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. 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. 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. 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. + + = 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. 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. 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. + + = 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. = = = = = = = = = =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. = = = =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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. = + = + += + + =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. = += + += + +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. + + =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. + + =+ + =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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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=