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Problem 5.32 In a measure to decrease costs, the
manufacturer of the fan described in Problem 5.31
proposes to support the fan with three equally spaced
legs instead of four. An engineer is assigned to analyze
the safety implications of the change. The weight of the
fan decreases to WD19.6 lb. The dimensions band h
are unchanged. What thrust Twill cause the fan to be
on the verge of tipping over in this case? Compare your
answer to the answer to Problem 5.31.
b
T
Solution: The free-body diagram is shown.
Problem 5.33 A force FD400 N acts on the bracket.
What are the reactions at Aand B?
B
A
F
320 mm
80 mm
Solution: The joint Ais a pinned joint; Bis a roller joint. The
pinned joint has two reaction forces AX,A
Y. The roller joint has one
reaction force BX. The sum of the forces is
AX
AY
80
F
288
c
2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they
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Problem 5.34 The sign’s weight WSD32 lb acts at
the point shown. The 10-lb weight of bar AD acts at the
midpoint of the bar. Determine the tension in the cable
AE and the reactions at D.
Ws
B
A
C
D
E
33 in
30 in
11 in 11 in
15⬚
30⬚
Solution: Treat the bar AD and sign as one single object. Let TAE
be the tension in the cable. The equilibium equations are
Solving yields
DyD34.0 lb
Problem 5.35 The device shown, called a swape or
W
25°
Solution:
F
441 N
883 N
Problem 5.36 This structure, called a truss, has a pin
support at Aand a roller support at Band is loaded by
two forces. Determine the reactions at the supports.
Solution:
MA:⊲4kN⊳p2b⊲2 kN cos 30°⊳3b
290
Problem 5.37 An Olympic gymnast is stationary in
the “iron cross” position. The weight of his left arm
and the weight of his body not including his arms are
shown. The distances are aDbD9 in and cD13 in.
Treat his shoulder Sas a fixed support, and determine
the magnitudes of the reactions at his shoulder. That
is, determine the force and couple his shoulder must
support.
144 lb 8 lb
ab c
S
Solution: The shoulder as a built-in joint has two-force and couple
reactions. The left hand must support the weight of the left arm and
half the weight of the body:
FH
FH
Problem 5.38 Determine the reactions at A.
6 ft 3 ft
300 lb
200 lb
5 ft
800 ft-lb
200 lb
A
Solution: The built-in support at Ais a two-force and couple reac-
tion support. The sum of the forces for the system is
from which
292
Problem 5.39 The car’s brakes keep the rear wheels
locked, and the front wheels are free to turn. Determine
the forces exerted on the front and rear wheels by the
road when the car is parked (a) on an up slope with
˛D15°; (b) on a down slope with ˛D15°.
3300 lb
36 in 70 in
y
20 in
x
α
Solution: The rear wheels are two force reaction support, and the
front wheels are a one force reaction support. Denote the rear wheels
by Aand the front wheels by B, and define the reactions as being
parallel to and normal to the road. The sum of forces:
Since the mass center of the vehicle is displaced above the point A,
a component of the weight (20Wsin ˛) produces a positive moment
about A, whereas the other component (36Wcos ˛) produces a negative
moment about A. The sum of the moments about A:
MAD36⊲3300 cos 15°⊳C20⊲3300 sin 15°⊳CBY⊲106⊳D0,
from which
BYDC97669
106 D921.4lb.
Substitute into the sum of forces equation to obtain AYD2266.1lb
(b) For the car parked down-slope the sum of the forces is
FXDAXC3300 sin 15°D0,
from which AXD854 lb
FYDAY3300 cos 15°CBYD0.
The component (20Wsin ˛) now produces a negative moment about
A. The sum of the moments about Ais
MAD3300⊲36⊳cos 15°3300⊲20⊳sin 15°C106BYD0,
from which
BYD131834
106 D1243.7lb.
Substitute into the sum of forces equation to obtain AYD1943.8lb
20 in.
α
Problem 5.40 The length of the bar is LD4 ft. Its
weight WD6 lb acts at the midpoint of the bar. The
floor and wall are smooth. The spring is unstretched
when the angle ˛= 0. If the bar is in equilibrium when
˛D40°, what is the spring constant k?
L
k
α
Solution: The free-body diagram is shown.
294
Problem 5.41 The weight Wof the bar acts at its
midpoint. The floor and wall are smooth. The spring is
unstretched when the angle ˛D0. Determine the angle
˛at which the bar is in equilibrium in terms of W,k,
and L.
L
k
α
Problem 5.42 The plate is supported by a pin in a
smooth slot at B. What are the reactions at the supports?
B
2 kN-m
6 kN-m
A
296
Problem 5.43 Determine the reactions at the fixed
support A.30 lb 40 lb
150 ft-lb
y
Solution: The free-body diagram is shown.
Problem 5.44 Suppose the you want to represent the
two forces and couple acting on the beam in Problem
5.43 by an equivalent force Fas shown. (a) Determine F
and the distance Dat which its line of action crosses the
xaxis. (b) Assume that Fis the only load acting on the
beam and determine the reactions at the fixed support A.
Compare your answers to answers to Problem 5.43.
F
y
x
D
A
Solution: The free-body diagram is shown.
Problem 5.45 The bicycle brake on the right is pinned
to the bicycle’s frame at A. Determine the force exerted
by the brake pad on the wheel rim at Bin terms of the
cable tension T.
T
40 mm
45 mm
40 mm
Brake pad
Wheel rim
35°
A
B
Solution: From the force balance equation for the cables: the force
on the brake mechanism TBin terms of the cable tension Tis
TB
298
Problem 5.46 The mass of each of the suspended
weights is 80 kg. Determine the reactions at the supports
at Aand E.
ABC
D
E
200 mm 200 mm
300 mm
Solution: From the free body diagram, the equations of
equilibrium for the rigid body are
FxDAXCEXD0,
AY
AX
y
x
A
0.2 m
0.2 m
Problem 5.47 The suspended weights in Problem 5.46
are each of mass m. The supports at Aand Ewill each
safely support a force of 6 kN magnitude. Based on this
criterion, what is the largest safe value of m?
Solution: Written with the mass value of 80 kg replaced by the
symbol m, the equations of equilibrium from Problem 5.46 are
FxDAXCEXD0,
Problem 5.48 The tension in cable BC is 100 lb.
Determine the reactions at the built-in support.
6 ft
B
C
A
300 ft-lb
FXDAXD0.
FYDAY200 D0,
from which AYD200 lb.
The sum of the moments about Ais
MDMA⊲3⊳⊲200⊳300 D0,
from which MAD900 ft lb
AX200 lb
3 ft
Problem 5.49 The tension in cable AB is 2 kN. What
are the reactions at Cin the two cases?
C
60°
ABC
60°
AB
300
Problem 5.50 Determine the reactions at the supports. 6 in
5 in
3 in
50 lb
100 in-lb
A
Solution: The reaction at Ais a two-force reaction. The reaction
BD100
⊲11 sin 60°6 cos 60°⊳D15.3lb.
Substitute into the force equations to obtain
AYDBsin 60°D13.3lb
and AXDBcos 60°C50 D57.7lb
Problem 5.51 The weight WD2 kN. Determine the
tension in the cable and the reactions at A.
0.6 m 0.6 m
A
W
30°
Solution: Equilibrium Eqns:
AY
Problem 5.52 The cable shown in Problem 5.51 will
safely support a tension of 6 kN. Based on this criterion,
what is the largest safe value of the weight W?
WD15.0kN
Problem 5.53 The blocks being compressed by the
clamp exert a 200-N force on the pin at Dthat points
from Atoward D. The threaded shaft BE exerts a force
on the pin at Ethat points from Btoward E.
pin support.
(b) Determine the reactions at C.
50 mm
125 mm
A
D
BE
125 mm 125 mm
302
Problem 5.54 Consider the clamp in Problem 5.53.
The blocks being compressed by the clamp exert a 200-
N force on the pin at Athat points from Dtoward A.
The threaded shaft BE exerts a force on the pin at Bthat
Fy: 200 N CCyD0
Solving we find
CxD500 N,C
yD200 N
Problem 5.55 Suppose that you want to design the
safety valve to open when the difference between the
pressure pin the circular pipe ⊲diameter D150 mm⊳
and the atmospheric pressure is 10 MPa (megapascals;
a pascal is 1 N/m2). The spring is compressed 20 mm
when the valve is closed. What should the value of the
spring constant be?
250 mm
150 mm
A
150 mm
k
p
Solution: The area of the valve is
22
The force at opening is
The force on the spring is found from the sum of the moments about
A,
Solving,
⊲0.4⊳L D0.15⊲1.7671 ð105⊳
⊲0.4⊳⊲0.02⊳
m
250
150
A
150 mm
k
A
F
m
m
k∆L
Problem 5.56 The 10-lb weight of the bar AB acts
at the midpoint of the bar. The length of the bar is 3 ft.
Determine the tension in the string BC and the reactions
at A.
30⬚
B
C
3 ft
304
Problem 5.57 The crane’s arm has a pin support at A.
The hydraulic cylinder BC exerts a force on the arm at
Cin the direction parallel to BC. The crane’s arm has a
mass of 200 kg, and its weight can be assumed to act at a
point 2 m to the right of A. If the mass of the suspended
box is 800 kg and the system is in equilibrium, what
is the magnitude of the force exerted by the hydraulic
cylinder?
1.8 m 1.2 m
7 m
C
A
1.2D63.4°
The equilibrium equations are
Fx:AxCFHcos D0
Problem 5.58 In Problem 5.57, what is the magnitude
of the force exerted on the crane’s arm by the pin support
at A?
Problem 5.59 A speaker system is suspended by the
cables attached at Dand E. The mass of the speaker
system is 130 kg, and its weight acts at G. Determine
the tensions in the cables and the reactions at Aand C.
0.5 m 0.5 m 0.5 m 0.5 m
1 m
G
A
BD
1 m
E
C
Solution: The weight of the speaker is WDmg D1275 N. The
equations of equilibrium for the entire assembly are
AYCY
1 m 1.5 m
306
Problem 5.60 The weight W1D1000 lb. Neglect the
weight of the bar AB. The cable goes over a pulley at
C. Determine the weight W2and the reactions at the pin
support A.
B
A
W1
W2
C
35°
50°
Solution: The strategy is to resolve the tensions at the end of bar
AB into x- and y-components, and then set the moment about Ato
zero. The angle between the cable and the positive xaxis is 35°.
The tension vector in the cable is
35°
T1
T2
rB
