Problem 3.1 Three forces act on a joint of a structure.
The joint’s weight is negligible and it is in equilibrium.
The force FA D 4 kN. Determine the force FB and the
vertical force FC .
y
FC
FB
15⬚
x
40⬚
FA
Solution:
FC
FA D 4 kN
FB
Fx : FA cos 40° FB cos 15° D 0
15°
Fy : FA sin 40° FC FB sin 15° D 0
Solving we find
40°
FB D 3.17 kN, FC D 1.75 kN
FA
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1
Problem 3.2 The mass of the ring is 2 kg. The y-axis
points upward. The angle ˛ D 45° .
(a)
(b)
y
F2
What is the ring’s weight in newtons?
Determine the forces F1 and F2 .
a
F1
30⬚
x
Solution:
F2
F1
(a)
(b)
W D mg D 2 kg9.81 m/s2 D 19.62 N
Fx : F1 cos 30° F2 cos 45° D 0
30°
45°
Fy : F1 sin 30° C F2 sin 45° 19.62 N D 0
Solving:
F1 D 14.36 N, F2 D 17.59 N
mg
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1
Problem 3.3 In Problem 3.2, suppose that you want to
choose the angle ˛ so that the force F2 is a minimum.
What is the required angle ˛ and the resulting value
of F2 ?
Strategy: Draw a vector diagram of the sum of the
forces acting on the ring.
Solution: The three forces must add to zero. The weight and force
F2
F1 have specified directions. For F2 to be minimum we must have
˛ C 30° D 90° ) ˛ D 60°
If that is the case then
mg
F2 D 19.62 cos 30° D 16.99 N
α
30°
F1
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1
Problem 3.4 The 200-kg engine block is suspended
by the cables AB and AC. The angle ˛ D 40° . The freebody diagram obtained by isolating the part of the system
within the dashed line is shown. Determine the forces
TAB and TAC .
y
TAB
B
TAC
C
a
A
a
A
x
(200 kg) (9.81 m/s2)
Solution:
TAB
TAC
˛ D 40°
Fx : TAC cos ˛ TAB cos ˛ D 0
α
α
Fy : TAC sin ˛ C TAB sin ˛ 1962 N D 0
Solving:
TAB D TAC D 1.526 kN
1962 Ν
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1
Problem 3.5 In Problem 3.4, suppose that you don’t
want either of the forces TAB or TAC to exceed 2 kN.
What is the smallest acceptable value of the angle ˛?
Solution:
TAB
TAC
Fx : TAC cos ˛ TAB cos ˛ D 0
Fy : TAC sin ˛ C TAB sin ˛ 1962 N D 0
α
α
Solving we find
TAB D TAC D
981 N
sin ˛
If we impose the limit:
2000 N D
981 N
981 N
) sin ˛ D
D 0.4905 ) ˛ D 29.4°
sin ˛
2000 N
1962 Ν
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1
Problem 3.6 A zoologist estimates that the jaw of a
predator, Martes, is subjected to a force P as large as
800 N. What forces T and M must be exerted by the
temporalis and masseter muscles to support this value
of P?
22°
T
P
M
36°
Solution: Resolve the forces into scalar components, and solve the
equilibrium equations. . .Express the forces in terms of horizontal and
vertical unit vectors:
T D jTji cos 22° C j sin 22° D jTj0.927i C 0.375j
P D 800i cos 270° C j sin 270° D 0i 800j
M D jMji cos 144° C j sin 144° D jMj0.809i C 0.588j
Apply the equilibrium conditions,
FD0DTCMCPD0
Collect like terms:
Fx D 0.927jTj 0.809jMji D 0
Fy D 0.375jTj C 0.588jMj 800j D 0
(1)
(2)
0.809
jMj D 0.873jMj
0.927
Substitute this value into the second equation, reduce algebraically,
and solve: jMj D 874 N, jTj D 763.3 N
Solve the first equation, jTj D
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1
Problem 3.7 The two springs are identical, with unstretched lengths 250 mm and spring constants k D
1200 N/m.
(a)
(b)
(c)
Draw the free-body diagram of block A.
Draw the free-body diagram of block B.
What are the masses of the two blocks?
300 mm
A
280 mm
B
Solution: The tension in the upper spring acts on block A in the
positive Y direction, Solve the spring force-deflection equation for
the tension in the upper spring. Apply the equilibrium conditions to
block A. Repeat the steps for block B.
300 mm
N
0.3 m 0.25 mj D 0i C 60j N
TUA D 0i C 1200
m
Similarly, the tension in the lower spring acts on block A in the negative Y direction
A
280 mm
N
0.28 m 0.25 mj D 0i 36j N
TLA D 0i 1200
m
B
The weight is WA D 0i jWA jj
The equilibrium conditions are
FD
Fx C
Fy D 0,
Tension,
upper spring
F D WA C TUA C TLA D 0
A
Collect and combine like terms in i, j
Solve
Fy D jWA j C 60 36j D 0
Tension,
lower
spring
Weight,
mass A
jWA j D 60 36 D 24 N
The mass of A is
mA D
jWL j
24 N
D 2.45 kg
D
jgj
9.81 m/s2
The free body diagram for block B is shown.
The tension in the lower spring TLB D 0i C 36j
The weight: WB D 0i jWB jj
Apply the equilibrium conditions to block B.
Tension,
lower spring
y
B
x
Weight,
mass B
F D WB C TLB D 0
Collect and combine like terms in i, j:
Solve:
Fy D jWB j C 36j D 0
jWB j D 36 N
The mass of B is given by mB D
jWB j
36 N
D 3.67 kg
D
jgj
9.81 m/s2
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1
Problem 3.8 The two springs in Problem 3.7 are identical, with unstretched lengths of 250 mm. Suppose that
their spring constant k is unknown and the sum of the
masses of blocks A and B is 10 kg. Determine the value
of k and the masses of the two blocks.
Solution: All of the forces are in the vertical direction so we will
use scalar equations. First, consider the upper spring supporting both
masses (10 kg total mass). The equation of equilibrium for block the
entire assembly supported by the upper spring is A is TUA mA C
mB g D 0, where TUA D kU 0.25 N. The equation of equilibrium
for block B is TUB mB g D 0, where TUB D kL 0.25 N. The
equation of equilibrium for block A alone is TUA C TLA mA g D 0
where TLA D TUB . Using g D 9.81 m/s2 , and solving simultaneously, we get k D 1962 N/m, mA D 4 kg, and mB D 6 kg .
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1
Problem 3.9 The inclined surface is smooth. The
two springs are identical, with unstretched lengths of
250 mm and spring constants k D 1200 N/m. What are
the masses of blocks A and B?
300 mm
A
280 mm
B
30⬚
Solution:
F1 D 1200 N/m0.3 0.25m D 60 N
mAg
F1
F2 D 1200 N/m0.28 0.25m D 36 N
F2
FB &: F2 C mB g sin 30° D 0
mBg
F2
FA &: F1 C F2 C mA g sin 30° D 0
NA
Solving: mA D 4.89 kg, mB D 7.34 kg
NB
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1
Problem 3.10 The mass of the crane is 20,000 kg. The
crane’s cable is attached to a caisson whose mass is
400 kg. The tension in the cable is 1 kN.
(a)
(b)
Determine the
friction forces
level ground.
Determine the
friction forces
level ground.
magnitudes of the normal and
exerted on the crane by the
45°
magnitudes of the normal and
exerted on the caisson by the
Strategy: To do part (a), draw the free-body diagram
of the crane and the part of its cable within the
dashed line.
Solution:
(a)
45°
Fy : Ncrane 196.2 kN 1 kN sin 45° D 0
196.2 kN
1 kN
Fx : Fcrane C 1 kN cos 45° D 0
y
Ncrane D 196.9 kN, Fcrane D 0.707 kN
(b)
Fy : Ncaisson 3.924 kN C 1 kN sin 45° D 0
x
Fcrane
Fx : 1 kN cos 45° C Fcaisson D 0
Ncrane
Ncaisson D 3.22 kN, Fcaisson D 0.707 kN
1 kN
3.924 kN
45°
Fcaisson
Ncaisson
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1
Problem 3.11 The inclined surface is smooth. The
100-kg crate is held stationary by a force T applied to
the cable.
Solution:
(a)
The FBD
T
(a)
(b)
Draw the free-body diagram of the crate.
Determine the force T.
981 Ν
T
Ν
60°
60⬚
(b)
F -: T 981 N sin 60° D 0
T D 850 N
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1
Problem 3.12
sloping road.
(a)
(b)
The 1200-kg car is stationary on the
If ˛ D 20° , what are the magnitudes of the total
normal and friction forces exerted on the car’s tires
by the road?
The car can remain stationary only if the total
friction force necessary for equilibrium is not
greater than 0.6 times the total normal force.
What is the largest angle ˛ for which the car can
remain stationary?
α
Solution:
11.772 kN
(a) ˛ D 20°
F% : N 11.772 kN cos ˛ D 0
F- : F 11.772 kN sin ˛ D 0
N D 11.06 kN, F D 4.03 kN
(b)
α
F D 0.6 N
F
F% : N 11.772 kN cos ˛ D 0
) ˛ D 31.0°
N
F- : F 11.772 kN sin ˛ D 0
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1
Problem 3.13 The crate is in equilibrium on the
smooth surface. (Remember that “smooth” means that
friction is negligible). The spring constant is k D
2500 N/m and the stretch of the spring is 0.055 m. What
is the mass of the crate?
20°
Solution:
K D 2500 N/m
y
Kδ
υ D 0.055 m
&C
%C
Fx D Kυ C m9.81 sin 20° D 0
x
Fy D N-m9.81 cos 20° D 0
N
20°
mg = 9.81 m
25000.055 C 3.355 m D 0
N 9.218 m D 0
Solving, m D 41.0 kg, N D 378 N
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1
Problem 3.14 A 600-lb box is held in place on the
smooth bed of the dump truck by the rope AB.
(a)
(b)
If ˛ D 25° , what is the tension in the rope?
If the rope will safely support a tension of 400 lb,
what is the maximum allowable value of ˛?
B
A
α
Solution: Isolate the box. Resolve the forces into scalar components, and solve the equilibrium equations.
A
B
The external forces are the weight, the tension in the rope, and the
normal force exerted by the surface. The angle between the x axis and
the weight vector is 90 ˛ (or 270 C ˛). The weight vector is
α
W D jWji sin ˛ j cos ˛ D 600i sin ˛ j cos ˛
The projections of the rope tension and the normal force are
y
T D jTx ji C 0j N D 0i C jNy jj
T
The equilibrium conditions are
x
FDWCNCTD0
Substitute, and collect like terms
N
W
α
Fx D 600 sin ˛ jTx ji D 0
Fy D 600 cos ˛ C jNy jj D 0
Solve for the unknown tension when
For ˛ D 25°
jTx j D 600 sin ˛ D 253.6 lb.
For a tension of 400 lb, (600 sin ˛ 400 D 0. Solve for the unknown
angle
sin ˛ D
400
D 0.667 or ˛ D 41.84°
600
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1
Problem 3.15 The 40-kg box is held in place on the
smooth inclined surface by the horizontal cable AB.
Determine the tension in the cable and the normal force
exerted on the box by the inclined surface.
A
B
50⬚
Solution:
Fx : T C N sin 50° D 0
392.4 N
T
y
Fy : N cos 50° 392.4 N D 0
Solving: N D 610 N, T D 468 N
x
50°
N
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1
Problem 3.16 The 1360-kg car and the 2100-kg tow
truck are stationary. The muddy surface on which the
car’s tires rest exerts negligible friction forces on them.
What is the tension in the tow cable?
18⬚
10⬚
26⬚
Solution: FBD of the car being towed
F- : T cos 8° 13.34 kN sin 26° D 0
13.34 kN
T
18°
T D 5.91 kN
26°
N
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1
Problem 3.17 In Problem 3.16, determine the magnitude of the total friction force exerted on the tow truck’s
tires. (This is the friction force the truck’s tires must
exert to prevent the truck and car from sliding down
the slope.)
Solution: Use the value of the tension from the previous problem
20.6 kN
F- : F 20.6 kN sin 10° T cos 8° D 0
Solving: F D 9.43 kN
18°
F
10°
T
N
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1
Problem 3.18 A 10-kg painting is hung with a wire
supported by a nail. The length of the wire is 1.3 m.
(a)
(b)
What is the tension in the wire?
What is the magnitude of the force exerted on the
nail by the wire?
1.2 m
Solution:
(a)
Fy : 98.1 N 2
98.1 N
5
TD0
13
T D 128 N
T
(b)
T
5
Force D 98.1 N
12
12
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1
Problem 3.19 A 10-kg painting is hung with a wire
supported by two nails. The length of the wire is 1.3 m.
(a)
(b)
What is the tension in the wire?
What is the magnitude of the force exerted on each
nail by the wire? (Assume that the tension is the
same in each part of the wire.)
0.4 m
0.4 m
0.4 m
Compare your answers to the answers to Problem 3.18.
T
Solution:
(a)
Examine the point on the left where the wire is attached to the
picture. This point supports half of the weight
R
27.3°
Fy : T sin 27.3° 49.05 N D 0
T D 107 N
(b)
49.05 N
Examine one of the nails
Fx : Rx T cos 27.3° C T D 0
RD
Ry
Fy : Ry T sin 27.3° D 0
Rx
27.3°
Rx 2 C Ry 2
T
T
R D 50.5 N
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1
Problem 3.20 Assume that the 150-lb climber is in
equilibrium. What are the tensions in the rope on the
left and right sides?
15°
14°
y
Solution:


Fx D TR cos15° TL cos14° D 0

Fy D TR sin15° C TL sin14° 150 D 0
14°
TR
TL
15°
Solving, we get TL D 299 lb, TR D 300 lb
x
150 lb
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1
Problem 3.21 If the mass of the climber shown in
Problem 3.20 is 80 kg, what are the tensions in the rope
on the left and right sides?
y
Solution:


Fx D TR cos15° TL cos14° D 0
 F D T sin15° C T sin14° mg D 0
y
R
L
TL
TR
14°
15°
Solving, we get
TL D 1.56 kN,
x
TR D 1.57 kN
mg = (80) (9.81) N
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1
Problem 3.22 A construction worker holds a 180-kg
crate in the position shown. What force must she exert
on the cable?
5°
30°
Solution: Eqns. of Equilibrium:

Fx D T2 cos 30° T1 sin 5° D 0



Fy D T1 cos 5° T2 sin 30° mg D 0



mg D 1809.81 N
Solving, we get
T1 D 1867 N T2 D 188 N
5°
y
T1
x
30°
T2
mg = (180) (9.81) N
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1
Problem 3.23 A construction worker on the moon
(acceleration due to gravity 1.62 m/s2 ) holds the same
crate described in Problem 3.22 in the position shown.
What force must she exert on the cable?
5°
30°
5°
Solution Eqns. of Equilibrium

Fx D T2 cos 30° T1 sin 5° D 0



Fy D T1 cos 5° T2 sin 30° mg D 0



mg D 1801.62 N
y
T1
Solving, we get
T1 D 308 N
T2 D 31.0 N
x
30°
T2
mg = (180) (1.62) N
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1
Problem 3.24 The person wants to cause the 200-lb
crate to start sliding toward the right. To achieve this,
the horizontal component of the force exerted on the
crate by the rope must equal 0.35 times the normal force
exerted on the crate by the floor. In Fig. a, the person
pulls on the rope in the direction shown. In Fig. b, the
person attaches the rope to a support as shown and pulls
upward on the rope. What is the magnitude of the force
he must exert on the rope in each case?
20°
(a)
10°
(b)
Solution:
(a)
T
200 lb
Fy : N 200 lb C T sin 20° D 0
20°
Also T cos 20° D 0.35 N
Solving: T D 66.1 lb
(b)
Ffr
N
The person exerts the force F
200 lb
Fy : N 200 lb D 0, T D 0.35 N D 70 lb
T
Fy : F T sin 10° D 0 ) F D 12.2 lb
Ffr
N
F
10°
T
T
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1
Problem 3.25 A traffic engineer wants to suspend a
200-lb traffic light above the center of the two right
lanes of a four-lane thoroughfare as shown. Determine
the tensions in the cables AB and BC.
Solution:
80 ft
20 ft
A
C
6
2
Fx : p TAB C p TBC D 0
37
5
1
1
Fy : p TAB C p TBC 200 lb D 0
37
5
Solving:
TAB D 304 lb, TBC D 335 lb
10 ft
B
TBC
30 ft
6
1
1
TAB
2
200 lb
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1
Problem 3.26 Cable AB is 3 m long and cable BC is
4 m long. The mass of the suspended object is 350 kg.
Determine the tensions in cables AB and BC.
5m
A
C
B
Solution:
TAB
TAC
3
4
Fx : TAB C TBC D 0
5
5
4
Fy :
4
3
TAB C TBC 3.43 kN D 0
5
5
4
3
3
TAB D 2.75 kN, TBC D 2.06 kN
3.43 kN
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1
Problem 3.27 In Problem 3.26, the length of cable AB
is adjustable. If you don’t want the tension in either cable
AB or cable BC to exceed 3 kN, what is the minimum
acceptable length of cable AB?
Solution: Consider the geometry:
x
5−x
We have the constraints
LAB 2 D x 2 C y 2 , 4 m2 D 5 m x2 C y 2
y
LAB
4m
These constraint imply
yD
LD
10 mx x 2 9 m2
TAB
TBC
10 mx 9 m2
Now draw the FBD and write the equations in terms of x
x
5x
Fx : p
TAB C
TBC D 0
4
10x 9
p
Fy :
10x x 2 9
p
TAB C
10x 9
p
y
4
y
x
5−x
10x x 2 9
TBC 3.43 kN D 0
4
If we set TAB D 3 kN and solve for x we find x D 1.535, TBC D
2.11 kN < 3 kN
3.43 kN
Using this value for x we find that LAB D 2.52 m
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1
Problem 3.28 What are the tensions in the upper and
lower cables? (Your answers will be in terms of W.
Neglect the weight of the pulley.)
45°
30°
W
Solution: Isolate the weight. The frictionless pulley changes the
direction but not the magnitude of the tension. The angle between the
right hand upper cable and the x axis is ˛, hence
TU
TU
β
TUR D jTU ji cos ˛ C j sin ˛.
α
y
The angle between the positive x and the left hand upper pulley is
180° ˇ, hence
TUL D jTU ji cos180 ˇ C j sin180 ˇ
TL
W
x
D jTU ji cos ˇ C j sin ˇ.
The lower cable exerts a force:
The weight:
TL D jTL ji C 0j
W D 0i jWjj
The equilibrium conditions are
F D W C TUL C TUR C TL D 0
Substitute and collect like terms,
Fx D jTU j cos ˇ C jTU j cos ˛ jTL ji D 0
Fy D jTU j sin ˛ C jTU j sin ˇ jWjj D 0.
Solve:
jTU j D
jWj
sin ˛ C sin ˇ
,
jTL j D jTU jcos ˛ cos ˇ.
From which
For
jTL j D jWj
cos ˛ cos ˇ
sin ˛ C sin ˇ
.
˛ D 30° and ˇ D 45°
jTU j D 0.828jWj,
jTL j D 0.132jWj
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1
Problem 3.29 Two tow trucks lift a motorcycle out of
a ravine following an accident. If the 100-kg motorcycle
is in equilibrium in the position shown, what are the
tensions in cables AB and AC?
(10, 9) m
y
(3, 8) m
C
B
(6, 4.5) m
A
x
Solution: We need to find unit vectors eAB and eAC . Then write
TAB D TAB eAB and TAC D TAC eAC . Finally, write and solve the
equations of equilibrium.
y
TAB
For the ring at A.
TAC
From the known locations of points A, B, and C,
eAB D
rAB
jrAB j
eAC D
rAC
jrAC j
rAB D 3i C 3.5j m jrAB j D 4.61 m
rAC D 4i C 4.5j m jrAC j D 6.02 m
A (6, 4, 5)
B (3, 8)
C (10, 9)
x
A
mg = (100) (9.81) N
eAB D 0.651i C 0.759j
eAC D 0.664i C 0.747j
TAB D 0.651TAB i C 0.759TAB j
TAC D 0.664TAC i C 0.747TAC j
W D mgj D 1009.81j N
For equilibrium,
TAB C TAC C W D 0
In component form, we have


Fx D 0.651TAB C 0.664TAC D 0

Fy D C0.759TAB C 0.747TAC 981 D 0
Solving, we get
TAB D 658 N, TAC D 645 N
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1
Problem 3.30 An astronaut candidate conducts experiments on an airbearing platform. While he carries out
calibrations, the platform is held in place by the horizontal tethers AB, AC, and AD. The forces exerted by
the tethers are the only horizontal forces acting on the
platform. If the tension in tether AC is 2 N, what are the
tensions in the other two tethers?
TOP VIEW
D
4.0 m
A
3.5 m
B
C
3.0 m
Solution: Isolate the platform. The angles ˛ and ˇ are
tan ˛ D
Also,
tan ˇ D
1.5
3.5
3.0
3.5
D 0.429,
˛ D 23.2° .
B
3.0 m
A
D 0.857,
C 3.5
m
ˇ D 40.6° .
4.0
m
y
B
x
TAB D jTAB ji cos180° ˇ C j sin180° ˇ
β
α
TAB D jTAB ji cos ˇ C j sin ˇ.
The angle between the tether AC and the positive x axis is 180° C ˛.
The tension is
D jTAC ji cos ˛ j sin ˛.
C
Solve:
jTAB j D
The tether AD is aligned with the positive x axis, TAD D jTAD ji C 0j.
The equilibrium condition:
F D TAD C TAB C TAC D 0.
Substitute and collect like terms,
D
A
TAC D jTAC ji cos180° C ˛ C j sin180° C ˛
D
1.5 m
The angle between the tether AB and the positive x axis is 180° ˇ,
hence
1.5 m
jTAD j D
sin ˛
sin ˇ
jTAC j,
jTAC j sin˛ C ˇ
sin ˇ
.
For jTAC j D 2 N, ˛ D 23.2° and ˇ D 40.6° ,
jTAB j D 1.21 N, jTAD j D 2.76 N
Fx D jTAB j cos ˇ jTAC j cos ˛ C jTAD ji D 0,
Fy D jTAB j sin ˇ jTAC j sin ˛j D 0.
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1
Problem 3.31 The forces exerted on the shoes and
back of the 72-kg climber by the walls of the “chimney”
are perpendicular to the walls exerting them. The tension
in the rope is 640 N. What is the magnitude of the force
exerted on his back?
10°
3°
4°
Solution: Draw a free body diagram of the climber-treating all
forces as if they act at a point. Write the forces in components and
then apply the conditions for particle equilibrium.

Fx D FFEET cos 4° FBACK cos 3° TROPE sin 10° D 0



Fy D FFEET sin 4° CFBACK sin 3° CTROPE cos 10° mg D 0



mg D 729.81 N, TROPE D 640 N
TROPE
10° y
FFEET
FBACK
3°
x
4°
Solving, we get
FBACK D 559 N, FFEET D 671 N
mg = (72) (9.81) N
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1
Problem 3.32 The slider A is in equilibrium and the
bar is smooth. What is the mass of the slider?
20°
200 N
A
45°
Solution: The pulley does not change the tension in the rope that
passes over it. There is no friction between the slider and the bar.
y
T = 200 N
20°
Eqns. of Equilibrium:


Fx D T sin 20° C N cos 45° D 0 T D 200 N

Fy D N sin 45° C T cos 20° mg D 0 g D 9.81 m/s2
Substituting for T and g, we have two eqns in two unknowns
(N and m).
Solving, we get N D 96.7 N, m D 12.2 kg.
x
N
45°
mg = (9.81) g
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1
Problem 3.33 The 20-kg mass is suspended from three
cables. Cable AC is equipped with a turnbuckle so that
its tension can be adjusted and a strain gauge that allows
its tension to be measured. If the tension in cable AC is
40 N, what are the tensions in cables AB and AD?
0.4 m
0.4 m
B
0.48 m
C
D
0.64 m
A
Solution:
TAC
TAB
5
TAC D 40 N
5
5
11
TAD D 0
Fx : p TAB C p TAC C p
89
89
185
TAD
8
8
8
11
5
8
8
8
TAD 196.2 N D 0
Fy : p TAB C p TAC C p
89
89
185
Solving: TAB D 144.1 N, TAD D 68.2 N
196.2 N
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1
Problem 3.34 In Problem 3.33, suppose that you want
to adjust the tension in cable AC so that the tensions
in cables AC and AD are equal. What is the necessary
tension in cable AC? What is the resulting tension in
cable AB?
Solution:
TAC
TAB
5
TAC D TAD
5
5
11
Fx : p TAB C p TAC C p
TAD D 0
89
89
185
8
8
8
TAD 196.2 N D 0
Fy : p TAB C p TAC C p
89
89
185
TAD
8
8
8
11
5
Solving: TAB D 138.5 N, TAC D TAD D 54.8 N
196.2 N
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1
Problem 3.35 The collar A slides on the smooth
vertical bar. The masses mA D 20 kg and mB D 10 kg.
When h D 0.1 m, the spring is unstretched. When the
system is in equilibrium, h D 0.3 m. Determine the
spring constant k.
0.25 m
h
A
B
k
Solution: The triangles formed by the rope segments and the horizontal line level with A can be used to determine the lengths Lu and
Ls . The equations are
Lu D
0.252 C 0.12 and Ls D
0.252 C 0.32 .
The stretch in the spring when in equilibrium is given by υ D Ls Lu .
Carrying out the calculations, we get Lu D 0.269 m, Ls D 0.391 m,
and υ D 0.121 m. The angle, , between the rope at A and the horizontal when the system is in equilibrium is given by tan D 0.3/0.25,
or D 50.2° . From the free body diagram for mass A, we get two
equilibrium equations. They are
and
T
NA
A
mA g
Fx D NA C T cos D 0
T
Fy D T sin mA g D 0.
We have two equations in two unknowns and can solve. We get NA D
163.5 N and T D 255.4 N. Now we go to the free body diagram for B,
where the equation of equilibrium is T mB g kυ D 0. This equation
has only one unknown. Solving, we get k D 1297 N/m
Lu
0.1 m
B
mBg
Kδ
0.25 m
Ls
Lu
0.3 m
0.25 m
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1
Problem 3.36* You are designing a cable system to
support a suspended object of weight W. The two wires
must be identical, and the dimension b is fixed. The ratio
of the tension T in each wire to its cross-sectional area
A must equal a specified value T/A D . The “cost” of
your design is pthe total volume of material in the two
wires, V D 2A b2 C h2 . Determine the value of h that
minimizes the cost.
b
b
h
W
Solution: From the equation
T
T
θ
θ
Fy D 2T sin W D 0,
we obtain T D
p
W
W b2 C h2
D
.
2 sin 2h
Since T/A D , A D
p
W b2 C h2
T
D
2h
W
p
Wb2 C h2 and the “cost” is V D 2A b2 C h2 D
.
h
To determine the value of h that minimizes V, we set
W
dV
b2 C h2 D
C2 D0
2
dh
h
and solve for h, obtaining h D b.
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1
Problem 3.37 The system of cables suspends a 1000lb bank of lights above a movie set. Determine the
tensions in cables AB, CD, and CE.
20 ft
18 ft
B
D
Solution: Isolate juncture A, and solve the equilibrium equations.
C
Repeat for the cable juncture C.
E
The angle between the cable AC and the positive x axis is ˛. The
tension in AC is TAC D jTAC ji cos ˛ C j sin ˛
45°
30°
A
The angle between the x axis and AB is 180° ˇ. The tension is
TAB D jTAB ji cos180 ˇ C j sin180 ˇ
TAB D i cos ˇ C j sin ˇ.
The weight is W D 0i jWjj.
The equilibrium conditions are
Solve: jTCE j D jTCA j cos ˛,
F D 0 D W C TAB C TAC D 0.
jTCD j D jTCA j sin ˛;
Substitute and collect like terms,
for jTCA j D 732 lb and ˛ D 30° ,
Fx D jTAC j cos ˛ jTAB j cos ˇi D 0
jTAB j D 896.6 lb,
Fy D jTAB j sin ˇ C jTAC j sin ˛ jWjj D 0.
jTCE j D 634 lb,
Solving, we get
jTAB j D
cos ˛
cos ˇ
and
jTAC j
jTAC j D
jWj cos ˇ
sin˛ C ˇ
jTCD j D 366 lb
,
jWj D 1000 lb, and ˛ D 30° , ˇ D 45°
jTAC j D 1000
jTAB j D 732
0.7071
0.9659
0.866
0.7071
B
C
A
β
α
D 732.05 lb
y
x
W
D 896.5 lb
Isolate juncture C. The angle between the positive x axis and the cable
CA is 180° ˛. The tension is
D
TCA D jTCA ji cos180° C ˛ C j sin180° C ˛,
C
90°
E
or TCA D jTCA ji cos ˛ j sin ˛.
The tension in the cable CE is
α
y
A
TCE D ijTCE j C 0j.
x
The tension in the cable CD is TCD D 0i C jjTCD j.
The equilibrium conditions are
F D 0 D TCA C TCE C TCD D 0
Substitute t and collect like terms,
Fx D jTCE j jTCA j cos ˛i D 0,
Fy D jTCD j jTCA j sin ˛j D 0.
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1
Problem 3.38 Consider the 1000-lb bank of lights in
Problem 3.37. A technician changes the position of the
lights by removing the cable CE. What is the tension in
cable AB after the change?
Solution: The original configuration in Problem 3.35 is used to
solve for the dimensions and the angles. Isolate the juncture A, and
solve the equilibrium conditions.
18 ft
20 ft
D
B
C
The lengths are calculated as follows: The vertical interior distance
in the triangle is 20 ft, since the angle is 45 deg. and the base and
altitude of a 45 deg triangle are equal. The length AB is given by
α
β
A
AB D
20 ft
D 28.284 ft.
cos 45°
The length AC is given by
AC D
18 ft
D 20.785 ft.
cos 30°
38
B
The altitude of the triangle for which AC is the hypotenuse is
18 tan 30° D 10.392 ft. The distance CD is given by 20 10.392 D
9.608 ft.
D
β
α
β
20.784+9.608
= 30.392
α
28.284
The distance AD is given by
A
AD D AC C CD D 20.784 C 9.608 D 30.392
B
The new angles are given by the cosine law
D
β
AB2 D 382 C AD2 238AD cos ˛.
A
α
Reduce and solve:
cos ˛ D
cos ˇ D
382 C 30.3922 28.2842
23830.392
y
28.2842 C 382 30.3922
228.28438
D 0.6787, ˛ D 47.23° .
D 0.6142, ˇ D 52.1° .
Isolate the juncture A. The angle between the cable AD and the positive
x axis is ˛. The tension is:
Solve:
jTAB j D
and
jTAD j D
TAD D jTAD ji cos ˛ C j sin ˛.
The angle between x and the cable AB is 180° ˇ. The tension is
TAB D jTAB ji cos ˇ C j sin ˇ.
The weight is W D 0i jWjj
F D 0 D W C TAB C TAD D 0.
cos ˛
cos ˇ
jTAD j,
jWj cos ˇ
sin˛ C ˇ
.
For jWj D 1000 lb, and ˛ D 51.2° , ˇ D 47.2°
jTAD j D 1000
The equilibrium conditions are
x
W
0.6142
0.989
jTAB j D 622.3
0.6787
0.6142
D 621.03 lb,
D 687.9 lb
Substitute and collect like terms,
Fx D jTAD j cos ˛ jTAB j cos ˇi D 0,
Fy D jTAB j sin ˇ C jTAD j sin ˛ jWjj D 0.
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1
Problem 3.39 While working on another exhibit, a
curator at the Smithsonian Institution pulls the suspended
Voyager aircraft to one side by attaching three horizontal
cables as shown. The mass of the aircraft is 1250 kg.
Determine the tensions in the cable segments AB, BC,
and CD.
D
C
B
30°
50°
Solution: Isolate each cable juncture, beginning with A and solve
the equilibrium equations at each juncture. The angle between the
cable AB and the positive x axis is ˛ D 70° ; the tension in cable AB
is TAB D jTAB ji cos ˛ C j sin ˛. The weight is W D 0i jWjj. The
tension in cable AT is T D jTji C 0j. The equilibrium conditions are
A
70°
F D W C T C TAB D 0.
Substitute and collect like terms
Fx jTAB j cos ˛ jTji D 0,
Fy D jTAB j sin ˛ jWjj D 0.
y
Solve: the tension in cable AB is jTAB j D
jWj
.
sin ˛
m
For jWj D 1250 kg 9.81 2 D 12262.5 N and ˛ D 70°
s
12262.5
jTAB j D
D 13049.5 N
0.94
B
x
α
A
T
W
Isolate juncture B. The angles are ˛ D 50° , ˇ D 70° , and the tension
cable BC is TBC D jTBC ji cos ˛ C j sin ˛. The angle between the
cable BA and the positive x axis is 180 C ˇ; the tension is
y
C
x
TBA D jTBA ji cos180 C ˇ C j sin180 C ˇ
The tension in the left horizontal cable is T D jTji C 0j. The equilibrium conditions are
β
A
F D TBA C TBC C T D 0.
Substitute and collect like terms
α
B
T
D jTBA ji cos ˇ j sin ˇ
y
T
Fy D jTBC j sin ˛ jTBA j sin ˇj D 0.
Solve: jTBC j D
sin ˇ
sin ˛
D
x
Fx D jTBC j cos ˛ jTBA j cos ˇ jTji D 0
α
C
β
jTBA j.
B
For jTBA j D 13049.5 N, and ˛ D 50° , ˇ D 70° ,
jTBC j D 13049.5
0.9397
0.7660
D 16007.6 N
Isolate the cable juncture C. The angles are ˛ D 30° , ˇ D 50° . By
symmetry with the cable juncture B above, the tension in cable CD is
jTCD j D
sin ˇ
sin ˛
jTCB j.
Substitute: jTCD j D 16007.6
0.7660
0.5
D 24525.0 N.
This completes the problem solution.
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1
Problem 3.40 A truck dealer wants to suspend a 4-Mg
(megagram) truck as shown for advertising. The distance
b D 15 m, and the sum of the lengths of the cables AB
and BC is 42 m. What are the tensions in the cables?
40 m
b
A
C
B
Solution: Determine the dimensions and angles of the cables. Isolate the cable juncture B, and solve the equilibrium conditions. The
dimensions of the triangles formed by the cables:
15 m
b
A
b D 15 m,
L D 25 m,
25 m
L
β
AB C BC D S D 42 m.
C
α
Subdivide into two right triangles with a common side of unknown
length. Let the unknown length of this common side be d, then by the
Pythagorean Theorem b2 C d2 D AB2 , L 2 C d2 D BC2 .
B
y
Subtract the first equation from the second to eliminate the unknown d,
L 2 b2 D BC2 AB2 .
A
α
B
β
C
Note that BC2 AB2 D BC ABBC C AB.
W
Substitute and reduce to the pair of simultaneous equations in the
unknowns
x
BC AB D
L 2 b2
S
Solve:
2
1
L b2
CS
2
S
BC D
D
,
BC C AB D S
Substitute and collect like terms
2
1
25 152
C 42 D 25.762 m
2
42
Fx D jTBC j cos ˛ jTBA j cos ˇi D 0,
Fy D jTBC j sin ˛ C jTBA j sin ˇ jWjj D 0
and AB D S BC D 42 25.762 D 16.238 m.
Solve:
jTBC j D
The interior angles are found from the cosine law:
cos ˛ D
C BC2
AB2
2L C bBC
cos ˇ D
L
C b2
L C b2 C AB2 BC2
2L C bAB
and jTBA j D
D 0.9704
˛ D 13.97°
cos ˇ
cos ˛
jTBA j,
jWj cos ˛
sin˛ C ˇ
.
For jWj D 40009.81 D 39240 N,
D 0.9238
ˇ D 22.52°
Isolate cable juncture B. The angle between BC and the positive x axis
is ˛; the tension is
and ˛ D 13.97° , ˇ D 22.52° ,
jTBA j D 64033 D 64 kN,
jTBC j D 60953 D 61 kN
TBC D jTBC ji cos ˛ C j sin ˛
The angle between BA and the positive x axis is 180° ˇ; the
tension is
TBA D jTBA ji cos180 ˇ C j sin180 ˇ
D jTBA ji cos ˇ C j sin ˇ.
The weight is W D 0i jWjj.
The equilibrium conditions are
F D W C TBA C TBC D 0.
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1
Problem 3.41 The distance h D 12 in, and the tension
in cable AD is 200 lb. What are the tensions in cables
AB and AC?
B
12 in.
A
D
C
12 in.
h
8 in.
12 in.
Solution: Isolated the cable juncture. From the sketch, the angles
are found from
tan ˛ D
tan ˇ D
8
12
4
12
D 0.667
y
B
12 in
α
˛ D 33.7°
8 in
A
D 0.333
8 in.
ˇ D 18.4°
β
D
4 in
C
The angle between the cable AB and the positive x axis is 180° ˛,
the tension in AB is:
x
TAB D jTAB ji cos180 ˛ C j sin180 ˛
TAB D jTAB ji cos ˛ C j sin ˛.
The angle between AC and the positive x axis is 180 C ˇ. The
tension is
TAC D jTAC ji cos180 C ˇ C j sin180 C ˇ
TAC D jTAC ji cos ˇ j sin ˇ.
The tension in the cable AD is
TAD D jTAD ji C 0j.
The equilibrium conditions are
F D TAC C TAB C TAD D 0.
Substitute and collect like terms,
Fx D jTAB j cos ˛ jTAC j cos ˇ C jTAD ji D 0
Fy D jTAB j sin ˛ jTAC j sin ˇj D 0.
Solve:
jTAB j D
and jTAC j D
sin ˇ
sin ˛
jTAC j,
sin ˛
sin˛ C ˇ
jTAD j.
For jTAD j D 200 lb, ˛ D 33.7° , ˇ D 18.4°
jTAC j D 140.6 lb,
jTAB j D 80.1 lb
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1
Problem 3.42 You are designing a cable system to
support a suspended object of weight W. Because your
design requires points A and B to be placed as shown,
you have no control over the angle ˛, but you can choose
the angle ˇ by placing point C wherever you wish. Show
that to minimize the tensions in cables AB and BC, you
must choose ˇ D ˛ if the angle ˛ ½ 45° .
B
β
α
A
Strategy: Draw a diagram of the sum of the forces
exerted by the three cables at A.
W
Solution: Draw the free body diagram of the knot at point A. Then
y
draw the force triangle involving the three forces. Remember that ˛ is
fixed and the force W has both fixed magnitude and direction. From
the force triangle, we see that the force TAC can be smaller than
TAB for a large range of values for ˇ. By inspection, we see that the
minimum simultaneous values for TAC and TAB occur when the two
forces are equal. This occurs when ˛ D ˇ. Note: this does not happen
when ˛ < 45° .
TAC
TAB
and
A
x
W
B
Possible locations
for C lie on line
C?
C?
α
TAB
Fx D TAB cos ˛ C TAC cos ˇ D 0
Fy D TAB sin ˛ C TAC sin ˇ W D 0.
B
α
In this case, we solved the problem without writing the equations of
equilibrium. For reference, these equations are:
C
Candidate β
W
Candidate values
for TAC
Fixed direction for
line AB
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1
Problem 3.43* The length of the cable ABC is 1.4 m.
The 2-kN force is applied to a small pulley. The system
is stationary. What is the tension in the cable?
1m
A
C
B
0.75 m
15⬚
Solution: Examine the geometry
h2 C 0.75 m2 C
tan ˛ D
)
0.75 m
2 kN
0.25 m
β
α
h2 C 0.25 m2 D 1.4 m
h
h
h
, tan ˇ D
0.75 m
0.25 m
h D 0.458 m, ˛ D 31.39° , ˇ D 61.35°
Now draw a FBD and solve for the tension. We can use either of the
equilibrium equations
Fx : T cos ˛ C T cos ˇ C 2 kN sin 15° D 0
T
T
β
α
T D 1.38 kN
2 kN
15°
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1
Problem 3.44 The masses m1 D 12 kg and m2 D 6 kg
are suspended by the cable system shown. The cable BC
is horizontal. Determine the angle ˛ and the tensions in
the cables AB, BC, and CD.
TAB
A
D
α
TBC
α
B
B
C
70⬚
m1
m2
117.7 N
Solution: We have 4 unknowns and 4 equations
TCD
FBx : TAB cos ˛ C TBC D 0
FBy : TAB sin ˛ 117.7 N D 0
FCx : TBC C TCD cos 70° D 0
70°
TBC
C
FCy : TCD sin 70° 58.86 N D 0
Solving we find
˛ D 79.7° , TAB D 119.7 N, TBC D 21.4 N, TCD D 62.6 N
58.86 N
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1
Problem 3.45 The weights W1 D 50 lb and W2 are
suspended by the cable system shown. Determine the
weight W2 and the tensions in the cables AB, BC,
and CD.
30 in
30 in
30 in
A
D
16 in
20 in
C
B
W2
W1
Solution: We have 4 unknowns and 4 equilibrium equations to use
3
15
FBx : p TAB C p
TBC D 0
229
13
TAB
2
15
2
3
2
2
FBy : p TAB C p
TBC 50 lb D 0
229
13
TBC
B
15
15
TBC C
TCD D 0
FCx : p
17
229
50 lb
2
8
TBC C
TCD W2 D 0
FCy : p
17
229
TCD
W2 D 25 lb, TAB D 75.1 lb
8
)
C
TBC D 63.1 lb, TCD D 70.8 lb
TBC
15
15
2
W2
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1
Problem 3.46 In the system shown in Problem 3.45,
assume that W2 D W1 /2. If you don’t want the tension
anywhere in the supporting cable to exceed 200 lb, what
is the largest acceptable value of W1 ?
TAB
Solution:
3
15
FBx : p TAB C p
TBC D 0
229
13
2
2
2
FBy : p TAB C p
TBC W1 D 0
229
13
15
2
3
TBC
B
15
15
TBC C
TCD D 0
FCx : p
17
229
W1
2
8
W1
TBC C
TCD D0
FCy : p
17
2
229
TCD
TAB D 1.502W1 , TBC D 1.262W1 , TCD D 1.417W1
8
C
AB is the critical cable
200 lb D 1.502W1 ) W1 D 133.2 lb
TBC
15
2
15
W2 = W1/2
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1
Problem 3.47 The hydraulic cylinder is subjected to
three forces. An 8-kN force is exerted on the cylinder
at B that is parallel to the cylinder and points from B
toward C. The link AC exerts a force at C that is parallel
to the line from A to C. The link CD exerts a force at
C that is parallel to the line from C to D.
(a)
(b)
Draw the free-body diagram of the cylinder. (The
cylinder’s weight is negligible).
Determine the magnitudes of the forces exerted by
the links AC and CD.
Solution: From the figure, if C is at the origin, then points A, B,
and D are located at
1m
D
C
Hydraulic
cylinder
1m
0.6 m
B
A
0.6 m
0.15 m
Scoop
y
A0.15, 0.6
B0.75, 0.6
D
FCD
D1.00, 0.4
and forces FCA , FBC , and FCD are parallel to CA, BC, and CD, respectively.
C
x
FBC
We need to write unit vectors in the three force directions and express
the forces in terms of magnitudes and unit vectors. The unit vectors
are given by
eCA D
rCA
D 0.243i 0.970j
jrCA j
eCB D
rCB
D 0.781i 0.625j
jrCB j
eCD D
rCD
D 0.928i C 0.371j
jrCD j
FCA
A
B
Now we write the forces in terms of magnitudes and unit vectors. We
can write FBC as FCB D 8eCB kN or as FCB D 8eCB kN (because
we were told it was directed from B toward C and had a magnitude
of 8 kN. Either way, we must end up with
FCB D 6.25i C 5.00j kN
Similarly,
FCA D 0.243FCA i 0.970FCA j
FCD D 0.928FCD i C 0.371FCD j
For equilibrium, FCA C FCB C FCD D 0
In component form, this gives


Fx D 0.243FCA C 0.928FCD 6.25 (kN) D 0

Fy D 0.970FCA C 0.371FCD C 5.00 (kN) D 0
Solving, we get
FCA D 7.02 kN, FCD D 4.89 kN
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1
Problem 3.48
surfaces.
(a)
(b)
The 50-lb cylinder rests on two smooth
Draw the free-body diagram of the cylinder.
If ˛ D 30° , what are the magnitudes of the forces
exerted on the cylinder by the left and right
surfaces?
Solution: Isolate the cylinder. (a) The free body diagram of the
isolated cylinder is shown. (b) The forces acting are the weight and the
normal forces exerted by the surfaces. The angle between the normal
force on the right and the x axis is 90 C ˇ. The normal force is
α
45°
y
β
α
NL
NR D jNR ji cos90 C ˇ C j sin90 C ˇ
NR
W
x
NR D jNR ji sin ˇ C j cos ˇ.
The angle between the positive x axis and the left hand force is normal
90 ˛; the normal force is NL D jNL ji sin ˛ C j cos ˛. The weight
is W D 0i jWjj. The equilibrium conditions are
F D W C NR C NL D 0.
Substitute and collect like terms,
Fx D jNR j sin ˇ C jNL j sin ˛i D 0,
Solve:
jNR j D
and jNL j D
sin ˛
sin ˇ
jNL j,
jWj sin ˇ
sin˛ C ˇ
.
For jWj D 50 lb, and ˛ D 30° , ˇ D 45° , the normal forces are
jNL j D 36.6 lb,
jNR j D 25.9 lb
Fy D jNR j cos ˇ C jNL j cos ˛ jWjj D 0.
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1
Problem 3.49 For the 50-lb cylinder in Problem 3.48,
obtain an equation for the force exerted on the cylinder
by the left surface in terms of the angle ˛ in two ways:
(a) using a coordinate system with the y axis vertical,
(b) using a coordinate system with the y axis parallel to
the right surface.
y
Solution: The solution for Part (a) is given in Problem 3.48 (see
free body diagram).
jNR j D
sin ˛
sin ˇ
jNL j
jNL j D
jWj sin ˇ
sin˛ C ˇ
.
Part (b): The isolated cylinder with the coordinate system is shown.
The angle between the right hand normal force and the positive x axis
is 180° . The normal force: NR D jNR ji C 0j. The angle between the
left hand normal force and the positive x is 180 ˛ C ˇ. The normal
force is NL D jNL ji cos˛ C ˇ C j sin˛ C ˇ.
The angle between the weight vector and the positive x axis is ˇ.
The weight vector is W D jWji cos ˇ j sin ˇ.
The equilibrium conditions are
β
α
NR
NL
W
x
Substitute and collect like terms,
Fx D jNR j jNL j cos˛ C ˇ C jWj cos ˇi D 0,
Fy D jNL j sin˛ C ˇ jWj sin ˇj D 0.
F D W C NR C NL D 0.
Solve:
jNL j D
jWj sin ˇ
sin˛ C ˇ
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1
Problem 3.50 The two springs are identical, with
unstretched length 0.4 m. When the 50-kg mass is
suspended at B, the length of each spring increases to
0.6 m. What is the spring constant k?
0.6 m
A
C
k
k
B
Solution:
F
F
F D k0.6 m 0.4 m
Fy : 2F sin 60° 490.5 N D 0
60°
60°
k D 1416 N/m
490.5 N
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1
Problem 3.51 The cable AB is 0.5 m in length. The
unstretched length of the spring is 0.4 m. When the
50-kg mass is suspended at B, the length of the spring
increases to 0.45 m. What is the spring constant k?
0.7 m
A
C
k
B
Solution: The Geometry
0.7 m
Law of Cosines and Law of Sines
2
2
φ
θ
2
0.7 D 0.5 C 0.45 20.50.45 cos ˇ
sin sin sin ˇ
D
D
0.45 m
0.5 m
0.7 m
0.5 m
0.45 m
β
ˇ D 94.8° , D 39.8° D 45.4°
Now do the statics
TAB
F
F D k0.45 m 0.4 m
Fx : TAB cos C F cos D 0
θ
φ
Fy : TAB sin C F sin 490.5 N D 0
Solving: k D 7560 N/m
490.5 N
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1
Problem 3.52 The 1440-kg car is moving at constant
speed on a road with the slope shown. The aerodynamic
forces on the car are the drag D D 530 N, which is
parallel to the road, and the lift L D 360 N, which is
perpendicular to the road. Determine the magnitudes of
the total normal and friction forces exerted on the car
by the road.
L
D
15°
Solution: From the free body diagram, the equations of equilib-
y
L
rium are
Fx D f D W sin 15° D 0
x
f
Fy D L C N W cos 15° D 0
D
W D mg D 14409.81 N
15°
L D 360 N, D D 530 N
W
N
m D 1440 kg, g D 9.81 m/s2
Solving, we get
15°
f D 4.19 kN, N D 13.29 kN
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1
Problem 3.53 The inclined surface is smooth. Determine the force T that must be exerted on the cable to
hold the 100-kg crate in equilibrium and compare your
answer to the answer of Problem 3.11.
T
60⬚
3T
Solution:
981 N
F- : 3 T 981 N sin 60° D 0
T D 283 N
N
60°
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1
Problem 3.54 The mass of each pulley of the system
is m and the mass of the suspended object A is mA .
Determine the force T necessary for the system to be in
equilibrium.
A
T
Solution: Draw free body diagrams of each pulley and the object
A. Each pulley and the object A must be in equilibrium. The weights of
the pulleys and object A are W D mg and WA D mA g. The equilibrium
equations for the lower pulley, middle pulley, and upper pulley are,
respectively, A 2T W D 0, B 2A W D 0, and C 2B W D
0. The equilibrium equation for the weight is T C A C B WA D 0.
Solving the first equation for A in terms of T and W, substituting for
A in the second equation and solving for B in terms of T and W,
we get A D 2T C W and B D 4T C 3W. Substituting for A and B in
the equilibrium equation for the weight, we get 7T D WA 4W D
mA g 4mg. Thus, the tension, T, in terms of masses and g is T D
g
mA 4m
7
C
B
B
B
W
A
A
T
A
T
W
W
T
A
B
WA
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1
Problem 3.55 The mass of each pulley of the system
is m and the mass of the suspended object A is mA .
Determine the force T necessary for the system to be in
equilibrium.
Solution: Draw free body diagrams of each pulley and the object
A. Each pulley and the object A must be in equilibrium. The weights
of the pulleys and object A are W D mg and WA D mA g. The equilibrium equations for the weight A, the lower pulley, second pulley, third
pulley, and the top pulley are, respectively, B WA D 0, 2C B W D 0, 2D C W D 0, 2T D W D 0, and FS 2T W D 0.
Begin with the first equation and solve for B, substitute for B in the
second equation and solve for C, substitute for C in the third equation
and solve for D, and substitute for D in the fourth equation and solve
for T, to get T in terms of W and WA . The result is
T
A
Fs
W
T
W
B D WA ,
DD
CD
WA
W
C ,
2
2
3W
WA
7W
WA
C
, and T D
C
,
4
4
8
8
or in terms of the masses,
TD
g
mA C 7m.
8
T
T
T
W
D
D
D
C
W
C
C
B
B
WA
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1
Problem 3.56 The suspended mass m1 D 50 kg. Neglecting the masses of the pulleys, determine the value of
the mass m2 necessary for the system to be in equilibrium.
A
B
C
m2
m1
Solution:
FC : T1 C 2m2 g m1 g D 0
T1
T
T
FB : T1 2m2 g D 0
m2 D
m1
D 12.5 kg
4
C
T1
m1 g
B
T = m2 g
T
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1
Problem 3.57 Consider the pulley system shown in
Problem 3.56. The suspended mass m1 D 50 kg. If the
pulleys A, B, and C each have a mass of 2 kg, what mass
m2 is necessary for the system to be in equilibrium?
Solution:
T1
T
T
FC : T1 C 2m2 g m1 g 2 kgg D 0
FB : T1 2 kgg 2m2 g D 0
C
m1
D 12.5 kg
m2 D
4
T1
(m1 + 2 kg) g
B
T = m2 g
(2 kg) g
T
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1
Problem 3.58 Pulley systems containing one, two, and
three pulleys are shown. Neglecting the weights of the
pulleys, determine the force T required to support the
weight W in each case.
T
T
T
W
(a) One pulley
W
(b) Two pulleys
W
(c) Three pulleys
Solution:
(a)
(b)
(b) For two pulleys
T
Fy : 2T W D 0 ) T D
T
W
2
Fupper : 2T T1 D 0
Flower : 2T1 W D 0
T1
TD
(c)
T1
W
4
Fupper : 2T T1 D 0
Fmiddle : 2T1 T2 D 0
W
Flower : 2T2 W D 0
(c) For three pulleys
TD
T
W
8
T
(a) For one pulleys
T
T
T1
T1
W
T2
T2
T2
W
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1
Problem 3.59 Problem 3.58 shows pulley systems
containing one, two, and three pulleys. The number of
pulleys in the type of system shown could obviously be
extended to an arbitrary number N.
(a)
(b)
Neglecting the weights of the pulleys, determine
the force T required to support the weight W as a
function of the number of pulleys N in the system.
Using the result of part (a), determine the force T
required to support the weight W for a system with
10 pulleys.
Solution: By extrapolation of the previous problem
(a)
TD
W
2N
(b)
TD
W
1024
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1
Problem 3.60 A 14,000-kg airplane is in steady flight
in the vertical plane. The flight path angle is D 10° ,
the angle of attack is ˛ D 4° , and the thrust force exerted
by the engine is T D 60 kN. What are the magnitudes
of the lift and drag forces acting on the airplane?
Solution: Let us draw a more detailed free body diagram to see
the angles involved more clearly. Then we will write the equations of
equilibrium and solve them.
y
L
W D mg D 14,0009.81 N
The equilibrium equations are
x


Fx D T cos ˛ D W sin D 0

Fy D T sin ˛ C L W cos D 0
α
α = 4°
γ = 10°
T
γ
D
W
T D 60 kN D 60000 N
Solving, we get
γ
D D 36.0 kN, L D 131.1 kN
y
Path
x
T
γ
L
α
D
Horizon
W
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1
Problem 3.61 An airplane is in steady flight, the angle
of attack ˛ D 0, the thrust-to-drag ratio T/D D 2, and
the lift-to-drag ratio L/D D 4. What is the flight path
angle ?
Solution: Use the same strategy as in Problem 3.52. The angle
between the thrust vector and the positive x axis is ˛,
T D jTji cos ˛ C j sin ˛
The lift vector: L D 0i C jLjj
The drag: D D jDji C 0j. The angle between the weight vector and
the positive x axis is 270 ;
W D jWji sin j cos .
The equilibrium conditions are
F D T C L C D C W D 0.
Substitute and collect like terms
and
Fx D jTj cos ˛ jDj jWj sin i D 0,
Fy D jTj sin ˛ C jLj jWj cos j D 0
Solve the equations for the terms in :
jWj sin D jTj cos ˛ jDj,
and jWj cos D jTj sin ˛ C jLj.
Take the ratio of the two equations
tan D
jTj cos ˛ jDj
jTj sin ˛ C jLj
.
Divide top and bottom on the right by jDj.
For ˛ D 0,
jLj
jTj
D 2,
D 4,
jDj
jDj
tan D
21
4
D
1
or D 14°
4
y
Path
T
L
α
D
x
γ
Horizontal
W
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1
Problem 3.62 An airplane glides in steady flight (T D
0), and its lift-to-drag ratio is L/D D 4.
(a)
(b)
What is the flight path angle ?
If the airplane glides from an altitude of 1000 m
to zero altitude, what horizontal distance does it
travel?
Solution: Use the same strategy as in Problem 3.52. The angle
y
between the thrust vector and the positive x axis is ˛:
Path
x
T
T D jTji cos ˛ C j sin ˛.
L
α
The lift vector: L D 0i C jLjj.
The drag: D D jDji C 0j. The angle between the weight vector and
the positive x axis is 270 :
γ
Horizontal
D
W
W D jWji sin j cos .
The equilibrium conditions are
F D T C L C D C W D 0.
γ
Substitute and collect like terms:
1 km
Fx D jTj cos ˛ jDj jWj sin i D 0
γ
Fy D jTj sin ˛ C jLj jWj cos j D 0
h
Solve the equations for the terms in ,
jWj sin D jTj cos ˛ jDj,
and jWj cos D jTj sin ˛ C jLj
Part (a): Take the ratio of the two equilibrium equations:
tan D
jTj cos ˛ jDj
jTj sin ˛ C jLj
.
Divide top and bottom on the right by jDj.
For ˛ D 0, jTj D 0,
jLj
D 4,
jDj
tan D
1
4
D 14°
Part (b): The flight path angle is a negative angle measured from the
horizontal, hence from the equality of opposite interior angles the angle
is also the positive elevation angle of the airplane measured at the
point of landing.
tan D
1
,
h
hD
1
1
D D 4 km
1
tan 4
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1
Problem 3.63 Four forces F1 , F2 , F3 , and F4 act on
an object in equilibrium. The force F1 D 50i (N).
The forces F2 , F3 , and F4 point in the directions of the
unit vectors
e2 D 0.485i C 0.485j 0.728k,
e3 D 0.557i C 0.743j C 0.371k,
e4 D 0.371i 0.743j C 0.557k.
Determine the magnitudes of F2 , F3 , and F4 .
y
F3
z
F4
F2
F1
x
Solution: The Forces:
F1 D 50 Ni
F2 D F2 0.485i C 0.485j 0.728k
F3 D F3 0.557i C 0.743j C 0.371k
F4 D F4 0.371i 0.743j C 0.557k
The equilibrium equations:
Fx : 50 N 0.485F2 0.557F3 0.371F3 D 0
Fy : 0.485F2 C 0.743F3 0.743 F D 0
Fz : 0.728F2 C 0.371F3 C 0.557 F D 0
Solving: F2 D 45.8 N, F3 D 17.99 N, F4 D 47.9 N
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1
y
Problem 3.64 The force F D 5i (kN) acts on point A
where the cables AB, AC, and AD are joined. What are
the tensions in the three cables?
D (0, 6, 0) m
A
Strategy: Isolate part of the cable system near point
A. See Example 3.5.
F
(12, 4, 2) m
C
B
(6, 0, 0) m
x
(0, 4, 6) m
z
Solution: Isolate the cable juncture A. Get the unit vectors parallel
y
to the cables using the coordinates of the end points. Express the
tensions in terms of these unit vectors, and solve the equilibrium conditions. The coordinates of points A, B, C, D are:
A12, 4, 2,
B6, 0, 0,
C0, 4, 6,
D
(0, 6, 0) m
A
C
(0, 4, 6) m
rB r A
6 12i C 0 4j C 0 2k
D
jrB rA j
6 122 C 42 C 22
F
(12, 4, 2) m
D0, 6, 0.
The unit vector eAB is, by definition,
eAB D
D
B
x
(6, 0, 0) m
z
4
2
6
i
j
k
7.483
7.483
7.483
eAB D 0.8018i 0.5345j 0.267k.
Similarly, the other unit vectors are
eAC D 0.9487i C 0j C 0.3163k,
eAD D 0.9733i C 0.1622j 0.1622k.
The tensions in the cables are expressed in terms of the unit vectors,
TAB D jTAB jeAB ,
TAC D jTAC jeAC ,
TAD D jTAD jeAD .
The external force acting on the juncture is, F D 5i C 0j C 0k. The
equilibrium conditions are
F D 0 D TAB C TAC C TAD C F D 0.
Substitute and collect like terms,
Fx D 0.8018jTAB j 0.9487jTAC j 0.9733jTAD j C 5i D 0
Fy D 0.5345jTAB j C 0jTAC j 0.1622jTAD jj D 0
Fz D 0.2673jTAB j 0.3163jTAC j 0.1622jTAD jk D 0.
A hand held calculator was used to solve these simultaneous equations.
The results are:
jTAB j D 0.7795 kN,
jTAC j D 1.9765 kN,
jTAD j D 2.5688 kN.
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1
Problem 3.65 An 80-lb chandelier is suspended from
three wires AB, AC, and AD of equal length. The wires
are attached at points B, C, and D on the ceiling. Points
B, C, and D lie on a circle of 3-ft radius and are equally
spaced. (That is, they are placed at 120° intervals around
the circle.) Point A is 4 ft below the ceiling. Determine
the tensions in the wires.
C
120⬚
D
120⬚
120⬚
B
A
Solution: By symmetry
T
TAB D TAC D TAD D T.
Geometry: 3:4:5 right triangle
4
T 80 lb D 0
Equilibrium:
Fy : 3
5
4
3
TAB D TAC D TAD D 33.3 lb
80 lb
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1
Problem 3.66 To support the tent, the tension in the
rope AB must be 40 lb. What are the tensions in the
ropes AC, AD, and AE?
y
(0, 5, 0) ft
C
Solution: Get the unit vectors parallel to the cables using the coor-
(0, 6, 6) ft
dinates of the end points. Express the tensions in terms of these unit
vectors, and solve the equilibrium conditions. The coordinates of points
A, B, C, D, E are:
A5, 4, 3,
B8, 4, 3,
C0, 5, 0,
D0, 6, 6,
rB D 8i C 4j C 3k,
rC D 0i C 5j C 0k,
rD D 0i C 6j C 6k,
(5, 4, 3) ft (8, 4, 3) ft
A
B
x
E
(3, 0, 3) ft
E3, 0, 3.
The vector locations of these points are,
rA D 5i C 4j C 3k,
D
z
C
A
B
rE D 3i C 0j C 3k.
D
The unit vector parallel to the tension acting between the points A, B
in the direction of B is by definition
eAB D
E
rB r A
.
jrB rA j
Perform this operation for each unit vector. We get
eAB D 1i C 0j C 0k
eAC D 0.8452i C 0.1690j 0.5071k
eAD D 0.8111i C 0.3244j C 0.4867k
eAE D 0.4472i 0.8944j C 0k
The tensions in the cables are,
TAB D jTAB jeAB D 40eAB ,
TAD D jTAD jeAD ,
TAC D jTAC jeAC ,
TAE D jTAE jeAE .
The equilibrium conditions are
F D 0 D TAB C TAC C TAD C TAE D 0.
Substitute the tensions,
Fx D 40 0.8452jTAC j 0.8111jTAD j 0.4472jTAE ji D 0
Fy D C0.1690jTAC j 0.3244jTAD j 0.8944jTAE jj D 0
Fz D 0.5071jTAC j 0.4867jTAD jk D 0.
This set of simultaneous equations in the unknown forces may be
solved using any of several standard algorithms.: The results are:
jTAE j D 11.7 lb,
jTAC j D 20.6 lb,
jTAD j D 21.4 lb.
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1
Problem 3.67 The bulldozer exerts a force F D 2i (kip)
at A. What are the tensions in cables AB, AC, and AD?
y
6 ft
C
8 ft
2 ft
B
A
3 ft
D
z
4 ft
8 ft
x
Solution: Isolate the cable juncture. Express the tensions in terms
of unit vectors. Solve the equilibrium equations. The coordinates of
points A, B, C, D are:
A8, 0, 0,
B0, 3, 8,
C0, 2, 6,
D0, 4, 0.
The radius vectors for these points are
rA D 8i C 0j C 0k,
rB D 0i C 3j C 8k,
rC D 0i C 2j 6k,
rD D 0i C 4j C 0k.
By definition, the unit vector parallel to the tension in cable AB is
eAB D
rB r A
.
jrB rA j
Carrying out the operations for each of the cables, the results are:
eAB D 0.6835i C 0.2563j C 0.6835k,
eAC D 0.7845i C 0.1961j 0.5883k,
eAD D 0.8944i 0.4472j C 0k.
The tensions in the cables are expressed in terms of the unit vectors,
TAB D jTAB jeAB ,
TAC D jTAC jeAC ,
TAD D jTAD jeAD .
The external force acting on the juncture is F D 2000i C 0j C 0k. The
equilibrium conditions are
F D 0 D TAB C TAC C TAD C F D 0.
Substitute the vectors into the equilibrium conditions:
Fx D 0.6835jTAB j 0.7845jTAC j 0.8944jTAD jC2000i D 0
Fy D 0.2563jTAB j C 0.1961jTAC j 0.4472jTAD jj D 0
Fz D 0.6835jTAB j 0.5883jTAC j C 0jTAD jk D 0
The commercial program TK Solver Plus was used to solve these
equations. The results are
jTAB j D 780.31 lb ,
jTAC j D 906.9 lb ,
jTAD j D 844.74 lb .
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1
y
Problem 3.68 Prior to its launch, a balloon carrying a
set of experiments to high altitude is held in place by
groups of student volunteers holding the tethers at B, C,
and D. The mass of the balloon, experiments package,
and the gas it contains is 90 kg, and the buoyancy force
on the balloon is 1000 N. The supervising professor
conservatively estimates that each student can exert at
least a 40-N tension on the tether for the necessary length
of time. Based on this estimate, what minimum numbers
of students are needed at B, C, and D?
A (0, 8, 0) m
C (10,0, –12) m
D
(–16, 0, 4) m
x
B (16, 0, 16) m
z
Solution:
1000 N
Fy D 1000 909.81 T D 0
T D 117.1 N
(90) g
A0, 8, 0
B16, 0, 16
T
C10, 0, 12
D16, 0, 4
We need to write unit vectors eAB , eAC , and eAD .
y
T
eAB D 0.667i 0.333j C 0.667k
(0, 8, 0)
eAC D 0.570i 0.456j 0.684k
A
FAC
eAD D 0.873i 0.436j C 0.218k
FAD
We now write the forces in terms of magnitudes and unit vectors

FAB D 0.667FAB i 0.333FAB j C 0.667FAB k




 FAC D 0.570FAC i 0.456FAC j 0.684FAC k

FAD D 0.873FAD i 0.436FAC j C 0.218FAC k




T D 117.1j (N)
C (10, 0, −12) m
D
x
(−16, 0, 4)
z
B (16, 0, 16) m
The equations of equilibrium are
Fx D 0.667FAB C 0.570FAC 0.873FAD D 0
Fy D 0.333FAB 0.456FAC 0.436FAC C 117.1 D 0
Fz D 0.667FAB 0.684FAC C 0.218FAC D 0
Solving, we get
FAB D 64.8 N ¾ 2 students
FAC D 99.8 N ¾ 3 students
FAD D 114.6 N ¾ 3 students
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1
Problem 3.69 The 20-kg mass is suspended by cables
attached to three vertical 2-m posts. Point A is at (0, 1.2,
0) m. Determine the tensions in cables AB, AC, and AD.
y
C
B
D
A
1m
1m
2m
0.3 m
x
z
Solution: Points A, B, C, and D are located at
A0, 1.2, 0,
B0.3, 2, 1,
C0, 2, 1,
D2, 2, 0
y
C
FAC
B
FAB
FAD
Write the unit vectors eAB , eAC , eAD
D
A
eAB D 0.228i C 0.608j C 0.760k
W
eAC D 0i C 0.625j 0.781k
eAD D 0.928i C 0.371j C 0k
z
(20) (9.81) N
x
The forces are
FAB D 0.228FAB i C 0.608FAB j C 0.760FAB k
FAC D 0FAC i C 0.625FAC j 0.781FAC k
FAD D 0.928FAD i C 0.371FAD j C 0k
W D 209.81j
The equations of equilibrium are


Fx D 0.228FAB C 0 C 0.928FAD D 0




Fy D 0.608FAB C 0.625FAC C 0.371FAD 209.81 D 0





Fz D 0.760FAB 0.781FAC C 0 D 0
We have 3 eqns in 3 unknowns solving, we get
FAB D 150.0 N
FAC D 146.1 N
FAD D 36.9 N
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1
Problem 3.70 The weight of the horizontal wall section is W D 20,000 lb. Determine the tensions in the
cables AB, AC, and AD.
Solution: Set the coordinate origin at A with axes as shown. The
upward force, T, at point A will be equal to the weight, W, since the
cable at A supports the entire wall. The upward force at A is T D W
k. From the figure, the coordinates of the points in feet are
A
D
10 ft
A4, 6, 10,
B0, 0, 0,
C12, 0, 0,
and
7 ft
D4, 14, 0.
The three unit vectors are of the form
6 ft
C
B
4 ft
xI xA i C yI yA j C zI zA k
,
eAI D xI xA 2 C yI yA 2 C zI zA 2
14 ft
8 ft
W
where I takes on the values B, C, and D. The denominators of the unit
vectors are the distances AB, AC, and AD, respectively. Substitution
of the coordinates of the points yields the following unit vectors:
T
z
eAB D 0.324i 0.487j 0.811k,
A
y
TD
eAC D 0.566i 0.424j 0.707k,
10 ft
TB
D
7 ft
TC
and eAD D 0i C 0.625j 0.781k.
6 ft
The forces are
TAB D TAB eAB ,
14 ft
4 ft
TAC D TAC eAC ,
and
TAD D TAD eAD .
C
B
X
8 ft
W
The equilibrium equation for the knot at point A is
T C TAB C TAC C TAD D 0.
From the vector equilibrium equation, write the scalar equilibrium
equations in the x, y, and z directions. We get three linear equations
in three unknowns. Solving these equations simultaneously, we get
TAB D 9393 lb, TAC D 5387 lb, and TAD D 10,977 lb
A
D
10 ft
6 ft
C
B
4 ft
8 ft
7 ft
14 ft
W
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1
Problem 3.71 The car and the pallet supporting it in
Fig. a weigh 3000 lb. They are suspended by four cables
AB, AC, AD, and AE. The locations of the cable attachment points on the pallet are shown in Fig. b. The
tensions in cables AB and AC are equal. Determine the
tensions in the four cables.
y
A
(0, 18, 0) ft
E
C
z
B
x
(a)
8 ft
C
D
5 ft
4 ft
x
5 ft
E
B
7 ft
5 ft
z
Solution: The Forces in the cables:
TAB D TAB
TAB D TAD
5i 18j C 5k
p
374
8i 18j 4k
p
404
, TAC D TAC
5i 18j 5k
p
374
, TAE D TAE
(b)
7i 18j C 5k
p
398
The equilibrium equations:
5
5
8
7
Fx : p
TAB C p
TAC p
TAD p
TAE D 0
398
374
374
404
18
18
18
18
Fy : p
TAB p
TAC p
TAD p
TAE
398
374
374
404
C 3000 lb D 0
Fz : p
5
374
5
4
5
TAB p
TAC p
TAD C p
TAE D 0
398
374
404
The extra equation:
TAB D TAC
Solution:
TAB D TAC D 970 lb, TAD D 741 lb, TAE D 588 lb
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1
Problem 3.72 The 680-kg load suspended from the
helicopter is in equilibrium. The aerodynamic drag force
on the load is horizontal. The y axis is vertical, and cable
OA lies in the x-y plane. Determine the magnitude of the
drag force and the tension in cable OA.
y
A
10°
O
x
B
C
D
y
Solution:
TOA
Fx D TOA sin 10° D D 0,
Fy D TOA cos 10° 6809.81 D 0.
Solving, we obtain D D 1176 N, TOA D 6774 N.
10°
D
x
(680) (9.81) N
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1
Problem 3.73 In Problem 3.72, the coordinates of the
three cable attachment points B, C, and D are (3.3,
4.5, 0) m, (1.1, 5.3, 1) m, and (1.6, 5.4, 1) m,
respectively. What are the tensions in cables OB, OC,
and OD?
Solution: The position vectors from O to pts B, C, and D are
rOB D 3.3i 4.5j (m),
rOC D 1.1i 5.3j C k (m),
rOD D 1.6i 5.4j k (m).
Dividing by the magnitudes, we obtain the unit vectors
eOB D 0.591i 0.806j,
eOC D 0.200i 0.963j C 0.182k,
eOD D 0.280i 0.944j 0.175k.
Using these unit vectors, we obtain the equilibrium equations
Fx D TOA sin 10° 0.591TOB C 0.200TOC C 0.280TOD D 0,
Fy D TOA cos 10° 0.806TOB 0.963TOC 0.944TOD D 0,
Fz D 0.182TOC 0.175TOD D 0.
From the solution of Problem 3.72, TOA D 6774 N. Solving these
equations, we obtain
TOB D 3.60 kN,
TOC D 1.94 kN,
TOD D 2.02 kN.
y
TOA
10°
x
TOB
TOC
TOD
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1
Problem 3.74 If the mass of the bar AB is negligible
compared to the mass of the suspended object E, the
bar exerts a force on the “ball” at B that points from A
toward B. The mass of the object E is 200 kg. The yaxis points upward. Determine the tensions in the cables
BC and CD.
y
(0, 4, ⫺3) m
C
B
(4, 3, 1) m
D
(0, 5, 5) m
Strategy: Draw a free-body diagram of the ball at B.
(The weight of the ball is negligible.)
x
A
E
z
Solution:
FAB D FAB
4i 3j k
p
26
, TBC D TBC
4i C j 4k
p
33
,
The forces
TBD D TBD
4i C 2j C 4k
6
, W D 200 kg9.81 m/s2 j
The equilibrium equations
4
4
4
Fx : p FAB p TBC TBD D 0
6
26
33
3
1
2
Fy : p FAB C p TBC C TBD 1962 N D 0
6
26
33
1
4
4
Fz : p FAB p TBC C TBD D 0
6
26
33
TBC D 1610 N
)
TBD D 1009 N
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1
Problem 3.75 The 1350-kg car is at rest on a plane
surface. The unit vector en D 0.231i C 0.923j C 0.308k
is perpendicular to the surface. The y axis points upward.
Determine the magnitudes of the normal and friction
forces the car’s wheels exert on the surface.
Solution: The weight force is
W D mgj D 13509.81j D 13240j N.
The component of W normal to the surface is
FN D W Ð e D Wx ex C Wy ey C Wz ez D Wy ey
y
en
D 132400.923 D 12220 N.
The component of W tangent to the surface (the friction force) can be
calculated from
FT D
W2 F2N D
132402 122202 D 5096 N.
Thus, FN D 12220 N and FT D 5096 N.
x
z
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1
y
Problem 3.76 The system shown anchors a stanchion
of a cable-suspended roof. If the tension in cable AB is
900 kN, what are the tensions in cables EF and EG?
G
(0, 1.4, –1.2) m
Solution: From the figure, the coordinates of the points (in
E
F
meters) are
B
(0, 1.4, 1.2) m
A3.4, 1, 0,
B1.8, 1, 0,
C2, 0, 1,
(3.4, 1, 0) m
(2, 1, 0) m
(1, 1.2, 0) m
A
D2, 0, 1,
(2.2, 0, –1) m
E0.9, 1.2, 0,
F0, 1.4, 1.2,
and
G0, 1.4, 1.2.
D
The unit vectors are of the form
eIK
xI xK i C yI yK j C zI zK k
D ,
xI xK 2 C yI yK 2 C zI zK 2
z
(2.2, 0, 1) m
C
y
x
(0, 1.4, –1.2) m
G
where IK takes on the values BA, BC, BD, BE, EF, and EG.
We need to find unit vectors eBA , eBC , eBD , eBE , eEF , and eEG .
E
F
(3.4, 1, 0) m
A
(2, 1, 0) m
Substitution of the coordinates of the points yields the following six
unit vectors:
(1, 1.2, 0) m
B
(0, 1.4, 1.2) m
(2, 0, –1) m
D
eBA D 1i C 0j C 0k,
x
C
eBC D 0.140i 0.707j C 0.707k,
(2, 0, 1) m
z
eBD D 0.140i 0.707j 0.707k,
y
eBE D 0.981i C 0.196j C 0k,
(0, 1.4, −1.2) m
G
eEF D 0.635i C 0.127j C 0.762k,
E
F
and eEG D 0.635i C 0.127j 0.762k.
(3.4, 1, 0) m
TBE (2, 1, 0) m
(1, 1.2, 0) m
The forces are of the form TIK D TIK eIK where IK takes on the same
values as above. The known force magnitude jTBA j D 900 kN. Thus,
A
TBD
TBC
D
C
TBA D TBA eBA D 9001i C 0j C 0k kN D 900i kN.
The vector equation of equilibrium at point B (see the first free body
diagram) is
B
(0, 1.4, 1.2) m
(2, 0, 1) m
y
(0, 1.4, −1.2) m
G
TEG
Use the unit vectors as TBA above to write this equation in component
form, and then solve the resulting linear equations for the three scalar
unknowns TBC , TBD , and TBE .
F
E −T (2, 1, 0) m (3.4, 1, 0) m
BE
TEF
(1, 1.2, 0) m
B
A
(0, 1.4, 1.2) m
The result is
D
TBD D 127.3 kN,
(2, 0, −1) m
x
z
TBA C TBC C TBD C TBE D 0.
TBC D 127.3 kN,
TBA
and
TBE D 917.8 kN.
Once we know TBE , we can use the second free body diagram and
the equilibrium equation at point E to solve for the tensions TEF and
TEG . The vector equilibrium equation at point E (see the second free
body diagram) is TBE C TEF C TEG D 0. Using the unit vectors as
above and solving for TEF and TEG , we get TEF D TEG D 708.7 kN.
C
z
(2, 0, −1) m
x
(2, 0, 1) m
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1
Problem 3.77* The cables of the system in Problem 3.76 will each safely support a tension of 1500 kN.
Based on this criterion, what is the largest safe value of
the tension in cable AB?
Solution: The largest load found in the solution of Problem 3.76
is TBE D 917.8 kN. The scale factor, scaling this force up to 1500 kN
is f D 1500/917.8 D 1.634. The largest safe value for the load in
cable AB is TAB max D TBA f D 9001.634 D 1471 kN.
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1
y
Problem 3.78 The 200-kg slider at A is held in place
on the smooth vertical bar by the cable AB.
2m
(a)
(b)
Determine the tension in the cable.
Determine the force exerted on the slider by
the bar.
B
A
5m
2m
x
2m
z
Solution: The coordinates of the points A, B are A2, 2, 0,
B0, 5, 2. The vector positions
rA D 2i C 2j C 0k,
T
rB D 0i C 5j C 2k
N
The equilibrium conditions are:
F D T C N C W D 0.
W
Eliminate the slider bar normal force as follows: The bar is parallel to
the y axis, hence the unit vector parallel to the bar is eB D 0i C 1j C
0k. The dot product of the unit vector and the normal force vanishes:
eB Ð N D 0. Take the dot product of eB with the equilibrium conditions:
eB Ð N D 0.
eB Ð F D eB Ð T C eB Ð W D 0.
The weight is
eB Ð W D 1j Ð jjWj D jWj D 2009.81 D 1962 N.
Note: For this specific configuration, the problem can be solved without eliminating the slider bar normal force, since it does not appear in
the y-component of the equilibrium equation (the slider bar is parallel
to the y-axis). However, in the general case, the slider bar will not be
parallel to an axis, and the unknown normal force will be projected
onto all components of the equilibrium equations (see Problem 3.79
below). In this general situation, it will be necessary to eliminate the
slider bar normal force by some procedure equivalent to that used
above. End Note.
The unit vector parallel to the cable is by definition,
eAB D
rB r A
.
jrB rA j
Substitute the vectors and carry out the operation:
eAB D 0.4851i C 0.7278j C 0.4851k.
(a)
The tension in the cable is T D jTjeAB . Substitute into the modified equilibrium condition
eB F D 0.7276jTj 1962 D 0.
Solve: jTj D 2696.5 N from which the tension vector is
T D jTjeAB D 1308i C 1962j C 1308k.
(b)
The equilibrium conditions are
F D 0 D T C N C W D 1308i C 1308k C N D 0.
Solve for the normal force: N D 1308i 1308k. The magnitude
is jNj D 1850 N.
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1
Problem 3.79 The 100-lb slider at A is held in place
on the smooth circular bar by the cable AB. The circular
bar is contained in the x-y plane.
(a)
(b)
y
3 ft
Determine the tension in the cable.
Determine the normal force exerted on the slider
by the bar.
B
Solution: Strategy: Develop the unit vectors (i) parallel to the
cable and (ii) parallel to the slider bar. Apply the equilibrium conditions. Eliminate the slider bar normal force by taking the dot product
of the slider bar unit vector with the equilibrium conditions. Solve for
the force parallel to the cable. Substitute this force into the equilibrium
condition to find the slider bar normal force.
Assume that the circular bar is a quarter circle, so that the slider is
located on a radius vector (4 ft). With this assumption the coordinates
of the points A, B are
A
4 ft
20°
4 ft
x
z
A4 cos ˛, 4 sin ˛, 0 D A3.76, 1.37, 0, B0, 4, 3.
T
N
The vector positions are
rA D 3.76i C 1.37j C 0k,
rB D 0i C 4j C 3k
The equilibrium conditions are:
F D T C N C W D 0.
The normal force is to be eliminated from the equilibrium equations.
The bar is normal to the radius vector at point A. Hence the unit vector
parallel to the bar is jTj D 137.1 lb.
W
The dot product with the normal force is zero, eB N D 0. Take the dot
product of the unit vector and the equilibrium condition:
eB F D eB T C eB W D 0.
The weight is
eB W D eB jjWj D 0.9397jWj D 0.9397100 D 94 lb.
The unit vector parallel to the cable is by definition,
eAB D
rB r A
.
jrB rA j
Substitute the vectors and carry out the operation
eAB D 0.6856i C 0.4801j C 0.5472k.
(a)
The tension in the cable is T D jTjeAB . Substitute into the modified equilibrium condition
eB F D 0.6854jTj 94 D 0.
Solve: jTj D 137.1 lb, from which the tension vector is
T D jTjeAB D 94i C 65.8j C 75k
(b)
Substitute T into the original equilibrium conditions,
F D 0 D T C N C W D 94i C 65.8j
C 75k C N 100j D 0.
Solve for the normal force exerted by the bar on the slider
N D 94i C 34.2j 75k (lb)
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1
y
Problem 3.80 The cable AB keeps the 8-kg collar A in
place on the smooth bar CD. The y axis points upward.
What is the tension in the cable?
0.15 m
0.4 m
B
Solution: The coordinates of points C and D are C (0.4, 0.3, 0),
and D (0.2, 0, 0.25). The unit vector from C toward D is given by
C
0.2 m
A
0.3 m
0.5 m
O
eCD D eCDx i C eCDy j C eCDz k D 0.456i 0.684j C 0.570k.
x
0.25 m
The location of point A is given by xA D xC C dCA eCDx , with similar
equations for yA and zA . From the figure, dCA D 0.2 m. From this,
we find the coordinates of A are A (0.309, 0.162, 0.114). From the
figure, the coordinates of B are B (0, 0.5, 0.15). The unit vector from
A toward B is then given by
D
0.2 m
z
y
0.15 m
eAB D eABx i C eABy j C eABz k D 0.674i C 0.735j C 0.079k.
0.4 m
W C
B
The tension force in the cable can now be written as
TAB
TAB D 0.674TAB i C 0.735TAB j C 0.079TAB k.
0.5 m
FN
From the free body diagram, the equilibrium equations are:
FNx C TAB eABx D 0,
FNy C TAB eABy mg D 0,
z
0.2 m
D
A
0.2 m 0.3 m
0.25 m
x
and FNz C TAB eABz D 0.
We have three equation in four unknowns. We get another equation
from the condition that the bar CD is smooth. This means that the
normal force has no component parallel to CD. Mathematically, this
can be stated as FN Ð eCD D 0. Expanding this, we get
FNx eCDx C FNy eCDy C FNz eCDz D 0.
We now have four equations in our four unknowns. Substituting in the
numbers and solving, we get
TAB D 57.7 N,
FNx D 38.9 N,
FNy D 36.1 N, and FNz D 4.53 N.
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1
Problem 3.81* In Problem 3.80, determine the magnitude of the normal force exerted on the collar A by the
smooth bar.
Solution: The solution to Problem 3.80 above provides the magnitudes of the components of the normal force exerted on the collar
at A.
jFN j D
FNx 2 C FNy 2 C FNz 2 .
Substituting in the values found in Problem 3.81, we get
jFN j D 53.2 N.
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1
y
Problem 3.82* The 10-kg collar A and 20-kg collar B
are held in place on the smooth bars by the 3-m cable
from A to B and the force F acting on A. The force F
is parallel to the bar. Determine F.
(0, 5, 0) m
(0, 3, 0) m
Solution: The geometry is the first part of the Problem. To ease
our work, let us name the points C, D, E, and G as shown in the
figure. The unit vectors from C to D and from E to G are essential to
the location of points A and B. The diagram shown contains two free
bodies plus the pertinent geometry. The unit vectors from C to D and
from E to G are given by
F
A
3m
B
(4, 0, 0) m
eCD D erCDx i C eCDy j C eCDz k,
(0, 0, 4) m
z
and eEG D erEGx i C eEGy j C eEGz k.
y
Using the coordinates of points C, D, E, and G from the picture, the
unit vectors are
D (0, 5, 0)
m
eCD D 0.625i C 0.781j C 0k,
yA D yC C CAeCDy ,
B
mB g
z
yB D yA C ABeABy ,
A
TAB
NB
and zA D zC C CAeCDz ,
xB D xA C ABeABx ,
NA
TAB
The location of point A is given by
where CA D 3 m. From these equations, we find that the location of
point A is given by A (2.13, 2.34, 0) m. Once we know the location
of point A, we can proceed to find the location of point B. We have
two ways to determine the location of B. First, B is 3 m from point A
along the line AB (which we do not know). Also, B lies on the line
EG. The equations for the location of point B based on line AB are:
F
G (0, 3, 0)
m
and eEG D 0i C 0.6j C 0.8k.
xA D xC C CAeCDx ,
mA g
3m
C (4, 0, 0) m
We now have two fewer equation than unknowns. Fortunately, there
are two conditions we have not yet invoked. The bars at A and B
are smooth. This means that the normal force on each bar can have
no component along that bar. This can be expressed by using the dot
product of the normal force and the unit vector along the bar. The two
conditions are
NA Ð eCD D NAx eCDx C NAy eCDy C NAz eCDz D 0
The equations based on line EG are:
for slider A and
yB D yE C EBeEGy ,
and zB D zE C EBeEGz .
We have six new equations in the three coordinates of B and the
distance EB. Some of the information in the equations is redundant.
However, we can solve for EB (and the coordinates of B). We get that
the length EB is 2.56 m and that point B is located at (0, 1.53, 1.96) m.
We next write equilibrium equations for bodies A and B. From the free
body diagram for A, we get
NAx C TAB eABx C FeCDx D 0,
x
E (0, 0, 4) m
and zB D zA C ABeABz .
xB D xE C EBeEGx ,
x
NB Ð eEG D NBx eEGx C NBy eEGy C NBz eEGz D 0.
Solving the eight equations in the eight unknowns, we obtain
F D 36.6 N .
Other values obtained in the solution are EB D 2.56 m,
NAx D 145 N,
NBx D 122 N,
NAy D 116 N,
NBy D 150 N,
NAz D 112 N,
and NBz D 112 N.
NAy C TAB eABy C FeCDy mA g D 0,
and NAz C TAB eABz C FeCDz D 0.
From the free body diagram for B, we get
NBx TAB eABx D 0,
Nby TAB eABy mB g D 0,
and NBz TAB eABz D 0.
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1
Problem 3.83 (a) Plot the tensions in cables AB and
AC for values of d from d D 0 to d D 1.8 m.
(b) Each cable will safely support a tension of 1 kN. Use
your graph to estimate the acceptable range of values
of d.
1m
1m
B
d
1m
C
A
50 kg
Solution: Isolate the cable juncture A. Find the interior angles ˛
and ˇ. Solve the equilibrium conditions in terms of distance d. Plot
the result.
y
x
B
C
β
The angle between the positive x axis and the tension TAC is ˛.
α
A
The tension: TAC D jTAC ji cos ˛ C j sin ˛
W
The angle between the positive x axis and AB is 180° ˇ.
The tension is TAB D jTAB ji cos ˇ C j sin ˇ.
B
1m
1m
d
The weight is W D 0i jWjj. The equilibrium conditions are
1m
C
β
F D W C TAB C TAC D 0.
α
A
Substitute the vectors and collect like terms,
Fx D jTAC j cos ˛ jTAB j cos ˇi D 0
Tensions vs d
1200
1100
Fy D jTAC j sin ˛ C jTAB j sin ˇ jWj.
Solve:
jTAC j D
jTAB j D
and jTAC j D
cos ˇ
cos ˛
jTAB j,
jWj cos ˛
sinˇ C ˛
jWj cos ˇ
sinˇ C ˛
,
.
m
The weight is jWj D 50 kg 9.81 2 D 490 N.
s
T 1000
e 900
n
s 800
i 700
o
n 600
s 500
,
400
N 300
200
TAB
TAC
0
.2
.4
.6
.8
1
d, meters
1.2
1.4
1.6
1.8
The angles ˛ and ˇ are to be determined. Subdivide the cable interior
into two right triangles as shown. From geometry,
tan ˇ D
1m
D 1,
1m
tan ˛ D
1d
1
ˇ D 45° ,
D 1 d,
˛ D tan1 1 d.
(Note that the argument 1 d is dimensionless, since it has been
divided by 1 m.) The commercial package TK Solver Plus was used to
produce a graph of the tensions vs. the distance d. From the intersection
of the tension line with the 1000 N line, the range of d for safe tension
is 0 d 1.31 m.
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1
Problem 3.84 The suspended traffic light weighs
100 lb. The cables AB, BC, AD, and DE are each 11 ft
long. Determine the smallest permissible length of the
cable BD if the tensions in the cables must not exceed
1000 lb.
40 ft
C
B
Strategy: Plot the tensions in the cables for a range of
lengths of the cable BD.
A
Solution: Isolate the cable juncture A. Find the interior angles ˛
The equilibrium conditions are
and ˇ. Solve the equilibrium conditions in terms of distance d. Repeat
for the cable junctures B and D.
Substitute and collect like terms.
From symmetry, the angles ˛ formed by the suspension cables are
equal. The two right triangles formed by the cables ABD have a base
length of one half BD. Denote this distance by d/2. Note AD D AB D
11 ft. The angle ˛ is given by
cos ˛ D
d
,
22
˛ D cos1
d
22
sin ˛
sin jTBD j D jTDA j
The angle formed by cable AB and the positive x axis is 180 ˛.
The equilibrium conditions are
sin˛ sin x
B
F D W C TAD C TAB D 0.
jTDA j,
y
The tension is TAB D jTAB ji cos ˛ C j sin ˛.
The weight is W D 0i jjWj D 100j.
and
F D TDE C TBD C TDA D 0.
Fy D jTDA j sin ˛ C jTDE j sin j D 0.
Solve: jTDE j D
The angle formed by cable AD and the positive x axis is ˛. The
tension is TAD D jTAD ji cos ˛ C j sin ˛.
Fx D jTDA j cos ˛ jTBD j C jTDE j cos i D 0
.
E
D
α
α
y
x
D
θ
B
A
α
E
D
Substitute and collect like terms
W
A
Fx D jTAD j cos ˛ jTAB j cos ˛i D 0
40
Fy D jTAD j sin ˛ C jTAB j sin ˛ jWjj D 0.
Solve:
jTAD j D jTAB j D
1
jWj
.
2
sin ˛
C
E
11
B
d
D
Isolate the cable juncture D.
Subdivide the upper cable system into two right triangles and a rectangle.
The base of each right triangle is
20 The distance DE is 11 ft, hence cos D
d
.
2
40 d
.
22
The tension in DE is TDE D jTDE ji cos C j sin .
The tension in BD is TBD D jTBD ji C 0j.
The cable AD makes an angle 180 C ˛ with the positive x axis:
TDA D jTDA ji cos ˛ j sin ˛.
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1
3.84 (Continued )
Tensions vs BD
T 1200
e 1000
n
s 800
i
o 600
n
s 400
, 200
l
b
0
18
TDE
TBD
TAD
18.1
Combine terms, jTBD j D
jTDE j D
jTAD j D
18.2
18.3
BD, ft
1
jWj
.
2
sin 1
jWj
2
sin ˛
D cos1
18.5
1
jWj sin˛ .
2
sin sin ˛
For jWj D 100 lb, ˛ D cos1
and
18.4
40 d
22
d
22
,
the commercial package TK Solver Plus was used to produce a graph
of the tensions vs. the distance d over the interval 18 < d 18.5.
The shortest length of BD for the maximum tension not to exceed
1000 lb is d D 18.028 ft. From the graph, it is clear that for lengths
near 18 ft, a few hundredths of a foot change in length can have
tremendous effect on the maximum cable tension.
2
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 3.85 The 2000-lb scoreboard A is suspended
above a sports arena by the cables AB and AC. Each
cable is 160 ft long. Suppose you want to move the
scoreboard out of the way for a tennis match by shortening cable AB while keeping the length of cable AC
constant.
(a)
300 ft
C
B
Plot the tension in cable AB as a function of its
length for values of the length from 142 ft to 160 ft.
Use your graph to estimate how much you can raise
the scoreboard relative to its original position if
you don’t want to subject the cable AB to a tension
greater than 6000 lb.
(b)
A
HOME
Solution: Isolate the scoreboard. Use the cosine law to find the
angles as a function of AB. Solve the equilibrium conditions, and plot
the result. The cosine law is:
d2 C 64400
d2 C 3002 1602
D
cos ˇ D
2300d
600d
cos ˛ D
y
x
β
B
300
α
d
C
160
A
3002 C 1602 d2
115600 d2
D
.
2300160
96000
W
The angle formed by cable AB and the positive x axis is 180° ˇ.
The tension is
TAB D jTAB ji cos ˇ C j sin ˇ.
The angle formed by cable C and the positive x axis is ˛. The tension is
TAC D jTAC ji cos ˛ C j sin ˛.
The weight is W D 0i jjWj D 0i 2000j.
The equilibrium conditions are
VISITOR
TIME
PERIOD
F D W C TAB C TAC D 0.
T 10000
e 9000
n 8000
s 7000
i 6000
o
5000
n
s 4000
, 3000
l 2000
b 1000
140
Tensions vs AB
145
150
AB Length, ft
155
160
Substitute:
Fx D jTAB j cos ˛ C jTAC j cos ˇi D 0
Fy D jTAB j sin ˛ C jTAC j sin ˇ jWjj D 0.
Solve:
jTAB j D jTAC j
and jTAC j D
cos ˇ
cos ˛
jWj cos ˛
sin˛ C ˇ
,
where
˛ D cos1
115600 d2
96000
,
ˇ D cos1
64400 C d2
600d
The commercial package TK Solver Plus was used to produce a graph
of the tensions vs. the length AB. Both cables have approximately the
same tension; the values are so close to one another that the difference
cannot be distinguished on the scale in the graph. (The tension in cable
AB is slightly higher than the tension in cable AC.)The cable AB is
144.29 ft long at the 6000 lb limit on the tension. At this point the drop
of the scoreboard from the ceiling is 144.26 sin ˇ D 25.32 ft, where
ˇ D 10.1° . The original drop was 160 sin ˇ D 55.68 ft, where ˇ D
20.4° . Thus the scoreboard can be raised 55.68 25.32 D 30.36 ft
before the tension limit is exceeded.
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1
Problem 3.86 The mass of the truck is 4000 kg. The
sum of the lengths of the cables AB and BC is 42 m.
(a)
Draw graphs of the tensions in cables AB and BC
for values of b from zero to 20 m.
Each cable will safely support a tension of 60 kN.
Use the results of part (a) to estimate the allowable
range of the distance b.
(b)
40 m
b
A
C
B
Solution: We have the geometry:
LAB D
b2 C h2 , LBC D
40 − b
b
40 b2 C h2
h
LAB
LAB C LBC D 42
LBC
Solving we find
hD
p
41 41 C 20b
841 20b
41 C 40b b2 , LAB D
, LBC D
21
21
21
Fx : Fy :
TBC
TAB
Equilibrium:
h
h
b
40 b
TAB C
TBC D 0
LAB
LBC
40 − b
b
h
h
TAB C
TBC 39.24 kN D 0
LAB
LBC
3.924 kN
Solving we find
TAB D
60
The plots
From the plot we see that AB reaches the critical value first.
Solving for the value of b that makes TAB D 60 kN we find
b < 10.01 m
Problem 3.86
70
50
Tensions
(a)
(b)
3.06440 b2.05 C b
128.8b 3.064b2
p
, TBC D p
41 b1 C b
41 b1 C b
40
TAB
TBC
30
20
10
0
0
2
4
6
8
10
b
12
14
16
18
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20
1
Problem 3.87 The two springs are identical, with
unstretched length 0.4 m and spring constant k D
900 N/m. A 50-kg mass is suspended at B. What is the
resulting tension in each spring?
Solution: Let h be the vertical distance from point B to the line
AC. Then we have 2 unknowns (F and h).
F D 900 N/m
0.3 m2 C h2 0.4 m
Fy : 2 h
0.3 m2 C h2
F 490.5 N D 0
0.6 m
Solving h D 0.634 m, F D 271 N
A
F
F
C
k
h
k
h
0.3
0.3
B
490.5 N
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1
Problem 3.88 The cable AB is 0.5 m in length and
the unstretched length of the spring BC is 0.4 m. The
spring constant k is 5200 N/m. When the 50-kg mass
is suspended at B, what is the resulting length of the
stretched spring?
0.7 m
A
C
k
B
Solution: Introduce the distances b and h.
Then we have 4 unknowns (F, TAB , b, h, LBC ).
0.7 m − b
b
We have the constraint and equilibrium equations
0.5 m D
LBC D
C
b2 C h2
0.7
m b2
h
A
C h2
B
F D 5200 N/mLBC 0.4 m
Fx : Fy :
b
0.7 m b
FD0
TAB C
0.5 m
LBC
h
h
F 490.5 N D 0
TAB C
0.5 m
LBC
TAB
F
0.5 m
h
b
h
0.7 m − b
Solving we find
h D 0.335 m, b D 0.371 m, LBC D 0.470 m,
F D 364 N, TAB D 343 N
LBC D 0.470 m
490.5 N
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1
Problem 3.89 The length of the cable ABC is 1.4 m.
The 2-kN force is applied to a small pulley. The system
is stationary. Determine the horizontal distance from A
to B and the tension in the cable.
1m
A
C
B
20⬚
Solution: From the geometry
1.4 m D
tan ˛ D
1−b
b
b2 C h2 C 1 m b2 C h2
2 kN
α
β
h
h
h
, tan ˇ D
b
1.0 m b
From equilibrium
Fx : T cos ˛ C T cos ˇ C 2 kN sin 20° D 0
Fy : T sin ˛ C T sin ˇ 2 kN cos 20° D 0
T
T
α
β
Solving we find
b D 0.823 m, h D 0.435 m, T D 1.349 kN
20°
2 kN
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1
Problem 3.90 Consider the tethered balloon in Problem 3.68. The mass of the balloon, experiments package,
and the gas it contains is 90 kg, and the buoyancy force
on the balloon is 1000 N. If the tethers AB, AC, and
AD will each safely support a tension of 500 N and the
coordinates of point A are (0, h, 0), What is the minimum
allowable height h?
y
Solution: See the solution to Problem 3.68. Solve the problem
by computer with values of h ranging from 1 to 4 meters. FAD is
always the largest force. At h D 1 m, FAD D 827 N and at h D 2 m,
FAD D 416 N. Now solve for values between 1 and 2 meters.
h (m)
FAD (N)
1.1
1.2
1.3
1.4
1.5
1.6
1.7
752
689
637
591
552
518
488
1.62
1.64
1.66
1.68
511
505
499
493
h¾
D 1.66 m
A (0, h, 0) m
C (10, 0, ⫺12) m
D
(⫺16, 0, 4) m
z
x
B (16, 0, 16) m
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1
Problem 3.91 The collar A slides on a smooth vertical
bar. The masses mA D 20 kg and mB D 10 kg, and the
spring constant k D 360 N/m. When h D 0.2 m, the
spring is unstretched. Determine the value of h when
the system is in equilibrium.
0.25 m
h
A
B
k
Solution: The diagram of the triangle shows that the amount of
stretch in the spring is given by υ D c c0 .
y
c0
C
From the free body diagrams of masses A and B, the equations of
equilibrium are:
For A:
and
and for B:
α
h
x
N
0.25 m
WA
WB k (c−c0)
Fx D T cos ˛ N D 0
Fy D T sin ˛ mA g D 0,
Fy D T kc c0 mB g D 0.
From the geometry, we know that
c0 D
T
T
h0
h02 C 0.252 ,
and that sin ˛ D
cD
h2 C 0.252 ,
h
.
c
Substituting in the known values, we get a set of equations which
must be solved either by iteration or by graphical methods. Using an
iterative solution, we get h D 0.218 m .
U 206
p 204
202
F 200
o 198
r 196
c 194
e 192
190
& 188
186
W 184
A 182
.2
.205
.21
.215
.22
h, vertical coord of slider − m
.225
.23
A graphical solution strategy can be easily employed. Once we know
a value for h, we can calculate the values of all of the forces in
the Problem. The only equation which will not be satisfied is the ydirection equilibrium equation for mass A. We see from the free body
diagram for A that the weight of A must be balanced by the vertical
component of T for equilibrium. We need only calculate the vertical
component of the force T acting on A and compare this to the weight
of A. The results of such a comparison are shown here. Note that the
sloping line (the vertical component of T) crosses the horizontal line
(the weight of A) at h ³ 0.217 m. This is very close to our previous
result.
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1
Problem 3.92* The cable AB keeps the 8-kg collar
A in place on the smooth bar CD. The y axis points
upward. Determine the distance s from C to the collar
A for which the tension in the cable is 150 N.
y
0.15 m
0.4 m
B
C
s
A
0.3 m
0.5 m
O
x
0.25 m
D
0.2 m
z
Solution: From the figure, the coordinates of the points (in
meters) are B0, 0.5, 0.15, C0.4, 0.3, 0, and D0.2, 0, 0.25.
0.15 m
The first unit vector is of the form,
B
y
0.4 m
W
xI xK i C yI yK j zI zK k
,
eIK D xI xK 2 C yI yK 2 C zI zK 2
TAB
FN
0.5 m
where IK takes on the value CD. The coordinates of point A are
given by
A
s
0.3 m
x
0.25 m
D
Ax D Cx C seCDx ,
C
Ay D Cy C seCDy ,
0.2 m
z
and Az D Cz C seCDz ,
220
where we do not know the value of s. The equations of equilibrium
for this problem are:
and
200
T
180
Fx D TAB eABx C FNx D 0,
i
n 160
Fy D TAB eABy C FNy W D 0,
A 140
B
− 120
N
Fz D TAB eABz C FNz D 0,
where TAB D 150 N.
100
80
.25
The weight of the collar is given by
.275
.3
.325
.35
.375
.4
.425
.45
.475
.5
Distance, s (m)
W D mg, or W D 89.81 D 78.48 N.
The condition that the force FN is perpendicular to CD is
FN Ð eCD D 0, or FN Ð eCD D FNx eCDx C FNy eCDy C FNz eCDz D 0.
We have three equilibrium equations plus the dot product equation
in the four unknowns, s and the three components of FN . Several
methods of solution are open to us. Any iterative algebraic solution
method should give the result s D 0.3046 m and that
Alternative Solution: The complication in the algebra in the solution
is because we do not know the location of point A. We can assume
the location of A is known (assume that we know the distance s) and
solve for the value of the tension in cable AB which corresponds to
that location for A. We can plot the value of the tension versus the
distance s and find the value of s at which the tension is 150 N. If we
do this, we get the plot shown. From the plot, s ¾
D 0.305 N.
FN D 80.7i 47.7j C 7.29k N.
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1
Problem 3.93* In Problem 3.92, determine the distance s from C to the collar A for which the magnitude
of the normal force exerted on the collar A by the smooth
bar is 50 N.
Solution: Use the solution for Problem 3.92. The magnitude of the
tension in AB is no longer to be equal to 150 N. Instead, the magnitude
of the normal force
jFN j D
F2Nx C F2Ny C F2Nz
must be 50 N.
From the plot, the correct value of s is s ¾
D 0.395 m .
An iterative solution of the equations four equations derived in Problem 3.92, with the values from this problem, gives s D 0.396 m .
160
140
120
F100
N
− 80
N
60
40
20
.25
.275
.3
.325
.35
.375
.4
Distance, s (m)
.425
.45
.475
.5
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1
y
Problem 3.94* The 10-kg collar A and 20-kg collar B
slide on the smooth bars. The cable from A to B is 3 m in
length. Determine the value of the distance s in the range
1 s 5 m for which the system is in equilibrium.
(0, 5, 0) m
(0, 3, 0) m
s
A
B
(4, 0, 0) m
(0, 0, 4) m
z
Solution: The geometry is the first part of the Problem. To ease
our work, let us name the points C, D, E, and G as shown in the
figure. The unit vectors from C to D and from E to G are essential to
the location of points A and B. The diagram shown contains two free
bodies plus the pertinent geometry. The unit vectors from C to D and
from E to G are given by
eCD D erCDx i C eCDy j C eCDz k,
and eEG D erEGx i C eEGy j C eEGz k.
Using the coordinates of points C, D, E, and G from the picture, the
unit vectors are
x
F
200
f
o
r
100
e
q
u
i
l
i
b
r
i
u
m
−100
0
−200
−300
−400
−500
−600
− −700
N −800
eCD D 0.625i C 0.781j C 0k,
2
2.25
2.5
2.75
3
3.25
3.5
3.75
4
4.25
4.5
4.75
Distance from C to A, s (k)
and eEG D 0i C 0.6j C 0.8k.
The location of point A is given by
xA D xC C CAeCDx ,
F
yA D yC C CAeCDy ,
e
q
u
i
l
i
b
r
i
u
m
and zA D zC C CAeCDz ,
y
D (0,5,0)
m
TAB
B
mBg
z
A
mAg
17.5
15
12.5
10
7.5
5
2.5
0
−2.5
−5
−7.5
−10
−12.5
N
2.56 2.58
NA
TAB
NB
−
F
G (0,3,0)
m
f
o
r
2.6
2.62 2.64 2.66 2.68
3m
C (4,0,0)
m
2.7
2.72 2.74 2.76 2.78
Distance from C to A, s (m)
x
where CA D s, the parameter we vary to find Fs.
E (0,0,4) m
From these equations, we can find that the location of point A for any
value of s. Once we know the location of point A, we can proceed
to find the location of point B. We have two ways to determine the
location of B. First, B is 3 m from point A along the line AB (which
we do not know). Also, B lies on the line EG. The equations for the
location of point B based on line AB are:
xB D xA C ABeABx ,
yB D yA C ABeABy ,
and zB D zA C ABeABz .
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1
3.94 (Continued )
The equations for the location of point B based on line EG are:
xB D xE C EBeEGx ,
yB D yE C EBeEGy ,
and zB D zE C EBeEGz .
We have six new equations in the three coordinates of B and the
distance EB. Some of the information in the equations is redundant.
However, we can solve for EB (and the coordinates of B). We next
write equilibrium equations for bodies A and B. From the free body
diagram for A, we get
NAx C TAB eABx C FeCDx D 0,
NAy C TAB eABy C FeCDy mA g D 0,
and NAz C TAB eABz C FeCDz D 0.
From the free body diagram for B, we get
NBx TAB eABx D 0,
Nby TAB eABy mB g D 0,
and NBz TAB eABz D 0.
We now have two fewer equation than unknowns. Fortunately, there
are two conditions we have not yet invoked. The bars at A and B
are smooth. This means that the normal force on each bar can have
no component along that bar. This can be expressed by using the dot
product of the normal force and the unit vector along the bar.
The two conditions are
NA Ð eCD D NAx eCDx C NAy eCDy C NAz eCDz D 0
for slider A and
NB Ð eEG D NBx eEGx C NBy eEGy C NBz eEGz D 0.
Solving the eight equations in the eight unknowns, we obtain Fs
for any given s. The plot of Fs vs s, found using TK Solver Plus
is shown below. We see from the plot that the force F goes to zero
somewhere between s D 2.5 m and s D 2.75 m. We can expand the
plot around the zero crossing to obtain a more exact result. The plot
right hand plot is such an expansion.
From the second plot, the value s D 2.65 m is necessary for the configuration to be in equilibrium.
2
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Problem 3.95 The 100-lb crate is held in place on the
smooth surface by the rope AB. Determine the tension in
the rope and the magnitude of the normal force exerted
on the crate by the surface.
A
45°
B
100 lb
30°
Solution: Isolate the crate, and solve the equilibrium conditions.
The weight is W D 0i 100j. The angle between the normal force
and the positive x axis is 90 30 D 60° . The normal force is
N D jNji cos 60 C j sin 60 D jNj0.5i C 0.866j.
The angle between the string tension and the positive x axis is 180° 45° D 135° , hence the tension is
T D jTji cos 135° C j sin 135° D jTj0.7071i C 0.7071j.
The equilibrium conditions are
F D W C N C T D 0.
Substituting, and collecting like terms
Fx D 0.5jNj 0.7071jTji D 0
Fy D 0.866jNj C 0.7071jTj 100j D 0
Solve: jTj D 51.8 lb, jNj D 73.2 lb
Check: Use a coordinate system with the x axis parallel to the inclined
surface. The equilibrium equation for the x-coordinate is
Fx jWj sin 30° jTj cos 15° D 0
from which jTj D
sin 30°
cos 15°
100 D 51.76 D 51.8 lb.
The equilibrium equation for the y-coordinate is
Fy D jNj W cos 30° C jTj sin 15° 0,
from which jNj D 73.2 lb. check.
y
T
α
N
W
β
x
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1
Problem 3.96 The system shown is called Russell’s
traction. If the sum of the downward forces exerted at
A and B by the patient’s leg is 32.2 lb, what is the
weight W?
y
Solution: Isolate the leg. Express the tensions at A and B in scalar
components. Solve the equilibrium conditions. The pulleys change the
direction but not the magnitude of the force jWj. The force at B is
60°
20°
FB D jWji cos 60° C j sin 60° .
25°
FB D jWj0.5i C 0.866j.
B
A
The angles at A relative to the positive x axis are: 180° and 180° 25° D 155° . The force at A is the sum of the two forces:
W
FA D jWji cos 180° C j sin 180° C jWji cos 155° C j sin 155° FA D jWj1.906i C 0.4226j.
x
The total force exerted by the patient’s leg is FP D FH i 32.2j, where
FH is an unknown component. The equilibrium conditions are
F D FA C FtB C FP D 0,
from which:
and
FX D 0.5jWj 1.906jWj C FH i D 0
FY D 0.866jWj C 0.4226jWj 32.2j D 0.
Solve for the weight: jWj D
32.2
D 25 lb .
1.2886
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1
Problem 3.97 A heavy rope used as a hawser for a
cruise ship sags as shown. If it weighs 200 lb, what are
the tensions in the rope at A and B?
55°
A
B
40°
Solution: Resolve the tensions at A and B into scalar components.
Solve the equilibrium equations. The tension at B is
TB D jTB ji cos 40° C j sin 40° TB D jTB j0.7660i C 0.6428j.
The angle at A relative to the positive x axis is 180° 55° D 125° .
The tension at A:
TA D jTA ji cos 125° C j sin 125° D jTA j0.5736i C 0.8192j.
The weight is: W D 0i 200j. The equilibrium conditions are
F D TA C TB C W D 0,
from which
Solve:
Fx D 0.766jTB j 0.5736jTA ji D 0
Fy D 0.6428jTB j 0.8192jTA j 200i D 0.
jTB j D 115.1 lb,
jTA j D 153.8 lb.
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1
Problem 3.98 The cable AB is horizontal, and the box
on the right weighs 100 lb. The surfaces are smooth.
(a)
(b)
A
What is the tension in the cable?
What is the weight of the box on the left?
B
20°
40°
Solution: Isolate the right hand box, resolve the forces into components, and solve the equilibrium conditions. Repeat for the box on
the left.
(a)
T
For right hand box. The weight is W D 0i 100j. The angle
between the normal force and the positive x axis is 90° 40° D
50° . The force:
N
40°
W
N D jNji cos 50° C j sin 50° D jNj0.6428i C 0.7660j.
The cable tension is T D jTji C 0j. The equilibrium conditions are
from which
and
y
T
F D T C N C W D 0,
Fx D 0.6428jNj jTji D 0
20°
W
N
x
Fy D 0.7660jNj 100j D 0
Solve: jTj D 83.9 lb
(b)
For left hand box: The weight W D 0i jWjj. The angle between
the normal force and the positive x axis is 90° C 20° D 110° .
The normal force:
N D jNj0.3420i C 0.9397j.
The cable tension is: T D jTji C 0j. The equilibrium conditions are:
F D W C N C T D 0,
from which:
and
Fx D 0.342jNj C 83.9i D 0
Fy D 0.940jNj jWjj D 0.
Solving for the weight of the box, we get jWj D 230.6 lb.
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1
Problem 3.99 A concrete bucket used at a construction site is supported by two cranes. The 100-kg bucket
contains 500 kg of concrete. Determine the tensions in
the cables AB and AC.
(1.5, 14) m
y
B
C
(3, 8) m
(5, 14) m
A
x
Solution: We need unit vectors eAB and eAC . The coordinates of
A, B, and C are
eAB D 0.243i C 0.970j
eAC D 0.316i C 0.949j
The forces are

T D 0.243TAB i C 0.970TAB j

 AB
TAC D 0.316TAC i C 0.949TAC j


W D 5886j N


Fx D 0.243TAB C 0.316TAC D 0

Fy D 0.970TAB C 0.949TAC 5886 D 0
Solving, TAB D 3.47 kN, TAC D 2.66 kN
TAC
TAB
A
(3,8)
W = –mg j
= (600)(9.81)N
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1
Problem 3.100 The mass of the suspended object A is
mA and the masses of the pulleys are negligible. Determine the force T necessary for the system to be in
equilibrium.
T
A
Solution: Break the system into four free body diagrams as shown.
Carefully label the forces to ensure that the tension in any single cord
is uniform. The equations of equilibrium for the four objects, starting
with the leftmost pulley and moving clockwise, are:
S 3T D 0,
R 3S D 0,
F
F 3R D 0,
R
and 2T C 2S C 2R mA g D 0.
We want to eliminate S, R, and F from our result and find T in
terms of mA and g. From the first two equations, we get S D 3T,
and R D 3S D 9T. Substituting these into the last equilibrium equation
results in 2T C 23T C 29T D mA g.
R
R
R
S
S
S
Solving, we get T D mA g/26 .
S
T
T
S
S
R
R
T
T
T
A
mAg
Note: We did not have to solve for F to find the appropriate value of
T. The final equation would give us the value of F in terms of mA and
g. We would get F D 27mA g/26. If we then drew a free body diagram
of the entire assembly, the equation of equilibrium would be F T mA g D 0. Substituting in the known values for T and F, we see that
this equation is also satisfied. Checking the equilibrium solution by
using the “extra” free body diagram is often a good procedure.
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1
Problem 3.101 The assembly A, including the pulley,
weighs 60 lb. What force F is necessary for the system
to be in equilibrium?
F
A
Solution: From the free body diagram of the assembly A, we have
3F 60 D 0, or F D 20 lb
F
F
F F
F
F
F
60 lb.
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1
Problem 3.102 The mass of block A is 42 kg, and the
mass of block B is 50 kg. The surfaces are smooth. If
the blocks are in equilibrium, what is the force F?
B
F
45°
A
20°
Solution: Isolate the top block. Solve the equilibrium equations.
The weight is. The angle between the normal force N1 and the positive x axis is. The normal force is. The force N2 is. The equilibrium
conditions are
from which
Solve:
y
B
N2
F D N1 C N2 C W D 0
N1
Fx D 0.7071jN1 j jN2 ji D 0
Fy D 0.7071jN1 j 490.5j D 0.
N1 D 693.7 N,
W
α
x
y
N1
jN2 j D 490.5 N
Isolate the bottom block. The weight is
F
β
α
A
W D 0i jWjj D 0i 429.81j D 0i 412.02j (N).
The angle between the normal force N1 and the positive x axis is
270° 45° D 225° .
x
N3
W
The normal force:
N1 D jN1 ji cos 225° C j sin 225° D jN1 j0.7071i 0.7071j.
The angle between the normal force N3 and the positive x-axis is
90° 20° D 70° .
The normal force is
N1 D jN3 ji cos 70° C j sin 70° D jN3 j0.3420i C 0.9397j.
The force is . . . F D jFji C 0j. The equilibrium conditions are
F D W C N1 C N3 C F D 0,
from which:
Fx D 0.7071jN1 j C 0.3420jN3 j C jFji D 0
Fy D 0.7071jN1 j C 0.9397jN3 j 412j D 0
For jN1 j D 693.7 N from above:
jFj D 162 N
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1
Problem 3.103 The climber A is being helped up
an icy slope by two friends. His mass is 80 kg, and
the direction cosines of the force exerted on him by
the slope are cos x D 0.286, cos y D 0.429, cos z D
0.857. The y axis is vertical. If the climber is in
equilibrium in the position shown, what are the tensions
in the ropes AB and AC and the magnitude of the force
exerted on him by the slope?
y
A
(3, 0, 4) m
z
Solution: Get the unit vectors parallel to the ropes using the coordinates of the end points. Express the tensions in terms of these unit
vectors, and solve the equilibrium conditions. The rope tensions, the
normal force, and the weight act on the climber. The coordinates
of points A, B, C are given by the problem, A3, 0, 4, B2, 2, 0,
C5, 2, 1.
B
rB D 2i C 2j C 0k,
x
C
A
The vector locations of the points A, B, C are:
rA D 3i C 0j C 4k,
C
(5, 2, –1) m
B
(2, 2, 0) m
W
N
rC D 5i C 2j 1k.
Substitute and collect like terms,
The unit vector parallel to the tension acting between the points A, B
in the direction of B is
eAB D
rB r A
jrB rA j
The unit vectors are
eAB D 0.2182i C 0.4364j 0.8729k,
Fx D 0.2182jTAB j C 0.3482jTAC j 0.286jNji D 0
Fy D 0.4364jTAB j C 0.3482jTAC j C 0.429jNj 784.8j D 0
Fz D 0.8729jTAB j C 0.8704jTAC j 0.857jNjk D 0
We have three linear equations in three unknowns. The solution is:
eAC D 0.3482i C 0.3482j 0.8704k,
jTAB j D 100.7 N ,
jTAC j D 889.0 N ,
jNj D 1005.5 N .
and eN D 0.286i C 0.429j C 0.857k.
where the last was given by the problem statement. The forces are
expressed in terms of the unit vectors,
TAB D jTAB jeAB ,
TAC D jTAC jeAC ,
N D jNjeN .
The weight is
W D 0i jWjj C 0k D 0i 809.81j C 0k 0i 784.8j C 0k.
The equilibrium conditions are
F D 0 D TAB C TAC C N C W D 0.
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1
Problem 3.104 Consider the climber A being helped
by his friends in Problem 3.103. To try to make the
tensions in the ropes more equal, the friend at B moves
to the position (4, 2, 0) m. What are the new tensions
in the ropes AB and AC and the magnitude of the force
exerted on the climber by the slope?
Solution: Get the unit vectors parallel to the ropes using the coordinates of the end points. Express the tensions in terms of these
unit vectors, and solve the equilibrium conditions. The coordinates
of points A, B, C are A3, 0, 4, B4, 2, 0, C5, 2, 1. The vector
locations of the points A, B, C are:
rA D 3i C 0j C 4k,
rB D 4i C 2j C 0k,
rC D 5i C 2j 1k.
The unit vectors are
eAB D C0.2182i C 0.4364j 0.8729k,
eAC D C0.3482i C 0.3482j 0.8704k,
eN D 0.286i C 0.429j C 0.857k.
where the last was given by the problem statement. The forces are
expressed in terms of the unit vectors,
TAB D jTAB jeAB ,
TAC D jTAC jeAC ,
N D jNjeN .
The weight is
W D 0i jWjj C 0k D 0i 809.81j C 0k 0i 784.8j C 0k.
The equilibrium conditions are
F D 0 D TAB C TAC C N C W D 0.
Substitute and collect like terms,
Fx D C0.281jTAB j C 0.3482jTAC j 0.286jNji D 0
Fy D 0.4364jTAB j C 0.3482jTAC j C 0.429jNj 784.8j D 0
Fz D 0.8729jTAB j C 0.8704jTAC j 0.857jNjk D 0
The HP-28S hand held calculator was used to solve these simultaneous
equations. The solution is:
jTAB j D 420.5 N ,
jTAC j D 532.5 N ,
jNj D 969.3 N .
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1
Problem 3.105 A climber helps his friend up an icy
slope. His friend is hauling a box of supplies. If the
mass of the friend is 90 kg and the mass of the supplies
is 22 kg, what are the tensions in the ropes AB and CD?
Assume that the slope is smooth.
A
20°
B
Solution: Isolate the box. The weight vector is
C
W2 D 229.81j D 215.8j (N).
The angle between the normal force and the positive x axis is
60° D 30° .
40°
90°
75°
D
The normal force is NB D jNB j0.866i 0.5j.
60°
The angle between the rope CD and the positive x axis is 180° 75° D 105° ; the tension is:
T2 D jT2 ji cos 105° C j sin 105° D jT2 j0.2588i C 0.9659j
T
The equilibrium conditions are
y
β
Fx D 0.866jNB j C 0.2588jT2 ji D 0,
Fy D 0.5jNB j C 0.9659jT2 j 215.8j D 0.
N
Solve:
NB D 57.8 N,
jT2 j D 193.5 N.
α
x
Isolate the friend. The weight is
W
W D 909.81j D 882.9j (N).
The angle between the normal force and the positive x axis is 90° 40° D 50° . The normal force is:
y
T1
20°
NF D jNF j0.6428i C 0.7660j.
40°
The angle between the lower rope and the x axis is 75° ; the tension is
75°
T2
N
W
x
T2 D jT2 j0.2588i C 0.9659j.
The angle between the tension in the upper rope and the positive x
axis is 180° 20° D 160° , the tension is
T1 D jT1 j0.9397i C 0.3420j.
The equilibrium conditions are
F D W C T1 C T2 C NF D 0.
From which:
Fx D 0.6428jNF j C 0.2588jT2 j 0.9397jT1 ji D 0
Fy D 0.7660jNF j 0.9659jT2 j C 0.3420jT1 j 882.9j D 0
Solve, for jT2 j D 193.5 N. The result:
jNF j D 1051.6 N ,
jT1 j D 772.6 N .
c 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1
Problem 3.106 The small sphere of mass m is attached
to a string of length L and rests on the smooth surface of
a sphere of radius R. Determine the tension in the string
in terms of m, L, h, and R.
L
h
m
R
Solution: Isolate the small sphere. Use the law of sines to determine the interior angles that appear in the solution.
The angle between the normal force and the positive x axis is 90° ˇ; the normal force is N D jNji sin ˇ C j cos ˇ.
(90°
The angle between the string tension and the positive x axis is
C
˛) (use the rule of equality of opposite interior angles from geometry),
hence the string tension is T D jTji sin ˛ C j cos ˛.
The equilibrium conditions are
from which:
Solve: jTj D
L
h α
The weight is W D 0i mgj.
y
T
R
N
β
R
W
x
F D T C N C W D 0,
Fx D jNj sin ˇ jTj sin ˛i D 0
Fy D jNj cos ˇ jTj cos ˛ mgj D 0
mg sin ˇ
sin˛ C ˇ
.
From the law of sines for the triangle with sides R, R C h, and L,
R C h
L
D
.
sin˛ C ˇ
sin ˇ
Substitute into the tension:
jTj D
mgL
R C h
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1
Problem 3.107 An engineer doing preliminary design
studies for a new radio telescope envisions a triangular
receiving platform suspended by cables from three equally spaced 40-m towers. The receiving platform has
a mass of 20 Mg (megagrams) and is 10 m below the
tops of the towers. What tension would the cables be
subjected to?
TOP VIEW
Solution: Isolate the platform. Choose a coordinate system with
the origin at the center of the platform, with the z axis vertical, and
the x,y axes as shown. Express the tensions in terms of unit vectors,
and solve the equilibrium conditions. The cable connections at the
platform are labeled a, b, c, and the cable connections at the towers
are labeled A, B, C. The horizontal distance from the origin (center of
the platform) to any tower is given by
LD
20 m
65 m
C
65
D 37.5 m.
2 sin60
z
c
The coordinates of points A, B, C are
b
a
A37.5, 0, 10,
B
A
y
x
B37.5 cos120° , 37.5 sin120° .10,
C37.5 cos240° , 37.5 sin240° , 10,
The vector locations are:
The tensions in the cables are expressed in terms of the unit vectors,
rA D 37.5i C 0j C 10k,
rB D 18.764i C 32.5j C 10k,
rC D 18.764i C 32.5j C 10k.
The distance from the origin to any cable connection on the platform is
dD
20
D 11.547 m.
2 sin60° The coordinates of the cable connections are
a11.547, 0, 0,
b11.547 cos120° , 11547 sin120° , 0,
TcC D jTcC jecC .
The weight is W D 0i 0j 200009.81k D 0i C 0j 196200k.
The equilibrium conditions are
F D 0 D TaA C TbB C TcC C W D 0,
from which:
c11.547 cos240° , 11.547 sin240° , 0.
The vector locations of these points are,
ra D 11.547i C 0j C 0k,
TbB D jTbB jebB ,
TaA D jTaA jeaA ,
Fx D 0.9333jTaA j 0.4666jTbB j 0.4666jTcC ji D 0
Fy D 0jTaA j C 0.8082jTbB j 0.8082jTcC jj D 0
Fz D 0.3592jTaA j 0.3592jTbB j
rb D 5.774i C 10j C 0k,
C 0.3592jTcC 196200jk D 0
rc D 5.774i C 10j C 0k.
The unit vector parallel to the tension acting between the points A, a
in the direction of A is by definition
eaA D
rA r a
.
jrA ra
Perform this operation for each of the unit vectors to obtain
eaA D C0.9333i C 0j 0.3592k
The commercial package TK Solver Plus was used to solve these
equations. The results:
jTaA j D 182.1 kN ,
jTcC j D 182.1 kN .
Check: For this geometry, where from symmetry all cable tensions may
be assumed to be the same, only the z-component of the equilibrium
equations is required:
Fz D 3jTj sin 196200 D 0,
ebB D 0.4667i C 0.8082j 0.3592k
where D tan1
ecC D 0.4667i C 0.8082j C 0.3592k
jTbB j D 182.1 kN ,
10
37.5 11.547
D 21.07° ,
from which each tension is jTj D 182.1 kN. check.
c 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1
Problem 3.108 The metal disk A weighs 10 lb. It is
held in place at the center of the smooth inclined surface
by the strings AB and AC. What are the tensions in the
strings?
y
B
(0, 6, 0) ft
C
(8, 4, 0) ft
2 ft
A
z
Solution: Isolate the disk, express the tensions in terms of the
unit vectors, and solve the equilibrium equations. The coordinates of
points A, B, C are: A5, 1, 4, B0, 6, 0, C8, 4, 0, where the coordinates of A are determined from the geometry of the inclined plane.
The radius vectors corresponding to these coordinates are
rA D 5i C 1j C 4k,
rB D 0i C 6j C 0k,
8 ft
10 ft
B
C
A
rC D 8i C 4j C 0k.
The unit vector eAB is, by definition,
eAB
x
N
W
rB r A
D
.
jrB rA j
Apply this to find the unit vectors parallel to the cables,
eAB D 0.6155i C 0.6155j 0.4924k,
eAC D 0.5145i C 0.5145j 0.6860k.
The weight is W D 0i 10j C 0k. The normal force acts normally to
the inclined surface,
N D jNj0i C j cos ˛ C k sin ˛
where tan ˛ D
2
0.25, ˛ D 14° .
8
The tensions in the cables are expressed in terms of the unit vectors,
TAB D jTAB jeAB ,
TAC D jTAC jeAC .
The equilibrium conditions are
F D 0 D TAB C TAC C W C N D 0.
From which
Fx D 0.6155jTAB j C 0.5145jTAC ji D 0
Fy D 0.6155jTAB j C 0.5145jTAC j C 0.9703jNj 10j D 0
The solution to this set of simultaneous equations was obtained using
the commercial program TK Solver 2. The result:
jNj D 8.35 lb ,
jTAB j D 1.54 lb ,
jTAC j D 1.85 lb .
c 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1
y
Problem 3.109 Cable AB is attached to the top of the
vertical 3-m post, and its tension is 50 kN. What are the
tensions in cables AO, AC, and AD?
5m
5m
C
D
Solution: Get the unit vectors parallel to the cables using the
coordinates of the end points. Express the tensions in terms of these
unit vectors, and solve the equilibrium conditions. The coordinates of
points A, B, C, D, O are found from the problem sketch: The coordinates of the points are A6, 2, 0, B12, 3, 0, C0, 8, 5, D0, 4, 5,
O0, 0, 0.
4m
8m
(6, 2, 0) m
The vector locations of these points are:
rA D 6i C 2j C 0k,
rB D 12i C 3j C 0k,
O
rC D 0i C 8j C 5k,
B
A
z
3m
12 m
rD D 0i C 4j 5k,
rO D 0i C 0j C 0k.
x
The unit vector parallel to the tension acting between the points A, B
in the direction of B is by definition
eAB
y
rB r A
D
.
jrB rA j
5m
5m
Perform this for each of the unit vectors
D
4m
C
eAB D C0.9864i C 0.1644j C 0k
eAC D 0.6092i C 0.6092j C 0.5077k
8m
O
(6, 2, 0) m
eAD D 0.7442i C 0.2481j 0.6202k
A
eAO D 0.9487i 0.3162j C 0k
The tensions in the cables are expressed in terms of the unit vectors,
TAB D jTAB jeAB D 50eAB ,
TAD D jTAD jeAD ,
TAC D jTAC jeAC ,
TAO D jTAO jeAO .
The equilibrium conditions are
F D 0 D TAB C TAC C TAD C TAO D 0.
Substitute and collect like terms,
Fx D 0.986450 0.6092jTAC j 0.7422jTAD j
0.9487jTAO ji D 0
Fy D 0.164450 C 0.6092jTAC j C 0.2481jTAD j
0.3162jTAO jj D 0
Fz D C0.5077jTAC j 0.6202jTAD jk D 0.
This set of simultaneous equations in the unknown forces may be
solved using any of several standard algorithms. The results are:
jTAO j D 43.3 kN,
jTAC j D 6.8 kN,
jTAD j D 5.5 kN.
c 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1
Problem 3.110* The 1350-kg car is at rest on a
plane surface with its brakes locked. The unit vector
en D 0.231i C 0.923j C 0.308k is perpendicular to the
surface. The y axis points upward. The direction cosines
of the cable from A to B are cos x D 0.816, cos y D
0.408, cos z D 0.408, and the tension in the cable is
1.2 kN. Determine the magnitudes of the normal and
friction forces the car’s wheels exert on the surface.
y
en
B
ep
x
z
Solution: Assume that all forces act at the center of mass of the
car. The vector equation of equilibrium for the car is
y
en
"car"
FN
TAB
FS C TAB C W D 0.
Writing these forces in terms of components, we have
x
W D mgj D 13509.81 D 13240j N,
FS D FSx i C FSy j C FSz k,
FS
F
W
z
and TAB D TAB eAB ,
where
eAB D cos x i C cos y j C cos z k D 0.816i C 0.408j 0.408k.
Substituting these values into the equations of equilibrium and solving
for the unknown components of FS , we get three scalar equations of
equilibrium. These are:
FSx TABx D 0,
FSy TABy W D 0,
and FSz TABz D 0.
Substituting in the numbers and solving, we get
FSx D 979.2 N,
FSy D 12, 754 N,
and FSz D 489.6 N.
The next step is to find the component of FS normal to the surface.
This component is given by
FN D FN Ð en D FSx eny C FSx eny C FSz enz .
Substitution yields
FN D 12149 N .
From its components, the magnitude of FS is FS D 12800 N. Using
the Pythagorean theorem, the friction force is
fD
F2S F2N D 4033 N.
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1
Problem 3.111* The brakes of the car in Problem 3.110 are released, and the car is held in place
on the plane surface by the cable AB. The car’s front
wheels are aligned so that the tires exert no friction
forces parallel to the car’s longitudinal axis. The unit
vector ep D 0.941i C 0.131j C 0.314k is parallel to the
plane surface and aligned with the car’s longitudinal
axis. What is the tension in the cable?
Solution: Only the cable and the car’s weight exert forces in the
direction parallel to ep . Therefore
ep Ð T mgj D 0:
0.941i C 0.131j C 0.314k
Ð [T0.816i C 0.408j 0.408k mgj] D 0,
0.9410.816T
C 0.1310.408T mg C 0.3140.408T D 0.
Solving, we obtain T D 2.50 kN.
c 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1