Vector mechanics for engineers statics and dynamics 11th edition beer solutions manual

Vector Mechanics for Engineers Statics and Dynamics 11th Edition
Beer SOLUTIONS MANUAL
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CHAPTER 2
Copyright © McGraw-Hill Education. Permission required for reproduction or display.

SOLUTION
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.
(a) Parallelogram law:
(b) Triangle rule:
We measure: R =1391 kN, α = 47.8° R =1391 N 47.8° W

SOLUTION
PROBLEM 2.2
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.
(b) Triangle rule:
We measure: R = 906 lb, α = 26.6° R = 906 lb 26.6° W

SOLUTION
(b) Triangle rule:
PROBLEM 2.3
Two structural members B and C are bolted to bracket A. Knowing that
both members are in tension and that P = 10 kN and Q = 15 kN,
determine graphically the magnitude and direction of the resultant force
exerted on the bracket using (a) the parallelogram law, (b) the triangle
rule.
We measure: R = 20.1 kN, α = 21.2° R = 20.1kN 21.2° W

SOLUTION
(b) Triangle rule:
PROBLEM 2.4
Two structural members B and C are bolted to bracket A. Knowing that
both members are in tension and that P = 6 kips and Q = 4 kips,
determine graphically the magnitude and direction of the resultant force
exerted on the bracket using (a) the parallelogram law, (b) the triangle
rule.
We measure: R = 8.03 kips, α = 3.8° R = 8.03 kips 3.8° W

SOLUTION
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.
Using the triangle rule and the law of sines:
(a)
120 N
sin30°
=
P
sin 25°
P = 101.4 N W
(b) 30° + β + 25° = 180°
β = 180° − 25° − 30°
= 125°
120 N
sin30°
=
R
sin125°
R = 196.6 N W

SOLUTION
PROBLEM 2.6
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.
(a) 75° + 40° + α = 180°
α = 180° − 75° − 40°
= 65°
800 lb
=
T2
T =853 lb W
sin 65° sin 75°
2
(b)
800 lb
=
R
R = 567 lb W
sin 65° sin 40°

SOLUTION
PROBLEM 2.7
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.
(a) 75° + 40° + β = 180°
β = 180° − 75° − 40°
= 65°
1000 lb
=
T1
T = 938 lb W
sin75° sin65°
1
(b)
1000 lb
=
R
sin75° sin 40°
R = 665 lb W

PROBLEM 2.8
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:
TAC
=
R
=
2.2 kN
sin30° sin125°
TAC = 2.603 kN
R = 4.264 kN
sin 25D
(a) TAC = 2.60 kN W
(b) R = 4.26 kN W

SOLUTION
PROBLEM 2.9
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.
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
sinα
=
sin 30°
3 kN 2.6643 kN
α = 34.3°
TAC = 2.66 kN 34.3° W

SOLUTION
PROBLEM 2.10
Two forces are applied as shown to a hook support. Knowing that the
magnitude of P is 35 N, determine by trigonometry (a) the required
angle α if the resultant R of the two forces applied to the support is to
be horizontal, (b) the corresponding magnitude of R.
Using the triangle rule and law of sines:
(a)
(b)
sin α
=
sin 25°
50 N 35 N
sin α = 0.60374
α = 37.138°
α + β + 25° = 180°
β = 180° − 25° − 37.138°
= 117.862°
α = 37.1° W
R
=
sin117.862°
35 N
sin 25°
R = 73.2 N W

SOLUTION
PROBLEM 2.11
A steel tank is to be positioned in an excavation. Knowing
that α = 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.
(a) β + 50° + 60° = 180°
β = 180° − 50° − 60°
= 70°
(b)
425 lb
=
sin 70°
425 lb
=
sin 70°
P
sin60°
R
sin50°
P = 392 lb W
R = 346 lb W

PROBLEM 2.12
A steel tank is to be positioned in an excavation. Knowing
that the magnitude of P is 500 lb, determine by
trigonometry (a) the required angle α if the resultant R of
the two forces applied at A is to be vertical, (b) the
SOLUTION
corresponding magnitude of R.
(a) (α + 30°) + 60° + β =180°
β =180° − (α + 30°) − 60°
β = 90° −α
sin(90° −α)
=
sin60°
425 lb 500 lb
90° −α = 47.402° α = 42.6° W
(b)
R
=
500 lb
R = 551 lb W
sin(42.598° + 30°) sin60°

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
SOLUTION
of the two forces applied at A is
vertical, (b) the
The smallest force P will be perpendicular to R.
(a)
(b)
P = (425 lb)cos30°
R = (425 lb)sin30°
P = 368 lb
R = 213 lb

PROBLEM 2.14
For the hook support of Prob. 2.10, determine by trigonometry (a) the
magnitude and direction of the smallest force P for which the resultant
R of the two forces applied to the support is horizontal, (b) the
SOLUTION
The smallest force P will be perpendicular to R.
(a) P = (50 N)sin 25° P = 21.1 N W
(b) R = (50 N)cos 25° R = 45.3 N W

SOLUTION
Using the law of cosines:
PROBLEM 2.15
For the hook support shown, determine by trigonometry the
magnitude and direction of the resultant of the two forces applied
to the support.
R2
= (200 lb)2
+ (300 lb)2
−2(200 lb)(300 lb)cos(45D
+ 65°)
R = 413.57 lb
sinα sin (45D
+65°)
=
300 lb 413.57 lb
α = 42.972°
β = 90D
+ 25D
− 42.972° R = 414 lb 72.0° W

PROBLEM 2.16
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,
SOLUTION
R2
= (900N)2
+ (600N)2
−2(900N)(600 N)cos(135°)
R =1390.57N
sin(α−30D
) sin (135°)
Using the law of sines: =
600N 1390.57N
α − 30D
= 17.7642°
α = 47.764D
R =1391N 47.8° W

SOLUTION
PROBLEM 2.17
Solve Problem 2.4 by trigonometry.
PROBLEM 2.4 Two structural members B and C are bolted to bracket
A. Knowing that both members are in tension and that P = 6 kips and
Q = 4 kips, determine graphically the magnitude and direction of the
resultant force exerted on the bracket using (a) the parallelogram law,
Using the force triangle and the laws of cosines and sines:
We have: γ =180° − (50° + 25°)
=105°
Then R2
= (4 kips)2
+ (6 kips)2
− 2(4 kips)(6 kips)cos105°
= 64.423 kips2
R = 8.0264 kips
And
4 kips
=
8.0264 kips
sin(25° +α) sin105°
sin(25° +α) = 0.48137
25° +α = 28.775°
α = 3.775°
R = 8.03 kips 3.8° W

SOLUTION
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.
Using the laws of cosines and sines:
P2
= (120 N)2
+ (160 N)2
− 2(120 N)(160 N)cos 25°
P = 72.096 N
And
sinα
=
sin 25°
120 N 72.096 N
sinα = 0.70343
α = 44.703°
P = 72.1 N 44.7° W

PROBLEM 2.19
Two forces P and Q are applied to the lid of a storage bin as shown.
Knowing that P = 48 N and Q = 60 N, determine by trigonometry
the magnitude and direction of the resultant of the two forces.
SOLUTION
We have
Then
and
γ =180° − (20° +10°)
=150°
R2
= (48 N)2
+ (60 N)2
−2(48 N)(60 N)cos150°
R =104.366 N
48 N
=
104.366 N
sinα sin150°
Hence:
sinα = 0.22996
α =13.2947°
φ =180° −α − 80°
=180° −13.2947° − 80°
= 86.705°
R =104.4 N 86.7° W

SOLUTION
PROBLEM 2.20
Two forces P and Q are applied to the lid of a storage bin as shown.
Knowing that P = 60 N and Q = 48 N, determine by trigonometry the
magnitude and direction of the resultant of the two forces.
We have
Then
and
γ =180° − (20° +10°)
=150°
R2
= (60 N)2
+ (48 N)2
−2(60 N)(48 N)cos150°
R =104.366 N
60 N
=
104.366 N
sinα sin150°
Hence:
sinα = 0.28745
α =16.7054°
φ =180° −α −180°
=180° −16.7054° − 80°
= 83.295°
R =104.4 N 83.3° W

SOLUTION
PROBLEM 2.21
Determine the x and y components of each of the forces shown.
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.
84
29-lb Force:
50-lb Force:
Fx = +(29 lb)
116
F = +(29 lb)
80
y
116
F = −(50 lb)
28
x
100
96
Fx = +21.0 lb W
Fy = +20.0 lb W
Fx = −14.00 lb W
Fy = +(50 lb)
100
48
Fy = +48.0 lb W
51-lb Force: Fx = +(51 lb)
102
F = −(51 lb)
90
y
102
Fx = +24.0 lb W
Fy = −45.0 lb W

SOLUTION
PROBLEM 2.22
OA = (600)2
+ (800)2
=1000 mm
OB = (560)2
+ (900)2
=1060 mm
OC = (480)2
+ (900)2
=1020 mm
800
800-N Force:
424-N Force:
408-N Force:
Fx = +(800 N)
1000
F = +(800 N)
600
y
1000
F = −(424 N)
560
x
1060
F = −(424 N)
900
y
1060
F = +(408 N)
480
x
1020
F = −(408 N)
900
y
1020
Fx = +640 N W
Fy = +480 N W
Fx = −224 N W
Fy = −360 N W
Fx = +192.0 N W
Fy = −360 N W

PROBLEM 2.23
SOLUTION
80-N Force: Fx = +(80 N)cos40° Fx = 61.3 N W
Fy = +(80 N)sin 40° Fy = 51.4 N W
120-N Force: Fx = +(120 N)cos70°
Fy = +(120 N)sin 70°
Fx = 41.0 N W
Fy =112.8 N W
150-N Force: Fx = −(150 N)cos35°
Fy = +(150 N)sin35°
Fx = −122. 9 N W
Fy = 86.0 N W

PROBLEM 2.24
SOLUTION
40-lb Force: Fx = +(40 lb)cos60° Fx = 20.0 lb W
Fy = −(40 lb)sin 60° Fy = −34.6 lb W
50-lb Force: Fx = −(50 lb)sin50°
Fy = −(50 lb)cos50°
Fx = −38.3 lb W
Fy = −32.1 lb W
60-lb Force: Fx = +(60 lb)cos25°
Fy = +(60 lb)sin 25°
Fx = 54.4 lb W
Fy = 25.4 lb W

⎝ ⎠
⎝ ⎠
⎝ ⎠
⎝ ⎠
⎝ ⎠
SOLUTION
PROBLEM 2.25
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.
BC = (650 mm)2
+ (720 mm)2
= 970 mm
(a)
or
Px = P
⎛ 650 ⎞
⎜
970
⎟
⎛ 970 ⎞
P = Px ⎜
650
⎟
= 325 N
⎛ 970 ⎞
⎜ 650 ⎟
(b)
= 485 N
P = P
⎛ 720 ⎞
y ⎜ 970 ⎟
= 485 N
⎛ 720 ⎞
⎜ 970 ⎟
P = 485 N W
= 360 N
Py = 970 N W

PROBLEM 2.26
Member BD exerts on member ABC a force P directed along line BD.
SOLUTION
Knowing that P must have a 300-lb horizontal component, determine (a) the
magnitude of the force P, (b) its vertical component.
(a)
(b) Vertical component
Psin35° = 300 lb
P =
300 lb
sin35°
Pv = Pcos35°
= (523 lb)cos35°
P = 523 lb W
Pv = 428 lb W

PROBLEM 2.27
The hydraulic cylinder BC exerts on member AB a force P
directed along line BC. Knowing that P must have a 600-N
component perpendicular to member AB, determine (a) the
magnitude of the force P, (b) its component along line AB.
SOLUTION
(a)
180° = 45° + α + 90° + 30°
α =180° − 45° − 90° − 30°
=15°
cosα =
Px
P
P =
Px
cosα
(b)
=
600 N
cos15°
= 621.17 N
Py
tanα =
Px
Py = Px tanα
= (600 N)tan15°
=160.770 N
P = 621 N W
Py =160.8 N W

SOLUTION
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.
(a)
Py
P =
cos 55°
=
350 lb
cos 55°
(b)
= 610.21 lb
Px = Psin 55°
= (610.21 lb)sin 55°
= 499.85 lb
P = 610 lb W
Px = 500 lb W

PROBLEM 2.29
The hydraulic cylinder BD exerts on member ABC a force P directed
along line BD. Knowing that P must have a 750-N component
SOLUTION
perpendicular to member ABC, determine (a) the magnitude of the force
P, (b) its component parallel to ABC.
(a)
(b)
750 N = Psin 20°
P = 2192.9 N
PABC = Pcos20°
= (2192.9 N)cos20°
P = 2190 N W
PABC = 2060 N W

SOLUTION
(a)
PROBLEM 2.30
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.
P =
37
P
12
x
=
37
(720 N)
12
= 2220 N
(b) P =
35
P
P = 2.22 kN W
y
12
x
=
35
(720 N)
12
= 2100 N
Py = 2.10 kN W

SOLUTION
PROBLEM 2.31
Determine the resultant of the three forces of Problem 2.21.
PROBLEM 2.21 Determine the x and y components of each of the
forces shown.
Components of the forces were determined in Problem 2.21:
Force x Comp. (lb) y Comp. (lb)
29 lb
50 lb
51 lb
+21.0
–14.00
+24.0
+20.0
+48.0
–45.0
Rx = +31.0 Ry = +23.0
R = Rxi + Ry j
= (31.0 lb)i + (23.0 lb)j
Ry
tanα =
Rx
=
23.0
31.0
α = 36.573°
R =
23.0 lb
sin (36.573°)
= 38.601 lb R = 38.6 lb 36.6° W

SOLUTION
PROBLEM 2.32
forces shown.
Force x Comp. (N) y Comp. (N)
80 N
120 N
150 N
+61.3
+41.0
–122.9
+51.4
+112.8
+86.0
Rx = −20.6 Ry = +250.2
R = Rxi + Ry j
= (−20.6 N)i + (250.2 N)j
Ry
tanα =
Rx
tanα =
250.2 N
20.6 N
tanα =12.1456
α = 85.293°
R =
250.2 N
sin85.293°
R = 251 N 85.3° W

SOLUTION
PROBLEM 2.33
PROBLEM 2.24 Determine the x and y components of each of
the forces shown.
Force x Comp. (lb) y Comp. (lb)
40 lb
50 lb
60 lb
+20.00
–38.30
+54.38
–34.64
–32.14
+25.36
Rx = +36.08 Ry = −41.42
R = Rxi + Ry j
= (+36.08 lb)i + (−41.42 lb)j
Ry
tanα =
Rx
tan =
41.42 lb
36.08 lb
tanα =1.14800
= 48.942°
R =
41.42 lb
sin 48.942°
R = 54.9 lb 48.9° W

SOLUTION
PROBLEM 2.34
forces shown.
800 lb
424 lb
408 lb
+640
–224
+192
+480
–360
–360
Rx = +608 Ry = −240
R = Rxi + Ry j
= (608 lb)i + (−240 lb)j
Ry
tanα =
Rx
=
240
608
α = 21.541°
R =
240 N
sin(21.541°)
= 653.65 N R = 654 N 21.5° W

SOLUTION
PROBLEM 2.35
Knowing that α = 35°, determine the resultant of the three
forces shown.
100-N Force:
150-N Force:
200-N Force:
Fx = +(100 N)cos35° = +81.915 N
Fy = −(100 N)sin35° = −57.358 N
Fx = +(150 N)cos65° = +63.393 N
Fy = −(150 N)sin65° = −135.946 N
Fx = −(200 N)cos35° = −163.830 N
Fy = −(200 N)sin35° = −114.715 N
100 N
150 N
200 N
+81.915
+63.393
−163.830
−57.358
−135.946
−114.715
Rx = −18.522 Ry = −308.02
R = Rxi + Ry j
= (−18.522 N)i + (−308.02 N)j
Ry
tanα =
Rx
=
308.02
18.522
α = 86.559°
R =
308.02 N
sin86.559
R = 309 N 86.6° W

N
SOLUTION
PROBLEM 2.36
Knowing that the tension in rope AC is 365 N, determine the
resultant of the three forces exerted at point C of post BC.
Determine force components:
960
Cable force AC: Fx = −(365 N)
1460
1100
= −240 N
Fy = −(365 N)
1460
= −275 N
24
500-N Force: Fx = (500 N)
25
7
= 480 N
Fy = (500 N)
25
= 140 N
200-N Force: Fx = (200 N)
4
= 160 N
5
3
and
Fy = −(200 N)
5
= −120 N
Rx = ΣFx = −240 N + 480 N +160 N = 400 N
Ry = ΣFy = −275 N +140 N −120 N = −255
R = R2
+ R2
x y
= (400 N)2
+ (−255 N)2
Further:
= 474.37 N
tanα =
255
400
α = 32.5°
R = 474 N 32.5° W

PROBLEM 2.37
forces shown.
SOLUTION
60-lb Force:
80-lb Force:
120-lb Force:
and
Fx = (60 lb)cos20° = 56.382 lb
Fy = (60 lb)sin 20° = 20.521lb
Fx = (80 lb)cos60° = 40.000 lb
Fy = (80 lb)sin60° = 69.282 lb
Fx = (120 lb)cos30° =103.923 lb
Fy = −(120 lb)sin30° = −60.000 lb
Rx = ΣFx = 200.305 lb
Ry = ΣFy = 29.803 lb
R = (200.305 lb)2
+ (29.803 lb)2
= 202.510 lb
Further: tanα =
29.803
200.305
α = tan−1 29.803
200.305
= 8.46° R = 203 lb 8.46° W

PROBLEM 2.38
forces shown.
SOLUTION
60-lb Force:
80-lb Force:
120-lb Force:
Then
and
Fx = (60 lb)cos 20° = 56.382 lb
Fy = (60 lb)sin 20° = 20.521 lb
Fx = (80 lb)cos 95° = −6.9725 lb
Fy = (80 lb)sin 95° = 79.696 lb
Fx = (120 lb)cos 5° =119.543 lb
Fy = (120 lb)sin 5° =10.459 lb
Rx = ΣFx =168.953 lb
Ry = ΣFy =110.676 lb
R = (168.953 lb)2
+ (110.676 lb)2
= 201.976 lb
tanα =
110.676
168.953
tanα = 0.65507
α = 33.228° R = 202 lb 33.2° W

PROBLEM 2.39
For the collar of Problem 2.35, determine (a) the required value
of α if the resultant of the three forces shown is to be vertical,
(b) the corresponding magnitude of the resultant.
SOLUTION
Rx = ΣFx
= (100 N)cosα + (150 N)cos(α + 30°) − (200 N)cosα
Rx = −(100 N)cosα + (150 N)cos(α + 30°)
Ry = ΣFy
= −(100 N)sinα − (150 N)sin(α + 30°) − (200 N)sinα
Ry = −(300 N)sinα − (150 N)sin(α + 30°)
(1)
(2)
(a) For R to be vertical, we must have Rx = 0. We make Rx = 0 in Eq. (1):
−100cosα +150cos(α + 30°) = 0
−100cosα +150(cosα cos 30° − sin α sin 30°) = 0
29.904cosα = 75sinα
tanα =
29.904
75
= 0.39872
α = 21.738° α = 21.7° W
(b) Substituting for α in Eq. (2):
Ry = −300sin 21.738° −150sin51.738°
= −228.89 N
R = |Ry | = 228.89 N R = 229 N W

T
T
PROBLEM 2.40
For the post of Prob. 2.36, determine (a) the required
tension in rope AC if the resultant of the three forces
exerted at point C is to be horizontal, (b) the corresponding
magnitude of the resultant.
SOLUTION
R = ΣF = −
960
T +
24
(500 N) +
4
(200 N)
x x
1460
AC
25 5
48
R = − + 640 N (1)
x
73
R = ΣF
AC
= −
1100
T +
7
(500 N) −
3
(200 N)
y y
1460
AC
25 5
55
R = − + 20 N (2)
y
73
AC
(a) For R to be horizontal, we must have Ry = 0.
Set Ry = 0 in Eq. (2): −
55
T + 20 N = 0
73
AC
TAC = 26.545 N TAC = 26.5 N W
(b) Substituting for TAC into Eq. (1) gives
R
48
(26.545 N) + 640 N
x = −
73
Rx = 622.55 N
R = Rx = 623 N R = 623 N W

SOLUTION
PROBLEM 2.41
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.
Using the x and y axes shown:
(a) Set Ry = 0 in Eq. (2):
Rx = ΣFx = TAC sin10° + (50 lb)cos35° + (75 lb)cos60°
= TAC sin10° + 78.458 lb
Ry = ΣFy = (50 lb)sin35° + (75 lb)sin60° −TAC cos10°
Ry = 93.631 lb −TAC cos10°
93.631 lb −TAC cos10° = 0
(1)
(2)
(b) Substituting for TAC in Eq. (1):
TAC = 95.075 lb
Rx = (95.075 lb)sin10° + 78.458 lb
= 94.968 lb
R = Rx
TAC = 95.1 lb W
R = 95.0 lb W

SOLUTION
Select the x axis to be along a a′.
Then
PROBLEM 2.42
For the block of Problems 2.37 and 2.38, determine (a) the
required value of α if the resultant of the three forces shown is
to be parallel to the incline, (b) the corresponding magnitude of
the resultant.
and
(a) Set Ry = 0 in Eq. (2).
Rx = ΣFx = (60 lb) +(80 lb)cosα + (120 lb)sinα
Ry = ΣFy = (80 lb)sinα − (120 lb)cosα
(80 lb)sinα − (120 lb)cosα = 0
(1)
(2)
Dividing each term by cosα gives:
(80 lb)tanα =120 lb
tanα =
120 lb
80 lb
α = 56.310° α = 56.3° W
(b) Substituting for α in Eq. (1) gives:
Rx = 60 lb + (80 lb)cos56.31° + (120lb)sin56.31° = 204.22 lb Rx = 204 lb W

SOLUTION
PROBLEM 2.43
Two cables are tied together at C and are loaded as shown.
Determine the tension (a) in cable AC, (b) in cable BC.
Free-Body Diagram Force Triangle
Law of sines:
TAC
=
TBC
=
400 lb
sin60° sin 40° sin80°
(a) T =
400 lb
(sin 60°) T = 352 lb W
AC
sin80°
AC
(b) T =
400 lb
(sin 40°) T = 261 lb W
BC
sin80°
BC

SOLUTION
PROBLEM 2.44
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.
Law of sines:
TAC
=
TBC
=
6 kN
sin60° sin35° sin85°
(a) T =
6 kN
(sin 60°) T = 5.22 kN W
AC
sin85°
AC
(b) T =
6 kN
(sin35°) T = 3.45 kN W
BC
sin85°
BC

SOLUTION
PROBLEM 2.45
Two cables are tied together at C and loaded as shown.
Determine the tension (a) in cable AC, (b) in cable BC.
tanα =
1.4
4.8
α =16.2602°
tan β =
1.6
3
β = 28.073°
Force Triangle
Free-Body Diagram
Law of sines:
TAC
=
TBC
=
1.98 kN
sin 61.927° sin 73.740° sin 44.333°
(a) T =
1.98 kN
sin61.927° T = 2.50 kN W
AC
sin 44.333°
AC
(b) T =
1.98 kN
sin73.740° T = 2.72 kN W
BC
sin 44.333°
BC

PROBLEM 2.46
SOLUTION
Free-Body Diagram
Knowing that P = 500 N and α = 60°, determine the tension in
(a) in cable AC, (b) in cable BC.
Force Triangle
Law of sines:
TAC
=
sin35°
TBC
sin75°
=
500 N
sin70°
(a) T =
500 N
sin35° T = 305 N W
AC
sin70°
AC
(b) T =
500 N
sin 75° T = 514 N W
BC
sin70°
BC

SOLUTION
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.
Law of sines:
W = mg
= (200 kg)(9.81 m/s2
)
=1962 N
TAC
sin 15°
=
TBC
sin 105°
=
1962 N
sin 60°
(a)
(b)
TAC =
T
(1962 N) sin 15°
sin 60°
(1962 N)sin 105°
TAC = 586 N W
T = 2190 N W
TBC =
sin 60°
BC

SOLUTION
PROBLEM 2.48
Knowing that α = 20°,
AC, (b) in rope BC.
determine the tension (a) in cable
Law of sines:
TAC
sin 110°
=
TBC
sin 5°
=
1200 lb
sin 65°
(a)
(b)
T =
1200 lb
sin 110°AC
sin 65°
T =
1200 lb
sin 5°BC
sin 65°
TAC =1244 lb W
TBC =115.4 lb W

CA CB
CA CB
PROBLEM 2.49
SOLUTION
Knowing that P = 300 N, determine the tension in cables AC and
BC.
Free-Body Diagram
ΣFx = 0 −T sin30D
+T sin30D
− Pcos 45° − 200N = 0
For P = 200N we have,
−0.5TCA + 0.5TCB + 212.13 − 200 = 0 (1)
ΣFy = 0 T cos30° −T cos30D
− Psin 45D
= 0
0. 6603TCA + 0.86603TCB − 212.13 = 0 (2)
Solving equations (1) and (2) simultaneously gives, TCA =134.6 N W
TCB =110.4 N W

CA
CB
CA CB
PROBLEM 2.50
SOLUTION
Determine the range of values of P for which both cables remain
taut.
Free-Body Diagram
ΣFx = 0 −T sin30D
+T sin30D
− Pcos 45° − 200N = 0
For TCA = 0 we have,
0.5TCB + 0.70711P − 200 = 0 (1)
ΣFy = 0 T cos30° −T cos30D
− Psin 45D
= 0 ; again setting TCA = 0 yields,
0.86603TCB −0.70711P = 0 (2)
Adding equations (1) and (2) gives, 1.36603TCB = 200 hence TCB =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 W

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 = 500 lb and Q = 650 lb, determine the
SOLUTION
magnitudes of the forces exerted on the rods A and B.
Resolving the forces into x- and y-directions:
R = P +Q + FA +FB = 0
Free-Body Diagram
Substituting components: R = −(500 lb)j + [(650 lb)cos50°]i
−[(650 lb)sin50°]j
+ FBi − (FA cos50°)i + (FA sin 50°)j = 0
In the y-direction (one unknown force):
−500 lb − (650 lb)sin50°+ FA sin50° = 0
500 lb + (650 lb)sin50°
Thus, FA =
sin50°
In the x-direction:
Thus,
=1302.70 lb
(650 lb)cos50°+ FB − FA cos50° = 0
FB = FA cos50° − (650 lb)cos50°
= (1302.70 lb)cos50° − (650 lb)cos50°
= 419.55 lb
FA =1303 lb W
FB = 420 lb W

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 magnitudes of the forces exerted
on rods A and B are FA = 750lb and FB = 400 lb,
SOLUTION
determine the magnitudes of P and Q.
Resolving the forces into x- and y-directions:
R = P +Q + FA + FB = 0
Free-Body Diagram
Substituting components: R = −Pj + Q cos 50°i − Q sin 50°j
−[(750 lb)cos 50°]i
+[(750 lb)sin 50°]j + (400 lb)i
In the x-direction (one unknown force):
Q cos 50° −[(750 lb)cos 50°] + 400 lb = 0
Q =
(750 lb)cos 50° − 400 lb
cos 50°
= 127.710 lb
In the y-direction: −P − Q sin 50° + (750 lb)sin 50° = 0
P = −Q sin 50° + (750 lb)sin 50°
= −(127.710 lb)sin 50° + (750 lb)sin 50°
= 476.70 lb P = 477 lb; Q =127.7 lb W

PROBLEM 2.53
A welded connection is in equilibrium under the action of
the four forces shown. Knowing that FA = 8 kN and
SOLUTION
FB =16 kN, determine the magnitudes of the other
forces.
two
Free-Body Diagram of Connection
ΣF = 0:
3
F − F −
3
F = 0
x
5
B C
5
A
With FA = 8 kN
FB =16 kN
F =
4
(16 kN) −
4
(8 kN) F = 6.40 kN W
C
5 5
C
3 3
Σ Fy = 0:
3
− FD +
5
FB −
5
FA = 0
3
With FA and FB as above: FD = (16 kN) −
5
(8 kN)
5
FD = 4.80 kN W

PROBLEM 2.54
A welded connection is in equilibrium under the action of the
four forces shown. Knowing that FA = 5 kN and FD = 6 kN,
SOLUTION
determine the magnitudes of the other two forces.
Free-Body Diagram of Connection
ΣFy = 0: −F −
3
F +
3
F = 0D
5
A
5
B
3
or FB = FD +
5
FA
With FA = 5 kN,
5 ⎡
FD =8 kN
3 ⎤
FB =
3 ⎢6 kN +
5
(5 kN)⎥ FB =15.00 kN W
⎣
ΣFx = 0:
4
⎦
−F +
4
F −
4
F = 0C
5
B
5
A
FC =
5
(FB − FA )
=
4
(15 kN − 5 kN)
5
FC = 8.00 kN W

PROBLEM 2.55
A sailor is being rescued using a boatswain’s chair
that is suspended from a pulley that can roll freely
on the support cable ACB and is pulled at a
constant speed by cable CD. Knowing that α = 30°
and β = 10° and that the combined weight of the
boatswain’s chair and the sailor is 200 lb,
SOLUTION
determine the tension (a) in the support cable ACB,
(b) in the traction cable CD.
Free-Body Diagram
ΣFx = 0:
ΣFy = 0:
TACB cos 10° −TACB cos 30° −TCD cos 30° = 0
TCD = 0.137158TACB
TACB sin 10° + TACB sin 30° + TCD sin 30° − 200 = 0
0.67365TACB + 0.5TCD = 200
(1)
(2)
(a) Substitute (1) into (2): 0.67365TACB + 0.5(0.137158TACB ) = 200
TACB = 269.46 lb TACB = 269 lb W
(b) From (1): TCD = 0.137158(269.46 lb) TCD = 37.0 lb W

SOLUTION
PROBLEM 2.56
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 β = 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.
Free-Body Diagram
ΣFx = 0: TACB cos 15°−TACB cos 25°− (20 lb)cos 25° = 0
TACB = 304.04 lb
ΣFy = 0: (304.04 lb)sin 15° + (304.04 lb)sin 25°
+(20 lb)sin 25° −W = 0
W = 215.64 lb
(a)
(b)
W = 216 lb W
TACB = 304 lb W

AC
BC
SOLUTION
PROBLEM 2.57
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.
(a) For a minimum tension in cable BC, set angle between cables to 90 degrees.
By inspection, α = 35.0D
W
T = (6 kN)cos35D
TAC = 4.91 kN W
T = (6 kN)sin35D
TBC = 3.44 kN W
(b) For equal tension in both cables, the force triangle will be an isosceles.
Therefore, by inspection, α = 55.0D
W
T = T = (1/ 2)
6 kN
T =T = 3.66 kN W
AC BC
cos35°
AC BC

SOLUTION
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 α.
(a) Law of cosines P2
= (600)2
+ (750)2
− 2(600)(750) cos(25° + 45°)
P = 784.02 N P = 784 N W
(b) Law of sines
sin β
600 N
=
sin(25° + 45°)
784.02 N
β = 46.0° ∴ α = 46.0° + 25° α = 71.0° W

SOLUTION
PROBLEM 2.59
For the situation described in Figure P2.48, determine (a)
the value of α for which the tension in rope BC is as small
as possible, (b) the corresponding value of the tension.
To be smallest, TBC
(a) Thus,
must be perpendicular to the direction of TAC.
α = 5.00° α = 5.00° W
(b) TBC = (1200 lb)sin 5° TBC =104.6 lb W

PROBLEM 2.60
Two cables tied together at C are loaded as shown. Determine the
range of values of Q for which the tension will not exceed 60 lb in
either cable.
SOLUTION
Free-Body Diagram
Requirement:
ΣFx = 0:
ΣFy = 0:
−TBC −Qcos60°+ 75 lb = 0
TBC = 75 lb −Q cos60°
TAC − Qsin 60° = 0
TAC = Qsin60°
TAC = 60 lb:
(1)
(2)
From Eq. (2): Qsin60° = 60 lb
Q = 69.3 lb
Requirement: TBC = 60 lb:
From Eq. (1): 75 lb − Qcos60° = 60 lb
Q = 30.0 lb 30.0 lb ≤ Q ≤ 69.3 lb W

SOLUTION
PROBLEM 2.61
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.
Free-Body Diagram
Isosceles Force Triangle
tanα =
h
(1)
0.6 m
From Eq. (1): tan16.2602° =
h
0.6 m
∴ h = 0.175000 m
Law of sines:
1
(2.8 kN)
sinα = 2
TAC
TAC = 5 kN
1
(2.8 kN)
sinα = 2
5 kN
α =16.2602°
Half-length of chain = AC = (0.6 m)2
+ (0.175 m)2
Total length:
= 0.625 m
= 2× 0.625 m 1.250 m W

h
1+
1+
600 mm
SOLUTION
Free-Body Diagram
PROBLEM 2.62
For W = 800 N, P = 200 N, and d = 600 mm,
determine the value of h consistent with
equilibrium.
TAC =TBC =800 N
AC = BC = (h2
+ d2
)
ΣFy = 0: 2(800 N)
2
h
h2
+ d2
− P = 0
800 =
P
2
⎛ d ⎞
⎜ ⎟
⎝ ⎠
Data: P = 200 N, d = 600 mm and solving for h
800 N =
200 N ⎛ ⎞
2
⎜ ⎟
2 ⎝ h ⎠
h = 75.6 mm W

PROBLEM 2.63
Collar A is connected as shown to a 50-lb load and can
slide on a frictionless horizontal rod. Determine the
magnitude of the force P required to maintain the
SOLUTION
equilibrium
x =15 in.
of the collar when (a) x = 4.5 in., (b)
(a) Free Body: Collar A Force Triangle
P
=
50 lb
P =10.98 lb W
4.5 20.5
(b) Free Body: Collar A Force Triangle
P
=
50 lb
P = 30.0 lb W
15 25

PROBLEM 2.64
Collar A is connected as shown to a 50-lb load and can
slide on a frictionless horizontal rod. Determine the
SOLUTION
distance x for which the collar is in equilibrium when
P = 48 lb.
Free Body: Collar A Force Triangle
N2
= (50)2
− (48)2
=196
N =14.00 lb
Similar Triangles
x
=
48 lb
20 in. 14 lb
x = 68.6 in. W

SOLUTION
PROBLEM 2.65
Three forces are applied to a bracket as shown. The directions of the two 150-N
forces may vary, but the angle between these forces is always 50°. Determine
the range of values of α for which the magnitude of the resultant of the forces
acting at A is less than 600 N.
Combine the two 150-N forces into a resultant force Q:
Q = 2(150 N)cos25°
= 271.89 N
Equivalent loading at A:
(600 N)2
= (500 N)2
+ (271.89 N)2
+ 2(500 N)(271.89 N)cos(55° +α)
cos(55° +α) = 0.132685
Two values for α : 55° +α = 82.375
α = 27.4°
55° +α = −82.375°
55° +α = 360° − 82.375°
or α = 222.6°
For R < 600 lb: 27.4° <α < 222.6D
W

⎜ ⎟
y ⎜ ⎟
y ⎜ ⎟
SOLUTION
PROBLEM 2.66
A 200-kg crate is to be supported by the rope-and-pulley arrangement shown.
Determine the magnitude and direction of the force P that must be exerted on
the free end of the rope to maintain equilibrium. (Hint: The tension in the rope is
the same on each side of a simple pulley. This can be proved by the methods of
Ch. 4.)
Free-Body Diagram: Pulley A
ΣFx = 0: − 2P
⎛ 5 ⎞
+ Pcosα = 0
⎝
cosα = 0.59655
α = ±53.377°
For α = +53.377°:
281 ⎠
ΣF = 0: 2P
⎛ 16 ⎞
+ P sin 53.377° −1962 N = 0
⎝ 281 ⎠
P = 724 N 53.4° W
For α = −53.377°:
ΣF = 0: 2P
⎛ 16 ⎞
+ P sin(−53.377°) −1962 N = 0
⎝ 281 ⎠
P =1773 53.4° W

SOLUTION
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.)
Free-Body Diagram of Pulley
(a)
(b)
(c)
(d)
(e)
ΣFy = 0: 2T − (600 lb) = 0
T =
1
(600 lb)
2
ΣFy = 0: 2T − (600 lb) = 0
T =
1
(600 lb)
2
ΣFy = 0: 3T − (600 lb) = 0
T =
1
(600 lb)
3
ΣFy = 0: 3T − (600 lb) = 0
T =
1
(600 lb)
3
ΣFy = 0: 4T − (600 lb) = 0
T =
1
(600 lb)
4
T = 300 lb W
T = 300 lb W
T = 200 lb W
T = 200 lb W
T =150.0 lb W

PROBLEM 2.68
Solve Parts b and d of Problem 2.67, assuming
that the free end of the rope is attached to the
crate.
PROBLEM 2.67 A 600-lb crate is supported
by several rope-and-pulley arrangements as
shown. Determine for each arrangement the
SOLUTION
tension in the rope. (See the hint for Problem
2.66.)
Free-Body Diagram of Pulley and Crate
(b)
(d)
ΣFy = 0:
ΣFy = 0:
3T − (600 lb) = 0
T =
1
(600 lb)
3
4T − (600 lb) = 0
T =
1
(600 lb)
4
T = 200 lb W
T =150.0 lb W

PROBLEM 2.69
A load Q is applied to the pulley C, which can roll on the
cable ACB. The pulley is held in the position shown by a
second cable CAD, which passes over the pulley A and
supports a load P. Knowing that P = 750 N, determine
(a) the tension in cable ACB, (b) the magnitude of load Q.
SOLUTION
Free-Body Diagram: Pulley C
(a) ΣFx = 0:
Hence:
TACB (cos25°− cos55°) − (750 N)cos55° = 0
TACB =1292.88 N
TACB =1293 N W
(b) ΣFy = 0: TACB (sin 25° + sin 55°) + (750 N)sin 55° − Q = 0
(1292.88 N)(sin 25° + sin 55°) + (750 N)sin 55° − Q = 0
or Q = 2219.8 N Q = 2220 N W

PROBLEM 2.70
An 1800-N load Q is applied to the pulley C, which can roll
on the cable ACB. The pulley is held in the position shown by
a second cable CAD, which passes over the pulley A and
supports a load P. Determine (a) the tension in cable ACB,
(b) the magnitude of load P.
SOLUTION
Free-Body Diagram: Pulley C
ΣFx = 0:
or
ΣFy = 0:
or
TACB (cos25°− cos55°) − Pcos55° = 0
P = 0.58010TACB
TACB (sin 25° + sin55°) + Psin55° −1800 N = 0
1.24177TACB + 0.81915P =1800 N
(1)
(2)
(a) Substitute Equation (1) into Equation (2):
1.24177TACB + 0.81915(0.58010TACB ) =1800 N
Hence:
(b) Using (1),
TACB =1048.37 N
TACB =1048 N W
P = 0.58010(1048.37 N) = 608.16 N
P = 608 N W

D
D
SOLUTION
PROBLEM 2.71
Determine (a) the x, y, and z components of the 600-N force,
(b) the angles θx, θy, and θz that the force forms with the
coordinate axes.
(a) Fx = (600 N)sin 25°cos30
Fx = 219.60 N
Fy = (600 N)cos25°
Fx = 220 N W
Fy = 543.78 N Fy = 544 N W
Fz = (380.36 N)sin 25°sin30
Fz =126.785 N Fz =126.8 N W
(b) cos
cos
θ =
Fx
=x
F
Fy
219.60 N
600 N
543.78 N
θx = 68.5° W
cos
θy =
F
Fz
=
600 N
126.785 N
θy = 25.0° W
θz = =
F 600 N
θz = 77.8° W

D
D
SOLUTION
PROBLEM 2.72
Determine (a) the x, y, and z components of the 450-N force,
(b) the angles θx, θy, and θz that the force forms with the
coordinate axes.
(a) Fx = −(450 N)cos 35°sin 40
Fx = −236.94 N
Fy = (450 N)sin 35°
Fy = 258.11 N
Fx = −237 N W
Fy = 258 N W
Fz = (450 N)cos35°cos40
Fz = 282.38 N
Fz = 282 N W
(b) cos
cos
cos
θx =
θy =
θz =
Fx
=
F
Fy
=
F
Fz
=
F
-236.94 N
450 N
258.11 N
450 N
282.38 N
450 N
θx =121.8° W
θy = 55.0° W
θz = 51.1° W
Note: From the given data, we could have computed directly
D D D
θy = 90 − 35 = 55 , which checks with the answer obtained.

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
(a)
F = 400 N
∴ FH = (400 N)cos40°
= 306.42 N
Fx = −FH sin35°
= −(306.42 N)sin35°
= −175.755 N
Fy = −F sin 40°
= −(400 N)sin 40°
= −257.12 N
Fz = +FH cos35°
= +(306.42 N)cos35°
= +251.00 N
Fx = −175.8 N W
Fy = −257 N W
Fz = +251N W
(b) cosθx =
Fx
=
F
−175.755 N
400 N
θx =116.1° W
cosθy =
Fy
=
F
−257.12 N
400 N
θy =130.0° W
cosθz =
Fz
=
F
251.00 N
400 N
θz = 51.1°W

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
(a)
F = 400 N
∴ FH = (400 N)cos 25°
= 362.52 N
Fx = +FH cos15°
= +(362.52 N)cos15°
= +350.17 N
Fy = −F sin 25°
= −(400 N)sin 25°
= −169.047 N
Fz = +FH sin15°
= +(362.52 N)sin15°
= +93.827 N
Fx = +350 N W
Fy = −169.0 N W
Fz = +93.8 N W
(b) cosθx =
Fx
=
F
+350.17 N
400 N
θx = 28.9° W
cos
cos
θy =
θz =
Fy
=
F
Fz
=
F
−169.047 N
400 N
+93.827 N
400 N
θy =115.0° W
θz = 76.4°W

SOLUTION
Fh = F cos 60°
= (50 lb)cos 60°
Fh = 25.0 lb
PROBLEM 2.75
The angle between spring AB and the post DA is 30°. Knowing
that the tension in the spring is 50 lb, determine (a) the x, y,
and z components of the force exerted on the circular plate at
B, (b) the angles θx, θy, and θz defining the direction of the
force at B.
D D
Fx = −Fh cos 35° Fy = Fsin60 Fz = −Fh sin35
D D D
Fx = (−25.0 lb)cos35
Fx = −20.479 lb
(a)
Fy = (50.0 lb)sin60
Fy = 43.301 lb
Fz = (−25.0 lb)sin35
Fz = −14.3394 lb
Fx = −20.5 lb W
Fy = 43.3 lb W
Fz = −14.33 lb W
(b) cos
cos
cos
θx =
θ y =
θz =
Fx
=
F
Fy
=
F
Fz
=
F
−20.479 lb
50 lb
43.301 lb
50 lb
-14.3394 lb
50 lb
θx =114.2° W
θy = 30.0° W
θz =106.7° W

SOLUTION
PROBLEM 2.76
The angle between spring AC and the post DA is
30°. Knowing that the tension in the spring is 40 lb,
determine (a) the x, y, and z components of the force
exerted on the circular plate at C, (b) the angles θx, θy,
and θz defining the direction of the force at C.
(a)
Fh = F cos 60°
= (40 lb)cos 60°
Fh = 20.0 lb
Fx = Fh cos 35°
= (20.0 lb)cos 35°
Fx =16.3830 lb
Fy = F sin 60°
= (40 lb)sin 60°
Fy = 34.641 lb
Fz = −Fh sin 35°
= −(20.0 lb)sin35°
Fz = −11.4715 lb
(b) cos
cos
cos
θx =
θ y =
θz =
Fx
F
Fy
F
Fz
F
=
16.3830 lb
40 lb
=
34.641 lb
40 lb
=
-11.4715 lb
40 lb
Fx =16.38 lb W
Fy = 34.6 lb W
Fz = −11.47 lb W
θx = 65.8° W θ
y = 30.0° W θz
=106.7° W

= −
z
PROBLEM 2.77
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
SOLUTION
cable on the anchor B, (b) the angles
direction of that force.
θx , θy , and θz defining the
From triangle AOB: cosθy =
56 ft
65 ft
(a)
= 0.86154
θy = 30.51°
Fx = −F sinθy cos 20°
= −(3900 lb)sin 30.51° cos 20°
Fy = +F cosθy = (3900 lb)(0.86154)
Fx = −1861 lb W
Fy = +3360 lb W
Fz = +(3900 lb)sin 30.51° sin 20° Fz = +677 lb W
(b) cosθ =
Fx 1861 lb
= −0.4771 θ =118.5° W
x
F 3900 lb
x
From above: θy = 30.51°
cosθ =
Fz
= +
677 lb
= +0.1736
θy = 30.5° W
θz =80.0° W
F 3900 lb

SOLUTION
PROBLEM 2.78
Cable AC is 70 ft long, and the tension in that cable is 5250 lb. Determine
(a) the x, y, and z components of the force exerted by the cable on the
anchor C, (b) the angles θx, θy, and θz defining the direction of that
force.
In triangle AOB: AC = 70 ft
OA = 56 ft
F = 5250 lb
cosθy =
56 ft
70 ft
(a)
θy = 36.870°
FH = F sinθy
= (5250 lb)sin36.870°
= 3150.0 lb
Fx = −FH sin50° = −(3150.0 lb)sin50° = −2413.0 lb
Fx = −2410 lb W
Fy = +F cosθy = +(5250 lb)cos36.870° = +4200.0 lb
Fy = +4200 lb W
Fz = −FH cos50° = −3150cos50° = −2024.8 lb Fz = −2025 lb W
(b) cosθ =
Fx
=x
F
−2413.0 lb
5250 lb
θx =117.4° W
From above: θy = 36.870° θy = 36.9° W
θz =
Fz
=
F
−2024.8 lb
5250 lb
θz =112.7° W

PROBLEM 2.79
Determine the magnitude and direction of the force F = (240 N)i – (270 N)j + (680 N)k.
SOLUTION
2 2 2
F = Fx + Fy + Fz
F = (240 N)2
+ (−270 N)2
+ (−680 N)2
F = 770 N W
cos
cos
θx =
Fx
F
Fy
=
240 N
770 N
−270 N
θx = 71.8° W
cos
θy =
F
Fz
=
770 N
680 N
θy =110.5° W
θy =
F
=
770 N
θz = 28.0° W

PROBLEM 2.80
Determine the magnitude and direction of the force F = (320 N)i + (400 N)j − (250 N)k.
SOLUTION
2 2 2
F = Fx + Fy + Fz
F = (320 N)2
+ (400 N)2
+ (−250 N)2
F = 570 N W
cos
cos
cos
θx =
θy =
Fx
F
Fy
F
Fz
=
320 N
570 N
=
400 N
570 N
−250 N
θx = 55.8° W
θy = 45.4° W
θy = =
F 570 N
θz =116.0° W

y
PROBLEM 2.81
A force acts at the origin of a coordinate system in a direction defined by the angles θx = 69.3° and θz
= 57.9°. Knowing that the y component of the force is –174.0 lb, determine (a) the angle θy, (b) the other
components and the magnitude of the force.
SOLUTION
cos2
θ + cos2
θ + cos2
θ = 1
x y z
cos2
(69.3°) + cos2
θ
(a) Since Fy < 0, we choose cosθy = −0.7699
+ cos2
(57.9°) =1
cosθy = ±0.7699
∴ θy =140.3° W
(b) Fy = F cosθy
−174.0 lb = F(−0.7699)
F = 226.0 lb
Fx = F cosθx = (226.0 lb)cos69.3°
Fz = F cosθz = (226.0 lb)cos57.9°
F = 226 lb W
Fx = 79.9 lb W
Fz =120.1lb W

z
PROBLEM 2.82
A force acts at the origin of a coordinate system in a direction defined by the angles θx = 70.9° and
θy = 144.9°. Knowing that the z component of the force is –52.0 lb, determine (a) the angle θz, (b) the
other components and the magnitude of the force.
SOLUTION
cos2
θ + cos2
θ + cos2
θ =1
x y z
cos2
70.9D
+ cos2
144.9° + cos2
θ ° =1
cosθz = ±0.47282
(a) Since Fz < 0, we choose cosθz = −0.47282 ∴ θz =118.2° W
(b) Fz = F cosθz
−52.0 lb = F(−0.47282)
F =110.0 lb
Fx = F cosθx = (110.0 lb)cos70.9°
Fy = F cosθy = (110.0 lb)cos144.9°
F =110.0 lb W
Fx = 36.0 lb W
Fy = −90.0 lb W

y
x y z
y
PROBLEM 2.83
A force F of magnitude 210 N acts at the origin of a coordinate system. Knowing that Fx = 80 N,
θz = 151.2°, and Fy < 0, determine (a) the components Fy and Fz, (b) the angles θx and θy.
SOLUTION
(a)
Then:
So:
Fz = F cosθz = (210 N)cos151.2°
= −184.024 N
F2
= F2
+ F2
+ F2
(210 N)2
= (80 N)2
+ (F )2
+ (184.024 N)2
Fz = −184.0 N W
Hence: F = − (210 N)2
− (80 N)2
− (184.024 N)2
(b)
= −61.929 N
cosθ =
Fx
=
80 N
= 0.38095
Fy = −62.0 lb W
θ = 67.6° W
x
F 210 N
x
Fy 61.929 N
cosθ = = = −0.29490y
F 210 N
θy =107.2° W

z
z z
x x
PROBLEM 2.84
A force F of magnitude 1200 N acts at the origin of a coordinate system. Knowing that
θx = 65°, θy = 40°, and Fz > 0, determine (a) the components of the force, (b) the angle θz.
SOLUTION
cos2
θ + cos2
θ + cos2
θ =1
x y z
cos2
65D
+ cos2
40° + cos2
θ ° =1
cosθz = ±0.48432
(b) Since Fz > 0,
(a)
we choose cosθ = 0.48432, or θ = 61.032D
F = 1200 N
∴ θz = 61.0° W
F = F cosθ = (1200 N) cos65D
Fx = 507 N W
Fy = F cosθy = (1200 N)cos40°
Fz = F cosθz = (1200 N)cos61.032°
Fy =919 N W
Fz = 582 N W

PROBLEM 2.85
A frame ABC is supported in part by cable DBE that
passes through a frictionless ring at B. Knowing that the
tension in the cable is 385 N, determine the components
of the force exerted by the cable on the support at D.
SOLUTION
JJJG
DB = (480 mm)i − (510 mm)j+ (320 mm)k
DB = (480 mm)2
+ (510 mm2
) + (320 mm)2
= 770 mm
F = FλDB
JJJG
= F
DB
DB
=
385 N
[(480 mm)i − (510 mm)j+ (320 mm)k]
770 mm
= (240 N)i − (255 N)j+ (160 N)k
Fx = +240 N, Fy = −255 N, Fz = +160.0 N W

PROBLEM 2.86
For the frame and cable of Problem 2.85, determine the
components of the force exerted by the cable on the support
at E.
PROBLEM 2.85 A frame ABC is supported in part by
cable DBE that passes through a frictionless ring at B.
Knowing that the tension in the cable is 385 N, determine
the components of the force exerted by the cable on the
support at D.
SOLUTION
JJJG
EB = (270 mm)i − (400 mm)j+ (600 mm)k
EB = (270 mm)2
+ (400 mm)2
+ (600 mm)2
= 770 mm
F = FλEB
JJJG
= F
EB
EB
=
385 N
[(270 mm)i − (400 mm)j+ (600 mm)k]
770 mm
F = (135 N)i − (200 N)j+ (300 N)k
Fx = +135.0 N, Fy = −200 N, Fz = +300 N W

SOLUTION
PROBLEM 2.87
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.
Cable AB:
JJJG
λ =
AB
=
(−46.765 ft)i +(45 ft)j +(36 ft)k
AB
AB 74.216 ft
−46.765i + 45j + 36k
TAB = TAB λAB =
74.216
(TAB )x = −1.260 kips W
(TAB )y = +1.213 kips W
(TAB )z = +0.970 kips W

SOLUTION
PROBLEM 2.88
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.
Cable AB:
JJJG
λ =
AC
=
(−46.765 ft)i +(55.8 ft)j +(−45 ft)k
AC
AC
T = T λ
85.590 ft
= (1.5 kips)
−46.765i +55.8j −45k
AC AC AC
85.590
(TAC )x = −0.820 kips W
(TAC )y = +0.978 kips W
(TAC )z = −0.789 kips W

i +
SOLUTION
PROBLEM 2.89
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.
We have:
Thus:
JJJG
BA = +(320 mm)i + (480 mm)j - (360 mm)k
JJJG
BA = 680 mm
F = T λ = T
BA
= T
⎛ 8 12
j-
9
k
⎞
B BA BA BA
BA
BA ⎜17 17 17 ⎟
⎝ ⎠
⎛ 8
T
⎞
i+
⎛12
T
⎞
j−
⎛ 9
T
⎞
k = 0
⎜
17
BA ⎟ ⎜
17
BA ⎟ ⎜
17
BA ⎟
Setting TBA = 408 N yields,
⎝ ⎠ ⎝ ⎠ ⎝ ⎠
Fx = +192.0 N, Fy = +288 N, Fz = −216 N W

SOLUTION
PROBLEM 2.90
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.
We have:
Thus:
JJJG
DA = −(250 mm)i + (480 mm)j + (360 mm)k
JJJG
DA = 650 mm
F = T λ = T
DA
= T
⎛
−
5
i +
48
j+
36
k
⎞
D DA DA DA
DA
DA ⎜ 13 65 65 ⎟
⎝ ⎠
−
⎛ 5
T
⎞
i+
⎛ 48
T
⎞
j+
⎛ 36
T
⎞
k = 0
⎜
13
DA ⎟ ⎜
65
DA ⎟ ⎜
65
DA ⎟
Setting TDA = 429 N yields,
⎝ ⎠ ⎝ ⎠ ⎝ ⎠
Fx = −165.0 N, Fy = +317 N, Fz = +238 N W

SOLUTION
PROBLEM 2.91
Find the magnitude and direction of the resultant of the two forces
shown knowing that P = 300 N and Q = 400 N.
P = (300 N)[−cos30°sin15°i + sin30°j + cos30°cos15°k]
= −(67.243 N)i + (150 N)j + (250.95 N)k
Q = (400 N)[cos50°cos20°i + sin50°j − cos50°sin 20°k]
= (400 N)[0.60402i + 0.76604j− 0.21985]
= (241.61 N)i + (306.42 N)j− (87.939 N)k
R = P + Q
= (174.367 N)i + (456.42 N)j+ (163.011 N)k
R = (174.367 N)2
+ (456.42 N)2
+ (163.011 N)2
= 515.07 N
cosθ =
Rx
=
174.367 N
= 0.33853
R = 515 N W
θ = 70.2° W
x
R 515.07 N
x
Ry 456.42 N
cosθ = = = 0.88613 θ = 27.6° W
y
R 515.07 N
y
cosθ =
Rz
=
163.011 N
= 0.31648 θ = 71.5° W
z
R 515.07 N
z

SOLUTION
PROBLEM 2.92
P = (400 N)[−cos30°sin15°i + sin30°j + cos30°cos15°k]
= −(89.678 N)i + (200 N)j+ (334.61 N)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
= 515.07 N
cosθ =
Rx
=
91.532 N
= 0.177708
R = 515 N W
θ = 79.8° W
x
R 515.07 N
x
Ry 429.81 N
cosθ = = = 0.83447 θ = 33.4° W
y
R 515.07 N
y
cosθ =
Rz
=
268.66 N
= 0.52160 θ = 58.6° W
z
R 515.07 N
z

PROBLEM 2.93
Knowing that the tension is 425 lb in cable AB and 510 lb
in cable AC, determine the magnitude and direction of the
resultant of the forces exerted at A by the two cables.
SOLUTION
JJJG
AB = (40 in.)i − (45 in.)j + (60 in.)k
AB =
JJJG
(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.
JJJG
T = T λ = T
AB
= (425 lb)
⎡(40in.)i −(45 in.)j +(60in.)k ⎤
AB AB AB AB
AB
⎢
85 in.
⎥
⎣ ⎦
TAB = (200 lb)i − (225 lb)j + (300 lb)k
JJJG
T = T λ = T
AC
= (510 lb)
⎡(100 in.)i −(45 in.)j +(60 in.)k ⎤
AC AC AC AC
AC
⎢
125 in.
⎥
⎣ ⎦
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
cosθ =
608 lb
= 0.66599
R = 913 lb W
θ = 48.2° W
x
912.92 lb
x
cosθ =
408.6 lb
= −0.44757 θ =116.6° W
y
912.92 lb
y
cosθ =
544.8 lb
= 0.59677 θ = 53.4° W
z
912.92 lb
z

PROBLEM 2.94
Knowing that the tension is 510 lb in cable AB and 425 lb
in cable AC, determine the magnitude and direction of the
resultant of the forces exerted at A by the two cables.
SOLUTION
JJJG
AB = (40 in.)i − (45 in.)j + (60 in.)k
AB =
JJJG
(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.
JJJG
T = T λ = T
AB
= (510 lb)
⎡(40in.)i −(45 in.)j +(60in.)k ⎤
AB AB AB AB
AB
⎢
85 in.
⎥
⎣ ⎦
TAB = (240 lb)i − (270 lb)j + (360 lb)k
JJJG
T = T λ = T
AC
= (425 lb)
⎡(100 in.)i −(45 in.)j +(60 in.)k ⎤
AC AC AC AC
AC
⎢
125 in.
⎥
⎣ ⎦
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
cosθ =
580 lb
= 0.63532
R = 913 lb W
θ = 50.6° W
x
cosθy
912.92 lb
=
−423 lb
= −0.46335
912.92 lb
x
θy =117.6° W
cosθ =
564 lb
= 0.61780 θ = 51.8° W
z
912.92 lb
z

PROBLEM 2.95
For the frame of Problem 2.85, determine the magnitude and
direction of the resultant of the forces exerted by the cable at B
knowing that the tension in the cable is 385 N.
PROBLEM 2.85 A frame ABC is supported in part by cable
DBE that passes through a frictionless ring at B. Knowing that
the tension in the cable is 385 N, determine the components of
the force exerted by the cable on the support at D.
SOLUTION
JJJG
BD = −(480 mm)i + (510 mm)j − (320 mm)k
BD = (480 mm)2
+ (510 mm)2
+ (320 mm)2
= 770 mm
JJJG
F = T λ = T
BD
BD BD BD BD
BD
=
(385 N)
[−(480 mm)i + (510 mm)j− (320 mm)k]
(770 mm)
= −(240 N)i + (255 N)j− (160 N)k
JJJG
BE = −(270 mm)i + (400 mm)j − (600 mm)k
BE = (270 mm)2
+ (400 mm)2
+ (600 mm)2
= 770 mm
JJJG
F = T λ = T
BE
BE BE BE BE
BE
=
(385 N)
[−(270 mm)i + (400 mm)j− (600 mm)k]
(770 mm)
= −(135 N)i + (200 N)j− (300 N)k
R = FBD + FBE = −(375 N)i + (455 N)j−(460 N)k
R = (375 N)2
+ (455 N)2
+ (460 N)2
= 747.83 N R = 748 N W
cos θx
=
−375 N
747.83 N
θx =120.1° W
cos θ =
455 N
θ = 52.5° W
y
cos θz
747.83 N
=
−460 N
747.83 N
y
θz =128.0° W

SOLUTION
PROBLEM 2.96
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.
We have:
Thus:
JJJG
AB = −(320 mm)i − (480 mm)j+ (360 mm)k
JJJG
AC = (450 mm)i − (480 mm)j+ (360 mm)k
JJJG
AD = (250 mm)i − (480 mm)j− (360 mm)k
JJJG
AB = 680 mm
AC = 750 mm
AD = 650 mm
TAB = TAB λAB = TAB
AB
AB
JJJG
=
TAB
(−320i − 480j + 360k)
680
T = T λ = T
AC
=
54
(450i − 480j + 360k)
AC AC AC AC
AC
JJJG
T = T λ = T
AD
AD AD AD AD
AD
Substituting into the Eq. R = ΣF and factoring i, j, k:
750
=
TAD
(250i − 480j − 360k)
650
R =
⎛
−
320
T + 32.40 +
250
T
⎞
i
⎜ 680
AB
650
AD ⎟
⎝ ⎠
+
⎛
−
480
T − 34.560 −
480
T
⎞
j
⎜ 680
AB
650
AD ⎟
⎝ ⎠
+
⎛ 360
T + 25.920 −
360
T
⎞
k
⎜ 680
AB
650
AD ⎟
⎝ ⎠

PROBLEM 2.96 (Continued)
Since R is vertical, the coefficients of i and k are zero:
i: −
320
T + 32.40 +
250
T = 0 (1)
680
AB
650
AD
k:
360
T + 25.920 −
360
T = 0 (2)
680
AB
650
AD
Multiply (1) by 3.6 and (2) by 2.5 then add:
−
252
T
680
AB +181.440 = 0
TAB = 489.60 N
Substitute into (2) and solve for TAD :
360
(489.60 N) + 25.920 −
360
T = 0
680 650
AD
TAB = 490 N W
TAD = 514.80 N
TAD = 515 N W

SOLUTION
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.
Cable AB: TAB =183lb
JJJG
T = T λ = T
AB
= (183 lb)
(−48 in.)i +(29 in.)j+(24 in.)k
AB AB AB AB
AB
TAB = −(144 lb)i + (87 lb)j+ (72 lb)k
JJJG
61in.
Cable AC: T = T λ = T
AC
= T
(−48 in.)i +(25 in.)j+(−36 in.)k
AC AC AC AC
AC
AC
65 in.
48 25 36
TAC = − TACi +
65
TAC j−
65
TACk
65
Load P: P = P j
For resultant to be directed along OA, i.e., x-axis
R = 0: ΣF = (72 lb) −
36
T′ = 0 T =130.0 lb W
z z
65
AC AC

SOLUTION
PROBLEM 2.98
For the boom and loading of Problem. 2.97, determine
the magnitude of the load P.
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.
See Problem 2.97. Since resultant must be directed along OA, i.e., the x-axis, we write
R = 0: ΣF = (87 lb) +
25
T − P = 0
y y
TAC =130.0 lb from Problem 2.97.
65
AC
Then (87 lb) +
25
(130.0 lb) − P = 0
65
P =137.0 lb W

⎝ ⎠
SOLUTION
PROBLEM 2.99
A container is supported by three cables that are attached to
a ceiling as shown. Determine the weight W of the container,
knowing that the tension in cable AB is 6 kN.
Free-Body Diagram at A:
The forces applied at A are: TAB , TAC , TAD , and W
where W =Wj. To express the other forces in terms of the unit vectors i, j, k, we write
JJJG
AB = −(450 mm)i + (600 mm)j
JJJG
AC = +(600 mm)j − (320 mm)k
JJJG
AD = +(500 mm)i + (600 mm)j + (360 mm)k
JJJG
AB = 750 mm
AC = 680 mm
AD = 860 mm
and T = λ T = T
AB
= T
(−450 mm)i + (600 mm)j
AB AB AB AB
AB
AB
⎛ 45 60
750 mm
⎞
JJJG
= ⎜−
75
i +
75
j⎟TAB
T = λ T = T
AC
= T
(600 mm)i −(320 mm)j
AC AC AC AC
AC
AC
680 mm
=
⎛ 60
j−
32
k
⎞
T⎜ 68 68 ⎟ AC
JJJG
⎝ ⎠
T = λ T = T
AD
= T
(500 mm)i +(600 mm)j+(360 mm)k
AD AD AD AD
AD
AD
860 mm

=
⎛ 50
i +
60
j +
36
k
⎞
T⎜ 86 86 86 ⎟ AD
⎝ ⎠

Equilibrium condition: ΣF = 0: ∴ TAB + TAC + TAD + W = 0
Substituting the expressions obtained for TAB , TAC , and TAD ; factoring i, j, and k; and equating each of
the coefficients to zero gives the following equations:
From i: −
45
T +
50
T = 0 (1)
75
AB
86
AD
From j:
60
T +
60
T +
60
T −W = 0 (2)
From k:
75
AB
68
AC
−
32
T
86
AD
+
36
T = 0 (3)
68
AC
86
AD
Setting TAB = 6 kN in (1) and (2), and solving the resulting set of equations gives
TAC = 6.1920 kN
TAC = 5.5080 kN W =13.98 kN W

SOLUTION
PROBLEM 2.100
A container is supported by three cables that are attached to
a ceiling as shown. Determine the weight W of the container,
knowing that the tension in cable AD is 4.3 kN.
See Problem 2.99 for the figure and analysis leading to the following set of linear algebraic equations:
−
45
T +
50
T = 0 (1)
75
AB
86
AD
60
T +
60
T +
60
T −W = 0 (2)
75
AB
68
AC
−
32
T
86
AD
+
36
T = 0 (3)
68
AC
86
AD
Setting TAD = 4.3 kN into the above equations gives
TAB = 4.1667 kN
TAC = 3.8250 kN W = 9.71kN W

AB
ACAC AC AC AC AC
SOLUTION
The forces applied at A are:
PROBLEM 2.101
Three cables are used to tether a balloon as shown. Determine
the vertical force P exerted by the balloon at A knowing that the
tension in cable AD is 481 N.
FREE-BODY DIAGRAM AT A
TAB , TAC , TAD , and P
where P = Pj. To express the other forces in terms of the unit vectors i, j, k, we write
JJJG
AB = −(4.20 m)i − (5.60 m)j
JJJG
AC = (2.40 m)i − (5.60 m)j + (4.20 m)k
JJJG
AD = −(5.60 m)j − (3.30 m)k
JJJG
AB
AB = 7.00 m
AC = 7.40 m
AD = 6.50 m
and TAB = TAB λAB = TAB = (−0.6i − 0.8j)TAB
JJJG
T = T λ = T
AC
= (0.32432i − 0.75676j + 0.56757k)T
JJJG
T = T λ = T
AD
= (−0.86154j − 0.50769k)TAD AD AD AD
AD
AD

Equilibrium condition: ΣF = 0: TAB + TAC + TAD + Pj = 0
Substituting the expressions obtained for TAB , TAC , and TAD and factoring i, j, and k:
(−0.6TAB + 0.32432TAC )i + (−0.8TAB − 0.75676TAC − 0.86154TAD + P)j
+(0.56757TAC − 0.50769TAD )k = 0
Equating to zero the coefficients of i, j, k:
Setting TAD = 481N
−0.6TAB + 0.32432TAC = 0
−0.8TAB − 0.75676TAC − 0.86154TAD + P = 0
0.56757TAC − 0.50769TAD = 0
in (2) and (3), and solving the resulting set of equations gives
TAC = 430.26 N
TAD = 232.57 N P = 926 N
(1)
(2)
(3)
W

SOLUTION
PROBLEM 2.102
Three cables are used to tether a balloon as shown. Knowing
that the balloon exerts an 800-N vertical force at A, determine
the tension in each cable.
See Problem 2.101 for the figure and analysis leading to the linear algebraic Equations (1), (2), and (3).
From Eq. (1):
From Eq. (3):
−0.6TAB + 0.32432TAC = 0
−0.8TAB − 0.75676TAC − 0.86154TAD + P = 0
0.56757TAC − 0.50769TAD = 0
TAB = 0.54053TAC
TAD = 1.11795TAC
(1)
(2)
(3)
Substituting for TAB and TAD in terms of TAC into Eq. (2) gives
−0.8(0.54053TAC ) − 0.75676TAC − 0.86154(1.11795TAC ) + P = 0
2.1523TAC = P; P = 800 N
T =
800 N
AC
2.1523
= 371.69 N
Substituting into expressions for TAB and TAD gives
TAB = 0.54053(371.69 N)
TAD =1.11795(371.69 N)
TAB = 201 N, TAC = 372 N, TAD = 416 N W

DA
DBDB DB DB DB DB
SOLUTION
By Symmetry TDB = TDC
The forces applied at D are:
PROBLEM 2.103
A 36-lb triangular plate is supported by three wires as shown. Determine
the tension in each wire, knowing that a = 6 in.
Free-Body Diagram of Point D:
TDB , TDC , TDA , and P
where P = Pj = (36 lb)j. To express the other forces in terms of the unit vectors i, j, k, we write
JJJG
DA =(16 in.)i − (24 in.)j
JJJG
DB = −(8 in.)i − (24 in.)j + (6 in.)k
JJJG
DC = −(8 in.)i −(24 in.)j − (6 in.)k
JJG
DA
DA = 28.844 in.
DB = 26.0 in.
DC = 26.0 in.
and TDA = TDA λDA = TDA = (0.55471i − 0.83206j)TDA
JJJG
T = T λ = T
DB
= (−0.30769i − 0.92308j + 0.23077k)T
JJJG
T = T λ = T
DC
= (−0.30769i −0.92308j − 0.23077k)TDC DC DC DC
DC
DC

Equilibrium condition: ΣF = 0: TDA + TDB + TDC + (36 lb)j = 0
Substituting the expressions obtained for TDA , TDB , and TDC and factoring i, j, and k:
(0.55471TDA −0.30769TDB −0.30769TDC )i + (−0.83206TDA − 0.92308TDB − 0.92308TDC + 36 lb)j
+(0.23077TDB − 0.23077TDC )k = 0
0.55471TDA −0.30769TDB − 0.30769TDC = 0
−0.83206TDA − 0.92308TDB − 0.92308TDC + 36 lb = 0
0.23077TDB − 0.23077TDC = 0
Equation (3) confirms thatTDB = TDC . Solving simultaneously gives,
(1)
(2)
(3)
TDA = 14.42 lb; TDB = TDC = 13.00 lb W

DA
DBDB DB DB DB DB
SOLUTION
By Symmetry TDB = TDC
PROBLEM 2.104
Solve Prob. 2.103, assuming that a = 8 in.
PROBLEM 2.103 A 36-lb triangular plate is supported by three wires
as shown. Determine the tension in each wire, knowing that a = 6 in.
The forces applied at D are: TDB , TDC , TDA , and P
where P = Pj = (36 lb)j. To express the other forces in terms of the unit vectors i, j, k, we write
JJJG
DA =(16 in.)i − (24 in.)j
JJJG
DB = −(8 in.)i − (24 in.)j + (8 in.)k
JJJG
DC = −(8 in.)i −(24 in.)j − (8 in.)k
JJG
DA
DA = 28.844 in.
DB = 26.533 in.
DC = 26.533 in.
and TDA = TDA λDA = TDA = (0.55471i − 0.83206j)TDA
JJJG
T = T λ = T
DB
= (−0.30151i − 0.90453j + 0.30151k)T
JJJG
T = T λ = T
DC
= (−0.30151i −0.90453j − 0.30151k)TDC DC DC DC
DC
DC

Equilibrium condition: ΣF = 0: TDA + TDB + TDC + (36 lb)j = 0
(0.55471TDA −0.30151TDB − 0.30151TDC )i + (−0.83206TDA − 0.90453TDB − 0.90453TDC + 36 lb)j
+(0.30151TDB − 0.30151TDC )k = 0
0.55471TDA −0.30151TDB − 0.30151TDC = 0
−0.83206TDA − 0.90453TDB − 0.90453TDC + 36 lb = 0
0.30151TDB − 0.30151TDC = 0
Equation (3) confirms thatTDB = TDC . Solving simultaneously gives,
(1)
(2)
(3)
TDA = 14.42 lb; TDB = TDC = 13.27 lb W

PROBLEM 2.105
A crate is supported by three cables as shown. Determine
the weight of the crate knowing that the tension in cable AC
is 544 lb.
Solution The forces applied at A are:
TAB , TAC , TAD and W
JJJG
AB = −(36 in.)i + (60 in.)j− (27 in.)k
AB = 75 in.
JJJG
AC = (60 in.)j+ (32 in.)k
AC = 68 in.
JJJG
AD = (40 in.)i + (60 in.)j− (27 in.)k
AD = 77 in.
and TAB = TAB λAB = TAB
JJJG
AB
AB
= (−0.48i + 0.8j − 0.36k)TAB
JJJG
T = T λ = T
AC
AC AC AC AC
AC
= (0.88235j + 0.47059k )TAC
JJJG
T = T λ = T
AD
AD AD AD AD
AD
= (0.51948i + 0.77922 j − 0.35065k )TAD
Equilibrium Condition with W = −Wj
ΣF = 0: TAB + TAC + TAD − Wj = 0

(−0.48TAB + 0.51948TAD )i + (0.8TAB + 0.88235TAC + 0.77922TAD −W)j
+(−0.36TAB + 0.47059TAC − 0.35065TAD )k = 0
−0.48TAB + 0.51948TAD = 0
0.8TAB + 0.88235TAC + 0.77922TAD −W = 0
−0.36TAB + 0.47059TAC − 0.35065TAD = 0
(1)
(2)
(3)
Substituting TAC = 544 lb in Equations (1), (2), and (3) above, and solving the resulting set of equations
using conventional algorithms, gives:
TAB = 374.27 lb
TAD = 345.82 lb W =1049 lb W

PROBLEM 2.106
A 1600-lb crate is supported by three cables as shown.
Determine the tension in each cable.
SOLUTION
The forces applied at A are:
TAB , TAC , TAD and W
JJJG
AB = −(36 in.)i + (60 in.)j− (27 in.)k
AB = 75 in.
JJJG
AC = (60 in.)j+ (32 in.)k
AC = 68 in.
JJJG
AD = (40 in.)i + (60 in.)j− (27 in.)k
AD = 77 in.
and TAB = TAB λAB = TAB
JJJG
AB
AB
= (−0.48i + 0.8j − 0.36k)TAB
JJJG
T = T λ = T
AC
AC AC AC AC
AC
= (0.88235j + 0.47059k )TAC
JJJG
T = T λ = T
AD
AD AD AD AD
AD
= (0.51948i + 0.77922 j − 0.35065k )TAD
Equilibrium Condition with W = −Wj
ΣF = 0: TAB + TAC + TAD − Wj = 0

(−0.48TAB + 0.51948TAD )i + (0.8TAB + 0.88235TAC + 0.77922TAD −W)j
+(−0.36TAB + 0.47059TAC − 0.35065TAD )k = 0
−0.48TAB + 0.51948TAD = 0
0.8TAB + 0.88235TAC + 0.77922TAD −W = 0
−0.36TAB + 0.47059TAC − 0.35065TAD = 0
(1)
(2)
(3)
Substituting W = 1600 lb in Equations (1), (2), and (3) above, and solving the resulting set of equations
using conventional algorithms gives,
TAB = 571 lb W
TAC = 830 lb W
TAD = 528 lb W

PROBLEM 2.107
Three cables are connected at A, where the forces P and Q are
SOLUTION
applied as shown. Knowing that Q = 0,
which the tension in cable AD is 305 N.
find the value of P for
ΣFA = 0:
JJJG
TAB + TAC + TAD + P = 0 where P = Pi
AB = −(960 mm)i − (240 mm)j + (380 mm)k
JJJG
AC = −(960 mm)i − (240 mm)j − (320 mm)k
JJJG
AD = −(960 mm)i + (720 mm)j − (220 mm)k
JJJG
AB =1060 mm
AC =1040 mm
AD =1220 mm
T = T λ = T
AB
= T
⎛
−
48
i −
12
j +
19
k
⎞
AB AB AB AB
AB AB ⎜ 53 53 53 ⎟
T = T λ
JJJG
= T
AC
= T
⎝ ⎠
⎛
−
12
i −
3
j−
4
k
⎞
AC AC AC AC
AC AC ⎜ 13 13 13 ⎟
T = T λ
⎝ ⎠
=
305 N
[(−960 mm)i + (720 mm)j − (220 mm)k]
AD AD AD
1220 mm
= −(240 N)i + (180 N)j − (55 N)k
Substituting into ΣFA = 0, factoring i, j, k, and setting each coefficient equal to φ gives:
i : P =
48
T +
12
T + 240 N (1)
53
AB
13
AC
j:
12
T +
3
T = 180 N (2)
53
AB
13
AC
k:
19
T
53
AB
−
4
T
13AC

= 55 N (3)
Solving the system of linear equations using conventional algorithms gives:
TAB = 446.71 N
TAC = 341.71 N P = 960 N W

SOLUTION
PROBLEM 2.108
Three cables are connected at A, where the forces P and Q are
applied as shown. Knowing that P =1200 N, determine the
values of Q for which cable AD is taut.
We assume that TAD = 0 and write
JJJG
ΣFA = 0: TAB + TAC + Qj + (1200 N)i = 0
AB = −(960 mm)i − (240 mm)j+ (380 mm)k
JJJG
AC = −(960 mm)i − (240 mm)j− (320 mm)k
JJJG
AB =1060 mm
AC =1040 mm
T = T λ = T
AB
=
⎛
−
48
i −
12
j+
19
k
⎞
T
AB AB AB AB
AB ⎜ 53 53 53 ⎟ AB
JJJG
⎝ ⎠
T = T λ = T
AC
=
⎛
−
12
i −
3
j−
4
k
⎞
T
AC AC AC AC
AC ⎜ 13 13 13 ⎟ AC
Substituting into ΣFA = 0,
⎝ ⎠
factoring i, j, k, and setting each coefficient equal to φ gives:
i: −
48
T −
12
T +1200 N = 0 (1)
53
AB
j: −
12
T
13
AC
−
3
T + Q = 0 (2)
53
AB
13
AC
k:
19
T −
4
T = 0 (3)
53
AB
13
AC
Solving the resulting system of linear equations using conventional algorithms gives:

605.71 N TAC =
705.71 N Q =
300.00 N
0 ≤ Q < 300 N W
Note: This solution assumes that Q is directed upward as shown (Q ≥ 0), if negative values of Q
are considered, cable AD remains taut, but AC becomes slack for Q = −460 N.

⎝
⎝
k = 0
PROBLEM 2.109
SOLUTION
Knowing that the tension in cable AC is 60 N, determine the
weight of the plate.
We note that the weight of the plate is equal in magnitude to the force P exerted by the support on
Point A.
We have:
JJJG
ΣF = 0: TAB + TAC + TAD + Pj = 0
Free Body A:
Thus:
AB = −(320 mm)i − (480 mm)j+ (360 mm)k
JJJG
AC = (450 mm)i − (480 mm)j+ (360 mm)k
JJJG
AD = (250 mm)i − (480 mm)j− (360 mm)k
JJJG
AB
=
⎛
−
8
AB = 680 mm
AC = 750 mm
AD = 650 mm
12 9 ⎞ Dimensions in mm
TAB = TAB λAB = TAB
AB
JJJG
AC
⎜ i − j+ k ⎟TAB
⎝ 17 17 17 ⎠
TAC = TAC λAC = TAC
AC
JJJG
= (0.6i − 0.64j+ 0.48k)TAC
AD
=
⎛ 5
i −
9.6
j−
7.2
k
⎞
TTAD = TAD λAD = TAD
AD
Substituting into the Eq. ΣF = 0 and factoring i, j, k:
⎜13 13 13
⎟ AD
⎠
⎛ 8 5 ⎞
⎜−
17
TAB + 0.6TAC +
13
TAD ⎟i
⎝
⎛ 12
⎠
9.6 ⎞
+ ⎜−
17
TAB − 0.64TAC −
13
TAD + P⎟ j
⎝
+
⎛ 9
T⎜
17
AB + 0.48TAC −
7.2
T
13
⎠
⎞
AD ⎟⎠

Setting the coefficient of i, j, k equal to zero:
i: −
8
T
17
AB + 0.6TAC +
5
T = 0
13
AD (1)
j: −
12
T − 0.64T −
9.6
T + P = 0 (2)
7
AB AC
13
AD
k:
9
T + 0.48T −
7.2
T = 0 (3)
Making TAC = 60 N in (1) and (3):
17
AB AC
13
AD
−
8
T + 36 N +
5
T = 0 (1′)
17
AB
13
AD
9
T + 28.8 N −
7.2
T = 0 (3′)
17
AB
Multiply (1′) by 9, (3′) by 8, and add:
13
AD
554.4 N −
12.6
T = 0 T = 572.0 N
Substitute into (1′) and solve for TAB :
13
AD AD
T =
17 ⎛
36 +
5
×572
⎞
T = 544.0 N
AB
8 ⎜ 13 ⎟ AB
⎝ ⎠
Substitute for the tensions in Eq. (2) and solve for P:
P =
12
(544N) + 0.64(60 N) +
9.6
(572 N)
17 13
= 844.8 N Weight of plate = P = 845 N W

SOLUTION
PROBLEM 2.110
Knowing that the tension in cable AD is 520 N, determine the
weight of the plate.
See Problem 2.109 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and
(3) below:
−
8
T
17
AB + 0.6TAC +
5
T = 0
13
AD (1)
−
12
T + 0.64T −
9.6
T + P = 0 (2)
17
AB AC
13
AD
9
T + 0.48T −
7.2
T = 0 (3)
17
AB AC
13
AD
Making TAD = 520 N in Eqs. (1) and (3):
−
8
T + 0.6T + 200 N = 0 (1′)
17
AB AC
9
T + 0.48T − 288 N = 0 (3′)
17
AB AC
Multiply (1′) by 9, (3′) by 8, and add:
9.24TAC − 504 N = 0
Substitute into (1′) and solve for TAB :
TAC = 54.5455 N
TAB =
17
(0.6 × 54.5455 + 200)
8
TAB = 494.545 N
Substitute for the tensions in Eq. (2) and solve for P:
P =
12
(494.545 N) + 0.64(54.5455 N) +
9.6
(520 N)
17 13
= 768.00 N

⎝ ⎠
⎝ ⎠
⎝ ⎠
SOLUTION
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.
ΣF = 0: TAB + TAC + TAD + Pj = 0 Free-Body Diagram at A:
JJJG
AB = −20i −100j+ 25k
JJJG
AC = 60i −100j+18k
JJJG
AD = −20i −100j− 74k
JJG
AB = 105 ft
AC = 118 ft
AD =126 ft
T = T λ = T
AB
We write AB AB AB AB
AB
⎛ 4 20 5 ⎞= ⎜−
21
i −
21
j+
21
k⎟TAB
JJJG
T = T λ = T
AC
AC AC AC AC
AC
⎛ 30 50 9 ⎞
= ⎜ 59
i −
59
j+
59
k⎟TAC
JJJG
T = T λ = T
AD
AD AD AD AD
AD
⎛ 10 50 37 ⎞
= ⎜−
63
i −
63
j−
63
k⎟TAD

⎝ ⎠
⎝ ⎠
⎛ 4 30 10 ⎞
⎜−
21
TAB +
59
TAC −
63
TAD ⎟i
⎛ 20 50 50 ⎞+ ⎜−
21
TAB −
59
TAC −
63
TAD + P⎟ j
+
⎛ 5
T +
9
T −
37
T
⎞
k = 0
⎜ 21
AB
59
AC
63
AD ⎟
⎝ ⎠
Setting the coefficients of i, j, k, equal to zero:
i: −
4
T +
30
T −
10
T = 0 (1)
21
AB
59
AC
63
AD
j: −
20
T −
50
T −
50
T + P = 0 (2)
21
AB
k:
5
T
59
AC
+
9
T
63
AD
−
37
T = 0 (3)
21
AB
Set TAB = 840 lb in Eqs. (1) – (3):
59
AC
63
AD
−160 lb +
30
T −
10
T = 0 (1′)
59
AC
63
AD
−800 lb −
50
T −
50
T + P = 0 (2′)
59
AC
63
AD
200 lb +
9
T
59
AC −
37
T = 0
63
AD (3′)
Solving, TAC = 458.12 lb TAD = 459.53 lb P =1552.94 lb P = 1553 lb W

⎝ ⎠
⎝ ⎠
⎝ ⎠
SOLUTION
PROBLEM 2.112
A transmission tower is held by three guy wires attached
to a pin at A and anchored by bolts at B, C, and D. If the
tension in wire AC is 590 lb, determine the vertical force
P exerted by the tower on the pin at A.
ΣF = 0: TAB + TAC + TAD + Pj = 0 Free-Body Diagram at A:
JJJG
AB = −20i −100j+ 25k
JJJG
AC = 60i −100j+18k
JJJG
AD = −20i −100j− 74k
JJG
AB = 105 ft
AC = 118 ft
AD =126 ft
T = T λ = T
AB
We write AB AB AB AB
AB
⎛ 4 20 5 ⎞= ⎜−
21
i −
21
j+
21
k⎟TAB
JJJG
T = T λ = T
AC
AC AC AC AC
AC
⎛ 30 50 9 ⎞
= ⎜ 59
i −
59
j+
59
k⎟TAC
JJJG
T = T λ = T
AD
AD AD AD AD
AD
⎛ 10 50 37 ⎞
= ⎜−
63
i −
63
j −
63
k⎟TAD

⎝ ⎠
⎝ ⎠
⎛ 4 30 10 ⎞
⎜−
21
TAB +
59
TAC −
63
TAD ⎟i
⎛ 20 50 50 ⎞+ ⎜−
21
TAB −
59
TAC −
63
TAD + P⎟ j
+
⎛ 5
T +
9
T −
37
T
⎞
k = 0
⎜ 21
AB
59
AC
63
AD ⎟
⎝ ⎠
Setting the coefficients of i, j, k, equal to zero:
i: −
4
T +
30
T −
10
T = 0 (1)
21
AB
59
AC
63
AD
j: −
20
T −
50
T −
50
T + P = 0 (2)
21
AB
k:
5
T
59
AC
+
9
T
63
AD
−
37
T = 0 (3)
Set TAC = 590 lb
21
AB
in Eqs. (1) – (3):
59
AC
63
AD
−
4
T + 300 lb −
10
T = 0 (1′)
21
AB
63
AD
−
20
T − 500 lb −
50
T + P = 0 (2′)
21
AB
63
AD
5
T + 90 lb −
37
T = 0 (3′)
21
AB
63
AD
Solving, TAB = 1081.82 lb TAD = 591.82 lb P = 2000 lb W

⎝ ⎠
⎝ ⎠
SOLUTION
Free-Body Diagram at A
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.
N = N
⎛ 16
i +
30
j
⎞
⎜ 34 34 ⎟
T = T λ
JJJG
= T
AC
= T
⎝ ⎠
and W = W j = −(175 lb)j
(−30 ft)i +(20 ft)j −(12 ft)k
AC AC AC AC
AC
AC
38 ft
⎛ 15 10 6 ⎞
JJJG
= TAC ⎜−
19
i +
19
j −
19
k⎟
T = T λ = T
AB
= T
(−30 ft)i +(24 ft)j +(32 ft)k
AB AB AB AB
AB
AB
50 ft
⎛ 15 12 16 ⎞
Equilibrium condition: ΣF = 0
= TAB ⎜−
25
i +
25
j+
25
k⎟
TAB + TAC + N + W = 0

Substituting the expressions obtained for TAB ,TAC ,N,
the coefficients to zero gives the following equations:
and W; factoring i, j, and k; and equating each of
From i: −
15
T −
15
T +
16
N = 0 (1)
25
AB
19
AC
34
From j:
12
T +
10
T +
30
N − (175 lb) = 0 (2)
25
AB
19
AC
34
From k:
16
T
25
AB −
6
T = 0
19
AC (3)
Solving the resulting set of equations gives:
TAB = 30.8 lb; TAC = 62.5 lb W

SOLUTION
PROBLEM 2.114
Solve Problem 2.113, assuming that a friend is helping the man
at A by pulling on him with a force P = −(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.
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
T −
15
T +
16
N = 0 (1)
25
AB
19
AC
34
12
T
25
AB +
10
T
19
AC +
30
N − (175 lb) = 0
34
(2)
16
T −
6
T − (45 lb) = 0 (3)
25
AB
19
AC
Solving the resulting set of equations simultaneously gives:
TAB = 81.3 lb W
TAC = 22.2 lb W

SOLUTION
PROBLEM 2.115
For the rectangular plate of Problems 2.109 and 2.110,
determine the tension in each of the three cables knowing that
the weight of the plate is 792 N.
(3) below. Setting P = 792 N gives:
−
8
T + 0.6T +
5
T = 0 (1)
17
AB AC
13
AD
−
12
T − 0.64T −
9.6
T + 792 N = 0 (2)
17
AB AC
13
AD
9
T + 0.48T −
7.2
T = 0 (3)
17
AB AC
13
AD
Solving Equations (1), (2), and (3) by conventional algorithms gives
TAB = 510.00 N
TAC = 56.250 N
TAD = 536.25 N
TAB = 510 N W
TAC = 56.2 N W
TAD = 536 N W

PROBLEM 2.116
For the cable system of Problems 2.107 and 2.108,
determine the tension in each cable knowing that
SOLUTION
P = 2880 N and Q = 0.
ΣFA = 0: TAB + TAC + TAD + P + Q = 0
Where P = Pi and Q = Qj
JJJG
AB = −(960 mm)i − (240 mm)j+ (380 mm)k
JJJG
AC = −(960 mm)i − (240 mm)j − (320 mm)k
JJJG
AD = −(960 mm)i + (720 mm)j − (220 mm)k
JJJG
AB = 1060 mm
AC = 1040 mm
AD =1220 mm
T = T λ = T
AB
= T
⎛
−
48
i −
12
j +
19
k
⎞
AB AB AB AB
AB AB ⎜ 53 53 53 ⎟
T = T λ
JJJG
= T
AC
= T
⎝ ⎠
⎛
−
12
i −
3
j −
4
k
⎞
AC AC AC AC
AC
AC ⎜ 13 13 13 ⎟
T = T λ
JJJG
= T
AD
= T
⎝ ⎠
⎛
−
48
i +
36
j −
11
k
⎞
AD AD AD AD
AD AD ⎜ 61 61 61 ⎟
Substituting into ΣFA = 0, setting P = (2880 N)i and Q = 0,
⎝ ⎠
and setting the coefficients of i, j, k equal to
0, we obtain the following three equilibrium equations:
i : −
48
T −
12
T −
48
T + 2880 N = 0 (1)
53
AB
j: −
12
T
13
AC
−
3
T
61
AD
+
36
T = 0 (2)

53
AB
13
AC
61
AD
k :
19
T −
4
T −
11
T = 0 (3)
53
AB
13
AC
61
AD

TAB = 1340.14 N
TAC = 1025.12 N
TAD = 915.03 N TAB = 1340 N W
TAC = 1025 N W
TAD = 915 N W

PROBLEM 2.117
SOLUTION
P = 2880 N and Q = 576 N.
See Problem 2.116 for the analysis leading to the linear algebraic Equations (1), (2), and (3) below:
−
48
T −
12
T −
48
T + P = 0 (1)
53
AB
13
AC
61
AD
−
12
T −
3
T +
36
T + Q = 0 (2)
53
AB
13
AC
61
AD
19
T −
4
T −
11
T = 0 (3)
53
AB
13
AC
61
AD
Setting P = 2880 N and Q = 576 N gives:
−
48
T −
12
T −
48
T + 2880 N = 0 (1′)
53
AB
13
AC
61
AD
−
12
T −
3
T +
36
T + 576 N = 0 (2′)
53
AB
13
AC
61
AD
19
T −
4
T −
11
T = 0 (3′)
53
AB
13
AC
61
AD
Solving the resulting set of equations using conventional algorithms gives:

TAB = 1431.00 N
TAC = 1560.00 N
TAD =183.010 N
TAB = 1431 N W
TAC = 1560 N W
TAD = 183.0 N W

PROBLEM 2.118
SOLUTION
P = 2880 N and Q = −576 N. (Q is directed downward).
See Problem 2.116 for the analysis leading to the linear algebraic Equations (1), (2), and (3) below:
−
48
T −
12
T −
48
T + P = 0 (1)
53
AB
13
AC
61
AD
−
12
T −
3
T +
36
T + Q = 0 (2)
53
AB
13
AC
61
AD
19
T −
4
T −
11
T = 0 (3)
53
AB
13
AC
61
AD
Setting P = 2880 N and Q = −576 N gives:
−
48
T −
12
T −
48
T + 2880 N = 0 (1′)
53
AB
13
AC
61
AD
−
12
T −
3
T +
36
T − 576 N = 0 (2′)
53
AB
13
19
T
AC
61
−
4
T
AD
−
11
T = 0 (3′)
53
AB
13
AC
61
AD
Solving the resulting set of equations using conventional algorithms gives:

1249.29 N TAC =
490.31 N TAD =
1646.97 N
TAB = 1249 N W
TAC = 490 N W
TAD = 1647 N W

SOLUTION
PROBLEM 2.119
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.
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.
(3) below:
i: −
4
T +
30
T −
10
T = 0 (1)
21
AB
59
AC
63
AD
j: −
20
T −
50
T −
50
T + P = 0 (2)
21
AB
k:
5
T
59
AC
+
9
T
63
AD
−
37
T = 0 (3)
21
AB
59
AC
63
AD
Substituting for P = 1800 lb in Equations (1), (2), and (3) above and solving the resulting set of equations
using conventional algorithms gives:
−
4
T +
30
T −
10
T = 0 (1′)
21
AB
59
AC
63
AD
−
20
T −
50
T −
50
T +1800 lb = 0 (2′)
21
AB
59
AC
63
AD
5
T +
9
T −
37
T = 0 (3′)
21
AB
59
AC
63
AD

= 973.64 lb TAC
= 531.00 lb TAD
= 532.64 lb
TAB = 974 lb W
TAC = 531 lb W
TAD = 533 lb W

SOLUTION
PROBLEM 2.120
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.
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
JJJG
DA = (18 in.)j+ (22 in.)k
DA = 28.425 in.
JJJG
DB = −(24 in.)i + (18 in.)j− (16 in.)k
DB = 34.0 in.
JJJG
DC = (24 in.)i + (18 in.)j − (16 in.)k
DC = 34.0 in.

JJJG
T = T λ = T
DA
and DA Da DA Da
DA
= (0.63324 j + 0.77397k )TDA
JJJG
T = T λ = T
DB
DB DB DB DB
DB
= (−0.70588i + 0.52941j − 0.47059k )TDB
JJJG
T = T λ = T
DC
Equilibrium Condition with
DC DC DC DC
DC
= (0.70588i + 0.52941j − 0.47059k )TDC
W = −Wj
ΣF = 0: TDA + TDB + TDC − Wj = 0
(−0.70588TDB + 0.70588TDC )i
(0.63324TDA + 0.52941TDB + 0.52941TDC −W)j
(0.77397TDA − 0.47059TDB − 0.47059TDC )k
−0.70588TDB + 0.70588TDC = 0
0.63324TDA + 0.52941TDB + 0.52941TDC −W = 0
0.77397TDA − 0.47059TDB − 0.47059TDC = 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,
TDA = 119.7 lb W
TDB = 98.4 lb W
TDC = 98.4 lb W

SOLUTION
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.)
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
JJJG
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
JJJG
T = Tλ = T
AB
and
AB AB AB
AB
=
TAB
[−(0.78 m)i + (1.6 m)j + (0 m)k]
1.78 m
TAB = TAB (−0.4382i + 0.8989j+ 0k)
JJJG
AC = (0)i + (1.6 m)j+ (1.2 m)k
AC = (0 m)2
+ (1.6 m)2
+ (1.2 m)2
= 2 m
JJJG
TAC = TλAC = TAC
AC
=
TAC
[(0)i + (1.6 m)j+ (1.2 m)k]
AC 2 m
and
TAC = TAC (0.8j+ 0.6k)
JJJG
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
JJJG
TAD = TλAD = TAD
AD
=
TAD
[(1.3 m)i + (1.6 m)j+ (0.4 m)k]
AD 2.1 m
TAD = TAD (0.6190i + 0.7619j + 0.1905k)

Finally,
JJJG
AE = −(0.4 m)i + (1.6 m)j− (0.86 m)k
AE = (−0.4 m)2
+ (1.6 m)2
+ (−0.86 m)2
=1.86 m
JJJG
T = Tλ = T
AE
With the weight of the container
AE AE AE
AE
=
TAE
[−(0.4 m)i + (1.6 m)j − (0.86 m)k]
1.86 m
TAE = TAE (−0.2151i + 0.8602j− 0.4624k)
W = −Wj, at A we have:
ΣF = 0: TAB + TAC + TAD −Wj = 0
Equating the factors of i, j, and k to zero, we obtain the following linear algebraic equations:
−0.4382TAB + 0.6190TAD − 0.2151TAE = 0
0.8989TAB + 0.8TAC + 0.7619TAD + 0.8602TAE −W = 0
0.6TAC + 0.1905TAD − 0.4624TAE = 0
(1)
(2)
(3)
Knowing that W = 1000 N and that because of the pulley system at BTAB = 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 W

PROBLEM 2.122
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.
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
SOLUTION
magnitude of P. (Hint: The tension is the same in
all portions of cable FBAD.)
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
TAB = TAD = P
and using the linear algebraic equations of Problem 2.131 with TAC = 150 N, we obtain
(a) P = 454 N W
(b) W = 1202 N W

SOLUTION
PROBLEM 2.123
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.
Free Body Diagram at A:
Since TBAC = tension in cable BAC, it follows that
TAB = TAC = TBAC
T λ
(−17.5 in.)i +(60 in.)j ⎛ −17.5
i
60
j
⎞
AB = TBAC AB = TBAC
62.5 in.
= TBAC ⎜ 62.5
+ ⎟
62.5
⎝ ⎠
T λ
(60 in.)i +(25 in.)k ⎛ 60
j
25
k
⎞
AC = TBAC AC = TBAC
65 in.
= TBAC ⎜ 65
+
65 ⎟
⎝ ⎠
T λ
(80 in.)i +(60 in.)j ⎛ 4
i
3
j
⎞
AD = TAD AD = TAD
100 in.
= TAD ⎜ 5
+
5 ⎟
⎝ ⎠
(60 in.)j −(45 in.)k ⎛ 4
j
3
k
⎞
TAE = TAE λAE = TAE
75 in.
= TAE ⎜ 5
−
5 ⎟
⎝ ⎠

⎝ ⎠
Substituting into ΣFA = 0, setting P = (−200 lb)j, and setting the coefficients of i, j, k equal to φ, we
obtain the following three equilibrium equations:
From i: −
17.5
T +
4
T = 0 (1)
62.5 BAC
5
AD
From j:
⎛ 60
+
60 ⎞
T +
3
T +
4
T − 200 lb = 0 (2)
⎜
62.5 65
⎟ BAC
5
AD
5
AE
From k:
25
T −
3
T = 0 (3)
65
BAC
5
AE
TBAC = 76.7 lb; TAD = 26.9 lb; TAE = 49.2 lb W

⎝ ⎠
SOLUTION
PROBLEM 2.124
Knowing that the tension in cable AE of Prob. 2.123 is 75 lb,
determine (a) the magnitude of the load P, (b) the tension in
cables BAC and AD.
PROBLEM 2.123 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.
Refer to the solution to Problem 2.123 for the figure and analysis leading to the following set of
equilibrium equations, Equation (2) being modified to include Pj as an unknown quantity:
−
17.5
T +
4
T = 0 (1)
62.5 BAC
5
AD
⎛ 60
+
60 ⎞
T +
3
T +
4
T − P = 0 (2)
⎜
62.5 65
⎟ BAC
5
AD
5
AE
25
T −
3
T = 0 (3)
65
BAC
5
AE
Substituting for TAE = 75 lb and solving simultaneously gives:
(a)
(b)
P = 305 lb W
TBAC =117.0 lb; TAD = 40.9 lb W

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
SOLUTION
of the force Q required to maintain the equilibrium of the
system.
For both Problems 2.125 and 2.126:
(AB)2
= x2
+ y2
+ z2
Free-Body Diagrams of Collars:
Here
or
(0.525 m)2
= (0.20 m)2
+ y2
+ z2
y2
+ z2
= 0.23563 m2
Thus, when y given, z is determined,
JJJG
AB
Now λAB =
AB
=
1
(0.20i − yj+ zk)m
0.525 m
= 0.38095i −1.90476yj +1.90476zk
Where y and z are in units of meters, m.
From the F.B. Diagram of collar A: ΣF = 0: Nxi + Nzk + Pj + TABλAB = 0
Setting the j coefficient to zero gives
With
P − (1.90476y)TAB = 0
P = 341 N
T =
341 N
AB
1.90476y
Now, from the free body diagram of collar B: ΣF = 0: Nxi + Ny j+ Qk −TAB λAB = 0
Setting the k coefficient to zero gives
And using the above result for TAB , we have
Q −TAB (1.90476z) = 0
Q = T z =
341 N
(1.90476z) =
AB
(1.90476)y

Then from the specifications of the problem,
and
y =155 mm = 0.155 m
z2
= 0.23563 m2
− (0.155 m)2
z = 0.46 m
(a) T =
341 N
AB
0.155(1.90476)
= 1155.00 N
or
and
TAB = 1155 N W
(b)
or
Q =
341 N(0.46 m)(0.866)
(0.155 m)
= (1012.00 N)
Q =1012 N W

PROBLEM 2.126
Solve Problem 2.125 assuming that y = 275 mm.
PROBLEM 2.125 Collars A and B are connected by a
525-mm-long wire and can slide freely on frictionless
rods. If a force P = (341 N)j is applied to collar A,
SOLUTION
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.
From the analysis of Problem 2.125, particularly the results:
y2
+ z2
= 0.23563 m2
T =
341 N
AB
1.90476y
Q =
341 N
z
y
With
and
(a)
y = 275 mm = 0.275 m, we obtain:
z2
= 0.23563 m2
− (0.275 m)2
z = 0.40 m
T =
341 N
= 651.00
or
and
(b)
or
AB
(1.90476)(0.275 m)
Q =
341 N(0.40 m)
(0.275 m)
TAB = 651 N W
Q = 496 N W

SOLUTION
PROBLEM 2.127
Two structural members A and B are bolted to a bracket as
shown. Knowing that both members are in compression and
that the force is 15 kN in member A and 10 kN in member B,
determine by trigonometry the magnitude and direction of the
resultant of the forces applied to the bracket by members A and
B.
Using the force triangle and the laws of cosines and sines,
we have
Then
and
γ = 180° − (40° + 20°)
= 120°
R2
= (15 kN)2
+ (10 kN)2
−2(15 kN)(10 kN)cos120°
= 475 kN2
R = 21.794 kN
10 kN
=
21.794 kN
sinα
sinα =
⎛
sin120°
10 kN ⎞
sin120°
⎜ 21.794 kN ⎟
⎝ ⎠
= 0.39737
α = 23.414
Hence: φ = α + 50° = 73.414 R = 21.8 kN 73.4° W

SOLUTION
PROBLEM 2.128
OA = (24 in.)2
+ (45 in.)2
= 51.0 in.
OB = (28 in.)2
+ (45 in.)2
= 53.0 in.
OC = (40 in.)2
+ (30 in.)2
102-lb Force:
= 50.0 in.
F = −102 lb
24 in.
F = −48.0 lb W
x
51.0 in.
x
F = +102 lb
45 in.
F = +90.0 lb W
106-lb Force:
y
51.0 in.
F = +106 lb
28 in.
x
53.0 in.
y
Fx = +56.0 lb W
F = +106 lb
45 in.
F = +90.0 lb W
y
53.0 in.
y
200-lb Force: F = −200 lb
40 in.
F = −160.0 lb W
x
50.0 in.
x
F = −200 lb
30 in.
F = −120.0 lb W
y
50.0 in.
y

SOLUTION
PROBLEM 2.129
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.
Rx = ΣFx = P + (200 lb)sin 40° − (400 lb)cos40°
Rx = P −177.860 lb
Ry = ΣFy = (200 lb)cos40° + (400 lb)sin 40°
Ry = 410.32 lb
(a) For R to be vertical, we must have Rx = 0.
(1)
(2)
Set
(b) Since R is to be vertical:
Rx = 0 in Eq. (1)
0 = P −177.860 lb
P =177.860 lb
R = Ry = 410 lb
P = 177.9 lb W
R = 410 lb W

SOLUTION
PROBLEM 2.130
Knowing that α = 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.
Law of sines:
(a)
FAC
=
sin 35°
TBC
sin 50°
300 lb
=
300 lb
sin 95°
(b)
FAC =
sin 95°
sin 35°
T =
300 lb
sin 50°BC
sin 95°
FAC =172.7 lb W
TBC = 231 lb W

PROBLEM 2.131
Two cables are tied together at C and loaded as shown.
SOLUTION
Knowing that P = 360 N, determine the tension (a) in cable
AC, (b) in cable BC.
Free Body: C
(a) ΣFx = 0: −
12
T +
4
(360 N) = 0
13
AC
5
TAC = 312 N W
(b) ΣF = 0:
5
(312 N) + T +
3
(360 N) − 480 N = 0
y
13
BC
5
TBC = 480 N −120 N − 216 N TBC =144.0 N W

PROBLEM 2.132
Two cables tied together at C are loaded as shown. Knowing that
the maximum allowable tension in each cable is 800 N,
SOLUTION
determine (a) the magnitude of the largest force P that can be
applied at C, (b) the corresponding value of α.
Free-Body Diagram: C Force Triangle
Force triangle is isosceles with
(a)
2β =180° − 85°
β = 47.5°
P = 2(800 N)cos 47.5° =1081 N
(b)
Since P > 0, the solution is correct.
α = 180° − 50° − 47.5° = 82.5°
P = 1081 N W
α = 82.5° W

SOLUTION
PROBLEM 2.133
The end of the coaxial cable AE is attached to the pole AB, which
is strengthened by the guy wires AC and AD. Knowing that the
tension in wire AC is 120 lb, determine (a) the components of
the force exerted by this wire on the pole, (b) the angles θx,
θy, and θz that the force forms with the coordinate axes.
(a) Fx = (120 lb) cos 60° cos 20°
Fx = 56.382 lb
Fy = −(120 lb)sin 60°
Fy = −103.923 lb
Fz = −(120 lb)cos 60° sin 20°
Fz = −20.521 lb
Fx = +56.4 lb W
Fy = −103.9 lb W
Fz = −20.5 lb W
(b) cos
cos
cos
θx =
θ y =
θz =
Fx
=
F
Fy
=
F
Fz
=
F
56.382 lb
120 lb
−103.923 lb
120 lb
−20.52 lb
120 lb
θx = 62.0° W
θy =150.0° W
θz = 99.8° W

PROBLEM 2.134
Knowing that the tension in cable AC is 2130 N, determine
the components of the force exerted on the plate at C.
SOLUTION
JJJG
CA = −(900 mm)i + (600 mm)j− (920 mm)k
CA = (900 mm)2
+ (600 mm)2
+ (920 mm)2
=1420 mm
TCA = TCAλCA
JJJG
= T
CA
TCA
CA
CA
=
2130 N
[−(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 W

SOLUTION
PROBLEM 2.135
P = (600 N)[sin 40°sin 25°i + cos40°j + sin 40°cos25°k]
= (162.992 N)i + (459.63 N)j + (349.54 N)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 R = 940 N W
cos
cos
θx =
Rx
R
Ry
=
386.52 N
940.22 N
828.25 N
θx = 65.7° W
θy =
cosθz =
R
Rz
R
=
940.22 N
=
220.49 N
940.22 N
θy = 28.2° W
θz = 76.4° W

Vector mechanics for engineers statics and dynamics 11th edition beer solutions manual

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