PROBLEM 3.79
If P = 20 lb, replace the three couples with a single
equivalent couple, specifying its magnitude and the
direction of its axis.
SOLUTION
From the solution to Problem. 3.78:
16-lb force:
M1 = −(480 lb ⋅ in.)k
40-lb force:
M 2 = 8 5[(10 lb ⋅ in.)i + (30 lb ⋅ in.) j + (15 lb ⋅ in.)k ]
P = 20 lb
M 3 = rC × P
= (30 in.)i × (20 lb)k
= (600 lb ⋅ in.) j
M = M1 + M 2 + M 3
= −(480)k + 8 5 (10i + 30 j + 15k ) + 600 j
= (178.885 lb ⋅ in.)i + (1136.66 lb ⋅ in.) j − (211.67 lb ⋅ in.)k
M = (178.885) 2 + (113.66) 2 + (211.67)2
M = 1170 lb ⋅ in. 
= 1169.96 lb ⋅ in.
M
= 0.152898i + 0.97154 j − 0.180921k
M
cos θ x = 0.152898
λ axis =
cos θ y = 0.97154
θ x = 81.2° θ y = 13.70° θ z = 100.4° 
cos θ z = −0.180921
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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240
PROBLEM 3.87
Three control rods attached to a lever ABC exert on it the forces
shown. (a) Replace the three forces with an equivalent force-couple
system at B. (b) Determine the single force that is equivalent to the
force-couple system obtained in part a, and specify its point of
application on the lever.
SOLUTION
(a)
First note that the two 90-N forces form a couple. Then
F = 216 N
where
and
θ
θ = 180° − (60° + 55°) = 65°
M = ΣM B
= (0.450 m)(216 N) cos 55° − (1.050 m)(90 N) cos 20°
= −33.049 N ⋅ m
The equivalent force-couple system at B is
F = 216 N
(b)
65.0°; M = 33.0 N ⋅ m

The single equivalent force F′ is equal to F. Further, since the sense of M is clockwise, F′ must be
applied between A and B. For equivalence,
ΣM B : M = aF ′ cos 55°
where a is the distance from B to the point of application of F′. Then
−33.049 N ⋅ m = − a(216 N) cos 55°
a = 0.26676 m
or
F ′ = 216 N
65.0° applied to the lever 267 mm to the left of B 
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 limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
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248
PROBLEM 3.93
An antenna is guyed by three cables as shown. Knowing
that the tension in cable AB is 288 lb, replace the force
exerted at A by cable AB with an equivalent force-couple
system at the center O of the base of the antenna.
SOLUTION
We have
d AB = (−64)2 + (−128) 2 + (16) 2 = 144 ft
Then
TAB =
Now
288 lb
( −64i − 128 j + 16k )
144
= (32 lb)( −4i − 8 j + k )
M = M O = rA / O × TAB
= 128 j × 32(−4i − 8 j + k )
= (4096 lb ⋅ ft)i + (16,384 lb ⋅ ft)k
The equivalent force-couple system at O is
F = −(128.0 lb)i − (256 lb) j + (32.0 lb)k 
M = (4.10 kip ⋅ ft)i + (16.38 kip ⋅ ft)k 
PROPRIETARY MATERIAL. © 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 limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
254