1. 1. SOLUTIONMANUAL
2. 2. CHAPTER 2
3. 3. PROBLEM 2.1Two forces are applied at point B of beam AB. Determinegraphically the magnitude and direction of their resultant using(a) the parallelogram law, (b) the triangle rule.PROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.3SOLUTION(a) Parallelogram law:(b) Triangle rule:We measure: R = 3.30 kN, = 66.6 R = 3.30 kN 66.6
4. 4. PROBLEM 2.2The cable stays AB and AD help support pole AC. Knowingthat the tension is 120 lb in AB and 40 lb in AD, determinegraphically the magnitude and direction of the resultant of theforces exerted by the stays at A using (a) the parallelogramlaw, (b) the triangle rule.PROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.4SOLUTIONWe measure: 51.359.0= = (a) Parallelogram law:(b) Triangle rule:We measure: R = 139.1 lb, = 67.0 R =139.1 lb 67.0
5. 5. PROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.5PROBLEM 2.3Two structural members B and C are bolted to bracket A. Knowing that bothmembers are in tension and that P = 10 kN and Q = 15 kN, determinegraphically the magnitude and direction of the resultant force exerted on thebracket using (a) the parallelogram law, (b) the triangle rule.SOLUTION(a) Parallelogram law:(b) Triangle rule:We measure: R = 20.1 kN, = 21.2 R = 20.1 kN 21.2
6. 6. PROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.6PROBLEM 2.4Two structural members B and C are bolted to bracket A. Knowing thatboth members are in tension and that P = 6 kips and Q = 4 kips, determinegraphically the magnitude and direction of the resultant force exerted onthe bracket using (a) the parallelogram law, (b) the triangle rule.SOLUTION(a) Parallelogram law:(b) Triangle rule:We measure: R = 8.03 kips, = 3.8 R = 8.03 kips 3.8
7. 7. + + = RPROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.7PROBLEM 2.5A stake is being pulled out of the ground by means of two ropes as shown.Knowing that = 30, determine by trigonometry (a) the magnitude of theforce P so that the resultant force exerted on the stake is vertical, (b) thecorresponding magnitude of the resultant.SOLUTIONUsing the triangle rule and the law of sines:(a)120 Nsin 30 sin 25P = P = 101.4 N (b) 30 25 180180 25 30125= = 120 Nsin 30 sin125= R = 196.6 N
8. 8. PROBLEM 2.6A trolley that moves along a horizontal beam is acted upon by twoforces as shown. (a) Knowing that = 25, determine by trigonometrythe magnitude of the force P so that the resultant force exerted on thetrolley is vertical. (b) What is the corresponding magnitude of theresultant? + + = PROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.8SOLUTIONUsing the triangle rule and the law of sines:(a)1600 N=P sin 25 sin 75P = 3660 N (b) 25 75 180180 25 7580= = 1600 N=R sin 25 sin80R = 3730 N
9. 9. PROBLEM 2.7A trolley that moves along a horizontal beam is acted upon by two forcesas shown. Determine by trigonometry the magnitude and direction of theforce P so that the resultant is a vertical force of 2500 N.= = PROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.9SOLUTIONUsing the law of cosines: 2 (1600 N)2 (2500 N)2 2(1600 N)(2500 N) cos 752596 NPP= + =Using the law of sines:sin sin 751600 N 2596 N36.5P is directed 90 36.5 or 53.5 below the horizontal. P = 2600 N 53.5
10. 10. + + = PROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.10PROBLEM 2.8A telephone cable is clamped at A to the pole AB. Knowing that the tensionin the left-hand portion of the cable is T1 = 800 lb, determine bytrigonometry (a) the required tension T2 in the right-hand portion if theresultant R of the forces exerted by the cable at A is to be vertical, (b) thecorresponding magnitude of R.SOLUTIONUsing the triangle rule and the law of sines:(a) 75 40 180180 75 4065= = 800 lb 2sin 65 sin 75T = 2 T = 853 lb (b)800 lbsin 65 sin 40R = R = 567 lb
11. 11. PROBLEM 2.9A telephone cable is clamped at A to the pole AB. Knowing that thetension in the right-hand portion of the cable is T2 = 1000 lb, determineby trigonometry (a) the required tension T1 in the left-hand portion ifthe resultant R of the forces exerted by the cable at A is to be vertical,(b) the corresponding magnitude of R.PROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.11SOLUTIONUsing the triangle rule and the law of sines:(a) 75 40 180180 75 4065 + + = = = 1000 lb =T 1sin 75 sin 651 T = 938 lb (b)1000 lb=R sin 75 sin 40R = 665 lb
12. 12. PROBLEM 2.10Two forces are applied as shown to a hook support. Knowing that themagnitude of P is 35 N, determine by trigonometry (a) the required angle if the resultant R of the two forces applied to the support is to behorizontal, (b) the corresponding magnitude of R.= = = 37.138 = 37.1 + + = R = PROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.12SOLUTIONUsing the triangle rule and law of sines:(a)sin sin 2550 N 35 Nsin 0.60374(b) 25 180180 25 37.138117.862 = = 35 Nsin117.862 sin 25R = 73.2 N
13. 13. PROBLEM 2.11A steel tank is to be positioned in an excavation. Knowing that = 20, determine by trigonometry (a) the required magnitudeof the force P if the resultant R of the two forces applied at A isto be vertical, (b) the corresponding magnitude of R.+ + = PROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.13SOLUTIONUsing the triangle rule and the law of sines:(a) 50 60 180180 50 6070= = 425 lbsin 70 sin 60P = P = 392 lb (b)425 lbsin 70 sin 50R = R = 346 lb
14. 14. PROBLEM 2.12A steel tank is to be positioned in an excavation. Knowing thatthe magnitude of P is 500 lb, determine by trigonometry (a) therequired angle if the resultant R of the two forces applied at Ais to be vertical, (b) the corresponding magnitude of R.+ + + = = + = = 90 = 47.402 = 42.6 R = + PROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.14SOLUTIONUsing the triangle rule and the law of sines:(a) ( 30 ) 60 180180 ( 30 ) 6090sin (90 ) sin 60425 lb 500 lb(b)500 lbsin (42.598 30 ) sin 60R = 551 lb
15. 15. PROBLEM 2.13A steel tank is to be positioned in an excavation. Determine bytrigonometry (a) the magnitude and direction of the smallestforce P for which the resultant R of the two forces applied at Ais vertical, (b) the corresponding magnitude of R.PROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.15SOLUTIONThe smallest force P will be perpendicular to R.(a) P = (425 lb) cos30 P = 368 lb (b) R = (425 lb)sin 30 R = 213 lb
16. 16. PROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.16PROBLEM 2.14For the hook support of Prob. 2.10, determine by trigonometry (a) themagnitude and direction of the smallest force P for which the resultant R ofthe two forces applied to the support is horizontal, (b) the correspondingmagnitude of R.SOLUTIONThe smallest force P will be perpendicular to R.(a) P = (50 N)sin 25 P = 21.1 N (b) R = (50 N) cos 25 R = 45.3 N
17. 17. PROBLEM 2.15Solve Problem 2.2 by trigonometry.PROBLEM 2.2 The cable stays AB and AD help support poleAC. Knowing that the tension is 120 lb in AB and 40 lb in AD,determine graphically the magnitude and direction of theresultant of the forces exerted by the stays at A using (a) theparallelogram law, (b) the triangle rule.== == + + = + + = = = = += + = R = 139.1 lb 67.0 PROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.17SOLUTION8tan1038.666tan1030.96Using the triangle rule: 18038.66 30.96 180110.38= Using the law of cosines: 2 (120 lb)2 (40 lb)2 2(120 lb)(40 lb) cos110.38139.08 lbRR= + =Using the law of sines:sin sin110.3840 lb 139.08 lb15.64(90 )(90 38.66 ) 15.6466.98
18. 18. PROBLEM 2.16Solve Problem 2.4 by trigonometry.PROBLEM 2.4 Two structural members B and C are bolted to bracket A.Knowing that both members are in tension and that P = 6 kips and Q = 4 kips,determine graphically the magnitude and direction of the resultant forceexerted on the bracket using (a) the parallelogram law, (b) the triangle rule.SOLUTIONUsing the force triangle and the laws of cosines and sines:We have: 180 (50 25 )= + = (4 kips) (6 kips) 2(4 kips)(6 kips) cos10564.423 kips8.0264 kips= + + = + = PROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.18105Then 2 2 22RR= + ==And4 kips 8.0264 kipssin(25 ) sin105sin(25 ) 0.4813725 28.7753.775= R = 8.03 kips 3.8
19. 19. PROBLEM 2.17For the stake of Prob. 2.5, knowing that the tension in one rope is 120 N,determine by trigonometry the magnitude and direction of the force P sothat the resultant is a vertical force of 160 N.PROBLEM 2.5 A stake is being pulled out of the ground by means of tworopes as shown. Knowing that = 30, determine by trigonometry (a) themagnitude of the force P so that the resultant force exerted on the stake isvertical, (b) the corresponding magnitude of the resultant.= == PROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.19SOLUTIONUsing the laws of cosines and sines:2 (120 N)2 (160 N)2 2(120 N)(160 N) cos2572.096 NPP= + =Andsin sin 25120 N 72.096 Nsin 0.7034344.703P = 72.1 N 44.7
20. 20. PROBLEM 2.18For the hook support of Prob. 2.10, knowing that P = 75 N and = 50,determine by trigonometry the magnitude and direction of the resultant ofthe two forces applied to the support.PROBLEM 2.10 Two forces are applied as shown to a hook support.Knowing that the magnitude of P is 35 N, determine by trigonometry (a) therequired angle if the resultant R of the two forces applied to the support isto be horizontal, (b) the corresponding magnitude of R.SOLUTIONUsing the force triangle and the laws of cosines and sines:We have 180 (50 25 )= + = (75 N) (50 N)2(75 N)(50 N) cos10510,066.1 N100.330 NPROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.20105Then 2 2 22 2RRR= + ==andsin sin10575 N 100.330 Nsin 0.7220646.225= == Hence: 25 = 46.225 25 = 21.225 R =100.3 N 21.2
21. 21. PROBLEM 2.19Two forces P and Q are applied to the lid of a storage bin as shown.Knowing that P = 48 N and Q = 60 N, determine by trigonometry themagnitude and direction of the resultant of the two forces.SOLUTIONUsing the force triangle and the laws of cosines and sines:We have 180 (20 10 )= + = = = = = PROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.21150Then 2 (48 N)2 (60 N)22(48 N)(60 N) cos150104.366 NRR= + =and48 N 104.366 Nsin sin150sin 0.2299613.2947== Hence: 180 80180 13.2947 8086.705R =104.4 N 86.7
22. 22. PROBLEM 2.20Two forces P and Q are applied to the lid of a storage bin as shown.Knowing that P = 60 N and Q = 48 N, determine by trigonometry themagnitude and direction of the resultant of the two forces.SOLUTIONUsing the force triangle and the laws of cosines and sines:We have 180 (20 10 )= + = = = = = PROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.22150Then 2 (60 N)2 (48 N)22(60 N)(48 N)cos 150104.366 NRR= + =and60 N 104.366 Nsin sin150sin 0.2874516.7054== Hence: 180 180180 16.7054 8083.295R =104.4 N 83.3
23. 23. PROBLEM 2.21Determine the x and y components of each of the forces shown.SOLUTION80-N Force: (80 N) cos 40 x F = + 61.3 N x F = Fy = +(80 N) sin 40 51.4 N Fy = 120-N Force: (120 N) cos 70 x F = + 41.0 N x F = (120 N) sin 70 Fy = + 112.8 N Fy = 150-N Force: (150 N) cos35 x F = 122. 9 N x F = (150 N) sin 35 Fy = + 86.0 N Fy = PROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and 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
24. 24. PROBLEM 2.22Determine the x and y components of each of the forces shown.SOLUTION40-lb Force: (40 lb) cos60 x F = + 20.0 lb x F = Fy = (40 lb)sin 60 34.6 lb Fy = 50-lb Force: (50 lb)sin 50 x F = 38.3 lb x F = (50 lb) cos50 Fy = 32.1 lb Fy = 60-lb Force: (60 lb) cos25 x F = + 54.4 lb x F = (60 lb)sin 25 Fy = + 25.4 lb Fy = PROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and 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
25. 25. PROBLEM 2.23Determine the x and y components of each of the forces shown.(600) (800)1000 mm(560) (900)1060 mm(480) (900)1020 mm1000 x F = + 640 N x F = + 1000 Fy = + Fy = +480 N 1060 x F = 224 N x F = 1060 y F = 360 N Fy = 1020 x F = + 192.0 N x F = + 1020 y F = 360 N Fy = PROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.25SOLUTIONCompute the following distances:2 22 22 2OAOBOC= +== +== +=800-N Force:800(800 N)600(800 N)424-N Force:560(424 N)900(424 N)408-N Force:480(408 N)900(408 N)
26. 26. PROBLEM 2.24Determine the x and y components of each of the forces shown.(24 in.) (45 in.)51.0 in.(28 in.) (45 in.)53.0 in.(40 in.) (30 in.)50.0 in.51.0 in. Fx = 48.0 lb x F = 51.0 in. Fy = + 90.0 lb Fy = + = + Fx 56.0 lb x F = + 53.0 in. Fy = + 90.0 lb Fy = + 50.0 in. Fx = 160.0 lb x F = 50.0 in. Fy = 120.0 lb Fy = PROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.26SOLUTIONCompute the following distances:2 22 22 2OAOBOC= +== +== +=102-lb Force:24 in.102 lb45 in.102 lb106-lb Force:28 in.106 lb53.0 in.45 in.106 lb200-lb Force:40 in.200 lb30 in.200 lb
27. 27. PROBLEM 2.25The hydraulic cylinder BD exerts on member ABC a force P directedalong line BD. Knowing that P must have a 750-N componentperpendicular to member ABC, determine (a) the magnitude of the forceP, (b) its component parallel to ABC.PROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.27SOLUTION(a) 750 N = Psin 20P = 2192.9 N P = 2190 N (b) cos 20 ABC P = P = (2192.9 N) cos 20 2060 N ABC P =
28. 28. PROBLEM 2.26Cable AC exerts on beam AB a force P directed along line AC. Knowing thatP must have a 350-lb vertical component, determine (a) the magnitude of theforce P, (b) its horizontal component.y P= = 500 lb x P = PROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.28SOLUTION(a)cos 55P =350 lbcos 55610.21 lb== P = 610 lb (b) sin 55 xP = P (610.21 lb)sin 55499.85 lb
29. 29. PROBLEM 2.27Member BC exerts on member AC a force P directed along line BC. Knowingthat P must have a 325-N horizontal component, determine (a) the magnitudeof the force P, (b) its vertical component.= = x P P = = y P P = PROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.29SOLUTION(650 mm)2 (720 mm)2970 mmBC= +=(a)650970 Px P or970650970325 N650485 N =P = 485 N (b)720970720485 N970360 N =970 N Py =
30. 30. PROBLEM 2.28Member BD exerts on member ABC a force P directed along line BD.Knowing that P must have a 240-lb vertical component, determine (a)the magnitude of the force P, (b) its horizontal component.y PP or P = 373 lb = =yPROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.30SOLUTION(a)240 lbsin 40 sin 40(b)240 lbtan 40 tan 40xPP= =or 286 lb x P =
31. 31. PROBLEM 2.29The guy wire BD exerts on the telephone pole AC a force P directed alongBD. Knowing that P must have a 720-N component perpendicular to thepole AC, determine (a) the magnitude of the force P, (b) its componentalong line AC.P PxPy = Px==PROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.31SOLUTION(a)371237(720 N)122220 N===P = 2.22 kN (b)351235(720 N)122100 N= 2.10 kN Py
32. 32. PROBLEM 2.30The hydraulic cylinder BC exerts on member AB a force P directed alongline BC. Knowing that P must have a 600-N component perpendicular tomember AB, determine (a) the magnitude of the force P, (b) its componentalong line AB. = + + + = = PxPPxPPy=x== =P PPROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.32SOLUTION180 45 90 30180 45 90 3015(a) coscos600 Ncos15621.17 NP====P = 621 N (b) tantan(600 N) tan15160.770 Ny x160.8 N Py =
33. 33. PROBLEM 2.31Determine the resultant of the three forces of Problem 2.23.PROBLEM 2.23 Determine the x and y components of each ofthe forces shown.SOLUTIONComponents of the forces were determined in Problem 2.23:Force x Comp. (N) y Comp. (N)800 lb +640 +480424 lb 224 360408 lb +192 360Rx = +608 Ry = 240R R i Rjx y(608 lb) ( 240 lb)yRRxPROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.33tan24060821.541240 Nsin(21.541)653.65 NR= += +=== ==i jR = 654 N 21.5
34. 34. PROBLEM 2.32Determine the resultant of the three forces of Problem 2.21.PROBLEM 2.21 Determine the x and y components of each of theforces shown.SOLUTIONComponents of the forces were determined in Problem 2.21:Force x Comp. (N) y Comp. (N)80 N +61.3 +51.4120 N +41.0 +112.8150 N 122.9 +86.0Rx = 20.6 Ry = +250.2R Rx y( 20.6 N) (250.2 N)RRyxPROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.34tan250.2 Ntan20.6 Ntan 12.145685.293250.2 Nsin85.293R= += +==== =R i ji jR = 251 N 85.3
35. 35. PROBLEM 2.33Determine the resultant of the three forces of Problem 2.22.PROBLEM 2.22 Determine the x and y components of each of theforces shown.R Rx yyRRxPROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.35SOLUTIONForce x Comp. (lb) y Comp. (lb)40 lb +20.00 34.6450 lb 38.30 32.1460 lb +54.38 +25.36Rx = +36.08 Ry = 41.42( 36.08 lb) ( 41.42 lb)tan41.42 lbtan36.08 lbtan 1.1480048.94241.42 lbsin 48.942R= += + + ==== =R i ji jR = 54.9 lb 48.9
36. 36. PROBLEM 2.34Determine the resultant of the three forces of Problem 2.24.PROBLEM 2.24 Determine the x and y components of each of theforces shown.SOLUTIONComponents of the forces were determined in Problem 2.24:Force x Comp. (lb) y Comp. (lb)102 lb 48.0 +90.0106 lb +56.0 +90.0200 lb 160.0 120.0Rx = 152.0 Ry = 60.0= += +=R R i Rjx y( 152 lb) (60.0 lb)yxRRPROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.36tan60.0 lbtan152.0 lbtan 0.3947421.541=== i j60.0 lbsin 21.541R =R = 163.4 lb 21.5
37. 37. PROBLEM 2.35Knowing that = 35, determine the resultant of the three forcesshown.FFFFFF= += + =R R i Rjx yyRRx== PROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.37SOLUTION100-N Force: (100 N) cos35 81.915 N(100 N)sin 35 57.358 Nxy= + = += = 150-N Force: (150 N)cos65 63.393 N(150 N) sin 65 135.946 Nxy= + = += = 200-N Force: (200 N)cos35 163.830 N(200 N)sin 35 114.715 Nxy= = = = Force x Comp. (N) y Comp. (N)100 N +81.915 57.358150 N +63.393 135.946200 N 163.830 114.715Rx = 18.522 Ry = 308.02( 18.522 N) ( 308.02 N)tan308.0218.52286.559i j308.02 Nsin86.559R = R = 309 N 86.6
38. 38. PROBLEM 2.36Knowing that the tension in rope AC is 365 N, determine theresultant of the three forces exerted at point C of post BC.= = = = FFFFFF= = + + == = + = = += +=240 N 480 N 160 N 400 N275 N 140 N 120 N 255 NR x FxR Fy yR R Rx yPROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.38SOLUTIONDetermine force components:Cable force AC:960(365 N) 240 N14601100(365 N) 275 N1460xy500-N Force:24(500 N) 480 N257(500 N) 140 N25xy= == =200-N Force:4(200 N) 160 N53(200 N) 120 N5xy= == = and2 22 2(400 N) ( 255 N)474.37 NFurther:255tan40032.5== R = 474 N 32.5
39. 39. PROBLEM 2.37Knowing that = 40, determine the resultant of the three forcesshown.SOLUTION60-lb Force: (60 lb)cos 20 56.382 lb(60 lb)sin 20 20.521 lbFFFFFF= == == +=200.305 lb29.803 lbR FR FRx xy y = = R = 203 lb 8.46 PROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.39xy= == =80-lb Force: (80 lb) cos60 40.000 lb(80 lb)sin60 69.282 lbxy= == =120-lb Force: (120 lb) cos30 103.923 lb(120 lb)sin 30 60.000 lbxy= == = and2 2(200.305 lb) (29.803 lb)202.510 lbFurther:29.803tan200.305 =1 29.803tan200.3058.46
40. 40. PROBLEM 2.38Knowing that = 75, determine the resultant of the three forcesshown.SOLUTION60-lb Force: (60 lb) cos 20 56.382 lb(60 lb) sin 20 20.521 lbFFFFFF= == =R FR FR= +=PROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.40xy= == =80-lb Force: (80 lb) cos 95 6.9725 lb(80 lb)sin 95 79.696 lbxy= = = =120-lb Force: (120 lb) cos 5 119.543 lb(120 lb)sin 5 10.459 lbxy= == =Then 168.953 lb110.676 lbx xy yand (168.953 lb)2 (110.676 lb)2201.976 lb110.676tan168.953tan 0.6550733.228=== R = 202 lb 33.2
41. 41. PROBLEM 2.39For the collar of Problem 2.35, determine (a) the required value of if the resultant of the three forces shown is to be vertical, (b) thecorresponding magnitude of the resultant.= = + + = + + (1)R Fx xR= = + = + (2)R Fy y + + = + = =29.904=== = 21.7 Ry = PROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.41SOLUTION(100 N) cos (150 N) cos ( 30 ) (200 N) cos(100 N)cos (150 N) cos ( 30 )x (100 N) sin (150 N)sin ( 30 ) (200 N)sin(300 N) sin (150 N)sin ( 30 )yR (a) For R to be vertical, we must have 0. x R = We make 0 x R = in Eq. (1):100cos 150cos ( 30 ) 0100cos 150(cos cos 30 sin sin 30 ) 029.904cos 75sintan750.3987221.738(b) Substituting for in Eq. (2):300sin 21.738 150sin 51.738228.89 N= | | 228.89 N R = Ry = R = 229 N
42. 42. PROBLEM 2.40For the post of Prob. 2.36, determine (a) the required tension inrope AC if the resultant of the three forces exerted at point C isto be horizontal, (b) the corresponding magnitude of theresultant.= = + +R F Tx x AC= + (1)R Tx AC= = + R F Ty y AC= + (2)R Ty AC73 AC T + == +== =RRR R R = 623 N PROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.42SOLUTION960 24 4(500 N) (200 N)1460 25 548640 N731100 7 3(500 N) (200 N)1460 25 55520 N73(a) For R to be horizontal, we must have Ry = 0.Set 0 Ry = in Eq. (2):5520 N 026.545 N AC T = 26.5 N AC T = (b) Substituting for AC T into Eq. (1) gives48(26.545 N) 640 N73622.55 N623 Nxxx
43. 43. PROBLEM 2.41A hoist trolley is subjected to the three forces shown. Knowing that = 40,determine (a) the required magnitude of the force P if the resultant ofthe three forces is to be vertical, (b) the corresponding magnitude ofthe resultant.PPROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.43SOLUTIONx R = (Fx = P + 200 lb)sin 40 (400 lb) cos 40177.860 lb x R = P (1)Ry = (200 lb) cos 40 (400 lb) sin 40 Fy = + 410.32 lb Ry = (2)(a) For R to be vertical, we must have 0. x R =Set 0 x R = in Eq. (1)0 177.860 lb177.860 lbP= = P = 177.9 lb (b) Since R is to be vertical:= = 410 lb R Ry R = 410 lb
44. 44. PROBLEM 2.42A hoist trolley is subjected to the three forces shown. Knowingthat P = 250 lb, determine (a) the required value of if theresultant of the three forces is to be vertical, (b) the correspondingmagnitude of the resultant. = + = + = + + (400 lb) cos (200 lb) sin 250 lb2cos sin 1.254cos sin 2.5sin 1.5625 = + +4(1 sin ) sin 2.5sin 1.5625= + PROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.44SOLUTIONx R = 250 lb Fx = + (200 lb)sin (400 lb)cos250 lb (200 lb)sin (400 lb)cos x R = + (1)Ry = (200 lb) cos (400 lb)sin Fy = + (a) For R to be vertical, we must have 0. x R =Set 0 x R = in Eq. (1)0 = 250 lb + (200 lb)sin (400 lb) cos2 22 220 5sin 2.5sin 2.4375Using the quadratic formula to solve for the roots givessin = 0.49162or = 29.447 = 29.4 (b) Since R is to be vertical:(200 lb) cos 29.447 (400 lb) sin 29.447 R = Ry = + R = 371 lb
45. 45. PROBLEM 2.43Two cables are tied together at C and are loaded as shown.Determine the tension (a) in cable AC, (b) in cable BC.TAC TBC = = sin108.492 AC T= sin108.492 BC T= PROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.45SOLUTIONFree-Body Diagram1100tan96048.888400tan96022.620== == Force TriangleLaw of sines:15.696 kNsin 22.620 sin 48.888 sin108.492(a)15.696 kN(sin 22.620 )6.37 kN AC T = (b)15.696 kN(sin 48.888 )12.47 kN BC T =
46. 46. PROBLEM 2.44Two cables are tied together at C and are loaded as shown. Determinethe tension (a) in cable AC, (b) in cable BC.TAC TBC = = sin 94.979 AC T= sin 94.979 BC T= PROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.46SOLUTION3tan2.2553.1301.4tan2.2531.891== == Free-Body DiagramLaw of sines: Force-Triangle660 Nsin 31.891 sin 53.130 sin 94.979(a)660 N(sin 31.891 )350 N AC T = (b)660 N(sin 53.130 )530 N BC T =
47. 47. PROBLEM 2.45Knowing that = 20, determine the tension (a) in cable AC,(b) in rope BC.AC BC T T = = sin 65 TAC = sin 65 TBC = PROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.47SOLUTIONFree-Body Diagram Force TriangleLaw of sines:1200 lbsin 110 sin 5 sin 65(a)1200 lbsin 1101244 lb AC T = (b)1200 lbsin 5115.4 lb BC T =
48. 48. PROBLEM 2.46Knowing that = 55 and that boom AC exerts on pin C a force directedalong line AC, determine (a) the magnitude of that force, (b) the tension incable BC.AC BC F T = = sin 95 FAC = sin 95 TBC = PROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.48SOLUTIONFree-Body Diagram Force TriangleLaw of sines:300 lbsin 35 sin 50 sin 95(a)300 lbsin 35172.7 lb AC F = (b)300 lbsin 50231 lb BC T =
49. 49. PROBLEM 2.47Two cables are tied together at C and loaded as shown.Determine the tension (a) in cable AC, (b) in cable BC.TAC TBC = = sin 44.333 AC T= sin 44.333 BC T= PROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.49SOLUTIONFree-Body Diagram1.4tan4.816.26021.6tan328.073== == Force TriangleLaw of sines:1.98 kNsin 61.927 sin 73.740 sin 44.333(a)1.98 kNsin 61.9272.50 kN AC T = (b)1.98 kNsin 73.7402.72 kN BC T =
50. 50. PROBLEM 2.48Two cables are tied together at C and are loaded as shown.Knowing that P = 500 N and = 60, determine the tension in(a) in cable AC, (b) in cable BC.AC BC T T = = sin 70 AC T = sin 70 BC T = PROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.50SOLUTIONFree-Body Diagram Force TriangleLaw of sines:500 Nsin 35 sin 75 sin 70(a)500 Nsin 35305 N AC T = (b)500 Nsin 75514 N BC T =
51. 51. PROBLEM 2.49Two forces of magnitude TA = 8 kips and TB = 15 kips areapplied as shown to a welded connection. Knowing that theconnection is in equilibrium, determine the magnitudes of theforces TC and TD.TT = 5.87 kips C T PROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.51SOLUTIONFree-Body Diagram0 1 Fx = 5 kips 8 kips TD cos 40 = 09.1379 kips D T =0 Fy = sin 40 0 D C T T =(9.1379 kips) sin 40 05.8737 kips ==CC9.14 kips D T =
52. 52. PROBLEM 2.50Two forces of magnitude TA = 6 kips and TC = 9 kips areapplied as shown to a welded connection. Knowing that theconnection is in equilibrium, determine the magnitudes of theforces TB and TD. = =PROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.52SOLUTIONFree-Body Diagram Fx = 0 6 kips cos 40 0 B D T T = (1)0 Fy = sin 40 9 kips 09 kipssin 4014.0015 kipsDDDTTT==Substituting for TD into Eq. (1) gives:6 kips (14.0015 kips) cos40 016.7258 kipsBBTT=16.73 kips B T = 14.00 kips D T =
53. 53. PROBLEM 2.51Two cables are tied together at C and loaded as shown. Knowingthat P = 360 N, determine the tension (a) in cable AC, (b) incable BC.13 5 x AC F = T + = 312 N AC T = 13 5 y BC F = + T + =PROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.53SOLUTIONFree Body: C(a)12 40: (360 N) 0(b)5 30: (312 N) (360 N) 480 N 0480 N 120 N 216 N BC T = 144 N BC T =
54. 54. PROBLEM 2.52Two cables are tied together at C and loaded as shown.Determine the range of values of P for which both cablesremain taut.13 5 Fx = TAC + P =13 5 y AC BC F = T + T + P = + + = 15 BCT = P (2)PROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.54SOLUTIONFree Body: C12 40: 01315 ACT = P (1)5 30: 480 N 0Substitute for AC T from (1):5 13 3480 N 013 15 5 P TBC P14480 NFrom (1), 0 AC Trequires P0.From (2), 0 BC Trequires14480 N, 514.29 N15P PAllowable range: 0 P514 N
55. 55. PROBLEM 2.53A sailor is being rescued using a boatswains chair that issuspended from a pulley that can roll freely on the supportcable ACB and is pulled at a constant speed by cable CD.Knowing that = 30 and =10 and that the combinedweight of the boatswains chair and the sailor is 900 N,determine the tension (a) in the support cable ACB, (b) in thetraction cable CD.PROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.55SOLUTIONFree-Body Diagram0: cos 10 cos 30 cos 30 0 x ACB ACB CD F = T T T =0.137158 CD ACB T = T (1)0: sin 10 sin 30 sin 30 900 0 Fy = TACB + TACB + TCD =0.67365 0.5 900 ACB CD T + T = (2)(a) Substitute (1) into (2): 0.67365 0.5(0.137158 ) 900 ACB ACB T + T =1212.56 N ACB T = 1213 N ACB T = (b) From (1): 0.137158(1212.56 N) CD T = 166.3 N CD T =
56. 56. PROBLEM 2.54A sailor is being rescued using a boatswains chair that issuspended from a pulley that can roll freely on the supportcable ACB and is pulled at a constant speed by cable CD.Knowing that = 25 and =15 and that the tension incable CD is 80 N, determine (a) the combined weight of theboatswains chair and the sailor, (b) in tension in the supportcable ACB.+ =PROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.56SOLUTIONFree-Body Diagram0: cos 15 cos 25 (80 N) cos 25 0 x ACB ACB F = T T =1216.15 N ACB T =0: (1216.15 N) sin 15 (1216.15 N)sin 25 Fy = + (80 N) sin 25 0862.54 NWW=(a) W = 863 N (b) 1216 N ACB T =
57. 57. PROBLEM 2.55Two forces P and Q are applied as shown to an aircraftconnection. Knowing that the connection is in equilibrium andthat P = 500 lb and Q = 650 lb, determine the magnitudes ofthe forces exerted on the rods A and B.= + + + =R j iji i j= + B A F = F = = 419.55 lb 420 lb B F = PROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.57SOLUTIONFree-Body DiagramResolving the forces into x- and y-directions:0 A B R = P +Q + F + F =Substituting components: (500 lb) [(650 lb) cos50 ][(650 lb) sin 50 ]( cos50) ( sin50) 0 B A A F F FIn the y-direction (one unknown force):500 lb (650 lb)sin 50 sin 50 0 A + F =Thus,500 lb (650 lb) sin 50sin 50 A F=1302.70 lb 1303 lb A F = In the x-direction: (650 lb)cos50 cos50 0 B A + F F =Thus, cos50 (650 lb) cos50(1302.70 lb) cos50 (650 lb) cos50
58. 58. PROBLEM 2.56Two forces P and Q are applied as shown to an aircraftconnection. Knowing that the connection is in equilibriumand that the magnitudes of the forces exerted on rods A and Bare 750 A F = lb and 400 B F = lb, determine the magnitudes ofP and Q.= P + Q Q + +R j i jij iQ= P = Q + = + = P = 477 lb; Q =127.7 lb PROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.58SOLUTIONFree-Body DiagramResolving the forces into x- and y-directions:R = P +Q + FA + FB = 0Substituting components: cos 50 sin 50[(750 lb)cos 50 ][(750 lb)sin 50 ] (400 lb)In the x-direction (one unknown force):Q cos 50 [(750 lb) cos 50] + 400 lb = 0(750 lb) cos 50 400 lbcos 50127.710 lb=In the y-direction: P Q sin 50 + (750 lb) sin 50 = 0sin 50 (750 lb) sin 50(127.710 lb)sin 50 (750 lb)sin 50476.70 lb
59. 59. PROBLEM 2.57Two cables tied together at C are loaded as shown. Knowing thatthe 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 .PROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.59SOLUTIONFree-Body Diagram: C Force TriangleForce triangle is isosceles with2 180 8547.5= = (a) P = 2(800 N)cos 47.5 = 1081 NSince P0, the solution is correct. P =1081 N (b) =180 50 47.5 = 82.5 = 82.5
60. 60. PROBLEM 2.58Two cables tied together at C are loaded as shown. Knowing thatthe maximum allowable tension is 1200 N in cable AC and 600 Nin cable BC, determine (a) the magnitude of the largest force Pthat can be applied at C, (b) the corresponding value of .PROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.60SOLUTIONFree-Body Diagram Force Triangle(a) Law of cosines: 2 (1200 N)2 (600 N)2 2(1200 N)(600 N) cos 851294.02 NPP= + =Since P1200 N, the solution is correct.P =1294 N (b) Law of sines:sin sin 851200 N 1294.02 N67.5180 50 67.5= = = = 62.5
61. 61. PROBLEM 2.59For the situation described in Figure P2.45, determine (a) thevalue of for which the tension in rope BC is as small aspossible, (b) the corresponding value of the tension.PROBLEM 2.45 Knowing that = 20, determine the tension(a) in cable AC, (b) in rope BC.PROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.61SOLUTIONFree-Body Diagram Force TriangleTo be smallest, BC T must be perpendicular to the direction of . AC T(a) Thus, = 5 = 5.00 (b) (1200 lb)sin 5 BC T = 104.6 lb BC T =
62. 62. PROBLEM 2.60For the structure and loading of Problem 2.46, determine (a) the value of forwhich the tension in cable BC is as small as possible, (b) the correspondingvalue of the tension.PROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.62SOLUTIONBC T must be perpendicular to AC F to be as small as possible.Free-Body Diagram: C Force Triangle is a right triangleTo be a minimum, BC T must be perpendicular to . AC F(a) We observe: = 90 30 = 60.0 (b) (300 lb)sin 50 BC T = or 229.81 lb BC T = 230 lb BC T =
63. 63. PROBLEM 2.61For the cables of Problem 2.48, it is known that the maximumallowable 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 . = + = 46.0 = 46.0 + 25 = 71.0 PROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.63SOLUTIONFree-Body Diagram Force Triangle(a) Law of cosines P2 = (600)2 + (750)2 2(600)(750) cos (25 + 45)P = 784.02 N P = 784 N (b) Law of sinessin sin (25 45 )600 N 784.02 N
64. 64. PROBLEM 2.62A movable bin and its contents have a combined weight of 2.8 kN.Determine the shortest chain sling ACB that can be used to lift theloaded bin if the tension in the chain is not to exceed 5 kN. = h=AC=== h = h == AC = +PROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.64SOLUTIONFree-Body Diagramtan0.6 m(1)Isosceles Force TriangleLaw of sines:1212(2.8 kN)sin5 kN(2.8 kN)sin5 kN16.2602ACTTFrom Eq. (1): tan16.2602 0.175000 m0.6 mHalf length of chain (0.6 m)2 (0.175 m)20.625 m=Total length: = 2 0.625 m 1.250 m
65. 65. PROBLEM 2.63Collar A is connected as shown to a 50-lb load and can slide ona frictionless horizontal rod. Determine the magnitude of theforce 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 TriangleP = P = 10.98 lb P = P = 30.0 lb PROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.6550 lb4.5 20.5(b) Free Body: Collar A Force Triangle50 lb15 25
66. 66. PROBLEM 2.64Collar A is connected as shown to a 50-lb load and can slide on africtionless horizontal rod. Determine the distance x for which thecollar is in equilibrium when P = 48 lb.x =PROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.66SOLUTIONFree Body: Collar A Force Triangle2 (50)2 (48)2 19614.00 lbNN= ==Similar Triangles48 lb20 in. 14 lbx = 68.6 in.
67. 67. PROBLEM 2.65Three forces are applied to a bracket as shown. The directions of the two 150-Nforces may vary, but the angle between these forces is always 50. Determine therange of values of for which the magnitude of the resultant of the forces acting at Ais less than 600 N.SOLUTIONCombine the two 150-N forces into a resultant force Q:Q= = + = + = PROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.672(150 N) cos25271.89 NEquivalent loading at A:Using the law of cosines:(600 N)2 (500 N)2 (271.89 N)2 2(500 N)(271.89 N) cos(55 )cos(55 ) 0.132685= + + + + =Two values for : 55 82.37527.4 + == or 55 82.37555 360 82.375222.6= For R600 lb: 27.4 222.6
68. 68. PROBLEM 2.66A 200-kg crate is to be supported by the rope-and-pulley arrangement shown.Determine the magnitude and direction of the force P that must be exerted on the freeend of the rope to maintain equilibrium. (Hint: The tension in the rope is the same oneach side of a simple pulley. This can be proved by the methods of Ch. 4.)xF P P = + = = + =PROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.68SOLUTIONFree-Body Diagram: Pulley A50: 2 cos 0281cos 0.5965553.377 = + = == For = +53.377:160: 2 sin 53.377 1962 N 0281 yF P P P = 724 N 53.4 For = 53.377:160: 2 sin( 53.377 ) 1962 N 0281 yF P P P =1773 53.4
69. 69. = = = = = = = = = =PROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.69PROBLEM 2.67A 600-lb crate is supported by several rope-and-pulleyarrangements as shown. Determine for eacharrangement the tension in the rope. (See the hintfor Problem 2.66.)SOLUTIONFree-Body Diagram of Pulley(a) 0: 2 (600 lb) 01(600 lb)2Fy TT=T = 300 lb (b) 0: 2 (600 lb) 01(600 lb)2Fy TT=T = 300 lb (c) 0: 3 (600 lb) 01(600 lb)3Fy TT=T = 200 lb (d) 0: 3 (600 lb) 01(600 lb)3Fy TT=T = 200 lb (e) 0: 4 (600 lb) 01(600 lb)4Fy TT= T =150.0 lb
70. 70. = = = =PROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.70PROBLEM 2.68Solve Parts b and d of Problem 2.67, assuming thatthe free end of the rope is attached to the crate.PROBLEM 2.67 A 600-lb crate is supported byseveral rope-and-pulley arrangements as shown.Determine for each arrangement the tension in therope. (See the hint for Problem 2.66.)SOLUTIONFree-Body Diagram of Pulley and Crate(b) 0: 3 (600 lb) 01(600 lb)3Fy TT=T = 200 lb (d) 0: 4 (600 lb) 01(600 lb)4Fy TT=T =150.0 lb
71. 71. PROBLEM 2.69A load Q is applied to the pulley C, which can roll on thecable ACB. The pulley is held in the position shown by asecond cable CAD, which passes over the pulley A andsupports a load P. Knowing that P = 750 N, determine(a) the tension in cable ACB, (b) the magnitude of load Q. = + + =Fy TACB Q + + =PROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.71SOLUTIONFree-Body Diagram: Pulley C(a) 0: (cos 25 cos55 ) (750 N)cos55 0 x ACB F = T =Hence: 1292.88 N ACB T =1293 N ACB T = (b) 0: (sin 25 sin55 ) (750 N)sin 55 0(1292.88 N)(sin 25 sin 55 ) (750 N) sin 55 Q0or Q = 2219.8 N Q = 2220 N
72. 72. PROBLEM 2.70An 1800-N load Q is applied to the pulley C, which can rollon the cable ACB. The pulley is held in the position shownby a second cable CAD, which passes over the pulley A andsupports a load P. Determine (a) the tension in cable ACB,(b) the magnitude of load P.PROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.72SOLUTIONFree-Body Diagram: Pulley C0: Fx = TACB (cos 25 cos55) Pcos55 = 0or 0.58010 ACB P = T (1)0: (sin 25 sin 55 ) sin 55 1800 N 0 Fy = TACB + + P =or 1.24177 0.81915 1800 N ACB T + P = (2)(a) Substitute Equation (1) into Equation (2):1.24177 0.81915(0.58010 ) 1800 N ACB ACB T + T =Hence: 1048.37 N ACB T =1048 N ACB T = (b) Using (1), P = 0.58010(1048.37 N) = 608.16 NP = 608 N
73. 73. PROBLEM 2.71Determine (a) the x, y, and z components of the 900-N force, (b) theangles x, y, and z that the force forms with the coordinate axes.F FFFx = Fh == 130.091 N, x F 130.1N x F = F FF= = = +F Fz hF 357 N z F = + = = 98.3 x = = = + 66.6 z PROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.73SOLUTIONcos 65(900 N) cos 65380.36 Nhh= = =(a) sin 20(380.36 N)sin 20sin 65(900 N) sin 65815.68 N,yy= == + 816 N y F = + cos 20(380.36 N)cos 20357.42 Nz(b)130.091 Ncos900 NxxFF815.68 Ncos900 NyyFF = = + 25.0 y = 357.42 Ncos900 NzzFF=
74. 74. PROBLEM 2.72Determine (a) the x, y, and z components of the 750-N force, (b) theangles x, y, and z that the force forms with the coordinate axes.F FF= =x h F FF FFF Fz hF = = + 58.7 x = = = + 76.0 z PROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.74SOLUTIONsin 35(750 N)sin 35430.18 Nhh= = =(a) cos 25(430.18 N) cos 25= +389.88 N, x F 390 N x F = + cos35(750 N) cos 35614.36 N,yy= == + 614 N Fy = + sin 25(430.18 N)sin 25181.802 Nz= = = +181.8 N z F = + (b)389.88 Ncos750 NxxFF614.36 Ncos750 NyyFF = = + 35.0 y = 181.802 Ncos750 NzzFF=
75. 75. PROBLEM 2.73A gun is aimed at a point A located 35 east of north. Knowing that the barrel of the gun forms an angle of 40with the horizontal and that the maximum recoil force is 400 N, determine (a) the x, y, and z components ofthat force, (b) the values of the angles x, y, and z defining the direction of the recoil force. (Assume that thex, y, and z axes are directed, respectively, east, up, and south.) FH = x H F = F = yF = F = = 257 N Fy = z H F = +F = + = + 251N z F = + = = 116.1 x = PROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.75SOLUTIONRecoil force F = 400 N(400 N)cos40306.42 N=(a) sin 35(306.42 N)sin 35= 175.755 N 175.8 N x F = sin 40(400 N)sin 40257.12 Ncos35(306.42 N)cos35251.00 N(b)175.755 Ncos400 NxxFF257.12 Ncos400 NyyFF = = 130.0 y = 251.00 Ncos400 NzzFF = = 51.1 z =
76. 76. PROBLEM 2.74Solve Problem 2.73, assuming that point A is located 15 north of west and that the barrel of the gun forms anangle of 25 with the horizontal.PROBLEM 2.73 A gun is aimed at a point A located 35 east of north. Knowing that the barrel of the gunforms an angle of 40 with the horizontal and that the maximum recoil force is 400 N, determine (a) the x, y,and z components of that force, (b) the values of the angles x, y, and z defining the direction of the recoilforce. (Assume that the x, y, and z axes are directed, respectively, east, up, and south.)H F = x H F = +F = + = +350.17 N 350 N x F = + yF = F = = 169.0 N Fy = z H F = +F = + = + 93.8 N z F = + = = + 28.9 x = = = + 76.4 z PROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.76SOLUTIONRecoil force F = 400 N(400 N)cos25362.52 N=(a) cos15(362.52 N) cos15sin 25(400 N)sin 25169.047 Nsin15(362.52 N)sin1593.827 N(b)350.17 Ncos400 NxxFF169.047 Ncos400 NyyFF = = 115.0 y = 93.827 Ncos400 NzzFF=
77. 77. PROBLEM 2.75Cable AB is 65 ft long, and the tension in that cable is 3900 lb.Determine (a) the x, y, and z components of the force exertedby the cable on the anchor B, (b) the angles , x , y and z defining the direction of that force.Fx = F y = PROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.77SOLUTIONFrom triangle AOB:56 ftcos65 ft0.8615430.51yy=== (a) sin cos 20(3900 lb)sin 30.51 cos 201861 lb x F = cos (3900 lb)(0.86154) Fy = +F y = 3360 lb Fy = + (3900 lb)sin 30.51 sin 20 z F = + 677 lb z F = + (b)1861 lbcos 0.47713900 lbxxFF = = = 118.5 x = From above: 30.51 y = 30.5 y = 677 lbcos 0.17363900 lbzzFF = =+ =+ 80.0 z =
78. 78. PROBLEM 2.76Cable AC is 70 ft long, and the tension in that cable is 5250 lb.Determine (a) the x, y, and z components of the force exerted bythe cable on the anchor C, (b) the angles x, y, and z defining thedirection of that force.== == = = = 117.4 x = = = 112.7 z PROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.78SOLUTIONIn triangle AOB: 70 ft56 ft5250 lbACOAF===56 ftcos70 ft36.870sin(5250 lb) sin 36.8703150.0 lbyyFH F y(a) sin 50 (3150.0 lb)sin 50 2413.04 lb x H F = F = = 2413 lb x F = cos (5250 lb) cos36.870 4200.0 lb Fy = +F y = + = + 4200 lb Fy = + cos50 3150cos50 2024.8 lb z H F = F = = 2025 lb z F = (b)2413.04 lbcos5250 lbxxFFFrom above:36.870 y = 36.9 y = 2024.8 lb5250 lbzzFF=
79. 79. PROBLEM 2.77The end of the coaxial cable AE is attached to the pole AB, which isstrengthened by the guy wires AC and AD. Knowing that the tensionin wire AC is 120 lb, determine (a) the components of the forceexerted by this wire on the pole, (b) the angles x, y, and z that theforce forms with the coordinate axes.SOLUTION(a) (120 lb) cos 60 cos 20 x F= 56.382 lb x F = 56.4 lb x F = + (120 lb)sin 60103.923 lbFFFF = = 99.8 z PROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.79yy= = Fy = 103.9 lb (120 lb) cos 60 sin 2020.521 lbzz= = 20.5 lb z F = (b)56.382 lbcos120 lbxxFF = = 62.0 x = 103.923 lbcos120 lbyyFF = = 150.0 y = 20.52 lbcos120 lbzzFF=
80. 80. PROBLEM 2.78The end of the coaxial cable AE is attached to the pole AB, which isstrengthened by the guy wires AC and AD. Knowing that the tensionin wire AD is 85 lb, determine (a) the components of the forceexerted by this wire on the pole, (b) the angles x, y, and z that theforce forms with the coordinate axes.SOLUTION(a) (85 lb)sin 36 sin 48 x F= = 37.129 lb 37.1 lb x F = F = (85 lb)cos 36y = 68.766 lbFy = 68.8 lb F= (85 lb)sin 36 cos 48z = 33.431 lbF = 33.4 lb z PROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and 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(b)37.129 lbcos85 lbxxFF = = 64.1 x = 68.766 lbcos85 lbyyFF = = 144.0 y = 33.431 lbcos85 lbzzFF = = 66.8 z =
81. 81. = + = + += + + =F i j k(690 N) (300 N) (580 N)x y z F F F F = = 127.6 z PROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.81PROBLEM 2.79Determine the magnitude and direction of the force F = (690 lb)i + (300 lb)j (580 lb)k.SOLUTION2 2 22 2 2(690 N) (300 N) ( 580 N)950 NF = 950 N 690 Ncos950 NxxFF = = 43.4 x = 300 Ncos950 NyyFF = = 71.6 y = 580 Ncos950 NzzFF=
82. 82. = += + += + +F (650 N) i (320 N) j (760 N)kF Fx Fy FzPROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.82PROBLEM 2.80Determine the magnitude and direction of the force F = (650 N)i (320 N)j + (760 N)k.SOLUTION2 2 22 2 2(650 N) ( 320 N) (760 N)F =1050 N 650 Ncos1050 NxxFF = = 51.8 x = 320 Ncos1050 NyyFF = = 107.7 y = 760 Ncos1050 NzzFF = = 43.6 z =
83. 83. + + =cos cos cos 1x y z + + == =F = 416.09 lby y F FPROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.83PROBLEM 2.81A force acts at the origin of a coordinate system in a direction defined by the angles x = 75 and z = 130.Knowing that the y component of the force is +300 lb, determine (a) the angle y, (b) the other componentsand the magnitude of the force.SOLUTION2 2 22 2 2cos (75 ) cos cos (130 ) 1cos 0.72100yy= (a) Since Fy 0, we choose cos 0.72100 y 43.9 y = (b) cos300 lb F(0.72100)F = 416 lb cos 416.09 lbcos75 x x F = F = 107.7 lb x F = + cos 416.09 lbcos130 z z F = F = 267 lb z F =
84. 84. + + =+ + =cos cos cos 1x y zcos cos 55 cos 45 1= x x F F = PROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.84PROBLEM 2.82A force acts at the origin of a coordinate system in a direction defined by the angles y = 55 and z = 45.Knowing that the x component of the force is 500 N, determine (a) the angle x, (b) the other componentsand the magnitude of the force.SOLUTION2 2 22 2 2cos 0.41353xx= (a) Since Fy 0, we choose cos 0.41353 x 114.4 x = (b) cos500 N F( 0.41353)F =1209.10 NF =1209.1N cos 1209.10 Ncos55 Fy = F y = 694 N Fy = + cos 1209.10 Ncos45 z z F = F = 855 N z F = +
85. 85. PROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.85PROBLEM 2.83A force F of magnitude 230 N acts at the origin of a coordinate system. Knowing that x = 32.5, Fy = 60 N,and Fz0, determine (a) the components Fx and Fz, (b) the angles y and z.SOLUTION(a) We haveFx = F cos x = (230 N) cos32.5 194.0 N x F = Then: 193.980 N x F =2 2 2 2F = Fx + Fy + FzSo: (230 N)2 (193.980 N)2 ( 60 N)2 2 z = + + FHence: (230 N)2 (193.980 N)2 ( 60 N)2 z F = + 108.0 N z F = (b) 108.036 N z F =60 Ncos 0.26087230 NyyFF = = = 105.1 y = 108.036 Ncos 0.46972230 NzzFF = = = 62.0 z =
86. 86. PROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.86PROBLEM 2.84A force F of magnitude 210 N acts at the origin of a coordinate system. Knowing that Fx = 80 N, z = 151.2,and Fy0, determine (a) the components Fy and Fz, (b) the angles x and y.SOLUTION(a) Fz = F cos z = (210 N)cos151.2= 184.024 N 184.0 N z F = Then: 2 2 2 2F = Fx + Fy + FzSo: (210 N)2 (80 N)2 ( )2 (184.024 N)2 = + Fy +Hence: (210 N)2 (80 N)2 (184.024 N)2 y F = = 61.929 N 62.0 lb Fy = (b)80 Ncos 0.38095210 NxxFF = = = 67.6 x = 61.929 Ncos 0.29490210 NyyFF = = = 107.2 y =
87. 87. PROBLEM 2.85In order to move a wrecked truck, two cables are attached at Aand pulled by winches B and C as shown. Knowing that thetension in cable AB is 2 kips, determine the components of theforce exerted at A by the cable.= = + +i j k ABABT= = + +PROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.87SOLUTIONCable AB:( 46.765 ft) (45 ft) (36 ft)74.216 ft46.765 45 3674.216ABAB AB ABi j kT ( ) 1.260 kips AB x T = ( ) 1.213 kips TAB y = + ( ) 0.970 kips AB z T = +
88. 88. PROBLEM 2.86In order to move a wrecked truck, two cables are attached at Aand pulled by winches B and C as shown. Knowing that thetension in cable AC is 1.5 kips, determine the components ofthe force exerted at A by the cable.= = + + i j k ACACT= = + PROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.88SOLUTIONCable AB:( 46.765 ft) (55.8 ft) ( 45 ft)85.590 ft46.765 55.8 45(1.5 kips)85.590ACAC AC ACi j kT ( ) 0.820 kips AC x T = ( ) 0.978 kips TAC y = + ( ) = 0.789 kips AC z T
89. 89. PROBLEM 2.87Knowing that the tension in cable AB is 1425 N, determine thecomponents of the force exerted on the plate at B.= + += + +===i j k(900 mm) (600 mm) (360 mm)(900 mm) (600 mm) (360 mm)1140 mmBABAT BA BA BABAPROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.89SOLUTION2 2 21425 N[ (900 mm) (600 mm) (360 mm) ]1140 mm(1125 N) (750 N) (450 N)BATBATBA= + += + +T i j ki j k( ) TBA x = 1125 N, (TBA )y = 750 N, (TBA )z = 450 N
90. 90. PROBLEM 2.88Knowing that the tension in cable AC is 2130 N, determine thecomponents of the force exerted on the plate at C.= + = + +===i j k(900 mm) (600 mm) (920 mm)(900 mm) (600 mm) (920 mm)1420 mmCACAT CA CA CACAPROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.90SOLUTION2 2 22130 N[ (900 mm) (600 mm) (920 mm) ]1420 mm(1350 N) (900 N) (1380 N)CATCATCA= + = + T i j ki j k( TCA )x = 1350 N, (TCA )y = 900 N, (TCA )z = 1380 N
91. 91. PROBLEM 2.89A frame ABC is supported in part by cable DBE that passesthrough a frictionless ring at B. Knowing that the tension in thecable is 385 N, determine the components of the force exerted bythe cable on the support at D.= += + +===i j k(480 mm) (510 mm) (320 mm)(480 mm) (510 mm ) (320 mm)770 mmDBF PROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.91SOLUTION2 2 2385 N[(480 mm) (510 mm) (320 mm) ]770 mm(240 N) (255 N) (160 N)DBDBFDBFDB= += +i j ki j kFx = +240 N, Fy = 255 N, Fz = +160.0 N
92. 92. PROBLEM 2.90For the frame and cable of Problem 2.89, determine the componentsof the force exerted by the cable on the support at E.PROBLEM 2.89 A frame ABC is supported in part by cable DBEthat passes through a frictionless ring at B. Knowing that the tensionin the cable is 385 N, determine the components of the force exertedby the cable on the support at D.= += + +===i j k(270 mm) (400 mm) (600 mm)(270 mm) (400 mm) (600 mm)770 mmEBF PROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.92SOLUTION2 2 2385 N[(270 mm) (400 mm) (600 mm) ]770 mm(135 N) (200 N) (300 N)EBEBFEBFEB= += +i j kF i j kFx = +135.0 N, Fy = 200 N, Fz = +300 N
93. 93. PROBLEM 2.91Find the magnitude and direction of the resultant of the two forcesshown knowing that P = 600 N and Q = 450 N.= + + = + += + = + = += + +=P i j kPROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.93SOLUTION(600 N)[sin 40 sin 25 cos 40 sin 40 cos 25 ](162.992 N) (459.63 N) (349.54 N)(450 N)[cos55 cos30 sin 55 cos55 sin 30 ](223.53 N) (368.62 N) (129.055 N)(386.52 N) (828.25 N) (220.49 N)R (3i j kQ i j ki j kR P Qi j k86.52 N)2 (828.25 N)2 (220.49 N)2940.22 N+ += R = 940 N 386.52 Ncos940.22 NxxRR = = 65.7 x = 828.25 Ncos940.22 NyyRR = = 28.2 y = 220.49 Ncos940.22 NzzRR = = 76.4 z =
94. 94. PROBLEM 2.92Find the magnitude and direction of the resultant of the two forcesshown knowing that P = 450 N and Q = 600 N.= + + = + += + = + = += + +=P i j kPROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.94SOLUTION(450 N)[sin 40 sin 25 cos40 sin 40 cos 25 ](122.244 N) (344.72 N) (262.154 N)(600 N)[cos55 cos30 sin 55 cos55 sin 30 ](298.04 N) (491.49 N) (172.073 N)(420.28 N) (836.21N) (90.081N)R (i j kQ i j ki j kR P Qi j k420.28 N)2 + (836.21N)2 +(90.081N)2940.21N= R = 940 N 420.28cos940.21xxRR = = 63.4 x = 836.21cos940.21yyRR = = 27.2 y = 90.081cos940.21zzRR = = 84.5 z =
95. 95. PROBLEM 2.93Knowing that the tension is 425 lb in cable AB and 510 lb incable AC, determine the magnitude and direction of the resultantof the forces exerted at A by the two cables.= (40 in.) i (45 in.) j +(60 in.)k= (40 in.) + (45 in.) + (60 in.) =85 in.= (100 in.) (45 in.) +(60 in.)= (100 in.) + (45 in.) + (60 in.) =125 in.= = = +ABABACACi j kABT TT i j k912.92 lb x = = 48.2 x = 912.92 lb y = = 116.6 y = 912.92 lb z = = 53.4 z PROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.95SOLUTION2 2 22 2 2(40 in.) (45 in.) (60 in.)(425 lb)AB AB AB ABAB85 in. i j kT (200 lb) (225 lb) (300 lb)(100 in.) (45 in.) (60 in.)(510 lb)125 in.(408 lb) (183.6 lb) (244.8 lb)(608) (408.6 lb) (544.8 lb)ABAC AC AC ACACAB ACACT TAC = + + = = = = += + = +i j kT T i j kR T T i j kThen: R = 912.92 lb R = 913 lb and608 lbcos 0.66599408.6 lbcos 0.44757544.8 lbcos 0.59677=
96. 96. PROBLEM 2.94Knowing that the tension is 510 lb in cable AB and 425 lb in cableAC, determine the magnitude and direction of the resultant of theforces exerted at A by the two cables.= (40 in.) i (45 in.) j +(60 in.)k= (40 in.) + (45 in.) + (60 in.) =85 in.= (100 in.) (45 in.) +(60 in.)= (100 in.) + (45 in.) + (60 in.) =125 in.= = = +ABABACACi j kABT TT i j k912.92 lb x = = 50.6 x = 912.92 lb y = = 117.6 y = 912.92 lb z = = 51.8 z PROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.96SOLUTION2 2 22 2 2(40 in.) (45 in.) (60 in.)(510 lb)AB AB AB ABAB85 in. i j kT (240 lb) (270 lb) (360 lb)(100 in.) (45 in.) (60 in.)(425 lb)125 in.(340 lb) (153 lb) (204 lb)(580 lb) (423 lb) (564 lb)ABAC AC AC ACACAB ACACT TAC = + + = = = = += + = +i j kT T i j kR T T i j kThen: R = 912.92 lb R = 913 lb and580 lbcos 0.63532423 lbcos 0.46335564 lbcos 0.61780=