42 in.)i − (32 in.)k BE = 52.802 in. TCD TBD TBE ! CD = TCD = TCD (0.6 j − 0.8k ) CD ! BD = TBD = TBD (−0.72414i + 0.41379 j − 0.55172k ) BD ! BE = TBE = TBE (0.79542i − 0.60604k ) BE ΣM A = 0: (rC × TCD ) + (rB × TBD ) + (rB × TBE ) + (rW × W) = 0 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 494 PROBLEM 4.123 (Continued) rC = −(42 in.)i + (32 in.)k Noting that rB = (32 in.)k rW = − ai + (32 in.)k and using determinants, we write i j k i j k −42 0 32 TCD + 0 0 32 TBD 0 0.6 −0.8 −0.72414 0.41379 −0.55172 i j k i j k + 0 0 32 TBE + −a 0 32 = 0 0.79542 0 −0.60604 0 −450 0 Equating to zero the coefficients of the unit vectors: i: −19.2TCD − 13.241TBD + 14400 = 0 (1) j: − 33.6TCD − 23.172TBD + 25.453TBE = 0 (2) k: −25.2TCD + 450a = 0 (3) Recalling that a = 21 in., Eq. (3) yields TCD = From (1): 450(21) = 375 lb 25.2 −19.2(375) − 13.241TBD + 14400 = 0 TBD = 543.77 lb From (2): TCD = 375 lb TBD = 544 lb −33.6(375) − 23.172(543.77) + 25.453TBE = 0 TBE = 990.07 lb ! ! TBE = 990 lb PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 495 ! PROBLEM 4.124 Solve Problem 4.123, assuming that the 450-lb load is applied at C. PROBLEM 4.123 The rigid L-shaped member ABC is supported by a ball-and-socket joint at A and by three cables. If a 450-lb load is applied at F, determine the tension in each cable. SOLUTION See solution of Problem 4.123 for free-body diagram and derivation of Eqs. (1), (2), and (3): −19.2TCD − 13.241TBD + 14400 = 0 (1) −33.6TCD − 23.172TBD + 25.453TBE = 0 (2) −25.2TCD + 450a = 0 (3) In this problem, the 450-lb load is applied at C and we have a = 42 in. Carrying into (3) and solving for TCD , TCD = 750 lb From (1): From (2): −19.2(750) − 13.241TBD + 14400 = 0 −33.6(750) − 0 + 25.453TBE = 0 ! TCD = 750 lb TBD = 0 TBE = 990 lb PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 496 ! PROBLEM 4.125 Frame ABCD is supported by a ball-and-socket joint at A and by three cables. For a = 150 mm, determine the tension in each cable and the reaction at A. SOLUTION First note TDG = λ DG TDG = −(0.48 m)i + (0.14 m)j (0.48) 2 + (0.14) 2 m TDG −0.48i + 0.14 j TDG 0.50 T = DG (24i + 7 j) 25 = TBE = λ BE TBE = −(0.48 m)i + (0.2 m)k (0.48) 2 + (0.2) 2 m TBE −0.48i + 0.2k TBE 0.52 T = BE (−12 j + 5k ) 13 = From f.b.d. of frame ABCD 7 ! ΣM x = 0: " TDG # (0.3 m) − (350 N)(0.15 m) = 0 25 $ % TDG = 625 N or 24 5 ! ! ΣM y = 0: " × 625 N # (0.3 m) − " TBE # (0.48 m) = 0 $ 25 % $ 13 % TBE = 975 N or 7 ! ΣM z = 0: TCF (0.14 m) + " × 625 N # (0.48 m) − (350 N)(0.48 m) = 0 $ 25 % TCF = 600 N or PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 497 PROBLEM 4.125 (Continued) ΣFx = 0: Ax + TCF + (TBE ) x + (TDG ) x = 0 12 24 ! ! Ax − 600 N − " × 975 N # − " × 625 N # = 0 $ 13 % $ 25 % Ax = 2100 N ΣFy = 0: Ay + (TDG ) y − 350 N = 0 7 ! Ay + " × 625 N # − 350 N = 0 $ 25 % Ay = 175.0 N ΣFz = 0: Az + (TBE ) z = 0 5 ! Az + " × 975 N # = 0 $ 13 % Az = −375 N A = (2100 N)i + (175.0 N) j − (375 N)k Therefore PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 498 PROBLEM 4.126 Frame ABCD is supported by a ball-and-socket joint at A and by three cables. Knowing that the 350-N load is applied at D (a = 300 mm), determine the tension in each cable and the reaction at A. SOLUTION First note TDG = λ DG TDG = −(0.48 m)i + (0.14 m)j (0.48) 2 + (0.14)2 m TDG −0.48i + 0.14 j TDG 0.50 T = DG (24i + 7 j) 25 = TBE = λ BE TBE = −(0.48 m)i + (0.2 m)k (0.48) 2 + (0.2) 2 m TBE −0.48i + 0.2k TBE 0.52 T = BE (−12i + 5k ) 13 = From f.b.d. of frame ABCD 7 ! ΣM x = 0: " TDG # (0.3 m) − (350 N)(0.3 m) = 0 $ 25 % TDG = 1250 N or 24 5 ! ! ΣM y = 0: " × 1250 N # (0.3 m) − " TBE # (0.48 m) = 0 25 13 $ % $ % TBE = 1950 N or 7 ! ΣM z = 0: TCF (0.14 m) + " × 1250 N # (0.48 m) − (350 N)(0.48 m) = 0 $ 25 % TCF = 0 or ! PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 499 ! PROBLEM 4.126 (Continued) ΣFx = 0: Ax + TCF + (TBE ) x + (TDG ) x = 0 12 24 ! ! Ax + 0 − " × 1950 N # − " × 1250 N # = 0 ! $ 13 % $ 25 % Ax = 3000 N ! ! ΣFy = 0: Ay + (TDG ) y − 350 N = 0 7 ! Ay + " × 1250 N # − 350 N = 0 $ 25 % Ay = 0 ΣFz = 0: Az + (TBE ) z = 0 5 ! Az + " × 1950 N # = 0 $ 13 % Az = −750 N A = (3000 N)i − (750 N)k Therefore PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 500 PROBLEM 4.127 Three rods are welded together to form a “corner” that is supported by three eyebolts. Neglecting friction, determine the reactions at A, B, and C when P = 240 lb, a = 12 in., b = 8 in., and c = 10 in. SOLUTION From f.b.d. of weldment ΣM O = 0: rA/O × A + rB/O × B + rC/O × C = 0 i 12 0 j 0 Ay k i 0 + 0 Az Bx j k i 8 0 + 0 0 Bz Cx j k 0 10 = 0 Cy 0 (−12 Az j + 12 Ay k ) + (8 Bz i − 8 Bx k ) + (−10 C y i + 10 C x j) = 0 From i-coefficient Bz = 1.25C y or j-coefficient k-coefficient or! (3) ΣF = 0: A + B + C − P = 0 ! ( Bx + C x )i + ( Ay + C y − 240 lb) j + ( Az + Bz )k = 0 ! From i-coefficient Bx + C x = 0 C x = − Bx or j-coefficient or (2) 12 Ay − 8 Bx = 0 Bx = 1.5 Ay or (1) −12 Az + 10 C x = 0 C x = 1.2 Az or ! 8 Bz − 10 C y = 0 (4) Ay + C y − 240 lb = 0 Ay + C y = 240 lb (5) PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 501 PROBLEM 4.127 (Continued) k-coefficient Az + Bz = 0 Az = − Bz or (6) Substituting C x from Equation (4) into Equation (2) − Bz = 1.2 Az (7) Using Equations (1), (6), and (7) Cy = Bz − Az 1 Bx ! Bx = = = 1.25 1.25 1.25 "$ 1.2 #% 1.5 (8) From Equations (3) and (8) Cy = 1.5 Ay 1.5 or C y = Ay and substituting into Equation (5) 2 Ay = 240 lb Ay = C y = 120 lb (9) Using Equation (1) and Equation (9) Bz = 1.25(120 lb) = 150.0 lb Using Equation (3) and Equation (9) Bx = 1.5(120 lb) = 180.0 lb From Equation (4) C x = −180.0 lb From Equation (6) Az = −150.0 lb Therefore A = (120.0 lb) j − (150.0 lb)k ! ! B = (180.0 lb)i + (150.0 lb)k ! ! C = −(180.0 lb)i + (120.0 lb) j ! PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 502 PROBLEM 4.128 Solve Problem 4.127, assuming that the force P is removed and is replaced by a couple M = +(600 lb ⋅ in.)j acting at B. PROBLEM 4.127 Three rods are welded together to form a “corner” that is supported by three eyebolts. Neglecting friction, determine the reactions at A, B, and C when P = 240 lb, a = 12 in., b = 8 in., and c = 10 in. SOLUTION From f.b.d. of weldment ΣM O = 0: rA/O × A + rB/O × B + rC/O × C + M = 0 i 12 0 j 0 Ay k i 0 + 0 Az Bx j k i 8 0 + 0 0 Bz Cx j k 0 10 + (600 lb ⋅ in.) j = 0 Cy 0 (−12 Az j + 12 Ay k ) + (8 Bz j − 8 Bx k ) + ( −10C y i + 10C x j) + (600 lb ⋅ in.) j = 0 From i-coefficient 8 Bz − 10 C y = 0 C y = 0.8Bz or j-coefficient (1) −12 Az + 10 C x + 600 = 0 C x = 1.2 Az − 60 or k-coefficient (2) 12 Ay − 8 Bx = 0 Bx = 1.5 Ay or (3) ! ΣF = 0: A + B + C = 0 ! ! ( Bx + C x )i + ( Ay + C y ) j + ( Az + Bz )k = 0 ! From i-coefficient C x = − Bx (4) j-coefficient C y = − Ay (5) k-coefficient Az = − Bz (6) PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 503 PROBLEM 4.128 (Continued) Substituting C x from Equation (4) into Equation (2) B ! Az = 50 − " x # $ 1.2 % (7) 2! C y = 0.8 Bz = − 0.8 Az = " # Bx − 40 $3% (8) Using Equations (1), (6), and (7) From Equations (3) and (8) C y = Ay − 40 Substituting into Equation (5) 2 Ay = 40 Ay = 20.0 lb From Equation (5) C y = −20.0 lb Equation (1) Bz = −25.0 lb Equation (3) Bx = 30.0 lb Equation (4) C x = −30.0 lb Equation (6) Az = 25.0 lb A = (20.0 lb) j + (25.0 lb)k ! ! ! B = (30.0 lb)i − (25.0 lb)k ! ! C = − (30.0 lb)i − (20.0 lb) j ! Therefore PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 504 PROBLEM 4.129 In order to clean the clogged drainpipe AE, a plumber has disconnected both ends of the pipe and inserted a power snake through the opening at A. The cutting head of the snake is connected by a heavy cable to an electric motor that rotates at a constant speed as the plumber forces the cable into the pipe. The forces exerted by the plumber and the motor on the end of the cable can be represented by the wrench F = −(48 N)k , M = −(90 N ⋅ m)k. Determine the additional reactions at B, C, and D caused by the cleaning operation. Assume that the reaction at each support consists of two force components perpendicular to the pipe. SOLUTION From f.b.d. of pipe assembly ABCD ΣFx = 0: Bx = 0 ΣM D ( x -axis) = 0: (48 N)(2.5 m) − Bz (2 m) = 0 Bz = 60.0 N and B = (60.0 N)k ΣM D ( z -axis) = 0: C y (3 m) − 90 N ⋅ m = 0 C y = 30.0 N ΣM D ( y -axis) = 0: −C z (3 m) − (60.0 N)(4 m) + (48 N)(4 m) = 0 C z = −16.00 N and C = (30.0 N) j − (16.00 N)k ΣFy = 0: Dy + 30.0 = 0 Dy = −30.0 N ΣFz = 0: Dz − 16.00 N + 60.0 N − 48 N = 0 Dz = 4.00 N and D = − (30.0 N) j + (4.00 N)k PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 505 ! PROBLEM 4.130 Solve Problem 4.129, assuming that the plumber exerts a force F = −(48 N)k and that the motor is turned off (M = 0). PROBLEM 4.129 In order to clean the clogged drainpipe AE, a plumber has disconnected both ends of the pipe and inserted a power snake through the opening at A. The cutting head of the snake is connected by a heavy cable to an electric motor that rotates at a constant speed as the plumber forces the cable into the pipe. The forces exerted by the plumber and the motor on the end of the cable can be represented by the wrench F = −(48 N)k , M = −(90 N ⋅ m)k. Determine the additional reactions at B, C, and D caused by the cleaning operation. Assume that the reaction at each support consists of two force components perpendicular to the pipe. SOLUTION From f.b.d. of pipe assembly ABCD ΣFx = 0: Bx = 0 ΣM D ( x -axis) = 0: (48 N)(2.5 m) − Bz (2 m) = 0 Bz = 60.0 N and B = (60.0 N)k ΣM D ( z -axis) = 0: C y (3 m) − Bx (2 m) = 0 Cy = 0 ΣM D ( y -axis) = 0: C z (3 m) − (60.0 N)(4 m) + (48 N)(4 m) = 0 C z = −16.00 N and C = − (16.00 N)k ΣFy = 0: Dy + C y = 0 Dy = 0 ΣFz = 0: Dz + Bz + C z − F = 0 Dz + 60.0 N − 16.00 N − 48 N = 0 Dz = 4.00 N and D = (4.00 N)k PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 506 ! PROBLEM 4.131 The assembly shown consists of an 80-mm rod AF that is welded to a cross consisting of four 200-mm arms. The assembly is supported by a ball-and-socket joint at F and by three short links, each of which forms an angle of 45° with the vertical. For the loading shown, determine (a) the tension in each link, (b) the reaction at F. SOLUTION rE/F = −200 i + 80 j TB = TB (i − j) / 2 rB/F = 80 j − 200k TC = TC (− j + k ) / 2 rC/F = 200i + 80 j TD = TD (− i + j) / 2 rD/E = 80 j + 200k ΣM F = 0: rB/F × TB + rC/F × TC + rD/F × TD + rE/F × (− Pj) = 0 i j k i j k i j k i j k Tc TB TD 0 80 −200 + 200 80 0 + 0 80 200 + −200 80 0 = 0 2 2 2 1 −1 0 0 −1 1 −1 −1 0 0 −P 0 Equate coefficients of unit vectors to zero and multiply each equation by 2. i : − 200 TB + 80 TC + 200 TD = 0 (1) j: − 200 TB − 200 TC − 200 TD = 0 (2) k : − 80 TB − 200 TC + 80 TD + 200 2 P = 0 80 (2): 200 − 80 TB − 80 TC − 80 TD = 0 Eqs. (3) + (4): −160TB − 280TC + 200 2 P = 0 Eqs. (1) + (2): −400TB − 120TC = 0 TB = − (3) (4) (5) 120 TC − 0.3TC 400 (6) PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 507 PROBLEM 4.131 (Continued) Eqs. (6) −160( −0.3TC ) − 280TC + 200 2 P = 0 (5): −232TC + 200 2 P = 0 TC = 1.2191P TC = 1.219 P From Eq. (6): TB = −0.3(1.2191P) = − 0.36574 = P From Eq. (2): − 200(− 0.3657 P) − 200(1.2191P) − 200Tθ D = 0 TB = −0.366 P TD = − 0.8534 P ΣF = 0: TD = − 0.853P F + TB + TC + TD − Pj = 0 i : Fx + (− 0.36574 P) 2 − Fx = − 0.3448P j: Fy − k : Fz + =0 2 Fx = − 0.345P (− 0.36574 P) 2 Fy = P ( − 0.8534 P) − (1.2191P) 2 − (− 0.8534 P) 2 − 200 = 0 Fy = P (1.2191P) 2 =0 Fz = − 0.8620 P Fz = − 0.862 P F = − 0.345 Pi + Pj − 0.862 Pk PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 508 ! PROBLEM 4.132 The uniform 10-kg rod AB is supported by a ball-and-socket joint at A and by the cord CG that is attached to the midpoint G of the rod. Knowing that the rod leans against a frictionless vertical wall at B, determine (a) the tension in the cord, (b) the reactions at A and B. SOLUTION Free-Body Diagram: Five unknowns and six Eqs. of equilibrium. But equilibrium is maintained (ΣMAB = 0) W = mg = (10 kg) 9.81m/s 2 W = 98.1 N ! GC = − 300i + 200 j − 225k GC = 425 mm ! GC T = T =T (− 300i + 200 j − 225k ) GC 425 rB/ A = − 600i + 400 j + 150 mm rG/ A = − 300i + 200 j + 75 mm ΣMA = 0: rB/ A × B + rG/ A × T + rG/ A × (− Wj) = 0 i j k i j k i j k T 75 − 600 400 150 + − 300 200 + − 300 200 75 425 B 0 0 − 300 200 − 225 0 − 98.1 0 Coefficient of i : (−105.88 − 35.29)T + 7357.5 = 0 T = 52.12 N T = 52.1 N ! ! PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 509 ! PROBLEM 4.132 (Continued) Coefficient of j : 150 B − (300 × 75 + 300 × 225) 52.12 =0 425 B = 73.58 N B = (73.6 N)i ΣF = 0: A + B + T − Wj = 0 Coefficient of i : Ax + 73.58 − 52.15 300 =0 425 Ax = 36.8 N Coefficient of j: Ay + 52.15 200 − 98.1 = 0 425 Ay = 73.6 N Coefficient of k : Az − 52.15 225 =0 425 Az = 27.6 N PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 510 ! PROBLEM 4.133 The bent rod ABDE is supported by ball-and-socket joints at A and E and by the cable DF. If a 60-lb load is applied at C as shown, determine the tension in the cable. SOLUTION Free-Body Diagram: ! DF = −16i + 11j − 8k DF = 21 in. ! DE T = (−16i + 11j − 8k ) T=T DF 21 rD/E = 16 i rC/E = 16i − 14k ! EA 7i − 24k = EA = 25 EA ΣM EA = 0: EA ⋅ (rB/E × T) + EA ⋅ (rC/E ⋅ (− 60 j)) = 0 7 0 −24 7 0 −24 T 1 16 0 0 + 16 0 −14 =0 21 × 25 25 0 −60 0 −16 11 −8 − −7 × 14 × 60 + 24 × 16 × 60 24 × 16 × 11 T+ =0 21 × 25 25 201.14 T + 17,160 = 0 T = 85.314 lb T = 85.3 lb PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 511 ! PROBLEM 4.134 Solve Problem 4.133, assuming that cable DF is replaced by a cable connecting B and F. SOLUTION Free-Body Diagram: rB/ A = 9i rC/ A = 9i + 10k ! BF = −16i + 11j + 16k BF = 25.16 in. ! BF T = (−16i + 11j + 16k ) T =T BF 25.16 ! AE 7i − 24k = AE = AE 25 ΣMAE = 0: AF ⋅ (rB/ A × T) + AE ⋅ (rC/ A ⋅ (− 60 j)) = 0 7 0 −24 7 0 −24 1 T +9 0 =0 9 0 0 10 25 × 25.16 25 −16 11 16 0 −60 0 − 24 × 9 × 11 24 × 9 × 60 + 7 × 10 × 60 T+ 25 × 25.16 25 94.436 T − 17,160 = 0 T = 181.7 lb PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 512 ! PROBLEM 4.135 The 50-kg plate ABCD is supported by hinges along edge AB and by wire CE. Knowing that the plate is uniform, determine the tension in the wire. SOLUTION Free-Body Diagram: W = mg = (50 kg)(9.81 m/s 2 ) W = 490.50 N ! CE = − 240i + 600 j − 400k CE = 760 mm ! CE T (− 240i + 600 j − 400k ) T =T = CE 760 ! AB 480i − 200 j 1 = = = (12i − 5 j) AB 520 13 AB ΣMAB = 0: AB ⋅ (rE/ A × T ) + AB ⋅ (rG/ A × − Wj) = 0 rE/ A = 240i + 400 j; rG/ A = 240i − 100 j + 200k −5 −5 12 0 12 0 T 1 + 240 −100 200 =0 240 400 0 13 × 20 13 − 240 600 − 400 −W 0 0 (−12 × 400 × 400 − 5 × 240 × 400) T + 12 × 200W = 0 760 T = 0.76W = 0.76(490.50 N) T = 373 N PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 513 ! PROBLEM 4.136 Solve Problem 4.135, assuming that wire CE is replaced by a wire connecting E and D. PROBLEM 4.135 The 50-kg plate ABCD is supported by hinges along edge AB and by wire CE. Knowing that the plate is uniform, determine the tension in the wire. SOLUTION Free-Body Diagram: Dimensions in mm W = mg = (50 kg)(9.81 m/s 2 ) W = 490.50 N ! DE = − 240i + 400 j − 400k DE = 614.5 mm ! DE T (240i + 400 j − 400k ) T =T = DE 614.5 ! AB 480i − 200 j 1 = = (12i − 5 j) AB = 520 13 AB rE/ A = 240i + 400 j; rG/ A = 240i − 100 j + 200k 12 −5 0 12 5 0 T 1 + 240 −100 200 =0 240 400 0 13 × 614.5 13 −W 240 400 − 400 0 0 (−12 × 400 × 400 − 5 × 240 × 400) T + 12 × 200 × W = 0 614.5 T = 0.6145W = 0.6145(490.50 N) T = 301 N PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 514 ! PROBLEM 4.137 Two rectangular plates are welded together to form the assembly shown. The assembly is supported by ball-and-socket joints at B and D and by a ball on a horizontal surface at C. For the loading shown, determine the reaction at C. SOLUTION λ BD = First note −(6 in.)i − (9 in.) j + (12 in.)k (6) 2 + (9) 2 + (12)2 in. 1 (− 6i − 9 j + 12k ) 16.1555 = − (6 in.)i = rA/B P = (80 lb)k rC/D = (8 in.)i C = (C ) j From the f.b.d. of the plates ΣM BD = 0: λ BD ⋅ (rA/B × P ) + λ BD ⋅ ( rC/D × C ) = 0 −6 −9 12 −6 −9 12 & 6(80) ' & C (8) ' −1 0 0 ( + 1 0 0 ( =0 ) *16.1555 + * 16.1555 )+ 0 0 1 0 1 0 ( −9)(6)(80) + (12)(8)C = 0 C = 45.0 lb or C = (45.0 lb) j PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 515 ! PROBLEM 4.138 Two 2 × 4-ft plywood panels, each of weight 12 lb, are nailed together as shown. The panels are supported by ball-and-socket joints at A and F and by the wire BH. Determine (a) the location of H in the xy plane if the tension in the wire is to be minimum, (b) the corresponding minimum tension. SOLUTION Free-Body Diagram: ! AF = 4i − 2 j − 4k AF rG1/A AF = 6 ft 1 = (2i − j − 2k ) 3 = 2i − j rG2 /A = 4i − j − 2k rB/A = 4i ΣM AF = 0: AF ⋅ (rG/A × ( −12 j) + AF ⋅ (rG 2/A × (−12 j)) + 2 −1 −2 2 −1 −2 1 1 + 4 −1 −2 + 2 −1 0 3 3 0 −12 0 0 −12 0 1 1 (2 × 2 × 12) + (−2 × 2 × 12 + 2 × 4 × 12) + 3 3 AF AF ⋅ (rB/A × T ) = 0 AF AF ⋅ (rB/A × T) = −32 or T ⋅ ( ⋅ (rB/A × T) = 0 ⋅ (rB/A × T) = 0 A/F × rB/A ) = −32 (1) PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 516 PROBLEM 4.138 (Continued) Projection of T on ( AF × rB/A ) is constant. Thus, Tmin is parallel to AF Corresponding unit vector is 1 5 1 1 × rB/A = (2i − j − 2k ) × 4i = (−8 j + 4k ) 3 3 (−2 j + k ) Tmin = T ( −2 j + k ) Eq. (1): 1 (2) 5 T &1 ' (−2 j + k ) ⋅ ( (2i − j − 2k ) × 4i ) = −32 3 5 * + T 1 (−2 j + k ) ⋅ ( −8 j + 4k ) = −32 3 5 T (16 + 4) = −32 3 5 T =− 3 5(32) = 4.8 5 20 T = 10.7331 lb Eq. (2) Tmin = T ( −2 j + k ) 1 5 = 4.8 5( −2 j + k ) 1 5 Tmin = −(9.6 lb)j + (4.8 lb k ) Since Tmin has no i component, wire BH is parallel to the yz plane, and x = 4 ft. (a) (b) x = 4.00 ft; y = 8.00 ft Tmin = 10.73 lb ! PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 517 ! PROBLEM 4.139 Solve Problem 4.138, subject to the restriction that H must lie on the y axis. PROBLEM 4.138 Two 2 × 4-ft plywood panels, each of weight 12 lb, are nailed together as shown. The panels are supported by ball-andsocket joints at A and F and by the wire BH. Determine (a) the location of H in the xy plane if the tension in the wire is to be minimum, (b) the corresponding minimum tension. SOLUTION Free-Body Diagram: ! AF = 4i − 2 j − 4k AF rG1/A 1 = (2i − j − 2k ) 3 = 2i − j rG2 /A = 4i − j − 2k rB/A = 4i ΣMAF = 0: AF ⋅ (rG/A × (−12 j) + AF ⋅ (rG2 /A × (−12 j)) + 2 −1 2 2 −1 −2 1 1 2 −1 0 + 4 −1 −2 + 3 3 0 −12 0 0 −12 0 1 1 (2 × 2 × 12) + (−2 × 2 × 12 + 2 × 4 × 12) + 3 3 AF AF AF AF ⋅ (rB/A × T ) = 0 ⋅ (rB/A × T) = 0 ⋅ (rB/A × T) = 0 ⋅ (rB/A × T) = −32 ! BH = −4i + yj − 4k BH = (32 + y 2 )1/2 ! BH −4i + yj − 4k =T T=T BH (32 + y 2 )1/ 2 (1) PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 518 PROBLEM 4.139 (Continued) Eq. (1): AF 2 −1 −2 T ⋅ (rB/A × T ) = 4 0 0 = −32 3(32 + y 2 )1/2 −4 y −4 (−16 − 8 y )T = −3 × 32(32 + y 2 )1/2 T = 96 (32 + y 2 )1/2 8 y + 16 (2) (8y +16) 12 (32 + y 2 ) −1/ 2 (2 y ) + (32 + y 2 )1/2 (8) dT = 0: 96 dy (8 y + 16) 2 Numerator = 0: (8 y + 16) y = (32 + y 2 )8 8 y 2 + 16 y = 32 × 8 + 8 y 2 Eq. (2): T = 96 (32 + 162 )1/ 2 = 11.3137 lb 8 × 16 + 16 y = 16.00 ft Tmin = 11.31 lb PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 519 ! PROBLEM 4.140 The pipe ACDE is supported by ball-and-socket joints at A and E and by the wire DF. Determine the tension in the wire when a 640-N load is applied at B as shown. SOLUTION Free-Body Diagram: Dimensions in mm ! AE = 480i + 160 j − 240k AE = 560 mm ! AE 480i + 160 j − 240k = = AE 560 AE 6i + 2 j − 3k AE = 7 rB/A = 200i rD/A = 480i + 160 j ! DF = −480i + 330 j − 240k ; DF = 630 mm ! DF −480i + 330 j − 240k −16i + 11j − 8k TDF = TDF = TDF = TDF 630 21 DF ΣM AE = AE ⋅ (rD/A × TDF ) + AE ⋅ (rB/A × (−600 j)) = 0 6 2 −3 6 2 −3 TDE 1 480 160 0 + 200 0 0 =0 21 × 7 7 0 −640 0 −16 11 −8 3 × 200 × 640 −6 × 160 × 8 + 2 × 480 × 8 − 3 × 480 × 11 − 3 × 160 × 16 TDF + =0 21 × 7 7 −1120TDF + 384 × 103 = 0 TDF = 342.86 N TDF = 343 N PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 520 ! PROBLEM 4.141 Solve Problem 4.140, assuming that wire DF is replaced by a wire connecting C and F. PROBLEM 4.140 The pipe ACDE is supported by ball-and-socket joints at A and E and by the wire DF. Determine the tension in the wire when a 640-N load is applied at B as shown. SOLUTION Free-Body Diagram: Dimensions in mm ! AE = 480i + 160 j − 240k AE = 560 mm ! AE 480i + 160 j − 240k = = AE 560 AE 6i + 2 j − 3k AE = 7 rB/A = 200i rC/A = 480i ! CF = −480i + 490 j − 240k ; CF = 726.70 mm ! CE −480i + 490 j − 240k TCF = TCF = CF 726.70 ΣMAE = 0: AE ⋅ (rC/A × TCF ) + AE ⋅ (rB/A × (−600 j)) = 0 6 2 6 2 −3 −3 TCF 1 480 0 0 0 0 =0 + 200 726.7 × 7 7 0 −640 0 −480 +490 −240 2 × 480 × 240 − 3 × 480 × 490 3 × 200 × 640 TCF + =0 726.7 × 7 7 −653.91TCF + 384 × 103 = 0 TCF = 587 N PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 521 ! PROBLEM 4.142 A hand truck is used to move two kegs, each of mass 40 kg. Neglecting the mass of the hand truck, determine (a) the vertical force P that should be applied to the handle to maintain equilibrium when α = 35°, (b) the corresponding reaction at each of the two wheels. SOLUTION Free-Body Diagram: W = mg = (40 kg)(9.81 m/s 2 ) = 392.40 N a1 = (300 mm)sinα − (80 mm)cosα a2 = (430 mm)cosα − (300 mm)sinα b = (930 mm)cosα From free-body diagram of hand truck Dimensions in mm ΣM B = 0: P(b) − W ( a2 ) + W (a1 ) = 0 (1) ΣFy = 0: P − 2W + 2 B = 0 (2) α = 35° For a1 = 300sin 35° − 80 cos 35° = 106.541 mm a2 = 430 cos 35° − 300sin 35° = 180.162 mm b = 930cos 35° = 761.81 mm (a) From Equation (1) P(761.81 mm) − 392.40 N(180.162 mm) + 392.40 N(106.54 mm) = 0 P = 37.921 N (b) or P = 37.9 N From Equation (2) 37.921 N − 2(392.40 N) + 2 B = 0 or B = 373 N PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 522 PROBLEM 4.143 Determine the reactions at A and C when (a) α = 0, (b) α = 30°. SOLUTION (a) α = 0° From f.b.d. of member ABC ΣM C = 0: (300 N)(0.2 m) + (300 N)(0.4 m) − A(0.8 m) = 0 A = 225 N or A = 225 N ΣFy = 0: C y + 225 N = 0 C y = −225 N or C y = 225 N ΣFx = 0: 300 N + 300 N + C x = 0 C x = −600 N or C x = 600 N Then C = Cx2 + C y2 = (600) 2 + (225) 2 = 640.80 N and θ = tan −1 " Cy ! −1 −225 ! # = tan " # = 20.556° $ −600 % $ Cx % or (b) C = 641 N 20.6° α = 30° From f.b.d. of member ABC ΣM C = 0: (300 N)(0.2 m) + (300 N)(0.4 m) − ( A cos 30°)(0.8 m) + ( A sin 30°)(20 in.) = 0 A = 365.24 N or A = 365 N 60.0° ΣFx = 0: 300 N + 300 N + (365.24 N) sin 30° + C x = 0 Cx = −782.62 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 523 PROBLEM 4.143 (Continued) ΣFy = 0: C y + (365.24 N) cos 30° = 0 C y = −316.31 N or C y = 316 N Then C = Cx2 + C y2 = (782.62) 2 + (316.31)2 = 884.12 N and θ = tan −1 " Cy ! −1 −316.31 ! # = tan " # = 22.007° $ −782.62 % $ Cx % C = 884 N or 22.0° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 524 PROBLEM 4.144 A lever AB is hinged at C and attached to a control cable at A. If the lever is subjected to a 75-lb vertical force at B, determine (a) the tension in the cable, (b) the reaction at C. SOLUTION Free-Body Diagram: Geometry: x AC = (10 in.) cos 20° = 9.3969 in. y AC = (10 in.) sin 20° = 3.4202 in. , yDA = 12 in. − 3.4202 in. = 8.5798 in. yDA ! −1 8.5798 ! # = tan " # = 42.397° $ 9.3969 % $ x AC % α = tan −1 " β = 90° − 20° − 42.397° = 27.603° Equilibrium for lever: ΣM C = 0: TAD cos 27.603°(10 in.) − (75 lb)[(15 in.)cos 20°] = 0 (a) TAD = 119.293 lb TAD = 119.3 lb ΣFx = 0: C x + (119.293 lb) cos 42.397° = 0 (b) Cx = −88.097 lb ΣFy = 0: C y − 75 lb − (119.293 lb) sin 42.397° = 0 C y = 155.435 Thus: C = Cx2 + C y2 = (−88.097) 2 + (155.435) 2 = 178.665 lb and θ = tan −1 Cy Cx = tan −1 155.435 = 60.456° 88.097 C = 178.7 lb 60.5° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 525 PROBLEM 4.145 Neglecting friction and the radius of the pulley, determine (a) the tension in cable ADB, (b) the reaction at C. SOLUTION Free-Body Diagram: Dimensions in mm Geometry: Distance AD = (0.36) 2 + (0.150) 2 = 0.39 m Distance BD = (0.2)2 + (0.15)2 = 0.25 m Equilibrium for beam: ΣM C = 0: (a) 0.15 ! 0.15 ! T # (0.36 m) − " T # (0.2 m) = 0 (120 N)(0.28 m) − " $ 0.39 % $ 0.25 % T = 130.000 N or T = 130.0 N 0.36 ! 0.2 ! ΣFx = 0: Cx + " (130.000 N) + " # # (130.000 N) = 0 $ 0.39 % $ 0.25 % (b) Cx = − 224.00 N 0.15 ! 0.15 ! ΣFy = 0: C y + " (130.00 N) + " # # (130.00 N) − 120 N = 0 $ 0.39 % $ 0.25 % C y = − 8.0000 N 2 Thus: C = Cx2 + C y2 = (−224) 2 + ( − 8 ) = 224.14 N and θ = tan −1 Cy Cx = tan −1 8 = 2.0454° 224 C = 224 N 2.05° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 526 ! PROBLEM 4.146 The T-shaped bracket shown is supported by a small wheel at E and pegs at C and D. Neglecting the effect of friction, determine the reactions at C, D, and E when θ = 30°. SOLUTION Free-Body Diagram: ΣFy = 0: E cos 30° − 20 − 40 = 0 E= 60 lb = 69.282 lb cos 30° E = 69.3 lb 60.0° ΣM D = 0: (20 lb)(4 in.) − (40 lb)(4 in.) − C (3 in.) + E sin 30°(3 in.) = 0 −80 − 3C + 69.282(0.5)(3) = 0 C = 7.9743 lb C = 7.97 lb ΣFx = 0: E sin 30° + C − D = 0 (69.282 lb)(0.5) + 7.9743 lb − D = 0 D = 42.615 lb D = 42.6 lb PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 527 ! PROBLEM 4.147 The T-shaped bracket shown is supported by a small wheel at E and pegs at C and D. Neglecting the effect of friction, determine (a) the smallest value of θ for which the equilibrium of the bracket is maintained, (b) the corresponding reactions at C, D, and E. SOLUTION Free-Body Diagram: ΣFy = 0: E cos θ − 20 − 40 = 0 E= 60 cos θ (1) ΣM D = 0: (20 lb)(4 in.) − (40 lb)(4 in.) − C (3 in.) ! 60 +" sin θ # 3 in. = 0 θ cos $ % 1 C = (180 tan θ − 80) 3 (a) For C = 0, 180 tan θ = 80 tan θ = Eq. (1) E= 4 θ = 23.962° 9 θ = 24.0° 60 = 65.659 cos 23.962° ΣFx = 0: −D + C + E sin θ = 0 D = (65.659) sin 23.962 = 26.666 lb (b) C = 0 D = 26.7 lb E = 65.71 lb 66.0° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 528 PROBLEM 4.148 For the frame and loading shown, determine the reactions at A and C. SOLUTION Since member AB is acted upon by two forces, A and B, they must be colinear, have the same magnitude, and be opposite in direction for AB to be in equilibrium. The force B acting at B of member BCD will be equal in magnitude but opposite in direction to force B acting on member AB. Member BCD is a three-force body with member forces intersecting at E. The f.b.d.’s of members AB and BCD illustrate the above conditions. The force triangle for member BCD is also shown. The angle β is found from the member dimensions: 6 in. ! # = 30.964° 10 $ in. % β = tan −1 " Applying of the law of sines to the force triangle for member BCD, 30 lb B C = = sin(45° − β ) sin β sin135° or 30 lb B C = = sin14.036° sin 30.964° sin135° A= B= (30 lb)sin 30.964° = 63.641 lb sin14.036° or and C= A = 63.6 lb 45.0° C = 87.5 lb 59.0° (30 lb) sin135° = 87.466 lb sin14.036° or PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 529 PROBLEM 4.149 Determine the reactions at A and B when β = 50°. SOLUTION Free-Body Diagram: (Three-force body) Reaction A must pass through Point D where 100-N force and B intersect In right ∆ BCD α = 90° − 75° = 15° BD = 250 tan 75° = 933.01 mm Dimensions in mm In right ∆ ABD tan γ = 150 mm AB = BD 933.01 mm γ = 9.13 Force Triangle Law of sines 100 N A B = = sin 9.13° sin15° sin155.87° A = 163.1 N; B = 257.6 N A = 163.1 N 74.1° B = 258 N 65.0° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 530 PROBLEM 4.150 The 6-m pole ABC is acted upon by a 455-N force as shown. The pole is held by a ball-and-socket joint at A and by two cables BD and BE. For a = 3 m, determine the tension in each cable and the reaction at A. SOLUTION Free-Body Diagram: Five unknowns and six Eqs. of equilibrium, but equilibrium is maintained (ΣM AC = 0) rB = 3j rC = 6 j ! CF = −3i − 6 j + 2k CF = 7 m ! BD = 1.5i − 3j − 3k BD = 4.5 m ! BE = 1.5i − 3j + 3k BE = 4.5 m ! CF P P=P = (−3i − 6 j + 2k ) CE 7 ! T BD TBD (1.5i − 3j − 3k ) = BD (i − 2 j − 2k ) TBD = TBD = BD 4.5 3 ! BE TBD TBE = TBE = = (i − 2 j + 2k ) BE 3 ΣM A = 0: rB × TBD + rB × TBE + rC × P = 0 i j k i j k i j k TBD TBE P + 0 3 0 + 0 6 0 =0 0 3 0 3 3 7 −3 −6 2 1 −2 −2 1 −2 2 Coefficient of i: −2TBD + 2TBE + 12 P=0 7 (1) Coefficient of k: −TBD − TBF + 18 P=0 7 (2) PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 531 PROBLEM 4.150 (Continued) Eq. (1) + 2 Eq. (2): Eq. (2): − 4TBD + − 48 12 P = 0 TBD = P 7 7 12 18 6 P − TBE + P = 0 TBE = P 7 7 7 P = 445 N TBD = Since TBE = 12 (455) 7 6 (455) 7 TBD = 780 N TBE = 390 N ΣF = 0: TBD + TBE + P + A = 0 780 390 455 (3) + Ax = 0 + − 3 3 7 Coefficient of i: 260 + 130 − 195 + Ax = 0 Coefficient of j: − 780 390 455 (2) − (2) − (6) + Ay = 0 3 3 7 −520 − 260 − 390 + Ay = 0 Coefficient of k: − Ax = 195.0 N Ay = 1170 N 780 390 455 (2) + (2) + (2) + Az = 0 3 3 7 −520 + 260 + 130 + Az = 0 Az = +130.0 N A = −(195.0 N)i + (1170 N) j + (130.0 N)k PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 532 ! PROBLEM 4.151 Solve Problem 4.150 for a = 1.5 m. PROBLEM 4.150 The 6-m pole ABC is acted upon by a 455-N force as shown. The pole is held by a ball-and-socket joint at A and by two cables BD and BE. For a = 3 m, determine the tension in each cable and the reaction at A. SOLUTION Free-Body Diagram: Five unknowns and six Eqs. of equilibrium but equilibrium is maintained (ΣM AC = 0) rB = 3j ΣM A = 0: rB × TBD rC = 6 j ! CF = −1.5i − 6 j + 2k CF = 6.5 m ! BD = 1.5i − 3j − 3k BD = 4.5 m ! BE = 1.5i − 3j + 3k BE = 4.5 m ! CF P P ( −1.5i − 6 j + 2k ) = (−3i − 12 j + 4k ) P=P = CE 6.5 13 ! T BD TBD (1.5i − 3j − 3k ) = BD (i − 2 j − 2k ) TBD = TBD = BD 4.5 3 ! BE TBD TBE = TBE = = (i − 2 j + 2k ) BE 3 + rB × TBE + rC × P = 0 i j k i j k i j k TBD TBE P + 0 3 0 + 0 =0 0 3 0 6 0 3 3 13 −3 −12 + 4 1 −2 −2 1 −2 2 Coefficient of i: −2TBD + 2TBE + 24 P=0 13 (1) Coefficient of k: −TBD − TBE + 18 P=0 13 (2) PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 533 PROBLEM 4.151 (Continued) Eq. (1) + 2 Eq. (2): Eq (2): −4TBD + − 60 15 P = 0 TBD = P 13 13 15 18 3 P − TBE + P = 0 TBE = P 13 13 13 P = 445 N TBD = Since TBE = 15 (455) 13 3 (455) 13 TBD = 525 N ! TBE = 105.0 N ! ΣF = 0: TBD + TBE + P + A = 0 525 105 455 (3) + Ax = 0 + − 3 3 13 Coefficient of i: 175 + 35 − 105 + Ax = 0 Coefficient of j: − 525 105 455 (2) − (2) − (12) + Ay = 0 3 3 13 −350 − 70 − 420 + Ay = 0 Coefficient of k: Ax = 105.0 N − Ay = 840 N 525 105 455 (2) + (2) + (4) + Az = 0 3 3 13 −350 + 70 + 140 + Az = 0 Az = 140.0 N A = −(105.0 N)i + (840 N) j + (140.0 N)k PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 534 PROBLEM 4.152 The rigid L-shaped member ABF is supported by a ball-and-socket joint at A and by three cables. For the loading shown, determine the tension in each cable and the reaction at A. SOLUTION Free-Body Diagram: rB/A = 12i rF/A = 12 j − 8k rD/A = 12i − 16k rE/A = 12i − 24k rF/A = 12i − 32k ! BG = −12i + 9k BG = 15 in. BG = −0.8i + 0.6k ! DH = −12i + 16 j; DH = 20 in.; λDH = −0.6i + 0.8 j ! FJ = −12i + 16 j; FJ = 20 in.; λFJ = −0.6i + 0.8 j ΣM A = 0: rB/A × TBG λBG + rDH × TDH λDH + rF/A × TFJ λFJ +rF/A × (−24 j) + rE/A × ( −24 j) = 0 i j k i j k i j k 12 0 0 TBG + 12 0 −16 TDH + 12 0 −32 TFJ −0.8 0 0.6 −0.6 0.8 0 −0.6 0.8 0 i j k i j k + 12 0 −8 + 12 0 −24 = 0 0 −24 0 0 −24 0 Coefficient of i: +12.8TDH + 25.6TFJ − 192 − 576 = 0 (1) Coefficient of k: +9.6TDH + 9.6TFJ − 288 − 288 = 0 (2) 3 4 Eq. (1) − Eq. (2): 9.6TFJ = 0 TFJ = 0 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 535 PROBLEM 4.152 (Continued) Eq. (1): 12.8TDH − 268 = 0 Coefficient of j: −7.2TBG + (16 × 0.6)(60.0 lb) = 0 ΣF = 0: TDH = 60 lb TBG = 80.0 lb A + TBG λ BG + TDH λ DH + TFJ − 24 j − 24 j = 0 Coefficient of i: Ax + (80)( −0.8) + (60.0)(−0.6) = 0 Ax = 100.0 lb Coefficient of j: Ay + (60.0)(0.8) − 24 − 24 = 0 Ay = 0 Coefficient of k: Az + (80.0)(+0.6) = 0 Az = −48.0 lb A = (100.0 lb)i − (48.0 lb) j Note: The value Ay = 0 Can be confirmed by considering ΣM BF = 0 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 536 PROBLEM 4.153 A force P is applied to a bent rod ABC, which may be supported in four different ways as shown. In each case, if possible, determine the reactions at the supports. SOLUTION (a) ΣM A = 0: − Pa + (C sin 45°)2a + (cos 45°)a = 0 3 C 2 =P C= P 3 2 2 ! 1 ΣFx = 0: Ax − " P " 3 ## 2 $ % C = 0.471P Ax = 2 ! 1 ΣFy = 0: Ay − P + " P " 3 ## 2 $ % P 3 Ay = 2P 3 A = 0.745P (b) 45° 63.4° ΣM C = 0: +Pa − ( A cos 30°)2a + ( A sin 30°)a = 0 A(1.732 − 0.5) = P A = 0.812 P A = 0.812P 60.0° ΣFx = 0: (0.812 P)sin 30° + Cx = 0 C x = −0.406 P ΣFy = 0: (0.812 P) cos 30° − P + C y = 0 C y = −0.297 P C = 0.503P 36.2° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 537 PROBLEM 4.153 (Continued) ΣM C = 0: + Pa − ( A cos 30°)2a + ( A sin 30°)a = 0 (c) A(1.732 + 0.5) = P A = 0.448P A = 0.448P ΣFx = 0: − (0.448P) sin 30° + Cx = 0 60.0° Cx = 0.224 P ΣFy = 0: (0.448 P) cos 30° − P + C y = 0 C y = 0.612 P C = 0.652P 69.9° ! (d) Force T exerted by wire and reactions A and C all intersect at Point D. ΣM D = 0: Pa = 0 Equilibrium not maintained Rod is improperly constrained PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 538 ! CHAPTER 5 PROBLEM 5.1 Locate the centroid of the plane area shown. SOLUTION Dimensions in mm Then A, mm 2 x , mm y , mm x A, mm3 y A, mm3 1 6300 105 15 0.66150 × 106 0.094500 × 106 2 9000 225 150 2.0250 × 106 1.35000 × 106 Σ 15300 2.6865 × 106 1.44450 × 106 X= Σ x A 2.6865 × 106 = 15300 ΣA Y = Σ y A 1.44450 × 106 = ΣA 15300 X = 175.6 mm Y = 94.4 mm PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 541 PROBLEM 5.2 Locate the centroid of the plane area shown. SOLUTION Dimensions in mm Then A, mm 2 x , mm y , mm x A, mm3 y A, mm3 1 1200 10 30 12000 36000 2 540 30 36 16200 19440 Σ 1740 28200 55440 X= Σ x A 28200 = ΣA 1740 X = 16.21 mm Y = Σ y A 55440 = ΣA 1740 Y = 31.9 mm PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 542 PROBLEM 5.3 Locate the centroid of the plane area shown. SOLUTION Dimensions in in. Then A, in.2 x , in. y , in. x A, in.3 y A, in.3 1 1 × 12 × 15 = 90 2 8 5 720 450 2 21 × 15 = 315 22.5 7.5 7087.5 2362.5 Σ 405.00 7807.5 2812.5 X= Σ x A 7807.5 = Σ A 405.00 X = 19.28 in. Y = Σ y A 2812.5 = Σ A 405.00 Y = 6.94 in. PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 543 PROBLEM 5.4 Locate the centroid of the plane area shown. SOLUTION Then A, in.2 x , in. y , in. x A, in.3 y A, in.3 1 1 (12)(6) = 36 2 4 4 144 144 2 (6)(3) = 18 9 7.5 162 135 Σ 54 306 279 XA = Σ xA X (54) = 306 X = 5.67 in. YA = Σ yA Y (54) = 279 Y = 5.17 in. PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 544 PROBLEM 5.5 Locate the centroid of the plane area shown. SOLUTION Then A, in.2 x , in. y , in. x A, in.3 y A, in.3 1 14 × 20 = 280 7 10 1960 2800 2 −π (4) 2 = −16π 6 12 –301.59 –603.19 Σ 229.73 1658.41 2196.8 X= Σ xA 1658.41 = ΣA 229.73 X = 7.22 in. Y = Σ y A 2196.8 = Σ A 229.73 Y = 9.56 in. PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 545 PROBLEM 5.6 Locate the centroid of the plane area shown. SOLUTION A, mm 2 x , mm y , mm x A, mm3 y A, mm3 1 1 (120)(75) = 4500 2 80 25 360 × 103 112.5 × 103 2 (75)(75) = 5625 157.5 37.5 885.94 × 103 210.94 × 103 163.169 43.169 −720.86 × 103 −190.716 × 103 525.08 × 103 132.724 × 103 3 Σ Then − π 4 (75) 2 = −4417.9 5707.1 XA = Σx A X (5707.1) = 525.08 × 103 X = 92.0 mm YA = Σ y A Y (5707.1) = 132.724 × 103 Y = 23.3 mm PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 546 PROBLEM 5.7 Locate the centroid of the plane area shown. SOLUTION A, in.2 x , in. y , in. x A, in.3 y A, in.3 (38)2 = 2268.2 0 16.1277 0 36581 2 −20 × 16 = −320 −10 8 3200 −2560 Σ 1948.23 3200 34021 1 Then π 2 X= Σ xA 3200 = Σ A 1948.23 X = 1.643 in. Y = Σ y A 34021 = Σ A 1948.23 Y = 17.46 in. PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 547 PROBLEM 5.8 Locate the centroid of the plane area shown. SOLUTION 1 2 Σ Then − A, in.2 x , in. y , in. x A, in.3 y A, in.3 30 × 50 = 1500 15 25 22500 37500 23.634 30 –8353.0 –10602.9 14147.0 26.897 π 2 (15) 2 = 353.43 1146.57 X= Σ x A 14147.0 = Σ A 1146.57 X = 12.34 in. Y = Σ y A 26897 = Σ A 1146.57 Y = 23.5 in. PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 548 PROBLEM 5.9 Locate the centroid of the plane area shown. SOLUTION 1 Σ Then x , mm y , mm x A, mm3 y A, mm3 (60)(120) = 7200 –30 60 −216 × 103 432 × 103 (60) 2 = 2827.4 25.465 95.435 72.000 × 103 269.83 × 103 (60) 2 = −2827.4 –25.465 25.465 72.000 × 103 −72.000 × 103 −72.000 × 103 629.83 × 103 π 2 3 A, mm 2 4 − π 4 7200 XA = Σ x A X (7200) = −72.000 × 103 YA = Σ y A Y (7200) = 629.83 × 103 X = −10.00 mm Y = 87.5 mm PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 549 PROBLEM 5.10 Locate the centroid of the plane area shown. SOLUTION Dimensions in mm A, mm 2 1 Then π 2 × 47 × 26 = 1919.51 2 1 × 94 × 70 = 3290 2 Σ 5209.5 x , mm y , mm x A, mm3 y A, mm3 0 11.0347 0 21181 −15.6667 −23.333 −51543 −76766 −51543 −55584 X= Σ x A −51543 = ΣA 5209.5 X = −9.89 mm Y = Σ y A −55584 = ΣA 5209.5 Y = −10.67 mm PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 550 PROBLEM 5.11 Locate the centroid of the plane area shown. SOLUTION X =0 First note that symmetry implies A, in.2 1 2 − π (8) 2 2 π (12) 2 2 yA, in.3 = −100.531 3.3953 –341.33 = 226.19 5.0930 1151.99 125.659 Σ Then y , in. Y = ! 810.66 Σ y A 810.66 in.3 = Σ A 125.66 in.2 or Y = 6.45 in. PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 551 PROBLEM 5.12 Locate the centroid of the plane area shown. SOLUTION Then A, mm 2 x , mm y , mm xA, mm3 yA, mm3 1 (15)(80) = 1200 40 7.5 48 × 103 9 × 103 2 1 (50)(80) = 1333.33 3 60 30 80 × 103 40 × 103 Σ 2533.3 128 × 103 49 × 103 X A = Σ xA X (2533.3) = 128 × 103 X = 50.5 mm YA = Σ yA Y (2533.3) = 49 × 103 Y = 19.34 mm PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 552 PROBLEM 5.13 Locate the centroid of the plane area shown. SOLUTION 1 2 Σ Then A, mm 2 x , mm y , mm x A, mm3 y A, mm3 1 × 30 × 20 = 200 3 9 15 1800 3000 12.7324 32.7324 9000.0 23137 10800 26137 π 4 (30)2 = 706.86 906.86 X= Σ x A 10800 = Σ A 906.86 X = 11.91 mm Y = Σ y A 26137 = Σ A 906.86 Y = 28.8 mm PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 553 PROBLEM 5.14 Locate the centroid of the plane area shown. SOLUTION Dimensions in in. Then A, in.2 x , in. y , in. x A, in 3 y A, in.3 1 2 800 × (20)(20) = 3 3 12 7.5 3200 2000 2 −1 −400 (20)(20) = 3 3 15 6.0 –2000 –800 Σ 400 3 1200 1200 X= Y = Σ x A 1200 = 400 ! ΣA " # $ 3 % Σ yA 1200 = 400 ! ΣA " # $ 3 % X = 9.00 in. Y = 9.00 in. PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 554 PROBLEM 5.15 Locate the centroid of the plane area shown. SOLUTION Then A, mm 2 x , mm y , mm x A, mm3 y A, mm3 1 2 (75)(120) = 6000 3 28.125 48 168750 288000 2 1 − (75)(60) = −2250 2 25 20 –56250 –45000 Σ 3750 112500 243000 X ΣA = ΣxA X (3750 mm2 ) = 112500 mm3 and or X = 30.0 mm Y ΣA = Σ yA Y (3750 mm 2 ) = 243000 or Y = 64.8 mm PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 555 PROBLEM 5.16 Determine the y coordinate of the centroid of the shaded area in terms of r1, r2, and α. SOLUTION First, determine the location of the centroid. y2 = From Figure 5.8A: Similarly = 2 cos α r2 π 3 ( 2 −α ) y1 = 2 cos α r1 3 ( π2 − α ) Σ yA = Then π 2 sin ( 2 − α ) r2 π 3 ( 2 −α ) = π ! A2 = " − α # r22 $2 % π ! A1 = " − α # r12 2 $ % 2 cos α & π ! ' 2 cos α & π ! 2' r2 − α # r22 ) − r1 π ( (" − α # r1 ) 3 ( π2 − α ) *"$ 2 3 2 α − % + % + ( 2 ) *$ 2 3 3 r2 − r1 cos α 3 ( ) π π ! ! ΣA = " − α # r22 − " − α # r12 $2 % $2 % and π ! = " − α # r22 − r12 2 $ % ( ) Y ΣA = Σ yA Now & π ' 2 ! Y (" − α # r22 − r12 ) = r23 − r13 cos α 2 % *$ + 3 ( ) ( ) Y = 2 r23 − r13 " 3 "$ r22 − r12 ! 2 cos α ! ## " # % $ π − 2α % PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 556 PROBLEM 5.17 Show that as r1 approaches r2, the location of the centroid approaches that for an arc of circle of radius (r1 + r2 )/2. SOLUTION First, determine the location of the centroid. y2 = From Figure 5.8A: Similarly = 2 cos α r2 π 3 ( 2 −α ) y1 = 2 cos α r1 3 ( π2 − α ) Σ yA = Then π 2 sin ( 2 − α ) r2 π 3 ( 2 −α ) = π ! A2 = " − α # r22 $2 % π ! A1 = " − α # r12 2 $ % 2 cos α & π ! ' 2 cos α & π ! 2' r2 − α # r22 ) − r1 π ( (" − α # r1 ) 3 ( π2 − α ) *"$ 2 3 2 − α % + % + ( 2 ) *$ 2 3 3 r2 − r1 cos α 3 ( ) π π ! ! ΣA = " − α # r22 − " − α # r12 $2 % $2 % and π ! = " − α # r22 − r12 2 $ % ( ) Y ΣA = Σ yA Now & π ' 2 ! Y (" − α # r22 − r12 ) = r23 − r13 cos α 2 % *$ + 3 ( ) Y = ( ) 2 r23 − r13 " 3 "$ r22 − r12 ! 2cos α ! ## " # % $ π − 2α % PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 557 PROBLEM 5.17 (Continued) 1 (r1 + r2 ) is 2 Using Figure 5.8B, Y of an arc of radius Y = = Now r23 − r13 r22 − r12 = = sin( π − α ) 1 (r1 + r2 ) π 2 2 ( 2 −α) 1 cos α (r1 + r2 ) π 2 ( 2 −α) ( (r2 − r1 ) r22 + r1r2 + r12 (1) ) (r2 − r1 )(r2 + r1 ) r22 + r1r2 + r12 r2 + r1 r2 = r + ∆ Let r1 = r − ∆ r= Then and In the limit as ∆ r23 − r13 r22 − r12 1 (r1 + r2 ) 2 = (r + ∆) 2 + (r + ∆)(r − ∆)( r − ∆) 2 (r + ∆ ) + (r − ∆ ) = 3r 2 + ∆ 2 2r 0 (i.e., r1 = r2 ), then r23 − r13 r22 − r12 So that = 3 r 2 = 3 1 × ( r1 + r2 ) 2 2 Y = 2 3 cos α × (r1 + r2 ) π −α 3 4 2 or Y = ( r1 + r2 ) cos α π − 2α Which agrees with Equation (1). PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 558 PROBLEM 5.18 For the area shown, determine the ratio a/b for which x = y . SOLUTION Then A x y xA yA 1 2 ab 3 3 a 8 3 b 5 a 2b 4 2ab2 5 2 1 − ab 2 1 a 3 2 b 3 Σ 1 ab 6 − a 2b 6 a 2b 12 − ab 2 3 ab 2 15 X Σ A = Σ xA 1 ! a 2b X " ab # = $ 6 % 12 or 1 a 2 Y ΣA = Σ y A X= 1 ! ab 2 Y " ab # = $ 6 % 15 2 b 5 or Y = Now X =Y , 1 2 a= b 2 5 or a 4 = b 5 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 559 PROBLEM 5.19 For the semiannular area of Problem 5.11, determine the ratio r2/r1 so that y = 3r1/4. SOLUTION A − 1 Σ 2 Let 4r1 3π 2 − r13 3 r22 4r2 3π 2 3 r2 3 (r 2 2 − r12 2 3 3 r2 − r1 3 ) ( ) Y Σ A = Σ yA Then or r12 2 2 π YA π π 2 Y 3 2 π r1 × r22 − r12 = r23 − r13 4 2 3 2 3 ' r ! 9π & r2 ! (" # − 1) = " 2 # − 1 16 ($ r1 % ) $ r1 % * + ( ) p= ( ) r2 r1 9π [( p + 1)( p − 1)] = ( p − 1)( p 2 + p + 1) 16 or 16 p 2 + (16 − 9π ) p + (16 − 9π ) = 0 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 560 PROBLEM 5.19 (Continued) −(16 − 9π ) ± (16 − 9π ) 2 − 4(16)(16 − 9π ) 2(16) Then p= or p = −0.5726 p = 1.3397 r2 = 1.340 r1 Taking the positive root PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 561 PROBLEM 5.20 A composite beam is constructed by bolting four plates to four 60 × 60 × 12-mm angles as shown. The bolts are equally spaced along the beam, and the beam supports a vertical load. As proved in mechanics of materials, the shearing forces exerted on the bolts at A and B are proportional to the first moments with respect to the centroidal x axis of the red shaded areas shown, respectively, in Parts a and b of the figure. Knowing that the force exerted on the bolt at A is 280 N, determine the force exerted on the bolt at B. SOLUTION From the problem statement: F is proportional to Qx . (Qx ) B FA (Qx ) A Therefore: FA FB , or = (Qx ) A (Qx ) B For the first moments: 12 ! (Qx ) A = " 225 + # (300 × 12) 2% $ FB = = 831600 mm3 12 ! (Qx ) B = (Qx ) A + 2 " 225 − # (48 × 12) + 2(225 − 30)(12 × 60) 2% $ = 1364688 mm3 Then FB = 1364688 (280 N) 831600 or FB = 459 N PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 562 PROBLEM 5.21 The horizontal x axis is drawn through the centroid C of the area shown, and it divides the area into two component areas A1 and A2 . Determine the first moment of each component area with respect to the x axis, and explain the results obtained. SOLUTION Note that Then and Qx = Σ y A 5 ! 1 ! (Qx )1 = " in. #" × 6 × 5 # in.2 $ 3 %$ 2 % or (Qx )1 = 25.0 in.3 2 ! 1 ! (Qx ) 2 = " − × 2.5 in. #" × 9 × 2.5 # in.2 $ 3 %$ 2 % 1 ! 1 ! + " − × 2.5 in. #" × 6 × 2.5 # in.2 3 2 $ %$ % or (Qx ) 2 = −25.0 in.3 Now Qx = (Qx )1 + (Qx )2 = 0 This result is expected since x is a centroidal axis (thus y = 0) and Qx = Σ yA = Y ΣA( y = 0 , Qx = 0) ! PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 563 PROBLEM 5.22 The horizontal x axis is drawn through the centroid C of the area shown, and it divides the area into two component areas A1 and A2 . Determine the first moment of each component area with respect to the x axis, and explain the results obtained. SOLUTION First determine the location of the centroid C. We have Then A, in.2 y ′, in. y ′A, in.3 I 1 ! 2 " × 2 × 1.5 # = 3 $2 % 0.5 1.5 II 1.5 × 5.5 = 8.25 2.75 22.6875 III 4.5 × 2 = 9 6.5 58.5 Σ 20.25 82.6875 Y ′ Σ A = Σ y ′A Y ′(20.25) = 82.6875 or Y ′ = 4.0833 in. Now Qx = Σ y1 A Then &1 ' (Qx )1 = ( (5.5 − 4.0833)in.) [(1.5)(5.5 − 4.0833)]in.2 *2 + + [(6.5 − 4.0833)in.][(4.5)(2)]in.2 and or (Qx )1 = 23.3 in.3 &1 ' (Qx ) 2 = − ( (4.0833 in.) ) [(1.5)(4.0833)]in.2 *2 + & 1 ' ! − [(4.0833 − 0.5)in.] × 2 (" × 2 × 1.5 # in.2 ) % *$ 2 + or (Qx ) 2 = −23.3 in.3 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 564 PROBLEM 5.22 (Continued) Now Qx = (Qx )1 + (Qx )2 = 0 This result is expected since x is a centroidal axis (thus Y = 0) and Qx = Σ y A = Y Σ A (Y = 0 , Qx = 0) ! PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 565 PROBLEM 5.23 The first moment of the shaded area with respect to the x axis is denoted by Qx. (a) Express Qx in terms of b, c, and the distance y from the base of the shaded area to the x axis. (b) For what value of y is Ox maximum, and what is that maximum value? SOLUTION Shaded area: A = b (c − y ) Qx = yA = Qx = (a) (b) For Qmax : For y = 0: 1 (c + y )[b(c − y )] 2 1 b (c 2 − y 2 ) 2 dQ = 0 or dy (Qx ) = 1 b(−2 y ) = 0 2 1 2 bc 2 y=0 (Qx ) = 1 2 bc 2 ! PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 566 PROBLEM 5.24 A thin, homogeneous wire is bent to form the perimeter of the figure indicated. Locate the center of gravity of the wire figure thus formed. SOLUTION Perimeter of Figure 5.1 Dimensions in mm L x y xL, mm 2 yL, mm 2 I 30 0 15 0 0.45 × 103 II 210 105 30 22.05 × 103 6.3 × 103 III 270 210 165 56.7 × 103 44.55 × 103 IV 30 225 300 6.75 × 103 9 × 103 V 300 240 150 72 × 103 45 × 103 VII 240 120 0 28.8 × 103 0 Σ 1080 186.3 × 103 105.3 × 103 X ΣL = Σ x L X (1080 mm) = 186.3 × 103 mm 2 X = 172.5 mm Y ΣL = Σ y L Y (1080 mm) = 105.3 × 103 mm 2 Y = 97.5 mm ! PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 567 PROBLEM 5.25 A thin, homogeneous wire is bent to form the perimeter of the figure indicated. Locate the center of gravity of the wire figure thus formed. SOLUTION First note that because wire is homogeneous, its center of gravity will coincide with the centroid of the corresponding line. Then L, mm x , mm y , mm xL, mm 2 yL, mm 2 1 20 10 0 200 0 2 24 20 12 480 288 3 30 35 24 1050 720 4 46.861 35 42 1640.14 1968.16 5 20 10 60 200 1200 6 60 0 30 0 1800 Σ 200.86 3570.1 5976.2 X ΣL = Σ x L X (200.86) = 3570.1 X = 17.77 mm Y ΣL = Σ y L Y (200.86) = 5976.2 Y = 29.8 mm PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 568 ! PROBLEM 5.26 A thin, homogeneous wire is bent to form the perimeter of the figure indicated. Locate the center of gravity of the wire figure thus formed. SOLUTION First note that because the wire is homogeneous, its center of gravity will coincide with the centroid of the corresponding line. Then L, in. x , in. y , in. xL, in.2 yL, in.2 1 33 16.5 0 544.5 0 2 15 33 7.5 495 112.5 3 21 22.5 15 472.5 315 4 122 + 152 = 19.2093 6 7.5 115.256 144.070 Σ 88.209 1627.26 571.57 X ΣL = Σx L X (88.209) = 1627.26 and or X = 18.45 in. Y ΣL = Σ y L Y (88.209) = 571.57 or Y = 6.48 in. PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 569 ! PROBLEM 5.27 A thin, homogeneous wire is bent to form the perimeter of the figure indicated. Locate the center of gravity of the wire figure thus formed. SOLUTION First note that because the wire is homogeneous, its center of gravity will coincide with the centroid of the corresponding line. Y6 = Then 2 π (38 in.) L, in. x , in. y , in. xL, in.2 yL, in.2 1 18 –29 0 –522 0 2 16 –20 8 –320 128 3 20 –10 16 –200 320 4 16 0 8 0 128 5 38 19 0 722 0 6 π (38) = 119.381 0 24.192 0 2888.1 Σ 227.38 –320 3464.1 X= Σx L −320 = 227.38 ΣL X = −1.407 in. Y = Σ y L 3464.1 = 227.38 ΣL Y = 15.23 in. PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 570 PROBLEM 5.28 A uniform circular rod of weight 8 lb and radius 10 in. is attached to a pin at C and to the cable AB. Determine (a) the tension in the cable, (b) the reaction at C. SOLUTION For quarter circle (a) r= 2r π 2r ! ΣM C = 0: W " # − Tr = 0 $π % 2! 2! T = W " # = (8 lb) " # $π % $π % (b) T = 5.09 lb ΣFx = 0: T − C x = 0 5.09 lb − Cx = 0 C x = 5.09 lb ΣFy = 0: C y − W = 0 C y = 8 lb C y − 8 lb = 0 C = 9.48 lb 57.5° ! PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 571 PROBLEM 5.29 Member ABCDE is a component of a mobile and is formed from a single piece of aluminum tubing. Knowing that the member is supported at C and that l = 2 m, determine the distance d so that portion BCD of the member is horizontal. SOLUTION First note that for equilibrium, the center of gravity of the component must lie on a vertical line through C. Further, because the tubing is uniform, the center of gravity of the component will coincide with the centroid of the corresponding line. Thus, X = 0 So that Σx L = 0 Then 0.75 ! cos 55° # m × (0.75 m) −" d − 2 $ % + (0.75 − d )m × (1.5 m) & 1 !' + ((1.5 − d )m − " × 2 m × cos 55° # ) × (2 m) = 0 $2 %+ * or &1 ' (0.75 + 1.5 + 2)d = ( (0.75) 2 − 2 ) cos 55° + (0.75)(1.5) + 3 2 * + or d = 0.739 m ! PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 572 PROBLEM 5.30 Member ABCDE is a component of a mobile and is formed from a single piece of aluminum tubing. Knowing that the member is supported at C and that d is 0.50 m, determine the length l of arm DE so that this portion of the member is horizontal. SOLUTION First note that for equilibrium, the center of gravity of the component must lie on a vertical line through C. Further, because the tubing is uniform, the center of gravity of the component will coincide with the centroid of the corresponding line. Thus, X =0 So that or Σx L = 0 0.75 ! −" sin 20° + 0.5sin 35° # m × (0.75 m) $ 2 % + (0.25 m × sin 35°) × (1.5 m) l! + "1.0 × sin 35° − # m × (l m) = 0 2 $ % or + ( sin 35° − 2l ) l = 0 −0.096193 !!!!!!!!" !!!!!!" ( xL)DE ( xL) AB +( xL)BD The equation implies that the center of gravity of DE must be to the right of C. Then l 2 − 1.14715l + 0.192386 = 0 or l= or l = 0.204 m 1.14715 ± (−1.14715)2 − 4(0.192386) 2 or l = 0.943 m Note that sin 35° − 12 l . 0 for both values of l so both values are acceptable.! PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 573 PROBLEM 5.31 The homogeneous wire ABC is bent into a semicircular arc and a straight section as shown and is attached to a hinge at A. Determine the value of θ for which the wire is in equilibrium for the indicated position. SOLUTION First note that for equilibrium, the center of gravity of the wire must lie on a vertical line through A. Further, because the wire is homogeneous, its center of gravity will coincide with the centroid of the corresponding line. Thus, X =0 So that Σx L = 0 Then 1 2r ! ! " − 2 r cos θ # (r ) + " π − r cos θ # (π r ) = 0 $ % $ % or cos θ = 4 1 + 2π = 0.54921 or θ = 56.7° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 574 ! PROBLEM 5.32 Determine the distance h for which the centroid of the shaded area is as far above line BB as possible when (a) k = 0.10, (b) k = 0.80. SOLUTION A y yA 1 1 ba 2 1 a 3 1 2 a b 6 2 1 − (kb)h 2 1 h 3 1 − kbh 2 6 Σ b (a − kh) 2 b 2 ( a − kh 2 ) 6 Y Σ A = Σ yA Then &b ' b Y ( ( a − kh) ) = (a 2 − kh 2 ) *2 + 6 Y = or (1) dY 1 −2kh(a − kh) − (a 2 − kh 2 )(− k ) = =0 dh 3 ( a − kh) 2 and or a 2 − kh 2 3(a − kh) 2h(a − kh) − a 2 + kh 2 = 0 (2) Simplifying Eq. (2) yields kh 2 − 2ah + a 2 = 0 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 575 PROBLEM 5.32 (Continued) Then 2a ± (−2a) 2 − 4( k )(a 2 ) 2k a = &*1 ± 1 − k '+ k h= Note that only the negative root is acceptable since h , a. Then (a) k = 0.10 h= (b) a & 1 − 1 − 0.10 ' + 0.10 * or h = 0.513a k = 0.80 h= a & 1 − 1 − 0.80 ' + 0.80 * or h = 0.691a ! PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 576 PROBLEM 5.33 Knowing that the distance h has been selected to maximize the distance y from line BB to the centroid of the shaded area, show that y = 2h/3. SOLUTION See solution to Problem 5.32 for analysis leading to the following equations: Y = a 2 − kh 2 3(a − kh) (1) 2h(a − kh) − a 2 + kh 2 = 0 (2) Rearranging Eq. (2) (which defines the value of h which maximizes Y ) yields a 2 − kh 2 = 2h(a − kh) Then substituting into Eq. (1) (which defines Y ) Y = 1 × 2h(a − kh) 3( a − kh) or Y = 2 h 3 ! PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 577 PROBLEM 5.34 Determine by direct integration the centroid of the area shown. Express your answer in terms of a and h. SOLUTION y h = x a h y= x a xEL = x 1 yEL = y 2 dA = ydx A= -x - EL dA - a 0 ydx = - = xydx = yEL dA = - a 0 1 "2 $ - a 0 a 0 1 h ! " a x # dx = 2 ah $ % a 1 h ! h & x3 ' x " x # dx = ( ) = ha 2 a * 3 +0 3 $a % 1 ! y # ydx = 2 % - a 0 2 1 h2 h ! " a x # dx = 2 2 b $ % a & x3 ' 1 2 ( ) = h a 3 * +0 6 1 ! 1 xA = xEL dA: x " ah # = ha 2 $2 % 3 x= 1 ! 1 yA = yEL dA: y " ah # = h 2 a $2 % 6 1 y= h 3 - - 2 a 3 ! PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 578 PROBLEM 5.35 Determine by direct integration the centroid of the area shown. Express your answer in terms of a and h. SOLUTION y1 : h = ka 2 At (a, h) k= or h a2 y2 : h = ma m= or xEL = x Now yEL = 1 ( y1 + y2 ) 2 h ' &h dA = ( y2 − y1 )dx = ( x − 2 x 2 ) dx a a * + h = 2 (ax − x 2 ) dx a and - A = dA = Then and h a - - a 0 a 1 ' 1 h h &a (ax − x 2 )dx = 2 ( x 2 − x3 ) = ah 3 +0 6 a2 a *2 a 1 ' 1 h &a &h x 2 (ax − x 2 ) dx = 2 ( x3 − x 4 ) = a 2 h 0 ( 4 + 0 12 a *3 *a 1 1 2 ( y1 + y2 )[( y2 − y1 ) dx] = yEL dA = y2 − y12 dx 2 2 xEL dA = - a - = 1 2 - ( - a 0 ) h 2 2 h2 4 ! "" 2 x − 4 x ##dx a $a % a 1 h2 & a 2 3 1 5 ' = ( x − x ) 2 a4 * 3 5 +0 1 = ah 2 15 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 579 PROBLEM 5.35 (Continued) 1 ! 1 xA = xEL dA: x " ah # = a 2 h $ 6 % 12 x= 1 a 2 1 ! 1 yA = yEL dA: y " ah # = ah 2 $ 6 % 15 y= 2 h 5 - - PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 580 ! PROBLEM 5.36 Determine by direct integration the centroid of the area shown. Express your answer in terms of a and h. SOLUTION For the element (EL) shown At x = a, Then x= xEL yEL h a3 a 1/3 y h1/3 a 1/3 y dy h1/3 1 1 a 1/3 y = x= 2 2 h1/ 3 =y - A = dA = Then Hence or k = dA = xdy = Now and y = h : h = ka3 - xEL dA = - yEL dA = - h 0 h 0 - h 0 3 a a 1/3 y dy = y 4/3 1/3 4 h1/3 h ( ) h = 0 3 ah 4 h 1 a 1/3 a 1/3 ! 1 a 3 5/3 ! 3 y " 1/3 y dy # = y # = a2h 2/3 " 2 h1/3 2 5 10 h h $ % $ %0 h a a 3 3 ! ! y " 1/3 y1/3 dy # = 1/3 " y 7/3 # = ah 2 $h % h $7 %0 7 - 3 ! 3 x " ah # = a 2 h $ 4 % 10 x= 2 a 5 ! - 3 ! 3 y " ah # = ah2 $4 % 7 y= 4 h 7 ! xA = xEL dA: yA = yEL dA: PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 581 PROBLEM 5.37 Determine by direct integration the centroid of the area shown. SOLUTION y= For the element (EL) shown b 2 a − x2 a dA = (b − y ) dx and xEL ) ( b a − a 2 − x 2 dx a =x = 1 ( y + b) 2 b = a + a 2 − x2 2a yEL = ( - A = dA = Then - To integrate, let x = a sin θ : Then A= = - π /2 0 a 0 ) ) ( b a − a 2 − x 2 dx a a 2 − x 2 = a cos θ , dx = a cos θ dθ b (a − a cos θ )(a cos θ dθ ) a b& 2 2θ 2 θ ( a sin θ − a " + sin a* 2 4 $ π /2 !' #) %+ 0 π! = ab "1 − # 4% $ and -x EL dA = - a 0 &b ' x ( a − a 2 − x 2 dx ) a * + ) ( π /2 b& a 1 !' = (" x 2 + (a 2 − x 2 )3/2 # ) a *$ 2 3 %+ 0 = 1 3 ab 6 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 582 PROBLEM 5.37 (Continued) -y EL dA = - a 0 b &b ' a + a 2 − x 2 ( a − a 2 − x 2 dx ) 2a a * + b2 = 2 2a = ) ( ( ) a - a 0 b 2 x3 ! ( x ) dx = 2 "" ## 2a $ 3 % 0 2 1 2 ab 6 - & π !' 1 x ( ab "1 − # ) = a 2 b 4 %+ 6 * $ or x = 2a 3(4 − π ) - & π !' 1 y ( ab "1 − # ) = ab 2 4 %+ 6 * $ or y = 2b 3(4 − π ) xA = xEL dA: yA = yEL dA: PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 583 ! PROBLEM 5.38 Determine by direct integration the centroid of the area shown. SOLUTION x =0 First note that symmetry implies For the element (EL) shown yEL = 2r (Figure 5.8B) π dA = π rdr - A = dA = Then and -y EL dA = - r2 r1 - r2 r1 r2 π r d r = π "" $ 2 r2 ! π 2 2 r2 − r1 ## = % r1 2 ( r2 1 ! 2 (π rdr ) = 2 " r 3 # = r23 − r13 π $3 %r 3 2r ( ) ) 1 So &π ' 2 yA = yEL dA: y ( r22 − r12 ) = r23 − r13 2 * + 3 - ( ) ( ) or y = 4 r23 − r13 3π r22 − r12 ! PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 584 PROBLEM 5.39 Determine by direct integration the centroid of the area shown. SOLUTION y =0 First note that symmetry implies xEL - A = dA = Then - = a[ x]0a − and - xEL dA = 1 a( adφ ) 2 2 = a cos φ 3 dA = dA = adx xEL = x - a 0 a 0 adx − α 1 - α 2 a dφ 2 − a2 α [φ ]α = a 2 (1 − α ) 2 x(adx) − α 2 1 ! - α 3 a cos φ "$ 2 a dφ #% 2 − a & x2 ' 1 = a ( ) − a3 [sin φ ]α−α 2 * +0 3 1 2 ! = a3 " − sin α # $2 3 % 1 2 ! xA = xEL dA: x [a 2 (1 − α )] = a3 " − sin α # $2 3 % - or x = 3 − 4sin α a 6(1 − α ) PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 585 ! PROBLEM 5.40 Determine by direct integration the centroid of the area shown. Express your answer in terms of a and b. SOLUTION x = 0, At y =b b = k (0 − a) 2 y= Now xEL = x - A = dA = Then and b ( x − a) 2 dx a2 - a 0 a 1 b b & 3' 2 ( ) − = − = ab x a dx x a ( ) 2 2 * + 0 3 a 3a &b ' b a 3 x ( 2 ( x − a) 2 dx ) = 2 x − 2ax 2 + a 2 x dx 0 *a 0 + a 4 2 b x a 2! 1 2 2 = 2 "" − ax3 + x ## = a b 2 a $ 4 3 % 12 - xEL dA = - yEL dA = = Hence y b = 2 ( x − a)2 2 2a dA = ydx = and b a2 b ( x − a )2 a2 Then yEL = or k = - a - a 0 - ( ) a 2 b &b ' b &1 ' ( x − a) 2 ( 2 ( x − a) 2 dx ) = 4 ( ( x − a)5 ) 2 5 2a *a + 2a * +0 1 2 ab 10 1 ! 1 xA = xEL dA: x " ab # = a 2 b $ 3 % 12 - 1 ! 1 yA = yEL dA: y " ab # = ab 2 $ 3 % 10 - x= y= 1 a 4 3 b 10 ! PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 586 PROBLEM 5.41 Determine by direct integration the centroid of the area shown. Express your answer in terms of a and b. SOLUTION b 2 x a2 b y2 = k2 x 4 but b = k2 a 4 y2 = 4 x 4 a b x4 ! dA = ( y2 − y1 )dx = 2 "" x 2 − 2 ## dx a $ a % xEL = x y1 = k1 x 2 yEL = = but b = k1a 2 y1 = 1 ( y1 + y2 ) 2 b 2a 2 x4 2 "" x + 2 a $ , A = dA = b a2 , a 0 ! ## % x4 2 "" x − 2 a $ ! ## dx % a b & x3 x5 ' = 2 ( − 2) a * 3 5a + 0 2 = ba 15 ,x EL dA ! ## dx % × b x4 2 x − 2 " a "$ a2 b a2 , x5 ! 3 "" x − 2 ## dx 0 a % $ = b a2 & x4 x6 ' − ( 2) * 4 6a + 0 = 1 2 a b 12 = , = a 0 a a PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 587 PROBLEM 5.41 (Continued) , yEL dA = = , a 0 b x4 ! b x4 ! 2 2 + − x x " # " # dx 2a 2 "$ a 2 #% a 2 "$ a 2 #% b2 2a 4 , a 0 x8 4 "" x − 4 a $ ! ## dx % a 2 2 b 2 & x5 x9 ' = 4 ( − 4) = ab 2a * 5 9a + 0 45 2 ! 1 xA = xEL dA: x " ba # = a 2b 15 $ % 12 5 x= a 8 2 ! 2 2 yA = yEL dA: y " ba # = ab ! 15 $ % 45 1 y= b 3 , ! , PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 588 ! PROBLEM 5.42 Determine by direct integration the centroid of the area shown. SOLUTION xEL = x We have ! ## % 2 ! x x dA = ydx = a ""1 − + 2 ## dx L L % $ yEL = 1 a x x2 y = ""1 − + 2 2 2$ L L , A = dA = Then , 2L 0 2L & x x2 ! x2 x3 ' a ""1 − + 2 ## dx = a ( x − + 2) L L % 2 L 3L + 0 $ * 8 = aL 3 and , xEL dA = = , yEL dA = = , 2L 0 & x x2 x ( a ""1 − + 2 L L (* $ 2L ! ' & x 2 x3 x4 ' + 2) ## dx ) = a ( − % )+ * 2 3L 4 L + 0 10 2 aL 3 , 2L 0 a2 2 a x x2 ! & x x2 ! ' ""1 − + 2 ## ( a ""1 − + 2 ## dx ) 2$ L L % *( $ L L % +) , EL 0 x x2 x3 x 4 ! ""1 − 2 + 3 2 − 2 3 + 4 ## dx L L L L % $ 2L a2 & x 2 x3 x4 x5 ' = + 2 − 3 + 4) (x − 2 * L L 2L 5L + 0 11 = a2 L 5 Hence 8 ! 10 xA = xEL dA: x " aL # = aL2 $3 % 3 , 1 ! 11 yA = yEL dA: y " a # = a 2 $8 % 5 , x= y= 5 L 4 33 a 40 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 589 ! PROBLEM 5.43 Determine by direct integration the centroid of the area shown. Express your answer in terms of a and b. SOLUTION x = a, y = b : a = kb2 For y2 at b Then y2 = Now xEL = x and for a 0# x# : 2 yEL = a y2 b x1/ 2 = 2 2 a x1/2 a dx a b x 1 x1/2 ! 1 # x # a : yEL = ( y1 + y2 ) = "" − + # 2 2 2$ a 2 a #% x1/2 dA = ( y2 − y1 )dx = b "" $ Then a b2 x1/2 dA = y2 dx = b For or k = , A = dA = , a/2 0 b a x1/2 a − dx + , x 1! + # dx a 2 #% x1/ 2 x 1 ! b "" − + ## dx a/2 $ a a 2% a a a/ 2 & 2 x3/2 x 2 1 ' b & 2 3/2 ' = + − + x) x b ( ) a (* 3 +0 * 3 a 2a 2 + a/2 = 3/ 2 3/ 2 2 b & a! a! ' (" # + (a)3/ 2 − " # ) 3 a *($ 2 % $ 2 % +) 2 a! ' 1& a ! ' /. /- 1 & 2 (a ) − $ % ! + (a) − $ % ! # + b "− & 2 ' !) 2 ( & 2 ' ) *, *+ 2a ( 13 ab = 24 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 590 PROBLEM 5.43 (Continued) and 1 xEL dA = 1 a/2 0 / x1/ 2 0 x $$ b dx %% + a & ' 1 - / x1/ 2 x 1 0 . x b $$ − + %% ! dx a/2 ( & a a 2 ' )! a a a/ 2 - 2 x5/2 x3 x 4 . b - 2 5/2 . = − + ! x ! +b a (5 )0 ( 5 a 3a 4 ) a/2 5/2 5/2 2 b -/ a 0 /a0 . 5/2 + − ( ) a $ % $ % ! 5 a (& 2 ' & 2 ' )! 3 2 *2 1 - 3 / a 0 . 1 - 2 / a 0 . *3 + b "− ( a) − $ % ! + (a) − $ % ! # 3a ( & 2 ' !) 4 ( & 2 ' !) ,* +* 71 2 a b = 240 = 1 yEL dA = 1 a/2 0 + 1 b x1/ 2 - x1/ 2 . b dx ! 2 a ( a ) b / x 1 x1/2 0 - / x1/ 2 x 1 0 . − + % dx ! $ − + % b$ a/ 2 2 $ a 2 a %' ( $& a a 2 %' )! & a a a/ 2 3 b2 - 1 2 . b 2 -/ x 2 1 / x 1 0 0 . $ = − $ − % %! x ! + 2a ( 2 ) 0 2 $& 2a 3a & a 2 ' %' ! ( ) a/2 = b -/ a 0 /a0 2 $ % + (a ) − $ % 4 a (& 2 ' &2' = 11 2 ab 48 2 Hence 2. b2 / a 1 0 !− $ − % !) 6a & 2 2 ' / 13 0 71 2 xA = xEL dA : x $ ab % = a b & 24 ' 240 1 / 13 0 11 2 yA = yEL dA: y $ ab % = ab & 24 ' 48 1 3 x= 17 a = 0.546a 130 y= 11 b = 0.423b 26 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 591 ! PROBLEM 5.44 Determine by direct integration the centroid of the area shown. Express your answer in terms of a and b. SOLUTION x = a, For y1 at y = 2b 2b = ka 2 or k = 2b a2 2b 2 x a2 Then y1 = By observation b x0 / y2 = − ( x + 2b) = b $ 2 − % a a' & Now xEL = x and for 0 # x # a : yEL = 1 b y1 = 2 x 2 2 a For a # x # 2a : yEL = 1 b/ x0 x0 / y2 = $ 2 − % and dA = y2 dx = b $ 2 − % dx 2 2& a' a' & 1 A = dA = Then 1 a 0 and dA = y1dx = 2b 2 x dx + a2 1 2a a 2b 2 x dx a2 x0 / b $ 2 − % dx a' & 2a a 2 - a/ 2b - x3 . 7 x0 . = 2 ! + b − $ 2 − % ! = ab 6 a ' )! a ( 3 )0 ( 2& 0 and 1x EL dA a / 2b 0 x $ 2 x 2 dx % + &a ' - / x0 . x b $ 2 − % dx ! a ' ) ( & 1 = - 2 x3 . 2b - x 4 . ! +b x − ! 2 3a ) 0 a ( 4 )0 ( 0 1 2a = a a 2a 1 2 1 - 2 2 3 (2a ) − (a )3 .) # a b + b " -((2a) 2 − (a) 2 .) + ( 2 3a + , 7 2 = a b 6 = PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 592 PROBLEM 5.44 (Continued) 1y EL dA = 1 a 0 b 2 - 2b 2 . x x dx ! + a2 ( a2 ) 1 2a 0 b/ x 0- / x0 . $ 2 − % b $ 2 − % dx ! 2& a '( & a' ) 2a a 3 2b 2 - x5 . b2 - a / x0 . = 4 − $2 − % ! ! + a ' )! a ( 5 )0 2 ( 3 & a 17 2 = ab 30 Hence /7 0 7 xA = xEL dA: x $ ab % = a 2b &6 ' 6 1 / 7 0 17 2 yA = yEL dA: y $ ab % = ab & 6 ' 30 1 x =a y= 17 b 35 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 593 ! PROBLEM 5.45 A homogeneous wire is bent into the shape shown. Determine by direct integration the x coordinate of its centroid. SOLUTION First note that because the wire is homogeneous, its center of gravity coincides with the centroid of the corresponding line xEL = a cos3 θ Now and dL = dx 2 + dy 2 x = a cos3 θ : dx = −3a cos 2 θ sin θ dθ Where y = a sin 3 θ : dy = 3a sin 2 θ cos θ dθ dL = [(−3a cos 2 θ sin θ dθ )2 + (3a sin 2 θ cos θ dθ )2 ]1/ 2 Then = 3a cos θ sin θ (cos 2 θ + sin 2 θ )1/ 2 dθ = 3a cos θ sin θ dθ 1 L = dL = and 1x EL dL = 3 a 2 = 1 π /2 0 1 π /2 0 π /2 3a cos θ sin θ dθ = 3a a cos3θ (3a cos θ sin θ dθ ) π /2 - 1 . = 3a 2 − cos5 θ ! ( 5 )0 Hence -1 2 . sin θ ! (2 )0 3 = a2 5 /3 0 3 xL = xEL dL : x $ a % = a 2 &2 ' 5 1 x= 2 a 5 Alternative Solution / x0 x = a cos3 θ 4 cos 2 θ = $ % &a' / y0 y = a sin 3 θ 4 sin 2 θ = $ % &a' /x0 $a% & ' 2/3 / y0 +$ % &a' 2/3 2/3 2/3 = 1 or y = (a 2/3 − x 2/3 )3/2 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 594 PROBLEM 5.45 (Continued) dy = (a 2/3 − x 2/3 )1/ 2 (− x −1/3 ) dx Then Now xEL = x and / dy 0 dL = 1 + $ % & dx ' 2 { dx = 1 + -(( a 2/3 − x 2/3 )1/2 (− x −1/3 ) .) 1 L = dL = Then and Hence 1 xEL dL = 1 a 0 1 a 0 2 1/2 } dx a 3 a1/3 - 3 2/3 . dx = a1/3 x ! = a 1/3 x (2 )0 2 a / a1/3 0 3 -3 . x $$ 1/3 dx %% = a1/3 x5/3 ! = a 2 (5 )0 5 &x ' /3 0 3 xL = xEL dL : x $ a % = a 2 &2 ' 5 1 x= 2 a 5 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 595 PROBLEM 5.46 A homogeneous wire is bent into the shape shown. Determine by direct integration the x coordinate of its centroid. SOLUTION First note that because the wire is homogeneous, its center of gravity coincides with the centroid of the corresponding line Now xEL = r cos θ Then L = dL = and 1 1 xEL dL = 7π / 4 1π /4 and dL = rdθ 7π /4 1π /4 3 rdθ = r[θ ]π7π/ 4/ 4 = π r 2 r cos θ ( rdθ ) = r 2 [sin θ ]π7π/4/ 4 / 1 1 0 = r2 $ − − % 2 2' & = −r 2 2 Thus /3 0 xL = xdL : x $ π r % = −r 2 2 &2 ' 1 x =− 2 2 r 3π PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 596 PROBLEM 5.47* A homogeneous wire is bent into the shape shown. Determine by direct integration the x coordinate of its centroid. Express your answer in terms of a. SOLUTION First note that because the wire is homogeneous, its center of gravity will coincide with the centroid of the corresponding line. x = a, We have at y=a a = ka 3/2 1 y= Then a or k = 1 a x3/ 2 dy 3 1/2 = x dx 2 a and Now xEL = x and / dy 0 dL = 1 + $ % dx & dx ' 2 1/2 2 - / 3 0 . x1/2 % ! = 1+ $ ' !) ( &2 a 1 4a + 9 x dx = 2 a 1 L = dL = Then 1 a 1 0 2 a dx 4a + 9 x dx a 1 -2 1 . × (4a + 9 x)3/ 2 ! 3 9 2 a( )0 a = [(13)3/2 − 8] 27 = 1.43971a = and 1x EL dL = 1 a 0 x - 1 (2 a . 4a + 9 x dx ! ) PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 597 PROBLEM 5.47* (Continued) Use integration by parts with u=x du = dx Then 1 xEL dL = dv = 4a + 9 x dx 2 (4a + 9 x)3/ 2 v= 27 a 1 2* 2 3/2 . x a x × + (4 9 ) " ! − 2 a *+ ( 27 )0 1 a 0 3* 2 (4a + 9 x)3/ 2 dx # 27 *, a = (13)3/2 2 1 -2 . a − (4a + 9 x)5/2 ! 27 45 27 a ( )0 = a2 2 2 3 3/ 2 5/2 "(13) − [(13) − 32]# 27 + 45 , = 0.78566a 2 1 xL = xEL dL : x (1.43971a ) = 0.78566a 2 or x = 0.546a PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 598 PROBLEM 5.48* Determine by direct integration the centroid of the area shown. SOLUTION We have xEL = x 1 πx a yEL = y = cos 2 2 2L and dA = ydx = a cos 1 A = dA = Then 1 πx dx 2L L/2 a cos 0 πx 2L dx L/ 2 πx. - 2L sin =a 2 L !) 0 (π 2 = and 1x EL dA π aL πx 0 / = x $ a cos dx 2 L %' & 1 u=x Use integration by parts with dv = cos du = dx Then πx 1 x cos 2L dx = 1 2L v= × sin 2L π πx − πx dx 2L sin πx 2L 2L 1π πx dx 2L 2L π x 2L πx0 2L / cos x sin = + π $& 2L π 2 L %' π xEL dA = a =a 2L - π ( x sin πx 2L + sin 2L π cos π x. L/ 2 2 L !) 0 2 L -/ L 2 0 2L . L %− + $$ % π ! π (& 2 2 π ' )! = 0.106374aL2 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 599 PROBLEM 5.48* (Continued) Also 1y EL dA = 1 L/ 2 0 a2 = 2 πx/ πx 0 a dx % cos $ a cos 2 2L & 2L ' -x L/ 2 sin π x . + 2π L ! L (2 )! 0 = a −2 / L L 0 $ + % 2 & 4 2π ' = 0.20458a 2L / 2 0 xA = xEL dA: x $ aL % = 0.106374aL2 $ π % & ' or x = 0.236 L / 2 0 yA = yEL dA: y $ aL % = 0.20458a 2L $ π % & ' or y = 0.454a 1 1 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 600 PROBLEM 5.49* Determine by direct integration the centroid of the area shown. SOLUTION 2 2 r cos θ = aeθ cos θ 3 3 2 2 θ = r sin θ = ae sin θ 3 3 xEL = We have yEL dA = and 1 1 ( r )(rdθ ) = a 2 e2θ dθ 2 2 1 A = dA = Then 1 π 0 π 1 2 2θ 1 -1 . a e dθ = a 2 e 2θ ! 2 2 (2 )0 1 2 2π a (e − 1) 4 = 133.623a 2 = and 1x EL dA = 1 π 0 2 θ /1 0 ae cos θ $ a 2 e2θ dθ % 3 &2 ' 1 = a3 3 1 π 0 e3θ cos θ dθ To proceed, use integration by parts, with u = e3θ du = 3e3θ dθ and dv = cos θ dθ Then 1e 3θ 1 then du = 3e3θ dθ dv = sin θ dθ , then Then 1e 3θ v = sin θ cos θ dθ = e3θ sin θ − sin θ (3e3θ dθ ) u = e3θ Now let and v = − cos θ sin θ dθ = e3θ sin θ − 3 - −e3θ cos θ − (− cos θ )(3e3θ dθ ) .! ( ) 1 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 601 PROBLEM 5.49* (Continued) So that 1 e3θ cos θ dθ = e3θ (sin θ + 3cos θ ) 10 π 1 . 1 - e3θ (sin θ + 3cos θ ) ! xEL dA = a 3 3 ( 10 )0 1y Also EL dA = a3 (−3e3π − 3) = −1239.26a3 30 = 1 π 0 2 θ /1 0 ae sin θ $ a 2 e 2θ dθ % 3 &2 ' 1 = a3 3 1 π 0 e3θ sin θ dθ Using integration by parts, as above, with u = e3θ and 1 dv = sin θ dθ Then 1e So that 1 3θ du = 3e3θ dθ and v = − cos θ 1 sin θ dθ = −e3θ cos θ − (− cos θ )(3e3θ dθ ) e3θ sin θ dθ = e3θ ( − cos θ + 3sin θ ) 10 π 1 . 1 - e3θ (− cos θ + 3sin θ ) ! yEL dA = a3 3 ( 10 )0 = Hence a 3 3π (e + 1) = 413.09a3 30 1 or x = −9.27a 1 or xA = xEL dA: x (133.623a 2 ) = −1239.26a 3 yA = yEL dA: y (133.623a 2 ) = 413.09a3 y = 3.09a PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 602 PROBLEM 5.50 Determine the centroid of the area shown when a = 2 in. SOLUTION xEL = x We have yEL = / 10 dA = ydx = $1 − % dx x' & and 1 A = dA = Then and 1 1/ 10 y = $1 − % 2 2& x' = 1 a EL dA = 1 a 1x 1y EL dA 1 1 1 a/ 1 1 0 dx 2 a $1 − x % 2 = [ x − ln x]1 = (a − ln a − 1) in. & ' a . / a2 -/ 1 0 . - x 2 10 − x ! = $$ − a + %% in.3 x $1 − % dx ! = 2' x' ) ( 2 (& )1 & 2 1 / 1 0 -/ 1 0 . 1 1 − % $1 − % dx ! = 2 $& x ' (& x' ) 2 1 a/ 1 2 1 0 $ 1 − x + 2 % dx x ' & a = 11. 1/ 10 x − 2ln x − ! = $ a − 2ln a − % in.3 2( x )1 2 & a' 1 xA = xEL dA: x = 1 yA = yEL dA: y = a2 2 −a+ 1 2 in. a − ln a − 1 a − 2ln a − 1a 2( a − ln a − 1) in. Find: x and y when a = 2 in. We have x= and y= 1 2 (2) 2 − 2 + 1 2 2 − ln 2 − 1 2 − 2ln 2 − 12 2(2 − ln 2 − 1) or x = 1.629 in. or y = 0.1853 in. PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 603 PROBLEM 5.51 Determine the value of a for which the ratio x / y is 9. SOLUTION xEL = x We have yEL = 1 1/ 10 y = $1 − % 2 2& x' / 10 dA = ydx = $1 − % dx x' & and 1 A = dA = Then 1 a/ 1 1 0 dx = [ x − ln x]1a $1 − % x' 2 & = (a − ln a − 1) in.2 and 1 xEL dA = 1 a 1 a . -/ 1 0 . - x 2 − x! x $1 − % dx ! = x' ) ( 2 (& )1 / a2 10 = $$ − a + %% in.3 2' & 2 1 yEL dA = 1 a 1 1 / 1 0 -/ 1 0 . 1 1 − % $1 − % dx ! = x ' (& x' ) 2 2 $& 1 a/ 1 2 1 0 $ 1 − x + 2 % dx x ' & a = 11. x − 2ln x − ! 2( x )1 = 1/ 10 a − 2ln a − % in.3 $ 2& a' 1 xA = xEL dA: x = 1 yA = yEL dA: y = a2 2 −a+ 1 2 in. a − ln a − 1 a − 2ln a − 1a 2( a − ln a − 1) in. PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 604 PROBLEM 5.51 (Continued) Find: a so that x =9 y x xA = = y yA We have Then or 1 2 1 2 a2 − a + 1 2 ( a − 2 ln a − 1a ) 1x 1y EL dA EL dA =9 a 3 − 11a 2 + a + 18a ln a + 9 = 0 Using trial and error or numerical methods and ignoring the trivial solution a = 1 in., find a = 1.901 in. and a = 3.74 in. PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 605 PROBLEM 5.52 Determine the volume and the surface area of the solid obtained by rotating the area of Problem 5.1 about (a) the line x = 240 mm, (b) the y axis. PROBLEM 5.1 Locate the centroid of the plane area shown. SOLUTION From the solution to Problem 5.1 we have A = 15.3 × 103 mm 2 Σ xA = 2.6865 × 106 mm3 Σ yA = 1.4445 × 106 mm3 Applying the theorems of Pappus-Guldinus we have (a) Rotation about the line x = 240 mm Volume = 2π (240 − x ) A = 2π (240 A − Σ xA) = 2π [240(15.3 × 103 ) − 2.6865 × 106 ] Volume = 6.19 × 106 mm3 Area = 2π X line L = 2π Σ( xline ) L = 2π ( x1 L1 + x3 L3 + x4 L4 + x5 L5 + x6 L6 ) Where x1 , , x6 are measured with respect to line x = 240 mm. Area = 2π [(120)(240) + (15)(30) + (30)(270) + (135)(210) + (240)(30)] (b) Area = 458 × 103 mm 2 Rotation about the y axis Volume = 2π X area A = 2π (Σ xA) = 2π (2.6865 × 106 mm3 ) Volume = 16.88 × 106 mm3 Area = 2π X line L = 2π Σ( xline ) L = 2π ( x1 L1 + x2 L2 + x3 L3 + x4 L4 + x5 L5 ) = 2π [(120)(240) + (240)(300) + (225)(30) + (210)(270) + (105)(210)] Area = 1.171 × 106 mm 2 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companie

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