Solucionario de Mecanica vectorial para ingenieros Beer Johnston Cap 2-5-estatica

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Solucionario de Mecanica vectorial para ingenieros Beer Johnston Cap 2-5-estatica

  1. 1. 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° W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
  2. 2. 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° W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
  3. 3. 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) α = 51.3° β = 59.0° Triangle rule: We measure: γ = 67.0° R = 139.1 lb, R = 139.1 lb 67.0° W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
  4. 4. 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° W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
  5. 5. 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° W Fbb′ = 336 lb W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
  6. 6. 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° W α + β + 60° = 180° β = 180° − 60° − 20.268° = 99.732° Faa′ 300 lb = sin 99.732° sin 60° Faa′ = 341 lb W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
  7. 7. 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° W α + β + 25° = 180° β = 180° − 25° − 37.138° = 117.86° R 35 N = sin117.86 sin 25° R = 73.2 N W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
  8. 8. 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 W α + 20° + 35° = 180° α = 180° − 20° − 35° = 125° R 75 N = sin125° sin 35° R = 107.1 N W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
  9. 9. 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 W 25° + β + 75° = 180° β = 180° − 25° − 75° = 80° 1600 N R = sin 25° sin 80° R = 3730 N W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
  10. 10. 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° W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
  11. 11. 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° 425 lb P = sin 70° sin 60° (b) P = 392 lb W 425 lb R = sin 70° sin 50° R = 346 lb W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
  12. 12. 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 = 500 lb 90° − α = 47.40° (b) α = 42.6° W R 500 lb = sin (42.6° + 30°) sin 60° R = 551 lb W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
  13. 13. 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) P = 21.1 N R = (50 N) cos 25° W R = 45.3 N W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
  14. 14. 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) P = 368 lb R = (425 lb)sin 30° W R = 213 lb W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
  15. 15. 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° W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
  16. 16. 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° W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
  17. 17. 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° W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
  18. 18. 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 ¹ = 0.39737 α = 23.414 φ = α + 50° = 73.414 R = 21.8 kN 73.4° W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
  19. 19. 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 ¹ = 0.59605 α = 36.588° φ = α + 50° = 86.588° R = 21.8 kN 86.6° W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
  20. 20. 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° W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
  21. 21. 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 Fx = +640 N W 600 1000 Fy = +480 N W Fx = −(424 N) 560 1060 Fx = −224 N W 900 1060 Fy = −360 N W Fx = + (408 N) 480 1020 Fx = +192.0 N W Fy = −(408 N) 408-N Force: 800 1000 Fy = −(424 N) 424-N Force: Fx = + (800 N) Fy = +(800 N) 800-N Force: 900 1020 Fy = −360 N W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
  22. 22. 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. Fx = +21.0 lb W 80 116 Fy = +20.0 lb W Fx = −(50 lb) 28 100 Fx = −14.00 lb W 96 100 Fy = + 48.0 lb W Fx = + (51 lb) 48 102 Fx = +24.0 lb W Fy = −(51 lb) 51-lb Force: 84 116 Fy = +(50 lb) 50-lb Force: Fx = + (29 lb) Fy = +(29 lb) 29-lb Force: 90 102 Fy = −45.0 lb W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
  23. 23. PROBLEM 2.23 Determine the x and y components of each of the forces shown. SOLUTION Fy = −34.6 lb W Fx = −(50 lb)sin 50° Fx = −38.3 lb W Fy = −32.1 lb W Fx = + (60 lb) cos 25° Fx = 54.4 lb W Fy = +(60 lb)sin 25° 60-lb Force: Fx = 20.0 lb W Fy = −(50 lb) cos 50° 50-lb Force: Fx = + (40 lb) cos 60° Fy = −(40 lb)sin 60° 40-lb Force: Fy = 25.4 lb W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
  24. 24. PROBLEM 2.24 Determine the x and y components of each of the forces shown. SOLUTION Fy = 51.4 N W Fx = + (120 N) cos 70° Fx = 41.0 N W Fy = 112.8 N W Fx = −(150 N) cos 35° Fx = −122. 9 N W Fy = +(150 N) sin 35° 150-N Force: Fx = 61.3 N W Fy = +(120 N) sin 70° 120-N Force: Fx = + (80 N) cos 40° Fy = + (80 N) sin 40° 80-N Force: Fy = 86.0 N W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
  25. 25. 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 W Pv = P cos 35° = (523 lb) cos 35° Pv = 428 lb W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
  26. 26. 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 W PABC = P cos 20° = (2193 N) cos 20° PABC = 2060 N W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
  27. 27. 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 W or Py = 153.6 N W 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
  28. 28. 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 W Px = Py tan 38° = (180 N) tan 38° = 140.63 N Px = 140.6 N W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
  29. 29. 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 W Px = Py tan 55° Px tan 55° 1200 N = tan 55° = 840.2 N Py = Py = 840 N W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
  30. 30. 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 W Px = P sin 55° = (610.2 lb) sin 55° = 499.8 lb Px = 500 lb W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
  31. 31. 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° W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
  32. 32. 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° W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
  33. 33. 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° W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
  34. 34. 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° W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
  35. 35. 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° W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
  36. 36. 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° W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
  37. 37. 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° W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
  38. 38. 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° W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
  39. 39. 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° W Substituting for α in Eq. (2): Ry = −300sin 21.74° − 150sin 51.74° = −228.9 N R = | Ry | = 228.9 N R = 229 N W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
  40. 40. 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 W Substituting for TBC in Eq. (2): 20 (580 N) − 700 N 29 Ry = −300 N Ry = R = | Ry | = 300 N R = 300 N W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
  41. 41. 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 W Substituting for TAC in Eq. (1): Rx = (95.07 lb) sin10° + 78.46 lb = 94.97 lb R = Rx R = 95.0 lb W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
  42. 42. 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° W 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 W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
  43. 43. 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 W (b) TBC = 1962 N sin 50° = 1735.49 N sin 60° TBC = 1.735 kN W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
  44. 44. 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 W (b) TBC = 500 N sin 40° = 326.35 N sin 80° TBC = 326 N W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
  45. 45. 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 W (b) TBC = 500 N sin 75° sin 70° TBC = 514 N W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
  46. 46. 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 W (b) TBC = (1962 N) sin 105° sin 60° TBC = 2190 N W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
  47. 47. 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 W (b) TBC = 1200 lb sin 5° sin 65° TBC = 115.4 lb W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
  48. 48. 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 W (b) TBC = 300 lb sin 50° sin 95° TBC = 231 lb W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
  49. 49. 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 W (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 W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
  50. 50. 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 W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
  51. 51. 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 W 3 3 FB − FA = 0 5 5 3 3 FD = (16 kN) − (8 kN) 5 5 FD = 4.80 kN W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
  52. 52. 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 = 5ª 3 º 6 kN + (5 kN) » 3« 5 ¬ ¼ ΣFx = 0: − FC + FB = 15.00 kN W 4 4 FB − FA = 0 5 5 4 ( FB − FA ) 5 4 = (15 kN − 5 kN) 5 FC = FC = 8.00 kN W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
  53. 53. 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 W P = 75 lb TCB = 75 lb − (60 lb)(0.50) or TCB = 45.0 lb W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
  54. 54. 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 Յ 60 lb: Requirement Q sin 60° Յ 60 lb From Eq. (2): Q Յ 69.3 lb TBC Յ 60 lb: Requirement From Eq. (1): 75 lb − Q sin 60° Յ 60 lb Q Ն 30.0 lb 30.0 lb Յ Q Յ 69.3 lb W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
  55. 55. 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 W TCD = 166.3 N W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
  56. 56. 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 W (b) TACB = 1216 N W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
  57. 57. 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 W sin β sin (25° + 45°) = 600 N 784.02 N β = 46.0° α = 46.0° + 25° = 71.0° W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
  58. 58. 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° Į = 5.00° TBC = (1200 lb) sin 5° W TBC = 104.6 lb W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
  59. 59. 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° W TBC = (300 lb)sin 50° (b) or TBC = 229.81 lb TBC = 230 lb W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
  60. 60. 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: BC 2.1 m = 870 N 630 N BC = 2.90 m L = 2( BC ) = 5.80 m L = 5.80 m W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
  61. 61. 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 W Since P Ͼ 0, the solution is correct. (b) α = 180° − 50° − 47.5° = 82.5° α = 82.5° W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
  62. 62. 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 Ͼ 1200 N, the solution is correct. P = 1294 N W (b) Law of sines: sin β sin 85° = 1200 N 1294 N β = 67.5° α = 180° − 50° − 67.5° α = 62.5° W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
  63. 63. 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 W Force Triangle P 50 lb = 15 25 P = 30.0 lb W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
  64. 64. 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. W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
  65. 65. 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° W α = −46.840 For ΣFy = 0: 2P cos 20° + P sin( −46.840°) − 1569.60 N = 0 or P = 1365 N 46.8° W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
  66. 66. 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° W (b) ΣFy = 0: P sin 40° + 2 P cos 22.52° − 1569.60 N = 0 P = 630 N W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
  67. 67. 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 W (b) ΣFy = 0: 2T − (600 lb) = 0 T= 1 (600 lb) 2 T = 300 lb W (c) ΣFy = 0: 3T − (600 lb) = 0 1 T = (600 lb) 3 T = 200 lb W (d) ΣFy = 0: 3T − (600 lb) = 0 1 T = (600 lb) 3 T = 200 lb W (e) ΣFy = 0: 4T − (600 lb) = 0 T= 1 (600 lb) 4 T = 150.0 lb W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
  68. 68. 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 W (d) ΣFy = 0: 4T − (600 lb) = 0 T= 1 (600 lb) 4 T = 150.0 lb W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
  69. 69. 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 W 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 W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
  70. 70. 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 W (b) P = 0.58010(1048.37 N) = 608.16 N Using (1), P = 608 N W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
  71. 71. 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 W θ x = 58.7° W θ y = 35.0° W θ z = 76.0° W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
  72. 72. 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 W θ x = 98.3° W +816 N 900 N θ y = 25.0° W Fz +357 N = 900 N F θ z = 66.6° W 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
  73. 73. 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 W Fx 110.3 N = = 0.38303 F 287.97 N θ x = 67.5° W Fy = F cos 30° = 249.39 cos θ y = Fy F = 249.39 N = 0.86603 287.97 N θ y = 30.0° W 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° W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
  74. 74. 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 W 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° W Fy = F cos 30° = 86.603 N cos θ y = Fy F = 86.603 N = 0.86603 100 N θ y = 30.0° W Fz = −32.14 N cos θ z = Fz −32.14 N = = −0.32140 F 100 N θ z = 108.7° W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
  75. 75. 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) Fx = −(60 lb) sin 30°cos 60° = −15 lb Fx = −15.00 lb W Fy = (60 lb) cos 30° = 51.96 lb Fz = (60 lb) sin 30° sin 60° = 25.98 lb (b) Fy = +52.0 lb W Fz = +26.0 lb W cos θ x = cos θ y = cos θ z = Fx −15.0 lb = = −0.25 60 lb F Fy θ x = 104.5° W 51.96 lb = 0.866 60 lb θ y = 30.0° W Fz 25.98 lb = = 0.433 F 60 lb θ z = 64.3° W 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
  76. 76. 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 W Fx −20 lb = = − 0.25 80 lb F θ x = 104.5° W Fy = (80 lb) cos 30° = 69.282 lb cos θ y = Fy F = 69.282 lb = 0.86615 80 lb θ y = 30.0° W Fz = (80 lb)sin 30° sin 60° = 34.641 lb cos θ z = Fz 34.641 = = 0.43301 80 F θ z = 64.3° W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
  77. 77. 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 W Fy = −(120 lb)sin 60° Fy = −103.923 lb Fy = −103.9 lb W 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 W 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° W θ y = 150.0° W θ z = 99.8° W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
  78. 78. 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 W Fy = −(85 lb) cos 36° = −68.766 lb Fy = −68.8 lb W Fz = (85 lb)sin 36° cos 48° Fz = 33.4 lb W = 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° W θ y = 144.0° W θ z = 66.8° W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
  79. 79. 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 W θ x = 55.8° W 400 N 570 N θ y = 45.4° W Fz −250 N = F 570 N θ z = 116.0° W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
  80. 80. 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 W θ x = 71.8° W −270 N 770 N Fz 680 N = F 770 N θ y = 110.5° W θ z = 28.0° W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
  81. 81. 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 Ÿ (cos θ y ) 2 = 1 − (cos θ y ) 2 − (cos θ z ) 2 Since Fz Ͻ 0 we must have cos θ z Ͻ 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° W Then: F= 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 W Fy = F cos θ y = (109.978 lb) cos144.9° and Fy = −90.0 lb W F = 110.0 lb W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
  82. 82. 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 Ÿ (cos θ y ) 2 = 1 − (cos θ y ) 2 − (cos θ z ) 2 Since Fx Ͻ 0 we must have cos θ x Ͻ 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° W Then: Fx 500 lb = = 1209.10 lb cos θ x 0.41353 F = 1209 lb W Fy = F cos θ y = (1209.10 lb) cos 55° Fy = 694 lb W Fz = F cos θ z = (1209.10 lb) cos 45° Fz = 855 lb W 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
  83. 83. 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 W cos θ x = cos θ y = Fy = −62.0 lb W Fx 80 N = = 0.38095 F 210 N Fy F = 61.929 N = −0.29490 210 N θ x = 67.6° W θ y = 107.2 W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
  84. 84. 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 W 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 W 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° W θ z = 62.0° W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
  85. 85. 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 JJJ G BA = (20 ft)i + (100 ft) j − (25 ft)k BA = (20 ft) 2 + (100 ft) 2 + (−25 ft)2 = 105 ft F = F Ȝ BA JJJ G 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 W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
  86. 86. 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 JJJ G DA = (20 ft)i + (100 ft) j + (70 ft)k DA = (20 ft) 2 + (100 ft) 2 + ( +70 ft) 2 = 126 ft F = F Ȝ DA JJJ G 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 W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
  87. 87. 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 JJJ G DB = (480 mm)i − (510 mm) j + (320 mm)k DB = (480 mm) 2 + (510 mm 2 ) + (320 mm) 2 = 770 mm F = F Ȝ DB JJJ G 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 W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
  88. 88. 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 JJJ G EB = (270 mm)i − (400 mm) j + (600 mm)k EB = (270 mm) 2 + (400 mm) 2 + (600 mm) 2 = 770 mm F = F Ȝ EB JJJ G 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 W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
  89. 89. 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 JJJ G BA = −(900 mm)i + (600 mm) j + (360 mm)k BA = (900 mm) 2 + (600 mm) 2 + (360 mm) 2 = 1140 mm TBA = TBA Ȝ BA JJJ G BA = TBA BA 1425 N TBA = [ −(900 mm)i + (600 mm) j + (360 mm)k ] 1140 mm = −(1125 N)i + (750 N) j + (450 N)k (TBA ) x = −1125 N, (TBA ) y = 750 N, (TBA ) z = 450 N W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
  90. 90. 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 JJJ G 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 JJJ G 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 W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
  91. 91. 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 W Rx 174.367 N = = 0.33853 515.07 N R Ry θ x = 70.2° W 456.42 N = 0.88613 515.07 N θ y = 27.6° W Rz 163.011 N = = 0.31648 R 515.07 N θ z = 71.5° W 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
  92. 92. 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 W Rx 91.532 N = = 0.177708 R 515.07 N Ry θ x = 79.8° W 429.81 N = 0.83447 515.07 N θ y = 33.4° W Rz 268.66 N = = 0.52160 R 515.07 N θ z = 58.6° W 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
  93. 93. 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 JJJ G AB = (40 in.)i − (45 in.) j + (60 in.)k AB = (40 in.) 2 + (45 in.) 2 + (60 in.) 2 = 85 in. JJJG AC = (100 in.)i − (45 in.) j + (60 in.)k AC = (100 in.) 2 + (45 in.) 2 + (60 in.) 2 = 125 in. JJJ G ª (40 in.)i − (45 in.) j + (60 in.)k º AB = (425 lb) « TAB = TAB ȜAB = TAB » 85 in. AB ¬ ¼ TAB = (200 lb)i − (225 lb) j + (300 lb)k JJJG ª (100 in.)i − (45 in.) j + (60 in.)k º AC = (510 lb) « TAC = TAC ȜAC = TAC » AC 125 in. ¬ ¼ 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: R = 912.92 lb R = 913 lb W 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° W cos θ z = and 544.8 lb = 0.59677 912.92 lb θ z = 53.4° W θ x = 48.2° W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
  94. 94. 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 JJJ G AB = (40 in.)i − (45 in.) j + (60 in.)k AB = (40 in.) 2 + (45 in.) 2 + (60 in.) 2 = 85 in. JJJG AC = (100 in.)i − (45 in.) j + (60 in.)k AC = (100 in.) 2 + (45 in.) 2 + (60 in.) 2 = 125 in. JJJ G ª (40 in.)i − (45 in.) j + (60 in.)k º AB = (510 lb) « TAB = TAB ȜAB = TAB » 85 in. AB ¬ ¼ TAB = (240 lb)i − (270 lb) j + (360 lb)k JJJG ª (100 in.)i − (45 in.) j + (60 in.)k º AC = (425 lb) « TAC = TAC ȜAC = TAC » AC 125 in. ¬ ¼ TAC = (340 lb)i − (153 lb) j + (204 lb)k R = TAB + TAC = (580 lb)i − (423 lb) j + (564 lb)k Then: R = 912.92 lb R = 913 lb W cos θ x = 580 lb = 0.63532 912.92 lb cos θ y = −423 lb = −0.46335 912.92 lb cos θ z = and 564 lb = 0.61780 912.92 lb θ x = 50.6° W θ y = 117.6° W θ z = 51.8° W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
  95. 95. 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 JJJ G BD = −(480 mm)i + (510 mm) j − (320 mm)k BD = (480 mm) 2 + (510 mm) 2 + (320 mm) 2 = 770 mm JJJ G BD FBD = TBD Ȝ 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 JJJ G BE = −(270 mm)i + (400 mm) j − (600 mm)k BE = (270 mm) 2 + (400 mm) 2 + (600 mm) 2 = 770 mm JJJ G BE FBE = TBE Ȝ 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 W cos θ x = −375 N 747.83 N θ x = 120.1° W cos θ y = 455 N 747.83 N θ y = 52.5° W cos θ z = −460 N 747.83 N θ z = 128.0° W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
  96. 96. 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 2 2 2 R = Rx + Ry + Rz = (+2475) 2 + (−1650) 2 + ( +930)2 = 3116.6 N cos θ x = cos θ y = cos θ z = R = 3120 N W Rx +2475 = R 3116.6 Ry θ x = 37.4° W −1650 3116.6 θ y = 122.0° W Rz + 930 = R 3116.6 θ z = 72.6° W 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
  97. 97. 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 W 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 W cos θ x = 98.966 lb 207.76 lb θ x = 61.6° W cos θ y = 182.668 lb 207.76 lb θ y = 151.6° W cos θ z = 0 θ z = 90.0° W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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

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