CHAPTER 2
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1
600 N
45°
900 N
30°
PROBLEM 2.1
Two forces are applied as shown to a hook. 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 = 1391 kN, a = 47.8
R = 1391 N
47.8 ◀
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2
PROBLEM 2.2
800 lb
Two forces are applied as shown to a bracket support. Determine
graphically the magnitude and direction of their resultant using
(a) the parallelogram law, (b) the triangle rule.
60°
35°
500 lb
SOLUTION
(a)
Parallelogram law:
(b)
Triangle rule:
We measure:
R = 906 lb, a = 26.6
R = 906 lb
26.6 ◀
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3
PROBLEM 2.3
Two forces P and Q are applied as shown at Point A of a hook support. Knowing that
P = 75 N and Q = 125 N, 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 = 179 N, a = 75.1
R = 179 N
75.1
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4
PROBLEM 2.4
Two forces P and Q are applied as shown at Point A of a hook support. Knowing that
P = 60 lb and Q = 25 lb, 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 = 77.1 lb, a = 85.4
R = 77.1 lb
85.4
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5
P
120 N
25°
α
PROBLEM 2.5
A stake is being pulled out of the ground by means of two ropes as shown.
Knowing that α = 30°, 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:
120 N
=
P
sin 25
(a)
sin30
(b)
30 + b + 25 = 180
P = 101.4 N ◀
b = 180 - 25 - 30
= 125
120 N
sin 30
=
R
sin125
R = 196.6 N ◀
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6
A
15°
PROBLEM 2.6
25°
T1
T2
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.
B
SOLUTION
Using the triangle rule and the law of sines:
(a)
75 + 40 + a = 180
a = 180 - 75 - 40
= 65
800 lb
sin 65
(b)
800 lb
sin 65
=
T2
sin 75
T 2 = 853 lb ◀
=
R
sin 40
R = 567 lb ◀
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7
A
15°
PROBLEM 2.7
25°
T1
T2
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.
B
SOLUTION
Using the triangle rule and the law of sines:
(a)
75 + 40 + b = 180
b = 180 - 75 - 40
= 65
(b)
T1
1000 lb
=
sin 75° sin 65
T1 = 938 lb ◀
1000 lb
R
=
sin 75° sin 40
R = 665 lb ◀
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8
PROBLEM 2.8
B
A
30°
α
C
A disabled automobile is pulled by means of two ropes as
shown. The tension in rope AB is 2.2 kN, and the angle α is
25. Knowing that the resultant of the two forces applied at
A is directed along the axis of the automobile, determine by
trigonometry (a) the tension in rope AC, (b) the magnitude
of the resultant of the two forces applied at A.
SOLUTION
Using the law of sines:
TA C
R
2.2 kN
=
=
sin 30° sin125 sin 25
TAC = 2.603 kN
R = 4.264 kN
(a)
T A C = 2.60 kN ◀
(b)
R = 4.26 kN ◀
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9
PROBLEM 2.9
B
A
30°
α
A disabled automobile is pulled by means of two ropes as
shown. Knowing that the tension in rope AB is 3 kN,
determine by trigonometry the tension in rope AC and the
value of α so that the resultant force exerted at A is a
4.8-kN force directed along the axis of the automobile.
C
SOLUTION
Using the law of cosines:
TAC 2 = (3 kN)2 + (4.8 kN)2 - 2(3 kN)(4.8 kN)cos 30°
TAC = 2.6643 kN
Using the law of sines:
sin a
sin 30
=
3 kN 2.6643 kN
a = 34.3
TA C = 2.66 kN
34.3 ◀
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10
PROBLEM 2.10
50 N
25°
α
P
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 a 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 a
=
sin 25
50 N
35 N
sin a = 0.60374
a = 37.138 
(b)
a = 37.1  ◀
a + b + 25 = 180
b = 180 - 25 - 37.138
= 117.862
R
35 N
=
sin117.862 sin 25
R = 73.2 N ◀
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11
30°
A
PROBLEM 2.11
P
425 lb
α
A steel tank is to be positioned in an excavation. Knowing
that a = 20°, 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)
b + 50 + 60 = 180
b = 180 - 50 - 60
= 70
(b)
425 lb
P
=
sin 70 sin 60
P = 392 lb ◀
425 lb
R
=
sin 70 sin 50
R = 346 lb ◀
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12
30°
A
PROBLEM 2.12
P
425 lb
α
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 a 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)
(a + 30) + 60 + b = 180
b = 180 - (a + 30) - 60
b = 90 - a
sin (90 - a )
425 lb
=
sin 60
500 lb
90 - a = 47.402
(b)
R
500 lb
=
sin (42.598 + 30) sin 60
a = 42.6 ◀
R = 551 lb ◀
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13
P
425 lb
30°
A
α
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) cos30
(b)
R = (425 lb)sin 30
P = 368 lb
◀
R = 213 lb ◀
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14
PROBLEM 2.14
50 N
25°
α
P
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
P = 21.1 N ◀
(b)
R = (50 N) cos 25
R = 45.3 N ◀
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15
PROBLEM 2.15
The barge B is pulled by two tugboats A and C. At a given
instant the tension in cable AB is 4500 lb and the tension
in cable BC is 2000 lb. Determine by trigonometry the
magnitude and direction of the resultant of the two forces
applied at B at that instant.
SOLUTION
Using the law of cosines:
b = 180 - 30 - 45
b = 105
R 2 = (4500 lb) + (2000 lb) - 2 (4500 lb)(2000 lb) cos 105
2
2
R = 5380 lb
Using the law of sines:
2000 lb
R
=
sin b sin (30 - a)
5380 lb
2000 lb
=
sin105 sin (30 - a)
a = 8.94 
R = 5380 lb
8.94 ◀
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16
PROBLEM 2.16
600 N
45°
900 N
30°
Solve Prob. 2.1 by trigonometry.
PROBLEM 2.1
Two forces are applied as shown to a hook. Determine graphically the
magnitude and direction of their resultant using (a) the parallelogram law,
(b) the triangle rule.
SOLUTION
Using the law of cosines:
R 2 = (900 N)2 + (600 N )2
- 2(900 N )(600 N)cos(135)
R = 1390.57N
Using the law of sines:
sin(a - 30) sin (135)
=
600N
1390.57N
a - 30 = 17.7642
a = 47.764
R = 1391N
47.8 ◀
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17
PROBLEM 2.17
Solve Problem 2.4 by trigonometry.
PROBLEM 2.4 Two forces P and Q are applied as shown at Point A of
a hook support. Knowing that P = 60 lb and Q = 25 lb, determine
graphically the magnitude and direction of their resultant using (a) the
parallelogram law, (b) the triangle rule.
SOLUTION
Using the triangle rule and the law of cosines:
20 + 35 + a = 180
a = 125
R 2 = P 2 + Q 2 - 2 PQ cos a
R 2 = (60 lb)2 + (25 lb)2
- 2(60 lb)(25 lb)cos125
R 2 = 3600 + 625 + 3000(0.5736)
R = 77.108 lb
Using the law of sines:
sin125
sin b
=
25 lb 77.108 lb
b = 15.402
70 + b = 85.402
R = 77.1 lb
85.4 ◀
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18
P
120 N
25°
α
PROBLEM 2.18
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°, 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)cos25
P = 72.096 N
And
sin a
sin 25
=
120 N 72.096 N
sin a = 0.70343
a = 44.703
P = 72.1 N
44.7 ◀
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19
PROBLEM 2.19
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 10 kN in member A and 15 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
g = 180 - (40 + 20)
= 120
R 2 = (10 kN)2 + (15 kN)2
Then
- 2(10 kN)(15 kN)cos120
= 475 kN2
R = 21.794 kN
15 kN 21.794 kN
=
sin a
sin120
æ 15 kN ö÷
sin a = çç
÷ sin120
çè 21.794 kN ÷÷ø
= 0.59605
a = 36.588
and
Hence:
f = a + 50 = 86.588
R = 21.8 kN
86.6°
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20
PROBLEM 2.20
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
g = 180 - (40 + 20)
= 120
R 2 = (15 kN)2 + (10 kN)2
Then
- 2(15 kN)(10 kN)cos120
= 475 kN2
R = 21.794 kN
10 kN 21.794 kN
=
sin a
sin120
æ 10 kN ö÷
sin a = çç
÷ sin120
çè 21.794 kN ø÷÷
= 0.39737
a = 23.414
and
Hence:
f = a + 50 = 73.414
R = 21.8 kN
73.4°
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21
PROBLEM 2.21
y
28 in.
96 in.
Determine the x and y components of each of the forces shown.
84 in.
50 lb
80 in.
29 lb
O
x
51 lb
90 in.
48 in.
SOLUTION
Compute the following distances:
OA = (84)2 + (80)2
= 116 in.
OB = (28)2 + (96)2
= 100 in.
OC = (48)2 + (90)2
= 102 in.
29-lb Force:
50-lb Force:
Fx = +(29 lb)
84
116
Fx = + 21.0 lb ◀
Fy = +(29 lb)
80
116
Fy = +20.0 lb ◀
Fx = -(50 lb)
28
100
Fx = - 14.00 lb ◀
96
100
Fy = + 48.0 lb ◀
Fx = +(51 lb)
48
102
Fx = + 24.0 lb ◀
Fy = -(51 lb)
90
102
Fy = -45.0 lb ◀
Fy = +(50 lb)
51-lb Force:
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22
PROBLEM 2.22
y
Determine the x and y components of each of the forces shown.
800
Dimensions
in mm
800 N
600
O
424 N
408 N
x
900
560
480
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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23
PROBLEM 2.23
Determine the x and y components of each of the forces shown.
SOLUTION
350-N Force:
800-N Force:
600-N Force:
Fx = +(350 N)cos25
Fx = + 317 N ◀
Fy = +(350 N)sin 25
Fy = +147.9 N ◀
Fx = +(800 N)cos70
Fx = + 274 N ◀
Fy = +(800 N)sin 70
Fy = +752 N ◀
Fx = -(600 N)cos60
Fx = - 300 N ◀
Fy = +(600 N)sin 60
Fy = +520 N ◀
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24
PROBLEM 2.24
Determine the x and y components of each of the forces shown.
SOLUTION
80-lb Force:
120-lb Force:
150-lb Force:
Fx = +(80 lb)cos30
Fx = + 69.3 lb ◀
Fy = -(80 lb)sin30
Fy = -40.0 lb ◀
Fx = +(120 lb)cos75
Fx = + 31.1 lb ◀
Fy = -(120 lb)sin 75
Fy = -115.9 lb ◀
Fx = -(150 lb)cos 40
Fx = - 114.9 lb ◀
Fy = -(150 lb)sin 40
Fy = -96.4 lb ◀
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25
PROBLEM 2.25
A
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.
C
720 mm
B
650 mm
SOLUTION
BC = (650 mm)2 + (720 mm)2
= 970 mm
æ 650 ö÷
Px = P çç
÷
çè 970 ÷÷ø
(a)
or
æ 970 ö÷
÷
P = Px çç
çè 650 ÷÷ø
æ 970 ö÷
= 325 N çç
÷
çè 650 ÷÷ø
= 485 N
P = 485 N ◀
(b)
æ 720 ö÷
Py = P çç
÷
èç 970 ÷ø÷
æ 720 ÷ö
= 485 N çç
÷
çè 970 ÷÷ø
= 360 N
Py = 970 N ◀
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26
D
C
35°
PROBLEM 2.26
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.
B
A
Q
SOLUTION
P sin 35 = 300 lb
(a)
P=
(b)
Vertical component
300 lb
sin 35
P = 523 lb ◀
Pv = P cos 35 
= (523 lb)cos35
Pv = 428 lb ◀
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27
PROBLEM 2.27
B
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.
M
45°
A
30°
C
SOLUTION
180 = 45 + a + 90 + 30
a = 180 - 45 - 90 - 30
= 15
(a)
Px
P
P
P= x
cos a
600 N
=
cos15
= 621.17 N
cos a =
P = 621 N ◀
(b)
tan a =
Py
Px
Py = Px tan a
= (600 N) tan15
= 160.770 N
Py = 160.8 N ◀
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28
C
PROBLEM 2.28
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.
55°
A
B
Q
SOLUTION
(a)
P=
=
Py
cos 55
350 lb
cos 55
= 610.21 lb
(b)
P = 610 lb ◀
Px = P sin 55
= (610.21 lb)sin 55
= 499.85 lb
Px = 500 lb ◀
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29
PROBLEM 2.29
A
B
Q
60°
C
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.
50°
D
SOLUTION
(a)
750 N = P sin 20
P = 2192.9 N
(b)
P = 2190 N ◀
PA B C = P cos 20 
= (2192.9 N)cos20
PA B C = 2060 N ◀
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30
PROBLEM 2.30
A
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.
B
7m
C
D
2.4 m
SOLUTION
(a)
37
P
12 x
37
= (720 N)
12
= 2220 N
P=
P = 2.22 kN ◀
(b)
35
P
12 x
35
= (720 N)
12
= 2100 N
Py =
Py = 2.10 kN ◀
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31
PROBLEM 2.31
y
28 in.
96 in.
Determine the resultant of the three forces of Problem 2.21.
84 in.
PROBLEM 2.21 Determine the x and y components of each of the
forces shown.
50 lb
80 in.
29 lb
O
x
51 lb
90 in.
48 in.
SOLUTION
Components of the forces were determined in Problem 2.21:
Force
x Comp. (lb)
y Comp. (lb)
29 lb
+21.0
+20.0
50 lb
-14.00
+48.0
51 lb
+24.0
-45.0
R x = + 31.0
R y = + 23.0
R = Rx i + Ry j
= (31.0 lb)i + (23.0 lb) j
Ry
tan a =
Rx
23.0
=
31.0
a = 36.573
R=
23.0 lb
sin (36.573)
R = 38.6 lb
= 38.601 lb
36.6 ◀
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32
PROBLEM 2.32
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)
350 N
+317
+147.9
800 N
+274
+752
600 N
-300
+520
R x = + 291
R y = +1419.9
R = R x i + Ry j
= (291 N)i + (1419.9 N) j
tan a =
Ry
Rx
1419.9 N
=
291 N
a = 78.418
R=
1419.9 N
sin (78.418)
R = 1449 N
= 1449 N
78.4 ◀
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33
PROBLEM 2.33
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)
80 lb
+69.3
-40.0
120 lb
+31.1
-115.9
150 lb
-114.9
-96.4
R x = - 14.50
Ry = -252.3
R = Rx i + Ry j
= -(14.50 lb)i - (252.3 lb) j
Rx
Ry
-14.50
=
-252.3
b = 3.289
tan b =
R=
252.3 lb
cos(3.289)
R = 253 lb
= 253 lb
86.7 ◀
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34
PROBLEM 2.34
y
Determine the resultant of the three forces of Problem 2.22.
800
Dimensions
in mm
PROBLEM 2.22 Determine the x and y components of each of the
forces shown.
800 N
600
O
424 N
408 N
x
900
560
480
SOLUTION
Components of the forces were determined in Problem 2.22:
Force
x Comp. (N)
y Comp. (N)
800 lb
+640
+480
424 lb
-224
-360
408 lb
+192
-360
R x = + 608
R y = - 240
R = Rx i + R y j
= (608 lb)i + (-240 lb) j
tan a =
Ry
Rx
240
=
608
a = 21.541
R=
240 N
sin(21.541°)
= 653.65 N
R = 654 N
21.5 ◀
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35
PROBLEM 2.35
Knowing that a = 35°, determine the resultant of the three
forces shown.
α
α
30°
100 N
200 N
150 N
SOLUTION
100-N Force:
Fx = +(100 N) cos35 = +81.915 N
Fy = -(100 N)sin 35 = -57.358 N
150-N Force:
Fx = +(150 N) cos65 = +63.393 N
Fy = -(150 N)sin 65 = -135.946 N
200-N Force:
Fx = -(200 N) cos35 = -163.830 N
Fy = -(200 N)sin 35 = -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
R y = -308.02
R x = - 18.522
R = Rx i + R y j
= (-18.522 N)i + (-308.02 N) j
Ry
tan a =
Rx
308.02
=
18.522
a = 86.559
R=
308.02 N
sin 86.559
R = 309 N
86.6 ◀
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36
PROBLEM 2.36
Knowing that the tension in cable BC is 725 N, determine the
resultant of the three forces exerted at Point B of beam AB.
SOLUTION
Cable BC Force:
500-N Force:
840
= -525 N
1160
840
Fy = (725 N)
= 500 N
1160
Fx = -(725 N)
3
Fx = -(500 N) = -300 N
5
4
Fy = -(500 N) = -400 N
5
12
= 720 N
13
5
Fy = -(780 N) = -300 N
13
780-N Force:
Fx = (780 N)
and
Rx = SFx = -105 N
Ry = SFy = -200 N
R = (-105 N)2 + (-200 N)2
= 225.89 N
Further:
tan a =
200
105
a = tan-1
= 62.3
200
105
R = 226 N
Thus:
62.3
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37
PROBLEM 2.37
80 lb
120 lb
α
α
60 lb
a'
Knowing that a = 40°, determine the resultant of the three
forces shown.
20°
a
SOLUTION
60-lb Force:
Fx = (60 lb) cos 20 = 56.382 lb
Fy = (60 lb)sin 20 = 20.521 lb
80-lb Force:
Fx = (80 lb) cos60 = 40.000 lb
Fy = (80 lb)sin 60 = 69.282 lb
120-lb Force:
Fx = (120 lb) cos30 = 103.923 lb
Fy = -(120 lb)sin 30 = -60.000 lb
and
R x = SFx = 200.305 lb
R y = SFy = 29.803 lb
R = (200.305 lb)2 + (29.803 lb)2
= 202.510 lb
Further:
tan a =
29.803
200.305
a = tan-1
= 8.46
29.803
200.305
R = 203 lb
8.46 ◀
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38
PROBLEM 2.38
80 lb
120 lb
α
α
Knowing that a = 75°, determine the resultant of the three
forces shown.
60 lb
a'
20°
a
SOLUTION
60-lb Force:
Fx = (60 lb) cos 20 = 56.382 lb
Fy = (60 lb)sin 20 = 20.521 lb
80-lb Force:
Fx = (80 lb) cos 95  = -6.9725 lb
Fy = (80 lb)sin 95 = 79.696 lb
120-lb Force:
Fx = (120 lb) cos 5  = 119.543 lb
Fy = (120 lb)sin 5 = 10.459 lb
Then
R x = SFx = 168.953 lb
R y = SFy = 110.676 lb
and
R = (168.953 lb)2 + (110.676 lb)2
= 201.976 lb
110.676
168.953
tan a = 0.65507
tan a =
a = 33.228
R = 202 lb
33.2 ◀
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39
PROBLEM 2.39
A collar that can slide on a vertical rod is subjected to
the three forces shown. Determine (a) the required
value of α if the resultant of the three forces is to be
horizontal, (b) the corresponding magnitude of the
resultant.
SOLUTION
(a)
Rx = SFx
= (85 N) cos a + (170 N)sin (a )
(1)
Ry = SFy
= +(110 N) + (85 N)sin (a ) - (170 N) cos a
(2)
For R to be horizontal, we must have R y = 0. We make R y = 0 in Eq. (2):
110 + 85sin a - 170 cos a = 0
22 + 17sin a - 34 cos a = 0
17sin a + 22 = -34 1 - sin 2 a
(
289sin 2 a + 748sin a + 484 = 1156 1 - sin 2 a
)
1445sin a + 748sin a - 672 = 0
2
Solving by use of the quadratic formula:
sin a = 0.47059
a = 28.072
(b)
a = 28.1 ◀
Since R = Rx using Eq. (1):
R = 85cos28.072 + 170sin 28.072
= 155.0 N
R = 155.0 N ◀
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40
PROBLEM 2.40
For the beam of Problem 2.36, determine (a) the required
tension in cable BC if the resultant of the three forces exerted at
Point B is to be vertical, (b) the corresponding magnitude of the
resultant.
SOLUTION
Rx = SFx = Rx = -
21
T + 420 N
29 BC
Ry = SFy =
Ry =
(a)
(1)
800
5
4
TBC - (780 N) - (500 N)
1160
13
5
20
T - 700 N
29 BC
(2)
For R to be vertical, we must have Rx = 0
Set Rx = 0 in Eq. (1)
(b)
840
12
3
TBC + (780 N) - (500 N)
1160
13
5
-
21
T + 420 N = 0
29 BC
TBC = 580 N
Substituting for TBC in Eq. (2):
20
(580 N) - 700 N
29
Ry = -300 N
Ry =
R = | Ry | = 300 N
R = 300 N
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41
PROBLEM 2.41
A
65°
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.
C
25°
35°
75 lb
50 lb
B
SOLUTION
Using the x and y axes shown:
(a)
R x = SFx = T A C sin10 + (50 lb) cos35 + (75 lb) cos60
= T A C sin10 + 78.458 lb
(1)
R y = SFy = (50 lb)sin 35 + (75 lb)sin 60 - T A C cos10
R y = 93.631 lb - T A C cos10
(2)
Set R y = 0 in Eq. (2):
93.631 lb - TA C cos10 = 0
T A C = 95.075 lb
(b)
T A C = 95.1 lb ◀
Substituting for T A C in Eq. (1):
R x = (95.075 lb)sin10 + 78.458 lb
= 94.968 lb
R = Rx
R = 95.0 lb ◀
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42
120 lb
α
80 lb
PROBLEM 2.42
α
For the block of Problems 2.37 and 2.38, determine (a) the
required value of a if the resultant of the three forces shown is
to be parallel to the incline, (b) the corresponding magnitude of
the resultant.
60 lb
a'
20°
a
SOLUTION
Select the x axis to be along a a¢.
Then
R x = S Fx = (60 lb) + (80 lb) cos a + (120 lb) sin a
(1)
R y = SFy = (80 lb)sin a - (120 lb) cos a
(2)
and
(a)
Set R y = 0 in Eq. (2).
(80 lb)sin a - (120 lb) cos a = 0
Dividing each term by cosa gives:
(80 lb) tan a = 120 lb
120 lb
tana =
80 lb
a = 56.310
(b)
a = 56.3 ◀
Substituting for a in Eq. (1) gives:
R x = 60 lb + (80 lb) cos 56.31 + (120 lb)sin 56.31 = 204.22 lb
R x = 204 lb ◀
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43
PROBLEM 2.43
A
50°
30°
B
Two cables are tied together at C and are loaded as shown.
Determine the tension (a) in cable AC, (b) in cable BC.
C
400 lb
SOLUTION
Free-Body Diagram
Force Triangle
Law of sines:
TAC
T
400 lb
= BC =
sin 60 sin 40 sin 80
(a)
TAC =
400 lb
(sin 60)
sin 80
TAC = 352 lb ◀
(b)
TBC =
400 lb
(sin 40)
sin80
T B C = 261 lb ◀
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44
PROBLEM 2.44
A
Two cables are tied together at C and are loaded as shown. Knowing that
α = 30°, determine the tension (a) in cable AC, (b) in cable BC.
6 kN
55°
C
α
B
SOLUTION
Free-Body Diagram
Force Triangle
Law of sines:
TA C
T
6 kN
= BC =
sin 60 sin 35 sin 85
(a)
TA C =
6 kN
(sin 60)
sin85
T A C = 5.22 kN ◀
(b)
TBC =
6 kN
(sin 35)
sin85
T BC = 3.45 kN ◀
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45
PROBLEM 2.45
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 = (120 kg) 9.81 m/s2 = 1177 N
Law of sines:
TAC
T
1.177 kN
= BC =
sin130 sin 30
sin 20
(a)
TAC =
1.177 kN
(sin130)
sin 20
T AC = 2.64 kN ◀
(b)
TBC =
1.177 kN
(sin 30)
sin 20
TBC = 1.721 kN ◀
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46
A
25°
45°
B
C
α
PROBLEM 2.46
Two cables are tied together at C and are loaded as shown.
Knowing that P = 500 N and a = 60°, determine the tension in
(a) in cable AC, (b) in cable BC.
P
SOLUTION
Free-Body Diagram
Law of sines:
Force Triangle
TA C
T
500 N
= BC =
sin 35 sin 75 sin 70°
(a)
TA C =
(b)
TBC =
500 N
sin 70
500 N
sin 70
sin 35
T A C = 305 N ◀
sin 75
T B C = 514 N ◀
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47
PROBLEM 2.47
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 = (50 kg) 9.81 m/s2 = 490 N
Law of sines:
TAC
T
490 N
= BC =
sin110 sin 30 sin 40°
(a)
TAC =
(b)
TBC =
490 N
sin 40
490 N
sin 40
sin110
T AC = 716 N ◀
sin 30
TBC = 381 N ◀
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48
A
PROBLEM 2.48
5°
Knowing that a = 20, determine the tension (a) in cable
AC, (b) in rope BC.
C
1200 lb
α
B
SOLUTION
Free-Body Diagram
Law of sines:
Force Triangle
TA C
T
1200 lb
= BC =
sin 110 sin 5 sin 65
(a)
TA C =
1200 lb
sin 110
sin 65
(b)
T BC =
1200 lb
sin 5
sin 65
T A C = 1244 lb ◀
T B C = 115.4 lb ◀
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49
A
B
Two cables are tied together at C and are loaded as shown. Knowing that
P = 300 N, determine the tension in cables AC and BC.
30° 30°
200 N
C
PROBLEM 2.49
45°
P
SOLUTION
Free-Body Diagram
SFx = 0
- TCA sin 30 + TCB sin 30 - P cos 45 - 200N = 0
For P = 200N we have,
- 0.5TCA + 0.5TCB + 212.13 - 200 = 0 (1)
SFy = 0
TCA cos30 - TCB cos30 - P sin 45 = 0
0.86603TCA + 0.86603TCB - 212.13 = 0
(2)
TCA = 134.6 N ◀
Solving equations (1) and (2) simultaneously gives,
TCB = 110.4 N ◀
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50
A
B
Two cables are tied together at C and are loaded as shown. Determine the
range of values of P for which both cables remain taut.
30° 30°
200 N
C
PROBLEM 2.50
45°
P
SOLUTION
Free-Body Diagram
SFx = 0
- TCA sin 30 + TCB sin 30 - P cos 45 - 200N = 0
For TCA = 0 we have,
0.5TCB + 0.70711P - 200 = 0 (1)
SFy = 0
TCA cos30 - TCB cos30 - P sin 45 = 0 ; again setting TCA = 0 yields,
0.86603TCB - 0.70711P = 0
Adding equations (1) and (2) gives,
(2)
1.36603TCB = 200 hence T CB = 146.410N and P = 179.315N
Substituting for TCB = 0 into the equilibrium equations and solving simultaneously gives,
- 0.5TCA + 0.70711P - 200 = 0
0.86603TCA - 0.70711P = 0
And TCA = 546.40N, P = 669.20N Thus for both cables to remain taut, load P must be within the
range of 179.315 N and 669.20 N.
179.3 N < P < 669 N ◀
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51
PROBLEM 2.51
Two forces P and Q are applied as shown to an aircraft
connection. Knowing that the connection is in
equilibrium and that P = 600 lb and Q = 800 lb,
determine the tension in rods A and B.
SOLUTION
Free-Body Diagram
Resolving the forces into x- and y-directions:
R = P + Q + TA + TB = 0
Substituting components:
R = -(600 lb) j - [(800 lb)cos30]i
+ [(800 lb)sin 30]j
+ TB i + (TA cos60)i + (TA sin 60) j = 0
Summing forces in the y-direction:
-600 lb + (800 lb)sin 30 + TA sin 60 = 0
TA = 230.94 lb
TA = 231 lb ◀
Summing forces in the x-direction:
-(800 lb)cos30 + TB + TA cos60 = 0
Thus,
TB = -(230.94 lb) cos60 + (800 lb) cos30
= 577.35 lb
TB = 577 lb ◀
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52
PROBLEM 2.52
Two forces P and Q are applied as shown to an
aircraft connection. Knowing that the connection is in
equilibrium and that the tensions in rods A and B are
TA = 240 lb and TB = 500 lb, determine the
magnitudes of P and Q.
SOLUTION
Free-Body Diagram
Resolving the forces into x- and y-directions:
R = P + Q + TA + TB = 0
Substituting components:
R = -Pj - Q cos 30i + Q sin 30 j
+ [(240 lb)cos 60]i
+ [(240 lb)sin 60]j + (500 lb)i
Summing forces in the x-direction:
-Q cos 30 + (240 lb)cos 60 + 500 lb = 0
Q = 715.91 lb
Summing forces in the y-direction:
-P + Q sin 30 + (240 lb)sin 60 = 0
P = Q sin 30 + (240 lb)sin 60
= (715.91 lb)sin 30 + (240 lb)sin 60
= 565.80 lb
P = 566 lb; Q = 716 lb ◀
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53
PROBLEM 2.53
FB
B
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.
C
FC
A
3
4
FA
D
FD
SOLUTION
Free-Body Diagram of Connection
SFx = 0:
3
3
FB - FC - FA = 0
5
5
FA = 8 kN
FB = 16 kN
With
FC =
4
4
(16 kN) - (8 kN)
5
5
FC = 6.40 kN ◀
3
3
S Fy = 0: - FD + FB - FA = 0
5
5
With FA and FB as above:
3
3
FD = (16 kN) - (8 kN)
5
5
FD = 4.80 kN ◀
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54
PROBLEM 2.54
FB
B
C
FC
A
3
A welded connection is in equilibrium under the action of the
four forces shown. Knowing that FA = 5 kN and FD = 6 kN,
determine the magnitudes of the other two forces.
4
FA
D
FD
SOLUTION
Free-Body Diagram of Connection
3
3
SFy = 0: -FD - FA + FB = 0
5
5
or
3
FB = FD + FA
5
With
FA = 5 kN, FD = 8 kN
FB =
ù
5é
3
ê6 kN + (5 kN)ú
ê
úû
3ë
5
FB = 15.00 kN ◀
4
4
SFx = 0: -FC + FB - FA = 0
5
5
4
FC = (FB - FA )
5
4
= (15 kN - 5 kN)
5
FC = 8.00 kN ◀
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55
PROBLEM 2.55
D
A
α
B
C
β
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 a = 30° and b = 10° and that the combined
weight of the boatswain’s chair and the sailor is 200 lb,
determine the tension (a) in the support cable ACB, (b) in
the traction cable CD.
SOLUTION
Free-Body Diagram
SFx = 0: T A CB cos 10 - T A CB cos 30 - TCD cos 30 = 0
TCD = 0.137158T A CB
(1)
SFy = 0: T A CB sin 10 + T A CB sin 30 + TCD sin 30 - 200 = 0
0.67365T A CB + 0.5T CD = 200
(a)
Substitute (1) into (2):
0.67365T A CB + 0.5(0.137158T A CB ) = 200
T A CB = 269.46 lb
(b)
From (1):
(2)
T A CB = 269 lb ◀
TCD = 0.137158(269.46 lb)
TCD = 37.0 lb ◀
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56
PROBLEM 2.56
D
A
α
C
B
β
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° and b = 15° and that the tension in
cable CD is 20 lb, determine (a) the combined weight of
the boatswain’s chair and the sailor, (b) the tension in the
support cable ACB.
SOLUTION
Free-Body Diagram
SFx = 0: T A CB cos 15 - T A CB cos 25 - (20 lb) cos 25 = 0
T A CB = 304.04 lb
SFy = 0: (304.04 lb)sin 15 + (304.04 lb)sin 25
+ (20 lb)sin 25 -W = 0
W = 215.64 lb
(a)
W = 216 lb ◀
(b) T A CB = 304 lb ◀
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57
PROBLEM 2.57
A
6 kN
55°
C
For the cables of prob. 2.44, find the value of α for which the tension is as small
as possible (a) in cable bc, (b) in both cables simultaneously. In each case
determine the tension in each cable.
α
B
SOLUTION
Free-Body Diagram
Force Triangle
(a) For a minimum tension in cable BC, set angle between cables to 90 degrees.
a = 35.0 ◀
By inspection,
T A C = (6 kN) cos 35 
T A C = 4.91 kN ◀
T B C = (6 kN) sin 35 
T B C = 3.44 kN ◀
(b) For equal tension in both cables, the force triangle will be an isosceles.
a = 55.0 ◀
Therefore, by inspection,
TA C = TBC = (1 / 2)
6 kN
cos35
T A C = T BC = 3.66 kN ◀
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58
A
25°
45°
α
C
B
PROBLEM 2.58
For the cables of Problem 2.46, 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 a.
P
SOLUTION
Free-Body Diagram
(a)
Law of cosines
Force Triangle
P 2 = (600)2 + (750)2 - 2(600)(750)cos(25 + 45)
P = 784.02 N
(b)
Law of sines
P = 784 N ◀
sin b
sin(25 + 45)
=
600 N
784.02 N
b = 46.0
\ a = 46.0 + 25
a = 71.0 ◀
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59
PROBLEM 2.59
A
For the situation described in Figure P2.48, determine (a)
the value of a for which the tension in rope BC is as small
as possible, (b) the corresponding value of the tension.
5°
C
1200 lb
α
B
SOLUTION
Free-Body Diagram
Force Triangle
To be smallest, T B C must be perpendicular to the direction of T A C .
(a)
(b)
Thus,
a = 5.00
a = 5.00
T BC = (1200 lb)sin 5
◀
T B C = 104.6 lb ◀
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60
PROBLEM 2.60
Two cables tied together at C are loaded as shown.
Determine the range of values of P for which both cables
remain taut.
SOLUTION
Free-Body Diagram
SFx = 0:
3
P -TBC - TAC sin 30 = 0
5
P=
SFy = 0:
(
5
T + TAC sin 30
3 BC
)
(1)
4
P + TAC cos30 - 120 = 0
5
P=
(
5
120 - TAC cos30
4
Requirement:
TAC = 0:
From Eq. (2):
P = 150.0 lb
Requirement:
TBC = 0:
(
(2)
)
From Eq. (1): P =
5
T sin 30
3 AC
From Eq. (2): P =
5
120 - TAC cos 30
4
(
)
)
Solving simultaneously yields:
TAC = 78.294 lb and
P = 65.2 lb
65.2 lb< P < 150.0 lb ◀
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61
PROBLEM 2.61
C
A
B
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.
0.7 m
1.2 m
SOLUTION
Free-Body Diagram
tan a =
h
0.6 m
(1)
Isosceles Force Triangle
Law of sines:
sin a =
1
2
(2.8 kN)
TA C
TA C = 5 kN
sin a =
1
2
(2.8 kN)
5 kN
a = 16.2602
From Eq. (1): tan16.2602 =
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 ´ 0.625 m
1.250 m ◀
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62
d
PROBLEM 2.62
d
h
For W = 800 N, P = 200 N, and d = 600 mm, determine
the value of h consistent with equilibrium.
P
W
SOLUTION
Free-Body Diagram
T A C = T B C = 800 N
(h
A C = BC =
SFy = 0: 2(800 N)
2
+d2
h
h +d2
2
)
-P = 0
æ d ö÷2
P
800 =
1 + çç ÷÷
çè h ÷ø
2
Data:
P = 200 N, d = 600 mm and solving for h
800 N =
æ 600 mm ö÷2
200 N
1 + çç
÷
çè h ÷÷ø
2
h = 75.6 mm ◀
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63
PROBLEM 2.63
x
B
C
20 in.
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.
50 lb
P
A
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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64
PROBLEM 2.64
x
B
C
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.
20 in.
50 lb
P
A
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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65
PROBLEM 2.65
A cable loop of length 1.5 m is placed around a crate. Knowing that the mass
of the crate is 300 kg, determine the tension in the cable for each of the
arrangements shown.
SOLUTION
Free-Body Diagram
(
Isosceles Force Triangle
)
W = (300 kg) 9.81 m/s2 = 2943.0 N
1
EB = (1500 mm - 400 mm - 300 mm - 300 mm)
2
EB = 250 mm
æ 200 mm ö÷
a = cos-1 çç
÷ = 36.87
çè 250 mm ø÷÷
TAE = TBE
1
TAE sin a = (2943.0 N)
2
1
TAE sin 36.87 = (2943.0 N)
2
TAE = 2452.5 N
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66
TAE = 2450 N ◀
(a)
Problem 2.65 (Continued)
Isosceles Force Triangle
Free-Body Diagram
1
(1500 mm - 300 mm - 400 mm - 400 mm)
2
EB = 250 mm
æ 150 mm ÷ö
÷ = 41.41
a = cos-1 çç
èç 200 mm ÷÷ø
EB =
TAE = TBE
1
TAE sin a = (2943.0 N)
2
1
TAE sin 41.41 = (2943.0 N)
2
TAE = 2224.7 N
TAE = 2220 N ◀
(b)
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67
PROBLEM 2.66
0.75 m
B
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.)
2.4 m
P
α
A
200 kg
SOLUTION
Free-Body Diagram: Pulley A
æ 5 ö÷
÷ + P cos a = 0
SFx = 0: - 2P ççç
çè 281 ÷÷ø
cos a = 0.59655
a = 53.377
For a = + 53.377:
æ 16 ö÷
÷ + P sin 53.377 - 1962 N = 0
SFy = 0: 2 P ççç
çè 281 ÷÷ø
P = 724 N
53.4 ◀
For a = - 53.377:
æ 16 ö÷
÷ + P sin(-53.377) - 1962 N = 0
SFy = 0: 2 P ççç
çè 281 ÷÷ø
P = 1773
53.4 ◀
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68
PROBLEM 2.67
T
T
A 600-lb crate is supported by several ropeand-pulley arrangements as shown. Determine
for each arrangement the tension in the rope.
(See the hint for Problem 2.66.)
T
T
T
(a)
(b)
(c)
(d)
(e)
SOLUTION
Free-Body Diagram of Pulley
(a)
SFy = 0: 2T - (600 lb) = 0
1
T = (600 lb)
2
T = 300 lb ◀
(b)
SFy = 0: 2T - (600 lb) = 0
1
T = (600 lb)
2
T = 300 lb ◀
(c)
SFy = 0: 3T - (600 lb) = 0
1
T = (600 lb)
3
T = 200 lb ◀
(d )
SFy = 0: 3T - (600 lb) = 0
1
T = (600 lb)
3
T = 200 lb ◀
(e)
SFy = 0: 4T - (600 lb) = 0
1
T = (600 lb)
4
T = 150.0 lb ◀
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69
PROBLEM 2.68
T
T
Solve Parts b and d of Problem 2.67, assuming
that the free end of the rope is attached to the
crate.
T
T
T
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.)
(a)
(b)
(c)
(d)
(e)
SOLUTION
Free-Body Diagram of Pulley and Crate
(b)
SFy = 0: 3T - (600 lb) = 0
1
T = (600 lb)
3
T = 200 lb ◀
(d )
SFy = 0: 4T - (600 lb) = 0
1
T = (600 lb)
4
T = 150.0 lb ◀
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70
PROBLEM 2.69
B
A
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.
25°
D 55°
P
C
Q
SOLUTION
Free-Body Diagram: Pulley C
(a)
SFx = 0: T A CB (cos 25 - cos 55) - (750 N) cos 55° = 0
T A CB = 1292.88 N
Hence:
T A CB = 1293 N ◀
(b)
SFy = 0: T A CB (sin 25 + sin 55) + (750 N)sin 55 - Q = 0
(1292.88 N)(sin 25 + sin 55) + (750 N)sin 55 - Q = 0
Q = 2219.8 N
or
Q = 2220 N ◀
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71
B
A
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.
25°
D 55°
P
PROBLEM 2.70
C
Q
SOLUTION
Free-Body Diagram: Pulley C
SFx = 0: T A CB (cos 25 - cos 55) - P cos 55 = 0
P = 0.58010 TACB (1)
or
SFy = 0: T A CB (sin 25 + sin 55) + P sin 55 - 1800 N = 0
1.24177T A CB + 0.81915P = 1800 N (2)
or
(a)
Substitute Equation (1) into Equation (2):
1.24177T A CB + 0.81915(0.58010T A CB ) = 1800 N
T A CB = 1048.37 N
Hence:
T A CB = 1048 N ◀
(b)
Using (1),
P = 0.58010(1048.37 N) = 608.16 N
P = 608 N ◀
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72
PROBLEM 2.71
Determine (a) the x, y, and z components of the 500-N force,
(b) the angles qx, qy, and qz that the force forms with the
coordinate axes.
SOLUTION
(a)
Fx = (500 N)sin 40 cos30
Fx = 278 N ◀
Fx = 278.34 N
Fy = (500 N)cos 40°
Fy = 383.02 N
Fy = 383 N ◀
Fz = (500 N)sin 40 sin30
Fz = 160.697 N
(b)
cos qx =
cos qy =
cos qz =
Fz = 160.7 N ◀
Fx
278.34 N
=
500 N
F
Fy
F
Fz
F
q x = 56.2  ◀
=
383.02 N
500 N
qy = 40.0 ◀
=
160.697 N
500 N
q z = 71.3  ◀
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73
PROBLEM 2.72
Determine (a) the x, y, and z components of the 800-N force,
(b) the angles qx, qy, and qz that the force forms with the
coordinate axes.
SOLUTION
(a)
Fx = -(800 N)cos 70 sin 25
Fx = - 115.635 N
Fx = -115.6 N ◀
Fy = (800 N)sin 70°
(b)
Fy = 751.75 N
Fy = 752 N ◀
Fz = (800 N)cos 70 cos 25
Fz = 247.98 N
Fz = 248 N ◀
Fx -115.635 N
=
800 N
F
q x = 98.3  ◀
cos qx =
cos qy =
cos qz =
Fy
F
Fz
F
=
751.75 N
800 N
qy = 20.0 ◀
=
247.98 N
800 N
q z = 71.9  ◀
Note: From the given data, we could have computed directly
qy = 90 - 35 = 55, which checks with the answer obtained.
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74
PROBLEM 2.73
A gun is aimed at a point A located 35° east of north. Knowing that the barrel of the gun forms an angle
of 40° with the horizontal and that the maximum recoil force is 400 N, determine (a) the x, y, and z
components of that force, (b) the values of the angles 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
= 306.42 N
(a)
Fx = -FH sin35
= -(306.42 N)sin35
= -175.755 N
Fy = -F sin 40
= -(400 N)sin 40
= -257.12 N
Fx = - 175.8 N ◀
Fy = -257 N ◀
Fz = +FH cos35
= +(306.42 N)cos35
Fz = +251 N ◀
= +251.00 N
(b)
cos qx =
cos qy =
cos qz =
Fx -175.755 N
=
400 N
F
Fy
F
=
-257.12 N
400 N
Fz 251.00 N
=
400 N
F
qx = 116.1 ◀
qy = 130.0 ◀
qz = 51.1  ◀
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75
PROBLEM 2.74
Solve Problem 2.73, assuming that point A is located 15° north of west and that the barrel of the gun forms
an angle of 25° with the horizontal.
PROBLEM 2.73 A gun is aimed at a point A located 35° east of north. Knowing that the barrel of the gun
forms an angle of 40° with the horizontal and that the maximum recoil force is 400 N, determine (a) the x, y,
and z components of that force, (b) the values of the angles 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
= 362.52 N
(a)
Fx = +FH cos15
= +(362.52 N)cos15
= +350.17 N
Fx = + 350 N ◀
Fy = -F sin 25
= -(400 N)sin 25
Fy = -169.0 N ◀
= -169.047 N
Fz = +FH sin15
= +(362.52 N)sin15
Fz = +93.8 N ◀
= +93.827 N
(b)
cos qx =
cos qy =
cos qz =
Fx +350.17 N
=
F
400 N
Fy
F
=
qx = 28.9  ◀
-169.047 N
400 N
Fz +93.827 N
=
400 N
F
qy = 115.0 ◀
qz = 76.4  ◀
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76
PROBLEM 2.75
The angle between the guy wire AB and the mast is 20°.
Knowing that the tension in AB is 300 lb, determine (a) the x,
y, and z components of the force exerted on the boat at B, (b)
the angles qx, qy, and qz defining the direction of the force
exerted at B.
SOLUTION
Fh = FAB sin 20
= (300 lb)sin 20
Fh = 102.606 lb
Fx = -Fh cos 40
Fy = FAB cos 20
Fx = (-102.606 lb) cos 40
Fy = (300 lb)cos20
Fx = -78.601 lb
Fy = 281.91 lb
Fz = -Fh sin 40
Fz = (-102.606 lb)sin 40
Fz = -65.954 lb
(a)
Fx = - 78.6 lb ◀
Fy = 282 lb ◀
Fz = - 66.0 lb ◀
(b)
Fx -78.601 lb
=
F
300 lb
Fy
281.91 lb
cos qy =
=
300 lb
F
Fz -65.954 lb
=
cos qz =
F
300 lb
cos qx =
q x = 105.2 ◀
qy = 20.0 ◀
q z = 102.7  ◀
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77
PROBLEM 2.76
The angle between the guy wire AC and the mast is 20°.
Knowing that the tension in AC is 300 lb, determine (a) the
x, y, and z components of the force exerted on the boat at C,
(b) the angles qx, qy, and qz defining the direction of the
force exerted at C.
SOLUTION
Fh = FAC sin 20
= (300 lb)sin 20
Fh = 102.606 lb
Fx = Fh cos 40
Fy = FAC cos 20
Fz = -Fh sin 40
Fx = (102.606 lb) cos 40
Fy = (300 lb)cos20
Fz = (-102.606 lb)sin 40
Fx = 78.601 lb
Fy = 281.91 lb
Fz = -65.954 lb
(a)
Fx = 78.6 lb ◀
Fy = 282 lb ◀
Fz = - 66.0 lb ◀
(b)
Fx
78.601 lb
=
F
300 lb
Fy
281.91 lb
cos qy =
=
300 lb
F
F
-65.954 lb
cos qz = z =
F
300 lb
cos qx =
q x = 74.8  ◀
qy = 20.0 ◀
q z = 102.7  ◀
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78
PROBLEM 2.77
y
A
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 qx , qy , and qz defining the direction of that
force.
56 ft
D
α
O
50°
z
B
20°
x
C
SOLUTION
56 ft
65 ft
= 0.86154
cos qy =
From triangle AOB:
qy = 30.51
Fx = -F sin qy cos20
(a)
= -(3900 lb)sin 30.51 cos20
Fx = - 1861 lb ◀
Fy = +F cos qy = (3900 lb)(0.86154) Fy = +3360 lb ◀
Fz = +(3900 lb)sin 30.51° sin 20°
(b)
cos qx =
From above:
Fx
1861 lb
== - 0.4771
3900 lb
F
qy = 30.51
cos qz =
Fz
677 lb
=+
= + 0.1736
F
3900 lb
Fz = + 677 lb ◀
qx = 118.5 ◀
qy = 30.5 ◀
qz = 80.0 ◀
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79
y
PROBLEM 2.78
A
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 qx, qy, and qz defining the direction of that force.
56 ft
D
α
O
50°
z
B
20°
C
x
SOLUTION
AC = 70 ft
In triangle AOB:
OA = 56 ft
F = 5250 lb
56 ft
70 ft
qy = 36.870
cos qy =
FH = F sin qy
= (5250 lb)sin 36.870
= 3150.0 lb
(a)
Fx = -FH sin 50 = -(3150.0 lb)sin 50 = -2413.0 lb
Fx = - 2410 lb ◀
Fy = +F cos qy = +(5250 lb)cos36.870 = +4200.0 lb
Fy = +4200 lb ◀
Fz = -FH cos50 = -3150 cos50 = -2024.8 lb
(b)
cos qx =
From above:
Fx -2413.0 lb
=
5250 lb
F
qy = 36.870
qz =
Fz -2024.8 lb
=
5250 lb
F
Fz = - 2025 lb ◀
qx = 117.4 ◀
qy = 36.9 ◀
qz = 112.7 ◀
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80
PROBLEM 2.79
Determine the magnitude and direction of the force F = (260 N)i – (320 N)j + (800 N)k.
SOLUTION
F = Fx2 + Fy2 + Fz2
F = (260 N)2 + (-320 N)2 + (800 N)2
cos qx =
cos qy =
cos qy =
Fx 260 N
=
900 N
F
Fy
F
Fz
F
=
- 320 N
900 N
=
800 N
900 N
F = 900 N ◀
q x = 73.2  ◀
qy = 110.8 ◀
q z = 27.3  ◀
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81
PROBLEM 2.80
Determine the magnitude and direction of the force F = (700 N)i – (820 N)j + (960 N)k.
SOLUTION
F = Fx2 + Fy2 + Fz2
F = (700 N)2 - (820 N)2 + (960 N)2
F = 1443.61 N
cos qx =
cos qy =
cos qy =
F = 1444 N ◀
Fx
700 N
=
F 1443.61 N
q x = 61.0  ◀
=
-820 N
1443.61 N
qy = 124.6 ◀
=
960 N
1443.61 N
Fy
F
Fz
F
qz = 48.3 ◀
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82
PROBLEM 2.81
A force F of magnitude 250 lb acts at the origin of a coordinate system. Knowing that qx =
65, qy = 40°, and Fz > 0, determine (a) the components of the force, (b) the angle qz.
SOLUTION
cos2 qx + cos2 qy + cos2 qz = 1
cos2 (65) + cos2 (40) + cos2 qz = 1
cos qz = 0.48432
(b)
\ qz = 61.0 ◀
Since Fz > 0, we choose cos qz = 0.48432
(a)
Fx = F cos qx = (250 lb)cos65
Fy = F cos qy = (250 lb)cos 40
Fx = 105.7 lb ◀
Fy = 191.5 lb ◀
Fz = F cos qz = (250 lb)cos 61
Fz = 121.2 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 qx = 70.9 and qy = 144.9.
Knowing that the z component of the force is s-52.0 lb, determine (a) the angle qz, (b) the other components
and the magnitude of the force.
SOLUTION
cos2 qx + cos2 qy + cos2 qz = 1
cos2 70.9 + cos2 144.9 + cos2 qz = 1
cos qz = 0.47282
(a)
Since Fz < 0, we choose cos qz = - 0.47282
(b)
Fz = F cos qz
-52.0 lb = F (-0.47282)
\ qz = 118.2 ◀
F = 110.0 lb ◀
F = 110.0 lb
Fx = F cos qx = (110.0 lb) cos 70.9
Fy = F cos qy = (110.0 lb)cos144.9
Fx = 36.0 lb ◀
Fy = -90.0 lb ◀
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84
PROBLEM 2.83
A force F of magnitude 210 N acts at the origin of a coordinate system. Knowing that Fx = 80 N,
qz= 151.2, and Fy< 0, determine (a) the components Fy and Fz, (b) the angles qx and qy.
SOLUTION
Fz = F cos qz = (210 N)cos151.2
(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 ◀
cos qx =
cos qy =
Fy = -62.0 lb ◀
Fx
80 N
=
= 0.38095
210 N
F
Fy
F
=
qx = 67.6  ◀
61.929 N
= -0.29490
210 N
qy = 107.2 ◀
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85
PROBLEM 2.84
A force acts at the origin of a coordinate system in a direction defined by the angles qy = 120 and qz
= 75. Knowing that the x component of the force is +40 N, determine (a) the angle qx, (b) the magnitude
of the force.
SOLUTION
cos2 qx + cos2 qy + cos2 qz = 1
cos2 qx + cos2 120 + cos2 75  = 1
cos qx = 0.82644
(b)
\ qx = 34.3 ◀
Since Fx > 0, we choose cos q x = 0.82644
(a)
Fx = F cos q x
40 N = F cos34.3
F = 48.4 N ◀
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86
PROBLEM 2.85
Two cables BG and BH are attached to frame ACD
as shown. Knowing that the tension in cable BG is
540 N, determine the components of the force
exerted by cable BG on the frame at B.
SOLUTION

BG = -(0.8 m)i + (1.48 m) j - (0.64 m)k
BG = (-0.8 m)2 + (1.48 m 2 ) + (-0.64 m)2
= 1.8 m
TBG = TBG λ BG

BG
= TBG
BG
540 N
=
[(-0.8 m)i + (1.48 m)j + (-0.64 m)k ]
1.8 m
TBG = (-240 N)i + (444 N) j - (192.0 N)k
Fx = -240 N, Fy = +444 N, Fz = +192.0 N ◀
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87
PROBLEM 2.86
Two cables BG and BH are attached to frame ACD
as shown. Knowing that the tension in cable BH is
750 N, determine the components of the force
exerted by cable BH on the frame at B.
SOLUTION

BH = (0.6 m)i + (1.2 m) j - (1.2 m)k
BH = (0.6 m)2 + (1.2 m)2 - (1.2 m)2
= 1.8 m
TBH = TBH λ BH

BH
= TBH
BH
750 N
=
[i + 2 j - 2 k ] (m)
3m
TBH = (250 N)i + (500 N) j - (500 N)k
Fx = +250 N, Fy = +500 N, Fz = -500 N ◀
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88
PROBLEM 2.87
C
28.8 ft
B
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.
45 ft
18 ft
36 ft
54 ft
A
30°
SOLUTION
Cable AB:
λAB
TAB

AB (-46.765 ft)i + (45 ft) j + (36 ft)k
=
=
AB
74.216 ft
-46.765i + 45 j + 36 k
= TAB λAB =
74.216
(TA B )x = -1.260 kips ◀
(TA B )y = +1.213 kips ◀
(TA B )z = +0.970 kips ◀
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89
PROBLEM 2.88
C
28.8 ft
B
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.
45 ft
18 ft
36 ft
54 ft
A
30°
SOLUTION
Cable AB:
λAC
TAC

AC (-46.765 ft)i + (55.8 ft) j + (-45 ft)k
=
=
AC
85.590 ft
-46.765i + 55.8 j - 45k
= TAC λAC = (1.5 kips)
85.590
(TA C )x = -0.820 kips ◀
(TAC )y = +0.978 kips ◀
(TA C )z = -0.789 kips ◀
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90
PROBLEM 2.89
y
A rectangular plate is supported by three cables as shown.
Knowing that the tension in cable AB is 408 N, determine the
components of the force exerted on the plate at B.
A
480
250
O
B
D
360
130
320
z
450
360
C
x
Dimensions in mm
SOLUTION
We have:

BA = +(320 mm)i + (480 mm)j - (360 mm)k
Thus:
FB = TBA λBA = TBA
BA = 680 mm

æ8
12
9 ö
BA
= TBA çç i + j - k ÷÷÷
ç
17
17 ÷ø
BA
è17
æ8
ö æ
ö æ
ö
çç T ÷÷ i + çç12 T ÷÷ j- çç 9 T ÷÷ k = 0
BA
BA
BA
÷
÷
èç17
ø÷ èç17
ø÷ èç17
ø÷÷
Setting T BA = 408 N yields,
Fx = +192.0 N, Fy = +288 N, Fz = -216 N ◀
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91
PROBLEM 2.90
y
A rectangular plate is supported by three cables as shown.
Knowing that the tension in cable AD is 429 N, determine the
components of the force exerted on the plate at D.
A
480
250
O
B
D
360
130
320
z
450
360
C
x
Dimensions in mm
SOLUTION
We have:

DA = -(250 mm)i + (480 mm)j + (360 mm)k
Thus:
FD = TDA λDA = TDA
DA = 650 mm

æ 5
48
36 ö÷
DA
j+
k÷
= TDA çç- i +
65
65 ÷ø÷
DA
èç 13
æ5
ö æ 48
ö æ 36
ö
-çç TDA ÷÷÷ i + çç TDA ÷÷÷ j + çç TDA ÷÷÷ k = 0
çè13
÷ø çè 65
÷ø èç 65
ø÷
Setting T DA = 429 N yields,
Fx = -165.0 N, Fy = +317 N, Fz = +238 N ◀
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92
PROBLEM 2.91
y
Q
P
Find the magnitude and direction of the resultant of the two forces
shown knowing that P = 300 N and Q = 400 N.
50°
20°
30°
x
15°
z
SOLUTION
P = (300 N)[- cos30 sin15i + sin 30 j + cos30 cos15k ]
= - (67.243 N)i + (150 N) j + (250.95 N)k
Q = (400 N)[cos50 cos20i + sin 50 j - cos50 sin 20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
R = 515 N ◀
= 515.07 N
cos qx =
cos qy =
cos qz =
Rx 174.367 N
=
= 0.33853
515.07 N
R
Ry
qx = 70.2  ◀
456.42 N
= 0.88613
515.07 N
qy = 27.6 ◀
Rz 163.011 N
=
= 0.31648
515.07 N
R
qz = 71.5  ◀
R
=
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93
PROBLEM 2.92
y
Q
P
Find the magnitude and direction of the resultant of the two forces
shown knowing that P = 400 N and Q = 300 N.
50°
20°
30°
x
15°
z
SOLUTION
P = (400 N)[- cos30 sin15i + sin 30 j + cos30 cos15k ]
= - (89.678 N)i + (200 N) j + (334.61 N)k
Q = (300 N)[cos 50 cos 20i + sin 50 j - cos 50 sin 20k ]
= (181.21 N)i + (229.81 N)j - (65.954 N)k
R= P+Q
= (91.532 N)i + (429.81 N) j + (268.66 N)k
R = (91.532 N)2 + (429.81 N)2 + (268.66 N)2
R = 515 N ◀
= 515.07 N
cos qx =
cos qy =
cos qz =
Rx 91.532 N
=
= 0.177708
515.07 N
R
Ry
qx = 79.8  ◀
429.81 N
= 0.83447
515.07 N
qy = 33.4 ◀
Rz
268.66 N
=
= 0.52160
515.07 N
R
qz = 58.6  ◀
R
=
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94
A
y
PROBLEM 2.93
40 in.
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.
45 in.
60 in.
O
D
B
z
C
60 in.
x
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.

é (40 in.)i - (45 in.) j + (60 in.)k ù
AB
ú
TAB = TAB λAB = TAB
= (425 lb) êê
ú
AB
85
in.
êë
úû
TAB = (200 lb)i - (225 lb) j + (300 lb)k

é (100 in.)i - (45 in.) j + (60 in.)k ù
AC
ú
TAC = TAC λAC = TAC
= (510 lb) êê
ú
125
in.
AC
êë
úû
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 qx =
608 lb
= 0.66599
912.92 lb
cos qy =
408.6 lb
= -0.44757
912.92 lb
cos qz =
544.8 lb
= 0.59677
912.92 lb
qx = 48.2  ◀
qy = 116.6 ◀
qz = 53.4  ◀
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95
A
y
PROBLEM 2.94
40 in.
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.
45 in.
60 in.
O
D
B
z
C
60 in.
x
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.

é (40 in.)i - (45 in.) j + (60 in.)k ù
AB
ú
TAB = TAB λAB = TAB
= (510 lb) êê
ú
AB
85 in.
êë
úû
TAB = (240 lb)i - (270 lb) j + (360 lb)k

é (100 in.)i - (45 in.) j + (60 in.)k ù
AC
ú
TAC = TAC λAC = TAC
= (425 lb) êê
ú
125 in.
AC
êë
úû
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 qx =
580 lb
= 0.63532
912.92 lb
cos qy =
-423 lb
= -0.46335
912.92 lb
cos qz =
564 lb
= 0.61780
912.92 lb
qx = 50.6  ◀
qy = 117.6 ◀
qz = 51.8  ◀
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96
PROBLEM 2.95
For the frame of Prob. 2.85, determine the magnitude and
direction of the resultant of the forces exerted by the cables
at B knowing that the tension is 540 N in cable BG and 750
N in cable BH.
PROBLEM 2.85 Two cables BG and BH are attached to
frame ACD as shown. Knowing that the tension in cable BG
is 540 N, determine the components of the force exerted by
cable BG on the frame at B.
SOLUTIO
ON
JJJG
BG = −(0.8 m)i + (1.448 m) j − (0.644 m)k
BG = (−0.8 m)2 + ((1.48 m 2 ) + (− 0.64 m)2 = 11.8 m
JJJG
BH = (00.6 m)i + (1.2 m) j − (1.2 m))k
BH = (0.6 m)2 + (1..2 m)2 − (1.2 m
m)2 = 1.8 m
TBG = TBG
B λ BG
JJJG
BG 540 N
= TBG
=
[(−0.8 m)i + (1.48 m)j + (−0.64 m)k ]
B
BG 1.8 m
= (−240 N)i + (4444 N) j − (1922.0 N)k
TBH = TBH
B λ BH
JJJG
BH 750 N
[i + 2 j − 2 k ] (m)
= TBH
=
B
BH
3m
TBH = (2250 N)i + (5000 N) j − (500 N
N)k
R = TBG
+
i
N) j − (692 N )k
+
T
=
(1
10
N)
(944
B
BH
Then:
and
R=
(10
2
+ 944 2 + (−692)
2
) = 1170.51 N
R = 1171 N ◀
cos θx =
10 N
1170.51 N
θ x = 89.5 ° ◀
cos θy =
944 N
1170.51 N
θy = 36.2° ◀
−692 N
1170.51
1
N
θ z = 126.2 ° ◀
cos θz =
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97
PROBLEM 2.96
y
For the plate of Prob. 2.89, determine the tensions in cables AB
and AD knowing that the tension in cable AC is 54 N and that
the resultant of the forces exerted by the three cables at A must
be vertical.
A
480
250
O
B
D
360
130
320
z
450
360
C
x
Dimensions in mm
SOLUTION
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
TAD

AB TAB
= TAB λAB = TAB
=
(-320i - 480 j + 360k)
AB
 680
54
AC
= TAC λAC = TAC
=
(450i - 480 j + 360k)
AC 750

AD TAD
= TAD λAD = TAD
=
(250i - 480 j - 360k)
AD 650
Substituting into the Eq. R = SF and factoring i, j, k :
æ 320
ö
250
TAB + 32.40 +
TAD ÷÷÷ i
R = çççè 680
÷ø
650
æ 480
ö
480
+ ççT - 34.560 TAD ÷÷÷ j
çè 680 AB
÷ø
650
æ 360
ö
360
+ çç
TAB + 25.920 TAD ÷÷÷ k
650
èç 680
ø÷
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written consent of McGraw-Hill Education.
98
PROBLEM 2.96 (Continued)
Since R is vertical, the coefficients of i and k are zero:
i:
-
320
250
TAB + 32.40 +
T =0
680
650 AD
(1)
360
360
TAB + 25.920 T =0
680
650 AD
(2)
k:
Multiply (1) by 3.6 and (2) by 2.5 then add:
-
252
T + 181.440 = 0
680 AB
T A B = 489.60 N
T A B = 490 N ◀
Substitute into (2) and solve for TAD :
360
360
(489.60 N) + 25.920 T =0
680
650 AD
T A D = 514.80 N
T A D = 515 N ◀
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99
PROBLEM 2.97
y
36 in.
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.
C
25 in.
24 in.
B
29 in.
O
48 in.
z
A
x
P
SOLUTION
Cable AB:
Cable AC:
Load P:
TA B = 183 lb

(-48 in.)i + (29 in.) j + (24 in.)k
AB
= (183 lb)
TAB = TAB λAB = TAB
61 in.
AB
TAB = -(144 lb)i + (87 lb) j + (72 lb)k

(-48 in.)i + (25 in.) j + (-36 in.)k
AC
= TAC
TAC = TAC λAC = TAC
AC
65 in.
48
25
36
TAC = - TAC i + TAC j - TAC k
65
65
65
P=P j
For resultant to be directed along OA, i.e., x-axis
Rz = 0: SFz = (72 lb) -
36 ¢
TAC = 0
65
T A C = 130.0 lb ◀
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100
PROBLEM 2.98
y
36 in.
C
For the boom and loading of Problem. 2.97, determine
the magnitude of the load P.
25 in.
24 in.
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.
B
29 in.
O
48 in.
z
A
x
P
SOLUTION
See Problem 2.97. Since resultant must be directed along OA, i.e., the x-axis, we write
Ry = 0: SFy = (87 lb) +
25
T -P = 0
65 AC
T A C = 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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written consent of McGraw-Hill Education.
101
y
PROBLEM 2.99
450 mm 500 mm
D
360 mm
B
320 mm
C
z
A
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.
600 mm
x
SOLUTION
Free-Body Diagram at A:
The forces applied at A are:
TA B , TA C , TA D , 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

(-450 mm)i + (600 mm) j
AB
and
TAB = λAB TAB = TAB
= TAB
750 mm
AB
æ 45
60 ÷ö
= çç- i +
j÷T
çè 75
75 ÷÷ø A B
TAC = λAC TAC = TAC
TAD = λADTAD = TAD

(600 mm)i - (320 mm) j
AC
= TAC
680 mm
AC
æ 60
ö÷
32
= çç j - k÷÷TAC
çè 68
68 ÷ø

AD
(500 mm)i + (600 mm) j + (360 mm)k
= TAD
AD
860 mm
æ 50
ö÷
60
36
j + k÷÷TAD
= çç i +
çè 86
86
86 ÷ø
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102
PROBLEM 2.99 (Continued)
Equilibrium condition:
SF = 0: \ TA B + TA C + TA D + W = 0
Substituting the expressions obtained for TA B , TA C , and TA D ; 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
TA B + TA D = 0
75
86
60
60
60
T + T + T -W = 0
75 A B 68 A C 86 A D
-
32
36
TA C + TA D = 0
68
86
(1)
(2)
(3)
Setting T A B = 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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103
y
PROBLEM 2.100
450 mm 500 mm
D
360 mm
B
320 mm
C
z
A
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.
600 mm
x
SOLUTION
See Problem 2.99 for the figure and analysis leading to the following set of linear algebraic equations:
-
45
50
TA B + TA D = 0
75
86
60
60
60
TA B + TA C + TA D - W = 0
75
68
86
-
32
36
T + T =0
68 A C 86 A D
(1)
(2)
(3)
Setting TA D = 4.3 kN into the above equations gives
TAB = 4.1667 kN
TAC = 3.8250 kN
W = 9.71 kN ◀
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104
PROBLEM 2.101
y
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.
A
5.60 m
B
D
4.20 m
O
C
z
3.30 m
4.20 m
2.40 m
SOLUTION
The forces applied at A are:
x
FREE-BODY DIAGRAM AT A
TA B , TA C , TA D , 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
AD = 6.50 m

AB
and
= (-0.6 i - 0.8 j)TAB
TAB = TAB λAB = TAB
AB

AC
= (0.32432 i - 0.75676 j + 0.56757k )TAC
TAC = TAC λAC = TAC
AC

AD
= (-0.86154 j - 0.50769k )TAD
TAD = TAD λAD = TAD
AD
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105
PROBLEM 2.101 (Continued)
Equilibrium condition:
SF = 0: TA B + TA C + TA D + Pj = 0
Substituting the expressions obtained for TA B , TA C , and TA D 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:
- 0.6T A B + 0.32432T A C = 0
- 0.8T A B - 0.75676T A C - 0.86154T A D + P = 0
0.56757T A C - 0.50769T A D = 0
(1)
(2)
(3)
Setting T A D = 481 N in (2) and (3), and solving the resulting set of equations gives
TAC = 430.26 N
TAD = 232.57 N
P = 926 N ◀
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106
PROBLEM 2.102
y
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.
A
5.60 m
B
D
4.20 m
O
C
z
2.40 m
3.30 m
4.20 m
x
SOLUTION
See Problem 2.101 for the figure and analysis leading to the linear algebraic Equations (1), (2), and (3).
-0.6T A B + 0.32432TA C = 0
-0.8T A B - 0.75676T A C - 0.86154T A D + P = 0
0.56757T A C - 0.50769T A D = 0
From Eq. (1):
TA B = 0.54053TA C
From Eq. (3):
TA D = 1.11795TA C
(1)
(2)
(3)
Substituting for TA B and TAD in terms of TAC into Eq. (2) gives
- 0.8(0.54053T A C ) - 0.75676TA C - 0.86154(1.11795T A C ) + P = 0
2.1523T A C = P ; P = 800 N
800 N
2.1523
= 371.69 N
and TAD gives
TA C =
Substituting into expressions for TA B
T A B = 0.54053(371.69 N)
T A D = 1.11795(371.69 N)
T A B = 201 N, T A C = 372 N, T A D = 416 N ◀
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107
PROBLEM 2.103
A crate is supported by three cables as shown. Determine
the weight W of the crate, knowing that the tension in cable
AD is 924 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 = (28 in.)i + (45 in.) j
AB = 53 in.

AC = (45 in.) j - (24 in.)k
AC = 51 in.

AD = -(26 in.)i + (45 in.) j + (18 in.)k
AD = 55 in.
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108
TAB
and
TAC
TAD
Equilibrium Condition with

AB
= TAB λAB = TAB
AB
= (0.5283i + 0.84906 j)TAB

AC
= TAC λAC = TAC
AC
= (0.88235 j - 0.47059k )TAC

AD
= TAD λAD = TAD
AD
= (-0.47273i + 0.81818 j + 0.32727k )TAD
W = - Wj
SF = 0: TAB + TAC + TAD - Wj = 0
PROBLEM 2.103 (Continued)
Substituting the expressions obtained for TAB , TAC , and TAD and factoring i, j, and k:
(0.5283TAB - 0.47273TAD )i + (0.84906TAB + 0.88235TAC + 0.81818TAD - W ) j
+ (-0.47059TAC + 0.32727TAD )k = 0
Equating to zero the coefficients of i, j, k:
0.5283TAB - 0.47273TAD = 0 (1)
0.84906TAB + 0.88235TAC + 0.81818TAD - W = 0 (2)
-0.47059TAC + 0.32727TAD = 0 (3)
Substituting TAD = 924 lb in Equations (1), (2), and (3) and solving the resulting set of equations, using
conventional algorithms for solving linear algebraic equations, gives:
TAB = 826.81 lb
TAC = 642.59 lb
W = 2030 lb ◀
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109
PROBLEM 2.104
A crate is supported by three cables as shown. Determine
the weight W of the crate, knowing that the tension in cable
AB is 1378 lb.
SOLUTION
See Problem 2.103 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3)
below:
0.5283TAB - 0.47273TAD = 0 (1)
0.84906TAB + 0.88235TAC + 0.81818TAD - W = 0 (2)
-0.47059TAC + 0.32727TAD = 0 (3)
Substituting TAB = 1378 lb in Equations (1), (2), and (3) above, and solving the resulting set of equations
using conventional algorithms, gives:
TAD = 1539.99 lb
TAC = 1070.98 lb
W = 3380 lb ◀
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110
PROBLEM 2.105
A 12-lb circular plate of 7-in. radius is supported as shown by three
wires, each of 25-in. length. Determine the tension in each wire,
knowing that α = 30°.
SOLUTION
Let q be angle between the vertical and any wire.
OA =
(25
2
)
- 72 = 24 in. thus cos q =
By symmetry TAB = TAC
24
25
SFx = 0:
-2(TAB sin q )(cos a ) + TAD (sin q ) = 0
For a = 30 :
TAD = 2 (cos30)TAB = 1.73205TAB
SFy = 0: 12 lb - 2TAB (cos q) - TAD (cos q) = 0
or
12 lb = (2TAB + TAD ) cos q
æ 24 ö
12 lb = (2TAB + 1.73205TAB )çç ÷÷÷
èç 25 ÷ø
TAB = TAC = 3.35 lb ◀
TAD = 5.80 lb ◀
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111
PROBLEM 2.106
Solve Prob. 2.105, knowing that α = 45°.
PROBLEM 2.105 A 12-lb circular plate of 7-in. radius is supported as
shown by three wires, each of 25-in. length. Determine the tension in each
wire, knowing that α = 30°.
SOLUTION
Let q be angle between the vertical and any wire.
OA =
(25
2
)
- 72 = 24 in. thus cos q =
By symmetry TAB = TAC
24
25
SFx = 0:
-2(TAB sin q )(cos a ) + TAD (sin q ) = 0
For a = 45 :
TAD = 2 (cos 45)TAB = 1.41421TAB
SFy = 0: 12 lb - 2TAB (cos q) - TAD (cos q) = 0
or
12 lb = (2TAB + TAD ) cos q
æ 24 ö
12 lb = (2TAB + 1.41421TAB )çç ÷÷÷
çè 25 ÷ø
TAB = TAC = 3.66 lb ◀
TAD = 5.18 lb ◀
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112
PROBLEM 2.107
y
220 mm
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.
D
960 mm
320 mm
C
380 mm
Q
O
240 mm
B
A
z
960 mm
P
x
SOLUTION
SFA = 0: TAB + TAC + TAD + P = 0 where P = P i

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

æ 48
AB
12
19 ö
TAB = TAB λAB = TAB
= TAB çç- i - j + k÷÷÷
ç
AB
53
53 ÷ø
è 53

æ 12
AC
3
4 ö
TAC = TAC λAC = TAC
= TAC çç- i - j - k÷÷÷
ç
AC
13
13 ø÷
è 13
305 N
[(-960 mm)i + (720 mm) j - (220 mm)k]
1220 mm
= -(240 N)i + (180 N) j - (55 N)k
TAD = TAD λAD =
Substituting into SFA = 0, factoring i, j, k, and setting each coefficient equal to f gives:
i: P=
j:
k:
48
12
TAB + TAC + 240 N
53
13
12
3
T + T = 180 N
53 A B 13 A C
19
4
T A B - TA C = 55 N
53
13
(1)
(2)
(3)
Solving the system of linear equations using conventional algorithms gives:
T A B = 446.71 N
P = 960 N ◀
T A C = 341.71 N
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113
PROBLEM 2.108
y
220 mm
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.
D
960 mm
320 mm
C
380 mm
Q
O
240 mm
B
A
z
960 mm
P
x
SOLUTION
We assume that T A D = 0 and write
SFA = 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

12
19 ö
AB æç 48
= ç- i - j + k ÷÷÷TAB
TAB = TAB λAB = TAB
ç
AB è 53
53
53 ÷ø

AC æç 12
3
4 ö
TAC = TAC λAC = TAC
= ç- i - j - k ÷÷÷TAC
ç
AC è 13
13
13 ø÷
Substituting into SFA = 0, factoring i, j, k, and setting each coefficient equal to f gives:
48
12
TAB - TAC + 1200 N = 0
53
13
12
3
j : - TAB - TAC + Q = 0
53
13
19
4
k:
TAB - TAC = 0
53
13
Solving the resulting system of linear equations using conventional algorithms gives:
i: -
TA B = 605.71 N
TA C = 705.71 N
Q = 300.00 N
(1)
(2)
(3)
0 £ Q < 300 N ◀
Note: This solution assumes that Q is directed upward as shown (Q ³ 0), if negative values of Q
are considered, cable AD remains taut, but AC becomes slack for Q = -460 N.
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114
PROBLEM 2.109
y
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.
A
480
250
O
B
D
360
130
320
z
450
x
360
C
Dimensions in mm
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 :
SF = 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
TAD

12
9 ö
AB æç 8
= TAB λAB = TAB
= ç- i - j + k ÷÷÷TAB Dimensions in mm
17
17 ÷ø
AB çè 17

AC
= TAC λAC = TAC
= (0.6 i - 0.64 j + 0.48k )TAC
AC

9.6
7.2 ö÷
AD æç 5
= TAD λAD = TAD
=ç ijk÷T
13
13 ÷÷ø AD
AD çè13
Substituting into the Eq. SF = 0 and factoring i, j, k :
æ 8
ö
çç- T + 0.6T + 5 T ÷÷ i
AB
AC
AD
÷ø÷
çè 17
13
æ 12
ö
9.6
+ çç- TAB - 0.64TAC TAD + P ÷÷÷ j
çè 17
÷ø
13
æ9
ö
7.2
+ çç TAB + 0.48TAC TAD ÷÷÷ k = 0
çè17
÷
13
ø
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115
PROBLEM 2.109 (Continued)
Setting the coefficient of i, j, k equal to zero:
i:
-
8
5
T + 0.6TA C + TA D = 0
17 A B
13
(1)
j:
-
12
9.6
TA B - 0.64TA C T +P=0
7
13 A D
(2)
9
7.2
TA B + 0.48T AC T =0
17
13 A D
(3)
8
5
T + 36 N + TA D = 0
17 A B
13
(1¢)
9
7.2
TA B + 28.8 N T =0
17
13 A D
(3¢)
k:
Making T A C = 60 N in (1) and (3):
-
Multiply (1¢) by 9, (3¢) by 8, and add:
554.4 N -
12.6
T = 0 TAD = 572.0 N
13 A D
Substitute into (1¢) and solve for TA B :
TA B =
ö
17 æç
5
çç36 + ´ 572÷÷÷÷
8è
13
ø
T A B = 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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116
PROBLEM 2.110
y
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.
A
480
250
O
B
D
360
130
320
z
x
360
450
C
Dimensions in mm
SOLUTION
See Problem 2.109 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and
(3) below:
8
5
TA B + 0.6TA C + TA D = 0
17
13
(1)
12
9.6
TAB + 0.64 TAC T +P =0
17
13 AD
(2)
-
9
7.2
TA B + 0.48T AC T =0
17
13 A D
(3)
Making T A D = 520 N in Eqs. (1) and (3):
8
T + 0.6TA C + 200 N = 0
17 A B
(1¢)
9
T + 0.48TA C - 288 N = 0
17 A B
(3¢)
-
Multiply (1¢) by 9, (3¢) by 8, and add:
9.24T A C - 504 N = 0 T A C = 54.5455 N
Substitute into (1¢) and solve for TA B :
TA B =
17
(0.6 ´ 54.5455 + 200) T A B = 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
= 768.00 N
P=
Weight of plate = P = 768 N ◀
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117
PROBLEM 2.111
y
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 840 lb, determine the vertical force
P exerted by the tower on the pin at A.
A
100 ft
D
20 ft
25 ft
B
O
74 ft
C
20 ft
60 ft
18 ft
z
x
SOLUTION
SF = 0: TAB + TAC + TAD + Pj = 0
Free-Body Diagram at A:

AB = -20 i - 100 j + 25k AB = 105 ft

AC = 60 i - 100 j + 18k AC = 118 ft

AD = -20i - 100 j - 74 k AD = 126 ft
We write

AB
TAB = TAB λ AB = TAB
AB
æ 4
20
5 ö
= çç- i - j + k÷÷÷TAB
çè 21
21
21 ÷ø

AC
TAC = TAC λAC = TAC
AC
æ 30
50
9 ö
= çç i j + k÷÷÷TAC
59
59 ø÷
èç 59

AD
TAD = TAD λAD = TAD
AD
æ 10
50
37 ö
ç
= ç- i - j - k÷÷÷TAD
63
63 ø÷
èç 63
Substituting into the Eq. SF = 0 and factoring i, j, k :
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written consent of McGraw-Hill Education.
118
PROBLEM 2.111 (Continued)
æ 4
ö
çç- T + 30 T - 10 T ÷÷ i
AB
AC
AD
÷
çè 21
÷ø
59
63
æ 20
ö
50
50
+ çç- TAB - TAC - TAD + P ÷÷÷ j
çè 21
59
63
ø÷
æ5
ö
9
37
+ çç TAB + TAC - TAD ÷÷÷ k = 0
çè 21
59
63
ø÷
Setting the coefficients of i, j, k, equal to zero:
i:
-
4
30
10
T + T - T =0
21 A B 59 A C 63 A D
(1)
j:
-
20
50
50
TA B - TA C - TA D + P = 0
21
59
63
(2)
5
9
37
TA B + TA C - TA D = 0
21
59
63
(3)
k:
Set TA B = 840 lb in Eqs. (1) - (3):
30
10
TA C - TA D = 0
59
63
(1¢)
50
50
T - T +P =0
59 A C 63 A D
(2¢)
9
37
TA C - TA D = 0
59
63
(3¢)
-160 lb +
-800 lb -
200 lb +
Solving,
T A C = 458.12 lb T A D = 459.53 lb P = 1552.94 lb
P = 1553 lb ◀
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written consent of McGraw-Hill Education.
119
PROBLEM 2.112
y
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 590 lb, determine the vertical force
P exerted by the tower on the pin at A.
A
100 ft
D
20 ft
25 ft
B
O
74 ft
C
20 ft
60 ft
18 ft
z
x
SOLUTION
SF = 0: TAB + TAC + TAD + Pj = 0
We write
Free-Body Diagram at A:

AB = -20 i - 100 j + 25k AB = 105 ft

AC = 60 i - 100 j + 18k AC = 118 ft

AD = -20i - 100 j - 74 k AD = 126 ft

AB
TAB = TAB λ AB = TAB
AB
æ 4
20
5 ö
= çç- i - j + k÷÷÷TAB
çè 21
21
21 ÷ø

AC
TAC = TAC λAC = TAC
AC
æ 30
50
9 ö
= çç i j + k÷÷÷TAC
çè 59
59
59 ÷ø

AD
TAD = TAD λAD = TAD
AD
æ 10
50
37 ö
= çç- i - j - k÷÷÷TAD
63
63 ø÷
èç 63
Substituting into the Eq. SF = 0 and factoring i, j, k :
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written consent of McGraw-Hill Education.
120
PROBLEM 2.112 (Continued)
æ 4
ö
çç- T + 30 T - 10 T ÷÷ i
AB
AC
AD
÷ø÷
çè 21
59
63
æ 20
ö
50
50
+ çç- TAB - TAC - TAD + P ÷÷÷ j
çè 21
÷ø
59
63
æ5
ö
9
37
+ çç TAB + TAC - TAD ÷÷÷ k = 0
çè 21
59
63
ø÷
Setting the coefficients of i, j, k, equal to zero:
i:
-
4
30
10
TA B + TA C - TA D = 0
21
59
63
(1)
j:
-
20
50
50
TA B - TA C - TA D + P = 0
21
59
63
(2)
5
9
37
T + T - T =0
21 A B 59 A C 63 A D
(3)
k:
Set T A C = 590 lb in Eqs. (1) – (3):
4
10
T A B + 300 lb - TA D = 0
21
63
(1¢)
20
50
TA B - 500 lb - TA D + P = 0
21
63
(2¢)
-
5
37
T + 90 lb - T A D = 0
21 A B
63
Solving,
TA B = 1081.82 lb TA D = 591.82 lb
(3¢)
P = 2000 lb ◀
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121
PROBLEM 2.113
y
O
B
C
4 ft
x
8 ft
z
16 ft
12 ft
32 ft
A
In trying to move across a slippery icy surface, a 175-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.
30 ft
SOLUTION
Free-Body Diagram at A
æ 16
30
N = N çç i +
çè 34
34
ö
j÷÷÷
÷ø
and W = W j = -(175 lb) j

(-30 ft)i + (20 ft) j - (12 ft)k
AC
TAC = TAC λ AC = TAC
= TAC
38 ft
AC
æ 15
10
6 ö
= TAC çç- i + j - k÷÷÷
çè 19
19
19 ø÷

(-30 ft)i + (24 ft) j + (32 ft)k
AB
= TAB
TAB = TAB λ AB = TAB
50 ft
AB
æ 15
12
16 ö
= TAB çç- i + j + k÷÷÷
çè 25
25
25 ÷ø
Equilibrium condition: SF = 0
TA B + TA C + N + W = 0
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122
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:
From i:
From j:
From k:
-
15
15
16
TA B - TA C + N = 0
25
19
34
12
10
30
TA B + TA C + N - (175 lb) = 0
25
19
34
16
6
T - T =0
25 A B 19 A C
(1)
(2)
(3)
Solving the resulting set of equations gives:
T A B = 30.8 lb; T A C = 62.5 lb ◀
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123
PROBLEM 2.114
y
O
B
C
4 ft
x
8 ft
z
16 ft
12 ft
32 ft
Solve Problem 2.113, assuming that a friend is helping the man
at A by pulling on him with a force P =-(45 lb)k.
PROBLEM 2.113 In trying to move across a slippery icy surface,
a 175-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.
A
30 ft
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 = (- 45 lb) k.
15
15
16
TA B - TA C + N = 0
25
19
34
(1)
12
10
30
T + T + N - (175 lb) = 0
25 A B 19 A C 34
(2)
16
6
TA B - TA C - (45 lb) = 0
25
19
(3)
-
Solving the resulting set of equations simultaneously gives:
T A B = 81.3 lb ◀
TA C = 22.2 lb ◀
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124
PROBLEM 2.115
y
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.
A
480
250
O
B
D
360
130
320
z
450
x
360
C
Dimensions in mm
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
TA B + 0.6TA C + TA D = 0
17
13
(1)
12
9.6
TA B - 0.64TA C T + 792 N = 0
17
13 A D
(2)
9
7.2
TA B + 0.48TA C T =0
17
13 A D
(3)
-
Solving Equations (1), (2), and (3) by conventional algorithms gives
TA B = 510.00 N
TA B = 510 N ◀
T A C = 56.250 N
TA C = 56.2 N ◀
T A D = 536.25 N
T A D = 536 N ◀
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125
PROBLEM 2.116
y
220 mm
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.
D
960 mm
320 mm
C
380 mm
Q
O
240 mm
B
A
z
P
960 mm
x
SOLUTION
SFA = 0: TAB + TAC + TAD + P + Q = 0
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

æ 48
12
19 ö
AB
TAB = TAB λAB = TAB
= TAB çç- i - j + k÷÷÷
ç
53
53 ÷ø
AB
è 53

æ 12
3
4 ö
AC
TAC = TAC λAC = TAC
= TAC çç- i - j - k÷÷÷
ç
13
13 ÷ø
AC
è 13

æ 48
36
11 ö
AD
TAD = TAD λAD = TAD
j - k÷÷÷
= TAD çç- i +
çè 61
61
61 ø÷
AD
Substituting into SFA = 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:
Where
i: -
48
12
48
T - T - T + 2880 N = 0
53 AB 13 AC 61 AD
(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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126
PROBLEM 2.116 (Continued)
Solving the system of linear equations using conventional algorithms gives:
T A B = 1340.14 N
TA C = 1025.12 N
TA B = 1340 N ◀
TA D = 915.03 N
TA C = 1025 N ◀
TA D = 915 N ◀
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127
PROBLEM 2.117
y
220 mm
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.
D
960 mm
320 mm
C
380 mm
Q
O
240 mm
B
A
z
P
960 mm
x
SOLUTION
See Problem 2.116 for the analysis leading to the linear algebraic Equations (1), (2), and (3) below:
-
48
12
48
TA B - TA C - TA D + P = 0
53
13
61
-
12
3
36
TA B - TA C + TA D + Q = 0
53
13
61
19
4
11
TA B - TA C - TA D = 0
53
13
61
(1)
(2)
(3)
Setting P = 2880 N and Q = 576 N gives:
-
48
12
48
TA B - TA C - TA D + 2880 N = 0
53
13
61
(1¢)
12
3
36
TA B - T A C + T A D + 576 N = 0
53
13
61
19
4
11
T - T - T =0
53 A B 13 A C 61 A D
(2¢)
-
(3¢)
Solving the resulting set of equations using conventional algorithms gives:
T A B = 1431.00 N
TA C = 1560.00 N
TA D = 183.010 N
TA B = 1431 N ◀
TA C = 1560 N ◀
T A D = 183.0 N ◀
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128
PROBLEM 2.118
Three cables are connected at D, where an upward force of
30 kN is applied. Determine the tension in each cable.
SOLUTION
Free-Body Diagram of Point D:
SFD = 0: TDA + TDB + TDC + P = 0
Where
P = Pj

DA = -(1 m)i DA = 1 m

DB = +(0.75 m)i - (3 m) j - (1 m)k DB = 3.25 m

DC = +(1 m)i - (3 m) j + (1.5 m)k DC = 3.5 m

DA
= TDA (-i)
TDA = TDA λDA = TDA
DA

æ3
DB
12
4 ö
= TDB çç i - j - k÷÷÷
TDB = TDB λDB = TDB
ç
DB
13
13 ø÷
è13

æ2
DC
6
3 ö
= TDC çç i - j + k÷÷÷
TDC = TDC λDC = TDC
ç
DC
7
7 ÷ø
è7
Substituting into SFD = 0, setting P = (30 kN) j and setting the coefficients of i, j, k equal to 0, we
obtain the following three equilibrium equations:
i : - TDA +
3
2
TDB + TDC = 0
13
7
(1)
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129
PROBLEM 2.118 (Continued)
j: -
12
6
TDB - TDC + 30 kN = 0 (2)
13
7
k: -
4
3
T + T =0
13 DB 7 DC
(3)
Solving the system of linear equations using conventional algorithms gives:
TDA = 8.50 kN ◀
TDB = 19.50 kN ◀
TDC = 14.00 kN ◀
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130
PROBLEM 2.119
y
For the transmission tower of Probs. 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
1800 lb.
A
100 ft
D
20 ft
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 840 lb, determine
the vertical force P exerted by the tower on the pin at A.
25 ft
B
O
74 ft
C
20 ft
60 ft
18 ft
z
x
SOLUTION
See Problem 2.111 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and
(3) below:
i:
-
4
30
10
TA B + TA C - TA D = 0
21
59
63
(1)
j:
-
20
50
50
TA B - TA C - TA D + P = 0
21
59
63
(2)
5
9
37
TA B + TA C - TA D = 0
21
59
63
(3)
k:
Substituting for P = 1800 lb in Equations (1), (2), and (3) above and solving the resulting set of
equations using conventional algorithms gives:
-
4
30
10
TA B + TA C - TA D = 0
21
59
63
20
50
50
T A B - T A C - T A D + 1800 lb = 0
21
59
63
5
9
37
TA B + TA C - TA D = 0
21
59
63
(1¢)
(2¢)
(3¢)
T A B = 973.64 lb
T A C = 531.00 lb
T A D = 532.64 lb
T A B = 974 lb ◀
TA C = 531 lb ◀
T A D = 533 lb ◀
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131
PROBLEM 2.120
24 in.
B
16 in.
24 in.
Three wires are connected at point D, which is
located 18 in. below the T-shaped pipe support
ABC. Determine the tension in each wire when
a 180-lb cylinder is suspended from point D as
shown.
22 in.
C
A
18 in.
D
180 lb
SOLUTION
Free-Body Diagram of Point D:
The forces applied at D are:
TDA , TDB , TDC and W
where W = - 180.0 lbj . To express the other forces in terms of the unit vectors i, j, k, we write

DA = (18 in.) j + (22 in.)k
DA = 28.425 in.

DB = -(24 in.)i + (18 in.) j - (16 in.)k
DB = 34.0 in.

DC = (24 in.)i + (18 in.) j - (16 in.)k
DC = 34.0 in.
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132
PROBLEM 2.120 (Continued)

DA
TDA = TDa λDA = TDa
DA
= (0.63324 j + 0.77397k )TDA

DB
TDB = TDB λDB = TDB
DB
= (-0.70588i + 0.52941j - 0.47059k )TDB

DC
TDC = TDC λDC = TDC
DC
= (0.70588i + 0.52941j - 0.47059k )TDC
and
Equilibrium Condition with
W = -W j
SF = 0: TDA + TDB + TDC - W j = 0
Substituting the expressions obtained for TDA , TDB , and TDC and factoring i, j, and k:
(-0.70588TDB + 0.70588TDC )i
(0.63324TDA + 0.52941TDB + 0.52941TDC - W ) j
(0.77397TDA - 0.47059TDB - 0.47059TDC )k
Equating to zero the coefficients of i, j, k:
-0.70588T DB + 0.70588T DC = 0
0.63324TDA + 0.52941T DB + 0.52941TDC - W = 0
0.77397T DA - 0.47059T DB - 0.47059T DC = 0
(1)
(2)
(3)
Substituting W = 180 lb in Equations (1), (2), and (3) above, and solving the resulting set of equations
using conventional algorithms gives,
T DA = 119.7 lb ◀
T DB = 98.4 lb ◀
T DC = 98.4 lb ◀
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133
PROBLEM 2.121
y
A container of weight W is suspended from ring
A, to which cables AC and AE are attached. A
force P is applied to the end F of a third cable
that passes over a pulley at B and through ring A
and that is attached to a support at D. Knowing
that W = 1000 N, determine the magnitude of P.
(Hint: The tension is the same in all portions of
cable FBAD.)
0.40 m
0.86 m
1.20 m
E
B
1.30 m
O
0.78 m
C
F
D
P
z
1.60 m
0.40 m
x
A
W
SOLUTION
The (vector) force in each cable can be written as the product of the (scalar) force and the unit vector
along the cable. That is, with

AB = -(0.78 m)i + (1.6 m) j + (0 m)k
AB = (-0.78 m)2 + (1.6 m)2 + (0)2
= 1.78 m
TAB = T λAB = TAB

AB
AB
TAB
[-(0.78 m)i + (1.6 m) j + (0 m)k ]
1.78 m
= TAB (-0.4382i + 0.8989 j + 0 k )
=
and
TAB

AC = (0)i + (1.6 m) j + (1.2 m)k
and
AC = (0 m)2 + (1.6 m)2 + (1.2 m)2 = 2 m

AC TAC
TAC = T λAC = TAC
=
[(0)i + (1.6 m) j + (1.2 m)k ]
AC 2 m
TAC = TAC (0.8 j + 0.6 k )

AD = (1.3 m)i + (1.6 m) j + (0.4 m)k
AD = (1.3 m)2 + (1.6 m)2 + (0.4 m)2 = 2.1 m

T
AD
TAD = T λAD = TAD
= AD [(1.3 m)i + (1.6 m) j + (0.4 m)k ]
AD 2.1 m
TAD = TAD (0.6190i + 0.7619 j + 0.1905k )
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134
PROBLEM 2.121 (Continued)

AE = -(0.4 m)i + (1.6 m) j - (0.86 m)k
Finally,
AE = (-0.4 m)2 + (1.6 m)2 + (-0.86 m)2 = 1.86 m

AE
TAE = T λAE = TAE
AE
TAE
=
[-(0.4 m)i + (1.6 m) j - (0.86 m)k ]
1.86 m
TAE = TAE (-0.2151i + 0.8602 j - 0.4624 k )
With the weight of the container
W = -W j, at A we have:
SF = 0: TA B + TA C + TA D - W j = 0
Equating the factors of i, j, and k to zero, we obtain the following linear algebraic equations:
-0.4382TA B + 0.6190T A D - 0.2151TA E = 0
0.8989TA B + 0.8T A C + 0.7619TA D + 0.8602T A E - W = 0
0.6T A C + 0.1905T A D - 0.4624TA E = 0
(1)
(2)
(3)
Knowing that W = 1000 N and that because of the pulley system at B TA B = TAD = P, where P is the
externally applied (unknown) force, we can solve the system of linear Equations (1), (2) and (3) uniquely
for P.
P = 378 N ◀
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135
PROBLEM 2.122
y
Knowing that the tension in cable AC of the
system described in Problem 2.121 is 150 N,
determine (a) the magnitude of the force P, (b)
the weight W of the container.
0.40 m
0.86 m
1.20 m
E
B
1.30 m
O
0.78 m
C
F
D
P
z
1.60 m
0.40 m
A
W
x
PROBLEM 2.121 A container of weight W is
suspended from ring A, to which cables AC and
AE are attached. A force P is applied to the end F
of a third cable that passes over a pulley at B and
through ring A and that is attached to a support at
D. Knowing that W = 1000 N, determine the
magnitude of P. (Hint: The tension is the same in
all portions of cable FBAD.)
SOLUTION
Here, as in Problem 2.121, the support of the container consists of the four cables AE, AC, AD, and AB,
with the condition that the force in cables AB and AD is equal to the externally applied force P. Thus,
with the condition
TA B = TA D = P
and using the linear algebraic equations of Problem 2.131 with T A C = 150 N, we obtain
(a)
P = 454 N ◀
(b)
W = 1202 N ◀
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136
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 = 270 lb,
determine P and Q. (Hint: The tension is the same in
both portions of cable BAC.)
SOLUTION
Free-Body Diagram of Point A:
TAB = T λAB

AB
=T
AB
(-48 in.)i + (72 in.) j - (16 in.)k
=T
88 in.
æ 6
9
2 ö
= T çç- i + j - k÷÷÷
11
11 ø÷
èç 11
TAC = T λAC

AC
=T
AC
(24 in.)i + (72 in.) j + (-13 in.)k
=T
77 in.
æ 24
72
13 ö
j - k÷÷÷
= T çç i +
çè 77
77
77 ÷ø
SF = 0: TAB + TAC + Q + P + W = 0
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137
PROBLEM 2.123 (continued)
Setting coefficients of i, j, k equal to zero:
Data:
i: -
6
24
T + T +P=0
11
77
æ 18 ö
- çç ÷÷÷T + P = 0
çè 77 ø÷
(1)
j: +
9
72
T + T -W = 0
11
77
æ135 ö÷
çç
÷T - W = 0
èç 77 ø÷÷
(2)
k: -
2
13
T - T +Q = 0
11
77
æ 27 ö
- çç ÷÷÷T + Q = 0
çè 77 ÷ø
(3)
æ 77 ÷ö
÷ 270 lb T = 154.0 lb
W = 270 lb T = çç
çè135 ÷÷ø
Substituting for T in Eqn . (1) and (2):
P = 36.0 lb ◀
Q = 54.0 lb ◀
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138
PROBLEM 2.124
For the system of Prob. 2.123, determine W and P knowing
that Q = 36 lb.
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 = 270 lb, 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). Setting Q = 36 lb
we have:
æ 77 ö
T = çç ÷÷÷ 36 lb
èç 27 ø÷
Eq. (3):
Eq. (1):
Eq. (2):
-
18
(102.667 lb) + P = 0
77
135
(102.667 lb) - W = 0
77
T = 102.667 lb
P = 24.0 lb ◀
W = 180.0 lb ◀
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written consent of McGraw-Hill Education.
139
PROBLEM 2.125
y
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.
P
A
y
200 mm
O
z
B
z
x
Q
SOLUTION
Free-Body Diagrams of Collars:
For both Problems 2.125 and 2.126:
( A B )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
(0.20 i - 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:
SF = 0: N x i + N z k + Pj + T A B lA B = 0
Setting the j coefficient to zero gives
P - (1.90476 y )T A B = 0
P = 341 N
With
TAB =
341 N
Now, from the free body diagram of collar B:
1.90476 y
SF = 0: N x i + N y j + Qk - T A B λA B = 0
Setting the k coefficient to zero gives
Q - TA B (1.90476z ) = 0
And using the above result for TAB , we have
Q = TA B z =
341 N
(1.90476) y
(1.90476z ) =
(341 N)(z )
y
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140
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
TA B =
(a)
TA B = 1155 N ◀
or
and
Q=
(b)
341 N(0.46 m)(0.866)
(0.155 m)
= (1012.00 N)
Q = 1012 N ◀
or
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141
PROBLEM 2.126
y
Solve Problem 2.125 assuming that y = 275 mm.
P
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.
A
y
200 mm
O
z
B
z
x
Q
SOLUTION
From the analysis of Problem 2.125, particularly the results:
y 2 + z 2 = 0.23563 m 2
341 N
1.90476 y
341 N
Q=
z
y
TAB =
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)
T A B = 651 N ◀
or
and
Q=
(b)
341 N(0.40 m)
(0.275 m)
Q = 496 N ◀
or
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142
PROBLEM 2.127
Q
P
85°
55°
25°
A
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
g = 180 - (20 + 10)
= 150
R 2 = (48 N)2 + (60 N)2
- 2(48 N)(60 N) cos150
R = 104.366 N
Then
48 N 104.366 N
=
sin a
sin150
sin a = 0.22996
and
a = 13.2947
Hence:
f = 180 - a - 80
= 180 - 13.2947 - 80
= 86.705
R = 104.4 N
86.7° ◀
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143
PROBLEM 2.128
y
120 N
Determine the x and y components of each of the forces shown.
80 N
150 N
30°
35°
40°
x
SOLUTION
80-N Force:
120-N Force:
150-N Force:
Fx = +(80 N) cos 40 
Fx = 61.3 N ◀
Fy = +(80 N)sin 40
Fy = 51.4 N ◀
Fx = +(120 N) cos 70 
Fx = 41.0 N ◀
Fy = +(120 N)sin 70
Fy = 112.8 N ◀
Fx = -(150 N) cos 35 
Fx = -122.9 N ◀
Fy = +(150 N)sin 35
Fy = 86.0 N ◀
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144
PROBLEM 2.129
P
α
400 lb
A hoist trolley is subjected to the three forces shown. Knowing that
α = 40°, 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.
α
200 lb
SOLUTION
Rx =
S Fx = P + (200 lb) sin 40  - (400 lb) cos 40 
R x = P - 177.860 lb
Ry =
SFy = (200 lb) cos 40 + (400 lb)sin 40
R y = 410.32 lb
(a)
(1)
(2)
For R to be vertical, we must have R x = 0.
Set
R x = 0 in Eq. (1)
0 = P - 177.860 lb
P = 177.9 lb ◀
P = 177.860 lb
(b)
Since R is to be vertical:
R = R y = 410 lb
R = 410 lb ◀
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145
PROBLEM 2.130
B
α
30°
Knowing that a = 55 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.
C
20°
300 lb
A
SOLUTION
Free-Body Diagram
Law of sines:
Force Triangle
FAC
T
300 lb
= BC =
sin 35 sin 50 sin 95
(a)
FAC =
300 lb
sin 35
sin 95
FA C = 172.7 lb ◀
(b)
TBC =
300 lb
sin 50
sin 95
T B C = 231 lb ◀
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146
PROBLEM 2.131
600 mm
A
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.
B
250 mm
P
C
3
4
Q = 480 N
SOLUTION
Free Body: C
(a)
SFx = 0: -
(b)
SFy = 0:
12
4
TAC + (360 N) = 0
13
5
TA C = 312 N ◀
5
3
(312 N) + TBC + (360 N) - 480 N = 0
13
5
T B C = 480 N - 120 N - 216 N
TBC = 144.0 N ◀
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147
A
B
35°
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 a.
50°
C
PROBLEM 2.132
α
P
SOLUTION
Free-Body Diagram: C
Force Triangle
Force triangle is isosceles with
2b = 180 - 85
b = 47.5
P = 2(800 N)cos 47.5° = 1081 N
(a)
Since P > 0, the solution is correct.
(b)
P = 1081 N ◀
a = 180 - 50 - 47.5 = 82.5
a = 82.5 ◀
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148
y
PROBLEM 2.133
A
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 qx, qy, and qz that the force forms with the coordinate
axes.
E
36°
B
60°
20°
48°
C
x
D
z
SOLUTION
(a)
Fx = (120 lb)cos 60 cos 20
Fx = 56.382 lb
Fx = + 56.4 lb ◀
Fy = -(120 lb)sin 60
Fy = -103.923 lb
Fz = -(120 lb) cos 60 sin 20
Fz = -20.521 lb
(b)
cos qx =
cos qy =
cos qz =
Fx 56.382 lb
=
120 lb
F
Fy
F
=
-103.923 lb
120 lb
Fz -20.52 lb
=
120 lb
F
Fy = -103.9 lb ◀
Fz = - 20.5 lb ◀
qx = 62.0  ◀
qy = 150.0 ◀
qz = 99.8  ◀
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149
PROBLEM 2.134
y
360 mm
Knowing that the tension in cable AC is 2130 N, determine
the components of the force exerted on the plate at C.
920 mm
A
O
B
x
600 mm D
z
900 mm
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 λCA

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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150
y
P
PROBLEM 2.135
Q
Find the magnitude and direction of the resultant of the two forces
shown knowing that P = 600 N and Q = 450 N.
40°
55°
30°
O
x
z 25°
SOLUTION
P = (600 N)[sin 40 sin 25i + cos 40 j + sin 40 cos 25k ]
= (162.992 N)i + (459.63 N) j + (349.54 N)k
Q = (450 N)[cos 55 cos30i + sin 55 j - cos 55 sin 30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
= 940.22 N
cos qx =
cos qy =
cos qz =
R = 940 N ◀
Rx 386.52 N
=
R
940.22 N
Ry
qx = 65.7  ◀
828.25 N
940.22 N
qy = 28.2 ◀
Rz 220.49 N
=
940.22 N
R
qz = 76.4  ◀
R
=
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151
PROBLEM 2.136
y
17.5 in.
25 in.
B
45 in.
E
O
D
C
x
z
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.
60 in.
A
80 in.
P
SOLUTION
Free Body Diagram at A:
Since T B A C = tension in cable BAC, it follows that
TA B = TA C = TBA C
TAB = TBAC λAB = TBAC
TAC = TBAC λAC = TBAC
TAD = TAD λAD = TAD
TAE = TAE λAE = TAE
(-17.5 in.)i + (60 in.) j
62.5 in.
(60 in.)i + (25 in.)k
65 in.
(80 in.)i + (60 in.) j
100 in.
(60 in.) j - (45 in.)k
75 in.
æ -17.5
60 ö÷
i+
j÷
= TBAC çç
çè 62.5
62.5 ø÷÷
æ 60
25 ö
= TBAC çç j + k÷÷÷
çè 65
65 ÷ø
æ4
3 ö
= TAD çç i + j÷÷÷
çè 5
5 ø÷
æ4
3 ö
= TAE çç j - k÷÷÷
çè 5
5 ÷ø
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152
PROBLEM 2.136 (Continued)
Substituting into S FA = 0, setting P = (-200 lb) j, and setting the coefficients of i, j, k equal to f, we
obtain the following three equilibrium equations:
17.5
4
TBAC + TAD = 0
62.5
5
From
i: -
From
æ 60
60 ö
3
4
j : çç
+ ÷÷÷TBAC + TAD + TAE - 200 lb = 0
çè 62.5 65 ÷ø
5
5
From
k:
(1)
25
3
TBAC - TAE = 0
65
5
(2)
(3)
Solving the system of linear equations using conventional algorithms gives:
TBA C = 76.7 lb; TA D = 26.9 lb; TA E = 49.2 lb ◀
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153
PROBLEM 2.137
y
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.
x
P
A
O
20 in.
Q
B
z
z
x
SOLUTION
Free-Body Diagrams of Collars:
A:
B:

AB -xi - (20 in.) j + zk
λAB =
=
AB
25 in.
SF = 0: Pi + N y j + N z k + TAB λ AB = 0
Collar A:
Substitute for λA B and set coefficient of i equal to zero:
TAB x
=0
25 in.
SF = 0: (60 lb)k + N x¢ i + N y¢ j - TAB λ AB = 0
P-
Collar B:
(1)
Substitute for λA B and set coefficient of k equal to zero:
60 lb x = 9 in.
(a)
TAB z
=0
25 in.
(2)
(9 in.)2 + (20 in.)2 + z 2 = (25 in.)2
z = 12 in.
From Eq. (2):
(b)
From Eq. (1):
60 lb - TAB (12 in.)
25 in.
P=
(125.0 lb)(9 in.)
25 in.
T A B = 125.0 lb ◀
P = 45.0 lb ◀
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154
PROBLEM 2.138
y
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.
x
P
A
O
20 in.
Q
B
z
z
x
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
T A B x = (25 in.)(20 lb)
(1¢)
From Eq. (2):
T A B z = (25 in.)(60 lb)
(2¢)
x
=2
z
Dividing Eq. (1¢) by (2¢),
Now write
(3)
x 2 + z 2 + (20 in.)2 = (25 in.)2
(4)
Solving (3) and (4) simultaneously,
4z 2 + z 2 + 400 = 625
z 2 = 45
z = 6.7082 in.
From Eq. (3):
x = 2z = 2(6.7082 in.)
= 13.4164 in.
x = 13.42 in., z = 6.71 in. ◀
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155
1100 mm
PROBLEM 2F1
400 mm
Two cables are tied together at C and loaded as shown. Draw
the free-body diagram needed to determine the tension in AC
and BC.
B
A
960 mm
C
1600 kg
SOLUTION
Free-Body Diagram of Point C:
W = (1600 kg)(9.81 m/s2 )
W = 15.6960(103 ) N
W = 15.696 kN
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156
TA
TB
40°
TD
PROBLEM 2.F2
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, draw the free-body diagram
needed to determine the magnitudes of the forces TC and TD.
TC
SOLUTION
Free-Body Diagram of Point E:
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157
PROBLEM 2.F3
15 in.
B
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.
C
65 lb
A
h
60 lb
SOLUTION
Free-Body Diagram of Point A:
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158
A
14 m
24 m
PROBLEM 2.F4
6m
8.25 m
B
E
10 m
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.
C
F
D
1.10 m
SOLUTION
Free-Body Diagram of Point B:
W E = 250 N + 765 N = 1015 N
8.25
qA B = tan-1
= 30.510
14
10
qBC = tan-1
= 22.620
24
Use this free body to determine TAB and TBC.
Free-Body Diagram of Point C:
qCD = tan-1
1.1
= 10.3889
6
Use this free body to determine TCD and WF.
Then weight of skier WS is found by
W S = W F - 250 N ◀
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159
PROBLEM 2.F5
y
Three cables are used to tether a balloon as shown. Knowing that the
tension in cable AC is 444 N, draw the free-body diagram needed to
determine the vertical force P exerted by the balloon at A.
A
5.60 m
B
D
4.20 m
O
C
z
2.40 m
3.30 m
4.20 m
x
SOLUTION
Free-Body Diagram of Point A:
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160
PROBLEM 2.F6
y
360 mm
C
D
450 mm
O
500 mm
B
A container of mass m = 120 kg is supported by three cables
as shown. Draw the free-body diagram needed to determine
the tension in each cable.
320 mm
z
x
A
600 mm
SOLUTION
Free-Body Diagram of Point A:
W = (120 kg)(9.81 m/s2 )
= 1177.2 N
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161
PROBLEM 2.F7
y
10.8 ft
A
A 150-lb cylinder is supported by two cables AC and BC that are
attached to the top of vertical posts. A horizontal force P, which
is 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.
10.8 ft
B
3.6 ft
C
P
7.2 ft
O
15 ft
x
z
SOLUTION
Free-Body Diagram of Point C:
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162
PROBLEM 2.F8
y
A
90 ft
20 ft
30 ft
B
A transmission tower is held by three guy wires attached
to a pin at A and anchored by bolts at B, C, and D.
Knowing that the tension in wire AB is 630 lb, draw the
free-body diagram needed to determine the vertical force
P exerted by the tower on the pin at A.
D
O
60 ft
45 ft
z
30 ft
C
65 ft
x
SOLUTION
Free-Body Diagram of point A:
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163