Solution 3.11
PROBLEM STATEMENT
The tailgate of a car is supported by the hydraulic lift BC. If the
lift exerts a 125-lb force directed along its centerline on the ball
and socket at B, determine the moment of the force about A.
First note
dCB  (12.0 in.)2  (2.33 in.)2
 12.2241 in.
Then
cos  
sin  
and
12.0 in.
12.2241 in.
2.33 in.
12.2241 in.
FCB  FCB cos  i  FCB sin  j

125 lb
12.2241 in.
[(12.0 in.) i  (2.33 in.) j]
Now
MA  rB/A  FCB
where
rB/A  (15.3 in.) i  (12.0 in.  2.33 in.) j
 (15.3 in.) i  (14.33 in.) j
Then
M A  [(15.3 in.)i  (14.33 in.) j]
125 lb
12.2241 in.
(12.0i  2.33 j)
 (1393.87 lb  in.)k
 (116.156 lb  ft)k
or M A  116.2 lb  ft
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Solution 3.14
PROBLEM STATEMENT
It is known that the connecting rod AB exerts on the crank BC a 2.5-kN force
directed down and to the left along the centerline of AB. Determine the moment of
the force about C.
Using (a):
MC  y1 (FAB ) x  x1 (FAB ) y
 24

7

 (0.056 m)   2500 N  (0.042 m)   2500 N
 25

 25

 61.6 N  m
(a)
MC  61.6 N  m
◀
Using (b):
MC  y2 (FAB ) x
7

 (0.088 m)   2500 N
 25

 61.6 N  m
(b)
MC  61.6 N  m
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Solution 3.21
PROBLEM STATEMENT
Before the trunk of a large tree is felled, cables AB and BC are
attached as shown. Knowing that the tensions in cables AB and
BC are 555 N and 660 N, respectively, determine the moment
about O of the resultant force exerted on the tree by the cables
at B.
dBA  (0.75 m)2  (7 m)2  (6 m)2  9.25 m
dBC  (4.25 m)2  (7 m)2  (1 m)2  8.25 m
Have

BA
555 N

(0.75 m i  7 m j  6 m k )
dB A 9.25 m
TB A  TB A
TBA  (45 N)i  (420 N)j  (360 N)k

660 N
BC
(4.25 m i  7 m j  k )

dBC 8.25 m
TBC  TBC
TBC  (340 N)i  (540 N) j  (80 N)k
R  TBA  TBC
 

 

 R  295 N i  980 N j  440 N k
MO  rB / O  R where rB / O  (7 m)j
i
MO  0
j
7
k
0 Nm
295 980 440

 

 3080 N  m i  2065 N  m k

 

MO  3080 N  m i  2070 N  m k ◀
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Solution 3.38
PROBLEM STATEMENT
Three cables are attached to the top of the tower at A.
Determine the angle formed by cables AD and AB.
First note:

AB  (16 m)i  (48 m) j  (12 m)k

AD  (14 m)i  (48 m) j
AB  (16 m)2  (48 m)2  (12 m)2
 52.0 m
AD  (14 m)2  (48 m)2
 50.0 m
By definition,
 
AB  AD  ( AB)( AD)cos BAD
 
AB  AD
cos BAD 
 AB AD
(16 m)i  (48 m) j  (12 m)k  (14 m)i  (48 m) j
52.0 m50.0 m
2
1614  48
cos BAD 
52.050.0
cos BAD 
cos BAD  0.800
  36.9 ◀
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Solution 3.55
PROBLEM STATEMENT
A force P of magnitude 520 lb acts on the frame shown at
point E. Determine the moment of P about a line joining
points O and D.
First develop expression for P in terms of its unit vector:

EH  (30 in.)i  (7.5 in.) j  (10 in.)k
EH  (30)2  (7.5)2  (10)2  32.50 in.

EH 30i  7.5 j  10k
λEH 

EH
32.50
λEH  0.92308i  0.23077 j  0.30769k
P  PλEH  520 lb 0.92308i  0.23077 j  0.30770 k
P  (480.0 lb)i  (120.0 lb) j  (160.0 lb)k
The moment of P about axis OD is given by the mixed triple product Eq. (3.43):
MOD  λOD  (rE / D  P )
rE / D  (30 in.)i  (7.5 in.) j
λOD
λOD
MOD

OD 30i  15 j  10k


OD
35.0
 0.85714 i  0.42857 j  0.28571k
0.85714 0.42857 0.28571
 λOD  (rE / D  P )  30.0 in. 7.50 in.
0
120 lb 160 lb
480 lb
MOD  0.85714[(7.50 in.)(160 lb)  0]
 0.42857[0  (30.0 in.)(160 lb)]
 0.28571[(30.0 in.)(120 lb)  (7.50)(480]
 3085.7 lb  in.
MOD  3090 lb  in.
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authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded,
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