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MANUAL
CHAPTER 2
PROBLEM 2.1
Two forces are applied at point B of beam AB. Determine
graphically the magnitude and direction of their resultant using
(a) the parallelogram law, (b) the triangle rule.
SOLUTION
(a)
Parallelogram law:
(b)
Triangle rule:
We measure:
R = 3.30 kN, α = 66.6&deg;
R = 3.30 kN
66.6&deg; 
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3
PROBLEM 2.2
The cable stays AB and AD help support pole AC. Knowing
that the tension is 120 lb in AB and 40 lb in AD, determine
graphically the magnitude and direction of the resultant of the
forces exerted by the stays at A using (a) the parallelogram
law, (b) the triangle rule.
SOLUTION
We measure:
(a)
Parallelogram law:
(b)
Triangle rule:
We measure:
α = 51.3&deg;
β = 59.0&deg;
R = 139.1 lb,
γ = 67.0&deg;
R = 139.1 lb
67.0&deg; 

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4
PROBLEM 2.3
Two structural members B and C are bolted to bracket A. Knowing that both
members are in tension and that P = 10 kN and Q = 15 kN, determine
graphically the magnitude and direction of the resultant force exerted on the
bracket using (a) the parallelogram law, (b) the triangle rule.
SOLUTION
(a)
Parallelogram law:
(b)
Triangle rule:
We measure:
α = 21.2&deg;
R = 20.1 kN,
R = 20.1 kN
21.2&deg; 
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5
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 force exerted on
the bracket using (a) the parallelogram law, (b) the triangle rule.
SOLUTION
(a)
Parallelogram law:
(b)
Triangle rule:
We measure:
R = 8.03 kips, α = 3.8&deg;
R = 8.03 kips
3.8&deg; 
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6
PROBLEM 2.5
A stake is being pulled out of the ground by means of two ropes as shown.
Knowing that α = 30&deg;, determine by trigonometry (a) the magnitude of the
force P so that the resultant force exerted on the stake is vertical, (b) the
corresponding magnitude of the resultant.
SOLUTION
Using the triangle rule and the law of sines:
(a)
(b)
120 N
P
=
sin 30&deg; sin 25&deg;
P = 101.4 N 
30&deg; + β + 25&deg; = 180&deg;
β = 180&deg; − 25&deg; − 30&deg;
= 125&deg;
120 N
R
=
sin 30&deg; sin125&deg;
R = 196.6 N 
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7
PROBLEM 2.6
A trolley that moves along a horizontal beam is acted upon by two
forces as shown. (a) Knowing that α = 25&deg;, determine by trigonometry
the magnitude of the force P so that the resultant force exerted on the
trolley is vertical. (b) What is the corresponding magnitude of the
resultant?
SOLUTION
Using the triangle rule and the law of sines:
(a)
(b)
1600 N
P
=
sin 25&deg; sin 75&deg;
P = 3660 N 
25&deg; + β + 75&deg; = 180&deg;
β = 180&deg; − 25&deg; − 75&deg;
= 80&deg;
1600 N
R
=
sin 25&deg; sin 80&deg;
R = 3730 N 
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8
PROBLEM 2.7
A trolley that moves along a horizontal beam is acted upon by two forces
as shown. Determine by trigonometry the magnitude and direction of the
force P so that the resultant is a vertical force of 2500 N.
SOLUTION
Using the law of cosines:
Using the law of sines:
P 2 = (1600 N)2 + (2500 N)2 − 2(1600 N)(2500 N) cos 75&deg;
P = 2596 N
sin α
sin 75&deg;
=
1600 N 2596 N
α = 36.5&deg;
P is directed 90&deg; − 36.5&deg; or 53.5&deg; below the horizontal.
P = 2600 N
53.5&deg; 
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9
PROBLEM 2.8
A telephone cable is clamped at A to the pole AB. Knowing that the tension
in the left-hand portion of the cable is T1 = 800 lb, determine by
trigonometry (a) the required tension T2 in the right-hand portion if the
resultant R of the forces exerted by the cable at A is to be vertical, (b) the
corresponding magnitude of R.
SOLUTION
Using the triangle rule and the law of sines:
(a)
75&deg; + 40&deg; + α = 180&deg;
α = 180&deg; − 75&deg; − 40&deg;
= 65&deg;
(b)
T2
800 lb
=
sin 65&deg; sin 75&deg;
T2 = 853 lb 
800 lb
R
=
sin 65&deg; sin 40&deg;
R = 567 lb 
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10
PROBLEM 2.9
A telephone cable is clamped at A to the pole AB. Knowing that the
tension in the right-hand portion of the cable is T2 = 1000 lb, determine
by trigonometry (a) the required tension T1 in the left-hand portion if
the resultant R of the forces exerted by the cable at A is to be vertical,
(b) the corresponding magnitude of R.
SOLUTION
Using the triangle rule and the law of sines:
(a)
75&deg; + 40&deg; + β = 180&deg;
β = 180&deg; − 75&deg; − 40&deg;
= 65&deg;
(b)
T1
1000 lb
=
sin 75&deg; sin 65&deg;
T1 = 938 lb 
1000 lb
R
=
sin 75&deg; sin 40&deg;
R = 665 lb 
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11
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) the required angle
α if the resultant R of the two forces applied to the support is to be
horizontal, (b) the corresponding magnitude of R.
SOLUTION
Using the triangle rule and law of sines:
(a)
sin α sin 25&deg;
=
50 N
35 N
sin α = 0.60374
α = 37.138&deg;
(b)
α = 37.1&deg; 
α + β + 25&deg; = 180&deg;
β = 180&deg; − 25&deg; − 37.138&deg;
= 117.862&deg;
R
35 N
=
sin117.862&deg; sin 25&deg;
R = 73.2 N 
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12
PROBLEM 2.11
A steel tank is to be positioned in an excavation. Knowing that
α = 20&deg;, determine by trigonometry (a) the required magnitude
of the force P if the resultant R of the two forces applied at A is
to be vertical, (b) the corresponding magnitude of R.
SOLUTION
Using the triangle rule and the law of sines:
(a)
β + 50&deg; + 60&deg; = 180&deg;
β = 180&deg; − 50&deg; − 60&deg;
= 70&deg;
(b)
425 lb
P
=
sin 70&deg; sin 60&deg;
P = 392 lb 
425 lb
R
=
sin 70&deg; sin 50&deg;
R = 346 lb 
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13
PROBLEM 2.12
A steel tank is to be positioned in an excavation. Knowing that
the magnitude of P is 500 lb, determine by trigonometry (a) the
required angle α if the resultant R of the two forces applied at A
is to be vertical, (b) the corresponding magnitude of R.
SOLUTION
Using the triangle rule and the law of sines:
(a)
(α + 30&deg;) + 60&deg; + β = 180&deg;
β = 180&deg; − (α + 30&deg;) − 60&deg;
β = 90&deg; − α
sin (90&deg; − α ) sin 60&deg;
425 lb
=
500 lb
90&deg; − α = 47.402&deg;
(b)
R
500 lb
=
sin (42.598&deg; + 30&deg;) sin 60&deg;
α = 42.6&deg; 
R = 551 lb 
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14
PROBLEM 2.13
A steel tank is to be positioned in an excavation. Determine by
trigonometry (a) the magnitude and direction of the smallest
force P for which the resultant R of the two forces applied at A
is vertical, (b) the corresponding magnitude of R.
SOLUTION
The smallest force P will be perpendicular to R.
(a)
P = (425 lb) cos 30&deg;
(b)
R = (425 lb)sin 30&deg;
P = 368 lb

R = 213 lb 
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15
PROBLEM 2.14
For the hook support of Prob. 2.10, determine by trigonometry (a) the
magnitude and direction of the smallest force P for which the resultant R of
the two forces applied to the support is horizontal, (b) the corresponding
magnitude of R.
SOLUTION
The smallest force P will be perpendicular to R.
(a)
P = (50 N)sin 25&deg;
(b)
R = (50 N) cos 25&deg;
P = 21.1 N

R = 45.3 N 
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16
PROBLEM 2.15
Solve Problem 2.2 by trigonometry.
PROBLEM 2.2 The cable stays AB and AD help support pole
AC. Knowing that the tension is 120 lb in AB and 40 lb in AD,
determine graphically the magnitude and direction of the
resultant of the forces exerted by the stays at A using (a) the
parallelogram law, (b) the triangle rule.
SOLUTION
8
10
α = 38.66&deg;
tan α =
6
10
β = 30.96&deg;
tan β =
Using the triangle rule:
Using the law of cosines:
α + β + ψ = 180&deg;
38.66&deg; + 30.96&deg; + ψ = 180&deg;
ψ = 110.38&deg;
R 2 = (120 lb)2 + (40 lb) 2 − 2(120 lb)(40 lb) cos110.38&deg;
R = 139.08 lb
Using the law of sines:
sin γ
sin110.38&deg;
=
40 lb
139.08 lb
γ = 15.64&deg;
φ = (90&deg; − α ) + γ
φ = (90&deg; − 38.66&deg;) + 15.64&deg;
φ = 66.98&deg;
R = 139.1 lb
67.0&deg; 
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17
PROBLEM 2.16
Solve 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 force
exerted on the bracket using (a) the parallelogram law, (b) the triangle rule.
SOLUTION
Using the force triangle and the laws of cosines and sines:
We have:
γ = 180&deg; − (50&deg; + 25&deg;)
= 105&deg;
Then
R 2 = (4 kips) 2 + (6 kips)2 − 2(4 kips)(6 kips) cos105&deg;
= 64.423 kips 2
R = 8.0264 kips
And
4 kips
8.0264 kips
=
sin(25&deg; + α )
sin105&deg;
sin(25&deg; + α ) = 0.48137
25&deg; + α = 28.775&deg;
α = 3.775&deg;
R = 8.03 kips
3.8&deg; 
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18
PROBLEM 2.17
For 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 so
that the resultant is a vertical force of 160 N.
PROBLEM 2.5 A stake is being pulled out of the ground by means of two
ropes as shown. Knowing that α = 30&deg;, determine by trigonometry (a) the
magnitude of the force P so that the resultant force exerted on the stake is
vertical, (b) the corresponding magnitude of the resultant.
SOLUTION
Using the laws of cosines and sines:
P 2 = (120 N) 2 + (160 N)2 − 2(120 N)(160 N) cos 25&deg;
P = 72.096 N
And
sin α
sin 25&deg;
=
120 N 72.096 N
sin α = 0.70343
α = 44.703&deg;
P = 72.1 N
44.7&deg; 
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19
PROBLEM 2.18
For the hook support of Prob. 2.10, knowing that P = 75 N and α = 50&deg;,
determine by trigonometry the magnitude and direction of the resultant of
the 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) the
required angle α if the resultant R of the two forces applied to the support is
to be horizontal, (b) the corresponding magnitude of R.
SOLUTION
Using the force triangle and the laws of cosines and sines:
We have
β = 180&deg; − (50&deg; + 25&deg;)
= 105&deg;
Then
R 2 = (75 N) 2 + (50 N) 2
− 2(75 N)(50 N) cos 105&deg;
R = 10, 066.1 N 2
R = 100.330 N
2
and
Hence:
sin γ
sin105&deg;
=
75 N 100.330 N
sin γ = 0.72206
γ = 46.225&deg;
γ − 25&deg; = 46.225&deg; − 25&deg; = 21.225&deg;
R = 100.3 N
21.2&deg; 
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20
PROBLEM 2.19
Two 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 the
magnitude and direction of the resultant of the two forces.
SOLUTION
Using the force triangle and the laws of cosines and sines:
We have
γ = 180&deg; − (20&deg; + 10&deg;)
= 150&deg;
Then
R 2 = (48 N)2 + (60 N)2
− 2(48 N)(60 N) cos150&deg;
R = 104.366 N
and
Hence:
48 N 104.366 N
=
sin α
sin150&deg;
sin α = 0.22996
α = 13.2947&deg;
φ = 180&deg; − α − 80&deg;
= 180&deg; − 13.2947&deg; − 80&deg;
= 86.705&deg;
R = 104.4 N
86.7&deg; 
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21
PROBLEM 2.20
Two 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 the
magnitude and direction of the resultant of the two forces.
SOLUTION
Using the force triangle and the laws of cosines and sines:
We have
γ = 180&deg; − (20&deg; + 10&deg;)
= 150&deg;
Then
R 2 = (60 N)2 + (48 N) 2
− 2(60 N)(48 N) cos 150&deg;
R = 104.366 N
and
Hence:
60 N 104.366 N
=
sin α
sin150&deg;
sin α = 0.28745
α = 16.7054&deg;
φ = 180&deg; − α − 180&deg;
= 180&deg; − 16.7054&deg; − 80&deg;
= 83.295&deg;
R = 104.4 N
83.3&deg; 
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22
PROBLEM 2.21
Determine the x and y components of each of the forces shown.
SOLUTION
80-N Force:
120-N Force:
150-N Force:
Fx = + (80 N) cos 40&deg;
Fx = 61.3 N 
Fy = + (80 N) sin 40&deg;
Fy = 51.4 N 
Fx = + (120 N) cos 70&deg;
Fx = 41.0 N 
Fy = +(120 N) sin 70&deg;
Fy = 112.8 N 
Fx = −(150 N) cos 35&deg;
Fx = −122. 9 N 
Fy = +(150 N) sin 35&deg;
Fy = 86.0 N 
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23
PROBLEM 2.22
Determine the x and y components of each of the forces shown.
SOLUTION
40-lb Force:
50-lb Force:
60-lb Force:
Fx = + (40 lb) cos 60&deg;
Fx = 20.0 lb 
Fy = −(40 lb)sin 60&deg;
Fy = −34.6 lb 
Fx = −(50 lb)sin 50&deg;
Fx = −38.3 lb 
Fy = −(50 lb) cos 50&deg;
Fy = −32.1 lb 
Fx = + (60 lb) cos 25&deg;
Fx = 54.4 lb 
Fy = +(60 lb)sin 25&deg;
Fy = 25.4 lb 
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24
PROBLEM 2.23
Determine the x and y components of each of the forces shown.
SOLUTION
Compute the following distances:
OA = (600) 2 + (800) 2
= 1000 mm
OB = (560)2 + (900) 2
= 1060 mm
OC = (480) 2 + (900)2
= 1020 mm
800-N Force:
424-N Force:
408-N Force:
Fx = + (800 N)
800
1000
Fx = +640 N 
Fy = +(800 N)
600
1000
Fy = +480 N 
Fx = −(424 N)
560
1060
Fx = −224 N 
Fy = −(424 N)
900
1060
Fy = −360 N 
Fx = + (408 N)
480
1020
Fx = +192.0 N 
Fy = −(408 N)
900
1020
Fy = −360 N 
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25
PROBLEM 2.24
Determine the x and y components of each of the forces shown.
SOLUTION
Compute the following distances:
OA = (24 in.)2 + (45 in.)2
= 51.0 in.
OB = (28 in.) 2 + (45 in.) 2
= 53.0 in.
OC = (40 in.) 2 + (30 in.) 2
= 50.0 in.
102-lb Force:
106-lb Force:
200-lb Force:
Fx = −102 lb
24 in.
51.0 in.
Fx = −48.0 lb 
Fy = +102 lb
45 in.
51.0 in.
Fy = +90.0 lb 
Fx = +106 lb
28 in.
53.0 in.
Fx = +56.0 lb 
Fy = +106 lb
45 in.
53.0 in.
Fy = +90.0 lb 
Fx = −200 lb
40 in.
50.0 in.
Fx = −160.0 lb 
Fy = −200 lb
30 in.
50.0 in.
Fy = −120.0 lb 
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26
PROBLEM 2.25
The hydraulic cylinder BD exerts on member ABC a force P directed
along line BD. Knowing that P must have a 750-N component
perpendicular to member ABC, determine (a) the magnitude of the force
P, (b) its component parallel to ABC.
SOLUTION
(a)
750 N = P sin 20&deg;
P = 2192.9 N
(b)
P = 2190 N 
PABC = P cos 20&deg;
= (2192.9 N) cos 20&deg;
PABC = 2060 N 
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27
PROBLEM 2.26
Cable AC exerts on beam AB a force P directed along line AC. Knowing that
P must have a 350-lb vertical component, determine (a) the magnitude of the
force P, (b) its horizontal component.
SOLUTION
(a)
P=
Py
cos 55&deg;
350 lb
cos 55&deg;
= 610.21 lb
=
(b)
P = 610 lb 
Px = P sin 55&deg;
= (610.21 lb)sin 55&deg;
= 499.85 lb
Px = 500 lb 
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28
PROBLEM 2.27
Member BC exerts on member AC a force P directed along line BC. Knowing
that P must have a 325-N horizontal component, determine (a) the magnitude
of the force P, (b) its vertical component.
SOLUTION
BC = (650 mm)2 + (720 mm) 2
= 970 mm
 650 
Px = P 

 970 
(a)
or
 970 
P = Px 

 650 
 970 
= 325 N 

 650 
= 485 N
P = 485 N 
(b)
 720 
Py = P 

 970 
 720 
= 485 N 

 970 
= 360 N
Py = 970 N 
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29
PROBLEM 2.28
Member 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.
SOLUTION
(a)
P=
(b)
Px =
Py
sin 40&deg;
=
Py
tan 40&deg;
240 lb
sin 40&deg;
=
240 lb
tan 40&deg;
or P = 373 lb 
or Px = 286 lb 
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30
PROBLEM 2.29
The guy wire BD exerts on the telephone pole AC a force P directed along
BD. Knowing that P must have a 720-N component perpendicular to the
pole AC, determine (a) the magnitude of the force P, (b) its component
along line AC.
SOLUTION
(a)
37
Px
12
37
=
(720 N)
12
= 2220 N
P=
P = 2.22 kN 
(b)
35
Px
12
35
= (720 N)
12
= 2100 N
Py =
Py = 2.10 kN 
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31
PROBLEM 2.30
The hydraulic cylinder BC exerts on member AB a force P directed along
line BC. Knowing that P must have a 600-N component perpendicular to
member AB, determine (a) the magnitude of the force P, (b) its component
along line AB.
SOLUTION
180&deg; = 45&deg; + α + 90&deg; + 30&deg;
α = 180&deg; − 45&deg; − 90&deg; − 30&deg;
= 15&deg;
(a)
Px
P
P
P= x
cos α
600 N
=
cos15&deg;
= 621.17 N
cos α =
P = 621 N 
(b)
tan α =
Py
Px
Py = Px tan α
= (600 N) tan15&deg;
= 160.770 N
Py = 160.8 N 
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32
PROBLEM 2.31
Determine the resultant of the three forces of Problem 2.23.
PROBLEM 2.23 Determine the x and y components of each of
the forces shown.
SOLUTION
Components of the forces were determined in Problem 2.23:
Force
x Comp. (N)
y Comp. (N)
800 lb
+640
+480
424 lb
–224
–360
408 lb
+192
–360
Rx = +608
Ry = −240
R = Rx i + Ry j
= (608 lb)i + (−240 lb) j
Ry
tan α =
Rx
240
608
α = 21.541&deg;
240 N
R=
sin(21.541&deg;)
= 653.65 N
=
R = 654 N
21.5&deg; 
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33
PROBLEM 2.32
Determine the resultant of the three forces of Problem 2.21.
PROBLEM 2.21 Determine the x and y components of each of the
forces shown.
SOLUTION
Components of the forces were determined in Problem 2.21:
Force
x Comp. (N)
y Comp. (N)
80 N
+61.3
+51.4
120 N
+41.0
+112.8
150 N
–122.9
+86.0
Rx = −20.6
Ry = + 250.2
R = Rx i + Ry j
= ( −20.6 N)i + (250.2 N) j
tan α =
Ry
Rx
250.2 N
20.6 N
tan α = 12.1456
tan α =
α = 85.293&deg;
R=
250.2 N
sin 85.293&deg;
R = 251 N
85.3&deg; 
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34
PROBLEM 2.33
Determine the resultant of the three forces of Problem 2.22.
PROBLEM 2.22 Determine the x and y components of each of the
forces shown.
SOLUTION
Force
x Comp. (lb)
y Comp. (lb)
40 lb
+20.00
–34.64
50 lb
–38.30
–32.14
60 lb
+54.38
+25.36
Rx = +36.08
Ry = −41.42
R = Rx i + Ry j
= ( +36.08 lb)i + (−41.42 lb) j
tan α =
Ry
Rx
41.42 lb
36.08 lb
tan α = 1.14800
tan α =
α = 48.942&deg;
R=
41.42 lb
sin 48.942&deg;
R = 54.9 lb
48.9&deg; 
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35
PROBLEM 2.34
Determine the resultant of the three forces of Problem 2.24.
PROBLEM 2.24 Determine the x and y components of each of the
forces shown.
SOLUTION
Components of the forces were determined in Problem 2.24:
Force
x Comp. (lb)
y Comp. (lb)
102 lb
−48.0
+90.0
106 lb
+56.0
+90.0
200 lb
−160.0
−120.0
Rx = −152.0
Ry = 60.0
R = Rx i + Ry j
= ( −152 lb)i + (60.0 lb) j
tan α =
Ry
Rx
60.0 lb
152.0 lb
tan α = 0.39474
tan α =
α = 21.541&deg;
R=
60.0 lb
sin 21.541&deg;
R = 163.4 lb
21.5&deg; 
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36
PROBLEM 2.35
Knowing that α = 35&deg;, determine the resultant of the three forces
shown.
SOLUTION
Fx = +(100 N) cos 35&deg; = +81.915 N
100-N Force:
Fy = −(100 N)sin 35&deg; = −57.358 N
Fx = +(150 N) cos 65&deg; = +63.393 N
150-N Force:
Fy = −(150 N) sin 65&deg; = −135.946 N
Fx = −(200 N) cos 35&deg; = −163.830 N
200-N Force:
Fy = −(200 N)sin 35&deg; = −114.715 N
Force
x Comp. (N)
y Comp. (N)
100 N
+81.915
−57.358
150 N
+63.393
−135.946
200 N
−163.830
−114.715
Rx = −18.522
Ry = −308.02
R = Rx i + Ry j
= (−18.522 N)i + (−308.02 N) j
tan α =
Ry
Rx
308.02
18.522
α = 86.559&deg;
=
R=
308.02 N
sin 86.559
R = 309 N
86.6&deg; 
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37
PROBLEM 2.36
Knowing that the tension in rope AC is 365 N, determine the
resultant of the three forces exerted at point C of post BC.
SOLUTION
Determine force components:
Cable force AC:
500-N Force:
200-N Force:
and
960
= −240 N
1460
1100
= −275 N
Fy = −(365 N)
1460
Fx = −(365 N)
24
= 480 N
25
7
Fy = (500 N)
= 140 N
25
Fx = (500 N)
4
= 160 N
5
3
Fy = −(200 N) = −120 N
5
Fx = (200 N)
Rx = ΣFx = −240 N + 480 N + 160 N = 400 N
Ry = ΣFy = −275 N + 140 N − 120 N = −255 N
R = Rx2 + Ry2
= (400 N) 2 + (−255 N) 2
= 474.37 N
Further:
255
400
α = 32.5&deg;
tan α =
R = 474 N
32.5&deg; 
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38
PROBLEM 2.37
Knowing that α = 40&deg;, determine the resultant of the three forces
shown.
SOLUTION
60-lb Force:
Fx = (60 lb) cos 20&deg; = 56.382 lb
Fy = (60 lb)sin 20&deg; = 20.521 lb
80-lb Force:
Fx = (80 lb) cos 60&deg; = 40.000 lb
Fy = (80 lb)sin 60&deg; = 69.282 lb
120-lb Force:
Fx = (120 lb) cos 30&deg; = 103.923 lb
Fy = −(120 lb)sin 30&deg; = −60.000 lb
and
Rx = ΣFx = 200.305 lb
Ry = ΣFy = 29.803 lb
R = (200.305 lb)2 + (29.803 lb) 2
= 202.510 lb
Further:
tan α =
29.803
200.305
α = tan −1
29.803
200.305
R = 203 lb
= 8.46&deg;
8.46&deg; 
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39
PROBLEM 2.38
Knowing that α = 75&deg;, determine the resultant of the three forces
shown.
SOLUTION
60-lb Force:
Fx = (60 lb) cos 20&deg; = 56.382 lb
Fy = (60 lb) sin 20&deg; = 20.521 lb
80-lb Force:
Fx = (80 lb) cos 95&deg; = −6.9725 lb
Fy = (80 lb)sin 95&deg; = 79.696 lb
120-lb Force:
Fx = (120 lb) cos 5 &deg; = 119.543 lb
Fy = (120 lb)sin 5&deg; = 10.459 lb
Then
Rx = ΣFx = 168.953 lb
Ry = ΣFy = 110.676 lb
and
R = (168.953 lb) 2 + (110.676 lb)2
= 201.976 lb
110.676
168.953
tan α = 0.65507
α = 33.228&deg;
tan α =
R = 202 lb
33.2&deg; 
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40
PROBLEM 2.39
For the collar of Problem 2.35, determine (a) the required value of
α if the resultant of the three forces shown is to be vertical, (b) the
corresponding magnitude of the resultant.
SOLUTION
Rx = ΣFx
= (100 N) cos α + (150 N) cos (α + 30&deg;) − (200 N) cos α
Rx = −(100 N) cos α + (150 N) cos (α + 30&deg;)
(1)
Ry = ΣFy
= −(100 N) sin α − (150 N)sin (α + 30&deg;) − (200 N)sin α
Ry = −(300 N) sin α − (150 N)sin (α + 30&deg;)
(a)
(2)
For R to be vertical, we must have Rx = 0. We make Rx = 0 in Eq. (1):
−100 cos α + 150cos (α + 30&deg;) = 0
−100cos α + 150 (cos α cos 30&deg; − sin α sin 30&deg;) = 0
29.904cos α = 75sin α
29.904
75
= 0.39872
tan α =
α = 21.738&deg;
(b)
α = 21.7&deg; 
Substituting for α in Eq. (2):
Ry = −300sin 21.738&deg; − 150sin 51.738&deg;
= −228.89 N
R = | Ry | = 228.89 N
R = 229 N 
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41
PROBLEM 2.40
For the post of Prob. 2.36, determine (a) the required tension in
rope AC if the resultant of the three forces exerted at point C is
to be horizontal, (b) the corresponding magnitude of the
resultant.
SOLUTION
Rx = ΣFx = −
Rx = −
48
TAC + 640 N
73
R y = ΣFy = −
Ry = −
(a)
960
24
4
TAC +
(500 N) + (200 N)
1460
25
5
1100
7
3
TAC +
(500 N) − (200 N)
1460
25
5
55
TAC + 20 N
73
(2)
For R to be horizontal, we must have R y = 0.
Set R y = 0 in Eq. (2):
−
55
TAC + 20 N = 0
73
TAC = 26.545 N
(b)
(1)
TAC = 26.5 N 
Substituting for TAC into Eq. (1) gives
48
(26.545 N) + 640 N
73
Rx = 622.55 N
Rx = −
R = Rx = 623 N
R = 623 N 
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42
PROBLEM 2.41
A hoist trolley is subjected to the three forces shown. Knowing that α = 40&deg;,
determine (a) the required magnitude of the force P if the resultant of
the three forces is to be vertical, (b) the corresponding magnitude of
the resultant.
SOLUTION
Rx =
ΣFx = P + (200 lb)sin 40&deg; − (400 lb) cos 40&deg;
Rx = P − 177.860 lb
Ry =
ΣFy = (200 lb) cos 40&deg; + (400 lb) sin 40&deg;
Ry = 410.32 lb
(a)
(2)
For R to be vertical, we must have Rx = 0.
Set
Rx = 0 in Eq. (1)
0 = P − 177.860 lb
P = 177.860 lb
(b)
(1)
P = 177.9 lb 
Since R is to be vertical:
R = Ry = 410 lb
R = 410 lb 
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43
PROBLEM 2.42
A hoist trolley is subjected to the three forces shown. Knowing
that P = 250 lb, determine (a) the required value of α if the
resultant of the three forces is to be vertical, (b) the corresponding
magnitude of the resultant.
SOLUTION
Rx =
ΣFx = 250 lb + (200 lb)sin α − (400 lb) cos α
Rx = 250 lb + (200 lb)sin α − (400 lb) cos α
Ry =
(a)
(1)
ΣFy = (200 lb) cos α + (400 lb)sin α
For R to be vertical, we must have Rx = 0.
Rx = 0 in Eq. (1)
Set
0 = 250 lb + (200 lb)sin α − (400 lb) cos α
(400 lb) cos α = (200 lb) sin α + 250 lb
2 cos α = sin α + 1.25
4cos 2 α = sin 2 α + 2.5sin α + 1.5625
4(1 − sin 2 α ) = sin 2 α + 2.5sin α + 1.5625
0 = 5sin 2 α + 2.5sin α − 2.4375
Using the quadratic formula to solve for the roots gives
sin α = 0.49162
α = 29.447&deg;
or
(b)
α = 29.4&deg; 
Since R is to be vertical:
R = Ry = (200 lb) cos 29.447&deg; + (400 lb) sin 29.447&deg;
R = 371 lb 
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44
PROBLEM 2.43
Two cables are tied together at C and are loaded as shown.
Determine the tension (a) in cable AC, (b) in cable BC.
SOLUTION
Free-Body Diagram
1100
960
α = 48.888&deg;
400
tan β =
960
β = 22.620&deg;
tan α =
Force Triangle
Law of sines:
TAC
TBC
15.696 kN
=
=
sin 22.620&deg; sin 48.888&deg; sin108.492&deg;
(a)
TAC =
15.696 kN
(sin 22.620&deg;)
sin108.492&deg;
TAC = 6.37 kN 
(b)
TBC =
15.696 kN
(sin 48.888&deg;)
sin108.492&deg;
TBC = 12.47 kN 
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45
PROBLEM 2.44
Two cables are tied together at C and are loaded as shown. Determine
the tension (a) in cable AC, (b) in cable BC.
SOLUTION
3
2.25
α = 53.130&deg;
1.4
tan β =
2.25
β = 31.891&deg;
tan α =
Free-Body Diagram
Law of sines:
Force-Triangle
TAC
TBC
660 N
=
=
sin 31.891&deg; sin 53.130&deg; sin 94.979&deg;
(a)
TAC =
660 N
(sin 31.891&deg;)
sin 94.979&deg;
TAC = 350 N 
(b)
TBC =
660 N
(sin 53.130&deg;)
sin 94.979&deg;
TBC = 530 N 
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46
PROBLEM 2.45
Knowing that α = 20&deg;, determine the tension (a) in cable AC,
(b) in rope BC.
SOLUTION
Free-Body Diagram
Law of sines:
Force Triangle
TAC
T
1200 lb
= BC =
sin 110&deg; sin 5&deg; sin 65&deg;
(a)
TAC =
1200 lb
sin 110&deg;
sin 65&deg;
TAC = 1244 lb 
(b)
TBC =
1200 lb
sin 5&deg;
sin 65&deg;
TBC = 115.4 lb 
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47
PROBLEM 2.46
Knowing that α = 55&deg; and that boom AC exerts on pin C a force directed
along line AC, determine (a) the magnitude of that force, (b) the tension in
cable BC.
SOLUTION
Free-Body Diagram
Law of sines:
Force Triangle
FAC
T
300 lb
= BC =
sin 35&deg; sin 50&deg; sin 95&deg;
(a)
FAC =
300 lb
sin 35&deg;
sin 95&deg;
(b)
TBC =
300 lb
sin 50&deg;
sin 95&deg;
FAC = 172.7 lb 
TBC = 231 lb 
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48
PROBLEM 2.47
Two cables are tied together at C and loaded as shown.
Determine the tension (a) in cable AC, (b) in cable BC.
SOLUTION
Free-Body Diagram
1.4
4.8
α = 16.2602&deg;
tan α =
1.6
3
β = 28.073&deg;
tan β =
Force Triangle
Law of sines:
TAC
TBC
1.98 kN
=
=
sin 61.927&deg; sin 73.740&deg; sin 44.333&deg;
(a)
TAC =
1.98 kN
sin 61.927&deg;
sin 44.333&deg;
TAC = 2.50 kN 
(b)
TBC =
1.98 kN
sin 73.740&deg;
sin 44.333&deg;
TBC = 2.72 kN 
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49
PROBLEM 2.48
Two cables are tied together at C and are loaded as shown.
Knowing that P = 500 N and α = 60&deg;, determine the tension in
(a) in cable AC, (b) in cable BC.
SOLUTION
Free-Body Diagram
Law of sines:
Force Triangle
TAC
T
500 N
= BC =
sin 35&deg; sin 75&deg; sin 70&deg;
(a)
TAC =
500 N
sin 35&deg;
sin 70&deg;
TAC = 305 N 
(b)
TBC =
500 N
sin 75&deg;
sin 70&deg;
TBC = 514 N 
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50
PROBLEM 2.49
Two forces of magnitude TA = 8 kips and TB = 15 kips are
applied as shown to a welded connection. Knowing that the
connection is in equilibrium, determine the magnitudes of the
forces TC and TD.
SOLUTION
Free-Body Diagram
ΣFx = 0
15 kips − 8 kips − TD cos 40&deg; = 0
TD = 9.1379 kips
ΣFy = 0
TD sin 40&deg; − TC = 0
(9.1379 kips) sin 40&deg; − TC = 0
TC = 5.8737 kips
TC = 5.87 kips 
TD = 9.14 kips 
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51
PROBLEM 2.50
Two forces of magnitude TA = 6 kips and TC = 9 kips are
applied as shown to a welded connection. Knowing that the
connection is in equilibrium, determine the magnitudes of the
forces TB and TD.
SOLUTION
Free-Body Diagram
Σ Fx = 0
TB − 6 kips − TD cos 40&deg; = 0
Σ Fy = 0
TD sin 40&deg; − 9 kips = 0
(1)
9 kips
sin 40&deg;
TD = 14.0015 kips
TD =
Substituting for TD into Eq. (1) gives:
TB − 6 kips − (14.0015 kips) cos 40&deg; = 0
TB = 16.7258 kips
TB = 16.73 kips 
TD = 14.00 kips 
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52
PROBLEM 2.51
Two cables are tied together at C and loaded as shown. Knowing
that P = 360 N, determine the tension (a) in cable AC, (b) in
cable BC.
SOLUTION
Free Body: C
(a)
ΣFx = 0: −
(b)
ΣFy = 0:
12
4
TAC + (360 N) = 0
13
5
TAC = 312 N 
5
3
(312 N) + TBC + (360 N) − 480 N = 0
13
5
TBC = 480 N − 120 N − 216 N
TBC = 144 N 
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53
PROBLEM 2.52
Two cables are tied together at C and loaded as shown.
Determine the range of values of P for which both cables
remain taut.
SOLUTION
Free Body: C
ΣFx = 0: −
12
4
TAC + P = 0
13
5
TAC =
ΣFy = 0:
Substitute for TAC from (1):
13
P
15
(1)
5
3
TAC + TBC + P − 480 N = 0
13
5
3
 5  13 
 13  15  P + TBC + 5 P − 480 N = 0
  
TBC = 480 N −
14
P
15
(2)
From (1), TAC ⬎ 0 requires P ⬎ 0.
From (2), TBC ⬎ 0 requires
14
P ⬍ 480 N, P ⬍ 514.29 N
15
0 ⬍ P ⬍ 514 N 
Allowable range:
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54
PROBLEM 2.53
A sailor is being rescued using a boatswain’s chair that is
suspended from a pulley that can roll freely on the support
cable ACB and is pulled at a constant speed by cable CD.
Knowing that α = 30&deg; and β = 10&deg; and that the combined
weight of the boatswain’s chair and the sailor is 900 N,
determine the tension (a) in the support cable ACB, (b) in the
traction cable CD.
SOLUTION
Free-Body Diagram
ΣFx = 0: TACB cos 10&deg; − TACB cos 30&deg; − TCD cos 30&deg; = 0
TCD = 0.137158TACB
(1)
ΣFy = 0: TACB sin 10&deg; + TACB sin 30&deg; + TCD sin 30&deg; − 900 = 0
0.67365TACB + 0.5TCD = 900
(a)
Substitute (1) into (2):
0.67365 TACB + 0.5(0.137158 TACB ) = 900
TACB = 1212.56 N
(b)
From (1):
(2)
TACB = 1213 N 
TCD = 0.137158(1212.56 N)
TCD = 166.3 N 
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55
PROBLEM 2.54
A sailor is being rescued using a boatswain’s chair that is
suspended from a pulley that can roll freely on the support
cable ACB and is pulled at a constant speed by cable CD.
Knowing that α = 25&deg; and β = 15&deg; and that the tension in
cable CD is 80 N, determine (a) the combined weight of the
boatswain’s chair and the sailor, (b) in tension in the support
cable ACB.
SOLUTION
Free-Body Diagram
ΣFx = 0: TACB cos 15&deg; − TACB cos 25&deg; − (80 N) cos 25&deg; = 0
TACB = 1216.15 N
ΣFy = 0: (1216.15 N) sin 15&deg; + (1216.15 N) sin 25&deg;
+ (80 N) sin 25&deg; − W = 0
W = 862.54 N
(a)
W = 863 N 
(b) TACB = 1216 N 
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56
PROBLEM 2.55
Two forces P and Q are applied as shown to an aircraft
connection. Knowing that the connection is in equilibrium and
that P = 500 lb and Q = 650 lb, determine the magnitudes of
the forces exerted on the rods A and B.
SOLUTION
Free-Body Diagram
Resolving the forces into x- and y-directions:
R = P + Q + FA + FB = 0
Substituting components:
R = −(500 lb) j + [(650 lb) cos 50&deg;]i
− [(650 lb) sin 50&deg;] j
+ FB i − ( FA cos 50&deg;)i + ( FA sin 50&deg;) j = 0
In the y-direction (one unknown force):
−500 lb − (650 lb)sin 50&deg; + FA sin 50&deg; = 0
Thus,
FA =
500 lb + (650 lb) sin 50&deg;
sin 50&deg;
= 1302.70 lb
In the x-direction:
Thus,
FA = 1303 lb 
(650 lb) cos 50&deg; + FB − FA cos 50&deg; = 0
FB = FA cos 50&deg; − (650 lb) cos50&deg;
= (1302.70 lb) cos 50&deg; − (650 lb) cos 50&deg;
= 419.55 lb
FB = 420 lb 
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57
PROBLEM 2.56
Two forces P and Q are applied as shown to an aircraft
connection. Knowing that the connection is in equilibrium
and that the magnitudes of the forces exerted on rods A and B
are FA = 750 lb and FB = 400 lb, determine the magnitudes of
P and Q.
SOLUTION
Free-Body Diagram
Resolving the forces into x- and y-directions:
R = P + Q + FA + FB = 0
Substituting components:
R = − Pj + Q cos 50&deg;i − Q sin 50&deg; j
− [(750 lb) cos 50&deg;]i
+ [(750 lb)sin 50&deg;] j + (400 lb)i
In the x-direction (one unknown force):
Q cos 50&deg; − [(750 lb) cos 50&deg;] + 400 lb = 0
(750 lb) cos 50&deg; − 400 lb
cos 50&deg;
= 127.710 lb
Q=
In the y-direction:
− P − Q sin 50&deg; + (750 lb) sin 50&deg; = 0
P = −Q sin 50&deg; + (750 lb) sin 50&deg;
= −(127.710 lb)sin 50&deg; + (750 lb) sin 50&deg;
= 476.70 lb
P = 477 lb; Q = 127.7 lb 
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58
PROBLEM 2.57
Two cables tied together at C are loaded as shown. Knowing that
the maximum allowable tension in each cable is 800 N, determine
(a) the magnitude of the largest force P that can be applied at C,
(b) the corresponding value of α.
SOLUTION
Free-Body Diagram: C
Force Triangle
Force triangle is isosceles with
2 β = 180&deg; − 85&deg;
β = 47.5&deg;
P = 2(800 N)cos 47.5&deg; = 1081 N
(a)
P = 1081 N 
Since P ⬎ 0, the solution is correct.
(b)
α = 180&deg; − 50&deg; − 47.5&deg; = 82.5&deg;
α = 82.5&deg; 
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59
PROBLEM 2.58
Two cables tied together at C are loaded as shown. Knowing that
the maximum allowable tension is 1200 N in cable AC and 600 N
in cable BC, determine (a) the magnitude of the largest force P
that can be applied at C, (b) the corresponding value of α.
SOLUTION
Free-Body Diagram
(a)
Law of cosines:
Force Triangle
P 2 = (1200 N) 2 + (600 N) 2 − 2(1200 N)(600 N) cos 85&deg;
P = 1294.02 N
Since P ⬎ 1200 N, the solution is correct.
P = 1294 N 
(b)
Law of sines:
sin β
sin 85&deg;
=
1200 N 1294.02 N
β = 67.5&deg;
α = 180&deg; − 50&deg; − 67.5&deg;
α = 62.5&deg; 
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60
PROBLEM 2.59
For the situation described in Figure P2.45, determine (a) the
value of α for which the tension in rope BC is as small as
possible, (b) the corresponding value of the tension.
PROBLEM 2.45 Knowing that α = 20&deg;, determine the tension
(a) in cable AC, (b) in rope BC.
SOLUTION
Free-Body Diagram
Force Triangle
To be smallest, TBC must be perpendicular to the direction of TAC .
(a)
(b)
Thus,
α = 5&deg;
α = 5.00&deg;
TBC = (1200 lb) sin 5&deg;

TBC = 104.6 lb 
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61
PROBLEM 2.60
For the structure and loading of Problem 2.46, determine (a) the value of α for
which the tension in cable BC is as small as possible, (b) the corresponding
value of the tension.
SOLUTION
TBC must be perpendicular to FAC to be as small as possible.
Free-Body Diagram: C
Force Triangle is a right triangle
To be a minimum, TBC must be perpendicular to FAC .
(a)
We observe:
α = 90&deg; − 30&deg;
α = 60.0&deg; 
TBC = (300 lb)sin 50&deg;
(b)
or
TBC = 229.81 lb
TBC = 230 lb 
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62
PROBLEM 2.61
For the cables of Problem 2.48, it is known that the maximum
allowable tension is 600 N in cable AC and 750 N in cable BC.
Determine (a) the maximum force P that can be applied at C,
(b) the corresponding value of α.
SOLUTION
Free-Body Diagram
(a)
Law of cosines
Force Triangle
P 2 = (600) 2 + (750)2 − 2(600)(750) cos (25&deg; + 45&deg;)
P = 784.02 N
(b)
Law of sines
P = 784 N 
sin β
sin (25&deg; + 45&deg;)
=
600 N
784.02 N
β = 46.0&deg;
∴ α = 46.0&deg; + 25&deg;
α = 71.0&deg; 
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63
PROBLEM 2.62
A 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 the
loaded bin if the tension in the chain is not to exceed 5 kN.
SOLUTION
Free-Body Diagram
tan α =
h
0.6 m
(1)
Isosceles Force Triangle
Law of sines: sin α =
1
2
(2.8 kN)
TAC
TAC = 5 kN
sin α =
1
2
(2.8 kN)
5 kN
α = 16.2602&deg;
From Eq. (1): tan16.2602&deg; =
h
0.6 m
∴ h = 0.175000 m
Half length of chain = AC = (0.6 m) 2 + (0.175 m)2
= 0.625 m
Total length:
= 2 &times; 0.625 m
1.250 m 
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64
PROBLEM 2.63
Collar A is connected as shown to a 50-lb load and can slide on
a frictionless horizontal rod. Determine the magnitude of the
force P required to maintain the equilibrium of the collar when
(a) x = 4.5 in., (b) x = 15 in.
SOLUTION
(a)
Free Body: Collar A
Force Triangle
P 50 lb
=
4.5 20.5
(b)
Free Body: Collar A
P = 10.98 lb 
Force Triangle
P 50 lb
=
15
25
P = 30.0 lb 

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65
PROBLEM 2.64
Collar A is connected as shown to a 50-lb load and can slide on a
frictionless horizontal rod. Determine the distance x for which the
collar is in equilibrium when P = 48 lb.
SOLUTION
Free Body: Collar A
Force Triangle
N 2 = (50) 2 − (48) 2 = 196
N = 14.00 lb
Similar Triangles
x
48 lb
=
20 in. 14 lb
x = 68.6 in. 
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66
PROBLEM 2.65
Three forces are applied to a bracket as shown. The directions of the two 150-N
forces may vary, but the angle between these forces is always 50&deg;. Determine the
range of values of α for which the magnitude of the resultant of the forces acting at A
is less than 600 N.
SOLUTION
Combine the two 150-N forces into a resultant force Q:
Q = 2(150 N) cos 25&deg;
= 271.89 N
Using the law of cosines:
(600 N) 2 = (500 N) 2 + (271.89 N)2 + 2(500 N)(271.89 N) cos(55&deg; + α )
cos(55&deg; + α ) = 0.132685
Two values for α :
55&deg; + α = 82.375
α = 27.4&deg;
or
55&deg; + α = −82.375&deg;
55&deg; + α = 360&deg; − 82.375&deg;
α = 222.6&deg;
27.4&deg; &lt; α &lt; 222.6 
For R &lt; 600 lb:
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67
PROBLEM 2.66
A 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 free
end of the rope to maintain equilibrium. (Hint: The tension in the rope is the same on
each side of a simple pulley. This can be proved by the methods of Ch. 4.)
SOLUTION
Free-Body Diagram: Pulley A
 5 
ΣFx = 0: − 2 P 
 + P cos α = 0
 281 
cos α = 0.59655
α = &plusmn;53.377&deg;
For α = +53.377&deg;:
 16 
ΣFy = 0: 2 P 
 + P sin 53.377&deg; − 1962 N = 0
 281 
P = 724 N
53.4&deg; 
For α = −53.377&deg;:
 16 
ΣFy = 0: 2 P 
 + P sin(−53.377&deg;) − 1962 N = 0
 281 
P = 1773
53.4&deg; 
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68
PROBLEM 2.67
A 600-lb crate is supported by several rope-andpulley arrangements as shown. Determine for each
arrangement the tension in the rope. (See the hint
for Problem 2.66.)
SOLUTION
Free-Body Diagram of Pulley
(a)
ΣFy = 0: 2T − (600 lb) = 0
T=
1
(600 lb)
2
T = 300 lb 
(b)
ΣFy = 0: 2T − (600 lb) = 0
T=
1
(600 lb)
2
T = 300 lb 
(c)
ΣFy = 0: 3T − (600 lb) = 0
1
T = (600 lb)
3
T = 200 lb 
(d)
ΣFy = 0: 3T − (600 lb) = 0
1
T = (600 lb)
3
T = 200 lb 
(e)
ΣFy = 0: 4T − (600 lb) = 0
T=
1
(600 lb)
4
T = 150.0 lb 

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69
PROBLEM 2.68
Solve Parts b and d of Problem 2.67, assuming that
the free end of the rope is attached to the crate.
PROBLEM 2.67 A 600-lb crate is supported by
several rope-and-pulley arrangements as shown.
Determine for each arrangement the tension in the
rope. (See the hint for Problem 2.66.)
SOLUTION
Free-Body Diagram of Pulley and Crate
(b)
ΣFy = 0: 3T − (600 lb) = 0
1
T = (600 lb)
3
T = 200 lb 
(d)
ΣFy = 0: 4T − (600 lb) = 0
T=
1
(600 lb)
4
T = 150.0 lb 

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70
PROBLEM 2.69
A load Q is applied to the pulley C, which can roll on the
cable ACB. The pulley is held in the position shown by a
second cable CAD, which passes over the pulley A and
supports a load P. Knowing that P = 750 N, determine
(a) the tension in cable ACB, (b) the magnitude of load Q.
SOLUTION
Free-Body Diagram: Pulley C
ΣFx = 0: TACB (cos 25&deg; − cos 55&deg;) − (750 N) cos 55&deg; = 0
(a)
TACB = 1292.88 N
Hence:
TACB = 1293 N 
ΣFy = 0: TACB (sin 25&deg; + sin 55&deg;) + (750 N) sin 55&deg; − Q = 0
(b)

(1292.88 N)(sin 25&deg; + sin 55&deg;) + (750 N) sin 55&deg; − Q = 0
Q = 2219.8 N
or
Q = 2220 N 
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71
PROBLEM 2.70
An 1800-N load Q is applied to the pulley C, which can roll
on the cable ACB. The pulley is held in the position shown
by a second cable CAD, which passes over the pulley A and
supports a load P. Determine (a) the tension in cable ACB,
(b) the magnitude of load P.
SOLUTION
Free-Body Diagram: Pulley C
ΣFx = 0: TACB (cos 25&deg; − cos 55&deg;) − P cos 55&deg; = 0
P = 0.58010TACB
or
(1)
ΣFy = 0: TACB (sin 25&deg; + sin 55&deg;) + P sin 55&deg; − 1800 N = 0
1.24177TACB + 0.81915 P = 1800 N
or
(a)
(2)
Substitute Equation (1) into Equation (2):
1.24177TACB + 0.81915(0.58010TACB ) = 1800 N
TACB = 1048.37 N
Hence:
TACB = 1048 N 
(b)
P = 0.58010(1048.37 N) = 608.16 N
Using (1),
P = 608 N 
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72
PROBLEM 2.71
Determine (a) the x, y, and z components of the 900-N force, (b) the
angles θx, θy, and θz that the force forms with the coordinate axes.
SOLUTION
Fh = F cos 65&deg;
= (900 N) cos 65&deg;
Fh = 380.36 N
(a)
Fx = Fh sin 20&deg;
= (380.36 N)sin 20&deg;
Fx = −130.091 N,
Fx = −130.1 N 
Fy = F sin 65&deg;
= (900 N)sin 65&deg;
Fy = +815.68 N,
Fy = +816 N 
Fz = Fh cos 20&deg;
= (380.36 N) cos 20&deg;
Fz = +357.42 N
(b)
cos θ x =
cos θ y =
cos θ z =
Fx −130.091 N
=
900 N
F
Fy
F
=
+815.68 N
900 N
Fz +357.42 N
=
900 N
F
Fz = +357 N 
θ x = 98.3&deg; 
θ y = 25.0&deg; 
θ z = 66.6&deg; 
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73
PROBLEM 2.72
Determine (a) the x, y, and z components of the 750-N force, (b) the
angles θx, θy, and θz that the force forms with the coordinate axes.
SOLUTION
Fh = F sin 35&deg;
= (750 N)sin 35&deg;
Fh = 430.18 N
(a)
Fx = Fh cos 25&deg;
= (430.18 N) cos 25&deg;
Fx = +389.88 N,
Fx = +390 N 
Fy = F cos 35&deg;

= (750 N) cos 35&deg;
Fy = +614.36 N,
Fy = +614 N 
Fz = Fh sin 25&deg;
= (430.18 N)sin 25&deg;
Fz = +181.802 N
(b)
cos θ x =
cos θ y =
cos θ z =
Fx +389.88 N
=
750 N
F
Fy
F
=
+614.36 N
750 N
Fz +181.802 N
=
750 N
F
Fz = +181.8 N 
θ x = 58.7&deg; 
θ y = 35.0&deg; 
θ z = 76.0&deg; 
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74
PROBLEM 2.73
A gun is aimed at a point A located 35&deg; east of north. Knowing that the barrel of the gun forms an angle of 40&deg;
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 θx, θy, and θz defining the direction of the recoil force. (Assume that the
x, y, and z axes are directed, respectively, east, up, and south.)
SOLUTION
Recoil force
F = 400 N
∴ FH = (400 N) cos 40&deg;
= 306.42 N
(a)
Fx = − FH sin 35&deg;
= −(306.42 N)sin 35&deg;
= −175.755 N
Fx = −175.8 N 
Fy = − F sin 40&deg;
= −(400 N)sin 40&deg;
= −257.12 N
Fy = −257 N 
Fz = + FH cos 35&deg;
= +(306.42 N) cos35&deg;
= +251.00 N
(b)
cos θ x =
Fx −175.755 N
=
400 N
F
−257.12 N
400 N
cos θ y =
Fy
cos θ z =
Fz 251.00 N
=
400 N
F
F
=
Fz = +251 N 
θ x = 116.1&deg; 
θ y = 130.0&deg; 
θ z = 51.1&deg; 
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75
PROBLEM 2.74
Solve Problem 2.73, assuming that point A is located 15&deg; north of west and that the barrel of the gun forms an
angle of 25&deg; with the horizontal.
PROBLEM 2.73 A gun is aimed at a point A located 35&deg; east of north. Knowing that the barrel of the gun
forms an angle of 40&deg; 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 θx, θy, and θz defining the direction of the recoil
force. (Assume that the x, y, and z axes are directed, respectively, east, up, and south.)
SOLUTION
Recoil force
F = 400 N
∴ FH = (400 N) cos 25&deg;
= 362.52 N
(a)
Fx = + FH cos15&deg;
= + (362.52 N) cos15&deg;
= +350.17 N
Fx = +350 N 
Fy = − F sin 25&deg;
= −(400 N)sin 25&deg;
= −169.047 N
Fy = −169.0 N 
Fz = + FH sin15&deg;
= +(362.52 N)sin15&deg;
= +93.827 N
(b)
cos θ x =
cos θ y =
cos θ z =
Fx +350.17 N
=
400 N
F
Fy
F
=
−169.047 N
400 N
Fz +93.827 N
=
400 N
F
Fz = +93.8 N 
θ x = 28.9&deg; 
θ y = 115.0&deg; 
θ z = 76.4&deg; 
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76
PROBLEM 2.75
Cable 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 exerted
by the cable on the anchor B, (b) the angles θ x , θ y , and θ z
defining the direction of that force.
SOLUTION
56 ft
65 ft
= 0.86154
cos θ y =
From triangle AOB:
θ y = 30.51&deg;
Fx = − F sin θ y cos 20&deg;
(a)
= −(3900 lb)sin 30.51&deg; cos 20&deg;
Fx = −1861 lb 
Fy = + F cos θ y = (3900 lb)(0.86154)

Fz = + (3900 lb)sin 30.51&deg; sin 20&deg;
(b)
cos θ x =
From above:
Fx
1861 lb
=−
= − 0.4771
3900 lb
F
θ y = 30.51&deg;
cos θ z =
Fz
677 lb
=+
= + 0.1736
3900 lb
F
Fy = +3360 lb 
Fz = +677 lb 
θ x = 118.5&deg; 
θ y = 30.5&deg; 
θ z = 80.0&deg; 
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77
PROBLEM 2.76
Cable 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 by
the cable on the anchor C, (b) the angles θx, θy, and θz defining the
direction of that force.
SOLUTION
AC = 70 ft
OA = 56 ft
F = 5250 lb
In triangle AOB:
cos θ y =

56 ft
70 ft
θ y = 36.870&deg;
FH = F sin θ y
= (5250 lb) sin 36.870&deg;
= 3150.0 lb
(a)
(b)
Fx = − FH sin 50&deg; = −(3150.0 lb)sin 50&deg; = −2413.04 lb
Fx = −2413 lb 
Fy = + F cos θ y = + (5250 lb) cos 36.870&deg; = +4200.0 lb
Fy = +4200 lb 
Fz = − FH cos 50&deg; = −3150cos50&deg; = −2024.8 lb
Fz = −2025 lb 
cos θ x =
From above:
Fx −2413.04 lb
=
5250 lb
F
θ y = 36.870&deg;
θz =
Fz −2024.8 lb
=
5250 lb
F
θ x = 117.4&deg; 
θ y = 36.9&deg; 
θ z = 112.7&deg; 
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78
PROBLEM 2.77
The end of the coaxial cable AE is attached to the pole AB, which is
strengthened by the guy wires AC and AD. Knowing that the tension
in wire AC is 120 lb, determine (a) the components of the force
exerted by this wire on the pole, (b) the angles θx, θy, and θz that the
force forms with the coordinate axes.
SOLUTION
(a)
Fx = (120 lb) cos 60&deg; cos 20&deg;
Fx = 56.382 lb
Fx = +56.4 lb 
Fy = −(120 lb)sin 60&deg;
Fy = −103.923 lb
Fy = −103.9 lb 
Fz = −(120 lb) cos 60&deg; sin 20&deg;
Fz = −20.521 lb
(b)
Fz = −20.5 lb 
cos θ x =
Fx 56.382 lb
=
F
120 lb
cos θ y =
Fy
cos θ z =
F
=
−103.923 lb
120 lb
Fz −20.52 lb
=
F
120 lb
θ x = 62.0&deg; 
θ y = 150.0&deg; 
θ z = 99.8&deg; 
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79
PROBLEM 2.78
The end of the coaxial cable AE is attached to the pole AB, which is
strengthened by the guy wires AC and AD. Knowing that the tension
in wire AD is 85 lb, determine (a) the components of the force
exerted by this wire on the pole, (b) the angles θx, θy, and θz that the
force forms with the coordinate axes.
SOLUTION
(a)
Fx = (85 lb)sin 36&deg; sin 48&deg;
= 37.129 lb
Fx = 37.1 lb 
Fy = −(85 lb) cos 36&deg;
= −68.766 lb
Fy = −68.8 lb 
Fz = (85 lb)sin 36&deg; cos 48&deg;
Fz = 33.4 lb 
= 33.431 lb
(b)
cos θ x =
Fx 37.129 lb
=
F
85 lb
cos θ y =
Fy
cos θ z =
Fz 33.431 lb
=
F
85 lb
F
=
−68.766 lb
85 lb
θ x = 64.1&deg; 
θ y = 144.0&deg; 
θ z = 66.8&deg; 
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80
PROBLEM 2.79
Determine the magnitude and direction of the force F = (690 lb)i + (300 lb)j – (580 lb)k.
SOLUTION
F = (690 N)i + (300 N) j − (580 N)k
F = Fx2 + Fy2 + Fz2
= (690 N) 2 + (300 N)2 + (−580 N) 2
F = 950 N 
= 950 N
cos θ x =
Fx 690 N
=
F 950 N
θ x = 43.4&deg; 
cos θ y =
Fy
θ y = 71.6&deg; 
cos θ z =
F
=
300 N
950 N
Fz −580 N
=
F
950 N
θ z = 127.6&deg; 
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81
PROBLEM 2.80
Determine the magnitude and direction of the force F = (650 N)i − (320 N)j + (760 N)k.
SOLUTION
F = (650 N)i − (320 N) j + (760 N)k
F = Fx2 + Fy2 + Fz2
= (650 N) 2 + (−320 N)2 + (760 N)2
F = 1050 N 
cos θ x =
Fx
650 N
=
F 1050 N
θ x = 51.8&deg; 
cos θ y =
Fy
−320 N
1050 N
θ y = 107.7&deg; 
cos θ z =
Fz
760 N
=
F 1050 N
θ z = 43.6&deg; 
F
=
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82
PROBLEM 2.81
A force acts at the origin of a coordinate system in a direction defined by the angles θx = 75&deg; and θz = 130&deg;.
Knowing that the y component of the force is +300 lb, determine (a) the angle θy, (b) the other components
and the magnitude of the force.
SOLUTION
cos 2 θ x + cos 2 θ y + cos 2 θ z = 1
cos 2 (75&deg;) + cos 2 θ y + cos 2 (130&deg;) = 1
cos θ y = &plusmn;0.72100
(a)
(b)
Since Fy ⬎ 0, we choose cos θ y ⫽⫹0.72100
∴ θ y = 43.9&deg; 
Fy = F cos θ y
300 lb = F (0.72100)
F = 416.09 lb
F = 416 lb 
Fx = F cos θ x = 416.09 lb cos 75&deg;
Fz = F cos θ z = 416.09 lb cos130&deg;
Fx = +107.7 lb 
Fz = −267 lb 
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83
PROBLEM 2.82
A force acts at the origin of a coordinate system in a direction defined by the angles θy = 55&deg; and θz = 45&deg;.
Knowing that the x component of the force is −500 N, determine (a) the angle θx, (b) the other components
and the magnitude of the force.
SOLUTION
cos 2 θ x + cos 2 θ y + cos 2 θ z = 1
cos 2 θ x + cos 2 55&deg; + cos 2 45&deg; = 1
cos θ x = &plusmn;0.41353
(a)
(b)
Since Fy ⬍ 0, we choose cos θ x ⫽⫺0.41353
∴ θ x = 114.4&deg; 
Fx = F cos θ x
−500 N = F (−0.41353)
F = 1209.10 N
F = 1209.1 N 
Fy = F cos θ y = 1209.10 N cos55&deg;
Fy = +694 N 
Fz = F cos θ z = 1209.10 N cos 45&deg;
Fz = +855 N 
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84
PROBLEM 2.83
A force F of magnitude 230 N acts at the origin of a coordinate system. Knowing that θx = 32.5&deg;, Fy = −60 N,
and Fz &gt; 0, determine (a) the components Fx and Fz, (b) the angles θy and θz.
SOLUTION
(a)
We have
Fx = F cos θ x = (230 N) cos 32.5&deg;
Then:
Fx = −194.0 N 
Fx = 193.980 N
F 2 = Fx2 + Fy2 + Fz2
So:
Hence:
(b)
(230 N) 2 = (193.980 N) 2 + (−60 N) 2 + Fz2
Fz = + (230 N)2 − (193.980 N)2 − (−60 N)2
Fz = 108.0 N 
Fz = 108.036 N
−60 N
=
= − 0.26087
F 230 N
F 108.036 N
cos θ z = z =
= 0.46972
230 N
F
cos θ y =
Fy
θ y = 105.1&deg; 
θ z = 62.0&deg; 
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85
PROBLEM 2.84
A force F of magnitude 210 N acts at the origin of a coordinate system. Knowing that Fx = 80 N, θz = 151.2&deg;,
and Fy &lt; 0, determine (a) the components Fy and Fz, (b) the angles θx and θy.
SOLUTION
Fz = F cos θ z = (210 N) cos151.2&deg;
(a)
= −184.024 N
Then:
So:
Hence:
F 2 = Fx2 + Fy2 + Fz2
(210 N) 2 = (80 N) 2 + ( Fy ) 2 + (184.024 N)2
Fy = − (210 N) 2 − (80 N) 2 − (184.024 N) 2
= −61.929 N
(b)
Fz = −184.0 N 
Fy = −62.0 lb 
cos θ x =
Fx
80 N
=
= 0.38095
F 210 N
θ x = 67.6&deg; 
cos θ y =
Fy
θ y = 107.2&deg; 
F
=
61.929 N
= −0.29490
210 N
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86
PROBLEM 2.85
In order to move a wrecked truck, two cables are attached at A
and pulled by winches B and C as shown. Knowing that the
tension in cable AB is 2 kips, determine the components of the
force exerted at A by the cable.
SOLUTION
Cable AB:
λAB

AB (−46.765 ft)i + (45 ft) j + (36 ft)k
=
=
74.216 ft
AB
TAB = TAB λAB =
−46.765i + 45 j + 36k
74.216
(TAB ) x = −1.260 kips 
(TAB ) y = +1.213 kips 
(TAB ) z = +0.970 kips 
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87
PROBLEM 2.86
In order to move a wrecked truck, two cables are attached at A
and pulled by winches B and C as shown. Knowing that the
tension in cable AC is 1.5 kips, determine the components of
the force exerted at A by the cable.
SOLUTION
Cable AB:
λAC

AC (−46.765 ft)i + (55.8 ft) j + (−45 ft)k
=
=
85.590 ft
AC
TAC = TAC λAC = (1.5 kips)
−46.765i + 55.8 j − 45k
85.590
(TAC ) x = −0.820 kips 
(TAC ) y = +0.978 kips 
(TAC ) z = −0.789 kips 
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88
PROBLEM 2.87
Knowing that the tension in cable AB is 1425 N, determine the
components of the force exerted on the plate at B.
SOLUTION

BA = −(900 mm)i + (600 mm) j + (360 mm)k
BA = (900 mm) 2 + (600 mm) 2 + (360 mm) 2
= 1140 mm
TBA = TBA λ BA

BA
= TBA
BA
1425 N
[ −(900 mm)i + (600 mm) j + (360 mm)k ]
TBA =
1140 mm
= −(1125 N)i + (750 N) j + (450 N)k
(TBA ) x = −1125 N, (TBA ) y = 750 N, (TBA ) z = 450 N 
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89
PROBLEM 2.88
Knowing that the tension in cable AC is 2130 N, determine the
components of the force exerted on the plate at C.
SOLUTION

CA = −(900 mm)i + (600 mm) j − (920 mm)k
CA = (900 mm)2 + (600 mm)2 + (920 mm) 2
= 1420 mm
TCA = TCA λ CA

CA
= TCA
CA
2130 N
TCA =
[−(900 mm)i + (600 mm) j − (920 mm)k ]
1420 mm
= −(1350 N)i + (900 N) j − (1380 N)k
(TCA ) x = −1350 N, (TCA ) y = 900 N, (TCA ) z = −1380 N 
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90
PROBLEM 2.89
A frame ABC is supported in part by cable DBE that passes
through a frictionless ring at B. Knowing that the tension in the
cable is 385 N, determine the components of the force exerted by
the cable on the support at D.
SOLUTION

DB = (480 mm)i − (510 mm) j + (320 mm)k
DB = (480 mm)2 + (510 mm 2 ) + (320 mm) 2
= 770 mm
F = F λ DB

DB
=F
DB
385 N
=
[(480 mm)i − (510 mm)j + (320 mm)k ]
770 mm
= (240 N)i − (255 N) j + (160 N)k
Fx = +240 N, Fy = −255 N, Fz = +160.0 N 
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91
PROBLEM 2.90
For the frame and cable of Problem 2.89, determine the components
of the force exerted by the cable on the support at E.
PROBLEM 2.89 A frame ABC is supported in part by cable DBE
that passes through a frictionless ring at B. Knowing that the tension
in the cable is 385 N, determine the components of the force exerted
by the cable on the support at D.
SOLUTION

EB = (270 mm)i − (400 mm) j + (600 mm)k
EB = (270 mm)2 + (400 mm) 2 + (600 mm)2
= 770 mm
F = F λ EB

EB
=F
EB
385 N
=
[(270 mm)i − (400 mm)j + (600 mm)k ]
770 mm
F = (135 N)i − (200 N) j + (300 N)k
Fx = +135.0 N, Fy = −200 N, Fz = +300 N 
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92
PROBLEM 2.91
Find the magnitude and direction of the resultant of the two forces
shown knowing that P = 600 N and Q = 450 N.
SOLUTION
P = (600 N)[sin 40&deg; sin 25&deg;i + cos 40&deg; j + sin 40&deg; cos 25&deg;k ]
= (162.992 N)i + (459.63 N) j + (349.54 N)k
Q = (450 N)[cos 55&deg; cos 30&deg;i + sin 55&deg; j − cos 55&deg; sin 30&deg;k ]
= (223.53 N)i + (368.62 N) j − (129.055 N)k
R =P+Q
= (386.52 N)i + (828.25 N) j + (220.49 N)k
R = (386.52 N)2 + (828.25 N)2 + (220.49 N) 2
R = 940 N 
= 940.22 N
cos θ x =
Rx 386.52 N
=
R 940.22 N
θ x = 65.7&deg; 
cos θ y =
Ry
θ y = 28.2&deg; 
cos θ z =
Rz 220.49 N
=
R 940.22 N
R
=
828.25 N
940.22 N
θ z = 76.4&deg; 
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93
PROBLEM 2.92
Find the magnitude and direction of the resultant of the two forces
shown knowing that P = 450 N and Q = 600 N.
SOLUTION
P = (450 N)[sin 40&deg; sin 25&deg;i + cos 40&deg; j + sin 40&deg; cos 25&deg;k ]
= (122.244 N)i + (344.72 N) j + (262.154 N)k
Q = (600 N)[cos 55&deg; cos 30&deg;i + sin 55&deg; j − cos 55&deg; sin 30&deg;k ]
= (298.04 N)i + (491.49 N)j − (172.073 N)k
R =P+Q
= (420.28 N)i + (836.21 N) j + (90.081 N)k
R = (420.28 N) 2 + (836.21 N) 2 + (90.081 N) 2
= 940.21 N
R = 940 N 
cos θ x =
Rx 420.28
=
R 940.21
θ x = 63.4&deg; 
cos θ y =
Ry
θ y = 27.2&deg; 
cos θ z =
Rz 90.081
=
R 940.21
R
=
836.21
940.21
θ z = 84.5&deg; 
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94
PROBLEM 2.93
Knowing that the tension is 425 lb in cable AB and 510 lb in
cable AC, determine the magnitude and direction of the resultant
of the forces exerted at A by the two cables.
SOLUTION

AB = (40 in.)i − (45 in.) j + (60 in.)k
AB = (40 in.)2 + (45 in.)2 + (60 in.) 2 = 85 in.

AC = (100 in.)i − (45 in.) j + (60 in.)k
AC = (100 in.)2 + (45 in.)2 + (60 in.)2 = 125 in.

 (40 in.)i − (45 in.) j + (60 in.)k 
AB
TAB = TAB λAB = TAB
= (425 lb) 

AB
85 in.


TAB = (200 lb)i − (225 lb) j + (300 lb)k

 (100 in.)i − (45 in.) j + (60 in.)k 
AC
TAC = TAC λAC = TAC
= (510 lb) 

125 in.
AC


TAC = (408 lb)i − (183.6 lb) j + (244.8 lb)k
R = TAB + TAC = (608)i − (408.6 lb) j + (544.8 lb)k
Then:
and
R = 912.92 lb
R = 913 lb 
cos θ x =
608 lb
= 0.66599
912.92 lb
cos θ y =
408.6 lb
= −0.44757
912.92 lb
cos θ z =
544.8 lb
= 0.59677
912.92 lb
θ x = 48.2&deg; 
θ y = 116.6&deg; 
θ z = 53.4&deg; 
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95
PROBLEM 2.94
Knowing that the tension is 510 lb in cable AB and 425 lb in cable
AC, determine the magnitude and direction of the resultant of the
forces exerted at A by the two cables.
SOLUTION

AB = (40 in.)i − (45 in.) j + (60 in.)k
AB = (40 in.)2 + (45 in.)2 + (60 in.) 2 = 85 in.

AC = (100 in.)i − (45 in.) j + (60 in.)k
AC = (100 in.)2 + (45 in.)2 + (60 in.)2 = 125 in.

 (40 in.)i − (45 in.) j + (60 in.)k 
AB
TAB = TAB λAB = TAB
= (510 lb) 

AB
85 in.


TAB = (240 lb)i − (270 lb) j + (360 lb)k

 (100 in.)i − (45 in.) j + (60 in.)k 
AC
TAC = TAC λAC = TAC
= (425 lb) 

125 in.
AC


TAC = (340 lb)i − (153 lb) j + (204 lb)k
R = TAB + TAC = (580 lb)i − (423 lb) j + (564 lb)k
Then:
and
R = 912.92 lb
R = 913 lb 
cos θ x =
580 lb
= 0.63532
912.92 lb
θ x = 50.6&deg; 
cos θ y =
−423 lb
= −0.46335
912.92 lb
θ y = 117.6&deg; 
cos θ z =
564 lb
= 0.61780
912.92 lb
θ z = 51.8&deg; 
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96
PROBLEM 2.95
For the frame of Problem 2.89, determine the magnitude and
direction of the resultant of the forces exerted by the cable at B
knowing that the tension in the cable is 385 N.
PROBLEM 2.89 A frame ABC is supported in part by cable
DBE that passes through a frictionless ring at B. Knowing that
the tension in the cable is 385 N, determine the components of
the force exerted by the cable on the support at D.
SOLUTION

BD = −(480 mm)i + (510 mm) j − (320 mm)k
BD = (480 mm) 2 + (510 mm) 2 + (320 mm) 2 = 770 mm

BD
FBD = TBD λ BD = TBD
BD
(385 N)
[−(480 mm)i + (510 mm) j − (320 mm)k ]
=
(770 mm)
= −(240 N)i + (255 N) j − (160 N)k

BE = −(270 mm)i + (400 mm) j − (600 mm)k
BE = (270 mm) 2 + (400 mm) 2 + (600 mm) 2 = 770 mm

BE
FBE = TBE λ BE = TBE
BE
(385 N)
[−(270 mm)i + (400 mm) j − (600 mm)k ]
=
(770 mm)
= −(135 N)i + (200 N) j − (300 N)k
R = FBD + FBE = −(375 N)i + (455 N) j − (460 N)k
R = (375 N)2 + (455 N)2 + (460 N)2 = 747.83 N
R = 748 N 
cos θ x =
−375 N
747.83 N
θ x = 120.1&deg; 
cos θ y =
455 N
747.83 N
θ y = 52.5&deg; 
cos θ z =
−460 N
747.83 N
θ z = 128.0&deg; 
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97
PROBLEM 2.96
For the cables of Problem 2.87, knowing that the tension is 1425 N
in cable AB and 2130 N in cable AC, determine the magnitude and
direction of the resultant of the forces exerted at A by the two
cables.
SOLUTION
TAB = −TBA
(use results of Problem 2.87)
(TAB ) x = +1125 N (TAB ) y = −750 N (TAB ) z = − 450 N
TAC = −TCA
(use results of Problem 2.88)
(TAC ) x = +1350 N (TAC ) y = −900 N (TAC ) z = +1380 N
Resultant:
Rx = ΣFx = +1125 + 1350 = +2475 N
Ry = ΣFy = −750 − 900 = −1650 N
Rz = ΣFz = −450 + 1380 = + 930 N
R = Rx2 + Ry2 + Rz2
= (+2475)2 + (−1650) 2 + (+930) 2
= 3116.6 N
cos θ x =
cos θ y =
cos θ z =
R = 3120 N 
Rx +2475
=
R 3116.6
θ x = 37.4&deg; 
−1650
3116.6
θ y = 122.0&deg; 
Ry
R
=
Rz
+ 930
=
R 3116.6
θ z = 72.6&deg; 
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98
PROBLEM 2.97
The boom OA carries a load P and is supported by two cables
as shown. Knowing that the tension in cable AB is 183 lb and
that the resultant of the load P and of the forces exerted at A
by the two cables must be directed along OA, determine the
tension in cable AC.
SOLUTION
Cable AB:
TAB = 183 lb
TAB
TAB
Cable AC:
TAC
TAC

AB
(−48 in.)i + (29 in.) j + (24 in.)k
= TAB λ AB = TAB
= (183 lb)
AB
61 in.
= −(144 lb)i + (87 lb) j + (72 lb)k

AC
(−48 in.)i + (25 in.) j + (−36 in.)k
= TAC λ AC = TAC
= TAC
AC
65 in.
48
25
36
= − TAC i + TAC j − TAC k
65
65
65
P = Pj
For resultant to be directed along OA, i.e., x-axis
Rz = 0: ΣFz = (72 lb) −
36
′ =0
TAC
65
TAC = 130.0 lb 
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99
PROBLEM 2.98
PROBLEM 2.97 The boom OA carries a load P and is
supported by two cables as shown. Knowing that the tension
in cable AB is 183 lb and that the resultant of the load P and
of the forces exerted at A by the two cables must be directed
along OA, determine the tension in cable AC.
SOLUTION
See Problem 2.97. Since resultant must be directed along OA, i.e., the x-axis, we write
Ry = 0: ΣFy = (87 lb) +
25
TAC − P = 0
65
TAC = 130.0 lb from Problem 2.97.
Then
(87 lb) +
25
(130.0 lb) − P = 0
65
P = 137.0 lb 
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100
PROBLEM 2.99
A container is supported by three cables that are attached to a
ceiling as shown. Determine the weight W of the container,
knowing that the tension in cable AB is 6 kN.
SOLUTION
Free-Body Diagram at A:
The forces applied at A are:
TAB , TAC , TAD , and W
where W = W j. To express the other forces in terms of the unit vectors i, j, k, we write

AB = − (450 mm)i + (600 mm) j
AB = 750 mm

AC = + (600 mm) j − (320 mm)k
AC = 680 mm

AD = + (500 mm)i + (600 mm) j + (360 mm)k AD = 860 mm

AB
(−450 mm)i + (600 mm) j
and
TAB = λ ABTAB = TAB
= TAB
AB
750 mm
 45 60 
j  TAB
= − i +
 75 75 
TAC = λ AC TAC = TAC

AC
(600 mm)i − (320 mm) j
= TAC
AC
680 mm
32 
 60
=  j − k  TAC
68
68 


(500 mm)i + (600 mm) j + (360 mm)k
860 mm
60
36 
 50
= i+
86
86 
 86
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101
PROBLEM 2.99 (Continued)
Equilibrium condition:
ΣF = 0: ∴ TAB + TAC + TAD + W = 0
Substituting the expressions obtained for TAB , TAC , and TAD ; factoring i, j, and k; and equating each of the
coefficients to zero gives the following equations:
From i:
From j:
From k:
45
50
75
86
(1)
60
60
60
TAB + TAC + TAD − W = 0
75
68
86
(2)
32
36
68
86
(3)
−
−
Setting TAB = 6 kN in (1) and (2), and solving the resulting set of equations gives
TAC = 6.1920 kN
TAC = 5.5080 kN
W = 13.98 kN 
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102
PROBLEM 2.100
A container is supported by three cables that are attached to a
ceiling as shown. Determine the weight W of the container,
knowing that the tension in cable AD is 4.3 kN.
SOLUTION
See Problem 2.99 for the figure and analysis leading to the following set of linear algebraic equations:
45
50
75
86
(1)
60
60
60
TAB + TAC + TAD − W = 0
75
68
86
(2)
32
36
68
86
(3)
−
−
Setting TAD = 4.3 kN into the above equations gives
TAB = 4.1667 kN
TAC = 3.8250 kN
W = 9.71 kN 
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103
PROBLEM 2.101
Three cables are used to tether a balloon as shown. Determine the
vertical force P exerted by the balloon at A knowing that the tension
in cable AD is 481 N.
SOLUTION
The forces applied at A are:
TAB , TAC , TAD , and P
where P = Pj. To express the other forces in terms of the unit vectors i, j, k, we write

AB = − (4.20 m)i − (5.60 m) j
AB = 7.00 m

AC = (2.40 m)i − (5.60 m) j + (4.20 m)k AC = 7.40 m

AD = − (5.60 m)j − (3.30 m)k

AB
and
TAB = TAB λ AB = TAB
= ( − 0.6i − 0.8 j)TAB
AB

AC
TAC = TAC λ AC = TAC
= (0.32432i − 0.75676 j + 0.56757k )TAC
AC

= (− 0.86154 j − 0.50769k )TAD
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104
PROBLEM 2.101 (Continued)
Equilibrium condition:
ΣF = 0: TAB + TAC + TAD + Pj = 0
Substituting the expressions obtained for TAB , TAC , and TAD and factoring i, j, and k:
(− 0.6TAB + 0.32432TAC )i + (−0.8TAB − 0.75676TAC − 0.86154TAD + P) j
+ (0.56757TAC − 0.50769TAD )k = 0
Equating to zero the coefficients of i, j, k:
− 0.6TAB + 0.32432TAC = 0
(1)
− 0.8TAB − 0.75676TAC − 0.86154TAD + P = 0
(2)
(3)
Setting TAD = 481 N in (2) and (3), and solving the resulting set of equations gives
TAC = 430.26 N
P = 926 N 
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105
PROBLEM 2.102
Three cables are used to tether a balloon as shown. Knowing that the
balloon exerts an 800-N vertical force at A, determine the tension in
each cable.
SOLUTION
See Problem 2.101 for the figure and analysis leading to the linear algebraic Equations (1), (2), and (3).
− 0.6TAB + 0.32432TAC = 0
(1)
− 0.8TAB − 0.75676TAC − 0.86154TAD + P = 0
(2)
(3)
From Eq. (1):
TAB = 0.54053TAC
From Eq. (3):
Substituting for TAB and TAD in terms of TAC into Eq. (2) gives
− 0.8(0.54053TAC ) − 0.75676TAC − 0.86154(1.11795TAC ) + P = 0
2.1523TAC = P ; P = 800 N
800 N
2.1523
= 371.69 N
TAC =
Substituting into expressions for TAB and TAD gives
TAB = 0.54053(371.69 N)
TAB = 201 N, TAC = 372 N, TAD = 416 N 
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106
PROBLEM 2.103
A crate is supported by three cables as shown. Determine
the weight of the crate knowing that the tension in cable AB
is 750 lb.
SOLUTION
The forces applied at A are:
TAB , TAC , TAD and W
where P = Pj. To express the other forces in terms of the unit vectors i, j, k, we write

AB = − (36 in.)i + (60 in.) j − (27 in.)k
AB = 75 in.

AC = (60 in.) j + (32 in.)k
AC = 68 in.

AD = (40 in.)i + (60 in.) j − (27 in.)k

AB
and
TAB = TAB λAB = TAB
AB
= (− 0.48i + 0.8 j − 0.36k )TAB

AC
TAC = TAC λAC = TAC
AC
= (0.88235 j + 0.47059k )TAC

= (0.51948i + 0.77922 j − 0.35065k )TAD
Equilibrium Condition with
W = − Wj
ΣF = 0: TAB + TAC + TAD − Wj = 0
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107
PROBLEM 2.103 (Continued)
Substituting the expressions obtained for TAB , TAC , and TAD and factoring i, j, and k:
(−0.48TAB + 0.51948TAD )i + (0.8TAB + 0.88235TAC + 0.77922TAD − W ) j
+ (−0.36TAB + 0.47059TAC − 0.35065TAD )k = 0
Equating to zero the coefficients of i, j, k:
0.8TAB + 0.88235TAC + 0.77922TAD − W = 0
−0.36TAB + 0.47059TAC − 0.35065TAD = 0
Substituting TAB = 750 lb in Equations (1), (2), and (3) and solving the resulting set of equations, using
conventional algorithms for solving linear algebraic equations, gives:
TAC = 1090.1 lb
W = 2100 lb 
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108
PROBLEM 2.104
A crate is supported by three cables as shown. Determine
the weight of the crate knowing that the tension in cable AD
is 616 lb.
SOLUTION
See Problem 2.103 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3)
below:
− 0.48TAB + 0.51948TAD = 0
(1)
0.8TAB + 0.88235TAC + 0.77922TAD − W = 0
(2)
− 0.36TAB + 0.47059TAC − 0.35065TAD = 0
(3)
Substituting TAD = 616 lb in Equations (1), (2), and (3) above, and solving the resulting set of equations using
conventional algorithms, gives:
TAB = 667.67 lb
TAC = 969.00 lb
W = 1868 lb 
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109
PROBLEM 2.105
A crate is supported by three cables as shown. Determine the weight
of the crate knowing that the tension in cable AC is 544 lb.
SOLUTION
See Problem 2.103 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3)
below:
− 0.48TAB + 0.51948TAD = 0
(1)
0.8TAB + 0.88235TAC + 0.77922TAD − W = 0
(2)
− 0.36TAB + 0.47059TAC − 0.35065TAD = 0
(3)
Substituting TAC = 544 lb in Equations (1), (2), and (3) above, and solving the resulting set of equations using
conventional algorithms, gives:
TAB = 374.27 lb
W = 1049 lb 
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110
PROBLEM 2.106
A 1600-lb crate is supported by three cables as shown. Determine
the tension in each cable.
SOLUTION
See Problem 2.103 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3)
below:
(1)
0.8TAB + 0.88235TAC + 0.77922TAD − W = 0
(2)
−0.36TAB + 0.47059TAC − 0.35065TAD = 0
(3)
Substituting W = 1600 lb in Equations (1), (2), and (3) above, and solving the resulting set of equations using
conventional algorithms, gives
TAB = 571 lb 
TAC = 830 lb 
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111
PROBLEM 2.107
Three cables are connected at A, where the forces P and Q are
applied as shown. Knowing that Q = 0, find the value of P for
which the tension in cable AD is 305 N.
SOLUTION

ΣFA = 0: TAB + TAC + TAD + P = 0 where P = Pi

AB = −(960 mm)i − (240 mm)j + (380 mm)k
AB = 1060 mm

AC = −(960 mm)i − (240 mm) j − (320 mm)k AC = 1040 mm 

AD = −(960 mm)i + (720 mm) j − (220 mm)k AD = 1220 mm

AB
19 
 48 12
= TAB  − i − j + k 
TAB = TAB λAB = TAB
AB
53
53
53 


AC
3
4 
 12
TAC = TAC λAC = TAC
= TAC  − i − j − k 
AC
 13 13 13 
305 N
[( −960 mm)i + (720 mm) j − (220 mm)k ]
1220 mm
= −(240 N)i + (180 N) j − (55 N)k
Substituting into ΣFA = 0, factoring i, j, k , and setting each coefficient equal to φ gives:
i: P =
48
12
TAB + TAC + 240 N
53
13
(1)
j:
12
3
TAB + TAC = 180 N
53
13
(2)
k:
19
4
TAB − TAC = 55 N
53
13
(3)
Solving the system of linear equations using conventional algorithms gives:
TAB = 446.71 N
TAC = 341.71 N
P = 960 N 
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112
PROBLEM 2.108
Three cables are connected at A, where the forces P and Q are
applied as shown. Knowing that P = 1200 N, determine the values
of Q for which cable AD is taut.
SOLUTION
We assume that TAD = 0 and write
ΣFA = 0: TAB + TAC + Qj + (1200 N)i = 0

AB = −(960 mm)i − (240 mm)j + (380 mm)k AB = 1060 mm

AC = −(960 mm)i − (240 mm) j − (320 mm)k AC = 1040 mm

AB  48 12
19 
TAB = TAB λAB = TAB
=  − i − j + k  TAB
AB  53 53
53 

AC  12
3
4 
TAC = TAC λAC = TAC
=  − i − j − k  TAC
AC  13 13 13 
Substituting into ΣFA = 0, factoring i, j, k , and setting each coefficient equal to φ gives:
i: −
48
12
TAB − TAC + 1200 N = 0
53
13
(1)
j: −
12
3
TAB − TAC + Q = 0
53
13
(2)
k:
19
4
TAB − TAC = 0
53
13
(3)
Solving the resulting system of linear equations using conventional algorithms gives:
TAB = 605.71 N
TAC = 705.71 N
Q = 300.00 N
0 ⱕ Q ⬍ 300 N 
Note: This solution assumes that Q is directed upward as shown (Q ⱖ 0), if negative values of Q
are considered, cable AD remains taut, but AC becomes slack for Q = −460 N. 
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113
PROBLEM 2.109
A rectangular plate is supported by three cables as shown.
Knowing that the tension in cable AC is 60 N, determine the
weight of the plate.
SOLUTION
We note that the weight of the plate is equal in magnitude to the force P exerted by the support on Point A.
Free Body A:
ΣF = 0: TAB + TAC + TAD + Pj = 0
We have:

AB = −(320 mm)i − (480 mm)j + (360 mm)k AB = 680 mm

AC = (450 mm)i − (480 mm) j + (360 mm)k AC = 750 mm

AD = (250 mm)i − (480 mm) j − ( 360 mm ) k AD = 650 mm
Thus:
TAB
TAC

AB  8
12
9 
= TAB λ AB = TAB
=  − i − j + k  TAB
AB  17 17 17 

AC
= TAC λAC = TAC
= ( 0.6i − 0.64 j + 0.48k ) TAC
AC

9.6
7.2 
=
i−
j−
13
13 
Dimensions in mm
Substituting into the Eq. ΣF = 0 and factoring i, j, k :
5
 8

 − TAB + 0.6TAC + TAD  i
13
 17

9.6
 12

+  − TAB − 0.64TAC −
13
 17

7.2
 9

+  TAB + 0.48TAC −
13
 17

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114
PROBLEM 2.109 (Continued)
Setting the coefficient of i, j, k equal to zero:
i:
−
8
5
TAB + 0.6TAC + TAD = 0
17
13
(1)
j:
−
12
9.6
TAB − 0.64TAC −
7
13
(2)
9
7.2
TAB + 0.48TAC −
17
13
(3)
8
5
TAB + 36 N + TAD = 0
17
13
(1′)
9
7.2
TAB + 28.8 N −
17
13
(3′)
k:
Making TAC = 60 N in (1) and (3):
−
Multiply (1′) by 9, (3′) by 8, and add:
554.4 N −
12.6
13
Substitute into (1′) and solve for TAB :
TAB =
17 
5

36 + &times; 572 

8 
13

TAB = 544.0 N
Substitute for the tensions in Eq. (2) and solve for P:
12
9.6
(544 N) + 0.64(60 N) +
(572 N)
17
13
= 844.8 N
P=
Weight of plate = P = 845 N 
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115
PROBLEM 2.110
A rectangular plate is supported by three cables as shown. Knowing
that the tension in cable AD is 520 N, determine the weight of the plate.
SOLUTION
See Problem 2.109 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and
(3) below:
8
5
TAB + 0.6TAC + TAD = 0
17
13
(1)
12
9.6
TAB + 0.64 TAC −
17
13
(2)
9
7.2
TAB + 0.48TAC −
17
13
(3)
−
−
Making TAD = 520 N in Eqs. (1) and (3):
8
TAB + 0.6TAC + 200 N = 0
17
(1′)
9
TAB + 0.48TAC − 288 N = 0
17
(3′)
−
Multiply (1′) by 9, (3′) by 8, and add:
9.24TAC − 504 N = 0 TAC = 54.5455 N
Substitute into (1′) and solve for TAB :
TAB =
17
(0.6 &times; 54.5455 + 200) TAB = 494.545 N
8
Substitute for the tensions in Eq. (2) and solve for P:
12
9.6
(494.545 N) + 0.64(54.5455 N) +
(520 N)
17
13
Weight of plate = P = 768 N 
= 768.00 N
P=
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116
PROBLEM 2.111
A transmission tower is held by three guy wires attached to a
pin at A and anchored by bolts at B, C, and D. If the tension in
wire AB is 630 lb, determine the vertical force P exerted by the
tower on the pin at A.
SOLUTION
Free Body A:
We write
ΣF = 0: TAB + TAC + TAD + Pj = 0

AB = −45i − 90 j + 30k AB = 105 ft

AC = 30i − 90 j + 65k AC = 115 ft

AD = 20i − 90 j − 60k AD = 110 ft

AB
TAB = TAB λ AB = TAB
AB
6
2 
 3
=  − i − j + k  TAB
7
7 
 7

AC
TAC = TAC λAC = TAC
AC
18
13 
 6
= i−
j + k  TAC
23
23
23 


9
6 
2
=  i − j − k  TAD
 11 11 11 
Substituting into the Eq. ΣF = 0 and factoring i, j, k :
6
2
 3

 − TAB + TAC + TAD  i
23
11
 7

6
18
9


+  − TAB − TAC − TAD + P  j
23
11
 7

13
6
2

+  TAB + TAC − TAD  k = 0
23
11
7

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117
PROBLEM 2.111 (Continued)
Setting the coefficients of i, j, k , equal to zero:
i:
3
6
2
− TAB + TAC + TAD = 0
7
23
11
(1)
j:
6
18
9
− TAB − TAC − TAD + P = 0
7
23
11
(2)
k:
2
13
6
TAB + TAC − TAD = 0
7
23
11
(3)
Set TAB = 630 lb in Eqs. (1) – (3):
6
2
23
11
(1′)
18
9
TAC − TAD + P = 0
23
11
(2′)
13
6
23
11
(3′)
−270 lb +
−540 lb −
180 lb +
Solving,
TAC = 467.42 lb TAD = 814.35 lb P = 1572.10 lb
P = 1572 lb 
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118
PROBLEM 2.112
A transmission tower is held by three guy wires attached to a
pin at A and anchored by bolts at B, C, and D. If the tension in
wire AC is 920 lb, determine the vertical force P exerted by the
tower on the pin at A.
SOLUTION
See Problem 2.111 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3)
below:
3
6
2
− TAB + TAC + TAD = 0
7
23
11
(1)
6
18
9
− TAB − TAC − TAD + P = 0
7
23
11
(2)
2
13
6
TAB + TAC − TAD = 0
7
23
11
(3)
Substituting for TAC = 920 lb in Equations (1), (2), and (3) above and solving the resulting set of equations
using conventional algorithms gives:
Solving,
3
2
− TAB + 240 lb + TAD = 0
7
11
(1′)
6
9
− TAB − 720 lb − TAD + P = 0
7
11
(2′)
2
6
TAB + 520 lb − TAD = 0
7
11
(3′)
TAB = 1240.00 lb
P = 3094.3 lb
P = 3090 lb 
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119
PROBLEM 2.113
In trying to move across a slippery icy surface, a 180-lb man uses two
ropes AB and AC. Knowing that the force exerted on the man by the
icy surface is perpendicular to that surface, determine the tension in
each rope.
SOLUTION
Free-Body Diagram at A
30 
 16
N= N i+
j
34 
 34
and W = W j = −(180 lb) j
TAC = TAC λ AC = TAC

AC
( −30 ft)i + (20 ft) j − (12 ft)k
= TAC
AC
38 ft
6 
 15 10
= TAC  − i + j − k 
 19 19 19 
TAB = TAB λ AB = TAB

AB
(−30 ft)i + (24 ft) j + (32 ft)k
= TAB
AB
50 ft
12
16 
 15
= TAB  − i +
j+ k
25
25 
 25
Equilibrium condition: ΣF = 0
TAB + TAC + N + W = 0
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120
PROBLEM 2.113 (Continued)
Substituting the expressions obtained for TAB , TAC , N, and W; factoring i, j, and k; and equating each of the
coefficients to zero gives the following equations:
15
15
16
TAB − TAC +
N =0
25
19
34
(1)
From j:
12
10
30
TAB + TAC +
N − (180 lb) = 0
25
19
34
(2)
From k:
16
6
TAB − TAC = 0
25
19
(3)
From i:
−
Solving the resulting set of equations gives:
TAB = 31.7 lb; TAC = 64.3 lb 
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121
PROBLEM 2.114
Solve Problem 2.113, assuming that a friend is helping the man
at A by pulling on him with a force P = −(60 lb)k.
PROBLEM 2.113 In trying to move across a slippery icy surface,
a 180-lb man uses two ropes AB and AC. Knowing that the force
exerted on the man by the icy surface is perpendicular to that
surface, determine the tension in each rope.
SOLUTION
Refer to Problem 2.113 for the figure and analysis leading to the following set of equations, Equation (3)
being modified to include the additional force P = ( −60 lb)k.
15
15
16
TAB − TAC +
N =0
25
19
34
(1)
12
10
30
TAB + TAC +
N − (180 lb) = 0
25
19
34
(2)
16
6
TAB − TAC − (60 lb) = 0
25
19
(3)
−
Solving the resulting set of equations simultaneously gives:
TAB = 99.0 lb 
TAC = 10.55 lb 
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122
PROBLEM 2.115
For the rectangular plate of Problems 2.109 and 2.110, determine
the tension in each of the three cables knowing that the weight of
the plate is 792 N.
SOLUTION
See Problem 2.109 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and
(3) below. Setting P = 792 N gives:
8
5
TAB + 0.6TAC + TAD = 0
17
13
(1)
12
9.6
TAB − 0.64TAC −
TAD + 792 N = 0
17
13
(2)
9
7.2
TAB + 0.48TAC −
17
13
(3)
−
−
Solving Equations (1), (2), and (3) by conventional algorithms gives
TAB = 510.00 N
TAB = 510 N 
TAC = 56.250 N
TAC = 56.2 N 
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123
PROBLEM 2.116
For the cable system of Problems 2.107 and 2.108, determine the
tension in each cable knowing that P = 2880 N and Q = 0.
SOLUTION
ΣFA = 0: TAB + TAC + TAD + P + Q = 0
Where
P = Pi and Q = Qj

AB = −(960 mm)i − (240 mm) j + (380 mm)k AB = 1060 mm

AC = −(960 mm)i − (240 mm) j − (320 mm)k AC = 1040 mm

AD = −(960 mm)i + (720 mm) j − (220 mm)k AD = 1220 mm

AB
19 
 48 12
TAB = TAB λAB = TAB
= TAB  − i − j + k 
AB
53
53 
 53

AC
3
4 
 12
TAC = TAC λAC = TAC
= TAC  − i − j − k 
AC
 13 13 13 

36
11 
 48
= TAD  − i +
j− k
61
61 
 61
Substituting into ΣFA = 0, setting P = (2880 N)i and Q = 0, and setting the coefficients of i, j, k equal to 0,
we obtain the following three equilibrium equations:
i: −
48
12
48
TAB − TAC − TAD + 2880 N = 0
53
13
61
(1)
j: −
12
3
36
TAB − TAC + TAD = 0
53
13
61
(2)
k:
19
4
11
TAB − TAC − TAD = 0
53
13
61
(3)
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124
PROBLEM 2.116 (Continued)
Solving the system of linear equations using conventional algorithms gives:
TAB = 1340.14 N
TAC = 1025.12 N
TAB = 1340 N 
TAC = 1025 N 
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125
PROBLEM 2.117
For the cable system of Problems 2.107 and 2.108, determine the
tension in each cable knowing that P = 2880 N and Q = 576 N.
SOLUTION
See Problem 2.116 for the analysis leading to the linear algebraic Equations (1), (2), and (3) below:
−
48
12
48
TAB − TAC − TAD + P = 0
53
13
61
(1)
−
12
3
36
TAB − TAC + TAD + Q = 0
53
13
61
(2)
19
4
11
TAB − TAC − TAD = 0
53
13
61
(3)
Setting P = 2880 N and Q = 576 N gives:
−
48
12
48
TAB − TAC − TAD + 2880 N = 0
53
13
61
(1′)
12
3
36
TAB − TAC + TAD + 576 N = 0
53
13
61
(2′)
19
4
11
TAB − TAC − TAD = 0
53
13
61
(3′)
−
Solving the resulting set of equations using conventional algorithms gives:
TAB = 1431.00 N
TAC = 1560.00 N
TAB = 1431 N 
TAC = 1560 N 
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126
PROBLEM 2.118
For the cable system of Problems 2.107 and 2.108, determine the
tension in each cable knowing that P = 2880 N and Q = −576 N.
(Q is directed downward).
SOLUTION
See Problem 2.116 for the analysis leading to the linear algebraic Equations (1), (2), and (3) below:

−
48
12
48
TAB − TAC − TAD + P = 0
53
13
61
(1)
−
12
3
36
TAB − TAC + TAD + Q = 0
53
13
61
(2)
19
4
11
TAB − TAC − TAD = 0
53
13
61
(3)
Setting P = 2880 N and Q = −576 N gives:
−
48
12
48
TAB − TAC − TAD + 2880 N = 0
53
13
61
(1′)
12
3
36
TAB − TAC + TAD − 576 N = 0
53
13
61
(2′)
19
4
11
TAB − TAC − TAD = 0
53
13
61
(3′)
−
Solving the resulting set of equations using conventional algorithms gives:
TAB = 1249.29 N
TAC = 490.31 N
TAB = 1249 N 
TAC = 490 N 
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127
PROBLEM 2.119
For the transmission tower of Problems 2.111 and 2.112, determine
the tension in each guy wire knowing that the tower exerts on the
pin at A an upward vertical force of 2100 lb.
SOLUTION
See Problem 2.111 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and
(3) below:
3
6
2
− TAB + TAC + TAD = 0
7
23
11
(1)
6
18
9
− TAB − TAC − TAD + P = 0
7
23
11
(2)
2
13
6
TAB + TAC − TAD = 0
7
23
11
(3)
Substituting for P = 2100 lb in Equations (1), (2), and (3) above and solving the resulting set of equations
using conventional algorithms gives:
3
6
2
− TAB + TAC + TAD = 0
7
23
11
(1′)
6
18
9
− TAB − TAC − TAD + 2100 lb = 0
7
23
11
(2′)
2
13
6
TAB + TAC − TAD = 0
7
23
11
(3′)
TAB = 841.55 lb
TAC = 624.38 lb
TAB = 842 lb 
TAC = 624 lb 
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128
PROBLEM 2.120
A horizontal circular plate weighing 60 lb is suspended as shown from
three wires that are attached to a support at D and form 30&deg; angles with
the vertical. Determine the tension in each wire.
SOLUTION
ΣFx = 0:
−TAD (sin 30&deg;)(sin 50&deg;) + TBD (sin 30&deg;)(cos 40&deg;) + TCD (sin 30&deg;)(cos 60&deg;) = 0
Dividing through by sin 30&deg; and evaluating:
−0.76604TAD + 0.76604TBD + 0.5TCD = 0
(1)
ΣFy = 0: −TAD (cos 30&deg;) − TBD (cos 30&deg;) − TCD (cos 30&deg;) + 60 lb = 0
TAD + TBD + TCD = 69.282 lb
or
(2)
ΣFz = 0: TAD sin 30&deg; cos 50&deg; + TBD sin 30&deg; sin 40&deg; − TCD sin 30&deg; sin 60&deg; = 0

or
0.64279TAD + 0.64279TBD − 0.86603TCD = 0
(3)
Solving Equations (1), (2), and (3) simultaneously:
TBD = 10.25 lb 



TCD = 29.5 lb 
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129
PROBLEM 2.121
Cable BAC passes through a frictionless ring A and is attached to
fixed supports at B and C, while cables AD and AE are both tied
to the ring and are attached, respectively, to supports at D and E.
Knowing that a 200-lb vertical load P is applied to ring A,
determine the tension in each of the three cables.
SOLUTION
Free Body Diagram at A:
Since TBAC = tension in cable BAC, it follows that
TAB = TAC = TBAC
TAB = TBAC λ AB = TBAC
(−17.5 in.)i + (60 in.) j
60 
 −17.5
= TBAC 
i+
j
62.5 in.
62.5 
 62.5
TAC = TBAC λ AC = TBAC
(60 in.)i + (25 in.)k
25 
 60
= TBAC  j + k 
65 in.
65 
 65
(80 in.)i + (60 in.) j
3 
4
= TAD  i + j 
100 in.
5 
5
TAE = TAE λ AE = TAE
(60 in.) j − (45 in.)k
3 
4
= TAE  j − k 
75 in.
5 
5
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130
PROBLEM 2.121 (Continued)
Substituting into ΣFA = 0, setting P = ( −200 lb) j, and setting the coefficients of i, j, k equal to φ , we obtain
the following three equilibrium equations:
17.5
4
62.5
5
From
i: −
From
3
4
 60 60 
j: 
+  TBAC + TAD + TAE − 200 lb = 0
62.5
65
5
5


From
k:
(1)
25
3
TBAC − TAE = 0
65
5
(2)
(3)
Solving the system of linear equations using convential acgorithms gives:
TBAC = 76.7 lb; TAD = 26.9 lb; TAE = 49.2 lb 
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131
PROBLEM 2.122
Knowing that the tension in cable AE of Prob. 2.121 is 75 lb,
determine (a) the magnitude of the load P, (b) the tension in
PROBLEM 2.121 Cable BAC passes through a frictionless ring A
and is attached to fixed supports at B and C, while cables AD and
AE are both tied to the ring and are attached, respectively, to
supports at D and E. Knowing that a 200-lb vertical load P is
applied to ring A, determine the tension in each of the three
cables.
SOLUTION
Refer to the solution to Problem 2.121 for the figure and analysis leading to the following set of equilibrium
equations, Equation (2) being modified to include Pj as an unknown quantity:
−
17.5
4
62.5
5
(1)
60 
3
4
 60
 62.5 + 65  TBAC + 5 TAD + 5 TAE − P = 0


25
3
TBAC − TAE = 0
65
5
(2)
(3)
Substituting for TAE = 75 lb and solving simultaneously gives:
P = 305 lb; TBAC = 117.0 lb; TAD = 40.9 lb 

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132
PROBLEM 2.123
A container of weight W is suspended from ring A.
Cable BAC passes through the ring and is attached to
fixed supports at B and C. Two forces P = Pi and
Q = Qk are applied to the ring to maintain the
container in the position shown. Knowing that
W = 376 N, determine P and Q. (Hint: The tension is
the same in both portions of cable BAC.)
SOLUTION
TAB = T λ AB

AB
=T
AB
(−130 mm)i + (400 mm) j + (160 mm)k
=T
450 mm
40
16 
 13
j+ k
=T − i +
45
45 
 45
Free-Body A:
TAC = T λ AC

AC
=T
AC
( −150 mm)i + (400 mm) j + (−240 mm)k
=T
490 mm
40
24 
 15
= T − i +
j− k
49
49 
 49
ΣF = 0: TAB + TAC + Q + P + W = 0
Setting coefficients of i, j, k equal to zero:
i: −
13
15
T − T +P=0
45
49
0.59501T = P
(1)
j: +
40
40
T + T −W = 0
45
49
1.70521T = W
(2)
k: +
16
24
T − T +Q =0
45
49
0.134240 T = Q
(3)
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133
PROBLEM 2.123 (Continued)
Data:
W = 376 N 1.70521T = 376 N T = 220.50 N
0.59501(220.50 N) = P
P = 131.2 N 
0.134240(220.50 N) = Q
Q = 29.6 N 
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134
PROBLEM 2.124
For the system of Problem 2.123, determine W and Q
knowing that P = 164 N.
PROBLEM 2.123 A container of weight W is suspended
from ring A. Cable BAC passes through the ring and is
attached to fixed supports at B and C. Two forces P = Pi
and Q = Qk are applied to the ring to maintain the container
in the position shown. Knowing that W = 376 N, determine
P and Q. (Hint: The tension is the same in both portions of
cable BAC.)
SOLUTION
Refer to Problem 2.123 for the figure and analysis resulting in Equations (1), (2), and (3) for P, W, and Q in
terms of T below. Setting P = 164 N we have:
Eq. (1):
0.59501T = 164 N
Eq. (2):
1.70521(275.63 N) = W
Eq. (3):
0.134240(275.63 N) = Q
T = 275.63 N
W = 470 N 
Q = 37.0 N 
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135
PROBLEM 2.125
Collars A and B are connected by a 525-mm-long wire and can
slide freely on frictionless rods. If a force P = (341 N)j is applied
to collar A, determine (a) the tension in the wire when
y = 155 mm, (b) the magnitude of the force Q required to
maintain the equilibrium of the system.
SOLUTION
For both Problems 2.125 and 2.126:
Free-Body Diagrams of Collars:
( AB) 2 = x 2 + y 2 + z 2
Here
(0.525 m) 2 = (0.20 m) 2 + y 2 + z 2
y 2 + z 2 = 0.23563 m 2
or
Thus, when y given, z is determined,
Now
λAB

AB
=
AB
1
(0.20i − yj + zk )m
0.525 m
= 0.38095i − 1.90476 yj + 1.90476 zk
=
Where y and z are in units of meters, m.
From the F.B. Diagram of collar A:
ΣF = 0: N x i + N z k + Pj + TAB λ AB = 0
Setting the j coefficient to zero gives
P − (1.90476 y )TAB = 0
P = 341 N
With
TAB =
341 N
1.90476 y
Now, from the free body diagram of collar B:
ΣF = 0: N x i + N y j + Qk − TAB λAB = 0
Setting the k coefficient to zero gives
Q − TAB (1.90476 z ) = 0
And using the above result for TAB , we have
Q = TAB z =
341 N
(341 N)( z )
(1.90476 z ) =
(1.90476) y
y
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136
PROBLEM 2.125 (Continued)
Then from the specifications of the problem, y = 155 mm = 0.155 m
z 2 = 0.23563 m 2 − (0.155 m) 2
z = 0.46 m
and
341 N
0.155(1.90476)
= 1155.00 N
TAB =
(a)
TAB = 1155 N 
or
and
341 N(0.46 m)(0.866)
(0.155 m)
= (1012.00 N)
Q=
(b)
Q = 1012 N 
or
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137
PROBLEM 2.126
Solve Problem 2.125 assuming that y = 275 mm.
PROBLEM 2.125 Collars A and B are connected by a
525-mm-long wire and can slide freely on frictionless
rods. If a force P = (341 N)j is applied to collar A,
determine (a) the tension in the wire when y = 155 mm,
(b) the magnitude of the force Q required to maintain the
equilibrium of the system.
SOLUTION
From the analysis of Problem 2.125, particularly the results:
y 2 + z 2 = 0.23563 m 2
341 N
TAB =
1.90476 y
341 N
Q=
z
y
With y = 275 mm = 0.275 m, we obtain:
z 2 = 0.23563 m 2 − (0.275 m) 2
z = 0.40 m
and
TAB =
(a)
341 N
= 651.00
(1.90476)(0.275 m)
TAB = 651 N 
or
and
Q=
(b)
341 N(0.40 m)
(0.275 m)
Q = 496 N 
or
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138
PROBLEM 2.127
Two structural members A and B are bolted to a bracket as shown. Knowing
that both members are in compression and that the force is 15 kN in
member A and 10 kN in member B, determine by trigonometry the
magnitude and direction of the resultant of the forces applied to the bracket
by members A and B.
SOLUTION
Using the force triangle and the laws of cosines and sines,
we have
γ = 180&deg; − (40&deg; + 20&deg;)
= 120&deg;
Then
R 2 = (15 kN) 2 + (10 kN)2
− 2(15 kN)(10 kN) cos120&deg;
= 475 kN 2
R = 21.794 kN
and
Hence:
10 kN 21.794 kN
=
sin α
sin120&deg;
 10 kN 
sin α = 
 sin120&deg;
 21.794 kN 
= 0.39737
α = 23.414
φ = α + 50&deg; = 73.414
R = 21.8 kN
73.4&deg; 
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139
PROBLEM 2.128
Member BD exerts on member ABC a force P directed along line BD.
Knowing that P must have a 300-lb horizontal component, determine (a) the
magnitude of the force P, (b) its vertical component.
SOLUTION
P sin 35&deg; = 300 lb
(a)
P=
(b)
Vertical component
300 lb
sin 35&deg;
P = 523 lb 
Pv = P cos 35&deg;
= (523 lb) cos 35&deg;
Pv = 428 lb 
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140
PROBLEM 2.129
Determine (a) the required tension in cable AC, knowing that the resultant
of the three forces exerted at Point C of boom BC must be directed along
BC, (b) the corresponding magnitude of the resultant.
SOLUTION
Using the x and y axes shown:
Rx = ΣFx = TAC sin10&deg; + (50 lb) cos 35&deg; + (75 lb) cos 60&deg;
= TAC sin10&deg; + 78.458 lb
(1)
Ry = ΣFy = (50 lb)sin 35&deg; + (75 lb)sin 60&deg; − TAC cos10&deg;
Ry = 93.631 lb − TAC cos10&deg;
(a)
(2)
Set Ry = 0 in Eq. (2):
93.631 lb − TAC cos10&deg; = 0
TAC = 95.075 lb
(b)
TAC = 95.1 lb 
Substituting for TAC in Eq. (1):
Rx = (95.075 lb)sin10&deg; + 78.458 lb
= 94.968 lb
R = Rx
R = 95.0 lb 
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141
PROBLEM 2.130
Two cables are tied together at C and are loaded as shown. Determine the
tension (a) in cable AC, (b) in cable BC.
SOLUTION
Free-Body Diagram
Force Triangle
W = mg
= (200 kg)(9.81 m/s 2 )
= 1962 N
Law of sines:
TAC
TBC
1962 N
=
=
sin 15&deg; sin 105&deg; sin 60&deg;
(a)
TAC =
(1962 N) sin 15&deg;
sin 60&deg;
TAC = 586 N 
(b)
TBC =
(1962 N) sin 105&deg;
sin 60&deg;
TBC = 2190 N 
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142
PROBLEM 2.131
A welded connection is in equilibrium under the action of the
four forces shown. Knowing that FA = 8 kN and FB = 16 kN,
determine the magnitudes of the other two forces.
SOLUTION
Free-Body Diagram of Connection
ΣFx = 0:
With
3
3
FB − FC − FA = 0
5
5
FA = 8 kN
FB = 16 kN
FC =
4
4
(16 kN) − (8 kN)
5
5
Σ Fy = 0: − FD +
With FA and FB as above:
FC = 6.40 kN 
3
3
FB − FA = 0 
5
5
3
3
FD = (16 kN) − (8 kN) 
5
5
FD = 4.80 kN 
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143
PROBLEM 2.132
Two cables tied together at C are loaded as shown. Determine
the range of values of Q for which the tension will not exceed
60 lb in either cable.
SOLUTION
ΣFx = 0: −TBC − Q cos 60&deg; + 75 lb = 0
Free-Body Diagram
TBC = 75 lb − Q cos 60&deg;
(1)
ΣFy = 0: TAC − Q sin 60&deg; = 0
TAC = Q sin 60&deg;
(2)
TAC ⱕ 60 lb:
Requirement:
Q sin 60&deg; ⱕ 60 lb
From Eq. (2):
Q ⱕ 69.3 lb
TBC ⱕ 60 lb:
Requirement:
From Eq. (1):
75 lb − Q sin 60&deg; ⱕ 60 lb
Q ⱖ 30.0 lb
30.0 lb ⱕ Q ⱕ 69.3 lb 
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144
PROBLEM 2.133
A horizontal circular plate is suspended as shown from three wires that
are attached to a support at D and form 30&deg; angles with the vertical.
Knowing that the x component of the force exerted by wire AD on the
plate is 110.3 N, determine (a) the tension in wire AD, (b) the angles
θ x, θ y, and θ z that the force exerted at A forms with the coordinate axes.
SOLUTION
(a)
Fx = F sin 30&deg; sin 50&deg; = 110.3 N (Given)
F=
(b)
cos θ x =
110.3 N
= 287.97 N
sin 30&deg; sin 50&deg;
F = 288 N 
Fx
110.3 N
=
= 0.38303
F 287.97 N
θ x = 67.5&deg; 
Fy = F cos 30&deg; = 249.39
cos θ y =
Fy
F
=
249.39 N
= 0.86603
287.97 N
θ y = 30.0&deg; 
Fz = − F sin 30&deg; cos 50&deg;
= −(287.97 N)sin 30&deg;cos 50&deg;
= −92.552 N
cos θ z =
Fz −92.552 N
=
= −0.32139
F
287.97 N
θ z = 108.7&deg; 
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145
PROBLEM 2.134
A force acts at the origin of a coordinate system in a direction defined by the angles θy = 55&deg; and θz = 45&deg;.
Knowing that the x component of the force is −500 lb, determine (a) the angle θx, (b) the other components
and the magnitude of the force.
SOLUTION
(a)
We have
(cos θ x ) 2 + (cos θ y )2 + (cos θ z ) 2 = 1  (cos θ y ) 2 = 1 − (cos θ y ) 2 − (cos θ z ) 2
Since Fx ⬍ 0, we must have cos θ x ⬍ 0.
Thus, taking the negative square root, from above, we have
cos θ x = − 1 − (cos 55) 2 − (cos 45) 2 = 0.41353
(b)
θ x = 114.4&deg; 
Then
Fx
500 lb
=
= 1209.10 lb
cos θ x 0.41353
F = 1209 lb 
Fy = F cos θ y = (1209.10 lb) cos 55&deg;
Fy = 694 lb 
Fz = F cos θ z = (1209.10 lb) cos 45&deg;
Fz = 855 lb 
F=
and
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146
PROBLEM 2.135
Find the magnitude and direction of the resultant of the two
forces shown knowing that P = 300 N and Q = 400 N.
SOLUTION
P = (300 N)[− cos 30&deg; sin15&deg;i + sin 30&deg; j + cos 30&deg; cos15&deg;k ]
= − (67.243 N)i + (150 N) j + (250.95 N)k
Q = (400 N)[cos 50&deg; cos 20&deg;i + sin 50&deg; j − cos 50&deg; sin 20&deg;k ]
= (400 N)[0.60402i + 0.76604 j − 0.21985]
= (241.61 N)i + (306.42 N) j − (87.939 N)k
R = P+Q
= (174.367 N)i + (456.42 N) j + (163.011 N)k
R = (174.367 N)2 + (456.42 N)2 + (163.011 N) 2
= 515.07 N
R = 515 N 
cos θ x =
Rx 174.367 N
=
= 0.33853
515.07 N
R
θ x = 70.2&deg; 
cos θ y =
Ry
θ y = 27.6&deg; 
cos θ z =
Rz 163.011 N
=
= 0.31648
R
515.07 N
R
=
456.42 N
= 0.88613
515.07 N
θ z = 71.5&deg; 
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147
PROBLEM 2.136
Three cables are used to tether a balloon as shown. Determine the
vertical force P exerted by the balloon at A knowing that the
tension in cable AC is 444 N.
SOLUTION
See Problem 2.101 for the figure and the analysis leading to the linear algebraic Equations (1), (2),
and (3) below:
− 0.6TAB + 0.32432TAC = 0
(1)
− 0.8TAB − 0.75676TAC − 0.86154TAD + P = 0
(2)
(3)
Substituting TAC = 444 N in Equations (1), (2), and (3) above, and solving the resulting set of
equations using conventional algorithms gives
TAB = 240 N
P = 956 N 
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148
PROBLEM 2.137
Collars A and B are connected by a 25-in.-long wire and can slide
freely on frictionless rods. If a 60-lb force Q is applied to collar B
as shown, determine (a) the tension in the wire when
x = 9 in., (b) the corresponding magnitude of the force P required
to maintain the equilibrium of the system.
SOLUTION
Free-Body Diagrams of Collars:
A:
B:
λAB

AB − xi − (20 in.) j + zk
=
=
AB
25 in.
ΣF = 0: Pi + N y j + N z k + TAB λ AB = 0
Collar A:
Substitute for λAB and set coefficient of i equal to zero:
P−
Collar B:
TAB x
=0
25 in.
(1)
ΣF = 0: (60 lb)k + N x′ i + N y′ j − TAB λ AB = 0
Substitute for λAB and set coefficient of k equal to zero:
60 lb −
x = 9 in.
(a)
From Eq. (2):
(b)
From Eq. (1):
TAB z
=0
25 in.
(2)
(9 in.)2 + (20 in.) 2 + z 2 = (25 in.) 2
z = 12 in.
60 lb − TAB (12 in.)
25 in.
P=
(125.0 lb)(9 in.)

25 in.
TAB = 125.0 lb 
P = 45.0 lb 
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149
PROBLEM 2.138
Collars A and B are connected by a 25-in.-long wire and can slide
freely on frictionless rods. Determine the distances x and z for
which the equilibrium of the system is maintained when P = 120 lb
and Q = 60 lb.
SOLUTION
See Problem 2.137 for the diagrams and analysis leading to Equations (1) and (2) below:
P=
TAB x
=0
25 in.
(1)
60 lb −
TAB z
=0
25 in.
(2)
For P = 120 lb, Eq. (1) yields
TAB x = (25 in.)(20 lb)
(1′)
From Eq. (2):
TAB z = (25 in.)(60 lb)
(2′)
x
=2
z
Dividing Eq. (1′) by (2′),
Now write
x 2 + z 2 + (20 in.) 2 = (25 in.) 2
(3)
(4)
Solving (3) and (4) simultaneously,
4 z 2 + z 2 + 400 = 625
z 2 = 45
z = 6.7082 in.
From Eq. (3):
x = 2 z = 2(6.7082 in.)
= 13.4164 in.
x = 13.42 in., z = 6.71 in. 
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150
PROBLEM 2F1
Two cables are tied together at C and loaded as shown. Draw the freebody diagram needed to determine the tension in AC and BC.
SOLUTION
Free-Body Diagram of Point C:

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151
PROBLEM 2.F2
A chairlift has been stopped in the position
shown. Knowing that each chair weighs 250 N
and that the skier in chair E weighs 765 N, draw
the free-body diagrams needed to determine the
weight of the skier in chair F.
SOLUTION
Free-Body Diagram of Point B:
WE = 250 N + 765 N = 1015 N
8.25
= 30.510&deg;
14
10
= tan −1
= 22.620&deg;
24
θ AB = tan −1
θ BC
Use this free body to determine TAB and TBC.
Free-Body Diagram of Point C:
θCD = tan −1
1.1
= 10.3889&deg;
6
Use this free body to determine TCD and WF.
Then weight of skier WS is found by
WS = WF − 250 N 

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152
PROBLEM 2.F3
Two cables are tied together at A and loaded as shown. Draw the freebody diagram needed to determine the tension in each cable.
SOLUTION
Free-Body Diagram of Point A:

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153
PROBLEM 2.F4
The 60-lb collar A can slide on a frictionless vertical rod and is connected
as shown to a 65-lb counterweight C. Draw the free-body diagram needed
to determine the value of h for which the system is in equilibrium.
SOLUTION
Free-Body Diagram of Point A:

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154
PROBLEM 2.F5
A 36-lb triangular plate is supported by three cables as shown. Draw the
free-body diagram needed to determine the tension in each wire.
SOLUTION
Free-Body Diagram of Point D:

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155
PROBLEM 2.F6
A 70-kg cylinder is supported by two cables AC and BC, which are
attached to the top of vertical posts. A horizontal force P,
perpendicular to the plane containing the posts, holds the cylinder in
the position shown. Draw the free-body diagram needed to
determine the magnitude of P and the force in each cable.
SOLUTION
Free-Body Diagram of Point C:
W = (70 kg)(9.81 m/s 2 )
= 686.7 N

W = −(686.7 N) j
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156
PROBLEM 2.F7
Three cables are connected at point D, which is
located 18 in. below the T-shaped pipe support ABC.
The cables support a 180-lb cylinder as shown. Draw
the free-body diagram needed to determine the tension
in each cable.
SOLUTION
Free-Body Diagram of Point D:


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157
PROBLEM 2.F8
A 100-kg container is suspended from ring A, to
which cables AC and AE are attached. A force P is
applied to end F of a third cable that passes over a
pulley at B and through ring A and then is attached
to a support at D. Draw the free-body diagram
needed to determine the magnitude of P. (Hint: The
tension is the same in all portions of cable FBAD.)
SOLUTION
Free-Body Diagram of Ring A:
W = (100 kg)(9.81 m/s 2 )
= 981 N

W = −(681 N) j

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158
CHAPTER 3
PROBLEM 3.1
A 20-lb force is applied to the control rod AB as shown. Knowing that the length of
the rod is 9 in. and that α = 25&deg;, determine the moment of the force about Point B
by resolving the force into horizontal and vertical components.
SOLUTION
Free-Body Diagram of Rod AB:
x = (9 in.) cos 65&deg;
= 3.8036 in.
y = (9 in.)sin 65&deg;
= 8.1568 in.
F = Fx i + Fy j
rA/B
= (20 lb cos 25&deg;)i + ( −20 lb sin 25&deg;) j
= (18.1262 lb)i − (8.4524 lb) j

= BA = ( −3.8036 in.)i + (8.1568 in.)j
M B = rA /B &times; F
= (−3.8036i + 8.1568 j) &times; (18.1262i − 8.4524 j)
= 32.150k − 147.852k
= −115.702 lb-in.
M B = 115.7 lb-in.

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161
PROBLEM 3.2
A 20-lb force is applied to the control rod AB as shown. Knowing that the length of
the rod is 9 in. and that α = 25&deg;, determine the moment of the force about Point B by
resolving the force into components along AB and in a direction perpendicular to AB.
SOLUTION
Free-Body Diagram of Rod AB:
θ = 90&deg; − (65&deg; − 25&deg;)
= 50&deg;
Q = (20 lb) cos 50&deg;
= 12.8558 lb
M B = Q (9 in.)
= (12.8558 lb)(9 in.)
= 115.702 lb-in.
M B = 115.7 lb-in.

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162
PROBLEM 3.3
A 20-lb force is applied to the control rod AB as shown. Knowing that the length of the
rod is 9 in. and that the moment of the force about B is 120 lb &middot; in. clockwise, determine
the value of α.
SOLUTION
Free-Body Diagram of Rod AB:
α = θ − 25&deg;
Q = (20 lb) cos θ
and
Therefore,
M B = (Q )(9 in.)
120 lb-in. = (20 lb)(cos θ )(9 in.)
120 lb-in.
cos θ =
180 lb-in.
or
θ = 48.190&deg;
Therefore,
α = 48.190&deg; − 25&deg;
α = 23.2&deg; 
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163
PROBLEM 3.4
A crate of mass 80 kg is held in the position shown. Determine
(a) the moment produced by the weight W of the crate about E,
(b) the smallest force applied at B that creates a moment of
equal magnitude and opposite sense about E.
SOLUTION
(a)
By definition,
We have
W = mg = 80 kg(9.81 m/s 2 ) = 784.8 N
ΣM E : M E = (784.8 N)(0.25 m)
M E = 196.2 N ⋅ m
(b)

For the force at B to be the smallest, resulting in a moment (ME) about E, the line of action of force FB
must be perpendicular to the line connecting E to B. The sense of FB must be such that the force
produces a counterclockwise moment about E.
Note:
We have
d = (0.85 m) 2 + (0.5 m) 2 = 0.98615 m
ΣM E : 196.2 N ⋅ m = FB (0.98615 m)
FB = 198.954 N
and
 0.85 m 
 = 59.534&deg;
 0.5 m 
θ = tan −1 
FB = 199.0 N
or
59.5&deg; 
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164
PROBLEM 3.5
A crate of mass 80 kg is held in the position shown. Determine
(a) the moment produced by the weight W of the crate about E,
(b) the smallest force applied at A that creates a moment of equal
magnitude and opposite sense about E, (c) the magnitude, sense,
and point of application on the bottom of the crate of the smallest
vertical force that creates a moment of equal magnitude and
SOLUTION
First note. . .
W = mg = (80 kg)(9.81 m/s 2 ) = 784.8 N
(a)
We have
M E = rH /EW = (0.25 m)(784.8 N) = 196.2 N ⋅ m
(b)
(c)
or M E = 196.2 N ⋅ m

For FA to be minimum, it must be perpendicular to the line
joining Points A and E. Then with FA directed as shown,
we have (− M E ) = rA/E ( FA )min .
Where
rA /E = (0.35 m)2 + (0.5 m)2 = 0.61033 m
then
196.2 N ⋅ m = (0.61033 m)( FA )min
or
( FA ) min = 321 N
Also
tan φ =
0.35 m
0.5 m
or
φ = 35.0&deg;
(FA ) min = 321 N
35.0&deg; 
For Fvertical to be minimum, the perpendicular distance from its line of action to Point E must be
maximum. Thus, apply (Fvertical)min at Point D, and then
(− M E ) = rD / E ( Fvertical ) min
or (Fvertical )min = 231 N
196.2 N ⋅ m = (0.85 m)( Fvertical ) min
at Point D 
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165
PROBLEM 3.6
A 300-N force P is applied at Point A of the bell crank shown.
(a) Compute the moment of the force P about O by resolving
it into horizontal and vertical components. (b) Using the
result of part (a), determine the perpendicular distance from
O to the line of action of P.
SOLUTION
x = (0.2 m) cos 40&deg;
= 0.153209 m
y = (0.2 m)sin 40&deg;
= 0.128558 m
∴ rA /O = (0.153209 m)i + (0.128558 m) j
(a)
Fx = (300 N)sin 30&deg;
= 150 N
Fy = (300 N) cos 30&deg;
= 259.81 N
F = (150 N)i + (259.81 N) j
M O = rA/ O &times; F
= (0.153209i + 0.128558 j) m &times; (150i + 259.81j) N
= (39.805k − 19.2837k ) N ⋅ m
= (20.521 N ⋅ m)k
(b)
M O = 20.5 N ⋅ m

M O = Fd
20.521 N ⋅ m = (300 N)(d )
d = 0.068403 m
d = 68.4 mm 
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166
PROBLEM 3.7
A 400-N force P is applied at Point A of the bell crank shown. (a) Compute
the moment of the force P about O by resolving it into components along
line OA and in a direction perpendicular to that line. (b) Determine the
magnitude and direction of the smallest force Q applied at B that has the
same moment as P about O.
SOLUTION
(a)
Portion OA of crank:
θ = 90&deg; − 30&deg; − 40&deg;
θ = 20&deg;
S = P sin θ
= (400 N) sin 20&deg;
= 136.81 N
M O = rO /A S
= (0.2 m)(136.81 N)
= 27.362 N ⋅ m
(b)
M O = 27.4 N ⋅ m

Smallest force Q must be perpendicular to OB.
Portion OB of crank:
M O = rO /B Q
M O = (0.120 m)Q
27.362 N ⋅ m = (0.120 m)Q
Q = 228 N
42.0&deg; 
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167
PROBLEM 3.8
It is known that a vertical force of 200 lb is required to remove the nail
at C from the board. As the nail first starts moving, determine (a) the
moment about B of the force exerted on the nail, (b) the magnitude of
the force P that creates the same moment about B if α = 10&deg;, (c) the
smallest force P that creates the same moment about B.
SOLUTION
(a)
M B = rC/B FN
We have
= (4 in.)(200 lb)
= 800 lb ⋅ in.
or MB = 800 lb ⋅ in. 
(b)
By definition,
M B = rA/B P sin θ
θ = 10&deg; + (180&deg; − 70&deg;)
= 120&deg;
Then
800 lb ⋅ in. = (18 in.) &times; P sin120&deg;
or P = 51.3 lb 
(c)
For P to be minimum, it must be perpendicular to the line joining
Points A and B. Thus, P must be directed as shown.
Thus
M B = dPmin
d = rA/B
or
or
800 lb ⋅ in. = (18 in.)Pmin
Pmin = 44.4 lb
Pmin = 44.4 lb
20&deg; 
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168
PROBLEM 3.9
It is known that the connecting rod AB exerts on the crank BC a 500-lb force
directed down and to the left along the centerline of AB. Determine the moment of
SOLUTION
Using (a):
M C = y1 ( FAB ) x + x1 ( FAB ) y
 7

 24

= (2.24 in.)  &times; 500 lb  + (1.68 in.)  &times; 500 lb 
 25

 25

= 1120 lb ⋅ in.
(a)
MC = 1.120 kip ⋅ in.

Using (b):
M C = y2 ( FAB ) x
 7

= (8 in.)  &times; 500 lb 
 25

= 1120 lb ⋅ in.
(b)
M C = 1.120 kip ⋅ in.

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169
PROBLEM 3.10
It is known that the connecting rod AB exerts on the crank BC a 500-lb force directed
down and to the left along the centerline of AB. Determine the moment of the force
SOLUTION
Using (a):
M C = − y1 ( FAB ) x + x1 ( FAB ) y
 7

 24

= −(2.24 in.)  &times; 500 lb  + (1.68 in.)  &times; 500 lb 
 25

 25

= +492.8 lb ⋅ in.
(a)
MC = 493 lb ⋅ in.

Using (b):
M C = y2 ( FAB ) x
 7

= (3.52 in.)  &times; 500 lb 
 25

= +492.8 lb ⋅ in.
(b)
M C = 493 lb ⋅ in.

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170
PROBLEM 3.11
A winch puller AB is used to straighten a fence post. Knowing that the tension in cable BC is 1040 N and
length d is 1.90 m, determine the moment about D of the force exerted by the cable at C by resolving that
force into horizontal and vertical components applied (a) at Point C, (b) at Point E.
SOLUTION
(a)
Slope of line:
Then
and
Then
EC =
0.875 m
5
=
1.90 m + 0.2 m 12
12
(TAB )
13
12
= (1040 N)
13
= 960 N
5
TABy = (1040 N)
13
= 400 N
TABx =
(a)
M D = TABx (0.875 m) − TABy (0.2 m)
= (960 N)(0.875 m) − (400 N)(0.2 m)
= 760 N ⋅ m
(b)
We have
or M D = 760 N ⋅ m
M D = TABx ( y ) + TABx ( x)
= (960 N)(0) + (400 N)(1.90 m)
= 760 N ⋅ m

(b)
or M D = 760 N ⋅ m

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171
PROBLEM 3.12
It is known that a force with a moment of 960 N &middot; m about D is required to straighten the fence post CD. If
d = 2.80 m, determine the tension that must be developed in the cable of winch puller AB to create the
SOLUTION
Slope of line:
EC =
0.875 m
7
=
2.80 m + 0.2 m 24
Then
TABx =
24
TAB
25
and
TABy =
7
TAB
25
We have
M D = TABx ( y ) + TABy ( x)
24
7
TAB (0) + TAB (2.80 m)
25
25
= 1224 N
960 N ⋅ m =
TAB
or TAB = 1224 N 
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172
PROBLEM 3.13
It is known that a force with a moment of 960 N &middot; m about D is required to straighten the fence post CD. If the
capacity of winch puller AB is 2400 N, determine the minimum value of distance d to create the specified
SOLUTION
The minimum value of d can be found based on the equation relating the moment of the force TAB about D:
M D = (TAB max ) y (d )
where
M D = 960 N ⋅ m
(TAB max ) y = TAB max sin θ = (2400 N) sin θ
Now
sin θ =
0.875 m
(d + 0.20)2 + (0.875)2 m

0.875
960 N ⋅ m = 2400 N 
 ( d + 0.20) 2 + (0.875) 2

or
(d + 0.20) 2 + (0.875) 2 = 2.1875d
or
(d + 0.20) 2 + (0.875) 2 = 4.7852d 2
or

 (d )


3.7852d 2 − 0.40d − 0.8056 = 0
Using the quadratic equation, the minimum values of d are 0.51719 m and − 0.41151 m.
Since only the positive value applies here, d = 0.51719 m
or d = 517 mm 
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173
PROBLEM 3.14
A mechanic uses a piece of pipe AB as a lever when tightening an
alternator belt. When he pushes down at A, a force of 485 N is
exerted on the alternator at B. Determine the moment of that force
about bolt C if its line of action passes through O.
SOLUTION
We have
M C = rB/C &times; FB
Noting the direction of the moment of each force component about C is
clockwise,
M C = xFBy + yFBx
where
and
x = 120 mm − 65 mm = 55 mm
y = 72 mm + 90 mm = 162 mm
FBx =
FBy =
65
(65) + (72) 2
2
72
(65) + (72) 2
2
(485 N) = 325 N
(485 N) = 360 N
M C = (55 mm)(360 N) + (162)(325 N)
= 72450 N ⋅ m
= 72.450 N ⋅ m
or M C = 72.5 N ⋅ m

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174
PROBLEM 3.15
Form the vector products B &times; C and B′ &times; C, where B = B′, and use the results
obtained to prove the identity
sin α cos β =
1
1
sin (α + β ) + sin (α − β ).
2
2
SOLUTION
Note:
B = B(cos β i + sin β j)
B′ = B(cos β i − sin β j)
C = C (cos α i + sin α j)
By definition,
Now
| B &times; C | = BC sin (α − β )
(1)
| B′ &times; C | = BC sin (α + β )
(2)
B &times; C = B(cos β i + sin β j) &times; C (cos α i + sin α j)
= BC (cos β sin α − sin β cos α )k
and
(3)
B′ &times; C = B(cos β i − sin β j) &times; C (cos α i + sin α j)
= BC (cos β sin α + sin β cos α ) k
(4)
Equating the magnitudes of B &times; C from Equations (1) and (3) yields:
BC sin(α − β ) = BC (cos β sin α − sin β cos α )
(5)
Similarly, equating the magnitudes of B′ &times; C from Equations (2) and (4) yields:
BC sin(α + β ) = BC (cos β sin α + sin β cos α )
(6)
Adding Equations (5) and (6) gives:
sin(α − β ) + sin(α + β ) = 2cos β sin α
or sin α cos β =
1
1
sin(α + β ) + sin(α − β ) 
2
2
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175
PROBLEM 3.16
The vectors P and Q are two adjacent sides of a parallelogram. Determine the area of the parallelogram when
(a) P = −7i + 3j − 3k and Q = 2i + 2j + 5k, (b) P = 6i − 5j − 2k and Q = −2i + 5j − k.
SOLUTION
(a)
We have
A = |P &times; Q|
where
P = −7i + 3j − 3k
Q = 2i + 2 j + 5k
Then
i j k
P &times; Q = −7 3 −3
2 2 5
= [(15 + 6)i + ( −6 + 35) j + ( −14 − 6)k ]
= (21)i + (29) j(−20)k
A = (20) 2 + (29) 2 + (−20) 2
(b)
We have
A = |P &times; Q|
where
P = 6i − 5 j − 2k
or A = 41.0 
Q = −2i + 5 j − 1k
Then
i
j k
P &times; Q = 6 −5 −2
−2 5 −1
= [(5 + 10)i + (4 + 6) j + (30 − 10)k ]
= (15)i + (10) j + (20)k
A = (15) 2 + (10) 2 + (20) 2
or A = 26.9 
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176
PROBLEM 3.17
A plane contains the vectors A and B. Determine the unit vector normal to the plane when A and B are equal
to, respectively, (a) i + 2j − 5k and 4i − 7j − 5k, (b) 3i − 3j + 2k and −2i + 6j − 4k.
SOLUTION
(a)
A&times;B
|A &times; B|
We have
λ=
where
A = 1i + 2 j − 5k
B = 4i − 7 j − 5k
Then
i j k
A &times; B = 1 +2 −5
4 −7 −5
= (−10 − 35)i + (20 + 5) j + (−7 − 8)k
= 15(3i − 1j − 1k )
and
|A &times; B | = 15 (−3)2 + (−1) 2 + (−1)2 = 15 11
λ=
(b)
15(−3i − 1j − 1k )
15 11
or λ =
1
11
(−3i − j − k ) 
A&times;B
|A &times; B|
We have
λ=
where
A = 3i − 3 j + 2k
B = −2i + 6 j − 4k
Then
i
j k
A &times; B = 3 −3 2
−2 6 −4
= (12 − 12)i + (−4 + 12) j + (18 − 6)k
= (8 j + 12k )
and
|A &times; B| = 4 (2) 2 + (3) 2 = 4 13
λ=
4(2 j + 3k )
4 13
or λ =
1
13
(2 j + 3k ) 
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177
PROBLEM 3.18
A line passes through the Points (20 m, 16 m) and (−1 m, −4 m). Determine the perpendicular distance d from
the line to the origin O of the system of coordinates.
SOLUTION
d AB = [20 m − ( −1 m)]2 + [16 m − ( −4 m)]2
= 29 m
Assume that a force F, or magnitude F(N), acts at Point A and is
directed from A to B.
Then
where
F = F λ AB
λ AB =
=
By definition,
where
Then
rB − rA
d AB
1
(21i + 20 j)
29
M O = | rA &times; F | = dF
rA = −(1 m)i − (4 m) j
M O = [ −(−1 m)i − (4 m) j] &times;
F
[(21 m)i + (20 m) j]
29 m
F
[−(20)k + (84)k ]
29
 64 
=  F k N ⋅ m
 29 
=
Finally,
 64 
 29 F  = d ( F )


64
d=
m
29
d = 2.21 m 
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178
PROBLEM 3.19
Determine the moment about the origin O of the force F = 4i − 3j + 5k that acts at a Point A. Assume that the
position vector of A is (a) r = 2i + 3j − 4k, (b) r = −8i + 6j − 10k, (c) r = 8i − 6j + 5k.
SOLUTION
MO = r &times; F
(a)
i j k
M O = 2 3 −4
4 −3 5
= (15 − 12)i + (−16 − 10) j + (−6 − 12)k
(b)
i
j
k
M O = −8 6 −10
4 −3 5
= (30 − 30)i + ( −40 + 40) j + (24 − 24)k
(c)
M O = 3i − 26 j − 18k 
MO = 0 
i j k
M O = 8 −6 5
4 −3 5
= (−30 + 15)i + (20 − 40) j + (−24 + 24)k
M O = −15i − 20 j 
Note: The answer to Part (b) could have been anticipated since the elements of the last two rows of the
determinant are proportional.
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179
PROBLEM 3.20
Determine the moment about the origin O of the force F = 2i + 3j − 4k that acts at a Point A. Assume that the
position vector of A is (a) r = 3i − 6j + 5k, (b) r = i − 4j − 2k, (c) r = 4i + 6j − 8k.
SOLUTION
MO = r &times; F
(a)
i j k
M O = 3 −6 5
2 3 −4
= (24 − 15)i + (10 + 12) j + (9 + 12)k
(b)
i j k
M O = 1 −4 −2
2 3 −4
= (16 + 6)i + (−4 + 4) j + (3 + 8)k
(c)
M O = 9i + 22 j + 21k 
M O = 22i + 11k 
i j k
M O = 4 6 −8
2 3 −4
= (−24 + 24)i + ( −16 + 16) j + (12 − 12)k
MO = 0 
Note: The answer to Part (c) could have been anticipated since the elements of the last two rows of the
determinant are proportional.
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180
PROBLEM 3.21
The wire AE is stretched between the corners A and E of a bent
plate. Knowing that the tension in the wire is 435 N, determine the
moment about O of the force exerted by the wire (a) on corner A,
(b) on corner E.
SOLUTION

AE = (0.21 m)i − (0.16 m) j + (0.12 m)k
(a)
AE = (0.21 m) 2 + (−0.16 m) 2 + (0.12 m) 2 = 0.29 m

AE
FA = FA λ AE = F
AE
0.21i − 0.16 j + 0.12k
= (435 N)
0.29
= (315 N)i − (240 N) j + (180 N)k
rA/O = −(0.09 m)i + (0.16 m) j
i
j
k
M O = −0.09 0.16
0
315 −240 180
= 28.8i + 16.20 j + (21.6 − 50.4)k
(b)
M O = (28.8 N ⋅ m)i + (16.20 N ⋅ m) j − (28.8 N ⋅ m)k 
FE = −FA = −(315 N)i + (240 N) j − (180 N)k
rE / O = (0.12 m)i + (0.12 m)k
i
j
k
M O = 0.12
0
0.12
−315 240 −180
= −28.8i + (−37.8 + 21.6) j + 28.8k
M O = −(28.8 N ⋅ m)i − (16.20 N ⋅ m) j + (28.8 N ⋅ m)k 
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181
PROBLEM 3.22
A small boat hangs from two davits, one of which is shown in the
figure. The tension in line ABAD is 82 lb. Determine the moment
about C of the resultant force RA exerted on the davit at A.
SOLUTION
We have
R A = 2FAB + FAD
where
and
FAB = −(82 lb) j

6i − 7.75 j − 3k
= (82 lb)
10.25
FAD = (48 lb)i − (62 lb) j − (24 lb)k
Thus
R A = 2FAB + FAD = (48 lb)i − (226 lb) j − (24 lb)k
Also
rA/C = (7.75 ft) j + (3 ft)k
Using Eq. (3.21):
i
j
k
M C = 0 7.75
3
48 − 226 −24
= (492 lb ⋅ ft)i + (144.0 lb ⋅ ft) j − (372 lb ⋅ ft)k
M C = (492 lb ⋅ ft)i + (144.0 lb ⋅ ft) j − (372 lb ⋅ ft)k 
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182
PROBLEM 3.23
A 6-ft-long fishing rod AB is securely anchored in the sand of a beach. After a fish takes the bait, the resulting
force in the line is 6 lb. Determine the moment about A of the force exerted by the line at B.
SOLUTION
We have
Then
Txz = (6 lb) cos 8&deg; = 5.9416 lb
Tx = Txz sin 30&deg; = 2.9708 lb
Ty = TBC sin 8&deg; = − 0.83504 lb
Tz = Txz cos 30&deg; = −5.1456 lb
Now
M A = rB/A &times; TBC
where
rB/A = (6sin 45&deg;) j − (6cos 45&deg;)k
=
Then
or
6 ft
2
(j − k)
i
j
k
MA =
0
1
−1
2
2.9708 −0.83504 −5.1456
6
6
6
(−5.1456 − 0.83504)i −
(2.9708) j −
(2.9708)k
=
2
2
2
6
M A = −(25.4 lb ⋅ ft)i − (12.60 lb ⋅ ft) j − (12.60 lb ⋅ ft)k

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183
PROBLEM 3.24
A precast concrete wall section is temporarily held by two
cables as shown. Knowing that the tension in cable BD is 900
N, determine the moment about Point O of the force exerted by
the cable at B.
SOLUTION
F=F

BD
BD
where F = 900 N

BD = −(1 m)i − (2 m) j + (2 m)k
BD = (−1 m) 2 + ( −2 m) 2 + (2 m) 2
=3m
−i − 2 j + 2k
3
= −(300 N)i − (600 N) j + (600 N)k
= (2.5 m)i + (2 m) j
F = (900 N)
rB /O
M O = rB /O &times; F
i
j
k
= 2.5
2
0
−300 −600 600
= 1200i − 1500 j + ( −1500 + 600)k
MO = (1200 N ⋅ m)i − (1500 N ⋅ m) j − (900 N ⋅ m)k 
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184
PROBLEM 3.25
A 200-N force is applied as shown to the bracket ABC. Determine
the moment of the force about A.
SOLUTION
We have
M A = rC/A &times; FC
where
rC/A = (0.06 m)i + (0.075 m) j
FC = −(200 N) cos 30&deg; j + (200 N)sin 30&deg;k
Then
i
j
k
M A = 200 0.06
0.075
0
0
− cos 30&deg; sin 30&deg;
= 200[(0.075sin 30&deg;)i − (0.06sin 30&deg;) j − (0.06 cos 30&deg;)k ]
or M A = (7.50 N ⋅ m)i − (6.00 N ⋅ m) j − (10.39 N ⋅ m)k 
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185
PROBLEM 3.26
The 6-m boom AB has a fixed end A. A steel cable is stretched from
the free end B of the boom to a Point C located on the vertical wall.
If the tension in the cable is 2.5 kN, determine the moment about A of
the force exerted by the cable at B.
SOLUTION
First note
d BC = (−6)2 + (2.4) 2 + (−4) 2
= 7.6 m
2.5 kN
(−6i + 2.4 j − 4k )
7.6
Then
TBC =
We have
M A = rB/A &times; TBC
where
rB/A = (6 m)i
Then
M A = (6 m)i &times;
2.5 kN
(−6i + 2.4 j − 4k )
7.6
or M A = (7.89 kN ⋅ m) j + (4.74 kN ⋅ m)k 
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186
PROBLEM 3.27
In Prob. 3.21, determine the perpendicular distance from point O to
wire AE.
PROBLEM 3.21 The wire AE is stretched between the corners A
and E of a bent plate. Knowing that the tension in the wire is 435 N,
determine the moment about O of the force exerted by the wire (a) on
corner A, (b) on corner E.
SOLUTION
From the solution to Prob. 3.21
M O = (28.8 N ⋅ m)i + (16.20 N ⋅ m) j − (28.8 N ⋅ m)k
M O = (28.8) 2 + (16.20) 2 + (28.8) 2
= 43.8329 N ⋅ m
But
M O = FA d
or
MO
FA
43.8329 N ⋅ m
d=
435 N
= 0.100765 m
d=
d = 100.8 mm 
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187
PROBLEM 3.28
In Prob. 3.21, determine the perpendicular distance from point B to
wire AE.
PROBLEM 3.21 The wire AE is stretched between the corners A
and E of a bent plate. Knowing that the tension in the wire is 435
N, determine the moment about O of the force exerted by the wire
(a) on corner A, (b) on corner E.
SOLUTION
From the solution to Prob. 3.21
FA = (315 N)i − (240 N) j + (180 N)k
rA/B = −(0.210 m)i
M B = rA /B &times; FA = −0.21i &times; (315i − 240 j + 180k )
= 50.4k + 37.8 j
M B = (50.4) 2 + (37.8) 2
= 63.0 N ⋅ m
M B = FA d
or
MB
FA
63.0 N ⋅ m
d=
435 N
= 0.144829 m
d=
d = 144.8 mm 
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188
PROBLEM 3.29
In Problem 3.22, determine the perpendicular distance from
PROBLEM 3.22 A small boat hangs from two davits, one of
which is shown in the figure. The tension in line ABAD is
82 lb. Determine the moment about C of the resultant force
RA exerted on the davit at A.
SOLUTION
First compute the moment about C of the force FDA exerted by the line on D:
From Problem 3.22:
= −(48 lb) i + (62 lb) j + (24 lb)k
M C = rD/C &times; FDA
= + (6 ft)i &times; [−(48 lb)i + (62 lb) j + (24 lb)k ]
= −(144 lb ⋅ ft) j + (372 lb ⋅ ft)k
M C = (144) 2 + (372)2
= 398.90 lb ⋅ ft
Then
M C = FDA d
Since
FDA = 82 lb
d=
=
MC
FDA
398.90 lb ⋅ ft
82 lb
d = 4.86 ft 
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189
PROBLEM 3.30
In Prob. 3.23, determine the perpendicular distance from point A to a line drawn through points B and C.
PROBLEM 3.23 A 6-ft-long fishing rod AB is securely anchored in the sand of a beach. After a fish takes the
bait, the resulting force in the line is 6 lb. Determine the moment about A of the force exerted by the line at B.
SOLUTION
From the solution to Prob. 3.23:
M A = −(25.4 lb ⋅ ft)i − (12.60 lb ⋅ ft) j − (12.60 lb ⋅ ft)k
M A = (−25.4) 2 + (−12.60) 2 + (−12.60)2
= 31.027 lb ⋅ ft
M A = TBC d
MA
TBC
31.027 lb ⋅ ft
=
6 lb
= 5.1712 ft
d=
or
d = 5.17 ft 
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190
PROBLEM 3.31
In Prob. 3.23, determine the perpendicular distance from point D to a line drawn through points B and C.
PROBLEM 3.23 A 6-ft-long fishing rod AB is securely anchored in the sand of a beach. After a fish takes the
bait, the resulting force in the line is 6 lb. Determine the moment about A of the force exerted by the line at B.
SOLUTION
AB = 6 ft
TBC = 6 lb
We have
| M D | = TBC d
where d = perpendicular distance from D to line BC.
M D = rB /D &times; TBC
rB /D = (6sin 45&deg; ft) j = (4.2426 ft)
TBC : (TBC ) x = (6 lb) cos8&deg; sin 30&deg; = 2.9708 lb
(TBC ) y = −(6 lb) sin 8&deg; = −0.83504 lb
(TBC ) z = −(6 lb) cos8&deg; cos 30&deg; = −5.1456 lb
TBC = (2.9708 lb)i − (0.83504 lb) j − (5.1456 lb)k
i
j
k
MD =
0
4.2426
0
2.9708 −0.83504 −5.1456
= −(21.831 lb ⋅ ft)i − (12.6039 lb ⋅ ft)
| M D | = ( −21.831) 2 + (−12.6039) 2 = 25.208 lb ⋅ ft
25.208 lb ⋅ ft = (6 lb)d
d = 4.20 ft 
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191
PROBLEM 3.32
In Prob. 3.24, determine the perpendicular distance from point O
to cable BD.
PROBLEM 3.24 A precast concrete wall section is temporarily
held by two cables as shown. Knowing that the tension in cable
BD is 900 N, determine the moment about Point O of the force
exerted by the cable at B.
SOLUTION
From the solution to Prob. 3.24 we have
M O = (1200 N ⋅ m)i − (1500 N ⋅ m) j − (900 N ⋅ m)k
M O = (1200) 2 + (−1500) 2 + ( −900)2 = 2121.3 N ⋅ m
M O = Fd
MO
F
2121.3 N ⋅ m
=
900 N
d=
d = 2.36 m 
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192
PROBLEM 3.33
In Prob. 3.24, determine the perpendicular distance from point C
to cable BD.
PROBLEM 3.24 A precast concrete wall section is
temporarily held by two cables as shown. Knowing that the
tension in cable BD is 900 N, determine the moment about
Point O of the force exerted by the cable at B.
SOLUTION
From the solution to Prob. 3.24 we have
F = −(300 N)i − (600 N) j + (600 N)k
rB /C = (2 m) j
M C = rB /C &times; F = (2 m) j &times; ( −300 Ni − 600 Nj + 600 Nk )
= (600 N ⋅ m)k + (1200 N ⋅ m)i
M C = (600)2 + (1200) 2 = 1341.64 N ⋅ m
M C = Fd
MC
F
1341.64 N ⋅ m
=
900 N
d=
d = 1.491 m 
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193
PROBLEM 3.34
Determine the value of a that minimizes the
perpendicular distance from Point C to a section
of pipeline that passes through Points A and B.
SOLUTION
Assuming a force F acts along AB,
|M C | = |rA / C &times; F| = F ( d )
d = perpendicular distance from C to line AB
where
F = λ AB F
=
(24 ft) i + (24 ft) j − (28 ft) k
(24) 2 + (24) 2 + (28) 2 ft
F
F
(6) i + (6) j − (7) k
11
= (3 ft)i − (10 ft) j − (a − 10 ft)k
=
rA/C
i
j
k
F
M C = 3 −10 10a
11
6 6
−7
= [(10 + 6a)i + (81 − 6a) j + 78 k ]
|M C | = |rA/C &times; F 2 |
Since
or
F
11
|rA/C &times; F 2 | = ( dF ) 2
1
(10 + 6a) 2 + (81 − 6a) 2 + (78)2 = d 2
121
Setting
d
da
(d 2 ) = 0 to find a to minimize d:
1
[2(6)(10 + 6a) + 2(−6)(81 − 6a)] = 0
121
Solving
a = 5.92 ft
or a = 5.92 ft 
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194
PROBLEM 3.35
Given the vectors P = 3i − j + 2k, Q = 4i + 5j − 3k, and S = −2i + 3j − k, compute the scalar products P &middot; Q,
P &middot; S, and Q &middot; S.
SOLUTION
P ⋅ Q = (3i − j + 2k ) ⋅ (4i + 5 j − 3k )
= (3)(4) + (−1)(5) + (2)(−3)
= 12 − 5 − 6
P ⋅ Q = +1 
P ⋅ S = (3i − j + 2k ) ⋅ (−2i + 3j − k )
= (3)(−2) + (−1)(3) + (2)( −1)
= −6 − 3 − 2
P ⋅ S = −11 
Q ⋅ S = (4i + 5 j − 3k ) ⋅ (−2i + 3j − k )
= (4)(−2) + (5)(3) + ( −3)(−1)
= −8 + 15 + 3
Q ⋅ S = +10 
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195
PROBLEM 3.36
Form the scalar product B &middot; C and use the result obtained to prove the
identity
cos (α − β) = cos α cos β + sin α sin β .
SOLUTION
B = B cos α i + B sin α j
(1)
C = C cos β i + C sin β j
(2)
By definition:
B ⋅ C = BC cos(α − β )
(3)
From (1) and (2):
B ⋅ C = ( B cos α i + B sin α j) ⋅ (C cos β i + C sin β j)
= BC (cos α cos β + sin α sin β )
(4)
Equating the right-hand members of (3) and (4),
cos(α − β ) = cos α cos β + sin α sin β 
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196
PROBLEM 3.37
Consider the volleyball net shown.
Determine the angle formed by guy
wires AB and AC.
SOLUTION
First note:
AB = (−6.5)2 + (−8)2 + (2) 2 = 10.5 ft
AC = (0) 2 + (−8) 2 + (6)2 = 10 ft
and
By definition,
or
or

AB = −(6.5 ft)i − (8 ft) j + (2 ft)k

AC = −(8 ft) j + (6 ft)k
 
AB ⋅ AC = ( AB )( AC ) cos θ
(−6.5i − 8 j + 2k ) ⋅ (−8 j + 6k ) = (10.5)(10) cos θ
(−6.5)(0) + ( −8)( −8) + (2)(6) = 105cos θ
cos θ = 0.72381
or θ = 43.6&deg; 
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197
PROBLEM 3.38
Consider the volleyball net shown.
Determine the angle formed by guy
SOLUTION
First note:
AC = (0)2 + (−8) 2 + (6) 2
= 10 ft
AD = (4) 2 + ( −8) 2 + (1) 2
and
By definition,
or
= 9 ft

AC = −(8 ft)j + (6 ft)k

AD = (4 ft)i − (8 ft) j + (1 ft)k
 
(−8 j + 6k ) ⋅ (4i − 8 j + k ) = (10)(9) cos θ
(0)(4) + ( −8)( −8) + (6)(1) = 90 cos θ
or
cos θ = 0.77778
or θ = 38.9&deg; 
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198
PROBLEM 3.39
Three cables are used to support a container as shown. Determine
the angle formed by cables AB and AD.
SOLUTION
First note:
AB = (450 mm)2 + (600 mm)2
= 750 mm
AD = (−500 mm) 2 + (600 mm) 2 + (360 mm) 2
and
By definition,
= 860 mm

AB = (450 mm)i + (600 mm) j

AD = (−500 mm)i + (600 mm) j + (360 mm)k
 
(450i + 600 j) ⋅ (−500i − 600 j + 360k ) = (750)(860) cos θ
(450)(−500) + (600)(600) + (0)(360) = (750)(860) cos θ
cos θ = 0.20930
or
θ = 77.9&deg; 
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199
PROBLEM 3.40
Three cables are used to support a container as shown. Determine
the angle formed by cables AC and AD.
SOLUTION
First note:
AC = (600 mm) 2 + (−320 mm) 2
= 680 mm
AD = (−500 mm) 2 + (600 mm) 2 + (360 mm) 2
and
By definition,
= 860 mm

AC = (600 mm)j + (−320 mm)k

AD = (−500 mm)i + (600 mm) j + (360 mm)k
 
(600 j − 320k ) ⋅ (−500i + 600 j + 360k ) = (680)(860) cos θ
0(−500) + (600)(600) + (−320)(360) = (680)(860) cos θ
cos θ = 0.41860
θ = 65.3&deg; 
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200
PROBLEM 3.41
The 20-in. tube AB can slide along a horizontal rod. The ends A and B
of the tube are connected by elastic cords to the fixed point C. For the
position corresponding to x = 11 in., determine the angle formed by
the two cords (a) using Eq. (3.32), (b) applying the law of cosines to
triangle ABC.
SOLUTION
(a)
Using Eq. (3.32):

CA = 11i − 12 j + 24k
CA = (11) 2 + ( −12) 2 + (24) 2 = 29 in.

CB = 31i − 12 j + 24k
CB = (31) 2 + ( −12) 2 + (24) 2 = 41 in.
 
CA ⋅ CB
cos θ =
(CA)(CB)
(11i − 12 j + 24k ) ⋅ (31i − 12 j + 24k )
=
(29)(41)
(11)(31) + (−12)(−12) + (24)(24)
=
(29)(41)
= 0.89235
(b)
θ = 26.8&deg; 
Law of cosines:
( AB) 2 = (CA) 2 + (CB) 2 − 2(CA)(CB ) cos θ
(20) 2 = (29)2 + (41) 2 − 2(29)(41) cos θ
cos θ = 0.89235
θ = 26.8&deg; 
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201
PROBLEM 3.42
Solve Prob. 3.41 for the position corresponding to x = 4 in.
PROBLEM 3.41 The 20-in. tube AB can slide along a horizontal
rod. The ends A and B of the tube are connected by elastic cords
to the fixed point C. For the position corresponding to x = 11 in.,
determine the angle formed by the two cords (a) using Eq. (3.32),
(b) applying the law of cosines to triangle ABC.
SOLUTION
(a)
Using Eq. (3.32):

CA = 4i − 12 j + 24k
CA = (4) 2 + (−12) 2 + (24) 2 = 27.129 in.

CB = 24i − 12 j + 24k
CB = (24) 2 + (−12)2 + (24)2 = 36 in.
 
CA ⋅ CB
cos θ =
(CA)(CB)
(4i − 12 j + 24k ) ⋅ (24i − 12 j + 24k )
=
(27.129)(36)
= 0.83551
(b)
θ = 33.3&deg; 
Law of cosines:
( AB) 2 = (CA) 2 + (CB )2 − 2(CA)(CB ) cos θ
(20) 2 = (27.129)2 + (36) 2 − 2(27.129)(36) cos θ
cos θ = 0.83551
θ = 33.3&deg; 
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202
PROBLEM 3.43
Ropes AB and BC are two of the ropes used to
support a tent. The two ropes are attached to a
stake at B. If the tension in rope AB is 540 N,
determine (a) the angle between rope AB and
the stake, (b) the projection on the stake of the
force exerted by rope AB at Point B.
SOLUTION
First note:
BA = (−3) 2 + (3)2 + (−1.5) 2 = 4.5 m
BD = (−0.08) 2 + (0.38)2 + (0.16) 2 = 0.42 m
λ BD
(a)
TBA
(−3i + 3j − 1.5k )
4.5
T
= BA (−2i + 2 j − k )
3

1
BD
( −0.08i + 0.38 j + 0.16k )
=
=
BD 0.42
1
= (−4i + 19 j + 8k )
21
TBA =
Then
We have
or
or
TBA ⋅ λ BD = TBA cos θ
TBA
1
( −2i + 2 j − k ) ⋅ (−4i + 19 j + 8k ) = TBA cos θ
3
21
1
[(−2)( −4) + (2)(19) + (−1)(8)]
63
= 0.60317
cos θ =
or θ = 52.9&deg; 
(b)
We have
(TBA ) BD = TBA ⋅ λ BD
= TBA cos θ
= (540 N)(0.60317)
or (TBA ) BD = 326 N 
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203
PROBLEM 3.44
Ropes AB and BC are two of the ropes used to
support a tent. The two ropes are attached to a
stake at B. If the tension in rope BC is 490 N,
determine (a) the angle between rope BC and
the stake, (b) the projection on the stake of the
force exerted by rope BC at Point B.
SOLUTION
First note:
BC = (1) 2 + (3) 2 + (−1.5) 2 = 3.5 m
BD = (−0.08) 2 + (0.38)2 + (0.16) 2 = 0.42 m
TBC
(i + 3j − 1.5k )
3.5
T
= BC (2i + 6 j − 3k )
7

1
BD
( −0.08i + 0.38 j + 0.16k )
=
=
BD 0.42
1
= (−4i + 19 j + 8k )
21
TBC =
λBD
(a)
TBC ⋅ λBD = TBC cos θ
TBC
1
(2i + 6 j − 3k ) ⋅ (−4i + 19 j + 8k ) = TBC cos θ
7
21
1
[(2)( −4) + (6)(19) + (−3)(8)]
147
= 0.55782
cos θ =
θ = 56.1&deg; 
(b)
(TBC ) BD = TBC ⋅ λBD
= TBC cos θ
= (490 N)(0.55782)
(TBC ) BD = 273 N 
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204
PROBLEM 3.45
Given the vectors P = 4i − 2 j + 3k , Q = 2i + 4 j − 5k , and S = S x i − j + 2k , determine the value of S x for
which the three vectors are coplanar.
SOLUTION
If P, Q, and S are coplanar, then P must be perpendicular to (Q &times; S).
P ⋅ (Q &times; S) = 0
(or, the volume of a parallelepiped defined by P, Q, and S is zero).
Then
or
4
2
Sx
−2 3
4 −5 = 0
−1 2
32 + 10S x − 6 − 20 + 8 − 12S x = 0
Sx = 7 
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205
PROBLEM 3.46
Determine the volume of the parallelepiped of Fig. 3.25 when
(a) P = 4i − 3j + 2k, Q = −2i − 5j + k, and S = 7i + j − k,
(b) P = 5i − j + 6k, Q = 2i + 3j + k, and S = −3i − 2j + 4k.
SOLUTION
Volume of a parallelepiped is found using the mixed triple product.
(a)
Vol. = P ⋅ (Q &times; S)
4 −3 2
= −2 −5 1 in.3
7
1 −1
= (20 − 21 − 4 + 70 + 6 − 4)
= 67
or Volume = 67.0 
(b)
Vol. = P ⋅ (Q &times; S)
5 −1 6
= 2
3 1 in.3
−3 −2 4
= (60 + 3 − 24 + 54 + 8 + 10)
= 111
or Volume = 111.0 
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206
PROBLEM 3.47
Knowing that the tension in cable AB is 570 N, determine the
moment about each of the coordinate axes of the force exerted
on the plate at B.
SOLUTION
BA = (−900 mm)i + (600 mm) j + (360 mm)k
BA = (−900) 2 + (600)2 + (360) 2 = 1140 mm

BA
FB = FB
BA
−900i + 600 j + 360k
= (570 N)
1140
= −(450 N)i + (300 N) j + (180 N)k
rB = (0.9 m)i
M O = rB &times; FB = 0.9i &times; (−450i + 300 j + 180k )
= 270k − 162 j
MO = M x i + M y j + M z k
= −(162 N ⋅ m) j + (270 N ⋅ m)k
M x = 0, M y = −162.0 N ⋅ m, M z = +270 N ⋅ m 
Therefore,
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207
PROBLEM 3.48
Knowing that the tension in cable AC is 1065 N, determine the
moment about each of the coordinate axes of the force exerted
on the plate at C.
SOLUTION

CA = (−900 mm)i + (600 mm) j + (−920 mm)k
CA = ( −900)2 + (600) 2 + (−920) 2 = 1420 mm

CA
FC = FC
CA
−900i + 600 j − 920k
= (1065 N)
1420
= −(675 N)i + (450 N) j − (690 N)k
rC = (0.9 m)i + (1.28 m)k
Using Eq. (3.19):
i
j
k
0.9
0 1.28
M O = rC &times; FC =
−675 450 −690
M O = −(576 N ⋅ m)i − (243 N ⋅ m) j + (405 N ⋅ m)k
But
Therefore,
MO = M x i + M y j + M z k
M x = −576 N ⋅ m, M y = −243 N ⋅ m, M z = +405 N ⋅ m 
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208
PROBLEM 3.49
A small boat hangs from two davits, one of which is shown in
the figure. It is known that the moment about the z-axis of the
resultant force RA exerted on the davit at A must not exceed
279 lb⋅ft in absolute value. Determine the largest allowable
tension in line ABAD when x = 6 ft.
SOLUTION
First note:
R A = 2TAB + TAD
Also note that only TAD will contribute to the moment about the z-axis.
Now
AD = (6) 2 + (−7.75) 2 + (−3) 2
= 10.25 ft

=T
T
(6i − 7.75 j − 3k )
=
10.25
Then
Now
M z = k ⋅ (rA/C &times; TAD )
where
rA/C = (7.75 ft) j + (3 ft)k
Then for Tmax ,
0
0
1
Tmax
279 =
0 7.75 3
10.25
6 −7.75 −3
=
Tmax
| − (1)(7.75)(6)|
10.25
or Tmax = 61.5 lb 
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209
PROBLEM 3.50
For the davit of Problem 3.49, determine the largest allowable
distance x when the tension in line ABAD is 60 lb.
SOLUTION
From the solution of Problem 3.49, TAD is now

=
60 lb
x 2 + ( −7.75) 2 + (−3)2
( xi − 7.75 j − 3k )
Then M z = k ⋅ (rA / C &times; TAD ) becomes
279 =
279 =
60
x 2 + (−7.75) 2 + ( −3) 2
60
x 2 + 69.0625
0
0
1
0 7.75 3
x −7.75 −3
| − (1)(7.75)( x) |
279 x 2 + 69.0625 = 465 x
0.6 x 2 + 69.0625 = x
Squaring both sides:
0.36 x 2 + 24.8625 = x 2
x 2 = 38.848
x = 6.23 ft 
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210
PROBLEM 3.51
A farmer uses cables and winch pullers B and E to plumb
one side of a small barn. If it is known that the sum of the
moments about the x-axis of the forces exerted by the
cables on the barn at Points A and D is equal to 4728 lb ⋅ ft,
determine the magnitude of TDE when TAB = 255 lb.
SOLUTION
The moment about the x-axis due to the two cable forces can be found using the z components of each
force acting at their intersection with the xy plane (A and D). The x components of the forces are parallel to
the x-axis, and the y components of the forces intersect the x-axis. Therefore, neither the x or y components
produce a moment about the x-axis.
We have
ΣM x : (TAB ) z ( y A ) + (TDE ) z ( yD ) = M x
where
(TAB ) z = k ⋅ TAB
= k ⋅ (TAB λ AB )

 − i − 12 j + 12k  
= k ⋅  255 lb 

17



= 180 lb
(TDE ) z = k ⋅ TDE
= k ⋅ (TDE λDE )

 1.5i − 14 j + 12k  
= k ⋅ TDE 

18.5



= 0.64865TDE
y A = 12 ft
yD = 14 ft
M x = 4728 lb ⋅ ft
(180 lb)(12 ft) + (0.64865TDE )(14 ft) = 4728 lb ⋅ ft
and
TDE = 282.79 lb
or TDE = 283 lb 
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211
PROBLEM 3.52
Solve Problem 3.51 when the tension in cable AB is 306 lb.
PROBLEM 3.51 A farmer uses cables and winch pullers B
and E to plumb one side of a small barn. If it is known that the
sum of the moments about the x-axis of the forces exerted by
the cables on the barn at Points A and D is equal to 4728 lb ⋅ ft,
determine the magnitude of TDE when TAB = 255 lb.
SOLUTION
The moment about the x-axis due to the two cable forces can be found using the z components of each force
acting at the intersection with the xy plane (A and D). The x components of the forces are parallel to the x-axis,
and the y components of the forces intersect the x-axis. Therefore, neither the x or y components produce a
We have
ΣM x : (TAB ) z ( y A ) + (TDE ) z ( yD ) = M x
Where
(TAB ) z = k ⋅ TAB
= k ⋅ (TAB λAB )

 − i − 12 j + 12k  
= k ⋅ 306 lb 

17



= 216 lb
(TDE ) z = k ⋅ TDE
= k ⋅ (TDE λDE )

 1.5i − 14 j + 12k  
= k ⋅ TDE 

18.5



= 0.64865TDE
y A = 12 ft
yD = 14 ft
M x = 4728 lb ⋅ ft
(216 lb)(12 ft) + (0.64865TDE )(14 ft) = 4728 lb ⋅ ft
and
or TDE = 235 lb 
TDE = 235.21 lb
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212
PROBLEM 3.53
A single force P acts at C in a direction perpendicular to the
handle BC of the crank shown. Knowing that Mx = +20 N &middot;
m and My = −8.75 N &middot; m, and Mz = −30 N &middot; m, determine the
magnitude of P and the values of φ and θ.
SOLUTION
rC = (0.25 m)i + (0.2 m) sin θ j + (0.2 m) cos θ k
P = − P sin φ j + P cos φ k
i
j
M O = rC &times; P = 0.25 0.2sin θ
0
− P sin φ
k
0.2 cos θ
P cos φ
Expanding the determinant, we find
M x = (0.2) P(sin θ cos φ + cos θ sin φ )
Dividing Eq. (3) by Eq. (2) gives:
M x = (0.2) P sin(θ + φ )
(1)
M y = −(0.25) P cos φ
(2)
M z = −(0.25) P sin φ
(3)
tan φ =
Mz
My
(4)
−30 N ⋅ m
−8.75 N ⋅ m
φ = 73.740
tan φ =
φ = 73.7&deg; 
Squaring Eqs. (2) and (3) and adding gives:
M y2 + M z2 = (0.25) 2 P 2
or
P = 4 M y2 + M z2
(5)
P = 4 (8.75) 2 + (30) 2
= 125.0 N
P = 125.0 N 
Substituting data into Eq. (1):
(+20 N ⋅ m) = 0.2 m(125.0 N) sin(θ + φ )
(θ + φ ) = 53.130&deg; and (θ + φ ) = 126.87&deg;
θ = −20.6&deg; and θ = 53.1&deg;
Q = 53.1&deg; 
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213
PROBLEM 3.54
A single force P acts at C in a direction perpendicular to the
handle BC of the crank shown. Determine the moment Mx of P
about the x-axis when θ = 65&deg;, knowing that My = −15 N &middot; m
and Mz = −36 N &middot; m.
SOLUTION
See the solution to Prob. 3.53 for the derivation of the following equations:
M x = (0.2) P sin(θ + φ )
M
tan φ = z
My
P = 4 M y2 + M z2
(1)
(4)
(5)
Substituting for known data gives:
tan φ =
−36 N ⋅ m
−15 N ⋅ m
φ = 67.380&deg;
P = 4 ( −15) 2 + (−36) 2
P = 156.0 N
M x = 0.2 m(156.0 N) sin(65&deg; + 67.380&deg;)
= 23.047 N ⋅ m
M x = 23.0 N ⋅ m 
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214
PROBLEM 3.55
The triangular plate ABC is supported by ball-and-socket joints
at B and D and is held in the position shown by cables AE and
CF. If the force exerted by cable AE at A is 55 N, determine the
moment of that force about the line joining Points D and B.
SOLUTION
First note:
TAE = TAE

AE
AE
AE = (0.9)2 + (−0.6) 2 + (0.2) 2 = 1.1 m
Then
Also,
Then
Now
where
Then
55 N
(0.9i − 0.6 j + 0.2k )
1.1
= 5[(9 N)i − (6 N) j + (2 N)k ]
TAE =
DB = (1.2) 2 + ( −0.35) 2 + (0)2
λ DB
= 1.25 m

DB
=
DB
1
(1.2i − 0.35 j)
=
1.25
1
(24i − 7 j)
=
25
M DB = λ DB ⋅ (rA/D &times; TAE )
rA/D = −(0.1 m) j + (0.2 m)k
M DB
24 −7
0
1
(5) 0 −0.1 0.2
=
25
9
2
−6
1
= (−4.8 − 12.6 + 28.8)
5
or M DB = 2.28 N ⋅ m 
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215
PROBLEM 3.56
The triangular plate ABC is supported by ball-and-socket joints
at B and D and is held in the position shown by cables AE and
CF. If the force exerted by cable CF at C is 33 N, determine the
moment of that force about the line joining Points D and B.
SOLUTION
First note:
TCF = TCF

CF
CF
CF = (0.6)2 + (−0.9) 2 + (−0.2) 2 = 1.1 m
33 N
(0.6i − 0.9 j + 0.2k )
1.1
= 3[(6 N)i − (9 N) j − (2 N)k ]
Then
TCF =
Also,
DB = (1.2) 2 + ( −0.35) 2 + (0)2
= 1.25 m

DB
=
DB
1
(1.2i − 0.35 j)
=
1.25
1
(24i − 7 j)
=
25
Then
λ DB
Now
M DB = λ DB ⋅ (rC/D &times; TCF )
where
rC/D = (0.2 m) j − (0.4 m)k
Then
M DB
24 −7
0
1
(3) 0 0.2 −0.4
=
25
6 −9 −2
=
3
(−9.6 + 16.8 − 86.4)
25
or M DB = −9.50 N ⋅ m 
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216
PROBLEM 3.57
The 23-in. vertical rod CD is welded to the midpoint C of the
50-in. rod AB. Determine the moment about AB of the 235-lb
force P.
SOLUTION

AB = (32 in.)i − (30 in.) j − (24 in.)k
AB = (32)2 + ( −30) 2 + ( −24) 2 = 50 in.

AB
= 0.64i − 0.60 − 0.48k
λ AB =
AB
We shall apply the force P at Point G:
rG /B = (5 in.)i + (30 in.)k

DG = (21 in.)i − (38 in.) j + (18 in.)k
DG = (21) 2 + (−38)2 + (18) 2 = 47 in.

DG
21i − 38 j + 18k
P=P
= (235 lb)
47
DG
P = (105 lb)i − (190 lb) j + (90 lb)k
The moment of P about AB is given by Eq. (3.46):
M AB
0.64
−0.60 −0.48
0
30 in.
= λ AB ⋅ (rG /B &times; P) = 5 in.
105 lb −190 lb 90 lb
M AB = 0.64[0 − (30 in.)(−190 lb)]
− 0.60[(30 in.)(105 lb) − (5 in.)(90 lb)]
− 0.48[(5 in.)( −190 lb) − 0]
= +2484 lb ⋅ in.
M AB = +207 lb ⋅ ft 
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217
PROBLEM 3.58
The 23-in. vertical rod CD is welded to the midpoint C of the
50-in. rod AB. Determine the moment about AB of the 174-lb
force Q.
SOLUTION

AB = (32 in.)i − (30 in.) j − (24 in.)k
AB = (32)2 + ( −30) 2 + ( −24) 2 = 50 in.

AB
= 0.64i − 0.60 j − 0.48k
λ AB =
AB
We shall apply the force Q at Point H:
rH /B = −(32 in.)i + (17 in.) j

DH = −(16 in.)i − (21 in.) j − (12 in.)k
DH = (16) 2 + (−21) 2 + (−12)2 = 29 in.

DH
−16i − 21j − 12k
Q=
= (174 lb)
DH
29
Q = −(96 lb)i − (126 lb) j − (72 lb)k
The moment of Q about AB is given by Eq. (3.46):
M AB
0.64
−0.60
−0.48
= λ AB ⋅ (rH /B &times; Q) = −32 in. 17 in.
0
−96 lb −126 lb −72 lb
M AB = 0.64[(17 in.)(−72 lb) − 0]
− 0.60[(0 − (−32 in.)( −72 lb)]
− 0.48[(−32 in.)(−126 lb) − (17 in.)(−96 lb)]
= −2119.7 lb ⋅ in.
M AB = 176.6 lb ⋅ ft 
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218
PROBLEM 3.59
The frame ACD is hinged at A and D and is supported by a
cable that passes through a ring at B and is attached to hooks
at G and H. Knowing that the tension in the cable is 450 N,
exerted on the frame by portion BH of the cable.
SOLUTION
Where
and
1
λ AD = (4i − 3k )
5
rB/A = (0.5 m)i
d BH = (0.375)2 + (0.75) 2 + (−0.75) 2
= 1.125 m
Then
Finally,
450 N
(0.375i + 0.75 j − 0.75k )
1.125
= (150 N)i + (300 N) j − (300 N)k
TBH =
4
0
−3
1
0
= 0.5 0
5
150 300 −300
1
= [(−3)(0.5)(300)]
5
or M AD = − 90.0 N ⋅ m 
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219
PROBLEM 3.60
of the force exerted on the frame by portion BG of the cable.
PROBLEM 3.59 The frame ACD is hinged at A and D and is
supported by a cable that passes through a ring at B and is
attached to hooks at G and H. Knowing that the tension in the
of the force exerted on the frame by portion BH of the cable.
SOLUTION
Where
and
1
λ AD = (4i − 3k )
5
rB/A = (0.5 m) j
BG = (−0.5) 2 + (0.925)2 + (−0.4)2
= 1.125 m
Then
Finally,
450 N
(−0.5i + 0.925 j − 0.4k )
1.125
= −(200 N)i + (370 N) j − (160 N)k
TBG =
4
0
−3
1
=
0.5
0
0
5
−200 370 −160
1
= [(−3)(0.5)(370)]
5
M AD = −111.0 N ⋅ m 
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220
PROBLEM 3.61
A regular tetrahedron has six edges of length a. A force P is
directed as shown along edge BC. Determine the moment of P
SOLUTION
We have
From triangle OBC:
M OA = λOA ⋅ (rC /O &times; P)
(OA) x =
a
2
(OA) z = (OA) x tan 30&deg; =
Since
a 1 
a

=
2 3  2 3
(OA) 2 = (OA) 2x + (OA) 2y + (OAz )2
2
or
 a 
a
a =   + (OA) 2y + 

2
2 3
2
2
(OA) y = a 2 −
2
a2 a2
−
=a
4 12
3
a
2
a
i+a
j+
k
2
3
2 3
Then
rA/O =
and
λOA = i +
1
2
2
1
j+
k
3
2 3
P = λ BC P =
(a sin 30&deg;)i − (a cos 30&deg;)k
P
( P) = (i − 3k )
a
2
rC /O = ai
M OA
1
2
=
1
2
3
0
2 3
P
(a)  
0
2
1
0
− 3
=
1
aP  2 
aP
 −
 (1)(− 3) =
2  3
2
M OA =
aP
2

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221
PROBLEM 3.62
A regular tetrahedron has six edges of length a. (a) Show that two
opposite edges, such as OA and BC, are perpendicular to each
other. (b) Use this property and the result obtained in Problem 3.61
to determine the perpendicular distance between edges OA and BC.
SOLUTION
(a)
For edge OA to be perpendicular to edge BC,
 
OA ⋅ BC = 0
From triangle OBC:
(OA) x =
a
2
a 1 
a

=
2 3  2 3
  a 
 a 
OA =   i + (OA) y j + 
k
2
2 3

BC = ( a sin 30&deg;) i − (a cos 30&deg;) k
(OA) z = (OA) x tan 30&deg; =
and
=
Then
or
so that
(b)
We have M OA
a
a 3
a
i−
k = (i − 3 k )
2
2
2
a
 a  
a
 i + (OA) y j + 
 k  ⋅ (i − 3k ) = 0
2
2 3 
2
a2
a2
+ (OA) y (0) −
=0
4
4
 
OA ⋅ BC = 0


OA is perpendicular to BC. 
 
= Pd , with P acting along BC and d the perpendicular distance from OA to BC.
From the results of Problem 3.57,
M OA =
Pa
2
Pa
2
= Pd
or d =
a
2

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222
PROBLEM 3.63
Two forces F1 and F2 in space have the same magnitude F. Prove that the moment of F1 about the line of
action of F2 is equal to the moment of F2 about the line of action of F1 .
SOLUTION
First note that
F1 = F1λ1 and F2 = F2 λ 2
Let M1 = moment of F2 about the line of action of F1 and M 2 = moment of F1 about the line of
action of F2 .
Now, by definition,
M1 = λ 1 ⋅ (rB /A &times; F2 )
= λ 1 ⋅ (rB /A &times; λ 2 ) F2
M 2 = λ 2 ⋅ (rA/B &times; F1 )
= λ 2 ⋅ (rA/B &times; λ 1 ) F1
Since
F1 = F2 = F
and rA /B = −rB /A
M1 = λ 1 ⋅ (rB /A &times; λ 2 ) F
M 2 = λ 2 ⋅ (−rB /A &times; λ 1 ) F
Using Equation (3.39):
so that
λ 1 ⋅ (rB /A &times; λ 2 ) = λ 2 ⋅ (−rB /A &times; λ 1 )
M 2 = λ 1⋅ (rB /A &times; λ 2 ) F 
M12 = M 21 
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223
PROBLEM 3.64
In Problem 3.55, determine the perpendicular distance between
cable AE and the line joining Points D and B.
PROBLEM 3.55 The triangular plate ABC is supported by
ball-and-socket joints at B and D and is held in the position
shown by cables AE and CF. If the force exerted by cable AE
at A is 55 N, determine the moment of that force about the line
joining Points D and B.
SOLUTION
From the solution to Problem 3.55:
ΤAE = 55 N
TAE = 5[(9 N)i − (6 N) j + (2 N)k ]
| M DB | = 2.28 N ⋅ m
λ DB =
1
(24i − 7 j)
25
Based on the discussion of Section 3.11,
it follows that only the perpendicular component of TAE will
contribute to the moment of TAE about line DB.
Now
(TAE )parallel = TAE ⋅ λ DB
= 5(9i − 6 j + 2k ) ⋅
1
(24i − 7 j)
25
1
= [(9)(24) + (−6)(−7)]
5
= 51.6 N
Also,
so that
TAE = (TAE ) parallel + (TAE ) perpendicular
(TAE )perpendicular = (55) 2 + (51.6)2 = 19.0379 N
Since λ DB and (TAE )perpendicular are perpendicular, it follows that
M DB = d (TAE ) perpendicular
or
2.28 N ⋅ m = d (19.0379 N)
d = 0.119761
d = 0.1198 m 
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224
PROBLEM 3.65
In Problem 3.56, determine the perpendicular distance
between cable CF and the line joining Points D and B.
PROBLEM 3.56 The triangular plate ABC is supported by
ball-and-socket joints at B and D and is held in the position
shown by cables AE and CF. If the force exerted by cable CF at
C is 33 N, determine the moment of that force about the line
joining Points D and B.
SOLUTION
ΤCF = 33 N
From the solution to Problem 3.56:
TCF = 3[(6 N)i − (9 N) j − (2 N)k ]
| M DB | = 9.50 N ⋅ m
λ DB =
1
(24i − 7 j)
25
Based on the discussion of Section 3.11,
it follows that only the perpendicular component of TCF will
contribute to the moment of TCF about line DB.
Now
(TCF )parallel = TCF ⋅ λ DB
= 3(6i − 9 j − 2k ) ⋅
1
(24i − 7 j)
25
3
[(6)(24) + (−9)( −7)]
25
= 24.84 N
=
Also,
so that
TCF = (TCF ) parallel + (TCF ) perpendicular
(TCF )perpendicular = (33) 2 − (24.84) 2
= 21.725 N
Since λ DB and (TCF )perpendicular are perpendicular, it follows that
| M DB | = d (TCF ) perpendicular
or
9.50 N ⋅ m = d &times; 21.725 N
or d = 0.437 m 
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225
PROBLEM 3.66
In Prob. 3.57, determine the perpendicular distance between
rod AB and the line of action of P.
PROBLEM 3.57 The 23-in. vertical rod CD is welded to the
midpoint C of the 50-in. rod AB. Determine the moment about
AB of the 235-lb force P.
SOLUTION

AB = (32 in.)i − (30 in.) j − (24 in.)k
AB = (32)2 + (−30) 2 + ( −24) 2 = 50 in.

AB
= 0.64i − 0.60 j − 0.48k
λ AB =
AB
λP =
P 105i − 190 j + 90k
=
P
235
Angle θ between AB and P:
cos θ = λ AB ⋅ λ P
= (0.64i − 0.60 j − 0.48k ) ⋅
105i − 190 j + 90k
235
= 0.58723
∴ θ = 54.039&deg;
The moment of P about AB may be obtained by multiplying the projection of P on a plane perpendicular to
AB by the perpendicular distance d from AB to P:
M AB = ( P sin θ )d
From the solution to Prob. 3.57: M AB = 207 lb ⋅ ft = 2484 lb ⋅ in.
We have
2484 lb ⋅ in. = (235 lb)(sin 54.039) d
d = 13.06 in. 
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226
PROBLEM 3.67
In Prob. 3.58, determine the perpendicular distance between
rod AB and the line of action of Q.
PROBLEM 3.58 The 23-in. vertical rod CD is welded to the
midpoint C of the 50-in. rod AB. Determine the moment about
AB of the 174-lb force Q.
SOLUTION

AB = (32 in.)i − (30 in.) j − (24 in.)k
AB = (32)2 + (−30) 2 + ( −24) 2 = 50 in.

AB
λ AB =
= 0.64i − 0.60 j − 0.48k
AB
λQ =
Q −96i − 126 j − 72k
=
Q
174
Angle θ between AB and Q:
cos θ = λ AB ⋅ λQ
= (0.64i − 0.60 j − 0.48k ) ⋅
(−96i − 126 j − 72k )
174
= 0.28000
∴ θ = 73.740&deg;
The moment of Q about AB may be obtained by multiplying the projection of Q on a plane perpendicular to
AB by the perpendicular distance d from AB to Q:
M AB = (Q sin θ )d
From the solution to Prob. 3.58: M AB = 176.6 lb ⋅ ft = 2119.2 lb ⋅ in.
2119.2 lb ⋅ in. = (174 lb)(sin 73.740&deg;) d
d = 12.69 in. 
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227
PROBLEM 3.68
In Problem 3.59, determine the perpendicular distance between
portion BH of the cable and the diagonal AD.
PROBLEM 3.59 The frame ACD is hinged at A and D and is
supported by a cable that passes through a ring at B and is
attached to hooks at G and H. Knowing that the tension in the
of the force exerted on the frame by portion BH of the cable.
SOLUTION
From the solution to Problem 3.59:
TBH = 450 N
TBH = (150 N)i + (300 N) j − (300 N)k
| M AD | = 90.0 N ⋅ m
1
λ AD = (4i − 3k )
5
Based on the discussion of Section 3.11, 
it follows that only the perpendicular component of TBH will
Now
(TBH )parallel = TBH ⋅ λ AD
1
= (150i + 300 j − 300k ) ⋅ (4i − 3k )
5
1
= [(150)(4) + (−300)(−3)]
5
= 300 N
Also,
so that
TBH = (TBH ) parallel + (TBH )perpendicular
(TBH )perpendicular = (450) 2 − (300) 2 = 335.41 N
Since λ AD and (TBH )perpendicular are perpendicular, it follows that
M AD = d (TBH ) perpendicular
or
90.0 N ⋅ m = d (335.41 N)
d = 0.26833 m
d = 0.268 m 
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228
PROBLEM 3.69
In Problem 3.60, determine the perpendicular distance between
portion BG of the cable and the diagonal AD.
PROBLEM 3.60 In Problem 3.59, determine the moment
about the diagonal AD of the force exerted on the frame by
portion BG of the cable.
SOLUTION
From the solution to Problem 3.60:
ΤBG = 450 N
TBG = −(200 N)i + (370 N) j − (160 N)k
| M AD | = 111 N ⋅ m
1
λ AD = (4i − 3k )
5
Based on the discussion of Section 3.11, 
it follows that only the perpendicular component of TBG will
Now
(TBG ) parallel = TBG ⋅ λ AD
1
= ( −200i + 370 j − 160k ) ⋅ (4i − 3k )
5
1
= [(−200)(4) + (−160)(−3)]
5
= −64 N
Also,
so that
TBG = (TBG ) parallel + (TBG )perpendicular
(TBG ) perpendicular = (450) 2 − ( −64) 2 = 445.43 N
Since λ AD and (TBG ) perpendicular are perpendicular, it follows that
M AD = d (TBG ) perpendicular
or
111 N ⋅ m = d (445.43 N)
d = 0.24920 m
d = 0.249 m 
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229
PROBLEM 3.70
A plate in the shape of a parallelogram is acted upon by two
couples. Determine (a) the moment of the couple formed by the
two 21-lb forces, (b) the perpendicular distance between the 12-lb
forces if the resultant of the two couples is zero, (c) the value of α
if the resultant couple is 72 lb ⋅ in. clockwise and d is 42 in.
SOLUTION
(a)
M1 = d1 F1
We have
d1 = 16 in.
where
F1 = 21 lb
M1 = (16 in.)(21 lb)
= 336 lb ⋅ in.
(b)
(c)
336 lb ⋅ in. − d 2 (12 lb) = 0
d 2 = 28.0 in. 
M total = M1 + M 2
We have
or

M1 + M 2 = 0
We have
or
or M1 = 336 lb ⋅ in.
−72 lb ⋅ in. = 336 lb ⋅ in. − (42 in.)(sin α )(12 lb)
sin α = 0.80952
α = 54.049&deg;
and
or α = 54.0&deg; 
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230
PROBLEM 3.71
Four 1-in.-diameter pegs are attached to a board as shown. Two
strings are passed around the pegs and pulled with the forces
indicated. (a) Determine the resultant couple acting on the board.
(b) If only one string is used, around which pegs should it pass
and in what directions should it be pulled to create the same
couple with the minimum tension in the string? (c) What is the
value of that minimum tension?
SOLUTION
M = (35 lb)(7 in.) + (25 lb)(9 in.)
= 245 lb ⋅ in. + 225 lb ⋅ in.
(a)
M = 470 lb ⋅ in.
(b)

With only one string, pegs A and D, or B and C should be used. We have
6
8
tan θ =
θ = 36.9&deg;
90&deg; − θ = 53.1&deg;
Direction of forces:
(c)
With pegs A and D:
θ = 53.1&deg; 
With pegs B and C:
θ = 53.1&deg; 
The distance between the centers of the two pegs is
82 + 62 = 10 in.
Therefore, the perpendicular distance d between the forces is
1 
d = 10 in. + 2  in. 
2 
= 11 in.
M = Fd
We must have
470 lb ⋅ in. = F (11 in.)
F = 42.7 lb 
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231
PROBLEM 3.72
Four pegs of the same diameter are attached to a board as shown.
Two strings are passed around the pegs and pulled with the
forces indicated. Determine the diameter of the pegs knowing
that the resultant couple applied to the board is 485 lb&middot;in.
counterclockwise.
SOLUTION
485 lb ⋅ in. = [(6 + d ) in.](35 lb) + [(8 + d ) in.](25 lb)
d = 1.250 in. 
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232
PROBLEM 3.73
A piece of plywood in which several holes are being drilled
successively has been secured to a workbench by means of two
nails. Knowing that the drill exerts a 12-N&middot;m couple on the piece of
plywood, determine the magnitude of the resulting forces applied to
the nails if they are located (a) at A and B, (b) at B and C, (c) at A
and C.
SOLUTION
(a)
M = Fd
12 N ⋅ m = F (0.45 m)
F = 26.7 N 
(b)
M = Fd
12 N ⋅ m = F (0.24 m)
F = 50.0 N 
(c)
M = Fd
d = (0.45 m)2 + (0.24 m)2
= 0.510 m
12 N ⋅ m = F (0.510 m)
F = 23.5 N 
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233
PROBLEM 3.74
Two parallel 40-N forces are applied to a lever as shown.
Determine the moment of the couple formed by the two forces
(a) by resolving each force into horizontal and vertical components
and adding the moments of the two resulting couples, (b) by using
the perpendicular distance between the two forces, (c) by summing
the moments of the two forces about Point A.
SOLUTION
(a)
We have
where
ΣM B : − d1C x + d 2 C y = M
d1 = (0.270 m) sin 55&deg;
= 0.22117 m
d 2 = (0.270 m) cos 55&deg;
= 0.154866 m
C x = (40 N) cos 20&deg;
= 37.588 N
C y = (40 N) sin 20&deg;
= 13.6808 N
M = −(0.22117 m)(37.588 N)k + (0.154866 m)(13.6808 N)k
= −(6.1946 N ⋅ m)k
(b)
We have
We have

or M = 6.19 N ⋅ m

or M = 6.19 N ⋅ m

M = Fd (−k )
= 40 N[(0.270 m)sin(55&deg; − 20&deg;)](−k )
= −(6.1946 N ⋅ m)k
(c)
or M = 6.19 N ⋅ m
ΣM A : Σ(rA &times; F ) = rB/A &times; FB + rC/A &times; FC = M
i
j
k
sin 55&deg; 0
M = (0.390 m)(40 N) cos 55&deg;
− cos 20&deg; − sin 20&deg; 0
i
j
k
+ (0.660 m)(40 N) cos 55&deg; sin 55&deg; 0
cos 20&deg; sin 20&deg; 0
= (8.9478 N ⋅ m − 15.1424 N ⋅ m)k
= −(6.1946 N ⋅ m)k
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234
PROBLEM 3.75
The two shafts of a speed-reducer unit are subjected to
couples of magnitude M1 = 15 lb&middot;ft and M2 = 3 lb&middot;ft,
respectively. Replace the two couples with a single equivalent
couple, specifying its magnitude and the direction of its axis.
SOLUTION
M1 = (15 lb ⋅ ft)k
M 2 = (3 lb ⋅ ft)i
M = M12 + M 22
= (15) 2 + (3) 2
= 15.30 lb ⋅ ft
15
tan θ x =
=5
3
θ x = 78.7&deg;
θ y = 90&deg;
θ z = 90&deg; − 78.7&deg;
= 11.30&deg;
M = 15.30 lb ⋅ ft; θ x = 78.7&deg;, θ y = 90.0&deg;, θ z = 11.30&deg; 
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235
PROBLEM 3.76
Replace the two couples shown with a single equivalent couple,
specifying its magnitude and the direction of its axis.
SOLUTION
Replace the couple in the ABCD plane with two couples P and Q shown:
P = (50 N)
CD
 160 mm 
= (50 N) 
 = 40 N
CG
 200 mm 
Q = (50 N)
CF
 120 mm 
= (50 N) 
 = 30 N
CG
 200 mm 
Couple vector M1 perpendicular to plane ABCD:
M1 = (40 N)(0.24 m) − (30 N)(0.16 m) = 4.80 N ⋅ m
Couple vector M2 in the xy plane:
M 2 = −(12.5 N)(0.192 m) = −2.40 N ⋅ m
144 mm
θ = 36.870&deg;
192 mm
M1 = (4.80 cos 36.870&deg;) j + (4.80 sin 36.870&deg;)k
= 3.84 j + 2.88k
tan θ =
M 2 = −2.40 j
M = M1 + M 2 = 1.44 j + 2.88k
M = 3.22 N ⋅ m; θ x = 90.0&deg;, θ y = 53.1&deg;, θ z = 36.9&deg; 
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236
PROBLEM 3.77
Solve Prob. 3.76, assuming that two 10-N vertical forces have
been added, one acting upward at C and the other downward at B.
PROBLEM 3.76 Replace the two couples shown with a single
equivalent couple, specifying its magnitude and the direction of
its axis.
SOLUTION
Replace the couple in the ABCD plane with two couples P and Q shown.
P = (50 N)
CD
 160 mm 
= (50 N) 
 = 40 N
CG
 200 mm 
Q = (50 N)
CF
 120 mm 
= (50 N) 
 = 30 N
CG
 200 mm 
Couple vector M1 perpendicular to plane ABCD.
M1 = (40 N)(0.24 m) − (30 N)(0.16 m) = 4.80 N ⋅ m
144 mm
θ = 36.870&deg;
192 mm
M1 = (4.80cos 36.870&deg;) j + (4.80sin 36.870&deg;)k
= 3.84 j + 2.88k
tan θ =
M 2 = −(12.5 N)(0.192 m) = −2.40 N ⋅ m
= −2.40 j
M 3 = rB /C &times; M 3 ; rB /C = (0.16 m)i + (0.144 m) j − (0.192 m)k
= (0.16 m)i + (0.144 m) j − (0.192 m)k &times; (−10 N) j
= −1.92i − 1.6k
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237
PROBLEM 3.77 (Continued)
M = M1 + M 2 + M 3 = (3.84 j + 2.88k ) − 2.40 j + (−1.92i − 1.6k )
= −(1.92 N ⋅ m)i + (1.44 N ⋅ m) j + (1.28 N ⋅ m)k
M = ( −1.92) 2 + (1.44) 2 + (1.28) 2 = 2.72 N ⋅ m
M = 2.72 N ⋅ m 
cos θ x = −1.92/2.72
cos θ y = 1.44/2.72
θ x = 134.9&deg; θ y = 58.0&deg; θ z = 61.9&deg; 
cos θ z = 1.28/2.72
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238
PROBLEM 3.78
If P = 0, replace the two remaining couples with a single
equivalent couple, specifying its magnitude and the direction
of its axis.
SOLUTION
M = M1 + M 2 ; F1 = 16 lb, F2 = 40 lb
M1 = rC &times; F1 = (30 in.)i &times; [−(16 lb) j] = −(480 lb ⋅ in.)k
M 2 = rE/B &times; F2 ; rE/B = (15 in.)i − (5 in.) j
d DE = (0) 2 + (5) 2 + (10) 2 = 5 5 in.
F2 =
40 lb
5 5
(5 j − 10k )
= 8 5[(1 lb) j − (2 lb)k ]
i
j k
M 2 = 8 5 15 −5 0
0 1 −2
= 8 5[(10 lb ⋅ in.)i + (30 lb ⋅ in.) j + (15 lb ⋅ in.)k ]
M = −(480 lb ⋅ in.)k + 8 5[(10 lb ⋅ in.)i + (30 lb ⋅ in.) j + (15 lb ⋅ in.)k ]
= (178.885 lb ⋅ in.)i + (536.66 lb ⋅ in.) j − (211.67 lb ⋅ in.)k
M = (178.885) 2 + (536.66) 2 + (−211.67) 2
= 603.99 lb ⋅ in
M = 604 lb ⋅ in. 
M
= 0.29617i + 0.88852 j − 0.35045k
M
cos θ x = 0.29617
λ axis =
cos θ y = 0.88852
θ x = 72.8&deg; θ y = 27.3&deg; θ z = 110.5&deg; 
cos θ z = −0.35045
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239
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 &times; P
= (30 in.)i &times; (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&deg; θ y = 13.70&deg; θ z = 100.4&deg; 
cos θ z = −0.180921
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240
PROBLEM 3.80
In a manufacturing operation, three holes are drilled
simultaneously in a workpiece. If the holes are perpendicular to the
surfaces of the workpiece, replace the couples applied to the drills
with a single equivalent couple, specifying its magnitude and the
direction
of
its axis.
SOLUTION
M = M1 + M 2 + M 3
= (1.5 N ⋅ m)(− cos 20&deg; j + sin 20&deg;k ) − (1.5 N ⋅ m) j
+ (1.75 N ⋅ m)(− cos 25&deg; j + sin 25&deg;k )
= −(4.4956 N ⋅ m) j + (0.22655 N ⋅ m)k
M = (0) 2 + (−4.4956) 2 + (0.22655) 2
= 4.5013 N ⋅ m
M = 4.50 N ⋅ m 
M
= −(0.99873j + 0.050330k )
M
cos θ x = 0
λaxis =
cos θ y = −0.99873
θ x = 90.0&deg;, θ y = 177.1&deg;, θ z = 87.1&deg; 
cos θ z = 0.050330
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241
PROBLEM 3.81
A 260-lb force is applied at A to the rolled-steel section shown. Replace that
force with an equivalent force-couple system at the center C of the section.
SOLUTION
AB = (2.5 in.) 2 + (6.0 in.)2 = 6.50 in.
2.5 in. 5
=
6.5 in. 13
6.0 in. 12
cos α =
=
6.5 in. 13
sin α =
α = 22.6&deg;
F = − F sin α i − F cos α j
5
12
= −(260 lb) i − (260 lb) j
13
13
= −(100.0 lb)i − (240 lb) j
M C = rA /C &times; F
= (2.5i + 4.0 j) &times; (−100.0i − 240 j)
= 400k − 600k
= −(200 lb ⋅ in.)k
F = 260 lb
67.4&deg;; M C = 200 lb ⋅ in.

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242
PROBLEM 3.82
A 30-lb vertical force P is applied at A to the bracket shown, which is held by
screws at B and C. (a) Replace P with an equivalent force-couple system at B.
(b) Find the two horizontal forces at B and C that are equivalent to the couple
obtained in part a.
SOLUTION
(a)
M B = (30 lb)(5 in.)
= 150.0 lb ⋅ in.
F = 30.0 lb , M B = 150.0 lb ⋅ in.
(b)
B=C =

150 lb ⋅ in.
= 50.0 lb
3.0 in.
B = 50.0 lb
; C = 50.0 lb

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243
PROBLEM 3.83
The force P has a magnitude of 250 N and is applied at the end C of
a 500-mm rod AC attached to a bracket at A and B. Assuming α = 30&deg;
and β = 60&deg;, replace P with (a) an equivalent force-couple system at B,
(b) an equivalent system formed by two parallel forces applied at
A and B.
SOLUTION
(a)
ΣF : F = P or F = 250 N
Equivalence requires
60&deg;
ΣM B : M = −(0.3 m)(250 N) = −75 N ⋅ m
The equivalent force-couple system at B is
FB = 250 N
(b)
M B = 75.0 N ⋅ m
60&deg;

We require
Equivalence then requires
ΣFx : 0 = FA cos φ + FB cos φ
FA = − FB
or cos φ = 0
ΣFy : − 250 = − FA sin φ − FB sin φ
Now if
FA = − FB  −250 = 0, reject.
cos φ = 0
or
and
Also,
φ = 90&deg;
FA + FB = 250
ΣM B : − (0.3 m)(250 N) = (0.2m) FA
or
FA = −375 N
and
FB = 625 N
FA = 375 N
60&deg;
FB = 625 N
60.0&deg; 
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244
PROBLEM 3.84
Solve Problem 3.83, assuming α = β = 25&deg;.
PROBLEM 3.83 The force P has a magnitude of 250 N and is applied
at the end C of a 500-mm rod AC attached to a bracket at A and B.
Assuming α = 30&deg; and β = 60&deg;, replace P with (a) an equivalent forcecouple system at B, (b) an equivalent system formed by two parallel
forces applied at A and B.
SOLUTION
(a)
Equivalence requires
ΣF : FB = P or FB = 250 N
25.0&deg;
ΣM B : M B = −(0.3 m)[(250 N)sin 50&deg;] = −57.453 N ⋅ m
The equivalent force-couple system at B is
FB = 250 N
(b)
25.0&deg;
M B = 57.5 N ⋅ m

We require
Equivalence requires
M B = d AE Q (0.3 m)[(250 N) sin 50&deg;]
= [(0.2 m) sin 50&deg;]Q
Q = 375 N
FA = 375 N
25.0&deg;
FB = 625 N
25.0&deg; 
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245
PROBLEM 3.85
The 80-N horizontal force P acts on a bell crank as shown.
(a) Replace P with an equivalent force-couple system at B.
(b) Find the two vertical forces at C and D that are equivalent
to the couple found in part a.
SOLUTION
(a)
ΣF : FB = F = 80 N
Based on
or
FB = 80.0 N

M B = 4.00 N ⋅ m

ΣM : M B = Fd B
= 80 N (0.05 m)
= 4.0000 N ⋅ m
or
(b)
If the two vertical forces are to be equivalent to MB, they must be
a couple. Further, the sense of the moment of this couple must be
counterclockwise.
Then with FC and FD acting as shown,
ΣM : M D = FC d
4.0000 N ⋅ m = FC (0.04 m)
FC = 100.000 N
or FC = 100.0 N 
ΣFy : 0 = FD − FC
FD = 100.000 N
or FD = 100.0 N 
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246
PROBLEM 3.86
A dirigible is tethered by a cable attached to its cabin at B. If
the tension in the cable is 1040 N, replace the force exerted by
the cable at B with an equivalent system formed by two
parallel forces applied at A and C.
SOLUTION
Require the equivalent forces acting at A and C be parallel and at an
angle of α with the vertical.
Then for equivalence,
ΣFx : (1040 N)sin 30&deg; = FA sin α + FB sin α
(1)
ΣFy : −(1040 N) cos 30&deg; = − FA cos α − FB cos α
(2)
Dividing Equation (1) by Equation (2),
( FA + FB ) sin α
(1040 N) sin 30&deg;
=
−(1040 N) cos 30&deg; −( FA + FB ) cos α
Simplifying yields α = 30&deg;.
Based on
ΣM C : [(1040 N) cos 30&deg;](4 m) = ( FA cos 30&deg;)(10.7 m)
FA = 388.79 N
FA = 389 N
or
60.0&deg; 
Based on
ΣM A : − [(1040 N) cos 30&deg;](6.7 m) = ( FC cos 30&deg;)(10.7 m)
FC = 651.21 N
FC = 651 N
or
60.0&deg; 
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247
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&deg; − (60&deg; + 55&deg;) = 65&deg;
M = ΣM B
= (0.450 m)(216 N) cos 55&deg; − (1.050 m)(90 N) cos 20&deg;
= −33.049 N ⋅ m
The equivalent force-couple system at B is
F = 216 N
(b)
65.0&deg;; 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&deg;
where a is the distance from B to the point of application of F′. Then
−33.049 N ⋅ m = − a(216 N) cos 55&deg;
a = 0.26676 m
or
F ′ = 216 N
65.0&deg; applied to the lever 267 mm to the left of B 
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248
PROBLEM 3.88
A hexagonal plate is acted upon by the force P and the couple shown.
Determine the magnitude and the direction of the smallest force P for
which this system can be replaced with a single force at E.
SOLUTION
From the statement of the problem, it follows that ΣM E = 0 for the given force-couple system. Further,
for Pmin, we must require that P be perpendicular to rB/E . Then
ΣM E : (0.2 sin 30&deg; + 0.2)m &times; 300 N
+ (0.2 m)sin 30&deg; &times; 300 N
− (0.4 m) Pmin = 0
or
Pmin = 300 N
Pmin = 300 N
30.0&deg; 
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249
PROBLEM 3.89
A force and couple act as shown on a square plate of side a = 25 in. Knowing
that P = 60 lb, Q = 40 lb, and α = 50&deg;, replace the given force and couple by
a single force applied at a point located (a) on line AB, (b) on line AC. In each
case determine the distance from A to the point of application of the force.
SOLUTION
Replace the given force-couple system with an equivalent forcecouple system at A.
Px = (60 lb)(cos 50&deg;) = 38.567 lb
Py = (60 lb)(sin 50&deg;) = 45.963 lb
M A = Py a − Qa
= (45.963 lb)(25 in.) − (40 lb)(25 in.)
= 149.075 lb ⋅ in.
(a)
149.075 lb ⋅ in. = (45.963 lb) x
x = 3.24 in.
P = 60.0 lb
(b)
50.0&deg;; 3.24 in. from A 
149.075 lb ⋅ in. = (38.567 lb) y
y = 3.87 in.
P = 60.0 lb
50.0&deg;; 3.87 in. below A 
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250
PROBLEM 3.90
The force and couple shown are to be replaced by an equivalent single force.
Knowing that P = 2Q, determine the required value of α if the line of action of
the single equivalent force is to pass through (a) Point A, (b) Point C.
SOLUTION
(a)
We must have M A = 0
( P sin α )a − Q (a ) = 0
sin α =
Q Q 1
=
=
P 2Q 2
α = 30.0&deg; 
(b)
MC = 0
We must have
( P sin α )a − ( P cos α ) a − Q (a) = 0
sin α − cos α =
Q Q 1
=
=
P 2Q 2
sin α = cos α +
1
2
(1)
1
4
1
1 − cos 2 α = cos 2 α + cos α +
4
sin 2 α = cos 2 α + cos α +
2 cos 2 α + cos α − 0.75 = 0
(2)
Solving the quadratic in cos α :
cos α =
−1 &plusmn; 7
4
α = 65.7&deg; or 155.7&deg;
Only the first value of α satisfies Eq. (1),
therefore α = 65.7&deg; 
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251
PROBLEM 3.91
The shearing forces exerted on the cross section of a steel channel can
be represented by a 900-N vertical force and two 250-N horizontal
forces as shown. Replace this force and couple with a single force F
applied at Point C, and determine the distance x from C to line BD.
(Point C is defined as the shear center of the section.)
SOLUTION
Replace the 250-N forces with a couple and move the 900-N force to Point C such that its moment about H is
equal to the moment of the couple
M H = (0.18)(250 N)
= 45 N ⋅ m
Then
or
M H = x(900 N)
45 N ⋅ m = x(900 N)
x = 0.05 m
F = 900 N
x = 50.0 mm 
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252
PROBLEM 3.92
A force and a couple are applied as shown to the end of a
cantilever beam. (a) Replace this system with a single force F
applied at Point C, and determine the distance d from C to a
line drawn through Points D and E. (b) Solve part a if the
directions of the two 360-N forces are reversed.
SOLUTION
(a)
We have
ΣF : F = (360 N) j − (360 N) j − (600 N)k
or F = −(600 N)k 
ΣM D : (360 N)(0.15 m) = (600 N)(d )
and
d = 0.09 m
or d = 90.0 mm below ED 
(b)
We have from part a:
F = −(600 N)k 
ΣM D : − (360 N)(0.15 m) = −(600 N)(d )
and
d = 0.09 m
or d = 90.0 mm above ED 
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253
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 &times; TAB
= 128 j &times; 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 
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254
PROBLEM 3.94
An antenna is guyed by three cables as shown. Knowing
that the tension in cable AD is 270 lb, replace the force
exerted at A by cable AD with an equivalent force-couple
system at the center O of the base of the antenna.
SOLUTION
We have
d AD = ( −64) 2 + (−128)2 + (−128) 2
= 192 ft
Then
Now
270 lb
(−64i − 128 j + 128k )
192
= (90 lb)(−i − 2 j − 2k )
M = M O = rA/O &times; TAD
= 128 j &times; 90(−i − 2 j − 2k )
= −(23, 040 lb ⋅ ft)i + (11,520 lb ⋅ ft)k
The equivalent force-couple system at O is
F = −(90.0 lb)i − (180.0 lb) j − (180.0 lb)k 
M = −(23.0 kip ⋅ ft)i + (11.52 kip ⋅ ft)k 
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255
PROBLEM 3.95
A 110-N force acting in a vertical plane parallel to the yz-plane
is applied to the 220-mm-long horizontal handle AB of a
socket wrench. Replace the force with an equivalent forcecouple system at the origin O of the coordinate system.
SOLUTION
We have
ΣF : PB = F
where
PB = 110 N[− (sin15&deg;) j + (cos15&deg;)k ]
= −(28.470 N) j + (106.252 N)k
or F = −(28.5 N) j + (106.3 N)k 
We have
where
ΣM O : rB/O &times; PB = M O
rB /O = [(0.22 cos 35&deg;)i + (0.15) j − (0.22sin 35&deg;)k ] m
= (0.180213 m)i + (0.15 m) j − (0.126187 m)k
i
j
k
0.180213 0.15 0.126187 N ⋅ m = M O
0
−28.5
106.3
M O = [(12.3487)i − (19.1566) j − (5.1361)k ] N ⋅ m
or M O = (12.35 N ⋅ m)i − (19.16 N ⋅ m)j − (5.13 N ⋅ m)k 
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256
PROBLEM 3.96
An eccentric, compressive 1220-N force P is applied to the end
of a cantilever beam. Replace P with an equivalent force-couple
system at G.
SOLUTION
We have
ΣF : − (1220 N)i = F
F = − (1220 N)i 
Also, we have
ΣM G : rA/G &times; P = M
i
j
k
1220 0 −0.1 −0.06 N ⋅ m = M
−1 0
0
M = (1220 N ⋅ m)[(−0.06)(−1) j − ( −0.1)( −1)k ]
or M = (73.2 N ⋅ m) j − (122 N ⋅ m)k 
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257
PROBLEM 3.97
To keep a door closed, a wooden stick is wedged between the
floor and the doorknob. The stick exerts at B a 175-N force directed
along line AB. Replace that force with an equivalent force-couple
system at C.
SOLUTION
We have
ΣF : PAB = FC
where
PAB = λAB PAB
=
(33 mm)i + (990 mm) j − (594 mm)k
(175 N)
1155.00 mm
or FC = (5.00 N)i + (150.0 N) j − (90.0 N)k 
We have
ΣM C : rB/C &times; PAB = M C
i
j
k
M C = 5 0.683 −0.860 0 N ⋅ m
1
30
−18
= (5){(− 0.860)(−18)i − (0.683)(−18) j
+ [(0.683)(30) − (0.860)(1)]k}
or M C = (77.4 N ⋅ m)i + (61.5 N ⋅ m) j + (106.8 N ⋅ m)k 
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258
PROBLEM 3.98
A 46-lb force F and a 2120-lb⋅in. couple M are
applied to corner A of the block shown. Replace
the given force-couple system with an equivalent
force-couple system at corner H.
SOLUTION
We have
Then
Also
d AJ = (18) 2 + (−14) 2 + (−3) 2 = 23 in.
46 lb
(18i − 14 j − 3k )
23
= (36 lb)i − (28 lb) j − (6 lb)k
F=
d AC = (−45) 2 + (0) 2 + (−28) 2 = 53 in.
2120 lb ⋅ in.
( −45i − 28k )
53
= −(1800 lb ⋅ in.)i − (1120 lb ⋅ in.)k
Then
M=
Now
M ′ = M + rA/H &times; F
where
Then
rA/H = (45 in.)i + (14 in.) j
i
j
k
M ′ = (−1800i − 1120k ) + 45 14
0
36 −28 −6
= (−1800i − 1120k ) + {[(14)(−6)]i + [−(45)( −6)]j + [(45)(−28) − (14)(36)]k}
= (−1800 − 84)i + (270) j + (−1120 − 1764)k
= −(1884 lb ⋅ in.)i + (270 lb ⋅ in.)j − (2884 lb ⋅ in.)k
= −(157 lb ⋅ ft)i + (22.5 lb ⋅ ft) j − (240 lb ⋅ ft)k
F′ = (36.0 lb)i − (28.0 lb) j − (6.00 lb)k 
The equivalent force-couple system at H is
M ′ = −(157.0 lb ⋅ ft)i + (22.5 lb ⋅ ft) j − (240 lb ⋅ ft)k 
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259
PROBLEM 3.99
A 77-N force F1 and a 31-N ⋅ m couple M1 are
applied to corner E of the bent plate shown. If
F1 and M1 are to be replaced with an equivalent
force-couple system (F2, M2) at corner B and if
(M2)z = 0, determine (a) the distance d, (b) F2
and M2.
SOLUTION
(a)
ΣM Bz : M 2 z = 0
We have
k ⋅ (rH /B &times; F1 ) + M 1z = 0
where
(1)
rH /B = (0.31 m)i − (0.0233) j
F1 = λ EH F1
(0.06 m)i + (0.06 m) j − (0.07 m)k
(77 N)
0.11 m
= (42 N)i + (42 N) j − (49 N)k
=
M1z = k ⋅ M1
M1 = λEJ M 1
=
− di + (0.03 m) j − (0.07 m)k
d 2 + 0.0058 m
(31 N ⋅ m)
Then from Equation (1),
0
0
1
( −0.07 m)(31 N ⋅ m)
=0
0.31 −0.0233 0 +
2
+
0.0058
d
−49
42
42
Solving for d, Equation (1) reduces to
(13.0200 + 0.9786) −
from which
2.17 N ⋅ m
d 2 + 0.0058
d = 0.1350 m
=0
or d = 135.0 mm 
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260
PROBLEM 3.99 (Continued)
(b)
F2 = F1 = (42i + 42 j − 49k ) N
or F2 = (42.0 N)i + (42.0 N) j − (49.0 N)k 
M 2 = rH /B &times; F1 + M1
i
j
k
(0.1350)i + 0.03j − 0.07k
(31 N ⋅ m)
= 0.31 −0.0233 0 +
0.155000
42
42
−49
= (1.14170i + 15.1900 j + 13.9986k ) N ⋅ m
+ (−27.000i + 6.0000 j − 14.0000k ) N ⋅ m
M 2 = − (25.858 N ⋅ m)i + (21.190 N ⋅ m) j
or M 2 = − (25.9 N ⋅ m)i + (21.2 N ⋅ m) j 
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261
PROBLEM 3.100
A 2.6-kip force is applied at Point D of the cast iron post shown.
Replace that force with an equivalent force-couple system at the
center A of the base section.
SOLUTION

DE = −(12 in.) j − (5 in.)k; DE = 13.00 in.

DE
F = (2.6 kips)
DE
F = (2.6 kips)
−12 j − 5k
13
F = −(2.40 kips) j − (1.000 kip)k 
M A = rD /A &times; F
where
rD /A = (6 in.)i + (12 in.) j
i
j
k
12 in.
0
M A = 6 in.
0
−2.4 kips −1.0 kips
M A = −(12.00 kip ⋅ in.)i + (6.00 kip ⋅ in.) j − (14.40 kip ⋅ in.)k 
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262
PROBLEM 3.101
SOLUTION
(a)
(a) We have
ΣFY : − 300 N − 200 N = Ra
or R a = 500 N 
and
ΣM A : − 400 N ⋅ m − (200 N)(3 m) = M a
or M a = 1000 N ⋅ m
(b) We have

ΣFY : 200 N + 300 N = Rb
or R b = 500 N 
and
ΣM A : − 400 N ⋅ m + (300 N)(3 m) = M b
or M b = 500 N ⋅ m
(c) We have

ΣFY : − 200 N − 300 N = Rc
or R c = 500 N 
and
ΣM A : 400 N ⋅ m − (300 N)(3 m) = M c
or M c = 500 N ⋅ m

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263
PROBLEM 3.101 (Continued)
(d) We have
ΣFY : − 500 N = Rd
or R d = 500 N 
and
ΣM A : 400 N ⋅ m − (500 N)(3 m) = M d
or M d = 1100 N ⋅ m
(e) We have

ΣFY : 300 N − 800 N = Re
or R e = 500 N 
and
ΣM A : 400 N ⋅ m + 1000 N ⋅ m − (800 N)(3 m) = M e
or M e = 1000 N ⋅ m
(f ) We have

ΣFY : − 300 N − 200 N = R f
or R f = 500 N 
and
ΣM A : 400 N ⋅ m − (200 N)(3 m) = M f
or M f = 200 N ⋅ m
(g) We have

ΣFY : − 800 N + 300 N = Rg
or R g = 500 N 
and
ΣM A : 1000 N ⋅ m + 400 N ⋅ m + (300 N)(3 m) = M g
or M g = 2300 N ⋅ m
(h) We have

ΣFY : − 250 N − 250 N = Rh
or R h = 500 N 
and
ΣM A : 1000 N ⋅ m + 400 N ⋅ m − (250 N)(3 m) = M h
or M h = 650 N ⋅ m
(b)

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264
PROBLEM 3.102
SOLUTION
We have
ΣFY : − 200 N − 300 N = R
R = 500 N
or
and
ΣM A : 500 N ⋅ m + 200 N ⋅ m − (300 N)(3 m) = M
M = 200 N ⋅ m
or
Problem 3.101 equivalent force-couples at A:
Case

R

M
(a)
500 N
1000 N⋅m
(b)
500 N
500 N⋅m
(c)
500 N
500 N⋅m
(d)
500 N
1100 N⋅m
(e)
500 N
1000 N⋅m
(f )
500 N
200 N⋅m
(g)
500 N
2300 N⋅m
(h)
500 N
650 N⋅m
Equivalent to case (f ) of Problem 3.101 
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265
PROBLEM 3.103
Determine the single equivalent force and the distance from Point A to its line of action for the beam and
SOLUTION
For equivalent single force at distance d from A:
(a)
We have
ΣFY : − 300 N − 200 N = R
or R = 500 N 
ΣM C : − 400 N ⋅ m + (300 N)(d )
and
− (200 N)(3 − d ) = 0
or d = 2.00 m 
(b)
We have
ΣFY : 200 N + 300 N = R
or R = 500 N 
ΣM C : − 400 N ⋅ m − (200 N)(d )
and
+ (300 N)(3 − d ) = 0
or d = 1.000 m 
(c)
We have
ΣFY : − 200 N − 300 N = R
or R = 500 N 
ΣM C : 500 N ⋅ m + 200 N ⋅ m
and
+ (200 N)( d ) − (300 N)(3 − d ) = 0
or d = 0.400 m 
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266
PROBLEM 3.104
Five separate force-couple systems act at the corners of a piece of sheet metal, which has been bent into the
shape shown. Determine which of these systems is equivalent to a force F = (10 lb)i and a couple of moment
M = (15 lb ⋅ ft)j + (15 lb ⋅ ft)k located at the origin.
SOLUTION
First note that the force-couple system at F cannot be equivalent because of the direction of the force [The
force of the other four systems is (10 lb)i]. Next, move each of the systems to the origin O; the forces remain
unchanged.
A: M A = ΣM O = (5 lb ⋅ ft) j + (15 lb ⋅ ft)k + (2 ft)k &times; (10 lb)i
= (25 lb ⋅ ft) j + (15 lb ⋅ ft)k
D : M D = ΣM O = −(5 lb ⋅ ft) j + (25 lb ⋅ ft)k
+ [(4.5 ft)i + (1 ft) j + (2 ft)k ] &times; 10 lb)i
= (15 lb ⋅ ft)i + (15 lb ⋅ ft)k
G : M G = ΣM O = (15 lb ⋅ ft)i + (15 lb ⋅ ft) j
I : M I = ΣM I = (15 lb ⋅ ft) j − (5 lb ⋅ ft)k
+ [(4.5 ft)i + (1 ft) j] &times; (10 lb) j
= (15 lb ⋅ ft) j − (15 lb ⋅ ft)k
The equivalent force-couple system is the system at corner D.

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267
PROBLEM 3.105
Three horizontal forces are applied as shown to a vertical cast iron arm.
Determine the resultant of the forces and the distance from the ground to
its line of action when (a) P = 200 N, (b) P = 2400 N, (c) P = 1000 N.
SOLUTION
(a)
RD = +200 N − 600 N − 400 N = −800 N
M D = −(200 N)(0.450 m) + (600 N)(0.300 m) + (400 N)(0.1500 m)
= +150.0 N ⋅ m
y=
M D 150 N ⋅ m
=
= 0.1875 m
R
800 N
R = 800 N
; y = 187.5 mm 
(b)
RD = +2400 N − 600 N − 400 N = +1400 N
M D = −(2400 N)(0.450 m) + (600 N)(0.300 m) + (400 N)(0.1500 m)
= −840 N ⋅ m
y=
M D 840 N ⋅ m
=
= 0.600 m
R
1400 N
R = 1400 N
; y = 600 mm 
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268
PROBLEM 3.105 (Continued)
(c)
RD = +1000 − 600 − 400 = 0
M D = −(1000 N)(0.450 m) + (600 N)(0.300 m) + (400 N)(0.1500 m)
= −210 N ⋅ m
∴ y = ∞ System reduces to a couple.
M D = 210 N ⋅ m

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269
PROBLEM 3.106
Three stage lights are mounted on a pipe as shown.
The lights at A and B each weigh 4.1 lb, while the one
at C weighs 3.5 lb. (a) If d = 25 in., determine the
distance from D to the line of action of the resultant
of the weights of the three lights. (b) Determine the
value of d so that the resultant of the weights passes
through the midpoint of the pipe.
SOLUTION
For equivalence,
ΣFy : − 4.1 − 4.1 − 3.5 = − R or R = 11.7 lb
ΣFD : − (10 in.)(4.1 lb) − (44 in.)(4.1 lb)
−[(4.4 + d ) in.](3.5 lb) = −( L in.)(11.7 lb)
375.4 + 3.5d = 11.7 L (d , L in in.)
or
d = 25 in.
(a)
We have
375.4 + 3.5(25) = 11.7 L or
L = 39.6 in.
The resultant passes through a point 39.6 in. to the right of D.

L = 42 in.
(b)
We have
375.4 + 3.5d = 11.7(42)
or d = 33.1 in. 
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270
PROBLEM 3.107
The weights of two children sitting at ends A and B of a seesaw
are 84 lb and 64 lb, respectively. Where should a third child sit
so that the resultant of the weights of the three children will
pass through C if she weighs (a) 60 lb, (b) 52 lb.
SOLUTION
(a)
For the resultant weight to act at C,
Then
ΣM C = 0 WC = 60 lb
(84 lb)(6 ft) − 60 lb(d ) − 64 lb(6 ft) = 0
d = 2.00 ft to the right of C 
(b)
For the resultant weight to act at C,
Then
ΣM C = 0 WC = 52 lb
(84 lb)(6 ft) − 52 lb(d ) − 64 lb(6 ft) = 0
d = 2.31 ft to the right of C 
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271
PROBLEM 3.108
A couple of magnitude M = 54 lb ⋅ in. and the three forces shown are
applied to an angle bracket. (a) Find the resultant of this system of
forces. (b) Locate the points where the line of action of the resultant
intersects line AB and line BC.
SOLUTION
(a)
We have
ΣF : R = (−10 j) + (30 cos 60&deg;)i
+ 30 sin 60&deg; j + (−45i )
= −(30 lb)i + (15.9808 lb) j
or R = 34.0 lb
(b)
28.0&deg; 
First reduce the given forces and couple to an equivalent force-couple system (R , M B ) at B.
We have
ΣM B : M B = (54 lb ⋅ in) + (12 in.)(10 lb) − (8 in.)(45 lb)
= −186 lb ⋅ in.
Then with R at D,
or
and with R at E,
or
ΣM B : −186 lb ⋅ in = a(15.9808 lb)
a = 11.64 in.
ΣM B : −186 lb ⋅ in = C (30 lb)
C = 6.2 in.
The line of action of R intersects line AB 11.64 in. to the left of B and intersects line BC 6.20 in.
below B.

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272
PROBLEM 3.109
A couple M and the three forces shown are applied to an angle bracket.
Find the moment of the couple if the line of action of the resultant of the
force system is to pass through (a) Point A, (b) Point B, (c) Point C.
SOLUTION
In each case, we must have M1R = 0
(a)
M AB = ΣM A = M + (12 in.)[(30 lb) sin 60&deg;] − (8 in.)(45 lb) = 0
M = +48.231 lb ⋅ in.
(b)

M = 240 lb ⋅ in.

M BR = ΣM B = M + (12 in.)(10 lb) − (8 in.)(45 lb) = 0
M = +240 lb ⋅ in.
(c)
M = 48.2 lb ⋅ in.
M CR = ΣM C = M + (12 in.)(10 lb) − (8 in.)[(30 lb) cos 60&deg;] = 0
M =0
M=0 
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273
PROBLEM 3.110
A 32-lb motor is mounted on the floor. Find the resultant of the
weight and the forces exerted on the belt, and determine where
the line of action of the resultant intersects the floor.
SOLUTION
We have
ΣF : (60 lb)i − (32 lb) j + (140 lb)(cos 30&deg;i + sin 30&deg; j) = R
R = (181.244 lb)i + (38.0 lb) j
or R = 185.2 lb
We have
11.84&deg; 
ΣM O : ΣM O = xRy
− [(140 lb) cos 30&deg;][(4 + 2 cos 30&deg;)in.] − [(140 lb) sin 30&deg;][(2 in.)sin 30&deg;]
− (60 lb)(2 in.) = x(38.0 lb)
x=
and
1
(− 694.97 − 70.0 − 120) in.
38.0
x = −23.289 in.
Or resultant intersects the base (x-axis) 23.3 in. to the left of
the vertical centerline (y-axis) of the motor.

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274
PROBLEM 3.111
A machine component is subjected to the forces and couples
shown. The component is to be held in place by a single rivet
that can resist a force but not a couple. For P = 0, determine the
location of the rivet hole if it is to be located (a) on line FG,
(b) on line GH.
SOLUTION
We have
First replace the applied forces and couples with an equivalent force-couple system at G.
ΣFx : 200cos 15&deg; − 120 cos 70&deg; + P = Rx
Thus,
Rx = (152.142 + P) N
or
ΣFy : − 200sin 15&deg; − 120sin 70&deg; − 80 = Ry
Ry = −244.53 N
or
ΣM G : − (0.47 m)(200 N) cos15&deg; + (0.05 m)(200 N)sin15&deg;
+ (0.47 m)(120 N) cos 70&deg; − (0.19 m)(120 N)sin 70&deg;
− (0.13 m)( P N) − (0.59 m)(80 N) + 42 N ⋅ m
+ 40 N ⋅ m = M G
M G = −(55.544 + 0.13P) N ⋅ m
or
(1)
Setting P = 0 in Eq. (1):
Now with R at I,
ΣM G : − 55.544 N ⋅ m = − a(244.53 N)
a = 0.227 m
or
and with R at J,
ΣM G : − 55.544 N ⋅ m = −b(152.142 N)
b = 0.365 m
or
(a)
The rivet hole is 0.365 m above G.

(b)
The rivet hole is 0.227 m to the right of G.

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275
PROBLEM 3.112
Solve Problem 3.111, assuming that P = 60 N.
PROBLEM 3.111 A machine component is subjected to the
forces and couples shown. The component is to be held in
place by a single rivet that can resist a force but not a couple.
For P = 0, determine the location of the rivet hole if it is to
be located (a) on line FG, (b) on line GH.
SOLUTION
See the solution to Problem 3.111 leading to the development of Equation (1):
M G = −(55.544 + 0.13P) N ⋅ m
Rx = (152.142 + P) N
and
P = 60 N
For
we have
Rx = (152.142 + 60)
= 212.14 N
M G = −[55.544 + 0.13(60)]
= −63.344 N ⋅ m
Then with R at I,
ΣM G : −63.344 N ⋅ m = −a(244.53 N)
a = 0.259 m
or
and with R at J,
ΣM G : −63.344 N ⋅ m = −b(212.14 N)
b = 0.299 m
or
(a)
The rivet hole is 0.299 m above G.

(b)
The rivet hole is 0.259 m to the right of G.

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276
PROBLEM 3.113
force acting on the truss and the point of intersection of its
line of action with a line drawn through Points A and G.
SOLUTION
We have
R = ΣF
R = (240 lb)(cos 70&deg;i − sin 70&deg; j) − (160 lb) j
+ (300 lb)(− cos 40&deg;i − sin 40&deg; j) − (180 lb) j
R = −(147.728 lb)i − (758.36 lb) j
R = Rx2 + Ry2
= (147.728) 2 + (758.36) 2
= 772.62 lb
 Ry 

 Rx 
 −758.36 
= tan −1 

 −147.728 
= 78.977&deg;
θ = tan −1 
or
We have
ΣM A = dRy
where
ΣM A = −[240 lb cos 70&deg;](6 ft) − [240 lbsin 70&deg;](4 ft)
R = 773 lb
79.0&deg; 
− (160 lb)(12 ft) + [300 lb cos 40&deg;](6 ft)
− [300 lb sin 40&deg;](20 ft) − (180 lb)(8 ft)
= −7232.5 lb ⋅ ft
−7232.5 lb ⋅ ft
−758.36 lb
= 9.5370 ft
d=
or
d = 9.54 ft to the right of A 
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277
PROBLEM 3.114
Four ropes are attached to a crate and exert the forces shown.
If the forces are to be replaced with a single equivalent force
applied at a point on line AB, determine (a) the equivalent
force and the distance from A to the point of application of
the force when α = 30&deg;, (b) the value of α so that the single
equivalent force is applied at Point B.
SOLUTION
We have
(a)
For equivalence,
ΣFx : −100 cos 30&deg; + 400 cos 65&deg; + 90 cos 65&deg; = Rx
or
Rx = 120.480 lb
ΣFy : 100 sin α + 160 + 400 sin 65&deg; + 90 sin 65&deg; = Ry
or
Ry = (604.09 + 100sin α ) lb
With α = 30&deg;,
Ry = 654.09 lb
Then
R = (120.480) 2 + (654.09) 2
= 665 lb
(1)
654.09
120.480
or θ = 79.6&deg;
tan θ =
ΣM A : (46 in.)(160 lb) + (66 in.)(400 lb) sin 65&deg;
Also
or
+ (26 in.)(400 lb) cos 65&deg; + (66 in.)(90 lb)sin 65&deg;
+ (36 in.)(90 lb) cos 65&deg; = d (654.09 lb)
ΣM A = 42, 435 lb ⋅ in. and d = 64.9 in.
79.6&deg; 

and R is applied 64.9 in. to the right of A.
(b)
R = 665 lb
We have d = 66 in.
Then
or
Using Eq. (1):
ΣM A : 42, 435 lb ⋅ in = (66 in.) Ry
Ry = 642.95 lb
642.95 = 604.09 + 100sin α
or α = 22.9&deg; 
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278
PROBLEM 3.115
Solve Prob. 3.114, assuming that the 90-lb force is removed.
PROBLEM 3.114 Four ropes are attached to a crate and exert
the forces shown. If the forces are to be replaced with a single
equivalent force applied at a point on line AB, determine
(a) the equivalent force and the distance from A to the point of
application of the force when α = 30&deg;, (b) the value of α so
that the single equivalent force is applied at Point B.
SOLUTION
(a)
For equivalence,
ΣFx : − (100 lb) cos 30&deg; + (400 lb)sin 25&deg; = Rx
or
Rx = 82.445 lb
ΣFy : 160 lb + (100 lb) sin 30&deg; + (400 lb) cos 25&deg; = Ry
or
Ry = 572.52 lb
R = (82.445)2 + (572.52) = 578.43 lb
tan θ =
572.52
82.445
or θ = 81.806&deg;
ΣM A : (46 in.)(160 lb) + (66 in.)(400 lb) cos 25&deg; + (26 in.)(400 lb)sin 25&deg;
= d (527.52 lb)
d = 62.3 in.
R = 578 lb
(b)
81.8&deg; and is applied 62.3 in. to the right of A.

We have d = 66.0 in. For R applied at B,
ΣM A : Ry (66 in.) = (160 lb)(46 in.) + (66 in.)(400 lb) cos 25&deg; + (26 in.)(400 lb)sin 25&deg;
Ry = 540.64 lb
ΣFY : 160 lb + (100 lb)sin α + (400 lb) cos 25&deg; = 540.64 lb
α = 10.44&deg; 
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279
PROBLEM 3.116
Four forces act on a 700 &times; 375-mm plate as shown. (a) Find
the resultant of these forces. (b) Locate the two points where
the line of action of the resultant intersects the edge of the
plate.
SOLUTION
(a)
R = ΣF
= (−400 N + 160 N − 760 N)i
+ (600 N + 300 N + 300 N) j
= −(1000 N)i + (1200 N) j
R = (1000 N) 2 + (1200 N)2
= 1562.09 N
 1200 N 
tan θ =  −

 1000 N 
= −1.20000
θ = −50.194&deg;
(b)
R = 1562 N
50.2&deg; 
M CR = Σr &times; F
= (0.5 m)i &times; (300 N + 300 N) j
= (300 N ⋅ m)k
(300 N ⋅ m)k = xi &times; (1200 N) j
x = 0.25000 m
x = 250 mm
(300 N ⋅ m) = yj &times; ( −1000 N)i
y = 0.30000 m
y = 300 mm
Intersection 250 mm to right of C and 300 mm above C 
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280
PROBLEM 3.117
Solve Problem 3.116, assuming that the 760-N force is directed
to the right.
PROBLEM 3.116 Four forces act on a 700 &times; 375-mm plate as
shown. (a) Find the resultant of these forces. (b) Locate the two
points where the line of action of the resultant intersects the edge
of the plate.
SOLUTION
R = ΣF
(a)
= ( −400N + 160 N + 760 N)i
+ (600 N + 300 N + 300 N) j
= (520 N)i + (1200 N) j
R = (520 N) 2 + (1200 N) 2 = 1307.82 N
 1200 N 
tan θ = 
 = 2.3077
 520 N 
θ = 66.5714&deg;
R = 1308 N
66.6&deg; 
M CR = Σr &times; F
(b)
= (0.5 m)i &times; (300 N + 300 N) j
= (300 N ⋅ m)k
(300 N ⋅ m)k = xi &times; (1200 N) j
x = 0.25000 m
or
x = 0.250 mm
(300 N ⋅ m)k = [ x′i + (0.375 m) j] &times; [(520 N)i + (1200 N) j]
= (1200 x′ − 195)k
x′ = 0.41250 m
or
x′ = 412.5 mm
Intersection 412 mm to the right of A and 250 mm to the right of C 
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281
PROBLEM 3.118
As follower AB rolls along the surface of member C, it exerts
a constant force F perpendicular to the surface. (a) Replace F
with an equivalent force-couple system at Point D obtained
by drawing the perpendicular from the point of contact to the
x-axis. (b) For a = 1 m and b = 2 m, determine the value of x
for which the moment of the equivalent force-couple system
at D is maximum.
SOLUTION
(a)
The slope of any tangent to the surface of member C is
dy d  
x2
=
b 1 − 2
dx dx   a
  −2b
  = 2 x
  a
Since the force F is perpendicular to the surface,
 dy 
tan α = −  
 dx 
−1
=
a2  1 
2b  x 
For equivalence,
ΣF : F = R
ΣM D : ( F cos α )( y A ) = M D
where
cos α =
2bx
(a ) + (2bx)2
2 2

x2 
y A = b 1 − 2 
 a 

x3 
2 Fb 2  x − 2 
a 

MD =
4
2 2
a + 4b x
Therefore, the equivalent force-couple system at D is
R=F
 a2 
tan −1 
 
 2bx 

x3 
2 Fb2  x − 2 
a 

M=

a 4 + 4b 2 x 2
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282
PROBLEM 3.118 (Continued)
(b)
To maximize M, the value of x must satisfy
dM
=0
dx
a = 1 m, b = 2 m
where for
M=
8F ( x − x3 )
1 + 16 x 2
dM
= 8F
dx
1

1 + 16 x 2 (1 − 3x 2 ) − ( x − x3 )  (32 x)(1 + 16 x 2 ) −1/ 2 
2
 =0
(1 + 16 x 2 )
(1 + 16 x 2 )(1 − 3x 2 ) − 16 x( x − x3 ) = 0
32 x 4 + 3x 2 − 1 = 0
or
x2 =
−3 &plusmn; 9 − 4(32)(−1)
= 0.136011 m 2
2(32)
Using the positive value of x2:
x = 0.36880 m
and − 0.22976 m 2
or x = 369 mm 
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283
PROBLEM 3.119
As plastic bushings are inserted into a 60-mm-diameter cylindrical sheet
metal enclosure, the insertion tools exert the forces shown on the enclosure.
Each of the forces is parallel to one of the coordinate axes. Replace these
forces with an equivalent force-couple system at C.
SOLUTION
For equivalence,
ΣF :
R = FA + FB + FC + FD
= −(17 N) j − (12 N) j − (16 N)k − (21 N)i
= −(21 N)i − (29 N) j − (16 N)k
ΣM C : M = rA /C &times; FA + rB /C &times; FB + rD /C &times; FD
M = [(0.11 m) j − (0.03 m)k ] &times; [−(17 N)] j
+ [(0.02 m)i + (0.11 m) j − (0.03 m)k ] &times; [ −(12 N)]j
+ [(0.03 m)i + (0.03 m) j − (0.03 m)k ] &times; [ −(21 N)]i
= −(0.51 N ⋅ m)i + [−(0.24 N ⋅ m)k − (0.36 N ⋅ m)i]
+ [(0.63 N ⋅ m)k + (0.63 N ⋅ m) j]
∴
The equivalent force-couple system at C is
R = −(21.0 N)i − (29.0 N) j − (16.00 N)k 
M = −(0.870 N ⋅ m)i + (0.630 N ⋅ m) j + (0.390 N ⋅ m)k 
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284
PROBLEM 3.120
Two 150-mm-diameter pulleys are mounted on
line shaft AD. The belts at B and C lie in
vertical planes parallel to the yz-plane. Replace
the belt forces shown with an equivalent forcecouple system at A.
SOLUTION
Equivalent force-couple at each pulley:
Pulley B:
R B = (145 N)(− cos 20&deg; j + sin 20&deg;k ) − 215 Nj
= − (351.26 N) j + (49.593 N)k
M B = − (215 N − 145 N)(0.075 m)i
= − (5.25 N ⋅ m)i
Pulley C:
R C = (155 N + 240 N)(− sin10&deg; j − cos10&deg;k )
= − (68.591 N) j − (389.00 N)k
M C = (240 N − 155 N)(0.075 m)i
= (6.3750 N ⋅ m)i
Then
R = R B + R C = − (419.85 N) j − (339.41)k
or R = (420 N) j − (339 N)k 
M A = M B + M C + rB/ A &times; R B + rC/ A &times; R C
i
j
k
0
0
N⋅m
= − (5.25 N ⋅ m)i + (6.3750 N ⋅ m)i + 0.225
0
−351.26 49.593
i
j
k
+ 0.45
0
0
N⋅m
0
−68.591 −389.00
= (1.12500 N ⋅ m)i + (163.892 N ⋅ m) j − (109.899 N ⋅ m)k
or M A = (1.125 N ⋅ m)i + (163.9 N ⋅ m) j − (109.9 N ⋅ m)k 
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285
PROBLEM 3.121
Four forces are applied to the machine component
ABDE as shown. Replace these forces with an
equivalent force-couple system at A.
SOLUTION
R = −(50 N) j − (300 N)i − (120 N)i − (250 N)k
R = −(420 N)i − (50 N)j − (250 N)k
rB = (0.2 m)i
rD = (0.2 m)i + (0.16 m)k
rE = (0.2 m)i − (0.1 m) j + (0.16 m)k
M RA = rB &times; [−(300 N)i − (50 N) j]
+ rD &times; (−250 N)k + r &times; ( − 120 N)i
i
j
k
i
j
k
= 0.2 m
0
0 + 0.2 m 0 0.16 m
−300 N −50 N 0
0
0 −250 N
i
j
k
+ 0.2 m −0.1 m 0.16 m
−120 N
0
0
= −(10 N ⋅ m)k + (50 N ⋅ m) j − (19.2 N ⋅ m) j − (12 N ⋅ m)k
Force-couple system at A is
R = −(420 N)i − (50 N) j − (250 N)k M RA = (30.8 N ⋅ m) j − (220 N ⋅ m)k 
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286
PROBLEM 3.122
While using a pencil sharpener, a student applies
the forces and couple shown. (a) Determine the
forces exerted at B and C knowing that these
forces and the couple are equivalent to a forcecouple system at A consisting of the force
R = (2.6 lb)i + Ry j − (0.7 lb)k and the couple
M RA = M x i + (1.0 lb &middot; ft)j − (0.72 lb &middot; ft)k.
(b) Find the corresponding values of Ry and M x .
SOLUTION
(a)
From the statement of the problem, equivalence requires
ΣF : B + C = R
or
ΣFx : Bx + C x = 2.6 lb
(1)
ΣFy : − C y = R y
(2)
ΣFz : − C z = −0.7 lb or C z = 0.7 lb
and
ΣM A : (rB/A &times; B + M B ) + rC/A &times; C = M AR
or
 1.75 
ΣM x : (1 lb ⋅ ft) + 
ft  (C y ) = M x
 12

(3)
 3.75 
 1.75 
 3.5 
ΣM y : 
ft  ( Bx ) + 
ft  (C x ) + 
ft  (0.7 lb) = 1 lb ⋅ ft
 12 
 12 
 12 
or
Using Eq. (1):
3.75Bx + 1.75C x = 9.55
3.75Bx + 1.75(2.6 Bx ) = 9.55
or
Bx = 2.5 lb
and
C x = 0.1 lb
 3.5 
ΣM z : − 
ft  (C y ) = −0.72 lb ⋅ ft
 12 
C y = 2.4686 lb
or
B = (2.50 lb)i C = (0.1000 lb)i − (2.47 lb) j − (0.700 lb)k 
(b)
Eq. (2) 
Using Eq. (3):
Ry = −2.47 lb 
 1.75 
1+ 
 (2.4686) = M x
 12 
or M x = 1.360 lb ⋅ ft 
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287
PROBLEM 3.123
A blade held in a brace is used to tighten a screw at A.
(a) Determine the forces exerted at B and C, knowing
that these forces are equivalent to a force-couple
system at A consisting of R = −(30 N)i + Ry j + Rz k
and M RA = − (12 N &middot; m)i. (b) Find the corresponding
values of Ry and Rz . (c) What is the orientation of the
slot in the head of the screw for which the blade is least
likely to slip when the brace is in the position shown?
SOLUTION
(a)
Equivalence requires
or
Equating the i coefficients:
Also,
or
Equating coefficients:
ΣF : R = B + C
−(30 N)i + Ry j + Rz k = − Bk + (−C x i + C y j + C z k )
i : − 30 N = −C x
or C x = 30 N
ΣM A : M RA = rB/A &times; B + rC/A &times; C
−(12 N ⋅ m)i = [(0.2 m)i + (0.15 m)j] &times; (− B)k
+(0.4 m)i &times; [−(30 N)i + C y j + C z k ]
i : − 12 N ⋅ m = −(0.15 m) B
k : 0 = (0.4 m)C y
or
or
B = 80 N
Cy = 0
j: 0 = (0.2 m)(80 N) − (0.4 m)C z or
C z = 40 N
B = −(80.0 N)k C = −(30.0 N)i + (40.0 N)k 
(b)
Now we have for the equivalence of forces
−(30 N)i + Ry j + Rz k = −(80 N)k + [(−30 N)i + (40 N)k ]
Equating coefficients:
j: R y = 0
Ry = 0 
k : Rz = −80 + 40
(c)
or
Rz = −40.0 N 
First note that R = −(30 N)i − (40 N)k. Thus, the screw is best able to resist the lateral force Rz
when the slot in the head of the screw is vertical.

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288
PROBLEM 3.124
In order to unscrew the tapped faucet A, a plumber uses two
pipe wrenches as shown. By exerting a 40-lb force on each
wrench, at a distance of 10 in. from the axis of the pipe and in a
direction perpendicular to the pipe and to the wrench, he
prevents the pipe from rotating, and thus avoids loosening or
further tightening the joint between the pipe and the tapped
elbow C. Determine (a) the angle θ that the wrench at A should
form with the vertical if elbow C is not to rotate about the
vertical, (b) the force-couple system at C equivalent to the two
40-lb forces when this condition is satisfied.
SOLUTION
We first reduce the given forces to force-couple systems at A and B, noting that
| M A | = | M B | = (40 lb)(10 in.)
= 400 lb ⋅ in.
We now determine the equivalent force-couple system at C.
R = (40 lb)(1 − cos θ )i − (40 lb) sin θ j
(1)
M CR = M A + M B + (15 in.)k &times; [−(40 lb) cos θ i − (40 lb)sin θ j]
+ (7.5 in.)k &times; (40 lb)i
= + 400 − 400 − 600cos θ j + 600sin θ i + 300 j
= (600 lb ⋅ in.)sin θ i + (300 lb ⋅ in.)(1 − 2 cos θ ) j
(a)
(2)
For no rotation about vertical, y component of M CR must be zero.
1 − 2cos θ = 0
cos θ = 1/2
θ = 60.0&deg; 
(b)
For θ = 60.0&deg; in Eqs. (1) and (2),
R = (20.0 lb)i − (34.641 lb) j; M CR = (519.62 lb ⋅ in.)i
R = (20.0 lb)i − (34.6 lb) j; M CR = (520 lb ⋅ in.)i 
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289
PROBLEM 3.125
Assuming θ = 60&deg; in Prob. 3.124, replace the two 40-lb forces
with an equivalent force-couple system at D and determine
whether the plumber’s action tends to tighten or loosen the
joint between (a) pipe CD and elbow D, (b) elbow D and
pipe DE. Assume all threads to be right-handed.
PROBLEM 3.124 In order to unscrew the tapped faucet A,
a plumber uses two pipe wrenches as shown. By exerting a
40-lb force on each wrench, at a distance of 10 in. from the
axis of the pipe and in a direction perpendicular to the pipe
and to the wrench, he prevents the pipe from rotating, and
thus avoids loosening or further tightening the joint between
the pipe and the tapped elbow C. Determine (a) the angle θ
that the wrench at A should form with the vertical if elbow C
is not to rotate about the vertical, (b) the force-couple system
at C equivalent to the two 40-lb forces when this condition
is satisfied.
SOLUTION
The equivalent force-couple system at C for θ = 60&deg; was obtained in the solution to Prob. 3.124:
R = (20.0 lb)i − (34.641 lb) j
M CR = (519.62 lb ⋅ in.)i
The equivalent force-couple system at D is made of R and M RD where
M RD = M CR + rC /D &times; R
= (519.62 lb ⋅ in.)i + (25.0 in.) j &times; [(20.0 lb)i − (34.641 lb) j]
= (519.62 lb ⋅ in.)i − (500 lb ⋅ in.)k
Equivalent force-couple at D:
R = (20.0 lb)i − (34.6 lb) j; M CR = (520 lb ⋅ in.)i − (500 lb ⋅ in.)k 
(a)
(b)
Since M RD has no component along the y-axis, the plumber’s action will neither loosen
nor tighten the joint between pipe CD and elbow.

, the plumber’s action will tend to tighten
Since the x component of M RD is
the joint between elbow and pipe DE.

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290
PROBLEM 3.126
As an adjustable brace BC is used to bring a wall into plumb, the
force-couple system shown is exerted on the wall. Replace this
force-couple system with an equivalent force-couple system at A
if R = 21.2 lb and M = 13.25 lb &middot; ft.
SOLUTION
We have
where
or
We have
where
ΣF : R = R A = RλBC
λBC =
(42 in.)i − (96 in.) j − (16 in.)k
106 in.
RA =
21.2 lb
(42i − 96 j − 16k )
106
R A = (8.40 lb)i − (19.20 lb) j − (3.20 lb)k

ΣM A : rC/A &times; R + M = M A
rC/A = (42 in.)i + (48 in.)k =
1
(42i + 48k )ft
12
= (3.5 ft)i + (4.0 ft)k
R = (8.40 lb)i − (19.50 lb) j − (3.20 lb)k
M = −λBC M
−42i + 96 j + 16k
(13.25 lb ⋅ ft)
106
= −(5.25 lb ⋅ ft)i + (12 lb ⋅ ft) j + (2 lb ⋅ ft)k
=
Then
i
j
k
3.5
0
4.0 lb ⋅ ft + (−5.25i + 12 j + 2k ) lb ⋅ ft = M A
8.40 −19.20 −3.20
M A = (71.55 lb ⋅ ft)i + (56.80 lb ⋅ ft)j − (65.20 lb ⋅ ft)k
or M A = (71.6 lb ⋅ ft)i + (56.8 lb ⋅ ft)j − (65.2 lb ⋅ ft)k 
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291
PROBLEM 3.127
Three children are standing on a 5 &times; 5-m raft. If the weights
of the children at Points A, B, and C are 375 N, 260 N, and
400 N, respectively, determine the magnitude and the point
of application of the resultant of the three weights.
SOLUTION
We have
ΣF : FA + FB + FC = R
−(375 N) j − (260 N) j − (400 N) j = R
−(1035 N) j = R
or
We have
R = 1035 N 
ΣM x : FA ( z A ) + FB ( z B ) + FC ( zC ) = R ( z D )
(375 N)(3 m) + (260 N)(0.5 m) + (400 N)(4.75 m) = (1035 N)(z D )
z D = 3.0483 m
We have
or
z D = 3.05 m 
ΣM z : FA ( x A ) + FB ( xB ) + FC ( xC ) = R ( xD )
375 N(1 m) + (260 N)(1.5 m) + (400 N)(4.75 m) = (1035 N)( xD )
xD = 2.5749 m
or
xD = 2.57 m 
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292
PROBLEM 3.128
Three children are standing on a 5 &times; 5-m raft. The weights of
the children at Points A, B, and C are 375 N, 260 N, and 400 N,
respectively. If a fourth child of weight 425 N climbs onto the
raft, determine where she should stand if the other children
remain in the positions shown and the line of action of the
resultant of the four weights is to pass through the center of
the raft.
SOLUTION
We have
ΣF : FA + FB + FC = R
−(375 N) j − (260 N) j − (400 N) j − (425 N) j = R
R = −(1460 N) j
We have
ΣM x : FA ( z A ) + FB ( z B ) + FC ( zC ) + FD ( z D ) = R ( z H )
(375 N)(3 m) + (260 N)(0.5 m) + (400 N)(4.75 m)
+ (425 N)(z D ) = (1460 N)(2.5 m)
z D = 1.16471 m
We have
or
z D = 1.165 m 
ΣM z : FA ( x A ) + FB ( xB ) + FC ( xC ) + FD ( xD ) = R ( xH )
(375 N)(1 m) + (260 N)(1.5 m) + (400 N)(4.75 m)
+ (425 N)(xD ) = (1460 N)(2.5 m)
xD = 2.3235 m
or
xD = 2.32 m 
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293
PROBLEM 3.129
Four signs are mounted on a frame spanning a highway,
and the magnitudes of the horizontal wind forces acting on
the signs are as shown. Determine the magnitude and the
point of application of the resultant of the four wind forces
when a = 1 ft and b = 12 ft.
SOLUTION
We have
Assume that the resultant R is applied at Point P whose coordinates are (x, y, 0).
Equivalence then requires
ΣFz : − 105 − 90 − 160 − 50 = − R
or R = 405 lb 
ΣM x : (5 ft)(105 lb) − (1 ft)(90 lb) + (3 ft)(160 lb)
+ (5.5 ft)(50 lb) = − y (405 lb)
or
y = −2.94 ft
ΣM y : (5.5 ft)(105 lb) + (12 ft)(90 lb) + (14.5 ft)(160 lb)
+ (22.5 ft)(50 lb) = − x(405 lb)
x = 12.60 ft
or
R acts 12.60 ft to the right of member AB and 2.94 ft below member BC.

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294
PROBLEM 3.130
Four signs are mounted on a frame spanning a highway,
and the magnitudes of the horizontal wind forces acting on
the signs are as shown. Determine a and b so that the point
of application of the resultant of the four forces is at G.
SOLUTION
Since R acts at G, equivalence then requires that ΣM G of the applied system of forces also be zero. Then at
G : ΣM x : − (a + 3) ft &times; (90 lb) + (2 ft)(105 lb)
+ (2.5 ft)(50 lb) = 0
or a = 0.722 ft 

ΣM y : − (9 ft)(105 ft) − (14.5 − b) ft &times; (90 lb)
+ (8 ft)(50 lb) = 0

or b = 20.6 ft 
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295
PROBLEM 3.131*
A group of students loads a 2 &times; 3.3-m flatbed trailer with two
0.66 &times; 0.66 &times; 0.66-m boxes and one 0.66 &times; 0.66 &times; 1.2-m box. Each
of the boxes at the rear of the trailer is positioned so that it is
aligned with both the back and a side of the trailer. Determine the
smallest load the students should place in a second 0.66 &times; 0.66 &times;
1.2-m box and where on the trailer they should secure it, without
any part of the box overhanging the sides of the trailer, if each
box is uniformly loaded and the line of action of the resultant of
the weights of the four boxes is to pass through the point of
intersection of the centerlines of the trailer and the axle.
(Hint: Keep in mind that the box may be placed either on its
side or on its end.)
SOLUTION
For the smallest weight on the trailer so that the resultant force of the four weights acts over the axle at the
intersection with the center line of the trailer, the added 0.66 &times; 0.66 &times; 1.2-m box should be placed adjacent to
one of the edges of the trailer with the 0.66 &times; 0.66-m side on the bottom. The edges to be considered are based
on the location of the resultant for the three given weights.
We have
ΣF : − (224 N) j − (392 N) j − (176 N) j = R
R = −(792 N) j
We have
ΣM z : − (224 N)(0.33 m) − (392 N)(1.67 m) − (176 N)(1.67 m) = ( −792 N)( x)
xR = 1.29101 m
We have
ΣM x : (224 N)(0.33 m) + (392 N)(0.6 m) + (176 N)(2.0 m) = (792 N)( z )
z R = 0.83475 m
From the statement of the problem, it is known that the resultant of R from the original loading and the
lightest load W passes through G, the point of intersection of the two center lines. Thus, ΣM G = 0.
Further, since the lightest load W is to be as small as possible, the fourth box should be placed as far from G
as possible without the box overhanging the trailer. These two requirements imply
(0.33 m ≤ x ≤ 1 m) (1.5 m ≤ z ≤ 2.97 m)
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296
PROBLEM 3.131* (Continued)
xL = 0.33 m
With
at
G : ΣM z : (1 − 0.33) m &times; WL − (1.29101 − 1) m &times; (792 N) = 0
WL = 344.00 N
or
Now we must check if this is physically possible,
at
G : ΣM x : ( z L − 1.5) m &times; 344 N) − (1.5 − 0.83475) m &times; (792 N) = 0
z L = 3.032 m
or
which is not acceptable.
z L = 2.97 m:
With
at
G : ΣM x : (2.97 − 1.5) m &times; WL − (1.5 − 0.83475) m &times; (792 N) = 0
WL = 358.42 N
or
Now check if this is physically possible,
at
G : ΣM z : (1 − xL ) m &times; (358.42 N) − (1.29101 − 1) m &times; (792 N) = 0
or
xL = 0.357 m ok!
WL = 358 N 
The minimum weight of the fourth box is
And it is placed on end A (0.66 &times; 0.66-m side down) along side AB with the center of the box 0.357 m

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297
PROBLEM 3.132*
Solve Problem 3.131 if the students want to place as much
weight as possible in the fourth box and at least one side of
the box must coincide with a side of the trailer.
PROBLEM 3.131* A group of students loads a 2 &times; 3.3-m
flatbed trailer with two 0.66 &times; 0.66 &times; 0.66-m boxes and one
0.66 &times; 0.66 &times; 1.2-m box. Each of the boxes at the rear of the
trailer is positioned so that it is aligned with both the back
and a side of the trailer. Determine the smallest load the
students should place in a second 0.66 &times; 0.66 &times; 1.2-m box and
where on the trailer they should secure it, without any part
of the box overhanging the sides of the trailer, if each box is
uniformly loaded and the line of action of the resultant of
the weights of the four boxes is to pass through the point of
intersection of the centerlines of the trailer and the axle.
(Hint: Keep in mind that the box may be placed either on its
side or on its end.)
SOLUTION
First replace the three known loads with a single equivalent force R applied at coordinate ( xR , 0, z R ).
Equivalence requires
ΣFy : − 224 − 392 − 176 = − R
or
R = 792 N
ΣM x : (0.33 m)(224 N) + (0.6 m)(392 N)
+ (2 m)(176 N) = z R (792 N)
or
z R = 0.83475 m
ΣM z : − (0.33 m)(224 N) − (1.67 m)(392 N)
− (1.67 m)(176 N) = xR (792 N)
or
xR = 1.29101 m
From the statement of the problem, it is known that the resultant of R and the heaviest loads WH passes
through G, the point of intersection of the two center lines. Thus,
ΣM G = 0
Further, since WH is to be as large as possible, the fourth box should be placed as close to G as possible while
keeping one of the sides of the box coincident with a side of the trailer. Thus, the two limiting cases are
xH = 0.6 m or
z H = 2.7 m
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298
PROBLEM 3.132* (Continued)
Now consider these two possibilities.
With xH = 0.6 m
at
G : ΣM z : (1 − 0.6) m &times; WH − (1.29101 − 1) m &times; (792 N) = 0
WH = 576.20 N
or
Checking if this is physically possible
at
or
G : ΣM x : ( z H − 1.5) m &times; (576.20 N) − (1.5 − 0.83475) m &times; (792 N) = 0
z H = 2.414 m
which is acceptable.
With z H = 2.7 m
at
or
G : ΣM x : (2.7 − 1.5) m &times; WH − (1.5 − 0.83475) m &times; (792 N) = 0
WH = 439 N
Since this is less than the first case, the maximum weight of the fourth box is
WH = 576 N 
and it is placed with a 0.66 &times; 1.2-m side down, a 0.66-m edge along side AD, and the center 2.41 m
from side DC.

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299
PROBLEM 3.133*
A piece of sheet metal is bent into the shape shown and is acted
upon by three forces. If the forces have the same magnitude P,
replace them with an equivalent wrench and determine (a) the
magnitude and the direction of the resultant force R, (b) the pitch
of the wrench, (c) the axis of the wrench.
SOLUTION
(
)
First reduce the given forces to an equivalent force-couple system R, M OR at the origin.
We have
ΣF : − Pj + Pj + Pk = R
R = Pk
or

5  
ΣM O : − (aP ) j +  −( aP )i +  aP  k  = M OR
2  

5 

M OR = aP  −i − j + k 
2 

or
(a)
Then for the wrench,
R=P 
and
λ axis =
R
=k
R
cos θ x = 0 cos θ y = 0 cos θ z = 1
or
(b)
θ x = 90&deg; θ y = 90&deg; θ z = 0&deg;

Now
M1 = λ axis ⋅ M OR
5 

= k ⋅ aP  −i − j + k 
2 

5
= aP
2
Then
P=
M1 25 aP
=
R
P
or P =
5
a 
2
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300
PROBLEM 3.133* (Continued)
(c)
The components of the wrench are (R , M1 ), where M1 = M1λ axis , and the axis of the wrench is
assumed to intersect the xy-plane at Point Q, whose coordinates are (x, y, 0). Thus, we require
M z = rQ &times; R R
where
M z = M O &times; M1
Then
5  5

aP  −i − j + k  − aPk = ( xi + yj) + Pk
2  2

Equating coefficients:
i : − aP = yP
or
j: − aP = − xP or
y = −a
x=a
The axis of the wrench is parallel to the z-axis and intersects the xy-plane at
x = a, y = −a. 
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301
PROBLEM 3.134*
Three forces of the same magnitude P act on a cube of side a as
shown. Replace the three forces by an equivalent wrench and
determine (a) the magnitude and direction of the resultant force
R, (b) the pitch of the wrench, (c) the axis of the wrench.
SOLUTION
Force-couple system at O:
R = Pi + Pj + Pk = P(i + j + k )
M OR = aj &times; Pi + ak &times; Pj + ai &times; Pk
= − Pak − Pai − Paj
M OR = − Pa (i + j + k )
Since R and M OR have the same direction, they form a wrench with M1 = M OR . Thus, the axis of the wrench
is the diagonal OA. We note that
cos θ x = cos θ y = cos θ z =
a
a 3
=
1
3
R = P 3 θ x = θ y = θ z = 54.7&deg;
M1 = M OR = − Pa 3
Pitch = p =
M 1 − Pa 3
=
= −a
R
P 3
(a)
R = P 3 θ x = θ y = θ z = 54.7&deg; 
(b)
– a
(c)
Axis of the wrench is diagonal OA.
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302
PROBLEM 3.135*
The forces and couples shown are applied to two screws as a piece
of sheet metal is fastened to a block of wood. Reduce the forces and
the couples to an equivalent wrench and determine (a) the resultant
force R, (b) the pitch of the wrench, (c) the point where the axis of
the wrench intersects the xz-plane.
SOLUTION
First, reduce the given force system to a force-couple at the origin.
We have
ΣF : − (10 lb) j − (11 lb) j = R
R = − (21 lb) j
We have
ΣM O : Σ(rO &times; F ) + ΣM C = M OR
M OR
i
j
k
i
j
k
= 0 0 20 lb ⋅ in. + 0 0 −15 lb ⋅ in. − (12 lb ⋅ in) j
0 −10 0
0 −11 0
= (35 lb ⋅ in.)i − (12 lb ⋅ in.) j
R = − (21 lb) j
(a)
(b)
We have
or R = − (21.0 lb) j 
R
R
= (− j) ⋅ [(35 lb ⋅ in.)i − (12 lb ⋅ in.) j]
M1 = λ R ⋅ M OR
λR =
= 12 lb ⋅ in. and M1 = −(12 lb ⋅ in.) j
and pitch
p=
M 1 12 lb ⋅ in.
=
= 0.57143 in.
R
21 lb
or
p = 0.571 in. 
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303
PROBLEM 3.135* (Continued)
(c)
We have
M OR = M1 + M 2
M 2 = M OR − M1 = (35 lb ⋅ in.)i
We require
M 2 = rQ/O &times; R
(35 lb ⋅ in.)i = ( xi + zk ) &times; [ −(21 lb) j]
35i = −(21x)k + (21z )i
From i:
From k:
35 = 21z
z = 1.66667 in.
0 = − 21x
z=0
The axis of the wrench is parallel to the y-axis and intersects the xz-plane at x = 0, z = 1.667 in.

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304
PROBLEM 3.136*
The forces and couples shown are applied to two screws as a piece
of sheet metal is fastened to a block of wood. Reduce the forces and
the couples to an equivalent wrench and determine (a) the resultant
force R, (b) the pitch of the wrench, (c) the point where the axis of
the wrench intersects the xz-plane.
SOLUTION
First, reduce the given force system to a force-couple system.
We have
ΣF : − (20 N)i − (15 N) j = R
We have
ΣM O : Σ(rO &times; F ) + ΣM C = M OR
R = 25 N
M OR = −20 N(0.1 m)j − (4 N ⋅ m)i − (1 N ⋅ m)j
= −(4 N ⋅ m)i − (3 N ⋅ m) j
R = −(20.0 N)i − (15.00 N)j 
(a)
(b)
We have
R
R
= (−0.8i − 0.6 j) ⋅ [−(4 N ⋅ m)i − (3 N ⋅ m)j]
M1 = λR ⋅ M OR
λ=
= 5 N⋅m
Pitch:
p=
M1 5 N ⋅ m
=
= 0.200 m
R
25 N
or p = 0.200 m 
(c)
From above, note that
M1 = M OR
Therefore, the axis of the wrench goes through the origin. The line of action of the wrench lies in the
xy-plane with a slope of
y=
3
x 
4
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305
PROBLEM 3.137*
Two bolts at A and B are tightened by applying the forces
and couples shown. Replace the two wrenches with a
single equivalent wrench and determine (a) the resultant
R, (b) the pitch of the single equivalent wrench, (c) the
point where the axis of the wrench intersects the xz-plane.
SOLUTION
First, reduce the given force system to a force-couple at the origin.
We have
ΣF : − (84 N) j − (80 N)k = R
and
ΣM O : Σ(rO &times; F ) + ΣM C = M OR
R = 116 N
i
j k
i
j k
0.6 0 0.1 + 0.4 0.3 0 + (−30 j − 32k ) N ⋅ m = M OR
0 84 0
0
0 80
M OR = − (15.6 N ⋅ m)i + (2 N ⋅ m) j − (82.4 N ⋅ m)k
R = − (84.0 N) j − (80.0 N)k 
(a)
(b)
We have
M1 = λ R ⋅ M OR
λR =
R
R
−84 j − 80k
⋅ [− (15.6 N ⋅ m)i + (2 N ⋅ m) j − (82.4 N ⋅ m)k ]
116
= 55.379 N ⋅ m
=−
and
Then pitch
M1 = M1λR = − (40.102 N ⋅ m) j − (38.192 N ⋅ m)k
p=
M 1 55.379 N ⋅ m
=
= 0.47741 m
R
116 N
or p = 0.477 m 
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306
PROBLEM 3.137* (Continued)
(c)
We have
M OR = M1 + M 2
M 2 = M OR − M1 = [(−15.6i + 2 j − 82.4k ) − (40.102 j − 38.192k )] N ⋅ m
= − (15.6 N ⋅ m)i + (42.102 N ⋅ m) j − (44.208 N ⋅ m)k
We require
M 2 = rQ/O &times; R
(−15.6i + 42.102 j − 44.208k ) = ( xi + zk ) &times; (84 j − 80k )
= (84 z )i + (80 x) j − (84 x)k
From i:
or
From k:
or
−15.6 = 84 z
z = − 0.185714 m
z = − 0.1857 m
−44.208 = −84 x
x = 0.52629 m
x = 0.526 m
The axis of the wrench intersects the xz-plane at
x = 0.526 m
y = 0 z = − 0.1857 m 
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307
PROBLEM 3.138*
Two bolts at A and B are tightened by applying the forces and
couples shown. Replace the two wrenches with a single equivalent
wrench and determine (a) the resultant R, (b) the pitch of the
single equivalent wrench, (c) the point where the axis of the
wrench intersects the xz-plane.
SOLUTION
First, reduce the given force system to a force-couple at the origin at B.
(a)
We have
15 
 8
ΣF : − (26.4 lb)k − (17 lb)  i + j  = R
 17 17 
R = − (8.00 lb)i − (15.00 lb) j − (26.4 lb)k 
and
We have
R = 31.4 lb
ΣM B : rA/B &times; FA + M A + M B = M BR
M RB
i
j
k
15 
 8
0 − 220k − 238  i + j  = 264i − 220k − 14(8i + 15 j)
= 0 −10
 17 17 
0 0 − 26.4
M RB = (152 lb ⋅ in.)i − (210 lb ⋅ in.)j − (220 lb ⋅ in.)k
(b)
We have
R
R
−8.00i − 15.00 j − 26.4k
=
⋅ [(152 lb ⋅ in.)i − (210 lb ⋅ in.) j − (220 lb ⋅ in.)k ]
31.4
= 246.56 lb ⋅ in.
M1 = λ R ⋅ M OR
λR =
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308
PROBLEM 3.138* (Continued)
and
Then pitch
(c)
We have
M1 = M1λR = − (62.818 lb ⋅ in.)i − (117.783 lb ⋅ in.) j − (207.30 lb ⋅ in.)k
p=
M 1 246.56 lb ⋅ in.
=
= 7.8522 in.
31.4 lb
R
or p = 7.85 in. 
M RB = M1 + M 2
M 2 = M RB − M1 = (152i − 210 j − 220k ) − ( − 62.818i − 117.783j − 207.30k )
= (214.82 lb ⋅ in.)i − (92.217 lb ⋅ in.) j − (12.7000 lb ⋅ in.)k
We require
M 2 = rQ/B &times; R
i
j
k
214.82i − 92.217 j − 12.7000k = x
0
z
−8 −15 −26.4
= (15 z )i − (8 z ) j + (26.4 x) j − (15 x)k
From i:
214.82 = 15 z
z = 14.3213 in.
From k:
−12.7000 = −15 x
x = 0.84667 in.
The axis of the wrench intersects the xz-plane at
x = 0.847 in. y = 0 z = 14.32 in. 
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309
PROBLEM 3.139*
A flagpole is guyed by three cables. If the tensions in the
cables have the same magnitude P, replace the forces exerted
on the pole with an equivalent wrench and determine (a) the
resultant force R, (b) the pitch of the wrench, (c) the point
where the axis of the wrench intersects the xz-plane.
SOLUTION
(a)
First reduce the given force system to a force-couple at the origin.
We have
ΣF : Pλ BA + P λDC + P λDE = R
 4
3  3
4   −9
4
12  
R = P  j − k  +  i − j  +  i − j + k  
5
5
5
5
25
5
25  
 
 

R=
R=
We have
3P
(2i − 20 j − k ) 
25
3P
27 5
P
(2) 2 + (20) 2 + (1)2 =
25
25
ΣM : Σ(rO &times; P) = M OR
3P 
4P 
4P
12 P 
 −4 P
 3P
 −9 P
(24a) j &times; 
j−
k  + (20a) j &times; 
i−
j  + (20a) j &times; 
i−
j+
k  = M OR
5
5
5
5
25
5
25






M OR =
24 Pa
( −i − k )
5
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310
PROBLEM 3.139* (Continued)
(b)
We have
M1 = λ R ⋅ M OR
where
λR =
Then
M1 =
25
1
R 3P
(2i − 20 j − k )
(2i − 20 j − k )
=
=
R 25
27 5 P 9 5
p=
and pitch
1
9 5
(2i − 20 j − k ) ⋅
M 1 −8Pa  25  −8a
=

=
R 15 5  27 5 P  81
M1 = M1λ R =
(c)
Then
M 2 = M OR − M1 =
24 Pa
−8 Pa
(−i − k ) =
5
15 5
or p = − 0.0988a 
−8Pa  1 
8 Pa
( −2i + 20 j + k )

 (2i − 20 j − k ) =
675
15 5  9 5 
24 Pa
8Pa
8 Pa
( −i − k ) −
(−2i + 20 j + k ) =
(−430i − 20 j − 406k )
5
675
675
M 2 = rQ/O &times; R
We require
 8Pa 
 3P 
 675  (−403i − 20 j − 406k ) = ( xi + zk ) &times;  25  (2i − 20 j − k )




 3P 
=
 [20 zi + ( x + 2 z ) j − 20 xk ]
 25 
From i:
8(− 403)
Pa
 3P 
= 20 z 

675
 25 
From k:
8(−406)
Pa
 3P 
= −20 x 
 x = 2.0049a
675
 25 
z = −1.99012a
The axis of the wrench intersects the xz-plane at
x = 2.00a, z = −1.990a 
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311
PROBLEM 3.140*
Two ropes attached at A and B are used to move the trunk of a
fallen tree. Replace the forces exerted by the ropes with an
equivalent wrench and determine (a) the resultant force R,
(b) the pitch of the wrench, (c) the point where the axis of the
wrench intersects the yz-plane.
SOLUTION
(a)
(
)
First replace the given forces with an equivalent force-couple system R, M OR at the origin.
We have
d AC = (6) 2 + (2) 2 + (9) 2 = 11 m
d BD = (14) 2 + (2)2 + (5) 2 = 15 m
Then
1650 N
= (6i + 2 j + 9k )
11
= (900 N)i + (300 N) j + (1350 N)k
TAC =
and
1500 N
= (14i + 2 j + 5k )
15
= (1400 N)i + (200 N) j + (500 N)k
TBD =
Equivalence then requires
ΣF : R = TAC + TBD
= (900i + 300 j + 1350k )
+(1400i + 200 j + 500k )
= (2300 N)i + (500 N) j + (1850 N)k
ΣM O : M OR = rA &times; TAC + rB &times; TBD
= (12 m)k &times; [(900 N)i + (300 N)j + (1350 N)k ]
+ (9 m)i &times; [(1400 N)i + (200 N)j + (500 N)k ]
= −(3600)i + (10,800 − 4500) j + (1800)k
= −(3600 N ⋅ m)i + (6300 N ⋅ m)j + (1800 N ⋅ m)k
The components of the wrench are (R , M1 ), where
R = (2300 N)i + (500 N) j + (1850 N)k

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312
PROBLEM 3.140* (Continued)

(b)
We have
R = 100 (23)2 + (5) 2 + (18.5) 2 = 2993.7 N
Let
λ axis =
Then
1
R
(23i + 5 j + 18.5k )
=
R 29.937
M1 = λ axis ⋅ M OR
1
(23i + 5 j + 18.5k ) ⋅ (−3600i + 6300 j + 1800k )
29.937
1
[(23)( −36) + (5)(63) + (18.5)(18)]
=
0.29937
= −601.26 N ⋅ m
=
Finally,
P=
M1 −601.26 N ⋅ m
=
R
2993.7 N
or P = − 0.201 m 
(c)
We have
M1 = M 1 λ axis
= (−601.26 N ⋅ m) &times;
1
(23i + 5 j + 18.5k )
29.937
or
M1 = −(461.93 N ⋅ m)i − (100.421 N ⋅ m) j − (371.56 N ⋅ m)k
Now
M 2 = M OR − M1
= (−3600i + 6300 j + 1800k )
− ( −461.93i − 100.421j − 371.56k )
= − (3138.1 N ⋅ m)i + (6400.4 N ⋅ m)j + (2171.6 N ⋅ m)k
For equivalence:
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313
PROBLEM 3.140* (Continued)
Thus, we require
M 2 = rP &times; R
r = ( yj + zk )
Substituting:
−3138.1i + 6400.4 j + 2171.6k =
i
j
k
0
y
z
2300 500 1850
Equating coefficients:
j : 6400.4 = 2300 z
or
k : 2171.6 = −2300 y or
The axis of the wrench intersects the yz-plane at
z = 2.78 m
y = − 0.944 m
y = −0.944 m
z = 2.78 m

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314
PROBLEM 3.141*
Determine whether the force-and-couple system shown can be
reduced to a single equivalent force R. If it can, determine R
and the point where the line of action of R intersects the
yz-plane. If it cannot be so reduced, replace the given system
with an equivalent wrench and determine its resultant, its pitch,
and the point where its axis intersects the yz-plane.
SOLUTION
First determine the resultant of the forces at D. We have
d DA = (−12) 2 + (9) 2 + (8) 2 = 17 in.
d ED = (−6) 2 + (0)2 + (−8)2 = 10 in.
Then
34 lb
= (−12i + 9 j + 8k )
17
= −(24 lb)i + (18 lb) j + (16 lb)k
FDA =
and
30 lb
= (−6i − 8k )
10
= −(18 lb)i − (24 lb)k
FED =
Then
ΣF : R = FDA + FED
= (−24i + 18 j + 16k + ( −18i − 24k )
= −(42 lb)i + (18 lb)j − (8 lb)k
For the applied couple
d AK = ( −6) 2 + (−6) 2 + (18)2 = 6 11 in.
Then
M=
160 lb ⋅ in.
( −6i − 6 j + 18k )
6 11
160
=
[−(1 lb ⋅ in.)i − (1 lb ⋅ in.)j + (3 lb ⋅ in.)k ]
11
To be able to reduce the original forces and couple to a single equivalent force, R and M
must be perpendicular. Thus
?
R ⋅ M =0
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315
PROBLEM 3.141* (Continued)
Substituting
(−42i + 18 j − 8k ) ⋅
160
or
11
160
11
?
( −i − j + 3k ) = 0
?
[(−42)(−1) + (18)(−1) + (−8)(3)] = 0

0 =0
or
R and M are perpendicular so that the given system can be reduced to the single equivalent force.
R = −(42.0 lb)i + (18.00 lb) j − (8.00 lb)k

Then for equivalence,
M = rP/D &times; R
Thus, we require
where
rP/D = −(12 in.)i + [( y − 3)in.] j + ( z in.)k
Substituting:
i
j
k
( −i − j + 3k ) = −12 ( y − 3) z
11
18
−42
−8
= [( y − 3)( −8) − ( z )(18)]i
160
+ [( z )(−42) − (−12)(−8)]j
+ [( −12)(18) − ( y − 3)(−42)]k
Equating coefficients:
j: −
k:
160
= − 42 z − 96
or
11
480
= −216 + 42( y − 3) or
11
The line of action of R intersects the yz-plane at
x=0
z = −1.137 in.
y = 11.59 in.
y = 11.59 in. z = −1.137 in.

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316
PROBLEM 3.142*
Determine whether the force-and-couple system shown can be
reduced to a single equivalent force R. If it can, determine R
and the point where the line of action of R intersects the
yz-plane. If it cannot be so reduced, replace the given system
with an equivalent wrench and determine its resultant, its
pitch, and the point where its axis intersects the yz-plane.
SOLUTION
First, reduce the given force system to a force-couple at the origin.
We have
ΣF : FA + FG = R
 (40 mm)i + (60 mm) j − (120 mm)k 
R = (50 N)k + 70 N 

140 mm


= (20 N)i + (30 N) j − (10 N)k
and
We have
R = 37.417 N
ΣM O : Σ(rO &times; F) + ΣM C = M OR
M OR = [(0.12 m) j &times; (50 N)k ] + {(0.16 m)i &times; [(20 N)i + (30 N) j − (60 N)k ]}
 (160 mm)i − (120 mm) j 
+ (10 N ⋅ m) 

200 mm


 (40 mm)i − (120 mm) j + (60 mm)k 
+ (14 N ⋅ m) 

140 mm


M 0R = (18 N ⋅ m)i − (8.4 N ⋅ m) j + (10.8 N ⋅ m)k
To be able to reduce the original forces and couples to a single equivalent force, R and M must be
perpendicular. Thus, R ⋅ M = 0.
Substituting
?
(20i + 30 j − 10k ) ⋅ (18i − 8.4 j + 10.8k ) = 0
?
or
(20)(18) + (30)(−8.4) + (−10)(10.8) = 0
or
0=0

R and M are perpendicular so that the given system can be reduced to the single equivalent force.
R = (20.0 N)i + (30.0 N) j − (10.00 N)k 
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317
PROBLEM 3.142* (Continued)

Then for equivalence,
Thus, we require
M OR = rp &times; R rp = yj + zk
Substituting:
i
j
k
18i − 8.4 j + 10.8k = 0 y
z
20 30 −10
Equating coefficients:
j: − 8.4 = 20 z
k:
or
z = −0.42 m
10.8 = −20 y or
y = −0.54 m
The line of action of R intersects the yz-plane at
x=0
y = −0.540 m z = −0.420 m

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318
PROBLEM 3.143*
Replace the wrench shown with an equivalent system consisting of two
forces perpendicular to the y-axis and applied respectively at A and B.
SOLUTION
Express the forces at A and B as
A = Ax i + Az k
B = Bx i + Bz k
Then, for equivalence to the given force system,
ΣFx : Ax + Bx = 0
(1)
ΣFz : Az + Bz = R
(2)
ΣM x : Az ( a) + Bz ( a + b) = 0
(3)
ΣM z : − Ax (a) − Bx (a + b) = M
(4)
Bx = − Ax
From Equation (1),
Substitute into Equation (4):
− Ax ( a) + Ax ( a + b) = M
M
M
and Bx = −
Ax =
b
b
From Equation (2),
Bz = R − Az
and Equation (3),
Az a + ( R − Az )(a + b) = 0
 a
Az = R 1 + 
 b
and
a

Bz = R − R 1 + 
 b
a
Bz = − R
b
M
A=
 b
Then
a


 i + R 1 + b  k 



M
B = −
 b
 a 
i −  b Rk 
 

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319
PROBLEM 3.144*
Show that, in general, a wrench can be replaced with two forces chosen in such a way that one force passes
through a given point while the other force lies in a given plane.
SOLUTION
First, choose a coordinate system so that the xy-plane coincides with the given plane. Also, position the
coordinate system so that the line of action of the wrench passes through the origin as shown in Figure a.
Since the orientation of the plane and the components (R, M) of the wrench are known, it follows that the
scalar components of R and M are known relative to the shown coordinate system.
A force system to be shown as equivalent is illustrated in Figure b. Let A be the force passing through the
given Point P and B be the force that lies in the given plane. Let b be the x-axis intercept of B.
The known components of the wrench can be expressed as
R = Rx i + Ry j + Rz k and M = M x i + M y j + M z k
while the unknown forces A and B can be expressed as
A = Ax i + Ay j + Az k and B = Bx i + Bz k
Since the position vector of Point P is given, it follows that the scalar components (x, y, z) of the
position vector rP are also known.
Then, for equivalence of the two systems,
ΣFx : Rx = Ax + Bx
(1)
ΣFy : Ry = Ay
(2)
ΣFz : Rz = Az + Bz
(3)
ΣM x : M x = yAz − zAy
(4)
ΣM y : M y = zAx − xAz − bBz
(5)
ΣM z : M z = xAy − yAx
(6)
Based on the above six independent equations for the six unknowns ( Ax , Ay , Az , Bx , Bz , b), there
exists a unique solution for A and B.
Ay = Ry 
From Equation (2),
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320
PROBLEM 3.144* (Continued)
1
Ax =   ( xRy − M z ) 
 y
Equation (6):
1
Bx = Rx −   ( xRy − M z ) 
 y
Equation (1):
Equation (4):
1
Az =   ( M x + zRy ) 
 y
Equation (3):
1
Bz = Rz −   ( M x + zRy ) 
 y
b=
Equation (5):
( xM x + yM y + zM z )
( M x − yRz + zRy )

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321
PROBLEM 3.145*
Show that a wrench can be replaced with two perpendicular forces, one of which is applied at a given point.
SOLUTION
First, observe that it is always possible to construct a line perpendicular to a given line so that the constructed
line also passes through a given point. Thus, it is possible to align one of the coordinate axes of a rectangular
coordinate system with the axis of the wrench while one of the other axes passes through the given point.
See Figures a and b.
We have
R = Rj and M = Mj
and are known.
The unknown forces A and B can be expressed as
A = Ax i + Ay j + Az k and B = Bx i + By j + Bz k
The distance a is known. It is assumed that force B intersects the xz-plane at (x, 0, z). Then for equivalence,
ΣFx :
0 = Ax + Bx
(1)
ΣFy :
R = Ay + By
(2)
ΣFz :
0 = Az + Bz
(3)
0 = − zBy
(4)
ΣM x :
ΣM y : M = − aAz − xBz + zBx
(5)
ΣM z :
(6)
0 = aAy + xBy
Since A and B are made perpendicular,
A ⋅ B = 0 or
There are eight unknowns:
Ax Bx + Ay B y + Az Bz = 0
(7)
Ax , Ay , Az , Bx , By , Bz , x, z
But only seven independent equations. Therefore, there exists an infinite number of solutions.
0 = − zBy
Next, consider Equation (4):
If By = 0, Equation (7) becomes
Ax Bx + Az Bz = 0
Ax2 + Az2 = 0
Using Equations (1) and (3), this equation becomes
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322
PROBLEM 3.145* (Continued)
Since the components of A must be real, a nontrivial solution is not possible. Thus, it is required that By ≠ 0,
so that from Equation (4), z = 0.
To obtain one possible solution, arbitrarily let Ax = 0.
(Note: Setting Ay , Az , or Bz equal to zero results in unacceptable solutions.)
The defining equations then become
0 = Bx
(1)′
R = Ay + By
(2)
0 = Az + Bz
(3)
M = − aAz − xBz
(5)′
0 = aAy + xBy
(6)
Ay By + Az Bz = 0
(7)′
Then Equation (2) can be written
Ay = R − By
Equation (3) can be written
Bz = − Az
Equation (6) can be written
x=−
aAy
By
Substituting into Equation (5)′,

R − By
M = − aAz −  − a

By

M
Az = −
By
aR
or

 ( − Az )


(8)
Substituting into Equation (7)′,
 M
 M

( R − By ) By +  −
By 
By  = 0
 aR  aR 
or
By =
a 2 R3
a2 R2 + M 2
Then from Equations (2), (8), and (3),
a2 R2
RM 2
= 2 2
2
2
a R +M
a R + M2

M  a 2 R3
aR 2 M
Az = −
=
−


aR  a 2 R 2 + M 2 
a2 R2 + M 2
Ay = R −
Bz =
2
aR 2 M
a2 R2 + M 2
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323
PROBLEM 3.145* (Continued)
In summary,
A=
RM
( Mj − aRk )
a R2 + M 2

B=
aR 2
(aRj + Mk )
a2 R2 + M 2

2
Which shows that it is possible to replace a wrench with two perpendicular forces, one of which is applied at
a given point.
Lastly, if R &gt; 0 and M &gt; 0, it follows from the equations found for A and B that Ay &gt; 0 and B y &gt; 0.
From Equation (6), x &lt; 0 (assuming a &gt; 0). Then, as a consequence of letting Ax = 0, force A lies in a plane
parallel to the yz-plane and to the right of the origin, while force B lies in a plane parallel to the yz-plane but
to the left to the origin, as shown in the figure below.

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324
PROBLEM 3.146*
Show that a wrench can be replaced with two forces, one of which has a prescribed line of action.
SOLUTION
First, choose a rectangular coordinate system where one axis coincides with the axis of the wrench and
another axis intersects the prescribed line of action (AA′). Note that it has been assumed that the line of action
of force B intersects the xz-plane at Point P(x, 0, z). Denoting the known direction of line AA′ by
λ A = λx i + λ y j + λz k
it follows that force A can be expressed as
A = Aλ A = A(λx i + λ y j + λz k )
Force B can be expressed as
B = Bx i + B y j + Bz k
Next, observe that since the axis of the wrench and the prescribed line of action AA′ are known, it follows that
the distance a can be determined. In the following solution, it is assumed that a is known.
Then for equivalence,
ΣFx : 0 = Aλx + Bx
(1)
ΣFy : R = Aλ y + By
(2)
ΣFz : 0 = Aλz + Bz
(3)
ΣM x : 0 = − zBy
(4)
ΣM y : M = − aAλz + zBx − xBz
(5)
ΣM x : 0 = − aAλ y + xB y
(6)
Since there are six unknowns (A, Bx, By, Bz, x, z) and six independent equations, it will be possible to
obtain a solution.
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325
PROBLEM 3.146* (Continued)
Case 1: Let z = 0 to satisfy Equation (4).
Now Equation (2):
Equation (3):
Equation (6):
Aλ y = R − B y
Bz = − Aλz
x=−
aAλ y
By
 a
= −
 By


 ( R − By )


Substitution into Equation (5):
  a
M = − aAλz −  − 
  By
1 M 
A=− 
B
λz  aR  y


 ( R − By )(− Aλz ) 



Substitution into Equation (2):
R=−
By =
Then
1 M 
B λ + By
λz  aR  y y
λz aR 2
λz aR − λ y M
MR
R
=
λz aR − λ y M λ − aR λ
y
z
M
λx MR
Bx = − Aλx =
λz aR − λ y M
A=−
Bz = − Aλz =
λz MR
λz aR − λ y M
A=
In summary,
B=
and

R 
x = a 1 −


By 


 λz aR − λ y M
= a 1 − R 
 λ aR 2

z

P
λA 
aR
λy −
λz
M
R
(λ M i + λz aRj + λz M k ) 
λz aR − λ y M x

 
 
or x =
λy M
λz R

Note that for this case, the lines of action of both A and B intersect the x-axis.
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326
PROBLEM 3.146* (Continued)
Case 2: Let By = 0 to satisfy Equation (4).
Now Equation (2):
A=
R
λy
Equation (1):
λ
Bx = − R  x
 λy





Equation (3):
λ
Bz = − R  z
 λy





Equation (6):
aAλ y = 0
which requires a = 0
Substitution into Equation (5):
 λ
M = z − R  x
  λ y
 λ

 − x − R  z

  λ y


M
  or λz x − λx z = 

 R


 λy

This last expression is the equation for the line of action of force B.
In summary,
 R
A=
 λy


 λA



 R
B=
 λy


 ( −λ x i − λx k )



Assuming that λx , λ y , λz ⬎ 0, the equivalent force system is as shown below.
Note that the component of A in the xz-plane is parallel to B.
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327
PROBLEM 3.147
A 300-N force is applied at A as shown. Determine (a) the moment
of the 300-N force about D, (b) the smallest force applied at B that
creates the same moment about D.
SOLUTION
(a)
Fx = (300 N) cos 25&deg;
= 271.89 N
Fy = (300 N) sin 25&deg;
= 126.785 N
F = (271.89 N)i + (126.785 N) j

r = DA = −(0.1 m)i − (0.2 m) j
MD = r &times; F
M D = [−(0.1 m)i − (0.2 m) j] &times; [(271.89 N)i + (126.785 N) j]
= −(12.6785 N ⋅ m)k + (54.378 N ⋅ m)k
= (41.700 N ⋅ m)k
M D = 41.7 N ⋅ m
(b)

The smallest force Q at B must be perpendicular to

DB at 45&deg;

M D = Q ( DB)
41.700 N ⋅ m = Q (0.28284 m)
Q = 147.4 N
45.0&deg; 
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328
PROBLEM 3.148
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.
SOLUTION
First note
dCB = (12.0 in.) 2 + (2.33 in.) 2
= 12.2241 in.
Then
and
12.0 in.
12.2241 in.
2.33 in.
sin θ =
12.2241 in.
cos θ =
FCB = FCB cos θ i − FCB sin θ j
=
125 lb
[(12.0 in.) i − (2.33 in.) j]
12.2241 in.
Now
M A = rB/A &times; 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] &times;
125 lb
(12.0i − 2.33j)
12.2241 in.
= (1393.87 lb ⋅ in.)k
= (116.156 lb ⋅ ft)k
or M A = 116.2 lb ⋅ ft

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329
PROBLEM 3.149
The ramp ABCD is supported by cables at corners C and D.
The tension in each of the cables is 810 N. Determine the
moment about A of the force exerted by (a) the cable at D,
(b) the cable at C.
SOLUTION
(a)
We have
M A = rE/A &times; TDE
where
rE/A = (2.3 m) j
TDE = λ DE TDE
=
(0.6 m)i + (3.3 m) j − (3 m)k
(0.6) 2 + (3.3)2 + (3) 2 m
(810 N)
= (108 N)i + (594 N) j − (540 N)k
i
j
k
MA = 0
2.3
0 N⋅m
108 594 −540
= −(1242 N ⋅ m)i − (248.4 N ⋅ m)k
or M A = −(1242 N ⋅ m)i − (248 N ⋅ m)k 
(b)
We have
M A = rG/A &times; TCG
where
rG/A = (2.7 m)i + (2.3 m) j
TCG = λ CGTCG
=
−(.6 m)i + (3.3 m) j − (3 m)k
(.6) 2 + (3.3) 2 + (3) 2 m
(810 N)
= −(108 N)i + (594 N) j − (540 N)k
i
j
k
M A = 2.7 2.3
0 N⋅m
−108 594 −540
= −(1242 N ⋅ m)i + (1458 N ⋅ m) j + (1852 N ⋅ m)k
or M A = −(1242 N ⋅ m)i + (1458 N ⋅ m) j + (1852 N ⋅ m)k 
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330
PROBLEM 3.150
Section AB of a pipeline lies in the yz-plane and forms an angle
of 37&deg; with the z-axis. Branch lines CD and EF join AB as
shown. Determine the angle formed by pipes AB and CD.
SOLUTION
First note

AB = AB(sin 37&deg; j − cos 37&deg;k )
CD = CD(− cos 40&deg; cos 55&deg; j + sin 40&deg; j − cos 40&deg; sin 55&deg;k )
Now
 
AB ⋅ CD = ( AB)(CD ) cos θ
or
AB(sin 37&deg; j − cos 37&deg;k ) ⋅ CD (− cos 40&deg; cos 55&deg;i + sin 40&deg; j − cos 40&deg; sin 55&deg;k )
= (AB)(CD) cos θ
or
cos θ = (sin 37&deg;)(sin 40&deg;) + (− cos 37&deg;)(− cos 40&deg; sin 55&deg;)
= 0.88799
or θ = 27.4&deg; 
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331
PROBLEM 3.151
To lift a heavy crate, a man uses a block and tackle attached to the
bottom of an I-beam at hook B. Knowing that the moments about the y
and the z axes of the force exerted at B by portion AB of the rope are,
respectively, 120 N ⋅ m and −460 N ⋅ m, determine the distance a.
SOLUTION
First note

BA = (2.2 m)i − (3.2 m) j − ( a m)k
Now
M D = rA/D &times; TBA
where
rA/D = (2.2 m)i + (1.6 m) j
TBA =
Then
i
j
k
TBA
MD =
2.2 1.6 0
d BA
2.2 −3.2 − a
=
Thus
TBA
(2.2i − 3.2 j − ak ) (N)
d BA
TBA
{−1.6a i + 2.2a j + [(2.2)(−3.2) − (1.6)(2.2)]k}
d BA
M y = 2.2
TBA
a
d BA
M z = −10.56
Then forming the ratio
TBA
d BA
(N ⋅ m)
(N ⋅ m)
My
Mz
T
2.2 dBA (N ⋅ m)
120 N ⋅ m
BA
=
−460 N ⋅ m −10.56 TdBA (N ⋅ m)
or a = 1.252 m 
BA
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332
PROBLEM 3.152
To loosen a frozen valve, a force F of magnitude 70 lb is
applied to the handle of the valve. Knowing that θ = 25&deg;,
Mx = −61 lb ⋅ ft, and M z = − 43 lb ⋅ ft, determine φ and d.
SOLUTION
We have
ΣM O : rA/O &times; F = M O
where
rA/O = −(4 in.)i + (11 in.) j − (d )k
F = F (cos θ cos φ i − sin θ j + cos θ sin φ k )
For
F = 70 lb, θ = 25&deg;
F = (70 lb)[(0.90631cos φ )i − 0.42262 j + (0.90631sin φ )k ]
i
M O = (70 lb)
−4
−0.90631cos φ
j
k
11
−d
in.
−0.42262 0.90631sin φ
= (70 lb)[(9.9694sin φ − 0.42262d ) i + (−0.90631d cos φ + 3.6252sin φ ) j
+ (1.69048 − 9.9694cos φ )k ] in.
and
M x = (70 lb)(9.9694sin φ − 0.42262d ) in. = −(61 lb ⋅ ft)(12 in./ft)
(1)
M y = (70 lb)(−0.90631d cos φ + 3.6252sin φ ) in.
(2)
M z = (70 lb)(1.69048 − 9.9694 cos φ ) in. = − 43 lb ⋅ ft(12 in./ft)
(3)
 634.33 
 = 24.636&deg;
 697.86 
From Equation (3):
φ = cos −1 
or
From Equation (1):
 1022.90 
d =
 = 34.577 in.
 29.583 
or d = 34.6 in. 
φ = 24.6&deg; 
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333
PROBLEM 3.153
The tension in the cable attached to the end C of an
adjustable boom ABC is 560 lb. Replace the force
exerted by the cable at C with an equivalent forcecouple system (a) at A, (b) at B.
SOLUTION
(a)
Based on
ΣF : FA = T = 560 lb
FA = 560 lb
or
20.0&deg; 
ΣM A : M A = (T sin 50&deg;)(d A )
= (560 lb)sin 50&deg;(18 ft)
= 7721.7 lb ⋅ ft
M A = 7720 lb ⋅ ft
or
(b)
Based on
ΣF : FB = T = 560 lb
FB = 560 lb
or

20.0&deg; 
ΣM B : M B = (T sin 50&deg;)(d B )
= (560 lb) sin 50&deg;(10 ft)
= 4289.8 lb ⋅ ft
M B = 4290 lb ⋅ ft
or

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334
PROBLEM 3.154
While tapping a hole, a machinist applies the horizontal forces
shown to the handle of the tap wrench. Show that these forces
are equivalent to a single force, and specify, if possible, the
point of application of the single force on the handle.
SOLUTION
Since the forces at A and B are parallel, the force at B can be replaced with the sum of two forces with one of
the forces equal in magnitude to the force at A except with an opposite sense, resulting in a force-couple.
We have FB = 2.9 lb − 2.65 lb = 0.25 lb, where the 2.65-lb force is part of the couple. Combining the two
parallel forces,
M couple = (2.65 lb)[(3.2 in. + 2.8 in.) cos 25&deg;]
= 14.4103 lb ⋅ in.
and
M couple = 14.4103 lb ⋅ in.
A single equivalent force will be located in the negative z direction.
Based on
ΣM B : −14.4103 lb ⋅ in. = [(0.25 lb) cos 25&deg;](a )
a = 63.600 in.
F′ = (0.25 lb)(cos 25&deg;i + sin 25&deg;k )
F′ = (0.227 lb)i + (0.1057 lb)k and is applied on an extension of handle BD at a
distance of 63.6 in. to the right of B.

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335
PROBLEM 3.155
Replace the 150-N force with an equivalent force-couple
system at A.
SOLUTION
Equivalence requires
ΣF : F = (150 N)(− cos 35&deg; j − sin 35&deg;k )
= −(122.873 N) j − (86.036 N)k
ΣMA : M = rD/A &times; F
where
Then
rD/A = (0.18 m)i − (0.12 m) j + (0.1 m)k
i
j
k
−0.12
0.1 N ⋅ m
M = 0.18
0
−122.873 −86.036
= [( −0.12)(−86.036) − (0.1)(−122.873)]i
+ [−(0.18)(−86.036)]j
+ [(0.18)(−122.873)]k
= (22.6 N ⋅ m)i + (15.49 N ⋅ m) j − (22.1 N ⋅ m)k
The equivalent force-couple system at A is
F = −(122.9 N) j − (86.0 N)k 
M = (22.6 N ⋅ m)i + (15.49 N ⋅ m) j − (22.1 N ⋅ m)k 
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336
PROBLEM 3.156
A beam supports three loads of given magnitude and a fourth load whose magnitude is a function of position.
If b = 1.5 m and the loads are to be replaced with a single equivalent force, determine (a) the value of a so that
the distance from support A to the line of action of the equivalent force is maximum, (b) the magnitude of the
equivalent force and its point of application on the beam.
SOLUTION
For equivalence,
ΣFy : −1300 + 400
a
− 400 − 600 = − R
b
a

R =  2300 − 400  N
b


or
ΣM A :
a
a
400  − a (400) − (a + b)(600) = − LR
2 
b
L=
or
Then with
(1)
1000a + 600b − 200
2300 − 400
b = 1.5 m
L=
a2
b
a
b
10a + 9 −
4 2
a
3
8
23 − a
3
(2)
where a, L are in m.
(a)
Find value of a to maximize L.
8 
8  
4  8 

10 − a  23 − a  − 10a + 9 − a 2  − 
dL 
3 
3  
3  3 
=
2
da
8 

 23 − 3 a 


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337
PROBLEM 3.156 (Continued)
230 −
or
16a 2 − 276a + 1143 = 0
276 &plusmn; (−276) 2 − 4(16)(1143)
2(16)
Then
a=
or
a = 10.3435 m and a = 6.9065 m
Since
(b)
184
80
64 2 80
32
a− a+
a + a + 24 − a 2 = 0
3
3
9
3
9
or
AB = 9 m, a must be less than 9 m
a = 6.91 m 
6.9065
1.5
or R = 458 N 
Using Eq. (1),
R = 2300 − 400
and using Eq. (2),
4
10(6.9065) + 9 − (6.9065)2
3
L=
= 3.16 m
8
23 − (6.9065)
3
R is applied 3.16 m to the right of A. 
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338
PROBLEM 3.157
A mechanic uses a crowfoot wrench to loosen a bolt at C. The
mechanic holds the socket wrench handle at Points A and B
and applies forces at these points. Knowing that these forces
are equivalent to a force-couple system at C consisting of the
force C = (8 lb)i + (4 lb)k and the couple M C = (360 lb &middot; in.)i,
determine the forces applied at A and at B when Az = 2 lb.
SOLUTION
We have
ΣF :
A+B=C
or
Fx : Ax + Bx = 8 lb
Bx = −( Ax + 8 lb)
(1)
ΣFy : Ay + By = 0
Ay = − By
or
(2)
ΣFz : 2 lb + Bz = 4 lb
Bz = 2 lb
or
(3)
ΣM C : rB/C &times; B + rA/C &times; A = M C
We have
i
8
Bx
j
0
By
k
i
2 + 8
2
Ax
j
0
Ay
k
8
2
lb ⋅ in. = (360 lb ⋅ in.)i
(2 By − 8 Ay )i + (2 Bx − 16 + 8 Ax − 16) j
or
+ (8By + 8 Ay )k = (360 lb ⋅ in.)i
From
i-coefficient:
j-coefficient:
k-coefficient:
2 By − 8 Ay = 360 lb ⋅ in.
−2 Bx + 8 Ax = 32 lb ⋅ in.
8 By + 8 Ay = 0
(4)
(5)
(6)
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339
PROBLEM 3.157 (Continued)
From Equations (2) and (4):
2 By − 8(− By ) = 360
By = 36 lb
From Equations (1) and (5):
Ay = 36 lb
2(− Ax − 8) + 8 Ax = 32
Ax = 1.6 lb
From Equation (1):
Bx = −(1.6 + 8) = −9.6 lb
A = (1.600 lb)i − (36.0 lb) j + (2.00 lb)k 
B = −(9.60 lb)i + (36.0 lb) j + (2.00 lb)k 
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340
PROBLEM 3.158
A concrete foundation mat in the shape of a regular hexagon of side
12 ft supports four column loads as shown. Determine the magnitudes
of the additional loads that must be applied at B and F if the resultant
of all six loads is to pass through the center of the mat.
SOLUTION
From the statement of the problem, it can be concluded that the six applied loads are equivalent to the
resultant R at O. It then follows that
ΣM O = 0 or ΣM x = 0 ΣM z = 0
Then
ΣM x = 0: (6 3 ft) FB + (6 3 ft)(10 kips) − (6 3 ft)(20 kips)
− (6 3 ft) FF = 0
FB − FF = 10
or
(1)
ΣM z = 0: (12 ft)(15 kips) + (6 ft) FB − (6 ft)(10 kips)
− (12 ft)(30 kips) − (6 ft)(20 kips) + (6 ft) FF = 0
FB + FF = 60
or
(2)
Then Eqs. (1) + (2) 
FB = 35.0 kips 
and
FF = 25.0 kips 
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341
CHAPTER 4
PROBLEM 4.1
Two crates, each of mass 350 kg, are placed as shown in the
bed of a 1400-kg pickup truck. Determine the reactions at each
of the two (a) rear wheels A, (b) front wheels B.
SOLUTION
Free-Body Diagram:
W = (350 kg)(9.81 m/s 2 ) = 3.4335 kN
Wt = (1400 kg)(9.81 m/s 2 ) = 13.7340 kN
(a)
Rear wheels:
ΣM B = 0: W (1.7 m + 2.05 m) + W (2.05 m) + Wt (1.2 m) − 2 A(3 m) = 0
(3.4335 kN)(3.75 m) + (3.4335 kN)(2.05 m)
+ (13.7340 kN)(1.2 m) − 2 A(3 m) = 0
A = +6.0659 kN
(b)
Front wheels:
A = 6.07 kN 
ΣFy = 0: − W − W − Wt + 2 A + 2 B = 0
−3.4335 kN − 3.4335 kN − 13.7340 kN + 2(6.0659 kN) + 2B = 0
B = +4.2346 kN
B = 4.23 kN 
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345
PROBLEM 4.2
Solve Problem 4.1, assuming that crate D is removed and that
the position of crate C is unchanged.
PROBLEM 4.1 Two crates, each of mass 350 kg, are placed
as shown in the bed of a 1400-kg pickup truck. Determine the
reactions at each of the two (a) rear wheels A, (b) front wheels B.
SOLUTION
Free-Body Diagram:
W = (350 kg)(9.81 m/s 2 ) = 3.4335 kN
Wt = (1400 kg)(9.81 m/s 2 ) = 13.7340 kN
(a)
Rear wheels:
ΣM B = 0: W (1.7 m + 2.05 m) + Wt (1.2 m) − 2 A(3 m) = 0
(3.4335 kN)(3.75 m) + (13.7340 kN)(1.2 m) − 2 A(3 m) = 0
A = + 4.8927 kN
(b)
Front wheels:
A = 4.89 kN 
ΣM y = 0: − W − Wt + 2 A + 2 B = 0
−3.4335 kN − 13.7340 kN + 2(4.8927 kN) + 2B = 0
B = +3.6911 kN
B = 3.69 kN 
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346
PROBLEM 4.3
A T-shaped bracket supports the four loads shown. Determine the
reactions at A and B (a) if a = 10 in., (b) if a = 7 in.
SOLUTION
Free-Body Diagram:
ΣFx = 0: Bx = 0
ΣM B = 0: (40 lb)(6 in.) − (30 lb)a − (10 lb)(a + 8 in.) + (12 in.) A = 0
(40a − 160)
12
A=
(1)
ΣM A = 0: − (40 lb)(6 in.) − (50 lb)(12 in.) − (30 lb)(a + 12 in.)
− (10 lb)(a + 20 in.) + (12 in.) B y = 0
By =
Bx = 0, B =
Since
(a)
(b)
(1400 + 40a)
12
(1400 + 40a )
12
(2)
For a = 10 in.,
Eq. (1):
A=
(40 &times; 10 − 160)
= +20.0 lb
12
Eq. (2):
B=
(1400 + 40 &times; 10)
= +150.0 lb
12
B = 150.0 lb 
Eq. (1):
A=
(40 &times; 7 − 160)
= +10.00 lb
12
A = 10.00 lb 
Eq. (2):
B=
(1400 + 40 &times; 7)
= +140.0 lb
12
B = 140.0 lb 
A = 20.0 lb 
For a = 7 in.,
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347
PROBLEM 4.4
distance a if the bracket is not to move.
PROBLEM 4.3 A T-shaped bracket supports the four loads shown.
Determine the reactions at A and B (a) if a = 10 in., (b) if a = 7 in.
SOLUTION
Free-Body Diagram:
For no motion, reaction at A must be downward or zero; smallest distance a for no motion
corresponds to A = 0.
ΣM B = 0: (40 lb)(6 in.) − (30 lb)a − (10 lb)(a + 8 in.) + (12 in.) A = 0
A=
(40a − 160)
12
A = 0: (40a − 160) = 0
a = 4.00 in. 
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348
PROBLEM 4.5
A hand truck is used to move two kegs, each of mass 40 kg.
Neglecting the mass of the hand truck, determine (a) the vertical
force P that should be applied to the handle to maintain
equilibrium when α = 35&deg;, (b) the corresponding reaction at each
of the two wheels.
SOLUTION
Free-Body Diagram:
W = mg = (40 kg)(9.81 m/s 2 ) = 392.40 N
a1 = (300 mm)sinα − (80 mm)cosα
a2 = (430 mm)cosα − (300 mm)sinα
b = (930 mm)cosα
From free-body diagram of hand truck,
Dimensions in mm
ΣM B = 0: P(b) − W ( a2 ) + W (a1 ) = 0
(1)
ΣFy = 0: P − 2W + 2 B = 0
(2)
α = 35&deg;
For
a1 = 300sin 35&deg; − 80 cos 35&deg; = 106.541 mm
a2 = 430 cos 35&deg; − 300sin 35&deg; = 180.162 mm
b = 930cos 35&deg; = 761.81 mm
(a)
From Equation (1):
P(761.81 mm) − 392.40 N(180.162 mm) + 392.40 N(106.54 mm) = 0
P = 37.921 N
(b)
or P = 37.9 N 
From Equation (2):
37.921 N − 2(392.40 N) + 2 B = 0
or B = 373 N 
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349
PROBLEM 4.6
Solve Problem 4.5 when α = 40&deg;.
PROBLEM 4.5 A hand truck is used to move two kegs, each of
mass 40 kg. Neglecting the mass of the hand truck, determine
(a) the vertical force P that should be applied to the handle to
maintain equilibrium when α = 35&deg;, (b) the corresponding reaction
at each of the two wheels.
SOLUTION
Free-Body Diagram:
W = mg = (40 kg)(9.81 m/s 2 )
W = 392.40 N
a1 = (300 mm)sinα − (80 mm)cosα
a2 = (430 mm)cosα − (300 mm)sinα
b = (930 mm)cosα
From F.B.D.:
ΣM B = 0: P(b) − W ( a2 ) + W (a1 ) = 0
P = W ( a2 − a1 )/b
(1)
ΣFy = 0: − W − W + P + 2 B = 0
B =W −
For
1
P
2
(2)
α = 40&deg;:
a1 = 300sin 40&deg; − 80 cos 40&deg; = 131.553 mm
a2 = 430 cos 40&deg; − 300sin 40&deg; = 136.563 mm
b = 930cos 40&deg; = 712.42 mm
(a)
From Equation (1):
P=
392.40 N (0.136563 m − 0.131553 m)
0.71242 m
P = 2.7595 N
(b)
From Equation (2):
B = 392.40 N −
P = 2.76 N 
1
(2.7595 N)
2
B = 391 N 
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350
PROBLEM 4.7
A 3200-lb forklift truck is used to lift a 1700-lb crate. Determine
the reaction at each of the two (a) front wheels A, (b) rear wheels B.
SOLUTION
Free-Body Diagram:
(a)
Front wheels:
ΣM B = 0: (1700 lb)(52 in.) + (3200 lb)(12 in.) − 2 A(36 in.) = 0
A = +1761.11 lb
(b)
Rear wheels:
A = 1761 lb 
ΣFy = 0: − 1700 lb − 3200 lb + 2(1761.11 lb) + 2 B = 0
B = +688.89 lb
B = 689 lb 
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351
PROBLEM 4.8
For the beam and loading shown, determine (a) the reaction at A,
(b) the tension in cable BC.
SOLUTION
Free-Body Diagram:
(a)
Reaction at A:
ΣFx = 0: Ax = 0
ΣMB = 0: (15 lb)(28 in.) + (20 lb)(22 in.) + (35 lb)(14 in.)
+ (20 lb)(6 in.) − Ay (6 in.) = 0
Ay = +245 lb
(b)
Tension in BC:
A = 245 lb 
ΣM A = 0: (15 lb)(22 in.) + (20 lb)(16 in.) + (35 lb)(8 in.)
− (15 lb)(6 in.) − FBC (6 in.) = 0
FBC = +140.0 lb
Check:
FBC = 140.0 lb 
ΣFy = 0: − 15 lb − 20 lb = 35 lb − 20 lb + A − FBC = 0
−105 lb + 245 lb − 140.0 = 0
0 = 0 (Checks)
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352
PROBLEM 4.9
distance a for which the reaction at B does not exceed 100 lb
downward or 200 lb upward.
SOLUTION
Assume B is positive when directed .
Sketch showing distance from D to forces.
ΣM D = 0: (300 lb)(8 in. − a ) − (300 lb)(a − 2 in.) − (50 lb)(4 in.) + 16 B = 0
−600a + 2800 + 16B = 0
(2800 + 16B)
600
(1)
[2800 + 16( −100)] 1200
=
= 2 in.
600
600
a ≥ 2.00 in. 
a=
For B = 100 lb = −100 lb, Eq. (1) yields:
a≥
For B = 200 = +200 lb, Eq. (1) yields:
a≤
Required range:
[2800 + 16(200)] 6000
=
= 10 in.
600
600
2.00 in. ≤ a ≤ 10.00 in.
a ≤ 10.00 in. 

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353
PROBLEM 4.10
The maximum allowable value of each of the reactions is
180 N. Neglecting the weight of the beam, determine the range
of the distance d for which the beam is safe.
SOLUTION
ΣFx = 0: Bx = 0
B = By
ΣM A = 0: (50 N) d − (100 N)(0.45 m − d ) − (150 N)(0.9 m − d ) + B(0.9 m − d ) = 0
50d − 45 + 100d − 135 + 150d + 0.9 B − Bd = 0
d=
180 N ⋅ m − (0.9 m) B
300 A − B
(1)
ΣM B = 0: (50 N)(0.9 m) − A(0.9 m − d ) + (100 N)(0.45 m) = 0
45 − 0.9 A + Ad + 45 = 0
(0.9 m) A − 90 N ⋅ m
A
(2)
d≥
180 − (0.9)180 18
=
= 0.15 m
300 − 180
120
d ≥ 150.0 mm 
d≤
(0.9)180 − 90 72
=
= 0.40 m
180
180
d=
Since B ≤ 180 N, Eq. (1) yields
Since A ≤ 180 N, Eq. (2) yields
Range:
150.0 mm ≤ d ≤ 400 mm
d ≤ 400 mm 

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354
PROBLEM 4.11
Three loads are applied as shown to a light beam supported by
cables attached at B and D. Neglecting the weight of the beam,
determine the range of values of Q for which neither cable
becomes slack when P = 0.
SOLUTION
ΣM B = 0: (3.00 kN)(0.500 m) + TD (2.25 m) − Q (3.00 m) = 0
Q = 0.500 kN + (0.750) TD
(1)
ΣM D = 0: (3.00 kN)(2.75 m) − TB (2.25 m) − Q(0.750 m) = 0
Q = 11.00 kN − (3.00) TB
(2)
For cable B not to be slack, TB ≥ 0, and from Eq. (2),
Q ≤ 11.00 kN
For cable D not to be slack, TD ≥ 0, and from Eq. (1),
Q ≥ 0.500 kN
For neither cable to be slack,
0.500 kN ≤ Q ≤ 11.00 kN 
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355
PROBLEM 4.12
Three loads are applied as shown to a light beam supported by
cables attached at B and D. Knowing that the maximum allowable
tension in each cable is 4 kN and neglecting the weight of the beam,
determine the range of values of Q for which the loading is safe
when P = 0.
SOLUTION
ΣM B = 0: (3.00 kN)(0.500 m) + TD (2.25 m) − Q (3.00 m) = 0
Q = 0.500 kN + (0.750) TD
(1)
ΣM D = 0: (3.00 kN)(2.75 m) − TB (2.25 m) − Q(0.750 m) = 0
Q = 11.00 kN − (3.00) TB
(2)
For TB ≤ 4.00 kN, Eq. (2) yields
Q ≥ 11.00 kN − 3.00(4.00 kN)
Q ≥ −1.000 kN
For TD ≤ 4.00 kN, Eq. (1) yields
Q ≤ 0.500 kN + 0.750(4.00 kN)
Q ≤ 3.50 kN
For loading to be safe, cables must also not be slack. Combining with the conditions obtained in Problem 4.11,
0.500 kN ≤ Q ≤ 3.50 kN 
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356
PROBLEM 4.13
For the beam of Problem 4.12, determine the range of values of Q
PROBLEM 4.12 Three loads are applied as shown to a light beam
supported by cables attached at B and D. Knowing that the maximum
allowable tension in each cable is 4 kN and neglecting the weight of
the beam, determine the range of values of Q for which the loading
is safe when P = 0.
SOLUTION
ΣM B = 0: (3.00 kN)(0.500 m) − (1.000 kN)(0.750 m) + TD (2.25 m) − Q(3.00 m) = 0
Q = 0.250 kN + 0.75 TD
(1)
ΣM D = 0: (3.00 kN)(2.75 m) + (1.000 kN)(1.50 m)
− TB (2.25 m) − Q (0.750 m) = 0
Q = 13.00 kN − 3.00 TB
(2)
For the loading to be safe, cables must not be slack and tension must not exceed 4.00 kN.
Making 0 ≤ TB ≤ 4.00 kN in Eq. (2), we have
13.00 kN − 3.00(4.00 kN) ≤ Q ≤ 13.00 kN − 3.00(0)
1.000 kN ≤ Q ≤ 13.00 kN
(3)
Making 0 ≤ TD ≤ 4.00 kN in Eq. (1), we have
0.250 kN + 0.750(0) ≤ Q ≤ 0.250 kN + 0.750(4.00 kN)
0.250 kN ≤ Q ≤ 3.25 kN
(4)
1.000 kN ≤ Q ≤ 3.25 kN 
Combining Eqs. (3) and (4),
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357
PROBLEM 4.14
For the beam of Sample Problem 4.2, determine the range of values
of P for which the beam will be safe, knowing that the maximum
allowable value of each of the reactions is 30 kips and that the
reaction at A must be directed upward.
SOLUTION
ΣFx = 0: Bx = 0
B = By
ΣM A = 0: − P(3 ft) + B(9 ft) − (6 kips)(11 ft) − (6 kips)(13 ft) = 0
P = 3B − 48 kips
(1)
ΣM B = 0: − A(9 ft) + P (6 ft) − (6 kips)(2 ft) − (6 kips)(4 ft) = 0
P = 1.5 A + 6 kips
(2)
Since B ≤ 30 kips, Eq. (1) yields
P ≤ (3)(30 kips) − 48 kips
P ≤ 42.0 kips 
Since 0 ≤ A ≤ 30 kips, Eq. (2) yields
0 + 6 kips ≤ P ≤ (1.5)(30 kips)1.6 kips
6.00 kips ≤ P ≤ 51.0 kips

Range of values of P for which beam will be safe:
6.00 kips ≤ P ≤ 42.0 kips

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358
PROBLEM 4.15
The bracket BCD is hinged at C and attached to a control
in the cable, (b) the reaction at C.
SOLUTION
At B:
Ty
Tx
=
0.18 m
0.24 m
3
Ty = Tx
4
(a)
(1)
ΣM C = 0: Tx (0.18 m) − (240 N)(0.4 m) − (240 N)(0.8 m) = 0
Tx = +1600 N
From Eq. (1):
Ty =
3
(1600 N) = 1200 N
4
T = Tx2 + Ty2 = 16002 + 12002 = 2000 N
(b)
T = 2.00 kN 
ΣFx = 0: Cx − Tx = 0
Cx − 1600 N = 0 C x = +1600 N
C x = 1600 N
ΣFy = 0: C y − Ty − 240 N − 240 N = 0
C y − 1200 N − 480 N = 0
C y = +1680 N
C y = 1680 N
α = 46.4&deg;
C = 2320 N
C = 2.32 kN
46.4&deg; 
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359
PROBLEM 4.16
Solve Problem 4.15, assuming that a = 0.32 m.
PROBLEM 4.15 The bracket BCD is hinged at C and
determine (a) the tension in the cable, (b) the reaction at C.
SOLUTION
At B:
Ty
Tx
=
0.32 m
0.24 m
4
Ty = Tx
3
ΣM C = 0: Tx (0.32 m) − (240 N)(0.4 m) − (240 N)(0.8 m) = 0
Tx = 900 N
From Eq. (1):
Ty =
4
(900 N) = 1200 N
3
T = Tx2 + Ty2 = 9002 + 12002 = 1500 N
T = 1.500 kN 
ΣFx = 0: C x − Tx = 0
C x − 900 N = 0 C x = +900 N
C x = 900 N
ΣFy = 0: C y − Ty − 240 N − 240 N = 0
C y − 1200 N − 480 N = 0
C y = +1680 N
C y = 1680 N
α = 61.8&deg;
C = 1906 N
C = 1.906 kN
61.8&deg; 
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360
PROBLEM 4.17
The lever BCD is hinged at C and attached to a control rod at B. If
P = 100 lb, determine (a) the tension in rod AB, (b) the reaction at C.
SOLUTION
Free-Body Diagram:
(a)
ΣM C = 0: T (5 in.) − (100 lb)(7.5 in.) = 0
T = 150.0 lb 
(b)
3
ΣFx = 0: C x + 100 lb + (150.0 lb) = 0
5
C x = −190 lb
C x = 190 lb
4
ΣFy = 0: C y + (150.0 lb) = 0
5
C y = −120 lb
C y = 120 lb
α = 32.3&deg;
C = 225 lb
C = 225 lb
32.3&deg; 
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361
PROBLEM 4.18
The lever BCD is hinged at C and attached to a control rod at B.
Determine the maximum force P that can be safely applied at D if the
maximum allowable value of the reaction at C is 250 lb.
SOLUTION
Free-Body Diagram:
ΣM C = 0: T (5 in.) − P (7.5 in.) = 0
T = 1.5P
3
ΣFx = 0: P + C x + (1.5P) = 0
5
C x = −1.9 P
ΣFy = 0: C y +
C x = 1.9 P
4
(1.5P) = 0
5
C y = −1.2 P
C y = 1.2 P
C = C x2 + C y2
= (1.9 P) 2 + (1.2 P) 2
C = 2.2472 P
For C = 250 lb,
250 lb = 2.2472P
P = 111.2 lb
P = 111.2 lb

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362
PROBLEM 4.19
Two links AB and DE are connected by a bell crank as
shown. Knowing that the tension in link AB is 720 N,
determine (a) the tension in link DE, (b) the reaction at C.
SOLUTION
Free-Body Diagram:
ΣM C = 0: FAB (100 mm) − FDE (120 mm) = 0
FDE =
(a)
For
(1)
FAB = 720 N
FDE =
(b)
5
FAB
6
5
(720 N)
6
FDE = 600 N 
3
ΣFx = 0: − (720 N) + C x = 0
5
C x = +432 N
4
ΣFy = 0: − (720 N) + C y − 600 N = 0
5
C y = +1176 N
C = 1252.84 N
α = 69.829&deg;
C = 1253 N
69.8&deg; 
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363
PROBLEM 4.20
Two links AB and DE are connected by a bell crank as
shown. Determine the maximum force that may be safely
exerted by link AB on the bell crank if the maximum
allowable value for the reaction at C is 1600 N.
SOLUTION
See solution to Problem 4.15 for F.B.D. and derivation of Eq. (1).
FDE =
5
FAB
6
(1)
3
ΣFx = 0: − FAB + C x = 0
5
ΣFy = 0: −
Cx =
3
FAB
5
4
FAB + C y − FDE = 0
5
4
5
− FAB + C y − FAB = 0
5
6
49
Cy =
FAB
30
C = C x2 + C y2
1
(49) 2 + (18) 2 FAB
30
C = 1.74005FAB
=
For C = 1600 N, 1600 N = 1.74005FAB
FAB = 920 N 
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364
PROBLEM 4.21
Determine the reactions at A and C when (a) α = 0, (b) α = 30&deg;.
SOLUTION
(a)
α =0
From F.B.D. of member ABC:
ΣM C = 0: (300 N)(0.2 m) + (300 N)(0.4 m) − A(0.8 m) = 0
A = 225 N
or
A = 225 N 
ΣFy = 0: C y + 225 N = 0
C y = −225 N or C y = 225 N
ΣFx = 0: 300 N + 300 N + C x = 0
C x = −600 N or C x = 600 N
Then
C = C x2 + C y2 = (600) 2 + (225) 2 = 640.80 N
and
θ = tan −1 
 Cy 
−1  −225 
 = tan 
 = 20.556&deg;
 −600 
 Cx 
or
(b)
C = 641 N
20.6&deg; 
α = 30&deg;
From F.B.D. of member ABC:
ΣM C = 0: (300 N)(0.2 m) + (300 N)(0.4 m) − ( A cos 30&deg;)(0.8 m)
+ ( A sin 30&deg;)(20 in.) = 0
A = 365.24 N
or
A = 365 N
60.0&deg; 
ΣFx = 0: 300 N + 300 N + (365.24 N) sin 30&deg; + C x = 0
C x = −782.62
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365
PROBLEM 4.21 (Continued)
ΣFy = 0: C y + (365.24 N) cos 30&deg; = 0
C y = −316.31 N or C y = 316 N
Then
C = C x2 + C y2 = (782.62) 2 + (316.31) 2 = 884.12 N
and
θ = tan −1 
 Cy 
−1  −316.31 
 = tan 
 = 22.007&deg;
 −782.62 
 Cx 
C = 884 N
or
22.0&deg; 
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366
PROBLEM 4.22
Determine the reactions at A and B when (a) α = 0, (b) α = 90&deg;,
(c) α = 30&deg;.
SOLUTION
(a)
α =0
ΣM A = 0: B(20 in.) − 75 lb(10 in.) = 0
B = 37.5 lb
ΣFx = 0: Ax = 0
+ ΣFy = 0: Ay − 75 lb + 37.5 lb = 0
Ay = 37.5 lb
A = B = 37.5 lb 
(b)
α = 90&deg;
ΣM A = 0: B(12 in.) − 75 lb(10 in.) = 0
B = 62.5 lb
ΣFx = 0: Ax − B = 0
Ax = 62.5 lb
ΣFy = 0: Ay − 75 lb = 0
Ay = 75 lb
A = Ax2 + Ay2
= (62.5 lb) 2 + (75 lb) 2
= 97.6 lb
75
62.5
θ = 50.2&deg;
tan θ =
A = 97.6 lb
50.2&deg;; B = 62.51 lb



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367
PROBLEM 4.22 (Continued)
(c)
α = 30&deg;
ΣM A = 0: ( B cos 30&deg;)(20 in.) + ( B sin 30&deg;)(12 in.)
− (75 lb)(10 in.) = 0
B = 32.161 lb
ΣFx = 0: Ax − (32.161) sin 30&deg; = 0
Ax = 16.0805 lb
ΣFy = 0: Ay + (32.161) cos 30&deg; − 75 = 0
Ay = 47.148 lb
A = Ax2 + Ay2
= (16.0805) 2 + (47.148) 2
= 49.8 lb
47.148
16.0805
θ = 71.2&deg;
tan θ =
A = 49.8 lb
71.2&deg;; B = 32.2 lb
60.0&deg; 
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368
PROBLEM 4.23
Determine the reactions at A and B when (a) h = 0,
(b) h = 200 mm.
SOLUTION
Free-Body Diagram:
ΣM A = 0: ( B cos 60&deg;)(0.5 m) − ( B sin 60&deg;)h − (150 N)(0.25 m) = 0
37.5
B=
0.25 − 0.866h
(a)
(1)
When h = 0,
B=
From Eq. (1):
37.5
= 150 N
0.25
B = 150.0 N
30.0&deg; 
ΣFy = 0: Ax − B sin 60&deg; = 0
Ax = (150)sin 60&deg; = 129.9 N
A x = 129.9 N
ΣFy = 0: Ay − 150 + B cos 60&deg; = 0
Ay = 150 − (150) cos 60&deg; = 75 N
A y = 75 N
α = 30&deg;
A = 150.0 N
(b)
A = 150.0 N
30.0&deg; 
When h = 200 mm = 0.2 m,
From Eq. (1):
B=
37.5
= 488.3 N
0.25 − 0.866(0.2)
B = 488 N
30.0&deg; 
ΣFx = 0: Ax − B sin 60&deg; = 0
Ax = (488.3) sin 60&deg; = 422.88 N
A x = 422.88 N
ΣFy = 0: Ay − 150 + B cos 60&deg; = 0
Ay = 150 − (488.3) cos 60&deg; = −94.15 N
A y = 94.15 N
α = 12.55&deg;
A = 433.2 N
A = 433 N
12.55&deg; 
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369
PROBLEM 4.24
A lever AB is hinged at C and attached to a control cable at A. If the
lever is subjected to a 75-lb vertical force at B, determine (a) the
tension in the cable, (b) the reaction at C.
SOLUTION
Free-Body Diagram:
Geometry:
x AC = (10 in.) cos 20&deg; = 9.3969 in.
y AC = (10 in.)sin 20&deg; = 3.4202 in.
 yDA = 12 in. − 3.4202 in. = 8.5798 in.
 yDA 
−1  8.5798 
 = tan 
 = 42.397&deg;
 9.3969 
 x AC 
α = tan −1 
β = 90&deg; − 20&deg; − 42.397&deg; = 27.603&deg;
Equilibrium for lever:
ΣM C = 0: TAD cos 27.603&deg;(10 in.) − (75 lb)[(15 in.)cos 20&deg;] = 0
(a)
ΣFx = 0: C x + (119.293 lb) cos 42.397&deg; = 0
(b)
C x = −88.097 lb
ΣFy = 0: C y − 75 lb − (119.293 lb) sin 42.397&deg; = 0
C y = 155.435
Thus,
C = C x2 + C y2 = (−88.097) 2 + (155.435) 2 = 178.665 lb
and
θ = tan −1
Cy
Cx
= tan −1
155.435
= 60.456&deg;
88.097
C = 178.7 lb
60.5&deg; 
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370
PROBLEM 4.25
For each of the plates and loadings shown, determine the reactions at A and B.
SOLUTION
(a)
Free-Body Diagram:
ΣM A = 0: B(20 in.) − (50 lb)(4 in.) − (40 lb)(10 in.) = 0
B = +30 lb
B = 30.0 lb 
ΣFx = 0: Ax + 40 lb = 0
Ax = −40 lb
A x = 40.0 lb
ΣFy = 0: Ay + B − 50 lb = 0
Ay + 30 lb − 50 lb = 0
Ay = +20 lb
A y = 20.0 lb
α = 26.56&deg;
A = 44.72 lb
A = 44.7 lb
26.6&deg; 



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371
PROBLEM 4.25 (Continued)
(b)
Free-Body Diagram:
ΣM A = 0: ( B cos 30&deg;)(20 in.) − (40 lb)(10 in.) − (50 lb)(4 in.) = 0
B = 34.64 lb
B = 34.6 lb
60.0&deg; 
ΣFx = 0: Ax − B sin 30&deg; + 40 lb
Ax − (34.64 lb) sin 30&deg; + 40 lb = 0
Ax = −22.68 lb
A x = 22.68 lb
ΣFy = 0: Ay + B cos 30&deg; − 50 lb = 0
Ay + (34.64 lb) cos 30&deg; − 50 lb = 0
Ay = +20 lb
A y = 20.0 lb
α = 41.4&deg;
A = 30.2 lb
A = 30.24 lb
41.4&deg; 
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372
PROBLEM 4.26
For each of the plates and loadings shown, determine the reactions at A and B.
SOLUTION
(a)
Free-Body Diagram:
ΣM B = 0: A(20 in.) + (50 lb)(16 in.) − (40 lb)(10 in.) = 0
A = +20 lb
A = 20.0 lb 
ΣFx = 0: 40 lb + Bx = 0
Bx = −40 lb
B x = 40 lb
ΣFy = 0: A + By − 50 lb = 0
20 lb + By − 50 lb = 0
By = +30 lb
α = 36.87&deg;
B = 50 lb
B y = 30 lb
B = 50.0 lb
36.9&deg; 
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373
PROBLEM 4.26 (Continued)
(b)
ΣM A = 0: − ( A cos 30&deg;)(20 in.) − (40 lb)(10 in.) + (50 lb)(16 in.) = 0
A = 23.09 lb
A = 23.1 lb
60.0&deg; 
ΣFx = 0: A sin 30&deg; + 40 lb + Bx = 0
(23.09 lb) sin 30&deg; + 40 lb + 8 x = 0
Bx = −51.55 lb
B x = 51.55 lb
ΣFy = 0: A cos 30&deg; + By − 50 lb = 0
(23.09 lb) cos 30&deg; + By − 50 lb = 0
By = +30 lb
B y = 30 lb
α = 30.2&deg;
B = 59.64 lb
B = 59.6 lb
30.2&deg; 
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374
PROBLEM 4.27
A rod AB hinged at A and attached at B to cable BD supports
the loads shown. Knowing that d = 200 mm, determine (a) the
tension in cable BD, (b) the reaction at A.
SOLUTION
Free-Body Diagram:
(a)
Move T along BD until it acts at Point D.
ΣM A = 0: (T sin 45&deg;)(0.2 m) + (90 N)(0.1 m) + (90 N)(0.2 m) = 0
T = 190.919 N
(b)
T = 190.9 N 
ΣFx = 0: Ax − (190.919 N) cos 45&deg; = 0
Ax = +135.0 N
A x = 135.0 N
ΣFy = 0: Ay − 90 N − 90 N + (190.919 N) sin 45&deg; = 0
Ay = +45.0 N
A y = 45.0 N
A = 142.3 N
18.43&deg;
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375
PROBLEM 4.28
A rod AB, hinged at A and attached at B to cable BD, supports
the loads shown. Knowing that d = 150 mm, determine (a) the
tension in cable BD, (b) the reaction at A.
SOLUTION
Free-Body Diagram:
tan α =
(a)
10
; α = 33.690&deg;
15
Move T along BD until it acts at Point D.
ΣM A = 0: (T sin 33.690&deg;)(0.15 m) − (90 N)(0.1 m) − (90 N)(0.2 m) = 0
T = 324.50 N
(b)
T = 324 N 
ΣFx = 0: Ax − (324.50 N) cos 33.690&deg; = 0
Ax = +270 N
A x = 270 N
ΣFy = 0: Ay − 90 N − 90 N + (324.50 N) sin 33.690&deg; = 0
Ay = 0
A = 270 N

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376
PROBLEM 4.29
A force P of magnitude 90 lb is applied to member ACDE, which is
supported by a frictionless pin at D and by the cable ABE. Since the
cable passes over a small pulley at B, the tension may be assumed to be
the same in portions AB and BE of the cable. For the case when a = 3 in.,
determine (a) the tension in the cable, (b) the reaction at D.
SOLUTION
Free-Body Diagram:
(a)
ΣM D = 0: (90 lb)(9 in.) −
5
12
T (9 in.) − T (7 in.) + T (3 in.) = 0
13
13
T = 117 lb
(b)
ΣFx = 0: Dx − 117 lb −
T = 117.0 lb 
5
(117 lb) + 90 = 0
13
Dx = +72 lb
ΣFy = 0: D y +
12
(117 lb) = 0
13
Dy = −108 lb
D = 129.8 lb
56.3&deg; 

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377
PROBLEM 4.30
Solve Problem 4.29 for a = 6 in.
PROBLEM 4.29 A force P of magnitude 90 lb is applied to
member ACDE, which is supported by a frictionless pin at D and
by the cable ABE. Since the cable passes over a small pulley at B,
the tension may be assumed to be the same in portions AB and BE
of the cable. For the case when a = 3 in., determine (a) the tension
in the cable, (b) the reaction at D.
SOLUTION
Free-Body Diagram:
(a)
ΣM D = 0: (90 lb)(6 in.) −
5
12
T (6 in.) − T (7 in.) + T (6 in.) = 0
13
13
T = 195 lb
(b)
T = 195.0 lb 
ΣFx = 0: Dx − 195 lb −
5
(195 lb) + 90 = 0
13
Dx = +180 lb
ΣFy = 0: D y +
12
(195 lb) = 0
13
Dy = −180 lb
D = 255 lb
45.0&deg; 
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378
PROBLEM 4.31
Neglecting friction, determine the tension in cable ABD and the
reaction at support C.
SOLUTION
Free-Body Diagram:
ΣM C = 0: T (0.25 m) − T (0.1 m) − (120 N)(0.1 m) = 0
ΣFx = 0: C x − 80 N = 0
C x = +80 N
ΣFy = 0: C y − 120 N + 80 N = 0
C y = +40 N
T = 80.0 N 
C x = 80.0 N
C y = 40.0 N
C = 89.4 N
26.6&deg; 
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379
PROBLEM 4.32
Neglecting friction and the radius of the pulley, determine
(a) the tension in cable ADB, (b) the reaction at C.
SOLUTION
Free-Body Diagram:
Dimensions in mm
Geometry:
Distance:
AD = (0.36) 2 + (0.150) 2 = 0.39 m
Distance:
BD = (0.2)2 + (0.15)2 = 0.25 m
Equilibrium for beam:
ΣM C = 0:
(a)
 0.15 
 0.15 
(120 N)(0.28 m) − 
T  (0.36 m) − 
T  (0.2 m) = 0
 0.39 
 0.25 
T = 130.000 N
or
T = 130.0 N 
 0.36 
 0.2 
ΣFx = 0: C x + 
 (130.000 N) +  0.25  (130.000 N) = 0
0.39




(b)
C x = − 224.00 N
 0.15 
 0.15 
(130.00 N) + 
ΣFy = 0: C y + 

 (130.00 N) − 120 N = 0
 0.39 
 0.25 
C y = − 8.0000 N
Thus,
C = C x2 + C y2 = (−224) 2 + (− 8) 2 = 224.14 N
and
θ = tan −1
Cy
Cx
= tan −1
8
= 2.0454&deg;
224
C = 224 N
2.05&deg; 
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380
PROBLEM 4.33
Rod ABC is bent in the shape of an arc of circle of radius R. Knowing the
θ = 30&deg;, determine the reaction (a) at B, (b) at C.
SOLUTION
Free-Body Diagram:
ΣM D = 0: C x ( R) − P( R) = 0
Cx = + P
ΣFx = 0: C x − B sin θ = 0
P − B sin θ = 0
B = P/sin θ
B=
P
sin θ
θ
ΣFy = 0: C y + B cos θ − P = 0
C y + ( P/sin θ ) cos θ − P = 0
1 

C y = P 1 −

tan θ 

For θ = 30&deg;,
(a)
(b)
B = P/sin 30&deg; = 2 P
Cx = + P
B = 2P
60.0&deg; 
Cx = P
C y = P(1 − 1/tan 30&deg;) = − 0.732/P
C y = 0.7321P
C = 1.239P
36.2&deg; 
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381
PROBLEM 4.34
Rod ABC is bent in the shape of an arc of circle of radius R. Knowing that θ = 60&deg;,
determine the reaction (a) at B, (b) at C.
SOLUTION
See the solution to Problem 4.33 for the free-body diagram and analysis leading to the following expressions:
Cx = + P
1 

C y = P 1 −
tan θ 

P
B=
sin θ
For θ = 60&deg;,
(a)
(b)
B = P/sin 60&deg; = 1.1547 P
Cx = + P
B = 1.155P
30.0&deg; 
Cx = P
C y = P(1 − 1/tan 60&deg;) = + 0.4226 P
C y = 0.4226 P
C = 1.086P
22.9&deg; 
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382
PROBLEM 4.35
A movable bracket is held at rest by a cable attached at C and by
frictionless rollers at A and B. For the loading shown, determine (a) the
tension in the cable, (b) the reactions at A and B.
SOLUTION
Free-Body Diagram:
(a)
ΣFy = 0: T − 600 N = 0
T = 600 N 
(b)
ΣFx = 0: B − A = 0
∴ B=A
Note that the forces shown form two couples.
ΣM = 0: (600 N)(600 mm) − A(90 mm) = 0
A = 4000 N
∴ B = 4000 N
A = 4.00 kN
; B = 4.00 kN

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383
PROBLEM 4.36
A light bar AB supports a 15-kg block at its midpoint C. Rollers at A
and B rest against frictionless surfaces, and a horizontal cable AD is
attached at A. Determine (a) the tension in cable AD, (b) the reactions
at A and B.
SOLUTION
Free-Body Diagram:
W = (15 kg)(9.81 m/s 2 )
= 147.150 N
(a)
ΣFx = 0: TAD − 105.107 N = 0
(b)
ΣFy = 0: A − W = 0
A − 147.150 N = 0
A = 147.2 N 
ΣM A = 0: B(350 mm) − (147.150 N) (250 mm) = 0
B = 105.107 N
B = 105.1 N

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384
PROBLEM 4.37
A light bar AD is suspended from a cable BE and supports a
50-lb block at C. The ends A and D of the bar are in contact
with frictionless vertical walls. Determine the tension in cable
BE and the reactions at A and D.
SOLUTION
Free-Body Diagram:
ΣFx = 0:
A= D
ΣFy = 0:
TBE = 50.0 lb 
We note that the forces shown form two couples.
ΣM = 0: A(8 in.) − (50 lb)(3 in.) = 0
A = 18.75 lb
A = 18.75 lb
D = 18.75 lb

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385
PROBLEM 4.38
A light rod AD is supported by frictionless pegs at B and C and
rests against a frictionless wall at A. A vertical 120-lb force is
applied at D. Determine the reactions at A, B, and C.
SOLUTION
Free-Body Diagram:
ΣFx = 0: A cos 30&deg; − (120 lb) cos 60&deg; = 0
A = 69.28 lb
A = 69.3 lb

ΣM B = 0: C (8 in.) − (120 lb)(16 in.) cos 30&deg;
+ (69.28 lb)(8 in.)sin 30&deg; = 0
C = 173.2 lb
C = 173.2 lb
60.0&deg; 
B = 34.6 lb
60.0&deg; 
ΣM C = 0: B(8 in.) − (120 lb)(8 in.) cos 30&deg;
+ (69.28 lb)(16 in.) sin 30&deg; = 0
B = 34.6 lb
Check:
ΣFy = 0: 173.2 − 34.6 − (69.28)sin 30&deg; − (120)sin 60&deg; = 0
0 = 0 (check) 
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386
PROBLEM 4.39
Bar AD is attached at A and C to collars that can move freely
on the rods shown. If the cord BE is vertical (α = 0), determine
the tension in the cord and the reactions at A and C.
SOLUTION
Free-Body Diagram:
ΣFy = 0: − T cos 30&deg; + (80 N) cos 30&deg; = 0
T = 80 N
T = 80.0 N 
ΣM C = 0: ( A sin 30&deg;)(0.4 m) − (80 N)(0.2 m) − (80 N)(0.2 m) = 0
A = + 160 N
A = 160.0 N
30.0&deg; 
C = 160.0 N
30.0&deg; 
ΣM A = 0: (80 N)(0.2 m) − (80 N)(0.6 m) + (C sin 30&deg;)(0.4 m) = 0
C = + 160 N
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387
PROBLEM 4.40
Solve Problem 4.39 if the cord BE is parallel to the rods (α = 30&deg;).
PROBLEM 4.39 Bar AD is attached at A and C to collars that
can move freely on the rods shown. If the cord BE is vertical
(α = 0), determine the tension in the cord and the reactions at A
and C.
SOLUTION
Free-Body Diagram:
ΣFy = 0: − T + (80 N) cos 30&deg; = 0
T = 69.282 N
T = 69.3 N 
ΣM C = 0: − (69.282 N) cos 30&deg;(0.2 m)
− (80 N)(0.2 m) + ( A sin 30&deg;)(0.4 m) = 0
A = + 140.000 N
A = 140.0 N
30.0&deg; 
C = 180.0 N
30.0&deg; 
ΣM A = 0: + (69.282 N) cos 30&deg;(0.2 m)
− (80 N)(0.6 m) + (C sin 30&deg;)(0.4 m) = 0
C = + 180.000 N
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388
PROBLEM 4.41
The T-shaped bracket shown is supported by a small wheel at E and pegs at C
and D. Neglecting the effect of friction, determine the reactions at C, D, and E
when θ = 30&deg;.
SOLUTION
Free-Body Diagram:
ΣFy = 0: E cos 30&deg; − 20 − 40 = 0
E=
60 lb
= 69.282 lb
cos 30&deg;
E = 69.3 lb
60.0&deg; 
ΣM D = 0: (20 lb)(4 in.) − (40 lb)(4 in.)
− C (3 in.) + E sin 30&deg;(3 in.) = 0
−80 − 3C + 69.282(0.5)(3) = 0
C = 7.9743 lb
C = 7.97 lb

D = 42.6 lb

ΣFx = 0: E sin 30&deg; + C − D = 0
(69.282 lb)(0.5) + 7.9743 lb − D = 0
D = 42.615 lb
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389
PROBLEM 4.42
The T-shaped bracket shown is supported by a small wheel at E and pegs at C
and D. Neglecting the effect of friction, determine (a) the smallest value of θ for
which the equilibrium of the bracket is maintained, (b) the corresponding reactions
at C, D, and E.
SOLUTION
Free-Body Diagram:
ΣFy = 0: E cos θ − 20 − 40 = 0
E=
60
cos θ
(1)
ΣM D = 0: (20 lb)(4 in.) − (40 lb)(4 in.) − C (3 in.)
 60

+
sin θ  3 in. = 0
 cos θ

1
C = (180 tan θ − 80)
3
(a)
For C = 0,
180 tan θ = 80
tan θ =
From Eq. (1):
E=
4
θ = 23.962&deg;
9
θ = 24.0&deg; 
60
= 65.659
cos 23.962&deg;
ΣFx = 0: −D + C + E sin θ = 0
D = (65.659) sin 23.962 = 26.666 lb
(b)
C = 0 D = 26.7 lb
E = 65.7 lb
66.0&deg; 
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390
PROBLEM 4.43
Beam AD carries the two 40-lb loads shown. The beam is held by a
fixed support at D and by the cable BE that is attached to the
counterweight W. Determine the reaction at D when (a) W = 100 lb,
(b) W = 90 lb.
SOLUTION
W = 100 lb
(a)
ΣFx = 0: Dx = 0
ΣFy = 0: D y − 40 lb − 40 lb + 100 lb = 0
Dy = −20.0 lb
or
D = 20.0 lb 
ΣM D = 0: M D − (100 lb)(5 ft) + (40 lb)(8 ft)
+ (40 lb)(4 ft) = 0
M D = 20.0 lb ⋅ ft
or M D = 20.0 lb ⋅ ft

W = 90 lb
(b)
ΣFx = 0: Dx = 0
ΣFy = 0: D y + 90 lb − 40 lb − 40 lb = 0
Dy = −10.00 lb
or
D = 10.00 lb 
ΣM D = 0: M D − (90 lb)(5 ft) + (40 lb)(8 ft)
+ (40 lb)(4 ft) = 0
M D = −30.0 lb ⋅ ft
or M D = 30.0 lb ⋅ ft

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391
PROBLEM 4.44
For the beam and loading shown, determine the range of values of W for
which the magnitude of the couple at D does not exceed 40 lb ⋅ ft.
SOLUTION
For Wmin ,
M D = − 40 lb ⋅ ft
ΣM D = 0: (40 lb)(8 ft) − Wmin (5 ft)
+ (40 lb)(4 ft) − 40 lb ⋅ ft = 0
Wmin = 88.0 lb
For Wmax ,
M D = 40 lb ⋅ ft
ΣM D = 0: (40 lb)(8 ft) − Wmax (5 ft)
+ (40 lb)(4 ft) + 40 lb ⋅ ft = 0
Wmax = 104.0 lb
or 88.0 lb ≤ W ≤ 104.0 lb 
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392
PROBLEM 4.45
An 8-kg mass can be supported in the three different ways shown. Knowing that the pulleys have a 100-mm
radius, determine the reaction at A in each case.
SOLUTION

W = mg = (8 kg)(9.81 m/s2 ) = 78.480 N

(a)
ΣFx = 0: Ax = 0
ΣFy = 0: Ay − W = 0

A y = 78.480 N
ΣM A = 0: M A − W (1.6 m) = 0

M A = + (78.480 N)(1.6 m)

M A = 125.568 N ⋅ m
A = 78.5 N


(b)
M A = 125.6 N ⋅ m
ΣFx = 0: Ax − W = 0
A x = 78.480
ΣFy = 0: Ay − W = 0
A y = 78.480
A = (78.480 N) 2 = 110.987 N

45&deg;
ΣM A = 0: M A − W (1.6 m) = 0

M A = + (78.480 N)(1.6 m)


A = 111.0 N


(c)
M A = 125.568 N ⋅ m
M A = 125.6 N ⋅ m
45&deg;

ΣFx = 0: Ax = 0
ΣFy = 0: Ay − 2W = 0


Ay = 2W = 2(78.480 N) = 156.960 N
ΣM A = 0: M A − 2W (1.6 m) = 0
M A = + 2(78.480 N)(1.6 m)
A = 157.0 N
M A = 251.14 N ⋅ m
M A = 251 N ⋅ m

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393
PROBLEM 4.46
A tension of 20 N is maintained in a tape as it passes through the
support system shown. Knowing that the radius of each pulley is
10 mm, determine the reaction at C.
SOLUTION
Free-Body Diagram:
ΣFx = 0: C x + (20 N) = 0
C x = −20 N
ΣFy = 0: C y − (20 N) = 0
C y = +20 N
C = 28.3 N
45.0&deg; 
ΣM C = 0: M C + (20 N)(0.160 m) + (20 N) (0.055 m) = 0
M C = −4.30 N ⋅ m
M C = 4.30 N ⋅ m

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394
PROBLEM 4.47
Solve Problem 4.46, assuming that 15-mm-radius pulleys are used.
PROBLEM 4.46 A tension of 20 N is maintained in a tape as it
passes through the support system shown. Knowing that the radius
of each pulley is 10 mm, determine the reaction at C.
SOLUTION
Free-Body Diagram:
ΣFx = 0: C x + (20 N) = 0
C x = −20 N
ΣFy = 0: C y − (20 N) = 0
C y = +20 N
C = 28.3 N
45.0&deg; 
ΣM C = 0: M C + (20 N) (0.165 m) + (20 N) (0.060 m) = 0
M C = −4.50 N ⋅ m
M C = 4.50 N ⋅ m

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395
PROBLEM 4.48
The rig shown consists of a 1200-lb horizontal member ABC and
a vertical member DBE welded together at B. The rig is being used
to raise a 3600-lb crate at a distance x = 12 ft from the vertical
member DBE. If the tension in the cable is 4 kips, determine the
reaction at E, assuming that the cable is (a) anchored at F as
shown in the figure, (b) attached to the vertical member at a point
located 1 ft above E.
SOLUTION
Free-Body Diagram:
M E = 0: M E + (3600 lb) x + (1200 lb) (6.5 ft) − T (3.75 ft) = 0
M E = 3.75T − 3600 x − 7800
(a)
(1)
For x = 12 ft and T = 4000 lbs,
M E = 3.75(4000) − 3600(12) − 7800
= 36, 000 lb ⋅ ft
ΣFx = 0 ∴ Ex = 0
ΣFy = 0:
E y − 3600 lb − 1200 lb − 4000 = 0
E y = 8800 lb
E = 8.80 kips ; M E = 36.0 kip ⋅ ft




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396
PROBLEM 4.48 (Continued)
ΣM E = 0: M E + (3600 lb)(12 ft) + (1200 lb)(6.5 ft) = 0
(b)



M E = −51, 000 lb ⋅ ft
ΣFx = 0 ∴ Ex = 0
ΣFy = 0:
E y − 3600 lb − 1200 lb = 0
E y = 4800 lb
E = 4.80 kips ; M E = 51.0 kip ⋅ ft

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397
PROBLEM 4.49
For the rig and crate of Prob. 4.48, and assuming that cable is
anchored at F as shown, determine (a) the required tension in cable
ADCF if the maximum value of the couple at E as x varies from 1.5
to 17.5 ft is to be as small as possible, (b) the corresponding
maximum value of the couple.
SOLUTION
Free-Body Diagram:
M E = 0: M E + (3600 lb) x + (1200 lb)(6.5 ft) − T (3.75 ft) = 0
M E = 3.75T − 3600 x − 7800
(1)
For x = 1.5 ft, Eq. (1) becomes
( M E )1 = 3.75T − 3600(1.5) − 7800
(2)
For x = 17.5 ft, Eq. (1) becomes
( M E ) 2 = 3.75T − 3600(17.5) − 7800
(a)
For smallest max value of |M E |, we set
( M E )1 = − ( M E )2
3.75T − 13, 200 = −3.75T + 70,800
(b)
T = 11.20 kips 
From Equation (2), then
M E = 3.75(11.20) − 13.20
|M E | = 28.8 kip ⋅ ft 
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398
PROBLEM 4.50
A 6-m telephone pole weighing 1600 N is used to support the ends of two
wires. The wires form the angles shown with the horizontal axis and
the tensions in the wires are, respectively, T1 = 600 N and T2 = 375 N.
Determine the reaction at the fixed end A.
SOLUTION
Free-Body Diagram:
ΣFx = 0: Ax + (375 N) cos 20&deg; − (600 N) cos10&deg; = 0
Ax = +238.50 N
ΣFy = 0: Ay − 1600 N − (600 N)sin10&deg; − (375 N) sin 20&deg; = 0
Ay = +1832.45 N
A = 238.502 + 1832.452
1832.45
θ = tan −1
238.50
A = 1848 N
82.6&deg; 
ΣM A = 0: M A + (600 N) cos10&deg;(6 m) − (375 N) cos 20&deg;(6 m) = 0
M A = −1431.00 N ⋅ m
M A = 1431 N ⋅ m

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399
PROBLEM 4.51
A vertical load P is applied at end B of rod BC. (a) Neglecting
the weight of the rod, express the angle θ corresponding to the
equilibrium position in terms of P, l, and the counterweight W.
(b) Determine the value of θ corresponding to equilibrium if
P = 2W.
SOLUTION
Free-Body Diagram:
(a)
Triangle ABC is isosceles. We have
θ 
θ 
CD = ( BC ) cos   = l cos  
2
2
θ

ΣM C = 0: P(l cos θ ) − W  l cos  = 0
2

Setting cos θ = 2 cos 2
θ
2
− 1:
θ
θ


Pl  2 cos 2 − 1 − Wl cos = 0
2 
2

cos 2
θ
θ 1
W 
−
 cos − = 0
2  2P 
2 2
cos
θ
2
=

1 W
W2
 &plusmn;

8
+

4 P
P2


1 W
θ = 2cos −1  
4 P
 
&plusmn;

W2
+ 8   
2

P

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400
PROBLEM 4.51 (Continued)
(b)
For P = 2W ,
cos
θ
cos
θ
2
2
θ
2
=
 1
11
1
+ 8  = 1 &plusmn; 33
 &plusmn;
 8
42
4

(
= 0.84307 and cos
= 32.534&deg;
θ = 65.1&deg;
θ
2
θ
2
)
= −0.59307
= 126.375&deg;
θ = 65.1&deg; 
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401
PROBLEM 4.52
A vertical load P is applied at end B of rod BC. (a) Neglecting the weight of
the rod, express the angle θ corresponding to the equilibrium position in
terms of P, l, and the counterweight W. (b) Determine the value of θ
corresponding to equilibrium if P = 2W.
SOLUTION
(a)
Triangle ABC is isosceles. We have
CD = ( BC ) cos
θ
2
= l cos
θ
2
θ

ΣM C = 0: W  l cos  − P(l sin θ ) = 0
2

Setting sin θ = 2sin
θ
θ
θ
θ
W − 2 P sin
(b)
For P = 2W ,
sin
θ
2
θ
2
or
θ
cos : Wl cos − 2 Pl sin cos = 0
2
2
2
2
2
θ
2
=
θ
2
=0
θ = 2sin −1 
W 
 
 2P 
W
W
=
= 0.25
2 P 4W
θ = 29.0&deg; 
= 14.5&deg;
= 165.5&deg; θ = 331&deg; (discard)
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402
PROBLEM 4.53
A slender rod AB, of weight W, is attached to blocks A and B,
which move freely in the guides shown. The blocks are connected
by an elastic cord that passes over a pulley at C. (a) Express the
tension in the cord in terms of W and θ. (b) Determine the value of
θ for which the tension in the cord is equal to 3W.
SOLUTION
(a)
From F.B.D. of rod AB:
 1 

ΣM C = 0: T (l sin θ ) + W   cos θ  − T (l cos θ ) = 0
 2 

T=
W cos θ
2(cosθ − sin θ )
Dividing both numerator and denominator by cos θ,
T=
(b)
For T = 3W ,
or
3W =
W
2
1


 1 − tan θ 


or T =
( W2 )
(1 − tan θ )

( W2 )
(1 − tan θ )
1
1 − tan θ =
6
5
 
θ = tan −1   = 39.806&deg;
6
or
θ = 39.8&deg; 
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403
PROBLEM 4.54
Rod AB is acted upon by a couple M and two forces, each of magnitude P.
(a) Derive an equation in θ, P, M, and l that must be satisfied when the rod
is in equilibrium. (b) Determine the value of θ corresponding to equilibrium
when M = 150 N &middot; m, P = 200 N, and l = 600 mm.
SOLUTION
Free-Body Diagram:
(a)
From free-body diagram of rod AB:
ΣM C = 0: P(l cos θ ) + P(l sin θ ) − M = 0
or sinθ + cosθ =
(b)
M

Pl
For M = 150 lb ⋅ in., P = 20 lb, and l = 6 in.,
sin θ + cos θ =
150 lb ⋅ in.
5
= = 1.25
(20 lb)(6 in.) 4
sin 2 θ + cos 2 θ = 1
Using identity
sin θ + (1 − sin 2 θ )1/2 = 1.25
(1 − sin 2 θ )1/2 = 1.25 − sin θ
1 − sin 2 θ = 1.5625 − 2.5sin θ + sin 2 θ
2sin 2 θ − 2.5sin θ + 0.5625 = 0
sin θ =
=
or
−( −2.5) &plusmn; (625) − 4(2)(0.5625)
2(2)
2.5 &plusmn; 1.75
4
sin θ = 0.95572 and sin θ = 0.29428
θ = 72.886&deg; and θ = 17.1144&deg;
or θ = 17.11&deg; and θ = 72.9&deg; 
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404
PROBLEM 4.55
Solve Sample Problem 4.5, assuming that the spring is unstretched
when θ = 90&deg;.
SOLUTION
First note:
T = tension in spring = ks
where
s = deformation of spring
= rβ
F = kr β
From F.B.D. of assembly:
or
ΣM 0 = 0: W (l cos β ) − F (r ) = 0
Wl cos β − kr 2 β = 0
cos β =
For
kr 2
β
Wl
k = 250 lb/in.
r = 3 in.
l = 8 in.
W = 400 lb
cos β =
or
(250 lb/in.)(3 in.)2
β
(400 lb)(8 in.)
cos β = 0.703125β
Solving numerically,
or
β = 51.134&deg;
Then
θ = 90&deg; + 51.134&deg; = 141.134&deg;
or θ = 141.1&deg; 
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405
PROBLEM 4.56
A slender rod AB, of weight W, is attached to blocks A and B that
move freely in the guides shown. The constant of the spring is k,
and the spring is unstretched when θ = 0. (a) Neglecting the weight
of the blocks, derive an equation in W, k, l, and θ that must be
satisfied when the rod is in equilibrium. (b) Determine the value
of θ when W = 75 lb, l = 30 in., and k = 3 lb/in.
SOLUTION
Free-Body Diagram:
Spring force:
Fs = ks = k (l − l cos θ ) = kl (1 − cos θ )
l

ΣM D = 0: Fs (l sin θ ) − W  cos θ  = 0
2

(a)
kl (1 − cos θ )l sin θ −
kl (1 − cos θ ) tan θ −
(b)
For given values of
W
l cos θ = 0
2
W
=0
2
or (1 − cos θ ) tan θ =
W

2kl
W = 75 lb
l = 30 in.
k = 3 lb/in.
(1 − cos θ ) tan θ = tan θ − sin θ
75 lb
=
2(3 lb/in.)(30 in.)
= 0.41667
Solving numerically,
θ = 49.710&deg;
or
θ = 49.7&deg; 
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406
PROBLEM 4.57
A vertical load P is applied at end B of rod BC. The constant of the
spring is k, and the spring is unstretched when θ = 60&deg;. (a) Neglecting
the weight of the rod, express the angle θ corresponding to the
equilibrium position terms of P, k, and l. (b) Determine the value of θ
1
corresponding to equilibrium if P = 4 kl.
SOLUTION
Free-Body Diagram:
(a)
Triangle ABC is isosceles. We have
θ 
θ 
AB = 2( AD ) = 2l sin   ; CD = l cos  
2
2
Elongation of spring:
x = ( AB)θ − ( AB )θ = 60&deg;
θ 
= 2l sin   − 2l sin 30&deg;
2
 θ 1
T = k x = 2kl  sin − 
2 2

θ

ΣM C = 0: T  l cos  − P(l sin θ ) = 0
2

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407
PROBLEM 4.57 (Continued)
θ
θ
θ
 θ 1

2kl  sin −  l cos − Pl  2sin cos  = 0
2 2
2
2
2


cos
θ
2
=0
2(kl − P ) sin
or
θ = 180&deg; (trivial)
sin
θ
2
θ
2
− kl = 0
=
1
2
kl
kl − P
1
2


θ = 2sin −1  kl /( kl − P)  
(b)
For P =
1
kl ,
4
sin
θ
2
θ
2
=
1
2
3
4
kl
kl
=
2
3
θ = 83.6&deg; 
= 41.8&deg;
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408
PROBLEM 4.58
A collar B of weight W can move freely along the vertical rod shown.
The constant of the spring is k, and the spring is unstretched when
θ = 0. (a) Derive an equation in θ, W, k, and l that must be satisfied
when the collar is in equilibrium. (b) Knowing that W = 300 N,
l = 500 mm, and k = 800 N/m, determine the value of θ corresponding
to equilibrium.
SOLUTION
First note:
T = ks
where
k = spring constant
s = elongation of spring
l
=
−l
cos θ
l
(1 − cos θ )
=
cos θ
kl
T=
(1 − cos θ )
cos θ
(a)
From F.B.D. of collar B:
or
(b)
For
ΣFy = 0: T sin θ − W = 0
kl
(1 − cos θ )sin θ − W = 0
cos θ
or tan θ − sin θ =
W

kl
W = 3 lb
l = 6 in.
k = 8 lb/ft
6 in.
l=
= 0.5 ft
12 in./ft
tan θ − sin θ =
Solving numerically,
3 lb
= 0.75
(8 lb/ft)(0.5 ft)
θ = 57.957&deg;
or
θ = 58.0&deg; 
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409
PROBLEM 4.59
Eight identical 500 &times; 750-mm rectangular plates, each of mass m = 40 kg, are held in a vertical plane as
shown. All connections consist of frictionless pins, rollers, or short links. In each case, determine whether
(a) the plate is completely, partially, or improperly constrained, (b) the reactions are statically determinate or
indeterminate, (c) the equilibrium of the plate is maintained in the position shown. Also, wherever possible,
compute the reactions.
SOLUTION
1.
Three non-concurrent, non-parallel reactions:
(a)
Plate: completely constrained
(b)
Reactions: determinate
(c)
Equilibrium maintained
A = C = 196.2 N
2.
Three non-concurrent, non-parallel reactions:
(a)
Plate: completely constrained
(b)
Reactions: determinate
(c)
Equilibrium maintained
B = 0, C = D = 196.2 N
3.
Four non-concurrent, non-parallel reactions:
(a)
Plate: completely constrained
(b)
Reactions: indeterminate
(c)
Equilibrium maintained
A x = 294 N
,
D x = 294 N
( A y + D y = 392 N )
4.
Three concurrent reactions (through D):
(a)
Plate: improperly constrained
(b)
Reactions: indeterminate
(c)
No equilibrium
(ΣM D ≠ 0)
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410
PROBLEM 4.59 (Continued)
5.
Two reactions:
(a)
Plate: partial constraint
(b)
Reactions: determinate
(c)
Equilibrium maintained
C = D = 196.2 N
6.
Three non-concurrent, non-parallel reactions:
(a)
Plate: completely constrained
(b)
Reactions: determinate
(c)
Equilibrium maintained
B = 294 N
7.
8.
, D = 491 N
53.1&deg;
Two reactions:
(a)
Plate: improperly constrained
(b)
Reactions determined by dynamics
(c)
No equilibrium
(ΣFy ≠ 0)
Four non-concurrent, non-parallel reactions:
(a)
Plate: completely constrained
(b)
Reactions: indeterminate
(c)
Equilibrium maintained
B = D y = 196.2 N
(C + D x = 0)
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411
PROBLEM 4.60
The bracket ABC can be supported in the eight different ways shown. All connections consist of smooth pins,
rollers, or short links. For each case, answer the questions listed in Problem 4.59, and, wherever possible,
compute the reactions, assuming that the magnitude of the force P is 100 lb.
SOLUTION
1.
Three non-concurrent, non-parallel reactions:
(a)
Bracket: complete constraint
(b)
Reactions: determinate
(c)
Equilibrium maintained
A = 120.2 lb
2.
3.
56.3&deg;, B = 66.7 lb
Four concurrent, reactions (through A):
(a)
Bracket: improper constraint
(b)
Reactions: indeterminate
(c)
No equilibrium
(ΣM A ≠ 0)
Two reactions:
(a)
Bracket: partial constraint
(b)
Reactions: indeterminate
(c)
Equilibrium maintained
A = 50 lb , C = 50 lb
4.
Three non-concurrent, non-parallel reactions:
(a)
Bracket: complete constraint
(b)
Reactions: determinate
(c)
Equilibrium maintained
A = 50 lb , B = 83.3 lb
36.9&deg;, C = 66.7 lb
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412
PROBLEM 4.60 (Continued)
5.
6.
Four non-concurrent, non-parallel reactions:
(a)
Bracket: complete constraint
(b)
Reactions: indeterminate
(c)
Equilibrium maintained
(ΣM C = 0) A y = 50 lb
Four non-concurrent, non-parallel reactions:
(a)
Bracket: complete constraint
(b)
Reactions: indeterminate
(c)
Equilibrium maintained
A x = 66.7 lb
B x = 66.7 lb
( A y + B y = 100 lb )
7.
Three non-concurrent, non-parallel reactions:
(a)
Bracket: complete constraint
(b)
Reactions: determinate
(c)
Equilibrium maintained
A = C = 50 lb
8.
Three concurrent, reactions (through A)
(a)
Bracket: improper constraint
(b)
Reactions: indeterminate
(c)
No equilibrium
(ΣM A ≠ 0)
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413
PROBLEM 4.61
Determine the reactions at A and B when a = 150 mm.
SOLUTION
Free-Body Diagram:
Force triangle
80 mm 80 mm
=
a
150 mm
β = 28.072&deg;
tan β =
A=
320 N
sin 28.072&deg;
B=
320 N
tan 28.072&deg;
A = 680 N
28.1&deg; 
B = 600 N


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414
PROBLEM 4.62
Determine the value of a for which the magnitude of the reaction at B
is equal to 800 N.
SOLUTION
Free-Body Diagram:
Force triangle
tan β =
80 mm
a
a=
80 mm
tan β
(1)
From force triangle:
tan β =
From Eq. (1):
a=
320 N
= 0.4
800 N
80 mm
0.4
a = 200 mm 
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415
PROBLEM 4.63
Using the method of Sec. 4.7, solve Problem 4.22b.
PROBLEM 4.22 Determine the reactions at A and B when (a) α = 0,
(b) α = 90&deg;, (c) α = 30&deg;.
SOLUTION
Free-Body Diagram:
(Three-force body)
The line of action at A must pass through C, where B and the 75-lb load intersect.
In triangle ACE:
Force triangle
tan θ =
10 in.
12 in.
θ = 39.806&deg;
B = (75 lb) tan 39.806&deg;
= 62.5 lb
75 lb
A=
= 97.6&deg;
cos 39.806&deg;
A = 97.6 lb
50.2&deg;; B = 62.5 lb


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416
PROBLEM 4.64
A 500-lb cylindrical tank, 8 ft in diameter, is to be raised over a 2-ft
obstruction. A cable is wrapped around the tank and pulled horizontally as
shown. Knowing that the corner of the obstruction at A is rough, find the
required tension in the cable and the reaction at A.
SOLUTION
Free-Body Diagram:
Force triangle
cos α =
GD 2 ft
=
= 0.5
AG 4 ft
α = 60&deg;
1
( β = 60&deg;)
2
T = (500 lb) tan 30&deg;
T = 289 lb
θ = α = 30&deg;
A=
500 lb
cos 30&deg;
A = 577 lb
60.0&deg; 
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417
PROBLEM 4.65
For the frame and loading shown, determine the reactions at A and C.
SOLUTION
Since member AB is acted upon by two forces, A and B, they must be colinear, have the same magnitude, and
be opposite in direction for AB to be in equilibrium. The force B acting at B of member BCD will be equal in
magnitude but opposite in direction to force B acting on member AB. Member BCD is a three-force body with
member forces intersecting at E. The F.B.D.’s of members AB and BCD illustrate the above conditions. The
force triangle for member BCD is also shown. The angle β is found from the member dimensions:
 6 in. 
 = 30.964&deg;
 10 in. 
β = tan −1 
Applying the law of sines to the force triangle for member BCD,
30 lb
B
C
=
=
sin(45&deg; − β ) sin β sin135&deg;
or
30 lb
B
C
=
=
sin14.036&deg; sin 30.964&deg; sin135&deg;
A= B=
(30 lb)sin 30.964&deg;
= 63.641 lb
sin14.036&deg;
or
and
C=
A = 63.6 lb
45.0&deg; 
C = 87.5 lb
59.0&deg; 
(30 lb) sin135&deg;
= 87.466 lb
sin14.036&deg;
or
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418
PROBLEM 4.66
For the frame and loading shown, determine the reactions at C and D.
SOLUTION
Since BD is a two-force member, the reaction at D must pass through Points B and D.
Free-Body Diagram:
(Three-force body)
Reaction at C must pass through E, where the reaction at D and the 150-lb load intersect.
Triangle CEF:
tan β =
4.5 ft
3 ft
β = 56.310&deg;
Triangle ABE:
tan γ =
1
2
γ = 26.565&deg;
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419
PROBLEM 4.66 (Continued)
Force Triangle
Law of sines:
150 lb
C
D
=
=
sin 29.745&deg; sin116.565&deg; sin 33.690&deg;
C = 270.42 lb,
D = 167.704 lb
C = 270 lb
56.3&deg;; D = 167.7 lb
26.6&deg; 
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420
PROBLEM 4.67
Determine the reactions at B and D when b = 60 mm.
SOLUTION
Since CD is a two-force member, the line of action of reaction at D must pass through Points C and D.
Free-Body Diagram:
(Three-force body)
Reaction at B must pass through E, where the reaction at D and the 80-N force intersect.
220 mm
250 mm
β = 41.348&deg;
tan β =
Force triangle
Law of sines:
80 N
B
D
=
=
sin 3.652&deg; sin 45&deg; sin131.348&deg;
B = 888.0 N
D = 942.8 N

B = 888 N
41.3&deg;
D = 943 N
45.0&deg; 
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421
PROBLEM 4.68
Determine the reactions at B and D when b = 120 mm.
SOLUTION
Since CD is a two-force member, line of action of reaction at D must pass through C and D
.
Free-Body Diagram:
(Three-force body)
Reaction at B must pass through E, where the reaction at D and the 80-N force intersect.
280 mm
250 mm
β = 48.24&deg;
tan β =
Force triangle
Law of sines:
80 N
B
D
=
=
sin 3.24&deg; sin135&deg; sin 41.76&deg;
B = 1000.9 N
D = 942.8 N 
B = 1001 N
48.2&deg; D = 943 N
45.0&deg; 
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422
PROBLEM 4.69
A T-shaped bracket supports a 300-N load as shown. Determine
the reactions at A and C when α = 45&deg;.
SOLUTION
Free-Body Diagram:
(Three-force body)
The line of action of C must pass through E, where A and the 300-N force intersect.
Triangle ABE is isosceles:
EA = AB = 400 mm
In triangle CEF:
tan θ =
CF
CF
150 mm
=
=
EF EA + AF 700 mm
θ = 12.0948&deg;
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423
PROBLEM 4.69 (Continued)
Force Triangle
Law of sines:
A
C
300 N
=
=
sin 32.905&deg; sin135&deg; sin12.0948&deg;
A = 778 N ;
C = 1012 N
77.9&deg; 
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424
PROBLEM 4.70
A T-shaped bracket supports a 300-N load as shown. Determine
the reactions at A and C when α = 60&deg;.
SOLUTION
Free-Body Diagram:
EA = (400 mm) tan 30&deg;
= 230.94 mm
In triangle CEF:
tan θ =
CF
CF
=
EF EA + AF
150
230.94 + 300
θ = 15.7759&deg;
tan θ =
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425
PROBLEM 4.70 (Continued)
Force Triangle
Law of sines:
A
C
300 N
=
=
sin 44.224&deg; sin120&deg; sin15.7759&deg;
A = 770 N
C = 956 N
A = 770 N ;
C = 956 N
74.2&deg; 
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426
PROBLEM 4.71
A 40-lb roller, of diameter 8 in., which is to be used on a tile floor, is resting
directly on the subflooring as shown. Knowing that the thickness of each
tile is 0.3 in., determine the force P required to move the roller onto the tiles
if the roller is (a) pushed to the left, (b) pulled to the right.
SOLUTION
Geometry: For each case as roller comes into contact with tile,
3.7 in.
4 in.
α = 22.332&deg;
α = cos −1

(a)

Roller pushed to left (three-force body):
Forces must pass through O.
Law of sines:
Force Triangle
40 lb
P
=
; P = 24.87 lb
sin 37.668&deg; sin 22.332&deg;
P = 24.9 lb
(b)
30.0&deg; 
Roller pulled to right (three-force body):
Forces must pass through O.
Law of sines:
40 lb
P
=
; P = 15.3361 lb
sin 97.668&deg; sin 22.332&deg;
P = 15.34 lb
30.0&deg; 



Force Triangle


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427
PROBLEM 4.72
One end of rod AB rests in the corner A and the other end is attached to
cord BD. If the rod supports a 40-lb load at its midpoint C, find the reaction
at A and the tension in the cord.
SOLUTION
Free-Body Diagram: (Three-force body)
The line of action of reaction at A must pass through E, where T and the 40-lb load intersect.
Force triangle
EF 23
=
AF 12
α = 62.447&deg;
5
EH
tan β =
=
DH 12
β = 22.620&deg;
tan α =
A
T
40 lb
=
=
sin 67.380&deg; sin 27.553&deg; sin 85.067&deg;
A = 37.1 lb
62.4&deg; 
T = 18.57 lb 
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428
PROBLEM 4.73
A 50-kg crate is attached to the trolley-beam system shown.
Knowing that a = 1.5 m, determine (a) the tension in cable CD,
(b) the reaction at B.
SOLUTION
Three-force body: W and TCD intersect at E.
0.7497 m
1.5 m
β = 26.556&deg;
tan β =
Three forces intersect at E.
W = (50 kg) 9.81 m/s 2
= 490.50 N
Law of sines:
Force triangle
TCD
490.50 N
B
=
=
sin 61.556&deg; sin 63.444&deg; sin 55&deg;
TCD = 498.99 N
B = 456.96 N
TCD = 499 N 
(a)
B = 457 N
(b)
26.6&deg; 
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429
PROBLEM 4.74
Solve Problem 4.73, assuming that a = 3 m.
PROBLEM 4.73 A 50-kg crate is attached to the trolley-beam
system shown. Knowing that a = 1.5 m, determine (a) the tension in
cable CD, (b) the reaction at B.
SOLUTION
W and TCD intersect at E.
Free-Body Diagram:
Three-Force Body
AE 0.301 m
=
AB
3m
β = 5.7295&deg;
tan β =
Three forces intersect at E.
Force Triangle
W = (50 kg) 9.81 m/s 2
= 490.50 N
Law of sines:
TCD
490.50 N
B
=
=
sin 29.271&deg; sin 95.730&deg; sin 55&deg;
TCD = 998.18 N
B = 821.76 N
TCD = 998 N 
(a)
B = 822 N
(b)
5.73&deg; 
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430
PROBLEM 4.75
Determine the reactions at A and B when β = 50&deg;.
SOLUTION
Free-Body Diagram: (Three-force body)
Reaction A must pass through Point D where the 100-N force
and B intersect.
In right Δ BCD:
α = 90&deg; − 75&deg; = 15&deg;
BD = 250 tan 75&deg; = 933.01 mm
In right Δ ABD:
Dimensions in mm
AB
150 mm
=
BD 933.01 mm
γ = 9.1333&deg;
tan γ =
Force Triangle
Law of sines:
100 N
A
B
=
=
sin 9.1333&deg; sin15&deg; sin155.867&deg;
A = 163.1 N; B = 257.6 N
A = 163.1 N
74.1&deg; B = 258 N
65.0&deg; 
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431
PROBLEM 4.76
Determine the reactions at A and B when β = 80&deg;.
SOLUTION
Free-Body Diagram:
(Three-force body)
Reaction A must pass through D where the 100-N force and B intersect.
In right triangle BCD:
α = 90&deg; − 75&deg; = 15&deg;
BD = BC tan 75&deg; = 250 tan75&deg;
BD = 933.01 mm
In right triangle ABD:
tan γ =
AB
150 mm
=
BD 933.01 mm
γ = 9.1333&deg;
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432
PROBLEM 4.76 (Continued)
Force Triangle
Law of sines:
100 N
A
B
=
=
sin 9.1333&deg; sin15&deg; sin155.867&deg;

A = 163.1 N
55.9&deg; 

B = 258 N
65.0&deg; 
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433
PROBLEM 4.77
Knowing that θ = 30&deg;, determine the reaction (a) at B, (b) at C.
SOLUTION
Free-Body Diagram:
(Three-force body)
Reaction at C must pass through D where force P and reaction at B intersect.
In Δ CDE:
( 3 − 1) R
R
= 3 −1
β = 36.2&deg;
tan β =
Force Triangle
Law of sines:
P
B
C
=
=
sin 23.8&deg; sin126.2&deg; sin 30&deg;
B = 2.00 P
C = 1.239 P
(a)
B = 2P
(b)
C = 1.239P
60.0&deg; 
36.2&deg; 
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434
PROBLEM 4.78
Knowing that θ = 60&deg;, determine the reaction (a) at B, (b) at C.
SOLUTION
Reaction at C must pass through D where force P and reaction at B intersect.
In ΔCDE:
tan β =
R−
=1−
R
3
Free-Body Diagram:
(Three-force body)
R
1
3
β = 22.9&deg;
Force Triangle
Law of sines:
P
B
C
=
=
sin 52.9&deg; sin 67.1&deg; sin 60&deg;
B = 1.155P
C = 1.086 P
(a)
B = 1.155P
30.0&deg; 
(b)
C = 1.086P
22.9&deg; 
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435
PROBLEM 4.79
Using the method of Section 4.7, solve Problem 4.23.
PROBLEM 4.23 Determine the reactions at A and B when
(a) h = 0, (b) h = 200 mm.
SOLUTION


Free-Body Diagram:
(a)
h=0
Reaction A must pass through C where the 150-N weight
and B interect.
Force triangle is equilateral.
(b)
A = 150.0 N
30.0&deg; 
B = 150.0 N
30.0&deg; 
h = 200 mm

55.662
250
β = 12.5521&deg;
tan β =

Law of sines:
A
B
150 N
=
=
sin17.4480&deg; sin 60&deg; sin102.552&deg;
A = 433.24 N
B = 488.31 N


Force Triangle
A = 433 N
B = 488 N
12.55&deg; 
30.0&deg; 

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436
PROBLEM 4.80
Using the method of Section 4.7, solve Problem 4.24.
PROBLEM 4.24 A lever AB is hinged at C and attached to a control
cable at A. If the lever is subjected to a 75-lb vertical force at B,
determine (a) the tension in the cable, (b) the reaction at C.
SOLUTION

Reaction at C must pass through E, where the 75-lb force
and T intersect.


9.3969 in.
8.5798 in.
α = 47.602&deg;
tan α =


14.0954 in.
24.870 in.
β = 29.543&deg;

tan β =


Force Triangle







Law of sines:



75 lb
T
C
=
=
sin18.0590&deg; sin 29.543&deg; sin132.398&deg;
Free-Body Diagram:
 Dimensions in in.
(a)

(b)
T = 119.3 lb 
C = 178.7 lb
60.5&deg; 
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437
PROBLEM 4.81
Member ABC is supported by a pin and bracket at B and by an
inextensible cord attached at A and C and passing over a frictionless
pulley at D. The tension may be assumed to be the same in portions AD
and CD of the cord. For the loading shown and neglecting the size of
the pulley, determine the tension in the cord and the reaction at B.
SOLUTION
Free-Body Diagram:
Reaction at B must pass through D.
7 in.
12 in.
α = 30.256&deg;
7 in.
tan β =
24 in.
β = 16.26&deg;
tan α =
Force Triangle
Law of sines:
T
T − 72 lb
B
=
=
sin 59.744&deg; sin13.996&deg; sin106.26
T (sin13.996&deg;) = (T − 72 lb)(sin 59.744&deg;)
T (0.24185) = (T − 72)(0.86378)
T = 100.00 lb
sin 106.26&deg;
sin 59.744&deg;
= 111.14 lb
T = 100.0 lb 
B = (100 lb)
B = 111.1 lb
30.3&deg; 
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438
PROBLEM 4.82
Member ABC is supported by a pin and bracket at B and by an
inextensible cord attached at A and C and passing over a frictionless
pulley at D. The tension may be assumed to be the same in portions
of the pulley, determine the tension in the cord and the reaction at B.
SOLUTION
Free-Body Diagram:
Force Triangle
Reaction at B must pass through D.
tan α =
120
; α = 36.9&deg;
160
T T − 75 N B
=
=
4
3
5
3T = 4T − 300; T = 300 N
5
5
B = T = (300 N) = 375 N
4
4
B = 375 N
36.9&deg;
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439
PROBLEM 4.83
A thin ring of mass 2 kg and radius r = 140 mm is held against a frictionless
wall by a 125-mm string AB. Determine (a) the distance d, (b) the tension in
the string, (c) the reaction at C.
SOLUTION
Free-Body Diagram:
(Three-force body)
The force T exerted at B must pass through the center G of the ring, since C and W intersect at that point.
Thus, points A, B, and G are in a straight line.
(a) From triangle ACG:
d = ( AG ) 2 − (CG )2
= (265 mm)2 − (140 mm) 2
= 225.00 mm
d = 225 mm 
Force Triangle
W = (2 kg)(9.81 m/s 2 ) = 19.6200 N
Law of sines:
T
C
19.6200 N
=
=
265 mm 140 mm 225.00 mm
T = 23.1 N 
(b)
C = 12.21 N
(c)

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440
PROBLEM 4.84
A uniform rod AB of length 2R rests inside a hemispherical bowl of radius
R as shown. Neglecting friction, determine the angle θ corresponding to
equilibrium.
SOLUTION
Based on the F.B.D., the uniform rod AB is a three-force body. Point E is the point of intersection of the three
forces. Since force A passes through O, the center of the circle, and since force C is perpendicular to the rod,
triangle ACE is a right triangle inscribed in the circle. Thus, E is a point on the circle.
Note that the angle α of triangle DOA is the central angle corresponding to the inscribed angle θ of
triangle DCA.
α = 2θ
The horizontal projections of AE , ( x AE ), and AG , ( x AG ), are equal.
x AE = x AG = x A
or
( AE ) cos 2θ = ( AG ) cos θ
and
(2 R) cos 2θ = R cos θ
Now
then
or
cos 2θ = 2 cos 2 θ − 1
4 cos 2 θ − 2 = cos θ
4 cos 2 θ − cos θ − 2 = 0
cos θ = 0.84307 and cos θ = − 0.59307
θ = 32.534&deg; and θ = 126.375&deg; (Discard)
or θ = 32.5&deg; 
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441
PROBLEM 4.85
A slender rod BC of length L and weight W is held by two cables as
shown. Knowing that cable AB is horizontal and that the rod forms an
angle of 40&deg; with the horizontal, determine (a) the angle θ that cable
CD forms with the horizontal, (b) the tension in each cable.
SOLUTION
Free-Body Diagram:
(Three-force body)
(a)
The line of action of TCD must pass through E, where TAB and W intersect.
CF
EF
L sin 40&deg;
= 1
L cos 40&deg;
2
tan θ =
= 2 tan 40&deg;
= 59.210&deg;
θ = 59.2&deg; 
(b)
Force Triangle
TAB = W tan 30.790&deg;
= 0.59588W
TAB = 0.596W 

W
cos 30.790&deg; 
= 1.16408W
TCD =
TCD = 1.164W 

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442
PROBLEM 4.86
A slender rod of length L and weight W is attached to a collar at A and is
fitted with a small wheel at B. Knowing that the wheel rolls freely along
a cylindrical surface of radius R, and neglecting friction, derive an
equation in θ, L, and R that must be satisfied when the rod is in
equilibrium.
SOLUTION
Free-Body Diagram (Three-force body)
Reaction B must pass through D where B and W intersect.
Note that ΔABC and ΔBGD are similar.
AC = AE = L cos θ
In Δ ABC:
(CE ) 2 + ( BE )2 = ( BC )2
(2 L cos θ ) 2 + ( L sin θ )2 = R 2
2
R
2
2
  = 4cos θ + sin θ
L
2
R
2
2
  = 4cos θ + 1 − cos θ
L
2
R
2
  = 3cos θ + 1
L
2

1
cos 2 θ =  R  − 1 
3  L 

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443
PROBLEM 4.87
Knowing that for the rod of Problem 4.86, L = 15 in., R = 20 in., and
W = 10 lb, determine (a) the angle θ corresponding to equilibrium,
(b) the reactions at A and B.
SOLUTION
See the solution to Problem 4.86 for the free-body diagram and analysis leading to the following equation:
2

1
cos 2 θ =  R  − 1
3  L 

For L = 15 in., R = 20 in., and W = 10 lb,
2

1  20 in. 
cos θ = 
 − 1 ; θ = 59.39&deg;
3  15 in. 



2
(a)
In Δ ABC:
θ = 59.4&deg; 
BE
L sin θ
1
=
= tan θ
CE 2 L cos θ 2
1
tan α = tan 59.39&deg; = 0.8452
2
α = 40.2&deg;
tan α =
Force Triangle
A = W tan α = (10 lb) tan 40.2&deg; = 8.45 lb
W
(10 lb)
B=
=
= 13.09 lb
cos α cos 40.2&deg;
A = 8.45 lb
(b)
B = 13.09 lb

49.8&deg; 
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444
PROBLEM 4.88
Rod AB is bent into the shape of an arc of circle and is lodged between two
pegs D and E. It supports a load P at end B. Neglecting friction and the
weight of the rod, determine the distance c corresponding to equilibrium
when a = 20 mm and R = 100 mm.
SOLUTION
Free-Body Diagram:
Since yED = xED = a,
slope of ED is
45&deg;;
slope of HC is
45&deg;.
Also
DE = 2 a
and
a
1
DH = HE =   DE =
2
2
 
For triangles DHC and EHC,
sin β =
a
2
R
=
a
2R
Now
c = R sin(45&deg; − β )
For
a = 20 mm and
sin β =
R = 100 mm
20 mm
2(100 mm)
= 0.141421
β = 8.1301&deg;
and
c = (100 mm) sin(45&deg; − 8.1301&deg;)
= 60.00 mm
or c = 60.0 mm 
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445
PROBLEM 4.89
A slender rod of length L is attached to collars that can slide freely along
the guides shown. Knowing that the rod is in equilibrium, derive an
expression for the angle θ in terms of the angle β.
SOLUTION
As shown in the free-body diagram of the slender rod AB, the three forces intersect at C. From the force
geometry:
Free-Body Diagram:
tan β =
xGB
y AB
where
y AB = L cos θ
and
xGB =
tan β =
1
L sin θ
2
1
2
L sin θ
L cos θ
1
= tan θ
2
or tan θ = 2 tan β 
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446
PROBLEM 4.90
An 8-kg slender rod of length L is attached to collars that can slide freely
along the guides shown. Knowing that the rod is in equilibrium and that
β = 30&deg;, determine (a) the angle θ that the rod forms with the vertical,
(b) the reactions at A and B.
SOLUTION
(a)
As shown in the free-body diagram of the slender rod AB, the three forces intersect at C. From the
geometry of the forces:
Free-Body Diagram:
tan β =
xCB
yBC
where
xCB =
1
L sin θ
2
and
yBC = L cos θ
tan β =
1
tan θ
2
or
tan θ = 2 tan β
For
β = 30&deg;
tan θ = 2 tan 30&deg;
= 1.15470
θ = 49.107&deg;
or
W = mg = (8 kg)(9.81 m/s2 ) = 78.480 N
(b)
From force triangle:
A = W tan β
= (78.480 N) tan 30&deg;
= 45.310 N
and
θ = 49.1&deg; 
B=
or
W
78.480 N
=
= 90.621 N
cos β
cos 30&deg;
A = 45.3 N
or B = 90.6 N

60.0&deg; 
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447
PROBLEM 4.91
A 200-mm lever and a 240-mm-diameter pulley are
welded to the axle BE that is supported by bearings at C
and D. If a 720-N vertical load is applied at A when the
lever is horizontal, determine (a) the tension in the cord,
(b) the reactions at C and D. Assume that the bearing
at D does not exert any axial thrust.
SOLUTION
Dimensions in mm
We have six unknowns and six equations of equilibrium. —OK
ΣM C = 0: (−120k ) &times; ( Dx i + Dy j) + (120 j − 160k ) &times; T i + (80k − 200i ) &times; (−720 j) = 0
−120 Dx j + 120 D y i − 120T k − 160Tj + 57.6 &times; 103 i + 144 &times; 103 k = 0
Equating to zero the coefficients of the unit vectors:
k:
−120T + 144 &times; 103 = 0
i:
120 Dy + 57.6 &times; 103 = 0
j: − 120 Dx − 160(1200 N) = 0
(b)
ΣFx = 0:
C x + Dx + T = 0
ΣFy = 0:
C y + Dy − 720 = 0
ΣFz = 0:
Cz = 0
(a) T = 1200 N 
Dy = −480 N
Dx = −1600 N
C x = 1600 − 1200 = 400 N
C y = 480 + 720 = 1200 N
C = (400 N)i + (1200 N) j; D = −(1600 N)i − (480 N) j 
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448
PROBLEM 4.92
Solve Problem 4.91, assuming that the axle has been
rotated clockwise in its bearings by 30&deg; and that the
PROBLEM 4.91 A 200-mm lever and a 240-mmdiameter pulley are welded to the axle BE that is
supported by bearings at C and D. If a 720-N vertical
load is applied at A when the lever is horizontal,
determine (a) the tension in the cord, (b) the reactions at
C and D. Assume that the bearing at D does not exert
any axial thrust.
SOLUTION
Dimensions in mm
We have six unknowns and six equations of equilibrium.
ΣM C = 0: (−120k ) &times; ( Dx i + Dy j) + (120 j − 160k ) &times; T i + (80k − 173.21i ) &times; (−720 j) = 0
−120 Dx j + 120 D y i −120T k −160T j + 57.6 &times; 103 i + 124.71 &times; 103 k = 0
Equating to zero the coefficients of the unit vectors,
k : − 120T + 124.71 &times; 103 = 0
i:
T = 1039 N 
120 Dy + 57.6 &times; 103 = 0 Dy = −480 N
j: − 120 Dx − 160(1039.2)
(b)
T = 1039.2 N
ΣFx = 0:
C x + Dx + T = 0
ΣFy = 0:
C y + Dy − 720 = 0
ΣFz = 0:
Cz = 0
Dx = −1385.6 N
C x = 1385.6 − 1039.2 = 346.4
C y = 480 + 720 = 1200 N
C = (346 N)i + (1200 N) j D = −(1386 N)i − (480 N) j 
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449
PROBLEM 4.93
A 4 &times; 8-ft sheet of plywood weighing 40 lb has been temporarily
propped against column CD. It rests at A and B on small wooden
blocks and against protruding nails. Neglecting friction at all
surfaces of contact, determine the reactions at A, B, and C.
SOLUTION
Free-Body Diagram:
We have five unknowns and six equations of equilibrium. Plywood sheet is free to move in x direction, but
equilibrium is maintained (ΣFx = 0).
ΣM A = 0: rB /A &times; ( B y j + Bz k ) + rC /A &times; C k + rG /A &times; ( −40 lb) j = 0
i j
5 0
0 By
k
i
j
k
i
j
k
0 + 4 4sin 60&deg; −4cos 60&deg; + 2 2sin 60&deg; −2 cos 60&deg; = 0
Bz 0
C
−40
0
0
0
(4C sin 60&deg; − 80 cos 60&deg;) i + (−5Bz − 4C ) j + (5B y − 80)k = 0
Equating the coefficients of the unit vectors to zero,
i:
4C sin 60&deg; − 80 cos 60&deg; = 0
j:
−5Bz − 4C = 0
k:
5B y − 80 = 0
ΣFy = 0:
Ay + B y − 40 = 0
ΣFz = 0:
Az + Bz + C = 0
C = 11.5470 lb
Bz = 9.2376 lb
B y = 16.0000 lb
Ay = 40 − 16.0000 = 24.000 lb
Az = 9.2376 − 11.5470 = −2.3094 lb
A = (24.0 lb)j − (2.31 lb)k ; B = (16.00 lb) j − (9.24 lb)k ; C = (11.55 lb)k 
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450
PROBLEM 4.94
Two tape spools are attached to an axle supported by bearings
at A and D. The radius of spool B is 1.5 in. and the radius of
spool C is 2 in. Knowing that TB = 20 lb and that the system
rotates at a constant rate, determine the reactions at A and D.
Assume that the bearing at A does not exert any axial thrust and
neglect the weights of the spools and axle.
SOLUTION
Free-Body Diagram:
We have six unknowns and six equations of equilibrium.
ΣM A = 0: (4.5i + 1.5k ) &times; (−20 j) + (10.5i + 2 j) &times; (−TC k ) + (15i) &times; ( Dx i + Dy j + Dz k ) = 0
−90k + 30i + 10.5TC j − 2TC i + 15D y k − 15Dz j = 0
Equate coefficients of unit vectors to zero:

i:
30 − 2TC = 0

j:
10.5TC − 15Dz = 0 10.5(15) − 15Dz = 0

k:
−90 + 15 Dy = 0
ΣFx = 0:
Dx = 0
ΣFy = 0:
Ay + D y − 20 lb = 0
Ay = 20 − 6 = 14 lb
ΣFz = 0:
Az + Dz − 15 lb = 0
Az = 15 − 10.5 = 4.5 lb
TC = 15 lb
Dz = 10.5 lb
Dy = 6 lb
A = (14.00 lb) j + (4.50 lb)k ; D = (6.00 lb) j + (10.50 lb)k 
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451
PROBLEM 4.95
Two transmission belts pass over a double-sheaved
pulley that is attached to an axle supported by bearings
at A and D. The radius of the inner sheave is 125 mm
and the radius of the outer sheave is 250 mm. Knowing
that when the system is at rest, the tension is 90 N in
both portions of belt B and 150 N in both portions of
belt C, determine the reactions at A and D. Assume that
the bearing at D does not exert any axial thrust.
SOLUTION
We replace TB and TB′ by their resultant (−180 N)j and TC and TC′ by their resultant (−300 N)k.
Dimensions in mm
We have five unknowns and six equations of equilibrium. Axle AD is free to rotate about the x-axis, but
equilibrium is maintained (ΣMx = 0).
ΣM A = 0: (150i ) &times; (−180 j) + (250i) &times; ( −300k ) + (450i ) &times; ( Dy j + Dz k ) = 0
−27 &times; 103 k + 75 &times; 103 j + 450 Dy k − 450 Dz j = 0
Equating coefficients of j and k to zero,
j:
75 &times; 103 − 450 Dz = 0
Dz = 166.7 N
k:
− 27 &times; 103 + 450 Dy = 0
Dy = 60.0 N
ΣFx = 0:
Ax = 0
ΣFy = 0:
Ay + Dy − 180 N = 0
ΣFz = 0: Az + Dz − 300 N = 0
Ay = 180 − 60 = 120.0 N
Az = 300 − 166.7 = 133.3 N
A = (120.0 N) j + (133.3 N)k ; D = (60.0 N) j + (166.7 N)k 
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452
PROBLEM 4.96
Solve Problem 4.95, assuming that the pulley rotates at a
constant rate and that TB = 104 N, T′B = 84 N, TC = 175 N.
PROBLEM 4.95 Two transmission belts pass over a
double-sheaved pulley that is attached to an axle supported
by bearings at A and D. The radius of the inner sheave is
125 mm and the radius of the outer sheave is 250 mm.
Knowing that when the system is at rest, the tension is 90 N
in both portions of belt B and 150 N in both portions of
belt C, determine the reactions at A and D. Assume that
the bearing at D does not exert any axial thrust.
SOLUTION
Dimensions in mm
We have six unknowns and six equations of equilibrium. —OK
ΣM A = 0: (150i + 250k ) &times; (−104 j) + (150i − 250k ) &times; ( −84 j)
+ (250i + 125 j) &times; (−175k ) + (250i − 125 j) &times; (−TC )
+ 450i &times; ( D y j + Dz k ) = 0
−150(104 + 84)k + 250(104 − 84)i + 250(175 + TC′ ) j − 125(175 − TC′ )
+ 450 D y k − 450 Dz j = 0
Equating the coefficients of the unit vectors to zero,
i : 250(104 − 84) − 125(175 − TC′ ) = 0
175 = TC′ = 40
j: 250(175 + 135) − 450 Dz = 0
Dz = 172.2 N
k : − 150(104 + 84) + 450 D y = 0
Dy = 62.7 N
TC′ = 135;
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453
PROBLEM 4.96 (Continued)
ΣFx = 0:
Ax = 0
ΣFy = 0:
Ay − 104 − 84 + 62.7 = 0
Ay = 125.3 N
ΣFz = 0:
Az − 175 − 135 + 172.2 = 0
Az = 137.8 N
A = (125.3 N) j + (137.8 N)k; D = (62.7 N) j + (172.2 N)k 
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454
PROBLEM 4.97
Two steel pipes AB and BC, each having a mass per unit
length of 8 kg/m, are welded together at B and supported by
three wires. Knowing that a = 0.4 m, determine the tension
in each wire.
SOLUTION
W1 = 0.6m′g
W2 = 1.2m′g
ΣM D = 0: rA/D &times; TA j + rE/D &times; (−W1 j) + rF/D &times; (−W2 j) + rC/D &times; TC j = 0
(−0.4i + 0.6k ) &times; TA j + (−0.4i + 0.3k ) &times; (−W1 j) + 0.2i &times; (−W2 j) + 0.8i &times; TC j = 0
−0.4TAk − 0.6TA i + 0.4W1k + 0.3W1i − 0.2W2 k + 0.8TC k = 0
Equate coefficients of unit vectors to zero:
1
1
i : − 0.6TA + 0.3W1 = 0; TA = W1 = 0.6m′g = 0.3m′g
2
2
k : − 0.4TA + 0.4W1 − 0.2W2 + 0.8TC = 0
−0.4(0.3m′g ) + 0.4(0.6m′g ) − 0.2(1.2m′g ) + 0.8TC = 0
TC =
(0.12 − 0.24 − 0.24)m′g
= 0.15m′g
0.8
ΣFy = 0: TA + TC + TD − W1 − W2 = 0
0.3m′g + 0.15m′g + TD − 0.6m′g − 1.2m′g = 0
TD = 1.35m′g
m′g = (8 kg/m)(9.81m/s 2 ) = 78.48 N/m
TA = 0.3m′g = 0.3 &times; 78.45
TA = 23.5 N 
TB = 0.15m′g = 0.15 &times; 78.45
TB = 11.77 N 
TC = 1.35m′g = 1.35 &times; 78.45
TC = 105.9 N 
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455
PROBLEM 4.98
For the pipe assembly of Problem 4.97, determine (a) the
largest permissible value of a if the assembly is not to tip,
(b) the corresponding tension in each wire.
SOLUTION
W1 = 0.6m′g
W2 = 1.2m′g
ΣM D = 0: rA/D &times; TA j + rE/D &times; (−W1 j) + rF/D &times; (−W2 j) + rC/D &times; TC j = 0
(− ai + 0.6k ) &times; TA j + (− ai + 0.3k ) &times; (−W1 j) + (0.6 − a)i &times; (−W2 j) + (1.2 − a)i &times; TC j = 0
−TA ak − 0.6TA i + W1ak + 0.3W1i − W2 (0.6 − a)k + TC (1.2 − a )k = 0
Equate coefficients of unit vectors to zero:
1
1
i : − 0.6TA + 0.3W1 = 0; TA = W1 = 0.6m′g = 0.3m′g
2
2
k : − TA a + W1a − W2 (0.6 − a ) + TC (1.2 − a) = 0
−0.3m′ga + 0.6m′ga − 1.2m′g (0.6 − a ) + TC (1.2 − a) = 0
TC =
(a)
0.3a − 0.6a + 1.2(0.6 − a)
1.2 − a
For maximum a and no tipping, TC = 0.
−0.3a + 1.2(0.6 − a) = 0
−0.3a + 0.72 − 1.2a = 0
1.5a = 0.72
a = 0.480 m 
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456
PROBLEM 4.98 (Continued)
(b)
Reactions:
m′g = (8 kg/m) 9.81 m/s 2 = 78.48 N/m
TA = 0.3m′g = 0.3 &times; 78.48 = 23.544 N
TA = 23.5 N 
ΣFy = 0: TA + TC + TD − W1 − W2 = 0
TA + 0 + TD − 0.6m′g − 1.2m′g = 0
TD = 1.8m′g − TA = 1.8 &times; 78.48 − 23.544 = 117.72
TD = 117.7 N 
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457
PROBLEM 4.99
The 45-lb square plate shown is supported by three vertical wires.
Determine the tension in each wire.
SOLUTION
Free-Body Diagram:
ΣM B = 0: rC /B &times; TC j + rA/B &times; TA j + rG /B &times; (−45 lb) j = 0

[−(20 in.)i + (15 in.)k ] &times; TC j + (20 in.)k &times; TA j
+ [−(10 in.)i + (10 in.)k ] &times; [−(45 lb)j] = 0
−20TC k − 15TC i − 20TAi + 450k + 450i = 0
Equating to zero the coefficients of the unit vectors,
k:
−20TC + 450 = 0
TC = 22.5 lb 
i:
−15(22.5) − 20TA + 450 = 0
TA = 5.625 lb 
ΣFy = 0:
TA + TB + TC − 45 lb = 0
5.625 lb + TB + 22.5 lb − 45 lb = 0
TB = 16.875 lb 
TA = 5.63 lb; TB = 16.88 lb; TC = 22.5 lb 
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458
PROBLEM 4.100
The table shown weighs 30 lb and has a diameter of 4 ft. It is supported by
three legs equally spaced around the edge. A vertical load P of magnitude
100 lb is applied to the top of the table at D. Determine the maximum
value of a if the table is not to tip over. Show, on a sketch, the area of the
table over which P can act without tipping the table.
SOLUTION
r = 2 ft b = r sin 30&deg; = 1 ft
We shall sum moments about AB.
(b + r )C + (a − b) P − bW = 0
(1 + 2)C + (a − 1)100 − (1)30 = 0
1
C = [30 − (a − 1)100]
3
If table is not to tip, C ≥ 0.
[30 − ( a − 1)100] ≥ 0
30 ≥ (a − 1)100
a − 1 ≤ 0.3 a ≤ 1.3 ft a = 1.300 ft
Only ⊥ distance from P to AB matters. Same condition must be satisfied for each leg. P must be located
in shaded area for no tipping.
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459
PROBLEM 4.101
An opening in a floor is covered by a 1 &times; 1.2-m sheet of
plywood of mass 18 kg. The sheet is hinged at A and B and
is maintained in a position slightly above the floor by a small
block C. Determine the vertical component of the reaction
(a) at A, (b) at B, (c) at C.
SOLUTION
rB/A = 0.6i
rC/A = 0.8i + 1.05k
rG/A = 0.3i + 0.6k
W = mg = (18 kg)9.81
W = 176.58 N
ΣM A = 0: rB/A &times; Bj + rC/A &times; Cj + rG/A &times; (−Wj) = 0
(0.6i ) &times; Bj + (0.8i + 1.05k ) &times; Cj + (0.3i + 0.6k ) &times; ( −Wj) = 0
0.6 Bk + 0.8Ck − 1.05Ci − 0.3Wk + 0.6Wi = 0
Equate coefficients of unit vectors to zero:
 0.6 
i : 1.05C + 0.6W = 0 C = 
176.58 N = 100.90 N
 1.05 
k : 0.6 B + 0.8C − 0.3W = 0
0.6 B + 0.8(100.90 N) − 0.3(176.58 N) = 0 B = −46.24 N
ΣFy = 0: A + B + C − W = 0
A − 46.24 N + 100.90 N + 176.58 N = 0
A = 121.92 N
(a ) A = 121.9 N (b) B = −46.2 N (c) C = 100.9 N 
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460
PROBLEM 4.102
Solve Problem 4.101, assuming that the small block C is
moved and placed under edge DE at a point 0.15 m from
corner E.
PROBLEM 4.101 An opening in a floor is covered by a
1 &times; 1.2-m sheet of plywood of mass 18 kg. The sheet is
hinged at A and B and is maintained in a position slightly
above the floor by a small block C. Determine the vertical
component of the reaction (a) at A, (b) at B, (c) at C.
SOLUTION
rB/A = 0.6i
rC/A = 0.65i + 1.2k
rG/A = 0.3i + 0.6k
W = mg = (18 kg) 9.81 m/s 2
W = 176.58 N
ΣM A = 0: rB/A &times; Bj + rC/A &times; Cj + rG/A &times; (−Wj) = 0
0.6i &times; Bj + (0.65i + 1.2k ) &times; Cj + (0.3i + 0.6k ) &times; (−Wj) = 0
0.6 Bk + 0.65Ck − 1.2Ci − 0.3Wk + 0.6Wi = 0
Equate coefficients of unit vectors to zero:
 0.6 
C =
176.58 N = 88.29 N
 1.2 
i : −1.2C + 0.6W = 0
k : 0.6 B + 0.65C − 0.3W = 0
0.6 B + 0.65(88.29 N) − 0.3(176.58 N) = 0 B = −7.36 N
ΣFy = 0: A + B + C − W = 0
A − 7.36 N + 88.29 N − 176.58 N = 0
A = 95.648 N
(a ) A = 95.6 N (b) B = − 7.36 N (c) C = 88.3 N 
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461
PROBLEM 4.103
The rectangular plate shown weighs 80 lb and is supported by three
vertical wires. Determine the tension in each wire.
SOLUTION
Free-Body Diagram:
ΣM B = 0: rA/B &times; TA j + rC/B &times; TC j + rG/B &times; (−80 lb) j = 0
(60 in.)k &times; TA j + [(60 in.)i + (15 in.)k ] &times; TC j + [(30 in.)i + (30 in.)k ] &times; ( −80 lb) j = 0
−60TAi + 60TC k − 15TC i − 2400k + 2400i = 0
Equating to zero the coefficients of the unit vectors,
i:
60TA − 15(40) + 2400 = 0
TA = 30.0 lb 
k:
60TC − 2400 = 0
TC = 40.0 lb 
ΣFy = 0:
TA + TB + TC − 80 lb = 0
30 lb + TB + 40 lb − 80 lb = 0
TB = 10.00 lb 
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462
PROBLEM 4.104
The rectangular plate shown weighs 80 lb and is supported by three
vertical wires. Determine the weight and location of the lightest block
that should be placed on the plate if the tensions in the three wires are
to be equal.
SOLUTION
Free-Body Diagram:
Let −Wb j be the weight of the block and x and z the block’s coordinates.
Since tensions in wires are equal, let
TA = TB = TC = T
ΣM 0 = 0: (rA &times; Tj) + (rB &times; Tj) + (rC &times; Tj) + rG &times; (−Wj) + ( xi + zk ) &times; ( −Wb j) = 0
or
(75 k ) &times; Tj + (15 k ) &times; Tj + (60i + 30k ) &times; Tj + (30i + 45k ) &times; (−Wj) + ( xi + zk ) &times; (−Wb j) = 0
or
−75T i − 15T i + 60T k − 30T i − 30W k + 45W i − Wb &times; k + Wb z i = 0
Equate coefficients of unit vectors to zero:
i:
−120T + 45W + Wb z = 0
(1)
k:
60T − 30W − Wb x = 0
(2)
ΣFy = 0:
3T − W − Wb = 0
(3)
Eq. (1) + 40 Eq. (3):
5W + ( z − 40)Wb = 0
(4)
Eq. (2) – 20 Eq. (3):
−10W − ( x − 20)Wb = 0
(5)
Also,
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463
PROBLEM 4.104 (Continued)
Solving Eqs. (4) and (5) for Wb /W and recalling that 0 ≤ x ≤ 60 in., 0 ≤ z ≤ 90 in.,
Eq. (4):
Wb
5
5
=
≥
= 0.125
W 40 − z 40 − 0
Eq. (5):
Wb
10
10
=
≥
= 0.5
W 20 − x 20 − 0
Thus, (Wb ) min = 0.5W = 0.5(80) = 40 lb
(Wb ) min = 40.0 lb 
Making Wb = 0.5W in Eqs. (4) and (5):
5W + ( z − 40)(0.5W ) = 0
z = 30.0 in. 
−10W − ( x − 20)(0.5W ) = 0
x = 0 in. 
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464
PROBLEM 4.105
A 2.4-m boom is held by a ball-and-socket joint at C and by two
cables AD and AE. Determine the tension in each cable and the
reaction at C.
SOLUTION
Free-Body Diagram:
(ΣMAC = 0).
Five unknowns and six equations of equilibrium, but equilibrium is maintained
rB = 1.2k
TAE
rA = 2.4k

AD = −0.8i + 0.6 j − 2.4k

AE = 0.8i + 1.2 j − 2.4k
AE = 2.8 m

=
=
(−0.8i + 0.6 j − 2.4k )

AE TAE
=
=
(0.8i + 1.2 j − 2.4k )
AE 2.8
ΣM C = 0: rA &times; TAD + rA &times; TAE + rB &times; (−3 kN) j = 0
i
j
k
i
j
k
T
0
0
2.4
0
2.4 AE + 1.2k &times; (−3.6 kN) j = 0
+ 0
2.6
2.8
0.8 1.2 −2.4
−0.8 0.6 −2.4
Equate coefficients of unit vectors to zero:
i : − 0.55385TAD − 1.02857TAE + 4.32 = 0
(1)
j : − 0.73846TAD + 0.68671TAE = 0
From Eq. (1):
(2)
−0.55385(0.92857)TAE − 1.02857TAE + 4.32 = 0
1.54286TAE = 4.32
TAE = 2.800 kN
TAE = 2.80 kN 
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465
PROBLEM 4.105 (Continued)
From Eq. (2):
TAD = 0.92857(2.80) = 2.600 kN
0.8
0.8
(2.6 kN) +
(2.8 kN) = 0
2.6
2.8
0.6
1.2
(2.6 kN) +
(2.8 kN) − (3.6 kN) = 0
ΣFy = 0: C y +
2.6
2.8
2.4
2.4
(2.6 kN) −
(2.8 kN) = 0
ΣFz = 0: C z −
2.6
2.8
ΣFx = 0: C x −
Cx = 0
C y = 1.800 kN
C z = 4.80 kN
C = (1.800 kN) j + (4.80 kN)k 
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466
PROBLEM 4.106
Solve Problem 4.105, assuming that the 3.6-kN load is applied at
Point A.
PROBLEM 4.105 A 2.4-m boom is held by a ball-and-socket
joint at C and by two cables AD and AE. Determine the tension in
each cable and the reaction at C.
SOLUTION
Free-Body Diagram:
(ΣMAC = 0).
Five unknowns and six equations of equilibrium, but equilibrium is maintained

AD = −0.8i + 0.6 j − 2.4k

AE = 0.8i + 1.2 j − 2.4k
AE = 2.8 m

(−0.8i + 0.6 j − 2.4k )
=

AE TAE
(0.8i + 1.2 j − 2.4k )
TAE =
=
AE 2.8
ΣM C = 0: rA &times; TAD + rA &times; TAE + rA &times; (−3.6 kN) j
Factor rA :
or
Coefficient of i:
rA &times; (TAD + TAE − (3.6 kN) j)
TAD + TAE − (3 kN) j = 0
−
(Forces concurrent at A)
T
(0.8) + AE (0.8) = 0
2.6
2.8
Coefficient of j:
2.6
TAE
2.8
(1)
T
(0.6) + AE (1.2) − 3.6 kN = 0
2.6
2.8
2.6
 0.6  1.2
TAE 
TAE − 3.6 kN = 0
+
2.8
 2.6  2.8
 0.6 + 1.2 
TAE 
 = 3.6 kN
 2.8 
TAE = 5.600 kN
TAE = 5.60 kN 
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467
PROBLEM 4.106 (Continued)
From Eq. (1):
2.6
(5.6) = 5.200 kN
2.8
0.8
0.8
(5.2 kN) +
(5.6 kN) = 0
2.6
2.8
0.6
1.2
ΣFy = 0: C y +
(5.2 kN) +
(5.6 kN) − 3.6 kN = 0
2.6
2.8
2.4
2.4
ΣFz = 0: Cz −
(5.2 kN) −
(5.6 kN) = 0
2.6
2.8
ΣFx = 0: C x −
Cx = 0
Cy = 0
Cz = 9.60 kN
C = (9.60 kN)k 
Note: Since the forces and reaction are concurrent at A, we could have used the methods of Chapter 2.
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468
PROBLEM 4.107
A 10-ft boom is acted upon by the 840-lb force shown. Determine the
tension in each cable and the reaction at the ball-and-socket joint at A.
SOLUTION
We have five unknowns and six equations of equilibrium, but equilibrium is maintained (ΣM x = 0).
Free-Body Diagram:

BD = (−6 ft)i + (7 ft) j + (6 ft)k BD = 11 ft

BE = (−6 ft)i + (7 ft) j − (6 ft)k BE = 11 ft

BD TBD
TBD = TBD
=
(−6i + 7 j + 6k )
BD 11

BE TBE
TBE = TBE
=
(−6i + 7 j − 6k )
BE 11
ΣM A = 0: rB &times; TBD + rB &times; TBE + rC &times; ( −840 j) = 0
6i &times;
TBD
T
(−6i + 7 j + 6k ) + 6i &times; BE (−6i + 7 j − 6k ) + 10i &times; (−840 j) = 0
11
11
42
36
42
36
TBD k − TBD j + TBE k + TBE j − 8400k = 0
11
11
11
11
Equate coefficients of unit vectors to zero:
i: −
k:
36
36
TBD + TBE = 0 TBE = TBD
11
11
42
42
TBD + TBE − 8400 = 0
11
11
 42

2  TBD  = 8400
 11

TBD = 1100 lb 
TBE = 1100 lb 
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469
PROBLEM 4.107 (Continued)
ΣFx = 0: Ax −
6
6
(1100 lb) − (1100 lb) = 0
11
11
Ax = 1200 lb
ΣFy = 0: Ay +
7
7
(1100 lb) + (1100 lb) − 840 lb = 0
11
11
Ay = −560 lb
ΣFz = 0: Az +
6
6
(1100 lb) − (1100 lb) = 0
11
11
Az = 0
A = (1200 lb)i − (560 lb) j 
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470
PROBLEM 4.108
A 12-m pole supports a horizontal cable CD and is held by a ball
and socket at A and two cables BE and BF. Knowing that the
tension in cable CD is 14 kN and assuming that CD is parallel to
the x-axis (φ = 0), determine the tension in cables BE and BF and
the reaction at A.
SOLUTION
Free-Body Diagram:
There are five unknowns and six equations of equilibrium. The pole is free to rotate about the y-axis, but


Resolve BE and BF into components:

BE = (7.5 m)i − (8 m) j + (6 m)k

BF = (7.5 m)i − (8 m) j − (6 m)k
BE = 12.5 m
BF = 12.5 m
Express TBE and TBF in terms of components:
TBE = TBE
TBF = TBF

BE
= TBE (0.60i − 0.64 j + 0.48k )
BE

BF
= TBF (0.60i − 0.64 j − 0.48k )
BF
(1)
(2)
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471
PROBLEM 4.108 (Continued)
ΣM A = 0: rB/A &times; TBE + rB/A &times; TBF + rC/A &times; (−14 kN)i = 0
8 j &times; TBE (0.60i − 0.64 j + 0.48k ) + 8 j &times; TBF (0.60i − 0.64 j − 0.48k ) + 12 j &times; (−14i ) = 0
−4.8 TBE k + 3.84 TBE i − 4.8TBF k − 3.84TBF i + 168k = 0
Equating the coefficients of the unit vectors to zero,
i:
3.84TBE − 3.84TBF = 0
TBE = TBF
k:
−4.8TBE − 4.8TBF + 168 = 0
ΣFx = 0:
Ax + 2(0.60)(17.50 kN) − 14 kN = 0
Ax = 7.00 kN
ΣFy = 0:
Ay − z (0.64) (17.50 kN) = 0
Ay = 22.4 kN
ΣFz = 0:
Az + 0 = 0
TBE = TBF = 17.50 kN 
Az = 0
A = −(7.00 kN)i + (22.4 kN) j 
Because of the symmetry, we could have noted at the outset that TBF = TBE and eliminated one unknown.
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472
PROBLEM 4.109
Solve Problem 4.108, assuming that cable CD forms an angle φ = 25&deg;
with the vertical xy plane.
PROBLEM 4.108 A 12-m pole supports a horizontal cable CD and
is held by a ball and socket at A and two cables BE and BF.
Knowing that the tension in cable CD is 14 kN and assuming that
CD is parallel to the x-axis (φ = 0), determine the tension in cables
BE and BF and the reaction at A.
SOLUTION
Free-Body Diagram:

BE = (7.5 m)i − (8 m) j + (6 m)k
BE = 12.5 m

BF = (7.5 m)i − (8 m)j − (6 m)k
BF = 12.5 m

BE
TBE = TBE
= TBE (0.60i − 0.64 j + 0.48k )
BE

BF
TBF = TBF
= TBF (0.60i − 0.64 j − 0.48k )
BF
ΣM A = 0: rB/A &times; TBE + rB/A &times; TBF + rC/A &times; TCD = 0
8 j &times; TBE (0.60i − 0.64 j + 0.48k ) + 8 j &times; TBF (0.60i − 0.64 j − 0.48k )
+ 12 j &times; (19 kN)(− cos 25&deg; i + sin 25&deg;k ) = 0
−4.8TBE k + 3.84TBE i − 4.8TBF k − 3.84TBF i + 152.6 k − 71.00 i = 0
Equating the coefficients of the unit vectors to zero,
Solving simultaneously,
i: 3.84TBE − 3.84TBF + 71.00 = 0;
TBF − TBE = 18.4896
k: − 4.8TBE − 4.8 TBF + 152.26 = 0;
TBF + TBE = 31.721
TBE = 6.6157 kN;
TBF = 25.105 kN
TBE = 6.62 kN; TBF = 25.1 kN 
ΣFx = 0: Ax + (0.60) (TBF + TBE ) − 14 cos 25&deg; = 0
Ax = 12.6883 − 0.60(31.7207)
Ax = −6.34 kN
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473
PROBLEM 4.109 (Continued)
ΣFy = 0: Ay − (0.64) (TBF + TBE ) = 0
Ay = 0.64(31.721)
Ay = 20.3 kN
ΣFz = 0: Az − 0.48(TBF − TBE ) + 14sin 25&deg; = 0
Az = 0.48(18.4893) − 5.9167
Az = 2.96 kN
A = −(6.34 kN)i + (20.3 kN) j + (2.96 kN) k 
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474
PROBLEM 4.110
A 48-in. boom is held by a ball-and-socket joint at C and
by two cables BF and DAE; cable DAE passes around a
the tension in each cable and the reaction at C.
SOLUTION
Free-Body Diagram:
Five unknowns and six equations of equilibrium, but equilibrium is
maintained (ΣMAC = 0).
T = Tension in both parts of cable DAE.
rB = 30k
rA = 48k


AE = 20 j − 48k
AE = 52 in.

BF = 16i − 30k
BF = 34 in.

T
= (−20i − 48k ) = ( −5i − 12k )
13

AE T
T
TAE = T
= (20 j − 48k ) = (5 j − 12k )
AE 52
13

T
BF TBF
TBF = TBF
=
(16i − 30k ) = BF (8i − 15k )
BF
34
17
ΣM C = 0: rA &times; TAD + rA &times; TAE + rB &times; TBF + rB &times; (−320 lb) j = 0
i j k
i j k
i j k
T
T
T
+ 0 0 48
+ 0 0 30 BF + (30k ) &times; (−320 j) = 0
0 0 48
13
13
17
−5 0 −12
0 5 −12
8 0 −15
Coefficient of i:
−
240
T + 9600 = 0
13
T = 520 lb
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475
PROBLEM 4.110 (Continued)
Coefficient of j:
−
240
240
T+
TBD = 0
13
17
TBD =
17
17
T = (520) TBD = 680 lb
13
13
ΣF = 0: TAD + TAE + TBF − 320 j + C = 0
Coefficient of i:
−
20
8
(520) + (680) + Cx = 0
52
17
−200 + 320 + Cx = 0
Coefficient of j:
20
(520) − 320 + C y = 0
52
200 − 320 + C y = 0
Coefficient of k:
Cx = −120 lb
−
C y = 120 lb
48
48
30
(520) − (520) − (680) + Cz = 0
52
52
34
−480 − 480 − 600 + Cz = 0
Cz = 1560 lb
TDAE = 520 lb 
TBD = 680 lb 
C = −(120.0 lb)i + (120.0 lb) j + (1560 lb)k 
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476
PROBLEM 4.111
Solve Problem 4.110, assuming that the 320-lb load is
applied at A.
PROBLEM 4.110 A 48-in. boom is held by a ball-andsocket joint at C and by two cables BF and DAE; cable
DAE passes around a frictionless pulley at A. For the
the reaction at C.
SOLUTION
Free-Body Diagram:
Five unknowns and six equations of equilibrium, but equilibrium is
maintained (ΣMAC = 0).
T = tension in both parts of cable DAE.
rB = 30k
rA = 48k


AE = 20 j − 48k
AE = 52 in.

BF = 16i − 30k
BF = 34 in.

T
= (−20i − 48k ) = ( −5i − 12k )
13

AE T
T
TAE = T
= (20 j − 48k ) = (5 j − 12k )
AE 52
13

T
BF TBF
TBF = TBF
=
(16i − 30k ) = BF (8i − 15k )
BF
34
17
ΣM C = 0: rA &times; TAD + rA &times; TAE + rB &times; TBF + rA &times; ( −320 lb) j = 0
i j k
i j k
i j k
T
T
T
+ 0 0 48
+ 0 0 30 BF + 48k &times; (−320 j) = 0
0 0 48
13
13
17
−5 0 −12
0 5 −12
8 0 −15
Coefficient of i:
−
240
T + 15,360 = 0
13
T = 832 lb
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477
PROBLEM 4.111 (Continued)
Coefficient of j:
−
240
240
T+
TBD = 0
13
17
TBD =
17
17
T = (832)
13
13
TBD = 1088 lb
ΣF = 0: TAD + TAE + TBF − 320 j + C = 0
−
Coefficient of i:
20
8
(832) + (1088) + C x = 0
52
17
−320 + 512 + Cx = 0
20
(832) − 320 + C y = 0
52
Coefficient of j:
320 − 320 + C y = 0
Coefficient of k:
−
Cy = 0
48
48
30
(832) − (852) − (1088) + Cz = 0
52
52
34
−768 − 768 − 960 + Cz = 0
Cx = −192 lb
TDAE = T
Cz = 2496 lb
TDAE = 832 lb 
TBD = 1088 lb 
C = −(192.0 lb)i + (2496 lb)k 
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478
PROBLEM 4.112
A 600-lb crate hangs from a cable that passes over a
pulley B and is attached to a support at H. The 200-lb
boom AB is supported by a ball-and-socket joint at A
and by two cables DE and DF. The center of gravity
of the boom is located at G. Determine (a) the tension
in cables DE and DF, (b) the reaction at A.
SOLUTION
Free-Body Diagram:
WC = 600 lb
WG = 200 lb
We have five unknowns (TDE , TDF , Ax , Ay , Az ) and five equilibrium equations. The boom is free to spin about
the AB axis, but equilibrium is maintained, since ΣM AB = 0.

BH = (30 ft)i − (22.5 ft) j
BH = 37.5 ft
We have

8.8
DE = (13.8 ft)i −
(22.5 ft) j + (6.6 ft)k
12
= (13.8 ft)i − (16.5 ft) j + (6.6 ft)k
DE = 22.5 ft

DF = (13.8 ft)i − (16.5 ft) j − (6.6 ft)k
DF = 22.5 ft
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479
PROBLEM 4.112 (Continued)
TBH = TBH
Thus:
TDE = TDE
TDF = TDF
(a)

BH
30i − 22.5 j
= (600 lb)
= (480 lb)i − (360 lb) j
BH
37.5

DE TDE
=
(13.8i − 16.5 j + 6.6k )
DE 22.5

DF TDE
=
(13.8i − 16.5 j − 6.6k )
DF 22.5
ΣM A = 0: (rJ &times; WC ) + (rK &times; WG ) + (rH &times; TBH ) + (rE &times; TDE ) + (rF &times; TDF ) = 0
− (12i ) &times; (−600 j) − (6i ) &times; (−200 j) + (18i ) &times; (480i − 360 j)
i
j
k
i
j
k
TDE
TDF
5
0
6.6 +
5
0
+
−6.6 = 0
22.5
22.5
13.8 −16.5 6.6
13.8 −16.5 −6.6
7200k + 1200k − 6480k + 4.84(TDE − TDF )i
or
+
58.08
82.5
(TDE − TDF ) j −
(TDE + TDF )k = 0
22.5
22.5
Equating to zero the coefficients of the unit vectors,
i or j:
TDE − TDF = 0
k : 7200 + 1200 − 6480 −
TDE = TDF *
82.5
(2TDE ) = 0
22.5
TDE = 261.82 lb
TDE = TDF = 262 lb 
(b)
 13.8 
ΣFx = 0: Ax + 480 + 2 
 (261.82) = 0
 22.5 
 16.5 
ΣFy = 0: Ay − 600 − 200 − 360 − 2 
 (261.82) = 0
 22.5 
ΣFz = 0: Az = 0
Ax = −801.17 lb
Ay = 1544.00 lb
A = −(801 lb)i + (1544 lb) j 
*Remark: The fact is that TDE = TDF could have been noted at the outset from the symmetry of structure with
respect to xy plane.
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480
PROBLEM 4.113
A 100-kg uniform rectangular plate is supported in the position
shown by hinges A and B and by cable DCE that passes over a
frictionless hook at C. Assuming that the tension is the same in
both parts of the cable, determine (a) the tension in the cable,
(b) the reactions at A and B. Assume that the hinge at B does
not exert any axial thrust.
SOLUTION
rB/A (960 − 180)i = 780i
Dimensions in mm
 960
 450
rG/A = 
− 90  i +
k
2
 2

= 390i + 225k
rC/A = 600i + 450k
T = Tension in cable DCE

CD = −690i + 675 j − 450k

CE = 270i + 675 j − 450k
CD = 1065 mm
CE = 855 mm
T
(−690i + 675 j − 450k )
1065
T
TCE =
(270i + 675 j − 450k )
855
W = −mgi = −(100 kg)(9.81 m/s 2 ) j = −(981 N) j
TCD =
ΣM A = 0: rC/A &times; TCD + rC/A &times; TCE + rG/A &times; (−Wj) + rB/A &times; B = 0
i
j
k
i
j
k
T
T
600
0
450
450
+ 600 0
1065
855
270 675 −450
−690 675 −450
i
+ 390
0
j
0
−981
k
i
225 + 780
0
0
j
0
k
0 =0
By
Bz
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481
PROBLEM 4.113 (Continued)
Coefficient of i:
−(450)(675)
T
T
− (450)(675)
+ 220.73 &times; 103 = 0
1065
855
T = 344.64 N
Coefficient of j:
(−690 &times; 450 + 600 &times; 450)
T = 345 N 
344.64
344.64
+ (270 &times; 450 + 600 &times; 450)
− 780 Bz = 0
1065
855
Bz = 185.516 N
Coefficient of k: (600)(675)
344.64
344.64
+ (600)(675)
− 382.59 &times; 103 + 780 By = 0 By = 113.178 N
1065
855
B = (113.2 N) j + (185.5 N)k 
ΣF = 0: A + B + TCD + TCE + W = 0
690
270
(344.64) +
(344.64) = 0
1065
855
Coefficient of i:
Ax −
Ax = 114.5 N
Coefficient of j:
Ay + 113.178 +
675
675
(344.64) +
(344.64) − 981 = 0
1065
855
Ay = 377 N
Coefficient of k:
Az + 185.516 −
450
450
(344.64) −
(344.64) = 0
1065
855
Az = 141.5 N
A = (114.5 N)i + (377 N) j + (144.5 N)k 
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482
PROBLEM 4.114
Solve Problem 4.113, assuming that cable DCE is replaced
by a cable attached to Point E and hook C.
PROBLEM 4.113 A 100-kg uniform rectangular plate is
supported in the position shown by hinges A and B and by
cable DCE that passes over a frictionless hook at C.
Assuming that the tension is the same in both parts of the
cable, determine (a) the tension in the cable, (b) the
reactions at A and B. Assume that the hinge at B does not
exert any axial thrust.
SOLUTION
CD = 1065 mm
CE = 855 mm
T
(−690i + 675 j − 450k )
1065
T
(270i + 675 j − 450k )
TCE =
855
W = −mgi = −(100 kg)(9.81 m/s 2 ) j = −(981 N)j
TCD =
Now,
ΣM A = 0: rC/A &times; TCE + rG/A &times; (−W j) + rB/A &times; B = 0
i
j
k
i
j
k
i
j
T
600 0
450
0
225 + 780 0
+ 390
855
270 675 −450
0 −981 0
0 By
Coefficient of i:
−(450)(675)
k
0 =0
Bz
T
+ 220.73 &times; 103 = 0
855
T = 621.31 N
Coefficient of j:
Coefficient of k:
(270 &times; 450 + 600 &times; 450)
(600)(675)
T = 621 N 
621.31
− 780 Bz = 0 Bz = 364.74 N
855
621.31
− 382.59 &times; 103 + 780 By = 0 By = 113.186 N
855
B = (113.2 N)j + (365 N)k 
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483
PROBLEM 4.114 (Continued)
ΣF = 0: A + B + TCE + W = 0
270
(621.31) = 0
855
Coefficient of i:
Ax +
Coefficient of j:
Ay + 113.186 +
Coefficient of k:
Az + 364.74 −

Ax = −196.2 N
675
(621.31) − 981 = 0
855
450
(621.31) = 0
855
Ay = 377.3 N
Az = −37.7 N 
A = −(196.2 N)i + (377 N)j − (37.7 N)k 

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484
PROBLEM 4.115
The rectangular plate shown weighs 75 lb and is held in the
position shown by hinges at A and B and by cable EF. Assuming
that the hinge at B does not exert any axial thrust, determine
(a) the tension in the cable, (b) the reactions at A and B.
SOLUTION
rB/A = (38 − 8)i = 30i
rE/A = (30 − 4)i + 20k
rG/A
= 26i + 20k
38
= i + 10k
2
= 19i + 10k

EF = 8i + 25 j − 20k
EF = 33 in.

AE T
T =T
= (8i + 25 j − 20k )
AE 33
ΣM A = 0: rE/A &times; T + rG/A &times; (−75 j) + rB/A &times; B = 0
i
j
k
i
j
k
i
j
T
+ 19 0 10 + 30 0
26 0 20
33
8 25 −20
0 −75 0
0 By
−(25)(20)
Coefficient of i:
Coefficient of j:
Coefficient of k:
(160 + 520)
(26)(25)
T
+ 750 = 0:
33
k
0 =0
Bz
T = 49.5 lb 
49.5
− 30 Bz = 0: Bz = 34 lb
33
49.5
− 1425 + 30 By = 0: By = 15 lb
33
B = (15.00 lb)j + (34.0 lb)k 




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485
PROBLEM 4.115 (Continued)

ΣF = 0: A + B + T − (75 lb)j = 0 
Ax +
Coefficient of i:
Coefficient of j:
Coefficient of k:
Ay + 15 +
8
(49.5) = 0
33
25
(49.5) − 75 = 0
33
Az + 34 −
20
(49.5) = 0
33
Ax = −12.00 lb
Ay = 22.5 lb
Az = −4.00 lb 
A = −(12.00 lb)i + (22.5 lb)j − (4.00 lb)k 

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486
PROBLEM 4.116
Solve Problem 4.115, assuming that cable EF is replaced by a
cable attached at points E and H.
PROBLEM 4.115 The rectangular plate shown weighs 75 lb and
is held in the position shown by hinges at A and B and by cable
EF. Assuming that the hinge at B does not exert any axial thrust,
determine (a) the tension in the cable, (b) the reactions at A and B.
SOLUTION
rB/A = (38 − 8)i = 30i
rE/A = (30 − 4)i + 20k
= 26i + 20k
38
i + 10k
2
= 19i + 10k
rG/A =

EH = −30i + 12 j − 20k
EH = 38 in.

EH T
T=T
= (−30i + 12 j − 20k )
EH 38
ΣM A = 0: rE/A &times; T + rG/A &times; (−75 j) + rB/A &times; B = 0
i
j
k
i
j
k
i
j
T
+ 19 0 10 + 30 0
26 0 20
38
−30 12 −20
0 −75 0
0 By
−(12)(20)
Coefficient of i:
Coefficient of j:
Coefficient of k:
(−600 + 520)
(26)(12)
T
+ 750 = 0
38
T = 118.75
k
0 =0
Bz
T = 118.8lb 
118.75
− 30 Bz = 0 Bz = −8.33lb
38
118.75
− 1425 + 30 By = 0 B y = 15.00 lb
38
B = (15.00 lb)j − (8.33 lb)k 
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487
PROBLEM 4.116 (Continued)
ΣF = 0:
Ax −
Coefficient of i:
Coefficient of j:
Coefficient of k:
Ay + 15 +
A + B + T − (75 lb)j = 0
30
(118.75) = 0
38
12
(118.75) − 75 = 0
38
Az − 8.33 −
20
(118.75) = 0
38
Ax = 93.75 lb
Ay = 22.5 lb
Az = 70.83 lb
A = (93.8 lb)i + (22.5 lb)j + (70.8 lb)k 

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488
PROBLEM 4.117
A 20-kg cover for a roof opening is hinged at corners A and B.
The roof forms an angle of 30&deg; with the horizontal, and the
cover is maintained in a horizontal position by the brace CE.
Determine (a) the magnitude of the force exerted by the brace,
(b) the reactions at the hinges. Assume that the hinge at A does
not exert any axial thrust.
SOLUTION
Force exerted by CE:
F = F (cos 75&deg;)i + F (sin 75&deg;) j
F = F (0.25882i + 0.96593j)
W = mg = 20 kg(9.81 m/s 2 ) = 196.2 N
rA /B = 0.6k
rC /B = 0.9i + 0.6k
rG /B = 0.45i + 0.3k
F = F (0.25882i + 0.96593j)
ΣM B = 0: rG/B &times; (−196.2 j) + rC/B &times; F + rA/B &times; A = 0
(a)
i
j
k
i
j
k
i
0.45
0
0.3 + 0.9
0
0.6 F + 0
0
−196.2 0
0.25882 +0.96593 0
Ax
j
0
Ay
k
0.6 = 0
0
Coefficient of i :
+58.86 − 0.57956 F − 0.6 Ay = 0
(1)
Coefficient of j:
+0.155292 F + 0.6 Ax = 0
(2)
Coefficient of k:
−88.29 + 0.86934 F = 0:
From Eq. (2):
+58.86 − 0.57956(101.56) − 0.6 Ay = 0
From Eq. (3):
+0.155292(101.56) + 0.6 Ax = 0
F = 101.56 N
Ay = 0
Ax = −26.286 N
F = (101.6 N) 
(b)
Coefficient of i:
Coefficient of j:
ΣF : A + B + F − Wj = 0
26.286 + Bx + 0.25882(101.56) = 0 Bx = 0
B y + 0.96593(101.56) − 196.2 = 0 B y = 98.1 N
Bz = 0
Coefficient of k:
A = −(26.3 N)i; B = (98.1 N) j 
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489
PROBLEM 4.118
The bent rod ABEF is supported by bearings at C and D and by wire
AH. Knowing that portion AB of the rod is 250 mm long, determine
(a) the tension in wire AH, (b) the reactions at C and D. Assume that
the bearing at D does not exert any axial thrust.
SOLUTION
ΔABH is equilateral.
Free-Body Diagram:
Dimensions in mm
rH/C = −50i + 250 j
rD/C = 300i
rF/C = 350i + 250k
T = T (sin 30&deg;) j − T (cos 30&deg;)k = T (0.5 j − 0.866k )
ΣM C = 0: rH/C &times; T + rD &times; D + rF/C &times; (−400 j) = 0
i
j
k
i
j
−50 250
0 T + 300 0
0
0.5 −0.866
0 Dy
Coefficient i:
k
i
j
k
0 + 350
0
250 = 0
Dz
0 −400 0
−216.5T + 100 &times; 103 = 0
T = 461.9 N
Coefficient of j:
T = 462 N 
−43.3T − 300 Dz = 0
−43.3(461.9) − 300 Dz = 0
Dz = −66.67 N
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490
PROBLEM 4.118 (Continued)
Coefficient of k:
−25T + 300 D y − 140 &times; 103 = 0
−25(461.9) + 300 Dy − 140 &times; 103 = 0
D y = 505.1 N
D = (505 N) j − (66.7 N)k 
ΣF = 0: C + D + T − 400 j = 0
Coefficient i:
Cx = 0
Cx = 0
Coefficient j:
C y + (461.9)0.5 + 505.1 − 400 = 0 C y = −336 N
Coefficient k:
Cz − (461.9)0.866 − 66.67 = 0
Cz = 467 N
C = −(336 N) j + (467 N)k 
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491
PROBLEM 4.119
Solve Problem 4.115, assuming that the hinge at B is removed and
that the hinge at A can exert couples about axes parallel to the y
and z axes.
PROBLEM 4.115 The rectangular plate shown weighs 75 lb and
is held in the position shown by hinges at A and B and by cable
EF. Assuming that the hinge at B does not exert any axial thrust,
determine (a) the tension in the cable, (b) the reactions at A and B.
SOLUTION
rE/A = (30 − 4)i + 20k = 26i + 20k
rG/A = (0.5 &times; 38)i + 10k = 19i + 10k

AE = 8i + 25 j − 20k
AE = 33 in.

AE T
T =T
= (8i + 25 j − 20k )
AE 33
ΣM A = 0: rE/A &times; T + rG/A &times; (−75 j) + ( M A ) y j + ( M A ) z k = 0
i
j
k
i
j
k
T
+ 19 0 10 + ( M A ) y j + ( M A ) z k = 0
26 0 20
33
8 25 −20
0 −75 0
−(20)(25)
Coefficient of i:
Coefficient of j:
Coefficient of k:
(160 + 520)
(26)(25)
T
+ 750 = 0
33
T = 49.5 lb 
49.5
+ ( M A ) y = 0 ( M A ) y = −1020 lb ⋅ in.
33
49.5
− 1425 + ( M A ) z = 0
33
( M A ) z = 450 lb ⋅ in.
ΣF = 0: A + T − 75 j = 0
Ax +
Coefficient of i:
8
(49.5) = 0
33
M A = −(1020 lb ⋅ in.)j + (450 lb ⋅ in.)k 
Ax = 12.00 lb
Coefficient of j:
Ay +
25
(49.5) − 75 = 0
33
Ay = 37.5 lb
Coefficient of k:
Az −
20
(49.5) = 0
33
Az = 30.0 lb
A = −(12.00 lb)i + (37.5 lb)j + (30.0 lb)k 
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492
PROBLEM 4.120
Solve Problem 4.118, assuming that the bearing at D is removed
and that the bearing at C can exert couples about axes parallel to
the y and z axes.
PROBLEM 4.118 The bent rod ABEF is supported by bearings
at C and D and by wire AH. Knowing that portion AB of the rod
is 250 mm long, determine (a) the tension in wire AH, (b) the
reactions at C and D. Assume that the bearing at D does not exert
any axial thrust.
SOLUTION
Free-Body Diagram:
ΔABH is equilateral.
Dimensions in mm
rH/C = −50i + 250 j
rF/C = 350i + 250k
T = T (sin 30&deg;) j − T (cos 30&deg;)k = T (0.5 j − 0.866k )
ΣM C = 0: rF/C &times; (−400 j) + rH/C &times; T + ( M C ) y j + ( M C ) z k = 0
i
j
k
i
j
k
350
0
250 + −50 250
0 T + (M C ) y j + (M C ) z k = 0
0 −400 0
0
0.5 −0.866
Coefficient of i:
Coefficient of j:
+100 &times; 103 − 216.5T = 0 T = 461.9 N
T = 462 N 
−43.3(461.9) + ( M C ) y = 0
( M C ) y = 20 &times; 103 N ⋅ mm
(M C ) y = 20.0 N ⋅ m
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493
PROBLEM 4.120 (Continued)
Coefficient of k:
−140 &times; 103 − 25(461.9) + ( M C ) z = 0
( M C ) z = 151.54 &times; 103 N ⋅ mm
(M C ) z = 151.5 N ⋅ m
ΣF = 0: C + T − 400 j = 0
M C = (20.0 N ⋅ m)j + (151.5 N ⋅ m)k 
Coefficient of i:
Coefficient of j:
Coefficient of k:
Cx = 0
C y + 0.5(461.9) − 400 = 0 C y = 169.1 N
C z − 0.866(461.9) = 0 C z = 400 N
C = (169.1 N)j + (400 N)k 

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494
PROBLEM 4.121
The assembly shown is welded to collar A that fits on the vertical
pin shown. The pin can exert couples about the x and z axes but
shown, determine the tension in each cable and the reaction at A.
SOLUTION
Free-Body Diagram:
First note:
−(0.08 m)i + (0.06 m) j
TCF = λ CF TCF =
(0.08) 2 + (0.06)2 m
TCF
= TCF (−0.8i + 0.6 j)
TDE = λ DE TDE =
(0.12 m)j − (0.09 m)k
(0.12) 2 + (0.09)2 m
TDE
= TDE (0.8 j − 0.6k )
(a)
From F.B.D. of assembly:
ΣFy = 0: 0.6TCF + 0.8TDE − 480 N = 0
0.6TCF + 0.8TDE = 480 N
or
(1)
ΣM y = 0: − (0.8TCF )(0.135 m) + (0.6TDE )(0.08 m) = 0
TDE = 2.25TCF
or
(2)
Substituting Equation (2) into Equation (1),
0.6TCF + 0.8[(2.25)TCF ] = 480 N
TCF = 200.00 N
TCF = 200 N 
or

and from Equation (2):
TDE = 2.25(200.00 N) = 450.00
TDE = 450 N 
or
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495
PROBLEM 4.121 (Continued)
(b)
From F.B.D. of assembly:
ΣFz = 0: Az − (0.6)(450.00 N) = 0
Az = 270.00 N
ΣFx = 0: Ax − (0.8)(200.00 N) = 0
Ax = 160.000 N
or A = (160.0 N)i + (270 N)k 
ΣM x = 0:
MAx + (480 N)(0.135 m) − [(200.00 N)(0.6)](0.135 m)
− [(450 N)(0.8)](0.09 m) = 0
M Ax = −16.2000 N ⋅ m
ΣM z = 0:
MAz − (480 N)(0.08 m) + [(200.00 N)(0.6)](0.08 m)
+ [(450 N)(0.8)](0.08 m) = 0
M Az = 0
or M A = −(16.20 N ⋅ m)i 
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496
PROBLEM 4.122
The assembly shown is used to control the tension T in a tape that
passes around a frictionless spool at E. Collar C is welded to rods ABC
and CDE. It can rotate about shaft FG but its motion along the shaft is
prevented by a washer S. For the loading shown, determine (a) the
tension T in the tape, (b) the reaction at C.
SOLUTION
Free-Body Diagram:
rA/C = 4.2 j + 2k
rE/C = 1.6i − 2.4 j
ΣM C = 0: rA/C &times; (−6 j) + rE/C &times; T (i + k ) + ( M C ) y j + ( M C ) z k = 0
(4.2 j + 2k ) &times; (−6 j) + (1.6i − 2.4 j) &times; T (i + k ) + ( M C ) y j + ( M C ) z k = 0
Coefficient of i:
Coefficient of j:
Coefficient of k:
12 − 2.4T = 0
T = 5.00 lb 
−1.6(5 lb) + (M C ) y = 0 ( M C ) y = 8 lb ⋅ in.
2.4(5 lb) + ( M C ) z = 0 ( M C ) z = −12 lb ⋅ in.
M C = (8.00 lb ⋅ in.)j − (12.00 lb ⋅ in.)k 
ΣF = 0: Cx i + C y j + C z k − (6 lb)j + (5 lb)i + (5 lb)k = 0
Equate coefficients of unit vectors to zero.
Cx = −5 lb C y = 6 lb Cz = −5 lb 
C = −(5.00 lb)i + (6.00 lb)j − (5.00 lb)k 

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497
PROBLEM 4.123
The rigid L-shaped member ABF is supported by a ball-and-socket
joint at A and by three cables. For the loading shown, determine the
tension in each cable and the reaction at A.
SOLUTION
Free-Body Diagram:
rB/A = 12i
rF/A = 12 j − 8k
rD/A = 12i − 16k
rE/A = 12i − 24k
rF/A = 12i − 32k

BG = −12i + 9k
BG = 15 in.
λ BG = −0.8i + 0.6k

DH = −12i + 16 j; DH = 20 in.; λDH = −0.6i + 0.8 j

FJ = −12i + 16 j; FJ = 20 in.; λFJ = −0.6i + 0.8 j
ΣM A = 0: rB/A &times; TBG λBG + rDH &times; TDH λDH + rF/A &times; TFJ λFJ
+rF/A &times; (−24 j) + rE/A &times; ( −24 j) = 0
i
j k
i
j
k
i
j
k
12 0 0 TBG + 12
0 −16 TDH + 12
0 −32 TFJ
−0.8 0 0.6
−0.6 0.8 0
−0.6 0.8 0
i
j
k
i
j
k
+ 12 0 −8 + 12 0 −24 = 0
0 −24 0
0 −24 0
Coefficient of i:
+12.8TDH + 25.6TFJ − 192 − 576 = 0
(1)
Coefficient of k:
+9.6TDH + 9.6TFJ − 288 − 288 = 0
(2)
9.6TFJ = 0
TFJ = 0 
3
4
Eq. (1) − Eq. (2):
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498
PROBLEM 4.123 (Continued)
12.8TDH − 268 = 0
TDH = 60 lb 
−7.2TBG + (16 &times; 0.6)(60.0 lb) = 0
TBG = 80.0 lb 
From Eq. (1):
Coefficient of j:
ΣF = 0:
A + TBG λ BG + TDH λ DH + TFJ − 24 j − 24 j = 0
Coefficient of i:
Ax + (80)( −0.8) + (60.0)(−0.6) = 0
Coefficient of j:
Ay + (60.0)(0.8) − 24 − 24 = 0
Coefficient of k:
Az + (80.0)(+0.6) = 0
Ax = 100.0 lb
Ay = 0
Az = −48.0 lb
A = (100.0 lb)i − (48.0 lb) j 
Note: The value Ay = 0 can be confirmed by considering ΣM BF = 0.
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499
PROBLEM 4.124
Solve Problem 4.123, assuming that the load at C has been removed.
PROBLEM 4.123 The rigid L-shaped member ABF is supported
by a ball-and-socket joint at A and by three cables. For the loading
shown, determine the tension in each cable and the reaction at A.
SOLUTION
Free-Body Diagram:
rB/A = 12i
rB/A = 12i − 16k
rE/A = 12i − 24k
rF /A = 12i − 32k

BG = −12i + 9k ; BG = 15 in.; λ BG = −0.8i + 0.6k

DH = −12i + 16 j; DH = 20 in.; λ DH = −0.6i + 0.8 j

FJ = −12i + 16 j; FJ = 20 in.; λ FJ = −0.6i + 0.8 j
ΣM A = 0: rB/A &times; TBG λ BG + rD/A &times; TDH λ DH + rF /A &times; TFJ λ FJ + rE/A &times; (−24 j) = 0
i
j k
i
j
k
i
j
k
i
j
k
12 0 0 TBG + 12
0 −16 TDH + 12
0 −32 TFJ + 12 0 −24 = 0
−0.8 0 0.6
−0.6 0.8 0
−0.6 0.8 0
0 −24 0
i:
+ 12.8TDH + 25.6TFJ − 576 = 0
(1)
k:
+9.6TDH + 9.6TFJ − 288 = 0
(2)
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500
PROBLEM 4.124 (Continued)
Multiply Eq. (1) by
3
4
and subtract Eq. (2):
From Eq. (1):
9.6TFJ − 144 = 0
TFJ = 15.00 lb 
12.8TDH + 25.6(15.00) − 576 = 0
TDH = 15.00 lb 
j:
−7.2TBG + (16)(0.6)(15) + (32)(0.6)(15) = 0
− 7.2TBG + 432 = 0
TBG = 60.0 lb 
ΣF = 0: A + TBG λ BG + TDAλ DH + TFJ λ FJ − 24 j = 0
i : Ax + (60)(−0.8) + (15)( −0.6) + (15)(−0.6) = 0
j:
Ay + (15)(0.8) + (15)(0.8) − 24 = 0
k:
Az + (60)(0.6) = 0
Ax = 66.0 lb
Ay = 0
Az = −36.0 lb
A = (66.0 lb)i − (36.0 lb)k 
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501
PROBLEM 4.125
The rigid L-shaped member ABC is supported by
a ball-and-socket joint at A and by three cables. If
a 1.8-kN load is applied at F, determine the tension in
each cable.
SOLUTION
Free-Body Diagram:
In this problem:
We have
Thus,
Dimensions in mm
a = 210 mm

CD = (240 mm)j − (320 mm)k CD = 400 mm

BD = −(420 mm)i + (240 mm)j − (320 mm)k BD = 580 mm

BE = (420 mm)i − (320 mm)k BE = 528.02 mm
TCD
TBD
TBE

CD
= TCD
= TCD (0.6 j − 0.8k )
CD

BD
= TBD
= TBD (−0.72414i + 0.41379 j − 0.55172k )
BD

BE
= TBE
= TBE (0.79542i − 0.60604k )
BE
ΣM A = 0: (rC &times; TCD ) + (rB &times; TBD ) + (rB &times; TBE ) + (rW &times; W) = 0
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502
PROBLEM 4.125 (Continued)
rC = −(420 mm)i + (320 mm)k
Noting that
rB = (320 mm)k
rW = − ai + (320 mm)k
and using determinants, we write
i
j
k
i
j
k
−420 0 320 TCD +
0
0
320 TBD
0
0.6 −0.8
−0.72414 0.41379 −0.55172
i
j
k
i
j
k
+
0
0
320 TBE + −a
0
320 = 0
0.79542 0 −0.60604
0 −1.8 0
Equating to zero the coefficients of the unit vectors,
i:
−192TCD − 132.413TBD + 576 = 0
(1)
j:
−336TCD − 231.72TBD + 254.53TBE = 0
(2)
k:
−252TCD + 1.8a = 0
(3)
Recalling that a = 210 mm, Eq. (3) yields
TCD =
From Eq. (1):
1.8(210)
= 1.500 kN
252
−192(1.5) − 132.413TBD + 576 = 0
TBD = 2.1751 kN
From Eq. (2):
TCD = 1.500 kN 
TBD = 2.18 kN 
−336(1.5) − 231.72(2.1751) + 254.53TBE = 0
TBE = 3.9603 kN 

TBE = 3.96 kN 
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503
PROBLEM 4.126
Solve Problem 4.125, assuming that the 1.8-kN load is
applied at C.
PROBLEM 4.125 The rigid L-shaped member ABC is
supported by a ball-and-socket joint at A and by three
cables. If a 1.8-kN load is applied at F, determine the
tension in each cable.
SOLUTION
See solution of Problem 4.125 for free-body diagram and derivation of Eqs. (1), (2), and (3):
−192TCD − 132.413TBD + 576 = 0
(1)
−336TCD − 231.72TBD + 254.53TBE = 0
(2)
−252TCD + 1.8a = 0
(3)
In this problem, the 1.8-kN load is applied at C and we have a = 420 mm. Carrying into Eq. (3) and
solving for TCD ,
TCD = 3.00
From Eq. (1):
From Eq. (2):
−(192)(3) − 132.413TBD + 576 = 0
−336(3) − 0 + 254.53TBE = 0 
TCD = 3.00 kN 
TBD = 0 
TBE = 3.96 kN 
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504
PROBLEM 4.127
The assembly shown consists of an 80-mm rod
AF that is welded to a cross consisting of four
200-mm arms. The assembly is supported by a
ball-and-socket joint at F and by three short
links, each of which forms an angle of 45&deg; with
(a) the tension in each link, (b) the reaction at F.
SOLUTION
rE/F = −200 i + 80 j
TB = TB (i − j) / 2
rB/F = 80 j − 200k
TC = TC (− j + k ) / 2 rC/F = 200i + 80 j
TD = TD (− i + j) / 2
rD/E = 80 j + 200k
ΣM F = 0: rB/F &times; TB + rC/F &times; TC + rD/F &times; TD + rE/F &times; (− Pj) = 0
i j
k
i
j k
i
j
k
i
j k
TC
TB
TD
0 80 −200
+ 200 80 0
+ 0 80 200
+ −200 80 0 = 0
2
2
2
1 −1
0
0 −1 1
0
−1 −1 0
−P 0
Equate coefficients of unit vectors to zero and multiply each equation by 2.
i:
−200 TB + 80 TC + 200 TD = 0
(1)
j:
−200 TB − 200 TC − 200 TD = 0
(2)
k:
−80 TB − 200 TC + 80 TD + 200 2 P = 0
(3)
−80 TB − 80 TC − 80 TD = 0
(4)
80
(2):
200
Eqs. (3) + (4):
−160TB − 280TC + 200 2 P = 0
Eqs. (1) + (2):
−400TB − 120TC = 0
TB = −
(5)
120
TC − 0.3TC
400
(6)
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505
PROBLEM 4.127 (Continued)
Eqs. (6)
−160( −0.3TC ) − 280TC + 200 2 P = 0
(5):
−232TC + 200 2 P = 0
TC = 1.2191P
TC = 1.219 P 
From Eq. (6):
TB = −0.3(1.2191P) = − 0.36574 = P
From Eq. (2):
− 200(− 0.3657 P) − 200(1.2191P) − 200Tθ D = 0
TD = − 0.8534 P
ΣF = 0:
TB = −0.366 P 
TD = − 0.853P 
F + TB + TC + TD − Pj = 0
i : Fx +
(− 0.36574 P)
2
−
Fx = − 0.3448P
j: Fy −
k : Fz +
=0
2
Fx = − 0.345P
(− 0.36574 P)
2
Fy = P
( − 0.8534 P)
−
(1.2191P)
2
−
(− 0.8534 P)
2
− 200 = 0
Fy = P
(1.2191P)
2
=0
Fz = − 0.8620 P Fz = − 0.862 P
F = − 0.345P i + Pj − 0.862Pk 
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506
PROBLEM 4.128
The uniform 10-kg rod AB is supported by a ball-and-socket joint at
A and by the cord CG that is attached to the midpoint G of the rod.
Knowing that the rod leans against a frictionless vertical wall at B,
determine (a) the tension in the cord, (b) the reactions at A and B.
SOLUTION
Free-Body Diagram:
Five unknowns and six equations of equilibrium, but equilibrium is
maintained (ΣMAB = 0).
W = mg
= (10 kg) 9.81m/s 2
W = 98.1 N

GC = − 300i + 200 j − 225k GC = 425 mm

GC
T
T =T
=
(− 300i + 200 j − 225k )
GC 425
rB/ A = − 600i + 400 j + 150 mm
rG/ A = − 300i + 200 j + 75 mm
ΣMA = 0: rB/ A &times; B + rG/ A &times; T + rG/ A &times; (− W j) = 0
i
j
k
i
j
k
i
j
k
T
− 600 400 150 + − 300 200
75
+ − 300 200 75
425
B
0
0
− 300 200 − 225
0
− 98.1 0
Coefficient of i : (−105.88 − 35.29)T + 7357.5 = 0
T = 52.12 N
T = 52.1 N 


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507
PROBLEM 4.128 (Continued)
Coefficient of j : 150 B − (300 &times; 75 + 300 &times; 225)
52.12
=0
425
B = 73.58 N
B = (73.6 N)i 
ΣF = 0: A + B + T − W j = 0
Coefficient of i :
Coefficient of j:
Coefficient of k :
300
=0
425
Ax = −36.8 N 
200
− 98.1 = 0
425
Ay = 73.6 N 
Ax + 73.58 − 52.15
Ay + 52.15
Az − 52.15
225
=0
425
Az = 27.6 N 
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508
PROBLEM 4.129
Three rods are welded together to form a “corner” that is supported
by three eyebolts. Neglecting friction, determine the reactions at
A, B, and C when P = 240 lb, a = 12 in., b = 8 in., and c = 10 in.
SOLUTION
From F.B.D. of weldment:
ΣM O = 0: rA/O &times; A + rB/O &times; B + rC/O &times; C = 0
i
12
0
j
0
Ay
k
i
0 + 0
Az Bx
j k
i
8 0 + 0
0 Bz
Cx
j k
0 10 = 0
Cy 0
(−12 Az j + 12 Ay k ) + (8 Bz i − 8 Bx k ) + (−10 C y i + 10 Cx j) = 0
From i-coefficient:
Bz = 1.25C y
or
j-coefficient:
k-coefficient:
or
(3)
ΣF = 0: A + B + C − P = 0 
( Bx + Cx )i + ( Ay + C y − 240 lb) j + ( Az + Bz )k = 0 
From i-coefficient:
Bx + Cx = 0
C x = − Bx
or
j-coefficient:
or
(2)
12 Ay − 8 Bx = 0
Bx = 1.5 Ay
or
(1)
−12 Az + 10 C x = 0
Cx = 1.2 Az
or

8 Bz − 10 C y = 0
(4)
Ay + C y − 240 lb = 0
Ay + C y = 240 lb
(5)
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509
PROBLEM 4.129 (Continued)
k-coefficient:
Az + Bz = 0
Az = − Bz
or
(6)
Substituting Cx from Equation (4) into Equation (2),
− Bz = 1.2 Az
(7)
Using Equations (1), (6), and (7),
Cy =
Bz
− Az
1  Bx  Bx
=
=
=
1.25 1.25 1.25  1.2  1.5
(8)
From Equations (3) and (8):
Cy =
1.5 Ay
1.5
or C y = Ay
and substituting into Equation (5),
2 Ay = 240 lb
Ay = C y = 120 lb
(9)
Using Equation (1) and Equation (9),
Bz = 1.25(120 lb) = 150.0 lb
Using Equation (3) and Equation (9),
Bx = 1.5(120 lb) = 180.0 lb
From Equation (4):
Cx = −180.0 lb
From Equation (6):
Az = −150.0 lb
Therefore,
A = (120.0 lb) j − (150.0 lb)k 

B = (180.0 lb)i + (150.0 lb)k 

C = −(180.0 lb)i + (120.0 lb) j 
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510
PROBLEM 4.130
Solve Problem 4.129, assuming that the force P is removed and is
replaced by a couple M = +(600 lb ⋅ in.)j acting at B.
PROBLEM 4.129 Three rods are welded together to form a “corner”
that is supported by three eyebolts. Neglecting friction, determine the
reactions at A, B, and C when P = 240 lb, a = 12 in., b = 8 in., and
c = 10 in.
SOLUTION
From F.B.D. of weldment:
ΣM O = 0: rA/O &times; A + rB/O &times; B + rC/O &times; C + M = 0
i
12
0
j
0
Ay
k
i
0 + 0
Az Bx
j k
i
8 0 + 0
0 Bz
Cx
j k
0 10 + (600 lb ⋅ in.) j = 0
Cy 0
(−12 Az j + 12 Ay k ) + (8 Bz j − 8 Bx k ) + ( −10C y i + 10C x j) + (600 lb ⋅ in.) j = 0
From i-coefficient:
8 Bz − 10 C y = 0
C y = 0.8Bz
or
j-coefficient:
(1)
−12 Az + 10 Cx + 600 = 0
Cx = 1.2 Az − 60
or
k-coefficient:
(2)
12 Ay − 8 Bx = 0
Bx = 1.5 Ay
or
(3)

ΣF = 0: A + B + C = 0 

( Bx + C x )i + ( Ay + C y ) j + ( Az + Bz )k = 0 
From i-coefficient:
C x = − Bx
(4)
j-coefficient:
C y = − Ay
(5)
k-coefficient:
Az = − Bz
(6)
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511
PROBLEM 4.130 (Continued)
Substituting Cx from Equation (4) into Equation (2),
B 
Az = 50 −  x 
 1.2 
(7)
2
C y = 0.8 Bz = − 0.8 Az =   Bx − 40
3
(8)
Using Equations (1), (6), and (7),
From Equations (3) and (8):
C y = Ay − 40
Substituting into Equation (5),
2 Ay = 40
Ay = 20.0 lb
From Equation (5):
C y = −20.0 lb
Equation (1):
Bz = −25.0 lb
Equation (3):
Bx = 30.0 lb
Equation (4):
Cx = −30.0 lb
Equation (6):
Az = 25.0 lb
A = (20.0 lb) j + (25.0 lb)k 
Therefore,

 B = (30.0 lb)i − (25.0 lb)k 

C = − (30.0 lb)i − (20.0 lb) j 
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512
PROBLEM 4.131
In order to clean the clogged drainpipe AE, a plumber has
disconnected both ends of the pipe and inserted a power
snake through the opening at A. The cutting head of the
snake is connected by a heavy cable to an electric motor that
rotates at a constant speed as the plumber forces the cable
into the pipe. The forces exerted by the plumber and
the motor on the end of the cable can be represented by
the wrench F = −(48 N)k , M = −(90 N ⋅ m)k. Determine the
additional reactions at B, C, and D caused by the cleaning
operation. Assume that the reaction at each support consists
of two force components perpendicular to the pipe.
SOLUTION
From F.B.D. of pipe assembly ABCD:
ΣFx = 0: Bx = 0
ΣM D ( x -axis) = 0: (48 N)(2.5 m) − Bz (2 m) = 0
Bz = 60.0 N
and B = (60.0 N)k 
ΣM D ( z -axis) = 0: C y (3 m) − 90 N ⋅ m = 0
C y = 30.0 N
ΣM D ( y -axis) = 0: − C z (3 m) − (60.0 N)(4 m) + (48 N)(4 m) = 0
Cz = −16.00 N
and C = (30.0 N) j − (16.00 N)k 
ΣFy = 0: Dy + 30.0 = 0
Dy = −30.0 N
ΣFz = 0: Dz − 16.00 N + 60.0 N − 48 N = 0
Dz = 4.00 N
and D = − (30.0 N) j + (4.00 N)k 
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513
PROBLEM 4.132
Solve Problem 4.131, assuming that the plumber exerts a force
F = −(48 N)k and that the motor is turned off (M = 0).
PROBLEM 4.131 In order to clean the clogged drainpipe AE, a
plumber has disconnected both ends of the pipe and inserted a
power snake through the opening at A. The cutting head of the snake
is connected by a heavy cable to an electric motor that rotates at a
constant speed as the plumber forces the cable into the pipe. The
forces exerted by the plumber and the motor on the end of the cable
can be represented by the wrench F = −(48 N)k , M = −(90 N ⋅ m)k.
Determine the additional reactions at B, C, and D caused by the
cleaning operation. Assume that the reaction at each support consists
of two force components perpendicular to the pipe.
SOLUTION
From F.B.D. of pipe assembly ABCD:
ΣFx = 0: Bx = 0
ΣM D ( x -axis) = 0: (48 N)(2.5 m) − Bz (2 m) = 0
Bz = 60.0 N
and B = (60.0 N)k 
ΣM D ( z -axis) = 0: C y (3 m) − Bx (2 m) = 0
Cy = 0
ΣM D ( y -axis) = 0: C z (3 m) − (60.0 N)(4 m) + (48 N)(4 m) = 0
Cz = −16.00 N
and C = − (16.00 N)k 
ΣFy = 0: Dy + C y = 0
Dy = 0
ΣFz = 0: Dz + Bz + C z − F = 0
Dz + 60.0 N − 16.00 N − 48 N = 0
Dz = 4.00 N
and D = (4.00 N)k 
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514
PROBLEM 4.133
The 50-kg plate ABCD is supported by hinges along edge AB
and by wire CE. Knowing that the plate is uniform, determine
the tension in the wire.
SOLUTION
Free-Body Diagram:
W = mg = (50 kg)(9.81 m/s 2 )
W = 490.50 N

CE = − 240i + 600 j − 400k
CE = 760 mm

CE
T
=
(− 240i + 600 j − 400k )
T =T
CE 760

AB 480i − 200 j 1
λ AB =
=
= (12i − 5 j)
AB
520
13
ΣMAB = 0: λ AB ⋅ (rE/ A &times; T ) + λ AB ⋅ (rG/ A &times; − W j) = 0
rE/ A = 240i + 400 j; rG/ A = 240i − 100 j + 200k
12
0
12
0
−5
−5
1
T
240 400
0
+ 240 −100 200
=0
13 &times; 20
13
− 240 600 − 400
0
−W
0
(−12 &times; 400 &times; 400 − 5 &times; 240 &times; 400)
T
+ 12 &times; 200W = 0
760
T = 0.76W = 0.76(490.50 N)
T = 373 N 
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515
PROBLEM 4.134
Solve Problem 4.133, assuming that wire CE is replaced by a
wire connecting E and D.
PROBLEM 4.133 The 50-kg plate ABCD is supported by
hinges along edge AB and by wire CE. Knowing that the plate
is uniform, determine the tension in the wire.
SOLUTION
Free-Body Diagram:
Dimensions in mm
W = mg = (50 kg)(9.81 m/s 2 )
W = 490.50 N

DE = − 240i + 400 j − 400k
DE = 614.5 mm

DE
T
=
(240i + 400 j − 400k )
T =T
DE 614.5

AB 480i − 200 j 1
λ AB =
=
= (12i − 5 j)
AB
520
13
rE/ A = 240i + 400 j; rG/ A = 240i − 100 j + 200k
12 −5
0
12
5
0
1
T
240 400
0
+ 240 −100 200
=0
13 &times; 614.5
13
240 400 − 400
0
−W
0
(−12 &times; 400 &times; 400 − 5 &times; 240 &times; 400)
T
+ 12 &times; 200 &times; W = 0
614.5
T = 0.6145W = 0.6145(490.50 N)
T = 301 N 
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516
PROBLEM 4.135
Two rectangular plates are welded together to form the assembly
shown. The assembly is supported by ball-and-socket joints at B
and D and by a ball on a horizontal surface at C. For the loading
shown, determine the reaction at C.
SOLUTION
λ BD =
First note:
−(6 in.)i − (9 in.) j + (12 in.)k
(6) 2 + (9) 2 + (12)2 in.
1
(− 6i − 9 j + 12k )
16.1555
= − (6 in.)i
=
rA/B
P = (80 lb)k
rC/D = (8 in.)i
C = (C ) j
From the F.B.D. of the plates:
ΣM BD = 0: λ BD ⋅ (rA/B &times; P ) + λ BD ⋅ ( rC/D &times; C ) = 0
−6 −9 12
−6 −9 12
 6(80) 
 C (8) 
−1 0 0 
+ 1 0 0 
=0

16.1555 
16.1555 
0 0 1
0 1 0
( −9)(6)(80) + (12)(8)C = 0
C = 45.0 lb
or C = (45.0 lb) j 
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517
PROBLEM 4.136
Two 2 &times; 4-ft plywood panels, each of weight 12 lb, are nailed
together as shown. The panels are supported by ball-and-socket
joints at A and F and by the wire BH. Determine (a) the location
of H in the xy plane if the tension in the wire is to be minimum,
(b) the corresponding minimum tension.
SOLUTION
Free-Body Diagram:

AF = 4i − 2 j − 4k
AF = 6 ft
1
λ AF = (2i − j − 2k )
3
rG1/A = 2i − j
rG2 /A = 4i − j − 2k
rB/A = 4i
ΣM AF = 0: λ AF ⋅ (rG1 /A &times; ( −12 j) + λ AF ⋅ (rG2 /A &times; (−12 j)) + λ AF ⋅ (rB /A &times; T ) = 0
2 −1 −2
2 −1 −2
1
1
+ 4 −1 −2 + λ AF ⋅ (rB/A &times; T) = 0
2 −1 0
3
3
0 −12 0
0 −12 0
1
1
(2 &times; 2 &times; 12) + (−2 &times; 2 &times; 12 + 2 &times; 4 &times; 12) + λ AF ⋅ (rB/A &times; T) = 0
3
3
λ AF ⋅ (rB/A &times; T) = −32 or T ⋅ (λ A/F &times; rB/A ) = −32
(1)
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518
PROBLEM 4.136 (Continued)
Projection of T on (λ AF &times; rB/A ) is constant. Thus, Tmin is parallel to
1
1
λ AF &times; rB/A = (2i − j − 2k ) &times; 4i = (−8 j + 4k )
3
3
Corresponding unit vector is
1
5
(−2 j + k ).
Tmin = T ( −2 j + k )
From Eq. (1):
1
(2)
5
T
1

(−2 j + k ) ⋅  (2i − j − 2k ) &times; 4i  = −32
5
3

T
1
(−2 j + k ) ⋅ ( −8 j + 4k ) = −32
3
5
T
(16 + 4) = −32
3 5
T =−
3 5(32)
= 4.8 5
20
T = 10.7331 lb
From Eq. (2):
Tmin = T ( −2 j + k )
1
5
= 4.8 5( −2 j + k )
Tmin
1
5
= −(9.6 lb)j + (4.8 lb k )
Since Tmin has no i component, wire BH is parallel to the yz plane, and x = 4 ft.
(a)
(b)
x = 4.00 ft;
y = 8.00 ft 
Tmin = 10.73 lb 
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519
PROBLEM 4.137
Solve Problem 4.136, subject to the restriction that H must lie on
the y-axis.
PROBLEM 4.136 Two 2 &times; 4-ft plywood panels, each of weight 12 lb,
are nailed together as shown. The panels are supported by ball-andsocket joints at A and F and by the wire BH. Determine (a) the
location of H in the xy plane if the tension in the wire is to be
minimum, (b) the corresponding minimum tension.
SOLUTION
Free-Body Diagram:

AF = 4i − 2 j − 4k
1
λ AF = (2i − j − 2k )
3
rG1/A = 2i − j
rG2 /A = 4i − j − 2k
rB/A = 4i
ΣMAF = 0: λ AF ⋅ (rG/A &times; (−12 j) + λ AF ⋅ (rG2 /A &times; (−12 j)) + λ AF ⋅ (rB/A &times; T ) = 0
2 −1 2
2 −1 −2
1
1
2 −1 0 + 4 −1 −2 + λ AF ⋅ (rB/A &times; T) = 0
3
3
0 −12 0
0 −12 0
1
1
(2 &times; 2 &times; 12) + (−2 &times; 2 &times; 12 + 2 &times; 4 &times; 12) + λ AF ⋅ (rB/A &times; T) = 0
3
3
λ AF ⋅ (rB/A &times; T) = −32

BH = −4i + yj − 4k
BH = (32 + y 2 )1/2

BH
−4i + yj − 4k
=T
T=T
BH
(32 + y 2 )1/ 2
(1)
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520
PROBLEM 4.137 (Continued)
From Eq. (1):
λ AF
2 −1 −2
T
⋅ (rB/A &times; T ) = 4 0 0
= −32
3(32 + y 2 )1/2
−4 y −4
(−16 − 8 y )T = −3 &times; 32(32 + y 2 )1/2
T = 96
(32 + y 2 )1/2
8 y + 16
(2)
(8y +16) 12 (32 + y 2 ) −1/ 2 (2 y ) + (32 + y 2 )1/2 (8)
dT
= 0: 96
dy
(8 y + 16) 2
Numerator = 0:
(8 y + 16) y = (32 + y 2 )8
8 y 2 + 16 y = 32 &times; 8 + 8 y 2
From Eq. (2):
T = 96
(32 + 162 )1/ 2
= 11.3137 lb
8 &times; 16 + 16
x = 0 ft; y = 16.00 ft 
Tmin = 11.31 lb 
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521
PROBLEM 4.138
The frame ACD is supported by ball-and-socket joints at A
and D and by a cable that passes through a ring at B and is
attached to hooks at G and H. Knowing that the frame
supports at Point C a load of magnitude P = 268 N, determine
the tension in the cable.
SOLUTION
Free-Body Diagram:
TBG
TBH

AD (1 m)i − (0.75 m)k
=
=
1.25 m
= 0.8i − 0.6k

BG
= TBG
BG
−0.5i + 0.925 j − 0.4k
= TBG
1.125

BH
= TBH
BH
0.375i + 0.75 j − 0.75k
= TBH
1.125
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522
PROBLEM 4.138 (Continued)
rB /A = (0.5 m)i; rC/A = (1 m)i; P = −(268 N)j
To eliminate the reactions at A and D, we shall write
λ AD ⋅ (rB /A &times; TBG ) + λ AD ⋅ (rB /A &times; TBH ) + λ AD ⋅ (rC /A &times; P) = 0
(1)
Substituting for terms in Eq. (1) and using determinants,
−0.6
−0.6
−0.6
0.8
0
0.8
0
0.8
0
TBG
TBH
+ 0.5
+ 1
0.5
0
0
0
0
0
0 =0
1.125
1.125
−0.5 0.925 −0.4
0.375 0.75 −0.75
0 −268
0
Multiplying all terms by (–1.125),
0.27750TBG + 0.22500TBH = 180.900
For this problem,
(2)
TBG = TBH = T
(0.27750 + 0.22500)T = 180.900
T = 360 N 
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523
PROBLEM 4.139
Solve Prob. 4.138, assuming that cable GBH is replaced by
a cable GB attached at G and B.
PROBLEM 4.138 The frame ACD is supported by ball-andsocket joints at A and D and by a cable that passes through a
ring at B and is attached to hooks at G and H. Knowing that the
frame supports at Point C a load of magnitude P = 268 N,
determine the tension in the cable.
SOLUTION
Free-Body Diagram:
TBG
TBH

AD (1 m)i − (0.75 m)k
=
=
1.25 m
= 0.8i − 0.6k

BG
= TBG
BG
−0.5i + 0.925 j − 0.4k
= TBG
1.125

BH
= TBH
BH
0.375i + 0.75 j − 0.75k
= TBH
1.125
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524
PROBLEM 4.139 (Continued)
rB /A = (0.5 m)i; rC/A = (1 m)i; P = −(268 N)j
To eliminate the reactions at A and D, we shall write
ΣM AD = 0: λ AD ⋅ (rB /A &times; TBG ) + λ AD ⋅ (rB /A &times; TBH ) + λ AD ⋅ (rC /A &times; P) = 0
(1)
Substituting for terms in Eq. (1) and using determinants,
−0.6
−0.6
−0.6
0.8
0
0.8
0
0.8
0
TBG
TBH
+ 0.5
+ 1
0.5
0
0
0
0
0
0 =0
1.125
1.125
−0.5 0.925 −0.4
0.375 0.75 −0.75
0 −268
0
Multiplying all terms by (–1.125),
0.27750TBG + 0.22500TBH = 180.900
(2)
For this problem, TBH = 0.
Thus, Eq. (2) reduces to
0.27750TBG = 180.900
TBG = 652 N 
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525
PROBLEM 4.140
The bent rod ABDE is supported by ball-and-socket joints at A
and E and by the cable DF. If a 60-lb load is applied at C as
shown, determine the tension in the cable.
SOLUTION
Free-Body Diagram:

DF = −16i + 11j − 8k
DF = 21 in.

DE T
T=T
= (−16i + 11j − 8k )
DF 21
rD/E = 16i
rC/E = 16i − 14k

EA 7i − 24k
=
λ EA =
EA
25
ΣM EA = 0: λ EA ⋅ (rB/E &times; T) + λ EA ⋅ (rC/E ⋅ (− 60 j)) = 0
7
0 −24
7
0 −24
1
T
16 0
0
+ 16 0 −14
=0
21 &times; 25
25
0 −60 0
−16 11 −8
−
24 &times; 16 &times; 11
−7 &times; 14 &times; 60 + 24 &times; 16 &times; 60
T+
=0
21 &times; 25
25
201.14 T + 17,160 = 0
T = 85.314 lb
T = 85.3 lb 
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526
PROBLEM 4.141
Solve Problem 4.140, assuming that cable DF is replaced by a
cable connecting B and F.
SOLUTION
Free-Body Diagram:
rB/ A = 9i
rC/ A = 9i + 10k

BF = −16i + 11j + 16k
BF = 25.16 in.

BF
T
=
(−16i + 11j + 16k )
T =T
BF 25.16

AE 7i − 24k
λ AE =
=
AE
25
ΣMAE = 0: λ AF ⋅ (rB/ A &times; T) + λ AE ⋅ (rC/ A ⋅ (− 60 j)) = 0
7
0 −24
7 0 −24
1
T
9
0
0
+9 0
10
=0
25 &times; 25.16
25
−16 11 16
0 −60 0
−
24 &times; 9 &times; 11
24 &times; 9 &times; 60 + 7 &times; 10 &times; 60
T+
=0
25 &times; 25.16
25
94.436 T − 17,160 = 0
T = 181.7 lb 
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527
PROBLEM 4.142
A gardener uses a 60-N wheelbarrow to transport a 250-N bag of
fertilizer. What force must she exert on each handle?
SOLUTION
Free-Body Diagram:

ΣM A = 0: (2 F )(1 m) − (60 N)(0.15 m) − (250 N)(0.3 m) = 0
F = 42.0 N 

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528
PROBLEM 4.143
The required tension in cable AB is 200 lb. Determine (a) the vertical
force P that must be applied to the pedal, (b) the corresponding
reaction at C.
SOLUTION
Free-Body Diagram:
BC = 7 in.
(a)
ΣM C = 0: P(15 in.) − (200 lb)(6.062 in.) = 0
P = 80.83 lb
(b)
P = 80.8 lb 
ΣFy = 0: C x − 200 lb = 0
C x = 200 lb
ΣFy = 0: C y − P = 0 C y − 80.83 lb = 0
C y = 80.83 lb
α = 22.0&deg;
C = 215.7 lb
C = 216 lb
22.0&deg; 
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529
PROBLEM 4.144
A lever AB is hinged at C and attached to a control cable at A. If
the lever is subjected to a 500-N horizontal force at B, determine
(a) the tension in the cable, (b) the reaction at C.
SOLUTION
Triangle ACD is isosceles with  C = 90&deg; + 30&deg; = 120&deg;  A =  D =
1
(180&deg; − 120&deg;) = 30&deg;.
2
Thus, DA forms angle of 60&deg; with the horizontal axis.
(a)
We resolve FAD into components along AB and perpendicular to AB.
ΣM C = 0: ( FAD sin 30&deg;)(250 mm) − (500 N)(100 mm) = 0
(b)
ΣFx = 0: − (400 N) cos 60&deg; + C x − 500 N = 0
C x = +300 N
ΣFy = 0: − (400 N) sin 60&deg; + C y = 0
C y = +346.4 N
C = 458 N
49.1&deg; 
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530
PROBLEM 4.145
A force P of magnitude 280 lb is applied to member ABCD, which is
supported by a frictionless pin at A and by the cable CED. Since the cable
passes over a small pulley at E, the tension may be assumed to be the
same in portions CE and ED of the cable. For the case when a = 3 in.,
determine (a) the tension in the cable, (b) the reaction at A.
SOLUTION
Free-Body Diagram:
(a)
ΣM A = 0: − (280 lb)(8 in.)
7
T (12 in.)
25
24
− T (8 in.) = 0
25
T (12 in.) −
(12 − 11.04)T = 840
(b)
ΣFx = 0:
T = 875 lb 
7
(875 lb) + 875 lb + Ax = 0
25
Ax = −1120
ΣFy = 0: Ay − 280 lb −
Ay = +1120
A x = 1120 lb
24
(875 lb) = 0
25
A y = 1120 lb
A = 1584 lb
45.0&deg; 
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531
PROBLEM 4.146
Two slots have been cut in plate DEF, and the plate has
been placed so that the slots fit two fixed, frictionless pins A
and B. Knowing that P = 15 lb, determine (a) the force each
pin exerts on the plate, (b) the reaction at F.
SOLUTION
Free-Body Diagram:
(a)
(b)
ΣFx = 0: 15 lb − B sin 30&deg; = 0
B = 30.0 lb
60.0&deg; 
ΣM A = 0: − (30 lb)(4 in.) + B sin 30&deg;(3 in.) + B cos 30&deg;(11 in.) − F (13 in.) = 0
−120 lb ⋅ in. + (30 lb) sin 30&deg;(3 in.) + (30 lb) cos 30&deg;(11 in.) − F (13 in.) = 0
F = + 16.2145 lb
(a)
F = 16.21 lb 
ΣFy = 0: A − 30 lb + B cos 30&deg; − F = 0
A − 30 lb + (30 lb) cos 30&deg; − 16.2145 lb = 0
A = + 20.23 lb
A = 20.2 lb 
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532
PROBLEM 4.147
Knowing that the tension in wire BD is 1300 N, determine the
reaction at the fixed support C of the frame shown.
SOLUTION
T = 1300 N
5
Tx = T
13
= 500 N
12
Ty = T
13
= 1200 N
ΣM x = 0: C x − 450 N + 500 N = 0
C x = −50 N
ΣFy = 0: C y − 750 N − 1200 N = 0 C y = +1950 N
C x = 50 N
C y = 1950 N
C = 1951 N
88.5&deg; 
ΣM C = 0: M C + (750 N)(0.5 m) + (4.50 N)(0.4 m)
− (1200 N)(0.4 m) = 0
M C = −75.0 N ⋅ m
M C = 75.0 N ⋅ m

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533
PROBLEM 4.148
The spanner shown is used to rotate a shaft. A pin fits in a
hole at A, while a flat, frictionless surface rests against the
shaft at B. If a 60-lb force P is exerted on the spanner at D,
find the reactions at A and B.
SOLUTION
Free-Body Diagram:
(Three-force body)
The line of action of A must pass through D, where B and P intersect.
3sin 50&deg;
3cos 50&deg; + 15
= 0.135756
α = 7.7310&deg;
tan α =
60 lb
sin 7.7310&deg;
= 446.02 lb
60 lb
B=
tan 7.7310&deg;
= 441.97 lb
A=
Force triangle
A = 446 lb
7.73&deg; 
B = 442 lb


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534
PROBLEM 4.149
Rod AB is supported by a pin and bracket at A and rests against a frictionless
peg at C. Determine the reactions at A and C when a 170-N vertical force is
applied at B.
SOLUTION
Free-Body Diagram:
(Three-force body)
The reaction at A must pass through D where C and the 170-N force intersect.
160 mm
300 mm
α = 28.07&deg;
tan α =
We note that triangle ABD is isosceles (since AC = BC) and, therefore,
 CAD = α = 28.07&deg;
Also, since CD ⊥ CB, reaction C forms angle α = 28.07&deg; with the horizontal axis.
Force triangle
We note that A forms angle 2α with the vertical axis. Thus, A and C form angle
180&deg; − (90&deg; − α ) − 2α = 90&deg; − α
Force triangle is isosceles, and we have
A = 170 N
C = 2(170 N)sin α
= 160.0 N
A = 170.0 N
33.9&deg;;
C = 160.0 N
28.1&deg; 
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535
PROBLEM 4.150
The 24-lb square plate shown is supported by three vertical
wires. Determine (a) the tension in each wire when a = 10 in.,
(b) the value of a for which the tension in each wire is 8 lb.
SOLUTION
rB/A = ai + 30k
rC/A = 30i + ak
rG/A = 15i + 15k
By symmetry, B = C.
ΣM A = 0: rB/A &times; Bj + rC &times; Cj + rG/A &times; (−W j) = 0
(ai + 30k ) &times; Bj + (30i + ak ) &times; Bj + (15i + 15k ) &times; (−W j) = 0
Bak − 30 Bi + 30 Bk − Bai − 15Wk + 15W i = 0
Equate coefficient of unit vector i to zero:
i : − 30 B − Ba + 15W = 0
B=
15W
30 + a
C=B=
15W
30 + a
(1)
ΣFy = 0: A + B + C − W = 0
 15W 
A+ 2
 − W = 0;
 30 + a 
(a)
For
a = 10 in.
From Eq. (1):
C=B=
From Eq. (2):
A=
A=
aW
30 + a
(2)
15(24 lb)
= 9.00 lb
30 + 10
10(24 lb)
= 6.00 lb
30 + 10
A = 6.00 lb; B = C = 9.00 lb 
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536
PROBLEM 4.150 (Continued)
(b)
For tension in each wire = 8 lb,
From Eq. (1):
8 lb =
15(24 lb)
30 + a
30 in. + a = 45
a = 15.00 in. 
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537
PROBLEM 4.151
Frame ABCD is supported by a ball-and-socket joint at A and
by three cables. For a = 150 mm, determine the tension in
each cable and the reaction at A.
SOLUTION
First note:
TDG = λ DG TDG =
−(0.48 m)i + (0.14 m)j
(0.48) 2 + (0.14) 2 m
TDG
−0.48i + 0.14 j
TDG
0.50
T
= DG (24i + 7 j)
25
=
TBE = λ BE TBE =
−(0.48 m)i + (0.2 m)k
(0.48)2 + (0.2)2 m
TBE
−0.48i + 0.2k
TBE
0.52
T
= BE (−12 j + 5k )
13
=
From F.B.D. of frame ABCD:
 7

ΣM x = 0:  TDG  (0.3 m) − (350 N)(0.15 m) = 0
 25

TDG = 625 N 
or
 24

 5

ΣM y = 0:  &times; 625 N  (0.3 m) −  TBE  (0.48 m) = 0
13
 25



TBE = 975 N 
or
 7

ΣM z = 0: TCF (0.14 m) +  &times; 625 N  (0.48 m) − (350 N)(0.48 m) = 0
 25

TCF = 600 N 
or
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538
PROBLEM 4.151 (Continued)
ΣFx = 0: Ax + TCF + (TBE ) x + (TDG ) x = 0
 12
  24

Ax − 600 N −  &times; 975 N  −  &times; 625 N  = 0
 13
  25

Ax = 2100 N
ΣFy = 0: Ay + (TDG ) y − 350 N = 0
 7

Ay +  &times; 625 N  − 350 N = 0
25


Ay = 175.0 N
ΣFz = 0: Az + (TBE ) z = 0
 5

Az +  &times; 975 N  = 0
13


Az = −375 N
A = (2100 N)i + (175.0 N) j − (375 N)k 
Therefore,
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539
PROBLEM 4.152
The pipe ACDE is supported by ball-and-socket joints at A and E and
by the wire DF. Determine the tension in the wire when a 640-N load is
applied at B as shown.
SOLUTION
Free-Body Diagram:
Dimensions in mm

AE = 480i + 160 j − 240k
AE = 560 mm

AE 480i + 160 j − 240k
λ AE =
=
AE
560
6i + 2 j − 3k
λ AE =
7
rB/A = 200i
rD/A = 480i + 160 j

DF = −480i + 330 j − 240k ;
DF = 630 mm

DF
−480i + 330 j − 240k
−16i + 11j − 8k
= TDF
= TDF
TDF = TDF
DF
630
21
ΣM AE = λ AE ⋅ (rD/A &times; TDF ) + λ AE ⋅ (rB/A &times; (−600 j)) = 0
6
2 −3
6
2
−3
TDF
1
480 160 0
0
0 =0
+ 200
21 &times; 7
7
0 −640 0
−16 11 −8
3 &times; 200 &times; 640
−6 &times; 160 &times; 8 + 2 &times; 480 &times; 8 − 3 &times; 480 &times; 11 − 3 &times; 160 &times; 16
TDF +
=0
21 &times; 7
7
−1120TDF + 384 &times; 103 = 0
TDF = 342.86 N
TDF = 343 N 
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540
PROBLEM 4.153
A force P is applied to a bent rod ABC, which may be supported in four different ways as shown. In each case,
if possible, determine the reactions at the supports.
SOLUTION
(a)
ΣM A = 0: − Pa + (C sin 45&deg;)2a + (cos 45&deg;)a = 0
3
C
2
=P
C=
2
P
3
 2  1
ΣFx = 0: Ax − 
P
 3  2


Ax =
P
3
 2  1
P
ΣFy = 0: Ay − P + 
 3  2


Ay =
2P
3
C = 0.471P
A = 0.745P
(b)
45&deg; 
63.4&deg; 
ΣM C = 0: +Pa − ( A cos 30&deg;)2a + ( A sin 30&deg;)a = 0
A(1.732 − 0.5) = P
A = 0.812 P
A = 0.812 P
ΣFx = 0: (0.812 P )sin 30&deg; + C x = 0
60.0&deg; 
Cx = −0.406 P
ΣFy = 0: (0.812 P) cos 30&deg; − P + C y = 0 C y = −0.297 P
C = 0.503P
36.2&deg; 
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541
PROBLEM 4.153 (Continued)
(c)
ΣM C = 0: + Pa − ( A cos 30&deg;)2a + ( A sin 30&deg;)a = 0
A(1.732 + 0.5) = P
A = 0.448P
A = 0.448P
ΣFx = 0: − (0.448P) sin 30&deg; + Cx = 0
60.0&deg; 
Cx = 0.224 P
ΣFy = 0: (0.448 P) cos 30&deg; − P + C y = 0 C y = 0.612 P
C = 0.652 P
69.9&deg; 

(d)
Force T exerted by wire and reactions A and C all intersect at Point D.
ΣM D = 0: Pa = 0
Equilibrium is not maintained.
Rod is improperly constrained. 
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542
PROBLEM 4.F1
needed to determine the reactions at A and E when α = 30&deg;.
SOLUTION
Free-Body Diagram of Frame:


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543
PROBLEM 4.F2
Neglecting friction, draw the free-body diagram needed to determine the
tension in cable ABD and the reaction at C when θ = 60&deg;.
SOLUTION
Free-Body Diagram of Member ACD:

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544
PROBLEM 4.F3
Bar AC supports two 400-N loads as shown. Rollers at A
and C rest against frictionless surfaces and a cable BD is
attached at B. Draw the free-body diagram needed to
determine the tension in cable BD and the reactions at
A and C.
SOLUTION
Free-Body Diagram of Bar AC:

Note: By similar triangles
yB
0.15 m
=
0.25 m 0.5 m
yB = 0.075 m 
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545
PROBLEM 4.F4
Draw the free-body diagram needed to determine the tension
in each cable and the reaction at D.
SOLUTION
Free-Body Diagram of Member ABCD:

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546
PROBLEM 4.F5
A 4 &times; 8-ft sheet of plywood weighing 34 lb has been temporarily
placed among three pipe supports. The lower edge of the sheet
rests on small collars at A and B and its upper edge leans against
pipe C. Neglecting friction on all surfaces, draw the free-body
diagram needed to determine the reactions at A, B, and C.
SOLUTION
Free-Body Diagram of Plywood sheet:

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547
PROBLEM 4.F6
Two transmission belts pass over sheaves welded to an
axle supported by bearings at B and D. The sheave at A
has a radius of 2.5 in. and the sheave at C has a radius
of 2 in. Knowing that the system rotates at a constant
rate, draw the free-body diagram needed to determine
the tension T and the reactions at B and D. Assume that
the bearing at D does not exert any axial thrust and
neglect the weights of the sheaves and axle.
SOLUTION
Free-Body Diagram of axle-sheave system:

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548
PROBLEM 4.F7
The 6-m pole ABC is acted upon by a 455-N force as shown.
The pole is held by a ball-and-socket joint at A and by two
cables BD and BE. Draw the free-body diagram needed to
determine the tension in each cable and the reaction at A.
SOLUTION
Free-Body Diagram of Pole:

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549
CHAPTER 5
PROBLEM 5.1
Locate the centroid of the plane area shown.
SOLUTION
A, in 2
x , in
y , in
x A,in 3
y A,in 3
1
8
0.5
4
4
32
2
3
2.5
2.5
7.5
7.5
Σ
11
11.5
39.5
X ΣA= xA
X (11 in 2 ) = 11.5 in 3
X = 1.045 in. 
Y ΣA=ΣyA
Y (11) = 39.5
Y = 3.59 in. 

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553
PROBLEM 5.2
Locate the centroid of the plane area shown.
SOLUTION
For the area as a whole, it can be concluded by observation that
Y =
2
(72 mm)
3
or Y = 48.0 mm 
Dimensions in mm
A, mm 2
x , mm
x A, mm3
1
1
&times; 30 &times; 72 = 1080
2
20
21,600
2
1
&times; 48 &times; 72 = 1728
2
46
79,488
Σ
Then X A = Σ x A
2808
101,088
X (2808) = 101, 088
or X = 36.0 mm 
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554
PROBLEM 5.3
Locate the centroid of the plane area shown.
SOLUTION
Then
A, mm 2
x , mm
y , mm
xA, mm3
yA, mm3
1
126 &times; 54 = 6804
9
27
61,236
183,708
2
1
&times; 126 &times; 30 = 1890
2
30
64
56,700
120,960
3
1
&times; 72 &times; 48 = 1728
2
48
−16
82,944
−27,648
Σ
10,422
200,880
277,020
X ΣA = Σ xA
X (10, 422 m 2 ) = 200,880 mm 2
and
or X = 19.27 mm 
Y Σ A = Σ yA
Y (10, 422 m 2 ) = 270, 020 mm3
or
Y = 26.6 mm 
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555
PROBLEM 5.4
Locate the centroid of the plane area shown.
SOLUTION
Then
A, in 2
x , in
y , in
x A, in 3
y A, in 3
1
1
(12)(6) = 36
2
4
4
144
144
2
(6)(3) = 18
9
7.5
162
135
Σ
54
306
279
XA = Σ xA
X (54) = 306
X = 5.67 in. 
YA = Σ yA
Y (54) = 279
Y = 5.17 in. 
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556
PROBLEM 5.5
Locate the centroid of the plane area shown.
SOLUTION
By symmetry, X = Y
A, in 2
Component
I
Quarter circle
II
Square
Σ
π
4
x , in.
x A, in 3
(10) 2 = 78.54
4.2441
333.33
−(5)2 = −25
2.5
−62.5
53.54
270.83
X Σ A = Σ x A: X (53.54 in 2 ) = 270.83 in 3
X = 5.0585 in.
X = Y = 5.06 in. 
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557
PROBLEM 5.6
Locate the centroid of the plane area shown.
SOLUTION
Then
A, in 2
x , in.
y , in.
x A, in 3
y A, in 3
1
14 &times; 20 = 280
7
10
1960
2800
2
−π (4) 2 = −16π
6
12
–301.59
–603.19
Σ
229.73
1658.41
2196.8
X=
Σ xA 1658.41
=
ΣA
229.73
X = 7.22 in. 
Y =
Σ y A 2196.8
=
Σ A 229.73
Y = 9.56 in. 
PROPRIETARY MATERIAL. &copy; 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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558
PROBLEM 5.7
Locate the centroid of the plane area shown.
SOLUTION
By symmetry, X = 0
Component
A, in 2
y , in.
y A, in 3
I
Rectangle
(3)(6) = 18
1.5
27.0
II
Semicircle
2.151
−13.51
Σ
−
π
2
(2) 2 = −6.28
13.49
11.72
Y ΣA=ΣyA
Y (11.72 in.2 ) = 13.49 in 3
Y = 1.151 in.
X =0
Y = 1.151 in. 
PROPRIETARY MATERIAL. &copy; 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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559
PROBLEM 5.8
Locate the centroid of the plane area shown.
SOLUTION
1
Σ
Then
x , mm
y , mm
x A, mm3
y A, mm3
(60)(120) = 7200
–30
60
−216 &times; 103
432 &times; 103
(60) 2 = 2827.4
25.465
95.435
72.000 &times; 103
269.83 &times; 103
(60) 2 = −2827.4
–25.465
25.465
72.000 &times; 103
−72.000 &times; 103
−72.000 &times; 103
629.83 &times; 103
π
2
3
A, mm 2
4
−
π
4
7200
XA = Σ x A
X (7200) = −72.000 &times; 103
YA = Σ y A
Y (7200) = 629.83 &times; 103
X = −10.00 mm 
Y = 87.5 mm 
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560
PROBLEM 5.9
Locate the centroid of the plane area shown.
SOLUTION
A, mm 2
x , mm
y , mm
x A, mm3
y A, mm3
1
1
(120)(75) = 4500
2
80
25
360 &times; 103
112.5 &times; 103
2
(75)(75) = 5625
157.5
37.5
885.94 &times; 103
210.94 &times; 103
163.169
43.169
−720.86 &times; 103
−190.716 &times; 103
525.08 &times; 103
132.724 &times; 103
3
Σ
Then
−
π
4
(75) 2 = −4417.9
5707.1
XA = Σx A
X (5707.1) = 525.08 &times; 103
X = 92.0 mm 
YA = Σ y A
Y (5707.1) = 132.724 &times; 103
Y = 23.3 mm 
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561
PROBLEM 5.10
Locate the centroid of the plane area shown.
SOLUTION
X = 0 
First note that symmetry implies
A, in 2
−
1
2
π (12) 2
2
2
Σ
Then
π (8) 2
y , in.
yA, in 3
= −100.531
3.3953
–341.33
= 226.19
5.0930
1151.99
125.659
Y =
810.66
Σ y A 810.66 in 3
=
Σ A 125.66 in 2
or Y = 6.45 in. 
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562
PROBLEM 5.11
Locate the centroid of the plane area shown.
SOLUTION
X =0 
First note that symmetry implies

A, m 2
4
&times; 4.5 &times; 3 = 18
3
1
2
y, m
−
π
Σ
2
(1.8) 2 = −5.0894
12.9106
yA, m3
1.2
21.6
0.76394
−3.8880
17.7120

Then
Y =
Σ y A 17.7120 m3
=
Σ A 12.9106 m 2
or Y = 1.372 m 
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563
PROBLEM 5.12
Locate the centroid of the plane area shown.
SOLUTION

Area mm2
x , mm
y , mm
xA, mm3
yA, mm3
1
1
(200)(480) = 32 &times; 103
3
360
60
11.52 &times; 106
1.92 &times; 106
2
1
− (50)(240) = 4 &times; 103
3
180
15
−0.72 &times; 106
−0.06 &times; 106
Σ
28 &times; 103
10.80 &times; 106
1.86 &times; 106

X Σ A = Σ x A:
X (28 &times; 103 mm 2 ) = 10.80 &times; 106 mm3
X = 385.7 mm
Y Σ A = Σ y A:
X = 386 mm 
Y (28 &times; 103 mm 2 ) = 1.86 &times; 106 mm3
Y = 66.43 mm
Y = 66.4 mm 
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564
PROBLEM 5.13
Locate the centroid of the plane area shown.
SOLUTION
Then
A, mm 2
x , mm
y , mm
xA, mm3
yA, mm3
1
(15)(80) = 1200
40
7.5
48 &times; 103
9 &times; 103
2
1
(50)(80) = 1333.33
3
60
30
80 &times; 103
40 &times; 103
Σ
2533.3
128 &times; 103
49 &times; 103
X A = Σ xA
X (2533.3) = 128 &times; 103
X = 50.5 mm 
YA = Σ yA
Y (2533.3) = 49 &times; 103
Y = 19.34 mm 
PROPRIETARY MATERIAL. &copy; 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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565
PROBLEM 5.14
Locate the centroid of the plane area shown.
SOLUTION
Dimensions in in.
Then
A, in 2
x , in.
y , in.
xA, in 3
yA, in 3
1
2
(4)(8) = 21.333
3
4.8
1.5
102.398
32.000
2
1
− (4)(8) = −16.0000
2
5.3333
1.33333
85.333
−21.333
Σ
5.3333
17.0650
10.6670
X ΣA = Σ xA
X (5.3333 in 2 ) = 17.0650 in 3
and
or X = 3.20 in. 
Y Σ A = Σ yA
Y (5.3333 in 2 ) = 10.6670 in 3
or
Y = 2.00 in. 
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566
PROBLEM 5.15
Locate the centroid of the plane area shown.
SOLUTION
Dimensions in mm
A, mm 2
1
Then
π
2
&times; 47 &times; 26 = 1919.51
2
1
&times; 94 &times; 70 = 3290
2
Σ
5209.5
x , mm
y , mm
x A, mm3
y A, mm3
0
11.0347
0
21,181
−15.6667
−23.333
−51,543
−76,766
−51,543
−55,584
X=
Σ x A −51,543
=
ΣA
5209.5
X = −9.89 mm 
Y =
Σ y A −55,584
=
ΣA
5209.5
Y = −10.67 mm 
PROPRIETARY MATERIAL. &copy; 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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567
PROBLEM 5.16
Determine the x coordinate of the centroid of the trapezoid shown in terms of
h1, h2, and a.
SOLUTION
A
x
xA
1
1
h1a
2
1
a
3
1
h1a 2
6
2
1
h2 a
2
2
a
3
2
h2 a2
6
Σ
1
a(h1 + h 2 )
2
X=
1 2
a ( h1 + 2h 2 )
6
1 2
Σ x A 6 a ( h1 + 2h 2 )
= 1
a (h1 + h 2 )
ΣA
2
1 h1 + 2h 2
X= a

3 h1 + h 2
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568
PROBLEM 5.17
For the plane area of Problem 5.5, determine the ratio a/r so that the
centroid of the area is located at point B.
SOLUTION
By symmetry, X = Y . For centroid to be at B, X = a.
Area
x
xA
I
Quarter circle
1 2
πr
4
4r
3π
1 3
r
3
II
Square
−a 2
1
a
2
1
− a3
2
π
Σ
4
r 2 − a2
X Σ A = Σ x A:
1
π
 1
X  r 2 − a 2  = r 3 − a3
2
4
 3
Set X = a :
1
π
 1
a  r 2 − a 2  = r 3 − a3
2
4
 3
1 3 1 3
r − a
3
2
1 3 π 2
1
a − r a + r3 = 0
2
4
3
Divide by
1 3
r :
2
3
a π a 2
r − 2 r + 3 =0
 
a
= 0.508 
r
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569
PROBLEM 5.18
Determine the y coordinate of the centroid of the shaded area in
terms of r1, r2, and α.
SOLUTION
First, determine the location of the centroid.
y2 =
From Figure 5.8A:
Similarly,
=
2
cos α
r2 π
3 ( 2 −α )
y1 =
2 cos α
r1
3 ( π2 − α )
Σ yA =
Then
π
2 sin ( 2 − α )
r2 π
3
( 2 −α )
=
π

A2 =  − α  r22
2

π

A1 =  − α  r12
2

2
cos α  π
  2 cos α  π
 2
− α  r22  − r1 π
r2

 − α  r1 
3 ( π2 − α )  2
3
2
α
 
 
( 2 − ) 
(
)
2 3 3
r2 − r1 cos α
3
π

π

ΣA =  − α  r22 −  − α  r12
2
2




and
π

=  − α  r22 − r12
2


(
)
Y ΣA = Σ yA
Now
 π
 2

Y  − α  r22 − r12  = r23 − r13 cos α

 2
 3
(
)
(
)
Y =
2  r23 − r13

3  r22 − r12
  2 cos α 
 
 
  π − 2α 
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570
PROBLEM 5.19
Show that as r1 approaches r2, the location of the centroid
approaches that for an arc of circle of radius (r1 + r2 )/2.
SOLUTION
First, determine the location of the centroid.
y2 =
From Figure 5.8A:
Similarly,
=
2
cos α
r2
3 ( π2 − α )
y1 =
2 cos α
r1
3 ( π2 − α )
Σ yA =
Then
π
2 sin ( 2 − α )
r2 π
3
( 2 −α )
=
π

A2 =  − α  r22
2

π

A1 =  − α  r12
2


2
cos α  π
  2 cos α  π
 2
r2
− α  r22  − r1 π

 − α  r1 
3 ( π2 − α )  2
3
2
 
 
( 2 − α ) 
(
)
2 3 3
r2 − r1 cos α
3
π

π

ΣA =  − α  r22 −  − α  r12
2
2




and
π

=  − α  r22 − r12
2


(
)
Y ΣA = Σ yA
Now
 π
 2

Y  − α  r22 − r12  = r23 − r13

 2
 3
2  r3 − r3
Y =  22 12
3  r2 − r1
(
)
(
) cos α
  2cos α 
 

  π − 2α 
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571
PROBLEM 5.19 (Continued)
Using Figure 5.8B, Y of an arc of radius
Y =
=
Now
r23 − r13
r22
−
r12
=
=
1
(r1 + r2 ) is
2
sin( π − α )
1
(r1 + r2 ) π 2
2
( 2 −α)
1
cos α
(r1 + r2 ) π
2
( 2 −α)
(
(r2 − r1 ) r22 + r1r2 + r12
(r2 − r1 )(r2 + r1 )
(1)
)
r22 + r1r2 + r12
r2 + r1
r2 = r + Δ
Let
r1 = r − Δ
r=
Then
and
In the limit as Δ
r23 − r13
r22 − r12
1
(r1 + r2 )
2
=
(r + Δ) 2 + (r + Δ)(r − Δ)( r − Δ) 2
(r + Δ ) + (r − Δ )
=
3r 2 + Δ 2
2r
0 (i.e., r1 = r2 ), then
r23 − r13
r22
So that
−
r12
=
3
r
2
=
3 1
&times; ( r1 + r2 )
2 2
Y =
2 3
cos α
&times; (r1 + r2 ) π
3 4
−α
2
or Y = ( r1 + r2 )
cos α

π − 2α
which agrees with Equation (1).
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572
PROBLEM 5.20
The horizontal x-axis is drawn through the centroid C of the area shown, and
it divides the area into two component areas A1 and A2. Determine the first
moment of each component area with respect to the x-axis, and explain the
results obtained.
SOLUTION
Length of BD:
BD = 0.48 in. + (1.44 in. − 0.48 in.)
0.84in.
= 0.48 + 0.56 = 1.04 in.
0.84 in. &times; 0.60 in.
Area above x-axis (consider two triangular areas):
1

1

Q1 = Σ y A = (0.28 in.)  (0.84 in.)(1.04 in.)  + (0.56 in.)  (0.84 in.)(0.48 in.) 
2
2




= 0.122304 in 3 + 0.112896 in 3
Q1 = 0.2352 in 3 
Area below x-axis:
1

1

Q2 = Σ yA = −(0.40 in.)  (0.60 in.)(1.44 in.)  − (0.20 in.)  (0.60 in.) 
2

2

= −0.1728 in 3 − 0.0624 in 3
Q2 = −0.2352 in 3 
|Q| = |Q2 |, since C is centroid and thus, Q = Σ y A = 0
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573
PROBLEM 5.21
The horizontal x-axis is drawn through the centroid C of the area shown,
and it divides the area into two component areas A1 and A2. Determine the
first moment of each component area with respect to the x-axis, and
explain the results obtained.
SOLUTION
Dimensions in mm
Area above x-axis (Area A1):
Q1 = Σ y A = (25)(20 &times; 80) + (7.5)(15 &times; 20)
= 40 &times; 103 + 2.25 &times; 103
Q1 = 42.3 &times; 103 mm3 
Area below x-axis (Area A2):
Q2 = Σ y A = (−32.5)(65 &times; 20)
Q2 = −42.3 &times; 103 mm3 
|Q1| = |Q2 |, since C is centroid and thus, Q = Σ y A = 0 
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574
PROBLEM 5.22
A composite beam is constructed by bolting four
plates to four 60 &times; 60 &times; 12-mm angles as shown.
The bolts are equally spaced along the beam, and
the beam supports a vertical load. As proved
in mechanics of materials, the shearing forces
exerted on the bolts at A and B are proportional to
the first moments with respect to the centroidal
x-axis of the red-shaded areas shown, respectively,
in parts a and b of the figure. Knowing that the
force exerted on the bolt at A is 280 N, determine
the force exerted on the bolt at B.
SOLUTION
From the problem statement, F is proportional to Qx .
(Qx ) B
FA
(Qx ) A
Therefore,
FA
FB
=
, or
(Qx ) A (Qx ) B
For the first moments,
12 

(Qx ) A =  225 +  (300 &times; 12)
2

FB =
= 831, 600 mm3
12 

(Qx ) B = (Qx ) A + 2  225 −  (48 &times; 12) + 2(225 − 30)(12 &times; 60)
2

= 1,364,688 mm3
Then
FB =
1,364, 688
(280 N)
831, 600
or FB = 459 N 
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575
PROBLEM 5.23
The first moment of the shaded area with respect to the x-axis is denoted by Qx.
(a) Express Qx in terms of b, c, and the distance y from the base of the shaded
area to the x-axis. (b) For what value of y is Qx maximum, and what is that
maximum value?
SOLUTION
A = b (c − y )
Qx = yA
=
Qx =
(a)
(b)
For Qmax ,
For y = 0,
1
(c + y )[b(c − y )]
2
1
b (c 2 − y 2 )
2
dQ
= 0 or
dy
(Qx ) =

1
b(−2 y ) = 0
2
1 2
bc
2
y=0 
(Qx ) =
1 2
bc 
2
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576
PROBLEM 5.24
A thin, homogeneous wire is bent to form the perimeter of the figure
indicated. Locate the center of gravity of the wire figure thus formed.
SOLUTION
First note that because the wire is homogeneous, its center of gravity will coincide with the centroid of the
corresponding line.
L, mm
Then
x , mm
y , mm
yL, mm 2
yL, mm 2
1
302 + 722 = 78
15
36
1170.0
2808.0
2
482 + 722 = 86.533
54
36
4672.8
3115.2
39
72
3042.0
5616.0
8884.8
11,539.2
3
78
Σ
242.53
X ΣL = Σx L
X (242.53) = 8884.8
or X = 36.6 mm 
and
Y ΣL = Σ yL
Y (242.53) = 11,539.2
or Y = 47.6 mm 
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577
PROBLEM 5.25
A thin, homogeneous wire is bent to form the perimeter of the figure
indicated. Locate the center of gravity of the wire figure thus formed.
SOLUTION
L, mm
1
Then
722 + 482 = 86.533
x , mm
y , mm
xL, mm 2
yL, mm 2
36
−24
3115.2
−2076.8
2
132
72
18
9504.0
2376.0
3
1262 + 302 = 129.522
9
69
1165.70
8937.0
4
54
−54
27
−2916.0
5
54
−27
0
−1458.0
Σ
456.06
9410.9
1458.0
0
10,694.2
X ΣL = Σx L
X (456.06) = 9410.9
or X = 20.6 mm 
Y ΣL = Σ y L
Y (456.06) = 10, 694.2
or Y = 23.4 mm 
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578
PROBLEM 5.26
A thin, homogeneous wire is bent to form the perimeter of the figure
indicated. Locate the center of gravity of the wire figure thus formed.
SOLUTION
Then
y , in.
xL, in 2
yL, in 2
3
80.498
40.249
7.5
36
22.5
9
9
54
54.0
3
6
7.5
18
22.5
5
6
3
6
18
36.0
6
6
0
3
0
18.0
Σ
37.416
206.50
193.249
L, in.
x , in.
1
122 + 62 = 13.4164
6
2
3
12
3
6
4
X ΣL = Σx L
Y ΣL =Σ y L
X (37.416) = 206.50
Y (37.416) = 193.249
X = 5.52 in. 
Y = 5.16 in. 
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579
PROBLEM 5.27
A thin, homogeneous wire is bent to form the perimeter of the figure
indicated. Locate the center of gravity of the wire figure thus formed.
SOLUTION
By symmetry, X = Y .
x , in.
L, in.
1
Then
1
π (10) = 15.7080
2
2(10)
π
= 6.3662
yL, in 2
100
2
5
0
0
3
5
2.5
12.5
4
5
5
25
5
5
7.5
37.5
Σ
35.708
X ΣL = Σx L
175
X (35.708) = 175
X = Y = 4.90 in. 
PROPRIETARY MATERIAL. &copy; 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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580
PROBLEM 5.28
The homogeneous wire ABCD is bent as shown and is attached to a
hinge at C. Determine the length L for which portion BCD of the
wire is horizontal.
SOLUTION
First note that for equilibrium, the center of gravity of the wire must lie on a vertical line through C. Further,
because the wire is homogeneous, the center of gravity of the wire will coincide with the centroid of the
corresponding line. Thus,
X = 0 so that Σ X L = 0
Then
L
+ (−40 mm)(80 mm) + (−40 mm)(100 mm) = 0
2
L2 = 14, 400 mm 2
L = 120.0 mm 
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581
PROBLEM 5.29
The homogeneous wire ABCD is bent as shown and is attached to a
hinge at C. Determine the length L for which portion AB of the wire
is horizontal.
SOLUTION
WI = 80w
WII = 100w
WIII = Lw
Σ M C = 0: (80 w)(32) + (100 w)(14) − ( Lw)(0.4 L) = 0
L2 = 9900
L = 99.5 mm 
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582
PROBLEM 5.30
The homogeneous wire ABC is bent into a semicircular arc and a straight section
as shown and is attached to a hinge at A. Determine the value of θ for which the
wire is in equilibrium for the indicated position.
SOLUTION
First note that for equilibrium, the center of gravity of the wire must lie on a vertical line through A. Further,
because the wire is homogeneous, its center of gravity will coincide with the centroid of the corresponding
line. Thus,
X =0
so that
Σx L = 0
Then
 1

 2r

 − 2 r cos θ  (r ) +  π − r cos θ  (π r ) = 0




or
cos θ =
4
1 + 2π
= 0.54921
or θ = 56.7&deg; 
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583
PROBLEM 5.31
A uniform circular rod of weight 8 lb and radius 10 in. is attached to a pin at C and
to the cable AB. Determine (a) the tension in the cable, (b) the reaction at C.
SOLUTION
For quarter circle,
(a)
r =
2r
π
 2r 
ΣM C = 0: W   − Tr = 0
π 
2
2
T = W   = (8 lb)  
π 
π 
(b)
T = 5.09 lb 
ΣFx = 0: T − C x = 0 5.09 lb − C x = 0
C x = 5.09 lb
ΣFy = 0: C y − W = 0
C y = 8 lb
C y − 8 lb = 0
C = 9.48 lb
57.5&deg; 
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584
PROBLEM 5.32
Determine the distance h for which the centroid of the shaded
area is as far above line BB′ as possible when (a) k = 0.10,
(b) k = 0.80.
SOLUTION
A
y
yA
1
1
ba
2
1
a
3
1 2
a b
6
2
1
− (kb)h
2
1
h
3
1
− kbh 2
6
Σ
b
(a − kh)
2
b 2
( a − kh 2 )
6
Y Σ A = Σ yA
Then
b
 b
Y  ( a − kh)  = (a 2 − kh 2 )
2

 6
Y =
or
(1)
dY 1 −2kh(a − kh) − (a 2 − kh 2 )(− k )
=
=0
dh 3
( a − kh) 2
and
or
a 2 − kh 2
3(a − kh)
2h(a − kh) − a 2 + kh 2 = 0
(2)
Simplifying Eq. (2) yields
kh 2 − 2ah + a 2 = 0
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585
PROBLEM 5.32 (Continued)
Then
2a &plusmn; (−2a) 2 − 4( k )(a 2 )
2k
a
= 1 &plusmn; 1 − k 
k
h=
Note that only the negative root is acceptable since h &lt; a. Then
(a)
k = 0.10
h=
(b)
a 
1 − 1 − 0.10 

0.10 
or h = 0.513a 
k = 0.80
h=
a 
1 − 1 − 0.80 

0.80 
or h = 0.691a 
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586
PROBLEM 5.33
Knowing that the distance h has been selected to maximize the
distance y from line BB′ to the centroid of the shaded area,
show that y = 2h/3.
SOLUTION
See solution to Problem 5.32 for analysis leading to the following equations:
Y =
a 2 − kh 2
3(a − kh)
(1)
2h(a − kh) − a 2 + kh 2 = 0
(2)
Rearranging Eq. (2) (which defines the value of h which maximizes Y ) yields
a 2 − kh 2 = 2h(a − kh)
Then substituting into Eq. (1) (which defines Y ),
Y =
1
&times; 2h(a − kh)
3( a − kh)
or Y =
2
h 
3
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587
PROBLEM 5.34
Determine by direct integration the centroid of the area shown. Express your
answer in terms of a and h.
SOLUTION
h
x
a
y=
We have
dA = (h − y )dx
and
x

= h 1 −  dx
 a
xEL = x
1
(h + y )
2
h
x
= 1 + 
2 a
yEL =

A = dA =
Then
and
x
y
a
=

EL dA =

EL dA
=
0
a
0
h2
2

a
0
a

x
x2 
1

h 1 −  dx = h  x −  = ah
a
2
a



0 2
a
 x 2 x3 
 
x 
1
x  h 1 −  dx  = h  −  = a 2 h
  a 
 2 3a  0 6
h
x  
x 
1 +   h 1 −  dx 

a   a  
2

a
0
x2
−
1

2
 a
a

h2 
x3 
1
=
−
= ah2
dx
x


2
2
3a  0 3


1  1
xA = xEL dA: x  ah  = a 2 h
2  6
x=
2
a 
3
1  1
y A = yEL dA: y  ah  = ah2
2  3
y=
2
h 
3


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588
PROBLEM 5.35
Determine by direct integration the centroid of the area shown. Express
SOLUTION
y = h(1 − kx3 )
For x = a, y = 0.
0 = h(1 − k a3 )
∴ k=
1
a3

x3 
y = h 1 − 3 
 a 
xEL = x,
yEL =

a
A = dA =
x

EL dA
=
yEL dA =
=

a
0


0
xydx =
1
y dA = ydx
2
ydx =

a
0

h2
2
0
a


x3 
x4 
3
h 1 − 3  dx = h  x − 3  = ah
4a  0 4
 a 

a

 x2
x4 
x5 
3
h  x − 3  dx = h  − 3  = a 2 h
a 

 2 5a  0 10
a
0
a
1 
1
 y  ydx =
2
2



a
0

x3 
h2
h 2 1 − 3  dx =
2
 a 

a
0
2 x3 x 6
1 − 3 + 6
a
a


 dx

a

x4
x7 
9 2
ah
x − 3 + 6  =
28
2a
7a  0

3  3
xA = xEL dA: x  ah  = a 2 h
 4  10
x=
2
a 
5
3  9 2
yA = yEL dA: y  ah  =
ah
 4  28
y=
3
h 
7


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589
PROBLEM 5.36
Determine by direct integration the centroid of the area shown. Express
SOLUTION
y1 : h = ka 2
At (a, h),
k=
or
h
a2
y2 : h = ma
m=
or
xEL = x
Now
yEL =
1
( y1 + y2 )
2
h 
h
dA = ( y2 − y1 )dx =  x − 2 x 2  dx
a
a


h
= 2 (ax − x 2 ) dx
a
and

A = dA =
Then
and
h
a



a
0
a
h
h a
1 
1
(ax − x 2 )dx = 2  x 2 − x3  = ah
2
3 0 6
a
a 2
a
h a
1 
1
h
xEL dA = x  2 (ax − x 2 ) dx = 2  x3 − x 4  = a 2 h
0 a
4  0 12
a 3
1
1 2
yEL dA =
y2 − y12 dx
( y1 + y2 )[( y2 − y1 ) dx] =
2
2

a
 (

=
1
2

)
a
0
h 2 2 h2 4 
 2 x − 4 x dx
a
a

a
1 h2  a 2 3 1 5 
=
 x − x 
2 a4  3
5 0
1
= ah 2
15
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590
PROBLEM 5.36 (Continued)
1  1
xA = xEL dA: x  ah  = a 2 h
 6  12
x=
1
a 
2
1  1
yA = yEL dA: y  ah  = ah 2
 6  15
y=
2
h 
5


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591
PROBLEM 5.37
Determine by direct integration the centroid of the area shown.
SOLUTION
x
,
k
y2 =
y1 = k x 2
a = k a 2 , thus, k =
But
y2 = a x ,
y1 =
1
a2
x2
a
xEL = x

x2 
dA = ( y2 − y1 ) dx =  ax −  dx
a 


A = dA =

a
x2 
 ax −  dx
0
a 

a
2
1
x3 
=
a x3/2 −  = a 2
3a  0 3
3
x
EL dA
=

a
0

x2 
x  a x −  dx =
a 


a
0
 a x

1  3 3
xA = xEL dA: x  a 2  =
a
 3  20

a
3/2
2
x3 
x4 
3 3
−  dx = 
a x5/2 −  =
a
a 
4a  0 20
5
x=
9a
20
y=x=
By symmetry,
9a

20
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592
PROBLEM 5.38
Determine by direct integration the centroid of the area shown.
SOLUTION
y=
For the element (EL) shown,
b 2
a − x2
a
dA = (b − y ) dx
and
xEL
)
(
b
a − a 2 − x 2 dx
a
=x
=
1
( y + b)
2
b
=
a + a 2 − x2
2a
yEL =
(

A = dA =
Then

To integrate, let
x = a sin θ :
Then
A=

π /2
0
a
0
)
)
(
b
a − a 2 − x 2 dx
a
a 2 − x 2 = a cos θ , dx = a cos θ dθ
b
(a − a cos θ )(a cos θ dθ )
a
π /2
=
b 2
2θ  
2 θ
 a sin θ − a  + sin

a
4   0
2
 π
= ab 1 − 
4

and
x
EL dA
a
)
(
b

x  a − a 2 − x 2 dx 
a


=

=
b  a 2 1 2
2 3/2  
 x + ( a − x )  
a  2
3
 0
=
1 3
ab
6
0
π /2
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593
PROBLEM 5.38 (Continued)
y
EL dA
=

a
0
b2
= 2
2a
=
) (
(
)
b
b

a + a 2 − x 2  a − a 2 − x 2 dx 
2a
a


a

a
0
b 2  x3 
( x ) dx = 2  
2a  3  0
2
1 2
ab
6

  π  1
x  ab 1 −   = a 2 b
4  6
 
or x =
2a

3(4 − π )

  π  1
y  ab 1 −   = ab 2
4  6
 
or y =
2b

3(4 − π )
xA = xEL dA:
yA = yEL dA:
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594
PROBLEM 5.39
Determine by direct integration the centroid of the area shown.
SOLUTION
x =0 
First note that symmetry implies
For the element (EL) shown,
yEL =
2r
(Figure 5.8B)
π
dA = π rdr

A = dA =
Then
and
So

yEL dA =

r2
r1

r2
r1
 r2
π rdr = π 
 2
r2

π 2 2
r2 − r1
 =
 r1 2
r2
(
2
1 
(π rdr ) = 2  r 3  = r23 − r13
π
 3  r1 3
2r
(
)
π
 2
yA = yEL dA: y  r22 − r12  = r23 − r13
2

 3

(
)
(
)
)
or y =
4 r23 − r13

3π r22 − r12
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595
PROBLEM 5.40
Determine by direct integration the centroid of the area shown. Express
SOLUTION
b 2
x
a2
b
y2 = k2 x 4 but b = k2 a 4 y2 = 4 x 4
a
4
b 
x 
dA = ( y2 − y1 )dx = 2  x 2 − 2  dx
a 
a 
xEL = x
y1 = k1 x 2
but b = k1a 2
1
( y1 + y2 )
2
b 
x4
= 2  x 2 + 2
2a 
a
y1 =
yEL =

A = dA =
b
a2




a
0
x4
2
−
x

a2


 dx

a

b  x3
x5 
= 2  − 2
a  3 5a  0
2
= ba
15
a
b 
x4 
xEL dA = x 2  x 2 − 2  dx
0
a 
a 
b a
x5 
= 2  x3 − 2  dx
a 0
a 


a
 x4
x6 
−

2
 4 6a  0
=
b
a2
=
1 2
a b
12
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596
PROBLEM 5.40 (Continued)

yEL dA =
=

a
0
b
2a 2
b2
2a 4

 2 x4  b
 x + 2  2
a a

 2 x4 
 x − 2  dx
a 

a
0
x8 
4
 x − 4  dx
a 

a
b 2  x5
x9 
2 2
= 4  − 4 =
ab
2a  5 9a  0 45
 2  1
xA = xEL dA: x  ba  = a 2b
 15  12
5
x= a 
8
 2  2 2
yA = yEL dA: y  ba  =
ab 
 15  45
1
y = b 
3



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597
PROBLEM 5.41
Determine by direct integration the centroid of the area shown. Express your
answer in terms of a and b.
SOLUTION
y =b 
First note that symmetry implies
x = a,
At
y=b
y1 : b = ka 2
y1 =
Then
k=
or
b
a2
b 2
x
a2
y2 : b = 2b − ca 2
c=
or

x2
y2 = b  2 − 2
a

Then



 
x2
dA = ( y2 − y1 )dx2 = b  2 − 2
a
 

x2 
= 2b 1 − 2  dx
 a 
Now
 b 2
 − 2 x  dx

 a
xEL = x
and
 
A = dA
Then
and
b
a2

a
0

x2
2b 1 − 2
 a
a


x3 
4
 dx = 2b  x − 2  = ab
3a  0 3


a
 
 x2
x2  
x4 
1
xEL dA = x  2b 1 − 2  dx  = 2b  − 2  = a 2b
0
  a  
 2 4a  0 2
4  1
xA = xEL dA: x  ab  = a 2b
3  2

a

3
x= a 
8
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598
PROBLEM 5.42
Determine by direct integration the centroid of the area shown.
SOLUTION
xEL = x
yEL =

L
A = dA =

xEL dA =

L
0

0
1
y dA = y dx
2
L


x
x2 
x 2 2 x3 
5
h 1 + − 2 2  dx = h  x +
−
= hL
2
L
2L 3 L 0 6
L 



x
x2 
xh 1 + − 2 2  dx = h
L
L 


L
0
x2
x3 
− 2 2  dx
 x +
L
L 

L
 x 2 1 x3 2 x 4 
1
= h +
−
= hL2
2 
 2 3 L 4 L 0 3
5  1
xA = xEL dA: x  hL  = hL2
6  3

A=

5
hL
6
yEL =

=
h2
2
h2
=
2


L
0
2
L 
5

x
x2 
y = h 1 + − 2 2 
L
L 

1
y
2
1 2
h2
yEL dA =
y dx =
2
2
x=
2
x
x2 
1 + − 2 2  dx
L
L 

L
0
x2
x4
x
x2
x3 
1 + 2 + 4 4 + 2 − 4 2 − 4 3  dx
L
L
L
L
L 

L

4
x3 4 x5 x 2 4 x3 x 4 
x
+
+ 4 +
− 2 − 3  = h2 L

2
10
L
3
5
3
L
L
L
L

0
5  4
yA = yEL dA: y  hL  = h 2 L
 6  10

y=
12
h 
25
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599
PROBLEM 5.43
Determine by direct integration the centroid of the area shown. Express
SOLUTION
For y1 at x = a,
y = 2b, 2b = ka 2 , or k =
Then
y1 =
By observation,
b
x

y2 = − ( x + 2b) = b  2 − 
a
a

2b
a2
2b 2
x
a2
Now
xEL = x
and for 0 ≤ x ≤ a,
yEL =
1
b
y1 = 2 x 2
2
a
For a ≤ x ≤ 2a,
yEL =
1
b
x
x

y2 =  2 −  and dA = y2 dx = b  2 −  dx
a
a
2
2



A = dA =
Then

a
0
and dA = y1 dx =
2b 2
x dx +
a2

2a
a
2b 2
x dx
a2
x

b  2 −  dx
a

2a
2
 a
2b  x3 
x 
7
= 2   + b  −  2 −   = ab
a  
6
a  3 0
 2 
0
a
and

xEL dA =

a
0
 2b

x  2 x 2 dx  +
a


2a
a
a
 
x 
x b  2 −  dx 
a 
 
2a

2b  x 4 
x3 
= 2   + b  x2 − 
3a  0
a  4 0

1 2
1  2


(2a ) − (a )3  
a b + b  (2a) 2 − (a) 2  +

2
3
a


7 2
= a b
6
=
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600
PROBLEM 5.43 (Continued)

yEL dA =

a
0
b 2  2b 2 
x
x dx  +
a 2  a 2


2a
0
b
x  
x 
 2 −  b  2 −  dx 
2
a  
a 
2a
3
2b 2  x5 
b2  a 
x 
= 4   +
−  2 −  
a  
a  5  0 2  3 
a
17 2
ab
=
30
a
Hence,
7  7
xA = xEL dA: x  ab  = a 2b
6  6

 7  17 2
yA = yEL dA: y  ab  =
ab
 6  30

x =a 
y=
17
b 
35
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601
PROBLEM 5.44
Determine by direct integration the centroid of the area shown. Express
SOLUTION
For y2 at x = a,
y = b, a = kb 2 , or k =
Then
y2 =
Now
xEL = x
and for 0 ≤ x ≤
a
,
2
yEL =
b
a
x1/2
y2 b x1/ 2
=
2 2 a
x1/2
dA = y2 dx = b
For
a
≤ x ≤ a,
2
yEL =
a
b2
a
dx
1
b  x 1 x1/ 2 
( y1 + y2 ) =  − +

2
2a 2
a 
 x1/2 x 1 
dA = ( y2 − y1 )dx = b 
− +  dx
 a a 2
Then

A = dA =

a /2
b
0
x1/ 2
a
dx +

 x1/2 x 1 
b 
− +  dx
a/ 2
 a a 2
a
a
a/ 2
 2 x3/2 x 2 1 
b  2 3/2 
=
x
+
b
−
+ x


a  3
0
 3 a 2a 2  a/2
3/ 2
3/ 2
2 b  a 
a 
3/ 2
=
  + ( a ) −   
3 a  2 
 2  
2
 1  2  a   1 
 a   
+ b −  (a ) −    +  (a) −    
2a 
 2   2 
 2   

13
ab
=
24
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602
PROBLEM 5.44 (Continued)
and

xEL dA =

a/2
0
 x1/ 2 
x  b
dx  +
a



  x1/ 2 x 1  
x b 
− +   dx
a/2
  a a 2  
a
a
a/ 2
 2 x5/2 x3 x 4 
b  2 5/2 
=
+
−
+ 
x
b


a  5
0
 5 a 3a 4  a/2
5/2
5/2
2 b  a 
a 
5/2
=
  + ( a ) −   
5 a  2 
 2  
3
2
 1 
a  1
 a   
+ b  −  ( a )3 −    +  ( a ) 2 −    
 2   4 
 2   
 3a 
71 2
a b
=
240

yEL dA =

a/2
0
+

b x1/ 2
2 a
 x1/ 2 
dx 
b
a 

b  x 1 x1/2    x1/ 2 x 1  
− +  dx 
 − +
 b 
a/ 2 2  a
2
a    a a 2  

a
a
a/ 2
3
b2  1 2 
b 2  x 2 1  x 1   



x
=
+
−
−
2a  2  0
2  2a 3a  a 2   

 a/2
Hence,
=
3
2
2
2
b  a 
a  b a 1
−
  + ( a ) 2 −    −


4a  2 
 2   6a  2 2 
=
11 2
ab
48
 13  71 2
xA = xEL dA : x  ab  =
a b
 24  240

 13  11 2
yA = yEL dA: y  ab  =
ab
 24  48

x=
17
a = 0.546a 
130
y=
11
b = 0.423b 
26
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603
PROBLEM 5.45
A homogeneous wire is bent into the shape shown. Determine by direct
integration the x coordinate of its centroid.
SOLUTION
First note that because the wire is homogeneous, its center of gravity coincides with the centroid of the
corresponding line.
Now
xEL = r cos θ
Then
L = dL =
and


xEL dL =
7π /4
π
/4
and dL = rdθ
7π / 4
π
/4
3
r dθ = r[θ ]π7π/ 4/ 4 = π r
2
r cos θ (rdθ )
= r 2 [sin θ ] π7π/4/ 4
 1
1 
= r2  −
−

2
2

= −r 2 2
Thus
3 
xL = x dL : x  π r  = − r 2 2
2 

x =−
2 2
r 
3π
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604
PROBLEM 5.46
A homogeneous wire is bent into the shape shown. Determine by direct
integration the x coordinate of its centroid.
SOLUTION
First note that because the wire is homogeneous, its center of gravity coincides with the centroid of the
corresponding line.
xEL = a cos3 θ
Now
and dL = dx 2 + dy 2
x = a cos3 θ : dx = −3a cos 2 θ sin θ dθ
where
y = a sin 3 θ : dy = 3a sin 2 θ cos θ dθ
dL = [(−3a cos 2 θ sin θ dθ ) 2 + (3a sin 2 θ cos θ dθ )2 ]1/ 2
Then
= 3a cos θ sin θ (cos 2 θ + sin 2 θ )1/ 2 dθ
= 3a cos θ sin θ dθ

L = dL =
and
x
EL dL
=
3
a
2
=

π /2
0

π /2
0
π /2
1

3a cos θ sin θ dθ = 3a  sin 2 θ 
2
0
a cos3θ (3a cos θ sin θ dθ )
π /2
 1

= 3a 2  − cos5 θ 
 5
0
Hence,
=
3 2
a
5
3  3
xL = xEL dL : x  a  = a 2
2  5

x=
2
a 
5
Alternative Solution:
 x
x = a cos3 θ  cos 2 θ =  
a
 y
y = a sin θ  sin θ =  
a
3
x
a
 
2/3
2/3
2/3
2
 y
+ 
a
2/3
= 1 or
y = (a 2/3 − x 2/3 )3/2
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605
PROBLEM 5.46 (Continued)
dy
= (a 2/3 − x 2/3 )1/ 2 (− x −1/3 )
dx
Then
Now
xEL = x
and
 dy 
dL = 1 +  
 dx 
2
{
dx = 1 + ( a 2/3 − x 2/3 )1/2 (− x −1/3 ) 

L = dL =
Then
and
Hence

xEL dL =

a
0

a
0
2
}
1/2
dx
a
a1/3
3
3

dx = a1/3  x 2/3  = a
1/3
x
2
0 2
a
 a1/3 
3
3

x  1/3 dx  = a1/3  x5/3  = a 2
5
0 5
x

3  3
xL = xEL dL : x  a  = a 2
2  5

x=
2
a 
5
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606
PROBLEM 5.47*
A homogeneous wire is bent into the shape shown. Determine by
direct integration the x coordinate of its centroid. Express your
SOLUTION
First note that because the wire is homogeneous, its center of gravity will coincide with the centroid of the
corresponding line.
We have at x = a,
y = a, a = ka3/2 , or k =
Then
y=
1
a
1
a
x3/ 2
dy
3 1/2
=
x
dx 2 a
and
Now
xEL = x
and
 dy 
dL = 1 +   dx
 dx 
2
1/2
2
  3
 
x1/2  
= 1 + 
  2 a
 
1
=
4a + 9 x dx
2 a

L = dL =
Then

a
1
0
2 a
dx
4a + 9 x dx
a
1 2 1

&times; (4a + 9 x)3/ 2 

3
9
2 a
0
a
= [(13)3/2 − 8]
27
= 1.43971a
=
and
x
EL dL
=

a
0
 1
x
2 a

4a + 9 x dx 

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607
PROBLEM 5.47* (Continued)
Use integration by parts with
u=x
du = dx
Then

dv = 4a + 9 x dx
2
v=
(4a + 9 x)3/ 2
27
a
1  
2
3/2 
xEL dL =
  x &times; (4a + 9 x)  −
2 a   27
0

a
0

2
(4a + 9 x)3/ 2 dx 
27

a
=
(13)3/2 2
1 2

(4a + 9 x)5/2 
a −
27
27 a  45
0
=
2
a2 

3/ 2
5/2
(13) − [(13) − 32]
27 
45

= 0.78566a 2

xL = xEL dL : x (1.43971a ) = 0.78566a 2
or x = 0.546a 
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608
PROBLEM 5.48
Determine by direct integration the centroid of the area shown.
SOLUTION
y = a sin
πx
L
1
y, dA = y dx
2
L /2
L /2
πx
sin
A=
y dx = a
dx
0
0
L
xEL = x,
yEL =


L /2
L
π x 
A = a   − cos

L  0
π
 
Setting u =
πx
L
, we have x =
Integrating by parts,
L /2

xEL dA =

L
u , dx =
L
π
0
π
xydx =

L /2
aL
=
π
xa sin
0
πx
L
du,

π /2 
L 
L 
L
 π u  a sin u  π du  = a  π 




 

xEL dA =

L 
xEL dA = a   [−u cos u ]π0 /2 +
π  

yEL dA =
0
2
=
dx

L /2
0
a2 L
2π 2
1 2
1
y dx = a 2
2
2

π /2 1
0
2

L /2
0

π /2
0
sin 2
2

π /2
0
u sin x du
2
 aL
cos u du = 2
 π
πx
L
dx =
a2 L
2π

π /2
0
sin 2 u du
π /2
(1 − cos 2u )du =
2
 aL  aL
xA = xEL dA: x 
=

2
π  π

 aL  1 2
yA = yEL dA: y 
= a L
π  8

a2 L 
1

u − sin 2u 

4π 
2
0
1
= a2 L
8
x=
y=
π
8
L
π

a 
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609
PROBLEM 5.49*
Determine by direct integration the centroid of the area shown.
SOLUTION
2
2
r cos θ = aeθ cos θ
3
3
2
2 θ
= r sin θ = ae sin θ
3
3
xEL =
We have
yEL
dA =
and
1
1
(r )(rdθ ) = a 2 e2θ d θ
2
2

A = dA =
Then

π
0
π
1 2 2θ
1 1

a e dθ = a 2  e 2θ 
2
2 2
0
1 2 2π
a (e − 1)
4
= 133.623a 2
=
and
x
EL dA =

π
0
2 θ
1

ae cos θ  a 2 e2θ dθ 
3
2

1
= a3
3

π
0
e3θ cos θ dθ
To proceed, use integration by parts, with
u = e3θ
du = 3e3θ dθ
and
dv = cos θ dθ
Then
e
3θ

then
du = 3e3θ dθ
dv = sin θ dθ , then
Then
e
3θ
v = sin θ
cos θ dθ = e3θ sin θ − sin θ (3e3θ dθ )
u = e3θ
Now let
and
v = − cos θ
cos θ dθ = e3θ sin θ − 3  −e3θ cos θ − ( − cos θ )(3e3θ dθ ) 



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610
PROBLEM 5.49* (Continued)
so that

e3θ cos θ dθ =

π

1  e3θ
(sin θ + 3cos θ ) 
xEL dA = a 3 
3  10
0
=

Also,
e3θ
(sin θ + 3cos θ )
10
yEL dA =
a3
(−3e3π − 3) = −1239.26a3
30

π
0
2 θ
1

ae sin θ  a 2 e2θ dθ 
3
2


1
= a3
3

π
0
e3θ sin θ dθ
Use integration by parts, as above, with
u = e3θ
and

dv = sin θ dθ
Then
e
so that

3θ
and
v = − cos θ

sin θ dθ = −e3θ cos θ − (− cos θ )(3e3θ dθ )
e3θ sin θ dθ =

e3θ
(− cos θ + 3sin θ )
10
π

1  e3θ
(− cos θ + 3sin θ ) 
yEL dA = a3 
3  10
0
=
Hence,
du = 3e3θ dθ
a 3 3π
(e + 1) = 413.09a 3
30

or x = −9.27a 

or
xA = xEL dA: x (133.623a 2 ) = −1239.26a 3
yA = yEL dA : y (133.623a 2 ) = 413.09a3
y = 3.09a 
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611
PROBLEM 5.50
Determine the centroid of the area shown when a = 2 in.
SOLUTION
xEL = x
We have
yEL =
 1
dA = ydx = 1 −  dx
x

and

A = dA =
Then
and
1
1 1
y = 1 − 
2
2
x
x

EL dA
=
yEL dA =

a

a
1
1

a
1
1  dx
2
a
1 − x  2 = [ x − ln x]1 = ( a − ln a − 1) in


a

 a2
 1    x 2
1
− a +  in 3
x 1 −  dx  =  − x  = 
x   2
2

1  2
1  1   1   1
1 −  1 −  dx  =
2 
x  
x  2

a
1
2 1 
1 − x + 2  dx
x 

a
=
1
1
1
1
x − 2ln x −  =  a − 2ln a −  in 3
2 
x 1 2 
a

xA = xEL dA: x =

yA = yEL dA: y =
a2
2
−a+
1
2
in.
a − ln a − 1
a − 2ln a − 1a
2(a − ln a − 1)
in.
Find x and y when a = 2 in.
We have
x=
and
y=
1
2
(2) 2 − 2 +
1
2
2 − ln 2 − 1
2 − 2ln 2 − 12
2(2 − ln 2 − 1)
or
x = 1.629 in. 
or y = 0.1853 in. 
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612
PROBLEM 5.51
Determine the value of a for which the ratio x / y is 9.
SOLUTION
xEL = x
We have
yEL =
1
1 1
y = 1 − 
2
2
x
 1
dA = y dx = 1 −  dx
x

and

A = dA =
Then

a
1
1  dx
= [ x − ln x]1a
1 − 
x 2

= (a − ln a − 1) in 2
and
x
EL dA
=

a
1
a

 1    x 2
x 1 −  dx  =  − x 
x   2

1
 a2
1
= 
− a +  in 3
2
 2

yEL dA =

a
1
1  1   1   1
1 −  1 −  dx  =
2 
x  
x  2

a
1
2 1 
1 − x + 2  dx
x 

a
=
1
1
x − 2ln x − 
2 
x 1
=
1
1
a − 2ln a −  in 3
2 
a

xA = xEL dA: x =

yA = yEL dA: y =
a2
2
−a+
1
2
in.
a − ln a − 1
a − 2ln a − 1a
2(a − ln a − 1)
in.
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613
PROBLEM 5.51 (Continued)
Find a so that
x
= 9.
y
x xA
=
=
y yA
We have
Then
or
1
2
1
2
a2 − a +
1
2
( a − 2 ln a − a1 )
x
y
EL dA
EL dA
=9
a 3 − 11a 2 + a + 18a ln a + 9 = 0
Using trial and error or numerical methods, and ignoring the trivial solution a = 1 in., we find
a = 1.901 in. and a = 3.74 in. 
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614
PROBLEM 5.52
Determine the volume and the surface area of the solid obtained by rotating
the area of Problem 5.1 about (a) the x-axis, (b) the y-axis.
SOLUTION
From the solution of Problem 5.1, we have
A = 11 in 2
Σ xA = 11.5 in 3
Σ yA = 39.5 in 3
Applying the theorems of Pappus-Guldinus, we have
(a)
Volume = 2π yarea A = 2π Σ yA
= 2π (39.5 in 3 )
or Volume = 248 in 3 
Area = 2π yline L = 2π Σ( yline ) L
= 2π ( y2 L2 + y3 L3 + y4 L4 + y5 L5 + y6 L6 + y7 L7 + y8 L8 )
= 2π [(1)(2) + (2)(3) + (2.5)(1) + (3)(3) + (5.5)(5) + (8)(1) + (4)(8)]
or Area = 547 in 2 
(b)
Volume = 2π xarea A = 2π Σ xA
= 2π (11.5 in 3 )
or Volume = 72.3 in 3 
Area = 2π xline L = 2π Σ( xline ) L
= 2π ( x1 L1 + x2 L2 + x3 L3 + x4 L4 + x5 L5 + x6 L6 + x7 L7 )
= 2π [(0.5)(1) + (1)(2) + (2.5)(3) + (4)(1) + (2.5)(3) + (1)(5) + (0.5)(1)]
or Area = 169.6 in 2 
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you are using it without permission.
615
PROBLEM 5.53
Determine the volume and the surface area of the solid obtained by rotating the
area of Problem 5.2 about (a) the line y = 72 mm, (b) the x-axis.
SOLUTION
From the solution of Problem 5.2, we have
A = 2808 mm 2
x = 36 mm
y = 48 mm
Applying the theorems of Pappus-Guldinus, we have
(a)
Rotation about the line y = 72 mm:
Volume = 2π (72 − y ) A
= 2π (72 − 48)(2808)
Volume = 423 &times; 103 mm3 
Area = 2π yline L
= 2π Σ( yline ) L
= 2π ( y1 L1 + y3 L3 )
where y1 and y3 are measured with respect to line y = 72 mm.
Area = 2π (36)

(
)
482 + 722 + (36)
(
)
302 + 722 

Area = 37.2 &times; 103 mm 2 
(b)
Volume = 2π yarea A
= 2π (48)(2808)
Volume = 847 &times; 103 mm3 
Area = 2π yline L = 2π Σ( yline ) L
= 2π ( y1 L1 + y2 L2 + y3 L3 )
= 2π (36)

(
)
482 + 722 + (72)(78) + (36)
(
)
302 + 722 

Area = 72.5 &times; 103 mm 2 
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616
PROBLEM 5.54
Determine the volume and the surface area of the solid obtained by rotating the
area of Problem 5.8 about (a) the line x = −60 mm, (b) the line y = 120 mm.
SOLUTION
From the solution of Problem 5.8, we have
A = 7200 mm 2
Σ x A = −72 &times; 103 mm3
Σ y A = 629.83 &times; 103 mm3
Applying the theorems of Pappus-Guldinus, we have
(a)
Rotation about line x = −60 mm:
Volume = 2π ( x + 60) A = 2π (Σ xA + 60 A)
= 2π [−72 &times; 103 + 60(7200)]
Volume = 2.26 &times; 106 mm3 
Area = 2π xline L = 2π Σ( xline ) L
= 2π ( x1 L1 + x2 L2 + x3 L3 )


2(60)  π (60)  
2(60)  π (60) 
= 2π  60 −
+  60 +
+ (60)(120) 




π  2  
π  2 


where x1 , x2 , x3 are measured with respect to line x = −60 mm.
(b)
Area = 116.3 &times; 103 mm 2 
Rotation about line y = 120 mm:
Volume = 2π (120 − y ) A = 2π (120 A −Σ yA)
= 2π [120(7200) − 629.83 &times; 103 ]
Volume = 1.471 &times; 106 mm3 
Area = 2π yline L = 2π Σ( yline ) L
= 2π ( y1 L1 + y2 L2 + y4 L4 )
where y1 , y2 , y4 are measured with respect to line y = 120 mm.


2(60)  π (60)   2(60)  π (60) 
Area = 2π 120 −
+
+ (60)(120) 




π  2   π  2 


Area = 116.3 &times; 103 mm 2 
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617
PROBLEM 5.55
Determine the volume of the solid generated by rotating the parabolic
area shown about (a) the x-axis, (b) the axis AA′.
SOLUTION
First, from Figure 5.8a, we have
4
ah
3
2
y= h
5
A=
Applying the second theorem of Pappus-Guldinus, we have
(a)
Volume = 2π yA
 2  4 
= 2π  h  ah 
 5  3 
(b)
or Volume =
16
π ah 2 
15
or Volume =
16 2
πa h 
3
Volume = 2π (2a) A
4 
= 2π (2a)  ah 
3 
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618
PROBLEM 5.56
Determine the volume and the surface area of the chain link
shown, which is made from a 6-mm-diameter bar, if R = 10 mm
and L = 30 mm.
SOLUTION
The area A and circumference C of the cross section of the bar are
A=
π
4
d 2 and C = π d .
Also, the semicircular ends of the link can be obtained by rotating the cross section through a horizontal
semicircular arc of radius R. Now, applying the theorems of Pappus-Guldinus, we have for the volume V,
V = 2(Vside ) + 2(Vend )
= 2( AL) + 2(π RA)
= 2( L + π R) A
or
π

V = 2[30 mm + π (10 mm)]  (6 mm) 2 
4

= 3470 mm3
For the area A,
or V = 3470 mm3 
A = 2( Aside ) + 2( Aend )
= 2(CL) + 2(π RC )
= 2( L + π R)C
or
A = 2[30 mm + π (10 mm)][π (6 mm)]
= 2320 mm 2
or A = 2320 mm 2 
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619
PROBLEM 5.57
Verify that the expressions for the volumes of the first four shapes in Figure 5.21 on Page 264 are correct.
SOLUTION
Following the second theorem of Pappus-Guldinus, in each case, a specific
generating area A will be rotated about the x-axis to produce the given
shape. Values of y are from Figure 5.8a.
(1)
Hemisphere: the generating area is a quarter circle.
We have
(2)
2
V = π a3 
3
 4a  π 
V = 2π y A = 2π 
 ha 
 3π  4 
2
or V = π a 2 h 
3
Paraboloid of revolution: the generating area is a quarter parabola.
We have
(4)
or
Semiellipsoid of revolution: the generating area is a quarter ellipse.
We have
(3)
 4a  π 2 
V = 2π y A = 2π 
 a 
 3π  4 
 3  2 
V = 2π y A = 2π  a  ah 
 8  3 
1
or V = π a 2 h 
2
Cone: the generating area is a triangle.
We have
 a  1 
V = 2π y A = 2π   ha 
 3  2 
1
or V = π a 2 h 
3
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620
PROBLEM 5.58
Determine the volume and weight of the solid brass knob shown, knowing
that the specific weight of brass is 0.306 lb/in3.
SOLUTION
Volume of knob is obtained by rotating area
of components shown below.
Area, in2
1
π
4
(0.75) 2 = 0.4418
y , in.
y A, in 3
0.8183
0.3615
2
(0.5)(0.75) = 0.375
0.25
0.0938
3
(1.25)(0.75) = 0.9375
0.625
0.5859
4
−π
(0.75) 2 = −0.4418
4
0.9317
Σ
−0.4116
0.6296
V = 2π Σ y A = 2π (0.6296 in 3 ) = 3.9559 in 3
V = 3.96 in 3 
W = γ V = (0.306 lb/in 3 )(3.9559 in 3 )
W = 1.211 lb 
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621
PROBLEM 5.59
Determine the total surface area of the solid brass knob shown.
SOLUTION
Area is obtained by rotating lines shown about the x-axis.
1
2
3
4
π
2
π
2
L, in.
y , in.
yL, in 2
0.5
0.25
0.1250
(0.75) = 1.1781
0.9775
1.1516
(0.75) = 1.1781
0.7725
0.9101
0.25
0.1250
0.5
Σ
2.3117
A = 2π Σ y L = 2π (2.3117 in 2 )
A = 14.52 in 2 
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622
PROBLEM 5.60
The aluminum shade for the small high-intensity lamp shown has a uniform thickness of 1 mm. Knowing that
the density of aluminum is 2800 kg/m3, determine the mass of the shade.
SOLUTION
The mass of the lamp shade is given by
m = ρV = ρ At
where A is the surface area and t is the thickness of the shade. The area can be generated by rotating the line
shown about the x-axis. Applying the first theorem of Pappus Guldinus, we have
A = 2π yL = 2π Σ yL
= 2π ( y1 L1 + y2 L2 + y3 L3 + y4 L4 )
or
13 mm
 13 + 16 
2
2
A = 2π 
(13 mm) + 
 mm &times; (32 mm) + (3 mm)
 2 
 2
 16 + 28 
2
2
+
 mm &times; (8 mm) + (12 mm)
 2 

 28 + 33 
mm &times; (28 mm) 2 + (5 mm) 2 
+

 2 

= 2π (84.5 + 466.03 + 317.29 + 867.51)
= 10,903.4 mm 2
Then
m = ρ At
= (2800 kg/m3 )(10.9034 &times; 10−3 m 2 )(0.001 m)
m = 0.0305 kg 
or
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623
PROBLEM 5.61
The escutcheon (a decorative plate placed on a pipe where the
pipe exits from a wall) shown is cast from brass. Knowing that
the density of brass is 8470 kg/m3, determine the mass of the
escutcheon.
SOLUTION
The mass of the escutcheon is given by m = (density)V , where V is the volume. V can be generated by
rotating the area A about the x-axis.
From the figure:
L1 = 752 − 12.52 = 73.9510 m
L2 =
37.5
= 76.8864 mm
tan 26&deg;
a = L2 − L1 = 2.9324 mm
12.5
= 9.5941&deg;
75
26&deg; − 9.5941&deg;
α=
2
φ = sin −1
Area A can be obtained by combining the following four areas:
Applying the second theorem of Pappus-Guldinus and using Figure 5.8a, we have
V = 2π yA = 2π Σ yA
Seg.
A, mm 2
y , mm
yA, mm3
1
1
(76.886)(37.5) = 1441.61
2
1
(37.5) = 12.5
3
18,020.1
2
−α (75) 2 = −805.32
2(75)sin α
sin (α + φ ) = 15.2303
3α
−12,265.3
3
1
− (73.951)(12.5) = −462.19
2
1
(12.5) = 4.1667
3
−1925.81
4
−(2.9354)(12.5) = −36.693
1
(12.5) = 6.25
2
−229.33
Σ
3599.7
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624
PROBLEM 5.61 (Continued)
Then
V = 2π Σ yA
= 2π (3599.7 mm3 )
= 22, 618 mm3
m = (density)V
= (8470 kg/m3 )(22.618 &times; 10−6 m3 )
= 0.191574 kg
or m = 0.1916 kg 
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625
PROBLEM 5.62
A 34 - in.-diameter hole is drilled in a piece of 1-in.-thick steel; the hole
is then countersunk as shown. Determine the volume of steel removed
during the countersinking process.
SOLUTION
The required volume can be generated by rotating the area shown about the y-axis. Applying the second
theorem of Pappus-Guldinus, we have
V = 2π x A
3 1  1   1 1
1

= 2π  +   in. &times;  &times; in. &times; in.
8
3
4
2
4
4






V = 0.0900 in 3 
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626
PROBLEM 5.63
Knowing that two equal caps have been removed from a 10-in.-diameter wooden
sphere, determine the total surface area of the remaining portion.
SOLUTION
The surface area can be generated by rotating the line shown about the y-axis. Applying the first theorem of
Pappus-Guldinus, we have
A = 2 π XL = 2 π Σ x L
= 2π (2 x1 L1 + x2 L2 )
Now
tan α =
4
3
or
α = 53.130&deg;
Then
x2 =
5 in. &times; sin 53.130&deg;
π
53.130&deg; &times; 180
&deg;
= 4.3136 in.
and
π 

L2 = 2  53.130&deg; &times;
 (5 in.)
180&deg; 

= 9.2729 in.
 3 

A = 2π  2  in.  (3 in.) + (4.3136 in.)(9.2729 in.) 
 2 

A = 308 in 2 
or
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627
PROBLEM 5.64
Determine the capacity, in liters, of the punch bowl
shown if R = 250 mm.
SOLUTION
The volume can be generated by rotating the triangle and circular sector shown about the y-axis. Applying the
second theorem of Pappus-Guldinus and using Figure 5.8a, we have
V = 2π xA = 2π Σ xA
= 2π ( x1 A1 + x2 A2 )
 1 1   1 1
3   2 R sin 30&deg;   π 2  
= 2π  &times; R   &times; R &times;
R + 
R 


2   3 &times; π6   6
 
 3 2   2 2
 R3
R3 
= 2π 
+

 16 3 2 3 
3 3
π R3
8
3 3
=
π (0.25 m)3
8
= 0.031883 m3
=
Since
103 l = 1 m3
V = 0.031883 m3 &times;
103 l
1 m3
V = 31.9 l 
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628
PROBLEM 5.65*
The shade for a wall-mounted light is formed from a thin sheet of
translucent plastic. Determine the surface area of the outside of
the shade, knowing that it has the parabolic cross section shown.
SOLUTION
First note that the required surface area A can be generated by rotating the parabolic cross section through π
radians about the y-axis. Applying the first theorem of Pappus-Guldinus, we have
A = π xL
Now at
x = 100 mm,
250 = k (100)
and
2
y = 250 mm
or k = 0.025 mm −1
xEL = x
2
 dy 
dL = 1 +   dx
 dx 
where
dy
= 2kx
dx
Then
dL = 1 + 4k 2 x 2 dx
We have
xL = xEL dL =


100
0
x
( 1 + 4k x dx )
2 2
100
1 1

(1 + 4k 2 x 2 )3/2 
xL = 
2
3
 4k
0
1
1
=
[1 + 4(0.025)2 (100) 2 ]3/2 − (1)3/2
12 (0.025) 2
{
}
= 17,543.3 mm 2
Finally,
A = π (17,543.3 mm 2 )
or A = 55.1 &times; 103 mm 2 
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629
PROBLEM 5.66
magnitude and location of the resultant of the distributed
load, (b) the reactions at the beam supports.
SOLUTION
(a)
1
RI = (1100 N/m) (6 m) = 2200 N
3
RII = (900 N/m) (6 m) = 5400 N
R = RI + RII = 2200 + 5400 = 7600 N
XR = Σ xR :
X (7600) = (2200)(1.5) + (5400)(3)
X = 2.5658 m
(b)
R = 7.60 kN ,
X = 2.57 m 
Σ M A = 0: B (6 m) − (7600 N)(2.5658 m) = 0
B = 3250.0 N
B = 3.25 kN 
Σ Fy = 0: A + 3250.0 N − 7600 N = 0
A = 4350.0 N
A = 4.35 kN 
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630
PROBLEM 5.67
and location of the resultant of the distributed load, (b) the
reactions at the beam supports.
SOLUTION
1
(150lb/ft)(9 ft) = 675 lb
2
1
RII = (120 lb/ft)(9ft) = 540lb
2
R = RI + RII = 675 + 540 = 1215 lb
RI =
XR = Σ x R: X (1215) = (3)(675) + (6)(540)
X = 4.3333 ft
R = 1215 lb
(a)
(b)
Reactions:
X = 4.33 ft 
Σ M A = 0: B (9 ft) − (1215 lb) (4.3333 ft) = 0
B = 585.00 lb
B = 585 lb 
Σ Fy = 0: A + 585.00 lb − 1215 lb = 0
A = 630.00 lb
A = 630 lb 
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631
PROBLEM 5.68
SOLUTION
defined by a linear relation between load and distance and have the same values at the end points.
1
(900 N/m) (1.5 m) = 675 N
2
1
R2 = (400 N/m) (1.5 m) = 300 N
2
R1 =
Σ M A = 0: − (675 N)(1.4 m) + (300 N)(0.9 m) + B(2.5 m) = C
B = 270 N
B = 270 N 
Σ Fy = 0: A − 675 N + 300 N + 270 N = 0
A = 105.0 N
A = 105.0 N 
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632
PROBLEM 5.69
SOLUTION
R I = (200 lb/ft)(15 ft)
R I = 3000 lb
1
(200 lb/ft)(6 ft)
2
R II = 600 lb
R II =
ΣM A = 0: − (3000 lb)(1.5 ft) − (600 lb)(9 ft + 2 ft) + B(15 ft) = 0
B = 740 lb
B = 740 lb 
ΣFy = 0: A + 740 lb − 3000 lb − 600 lb = 0
A = 2860 lb
A = 2860 lb 
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633
PROBLEM 5.70
SOLUTION
R I = (200 lb/ft)(4 ft) = 800 lb
R II =
1
(150 lb/ft)(3 ft) = 225 lb
2
ΣFy = 0: A − 800 lb + 225 lb = 0
A = 575 lb 
ΣM A = 0: M A − (800 lb)(2 ft) + (225 lb)(5 ft) = 0
M A = 475 lb ⋅ ft

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634
PROBLEM 5.71
Determine the reactions at the beam supports for the given
SOLUTION
1
(4 kN/m)(6 m)
2
= 12 kN
RII = (2 kN/m)(10 m)
RI =
= 20 kN
ΣFy = 0: A − 12 kN − 20 kN = 0
A = 32.0 kN 
ΣM A = 0: M A − (12 kN)(2 m) − (20 kN)(5 m) = 0
M A = 124.0 kN ⋅ m

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635
PROBLEM 5.72
SOLUTION
are defined by a parabolic relation between load and distance and the values at the end points are the same.
We have
RI = (6 m)(300 N/m) = 1800 N
RII =
Then
2
(6 m)(1200 N/m) = 4800 N
3
ΣFx = 0: Ax = 0
ΣFy = 0: Ay + 1800 N − 4800 N = 0
or
Ay = 3000 N
A = 3.00 kN 
 15 
ΣM A = 0: M A + (3 m)(1800 N) −  m  (4800 N) = 0
 4 
or
M A = 12.6 kN ⋅ m
M A = 12.60 kN ⋅ m

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636
PROBLEM 5.73
SOLUTION
We have
Then
1
RI = (12 ft)(200 lb/ft) = 800 lb
3
1
RII = (6 ft)(100 lb/ft) = 200 lb
3
ΣFx = 0: Ax = 0
ΣFy = 0: Ay − 800 lb − 200 lb = 0
or
Ay = 1000 lb
A = 1000 lb 
ΣM A = 0: M A − (3 ft)(800 lb) − (16.5 ft)(200 lb) = 0
or
M A = 5700 lb ⋅ ft
M A = 5700 lb ⋅ ft

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637
PROBLEM 5.74
Determine the reactions at the beam supports for the given
SOLUTION
1
1
wO (12 ft) = (400 lb/ft)(12 ft) = 2400 lb
2
2
1
RII = (300 lb/ft)(12 ft) = 1800 lb
2
RI =
ΣM B = 0: (2400 lb)(1 ft) − (1800 lb)(3 ft) + C (7 ft) = 0
C = 428.57 lb
C = 429 lb 
ΣFy = 0: B + 428.57 lb − 2400 lb − 1800 lb = 0
B = 3771 lb
B = 3770 lb 
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638
PROBLEM 5.75
Determine (a) the distributed load wO at the end A of the beam
ABC for which the reaction at C is zero, (b) the corresponding
reaction at B.
SOLUTION
For wO ,
1
wO (12 ft) = 6 wO
2
1
RII = (300 lb/ft)(12 ft) = 1800 lb
2
RI =
(a)
For C = 0,
ΣM B = 0: (6 wO )(1 ft) − (1800 lb)(3 ft) = 0
(b)
Corresponding value of R I :
wO = 900 lb/ft 
RI = 6(900) = 5400 lb
ΣFy = 0: B − 5400 lb − 1800 lb = 0
B = 7200 lb 
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639
PROBLEM 5.76
Determine (a) the distance a so that the vertical reactions at
supports A and B are equal, (b) the corresponding reactions at
the supports.
SOLUTION
(a)
We have
Then
1
(a m)(1800 N/m) = 900a N
2
1
RII = [(4 − a ) m](600 N/m) = 300(4 − a) N
2
RI =
ΣFy = 0: Ay − 900a − 300(4 − a ) + B y = 0
or
Ay + B y = 1200 + 600a
Now
Ay = B y  Ay = B y = 600 + 300a (N)
Also,
(1)

a 
ΣM B = 0: − (4 m) Ay +  4 −  m  [(900a) N]
3 

1

+  (4 − a ) m  [300(4 − a) N] = 0
3

or
Ay = 400 + 700a − 50a 2
Equating Eqs. (1) and (2),
600 + 300a = 400 + 700a − 50a 2
or
a 2 − 8a + 4 = 0
(2)
8 &plusmn; (−8) 2 − 4(1)(4)
2
Then
a=
or
a = 0.53590 m
Now
a≤4m
a = 7.4641 m
a = 0.536 m 
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640
PROBLEM 5.76 (Continued)
(b)
We have
From Eq. (1):
ΣFx = 0: Ax = 0
Ay = By
= 600 + 300(0.53590)
= 761 N
A = B = 761 N 
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641
PROBLEM 5.77
Determine (a) the distance a so that the reaction at support B is
minimum, (b) the corresponding reactions at the supports.
SOLUTION
(a)
We have
Then
or
Then
(b)
From Eq. (1):
and
1
(a m)(1800 N/m) = 900a N
2
1
RII = [(4 − a )m](600 N/m) = 300(4 − a) N
2
RI =
a 
8+ a 
ΣM A = 0: −  m  (900a N) − 
m  [300(4 − a)N] + (4 m) By = 0
3


 3

By = 50a 2 − 100a + 800
dBy
da
= 100a − 100 = 0
By = 50(1)2 − 100(1) + 800 = 750 N
(1)
or a = 1.000 m 
B = 750 N

ΣFx = 0: Ax = 0
ΣFy = 0: Ay − 900(1) N − 300(4 − 1) N + 750 N = 0
or
Ay = 1050 N
A = 1050 N 
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642
PROBLEM 5.78
A beam is subjected to a linearly distributed downward
load and rests on two wide supports BC and DE, which
exert uniformly distributed upward loads as shown.
Determine the values of wBC and wDE corresponding to
equilibrium when wA = 600 N/m.
SOLUTION
We have
1
(6 m)(600 N/m) = 1800 N
2
1
RII = (6 m)(1200 N/m) = 3600 N
2
RBC = (0.8 m) (wBC N/m) = (0.8wBC ) N
RI =
RDE = (1.0 m) (wDE N/m) = ( wDE ) N
Then
ΣM G = 0: − (1 m)(1800 N) − (3 m)(3600 N) + (4 m)( wDE N) = 0
wDE = 3150 N/m 
or
and
or
ΣFy = 0: (0.8wBC ) N − 1800 N − 3600 N + 3150 N = 0
wBC = 2812.5 N/m
wBC = 2810 N/m 
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643
PROBLEM 5.79
A beam is subjected to a linearly distributed downward load
and rests on two wide supports BC and DE, which exert
uniformly distributed upward loads as shown. Determine (a)
the value of wA so that wBC = wDE, (b) the corresponding
values of wBC and wDE.
SOLUTION
(a)
1
(6 m)( wA N/m) ⋅ (3 wA ) N
2
1
RII = (6 m)(1200 N/m) = 3600 N
2
RBC = (0.8 m) (wBC N/m) = (0.8wBC ) N
RI =
We have
RDE = (1 m) (wDE N/m) = ( wDE ) N
ΣFy = 0: (0.8wBC ) N − (3wA ) N − 3600 N + ( wDE ) N = 0
Then
or
0.8wBC + wDE = 3600 + 3wA
5
wBC = wDE  wBC = wDE = 2000 + wA
3
Now
(1)
ΣM G = 0: −(1 m)(3wA N) − (3 m)(3600 N) + (4 m)( wDE N) = 0
Also,
wDE = 2700 +
or
3
wA
4
(2)
Equating Eqs. (1) and (2),
5
3
2000 + wA = 2700 + wA
3
4
or
(b) Eq. (1) 
wA =
8400
N/m
11
wA = 764 N/m 
5  8400 
wBC = wDE = 2000 + 
3  11 
or wBC = wDE = 3270 N/m 
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644
PROBLEM 5.80
The cross section of a concrete dam is as shown. For a 1-m-wide
dam section, determine (a) the resultant of the reaction forces
exerted by the ground on the base AB of the dam, (b) the point of
application of the resultant of part a, (c) the resultant of the
pressure forces exerted by the water on the face BC of the dam.
SOLUTION
(a)
Consider free body made of dam and section BDE of water. (Thickness = 1 m)
p = (3 m)(10 kg/m3 )(9.81 m/s 2 )
W1 = (1.5 m)(4 m)(1 m)(2.4 &times; 103 kg/m3 )(9.81 m/s 2 ) = 144.26 kN
1
W2 = (2 m)(3 m)(1 m)(2.4 &times; 103 kg/m3 )(9.81 m/s 2 ) = 47.09 kN
3
2
W3 = (2 m)(3 m)(1 m)(103 kg/m3 )(9.81 m/s 2 ) = 39.24 kN
3
1
1
P = Ap = (3 m)(1 m)(3 m)(103 kg/m3 )(9.81 m/s 2 ) = 44.145 kN
2
2
ΣFx = 0: H − 44.145 kN = 0
H = 44.145 kN
H = 44.1 kN

ΣFy = 0: V − 141.26 − 47.09 − 39.24 = 0
V = 227.6 kN
V = 228 kN

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645
PROBLEM 5.80 (Continued)
1
(1.5 m) = 0.75 m
2
1
x2 = 1.5 m + (2 m) = 2 m
4
5
x3 = 1.5 m + (2 m) = 2.75 m
8
x1 =
ΣM A = 0: xV − Σ xW + P (1 m) = 0
x(227.6 kN) − (141.26 kN)(0.75 m) − (47.09 kN)(2 m)
− (39.24 kN)(2.75 m) + (44.145 kN)(1 m) = 0
x(227.6 kN) − 105.9 − 94.2 − 107.9 + 44.145 = 0
x(227.6) − 263.9 = 0
x = 1.159 m (to right of A) 
(b)
Resultant of face BC:
Consider free body of section BDE of water.
− R = 59.1 kN
R = 59.1 kN
41.6&deg;
41.6&deg; 
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646
PROBLEM 5.81
The cross section of a concrete dam is as shown. For a 1-m-wide
dam section, determine (a) the resultant of the reaction forces
exerted by the ground on the base AB of the dam, (b) the point of
application of the resultant of part a, (c) the resultant of the
pressure forces exerted by the water on the face BC of the dam.
SOLUTION
(a)
Consider free body made of dam and triangular section of water shown. (Thickness = 1 m.)
p = (7.2 m)(103 kg/m3 )(9.81m/s 2 )
2
(4.8 m)(7.2 m)(1 m)(2.4 &times; 103 kg/m3 )(9.81 m/s 2 )
3
= 542.5 kN
1
W2 = (2.4 m)(7.2 m)(1 m)(2.4 &times; 103 kg/m3 )(9.81 m/s 2 )
2
= 203.4 kN
1
W3 = (2.4 m)(7.2 m)(1 m)(103 kg/m3 )(9.81 m/s 2 )
2
= 84.8 kN
1
1
P = Ap = (7.2 m)(1 m)(7.2 m)(103 kg/m3 )(9.81 m/s 2 )
2
2
= 254.3 kN
W1 =
ΣFx = 0: H − 254.3 kN = 0
H = 254 kN

ΣFy = 0: V − 542.5 − 203.4 − 84.8 = 0
V = 830.7 kN
V = 831 kN 
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647
PROBLEM 5.81 (Continued)
5
x1 = (4.8 m) = 3 m
8
1
x2 = 4.8 + (2.4) = 5.6 m
3
2
x3 = 4.8 + (2.4) = 6.4 m
3
(b)
ΣM A = 0: xV − Σ xW + P(2.4 m) = 0
x(830.7 kN) − (3 m)(542.5 kN) − (5.6 m)(203.4 kN)
− (6.4 m)(84.8 kN) + (2.4 m)(254.3 kN) = 0
x(830.7) − 1627.5 − 1139.0 − 542.7 + 610.3 = 0
x(830.7) − 2698.9 = 0
x = 3.25 m (to right of A) 
(c)
Resultant on face BC:
Direct computation:
P = ρ gh = (103 kg/m3 )(9.81 m/s 2 )(7.2 m)
P = 70.63 kN/m 2
BC = (2.4) 2 + (7.2) 2
= 7.589 m
θ = 18.43&deg;
1
PA
2
1
= (70.63 kN/m 2 )(7.589 m)(1 m)
2
R=
R = 268 kN
18.43&deg; 
Alternate computation: Use free body of water section BCD.
− R = 268 kN
R = 268 kN
18.43&deg;
18.43&deg; 
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648
PROBLEM 5.82
An automatic valve consists of a 9 &times; 9-in. square plate that is pivoted about a
horizontal axis through A located at a distance h = 3.6 in. above the lower edge.
Determine the depth of water d for which the valve will open.
SOLUTION
Since valve is 9 in. wide, w = 9 p = 9γ h, where all dimensions are in inches.
w1 = 9γ (d − 9), w2 = 9γ d
1
1
(9 in.) w1 = (9)(9γ )(d − 9)
2
2
1
1
PII = (9 in.) w2 = (9)(9γ d )
2
2
PI =
Valve opens when B = 0.
ΣM A = 0: PI (6 in. − 3.6 in.) − PII (3.6 in. − 3 in.) = 0
1

1

 2 (9)(9γ )( d − 9)  (2.4) −  2 (9)(9γ d )  (0.6) = 0




(d − 9)(2.4) − d (0.6) = 0
1.8d − 21.6 = 0
d = 12.00 in. 
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649
PROBLEM 5.83
An automatic valve consists of a 9 &times; 9-in. square plate that is pivoted about
a horizontal axis through A. If the valve is to open when the depth of water
is d = 18 in., determine the distance h from the bottom of the valve to the
pivot A.
SOLUTION
Since valve is 9 in. wide, w = 9 p = 9γ h, where all dimensions are in inches.
w1 = 9γ ( d − 9)
w2 = 9γ d
For d = 18 in.,
w1 = 9γ (18 − 9) = 81γ
w2 = 9γ (18) = 162γ
1
1
(9)(9γ )(18 − 9) = (729γ )
2
2
1
PII = (9)(9γ )(18) = 729γ
2
PI =
Valve opens when B = 0.
ΣM A = 0: P1 (6 − h) − PII ( h − 3) = 0
1
729γ (6 − h) − 729( h − 3) = 0
2
1
3− h −h +3= 0
2
6 − 1.5h = 0
h = 4.00 in. 
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650
PROBLEM 5.84
The 3 &times; 4-m side AB of a tank is hinged at its bottom A and is held in place by a
thin rod BC. The maximum tensile force the rod can withstand without breaking
is 200 kN, and the design specifications require the force in the rod not to exceed
20 percent of this value. If the tank is slowly filled with water, determine the
maximum allowable depth of water d in the tank.
SOLUTION
Consider the free-body diagram of the side.
We have
Now
where
P=
1
1
Ap = A( ρ gd )
2
2
ΣM A = 0: hT −
d
P=0
3
h=3m
Then for d max ,
(3 m)(0.2 &times; 200 &times; 103 N) −
d max
3
1

3
3
2
 2 (4 m &times; d max ) &times; (10 kg/m &times; 9.81 m/s &times; d max )  = 0


3
120 N ⋅ m − 6.54d max
N/m 2 = 0
or
d max = 2.64 m 
or
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651
PROBLEM 5.85
The 3 &times; 4-m side of an open tank is hinged at its bottom A and is held in
place by a thin rod BC. The tank is to be filled with glycerine, whose density
is 1263 kg/m3. Determine the force T in the rod and the reactions at the hinge
after the tank is filled to a depth of 2.9 m.
SOLUTION
Consider the free-body diagram of the side.
We have
Then
1
1
Ap = A( ρ gd )
2
2
1
= [(2.9 m)(4 m)] [(1263 kg/m3 )(9.81 m/s 2 )(2.9 m)]
2
= 208.40 kN
P=
ΣFy = 0: Ay = 0
 2.9 
m  (208.4 kN) = 0
ΣM A = 0: (3 m)T − 
 3 
T = 67.151 kN
or
ΣFx = 0:
T = 67.2 kN

Ax + 208.40 kN − 67.151 kN = 0
Ax = −141.249 kN
or
A = 141.2 kN

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652
PROBLEM 5.86
The friction force between a 6 &times; 6-ft square sluice gate AB and its guides is equal
to 10 percent of the resultant of the pressure forces exerted by the water on the face
of the gate. Determine the initial force needed to lift the gate if it weighs 1000 lb.
SOLUTION
Consider the free-body diagram of the gate.
Now
1
1
ApI = [(6 &times; 6) ft 2 ][(62.4 lb/ft 3 )(9 ft)]
2
2
= 10,108.8 lb
PI =
1
1
ApII = [(6 &times; 6) ft 2 ][(62.4 lb/ft 3 )(15 ft)]
2
2
= 16,848 lb
PII =
Then
F = 0.1P = 0.1( PI + PII )
= 0.1(10,108.8 + 16,848) lb
= 2695.7 lb
Finally
ΣFy = 0: T − 2695.7 lb − 1000 lb = 0
or T = 3.70 kips 
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653
PROBLEM 5.87
A tank is divided into two sections by a 1 &times; 1-m square gate that is
hinged at A. A couple of magnitude 490 N &middot; m is required for the
gate to rotate. If one side of the tank is filled with water at the rate
of 0.1 m3/min and the other side is filled simultaneously with
methyl alcohol (density ρma = 789 kg/m3) at the rate of 0.2 m3/min,
determine at what time and in which direction the gate will rotate.
SOLUTION
Consider the free-body diagram of the gate.
First note V = A base d and V = rt.
Then
dW =
d MA =
Now
P=
0.1 m3 / min &times; t (min)
= 0.25t (m)
(0.4 m)(1 m)
0.2 m3 / min &times; t (min)
= t (m)
(0.2 m)(1 m)
1
1
Ap = A( ρ gh) so that
2
2
1
PW = [(0.25t ) m &times; (1 m)][(103 kg/m3 )(9.81 m/s 2 )(0.25t ) m]
2
= 306.56t 2 N
1
PMA = [(t ) m &times; (1 m)][(789 kg/m3 )(9.81 m/s 2 )(t ) m]
2
= 3870t 2 N
Now assume that the gate will rotate clockwise and when d MA ≤ 0.6 m. When rotation of the gate is
impending, we require
1
1



ΣM A : M R =  0.6 m − d MA  PMA −  0.6 m − dW
3
3




 PW

Substituting
1 
1



490 N ⋅ m =  0.6 − t  m &times; (3870t 2 ) N −  0.6 − &times; 0.25t  m &times; (306.56t 2 ) N
3
3




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654
PROBLEM 5.87 (Continued)
Simplifying
1264.45t 3 − 2138.1t 2 + 490 = 0
Solving (positive roots only)
t = 0.59451 min and t = 1.52411 min
Now check assumption using the smaller root. We have
d MA = (t ) m = 0.59451 m &lt; 0.6 m
∴ t = 0.59451min = 35.7 s 
and the gate rotates clockwise. 
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655
PROBLEM 5.88
A prismatically shaped gate placed at the end of a freshwater channel is
supported by a pin and bracket at A and rests on a frictionless support at B.
The pin is located at a distance h = 0.10 m below the center of gravity C
of the gate. Determine the depth of water d for which the gate will open.
SOLUTION
First note that when the gate is about to open (clockwise rotation is impending), By
0 and the line of
action of the resultant P of the pressure forces passes through the pin at A. In addition, if it is assumed that the
gate is homogeneous, then its center of gravity C coincides with the centroid of the triangular area. Then
a=
d
− (0.25 − h)
3
and
b=
2
8 d 
(0.4) −  
3
15  3 
Now
a 8
=
b 15
so that
d
− (0.25 − h)
3
2
(0.4) − 158 d3
3
( )
=
8
15
Simplifying yields
289
70.6
d + 15h =
45
12
(1)
Alternative solution:
Consider a free body consisting of a 1-m thick section of the gate and the triangular section BDE of water
above the gate.
Now
1
1
Ap′ = (d &times; 1 m)( ρ gd )
2
2
1
= ρ gd 2 (N)
2
1 8

W ′ = ρ gV = ρ g  &times; d &times; d &times; 1 m 
2
15


4
= ρ gd 2 (N)
15
P′ =
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656
PROBLEM 5.88 (Continued)
Then with By = 0 (as explained above), we have
2
1  8   4
 d
 1

ΣM A = 0:  (0.4) −  d    ρ gd 2  −  − (0.25 − h)   ρ gd 2  = 0
3
3
15
15
3
2

 
 



Simplifying yields
289
70.6
d + 15h =
45
12
as above.
Find d:
Substituting into Eq. (1),
h = 0.10 m
289
70.6
d + 15(0.10) =
45
12
or d = 0.683 m 
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657
PROBLEM 5.89
A prismatically shaped gate placed at the end of a freshwater channel
is supported by a pin and bracket at A and rests on a frictionless support
at B. Determine the distance h if the gate is to open when d = 0.75 m.
SOLUTION
First note that when the gate is about to open (clockwise rotation is impending), By
0 and the line of
action of the resultant P of the pressure forces passes through the pin at A. In addition, if it is assumed that the
gate is homogeneous, then its center of gravity C coincides with the centroid of the triangular area. Then
a=
d
− (0.25 − h)
3
and
b=
2
8 d
(0.4) −  
3
15  3 
Now
a 8
=
b 15
so that
d
− (0.25 − h)
3
2
(0.4) − 158 d3
3
( )
=
8
15
Simplifying yields
289
70.6
d + 15h =
45
12
(1)
Alternative solution:
Consider a free body consisting of a 1-m thick section of the gate and the triangular section BDE of water
above the gate.
Now
1
1
Ap′ = (d &times; 1 m)( ρ gd )
2
2
1
2
= ρ gd (N)
2
1 8

W ′ = ρ gV = ρ g  &times; d &times; d &times; 1 m 
 2 15

4
= ρ gd 2 (N)
15
P′ =
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658
PROBLEM 5.89 (Continued)
Then with By = 0 (as explained above), we have
2
1  8   4
 d
 1

ΣM A = 0:  (0.4) −  d    ρ gd 2  −  − (0.25 − h)   ρ gd 2  = 0
3  15    15
 3
 2

3
Simplifying yields
289
70.6
d + 15h =
45
12
as above.
Find h:
Substituting into Eq. (1),
d = 0.75 m
289
70.6
(0.75) + 15h =
45
12
or h = 0.0711 m 
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659
PROBLEM 5.90
The square gate AB is held in the position shown by hinges along its top edge A
and by a shear pin at B. For a depth of water d = 3.5 ft, determine the force
exerted on the gate by the shear pin.
SOLUTION
First consider the force of the water on the gate. We have
1
Ap
2
1
= A(γ h)
2
P=
Then
1
(1.8 ft) 2 (62.4 lb/ft 3 )(1.7 ft)
2
= 171.850 lb
PI =
1
(1.8 ft)2 (62.4 lb/ft 3 ) &times; (1.7 + 1.8cos 30&deg;) ft
2
= 329.43 lb
PII =
Now
or
1

2

ΣM A = 0:  LAB  PI +  LAB  PII − LAB FB = 0
3

3

1
2
(171.850 lb) + (329.43 lb) − FB = 0
3
3
FB = 276.90 lb
or
FB = 277 lb
30.0&deg; 
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660
PROBLEM 5.91
A long trough is supported by a continuous hinge along its
lower edge and by a series of horizontal cables attached to
its upper edge. Determine the tension in each of the cables,
at a time when the trough is completely full of water.
SOLUTION
Consider free body consisting of 20-in. length of the trough and water.
l = 20-in. length of free body
π

W = γ v = γ  r 2l 
4


PA = γ r
P=
1
1
1
PA rl = (γ r )rl = γ r 2l
2
2
2
1 
ΣM A = 0: Tr − Wr − P  r  = 0
3 
 π
 4r
Tr −  γ r 2 l 
 4
 3π
  1 2  1 
 −  2 γ r l  3 r  = 0
 
 
1
1
1
T = γ r 2l + γ r 2l = γ r 2l
3
6
2
Data:
γ = 62.4 lb/ft 3 r =
Then
T=
24
20
ft = 2 ft l =
ft
12
12
1
 20 
(62.4 lb/ft 3 )(2 ft)2 
ft 
2
 12 
= 208.00 lb
T = 208 lb 
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661
PROBLEM 5.92
A 0.5 &times; 0.8-m gate AB is located at the bottom of a tank filled
with water. The gate is hinged along its top edge A and rests on a
frictionless stop at B. Determine the reactions at A and B when
cable BCD is slack.
SOLUTION
First consider the force of the water on the gate.
We have
P=
1
1
Ap = A( ρ gh)
2
2
1
PI = [(0.5 m)(0.8 m)] &times; [(103 kg/m3 )(9.81 m/s 2 )(0.45 m)]
2
= 882.9 N
so that
1
PII = [(0.5 m)(0.8 m)] &times; [(103 kg/m3 )(9.81 m/s 2 )(0.93 m)]
2
= 1824.66 N
Reactions at A and B when T = 0:
We have
ΣM A = 0:
1
2
(0.8 m)(882.9 N) + (0.8 m)(1824.66 N) − (0.8 m)B = 0
3
3
B = 1510.74 N
or
or
B = 1511 N
53.1&deg; 
A = 1197 N
53.1&deg; 
ΣF = 0: A + 1510.74 N − 882.9 N − 1824.66 N = 0
or
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662
PROBLEM 5.93
A 0.5 &times; 0.8-m gate AB is located at the bottom of a tank filled with
water. The gate is hinged along its top edge A and rests on a frictionless
stop at B. Determine the minimum tension required in cable BCD to
open the gate.
SOLUTION
First consider the force of the water on the gate.
We have
so that
P=
1
1
Ap = A( ρ gh)
2
2
1
PI = [(0.5 m)(0.8 m)] &times; [(103 kg/m3 )(9.81 m/s 2 )(0.45 m)]
2
= 882.9 N
1
PII = [(0.5 m)(0.8 m)] &times; [(103 kg/m3 )(9.81 m/s 2 )(0.93 m)]
2
= 1824.66 N
T to open gate:
First note that when the gate begins to open, the reaction at B
Then
or
ΣM A = 0:
0.
1
2
(0.8 m)(882.9 N) + (0.8 m)(1824.66 N)
3
3
 8 
− (0.45 + 0.27)m &times;  T  = 0
 17 
235.44 + 973.152 − 0.33882 T = 0
or T = 3570 N 
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663
PROBLEM 5.94
A 4 &times; 2-ft gate is hinged at A and is held in position by rod CD.
End D rests against a spring whose constant is 828 lb/ft. The spring
is undeformed when the gate is vertical. Assuming that the force
exerted by rod CD on the gate remains horizontal, determine the
minimum depth of water d for which the bottom B of the gate will
move to the end of the cylindrical portion of the floor.
SOLUTION
First determine the forces exerted on the gate by the spring and the water when B is at the end of the
cylindrical portion of the floor.
sin θ =
We have
2
θ = 30&deg;
4
Then
xSP = (3 ft) tan 30&deg;
and
FSP = kxSP
= 828 lb/ft &times; 3 ft &times; tan30&deg;
= 1434.14 lb
Assume
d ≥ 4 ft
We have
P=
1
1
Ap = A(γ h)
2
2
1
PI = [(4 ft)(2 ft)] &times; [(62.4 lb/ft 3 )( d − 4) ft]
2
= 249.6(d − 4) lb
Then
1
PII = [(4 ft)(2 ft)] &times; [(62.4 lb/ft 3 )(d − 4 + 4cos 30&deg;)]
2
= 249.6(d − 0.53590&deg;) lb
For d min so that the gate opens, W = 0
Using the above free-body diagrams of the gate, we have
4 
8 
ΣM A = 0:  ft  [249.6(d − 4) lb] +  ft  [249.6( d − 0.53590) lb]
3 
3 
−(3 ft)(1434.14 lb) = 0
or
(332.8d − 1331.2) + (665.6d − 356.70) − 4302.4 = 0
d = 6.00 ft
or
d ≥ 4 ft  assumption correct
d = 6.00 ft 
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664
PROBLEM 5.95
Solve Problem 5.94 if the gate weighs 1000 lb.
PROBLEM 5.94 A 4 &times; 2-ft gate is hinged at A and is held in
position by rod CD. End D rests against a spring whose constant is
828 lb/ft. The spring is undeformed when the gate is vertical.
Assuming that the force exerted by rod CD on the gate remains
horizontal, determine the minimum depth of water d for which the
bottom B of the gate will move to the end of the cylindrical portion
of the floor.
SOLUTION
First determine the forces exerted on the gate by the spring and the water when B is at the end of the
cylindrical portion of the floor.
We have
sin θ =
2
θ = 30&deg;
4
Then
xSP = (3 ft) tan 30&deg;
and
FSP = kxSP = 828 lb/ft &times; 3 ft &times; tan30&deg;
= 1434.14 lb
Assume
d ≥ 4 ft
We have
P=
1
1
Ap = A(γ h)
2
2
1
PI = [(4 ft)(2 ft)] &times; [(62.4 lb/ft 3 )( d − 4) ft]
2
= 249.6(d − 4) lb
1
PII = [(4 ft)(2 ft)] &times; [(62.4 lb/ft 3 )(d − 4 + 4cos 30&deg;)]
2
= 249.6(d − 0.53590&deg;) lb
Then
For d min so that the gate opens, W = 1000 lb
Using the above free-body diagrams of the gate, we have
4 
8 
ΣM A = 0:  ft  [249.6( d − 4) lb] +  ft  [249.6( d − 0.53590) lb]
3 
3 
− (3 ft)(1434.14 lb) − (1 ft)(1000 lb) = 0
or
(332.8d − 1331.2) + (665.6d − 356.70) − 4302.4 − 1000 = 0
d = 7.00 ft
or
d ≥ 4 ft  assumption correct
d = 7.00 ft 
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665
PROBLEM 5.96
A hemisphere and a cone are attached as shown. Determine the location of the
centroid of the composite body when (a) h = 1.5a, (b) h = 2a.
SOLUTION
V
y
yV
Cone I
1 2
πa h
3
h
4
1
π a 2 h2
12
Hemisphere II
2 3
πa
3
−
3a
8
1
− π a4
4
1
V = π a 2 ( h + 2a )
3
1
Σ yV = π a 2 (h 2 − 3a 2 )
12
(a)
For h = 1.5a,
1
7
V = π a 2 (1.5a + 2a ) = π a 2
3
6
Σ yV =
1
1
π a 2 [(1.5a) 2 − 3a 2 ] = − π a 4
12
16
1
3
7

Y V = Σ yV : Y  π a3  = − π a 4 Y = − a
16
56
6

Centroid is 0.0536a below base of cone. 
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666
PROBLEM 5.96 (Continued)
(b)
For h = 2a,
1
4
V = π a 2 (2a + 2a) = π a 3
3
3
Σ yV =
1
1
π a 2 [(2a) 2 − 3a 2 ] = π a 4
12
12
1
4
 1
Y V = Σ yV : Y  π a 3  = π a 4 Y = a
16
3
 12
Centroid is 0.0625a above base of cone. 
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667
PROBLEM 5.97
Consider the composite body shown. Determine (a) the value
of x when h = L/2, (b) the ratio h/L for which x = L.
SOLUTION
V
x
xV
Rectangular prism
Lab
1
L
2
1 2
L ab
2
Pyramid
1 b
a
h
3  2 
1
h
4
1
1 

abh  L + h 
6
4 

1 

ΣV = ab  L + h 
6 

1 
1 

Σ xV = ab 3L2 + h  L + h  
6 
4 

Then
X ΣV = Σ xV
Now
so that
 
1  1 
1 
X  ab  L + h   = ab  3L2 + hL + h 2 
6
6
4 


 
h 1 h2 
 1 h 1 
X 1 +
L
3
=
+
+



L 4 L2 
 6 L  6 
or
(a)
L+
X = ? when h =
Substituting
(1)
1
L.
2
h 1
= into Eq. (1),
L 2
2
 1  1  1   1  1  1  
X 1 +    = L 3 +   +   
 6  2   6   2  4  2  
or
X=
57
L
104
X = 0.548L 
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668
PROBLEM 5.97 (Continued)
(b)
h
= ? when X = L.
L
Substituting into Eq. (1),
or
or
h 1 h2 
 1 h 1 
L 1 +
=
L
3
+
+



L 4 L2 
 6 L  6 
1+
1 h 1 1 h 1 h2
= +
+
6 L 2 6 L 24 L2
h2
= 12
L2
h
=2 3 
L
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669
PROBLEM 5.98
Determine the y coordinate of the centroid of the body shown.
SOLUTION
First note that the values of Y will be the same for the given body and the body shown below. Then
V
y
Cone
1 2
πa h
3
1
− h
4
2
Cylinder
1
a
−π   b = − π a 2 b
4
2
1
− b
2
Σ
We have
Then
π
12
a 2 (4 h − 3b)
yV
−
1
π a2 h2
12
1 2 2
πa b
8
−
π
24
a 2 (2 h 2 − 3b 2 )
Y ΣV = Σ yV
π
π

Y  a 2 (4h − 3b)  = − a 2 (2h 2 − 3b 2 )
24
 12

or Y = −
2h 2 − 3b 2

2(4h − 3b)
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670
PROBLEM 5.99
Determine the z coordinate of the centroid of the body shown. (Hint: Use
the result of Sample Problem 5.13.)
SOLUTION
First note that the body can be formed by removing a half cylinder from a half cone, as shown.
V
1 2
πa h
6
Half cone
Half cylinder
Σ
−
π a
2
24
π
1
− a3h
6
4 a
2a
=−


3π  2 
3π
1 3
ab
12
−
π
b = − a 2b
2  2 
8
π
zV
z
−
a
−
a 2 (4h − 3b)
1 3
a (2h − b)
12
From Sample Problem 5.13:
We have
Then
Z ΣV = Σ zV
1
π

Z  a 2 (4h − 3b)  = − a3 (2h − b)
12
 24

or Z = −
a  4h − 2b 

π  4h − 3b 
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671
PROBLEM 5.100
For the machine element shown, locate the y coordinate of the
center of gravity.
SOLUTION
For half-cylindrical hole,
r = 1.25 in.
4(1.25)
3π
= 1.470 in.
yIII = 2 −
For half-cylindrical plate,
r = 2 in.
4(2)
zIV = 7 +
= 7.85 in.
3π
V, in 3
y , in.
z , in.
y V , in 4
z V , in 4
3.5
–7.875
73.50
I
Rectangular plate
(7)(4)(0.75) = 21.0
–0.375
II
Rectangular plate
(4)(2)(1) = 8.0
1.0
2
8.000
16.00
III
–(Half cylinder)
(1.25) 2 (1) = 2.454
1.470
2
–3.607
–4.908
IV
Half cylinder
(2) 2 (0.75) = 4.712
–0.375
–1.767
36.99
V
–(Cylinder)
−π (1.25) 2 (0.75) = −3.682
1.381
–25.77
Σ
27.58
–3.868
95.81
−
π
2
π
2
–0.375
–7.85
7
Y ΣV = Σ yV
Y (27.58 in 3 ) = −3.868 in 4
Y = −0.1403 in. 
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672
PROBLEM 5.101
For the machine element shown, locate the y coordinate of the
center of gravity.
SOLUTION
First assume that the machine element is homogeneous so that its center of gravity will coincide with the
centroid of the corresponding volume.
V , mm3
x , mm
y , mm
x V, mm 4
y V, mm 4
I
(100)(18)(90) = 162, 000
50
9
8,100,000
1,458,000
II
(16)(60)(50) = 48, 000
92
48
4,416,000
2,304,000
III
π (12)2 (10) = 4523.9
105
54
475,010
244,290
IV
−π (13) 2 (18) = −9556.7
28
9
–267,590
–86,010
Σ
204,967.2
12,723,420
3,920,280
We have
Y ΣV = Σ yV
Y (204,967.2 mm3 ) = 3,920, 280 mm 4
or Y = 19.13 mm 
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673
PROBLEM 5.102
For the machine element shown, locate the x coordinate of
the center of gravity.
SOLUTION
First assume that the machine element is homogeneous so that its center of gravity will coincide with the
centroid of the corresponding volume.
V , mm3
x , mm
y , mm
xV , mm 4
yV, mm 4
I
(100)(18)(90) = 162, 000
50
9
8,100,000
1,458,000
II
(16)(60)(50) = 48, 000
92
48
4,416,000
2,304,000
III
π (12)2 (10) = 4523.9
105
54
475,010
244,290
IV
−π (13) 2 (18) = −9556.7
28
9
–267,590
–86,010
Σ
204,967.2
12,723,420
3,920,280
We have
X ΣV = Σ xV
X (204,967.2 mm3 ) = 12, 723, 420 mm 4
X = 62.1 mm 
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674
PROBLEM 5.103
For the machine element shown, locate the z coordinate of the
center of gravity.
SOLUTION
For half-cylindrical hole,
r = 1.25 in.
4(1.25)
3π
= 1.470 in.
yIII = 2 −
For half-cylindrical plate,
r = 2 in.
4(2)
zIV = 7 +
= 7.85 in.
3π
V, in 3
Now
y , in.
z , in.
yV , in 4
z V , in 4
3.5
–7.875
73.50
I
Rectangular plate
(7)(4)(0.75) = 21.0
–0.375
II
Rectangular plate
(4)(2)(1) = 8.0
1.0
2
8.000
16.00
III
–(Half cylinder)
(1.25) 2 (1) = 2.454
1.470
2
–3.607
–4.908
IV
Half cylinder
(2) 2 (0.75) = 4.712
–0.375
–1.767
36.99
V
–(Cylinder)
−π (1.25) 2 (0.75) = −3.682
1.381
–25.77
Σ
27.58
–3.868
95.81
−
π
2
π
2
–0.375
–7.85
7
Z ΣV = zV
Z (27.58 in 3 ) = 95.81 in 4
Z = 3.47 in. 
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675
PROBLEM 5.104
For the machine element shown, locate the x coordinate of the
center of gravity.
SOLUTION
Dimensions in mm
V, mm3
x , mm z , mm
x V , mm 4
z V , mm 4
19
45
649.8 &times; 103
1539 &times; 103
46.5
20
292.17 &times; 103
125.664 &times; 103
−π (12)2 (10) = −4.5239 &times; 103
38
20
Rectangular prism
(30)(10)(24) = 7.2 &times; 103
5
78
36 &times; 103
561.6 &times; 103
Triangular prism
1
(30)(9)(24) = 3.24 &times; 103
2
13
78
42.12 &times; 103
252.72 &times; 103
Σ
46.399 &times; 103
848.18 &times; 103
2388.5 &times; 103
I
Rectangular plate
II
Half cylinder
III
–(Cylinder)
IV
V
(10)(90)(38) = 34.2 &times; 103
π
2
(20)2 (10) = 6.2832 &times; 103
−171.908 &times; 103 −90.478 &times; 103
X ΣV = Σ xV
X=
Σ xV 848.18 &times; 103 mm4
=
ΣV
46.399 &times; 103 mm3
X = 18.28 mm 
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676
PROBLEM 5.105
For the machine element shown, locate the z coordinate of the
center of gravity.
SOLUTION
Dimensions in mm
V, mm3
x V , mm 4
z V , mm 4
19
45
649.8 &times; 103
1539 &times; 103
46.5
20
292.17 &times; 103
125.664 &times; 103
−π (12)2 (10) = −4.5239 &times; 103
38
20
(30)(10)(24) = 7.2 &times; 103
5
78
36 &times; 103
561.6 &times; 103
Triangular prism
1
(30)(9)(24) = 3.24 &times; 103
2
13
78
42.12 &times; 103
252.72 &times; 103
Σ
46.399 &times; 103
848.18 &times; 103
2388.5 &times; 103
I
Rectangular plate
II
Half cylinder
III
–(Cylinder)
IV Rectangular prism
V
x , mm z , mm
(10)(90)(38) = 34.2 &times; 103
π
2
(20)2 (10) = 6.2832 &times; 103
−171.908 &times; 103 −90.478 &times; 103
Z ΣV = Σ zV
Z=
Σ zV 2388.5 &times; 103 mm4
=
ΣV
46.399 &times; 103 mm3
Z = 51.5 mm 
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677
PROBLEM 5.106
Locate the center of gravity of the sheet-metal form shown.
SOLUTION
Y = 80.0 mm 
By symmetry,
4(80)
= 33.953 mm
3π
2(60)
zII = −
= −38.197 mm
zI =
π
A, mm 2
I
π
2
(80) 2 = 10, 053
II
π (60)(160) = 30,159
Σ
40,212
x , mm
z , mm
xA, mm3
zA, mm3
0
33.953
0
341.33 &times; 103
60
−38.197
1809.54 &times; 103 −1151.98 &times; 103
1809.54 &times; 103
−810.65 &times; 103
X Σ A = Σ xA: X (40, 212) = 1809.54 &times; 103
X = 45.0 mm 
Z Σ A = Σ zA: Z (40, 212) = −810.65 &times; 103
Z = −20.2 mm 
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678
PROBLEM 5.107
Locate the center of gravity of the sheet-metal form shown.
SOLUTION
First assume that the sheet metal is homogeneous so that the center of gravity of the form will coincide with
the centroid of the corresponding area.
1
yI = 0.18 + (0.12) = 0.22 m
3
1
zI = (0.2 m)
3
2 &times; 0.18 0.36
=
m
xII = yII =
π
xIV
I
II
IV
Σ
A, m 2
x, m
y, m
z, m
x A, m3
yA, m3
1
(0.2)(0.12) = 0.012
2
0
0.22
0.2
3
0
0.00264
0.0008
0.36
0.36
π
π
0.1
0.00648
0.00648
0.005655
0.26
0
0.1
0.00832
0
0.0032
0.31878
0
0.1
–0.001258
0
–0.000393
0.013542
0.00912
0.009262
π
(0.18)(0.2) = 0.018π
2
(0.16)(0.2) = 0.032
III
−
π
4 &times; 0.05
= 0.34 −
3π
= 0.31878 m
π
2
(0.05) 2 = −0.00125π
0.096622
z A, m3
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679
PROBLEM 5.107 (Continued)
We have
X ΣV = Σ xV : X (0.096622 m 2 ) = 0.013542 m3
or X = 0.1402 m 
Y ΣV = Σ yV : Y (0.096622 m 2 ) = 0.00912 m3
or Y = 0.0944 m 
Z ΣV = Σ zV : Z (0.096622 m 2 ) = 0.009262 m3
or Z = 0.0959 m 
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680
PROBLEM 5.108
A window awning is fabricated from sheet metal of uniform
thickness. Locate the center of gravity of the awning.
SOLUTION
First, assume that the sheet metal is homogeneous so that the center of gravity of the awning coincides with
the centroid of the corresponding area.
(4)(25)
= 14.6103 in.
3π
(4)(25) 100
in.
zII = zVI =
=
3π
3π
(2)(25)
yIV = 4 +
= 19.9155 in.
yII = yVI = 4 +
π
zIV =
(2)(25)
π
AII = AVI =
AIV =
A, in
2
I
(4)(25) = 100
II
490.87
III
(4)(34) = 136
IV
1335.18
V
(4)(25) = 100
VI
Σ
490.87
y , in.
2
π
2
19.9155
in.
(25) 2 = 490.87 in 2
(25)(34) = 1335.18 in 2
zA, in 3
12.5
200
1250
100
3π
14.6103
π
yA, in 3
π
12.5
2
4
50
z , in.
100
3π
25
50
14.6103
2
π
=
7171.8
272
3400
26,591
21,250
200
1250
7171.8
41,607
2652.9
5208.3
5208.3
37,567
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681
PROBLEM 5.108 (Continued)
X = 17.00 in. 
Now, symmetry implies
and
Y Σ A = Σ yA: Y (2652.9 in 2 ) = 41,607 in 3
or Y = 15.68 in. 
Z Σ A = Σ zA: Z (2652.9 in 2 ) = 37,567 in 3
or Z = 14.16 in. 
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682
PROBLEM 5.109
A thin sheet of plastic of uniform thickness is bent to
form a desk organizer. Locate the center of gravity of the
organizer.
SOLUTION
First assume that the plastic is homogeneous so that the center of gravity of the organizer will coincide with
the centroid of the corresponding area. Now note that symmetry implies
Z = 30.0 mm 
x2 = 6 −
2&times; 6
x4 = 36 +
x8 = 58 −
π
= 2.1803 mm
2&times;6
π
2&times; 6
π
x10 = 133 +
= 39.820 mm
= 54.180 mm
2&times;6
π
= 136.820 mm
y2 = y4 = y8 = y10 = 6 −
y6 = 75 +
2&times;5
π
2&times;6
π
= 2.1803 mm
= 78.183 mm
A2 = A4 = A8 = A10 =
π
&times; 6 &times; 60 = 565.49 mm 2
2
A6 = π &times; 5 &times; 60 = 942.48 mm 2
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683
PROBLEM 5.109 (Continued)
A, mm 2
We have
x , mm
1
(74)(60) = 4440
2
565.49
3
(30)(60) = 1800
4
565.49
5
(69)(60) = 4140
42
6
942.48
7
(69)(60) = 4140
8
565.49
9
y , mm
0
xA, mm3
yA, mm3
0
190,920
1233
1233
43
2.1803
2.1803
21
0
37,800
0
39.820
2.1803
22,518
1233
40.5
173,880
167,670
47
78.183
44,297
73,686
52
40.5
215,280
167,670
54.180
2.1803
30,638
1233
(75)(60) = 4500
95.5
0
429,750
0
10
565.49
136.820
2.1803
77,370
1233
Σ
22,224.44
1,032,766
604,878
X Σ A = Σ xA: X (22, 224.44 mm 2 ) = 1,032, 766 mm3
or X = 46.5 mm 
Y Σ A = Σ yA: Y (22, 224.44 mm 2 ) = 604,878 mm3
or Y = 27.2 mm 
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684
PROBLEM 5.110
A wastebasket, designed to fit in the corner of a room, is 16 in. high
and has a base in the shape of a quarter circle of radius 10 in. Locate
the center of gravity of the wastebasket, knowing that it is made of
sheet metal of uniform thickness.
SOLUTION
By symmetry,
X =Z
For III (Cylindrical surface),
x=
A=
For IV (Quarter-circle bottom),
x=
A=
2r
π
π
2
=
2(10)
rh =
π
π
2
= 6.3662 in.
(10)(16) = 251.33 in 2
4r 4(10)
=
= 4.2441 in.
3π
3π
π
A, in 2
4
r2 =
π
4
(10) 2 = 78.540 in 2
x , in.
x , in.
xA, in 3
yA, in 3
I
(10)(16) = 160
5
8
800
1280
II
(10)(16) = 160
0
8
0
1280
III
251.33
6.3662
8
4.2441
0
IV
Σ
X Σ A = Σ xA:
78.540
649.87
333.33
2733.3
2010.6
0
4570.6
X (649.87 in 2 ) = 2733.3 in 3
X = 4.2059 in.
Y Σ A = Σ yA:
1600.0
X = Z = 4.21 in. 
Y (649.87 in 2 ) = 4570.6 in 3
Y = 7.0331 in.
Y = 7.03 in. 
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685
PROBLEM 5.111
A mounting bracket for electronic components is formed from
sheet metal of uniform thickness. Locate the center of gravity
of the bracket.
SOLUTION
First, assume that the sheet metal is homogeneous so that the center of gravity of the bracket coincides with
the centroid of the corresponding area. Then (see diagram)
4(0.625)
3π
= 1.98474 in.
zV = 2.25 −
AV = −
π
(0.625) 2
2
= −0.61359 in 2
A, in 2
x , in.
I
(2.5)(6) = 15
1.25
II
(1.25)(6) = 7.5
2.5
III
(0.75)(6) = 4.5
2.875
IV
5
−   (3) = − 3.75
4
1.0
0
3.75
V
− 0.61359
1.0
0
1.98474 0.61359
Σ
22.6364
We have
y , in.
z , in.
0
xA, in 3
yA, in 3
zA, in 3
0
45
3
18.75
–0.625
3
18.75
–4.6875
22.5
–1.25
3
12.9375
–5.625
13.5
3.75
46.0739
0
–14.0625
0
–1.21782
10.3125
65.7197
X ΣA = Σ xA
X (22.6364 in 2 ) = 46.0739 in 3
or
X = 2.04 in. 
Y ΣA = Σ yA
Y (22.6364 in 2 ) = −10.3125 in 3
or Y = − 0.456 in. 
Z ΣA = Σ zA
Z (22.6364 in 2 ) = 65.7197 in 3
or
Z = 2.90 in. 
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686
PROBLEM 5.112
An 8-in.-diameter cylindrical duct and a 4 &times; 8-in. rectangular
duct are to be joined as indicated. Knowing that the ducts were
fabricated from the same sheet metal, which is of uniform
thickness, locate the center of gravity of the assembly.
SOLUTION
Assume that the body is homogeneous so that its center of gravity coincides with the centroid of the area.
By symmetry, z = 0.
1
2
A, in 2
x , in.
y , in.
xA, in 3
yA, in 3
π (8)(12) = 96π
0
6
0
576π
10
−128
−160π
4(4)
16
=−
3π
3π
12
− 42.667
96π
6
12
576
1152
6
8
576
768
4(4) 16
=
3π
3π
8
− 42.667
− 64π
6
10
288
480
6
10
288
480
1514.6
4287.4
−
π
2
π
3
Then
(4) 2 = 8π
4
2
(8)(12) = 96
5
(8)(12) = 96
6
−
2(4)
(8)(4) = −16π
π
(4) 2 = − 8π
7
2
(4)(12) = 48
8
(4)(12) = 48
Σ
539.33
π
−
=
8
π
X =
Σ xA 1514.67
=
in.
539.33
ΣA
or X = 2.81 in. 
Y =
Σ yA 4287.4
=
in.
539.33
ΣA
or Y = 7.95 in. 
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687
PROBLEM 5.113
An elbow for the duct of a ventilating system is made of sheet
metal of uniform thickness. Locate the center of gravity of the
elbow.
SOLUTION
First, assume that the sheet metal is homogeneous so that the center of gravity of the duct coincides with the
centroid of the corresponding area. Also, note that the shape of the duct implies
Y = 38.0 mm 
Note that
xI = zI = 400 −
xII = 400 −
zII = 300 −
2
π
2
π
2
π
(400) = 145.352 mm
(200) = 272.68 mm
(200) = 172.676 mm
xIV = zIV = 400 −
4
(400) = 230.23 mm
3π
4
(200) = 315.12 mm
3π
4
zV = 300 −
(200) = 215.12 mm
3π
xV = 400 −
Also note that the corresponding top and bottom areas will contribute equally when determining x and z .
Thus,
A, mm 2
x , mm
z , mm
xA, mm3
zA, mm3
(400)(76) = 47,752
145.352
145.352
6,940,850
6,940,850
(200)(76) = 23,876
272.68
172.676
6,510,510
4,122,810
III
100(76) = 7600
200
350
1,520,000
2,660,000
IV
π 
2   (400)2 = 251,327
4
230.23
230.23
57,863,020
57,863,020
V
π 
−2   (200)2 = −62,832
4
315.12
215.12
–19,799,620
–13,516,420
VI
−2(100)(200) = −40,000
300
350
–12,000,000
–14,000,000
Σ
227,723
41,034,760
44,070,260
I
II
π
2
π
2
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688
PROBLEM 5.113 (Continued)
We have
X Σ A = Σ xA: X (227, 723 mm 2 ) = 41,034, 760 mm3
or X = 180.2 mm 
Z Σ A = Σ zA: Z (227, 723 mm 2 ) = 44,070, 260 mm3
or Z = 193.5 mm 
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689
PROBLEM 5.114
Locate the center of gravity of the figure shown, knowing that it is
made of thin brass rods of uniform diameter.
SOLUTION
X =0 
By symmetry,
L, in.
y , in.
z , in.
yL, in 2
zL, in 2
AB
302 + 162 = 34
15
0
510
0
302 + 162 = 34
15
8
510
272
AE
302 + 162 = 34
15
0
510
0
BDE
π (16) = 50.265
0
0
512
Σ
152.265
1530
784
2(16)
π
= 10.186
Y ΣL = Σ y L : Y (152.265 in.) = 1530 in 2
Y = 10.048 in.

Y = 10.05 in. 
Z ΣL = Σ z L : Z (152.265 in.) = 784 in 2 
Z = 5.149 in. 

Z = 5.15 in. 
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690
PROBLEM 5.115
Locate the center of gravity of the figure shown, knowing that it is
made of thin brass rods of uniform diameter.
SOLUTION
Uniform rod:
AB 2 = (1 m) 2 + (0.6 m) 2 + (1.5 m) 2
AB = 1.9 m
L, m
x, m
y, m
z, m
xL, m 2
yL, m 2
ΣL , m
AB
1.9
0.5
0.75
0.3
0.95
1.425
0.57
BD
0.6
1.0
0
0.3
0.60
0
0.18
DO
1.0
0.5
0
0
0.50
0
0
OA
1.5
0
0.75
0
0
1.125
0
Σ
5.0
2.05
2.550
0.75
X Σ L = Σ x L : X (5.0 m) = 2.05 m 2
X = 0.410 m 
Y Σ L = Σ y L : Y (5.0 m) = 2.55 m 2
Y = 0.510 m 
Z Σ L = Σ z L : Z (5.0 m) = 0.75 m 2
Z = 0.1500 m 
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691
PROBLEM 5.116
A thin steel wire of uniform cross section is bent into the shape
shown. Locate its center of gravity.
SOLUTION
First assume that the wire is homogeneous so that its center of gravity
will coincide with the centroid of the corresponding line.
x2 = z2 =
1
2
We have
π
2
2 &times; 2.4
π
=
4.8
π
m
L, m
x, m
y, m
z, m
xL, m 2
yL, m 2
zL, m 2
2.6
1.2
0.5
0
3.12
1.3
0
π
5.76
0
5.76
&times; 2.4 = 1.2π
4.8
π
0
4.8
3
2.4
0
0
1.2
0
0
2.88
4
1.0
0
0.5
0
0
0.5
0
Σ
9.7699
8.88
1.8
8.64
X Σ L = Σ x L : X (9.7699 m) = 8.88 m 2
or X = 0.909 m 
Y Σ L = Σ y L : Y (9.7699 m) = 1.8 m 2
or Y = 0.1842 m 
Z Σ L = Σ z L : Z (9.7699 m) = 8.64 m 2
or Z = 0.884 m 
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692
PROBLEM 5.117
The frame of a greenhouse is constructed from uniform aluminum
channels. Locate the center of gravity of the portion of the frame
shown.
SOLUTION
First assume that the channels are homogeneous so that the center of gravity of the frame
will coincide with the centroid of the corresponding line.
2&times;3
x8 = x9 =
π
y8 = y9 = 5 +
L, ft
6
π
2&times;3
π
y , ft
ft
= 6.9099 ft
z , ft
xL, ft 2
yL, ft 2
zL, ft 2
1
2
3
0
1
6
0
2
2
3
1.5
0
2
4.5
0
6
3
5
3
2.5
0
15
12.5
0
4
5
3
2.5
2
15
12.5
10
5
8
0
4
2
0
32
16
6
2
3
5
1
6
10
2
7
3
1.5
5
2
4.5
15
6
6.9099
0
9
32.562
0
π
6.9099
2
9
32.562
9.4248
0
8
1
0
16
2
8
9
10
Σ
We have
x , ft
=
π
2
π
2
&times; 3 = 4.7124
&times; 3 = 4.7124
2
6
π
6
39.4248
69
163.124
53.4248
X Σ L = Σ x L : X (39.4248 ft) = 69 ft 2
or X = 1.750 ft 
Y Σ L = Σ y L : Y (39.4248 ft) = 163.124 ft 2
or
Z Σ L = Σ z L : Z (39.4248 ft) = 53.4248 ft 2
or Z = 1.355 ft 
Y = 4.14 ft 
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693
PROBLEM 5.118
Three brass plates are brazed to a steel pipe to form the flagpole
base shown. Knowing that the pipe has a wall thickness of 8 mm
and that each plate is 6 mm thick, determine the location of the
center of gravity of the base. (Densities: brass = 8470 kg/m3;
steel = 7860 kg/m3.)
SOLUTION
Since brass plates are equally spaced, we note that
the center of gravity lies on the y-axis.
x =z =0 
Thus,
V=
π
[(0.064 m) 2 − (0.048 m) 2 ](0.192 m)
4
= 270.22 &times; 10−6 m3
Steel pipe:
m = ρ V = (7860 kg/m3 )(270.22 &times; 10−6 m3 )
= 2.1239 kg
Each brass plate:
1
(0.096 m)(0.192 m)(0.006 m) = 55.296 &times; 10−6 m3
2
m = ρ V = (8470 kg/m3 )(55.296 &times; 10−6 m3 ) = 0.46836 kg
V=
Flagpole base:
Σm = 2.1239 kg + 3(0.46836 kg) = 3.5290 kg
Σ y m = (0.096 m)(2.1239 kg) + 3[(0.064 m)(0.46836 kg)] = 0.29382 kg ⋅ m
Y Σm = Σ y m :
Y (3.5290 kg) = 0.29382 kg ⋅ m
Y = 0.083259 m
Y = 83.3 mm above the base 
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694
PROBLEM 5.119
A brass collar, of length 2.5 in., is mounted on an aluminum rod
of length 4 in. Locate the center of gravity of the composite body.
(Specific weights: brass = 0.306 lb/in3, aluminum = 0.101 lb/in3)
SOLUTION
Aluminum rod:
W = γV
π

= (0.101 lb/in 3 )  (1.6 in.)2 (4 in.) 
4

= 0.81229 lb
Brass collar:
W = γV
π
= (0.306 lb/in.3 ) [(3 in.) 2 − (1.6 in.) 2 ](2.5 in.)
4
= 3.8693 lb
yW (lb ⋅ in.)
Component
W(lb)
Rod
0.81229
2
1.62458
Collar
3.8693
1.25
4.8366
Σ
4.6816
y (in.)
6.4612
Y ΣW = Σ y W : Y (4.6816 lb) = 6.4612 lb ⋅ in.
Y = 1.38013 in.
Y = 1.380 in. 
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695
PROBLEM 5.120
A bronze bushing is mounted inside a steel sleeve. Knowing that the
specific weight of bronze is 0.318 lb/in3 and of steel is 0.284 lb/in3,
determine the location of the center of gravity of the assembly.
SOLUTION
X =Z =0 
First, note that symmetry implies
Now
W = ( ρ g )V
 π 

yI = 0.20 in. WI = (0.284 lb/in 3 )   [(1.82 − 0.752 ) in 2 ](0.4 in.)  = 0.23889 lb
 4 

 π 

yII = 0.90 in. WII = (0.284 lb/in 3 )   [(1.1252 − 0.752 ) in 2 ](1 in.)  = 0.156834 lb
4
 

 π 

yIII = 0.70 in. WIII = (0.318 lb/in 3 )   [(0.752 − 0.52 ) in 2 ](1.4 in.)  = 0.109269 lb
 4 

We have
Y ΣW = Σ yW
(0.20 in.)(0.23889 lb) + (0.90 in.)(0.156834 lb) + (0.70 in.)(0.109269 lb)
Y =
0.23889 lb + 0.156834 lb + 0.109269 lb
Y = 0.526 in. 
or
(above base)
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696
PROBLEM 5.121
A scratch awl has a plastic handle and a steel blade and shank. Knowing that the density of plastic is 1030 kg/m3
and of steel is 7860 kg/m3, locate the center of gravity of the awl.
SOLUTION
Y = Z = 0 
First, note that symmetry implies
5
xI = (12.5 mm) = 7.8125 mm
8
 2π 
3
WI = (1030 kg/m3 ) 
 (0.0125 m)
3


= 4.2133 &times; 10−3 kg
xII = 52.5 mm
π 
WII = (1030 kg/m3 )   (0.025 m) 2 (0.08 m)
4
−3
= 40.448 &times; 10 kg
xIII = 92.5 mm − 25 mm = 67.5 mm
π 
WIII = −(1030 kg/m3 )   (0.0035 m) 2 (0.05 m)
4
−3
= −0.49549 &times; 10 kg
xIV = 182.5 mm − 70 mm = 112.5 mm
π 
WIV = (7860 kg/m3 )   (0.0035 m) 2 (0.14 m) 2 = 10.5871 &times; 10−3 kg
4
1
xV = 182.5 mm + (10 mm) = 185 mm
4
π 
WV = (7860 kg/m3 )   (0.00175 m) 2 (0.01 m) = 0.25207 &times; 10−3 kg
3
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697
PROBLEM 5.121 (Continued)
W, kg
We have
x, mm
xW, kg ⋅ mm
I
4.123 &times; 10−3
II
40.948 &times; 10−3
52.5
2123.5 &times; 10−3
III
−0.49549 &times; 10−3
67.5
−33.447 &times; 10−3
IV
10.5871 &times; 10−3
112.5
1191.05 &times; 10−3
V
0.25207 &times; 10−3
185
46.633 &times; 10−3
Σ
55.005 &times; 10−3
7.8125
32.916 &times; 10−3
3360.7 &times; 10−3
X ΣW = Σ xW : X (55.005 &times; 10−3 kg) = 3360.7 &times; 10−3 kg ⋅ mm
X = 61.1 mm 
or
(from the end of the handle)
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698
PROBLEM 5.122
Determine by direct integration the values of x for the two volumes obtained by passing a vertical cutting
plane through the given shape of Figure 5.21. The cutting plane is parallel to the base of the given shape and
divides the shape into two volumes of equal height.
A hemisphere
SOLUTION
Choose as the element of volume a disk of radius r and thickness dx. Then
dV = π r 2 dx, xEL = x
The equation of the generating curve is x 2 + y 2 = a 2 so that r 2 = a 2 − x 2 and then
dV = π ( a 2 − x 2 ) dx
V1 =
Component 1:
=
and

1
xEL dV =

a/2
0
a/2

x3 
π (a − x )dx = π  a 2 x − 
3 0

2
2
11 3
πa
24

a/2
0
x π (a 2 − x 2 ) dx 
a/2
 x2 x4 
= π a2
− 
2
4 0

7
= π a4
64
Now
x1V1 =
x
1
EL dV :
 11
 7
x1  π a 3  = π a 4
 24
 64
or
x1 =
21
a 
88
Component 2:
a

x3 
π (a − x )dx = π  a 2 x − 
V2 =
a /2
3  a/2


a
2
2


a 3 
a3   2 a  ( 2 )  
 2
= π   a (a ) −  − a   −

3   2
3 




5
= π a3
24
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699
PROBLEM 5.122 (Continued)
and

2
xEL dV =

a
 x 2 x4 
− 
x π ( a 2 − x 2 )dx  = π  a 2
a/2
2
4  a/2

a
2
4 


a
a
2
(
(a ) 4   2 ( 2 )
  2 (a)
2 ) 
= π a
−
−

− a
2
4  
2
4 





9
= π a4
64
Now
x2V2 =

2
 5
 9
xEL dV : x2  π a3  = π a 4
 24
 64
or
x2 =
27
a 
40
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700
PROBLEM 5.123
Determine by direct integration the values of x for the two volumes obtained by passing a vertical cutting
plane through the given shape of Figure 5.21. The cutting plane is parallel to the base of the given shape and
divides the shape into two volumes of equal height.
A semiellipsoid of revolution
SOLUTION
Choose as the element of volume a disk of radius r and thickness dx. Then
dV = π r 2 dx, xEL = x
x2 y2
+
= 1 so that
h2 a 2
The equation of the generating curve is
r2 =
a2 2
(h − x 2 )dx
2
h
dV = π
and then
V1 =
Component 1:
=
and
a2 2
(h − x 2 )
2
h

1
xEL dV =

h/2
0
h/2
a2
a2 
x3 
π 2 (h 2 − x 2 )dx = π 2  h 2 x − 
3 0
h
h 
11 2
πa h
24

h/2
0
 a2

x π 2 (h 2 − x 2 ) dx 
 h

a2
=π 2
h
=
Now
Component 2:
x1V1 =
V2 =
h/2
 2 x2 x4 
− 
h
4 0
 2
7
π a 2 h2
64
x
1

EL dV :
h
h/2
 11
 7
x1  π a 2 h  = π a 2 h 2
24

 64
or
x1 =
21
h 
88
h
π
a2 2
a2 
x3 
(h − x 2 )dx = π 2  h 2 x − 
2
3  h/2
h
h 


h 3 
a 2  2
( h)3   2  h  ( 2 )  
= π 2 h ( h ) −

− h  −
3   2
3 
h 



5
= π a2h
24
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701
PROBLEM 5.123 (Continued)
and

2
xEL dV =

 a2

x π 2 (h 2 − x 2 ) dx 
h/2
 h

h
a2
=π 2
h
h
 2 x2 x4 
− 
h
4  h/2
 2


h 2
h 4 
(
a 2   2 ( h) 2 ( h) 4   2 ( 2 )
2 ) 
= π 2 h
−
−

 − h
2
4 
2
4 
h 



9
= π a 2 h2
64
Now
x2V2 =

 5
 9
xEL dV : x2  π a 2 h  = π a 2 h 2
2
 24
 64
or
x2 =
27
h 
40
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702
PROBLEM 5.124
Determine by direct integration the values of x for the two volumes obtained by passing a vertical cutting
plane through the given shape of Figure 5.21. The cutting plane is parallel to the base of the given shape and
divides the shape into two volumes of equal height.
A paraboloid of revolution
SOLUTION
Choose as the element of volume a disk of radius r and thickness dx. Then
dV = π r 2 dx, xEL = x
The equation of the generating curve is x = h −
h 2
a2
y so that r 2 =
(h − x).
2
h
a
a2
(h − x)dx
h
dV = π
and then
V1 =
Component 1:

h/2
0
π
a2
=π
h
a2
( h − x) dx
h
h/2

x2 
hx
−


2 0

3
= π a2 h
8
and

1
xEL dV =

h/2
0
 a2

x π
(h − x)dx 
 h

h/2
a 2  x 2 x3 
1
2 2
=π
h −  = π a h
h  2
3 0
12
Now
x1V1 =
x
1
EL dV :
3
 1
x1  π a 2 h  = π a 2 h 2
8
 12
or
x1 =
2
h 
9
Component 2:
h
a2
a2 
x2 
V2 =
hx
π (h − x)dx = π
−


h/2
h
h 
2  h/2

h


h 2 
( h) 2    h  ( 2 )  
a2 
=π
  h( h ) −

 − h  −
h 
2   2
2 



1 2
= πa h
8
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703
PROBLEM 5.124 (Continued)
and

2
xEL dV =

a2
=π
h
=
Now
x2V2 =
h
 a2

a 2  x 2 x3 
x π
( h − x)dx  = π
h − 
h/2
h  2
3  h/2
 h

h

 h 2 ( h )3  
2
( h )3   ( 2 )
  ( h)

−
− 2 
h
− h


2
3
2
3
 

 

1
π a 2 h2
12

2
1
 1
xEL dV : x2  π a 2 h  = π a 2 h 2
8

 12
or
x2 =
2
h 
3
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704
PROBLEM 5.125
Locate the centroid of the volume obtained by rotating the shaded area
SOLUTION
y =0 
First note that symmetry implies
z =0 
Choose as the element of volume a disk of radius r and thickness dx. Then
dV = π r 2 dx, xEL = x
r = kx1/3
Now
dV = π k 2 x 2/3 dx
so that
at x = h, y = a,
a = kh1/3
or
k=
a
h1/3
a 2 2/3
x dx
h 2/3
dV = π
Then
V=
and

h
0
π
a 2 2/3
x dx
h 2/3
a2
h 2/3
=π
h
 3 5/3 
5 x 

0
3
= π a2 h
5
Also
x
EL dV
=

h
0
h
 a 2 2/3 
a 2  3 8/3 
x  π 2/3 x dx  = π 2/3  x 
h 8
0
 h

3
= π a 2 h2
8
Now

xV = xdV :
3
 3
x  π a 2 h  = π a 2 h2
5
 8
5
or x = h 
8
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705
PROBLEM 5.126
Locate the centroid of the volume obtained by rotating the shaded area about
the x-axis.
SOLUTION
y =0 
First, note that symmetry implies
z =0 
Choose as the element of volume a disk of radius r and thickness dx.
dV = π r 2 dx, xEL = x
Then
Now r = 1 −
2
 1
dV = π 1 −  dx
x

 2 1 
= π 1 − + 2  dx
x x 

1
so that
x
V=
Then
3

1


π 1 −
3
2 1 
1

+  dx = π  x − 2 ln x − 
x x2 
x 1


1 
1 
= π  3 − 2ln 3 −  − 1 − 2 ln1 −  
3 
1 

= (0.46944π ) m3
and
x
EL dV
=

3
1
3
 x2

  2 1  
x π 1 − + 2  dx  = π  − 2 x + ln x 
x x  
 
2
1
  32
 13
 
= π   − 2(3) + ln 3 −  − 2(1) + ln1 
  2
 2
 
= (1.09861π ) m
Now

xV = xEL dV : x (0.46944π m3 ) = 1.09861π m 4
or x = 2.34 m 
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706
PROBLEM 5.127
Locate the centroid of the volume obtained by rotating the shaded
area about the line x = h.
SOLUTION
x =h 
First, note that symmetry implies
z =0 
Choose as the element of volume a disk of radius r and thickness dx. Then
dV = π r 2 dy, yEL = y
Now x 2 =
Then
and
Let
Then
h2 2
h 2
(a − y 2 ) so that r = h −
a − y2 .
2
a
a
dV = π
V=

(
h2
a − a2 − y 2
a2
a
0
π
) dy
2
(
h2
a − a2 − y2
a2
) dy
2
y = a sin θ  dy = a cos θ dθ
h2
a2
h2
=π 2
a

= π ah 2

V =π

π /2
0
π /2
0
π /2
0
(
a − a 2 − a 2 sin 2 θ
) a cosθ dθ
2
 a 2 − 2a (a cos θ ) + (a 2 − a 2 sin 2 θ )  a cos θ dθ


(2cos θ − 2 cos 2 θ − sin 2 θ cos θ ) dθ
π /2

 θ sin 2θ  1 3 
= π ah  2sin θ − 2  +
− sin θ 
4  3
2

0
2

 π  1
= π ah 2  2 − 2  2  − 

 2  3 
= 0.095870π ah 2
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707
PROBLEM 5.127 (Continued)
and

yEL dV =
0
(
 h2
y π 2 a − a 2 − y 2
 a
( 2a y − 2ay
) dy 
2
)
=π
h2
a2

=π
h2
a2
1 4
 2 2 2
2
2 3/2
 a y + 3 a (a − y ) − 4 y 

0
=π
h2
a2
 2 2 1 4   2
2 3/2  
  a (a ) − a  −  a( a )  
4  3


=
Now

a
a
0
2
a 2 − y 2 − y 3 dy
a
1
π a 2 h2
12

y (0.095870π ah2 ) =
yV = yEL dV :
1
π a2 h2
12
or y = 0.869a 
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708
PROBLEM 5.128*
Locate the centroid of the volume generated by revolving the portion
of the sine curve shown about the x-axis.
SOLUTION
y =0 
First, note that symmetry implies
z =0 
Choose as the element of volume a disk of radius r and thickness dx.
dV = π r 2 dx, xEL = x
Then
r = b sin
Now
πx
2a
dV = π b 2 sin 2
so that
V=
Then

2a
a
πx
2a
dx
π b 2 sin 2
πx
2a
dx
2a
 x sin π x 
= π b  − π2 a 
2 a 
 2
a
2
= π b 2 ( 22a ) − ( a2 ) 
1
= π ab 2
2
and
x
EL dV
=

2a
a
πx 

x  π b 2 sin 2
dx
2a 

Use integration by parts with
u=x
dV = sin 2
du = dx
V=
x
−
2
πx
2a
sin πax
2π
a
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709
PROBLEM 5.128* (Continued)
Then

2a
 

2a  x
sin π x  
x sin πax  

xEL dV = π b   x  − 2π   −
− 2π a  dx 




a 2
   2
a
a
 a

 
2a

2
π x  
 2a 
 a   1 2 a
2 
= π b   2a   − a    −  x + 2 cos  
a a 
2π
 2   4
   2 


2
 3   1
a2
1
a 2  
= π b 2  a 2  −  (2a) 2 + 2 − (a) 2 + 2  
4
2π
2π  
 2   4
3 1 
= π a 2b 2  − 2 
4 π 
= 0.64868π a 2b 2
Now
1

xV = xEL dV : x  π ab 2  = 0.64868π a 2b 2
2



or x = 1.297a 
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710
PROBLEM 5.129*
Locate the centroid of the volume generated by revolving the
portion of the sine curve shown about the y-axis. (Hint: Use a
thin cylindrical shell of radius r and thickness dr as the element
of volume.)
SOLUTION
x =0 
First note that symmetry implies
z =0
Choose as the element of volume a cylindrical shell of radius r and thickness dr.
Then
Now
so that
Then
dV = (2π r )( y )(dr ),
y = b sin

2a
a
1
y
2
πr
2a
dV = 2π br sin
V=
yEL =
πr
2a
dr
2π br sin
πr
2a
dr
Use integration by parts with
u = rd
dv = sin
dr
2a
2a
πr
v = − cos
2a
π
du = dr
Then
πr
2a
   2a
π r 
V = 2π b  ( r )  − cos   −
π
2a   a
  

2a
π r  
 π cos 2a  dr 

 
2a 
a
2a
 2a
 4a 2
π r  

= 2π b  − [ (2a)(−1) ] +  2 sin  
2a  a 
 π
π

 4a 2 4 a 2
− 2
V = 2π b 
π
 π
1

= 8a 2 b  1 − 
π





= 5.4535a 2b
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711
PROBLEM 5.129* (Continued)
Also
y
EL dV
π r 
πr 
1
2π br sin
b sin
dr 


2
a 
2a 
2a 
2a
πr
r sin 2
dr
= π b2
a
2a
=

2a 

Use integration by parts with
u=r
dv = sin 2
du = dr
Then

v=
r
−
2
πr
dr
2a
sin πar
2π
a
2a
 

2a  r
sin π r  
r sin πar  

yEL dV = π b  ( r )  − 2π   −
− 2π a  dr 




a 2
   2
a
a
 a

 
2a

2
π r  
a2
 2a 
 a   r
2 
= π b  (2a)   − ( a)    −  + 2 cos  
a a 
 2 
 2    4 2π
 


2
 3
 (2a) 2
(a ) 2
a2
a 2  
= π b2  a 2 − 
+ 2−
+ 2 
4
2π
2π  
 2
 4
3 1 
= π a 2b 2  − 2 
4 π 
= 2.0379a 2 b 2
Now

yV = yEL dV : y (5.4535a 2b) = 2.0379a 2b 2
or y = 0.374b 
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712
PROBLEM 5.130*
Show that for a regular pyramid of height h and n sides (n = 3, 4, ) the centroid of the volume of the pyramid
is located at a distance h/4 above the base.
SOLUTION
Choose as the element of a horizontal slice of thickness dy. For any number N of sides, the area of the base of
the pyramid is given by
Abase = kb 2
where k = k ( N ); see note below. Using similar triangles, we have
s h− y
=
b
h
s=
or
b
(h − y )
h
dV = Aslice dy = ks 2 dy = k
Then
V=
and
h

k
0
b2
(h − y ) 2 dy
h2
b2
b2
2
(
h
y
)
dy
k
−
=
h2
h2
h
 1
3
 − 3 (h − y ) 

0
1
= kb 2 h
3
yEL = y
Also,
so that

yEL dV =

h
0
=k
Now
Note:
 b2

b2
y  k 2 ( h − y ) 2 dy  = k 2
h
 h

b2
h2

h
0
( h 2 y − 2hy 2 + y 3 ) dy
h
1 2 2
1 2 2 2 3 1 4 
 2 h y − 3 hy + 4 y  = 12 kb h

0
1
 1
yV = yEL dV : y  kb 2 h  = kb 2 h 2
3
 12

1
Abase = N  &times; b &times;
2
N
=
b2
4 tan πN
b
2
tan πN
or y =
1
h Q.E.D. 
4

 

= k ( N )b 2
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713
PROBLEM 5.131
Determine by direct integration the location of the centroid of one-half of a
thin, uniform hemispherical shell of radius R.
SOLUTION
x = 0
First note that symmetry implies
The element of area dA of the shell shown is obtained by cutting the shell with two planes parallel to the xy
plane. Now
dA = (π r )( Rdθ )
2r
yEL = −
π
r = R sin θ
where
dA = π R 2 sin θ dθ
2R
yEL = −
sin θ
so that
π
A=
Then

π /2
0
π R 2 sin θ dθ = π R 2 [− cos θ ]π0 /2
= π R2
and
y
EL dA
=

π /2 
0
2R

sin θ  (π R 2 sin θ dθ )
−
π


π /2
 θ sin 2θ 
= −2 R3  −
4  0
2
=−
Now
Symmetry implies
π
2
R3

yA = yEL dA: y (π R 2 ) = −
π
2
R3
1
or y = − R 
2
1
z = − R 
2
z=y
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714
PROBLEM 5.132
The sides and the base of a punch bowl are of
uniform thickness t. If t &lt; &lt; R and R = 250 mm,
determine the location of the center of gravity of
(a) the bowl, (b) the punch.
SOLUTION
(a)
Bowl:
x =0 
First note that symmetry implies
z =0 
for the coordinate axes shown below. Now assume that the bowl may be treated as a shell; the center of
gravity of the bowl will coincide with the centroid of the shell. For the walls of the bowl, an element of
area is obtained by rotating the arc ds about the y-axis. Then
dAwall = (2π R sin θ )( R dθ )
( yEL ) wall = − R cos θ
and
Awall =
Then
π /2
π
/6
2π R 2 sin θ dθ
π /2
= 2π R 2 [ − cos θ ]π /6
= π 3R2

π
=
π
ywall Awall = ( yEL ) wall dA
and
/2
/6
(− R cos θ )(2π R 2 sin θ dθ )
= π R3 [cos 2 θ ]ππ /2
/6
3
= − π R3
4
π
By observation,
Abase =
Now
y ΣA = Σ yA
or
or
4
R2 ,
ybase = −
3
R
2

π
3
π
3 


y  π 3R 2 + R 2  = − π R3 + R 2  −
R


4
4
4


 2 
y = − 0.48763R R = 250 mm
y = −121.9 mm 
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715
PROBLEM 5.132 (Continued)
(b)
Punch:
x =0 
First note that symmetry implies
z =0 
and that because the punch is homogeneous, its center of gravity will coincide with the centroid of
the corresponding volume. Choose as the element of volume a disk of radius x and thickness dy. Then
dV = π x 2 dy,
Now
yEL = y
x2 + y 2 = R2
dV = π ( R 2 − y 2 )dy
so that
V=
Then

0
− 3/2 R
π ( R 2 − y 2 ) dy
0
1 

= π  R 2 y − y3 
3 −

3/2 R
3
 
3  1
3   3
2

R − −
R  = π 3R3
= −π R  −



  2  3  2   8


and
y
EL dV
=

0
− 3/2 R
(
)
( y ) π R 2 − y 2 dy 


0
1 
1
= π  R2 y 2 − y4 
2
4 −

3/2 R
4


 1
 
1
3
3
15
2
R − −
R   = − π R4
= −π  R  −




2  2 
4 2  
64


2
Now
15
3

yV = yEL dV : y  π 3 R3  = − π R 4
64
8


y =−
or
5
8 3
R R = 250 mm
y = − 90.2 mm 
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716
PROBLEM 5.133
Locate the centroid of the section shown, which was cut from a thin circular
pipe by two oblique planes.
SOLUTION
x =0 
First note that symmetry implies
Assume that the pipe has a uniform wall thickness t and choose as the element of volume a vertical strip of
width adθ and height ( y2 − y1 ). Then
dV = ( y2 − y1 )ta dθ ,
Now
y1 =
=
and
h
3
2a
z+
h
6
yEL =
1
( y1 + y2 ) z EL = z
2
y2 = −
h
( z + a)
6a
=
2h
3
2a
z+
2
h
3
h
( − z + 2a )
3a
z = a cos θ
h
h
(−a cos θ + 2a) −
(a cos θ + a )
3a
6a
h
= (1 − cos θ )
2
Then
( y2 − y1 ) =
and
( y1 + y2 ) =
h
h
(a cos θ + a) + (− a cos θ + 2a)
6a
3a
h
= (5 − cos θ )
6
aht
h
dV =
(1 − cos θ )dθ yEL = (5 − cos θ ), z EL = a cos θ
2
12
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717
PROBLEM 5.133 (Continued)
and
V =2

π
=2

π
aht
(1 − cos θ )dθ = aht[θ − sin θ ]π0
0 2
= π aht
Then
y
EL dV
h
 aht

(5 − cos θ ) 
(1 − cos θ ) dθ 
12
2


0
=
ah 2 t
12
=
ah 2 t 
θ sin 2θ 
5θ − 6sin θ + +
12 
2
4  0
=
11
π ah2 t
24

π
0
(5 − 6 cos θ + cos 2 θ ) dθ
π
z
EL dV
=2

π
0
 aht

a cos θ 
(1 − cos θ ) dθ 
 2

π
θ sin 2θ 

= a ht sin θ − −
2
4  0

1
= − π a 2 ht
2
2

11
π ah 2t
24
Now
yV = yEL dV : y (π aht ) =
and
1
zV = zEL dV : z (π aht ) = − π a 2 ht
2

or y =
11
h 
24
1
or z = − a 
2
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718
PROBLEM 5.134*
Locate the centroid of the section shown, which was cut from an elliptical
cylinder by an oblique plane.
SOLUTION
x=0 
First note that symmetry implies
Choose as the element of volume a vertical slice of width zx, thickness dz, and height y. Then
dV = 2 xy dz ,
x=
and
y=−
Then
V=
Then

h/2
h h
z+ =
(b − z )
b
2 2b
 a 2
 h

b − z 2   (b − z )  dz
2

−b  b
  2b

b
z = b sin θ
V=
1
, z EL = z
24
a 2
b − z2
b
Now
Let
yEL =
ah
b2
π /2
π
= abh
/2
dz = b cos θ dθ
(b cos θ )[b(1 − sin θ )] b cos θ dθ
π /2
π
− /2
(cos 2 θ − sin θ cos 2 θ ) dθ
π /2
 θ sin 2θ 1

= abh  +
+ cos3 θ 
4
3
2
 −π /2
1
V = π abh
2
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719
PROBLEM 5.134* (Continued)
and
y
EL dV
Now
so that
Then
Also,

=
1 ah 2
4 b3

b
−b
z = b sin θ
Let
Then
1 h
  a 2
 h
 
b − z 2   (b − z )  dz 
&times; (b − z )   2

−b  2
2b
  b
  2b
 
b
=

dz = b cos θ dθ
1 ah 2
4 b3
1
= abh 2
4
sin 2 θ cos 2 θ =
y
EL dV
z
EL dV
π /2
π
yEL dV =
sin 2 θ =
(b − z ) 2 b 2 − z 2 dz
− /2
[b(1 − sin θ )]2 (b cos θ ) &times; (b cos θ dθ )
π /2
π
− /2
(cos 2 θ − 2sin θ cos 2 θ + sin 2 θ cos 2 θ ) dθ
1
1
(1 − cos 2θ ) cos 2 θ = (1 + cos 2θ )
2
2
1
(1 − cos 2 2θ )
4
π /2

=
1
abh 2
4
=
 θ sin 2θ
1
abh 2  +
4
4
 2
=
5
π abh2
32
π
− /2 

cos 2 θ − 2sin θ cos 2 θ +
1

(1 − cos 2 2θ )  dθ
4

1
1  θ sin 4θ
 1
3
 + 3 cos θ + 4 θ − 4  2 + 8


π /2


  −π /2
 a 2
h
 
z 2
a − z 2  (b − z )  dz 
−b  b
 2b
 
ah b
z (b − z ) b 2 − z 2 dz
= 2
b −b
=

b

z = b sin θ
Let
Then
z
EL dV
=
ah
b2
π /2
π
− /2
= ab 2 h
Using
sin 2 θ cos 2 θ =
z
EL dV
dz = b cos θ dθ
(b sin θ )[b(1 − sin θ )](b cos θ ) &times; (b cos θ dθ )
π /2
π
− /2
(sin θ cos 2 θ − sin 2 θ cos 2 θ ) dθ
1
(1 − cos 2 2θ ) from above,
4
= ab 2 h
π /2
π

− /2 

1

sin θ cos 2 θ − (1 − cos 2 2θ )  dθ
4

 1
1
1  θ sin 4θ
= ab 2 h  − cos3 θ − θ +  +
4
4 2
8
 3
π /2
1

= − π ab 2 h

8
  −π /2
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720
PROBLEM 5.134* (Continued)
Now
1
 5
yV = yEL dV : y  π abh  = π abh 2
2

 32
or y =
and
1
1

z V = z EL dV : z  π abh  = − π ab 2 h
8
2

1
or z = − b 
4


5
h 
16
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721
PROBLEM 5.135
After grading a lot, a builder places four stakes to designate the
corners of the slab for a house. To provide a firm, level base
for the slab, the builder places a minimum of 3 in. of gravel
beneath the slab. Determine the volume of gravel needed and
the x coordinate of the centroid of the volume of the gravel.
(Hint: The bottom surface of the gravel is an oblique plane,
which can be represented by the equation y = a + bx + cz.)
SOLUTION
The centroid can be found by integration. The equation for the bottom of the gravel is y = a + bx + cz , where
the constants a, b, and c can be determined as follows:
For x = 0 and z = 0,
y = − 3 in., and therefore,
−
For x = 30 ft and z = 0,
y = − 5 in., and therefore,
−
For x = 0 and z = 50 ft,
3
1
ft = a, or a = − ft
12
4
5
1
1
ft = − ft + b(30 ft), or b = −
12
4
180
y = − 6 in., and therefore,
−
6
1
1
ft = − ft + c(50 ft), or c = −
12
4
200
Therefore,
y=−
Now
x=
1
1
1
ft −
x−
z
4
180
200
x
EL dV
V
A volume element can be chosen as
dV = | y| dx dz
1
1
1 
1+
x+
z dx dz
4 
45
50 
or
dV =
and
xEL = x
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722
PROBLEM 5.135 (Continued)
Then
x
EL dV
=
50
30
0
0
 
x
1
1 
1+
x+
z dx dz
4  45
50 
30
50  x 2
z 2
1 3
x +
x  dz
 +
100  0
 2 135
1
=
4

=
1
4

=
1
9 
650 z + z 2 
4 
2 0
0
50
0
(650 + 9 z ) dz
50
= 10937.5 ft 4
The volume is

V dV =
50
30
0
0
 
=
1
4

=
1
4

1
1
1 
1+
x+
z dx dz

4
45
50 
30
50 
0
1 2 z 
 x + 90 x + 50 x  dz

0
50 
0
3 
 40 + z  dz
5 

50
1
3 
=  40 z + z 2 
4
10  0
= 687.50 ft 3
Then
x =
x
EL dV
V
=
10937.5ft 4
= 15.9091 ft
687.5 ft 3
V = 688 ft 3 
Therefore,
x = 15.91 ft 
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723
PROBLEM 5.136
Determine by direct integration the location of the centroid of
the volume between the xz plane and the portion shown of the
surface y = 16h(ax − x2)(bz − z2)/a2b2.
SOLUTION
First note that symmetry implies
x=
a

2
z=
b

2
Choose as the element of volume a filament of base dx &times; dz and height y. Then
dV = y dx dz ,
or
Then
dV =
yEL =
1
y
2
16h
(ax − x 2 )(bz − z 2 ) dx dz
2 2
a b
b
a
0
0
V=

V=
16h
a 2b2
16h
(ax − x 2 )(bz − z 2 ) dx dz
a 2b2

a
1 
a
(bz − z 2 )  x 2 − x3  dz
0
3 0
z
b
b
16h  a
1
1 
 b
= 2 2  ( a 2 ) − ( a )3   z 2 − z 3 
3
3 0
a b 2
 2
8ah  b 2 1 3 
(b) − (b) 
3
3b 2  2

4
= abh
9
=
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724
PROBLEM 5.136 (Continued)
and
y
EL dV
b
a
0
0
=

=
128h 2
a 4b 4
1  16h
  16h

( ax − x 2 )(bz − z 2 )   2 2 (ax − x 2 )(bz − z 2 ) dx dz 

2 2
2 a b
a b

128h 2
= 2 4
a b
Now
b
a
0
0


(a 2 x 2 − 2ax3 + x 4 )(b 2 z 2 − 2bz 3 + z 4 )dx dz
a
 a2
a
1 
(b z − 2bz + z )  x3 − x 4 + x5  dz
0
2
5 0
3
b
2 2
3
4
b
=
128h 2
a 4b4
 a2
a 4 1 5   b2 3 b 4 1 5 
3
−
(
a
)
(a) + (a)   z − z + z 

2
5
z
5 0
3
 3
=
64ah 2
15b 4
 b3 3 b 4 1 5  32
abh 2
 (b) − (b) + (b)  =
3
2
5
225


4
 32
yV = yEL dV : y  abh  =
abh 2
9
 225

8
or y = 25 h 
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725
PROBLEM 5.137
Locate the centroid of the plane area shown.
SOLUTION
A, in 2
1
Then
π
2
(38)2 = 2268.2
2
−20 &times; 16 = 320
Σ
1948.23
x , in.
y , in.
x A, in 3
y A, in 3
0
16.1277
0
36,581
3200
−2560
3200
34,021
−10
8
X=
Σ xA
3200
=
Σ A 1948.23
X = 1.643 in. 
Y =
Σ y A 34, 021
=
Σ A 1948.23
Y = 17.46 in. 
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726
PROBLEM 5.138
Locate the centroid of the plane area shown.
SOLUTION
Then
A, mm 2
x , mm
y , mm
x A, mm3
y A, mm3
1
2
(75)(120) = 6000
3
28.125
48
168,750
288,000
2
1
− (75)(60) = −2250
2
25
20
–56,250
–45,000
Σ
3750
112,500
243,000
X ΣA = ΣxA
X (3750 mm 2 ) = 112,500 mm3
and
or X = 30.0 mm 
Y ΣA = Σ yA
Y (3750 mm 2 ) = 243, 000 mm3
or Y = 64.8 mm 
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727
PROBLEM 5.139
The frame for a sign is fabricated from thin, flat steel bar stock of mass
per unit length 4.73 kg/m. The frame is supported by a pin at C and by a
cable AB. Determine (a) the tension in the cable, (b) the reaction at C.
SOLUTION
First note that because the frame is fabricated from uniform bar stock, its center of gravity will coincide with
the centroid of the corresponding line.
L, m
x, m
xL, m 2
1
1.35
0.675
0.91125
2
0.6
0.3
0.18
3
0.75
0
0
4
0.75
0.2
0.15
1.07746
1.26936
5
Σ
Then
π
2
(0.75) = 1.17810
4.62810
2.5106
X ΣL = Σx L
X (4.62810) = 2.5106
or
X = 0.54247 m
The free-body diagram of the frame is then
where
W = (m′Σ L) g
= 4.73 kg/m &times; 4.62810 m &times; 9.81 m/s 2
= 214.75 N
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728
PROBLEM 5.139 (Continued)
Equilibrium then requires
3

ΣM C = 0: (1.55 m)  TBA  − (0.54247 m)(214.75 N) = 0
5

(a)
TBA = 125.264 N
or
or TBA = 125.3 N 
3
ΣFx = 0: C x − (125.264 N) = 0
5
(b)
C x = 75.158 N
or
ΣFy = 0: C y +
4
(125.264 N) − (214.75 N) = 0
5
C y = 114.539 N
or
C = 137.0 N
Then
56.7&deg; 
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729
PROBLEM 5.140
Determine by direct integration the centroid of the area shown. Express your
answer in terms of a and h.
SOLUTION
For the element (EL) shown,
at x = a, y = h, h = ka 3
or k =
x=
Then
xEL
yEL
a 1/3
y dy
h1/3
1
1 a 1/3
= x=
y
2
2 h1/3
=y

A = dA =
Then
Hence
a 1/3
y
h1/3
dA = x dy =
Now
and
h
a3

xEL dA =

yEL dA =


h
0
h
0

h
0
( )
a 1/3
3 a
y dy =
y 4/3
1/3
1/3
4h
h
h
=
0
3
ah
4
h
1 a 1/3  a 1/3  1 a  3 5/3 
3
y  1/3 y dy  =
y  = a2h
1/3
2/3 
2h
h
 2 h 5
 0 10
h
3
a 3
 a


y  1/3 y1/3 dy  = 1/3  y 7/3  = ah 2
7
7
h
 h 
0

3  3
x  ah  = a 2 h
 4  10
x=
2
a 
5

3  3
y  ah  = ah2
4  7
y=
4
h 
7
xA = xEL dA:
yA = yEL dA:
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730
PROBLEM 5.141
Determine by direct integration the centroid of the area shown.
SOLUTION
xEL = x
We have
a
x x2 
1
y = 1 − + 2 
L L 
2
2

x x2 
dA = y dx = a 1 − + 2  dx
L L 

yEL =

A = dA =
Then

2L
0
2L


x x2 
x2
x3 
a 1 − + 2  dx = a  x −
+ 2
2 L 3L  0
L L 


8
= aL
3
and
x

Hence,
EL dA
2L
2L
 
 x 2 x3
x x2  
x4 
x  a 1 − + 2  dx  = a  −
+ 2
L L  
 2 3L 4 L  0
 
=

=
10 2
aL
3
yEL dA =

0
2L
0
a
x x2   
x x2  
1 − + 2   a 1 − + 2  dx 
L L   
L L  
2
EL 
x
x2
x3 x 4 
1 − 2 + 3 2 − 2 3 + 4  dx
L
L
L
L 

=
a2
2

=
a2
2

x 2 x3
x4
x5 
+ 2 − 3 + 4
x −
L L 2L 5L 0

=
11 2
a L
5
0
2L
 8  10
xA = xEL dA: x  aL  = aL2
3  3

 1  11
yA = yEL dA: y  a  = a 2
8  5

x=
y=
5
L 
4
33
a 
40
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731
PROBLEM 5.142
Three different drive belt profiles are to be
studied. If at any given time each belt makes
contact with one-half of the circumference of
its pulley, determine the contact area between
the belt and the pulley for each design.
SOLUTION
SOLUTION
Applying the first theorem of Pappus-Guldinus, the contact area AC of a belt
is given by
AC = π yL = π Σ yL
where the individual lengths are the lengths of the belt cross section that are
in contact with the pulley.
AC = π [2( y1 L1 ) + y2 L2 ]
(a)
 

0.125    0.125 in. 
= π  2  3 −
+ [(3 − 0.125) in.](0.625 in.) 
in. 


2    cos 20&deg; 
 

AC = 8.10 in 2 
or
AC = π [2( y1 L1 )]
(b)

0.375    0.375 in. 
= 2π  3 − 0.08 −
in.
2    cos 20&deg; 

AC = 6.85 in 2 
or
AC = π [2( y1 L1 )]
(c)

2(0.25)  
= π  3 −
in. [π (0.25 in.)]
π  

AC = 7.01 in 2 
or
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732
PROBLEM 5.143
Determine the reactions at the beam supports for the given
SOLUTION
We have
Then
1
(3ft)(480 lb/ft) = 720 lb
2
1
R II = (6 ft)(600 lb/ft) = 1800 lb
2
R III = (2 ft)(600 lb/ft) = 1200 lb
R I=
ΣFx = 0: Bx = 0
ΣM B = 0: (2 ft)(720 lb) − (4 ft)(1800 lb) + (6 ft)C y − (7 ft)(1200 lb) = 0
or
C y = 2360 lb
C = 2360 lb 
ΣFy = 0: −720 lb + B y − 1800 lb + 2360 lb − 1200 lb = 0
or
B y = 1360 lb
B = 1360 lb 
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733
PROBLEM 5.144
The beam AB supports two concentrated loads and rests on soil that
exerts a linearly distributed upward load as shown. Determine the
values of ω A and ω B corresponding to equilibrium.
SOLUTION
1
RI = ω A (1.8 m) = 0.9ω A
2
1
RII = ωB (1.8 m) = 0.9ωB
2
ΣM D = 0: (24 kN)(1.2 − a) − (30 kN)(0.3 m) − (0.9ω A )(0.6 m) = 0
(1)
24(1.2 − 0.6) − (30)(0.3) − 0.54ωa = 0
For a = 0.6 m,
14.4 − 9 − 0.54ω A = 0
ΣFy = 0: − 24 kN − 30 kN + 0.9(10 kN/m) + 0.9ωB = 0
ω A = 10.00 kN/m 
ωB = 50.0 kN/m 
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734
PROBLEM 5.145
The base of a dam for a lake is designed to resist up to 120 percent of
the horizontal force of the water. After construction, it is found that
silt (that is equivalent to a liquid of density ρ s = 1.76 &times; 103 kg/m3 ) is
settling on the lake bottom at the rate of 12 mm/year. Considering a
1-m-wide section of dam, determine the number of years until the
dam becomes unsafe.
SOLUTION
First determine force on dam without the silt,
1
1
Apw = A( ρ gh)
2
2
1
= [(6.6 m)(1 m)][(103 kg/m3 )(9.81 m/s 2 )(6.6 m)]
2
= 213.66 kN
= 1.2 Pw = (1.5)(213.66 kN) = 256.39 kN
Pw =
Pallow
Next determine the force P′ on the dam face after a depth d of silt has settled.
We have
1
Pw′ = [(6.6 − d ) m &times; (1 m)][(103 kg/m3 )(9.81 m/s 2 )(6.6 − d ) m]
2
= 4.905(6.6 − d ) 2 kN
( Ps ) I = [d (1 m)][(103 kg/m3 )(9.81 m/s 2 )(6.6 − d ) m]
= 9.81(6.6d − d 2 ) kN
1
( Ps )II = [ d (1 m)][(1.76 &times; 103 kg/m3 )(9.81 m/s 2 )(d ) m]
2
= 8.6328d 2 kN
P′ = Pw′ + ( Ps ) I + ( Ps ) II = [4.905(43.560 − 13.2000d + d 2 )
+ 9.81(6.6d − d 2 ) + 8.6328d 2 ] kN
= [3.7278d 2 + 213.66] kN
Now it’s required that P′ = Pallow to determine the maximum value of d.
(3.7278d 2 + 213.66) kN = 256.39 kN
or
Finally,
d = 3.3856 m
3.3856 m = 12 &times; 10−3
m
&times;N
year
or N = 282 years 
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735
PROBLEM 5.146
Determine the location of the centroid of the composite body shown
when (a) h = 2b, (b) h = 2.5b.
SOLUTION
V
x
xV
Cylinder I
π a 2b
1
b
2
1 2 2
πa b
2
Cone II
1 2
πa h
3
b+
1
h
4
1 2 
1 
π a hb + h
3
4 

1 

V = π a2  b + h 
3 

1
1
1

Σ xV = π a 2  b 2 + hb + h 2 
2
3
12


(a)
For h = 2b,
1

 5
V = π a 2 b + (2b)  = π a 2 b
3

 3
1
1
1

Σ xV = π a 2  b 2 + (2b)b + (2b) 2 
3
12
2

1 2 1 3
= π a 2b 2  + +  = π a 2 b2
 2 3 3 2
5
 3
XV = Σ xV : X  π a 2b  = π a 2 b 2
3
 2
X=
9
b
10
Centroid is
1
b
10
to left of base of cone. 
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you are using it without permission.
736
PROBLEM 5.146 (Continued)
(b)
For h = 2.5b,
1


V = π a 2 b + (2.5b)  = 1.8333π a 2 b
3


1
1
1

Σ xV = π a 2  b 2 + (2.5b)b + (2.5b) 2 
3
12
2

= π a 2b 2 [0.5 + 0.8333 + 0.52083]
= 1.85416π a 2b 2
XV = Σ xV : X (1.8333π a 2 b) = 1.85416π a 2b 2
X = 1.01136b
Centroid is 0.01136b to right of base of cone. 
Note: Centroid is at base of cone for h = 6b = 2.449b. 
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737
PROBLEM 5.147
Locate the center of gravity of the sheet-metal form shown.
SOLUTION
First, assume that the sheet metal is homogeneous so that the center of gravity of the form will coincide with
the centroid of the corresponding area.
1
yI = − (1.2) = −0.4 m
3
1
zI = (3.6) = 1.2 m
3
4(1.8)
2.4
xIII = −
=−
m
π
3π
A, m 2
x, m
y, m
z, m
xA, m3
I
1
(3.6)(1.2) = 2.16
2
1.5
−0.4
1.2
II
(3.6)(1.7) = 6.12
0.75
0.4
0.8
III
Σ
We have
π
2
(1.8)2 = 5.0894
−
2.4
π
13.3694
yA, m3
zA, m3
3.24
−0.864
2.592
1.8
4.59
2.448
11.016
1.8
−3.888
4.0715
3.942
5.6555
9.1609
22.769
X ΣV = Σ xV : X (13.3694 m 2 ) = 3.942 m3
or X = 0.295 m 
Y ΣV = Σ yV : Y (13.3694 m 2 ) = 5.6555 m3
or Y = 0.423 m 
Z ΣV = Σ zV : Z (13.3694 m 2 ) = 22.769 m3
or Z = 1.703 m 
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738
PROBLEM 5.148
Locate the centroid of the volume obtained by rotating the shaded area
SOLUTION
First note that symmetry implies
y =0 
and
z =0 
We have
y = k ( X − h) 2
At x = 0, y = a,
a = k ( − h) 2
or
k=
a
h2
Choose as the element of volume a disk of radius r and thickness dx. Then
dV = π r 2 dx, X EL = x
r=
Now
a
( x − h) 2
2
h
so that
dV = π
Then
V=

a2
( x − h) 4 dx
h4
h
0
π
a2
π a2
4
x
h
dx
(
−
)
=
[( x − h)5 ]0h
5 h4
h4
1
= π a2 h
5
and

 a2

x π 4 ( x − h) 4 dx 
0
 h

2
a h 5
( x − 4hx 4 + 6h 2 x3 − 4h3 x 2 + h4 x)dx
=π 4
h 0
xEL dV =

h

h
=π
=
Now
a2  1 6 4 5 3 2 4 4 3 3 1 4 2 
x − hx + h x − h x + h x 
5
2
3
2
h4  6
0
1
π a 2 h2
30
π
 π 2 2
xV = xEL dV : x  a 2 h  =
a h
5
 30

or
x=
1
h 
6
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739
CHAPTER 6
PROBLEM 6.1
Using the method of joints, determine the force in each member of the truss
shown. State whether each member is in tension or compression.
SOLUTION
Free body: Entire truss:
ΣFy = 0: By = 0
By = 0
ΣM C = 0: − Bx (3.2 m) − (48 kN)(7.2 m) = 0
Bx = −108 kN B x = 108 kN
ΣFx = 0: C − 108 kN + 48 kN = 0
C = 60 kN
C = 60 kN
Free body: Joint B:
FAB FBC 108 kN
=
=
5
4
3
FAB = 180.0 kN T 
FBC = 144.0 kN T 
Free body: Joint C:
FAC FBC 60 kN
=
=
13
12
5
FAC = 156.0 kN C 
FBC = 144 kN (checks)
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743
PROBLEM 6.2
Using the method of joints, determine the force in each member of the
truss shown. State whether each member is in tension or compression.
SOLUTION
Reactions:
ΣM A = 0: C = 1260 lb
ΣFx = 0: A x = 0
ΣFy = 0: A y = 960 lb
Joint B:
FAB FBC 300 lb
=
=
12
13
5
FAB = 720 lb T 
FBC = 780 lb C 
Joint A:
ΣFy = 0: − 960 lb −
4
FAC = 0
5
FAC = 1200 lb
FAC = 1200 lb C 
3
ΣFx = 0: 720 lb − (1200 lb) = 0 (checks)
5
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744
PROBLEM 6.3
Using the method of joints, determine the force in each member of the truss
shown. State whether each member is in tension or compression.
SOLUTION
AB = 32 + 1.252 = 3.25 m
BC = 32 + 42 = 5 m
Reactions:
ΣM A = 0: (84 kN)(3 m) − C (5.25 m) = 0
C = 48 kN
ΣFx = 0: Ax − C = 0
A x = 48 kN
ΣFy = 0: Ay = 84 kN = 0
A y = 84 kN
Joint A:
ΣFx = 0: 48 kN −
12
FAB = 0
13
FAB = +52 kN
ΣFy = 0: 84 kN −
FAB = 52.0 kN T 
5
(52 kN) − FAC = 0
13
FAC = +64.0 kN
FAC = 64.0 kN T 
Joint C:
FBC 48 kN
=
5
3
FBC = 80.0 kN C 
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745
PROBLEM 6.4
Using the method of joints, determine the force in each member
of the truss shown. State whether each member is in tension or
compression.
SOLUTION
Free body: Truss:
C = D = 600 lb
Free body: Joint B:
FAB
5
=
FBC 300 lb
=
2
1
FAB = 671 lb T
FBC = 600 lb C 
Free body: Joint C:
ΣFy = 0:
3
FAC + 600 lb = 0
5
FAC = −1000 lb
ΣFx = 0:
4
( −1000 lb) + 600 lb + FCD = 0
5
FAC = 1000 lb C 
FCD = 200 lb T 
From symmetry:
FAD = FAC = 1000 lb C , FAE = FAB = 671 lb T , FDE = FBC = 600 lb C 
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746
PROBLEM 6.5
Using the method of joints, determine the force in each
member of the truss shown. State whether each member is
in tension or compression.
SOLUTION
Reactions:
ΣM D = 0: Fy (24) − (4 + 2.4)(12) − (1)(24) = 0
Fy = 4.2 kips
ΣFx = 0: Fx = 0
ΣFy = 0: D − (1 + 4 + 1 + 2.4) + 4.2 = 0
D = 4.2 kips
Joint A:
ΣFx = 0: FAB = 0
FAB = 0 
ΣFy = 0 : −1 − FAD = 0
Joint D:
ΣFy = 0: − 1 + 4.2 +
ΣFx = 0:
8
FBD = 0
17
FBD = −6.8 kips
15
(−6.8) + FDE = 0
17
FDE = +6 kips
FAD = 1.000 kip C 
FBD = 6.80 kips C 
FDE = 6.00 kips T 
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747
PROBLEM 6.5 (Continued)
Joint E:
ΣFy = 0 : FBE − 2.4 = 0
FBE = +2.4 kips
FBE = 2.40 kips T 
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748
PROBLEM 6.6
Using the method of joints, determine the force in each member of
the truss shown. State whether each member is in tension or
compression.
SOLUTION
AD = 52 + 122 = 13 ft
BCD = 122 + 162 = 20 ft
Reactions:
ΣFx = 0: Dx = 0
ΣM E = 0: D y (21 ft) − (693 lb)(5 ft) = 0
ΣFy = 0: 165 lb − 693 lb + E = 0
Joint D:
D y = 165 lb
E = 528 lb
ΣFx = 0:
5
4
13
5
(1)
ΣFy = 0:
12
3
FAD + FDC + 165 lb = 0
13
5
(2)
Solving Eqs. (1) and (2) simultaneously,
FAD = 260 lb C 
FDC = +125 lb
FDC = 125 lb T 
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749
PROBLEM 6.6 (Continued)
Joint E:
ΣFx = 0:
5
4
FBE + FCE = 0
13
5
(3)
ΣFy = 0:
12
3
FBE + FCE + 528 lb = 0
13
5
(4)
Solving Eqs. (3) and (4) simultaneously,
FBE = −832 lb
FBE = 832 lb C 
FCE = +400 lb
FCE = 400 lb T 
Joint C:
Force polygon is a parallelogram (see Fig. 6.11, p. 209).
FAC = 400 lb T 
FBC = 125.0 lb T 
Joint A:
ΣFx = 0:
5
4
(260 lb) + (400 lb) + FAB = 0
13
5
FAB = −420 lb
ΣFy = 0:
FAB = 420 lb C 
12
3
(260 lb) − (400 lb) = 0
13
5
0 = 0 (Checks)
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750
PROBLEM 6.7
Using the method of joints, determine the force in each member of the truss shown.
State whether each member is in tension or compression.
SOLUTION
Free body: Entire truss:
ΣFx = 0: Cx + 2(5 kN) = 0
Cx = −10 kN

C x = 10 kN
ΣM C = 0: D(2 m) − (5 kN)(8 m) − (5 kN)(4 m) = 0

D = +30 kN
D = 30 kN

ΣFy = 0: C y + 30 kN = 0 C y = −30 kN C y = 30 kN
Free body: Joint A:
=
=
4
1
17
FAB = 20.0 kN T 


FAD = 20.6 kN C 
Free body: Joint B:
ΣFx = 0: 5 kN +
1
5
FBD = 0
FBD = −5 5 kN
ΣFy = 0: 20 kN − FBC −
2
5
FBD = 11.18 kN C 
(−5 5 kN) = 0
FBC = +30 kN
FBC = 30.0 kN T 

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751
PROBLEM 6.7 (Continued)
Free body: Joint C:
ΣFx = 0: FCD − 10 kN = 0
FCD = +10 kN
FCD = 10.00 kN T 
ΣFy = 0: 30 kN − 30 kN = 0 (checks)
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752
PROBLEM 6.8
Using the method of joints, determine the force in each member of the
truss shown. State whether each member is in tension or compression.
SOLUTION
Reactions:
ΣM C = 0: A x = 16 kN
ΣFy = 0: A y = 9 kN
ΣFx = 0:
C = 16 kN
Joint E:
FBE FDE 3 kN
=
=
5
4
3
FBE = 5.00 kN T 
FDE = 4.00 kN C 
Joint B:
ΣFx = 0:
4
(5 kN) − FAB = 0
5
FAB = +4 kN
FAB = 4.00 kN T 
3
ΣFy = 0: − 6 kN − (5 kN) − FBD = 0
5
FBD = −9 kN
FBD = 9.00 kN C 
Joint D:
ΣFy = 0: − 9 kN +
3
5
FAD = 15.00 kN T 
4
ΣFx = 0: − 4 kN − (15 kN) − FCD = 0
5
FCD = −16 kN
FCD = 16.00 kN C 
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753
PROBLEM 6.9
Determine the force in each member of the Gambrel roof
truss shown. State whether each member is in tension or
compression.
SOLUTION
Free body: Truss:
ΣFx = 0: H x = 0
A = Hy =
1
2
A = H y = 1200 lb
Free body: Joint A:
FAB FAC 900 lb
=
=
5
4
3
FAB = 1500 lb C 
FAC = 1200 lb T 
Free body: Joint C:
BC is a zero-force member.
FBC = 0
FCE = 1200 lb T 
Free body: Joint B:
ΣFx = 0:
24
4
4
FBD + FBE + (1500 lb) = 0
25
5
5
24 FBD + 20 FBE = −30, 000 lb
or
ΣFy = 0:
(1)
7
3
3
FBD − FBE + (1500) − 600 = 0
25
5
5
7 FBD − 15FBE = −7,500 lb
or
(2)
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754
PROBLEM 6.9 (Continued)
Multiply Eq. (1) by 3, Eq. (2) by 4, and add:
100 FBD = −120, 000 lb
FBD = 1200 lb C 
Multiply Eq. (1) by 7, Eq. (2) by –24, and add:
500 FBE = −30,000 lb
FBE = 60.0 lb C 
Free body: Joint D:
24
24
FDF = 0
(1200 lb) +
25
25
ΣFx = 0:
FDF = −1200 lb
FDF = 1200 lb C 
7
7
(1200 lb) − (−1200 lb) − 600 lb − FDE = 0
25
25
ΣFy = 0:
FDE = 72.0 lb
FDE = 72.0 lb T 
Because of the symmetry of the truss and loading, we deduce that
FEF = FBE
FEF = 60.0 lb C 
FEG = FCE
FEG = 1200 lb T 
FFG = FBC
FFG = 0 
FFH = FAB
FFH = 1500 lb C 
FGH = FAC
FGH = 1200 lb T 
Note: Compare results with those of Problem 6.11.
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755
PROBLEM 6.10
Determine the force in each member of the Howe roof
truss shown. State whether each member is in tension or
compression.
SOLUTION
Free body: Truss:
ΣFx = 0: H x = 0
A = Hy =
1
2
A = H y = 1200 lb
Free body: Joint A:
FAB FAC 900 lb
=
=
5
4
3
FAB = 1500 lb C 
FAC = 1200 lb T 
Free body: Joint C:
BC is a zero-force member.
FBC = 0
FCE = 1200 lb T 
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756
PROBLEM 6.10 (Continued)
Free body: Joint B:
ΣFx = 0:
or
FBD + FBE = −1500 lb
ΣFy = 0:
or
4
4
4
FBD + FBC + (1500 lb) = 0
5
5
5
(1)
3
3
3
FBD − FBE + (1500 lb) − 600 lb = 0
5
5
5
FBD − FBE = −500 lb
(2)
2 FBD = −2000 lb
FBD = 1000 lb C 
Subtract Eq. (2) from Eq. (1):
2 FBE = −1000 lb
FBE = 500 lb C 
Free Body: Joint D:
4
4
(1000 lb) + FDF = 0
5
5
ΣFx = 0:
FDF = −1000 lb
FDF = 1000 lb C 
3
3
(1000 lb) − ( −1000 lb) − 600 lb − FDE = 0
5
5
ΣFy = 0:
FDE = +600 lb
FDE = 600 lb T 
Because of the symmetry of the truss and loading, we deduce that
FEF = FBE
FEF = 500 lb C 
FEG = FCE
FEG = 1200 lb T 
FFG = FBC
FFG = 0 
FFH = FAB
FFH = 1500 lb C 
FGH = FAC
FGH = 1200 lb T 
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757
PROBLEM 6.11
Determine the force in each member of the Pratt roof
truss shown. State whether each member is in tension
or compression.
SOLUTION
Free body: Truss:
ΣFx = 0: Ax = 0
Due to symmetry of truss and load,
Ay = H =
1
2
Free body: Joint A:
FAB FAC 15.3 kN
=
=
37
35
12
FAB = 47.175 kN FAC = 44.625 kN
FAB = 47.2 kN C 
FAC = 44.6 kN T 
Free body: Joint B:
From force polygon:
FBD = 47.175 kN, FBC = 10.5 kN
FBC = 10.50 kN C 
FBD = 47.2 kN C 
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758
PROBLEM 6.11 (Continued)
Free body: Joint C:
ΣFy = 0:
3
FCD − 10.5 = 0
5
ΣFx = 0: FCE +
FCD = 17.50 kN T 
4
(17.50) − 44.625 = 0
5
FCE = 30.625 kN
Free body: Joint E:
DE is a zero-force member.
FCE = 30.6 kN T 
FDE = 0 
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759
PROBLEM 6.12
Determine the force in each member of the Fink roof truss shown.
State whether each member is in tension or compression.
SOLUTION
Free body: Truss:
ΣFx = 0: A x = 0
Ay = G =
1
2
A y = G = 6.00 kN
Free body: Joint A:
F
FAB
4.50 kN
= AC =
2.462 2.25
1
FAB = 11.08 kN C 
FAC = 10.125 kN
FAC = 10.13 kN T 
Free body: Joint B:
ΣFx = 0:
3
2.25
2.25
FBC +
FBD +
(11.08 kN) = 0
5
2.462
2.462
(1)
F
4
11.08 kN
FBC + BD +
− 3 kN = 0
5
2.462
2.462
(2)
ΣFy = 0: −
Multiply Eq. (2) by –2.25 and add to Eq. (1):
12
FBC + 6.75 kN = 0
5
FBC = −2.8125
FBC = 2.81 kN C 
Multiply Eq. (1) by 4, Eq. (2) by 3, and add:
12
12
FBD +
(11.08 kN) − 9 kN = 0
2.462
2.462
FBD = −9.2335 kN
FBD = 9.23 kN C 
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760
PROBLEM 6.12 (Continued)
Free body: Joint C:
ΣFy = 0:
4
4
FCD − (2.8125 kN) = 0
5
5
FCD = 2.8125 kN,
FCD = 2.81 kN T 
3
3
ΣFx = 0: FCE − 10.125 kN + (2.8125 kN) + (2.8125 kN) = 0
5
5
FCE = +6.7500 kN
FCE = 6.75 kN T 
Because of the symmetry of the truss and loading, we deduce that
FDE = FCD
FCD = 2.81 kN T 
FDF = FBD
FDF = 9.23 kN C 
FEF = FBC
FEF = 2.81 kN C 
FEG = FAC
FEG = 10.13 kN T 
FFG = FAB
FFG = 11.08 kN C 
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761
PROBLEM 6.13
Using the method of joints, determine the force in each member
of the double-pitch roof truss shown. State whether each member
is in tension or compression.
SOLUTION
Free body: Truss:
ΣM A = 0: H (18 m) − (2 kN)(4 m) − (2 kN)(8 m) − (1.75 kN)(12 m)
− (1.5 kN)(15 m) − (0.75 kN)(18 m) = 0
H = 4.50 kN
ΣFx = 0: Ax = 0
ΣFy = 0: Ay + H − 9 = 0
Ay = 9 − 4.50
A y = 4.50 kN
Free body: Joint A:
FAB
FAC 3.50 kN
=
2
1
5
FAB = 7.8262 kN C
=
FAB = 7.83 kN C 
FAC = 7.00 kN T 
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762
PROBLEM 6.13 (Continued)
Free body: Joint B:
ΣFx = 0:
2
5
FBD +
2
5
(7.8262 kN) +
1
2
FBC = 0
FBD + 0.79057 FBC = −7.8262 kN
or
ΣFy = 0:
1
5
FBD +
1
5
(7.8262 kN) −
1
2
(1)
FBC − 2 kN = 0
FBD − 1.58114 FBC = −3.3541
or
(2)
Multiply Eq. (1) by 2 and add Eq. (2):
3FBD = −19.0065
FBD = −6.3355 kN
FBD = 6.34 kN C 
Subtract Eq. (2) from Eq. (1):
2.37111FBC = −4.4721
FBC = 1.886 kN C 
FBC = −1.8861 kN
Free body: Joint C:
2
ΣFy = 0:
5
FCD −
1
2
(1.8861 kN) = 0
FCD = +1.4911 kN
ΣFx = 0: FCE − 7.00 kN +
FCE
1
FCD = 1.491 kN T 
(1.8861 kN) +
2
= +5.000 kN
1
5
(1.4911 kN) = 0
FCE = 5.00 kN T 
Free body: Joint D:
ΣFx = 0:
2
5
FDF +
1
2
FDE +
2
5
(6.3355 kN) −
1
5
(1.4911 kN) = 0
FDF + 0.79057 FDE = −5.5900 kN
or
ΣFy = 0:
or
1
5
FDF −
1
2
FDE +
1
5
(6.3355 kN) −
FDF − 0.79057 FDE = −1.1188 kN
2
5
(1.4911 kN) − 2 kN = 0
(2)
2 FDF = −6.7088 kN
FDF = −3.3544 kN
Subtract Eq. (2) from Eq. (1):
(1)
FDF = 3.35 kN C 
1.58114 FDE = −4.4712 kN
FDE = −2.8278 kN
FDE = 2.83 kN C 
PROPRIETARY MATERIAL. &copy; 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
763
PROBLEM 6.13 (Continued)
Free body: Joint F:
ΣFx = 0:
1
FFG +
2
2
(3.3544 kN) = 0
5
FFG = −4.243 kN
ΣFy = 0: −FEF − 1.75 kN +
1
5
FFG = 4.24 kN C 
(3.3544 kN) −
1
2
FEF = 2.750 kN
(−4.243 kN) = 0
FEF = 2.75 kN T 
Free body: Joint G:
ΣFx = 0:
1
FGH −
2
ΣFy = 0: −
1
2
1
2
(4.243 kN) = 0
FGH −
1
2
FEG −
1
2
(1)
(4.243 kN) − 1.5 kN = 0
FGH + FEG = −6.364 kN
or
(2)
2 FGH = −10.607
FGH = −5.303
Subtract Eq. (1) from Eq. (2):
2
FEG +
FGH − FEG = −4.243 kN
or
1
FGH = 5.30 kN C 
2 FEG = −2.121 kN
FEG = −1.0605 kN
FEG = 1.061 kN C 
Free body: Joint H:
FEH 3.75 kN
=
1
1
We can also write
FGH
2
=
FEH = 3.75 kN T 
3.75 kN
1
FGH = 5.30 kN C (Checks)
PROPRIETARY MATERIAL. &copy; 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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764
PROBLEM 6.14
The truss shown is one of several supporting an advertising panel. Determine
the force in each member of the truss for a wind load equivalent to the two
forces shown. State whether each member is in tension or compression.
SOLUTION
Free body: Entire truss:
ΣM F = 0: (800 N)(7.5 m) + (800 N)(3.75 m) − A(2 m) = 0
A = +2250 N
A = 2250 N
ΣFy = 0: 2250 N + Fy = 0
Fy = −2250 N Fy = 2250 N
ΣFx = 0: −800 N − 800 N + Fx = 0
Fx = +1600 N Fx = 1600 N
Joint D:
800 N FDE FBD
=
=
8
15
17
FBD = 1700 N C 
FDE = 1500 N T 
Joint A:
2250 N FAB FAC
=
=
15
17
8
FAB = 2250 N C 
FAC = 1200 N T 
Joint F:
ΣFx = 0: 1600 N − FCF = 0
FCF = +1600 N
FCF = 1600 N T 
ΣFy = 0: FEF − 2250 N = 0
FEF = +2250 N
FEF = 2250 N T 
PROPRIETARY MATERIAL. &copy; 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
765
PROBLEM 6.14 (Continued)
Joint C:
ΣFx = 0:
8
FCE − 1200 N + 1600 N = 0
17
FCE = −850 N
ΣFy = 0: FBC +
15
FCE = 0
17
FBC = −
15
15
FCE = − ( −850 N)
17
17
FBC = +750 N
Joint E:
FCE = 850 N C 
ΣFx = 0: − FBE − 800 N +
FBC = 750 N T 
8
(850 N) = 0
17
FBE = −400 N
ΣFy = 0: 1500 N − 2250 N +
FBE = 400 N C 
15
(850 N) = 0
17
0 = 0 (checks)
PROPRIETARY MATERIAL. &copy; 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
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you are using it without permission.
766
PROBLEM 6.15
Determine the force in each of the members located to the left
of line FGH for the studio roof truss shown. State whether each
member is in tension or compression.
SOLUTION
Free body: Truss:
ΣFx = 0: A x = 0
Ay = L =
1
2
A y = L = 1200 lb
Zero-Force Members: Examining joints C and H, we conclude that
BC, EH, and GH are zero-force members. Thus,
FBC = FEH = 0

Also,
FCE = FAC
(1)
Free body: Joint A:
FAB
FAC 1000 lb
=
2
1
5
FAB = 2236 lb C
=
FAB = 2240 lb C 
FAC = 2000 lb T 
FCE = 2000 lb T 
From Eq. (1):
Free body: Joint B:
ΣFx = 0:
2
5
FBD +
2
5
FBE +
2
5
(2236 lb) = 0
FBD + FBE = −2236 lb
or
ΣFy = 0:
1
5
FBD −
1
5
FBE +
1
FBD − FBE = −1342 lb
or
(2)
5
(2236 lb) − 400 lb = 0
(3)
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767
PROBLEM 6.15 (Continued)
2 FBD = −3578 lb
FBD = 1789 lb C 
Subtract Eq. (3) from Eq. (1): 2 FBE = − 894 lb
FBE = 447 lb C 
Free body: Joint E:
ΣFx = 0:
2
5
FEG +
2
5
(447 lb) − 2000 lb = 0
FEG = 1789 lb T 
ΣFy = 0: FDE +
1
5
(1789 lb) −
1
5
FDE = − 600 lb
(447 lb) = 0
FDE = 600 lb C 
Free body: Joint D:
ΣFx = 0:
2
5
FDF +
2
5
FDG +
2
5
(1789 lb) = 0
FDF + FDG = −1789 lb
or
ΣFy = 0:
1
FDF −
1
FDG +
(4)
1
5
5
5
+ 600 lb − 400 lb = 0
FDF − FDG = −2236 lb
or
(1789 lb)
(5)
2 FDF = − 4025 lb FDF = 2010 lb C 
Subtract Eq. (5) from Eq. (4): 2 FDG = 447 lb
FDG = 224 lb T 
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768
PROBLEM 6.16
Determine the force in member FG and in each of the members
located to the right of FG for the studio roof truss shown. State
whether each member is in tension or compression.
SOLUTION
L=
1
2
L = 1200 lb
(See F.B. diagram to the left for more details.)
Free body: Joint L:
9
= 26.57&deg;
18
3
β = tan −1 = 9.46&deg;
18
FJL
FKL
1000 lb
=
=
sin 63.43&deg; sin 99.46&deg; sin17.11&deg;
α = tan −1
FKL = 3352.7 lb C
FJL = 3040 lb T 
FKL = 3350 lb C 
Free body: Joint K:
ΣFx = 0: −
or
2
5
FIK −
2
5
2
FJK −
5
(3352.7 lb) = 0
FIK + FJK = − 3352.7 lb
ΣFy = 0:
or
1
5
FIK −
1
5
FJK +
(1)
1
5
(3352.7) − 400 = 0
FIK − FJK = − 2458.3 lb
(2)
2 FIK = − 5811.0
FIK = − 2905.5 lb
FIK = 2910 lb C 
Subtract Eq. (2) from Eq. (1): 2 FJK = − 894.4
FJK = − 447.2 lb
FJK = 447 lb C 
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769
PROBLEM 6.16 (Continued)
Free body: Joint J:
2
ΣFx = 0: −
13
3
ΣFy = 0:
FIJ +
13
6
FIJ −
37
1
37
6
FGJ +
FGJ −
37
1
37
2
(3040 lb) −
(3040 lb) −
5
1
5
(447.2) = 0
(447.2) = 0
(3)
(4)
Multiply Eq. (4) by 6 and add to Eq. (3):
16
13
FIJ −
8
(447.2) = 0
5
FIJ = 360.54 lb
FIJ = 361 lb T 
Multiply Eq. (3) by 3, Eq. (4) by 2, and add:
−
16
37
( FGJ − 3040) −
8
(447.2) = 0
5
FGJ = 2431.7 lb
FGJ = 2430 lb T 
Free body: Joint I:
2
ΣFx = 0: −
5
2
FFI −
5
2
FGI −
5
2
(2905.5) +
13
(360.54) = 0
FFI + FGI = − 2681.9 lb
or
ΣFy = 0:
1
5
FFI −
1
5
FGI +
1
5
(5)
(2905.5) −
3
13
(360.54) − 400 = 0
FFI − FGI = −1340.3 lb
or
(6)
2 FFI = −4022.2
Subtract Eq. (6) from Eq. (5):
FFI = −2011.1 lb
FFI = 2010 lb C 
2 FGI = −1341.6 lb
FGI = 671 lb C 
Free body: Joint F:
From
ΣFx = 0: FDF = FFI = 2011.1 lb C
 1

2011.1 lb  = 0
ΣFy = 0: FFG − 400 lb + 2 
 5

FFG = + 1400 lb
FFG = 1400 lb T 
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770
PROBLEM 6.17
Determine the force in each of the members located to the
left of FG for the scissors roof truss shown. State whether
each member is in tension or compression.
SOLUTION
Free Body: Truss:
ΣFx = 0: A x = 0
ΣM L = 0: (1 kN)(12 m) + (2 kN)(10 m) + (2 kN)(8 m) + (1 kN)(6 m) − Ay (12 m) = 0
A y = 4.50 kN
FBC = 0 
We note that BC is a zero-force member:
Also,
FCE = FAC
Free body: Joint A:
ΣFx = 0:
1
ΣFy = 0:
1
(1)
2
2
FAB +
2
FAB +
1
5
5
FAC = 0
(2)
FAC + 3.50 kN = 0
(3)
Multiply Eq. (3) by –2 and add Eq. (2):
−
1
2
FAB − 7 kN = 0
FAB = 9.90 kN C 
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771
PROBLEM 6.17 (Continued)
Subtract Eq. (3) from Eq. (2):
1
5
FAC − 3.50 kN = 0 FAC = 7.826 kN
FCE = FAC = 7.826 kN
From Eq. (1):
FAC = 7.83 kN T 
FCE = 7.83 kN T 
Free body: Joint B:
ΣFy = 0:
1
2
FBD +
1
2
(9.90 kN) − 2 kN = 0
FBD = −7.071 kN
1
ΣFx = 0: FBE +
2
(9.90 − 7.071) kN = 0
FBE = −2.000 kN
Free body: Joint E:
ΣFx = 0:
2
5
FBD = 7.07 kN C 
FBE = 2.00 kN C 
( FEG − 7.826 kN) + 2.00 kN = 0
FEG = 5.590 kN
ΣFy = 0: FDE −
1
5
FEG = 5.59 kN T 
(7.826 − 5.590) kN = 0
FDE = 1.000 kN
FDE = 1.000 kN T 
Free body: Joint D:
ΣFx = 0:
2
5
( FDF + FDG ) +
1
2
(7.071 kN)
FDF + FDG = −5.590 kN
or
ΣFy = 0:
or
1
5
( FDF − FDG ) +
1
2
(4)
(7.071 kN) = 2 kN − 1 kN = 0
FDE − FDG = −4.472
(5)
2 FDF = −10.062 kN
FDF = 5.03 kN C 
Subtract Eq. (5) from Eq. (4):
2 FDG = −1.1180 kN
FDG = 0.559 kN C 
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772
PROBLEM 6.18
Determine the force in member FG and in each of the
members located to the right of FG for the scissors roof
truss shown. State whether each member is in tension or
compression.
SOLUTION
Free body: Truss:
ΣM A = 0: L(12 m) − (2 kN)(2 m) − (2 kN)(4 m) − (1 kN)(6 m) = 0
L = 1.500 kN
Angles:
tan α = 1
tan β =
1
2
α = 45&deg;
β = 26.57&deg;
Zero-force members:
Examining successively joints K, J, and I, we note that the following members to the right of FG are zeroforce members: JK, IJ, and HI.
FHI = FIJ = FJK = 0 
Thus,
We also note that
and
FGI = FIK = FKL
(1)
FHJ = FJL
(2)
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773
PROBLEM 6.18 (Continued)
Free body: Joint L:
FJL
F
1.500 kN
= KL =
sin116.57&deg; sin 45&deg; sin18.43&deg;
FJL = 4.2436 kN
FJL = 4.24 kN C 
FKL = 3.35 kN T 
From Eq. (1):
FGI = FIK = FKL
FGI = FIK = 3.35 kN T 
From Eq. (2):
FHJ = FJL = 4.2436 kN
FHJ = 4.24 kN C 
FGH
FFH
4.2436
=
=
sin108.43&deg; sin18.43&deg; sin 53.14&deg;
FFH = 5.03 kN C 
Free body: Joint H:
FGH = 1.677 kN T 
Free body: Joint F:
ΣFx = 0: − FDF cos 26.57&deg; − (5.03 kN) cos 26.57&deg; = 0
FDF = −5.03 kN
ΣFy = 0: − FFG − 1 kN + (5.03 kN) sin 26.57&deg; − ( −5.03 kN)sin 26.57&deg; = 0
FFG = 3.500 kN
FFG = 3.50 kN T 
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774
PROBLEM 6.19
Determine the force in each member of the Warren bridge truss
shown. State whether each member is in tension or compression.
SOLUTION
ΣFx = 0: Ax = 0
Free body: Truss:
Ay = G =
Free body: Joint A:
1
2
FAB FAC 6 kips
=
=
5
3
4
FAB = 7.50 kips C 
FAC = 4.50 kips T 

Free body: Joint B:
FBC FBD 7.5 kips
=
=
5
6
5
FBC = 7.50 kips T 
FBD = 9.00 kips C 
Free body: Joint C:
ΣFy = 0:
4
4
(7.5) + FCD − 6 = 0
5
5
FCD = 0 
3
ΣFx = 0: FCE − 4.5 − (7.5) = 0
5
FCE = +9 kips
FCE = 9.00 kips T 
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775
PROBLEM 6.20
Solve Problem 6.19 assuming that the load applied at E has
been removed.
PROBLEM 6.19 Determine the force in each member of the
Warren bridge truss shown. State whether each member is in
tension or compression.
SOLUTION
Free body: Truss:
ΣFx = 0: Ax = 0
ΣM G = 0: 6(36) − Ay (54) = 0 A y = 4 kips
ΣFy = 0: 4 − 6 + G = 0 G = 2 kips
Free body: Joint A:
FAB FAC 4 kips
=
=
5
3
4
FAB = 5.00 kips C 
FAC = 3.00 kips T 
Free body Joint B:
FBC FBD 5 kips
=
=
5
6
5
FBC = 5.00 kips T 
FBD = 6.00 kips C 
Free body Joint C:
ΣM y = 0:
4
4
(5) + FCD − 6 = 0
5
5
3
3
ΣFx = 0: FCE + (2.5) − (5) − 3 = 0
5
5
FCD = 2.50 kips T 
FCE = 4.50 kips T 
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776
PROBLEM 6.20 (Continued)
Free body: Joint D:
4
4
ΣFy = 0: − (2.5) − FDE = 0
5
5
FDE = −2.5 kips
FDE = 2.50 kips C 
3
3
ΣFx = 0: FDF + 6 − (2.5) − (2.5) = 0
5
5
Free body: Joint F:
FDF = −3 kips
FDF = 3.00 kips C 
FEF FFG 3 kips
=
=
5
5
6
FEF = 2.50 kips T 
FFG = 2.50 kips C 

Free body: Joint G:
FEG 2 kips
=
3
4
Also,
FFG 2 kips
=
5
4
FEG = 1.500 kips T 
FFG = 2.50 kips C (Checks)
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777
PROBLEM 6.21
Determine the force in each member of the Pratt bridge truss
shown. State whether each member is in tension or compression.
SOLUTION
Free body: Truss:
ΣFz = 0: A x = 0
ΣM A = 0 : H (36 ft) − (4 kips)(9 ft)
− (4 kips)(18 ft) − (4 kips)(27 ft) = 0
H = 6 kips
ΣFy = 0: Ay + 6 kips − 12 kips = 0
A y = 6 kips
Free body: Joint A:
FAB FAC 6 kips
=
=
5
3
4
FAB = 7.50 kips C 
FAC = 4.50 kips T 
Free body: Joint C:
ΣFx = 0:
FCE = 4.50 kips T 
ΣFy = 0:
FBC = 4.00 kips T 
Free body: Joint B:
ΣFy = 0: −
4
4
FBE + (7.50 kips) − 4.00 kips = 0 
5
5
FBE = 2.50 kips T 
ΣFx = 0:
8
3
(7.50 kips) + (2.50 kips) + FBD = 0 
5
5
FBD = −6.00 kips
FBD = 6.00 kips C 
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778
PROBLEM 6.21 (Continued)
Free body: Joint D:
We note that DE is a zero-force member:
FDE = 0 
FDF = 6.00 kips C 
Also,
From symmetry:
FFE = FBE
FEF = 2.50 kips T 
FEG = FCE
FEG = 4.50 kips T 
FFG = FBC
FFG = 4.00 kips T 
FFH = FAB
FFH = 7.50 kips C 
FGH = FAC
FGH = 4.50 kips T 
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779
PROBLEM 6.22
Solve Problem 6.21 assuming that the load applied at G has
been removed.
PROBLEM 6.21 Determine the force in each member of the
Pratt bridge truss shown. State whether each member is in
tension or compression.
SOLUTION
Free body: Truss:
ΣFx = 0: A x = 0
ΣM A = 0: H (36 ft) − (4 kips)(9 ft) − (4 kips)(18 ft) = 0
H = 3.00 kips
ΣFy = 0: A y + 5.00 kips
We note that DE and FG are zero-force members.
FDE = 0,
Therefore,
FFG = 0 
Also,
FBD = FDF
(1)
and
FEG = FGH
(2)
Free body: Joint A:
FAB FAC 5 kips
=
=
5
3
4
FAB = 6.25 kips C 
FAC = 3.75 kips T 

Free body: Joint C:
ΣFx = 0:
FCE = 3.75 kips T 
ΣFy = 0:
FBC = 4.00 kips T 
Free body: Joint B:
ΣFx = 0:
4
4
(6.25 kips) − 4.00 kips − FBE = 0
5
5
FBE = 1.250 kips T 
3
3
ΣFx = 0: FBD + (6.25 kips) + (1.250 kips) = 0 
5
5
FBD = −4.50 kips
FBD = 4.50 kips C 
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780
PROBLEM 6.22 (Continued)
Free body: Joint F:
We recall that FFG = 0, and from Eq. (1) that
FDF = FBD
FDF = 4.50 kips C 
FEF FFH 4.50 kips
=
=
5
5
6
FEF = 3.75 kips T 
FFH = 3.75 kips C 
Free body: Joint H:
FGH 3.00 kips
=
3
4
FGH = 2.25 kips T 
Also,
FFH 3.00 kips
=
5
4
FFH = 3.75 kips C (checks)
From Eq. (2):
FEG = FGH
FEG = 2.25 kips T 
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781
PROBLEM 6.23
The portion of truss shown represents the upper part of
determine the force in each of the members located
above HJ. State whether each member is in tension or
compression.
SOLUTION
Free body: Joint A:
F
FAB
1.2 kN
= AC =
2.29 2.29
1.2
FAB = 2.29 kN T 
FAC = 2.29 kN C 
Free body: Joint F:
FDF
F
1.2 kN
= EF =
2.29 2.29
2.1
FDF = 2.29 kN T 
FEF = 2.29 kN C 
Free body: Joint D:
FBD FDE 2.29 kN
=
=
2.21 0.6
2.29
FBD = 2.21 kN T 
FDE = 0.600 kN C 
Free body: Joint B:
ΣFx = 0:
4
2.21
(2.29 kN) = 0
FBE + 2.21 kN −
5
2.29
FBE = 0 
3
0.6
(2.29 kN) = 0
ΣFy = 0: − FBC − (0) −
5
2.29
FBC = − 0.600 kN
FBC = 0.600 kN C 
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782
PROBLEM 6.23 (Continued)
Free body: Joint C:
ΣFx = 0: FCE +
2.21
(2.29 kN) = 0
2.29
FCE = − 2.21 kN
ΣFy = 0: − FCH − 0.600 kN −
FCH = −1.200 kN
FCE = 2.21 kN C 
0.6
(2.29 kN) = 0
2.29
FCH = 1.200 kN C 
Free body: Joint E:
ΣFx = 0: 2.21 kN −
2.21
4
(2.29 kN) − FEH = 0
2.29
5
FEH = 0 
ΣFy = 0: − FEJ − 0.600 kN −
FEJ = −1.200 kN
0.6
(2.29 kN) − 0 = 0
2.29
FEJ = 1.200 kN C 
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783
PROBLEM 6.24
FCH = FEJ = 1.2 kN C and FEH = 0, determine the force in
member HJ and in each of the members located between HJ and
NO. State whether each member is in tension or compression.
PROBLEM 6.23 The portion of truss shown represents the
upper part of a power transmission line tower. For the given
above HJ. State whether each member is in tension or
compression.
SOLUTION
Free body: Joint G:
FGH
F
1.2 kN
= GI =
3.03 3.03
1.2
FGH = 3.03 kN T 
FGI = 3.03 kN C 
Free body: Joint L:
FJL
F
1.2 kN
= KL =
3.03 3.03
1.2
FJL = 3.03 kN T 
FKL = 3.03 kN C 
Free body: Joint J:
ΣFx = 0: − FHJ +
2.97
(3.03 kN) = 0
3.03
FHJ = 2.97 kN T 
0.6
(3.03 kN) = 0
3.03
= −1.800 kN
FJK = 1.800 kN C 
Fy = 0: − FJK − 1.2 kN −
FJK
Free body: Joint H:
ΣFx = 0:
4
2.97
(3.03 kN) = 0
FHK + 2.97 kN −
5
3.03
FHK = 0 
0.6
3
(3.03) kN − (0) = 0
3.03
5
= −1.800 kN
FHI = 1.800 kN C 
ΣFy = 0: − FHI − 1.2 kN −
FHI
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784
PROBLEM 6.24 (Continued)
Free body: Joint I:
ΣFx = 0: FIK +
2.97
(3.03 kN) = 0
3.03
FIK = −2.97 kN
ΣFy = 0: − FIN − 1.800 kN −
FIN = −2.40 kN
FIK = 2.97 kN C 
0.6
(3.03 kN) = 0
3.03
FIN = 2.40 kN C 
Free body: Joint K:
ΣFx = 0: −
4
2.97
(3.03 kN) = 0
FKN + 2.97 kN −
5
3.03
FKN = 0 
ΣFy = 0: − FKO −
0.6
3
(3.03 kN) − 1.800 kN − (0) = 0
3.03
5
FKO = −2.40 kN
FKO = 2.40 kN C 
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785
PROBLEM 6.25
Solve Problem 6.23 assuming that the cables hanging from the
right side of the tower have fallen to the ground.
PROBLEM 6.23 The portion of truss shown represents the
upper part of a power transmission line tower. For the given
above HJ. State whether each member is in tension or
compression.
SOLUTION
Zero-Force Members:
Considering joint F, we note that DF and EF are zero-force members:
FDF = FEF = 0 
Considering next joint D, we note that BD and DE are zero-force
members:
FBD = FDE = 0 
Free body: Joint A:
F
FAB
1.2 kN
= AC =
2.29 2.29
1.2
FAB = 2.29 kN T 
FAC = 2.29 kN C 
Free body: Joint B:
ΣFx = 0:
4
2.21
(2.29 kN) = 0
FBE −
5
2.29
FBE = 2.7625 kN
ΣFy = 0: − FBC −
FBE = 2.76 kN T 
0.6
3
(2.29 kN) − (2.7625 kN) = 0
2.29
5
FBC = −2.2575 kN
FBC = 2.26 kN C 
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786
PROBLEM 6.25 (Continued)
Free body: Joint C:
ΣFx = 0: FCE +
2.21
(2.29 kN) = 0
2.29
FCE = 2.21 kN C 
ΣFy = 0: − FCH − 2.2575 kN −
FCH = −2.8575 kN
0.6
(2.29 kN) = 0
2.29
FCH = 2.86 kN C 
Free body: Joint E:
ΣFx = 0: −
4
4
FEH − (2.7625 kN) + 2.21 kN = 0
5
5
FEH = 0 
3
3
ΣFy = 0: − FEJ + (2.7625 kN) − (0) = 0
5
5
FEJ = +1.6575 kN
FEJ = 1.658 kN T 
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787
PROBLEM 6.26
Determine the force in each of the members connecting joints A
through F of the vaulted roof truss shown. State whether each
member is in tension or compression.
SOLUTION
Free body: Truss:
ΣFx = 0: A x = 0
ΣM K = 0: (1.2 kN)6a + (2.4 kN)5a + (2.4 kN)4a + (1.2 kN)3a
− Ay (6a) = 0 A y = 5.40 kN
Free body: Joint A:
FAB
FAC 4.20 kN
=
2
1
5
FAB = 9.3915 kN
=
FAB = 9.39 kN C 
FAC = 8.40 kN T 
Free body: Joint B:
ΣFx = 0:
2
ΣFy = 0:
1
5
5
FBD +
1
FBD −
1
2
2
FBC +
2
FBC +
1
5
5
(9.3915) = 0
(1)
(9.3915) − 2.4 = 0
(2)
3
5
FBD +
3
5
(9.3915 kN) − 2.4 kN = 0
FBD = − 7.6026 kN
FBD = 7.60 kN C 
Multiply Eq. (2) by –2 and add Eq. (1):
3
2
FB + 4.8 kN = 0
FBC = −2.2627 kN
FBC = 2.26 kN C 
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788
PROBLEM 6.26 (Continued)
Free body: Joint C:
ΣFx = 0:
1
ΣFy = 0:
2
5
5
4
FCD +
17
1
FCD +
17
FCE +
1
FCE −
1
2
2
(2.2627) − 8.40 = 0
(3)
(2.2627) = 0
(4)
Multiply Eq. (4) by −4 and add Eq. (1):
−
7
5
FCD +
5
(2.2627) − 8.40 = 0
2
FCD = − 0.1278 kN
FCD = 0.128 kN C 
Multiply Eq. (1) by 2 and subtract Eq. (2):
7
17
FCE +
3
2
(2.2627) − 2(8.40) = 0
FCE = 7.068 kN
FCE = 7.07 kN T 
Free body: Joint D:
ΣFx = 0:
2
1
+
ΣFy = 0:
FDF +
5
5
1
5
+
(0.1278) = 0
FDF −
2
5
1
2
FDE +
(7.6026)
1.524
5
(5)
1.15
1
FDE +
(7.6026)
1.524
5
(0.1278) − 2.4 = 0
(6)
Multiply Eq. (5) by 1.15 and add Eq. (6):
3.30
5
FDF +
3.30
5
(7.6026) +
3.15
5
(0.1278) − 2.4 = 0
FDF = − 6.098 kN
FDF = 6.10 kN C 
Multiply Eq. (6) by –2 and add Eq. (5):
3.30
3
FDE −
(0.1278) + 4.8 = 0
1.524
5
FDE = − 2.138 kN
FDE = 2.14 kN C 
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789
PROBLEM 6.26 (Continued)
Free body: Joint E:
ΣFx = 0:
0.6
4
1
FEF +
( FEH − FCE ) +
(2.138) = 0
2.04
1.524
17
(7)
ΣFy = 0:
1.95
1
1.15
FEF +
( FEH − FCE ) −
(2.138) = 0
2.04
1.524
17
(8)
Multiply Eq. (8) by 4 and subtract Eq. (7):
7.2
FEF − 7.856 kN = 0
2.04
FEF = 2.23 kN T 
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790
PROBLEM 6.27
Determine the force in each member of the truss shown. State whether
each member is in tension or compression.
SOLUTION
Free body: Truss:
ΣM F = 0: G (20 ft) − (15 kips)(16 ft) − (40 kips)(15 ft) = 0
G = 42 kips
ΣFx = 0: Fx + 15 kips = 0
Fx = 15 kips
ΣFy = 0: Fy − 40 kips + 42 kips = 0
Fy = 2 kips
Free body: Joint F:
ΣFx = 0:
1
5
FDF − 15 kips = 0
FDF = 33.54 kips
ΣFy = 0: FBF − 2 kips +
2
5
FDF = 33.5 kips T 
(33.54 kips) = 0
FBF = − 28.00 kips
FBF = 28.0 kips C 
Free body: Joint B:
ΣFx = 0:
5
ΣFy = 0:
2
29
29
FAB +
5
FAB −
6
61
61
FBD + 15 kips = 0
(1)
FBD + 28 kips = 0
(2)
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791
PROBLEM 6.27 (Continued)
Multiply Eq. (1) by 6, Eq. (2) by 5, and add:
40
29
FAB + 230 kips = 0
FAB = −30.96 kips
FAB = 31.0 kips C 
Multiply Eq. (1) by 2, Eq. (2) by –5, and add:
40
61
FBD − 110 kips = 0
FBD = 21.48 kips
FBD = 21.5 kips T 
Free body: Joint D:
2
ΣFy = 0:
5
2
5
(33.54) +
6
61
(21.48) = 0
FAD = 15.09 kips T 
1
ΣFx = 0: FDE +
5
(15.09 − 33.54) −
5
(21.48) = 0
61
FDE = 22.0 kips T 
Free body: Joint A:
5
ΣFx = 0:
29
ΣFy = 0: −
1
FAC +
2
29
5
FAC −
FAE +
2
5
5
FAE +
29
(30.36) −
2
29
1
5
(15.09) = 0
2
(30.96) −
5
(15.09) = 0
(3)
(4)
Multiply Eq. (3) by 2 and add Eq. (4):
8
29
FAC +
12
29
(30.96) −
4
5
FAC = −28.27 kips,
(15.09) = 0
FAC = 28.3 kips C 
Multiply Eq. (3) by 2, Eq. (4) by 5, and add:
−
8
5
FAE +
20
29
(30.96) −
12
5
(15.09) = 0
FAE = 9.50 kips T 
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792
PROBLEM 6.27 (Continued)
Free body: Joint C:
From force triangle:
FCE
61
=
FCG 28.27 kips
=
8
29
FCE = 41.0 kips T 
FCG = 42.0 kips C 
Free body: Joint G:
ΣFx = 0:
ΣFy = 0: 42 kips − 42 kips = 0
FEG = 0 
(Checks) 
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793
PROBLEM 6.28
Determine the force in each member of the truss shown. State
whether each member is in tension or compression.
SOLUTION
Reactions:
ΣFx = 0:
Ex = 0
ΣM F = 0:
E y = 45 kips
ΣFy = 0:
F = 60 kips
Joint D:
FCD FDH 15 kips
=
=
12
13
5
FCD = 36.0 kips T 
FDH = 39.0 kips C 
Joint H:
ΣF = 0:
FCH = 0 
ΣF = 0:
FGH = 39.0 kips C 
ΣF = 0:
FCG = 0 
ΣF = 0:
FBC = 36.0 kips T 
ΣF = 0:
FBG = 0 
ΣF = 0:
FFG = 39.0 kips C 
ΣF = 0:
FBF = 0 
Joint C:
Joint G:
Joint B:
ΣF = 0:
FAB = 36.0 kips T 
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794
PROBLEM 6.28 (Continued)
Joint A:
AE = 122 + 152 = 19.21 ft
tan 38.7&deg; =
36 kips
FAF
FAF = 45.0 kips C 
sin 38.7&deg; =
36 kips
FAE
FAE = 57.6 kips T 
Joint E:
ΣFx = 0: + (57.6 kips) sin 38.7&deg; + FEF = 0
FEF = −36.0 kips
FEF = 36.0 kips C 
ΣFy = 0: (57.6 kips) cos 38.7&deg; − 45 kips = 0 (Checks)
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795
PROBLEM 6.29
Determine whether the trusses of Problems 6.31a, 6.32a, and
6.33a are simple trusses.
SOLUTION
Truss of Problem 6.31a:
Starting with triangle HDI and adding two members at a time, we obtain
successively joints A, E, J, and B, but cannot go further. Thus, this truss
is not a simple truss. 
Truss of Problem 6.32a:
Starting with triangle ABC and adding two members at a time, we obtain joints
D, E, G, F, and H, but cannot go further. Thus, this truss
is not a simple truss. 
Truss of Problem 6.33a:
Starting with triangle ABD and adding two members at a time, we obtain
successively joints H, G, F, E, I, C, and J, thus completing the truss.
Therefore, this is a simple truss. 
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796
PROBLEM 6.30
Determine whether the trusses of Problems 6.31b, 6.32b, and 6.33b are simple trusses.
SOLUTION
Truss of Problem 6.31b:
Starting with triangle CGM and adding two members at a time, we obtain
successively joints B, L, F, A, K, J, then H, D, N, I, E, O, and P, thus completing
the truss.
Therefore, this truss is a simple truss. 
Truss of Problem 6.32b:
Starting with triangle ABC and adding two members at a time, we obtain
successively joints E, D, F, G, and H, but cannot go further. Thus, this truss
is not a simple truss. 
Truss of Problem 6.33b:
Starting with triangle GFH and adding two members at a time, we obtain
successively joints D, E, C, A, and B, thus completing the truss.
Therefore, this is a simple truss. 
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797
PROBLEM 6.31
each of the two trusses shown.
SOLUTION
Truss (a):
FB: Joint B: FBJ = 0
FB: Joint D: FDI = 0
FB: Joint E: FEI = 0
FB: Joint I : FAI = 0
(a)
FB: Joint F : FFK = 0
FB: Joint G: FGK = 0
FB: Joint K : FCK = 0
AI , BJ , CK , DI , EI , FK , GK 
The zero-force members, therefore, are
Truss (b):
FB: Joint K : FFK = 0
FB: Joint O : FIO = 0
(b)
The zero-force members, therefore, are
FK and IO 
All other members are either in tension or compression.
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798
PROBLEM 6.32
of the two trusses shown.
SOLUTION
Truss (a):
FB: Joint B: FBC = 0
FB: Joint C: FCD = 0
FB: Joint J : FIJ = 0
FB: Joint I : FIL = 0
FB: Joint N : FMN = 0
(a)
FB: Joint M : FLM = 0
BC , CD, IJ , IL, LM , MN 
The zero-force members, therefore, are
Truss (b):
FB: Joint C: FBC = 0
FB: Joint B: FBE = 0
FB: Joint G: FFG = 0
FB: Joint F : FEF = 0
FB: Joint E: FDE = 0
FB: Joint I : FIJ = 0
(b)
FB: Joint M : FMN = 0
FB: Joint N : FKN = 0
BC , BE , DE , EF , FG , IJ , KN , MN 
The zero-force members, therefore, are
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799
PROBLEM 6.33
of the two trusses shown.
SOLUTION
Truss (a):
Note: Reaction at F is vertical ( Fx = 0).
Joint G:
ΣF = 0,
FDG = 0 
Joint D:
ΣF = 0,
FDB = 0 
Joint F :
ΣF = 0,
FFG = 0 
Joint G:
ΣF = 0,
FGH = 0 
Joint J :
ΣF = 0,
FIJ = 0 
Joint I :
ΣF = 0,
FHI = 0 
PROPRIETARY MATERIAL. &copy; 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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800
PROBLEM 6.33 (Continued)
Truss (b):
Joint A:
ΣF = 0,
FAC = 0 
Joint C:
ΣF = 0,
FCE = 0 
Joint E:
ΣF = 0,
FEF = 0 
Joint F :
ΣF = 0,
FFG = 0 
Joint G:
ΣF = 0,
FGH = 0 
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801
PROBLEM 6.34
Determine the zero-force members in the truss of (a) Problem 6.26, (b) Problem 6.28.
SOLUTION
(a)
Truss of Problem 6.26:
FB : Joint I : FIJ = 0
FB : Joint J : FGJ = 0
FB : Joint G: FGH = 0
GH , GJ , IJ 
The zero-force members, therefore, are
(b)
Truss of Problem 6.28:
FB : Joint B: FBF = 0
FB : Joint B: FBG = 0
FB : Joint C: FCG = 0
FB : Joint C: FCH = 0
BF , BG, CG, CH 
The zero-force members, therefore, are
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802
PROBLEM 6.35*
The truss shown consists of six members and is supported
by a short link at A, two short links at B, and a ball and
socket at D. Determine the force in each of the members for
SOLUTION
Free body: Truss:
From symmetry:
Dx = Bx
and
Dy = B y
ΣM z = 0: − A(10 ft) − (400 lb)(24 ft) = 0
A = −960 lb
ΣFx = 0: Bx + Dx + A = 0
2 Bx − 960 lb = 0, Bx = 480 lb
ΣFy = 0: B y + Dy − 400 lb = 0
2 By = 400 lb
By = +200 lb
B = (480 lb)i + (200 lb) j 
Thus,
Free body: C:
FCA = FAC
FCB = FBC
FCD = FCD

CA FAC
=
( −24i + 10 j)
CA 26

CB FBC
=
(−24i + 7k )
CB
25

CD FCD
=
(−24i − 7k )
CD
25
ΣF = 0: FCA + FCB + FCD − (400 lb) j = 0
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803
PROBLEM 6.35* (Continued)
Substituting for FCA , FCB , FCD , and equating to zero the coefficients of i, j, k :
24
24
FAC − ( FBC + FCD ) = 0
26
25
i:
−
j:
10
FAC − 400 lb = 0
26
k:
7
( FBC − FCD ) = 0 FCD = FBC
25
(1)
FAC = 1040 lb T 
Substitute for FAC and FCD in Eq. (1):
−
24
24
(10.40 lb) − (2 FBC ) = 0 FBC = −500 lb
26
25
FBC = FCD = 500 lb C 
Free body: B:
FBC
FBA
FBD

CB
= (500 lb)
= −(480 lb)i + (140 lb)k
CB

BA FAB
= FAB
=
(10 j − 7k )
BA 12.21
= − FBD k
ΣF = 0: FBA + FBD + FBC + (480 lb)i + (200 lb) j = 0
Substituting for FBA , FBD , FBC and equating to zero the coefficients of j and k:
j:
10
FAB + 200 lb = 0 FAB = −244.2 lb
12.21
k: −
7
FAB − FBD + 140 lb = 0
12.21
FBD = −
From symmetry:
7
(−244.2 lb) + 140 lb = +280 lb
12.21
FAB = 244 lb C 
FBD = 280 lb T 
FAD = 244 lb C 
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804
PROBLEM 6.36*
The truss shown consists of six members and is supported by a
ball and socket at B, a short link at C, and two short links at D.
Determine the force in each of the members for P = (−2184 N)j
and Q = 0.
SOLUTION
Free body: Truss:
From symmetry:
Dx = Bx and Dy = By
ΣFx = 0: 2 Bx = 0
Bx = Dx = 0
ΣFz = 0: Bz = 0
ΣM c z = 0: − 2 By (2.8 m) + (2184 N)(2 m) = 0
By = 780 N
B = (780 N) j 
Thus,
Free body: A:

AB FAB
FAB = FAB
=
( −0.8i − 4.8 j + 2.1k )
AB 5.30

AC FAC
=
FAC = FAC
(2i − 4.8 j)
AC 5.20

=
(+0.8i − 4.8 j − 2.1k )
ΣF = 0: FAB + FAC + FAD − (2184 N) j = 0
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805
PROBLEM 6.36* (Continued)
Substituting for FAB , FAC , FAD , and equating to zero the coefficients of i, j, k :
i:
−
0.8
2
( FAB + FAD ) +
FAC = 0
5.30
5.20
(1)
j:
−
4.8
4.8
( FAB + FAD ) −
FAC − 2184 N = 0
5.30
5.20
(2)
2.1
( FAB − FAD ) = 0
5.30
k:
Multiply Eq. (1) by –6 and add Eq. (2):
 16.8 
−
 FAC − 2184 N = 0, FAC = −676 N
 5.20 
FAC = 676 N C 
Substitute for FAC and FAD in Eq. (1):
 0.8 
 2 
−
 2 FAB + 
 (−676 N) = 0, FAB = −861.25 N
 5.30 
 5.20 
Free body: B:
FAB = FAD = 861 N C 

AB
= −(130 N)i − (780 N) j + (341.25 N)k
FAB = (861.25 N)
AB
 2.8i − 2.1k 
FBC = FBC 
 = FBC (0.8i − 0.6k )
3.5


FBD = − FBD k
ΣF = 0: FAB + FBC + FBD + (780 N) j = 0
Substituting for FAB , FBC , FBD and equating to zero the coefficients of i and k ,
i:
k:
−130 N + 0.8FBC = 0 FBC = +162.5 N
341.25 N − 0.6 FBC − FBD = 0
FBD = 341.25 − 0.6(162.5) = +243.75 N
From symmetry:
FBC = 162.5 N T 
FCD = FBC
FBD = 244 N T 
FCD = 162.5 N T 
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806
PROBLEM 6.37*
The truss shown consists of six members and is supported
by a ball and socket at B, a short link at C, and two short
links at D. Determine the force in each of the members for
P = 0 and Q = (2968 N)i.
SOLUTION
Free body: Truss:
From symmetry:
Dx = Bx and Dy = By
ΣFx = 0: 2 Bx + 2968 N = 0
Bx = Dx = −1484 N
ΣM cz ′ = 0: − 2 By (2.8 m) − (2968 N)(4.8 m) = 0
By = −2544 N
B = −(1484 N)i − (2544 N) j 
Thus,
Free body: A:
FAB = FAB

AB
AB
FAB
(−0.8i − 4.8 j + 2.1k )
5.30

AC FAC
= FAC
=
(2i − 4.8 j)
AC 5.20

=
(−0.8i − 4.8 j − 2.1k )
5.30
=
FAC
ΣF = 0: FAB + FAC + FAD + (2968 N)i = 0
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807
PROBLEM 6.37* (Continued)
Substituting for FAB , FAC , FAD , and equating to zero the coefficients of i, j, k ,
0.8
2
FAC + 2968 N = 0
( FAB + FAD ) +
5.30
5.20
(1)
4.8
4.8
FAC = 0
( FAB + FAD ) −
5.30
5.20
(2)
i:
−
j:
−
k:
2.1
( FAB − FAD ) = 0
5.30
Multiply Eq. (1) by –6 and add Eq. (2):
 16.8 
−
 FAC − 6(2968 N) = 0, FAC = −5512 N
 5.20 
FAC = 5510 N C 
Substitute for FAC and FAD in Eq. (2):
 4.8 
 4.8 
−
 2 FAB − 
 (−5512 N) = 0, FAB = +2809 N
 5.30 
 5.20 
FAB = FAD = 2810 N T 
Free body: B:
FAB
FBC
FBD

BA
= (2809 N)
= (424 N)i + (2544 N) j − (1113 N)k
BA
 2.8 i − 2.1k 
= FBC 
 = FBC (0.8i − 0.6k )
3.5


= − FBD k
ΣF = 0: FAB + FBC + FBD − (1484 N)i − (2544 N) j = 0
Substituting for FAB , FBC , FBD and equating to zero the coefficients of i and k ,
i:
+24 N + 0.8FBC − 1484 N = 0, FBC = +1325 N
k:
−1113 N − 0.6 FBC − FBD = 0
From symmetry:
FBC = 1325 N T 
FBD = −1113 N − 0.6(1325 N) = −1908 N,
FBD = 1908 N C 
FCD = FBC
FCD = 1325 N T 
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808
PROBLEM 6.38*
The truss shown consists of nine members and is
supported by a ball and socket at A, two short links
at B, and a short link at C. Determine the force in
SOLUTION
Free body: Truss:
From symmetry:
Az = Bz = 0
ΣFx = 0: Ax = 0
ΣM BC = 0: Ay (6 ft) + (1600 lb)(7.5 ft) = 0
Ay = −2000 lb
A = −(2000 lb)j 
By = C
From symmetry:
ΣFy = 0: 2 By − 2000 lb − 1600 lb = 0
By = 1800 lb
B = (1800 lb) j 
ΣF = 0: FAB + FAC + FAD − (2000 lb) j = 0
Free body: A:
FAB
i+k
+ FAC
2
i−k
2
+ FAD (0.6 i + 0.8 j) − (2000 lb) j = 0
Factoring i, j, k and equating their coefficient to zero,
1
2
FAB +
1
FAC + 0.6 FAD = 0
(1)
0.8FAD − 2000 lb = 0
FAD = 2500 lb T 
2
1
2
FAB −
1
2
FAC = 0
FAC = FAB
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809
PROBLEM 6.38* (Continued)
Substitute for FAD and FAC into Eq. (1):
2
2
Free body: B:
FAB + 0.6(2500 lb) = 0, FAB = −1060.7 lb,
FBA
FBC
FAB = FAC = 1061 lb C 

BA
i+k
= FAB
= + (1060.7 lb)
= (750 lb)(i + k )
BA
2
= − FBC k
FBD = FBD (0.8 j − 0.6k )

BE FBE
(7.5i + 8 j − 6k )
FBE = FBE
=
BE 12.5
ΣF = 0: FBA + FBC + FBD + FBE + (1800 lb) j = 0
Substituting for FBA , FBC , FBD , and FBE and equating to zero the coefficients of i, j, k ,
i:
 7.5 
750 lb + 
 FBE = 0, FBE = −1250 lb
 12.5 
FBE = 1250 lb C 
j:
 8 
0.8 FBD + 
 (−1250 lb) + 1800 lb = 0
 12.5 
FBD = 1250 lb C 
k:
750 lb − FBC − 0.6( −1250 lb) −
6
( −1250 lb) = 0
12.5
FBC = 2100 lb T 
FBD = FCD = 1250 lb C 
From symmetry:
Free body: D:
ΣF = 0: FDA + FDB + FDC + FDE i = 0
We now substitute for FDA , FDB , FDC and equate to zero the coefficient of i. Only FDA contains i and its
coefficient is
−0.6 FAD = −0.6(2500 lb) = −1500 lb
i:
−1500 lb + FDE = 0
FDE = 1500 lb T 
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810
PROBLEM 6.39*
The truss shown consists of nine members and is supported by a ball
and socket at B, a short link at C, and two short links at D. (a) Check
that this truss is a simple truss, that it is completely constrained,
and that the reactions at its supports are statically determinate.
(b) Determine the force in each member for P = (−1200 N)j and Q = 0.
SOLUTION
Free body: Truss:
ΣM B = 0: 1.8i &times; Cj + (1.8i − 3k ) &times; ( D y j + Dk )
+ (0.6i − 0.75k ) &times; (−1200 j) = 0
−1.8 Ck + 1.8 D y k − 1.8 Dz j
+ 3D y i − 720k − 900i = 0
Equate to zero the coefficients of i, j, k :
i:
3Dy − 900 = 0, D y = 300 N
j:
Dz = 0,
D = (300 N) j 
k:
1.8C + 1.8(300) − 720 = 0
C = (100 N) j 
ΣF = 0: B + 300 j + 100 j − 1200 j = 0
B = (800 N) j 
Free body: B:
ΣF = 0: FBA + FBC + FBE + (800 N) j = 0, with

BA FAB
(0.6i + 3j − 0.75k )
FBA = FAB
=
BA 3.15
FBC = FBC i
FBE = − FBE k
Substitute and equate to zero the coefficient of j, i, k :
j:
 3 

 FAB + 800 N = 0, FAB = −840 N,
 3.315 
i:
 0.6 

 (−840 N) + FBC = 0
 3.15 
FBC = 160.0 N T 
k:
 0.75 
−
 (−840 N) − FBE = 0
 3.15 
FBE = 200 N T 
FAB = 840 N C 
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811
PROBLEM 6.39* (Continued)
Free body: C:
ΣF = 0: FCA + FCB + FCD + FCE + (100 N) j = 0, with

F
CA
FCA = FAC
= AC (−1.2i + 3j − 0.75 k )
CA 3.317
FCB = −(160 N) i
FCD = − FCD k FCE = FCE

F
CE
= CE (−1.8i − 3k )
CE 3, 499
Substitute and equate to zero the coefficient of j, i, k :
 3 

 FAC + 100 N = 0, FAC = −110.57 N
 3.317 
j:
−
i:
k:
−
FAC = 110.6 N C 
1.2
1.8
FCE = 0, FCE = −233.3
(−110.57) − 160 −
3.317
3.499
FCE = 233 N C 
0.75
3
(−110.57) − FCD −
(−233.3) = 0
3.317
3.499
FCD = 225 N T 
Free body: D:
ΣF = 0: FDA + FDC + FDE + (300 N) j = 0, with

F
DA
= AD (−1.2i + 3j + 2.25k )
DA 3.937
FDC = FCD k = (225 N)k
FDE = − FDE i
Substitute and equate to zero the coefficient of j, i, k :
j:
 3 

 FAD + 300 N = 0,
 3.937 
i:
 1.2 
−
 ( −393.7 N) − FDE = 0
 3.937 
k:
 2.25 

 (−393.7 N) + 225 N = 0 (Checks)
 3.937 
FAD = 394 N C 
FDE = 120.0 N T 
Free body: E:
Member AE is the only member at E which does not lie in the xz plane.
Therefore, it is a zero-force member.
FAE = 0 
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812
PROBLEM 6.40*
Solve Problem 6.39 for P = 0 and Q = (−900 N)k.
PROBLEM 6.39* The truss shown consists of nine members
and is supported by a ball and socket at B, a short link at C,
and two short links at D. (a) Check that this truss is a simple
truss, that it is completely constrained, and that the reactions
at its supports are statically determinate. (b) Determine the
force in each member for P = (−1200 N)j and Q = 0.
SOLUTION
Free body: Truss:
ΣM B = 0: 1.8i &times; Cj + (1.8i − 3k ) &times; ( D y j + Dz k )
+ (0.6i + 3j − 0.75k ) &times; (−900N)k = 0
1.8 Ck + 1.8D y k − 1.8Dz j
+ 3D y i + 540 j − 2700i = 0
Equate to zero the coefficient of i, j, k :
3Dy − 2700 = 0
Dy = 900 N
−1.8Dz + 540 = 0
Dz = 300 N
1.8C + 1.8 Dy = 0
Thus,
C = − D y = −900 N
C = −(900 N) j D = (900 N) j + (300 N)k
ΣF = 0: B − 900 j + 900 j + 300k − 900k = 0

B = (600 N)k 
Free body: B:
Since B is aligned with member BE,
FAB = FBC = 0,
FBE = 600 N T 
Free body: C:

ΣF = 0: FCA + FCD + FCE − (900 N) j = 0, with
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813
PROBLEM 6.40* (Continued)

FCA
FCD

F
CA
= FAC
= AC ( −1.2i + 3j − 0.75k ) 
CA 3.317

F
CE
= − FCD k FCE = FCE
= CE (−1.8i − 3k )
CE 3.499
Substitute and equate to zero the coefficient of j, i, k:
j:
 3 

 FAC − 900 N = 0,
 3.317 
FAC = 995.1 N
FAC = 995 N T 
FCE = −699.8 N
FCE = 700 N C 
i:
−
1.2
1.8
FCE = 0,
(995.1) −
3.317
3.499
k:
−
0.75
3
(995.1) − FCD −
( −699.8) = 0
3.317
3.499
FCD = 375 N T 
Free body: D:
ΣF = 0: FDA + FDE + (375 N)k +(900 N)j + (300 N) k = 0

F
DA
= AD (−1.2i + 3j + 2.25k )
DA 3.937
with
FDE = − FDE i
and
Substitute and equate to zero the coefficient j, i, k:
j:
 3 

 3.937 
i:
 1.2 
−
 (−1181.1 N) − FDE = 0
 3.937 
k:
FAD = 1181 N C 
FDE = 360 N T 
 2.25 

 (−1181.1 N + 375 N + 300 N = 0)
 3.937 
(Checks)
Free body: E:
Member AE is the only member at E which
does not lie in the xz plane. Therefore, it is
a zero-force member.
FAE = 0 
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814
PROBLEM 6.41*
The truss shown consists of 18 members and is supported by a
ball and socket at A, two short links at B, and one short link at G.
(a) Check that this truss is a simple truss, that it is completely
constrained, and that the reactions at its supports are statically
each of the six members joined at E.
SOLUTION
(a)
Check simple truss.
(2) Add members BD, ED, GD joining at D.
(3) Add members BA, DA, EA joining at A.
(4) Add members DH, EH, GH joining at H.
(5) Add members BC, DC, GC joining at C.
Truss has been completed: It is a simple truss.
Free body: Truss:
Check constraints and reactions.
Six unknown reactions—ok; however, supports at A and
B constrain truss to rotate about AB and support at G
prevents such a rotation. Thus,
Truss is completely constrained and reactions are statically determinate.
Determination of reactions:
ΣM A = 0: 11i &times; ( By j + Bz k ) + (11i − 9.6k ) &times; G j
+ (10.08 j − 9.6k ) &times; (275i + 240k ) = 0
11By k − 11By j + 11Gk + 9.6Gi − (10.08)(275)k
+ (10.08)(240)i − (9.6)(275) j = 0
Equate to zero the coefficient of i, j, k:
i : 9.6G + (10.08)(240) = 0 G = −252 lb
G = (−252 lb) j 
j: − 11Bz − (9.6)(275) = 0 Bz = −240 lb
k : 11By + 11(−252) − (10.08)(275) = 0, By = 504 lb
B = (504 lb) j − (240 lb)k 
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815
PROBLEM 6.41* (Continued)
ΣF = 0: A + (504 lb) j − (240 lb)k − (252 lb) j
+ (275 lb)i + (240 lb)k = 0
A = −(275 lb)i − (252 lb) j 
Zero-force members.
The determination of these members will facilitate our solution.
FB: C: Writing
ΣFx = 0, ΣFy = 0, ΣFz = 0
yields FBC = FCD = FCG = 0 
FB: F: Writing
ΣFx = 0, ΣFy = 0, ΣFz = 0
yields FBF = FEF = FFG = 0 
FB: A: Since
FB: H: Writing
Az = 0, writing ΣFz = 0
ΣFy = 0
yields FDH = 0 
FB: D: Since FAD = FCD = FDH = 0, we need
consider only members DB, DE, and DG.
Since FDE is the only force not contained in plane BDG, it must be
zero. Simple reasonings show that the other two forces are also zero.
FBD = FDE = FDG = 0 
The results obtained for the reactions at the supports and for the zero-force members are shown on the
figure below. Zero-force members are indicated by a zero (“0”).
(b)
Force in each of the members joined at E.
FDE = FEF = 0 
Free body: A:
ΣFy = 0
yields FAE = 252 lb T 
Free body: H:
ΣFz = 0
yields FEH = 240 lb C 
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816
PROBLEM 6.41* (Continued)
ΣF = 0: FEB + FEG + (240 lb)k − (252 lb) j = 0
Free body: E:
F
FBE
(11i − 10.08 j) + EG (11i − 9.6k ) + 240k − 252 j = 0
14.92
14.6
Equate to zero the coefficient of y and k:
 10.08 
j: − 
 FBE − 252 = 0
 14.92 
FBE = 373 lb C 
 9.6 
−
 FEG + 240 = 0
 14.6 
FEG = 365 lb T 
k:
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817
PROBLEM 6.42*
The truss shown consists of 18 members and is supported by a
ball and socket at A, two short links at B, and one short link at G.
(a) Check that this truss is a simple truss, that it is completely
constrained, and that the reactions at its supports are statically
of the six members joined at G.
SOLUTION
See solution to Problem 6.41 for part (a) and for reactions and zero-force members.
(b)
Force in each of the members joined at G.
FCG = FDG = FFG = 0 
Free body: H:
Free body: G:
ΣFx = 0
yields FGH = 275 lb C 
ΣF = 0: FGB + FGE + (275 lb)i − (252 lb) j = 0
FBG
F
(−10.08 j + 9.6k ) + EG (−11i + 9.6k ) + 275i − 252 j = 0
13.92
14.6
Equate to zero the coefficient of i, j, k:
 11 
i: − 
 FEG + 275 = 0
 14.6 
FEG = 365 lb T 
 10.08 
j: − 
 FBG − 252 = 0
 13.92 
FBG = 348 lb C 
 9.6 
 9.6 
k: 
 (−348) + 
 (365) = 0
 13.92 
 14.6 
(Checks)
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818
PROBLEM 6.43
Determine the force in members CD and DF of the truss shown.
SOLUTION
Reactions:
ΣM J = 0: (12 kN)(4.8 m) + (12 kN)(2.4 m) − B(9.6 m) = 0
B = 9.00 kN
ΣFy = 0: 9.00 kN − 12.00 kN − 12.00 kN + J = 0
J = 15.00 kN
Member CD:
ΣFy = 0: 9.00 kN + FCD = 0
FCD = 9.00 kN C 
Member DF:
ΣM C = 0: FDF (1.8 m) − (9.00 kN)(2.4 m) = 0
FDF = 12.00 kN T 
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819
PROBLEM 6.44
Determine the force in members FG and FH of the truss shown.
SOLUTION
Reactions:
ΣM J = 0: (12 kN)(4.8 m) + (12 kN)(2.4 m) − B (9.6 m) = 0
B = 9.00 kN
ΣFy = 0: 9.00 kN − 12.00 kN − 12.00 kN + J = 0
J = 15.00 kN
Member FG:
3
ΣFy = 0: − FFG − 12.00 kN + 15.00 kN = 0
5
FFG = 5.00 kN T 
Member FH:
ΣM G = 0: (15.00 kN)(2.4 m) − FFH (1.8 m) = 0
FFH = 20.0 kN T 
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820
PROBLEM 6.45
A Warren bridge truss is loaded as shown. Determine the force in
members CE, DE, and DF.
SOLUTION
Free body: Truss:
ΣFx = 0: k x = 0
ΣM A = 0: k y (62.5 ft) − (6000 lb)(12.5 ft)
− (6000 lb)(25 ft) = 0
k = k y = 3600 lb 
ΣFy = 0: A + 3600 lb − 6000 lb − 6000 lb = 0
A = 8400 lb 
We pass a section through members CE, DE, and DF and use the
free body shown.
ΣM D = 0: FCE (15 ft) − (8400 lb)(18.75 ft)
+ (6000 lb)(6.25 ft) = 0
FCE = +8000 lb
ΣFy = 0: 8400 lb − 6000 lb −
FCE = 8000 lb T 
15
FDE = 0
16.25
FDE = +2600 lb
FDE = 2600 lb T 
ΣM E = 0: 6000 lb (12.5 ft) − (8400 lb)(25 ft)
− FDF (15 ft) = 0
FDF = −9000 lb
FDF = 9000 lb C 
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821
PROBLEM 6.46
A Warren bridge truss is loaded as shown. Determine the force in
members EG, FG, and FH.
SOLUTION
See solution of Problem 6.45 for free-body diagram of truss and determination of reactions:
A = 8400 lb
k = 3600 lb 
We pass a section through members EG, FG, and FH, and use the free body shown.
ΣM F = 0: (3600 lb)(31.25 ft) − FEG (15 ft) = 0
FEG = +7500 lb
ΣFy = 0:
FEG = 7500 lb T 
15
FFG + 3600 lb = 0
16.25
FFG = −3900 lb
FFG = 3900 lb C 
ΣM G = 0: FFH (15 ft) + (3600 lb)(25 ft) = 0
FFH = −6000 lb
FFH = 6000 lb C 
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822
PROBLEM 6.47
Determine the force in members DF, EF, and EG of the
truss shown.
SOLUTION
tan β =
3
4
Reactions:
A=N=0
Member DF:
3
ΣM E = 0: + (16 kN)(6 m) − FDF (4 m) = 0
5
FDF = + 40 kN
Member EF:
+ ΣF = 0: (16 kN) sin β − FEF cos β = 0
FEF = 16 tan β = 16(0.75) = 12 kN
Member EG:
FDF = 40.0 kN T 
ΣM F = 0: (16 kN)(9 m) +
FEF = 12.00 kN T 
4
FEG (3 m) = 0
5
FEG = −60 kN
FEG = 60.0 kN C 
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823
PROBLEM 6.48
Determine the force in members GI, GJ, and HI of the
truss shown.
SOLUTION
Reactions:
A=N=0
Member GI:
+ ΣF = 0: (16 kN)sin β + FGI sin β = 0
FGI = −16 kN
Member GJ:
ΣM I = 0: − (16 kN)(9 m) −
4
FGJ (3 m) = 0
5
FGJ = −60 kN
Member HI:
ΣM G = 0: − (16 kN)(9 m) +
FGI = 16.00 kN C 
FGJ = 60.0 kN C 
3
FHI (4 m) = 0
5
FHI = +60 kN
FHI = 60.0 kN T 
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824
PROBLEM 6.49
Determine the force in members AD, CD, and CE of the
truss shown.
SOLUTION
Reactions:
ΣM k = 0: 36(2.4) − B(13.5) + 20(9) + 20(4.5) = 0
ΣFx = 0: −36 + K x = 0
B = 26.4 kN
K x = 36 kN
ΣFy = 0: 26.4 − 20 − 20 + K y = 0 K y = 13.6 kN
ΣM C = 0: 36(1.2) − 26.4(2.25) − FAD (1.2) = 0
 8

ΣM A = 0:  FCD  (4.5) = 0
17


FAD = 13.5 kN C 
FCD = 0 
 15

ΣM D = 0:  FCE  (2.4) − 26.4(4.5) = 0
 17

FCE = +56.1 kN
FCE = 56.1 kN T 
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825
PROBLEM 6.50
Determine the force in members DG, FG, and FH of the
truss shown.
SOLUTION
See the solution to Problem 6.49 for free-body diagram and analysis to determine the reactions at the supports
B and K.
B = 26.4 kN ;
K x = 36.0 kN
;
K y = 13.60 kN
ΣM F = 0: 36(1.2) − 26.4(6.75) + 20(2.25) − FDG (1.2) = 0
FDG = −75 kN
FDG = 75.0 kN C 
 8

ΣM D = 0: − 26.4(4.5) +  FFG  (4.5) = 0
17


FFG = +56.1 kN

 15
ΣM G = 0: 20(4.5) − 26.4(9) +  FFH
 17

 (2.4) = 0 

FFH = +69.7 kN 

FFG = 56.1 kN T 
FFH = 69.7 kN T 
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826
PROBLEM 6.51
in members AB, AG, and FG.
SOLUTION
We pass a section through members AB, AG, and FG, and use the free body shown.
ΣM G = 0:
 40

 41 FAB  (6.3 ft) − (1.8 kips)(14 ft)


− (0.9 kips)(28 ft) = 0
FAB = 8.20 kips T 
FAB = +8.20 kips
3

ΣM D = 0: −  FAG  (28 ft) + (1.8 kips)(28 ft)
5

+ (1.8 kips)(14 ft) = 0
FAG = 4.50 kips T 
FAG = +4.50 kips
ΣM A = 0: − FFG (9 ft) − (1.8 kips)(12 ft) − (1.8 kips)(26 ft)
− (0.9 kips)(40 ft) = 0
FFG = 11.60 kips C 
FFG = −11.60 kips
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827
PROBLEM 6.52
in members AE, EF, and FJ.
SOLUTION
We pass a section through members AE, EF, and FJ, and use the free body shown.


8
ΣM F = 0: 
FAE  (9 ft) − (1.8 kips)(12 ft) − (1.8 kips)(26 ft) − (0.9 kips)(40 ft) = 0
 82 + 92



FAE = +17.46 kips
FAE = 17.46 kips T 
ΣM A = 0: − FEF (9 ft) − (1.8 kips)(12 ft) − (1.8 kips)(26 ft) − (0.9 kips)(40 ft) = 0
FEF = −11.60 kips
FEF = 11.60 kips C 
ΣM E = 0: − FFJ (8 ft) − (0.9 kips)(8 ft) − (1.8 kips)(20 ft) − (1.8 kips)(34 ft) − (0.9 kips)(48 ft) = 0
FFJ = −18.45 kips
FFJ = 18.45 kips C 
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828
PROBLEM 6.53
Determine the force in members CD and DF of the truss shown.
SOLUTION
tan α =
5
α = 22.62&deg;
12
sin α =
5
12
cos α =
13
13
Member CD:
ΣM I = 0: FCD (9 m) + (10 kN)(9 m) + (10 kN)(6 m) + (10 kN)(3 m) = 0
FCD = 20.0 kN C 
FCD = −20 kN
Member DF:
ΣM C = 0: ( FDF cos α )(3.75 m) + (10 kN)(3 m) + (10 kN)(6 m) + (10 kN)(9 m) = 0
FDF cos α = −48 kN
 12 
FDF   = −48 kN
 13 
FDF = −52.0 kN
FDF = 52.0 kN C 
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829
PROBLEM 6.54
Determine the force in members CE and EF of the truss shown.
SOLUTION
Member CE:
ΣM F = 0: FCE (2.5 m) − (10 kN)(3 m) − (10 kN)(6 m) = 0
FCE = 36.0 kN T 
FCE = +36 kN
Member EF:
ΣM I = 0: FEF (6 m) + (10 kN)(6 m) + (10 kN)(3 m) = 0
FEF = 15.00 kN C 
FEF = −15 kN
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830
PROBLEM 6.55
The truss shown was designed to support the roof of
force in members FG, EG, and EH.
SOLUTION
Ax = 0,
Ay = O =
1
2
A = O = 4.48 kN 
We pass a section through members FG, EG, and EH, and use the free body shown.
Slope FG = Slope FI =
Slope EG =
1.75 m
6m
5.50 m
2.4 m
ΣM E = 0: (0.6 kN)(7.44 m) + (1.24 kN)(3.84 m)
− (4.48 kN)(7.44 m)
 6

−
FFG  (4.80 m) = 0
6.25


FFG = −5.231 kN
FFG = 5.23 kN C 
ΣM G = 0: FEH (5.50 m) + (0.6 kN)(9.84 m)
+ (1.24 kN)(6.24 m) + (1.04 kN)(2.4 m)
−(4.48 kN)(9.84 m) = 0
ΣFy = 0:
FEH = 5.08 kN T 
5.50
1.75
(−5.231 kN) + 4.48 kN − 0.6 kN − 1.24 kN − 1.04 kN = 0
FEG +
6.001
6.25
FEG = −0.1476 kN
FEG = 0.1476 kN C 
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831
PROBLEM 6.56
The truss shown was designed to support the roof of
force in members KM, LM, and LN.
SOLUTION
O=
1
2
O = 4.48 kN 
We pass a section through KM, LM, LN, and use free body shown.
 3.84

ΣM M = 0: 
FLN  (3.68 m)
 4

+ (4.48 kN − 0.6 kN)(3.6 m) = 0
FLN = −3.954 kN
FLN = 3.95 kN C 
ΣM L = 0: − FKM (4.80 m) − (1.24 kN)(3.84 m)
+ (4.48 kN − 0.6 kN)(7.44 m) = 0
FKM = +5.022 kN
ΣFy = 0:
FKM = 5.02 kN T 
4.80
1.12
(−3.954 kN) − 1.24 kN − 0.6 kN + 4.48 kN = 0
FLM +
6.147
4
FLM = −1.963 kN
FLM = 1.963 kN C 
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832
PROBLEM 6.57
A Polynesian, or duopitch, roof truss is loaded as
shown. Determine the force in members DF, EF,
and EG.
SOLUTION
Free body: Truss:
ΣFx = 0: N x = 0
ΣM N = 0: (200 lb)(8a ) + (400 lb)(7a + 6a + 5a )+(350 lb)(4a ) + (300 lb)(3a + 2a + a ) − A(8a ) = 0
A = 1500 lb 
ΣFy = 0: 1500 lb − 200 lb − 3(400 lb) − 350 lb − 3(300 lb) − 150 lb + N y = 0
N y = 1300 lb N = 1300 lb 
We pass a section through DF, EF, and EG, and use the free body shown.
(We apply FDF at F.)
ΣM E = 0: (200 lb)(18 ft) + (400 lb)(12 ft) + (400 lb)(6 ft) − (1500 lb)(18 ft)


18
FDF  (4.5 ft) = 0
−
 182 + 4.52



FDF = −3711 lb
FDF = 3710 lb C 
ΣM A = 0: FEF (18 ft) − (400 lb)(6 ft) − (400 lb)(12 ft) = 0
FEF = 400 lb T 
FEF = +400 lb
ΣM F = 0: FEG (4.5 ft) − (1500 lb)(18 ft)+(200 lb)(18 ft) + (400 lb)(12 ft) + (400 lb)(6 ft) = 0
FEG = +3600 lb
FEG = 3600 lb T 
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833
PROBLEM 6.58
A Polynesian, or duopitch, roof truss is loaded as
shown. Determine the force in members HI, GI,
and GJ.
SOLUTION
See solution of Problem 6.57 for reactions:
A = 1500 lb ,
N = 1300 lb 
We pass a section through HI, GI, and GJ, and use the free body shown.
(We apply FHI at H.)


6
ΣM G = 0: 
FHI  (8.5 ft) + (1300 lb)(24 ft) − (300 lb)(6 ft)
 2

2
 6 +4

− (300 lb)(12 ft) − (300 lb)(18 ft) − (150 lb)(24 ft) = 0
FHI = −2375.4 lb
FHI = 2375 lb C 
ΣM I = 0: (1300 lb)(18 ft) − (300 lb)(6 ft) − (300 lb)(12 ft)
− (150 lb)(18 ft) − FGJ (4.5 ft) = 0
FGJ = +3400 lb
ΣFx = 0: −
FGJ = 3400 lb T 
4
6
FGI −
(−2375.4 lb) − 3400 lb = 0
2
5
6 + 42
FGI = −1779.4 lb
FGI = 1779 lb C 
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834
PROBLEM 6.59
Determine the force in members DE and DF of the truss shown
when P = 20 kips.
SOLUTION
Reactions:
C = K = 2.5P
7.5
18
β = 22.62&deg;
tan β =
Member DE:
ΣM A = 0: (2.5 P )(6 ft) − FDE (12 ft) = 0
FDE = +1.25 P
For P = 20 kips,
FDE = +1.25(20) = +25 kips
FDE = 25.0 kips T 
Member DF:
ΣM E = 0: P (12 ft) − (2.5 P )(6 ft) − FDF cos β (5 ft) = 0
12 P − 15 P − FDF cos 22.62&deg;(5 ft) = 0
FDF = −0.65 P
For P = 20 kips,
FDF = −0.65(20) = −13 kips
FDF = 13.00 kips C 
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835
PROBLEM 6.60
Determine the force in members EG and EF of the truss shown
when P = 20 kips.
SOLUTION
Reactions:
C = K = 2.5P
7.5
tan α =
6
α = 51.34&deg;
Member EG:
ΣM F = 0: P (18 ft) − 2.5 P (12 ft) − P (6 ft) + FEG (7.5 ft) = 0
FEG = +0.8 P;
For P = 20 kips,
FEG = 16.00 kips T 
FEG = 0.8(20) = +16 kips
Member EF:
ΣM A = 0: 2.5 P (6 ft) − P (12 ft) + FEF sin 51.34&deg;(12 ft) = 0
FEF = −0.320 P;
For P = 20 kips,
FEF = −0.320(20) = −6.4 kips
FEF = 6.40 kips C 
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836
PROBLEM 6.61
Determine the force in members EH and GI of the
truss shown. (Hint: Use section aa.)
SOLUTION
Reactions:
ΣFx = 0: Ax = 0
ΣM P = 0: 12(45) + 12(30) + 12(15) − Ay (90) = 0
A y = 12 kips
ΣFy = 0: 12 − 12 − 12 − 12 + P = 0
P = 24 kips
ΣM G = 0: − (12 kips)(30 ft) − FEH (16 ft) = 0
FEH = −22.5 kips
ΣFx = 0: FGI − 22.5 kips = 0
FEH = 22.5 kips C 
FGI = 22.5 kips T 
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837
PROBLEM 6.62
Determine the force in members HJ and IL of the truss
shown. (Hint: Use section bb.)
SOLUTION
See the solution to Problem 6.61 for free body diagram and analysis to determine the reactions at supports A
and P.
A x = 0; A y = 12.00 kips ;
P = 24.0 kips
ΣM L = 0: FHJ (16 ft) − (12 kips)(15 ft) + (24 kips)(30 ft) = 0
FHJ = −33.75 kips
FHJ = 33.8 kips C 
ΣFx = 0: 33.75 kips − FIL = 0
FIL = +33.75 kips
FIL = 33.8 kips T 
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838
PROBLEM 6.63
Determine the force in members DG and FI of the truss shown. (Hint: Use
section aa.)
SOLUTION
ΣM F = 0: FDG (4 m) − (5 kN)(3 m) = 0
FDG = +3.75 kN
FDG = 3.75 kN T 
ΣFy = 0: − 3.75 kN − FFI = 0
FFI = −3.75 kN
FFI = 3.75 kN C 
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839
PROBLEM 6.64
Determine the force in members GJ and IK of the truss shown. (Hint: Use
section bb.)
SOLUTION
ΣM I = 0: FGJ (4 m) − (5 kN)(6 m) − (5 kN)(3 m) = 0
FGJ = +11.25 kN
FGJ = 11.25 kN T 
ΣFy = 0: − 11.25 kN − FIK = 0
FIK = −11.25 kN
FIK = 11.25 kN C 
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840
PROBLEM 6.65
The diagonal members in the center panels of the truss shown
are very slender and can act only in tension; such members are
known as counters. Determine the forces in the counters that
SOLUTION
Free body: Truss:
ΣFx = 0: Fx = 0
ΣM H = 0: 4.8(3a) + 4.8(2a) + 4.8a − 2.4a − Fy (2a) = 0
Fy = +13.20 kips
F = 13.20 kips 
ΣFy = 0: H + 13.20 kips − 3(4.8 kips) − 2(2.4 kips) = 0
H = +6.00 kips
H = 6.00 kips 
Free body: ABF:
We assume that counter BG is acting.
ΣFy = 0: −
9.6
FBG + 13.20 − 2(4.8) = 0
14.6
FBG = +5.475
FBG = 5.48 kips T 
Since BG is in tension, our assumption was correct.
Free body: DEH:
We assume that counter DG is acting.
ΣFy = 0: −
9.6
FDG + 6.00 − 2(2.4) = 0
14.6
FDG = 1.825 kips T 
FDG = +1.825
Since DG is in tension, O.K.
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841
PROBLEM 6.66
The diagonal members in the center panels of the truss shown
are very slender and can act only in tension; such members are
known as counters. Determine the forces in the counters that
SOLUTION
Free body: Truss:
ΣFx = 0: Fx = 0
ΣM G = 0: − Fy a + 4.8(2a) + 4.8a − 2.4a − 2.4(2a) = 0
Fy = 7.20
F = 7.20 kips 
Free body: ABF:
We assume that counter CF is acting.
ΣFy = 0:
9.6
FCF + 7.20 − 2(4.8) = 0
14.6
FCF = +3.65
FCF = 3.65 kips T 
Since CF is in tension, O.K.
Free body: DEH:
We assume that counter CH is acting.
ΣFy = 0:
9.6
FCH − 2(2.4 kips) = 0
14.6
FCH = +7.30
FCH = 7.30 kips T 
Since CH is in tension, O.K.
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842
PROBLEM 6.67
The diagonal members in the center panels of
the power transmission line tower shown are
very slender and can act only in tension; such
members are known as counters. For the given
counters listed below is acting, (b) the force
in that counter.
Counters CJ and HE
SOLUTION
Free body: Portion ABDFEC of tower.
We assume that counter CJ is acting and show the forces exerted by that counter and by members CH and EJ.
ΣFx = 0:
4
FCJ − 2(1.2 kN)sin 20&deg; = 0 FCJ = +1.026 kN
5
Since CJ is found to be in tension, our assumption was correct. Thus, the answers are
(a) CJ

(b) 1.026 kN T 
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843
PROBLEM 6.68
The diagonal members in the center panels of
the power transmission line tower shown are
very slender and can act only in tension; such
members are known as counters. For the given
counters listed below is acting, (b) the force
in that counter.
Counters IO and KN
SOLUTION
Free body: Portion of tower shown.
We assume that counter IO is acting and show the forces exerted by that counter and by members IN and KO.
ΣFx = 0 :
4
FIO − 4(1.2 kN)sin 20&deg; = 0 FIO = +2.05 kN
5
Since IO is found to be in tension, our assumption was correct. Thus, the answers are
(a) IO

(b) 2.05 kN T 
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844
PROBLEM 6.69
Classify each of the structures shown as completely, partially, or improperly constrained; if completely
constrained, further classify as determinate or indeterminate. (All members can act both in tension and in
compression.)
SOLUTION
Structure (a)
Number of members:
m = 16
Number of joints:
n = 10
Reaction components:
r=4
m + r = 20, 2 n = 20
(a)
m + r = 2n 
Thus,
To determine whether the structure is actually completely constrained and determinate, we must try to find the
reactions at the supports. We divide the structure into two simple trusses and draw the free-body diagram of
each truss.
This is a properly supported simple truss – O.K.
This is an improperly supported simple
truss. (Reaction at C passes through B. Thus,
Eq. ΣM B = 0 cannot be satisfied.)
Structure is improperly constrained. 
Structure (b)
m = 16
n = 10
r=4
m + r = 20, 2n = 20
(b)
m + r = 2n 
Thus,
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845
PROBLEM 6.69 (Continued)
We must again try to find the reactions at the supports dividing the structure as shown.
Both portions are simply supported simple trusses.
Structure is completely constrained and determinate. 
Structure (c)
m = 17
n = 10
r=4
m + r = 21, 2n = 20
(c)
m + r &gt; 2n 
Thus,
This is a simple truss with an extra support which causes reactions (and forces in members) to be
indeterminate.
Structure is completely constrained and indeterminate. 
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846
PROBLEM 6.70
Classify each of the structures shown as completely, partially, or improperly constrained; if completely
constrained, further classify as determinate or indeterminate. (All members can act both in tension and in
compression.)
SOLUTION
Structure (a):
Nonsimple truss with r = 4, m = 16, n = 10,
so m + r = 20 = 2n, but we must examine further.
FBD Sections:
FBD
I:
ΣM A = 0 
T1
II:
ΣFx = 0 
T2
I:
ΣFx = 0

Ax
I:
ΣFy = 0

Ay
II:
ΣM E = 0 
Cy
II:
ΣFy = 0 
Ey
Since each section is a simple truss with reactions determined,
structure is completely constrained and determinate. 
Structure (b):
Nonsimple truss with r = 3, m = 16, n = 10,
so
m + r = 19 &lt; 2n = 20
Structure is partially constrained. 
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847
PROBLEM 6.70 (Continued)
Structure (c):
Simple truss with r = 3, m = 17, n = 10, m + r = 20 = 2n, but the horizontal reaction forces Ax and E x
are collinear and no equilibrium equation will resolve them, so the structure is improperly constrained
and indeterminate.

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848
PROBLEM 6.71
Classify each of the structures shown as completely, partially, or improperly constrained; if completely
constrained, further classify as determinate or indeterminate. (All members can act both in tension and in
compression.)
SOLUTION
Structure (a):
Nonsimple truss with r = 4, m = 12, n = 8 so r + m = 16 = 2n.
Check for determinacy:
One can solve joint F for forces in EF, FG and then solve joint
E for E y and force in DE.
This leaves a simple truss ABCDGH with
r = 3, m = 9, n = 6 so r + m = 12 = 2n
Structure is completely constrained and determinate. 
Structure (b):
Simple truss (start with ABC and add joints alphabetically to complete truss) with r = 4, m = 13, n = 8
so
r + m = 17 &gt; 2n = 16
Constrained but indeterminate 
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849
PROBLEM 6.71 (Continued)
Structure (c):
Nonsimple truss with r = 3, m = 13, n = 8 so r + m = 16 = 2n. To further examine, follow procedure in
part (a) above to get truss at left.
Since F1 ≠ 0 (from solution of joint F),
ΣM A = aF1
≠ 0 and there is no equilibrium.
Structure is improperly constrained. 
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850
PROBLEM 6.72
Classify each of the structures shown as completely, partially, or improperly constrained; if completely
constrained, further classify as determinate or indeterminate. (All members can act both in tension and in
compression.)
SOLUTION
Structure (a)
Number of members:
m = 12
Number of joints:
n=8
Reaction components:
Thus,
r =3
m + r = 15,
2 n = 16
m + r &lt; 2n

Structure is partially constrained. 
Structure (b)
m = 13, n = 8
r =3
m + r = 16, 2n = 16
Thus,
m + r = 2n

To verify that the structure is actually completely constrained and determinate, we observe that it is a simple
truss (follow lettering to check this) and that it is simply supported by a pin-and-bracket and a roller. Thus,
structure is completely constrained and determinate. 
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851
PROBLEM 6.72 (Continued)
Structure (c)
m = 13, n = 8
r=4
m + r = 17, 2n = 16
Thus,
m + r &gt; 2n

Structure is completely constrained and indeterminate. 
This result can be verified by observing that the structure is a simple truss (follow lettering to check this),
therefore it is rigid, and that its supports involve four unknowns.
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852
PROBLEM 6.73
Classify each of the structures shown as completely, partially, or improperly constrained; if completely
constrained, further classify as determinate or indeterminate. (All members can act both in tension and in
compression.)
SOLUTION
Structure (a):
Rigid truss with r = 3, m = 14, n = 8,
so
r + m = 17 &gt; 2n = 16
so completely constrained but indeterminate 
Structure (b):
r = 3, m = 13, n = 8, so r + m = 16 = 2n
so completely constrained and determinate 
Structure (c):
Simple truss with r = 3, m = 13, n = 8, so r + m = 16 = 2n, but horizontal reactions ( Ax and Dx ) are collinear,
so cannot be resolved by any equilibrium equation.
Structure is improperly constrained. 
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853
PROBLEM 6.74
Classify each of the structures shown as completely, partially, or improperly constrained; if completely
constrained, further classify as determinate or indeterminate. (All members can act both in tension and in
compression.)
SOLUTION
Structure (a):
m = 12
No. of members
No. of joints
n = 8 m + r = 16 = 2n
No. of reaction components
r = 4 unknows = equations
FBD of EH:
ΣM H = 0
FDE ; ΣFx = 0
FGH ; ΣFy = 0
Hy
Then ABCDGF is a simple truss and all forces can be determined.
This example is completely constrained and determinate. 
Structure (b):
No. of members
m = 12
No. of joints
n = 8 m + r = 15 &lt; 2n = 16
No. of reaction components
r = 3 unknows &lt; equations
partially constrained 
Note: Quadrilateral DEHG can collapse with joint D moving downward; in (a), the roller at F prevents this
action.
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854
PROBLEM 6.74 (Continued)
Structure (c):
No. of members
m = 13
No. of joints
n = 8 m + r = 17 &gt; 2n = 16
No. of reaction components
r = 4 unknows &gt; equations
completely constrained but indeterminate 
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855
PROBLEM 6.75
Determine the force in member BD and the components of the
reaction at C.
SOLUTION
We note that BD is a two-force member. The force it exerts on ABC, therefore, is directed along line BD.
Free body: ABC:
BD = (24) 2 + (10) 2 = 26 in.
 10

ΣM C = 0: (160 lb)(30 in.) − 
FBD  (16 in.) = 0
 26

FBD = +780 lb
ΣM x = 0: C x +
FBD = 780 lb T 
24
(780 lb) = 0
26
C x = −720 lb
ΣFy = 0: C y − 160 lb +
C x = 720 lb

10
(780 lb) = 0
26
C y = −140.0 lb
C y = 140.0 lb 
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856
PROBLEM 6.76
Determine the force in member BD and the components of the
reaction at C.
SOLUTION
We note that BD is a two-force member. The force it exerts on ABC, therefore, is directed along line BD.
Free body: ABC:
Attaching FBD at D and resolving it into components, we write
ΣM C = 0:
 450

FBD  (240 mm) = 0
(400 N)(135 mm) + 
 510

FBD = −255 N
ΣFx = 0: C x +
FBD = 255 N C 
240
(−255 N) = 0
510
C x = +120.0 N
ΣFy = 0: C y − 400 N +
C x = 120.0 N

450
(−255 N) = 0
510
C y = +625 N
C y = 625 N 
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857
PROBLEM 6.77
Determine the components of all forces acting on member ABCD of the
assembly shown.
SOLUTION
Free body: Entire assembly:
ΣM B = 0: D (120 mm) − (480 N)(80 mm) = 0
D = 320 N 
ΣFx = 0: Bx + 480 N = 0
B x = 480 N

ΣFy = 0: By + 320 N = 0
B y = 320 N 
Free body: Member ABCD:
ΣM A = 0: (320 N)(200 mm) − C (160 mm) − (320 N)(80 mm)
− (480 N)(40 mm) = 0
C = 120.0 N 
ΣFx = 0: Ax − 480 N = 0
A x = 480 N

ΣFy = 0: Ay − 320 N − 120 N + 320 N = 0
A y = 120.0 N 
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858
PROBLEM 6.78
Determine the components of all forces acting on member ABD
of the frame shown.
SOLUTION
Free body: Entire frame:
ΣM D = 0: (300 lb) − (12 ft) + (450 lb)(4 ft) − E (6 ft) = 0
E = +900 lb
E = 900 lb

D x = 900 lb

ΣFx = 0: Dx − 900 lb = 0
ΣFy = 0: Dy − 300 lb − 450 lb = 0
D y = 750 lb 
Free body: Member ABD:
We note that BC is a two-force member and that B is directed along BC.
ΣM A = 0: (750 lb)(16 ft) − (900 lb)(6 ft) − B (8 ft) = 0
B = +825 lb
B = 825 lb 
ΣFx = 0: Ax + 900 lb = 0
Ax = −900 lb
A x = 900 lb

ΣFy = 0: Ay + 750 lb − 825 lb = 0
Ay = +75 lb
A y = 75.0 lb 
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859
PROBLEM 6.79
forces acting on member ABC.
SOLUTION
Free body: Entire frame:
ΣM E = 0: − Ax (4) − (20 kips)(5) = 0
Ax = −25 kips,
A x = 25.0 kips

ΣFy = 0: Ay − 20 kips = 0
Ay = 20 kips
A y = 20.0 kips 
Free body: Member ABC:
Note: BE is a two-force member, thus B is directed along line BE and By =
2
Bx .
5
ΣM C = 0: (25 kips)(4 ft) − (20 kips)(10 ft) + Bx (2 ft) + By (5 ft) = 0
−100 kip ⋅ ft + Bx (2 ft) +
2
Bx (5 ft) = 0
5
Bx = 25 kips
By =
B x = 25.0 kips
2
2
( Bx ) = (25) = 10 kips
5
5

B y = 10.00 kips 
ΣFx = 0: C x − 25 kips − 25 kips = 0
C x = 50 kips
C x = 50.0 kips

ΣFy = 0: C y + 20 kips − 10 kips = 0
C y = −10 kips
C y = 10.00 kips 
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860
PROBLEM 6.80
Solve Problem 6.79 assuming that the 20-kip load is replaced by a
clockwise couple of magnitude 100 kip ⋅ ft applied to member EDC
at Point D.
components of all forces acting on member ABC.
SOLUTION
Free body: Entire frame:
ΣFy = 0: Ay = 0
ΣM E = 0: − Ax (4 ft) − 100 kip ⋅ ft = 0
Ax = −25 kips
A x = 25.0 kips
A = 25.0 kips

B x = 25.0 kips

Free body: Member ABC:
Note: BE is a two-force member, thus B is directed along line BE and By =
2
Bx .
5
ΣM C = 0: (25 kips)(4 ft) + Bx (2 ft) + By (5 ft) = 0
100 kip ⋅ ft + Bx (2 ft) +
2
Bx (5 ft) = 0
5
Bx = −25 kips
By =
2
2
Bx = ( −25) = −10 kips;
5
5
ΣFx = 0: − 25 kips + 25 kips + C x = 0
B y = 10.00 kips 
Cx = 0
ΣFy = 0: +10 kips + C y = 0
C y = −10 kips
C y = 10 kips
C = 10.00 kips 
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861
PROBLEM 6.81
Determine the components of all forces acting on member
ABCD when θ = 0.
SOLUTION
Free body: Entire assembly:
ΣM B = 0: A(200) − (150 N)(500) = 0
A = +375 N
A = 375 N

B x = 375 N

ΣFx = 0: Bx + 375 N = 0
Bx = −375 N
ΣFy = 0: By − 150 N = 0
By = +150 N
B y = 150 N 
Free body: Member ABCD:
We note that D is directed along DE, since DE is a two-force member.
ΣM C = 0: D (300) − (150 N)(100) + (375 N)(200) = 0
D = −200 N
D = 200 N 
ΣFx = 0: C x + 375 − 375 = 0
Cx = 0
ΣFy = 0: C y + 150 − 200 = 0
C y = +50.0 N
C = 50.0 N 
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862
PROBLEM 6.82
Determine the components of all forces acting on member
ABCD when θ = 90&deg;.
SOLUTION
Free body: Entire assembly:
ΣM B = 0: A(200) − (150 N)(200) = 0
A = +150.0 N
A = 150.0 N

ΣFx = 0: Bx + 150 − 150 = 0
Bx = 0
ΣFy = 0: By = 0
B = 0 
Free body: Member ABCD:
We note that D is directed along DE, since DE is a two-force member.
ΣM C = 0: D (300) + (150 N)(200) = 0
D = −100.0 N
D = 100.0 N 
ΣFx = 0: C x + 150 N = 0
C x = −150 N
Cx = 150.0 N

ΣFy = 0: C y − 100 N = 0
C y = +100.0 N
C y = 100.0 N 
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863
PROBLEM 6.83
Determine the components of the reactions at A and E if a 750-N force
directed vertically downward is applied (a) at B, (b) at D.
SOLUTION
Free body: Entire frame:
The following analysis is valid for both parts (a) and (b) since position of load on its line of action is immaterial.
ΣM E = 0: − (750 N)(80 mm) − Ax (200 mm) = 0
Ax = − 300 N
A x = 300 N
ΣFx = 0: E x − 300 N = 0 E x = 300 N E x = 300 N
ΣFy = 0: Ay + E y − 750 N = 0
(a)
(1)
Free body: Member CE:
CE is a two-force member. Thus, the reaction at E must be directed along CE.
Ey
300 N
From Eq. (1):
=
75 mm
250 mm
E y = 90 N
Ay + 90 N − 750 N = 0
Ay = 660 N
Thus, reactions are
(b)
A x = 300 N
,
A y = 660 N 
E x = 300 N
,
E y = 90.0 N 
Free body: Member AC:
AC is a two-force member. Thus, the reaction at A must be directed along AC.
Ay
300 N
=
125 mm
250 mm
Ay = 150 N
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864
PROBLEM 6.83 (Continued)
Ay + E y − 750 N = 0
From Eq. (1):
150 N + E y − 750 N = 0
E y = 600 N E y = 600 N
Thus, reactions are
A x = 300 N
,
E x = 300 N
,
A y = 150.0 N 
E y = 600 N 
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865
PROBLEM 6.84
Determine the components of the reactions at A and E if a 750-N force
directed vertically downward is applied (a) at B, (b) at D.
SOLUTION
Free-body: Entire frame:
The following analysis is valid for both parts (a) and (b) since position of load on its line of action is immaterial.
ΣM E = 0: − (750 N)(240 mm) − Ax (400 mm) = 0
Ax = −450 N
A x = 450 N
ΣFx = 0: E x − 450 N = 0 E x = 450 N E x = 450 N
ΣFy = 0: Ay + E y − 750 N = 0
(a)
(1)
Free body: Member CE:
CE is a two-force member. Thus, the reaction at E must be directed along CE.
Ey
450 N
From Eq. (1):
=
240 mm
; E y = 225 N
480 mm
Ay + 225 − 750 = 0; A y = 525 N
Thus, reactions are
A x = 450 N
,
,
E y = 225 N 
A x = 450 N
,
A y = 150.0 N 
E x = 450 N
,
E x = 450N
(b)
A y = 525 N 
Free body: Member AC:
AC is a two-force member. Thus, the reaction at A must be directed along AC.
Ay
450 N
From Eq. (1):
=
160 mm
480 mm
A y = 150.0 N
Ay + E y − 750 N = 0
150 N + E y − 750 N = 0
E y = 600 N E y = 600 N
Thus, reactions are
E y = 600 N 
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866
PROBLEM 6.85
Determine the components of the reactions at A and E if the frame is
loaded by a clockwise couple of magnitude 36 N &middot; m applied (a) at B,
(b) at D.
SOLUTION
Free body: Entire frame:
The following analysis is valid for both parts (a) and (b) since the point of application of the couple is
immaterial.
ΣM E = 0: −36 N ⋅ m − Ax (0.2 m) = 0
Ax = − 180 N
A x = 180 N
ΣFx = 0: − 180 N + E x = 0
E x = 180 N E x = 180 N
ΣFy = 0: Ay + E y = 0
(a)
(1)
Couple applied at B.
Free body: Member CE:
AC is a two-force member. Thus, the reaction at E must be directed along EC.
Ey
180 N
From Eq. (1):
=
0.075 m
0.25 m
E y = 54 N
Ay + 54 N = 0
Ay = − 54 N A y = 54.0 N
Thus, reactions are
(b)
A x = 180.0 N
,
A y = 54.0 N 
E x = 180.0 N
,
E y = 54.0 N 
Couple applied at D.
Free body: Member AC:
AC is a two-force member. Thus, the reaction at A must be
directed along EC.
Ay
180 N
=
0.125 m
0.25 m
Ay = 90 N
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867
PROBLEM 6.85 (Continued)
From Eq. (1):
Ay + E y = 0
90 N + E y = 0
E y = − 90 N E y = 90 N
Thus, reactions are
A x = 180.0 N
E x = −180.0 N
A y = 90.0 N 
,
,
E y = 90.0 N 
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868
PROBLEM 6.86
Determine the components of the reactions at A and E if the frame is loaded
by a clockwise couple of magnitude 36 N &middot; m applied (a) at B, (b) at D.
SOLUTION
Free body: Entire frame:
The following analysis is valid for both parts (a) and (b) since the point of application of the couple is
immaterial.
ΣM E = 0: −36 N ⋅ m − Ax (0.4 m) = 0
Ax = − 90 N
A x = 90.0 N
ΣFx = 0: −90 +E x = 0
E x = 90 N
E x = 90.0 N
ΣFy = 0: Ay + E y = 0
(a)
(1)
Couple applied at B.
Free body: Member CE:
AC is a two-force member. Thus, the reaction at E must be directed along EC.
Ey
90 N
From Eq. (1):
=
0.24 m
; E y = 45.0 N
0.48 m
Ay + 45 N = 0
Ay = − 45 N A y = 45.0 N
Thus, reactions are
(b)
A x = 90.0 N
,
A y = 45.0 N 
E x = 90.0 N
,
E y = 45.0 N 
Couple applied at D.
Free body: Member AC:
AC is a two-force member. Thus, the reaction at A must be directed along AC.
Ay
90 N
=
0.16 m
; A y = 30 N
0.48 m
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869
PROBLEM 6.86 (Continued)
From Eq. (1):
Ay + E y = 0
30 N + E y = 0
E y = − 30 N E y = 30 N
A x = 90.0 N
Thus, reactions are
E x = − 90.0 N
A y = 30.0 N 
,
,
E y = 30.0 N 
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870
PROBLEM 6.87
Determine all the forces exerted on member AI if the frame is loaded
by a clockwise couple of magnitude 1200 lb &middot; in. applied (a) at Point D,
(b) at Point E.
SOLUTION
Free body: Entire frame:
Location of couple is immaterial.
ΣM H = 0: I (48 in.) − 1200 lb ⋅ in. = 0
I = +25.0 lb
(a) and (b)
I = 25.0 lb 
We note that AB, BC, and FG are two-force members.
Free body: Member AI:
tan α =
(a)
20 5
=
48 12
α = 22.6&deg;
Couple applied at D.
ΣFy = 0: −
ΣM G = 0:
5
A + 25 lb = 0
13
A = +65.0 lb
A = 65.0 lb
12
(65 lb)(40 in.) − C (20 in.) = 0
13
C = +120 lb
ΣFx = 0: −
12
(65 lb) + 120 lb + G = 0
13
G = −60.0 lb
22.6&deg; 
C = 120 lb
G = 60 lb


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871
PROBLEM 6.87 (Continued)
(b)
Couple applied at E.
ΣFy = 0: −
ΣM G = 0:
5
A + 25 lb = 0
13
A = +65.0 lb
A = 65.0 lb
22.6&deg; 
12
(65 lb) + (40 in.) − C (20 in.) − 1200 lb ⋅ in. = 0
13
C = +60.0 lb
C = 60.0 lb
ΣFx = 0: −
12
(65 lb) + 60 lb + G = 0
13

G=0 
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872
PROBLEM 6.88
Determine all the forces exerted on member AI if the frame is
loaded by a 40-lb force directed horizontally to the right and applied
(a) at Point D, (b) at Point E.
SOLUTION
Free body: Entire frame:
Location of 40-lb force on its line of action DE is immaterial.
ΣM H = 0: I (48 in.) − (40 lb)(30 in.) = 0
I = +25.0 lb
(a) and (b)
I = 25.0 lb 
We note that AB, BC, and FG are two-force members.
Free body: Member AI:
tan α =
(a)
20 5
=
48 12
α = 22.6&deg;
Force applied at D.
ΣFy = 0: −
ΣM G = 0:
5
A + 25 lb = 0
13
A = +65.0 lb
12
(65 lb)(40 in.) − C (20 in.) = 0
13
C = +120.0 lb
ΣFx = 0: −
12
(65 lb) + 120 lb + G = 0
13
G = −60.0 lb
A = 65.0 lb
22.6&deg; 
C = 120.0 lb

G = 60.0 lb

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873
PROBLEM 6.88 (Continued)
(b)
Force applied at E.
ΣFy = 0: −
ΣM G = 0:
5
A + 25 lb = 0
13
A = +65.0 lb
A = 65.0 lb
22.6&deg; 
12
(65 lb)(40 in.) − C (20 in.) − (40 lb)(10 in.) = 0
13
C = +100.0 lb
C = 100.0 lb
ΣFx = 0: −

12
(65 lb) + 100 lb + 40 lb + G = 0
13
G = −80.0 lb
G = 80.0 lb

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874
PROBLEM 6.89
Determine the components of the reactions at A and B, (a) if the 500-N load is
applied as shown, (b) if the 500-N load is moved along its line of action and is
applied at Point F.
SOLUTION
Free body: Entire frame:
Analysis is valid for either parts (a) or (b), since position of 100-lb load on its line of action is immaterial.
ΣM A = 0: By (10) − (100 lb)(6) = 0 By = + 60 lb
ΣFy = 0: Ay + 60 − 100 = 0
Ay = + 40 lb
ΣFx = 0: Ax + Bx = 0
(a)
(1)
Free body: Member AC:
Since AC is a two-force member, the reaction at A must be directed along CA. We have
Ax
40 lb
=
10 in. 5 in.
From Eq. (1):
A x = 80.0 lb
,
A y = 40.0 lb 
B x = 80.0 lb
,
B y = 60.0 lb 
− 80 + Bx = 0 Bx = + 80 lb
Thus,
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875
PROBLEM 6.89 (Continued)
(b)
Free body: Member BCD:
Since BCD is a two-force member (with forces applied at B and C only),
the reaction at B must be directed along CB. We have, therefore,
Bx = 0
The reaction at B is
From Eq. (1):
The reaction at A is
B y = 60.0 lb 
Bx = 0
Ax + 0 = 0
Ax = 0
A y = 40.0 lb 
Ax = 0
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reproduced or distributed in any form or by any means, without the prior written permission of the publisher, ```