Problem 1.1 Express the fractions 13 and 23 to three
significant digits.
Solution:
1/3 D 0.3333. . D 0.333
2/3 D 0.6666. . D 0.667
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1
Problem 1.2 The base of natural logarithms is e D
2.718281828 . . .
(a)
(b)
(c)
Express e to five significant digits.
Determine the value of e2 to five significant digits.
Use the value of e you obtained in part (a) to determine the value of e2 to five significant digits.
Solution: The value of e is: e D 2.718281828
(a)
To five significant figures e D 2.7183
(b)
e2 to five significant figures is e2 D 7.3891
(c)
Using the value from part (a) we find e2 D 7.3892 which is
not correct in the fifth digit.
[Part (c) demonstrates the hazard of using rounded-off
values in calculations.]
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1
Problem 1.3 A machinist drills a circular hole in a
panel with radius r D 5 mm. Determine the circumference C and area A of the hole to four significant digits.
Solution:
C D 2r D 10 D 31.42 mm
A D r 2 D 25 D 78.54 mm2
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1
Problem 1.4 The opening in a soccer goal is 24 ft
wide and 8 ft high. Use these values to determine its
dimensions in meters to three significant digits.
Solution: The conversion between feet and meters, found inside
the front cover of the textbook, is 1 m D 3.281 ft. The goal width,
w D 24 ft
1m
3.281 ft
D 7.3148 m D 7.31 m.
The goal height is given by
h D 8 ft
1m
3.281 ft
D 2.438 m D 2.44 m.
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1
Problem 1.5 The coordinates (in meters) of point A
are xA D 3, yA D 7, and the coordinates of point B are
xB D 10, yB D 2. Determine the length of the straight
line from A to B to three significant digits.
y
Solution: The length is
LD
10 32 C 2 72 m D
p
74 m D 8.602325 m
to three significant figures this is
L D 8.60 m
A
B
x
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1
Problem 1.6 Suppose that you have just purchased
a Ferrari F355 coupe and you want to know whether
you can use your set of SAE (U.S. Customary Units)
wrenches to work on it. You have wrenches with widths
w D 1/4 in, 1/2 in, 3/4 in, and 1 in, and the car has nuts
with dimensions n D 5 mm, 10 mm, 15 mm, 20 mm,
and 25 mm. Defining a wrench to fit if w is no more
than 2% larger than n, which of your wrenches can you
use?
Solution: Convert the metric size n to inches, and compute the
percentage difference between the metric sized nut and the SAE
wrench. The results are:
5 mm
1 inch
25.4 mm
D 0.19685.. in,
0.19685 0.25
0.19685
100
D 27.0%
10 mm
15 mm
n
20 mm
25 mm
1 inch
25.4 mm
1 inch
25.4 mm
1 inch
25.4 mm
1 inch
25.4 mm
D 0.3937.. in,
D 0.5905.. in,
D 0.7874.. in,
D 0.9843.. in,
0.3937 0.5
0.3937
0.5905 0.5
0.5905
100 D 27.0%
100 D C15.3%
0.7874 0.75
0.7874
0.9843 1.0
0.9843
100 D C4.7%
100 D 1.6%
A negative percentage implies that the metric nut is smaller than the
SAE wrench; a positive percentage means that the nut is larger then
the wrench. Thus within the definition of the 2% fit, the 1 in wrench
will fit the 25 mm nut. The other wrenches cannot be used.
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1
Problem 1.7 On August 20, 1974, Nolan Ryan threw
the first baseball pitch measured at over 100 mi/h. The
measured speed was 100.9 mi/h. Determine the speed of
the pitch to four significant digits (a) in ft/s; (b) in km/h.
Solution:
(a)
v D 100.9
mi
h
v D 100.9
(b)
mi
h
5280 ft
1 mi
5280 ft
mi
1h
3600 s
D 148.0
0.3048 m
ft
ft
s
1 km
1000 m
D 162.4 km/h
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1
Problem 1.8 On March 18, 1999, an experimental
Maglev (magnetic levitation) train in Japan reached a
maximum speed of 552 km/h. What was its velocity in
mi/h to three significant digits?
Solution:
v D 552
km
h
1000 m
km
1 ft
0.3048 m
1 mi
5280 ft
D 343 mi/h
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1
Problem 1.9 In May, 1963, in the last flight of
Project Mercury, Astronaut L. Gordon Cooper traveled
a distance of 546,167 miles in 1 day, 10 hours, 19
minutes, and 49 seconds. Determine his average speed
(the distance traveled divided by the time required) to
three significant digits (a) in mi/h; (b) in km/h.
Solution:
(a)
vD 546167 mi
D 15,900 mi/h
49
19
C
34 C
h
60
3600
v D 15,900
(b)
mi
h
5280 ft
1 mi
0.3048 m
ft
1 km
1000 m
D 25,600 km/h
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1
Problem 1.10 Engineers who study shock waves
sometimes express velocity in millimeters per microsecond (mm/µs). Suppose the velocity of a wavefront
is measured and determined to be 5 mm/µs. Determine
its velocity: (a) in m/s; (b) in mi/s.
Solution: Convert units using Tables 1.1 and 1.2. The results:
(a)
5
mm
s
1m
1000 mm
106 s
1s
D 5000
m
s
.
Next, use this result to get (b):
(b)
5000
m
s
1 ft
0.3048 m
1 mi
5280 ft
D 3.10685 . . .
D 3.11
mi
s
mi
s
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1
Problem 1.11 The kinetic energy of a particle of mass
m is defined to be 12 mv2 , where v is the magnitude of the
particle’s velocity. If the value of the kinetic energy of
a particle at a given time is 200 when m is in kilograms
and v is in meters per second, what is the value when m
is in slugs and v is in feet per second?
Solution:
200
kg-m2
s2
0.0685 slug
1 kg
D 147.46 D 147
1 ft
0.3048 m
2
slug-ft2
s2
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1
Problem 1.12 The acceleration due to gravity at sea
level in SI units is g D 9.81 m/s2 . By converting units,
use this value to determine the acceleration due to gravity
at sea level in U.S. Customary units.
Solution: Use Table 1.2. The result is:
g D 9.81
m
s2
1 ft
0.3048 m
D 32.185 . . .
ft
s2
D 32.2
ft
s2
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1
Problem 1.13 A furlong per fortnight is a facetious
unit of velocity, perhaps made up by a student as a
satirical comment on the bewildering variety of units
engineers must deal with. A furlong is 660 ft (1/8 mile).
A fortnight is 2 weeks (14 days). If you walk to class
at 2 m/s, what is your speed in furlongs per fortnight to
three significant digits?
Solution: Convert the units using the given conversions. Record
the first three digits on the left, and add zeros as required by the number
of tens in the exponent. The result is:
5
ft
s
1 furlong
660 ft
D
9160
furlongs
fortnight
3600 s
1 hr
24 hr
1 day
14 day
1 fortnight
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1
Problem 1.14 The cross-sectional area of a beam is
480 in2 . What is its cross-section in m2 ?
Solution: Convert units using Table 1.2. The result:
480 in2
1 ft
12 in
2 0.3048 m
1 ft
2
D 0.30967 . . . m2 D 0.310 m2
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1
Problem 1.15 The cross-sectional area of the C12ð30
American Standard Channel steel beam is A D 8.81 in2 .
What is its cross-sectional area in mm2 ?
y
A
Solution:
A D 8.81 in2
25.4 mm
1 in
2
D 5680 mm2
x
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1
Problem 1.16 A pressure transducer measures a value
of 300 lb/in2 . Determine the value of the pressure in
pascals. A pascal (Pa) is one newton per meter squared.
Solution: Convert the units using Table 1.2 and the definition of
the Pascal unit. The result:
300
lb
in2
4.448 N
1 lb
D 2.0683 . . . 106 12 in
1 ft
N
m2
2 1 ft
0.3048 m
2
D 2.07106 Pa
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1
Problem 1.17 A horsepower is 550 ft-lb/s. A watt is
1 N-m/s. Determine the number of watts generated by
(a) the Wright brothers’ 1903 airplane, which had a 12horsepower engine; (b) a modern passenger jet with a
power of 100,000 horsepower at cruising speed.
Solution: Convert units using inside front cover of textbook derive
the conversion between horsepower and watts. The result
(a)
12 hp
(b)
105 hp
746 watt
1 hp
746 watt
1 hp
D 8950 watt
Boeing 747
Wright
Brothers' Flier
(shown to scale)
D 7.46107 watt
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1
Problem 1.18 In SI units, the universal gravitational
constant G D 6.67 ð 1011 N-m2 /kg2 . Determine the
value of G in U.S. Customary units.
Solution: Convert units using Table 1.2. The result:
6.671011 N-m2
kg2
1 lb
4.448 N
D 3.43590 . . . 108 lb-ft2
slug2
1 ft
0.3048 m
2 D 3.44108 14.59 kg
1 slug
lb-ft2
slug2
2
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1
Problem 1.19 The moment of inertia of the rectangular area about the x axis is given by the equation
I D 13 bh3 .
The dimensions of the area are b D 200 mm and h D
100 mm. Determine the value of I to four significant
digits in terms of (a) mm4 ; (b) m4 ; (c) in4 .
Solution:
1
200 mm100 mm3 D 66.7 ð 106 mm4
3
(a)
ID
(b)
I D 66.7 ð 106 mm4
(c)
I D 66.7 ð 106 mm4
y
1m
1000 mm
1 in
25.4 mm
4
D 66.7 ð 106 m4
4
D 160 in4
h
x
b
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1
Problem 1.20
In the equation
T D 12 Iω2 ,
Solution: For (a), substitute the units into the expression for T:
(a)
the term I is in kg-m2 and ω is in s1 .
(a)
(b)
What are the SI units of T?
If the value of T is 100 when I is in kg-m2 and ω is
in s1 , what is the value of T when it is expressed
in U.S. Customary base units?
TD
kg-m2
1
I kg-m2 ωs1 2 D
2
s2
For (b), convert units using Table 1.2. The result:
(b)
100
kg-m2
s2
1 slug
14.59 kg
D 73.7759 . . .
slug-ft2
s2
1 ft
0.3048 m
D 73.8
2
slug-ft2
s2
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1
Problem 1.21
The equation
D
My
I
Solution:
(a)
is used in the mechanics of materials to determine
normal stresses in beams.
(a)
(b)
When this equation is expressed in terms of SI base
units, M is in newton-meters (N-m), y is in meters
(m), and I is in meters to the fourth power (m4 ).
What are the SI units of ?
If M D 2000 N-m, y D 0.1 m, and I D 7 ð
105 m4 , what is the value of in U.S. Customary
base units?
(b)
D
(N-m)m
My
N
D
D 2
I
m4
m
D
2000 N-m0.1 m
My
D
I
7 ð 105 m4
D 59,700
1 lb
4.448 N
0.3048 m
ft
2
lb
ft2
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1
Problem 1.22 Let W be your weight at sea level in
pounds. (a) What is your weight at sea level in newtons?
(b) What is your mass in kilograms?
Solution:
(a)
W (lb)
(b)
mD
4.448 N
lb
D 4.448W N
4.448W N
D 0.453W kg
9.81 m/s2
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1
Problem 1.23 The acceleration due to gravity is
1.62 m/s2 on the surface of the moon and 9.81 m/s2
on the surface of the earth. A female astronaut’s mass
is 57 kg. What is the maximum allowable mass of her
spacesuit and equipment if the engineers don’t want the
total weight on the moon of the woman, her spacesuit
and equipment to exceed 180 N?
Solution: Find the mass which weighs 180 N on the moon.
mD
180 N-s2
w
D
D 111.1 kg
g
1.62 m
This is the total allowable mass. Thus, the suit & equipment can have
mass of
mS/E D 111.1 kg 57 kg D 54.1 kg
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1
Problem 1.24
(a)
(b)
A person has a mass of 50 kg.
The acceleration due to gravity at sea level is g D
9.81 m/s2 . What is the person’s weight at sea level?
The acceleration due to gravity on the moon is g D
1.62 m/s2 . What would the person weigh on the
moon?
Solution: Use Eq (1.6).
(a)
m
We D 50 kg 9.81 2 D 490.5 N D 491 N, and
s
(b)
m
Wmoon D 50 kg 1.62 2 D 81 N.
s
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1
Problem 1.25 The acceleration due to gravity at
sea level is g D 9.81 m/s2 . The radius of the earth
is 6370 km. The universal gravitational constant is
G D 6.67 ð 1011 N-m2 /kg2 . Use this information to
determine the mass of the earth.
Solution: Use Eq (1.3) a D
GmE
. Solve for the mass,
R2
m 2
9.81 m/s2 6370 km2 103
gR2
km
mE D
D
G
N-m2
6.671011 kg2
D 5.9679 . . . 1024 kg D 5.971024 kg
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1
Problem 1.26 A person weighs 180 lb at sea level. The
radius of the earth is 3960 mi. What force is exerted on
the person by the gravitational attraction of the earth if
he is in a space station in orbit 200 mi above the surface
of the earth?
Solution: Use Eq (1.5).
W D mg
RE
r
2
D
WE
g
g
RE
RE C H
2
D WE
3960
3960 C 200
2
D 1800.90616 D 163 lb
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1
Problem 1.27 The acceleration due to gravity on the
surface of the moon is 1.62 m/s2 . The radius of the moon
is RM D 1738 km. Determine the acceleration due to
gravity of the moon at a point 1738 km above its surface.
Strategy: Write an equation equivalent to Eq. (1.4) for
the acceleration due to gravity of the moon.
Use Eq (1.4), rewritten to apply to the Moon. . . a D
Solution:
gM
RM
r
2
a D 1.62 m/s2 RM
RM CRM
2
D 1.62 m/s2 2
1
D 0.405 m/s2
2
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1
Problem 1.28 If an object is near the surface of the
earth, the variation of its weight with distance from the
center of the earth can often be neglected. The acceleration due to gravity at sea level is g D 9.81 m/s2 . The
radius of the earth is 6370 km. The weight of an object
at sea level is mg, where m is its mass. At what height
above the earth does the weight of the object decrease
to 0.99 mg?
Solution: Use a variation of Eq (1.5).
W D mg
RE
RE C h
2
D 0.99 mg
Solve for the radial height,
h D RE
1
p
1 D 63701.0050378 1.0
0.99
D 32.09 . . . km D 32,100 m D 32.1 km
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1
Problem 1.29 The centers of two oranges are 1 m
apart. The mass of each orange is 0.2 kg. What
gravitational force do they exert on each other? (The
universal gravitational constant is G D 6.67 ð 1011 Nm2 /kg2 .)
Solution: Use Eq (1.1) F D
FD
Gm1 m2
. Substitute:
r2
6.671011 0.20.2
D 2.6681012 N
12
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1
Problem 1.30 At a point between the earth and the
moon, the magnitude of the force exerted on an object
by the earth’s gravity equals the magnitude of the force
exerted on the object by the moon’s gravity. What is
the distance from the center of the earth to that point
to three significant digits? The distance from the center
of the earth to the center of the moon is 383,000 km,
and the radius of the earth is 6370 km. The radius of the
moon is 1738 km, and the acceleration due to gravity at
its surface is 1.62 m/s2 .
Solution: Let rEp be the distance from the Earth to the point where
the gravitational accelerations are the same and let rMp be the distance
from the Moon to that point. Then, rEp C rMp D rEM D 383,000 km.
The fact that the gravitational attractions by the Earth and the Moon
at this point are equal leads to the equation
gE
RE
rEp
2
D gM
RM
rMp
2
,
where rEM D 383,000 km. Substituting the correct numerical values
leads to the equation
9.81
m 6370 km 2
s2
rEp
D 1.62
m 1738 km 2
s2
rEM rEp
,
where rEp is the only unknown. Solving, we get rEp D 344,770 km D
345,000 km.
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1
Problem 2.1 The magnitudes jFA j D 60 N and jFB j D
80 N. The angle ˛ D 45° . Graphically determine the
magnitude of the vector FA C FB and the angle between
the vectors FB and FA C FB .
FB
FC
␤
a
FA
Strategy: Construct the parallelogram for determining
the sum of the forces, drawing the lengths of FA
and FB proportional to their magnitudes and accurately
measuring the angle ˛, as we did in Example 2.1. Then
you can measure the magnitude of FA C FB and the angle
between FB and FA C FB .
Solution: Draw the vectors to scale and measure the magnitude
and angle of the resultant
FA
Measuring we find
jFA C FB j D 130 N
˛ D 19°
FB
|F A
|
+ FB
80 N
α
45°
60 N
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1
Problem 2.2 The magnitudes jFA j D 40 N, jFB j D
50 N, and jFA C FB j D 80 N. Assume that 0 < ˛ < 90° .
Graphically determine the angle ˛.
Solution: Draw the vectors to scale and measure the angle ˛ using
a protractor.
˛ D 55°
|=
N
80
N
B
F
B =
50
|F A
+F
α
FA = 40 N
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1
Problem 2.3 The magnitudes jFA j D 40 N, jFB j D
50 N, and jFA C FB j D 80 N. Assume that 0 < ˛ < 90° .
Use trigonometry to determine the angle ˛.
Solution: Use the figure from Problem 2.2. The law of cosines:
802 D 502 C 402 25040 cos180° ˛ ) ˛ D 54.9°
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1
Problem 2.4 The magnitudes jFA j D 40 N, jFB j D
50 N, and jFC j D 40 N. The angles ˛ D 50° and ˇ D
80° . Graphically determine the magnitude of FA C FB C
FC .
Solution: Drawing everything to scale we can measure the magnitude as
R D jFA C FB C FC j D 83 N
=
FC
130°
F
B =
50
R=
|F
A+
F
B+
F
c|
40
50°
FA = 40
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1
Problem 2.5 The magnitudes jFA j D 40 N, jFB j D
50 N, and jFC j D 40 N. The angles ˛ D 50° and ˇ D
80° . Use trigonometry to determine the magnitude of
FA C FB C FC .
Solution: We have
RD
40 C 50 cos 50° C 40 cos 130° 2 C 50 sin 50° C 40 sin 130° 2 N
D 83.1 N
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1
Problem 2.6 If the magnitude of the vector rAC is
195 mm, what is the angle ?
150 mm
60 mm
B
rAB
rBC
C
θ
A
Solution: From the law of cosines
1502 D 602 C 1952 260195 cos ) D 35.2°
m
0m
r AB
rAC
rB
C=1
=6
50 m
m
θ
rAC = 195 mm
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1
Problem 2.7 The vectors FA and FB represent the
forces exerted on the pulley by the belt. Their magnitudes are jFA j D 80 N and jFB j D 60 N. What is the
magnitude jFA C FB j of the total force the belt exerts
on the pulley?
Solution:
FB
FB
+F
FA
45°
FB
35°
45°
B
FA
10°
5°
14
45°
β
35°
F
10° A
Law of cosines
FA
jFA C FB j2 D 802 C 602 28060 cos 145°
10°
jFA C FB j D 133.66 ³ 134 N
Law of sines
jFB j
jFA C FB j
60
133.66
D
)
D
ˇ D 14.92°
sin ˇ
sin 145
sin ˇ
sin 145
∴ jFA C FB j D 134 N
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1
Problem 2.8 The sum of the forces FA C FB C FC D
0. The magnitude jFA j D 100 N and the angle ˛ D 60° .
Determine jFB j and jFC j.
FB
30°
FA
a
FC
Solution: Using the Law of sines twice we find
100 N
FB
FC
D
D
) FB D 86.6 N,
sin 90°
sin 60°
sin 30°
FB
FC D 50 N
30°
FA = 100 N
α = 60°
FC
⇒
FC
90°
FB
60°
30°
100 N
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1
Problem 2.9 In Problem 2.8, the sum of the forces
FA C FB C FC D 0. If the magnitudes jFA j D 100 N and
jFB j D 80 N, what are jFC j and the angle ˛?
Solution: Using the Law of Cosines
F
B=
F2C D 100 N2 C 80 N2 2100 N80 N cos 30° ) FC
80
N
FA = 100 N
30°
D 50.4 N
α
Using the Law of Sines
FC
FC
80 N
D
) ˛ D 52.5°
sin 30°
sin ˛
β
FC
80
α
N
30°
100 N
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1
Problem 2.10 The forces acting on the sailplane are
represented by three vectors. The lift L and drag D are
perpendicular, the magnitude of the weight W is 3500 N,
and W C L C D D 0. What are the magnitudes of the lift
and drag?
L
25°
D
W
Solution: Draw the force triangle and then use the geometry plus
cos 25° D
sin 25° D
jLj
jWj
L
25°
W
jDj
jWj
jWj D 3500 N
65°
25°
D
jLj D 3500 cos 25°
jDj D 3500 sin 25°
jLj D 3170 N
jDj D 1480 N
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1
Problem 2.11 A spherical storage tank is supported by
cables. The tank is subjected to three forces: the forces
FA and FB exerted by the cables and the weight W.
The weight of the tank is jWj D 600 lb. The vector sum
of the forces acting on the tank equals zero. Determine
the magnitudes of FA and FB (a) graphically and (b) by
using trigonometry.
Solution: The vector construction is shown.
(a) The graphical solution is obtained from the construction by the
recognition that since the opposite interior angles of the triangle are
equal, the sides (magnitudes of the forces exerted by the cables) are
equal. A measurement determines the magnitudes. (b) The trigonometric solution is obtained from the law of sines:
jFA j
jFB j
jWj
D
D
sin 140°
sin 20°
sin 20°
Solving:
40°
FB
FA
20°
20°
jFA j D jFB j D jWj
sin 20
sin 140
D 319.25 . . . D 319.3 lb
FB
20
W
20
140
W
20
20
FA
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1
Problem 2.12 The rope ABC exerts forces FBA and
FBC on the block at B. Their magnitudes are equal:
jFBA j D jFBC j. The magnitude of the total force exerted
on the block at B by the rope is jFBA C FBC j D
920 N. Determine jFBA j (a) graphically and (b) by using
trigonometry.
FBC
C
20°
B
B
FBA
A
FBC
Solution: Law of Sines
70°
FBA
920 N
D
) FBA D FBC D 802 N
sin 55°
sin 70°
55°
55°
55°
FBA
55°
70°
|FBA + FBC| = 920 N
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1
Problem 2.13 Two snowcats tow a housing unit to a
new location at McMurdo Base, Antarctica. (The top
view is shown. The cables are horizontal.) The sum of
the forces FA and FB exerted on the unit is parallel to the
line L, and jFA j D 1000 lb. Determine jFB j and jFA C
FB j (a) graphically and (b) by using trigonometry.
L
FA
50°
30°
FB
TOP VIEW
Solution: The graphical construction is shown. The sum of the
interior angles must be 180° . (a) The magnitudes of jFB j and jFA C FB j
are determined from measurements. (b) The trigonometric solution is
obtained from the law of sines:
jFA C FB j D jFA j
sin 50
sin 30
FB
156°
sin 100
sin 30
FA + FB
38°
38°
jFA j
jFB j
jFA C FB j
D
D
sin 100
sin 30
sin 50
from which jFB j D jFA j
FB
38°
D 10001.532
D 1532 lb
50°
50°
FA
D 10001.9696
D 1970 lb
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1
Problem 2.14 A surveyor determines that the horizontal distance from A to B is 400 m and that the horizontal
distance from A to C is 600 m. Determine the magnitude
of the horizontal vector rBC from B to C and the angle
˛ (a) graphically and (b) by using trigonometry.
North
B
α
rBC
C
60°
20°
East
A
Solution: (a) The graphical solution is obtained by drawing the
figure to scale and measuring the unknowns. (b) The trigonometric
solution is obtained by breaking the figure into three separate right
triangles. The magnitude jrBC j is obtained by the cosine law:
B
F
α
C
jrBC j2 D 4002 C 6002 2400600 cos 40°
or jrBC j D 390.25 D 390.3 m
60°
20°
The three right triangles are shown. The distance BD is BD D
400 sin 60° D 346.41 m. The distance CE is CE D 600 sin 20° D
205.2 m. The distance FC is FC D 346.4 205.2 D 141.2 m.
The angle ˛ is sin ˛ D
A
D
E
141.2
D 0.36177 . . ., or ˛ D 21.2°
390.3
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1
Problem 2.15 The vector r extends from point A to
the midpoint between points B and C. Prove that
C
r D 12 rAB C rAC .
rAC
r
rAB
A
B
Solution: The proof is straightforward:
C
r D rAB C rBM , and r D rAC C rCM .
rAC
r
Add the two equations and note that rBM C rCM D 0, since the two
vectors are equal and opposite in direction.
Thus 2r D rAC C rAB , or r D
1
2
rAC C rAB A
M
B
rAB
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1
Problem 2.16
explain why
By drawing sketches of the vectors,
U C V C W D U C V C W.
Solution: Additive associativity for vectors is usually given as an
axiom in the theory of vector algebra, and of course axioms are not
subject to proof. However we can by sketches show that associativity
for vector addition is intuitively reasonable: Given the three vectors to
be added, (a) shows the addition first of V C W, and then the addition
of U. The result is the vector U C V C W.
V
U
V+W
(a)
W
U+[V+W]
V
(b) shows the addition of U C V, and then the addition of W, leading
to the result U C V C W.
The final vector in the two sketches is the same vector, illustrating that
associativity of vector addition is intuitively reasonable.
U
U+V
(b)
W
[U+V]+W
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1
Problem 2.17 A force F D 40 i 20 j N. What is
its magnitude jFj?
p
Solution: jFj D 402 C 202 D 44.7 N
Strategy: The magnitude of a vector in terms of its
components is given by Eq. (2.8).
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1
Problem 2.18 An engineer estimating the components
of a force F D Fx i C Fy j acting on a bridge abutment
has determined that Fx D 130 MN, jFj D 165 MN, and
Fy is negative. What is Fy ?
Solution:
jFj D
Thus jFy j D
jFy j D
jFx j2 C jFy j2
jFj2 jFx j2 mN
p
1652 1302 mN
jFy j D 101.6 mN
Fy D 102 mN
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1
Problem 2.19 A support is subjected to a force F D
Fx i C 80j (N). If the support will safely support a force
of 100 N, what is the allowable range of values of the
component Fx ?
Solution: Use the definition of magnitude in Eq. (2.8) and reduce
algebraically.
100 ½
Fx 2 C 802 , from which 1002 802 ½ Fx 2 .
Thus jFx j p
3600, or 60 Fx C60 (N)
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1
Problem 2.20 If FA D 600i 800j (kip) and FB D
200i 200j (kip), what is the magnitude of the force
F D FA 2FB ?
Solution: Take the scalar multiple of FB , add the components of
the two forces as in Eq. (2.9), and use the definition of the magnitude.
F D 600 2200i C 800 2200j D 200i 400j
jFj D
2002 C 4002 D 447.2 kip
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1
Problem 2.21 If FA D i 4.5j (kN) and FB D 2i 2j (kN), what is the magnitude of the force F D 6FA C
4FB ?
Solution: Take the scalar multiples and add the components.
F D 6 C 42i C 64.5 C 42j D 2i 35j, and
jFj D
22 C 352 D 35.1 kN
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1
Problem 2.22 Two perpendicular vectors U and V lie
in the x-y plane. The vector U D 6i 8j and jVj D 20.
What are the components of V? (Notice that this problem
has two answers.)
Solution: The two possible values of V are shown in the sketch.
The strategy is to (a) determine the unit vector associated with U,
(b) express this vector in terms of an angle, (c) add š90° to this
angle, (d) determine the two unit vectors perpendicular to U, and
(e) calculate the components of the two possible values of V. The
unit vector parallel to U is
eU D 6i
62 C 82
8j
62 C 82
D 0.6i 0.8j
y
V2
6
V1
U
x
8
Expressed in terms of an angle,
eU D i cos ˛ j sin ˛ D i cos53.1° j sin53.1° Add š90° to find the two unit vectors that are perpendicular to this
unit vector:
ep1 D i cos143.1° j sin143.1° D 0.8i 0.6j
ep2 D i cos36.9° j sin36.9° D 0.8i C 0.6j
Take the scalar multiple of these unit vectors to find the two vectors
perpendicular to U.
V1 D jVj0.8i 0.6j D 16i 12j.
The components are Vx D 16, Vy D 12
V2 D jVj0.8i C 0.6j D 16i C 12j.
The components are Vx D 16, Vy D 12
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1
Problem 2.23 A fish exerts a 40-N force on the line
that is represented by the vector F. Express F in terms
of components using the coordinate system shown.
Solution:
Fx D jFj cos 60° D 400.5 D 20 N
Fy D jFj sin 60° D 400.866 D 34.6 N
y
F D 20i 34.6j N
60°
F
x
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1
Problem 2.24 A person exerts a 60-lb force F to push
a crate onto a truck. Express F in terms of components.
Solution: The strategy is to express the force F in terms of the
angle. Thus
F D ijFj cos20° C jjFj sin20° y
F D 600.9397i C 0.342j or F D 56.4i C 20.5j (lb)
F
20°
x
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1
Problem 2.25 The missile’s engine exerts a 260-kN
force F. Express F in terms of components using the
coordinate system shown.
y
F
40°
x
Solution:
Fx D jFj cos 40°
Fx D 199 N
y
F
40°
Fy D jFj sin 40°
Fy D 167 N
F D 199i C 167j N
x
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1
Problem 2.26 For the truss shown, express the position vector rAD from point A to point D in terms of
components. Use your result to determine the distance
from point A to point D.
y
Solution: Coordinates A(1.8, 0.7) m, D(0, 0.4) m
rAD D 0 1.8 mi C 0.4 m 0.7 mj D 1.8i 0.3j m
rAD D 1.8 m2 C 0.3 m2 D 1.825 m
B
A
0.6 m
D
0.7 m
0.4 m
C
0.6 m
x
1.2 m
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1
Problem 2.27 The points A, B, . . . are the joints of
the hexagonal structural element. Let rAB be the position
vector from joint A to joint B, rAC the position vector
from joint A to joint C, and so forth. Determine the
components of the vectors rAC and rAF .
Solution: Use the xy coordinate system shown and find the locations of C and F in those coordinates. The coordinates of the points
in this system are the scalar components of the vectors rAC and rAF .
For rAC , we have
rAC D rAB C rBC D xB xA i C yB yA j
y
C xC xB i C yC yB j
E
D
2m
rAC D 2m 0i C 0 0j C 2m cos 60° 0i
or
C 2m cos 60° 0j,
F
C
giving
rAC D 2m C 2m cos 60° i C 2m sin 60° j. For rAF , we have
A
B
x
rAF D xF xA i C yF yA j
D 2m cos 60° xF 0i C 2m sin 60° 0j.
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1
Problem 2.28 For the hexagonal structural element in
Problem 2.27, determine the components of the vector
rAB rBC .
Solution: rAB rBC .
The angle between BC and the x-axis is 60° .
rBC D 2 cos60° i C 2sin60° j m
rBC D 1i C 1.73j m
rAB rBC D 2i 1i 1.73j m
rAB rBC D 1i 1.73j m
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1
Problem 2.29 The coordinates of point A are (1.8,
3.0) m. The y coordinate of point B is 0.6 m and the
magnitude of the vector rAB is 3.0 m. What are the components of rAB ?
y
A
rAB
B
x
Solution: Let the x-component of point B be xB . The vector from
A to B can be written as
rAB D xB xA i C yB yA j m
or rAB D xB 1.8i C 0.6 3.0j m
rAB D xB 1.8i 2.4j m
We also know jrAB j D 3.0 m. Thus
32 D xB 1.802 C 2.42
Solving, xB D 3.60. Thus
rAB D 1.80i 2.40j m
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1
Problem 2.30 (a) Express the position vector from
point A of the front-end loader to point B in terms of
components.
y
98 in.
45 in.
C
(b) Express the position vector from point B to point C
in terms of components.
A
55 in.
(c) Use the results of (a) and (b) to determine the distance
from point A to point C.
B
50 in.
35 in.
x
50 in.
Solution: The coordinates are A(50, 35); B(98, 50); C(45, 55).
(a)
The vector from point A to B:
rAB D 98 50i C 50 35j D 48i C 15j (in)
(b)
The vector from point B to C is
rBC D 45 98i C 55 50j D 53i C 5j (in).
(c)
The distance from A to C is the magnitude of the sum of the
vectors,
rAC D rAB C rBC D 48 53i C 15 C 5j D 5i C 20j.
The distance from A to C is
jrAC j D
52 C 202 D 20.62 in
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1
Problem 2.31 Five identical cylinders with radius R D
0.2 m are stacked as shown. Determine the components
of the position vectors (a) from point A to point B and
(b) from point B to point E.
y
D
B
R
C
A
E
x
Solution: The coordinates are (R D 0.2 m)
y
A0.2, 0.2 m
D
B
B0.4, 0.2 C 0.4 sin 60° m
E1.0, 0.2 m
Thus
A
C
E
rAB D 0.4 m 0.2 mi C 0.2 m C 0.4 m sin 60° 0.2 mj
(a)
D 0.2i C 0.346j m
x
rBE D 1.0 m 0.4 mi C 0.2 m 0.2 m 0.4 m sin 60° j
(b)
D 0.6i 0.346j m
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1
Problem 2.32 Determine the position vector rAB in
terms of its components if: (a) D 30° , (b) D 225° .
y
150 mm
60 mm
B
rAB
rBC
C
θ
x
A
Solution:
(a)
rAB D 60 cos30° i C 60 sin30° j, or
y
150
mm
60
mm
rAB D 51.96i C 30j mm. And
B
(b)
FAB
rAB D 60 cos225° i C 60 sin225° j or
rAB D 42.4i 42.4j mm.
A
θ
FBC
C
F
x
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1
Problem 2.33 In Problem 2.32 determine the position
vector rBC in terms of its components if: (a) D 30° ,
(b) D 225° .
Solution:
(a)
From Problem 2.32, rAB D 51.96i C 30j mm. Thus, the coordinates of point B are (51.96, 30) mm. The vector rBC is given by
rBC D xC xB i C yC yB j, whereyC D 0. The magnitude of
the vector rBC is 150 mm. Using these facts, we find that yBC D
30 mm, and xBC D 146.97 mm.
(b)
rAB D 60 cos225° i C 60 sin225° j or
rAB D 42.4i 42.4j mm.
From Problem 2.32, rAB D 42.4i 42.4j mm. Thus, the
coordinates of point B are (42.4, 42.4) mm. The vector rBC
is given by rBC D xC xB i C yC yB j, where yC D 0. The
magnitude of the vector rBC is 150 mm. Using these facts, we
find that yBC D 42.4 mm, and xBC D 143.9 mm.
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1
Problem 2.34 A surveyor measures the location of
point A and determines that rOA D 400i C 800j (m). He
wants to determine the location of a point B so that
jrAB j D 400 m and jrOA C rAB j D 1200 m. What are the
cartesian coordinates of point B?
y
B
A
N
rAB
rOA
Proposed
roadway
x
O
Solution: Two possibilities are: The point B lies west of point A,
or point B lies east of point A, as shown. The strategy is to determine
the unknown angles ˛, ˇ, and . The magnitude of OA is
jrOA j D
B
4002 C 8002 D 894.4.
α
The angle ˇ is determined by
tan ˇ D
800
D 2, ˇ D 63.4° .
400
A
B
y
α
θ
β
0
x
The angle ˛ is determined from the cosine law:
cos ˛ D
894.42 C 12002 4002
D 0.9689.
2894.41200
˛ D 14.3° . The angle is D ˇ š ˛ D 49.12° , 77.74° .
The two possible sets of coordinates of point B are
rOB D 1200i cos 77.7 C j sin 77.7 D 254.67i C 1172.66j (m)
rOB D 1200i cos 49.1 C j sin 49.1 D 785.33i C 907.34j (m)
The two possibilities lead to B(254.7 m, 1172.7 m) or B(785.3 m,
907.3 m)
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1
Problem 2.35 The magnitude of the position vector
rBA from point B to point A is 6 m and the magnitude of
the position vector rCA from point C to point A is 4 m.
What are the components of rBA ?
y
Thus
rBA D xA 0i C yA 0j ) 6 m2 D xA 2 C yA 2
rCA D xA 3 mi C yA 0j ) 4 m2 D xA 3 m2 C yA 2
3m
B
Solution: The coordinates are: AxA , yA , B0, 0, C3 m, 0
x
C
Solving these two equations, we find xA D 4.833 m, yA D š3.555 m.
We choose the “-” sign and find
rBA D 4.83i 3.56j m
A
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1
Problem 2.36 In Problem 2.35, determine the components of a unit vector eCA that points from point C toward
point A.
Strategy: Determine the components of rCA and then
divide the vector rCA by its magnitude.
Solution: From the previous problem we have
rCA D 1.83i 3.56j m,
rCA D
1.832 C 3.562 m D 3.56 m
Thus
eCA D
rCA
D 0.458i 0.889j
rCA
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1
Problem 2.37 The x and y coordinates of points A, B,
and C of the sailboat are shown.
(a)
(b)
Determine the components of a unit vector that
is parallel to the forestay AB and points from A
toward B.
Determine the components of a unit vector that
is parallel to the backstay BC and points from C
toward B.
Solution:
rAB D xB xA i C yB yA j
rCB D xB xC i C yC yB j
Points are: A (0, 1.2), B (4, 13) and C (9, 1)
Substituting, we get
rAB D 4i C 11.8j m, jrAB j D 12.46 m
y
B (4, 13) m
rCB D 5i C 12j m, jrCB j D 13 m
The unit vectors are given by
eAB D
rAB
rCB
and eCB D
jrAB j
jrCB j
Substituting, we get
eAB D 0.321i C 0.947j
eCB D 0.385i C 0.923j
A
(0, 1.2) m
C
(9, 1) m
x
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1
Problem 2.38 The length of the bar AB is 0.6 m.
Determine the components of a unit vector eAB that
points from point A toward point B.
y
B
0.4 m
A
0.3 m
x
Solution: We need to find the coordinates of point Bx, y
B
We have the two equations
2
2
x C y D 0.4 m
m
y
m
0.6
0.4
0.3 m C x2 C y 2 D 0.6 m2
2
Solving we find
x D 0.183 m,
y D 0.356 m
A
0.3 m
O
x
Thus
eAB D
rAB
0.183 m [0.3 m]i C 0.356 mj
D rAB
0.183 m C 0.3 m2 C 0.356 m2
D 0.806i C 0.593j
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1
y
Problem 2.39 Determine the components of a unit
vector that is parallel to the hydraulic actuator BC and
points from B toward C.
1m
D
C
1m
0.6 m
B
A
0.15 m
x
0.6 m
Scoop
Solution: Point B is at (0.75, 0) and point C is at (0, 0.6). The
vector
rBC D xC xB i C yC yB j
rBC D 0 0.75i C 0.6 0j m
rBC D 0.75i C 0.6j m
jrBC j D
eBC D
0.752 C 0.62 D 0.960 m
rBC
0.75
0.6
D
iC
j
jrBC j
0.96
0.96
eBC D 0.781i C 0.625j
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1
Problem 2.40 The hydraulic actuator BC in Problem
2.39 exerts a 1.2-kN force F on the joint at C that is
parallel to the actuator and points from B toward C.
Determine the components of F.
Solution: From the solution to Problem 2.39,
eBC D 0.781i C 0.625j
The vector F is given by F D jFjeBC
F D 1.20.781i C 0.625j k Ð N
F D 937i C 750j N
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1
Problem 2.41 A surveyor finds that the length of the
line OA is 1500 m and the length of line OB is 2000 m.
(a)
(b)
y
N
Determine the components of the position vector
from point A to point B.
Determine the components of a unit vector that
points from point A toward point B.
A
Proposed bridge
B
60°
30°
O
River
x
Solution: We need to find the coordinates of points A and B
rOA D 1500 cos 60° i C 1500 sin 60° j
rOA D 750i C 1299j m
Point A is at (750, 1299) (m)
rOB D 2000 cos 30° i C 2000 sin 30° j m
rOB D 1732i C 1000j m
Point B is at (1732, 1000) (m)
(a)
The vector from A to B is
rAB D xB xA i C yB yA j
rAB D 982i 299j m
(b)
The unit vector eAB is
eAB D
982i 299j
rAB
D
jrAB j
1026.6
eAB D 0.957i 0.291j
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1
Problem 2.42 The positions at a given time of the Sun
(S) and the planets Mercury (M), Venus (V), and Earth
(E) are shown. The approximate distance from the Sun
to Mercury is 57 ð 106 km, the distance from the Sun to
Venus is 108 ð 106 km, and the distance from the Sun
to the Earth is 150 ð 106 km. Assume that the Sun and
planets lie in the x y plane. Determine the components of a unit vector that points from the Earth toward
Mercury.
E
y
20°
S
M
x
40°
V
Solution: We need to find rE and rM in the coordinates shown
rE D jrE j sin 20° i C jrE jcos 20° j km
rM D jrM j cos 0° i km
rE D 51.3 ð 106 i C 141 ð 106 j km
rM D 57 ð 106 i km
rEM D xM xE i C yM yE j km
rEM D 108.3 ð 106 i 141 ð 106 j km
jrEM j D 177.8 ð 106 km
eEM D
rEM
D C0.609i 0.793j
jrEM j
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1
Problem 2.43 For the positions described in Problem
2.42, determine the components of a unit vector that
points from Earth toward Venus.
Solution: From the solution to Problem 2.42,
rEV D xV xE i C yV yE j km
rE D 51.3 ð 106 i C 141 ð 106 j km
rEV D 31.4 ð 106 i 210.4 ð 106 j km
The position of Venus is
jrEV j D 212.7 ð 106 km
rV D jrV j cos 40° i jrV j sin 40° j km
eEV D
rEV
jrEV j
rV D 82.7 ð 106 i 69.4 ð 106 j km
eEV D 0.148i 0.989j
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1
Problem 2.44 The rope ABC exerts forces FBA and
FBC on the block at B. Their magnitudes are equal:
jFBA j D jFBC j. The magnitude of the total force exerted
on the block at B by the rope is jFBA C FBC j D 920 N.
Determine jFBA j by expressing the forces FBA and FBC
in terms of components and compare your answer to the
answer of Problem 2.12.
FBC
C
20°
B
B
FBA
A
Solution:
FBC
FBC D Fcos 20° i C sin 20° j
20°
FBA D Fj
FBC C FBA D Fcos 20° i C [sin 20° 1]j
Therefore
920 N2 D F2 cos2 20° C [sin 20° 1]2 ) F D 802 N
FBA
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1
Problem 2.45 The magnitude of the horizontal force
F1 is 5 kN and F1 C F2 C F3 D 0. What are the magnitudes of F2 and F3 ?
y
F3
30˚
F1
45˚
F2
x
Solution: Using components we have
Fx : 5 kN C F2 cos 45° F3 cos 30° D 0
Fy : F2 sin 45° C F3 sin 30° D 0
Solving simultaneously yields:
) F2 D 9.66 kN,
F3 D 13.66 kN
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1
Problem 2.46 Four groups engage in a tug-of-war. The
magnitudes of the forces exerted by groups B, C, and D
are jFB j D 800 lb, jFC j D 1000 lb, jFD j D 900 lb. If the
vector sum of the four forces equals zero, what are the
magnitude of FA and the angle ˛?
y
FB
FC
70°
30°
20°
α
FD
FA
x
Solution: The strategy is to use the angles and magnitudes to
determine the force vector components, to solve for the unknown force
FA and then take its magnitude. The force vectors are
FB D 800i cos 110° C j sin 110° D 273.6i C 751.75j
FC D 1000i cos 30° C j sin 30° D 866i C 500j
FD D 900i cos20° C j sin20° D 845.72i 307.8j
FA D jFA ji cos180 C ˛ C j sin180 C ˛
D jFA ji cos ˛ j sin ˛
The sum vanishes:
FA C FB C FC C FD D i1438.1 jFA j cos ˛
C j944 jFA j sin ˛ D 0
From which FA D 1438.1i C 944j. The magnitude is
jFA j D
14382 C 9442 D 1720 lb
The angle is: tan ˛ D
944
D 0.6565, or ˛ D 33.3°
1438
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1
Problem 2.47 The two vernier engines of the launch
vehicle exert thrusts (forces) that control the vehicle’s
attitude, or angular position. Each engine exerts a 5000lb thrust. At the present instant, the thrusts are in the
directions shown. (a) What is the x component of the
force exerted on the vehicle by the vernier engines?
(b) If the launch vehicle’s main engines exert a 200,000lb thrust parallel to the y axis, what is the y component
of the total force on the launch vehicle?
y
Solution:
(a)
x
Fx : 5000 lb sin 30° 5000 lb sin 15° D 1210 lb
Vernier
engines
Fy : 5000 lb cos 30° C 5000 lb cos 15° C 200,000 lb
(b)
D 209,160 lb
30°
15°
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1
Problem 2.48 The bracket must support the two forces
shown, where jF1 j D jF2 j D 2 kN. An engineer determines that the bracket will safely support a total force
of magnitude 3.5 kN in any direction. Assume that 0 ˛ 90° . What is the safe range of the angle ˛?
F2
α
F1
F2
Solution:
Fx : 2 kN C 2 kN cos ˛ D 2 kN1 C cos ˛
F1
α
Fy : 2 kN sin ˛
β
α
F1 + F2
Thus the total force has a magnitude given by
F D 2 kN
p
1 C cos ˛2 C sin ˛2 D 2 kN 2 C 2 cos ˛ D 3.5 kN
Thus when we are at the limits we have
2 C 2 cos ˛ D
3.5 kN
2 kN
2
D
17
49
) cos ˛ D
) ˛ D 57.9°
16
32
In order to be safe we must have
57.9° ˛ 90°
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1
Problem 2.49 The figure shows three forces acting on
a joint of a structure. The magnitude of Fc is 60 kN, and
FA C FB C FC D 0. What are the magnitudes of FA and
FB ?
y
FC
FB
15°
x
40°
FA
Solution: We need to write each force in terms of its components.
FA
195°
FA D jFA j cos 40i C jFA j sin 40j kN
40°
x
FB D jFB j cos 195° i C jFB j sin 195j kN
FC D jFC j cos 270° i C jFC j sin 270° j kN
FB
Thus FC D 60j kN
Since FA C FB C FC D 0, their components in each direction must also
sum to zero.
270°
FC
FAx C FBx C FCx D 0
FAy C FBy C FCy D 0
Thus,
jFA j cos 40° C jFB j cos 195° C 0 D 0
jFA j sin 40° C jFB j sin 195° 60 kN D 0
Solving for jFA j and jFB j, we get
jFA j D 137 kN, jFB j D 109 kN
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1
Problem 2.50 Four forces act on a beam. The vector
sum of the forces is zero. The magnitudes jFB j D
10 kN and jFC j D 5 kN. Determine the magnitudes of
FA and FD .
FD
30°
FA
FB
FC
Solution: Use the angles and magnitudes to determine the vectors,
and then solve for the unknowns. The vectors are:
FA D jFA ji cos 30° C j sin 30° D 0.866jFA ji C 0.5jFA jj
FB D 0i 10j, FC D 0i C 5j, FD D jFD ji C 0j.
Take the sum of each component in the x- and y-directions:
and
Fx D 0.866jFA j jFD ji D 0
Fy D 0.5jFA j 10 5j D 0.
From the second equation we get jFA j D 10 kN . Using this value in
the first equation, we get jFD j D 8.7 kN
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1
Problem 2.51 Six forces act on a beam that forms part
of a building’s frame. The vector sum of the forces
is zero. The magnitudes jFB j D jFE j D 20 kN, jFC j D
16 kN, and jFD j D 9 kN. Determine the magnitudes of
FA and FG .
FA
70°
FC
40°
50°
40°
FB
Solution: Write each force in terms of its magnitude and direction
FG
FD
FE
y
as
F D jFj cos i C jFj sin j
where is measured counterclockwise from the Cx-axis.
Thus, (all forces in kN)
θ
FA D jFA j cos 110° i C jFA j sin 110° j kN
FB D 20 cos 270° i C 20 sin 270° j kN
x
FC D 16 cos 140° i C 16 sin 140° j kN
FD D 9 cos 40° i C 9 sin 40° j kN
FE D 20 cos 270° i C 20 sin 270° j kN
FG D jFG j cos 50° i C jFG j sin 50° j kN
We know that the x components and y components of the forces must
add separately to zero.
Thus
FAx C FBx C FCx C FDx C FEx C FGx D 0
FAy C FBy C FCy C FDy C FEy C FGy D 0
jFA j cos 110° C 0 12.26 C 6.89 C 0 C jFG j cos 50° D 0
jFA j sin 110° 20 C 10.28 C 5.79 20 C jFG j sin 50° D 0
Solving, we get
jFA j D 13.0 kN
jFG j D 15.3 kN
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1
Problem 2.52 The total weight of the man and parasail
is jWj D 230 lb. The drag force D is perpendicular to
the lift force L. If the vector sum of the three forces is
zero, what are the magnitudes of L and D?
y
20°
Solution: Three forces in equilibrium form a closed triangle. In
this instance it is a right triangle. The law of sines is
jLj
jDj
jWj
D
D
sin 90°
sin 70°
sin 20°
From which:
L
jLj D jWj sin 70° D 2300.9397 D 216.1 lb
jDj D jWj sin 20° D 2300.3420 D 78.66 lb
D
y
20°
L
D
x
W
x
W
D
W
20°
L
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1
Problem 2.53 The three forces acting on the car are
shown. The force T is parallel to the x axis and the
magnitude of the force W is 14 kN. If T C W C N D 0,
what are the magnitudes of the forces T and N?
Solution:
Fx : T N sin 20° D 0
Fy : N cos 20° 14 kN D 0
Solving we find
N D 14.90 N,
T D 5.10 N
20⬚
y
T
W
x
20⬚
N
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1
Problem 2.54 The cables A, B, and C help support a
pillar that forms part of the supports of a structure. The
magnitudes of the forces exerted by the cables are equal:
jFA j D jFB j D jFC j. The magnitude of the vector sum of
the three forces is 200 kN. What is jFA j?
Solution: Use the angles and magnitudes to determine the vector
components, take the sum, and solve for the unknown. The angles
between each cable and the pillar are:
A D tan1
B D tan1
FC
FA
FB
D 33.7° ,
8
D 53.1°
6
C D tan1
4m
6m
12
6
D 63.4° .
6m
A
B
Measure the angles counterclockwise form the x-axis. The force vectors acting along the cables are:
C
FA D jFA ji cos 303.7° C j sin 303.7° D 0.5548jFA ji 0.8319jFA jj
4m
4m
4m
FB D jFB ji cos 323.1° C j sin 323.1° D 0.7997jFB ji 0.6004jFB jj
FC D jFC ji cos 333.4° C j sin 333.4° D 0.8944jFC ji0.4472jFC jj
The sum of the forces are, noting that each is equal in magnitude, is
F D 2.2489jFA ji 1.8795jFA jj.
The magnitude of the sum is given by the problem:
200 D jFA j
2.24892 C 1.87952 D 2.931jFA j,
from which jFA j D 68.24 kN
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1
Problem 2.55 The total force exerted on the top of the
mast B by the sailboat’s forestay AB and backstay BC is
180i 820j (N). What are the magnitudes of the forces
exerted at B by the cables AB and BC ?
Solution: We first identify the forces:
4.0 mi 11.8 mj
FAB D TAB 4.0 m2 C 11.8 m2
FBC D TBC y
B (4, 13) m
5.0 mi 12.0 mj
5.0 m2 C 12.0 m2
Then if we add the force we find
4
5
TAB C p
TBC D 180 N
Fx : p
155.24
169
11.8
12
TAB p
TBC D 820 N
Fy : p
169
155.24
Solving simultaneously yields:
) TAB D 226 N,
A
(0, 1.2) m
C
(9, 1) m
TAC D 657 N
x
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1
Problem 2.56 The structure shown forms part of a
truss designed by an architectural engineer to support
the roof of an orchestra shell. The members AB, AC,
and AD exert forces FAB , FAC , and FAD on the joint A.
The magnitude jFAB j D 4 kN. If the vector sum of the
three forces equals zero, what are the magnitudes of FAC
and FAD ?
y
B
(– 4, 1) m
FAB
FAC
C
x
A
FAD
(4, 2) m
D
(–2, – 3) m
Solution: Determine the unit vectors parallel to each force:
B
eAD D p
2
22 C 32
3
iC p
j D 0.5547i 0.8320j
22 C 32
C
A
eAC D p
eAB D p
4
42 C 12
4
42 C 22
1
iC p
j D 0.9701i C 0.2425j
42 C 12
D
2
iC p
j D 0.89443i C 0.4472j
42 C 22
The forces are FAD D jFAD jeAD , FAC D jFAC jeAC ,
FAB D jFAB jeAB D 3.578i C 1.789j. Since the vector sum of the forces
vanishes, the x- and y-components vanish separately:
Fx D 0.5547jFAD j 0.9701jFAC j C 3.578i D 0, and
Fy D 0.8320jFAD j C 0.2425jFAC j C 1.789j D 0
These simultaneous equations in two unknowns can be solved by any
standard procedure. An HP-28S hand held calculator was used here:
The results: jFAC j D 2.108 kN , jFAD j D 2.764 kN
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1
Problem 2.57
Solution: The unit vector from B to A is the vector from B to A
The distance s D 45 in.
divided by its magnitude. The vector from B to A is given by
(a)
(b)
Determine the unit vector eBA that points from B
toward A.
Use the unit vector you obtained in (a) to determine
the coordinates of the collar C.
y
A
rBA D xA xB i C yA yB j or rBA D 14 75i C 45 12j in.
Hence, vector from B to A is given by rBA D 61i C 33j in. The
magnitude of the vector from B to A is 69.4 in and the unit vector
from B toward A is eBA D 0.880i C 0.476j.
(14, 45) in
C
s
B
(75, 12) in
x
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1
Problem 2.58 In Problem 2.57, determine the x and y
coordinates of the collar C as functions of the distance s.
Solution: The coordinates of the point C are given by
xC D xB C s0.880 and yC D yB C s0.476.
Thus, the coordinates of point C are xC D 75 0.880s in and yC D
12 C 0.476s in. Note from the solution of Problem 2.57 above, 0 s 69.4 in.
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1
Problem 2.59 The position vector r goes from point
A to a point on the straight line between B and C. Its
magnitude is jrj D 6 ft. Express r in terms of scalar
components.
y
B
(7, 9) ft
r
A (3, 5) ft
C
(12, 3) ft
x
Solution: Determine the perpendicular vector to the line BC from
point A, and then use this perpendicular to determine the angular orientation of the vector r. The vectors are
y
B[7,9]
P
rAB D 7 3i C 9 5j D 4i C 4j,
jrAB j D 5.6568
rAC D 12 3i C 3 5j D 9i 2j,
jrAC j D 9.2195
rBC D 12 7i C 3 9j D 5i 6j,
jrBC j D 7.8102
r
A[3,5]
C[12,3]
x
The unit vector parallel to BC is
eBC D
rBC
D 0.6402i 0.7682j D i cos 50.19° j sin 50.19° .
jrBC j
Add š90° to the angle to find the two possible perpendicular vectors:
eAP1 D i cos 140.19° j sin 140.19° , or
eAP2 D i cos 39.8° C j sin 39.8° .
Choose the latter, since it points from A to the line.
Given the triangle defined by vertices A, B, C, then the magnitude of
the perpendicular corresponds to the altitude when the base is the line
2area
. From geometry, the area of
BC. The altitude is given by h D
base
a triangle with known sides is given by
area D
p
ss jrBC js jrAC js jrAB j,
where s is the semiperimeter, s D 12 jrAC j C jrAB j C jrBC j. Substituting values, s D 11.343, and area D 22.0 and the magnitude of the
222
D 5.6333. The angle between the
perpendicular is jrAP j D
7.8102
5.6333
D 20.1° . Thus
vector r and the perpendicular rAP is ˇ D cos1
6
the angle between the vector r and the x-axis is ˛ D 39.8 š 20.1 D
59.1° or 19.7° . The first angle is ruled out because it causes the vector
r to lie above the vector rAB , which is at a 45° angle relative to the
x-axis. Thus:
r D 6i cos 19.7° C j sin 19.7° D 5.65i C 2.02j
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1
Problem 2.60 Let r be the position vector from point
C to the point that is a distance s meters from point
A along the straight line between A and B. Express r
in terms of scalar components. (Your answer will be in
terms of s.)
y
B
(10, 9) m
s
r
A (3, 4) m
C (9, 3) m
x
Solution: Determine the ratio of the parts of the line AB and use
this value to determine r. The vectors are:
Check: An alternate solution: Find the angle of the line AB:
D tan1
rAB D 10 3i C 9 4j D 7i C 5j,
jrAB j D 8.602
rCA D 3 9i C 4 3 D 6i C 1j,
jrCA j D 6.0828
rCB D 10 9i C 9 3j D 1i C 6j,
jrCB j D 6.0828
5
7
D 35.54° .
The components of s,
s D jsji cos C j sin D jsj0.8138i C 0.5812j.
The ratio of the magnitudes of the two parts of the line is
s
jrBP j
DRD
jrPA j
jrBC j s
Since the ratio is a scalar, then rBP D RrPA , from which r rCA D
RrCB r.
RrCB C rCA
Solve for the vector r, r D
. Substitute the values of the
1CR
s
, and reduce algebraically:
vectors, note that R D
8.602 s
The coordinates of point P 3 C 0.8138jsj, 4 C 0.5812jsj. Subtract
coordinates of point C to get
r D 0.8135jsj 6i C 0.5812jsj C 1j .
check .
B[10,9] m
y
P
r D 0.8138s 6i C 0.5813s C 1j (m) :
8
r
C[9,3] m
A[3,4] m
x
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1
Problem 2.61
magnitude?
A vector U D 3i 4j 12k. What is its
Strategy: The magnitude of a vector is given in terms
of its components by Eq. (2.14).
Solution: Use definition given in Eq. (14). The vector magnitude is
jUj D
32 C 42 C 122 D 13
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1
Problem 2.62 The vector e D 13 i C 23 j C ez k is a unit
vector. Determine the component ez . (Notice that there
are two answers.)
Solution:
eD
2
1
i C j C ez k )
3
3
2 2
4
1
2
C
C ez 2 D 1 ) e2 D
3
3
9
Thus
ez D
2
3
or ez D 2
3
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1
Problem 2.63 An engineer determines that an attachment point will be subjected to a force F D 20i C Fy j 45k kN. If the attachment point will safely support a
force of 80-kN magnitude in any direction, what is the
acceptable range of values for Fy ?
y
Solution:
802 ½ Fx2 C Fy2 C F2z
802 ½ 202 C Fy2 C 452
To find limits, use equality.
Fy2LIMIT D 802 202 452
Fy2LIMIT D 3975
F
Fy LIMIT D C63.0, 63.0 kN
jFy LIMIT j 63.0 kN 63.0 kN Fy 63.0 kN
z
x
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1
Problem 2.64 A vector U D Ux i C Uy j C Uz k. Its
magnitude is jUj D 30. Its components are related by
the equations Uy D 2Ux and Uz D 4Uy . Determine the
components. (Notice that there are two answers.)
Solution: Substitute the relations between the components, determine the magnitude, and solve for the unknowns. Thus
U D C3.61i C 23.61j C 423.61k
D 3.61i 7.22j 28.9k
U D Ux i C 2Ux j C 42Ux k D Ux 1i 2j 8k
where Ux can be factored out since it is a scalar. Take the magnitude,
noting that the absolute value of jUx j must be taken:
p
30 D jUx j 12 C 22 C 82 D jUx j8.31.
U D 3.61i C 23.61j
C 423.61k D 3.61i C 7.22j C 28.9k
Solving, we get jUx j D 3.612, or Ux D š3.61. The two possible
vectors are
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1
Problem 2.65 An object is acted upon by two
forces F1 D 20i C 30j 24k (kN) and F2 D 60i C
20j C 40k (kN). What is the magnitude of the total force
acting on the object?
Solution:
F1 D 20i C 30j 24k kN
F2 D 60i C 20j C 40k kN
F D F1 C F2 D 40i C 50j C 16k kN
Thus
FD
40 kN2 C 50 kN2 C 16 kN2 D 66 kN
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1
Problem 2.66 Two vectors U D 3i 2j C 6k and V D
4i C 12j 3k.
(a)
(b)
Determine the magnitudes of U and V.
Determine the magnitude of the vector 3U C 2V.
Solution: The magnitudes:
(a)
jUj D
p
p
32 C 22 C 62 D 7 and jVj D 42 C 122 C 32 D 13
The resultant vector
3U C 2V D 9 C 8i C 6 C 24j C 18 6k
D 17i C 18j C 12k
(b)
The magnitude j3U C 2Vj D
p
172 C 182 C 122 D 27.51
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1
Problem 2.67
(a)
(b)
A vector U D 40i 70j 40k.
What is its magnitude?
What are the angles x , y , and z between U and
the positive coordinate axes?
Strategy: Since you know the components of U, you
can determine the angles x , y , and z from Eqs. (2.15).
Solution: The magnitude:
(a) jUj D
(b)
p
402 C 702 C 402 D 90
The direction cosines:
U D 90
40
70
40
i
j
90
90
90
D 900.4444i 0.7777j 0.4444k
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1
Problem 2.68 A force vector is given in terms of its
components by F D 10i 20j 20k (N).
Solution:
F D 10i 20j 20k N
(a)
(b)
What are the direction cosines of F?
Determine the components of a unit vector e that
has the same direction as F.
FD
10 N2 C 20 N2 C 20 N2 D 30 N
cos x D
(a)
10 N
D 0.333,
30 N
cos z D
(b)
cos y D
20 N
D 0.667,
30 N
20 N
D 0.667
30 N
e D 0.333i 0.667j 0.667k
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1
Problem 2.69 The cable exerts a force F on the hook
at O whose magnitude is 200 N. The angle between the
vector F and the x axis is 40° , and the angle between
the vector F and the y axis is 70° .
(a)
(b)
y
70°
What is the angle between the vector F and the
z axis?
Express F in terms of components.
Strategy: (a) Because you know the angles between
the vector F and the x and y axes, you can use Eq. (2.16)
to determine the angle between F and the z axis.
(Observe from the figure that the angle between F and
the z axis is clearly within the range 0 < z < 180° .) (b)
The components of F can be obtained with Eqs. (2.15).
F
40°
O
x
z
Solution:
(a)
cos 40° 2 C cos 70° 2 C cos z 2 D 1 ) z D 57.0°
F D 200 Ncos 40° i C cos 70° j C cos 57.0° k
(b)
F D 153.2i C 68.4j C 108.8k N
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1
Problem 2.70 A unit vector has direction cosines
cos x D 0.5 and cos y D 0.2. Its z component is positive. Express it in terms of components.
Solution: Use Eq. (2.15) and (2.16). The third direction cosine is
cos z D š
1 0.52 0.22 D C0.8426.
The unit vector is
u D 0.5i C 0.2j C 0.8426k
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1
Problem 2.71 The airplane’s engines exert a total thrust
force T of 200-kN magnitude. The angle between T and
the x axis is 120° , and the angle between T and the y axis
is 130° . The z component of T is positive.
(a)
(b)
What is the angle between T and the z axis?
Express T in terms of components.
l D cos 120° D 0.5, m D cos 130° D 0.6428
from which the z-direction cosine is
n D cosz D š
1 0.52 0.64282 D C0.5804.
Thus the angle between T and the z-axis is
y
y
Solution: The x- and y-direction cosines are
(a)
z D cos1 0.5804 D 54.5° , and the thrust is
T D 2000.5i 0.6428j C 0.5804k, or:
130°
x
x
(b)
T D 100i 128.6j C 116.1k (kN)
120°
T
z
z
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1
Problem 2.72 Determine the components of the position vector rBD from point B to point D. Use your result
to determine the distance from B to D.
Solution: We have the following coordinates: A0, 0, 0,
B5, 0, 3 m, C6, 0, 0 m, D4, 3, 1 m
rBD D 4 m 5 mi C 3 m 0j C 1 m 3 mk
y
D i C 3j 2k m
rBD D 1 m2 C 3 m2 C 2 m2 D 3.74 m
D (4, 3, 1) m
A
C (6, 0, 0) m
x
z
B (5, 0, 3) m
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1
Problem 2.73 What are the direction cosines of the
position vector rBD from point B to point D?
Solution:
cos x D
1 m
D 0.267,
3.74 m
cos z D
cos y D
3m
D 0.802,
3.74 m
2 m
D 0.535
3.74 m
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1
Problem 2.74 Determine the components of the unit
vector eCD that points from point C toward point D.
Solution: We have the following
B5, 0, 3 m, C6, 0, 0 m, D4, 3, 1 m
coordinates:
A0, 0, 0,
rCD D 4 m 6 mi C 3 m 0j C 1 m 0k D 2i C 3j C 1k
rCD D
2 m2 C 3 m2 C 1 m2 D 3.74 m
Thus
eCD D
1
2i C 3j C k m D 0.535i C 0.802j C 0.267k
3.74 m
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1
Problem 2.75 What are the direction cosines of the
unit vector eCD that points from point C toward point D?
Solution: Using Problem 2.74
cos x D 0.535,
cos y D 0.802,
cos z D 0.267
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1
Problem 2.76 The bar CD exerts a force F on the
joint at point D that points from point C toward point
D. Its magnitude is jFj D 40 kN. Express F in terms
of components.
Solution: Using Problem 2.74
F D 40 kNeCD D 40 kN0.535i C 0.802j C 0.267k
F D 21.4i C 32.1j C 10.69k kN
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1
Problem 2.77 Astronauts on the space shuttle use radar
to determine the magnitudes and direction cosines of the
position vectors of two satellites A and B. The vector rA
from the shuttle to satellite A has magnitude 2 km, and
direction cosines cos x D 0.768, cos y D 0.384, cos z D
0.512. The vector rB from the shuttle to satellite B has
magnitude 4 km and direction cosines cos x D 0.743,
cos y D 0.557, cos z D 0.371. What is the distance
between the satellites?
B
rB
x
A
y
rA
z
Solution: The two position vectors are:
rA D 20.768iC 0.384jC 0.512k D 1.536i C 0.768j C 1.024k (km)
rB D 40.743iC 0.557j 0.371k D 2.972i C 2.228j 1.484k (km)
The distance is the magnitude of the difference:
jrA rB j
D
1.5362.9272 C 0.7682.2282 C 1.0241.4842
D 3.24 (km)
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1
y
Problem 2.78 Archaeologists measure a pre-Columbian ceremonial structure and obtain the dimensions
shown. Determine (a) the magnitude and (b) the direction cosines of the position vector from point A to
point B.
10 m
4m
4m
A
10 m
8m
B
b
8m
z
C
x
Solution: The coordinates are A(0, 16, 14), and B(10, 8, 4). The
vector from A to B is
rAB D 10 0i C 8 16j C 4 14k D 10i 8j 10k.
The magnitude is
p
(a)
jrAB j D
(b)
The direction cosines are
102 C 82 C 102 D 16.2 m , and
cos x D
10
D 0.6155,
16.2
cos y D
8
D 0.4938,
16.2
and cos z D
10
D 0.6155 .
16.2
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1
Problem 2.79 Consider the structure described in
Problem 2.78. After returning to the United States,
an archaeologist discovers that a graduate student has
erased the only data file containing the dimension b.
But from recorded GPS data he is able to calculate that
the distance from point B to point C is 16.61 m.
(a)
(b)
What is the distance b?
Determine the direction cosines of the position
vector from B to C.
Solution: We have the coordinates B10 m, 8 m, 4 m,
C10 m C b, 0, 18 m
rBC D 10 m C b 10 mi C 0 8 mj C 18 m 4 mk
rBC D bi C 8 mj C 14 mk
(a)
Now we have
16.61 m2 D b2 C 8 m2 C 14 m2 ) b D 3.99 m
cos x D
(b)
3.99 m
D 0.240,
16.61 m
cos z D
cos y D
8 m
D 0.482,
16.61 m
14 m
D 0.843
16.61 m
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1
Problem 2.80 Observers at A and B use theodolites to
measure the direction from their positions to a rocket
in flight. If the coordinates of the rocket’s position at a
given instant are (4, 4, 2) km, determine the direction
cosines of the vectors rAR and rBR that the observers
would measure at that instant.
y
rAR
rBR
A
x
z
B (5,0,2) km
Solution: The vector rAR is given by
rAR D 4i C 4j C 2k km
and the magnitude of rAR is given by
jrAR j D
42 C 42 C 22 km D 6 km.
The unit vector along AR is given by
uAR D rAR /jrAR j.
Thus, uAR D 0.667i C 0.667j C 0.333k
and the direction cosines are
cos x D 0.667, cos y D 0.667, and cos z D 0.333.
The vector rBR is given by
rBR D xR xB i C yR yB j C zR zB k km
D 4 5i C 4 0j C 2 2k km
and the magnitude of rBR is given by
jrBR j D
12 C 42 C 02 km D 4.12 km.
The unit vector along BR is given by
eBR D rBR /jrBR j.
Thus, uBR D 0.242i C 0.970j C 0k
and the direction cosines are
cos x D 0.242, cos y D 0.970, and cos z D 0.0.
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1
Problem 2.81 In Problem 2.80, suppose that the coordinates of the rocket’s position are unknown. At a given
instant, the person at A determines that the direction
cosines of rAR are cos x D 0.535, cos y D 0.802, and
cos z D 0.267, and the person at B determines that the
direction cosines of rBR are cos x D 0.576, cos y D
0.798, and cos z D 0.177. What are the coordinates of
the rocket’s position at that instant.
Solution: The vector from A to B is given by
rAB D xB xA i C yB yA j C zB zA k or
Similarly, the vector along BR, uBR D 0.576i C 0.798 0.177k.
From the diagram in the problem statement, we see that rAR D
rAB C rBR . Using the unit vectors, the vectors rAR and rBR can be
written as
rAB D 5 0i C 0 0j C 2 0k D 5i C 2k km.
The magnitude of rAB is given by jrAB j D 52 C 22 D 5.39 km.
The unit vector along AB, uAB , is given by
uAB D rAB /jrAB j D 0.928i C 0j C 0.371k km.
The unit vector along the line AR,
uAR D cos x i C cos y j C cos z k D 0.535i C 0.802j C 0.267k.
rAR D 0.535rAR i C 0.802rAR j C 0.267rAR k, and
rBR D 0.576rBR i C 0.798rBR j 0.177rBR k.
Substituting into the vector addition rAR D rAB C rBR and equating
components, we get, in the x direction, 0.535rAR D 0.576rBR , and
in the y direction, 0.802rAR D 0.798rBR . Solving, we get that rAR D
4.489 km. Calculating the components, we get
rAR D rAR eAR D 0.5354.489i C 0.8024.489j C 0.2674.489k.
Hence, the coordinates of the rocket, R, are (2.40, 3.60, 1.20) km.
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1
Problem 2.82* The height of Mount Everest was originally measured by a surveyor in the following way. He
first measured the altitudes of two points and the horizontal distance between them. For example, suppose that
the points A and B are 3000 m above sea level and are
10,000 m apart. He then used a theodolite to measure the
direction cosines of the vector rAP from point A to the
top of the mountain P and the vector rBP from point B to
P. Suppose that the direction cosines of rAP are cos x D
0.5179, cos y D 0.6906, and cos z D 0.5048, and the
direction cosines of rBP are cos x D 0.3743, cos y D
0.7486, and cos z D 0.5472. Using this data, determine
the height of Mount Everest above sea level.
z
P
y
B
x
A
Solution: We have the following coordinates A0, 0, 3000 m,
B10, 000, 0, 3000 m, Px, y, z
Then
rAP D xi C yj C z 3000 mk D rAP 0.5179i C 0.6906j C 0.5048k
rBP D x 10,000 mi C yj C z 3000 mk
D rBP 0.3743i C 0.7486j C 0.5472k
Equating components gives us five equations (one redundant) which
we can solve for the five unknowns.
x D rAP 0.5179
y D rAP 0.6906
z 3000 m D rAP 0.5048
) z D 8848 m
x 10000 m D rBP 0.7486
y D rBP 0.5472
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1
y
Problem 2.83 The distance from point O to point A is
20 ft. The straight line AB is parallel to the y axis, and
point B is in the x-z plane. Express the vector rOA in
terms of scalar components.
A
rOA
Strategy: You can resolve rOA into a vector from O to
B and a vector from B to A. You can then resolve the
vector form O to B into vector components parallel to
the x and z axes. See Example 2.9.
O
x
30°
60°
B
z
Solution: See Example 2.10. The length BA is, from the right
The vector rOA is given by rOA D rOB C rBA , from which
triangle OAB,
jrAB j D jrOA j sin 30° D 200.5 D 10 ft.
rOA D 15i C 10j C 8.66k (ft)
Similarly, the length OB is
A
jrOB j D jrOA j cos 30° D 200.866 D 17.32 ft
The vector rOB can be resolved into components along the axes by the
right triangles OBP and OBQ and the condition that it lies in the x-z
plane.
Hence,
rOA
y
30°
O
x
Q
z
P
60°
B
rOB D jrOB ji cos 30° C j cos 90° C k cos 60° or
rOB D 15i C 0j C 8.66k.
The vector rBA can be resolved into components from the condition
that it is parallel to the y-axis. This vector is
rBA D jrBA ji cos 90° C j cos 0° C k cos 90° D 0i C 10j C 0k.
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1
Problem 2.84 The pole supporting the sign is parallel
to the x axis and is 2 m long. Point A is contained in the
y-z plane. Express the position vector r from the origin
to the end of the pole in terms of components.
y
A
Bedford
Falls
r
Solution: The vector can be written in two ways.
45°
60°
r D 2 mi C yj C zk
O
D rsin 45° i C cos 45° sin 60° j C cos 45° cos 60° k
Equating components we find

2 m D r sin 45°



y D r cos 45° sin 60° ) r D 2.83 m,



z D r cos 45° cos 60°
y D 1.732 m,
x
zD1m
z
Thus the vector is
r D 2.00i C 1.732j C 1.000k m
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1
y
Problem 2.85 The straight line from the head of F to
point A is parallel to the y axis, and point A is contained
in the x-z plane. The x component of F is Fx D 100 N.
(a)
(b)
What is the magnitude of F?
Determine the angles x , y , and z between F and
the positive coordinate axes.
F
Solution: The triangle OpA is a right triangle, since OA lies in
the x-z plane, and Ap is parallel to the y-axis. Thus the magnitudes
are given by the sine law:
x
20°
O
60°
A
jFj
jrOA j
jrAp j
D
D
,
sin 20°
sin 90°
sin 70°
thus jrAp j D jFj0.342 and jrOA j D jFj0.9397. The components of
the two vectors are from the geometry
rOA D jrOA ji cos 30° C j cos 90° C k cos 60° z
(b) x D cos1 0.8138 D 35.5° ,
D jrOA j0.866i C 0j C 0.5k and
rAp D jrAp ji cos 90° C j cos 0° C k cos 90° D jrAp j0i C 1j C 0k
y D cos1 0.342 D 70° and z D cos1 0.4699 D 62°
Noting F D rOA C rAp , then from above
y
F D jFj0.34200i C 1j C 0k C jFj0.93970.866i C 0j C 0.5k
P
F
F D jFj0.8138i C 0.342j C 0.4699k
The x-component is given to be 100 N. Thus,
(a) jFj D
100
D 122.9 N The angles are given by
0.8138
O
q
z
20°
x
60°
A
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1
Problem 2.86 The position of a point P on the surface
of the earth is specified by the longitude , measured
from the point G on the equator directly south of Greenwich, England, and the latitude L measured from the
equator. Longitude is given as west (W) longitude or east
(E) longitude, indicating whether the angle is measured
west or east from point G. Latitude is given as north (N)
latitude or south (S) latitude, indicating whether the angle
is measured north or south from the equator. Suppose
that P is at longitude 30° W and latitude 45° N. Let RE
be the radius of the earth. Using the coordinate system
shown, determine the components of the position vector
of P relative to the center of the earth. (Your answer will
be in terms of RE .)
y
N
P
L
z
O
λ
G
Equator
x
Solution: Drop a vertical line from point P to the equatorial plane.
Let the intercept be B (see figure). The vector position of P is the sum
of the two vectors: P D rOB C rBP . The vector rOB D jrOB ji cos C
0j C k sin . From geometry, the magnitude is jrOB j D RE cos .
The vector rBP D jrBP j0i C 1j C 0k. From geometry, the magnitude
is jrBP j D RE sin P . Substitute: P D rOB C rBP D RE i cos cos C
j sin C k sin cos . Substitute from the problem statement: D
C30° , D 45° . Hence P D RE 0.6124i C 0.707j C 0.3536k
y
P
z
O
θ
B
λ
x
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1
Problem 2.87 An engineer calculates that the magnitude of the axial force in one of the beams of a geodesic
dome is jP D 7.65 kN. The cartesian coordinates of the
endpoints A and B of the straight beam are (12.4,
22.0, 18.4) m and (9.2, 24.4, 15.6) m, respectively.
Express the force P in terms of scalar components.
Solution: The components of the position vector from B to A are
rBA D xA xB i C yA yB j C zA zB k
D 12.4 C 9.2i C 22.0 24.4j
C 18.4 C 15.6k
B
D 3.2i 2.4j 2.8k m.
Dividing this vector by its magnitude, we obtain a unit vector that
points from B toward A:
P
A
eBA D 0.655i 0.492j 0.573k.
Therefore
P D jPjeBA
D 7.65 eBA
D 5.01i 3.76j 4.39k kN.
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1
y
Problem 2.88 The cable BC exerts an 8-kN force F
on the bar AB at B.
(a)
(b)
B (5, 6, 1) m
Determine the components of a unit vector that
points from B toward point C.
Express F in terms of components.
F
A
x
C (3, 0, 4) m
z
Solution:
(a)
eBC D
xC xB i C yC yB j C zC zB k
rBC
D jrBC j
xC xB 2 C yC yB 2 C zC zB 2
6
3
2i 6j C 3k
2
eBC D p
D i jC k
7
7
7
22 C 62 C 32
eBC D 0.286i 0.857j C 0.429k
(b)
F
D jFjeBC D 8eBC D 2.29i 6.86j C 3.43k kN
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1
Problem 2.89 A cable extends from point C to point
E. It exerts a 50-lb force T on plate C that is directed
along the line from C to E. Express T in terms of scalar
components.
y
6 ft
A
E
D
z
4 ft
T
2 ft
B
20°
x
C
4 ft
y
Solution: Find the unit vector eCE and multiply it times the magnitude of the force to get the vector in component form,
eCE D
6 ft
rCE
xE xC i C yE yC j C zE zC k
D jrCE j
xE xC 2 C yE yC 2 C zE zC 2
A
E
The coordinates of point C are 4, 4 sin 20° , 4 cos 20° or
4, 1.37, 3.76 (ft) The coordinates of point E are (0, 2, 6) (ft)
D
T
2 ft
0 4i C 2 1.37j C 6 3.76k
p
eCE D
42 C 3.372 C 2.242
x
4 ft
T
z
B
20°
C
4 ft
eCE D 0.703i C 0.592j C 0.394k
T D 50eCE lb
T D 35.2i C 29.6j C 19.7k lb
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1
Problem 2.90 What are the direction cosines of the
force T in Problem 2.89?
Solution: From the solution to Problem 2.89,
eCE D 0.703i C 0.592j C 0.394k
However
eCE D cos x i C cos y j C cos z k
Hence,
cos x D 0.703
cos y D 0.592
cos z D 0.394
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1
y
Problem 2.91 The cable AB exerts a 200-lb force FAB
at point A that is directed along the line from A to B.
Express FAB in terms of scalar components.
8 ft
C
8 ft
6 ft
B
x
FAB
Solution: The coordinates of B are B(0,6,8). The position vector
from A to B is
z
FAC
A (6, 0, 10) ft
rAB D 0 6i C 6 0j C 8 10k D 6i C 6j 2k
The magnitude is jrAB j D
p
62 C 62 C 22 D 8.718 ft.
The unit vector is
uAB D
6
2
6
iC
j
k
8.718
8.718
8.718
or
uAB D 0.6882i C 0.6882j 0.2294k.
FAB D jFAB juAB D 2000.6882i C 0.6882j 0.2294k
The components of the force are
FAB D jFAB juAB D 2000.6882i C 0.6882j 0.2294k or
FAB D 137.6i C 137.6j 45.9k
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1
Problem 2.92 Consider the cables and wall described
in Problem 2.91. Cable AB exerts a 200-lb force FAB
at point A that is directed along the line from A to B.
The cable AC exerts a 100-lb force FAC at point A that
is directed along the line from A to C. Determine the
magnitude of the total force exerted at point A by the
two cables.
Solution: Refer to the figure in Problem 2.91. From Problem 2.91
the force FAB is
The force is
FAC D jFAC juAC D 100uAC D 16.9i C 50.7j 84.5k.
FAB D 137.6i C 137.6j 45.9k
The resultant of the two forces is
The coordinates of C are C(8,6,0). The position vector from A to C is
FR D FAB C FAC D 137.6 C 16.9i C 137.6 C 50.7j
rAC D 8 6i C 6 0j C 0 10k D 2i C 6j 10k.
The magnitude is jrAC j D
The unit vector is
uAC D
p
C 84.5 45.9k.
22 C 62 C 102 D 11.83 ft.
6
10
2
iC
j
k D 0.1691i C 0.5072j 0.8453k.
11.83
11.83
11.83
FR D 120.7i C 188.3j 130.4k.
The magnitude is
p
jFR j D
120.72 C 188.32 C 130.42 D 258.9 lb
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1
Problem 2.93 The 70-m-tall tower is supported by
three cables that exert forces FAB , FAC , and FAD on it.
The magnitude of each force is 2 kN. Express the total
force exerted on the tower by the three cables in terms
of scalar components.
A
y
FAD
A
FAB
FAC
D
60 m
60 m
B
x
40 m
C
40 m
40 m
z
Solution: The coordinates of the points are A (0, 70, 0), B (40, 0,
0), C (40, 0, 40) D (60, 0, 60).
The position vectors corresponding to the cables are:
rAD D 60 0i C 0 70j C 60 0k
rAD D 60i 70k 60k
rAC D 40 0i C 0 70j C 40 0k
rAC D 40i 70j C 40k
rAB D 40 0i C 0 70j C 0 0k
rAB D 40i 70j C 0k
The unit vectors corresponding to these position vectors are:
uAD D
60
70
60
rAD
D
i
j
k
jrAD j
110
110
110
D 0.5455i 0.6364j 0.5455k
uAC D
40
70
40
rAC
D i
jC
k
jrAC j
90
90
90
D 0.4444i 0.7778j C 0.4444k
uAB D
40
70
rAB
D
i
j C 0k D 0.4963i 0.8685j C 0k
jrAB j
80.6
80.6
The forces are:
FAB D jFAB juAB D 0.9926i 1.737j C 0k
FAC D jFAC juAC D 0.8888i 1.5556j C 0.8888
FAD D jFAD juAD D 1.0910i 1.2728j 1.0910k
The resultant force exerted on the tower by the cables is:
FR D FAB C FAC C FAD D 0.9872i 4.5654j 0.2022k kN
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1
Problem 2.94 Consider the tower described in Problem 2.93. The magnitude of the force FAB is 2 kN. The
x and z components of the vector sum of the forces
exerted on the tower by the three cables are zero. What
are the magnitudes of FAC and FAD ?
Solution: From the solution of Problem 2.93, the unit vectors are:
uAC D
40
70
40
rAC
D i
jC
k
jrAC j
90
90
90
D 0.4444i 0.7778j C 0.4444k
uAD D
60
70
60
rAD
D
i
j
jrAD j
110
110
110
Taking the sum of the forces:
FR D FAB C FAC C FAD D 0.9926 0.4444jFAC j 0.5455jFAD ji
C 1.737 0.7778jFAC j 0.6364jFAD jj
C 0.4444jFAC j 0.5455jFAD jk
The sum of the x- and z-components vanishes, hence the set of simultaneous equations:
D 0.5455i 0.6364j 0.5455k
From the solution of Problem 2.93 the force FAB is
FAB D jFAB juAB D 0.9926i 1.737j C 0k
The forces FAC and FAD are:
FAC D jFAC juAC D jFAC j0.4444i 0.7778j C 0.4444k
FAD D jFAD juAD D jFAD j0.5455i 0.6364j 0.5455k
0.4444jFAC j C 0.5455jFAD j D 0.9926 and
0.4444jFAC j 0.5455jFAD j D 0
These can be solved by means of standard algorithms, or by the use of
commercial packages such as TK Solver Plus  or Mathcad. Here
a hand held calculator was used to obtain the solution:
jFAC j D 1.1168 kN
jFAD j D 0.9098 kN
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1
Problem 2.95 Express the position vector from point
O to the collar at A in terms of scalar components.
y
T
Solution: The vector from O to A can be expressed as the sum of
the vectors rOT from O to the top of the slider bar, and rTA from the
top of the slider bar to A. The coordinates of the top and base of the
slider bar are: T (0, 7, 0), B (4, 0, 4). The position vector of the top
of the bar is: rOT D 0i C 7j C 0k. The position vector from the top of
the bar to the base is:
6 ft
7 ft
A
rTB D 4 0i C 0 7j C 4 0k. or
x
O
rTB D 4i 7j C 4k. The unit vector pointing from the top of the bar
to the base is
uTB D
rTB
4
7
4
D i j C k D 0.4444i 0.7778j C 0.4444k.
jrTB j
9
9
9
4 ft
z
4 ft
B
The collar position is
rTA D jrTA juTB D 60.4444i 0.7778j C 0.4444k
D 2.6667i 4.6667j C 2.6667,
measured along the bar. The sum of the two vectors is the position
vector of A from origin O:
rOA D 2.6667 C 0i C 4.6667 C 7j C 2.6667 C 0k
D 2.67i C 2.33j C 2.67k ft
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1
y
Problem 2.96 The cable AB exerts a 32-lb force T on
the collar at A. Express T in terms of scalar components.
4 ft
B
T
6 ft
Solution: The coordinates of point B are B (0, 7, 4). The vector
position of B is rOB D 0i C 7j C 4k.
7 ft
A
x
The vector from point A to point B is given by
4 ft
rAB D rOB rOA .
From Problem 2.95, rOA D 2.67i C 2.33j C 2.67k. Thus
4 ft
z
rAB D 0 2.67i C 7 2.33j C 4 2.67j
rAB D 2.67i C 4.67j C 1.33k.
The magnitude is
p
jrAB j D
2.672 C 4.672 C 1.332 D 5.54 ft.
The unit vector pointing from A to B is
uAB D
rAB
D 0.4819i C 0.8429j C 0.2401k
jrAB j
The force T is given by
TAB D jTAB juAB D 32uAB D 15.4i C 27.0j C 7.7k (lb)
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1
Problem 2.97 The circular bar has a 4-m radius and
lies in the x-y plane. Express the position vector from
point B to the collar at A in terms of scalar components.
y
Solution: From the figure, the point B is at (0, 4, 3) m. The coordinates of point A are determined by the radius of the circular bar
and the angle shown in the figure. The vector from the origin to A
is rOA D 4 cos20° i C 4 sin20° j m. Thus, the coordinates of point A
are (3.76, 1.37, 0) m. The vector from B to A is given by rBA D xA xB i C yA yB j C zA zB k D 3.76i 2.63j 3k m. Finally, the
scalar components of the vector from B to A are (3.76, 2.63, 3) m.
3m
B
A
4m
20°
4m
x
z
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1
Problem 2.98 The cable AB in Problem 2.97 exerts a
60-N force T on the collar at A that is directed along
the line from A toward B. Express T in terms of scalar
components.
Solution: We know rBA D 3.76i 2.63j 3k m from Problem
2.97. The unit vector uAB D rBA /jrBA j. The unit vector is uAB D
0.686i C 0.480j C 0.547k. Hence, the force vector T is given by
T D jTj0.686iC 0.480jC 0.547k N D 41.1i C 28.8j C 32.8k N
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1
Problem 2.99 Determine the dot product U Ð V of the
vectors U D 4i C 6j 10k and V D 8i C 12j C 2k.
Solution:
U D 4i C 6j 10k,
Strategy: The vectors are expressed in terms of their
components, so you can use Eq. (2.23) to determine their
dot product.
V D 8i C 12j C 2k
U Ð V D 48 C 612 C 102 D 20
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1
Problem 2.100 Determine the dot product U Ð V of the
vectors U D 40i C 20j C 60k and V D 30i C 15k.
Solution: Use Eq. 2.23.
U Ð V D 4030 C 200 C 1560 D 300
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1
Problem 2.101 What is the dot product of the position
vector r D 10i C 25j (m) and the force vector
Solution: Use Eq. (2.23).
F Ð r D 30010 C 25025 C 3000 D 3250 N-m
F D 300i C 250j C 300k N?
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1
Problem 2.102 Suppose that the dot product of two
vectors U and V is U Ð V D 0. If jUj 6D 0, what do you
know about the vector V?
Solution:
Either jVj D 0 or V ? U
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1
Problem 2.103 Two perpendicular vectors are given
in terms of their components by
U D Ux i 4j C 6k
and V D 3i C 2j 3k.
Use the dot product to determine the component Ux .
Solution: When the vectors are perpendicular, U Ð V 0.
Thus
U Ð V D Ux Vx C Uy Vy C Uz Vz D 0
D 3Ux C 42 C 63 D 0
3Ux D 26
Ux D 8.67
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1
Problem 2.104
Three vectors
U D Ux i C 3j C 2k
V D 3i C Vy j C 3k
W D 2i C 4j C Wz k
are mutually perpendicular. Use the dot product to determine the components Ux , Vy , and Wz
Solution: For mutually perpendicular vectors, we have three
equations, i.e.,
UÐVD0
UÐWD0
VÐWD0
Thus

3Ux C 3Vy C 6 D 0

3 Eqns
2Ux C 12 C 2Wz D 0

 3 Unknowns
C6 C 4Vy C 3Wz D 0
Solving, we get
Ux
Vy
Wz
D 2.857
D 0.857
D 3.143
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1
Problem 2.105
20.
(a)
(b)
The magnitudes jUj D 10 and jVj D
y
V
Use the definition of the dot product to determine
U Ð V.
Use Eq. (2.23) to obtain U Ð V.
U
45°
30°
x
Solution:
(a)
The definition of the dot product (Eq. (2.18)) is
U Ð V D jUjjVj cos . Thus
U Ð V D 1020 cos45° 30° D 193.2
(b)
The components of U and V are
U D 10i cos 45° C j sin 45° D 7.07i C 7.07j
V D 20i cos 30° C j sin 30° D 17.32i C 10j
From Eq. (2.23)
U Ð V D 7.0717.32 C 7.0710 D 193.2
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1
Problem 2.106 By evaluating the dot product U Ð V,
prove the identity cos1 2 D cos 1 cos 2 C sin 1
sin 2 .
Strategy: Evaluate the dot product both by using the
definition and by using Eq. (2.23).
y
U
V
θ1
θ2
x
Solution: The strategy is to use the definition Eq. (2.18) and the
Eq. (2.23). From Eq. (2.18) and the figure,
U Ð V D jUjjVj cos1 2 . From Eq. (2.23) and the figure,
U D jUji cos 1 C j sin 2 , V D jVji cos 2 C j sin 2 ,
and the dot product is U Ð V D jUjjVjcos 1 cos 2 C sin 1 sin 2 .
Equating the two results:
U Ð V D jUjjVj cos1 2 D jUjjVjcos 1 cos 2 C sin 1 sin 2 ,
from which if jUj 6D 0 and jVj 6D 0, it follows that
cos1 2 D cos 1 cos 2 C sin 1 sin 2 ,
Q.E.D.
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1
Problem 2.107 Use the dot product to determine the
angle between the forestay (cable AB) and the backstay
(cable BC).
y
B (4, 13) m
A
(0, 1.2) m
C
(9, 1) m
x
Solution: The unit vector from B to A is
eBA D
rBA
D 0.321i 0.947j
jrBA j
The unit vector from B to C is
eBC D
rBC
D 0.385i 0.923j
jrBC j
From the definition of the dot product, eBA Ð eBC D 1 Ð 1 Ð cos , where
is the angle between BA and BC. Thus
cos D 0.3210.385 C 0.9470.923
cos D 0.750
D 41.3°
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1
y
Problem 2.108 Determine the angle between the
lines AB and AC (a) by using the law of cosines (see
Appendix A); (b) by using the dot product.
B
(4, 3, ⫺1) m
Solution:
(a)
We have the distances:
AB D
AC D
BC D
42 C 32 C 12 m D
52 C 12 C 32 m D
A
p
26 m
x
u
p
35 m
z
5 42 C 1 32 C 3 C 12 m D
p
33 m
(5, ⫺1, 3) m
C
The law of cosines gives
BC2 D AB2 C AC2 2ABAC cos cos D
(b)
AB2 C AC2 BC2
D 0.464
2ABAC
) D 62.3°
Using the dot product
rAB D 4i C 3j k m,
rAC D 5i j C 3k m
rAB Ð rAC D 4 m5 m C 3 m1 m C 1 m3 m D 14 m2
rAB Ð rAC D ABAC cos Therefore
cos D p
14 m2
p
D 0.464 ) D 62.3°
26 m 35 m
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1
Problem 2.109 The ship O measures the positions of
the ship A and the airplane B and obtains the coordinates
shown. What is the angle between the lines of sight
OA and OB?
y
B
(4, 4, –4) km
θ
x
O
A
z
(6, 0, 3) km
Solution: From the coordinates, the position vectors are:
rOA D 6i C 0j C 3k and rOB D 4i C 4j 4k
The dot product: rOA Ð rOB D 64 C 04 C 34 D 12
p
The magnitudes: jrOA j D
p
jrOA j D
62 C 02 C 32 D 6.71 km and
42 C 42 C 42 D 6.93 km.
rOA Ð rOB
D 0.2581, from which D š75° .
jrOA jjrOB j
From the problem and the construction, only the positive angle makes
sense, hence D 75°
From Eq. (2.24) cos D
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1
Problem 2.110 Astronauts on the space shuttle use
radar to determine the magnitudes and direction cosines
of the position vectors of two satellites A and B. The
vector rA from the shuttle to satellite A has magnitude 2 km and direction cosines cos x D 0.768, cos y D
0.384, cos z D 0.512. The vector rB from the shuttle to
satellite B has magnitude 4 km and direction cosines
cos x D 0.743, cos y D 0.557, cos z D 0.371. What
is the angle between the vectors rA and rB ?
B
rB
x
θ
A
y
rA
z
Solution: The direction cosines of the vectors along rA and rB
are the components of the unit vectors in these directions (i.e., uA D
cos x i C cos y j C cos z k, where the direction cosines are those for
rA ). Thus, through the definition of the dot product, we can find an
expression for the cosine of the angle between rA and rB .
cos D cos xA cos xB C cos yA cos yB C cos zA cos zB .
Evaluation of the relation yields
cos D 0.594 ) D 53.5° .
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1
y
Problem 2.111 The cable BC exerts an 800-N force
F on the bar AB at B. Use Eq. (2.26) to determine the
vector component of F parallel to the bar.
B (5, 6, 1) m
F
Solution: Eqn. 2.26 is UP D e Ð Ue where U is the vector for
which you want the component parallel to the direction indicated by
the unit vector e.
A
x
For the problem at hand, we must find two unit vectors. We need eBC
to be able to write the force FF D jFjeBC and eBA ¾ the direction
parallel to the bar.
eBC D
xC xB i C yC yB j C zC zB k
rBC
D jrBC j
xC xB 2 C yC yB 2 C zC zB 2
C (3, 0, 4) m
z
FP D F Ð eBA eBA
3 5i C 0 6j C 4 1k
p
eBC D
22 C 62 C 32
6
3
2
eBC D i j C k
7
7
7
FP D 624.1eBA
FP D 396.3i 475.6j 79.3k N
Similarly
5i 6j 1k
eBA D p
52 C 62 C 12
eBA D 0.635i 0.762j 0.127k
Now F D jFjeBC D 800 eBC
F D 228.6i 685.7j C 342.9k N
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1
Problem 2.112 The force F D 21i C 14j (kN). Resolve it into vector components parallel and normal to the
line OA.
y
F
O
Solution: The position vector of point A is
x
rA D 6i 2j C 3k
p
The magnitude is jrA j D 62 C 22 C 32 D 7. The unit vector parallel
rA
6
2
3
to OA is eOA D
D i jC k
jrA j
7
7
7
(a)
z
A (6, – 2, 3) m
The component of F parallel to OA is
F Ð eOA eOA D 36 C 22
1
7
6i 2j C 3k
FP D 12i 4j C 6k (kN)
(b)
The component of F normal to OA is
FN D F Fp D 21 12i C 14 4j C 0 6k
D 9i C 18j 6k (kN)
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1
Problem 2.113 At the instant shown, the Harrier’s
thrust vector is T D 17,000i C 68,000j 8,000k (N)
and its velocity vector is v D 7.3i C 1.8j 0.6k (m/s).
The quantity P D jTp jjvj, where Tp is the vector
component of T parallel to v, is the power currently
being transferred to the airplane by its engine. Determine
the value of P.
y
v
T
x
Solution:
T D 17,000i C 68,000j 8,000k N
v D 7.3i C 1.8j 0.6k m/s
Power D T Ð v D 17,000 N7.3 m/s C 68,000 N1.8 m/s
C 8,000 N0.6 m/s
Power D 251,000 Nm/s D 251 kW
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1
Problem 2.114 Cables extend from A to B and from
A to C. The cable AC exerts a 1000-lb force F at A.
(a)
(b)
y
A
What is the angle between the cables AB and AC?
Determine the vector component of F parallel to
the cable AB.
(0, 7, 0) ft
F
x
Solution: Use Eq. (2.24) to solve.
(a)
From the coordinates of the points, the position vectors are:
rAB D 0 0i C 0 7j C 10 0k
B
(0, 0, 10) ft
z
C
(14, 0, 14) ft
rAB D 0i 7j C 10k
rAC D 14 0i C 0 7j C 14 0k
rAC D 14i 7j C 14k
The magnitudes are:
jrAB j D
p
72 C 102 D 12.2 (ft) and
jrAB j D
p
142 C 72 C 142 D 21.
The dot product is given by
rAB Ð rAC D 140 C 77 C 1014 D 189.
The angle is given by
cos D
189
D 0.7377,
12.221
from which D š42.5° . From the construction: D C42.5°
(b)
The unit vector associated with AB is
eAB D
rAB
D 0i 0.5738j C 0.8197k.
jrAB j
The unit vector associated with AC is
eAC D jrrAC
D 0.6667i 0.3333j C 0.6667k.
AC j
Thus the force vector along AC is
FAC D jFjeAC D 666.7i 333.3j C 666.7k.
The component of this force parallel to AB is
FAC Ð eAB eAB D 737.5eAB D 0i 423.2j C 604.5k (lb)
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1
Problem 2.115 Consider the cables AB and AC shown
in Problem 2.114. Let rAB be the position vector from
point A to point B. Determine the vector component of
rAB parallel to the cable AC.
Solution: From Problem 2.114, rAB D 0i 7j C 10k, and eAC D
0.6667i 0.3333j C 0.6667k. Thus rAB Ð eAC D 9, and rAB Ð eAC eAC
D 6i 3j C 6k
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1
Problem 2.116 The force F D 10i C 12j 6k N.
Determine the vector components of F parallel and normal to line OA.
y
A
(0, 6, 4) m
Solution:
F
O
rOA
Find eOA D
jrOA j
x
Then
z
FP D F Ð eOA eOA
and FN D F FP
eOA D
0i C 6j C 4k
6j C 4k
p
D p
52
62 C 42
eOA D
4
6
jC
k D 0.832j C 0.555k
7.21
7.21
FP D [10i C 12j 6k Ð 0.832j C 0.555k]eOA
FP D [6.656]eOA D 0i C 5.54j C 3.69k N
FN D F FP
FN D 10i C 12 5.54j C 6 3.69k
FN D 10i C 6.46j 9.69k N
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1
y
Problem 2.117 The rope AB exerts a 50-N force T on
collar A. Determine the vector component of T parallel
to bar CD.
0.15 m
0.4 m
Solution: The vector from C to D is rCD D xD xC i C yD B
C
yC j C zD zC k. The magnitude of the vector
jrCD j D
T
xD xC 2 C yD yC 2 C zD zC 2 .
0.2 m
0.3 m
A
The components of the unit vector along CD are given by uCDx D
xD xC /jrCD j, uCDy D yD yC /jrCD j, etc. Numerical values are
jrCD j D 0.439 m, uCDx D 0.456, uCDy D 0.684, and uCDz D 0.570.
The coordinates of point A are given by xA D xC C jrCA jeCDx , yA D
yC C jrCA juCDy , etc. The coordinates of point A are (0.309, 0.163,
0.114) m. The vector from A to B and the corresponding unit vector
are found in the same manner as from C to D above. The results are
jrAB j D 0.458 m, uABx D 0.674, uABy D 0.735, and uABz D 0.079.
The force T is given by T D jTjuAB . The result is T D 33.7i C
36.7j C 3.93k N.
0.5 m
O
x
D
0.25 m
0.2 m
z
The component of T parallel to CD is given
Tparallel D T ž uCD D 7.52 N.
The negative sign means that the component of T parallel to CD points
from D toward C (opposite to the direction of the unit vector from C
to D).
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1
Problem 2.118 In Problem 2.117, determine the vector
component of T normal to the bar CD.
Solution: From the solution of Problem 2.117, jTj D 50 N, and
the component of T parallel to bar CD is Tparallel D 7.52 N. The
component of T normal to bar CD is given by
Tnormal D
jTj2 Tparallel 2 D 49.4 N.
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1
y
Problem 2.119 The disk A is at the midpoint of the
sloped surface. The string from A to B exerts a 0.2lb force F on the disk. If you resolve F into vector
components parallel and normal to the sloped surface,
what is the component normal to the surface?
B
(0, 6, 0) ft
F
2 ft
A
x
8 ft
10 ft
z
2
Solution: Consider a line on the sloped surface from A perpendicular to the surface. (see the diagram above) By SIMILAR triangles we
see that one such vector is rN D 8j C 2k. Let us find the component
of F parallel to this line.
The unit vector in the direction normal to the surface is
eN D
y
8
8j C 2k
rN
D p
D 0.970j C 0.243k
jrN j
82 C 22
2
The unit vector eAB can be found by
z
xB xA i C yB yA j C zB zA h
eAB D xB xA 2 C yB yA 2 C zB zA 2
8
Point B is at (0, 6, 0) (ft) and A is at (5, 1, 4) (ft).
Substituting, we get
eAB D 0.615i C 0.615j 0.492k
Now F D jFjeAB D 0.2eAB
F D 0.123i C 0.123j 0.0984k lb
The component of F normal to the surface is the component parallel
to the unit vector eN .
FNORMAL D F Ð eN eN D 0.955eN
FNORMAL D 0i C 0.0927j C 0.0232k lb
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1
Problem 2.120 In Problem 2.119, what is the vector
component of F parallel to the surface?
Solution: From the solution to Problem 2.119,
Thus
F D 0.123i C 0.123j 0.0984k lb and
Fparallel D F FNORMAL
FNORMAL D 0i C 0.0927j C 0.0232k lb
Substituting, we get
The component parallel to the surface and the component normal to
the surface add to give FF D FNORMAL C Fparallel .
Fparallel D 0.1231i C 0.0304j 0.1216k lb
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1
Problem 2.121 An astronaut in a maneuvering unit
approaches a space station. At the present instant, the
station informs him that his position relative to the origin
of the station’s coordinate system is rG D 50i C 80j C
180k (m) and his velocity is v D 2.2j 3.6k (m/s).
The position of the airlock is rA D 12i C 20k (m).
Determine the angle between his velocity vector and the
line from his position to the airlock’s position.
Solution: Points G and A are located at G: (50, 80, 180) m
and A: (12, 0, 20) m. The vector rGA is rGA D xA xG i C yA yG j C zA zG k D 12 50i C 0 80j C 20 180k m. The
dot product between v and rGA is v ž rGA D jvjjrGA j cos D vx xGA C
vy yGA C vz zGA , where is the angle between v and rGA . Substituting
in the numerical values, we get D 19.7° .
y
G
A
z
x
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1
Problem 2.122 In Problem 2.121, determine the vector component of the astronaut’s velocity parallel to the
line from his position to the airlock’s position.
Solution: The dot product v ž rGA D vx xGA C vy yGA C vz zGA D
752 m/s2 and the component of v parallel to GA is vparallel D jvj cos where is defined as in Problem 2.121 above.
vparallel D 4.220.941 D 3.96 m/s
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1
Problem 2.123 Point P is at longitude 30° W and latitude 45° N on the Atlantic Ocean between Nova
Scotia and France. (See Problem 2.86.) Point Q is at
longitude 60° E and latitude 20° N in the Arabian Sea.
Use the dot product to determine the shortest distance
along the surface of the earth from P to Q in terms of
the radius of the earth RE .
Strategy: Use the dot product to detrmine the angle
between the lines OP and OQ; then use the definition of
an angle in radians to determine the distance along the
surface of the earth from P to Q.
y
N
P
Q
45°
z
20°
O
30°
60°
G
Equator
x
Solution: The distance is the product of the angle and the radius of
the sphere, d D RE , where is in radian measure. From Eqs. (2.18)
and (2.24), the angular separation of P and Q is given by
cos D
PÐQ
jPjjQj
.
The strategy is to determine the angle in terms of the latitude and
longitude of the two points. Drop a vertical line from each point P and
Q to b and c on the equatorial plane. The vector position of P is the sum
of the two vectors: P D rOB C rBP . The vector rOB D jrOB ji cos P C
0j C k sin P . From geometry, the magnitude is jrOB j D RE cos P .
The vector rBP D jrBP j0i C 1j C 0k. From geometry, the magnitude
is jrBP j D RE sin P . Substitute and reduce to obtain:
The dot product is
P Ð Q D RE2 cosP Q cos P cos Q C sin P sin Q Substitute:
cos D
PÐQ
D cosP Q cos P cos Q C sin P sin Q
jPjjQj
Substitute P D C30° , Q D 60° , p D C45° , Q D C20° , to obtain
cos D 0.2418, or D 1.326 radians. Thus the distance is d D
1.326RE
y
P D rOB C rBP D RE i cos P cos P C j sin P C k sin P cos P .
N
P
θ
A similar argument for the point Q yields
45°
Q D rOC C rCQ D RE i cos Q cos Q C j sin Q C k sin Q cos Q b
30°
Using the identity cos2 ˇ C sin2 ˇ D 1, the magnitudes are
Q
RE
60°
x
20°
c
G
jPj D jQj D RE
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1
Problem 2.124 (a) Determine the cross product U ð V
of the vectors U D 4i C 6j 10k and V D 8i C 12j C
2k. (b) Use the dot product to prove that the vector U ð
V is perpendicular to U and perpendicular to V.
Strategy: The vectors are expressed in terms of their
components, so you can use Eq. (2.34) to determine their
cross product.
Solution:
(a)
U D 4i C 6j 10k, V D 8i C 12j C 2k
i
j
k 6 10 D 132i C 72j C 96k
UðVD 4
8 12
2 U Ð U ð V D 4132 C 672 C 1096 D 0
(b)
V Ð U ð V D 8132 C 1272 C 296 D 0
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1
Problem 2.125
C 4j.
(a)
(b)
Two vectors U D 3i C 2j and V D 2i
What is the cross product U ð V?
What is the cross product V ð U?
Solution: Use Eq. (2.34) and expand into 2 by 2 determinants.
i j k
U ð V D 3 2 0 D i20 40 j30 20
2 4 0
C k34 22 D 8k
i j k
V ð U D 2 4 0 D i40 20 j20 30
3 2 0
C k22 34 D 8k
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1
Problem 2.126 What is the cross product r ð F of the
position vector r D 2i C 2j C 2k (m) and the force
F D 20i 40k (N)?
Solution: Use Eq. (2.34) and expand into 2 by 2 determinants.
i
r ð F D 2
20
j
k 2
2 D i240 02 j240
0 40 202 C k20 220
r ð F D 80i C 120j 40k (N-m)
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1
Problem 2.127
Determine the cross product r ð F of
the position vector r D 4i 12j C 3k (m) and the force
F D 16i 22j 10k N.
Solution:
i
j
r ð F D 4 12
16 22
k 3 10 r ð F D 120 66i C 48 40j
C 88 192k N-m
r ð F D 186i C 88j C 104k N-m
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1
Problem 2.128 Suppose that the cross product of two
vectors U and V is U ð V D 0. If jUj 6D 0, what do you
know about the vector V?
Solution:
Either V D 0 or
VjjU
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1
Problem 2.129 The cross product of two vectors U
and V is U ð V D 30i C 40k. The vector V D 4i 2j C 3k. Determine the components of U.
Solution: We know
i
U ð V D Ux
4
j
Uy
2
Equating components of (1) and (2), we get
k Uz 3
3Uy C 2Uz D 30
4Uz 3Ux D 0
U ð V D 3Uy C 2Uz i C 4Uz 3Ux j C 2Ux 4Uy k (1)
We also know
U ð V D 30i C 0j C 40k
2Ux 4Uy D 40
Setting Ux D 4 and solving, we get
(2)
U D 4i 12j C 3k
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1
Problem 2.130
20.
(a)
The magnitudes jUj D 10 and jVj D
V
(c)
Use the definition of the cross product to determine
U ð V.
Use the definition of the cross product to determine
V ð U.
Use Eq. (2.34) to determine U ð V.
(d)
Use Eq. (2.34) to determine V ð U.
(b)
y
U
30°
45°
x
Solution: From Eq. (228) U ð V D jUjjVj sin e. From the sketch,
the positive z-axis is out of the paper. For U ð V, e D 1k (points into
the paper); for V ð U, e D C1k (points out of the paper). The angle
D 15° , hence (a) U ð V D 10200.2588e D 51.8e D 51.8k.
Similarly, (b) V ð U D 51.8e D 51.8k (c) The two vectors are:
U D 10i cos 45° C j sin 45 D 7.07i C 0.707j,
V D 20i cos 30° C j sin 30° D 17.32i C 10j
i
U ð V D 7.07
17.32
j
7.07
10
k 0 D i0 j0 C k70.7 122.45
0
D 51.8k
i
(d) V ð U D 17.32
7.07
j
10
7.07
k 0 D i0 j0 C k122.45 70.7
0
D 51.8k
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1
Problem 2.131 The force F D 10i 4j (N). Determine the cross product rAB ð F.
y
(6, 3, 0) m
A
rAB
x
z
(6, 0, 4) m
B
F
Solution: The position vector is
y
A (6, 3, 0)
rAB D 6 6i C 0 3j C 4 0k D 0i 3j C 4k
The cross product:
i
rAB ð F D 0
10
j k 3 4 D i16 j40 C k30
4 0 rAB
x
z
B (6, 0, 4)
F
D 16i C 40j C 30k (N-m)
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1
Problem 2.132 By evaluating the cross product U ð
V, prove the identity sin1 2 D sin 1 cos 2 cos 1 sin 2 .
y
U
V
θ1
θ2
Solution: Assume that both U and V lie in the x-y plane. The
strategy is to use the definition of the cross product (Eq. 2.28) and the
Eq. (2.34), and equate the two. From Eq. (2.28) U ð V D jUjjVj sin1
2 e. Since the positive z-axis is out of the paper, and e points into
the paper, then e D k. Take the dot product of both sides with e, and
note that k Ð k D 1. Thus
sin1 2 D U ð V Ð k
jUjjVj
x
y
U
V
θ1
θ2
x
The vectors are:
U D jUji cos 1 C j sin 2 , and V D jVji cos 2 C j sin 2 .
The cross product is
i
U ð V D jUj cos 1
jVj cos 2
j
jUj sin 1
jVj sin 2
k 0 0
D i0 j0 C kjUjjVjcos 1 sin 2 cos 2 sin 1 Substitute into the definition to obtain: sin1 2 D sin 1 cos 2 cos 1 sin 2 . Q.E.D.
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1
Problem 2.133 Use the cross product to determine the
components of a unit vector e that is normal to both
of the vectors U D 8i 6j C 4k and V D 3i C 7j C 9k.
(Notice that there are two answers.)
Solution: First, find U ð V D R
i
R D U ð V D 8
3
j k 6 4 7 9
R D 54 28i C 12 72j C 56 18 k
R D 82i 60j C 74k
eR D š
R
Dš
jRj
82i 60j C 74k
125.7
er D š0.652i 0.477j C 0.589k
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1
Problem 2.134 (a) What is the cross product rOA ð
rOB ? (b) Determine a unit vector e that is perpendicular
to rOA and rOB .
y
B ( 4, 4, –4) m
Solution: The two radius vectors are
rOB
rOB D 4i C 4j 4k, rOA D 6i 2j C 3k
(a)
The cross product is
i
rOA ð rOB D 6
4
O
j
k 2 3 D i8 12 j24 12
4 4 x
rOA
z
A (6, –2, 3) m
C k24 C 8
D 4i C 36j C 32k m2 The magnitude is
jrOA ð rOB j D
(b)
p
42 C 362 C 322 D 48.33 m2
The unit vector is
eDš
rOA ð rOB
jrOA ð rOB j
D š0.0828i C 0.7448j C 0.6621k
(Two vectors.)
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1
Problem 2.135 For the points O, A, and B in Problem 2.134, use the cross product to determine the length
of the shortest straight line from point B to the straight
line that passes through points O and A.
Solution:
(The magnitude of C is 338.3)
rOA D 6i 2j C 3k (m)
rOB D 4i C 4j 4k m
We now want to find the length of the projection, P, of line OB in
direction ec .
P D rOB Ð eC
rOA ð rOB D C
D 4i C 4j 4k Ð eC
(C is ? to both rOA and rOB )
i
j
C D 6 2
4
4
C8 12i
k 3 D C12 C 24j
C24 C 8k
4 P D 6.90 m
y
B ( 4, 4, –4) m
C D 4i C 36j C 32k
C is ? to both rOA and rOB . Any line ? to the plane formed by C and
rOA will be parallel to the line BP on the diagram. C ð rOA is such a
line. We then need to find the component of rOB in this direction and
compute its magnitude.
i
C ð rOA D 4
6
j
C36
2
rOB
k 32 3
C D 172i C 204j 208k
O
x
rOA
P
z
A(6, –2, 3) m
The unit vector in the direction of C is
eC D
C
D 0.508i C 0.603j 0.614k
jCj
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1
Problem 2.136 The cable BC exerts a 1000-lb force F
on the hook at B. Determine rAB ð F.
y
Solution: The coordinates of points A, B, and C are A (16, 0, 12),
B (4, 6, 0), C (4, 0, 8). The position vectors are
B
F
6 ft
rAB
rOA D 16i C 0j C 12k, rOB D 4i C 6j C 0k, rOC D 4i C 0j C 8k.
x
8 ft
The force F acts along the unit vector
eBC D
jrOC rOB j D
rAC
4 ft
rOC rOB
rAB
rBC
D
D
jrBC j
jrOC rOB j
jrAB j
Noting rOC rOB D 4 4i C 0 6j C 8 0k D 0i 6j C 8k
C
4 ft
z
p
62 C 82 D 10. Thus
y
eBC D 0i 0.6j C 0.8k, and F D jFjeBC D 0i 600j C 800k (lb).
B
The vector
6 ft
r
x
rAB D 4 16i C 6 0j C 0 12k D 12i C 6j 12k
8 ft
C
Thus the cross product is
i
j
6
rAB ð F D 12
0
600
A
12 ft
k 12 D 2400i C 9600j C 7200k (ft-lb)
800 4 ft
4 ft
12 ft
A
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1
Problem 2.137 The force vector F points along the
straight line from point A to point B. Its magnitude
is jFj D 20 N. The coordinates of points A and B
are xA D 6 m, yA D 8 m, zA D 4 m and xB D 8 m, yB D
1 m, zB D 2 m.
(a)
(b)
y
A
F
B
rA
Express the vector F in terms of its components.
Use Eq. (2.34) to determine the cross products
rA ð F and rB ð F.
rB
x
z
Solution: We have rA D 6i C 8j C 4k m, rB D 8i C j 2k m,
F D 20 N
(a)
8 6 mi C 1 8 mj C 2 4 mk
2 m2 C 7 m2 C 6 m2
20 N
D p 2i 7j 6k
89
i
j
k 20 N rA ð F D p 6 m 8 m 4 m 89 2
7 6 (b)
D 42.4i C 93.3j 123.0k Nm
i
j
k 20 N rB ð F D p 8 m 1 m 2 m 89 2
7
6 D 42.4i C 93.3j 123.0k Nm
Note that both cross products give the same result (as they must).
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1
Problem 2.138 The rope AB exerts a 50-N force T on
the collar at A. Let rCA be the position vector from point
C to point A. Determine the cross product rCA ð T.
y
0.15 m
0.4 m
B
C
T
0.2 m
Solution: The vector from C to D is rCD D xD xC i C yD jrCD j D
0.5 m
O
xD xC 2 C yD yC 2 C zD zC 2 .
x
D
The components of the unit vector along CD are given by uCDx D
xD xC /jrCD j, uCDy D yD yC /jrCD j, etc. Numerical values
are jrCD j D 0.439 m, uCDx D 0.456, uCDy D 0.684, and uCDz D
0.570. The coordinates of point A are given by xA D xC C jrCA juCDx ,
yA D yC C jrCA juCDy , etc. The coordinates of point A are (0.309,
0.162, 0.114) m. The vector rCA is given by rCA D xA xC i C yA yC j C zA zC k. The vector rCA is rCA D 0.091i C 0.137j C
0.114k m. The vector from A to B and the corresponding unit
vector are found in the same manner as from C to D above.
The results are jrAB j D 0.458 m, uABx D 0.674, uABy D 0.735, and
uABz D 0.079. The force T is given by T D jTjuAB . The result is
T D 33.7i C 36.7j C 3.93k N.
0.3 m
A
yC j C zD zC k. The magnitude of the vector
0.25 m
0.2 m
z
The cross product rCA ð T can now be calculated.
i
j
rCA ð T D 0.091 0.138
33.7
36.7
k 0.114 3.93 D 4.65i C 3.53j C 7.98k N-m
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1
Problem 2.139 The straight line L is collinear with the
force vector F. Let D be the perpendicular distance from
an arbitrary point P to L. Prove that
L
P
F
DjFj D jr ð Fj,
where r is a position vector from point P to any point
on L.
D
Solution: By definition
F
r
jr ð Fj D jrjjFj sin From the figure we see that
D
θ
θ
D D jrj sin Hence
jr ð Fj D DjFj
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1
Problem 2.140 The bar AB is 6 m long and is perpendicular to the bars AC and AD. Use the cross product to
determine the coordinates xB , yB , zB of point B.
y
B
Solution: The strategy is to determine the unit vector perpendicular to both AC and AD, and then determine the coordinates that will
agree with the magnitude of AB. The position vectors are:
(0, 3, 0) m
(xB, yB, zB)
A
rOA D 0i C 3j C 0k, rOD D 0i C 0j C 3k, and
rOC D 4i C 0j C 0k. The vectors collinear with the bars are:
rAD D 0 0i C 0 3j C 3 0k D 0i 3j C 3k,
rAC D 4 0i C 0 3j C 0 0k D 4i 3j C 0k.
The vector collinear with rAB is
i
j k R D rAD ð rAC D 0 3 3 D 9i C 12j C 12k
4 3 0 D
z
C
(0, 0, 3) m
(4, 0, 0) m
x
The magnitude jRj D 19.21 (m). The unit vector is
eAB D
R
D 0.4685i C 0.6247j C 0.6247k.
jRj
Thus the vector collinear with AB is
rAB D 6eAB D C2.811i C 3.75j C 3.75k.
Using the coordinates of point A:
xB D 2.81 C 0 D 2.81 (m)
yB D 3.75 C 3 D 6.75 (m)
zB D 3.75 C 0 D 3.75 (m)
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1
Problem 2.141* Determine the minimum distance from
point P to the plane defined by the three points A, B,
and C.
y
B
(0, 5, 0) m
P
(9, 6, 5) m
Solution: The strategy is to find the unit vector perpendicular to
the plane. The projection of this unit vector on the vector OP: rOP Ð e is
the distance from the origin to P along the perpendicular to the plane.
The projection on e of any vector into the plane (rOA Ð e, rOB Ð e, or
rOC Ð e) is the distance from the origin to the plane along this same
perpendicular. Thus the distance of P from the plane is
A
(3, 0, 0) m
C
d D rOP Ð e rOA Ð e.
The position vectors are: rOA D 3i, rOB D 5j, rOC D 4k and rOP D
9i C 6j C 5k. The unit vector perpendicular to the plane is found
from the cross product of any two vectors lying in the plane. Noting:
rBC D rOC rOB D 5j C 4k, and rBA D rOA rOB D 3i 5j. The
cross product:
(0, 0, 4) m
z
y
P[9,6,5]
i
j k rBC ð rBA D 0 5 4 D 20i C 12j C 15k.
3 5 0 The magnitude is jrBC ð rBA j D 27.73, thus the unit vector is e D
0.7212i C 0.4327j C 0.5409k. The distance of point P from the plane
is d D rOP Ð e rOA Ð e D 11.792 2.164 D 9.63 m. The second term
is the distance of the plane from the origin; the vectors rOB , or rOC
could have been used instead of rOA .
x
B[0,5,0]
x
O
A[3,0,0]
z
C[0,0,4]
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1
Problem 2.142* The force vector F points along
the straight line from point A to point B. Use
Eqs. (2.28)–(2.31) to prove that
y
A
F
rB ð F D rA ð F.
rA
Strategy: Let rAB be the position vector from point A
to point B. Express rB in terms of of rA and rAB . Notice
that the vectors rAB and F are parallel.
B
rB
x
z
Solution: We have
rB D rA C rAB .
Therefore
rB ð F D rA C rAB ð F D rA ð F C rAB ð F
The last term is zero since rAB jjF.
Therefore
rB ð F D rA ð F
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1
Problem 2.143 For the vectors U D 6i C 2j 4k,
V D 2i C 7j, and W D 3i C 2k, evaluate the following
mixed triple products: (a) U Ð V ð W; (b) W Ð V ð
U; (c) V Ð W ð U.
Solution: Use Eq. (2.36).
6 2 4 (a) U Ð V ð W D 2 7
0 3 0
2
D 614 24 C 421 D 160
3
(b) W Ð V ð U D 2
6
0
2 7
0 2 4 D 328 0 C 24 42 D 160
2
(c) V Ð W ð U D 3
6
7
0 0
2 2 4 D 24 712 12 C 0 D 160
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1
Problem 2.144 Use the mixed triple product to calculate the volume of the parallelepiped.
y
(140, 90, 30) mm
(200, 0, 0) mm
x
(160, 0, 100) mm
z
Solution: We are given the coordinates of point D. From the geometry, we need to locate points A and C. The key to doing this is to note
that the length of side OD is 200 mm and that side OD is the x axis.
Sides OD, AE, and CG are parallel to the x axis and the coordinates
of the point pairs (O and D), (A and E), and (C and D) differ only by
200 mm in the x coordinate. Thus, the coordinates of point A are (60,
90, 30) mm and the coordinates of point C are (40, 0, 100) mm.
Thus, the vectors rOA , rOD , and rOC are rOD D 200i mm, rOA D
60i C 90j C 30k mm, and rOC D 40i C 0j C 100k mm. The mixed
triple product of the three vectors is the volume of the parallelepiped.
The volume is
60
rOA Ð rOC ð rOD D 40
200
90 30 0 100 0
0 y
(140, 90, 30)
mm
E
A
B
F
O
D
x
(200, 0, 0)
mm
(160, 0, 100)
mm
G
C
z
D 600 C 90200100 C 300 mm3
D 1,800,000 mm3
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1
Problem 2.145
that
By using Eqs. (2.23) and (2.34), show
Ux
U Ð V ð W D Vx
W
x
Uy
Vy
Wy
Uz Vz Wz .
Solution: One strategy is to expand the determinant in terms of
its components, take the dot product, and then collapse the expansion.
Eq. (2.23) is an expansion of the dot product: Eq. (2.23): U Ð V D
UX VX C UY VY C UZ VZ . Eq. (2.34) is the determinant representation
of the cross product:
i
Eq. (2.34) U ð V D UX
VX
j
UY
VY
k UZ VZ U
Q Ð P D QX Y
VY
For notational convenience, write P D U ð V. Expand the determinant about its first row:
U
P D i Y
VY
UX
UZ j
VX
VZ UX
UZ C
k
VX
VZ Since the two-by-two determinants are scalars, this can be written in
the form: P D iPX C jPY C kPZ where the scalars PX , PY , and PZ are
the two-by-two determinants. Apply Eq. (2.23) to the dot product of
a vector Q with P. Thus Q Ð P D QX PX C QY PY C QZ PZ . Substitute
PX , PY , and PZ into this dot product
UZ VZ UX
UZ Q
Y
VX
VZ UX
UZ C
Q
z
VX
VZ UZ VZ But this expression can be collapsed into a three-by-three determinant
directly, thus:
QX
Q Ð U ð V D UX
VX
QY
UY
VY
QZ UZ . This completes the demonstration.
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1
Problem 2.146 The vectors U D i C UY j C 4k, V D
2i C j 2k, and W D 3i C j 2k are coplanar (they
lie in the same plane). What is the component Uy ?
Solution: Since the non-zero vectors are coplanar, the cross product of any two will produce a vector perpendicular to the plane, and
the dot product with the third will vanish, by definition of the dot
product. Thus U Ð V ð W D 0, for example.
1 UY
1
U Ð V ð W D 2
3 1
4 2 2 D 12 C 2 UY 4 6 C 42 C 3
D C10UY C 20 D 0
Thus UY D 2
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1
Problem 2.147 The magnitude of F is 8 kN. Express
F in terms of scalar components.
Solution: The unit vector collinear with the force F is developed
as follows: The collinear vector is r D 7 3i C 2 7j D 4i 5j
y
The magnitude: jrj D
(3, 7) m
eD
F
p
42 C 52 D 6.403 m. The unit vector is
r
D 0.6247i 0.7809j. The force vector is
jrj
F D jFje D 4.997i 6.247j D 5i 6.25j (kN)
(7, 2) m
x
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1
Problem 2.148 The magnitude of the vertical force W
is 600 lb, and the magnitude of the force B is 1500 lb.
Given that A C B C W D 0, determine the magnitude of
the force A and the angle ˛.
B
W
50°
α
A
Solution: The strategy is to use the condition of force balance to
determine the unknowns. The weight vector is W D 600j. The vector
B is
B D 1500i cos 50° C j sin 50° D 964.2i C 1149.1j
The vector A is A D jAji cos180 C ˛ C j sin180 C ˛
A D jAji cos ˛ j sin ˛. The forces balance, hence A C B C W D
0, or 964.2 jAj cos ˛i D 0, and 1149.1 600 jAj sin ˛j D 0.
Thus jAj cos ˛ D 964.2, and jAj sin ˛ D 549.1. Take the ratio of the
two equations to obtain tan ˛ D 0.5695, or ˛ D 29.7° . Substitute this
angle to solve: jAj D 1110 lb
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1
Problem 2.149 The magnitude of the vertical force
vector A is 200 lb. If A C B C C D 0, what are the magnitudes of the force vectors B and C?
70 in.
50 in.
100 in.
C
E
B
D
A
F
Solution: The strategy is to express the forces in terms of scalar
components, and then solve the force balance equations for the unknowns. C D jCji cos ˛ j sin ˛, where
tan ˛ D
50
D 0.7143, or ˛ D 35.5° .
70
Thus C D jCj0.8137i 0.5812j. Similarly, B D CjBji, and A D
C200j. The force balance equation is A C B C C D 0. Substituting,
0.8137jCj C jBji D 0, and 0.5812jCj C 200j D 0. Solving,
jCj D 344.1 lb, jBj D 280 lb
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1
Problem 2.150 The magnitude of the horizontal force
vector D in Problem 2.149 is 280 lb. If D C E C F D 0,
what are the magnitudes of the force vectors E and F?
Solution: The strategy is to express the force vectors in terms of
scalar components, and then solve the force balance equation for the
unknowns. The force vectors are:
E D jEji cos ˇ j sin ˇ, where tan ˇ D
50
D 0.5, or ˇ D 26.6° .
100
Thus
E D jEj0.8944i 0.4472j
D D 280i, and F D jFjj.
The force balance equation is D C E C F D 0. Substitute and resolve
into two equations:
0.8944jEj 280i D 0, and 0.4472jEj C jFjj D 0.
Solve: jEj D 313.1 lb, jFj D 140 lb
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1
Problem 2.151
y
What are the direction cosines of F?
F = 20i + 10j – 10k (lb)
Refer to this diagram when solving Problems 2.151–2.157.
A
(4, 4, 2) ft
θ
B (8, 1, – 2) ft
x
z
Solution: Use the definition of the direction cosines and the
ensuing discussion.
The magnitude of F: jFj D
p
202 C 102 C 102 D 24.5.
The direction cosines are cos x D
cos y D
10
Fy
D
D 0.4082
jFj
24.5
cos z D
10
Fz
D
D 0.4082
jFj
24.5
20
Fx
D
D 0.8165,
jFj
24.5
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1
Problem 2.152 Determine the scalar components of
a unit vector parallel to line AB that points from A
toward B.
Solution: Use the definition of the unit vector, we get
The position vectors are: rA D 4i C 4j C 2k, rB D 8i C 1j 2k. The
vector from A to B is rAB D 8 p4i C 1 4j C 2 2k D
4i 3j 4k. The magnitude: jrAB j D 42 C 32 C 42 D 6.4. The unit
vector is
eAB D
4
3
4
rAB
D
i
j
k D 0.6247i 0.4688j 0.6247k
jrAB j
6.4
6.4
6.4
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1
Problem 2.153 What is the angle between the line
AB and the force F?
Solution: Use the definition of the dot product Eq. (2.18), and
Eq. (2.24):
cos D
rAB Ð F
.
jrAB jjFj
From the solution to Problem 2.130, the vector parallel to AB is rAB D
4i 3j 4k, with a magnitude jrAB j D 6.4. From Problem 2.151, the
force is F D 20i C 10j 10k, with a magnitude of jFj D 24.5. The dot
product is rAB Ð F D 420 C 310 C 410 D 90. Substi90
D 0.574, D 55°
tuting, cos D
6.424.5
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1
Problem 2.154 Determine the vector component of F
that is parallel to the line AB.
Solution: Use the definition in Eq. (2.26): UP D e Ð Ue, where e
is parallel to a line L. From Problem 2.152 the unit vector parallel to
line AB is eAB D 0.6247i 0.4688j 0.6247k. The dot product is
e Ð F D 0.624720 C 0.468810 C 0.624710 D 14.053.
The parallel vector is
e Ð Fe D 14.053e D 8.78i 6.59j 8.78k (lb)
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1
Problem 2.155 Determine the vector component of F
that is normal to the line AB.
Solution: Use the Eq. (2.27) and the solution to Problem 2.154.
FN D F FP D 20 8.78i C 10 C 6.59j C 10 C 8.78k
D 11.22i C 16.59j 1.22k (lb)
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1
Problem 2.156 Determine the vector rBA ð F, where
rBA is the position vector from B to A.
Solution: Use the definition in Eq. (2.34). Noting rBA D rAB ,
from Problem 2.155 rBA D 4i C 3j C 4k. The cross product is
i
j
rBA ð F D 4 3
20 10
k 4 D 30 40i 40 80j
10 C 40 60
D 70i C 40j 100k (ft-lb)
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1
Problem 2.157 (a) Write the position vector rAB from
point A to point B in terms of scalar components.
(b) The vector F has magnitude jFj D 200 N and is
parallel to the line from A to B. Write F in terms of
scalar components.
y
Solution:
(a)
A (1,5,−1) m
rAB D xB xA i C yB yA j C zB zA k
D 8 1i C 1 5j C 1 C 1k
D 7i 4j C 2k m.
(b)
By dividing rAB by its magnitude, we obtain a unit vector parallel
to F:
B (8,1,1) m
F
rAB
D 0.843i 0.482j C 0.241k.
eAB D
jrAB j
x
Then
z
F D 200eAB D 169i 93.3j C 48.2k N.
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1
y
Problem 2.158 The rope exerts a force of magnitude
jFj D 200 lb on the top of the pole at B.
(a)
(b)
Determine the vector rAB ð F, where rAB is the
position vector from A to B.
Determine the vector rAC ð F, where rAC is the
position vector from A to C.
B (5, 6, 1) ft
F
A
x
C (3, 0, 4) ft
z
Solution: The strategy is to define the unit vector pointing from B
to A, express the force in terms of this unit vector, and take the cross
product of the position vectors with this force. The position vectors
rAB D 5i C 6j C 1k, rAC D 3i C 0j C 4k,
rBC D 3 5i C 0 6j C 4 1k D 2i 6j C 3k.
The magnitude jrBC j D
eBC D
p
22 C 62 C 32 D 7. The unit vector is
rBC
D 0.2857i 0.8571j C 0.4286k.
jrBC j
The force vector is
F D jFjeBC D 200eBC D 57.14i 171.42j C 85.72k.
The cross products:
i
rAB ð F D 5
57.14
j
6
171.42
k 1 85.72 D 685.74i 485.74j 514.26k
D 685.7i 485.7j 514.3k (ft-lb)
i
rAC ð F D 3
57.14
j
0
171.42
k 4 85.72 D 685.68i 485.72j 514.26k
D 685.7i 485.7j 514.3k (ft-lb)
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1
Problem 2.159 The magnitude of FB is 400 N and
jFA C FB j D 900 N. Determine the components of FA .
y
FB
FA
60°
30°
40°
x
50°
z
Solution:
Setting
jFB j D 400 N
900 N D jFA C FB j
We need to write each vector in terms of its known or unknown components. From the diagram
D [0.587FA 1002 C 0.643FA C 3462
C 0.492FA C 1732 ]1/2
FAx D jFA j cos 40° cos 40° D 0.587
FAz D jFA j cos 40° cos 50° D 0.492
and solving, we obtain FA D 595 N. Substituting this result into
Eq. (1),
FAy D jFA j sin 40° D 0.642
FA D 349i C 382j C 293k N.
FBx D 400 cos 60° cos 60°
FBz D 400 cos 60° cos 30°
FBy D 400 sin 60°
Let FA D jFA j and FB D jFB j D 400 N.
The components of the vectors are
FA D FA cos 40° sin 50° i C FA sin 40° j C FA cos 40° cos 50° k
D FA 0.587i C 0.643j C 0.492k,
(1)
FB D FB cos 60° sin 30° i C FB sin 60° j C FB cos 60° cos 30° k
D 100i C 346j C 173k N.
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1
Problem 2.160 Suppose that the forces FA and FB
shown in Problem 2.159 have the same magnitude and
FA Ð FB D 600 N2 . What are FA and FB ?
Solution: From Problem 2.159, the forces are:
FA D jFA ji cos 40° sin 50° C j sin 40° C k cos 40° cos 50° D jFA j0.5868i C 0.6428j C 0.4924k
FB D jFB ji cos 60° sin 30° C j sin 60° C k cos 60° cos 30° D jFB j0.25i C 0.866j C 0.433k
The dot product: FA Ð FB D jFA jjFB j0.6233 D 600 N2 , from
jFA j D jFB j D
600
D 31.03 N,
0.6233
and
FA D 18.21i C 19.95j C 15.28k (N)
FB D 7.76i C 26.87j C 13.44k (N)
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1
y
Problem 2.161 The magnitude of the force vector FB
is 2 kN. Express it in terms of scalar components.
F
D
(4, 3, 1) m
FC
FA
FB
A
z
C
x
(6, 0, 0) m
B
Solution: The strategy is to determine the unit vector collinear
with FB and then express the force in terms of this unit vector.
F
y
The radius vector collinear with FB is
D (4,3,1)
rBD D 4 5i C 3 0j C 1 3k or rBD D 1i C 3j 2k.
FA
The magnitude is
p
jrBD j D 12 C 32 C 22 D 3.74.
The unit vector is
(5, 0, 3) m
FC
A
z
x
C(6,0,0)
FB
B (5,0,3)
rBD
D 0.2673i C 0.8018j 0.5345k
eBD D
jrBD j
The force is
FB D jFB jeBD D 2eBD (kN) FB D 0.5345i C 1.6036j 1.0693k
D 0.53i C 1.60j 1.07k (kN)
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1
Problem 2.162 The magnitude of the vertical force
vector F in Problem 2.161 is 6 kN. Determine the vector
components of F parallel and normal to the line from B
to D.
Solution: The projection of the force F onto the line from B
to D is FP D F Ð eBD eBD . The vertical force has the component
F D 6j (kN). From Problem 2.139, the unit vector pointing from
B to D is eBD D 0.2673i C 0.8018j 0.5345k. The dot product is
F Ð eBD D 4.813. Thus the component parallel to the line BD is FP D
4.813eBD D C1.29i 3.86j C 2.57k (kN). The component perpendicular to the line is: FN D F FP . Thus FN D 1.29i 2.14j 2.57k (kN)
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1
Problem 2.163 The magnitude of the vertical force
vector F in Problem 2.161 is 6 kN. Given that F C FA C
FB C FC D 0, what are the magnitudes of FA , FB , and
FC ?
Solution: The strategy is to expand the forces into scalar components, and then use the force balance equation to solve for the unknowns. The unit vectors are used to expand the forces into scalar
components. The position vectors, magnitudes, and unit vectors are:
rAD D 4i C 3j C 1k, jrAD j D
p
26 D 5.1,
The forces are:
FA D jFA jeAD , FB D jFB jeBD , FC D jFC jeCD , F D 6j (kN).
Substituting into the force balance equation
F C FA C FB C FC D 0,
eAD D 0.7845i C 0.5883j C 0.1961k.
rBD D 1i C 3j 2k, jrBD j D
p
0.7843jFA j 0.2674jFB j 0.5348jFC ji D 0
14 D 3.74,
0.5882jFA j C 0.8021jFB j C 0.8021jFC j 6j
eBD D 0.2673i C 0.8018j 0.5345k.
rCD D 2i C 3j C 1k, jrCD j D
p
14 D 3.74,
eCD D 0.5345i C 0.8018j C 0.2673k
D 00.1961jFA j 0.5348jFB j C 0.2674jFC jk D 0
These simple simultaneous equations can be solved a standard method
(e.g., Gauss elimination) or, conveniently, by using a commercial
package, such as TK Solver, Mathcad, or other. An HP-28S hand
held calculator was used here: jFA j D 2.83 (kN), jFB j D 2.49 (kN),
jFC j D 2.91 (kN)
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1
Problem 2.164 The magnitude of the vertical force W
is 160 N. The direction cosines of the position vector from
A to B are cos x D 0.500, cos y D 0.866, and cos z D
0, and the direction cosines of the position vector from
B to C are cos x D 0.707, cos y D 0.619, and cos z D
0.342. Point G is the midpoint of the line from B to C.
Determine the vector rAG ð W, where rAG is the position
vector from A to G.
Solution: Express the position vectors in terms of scalar components, calculate rAG , and take the cross product. The position vectors
are: rAB D 0.6.5i C 0.866j C 0k rAB D 0.3i C 0.5196j C 0k,
rBG D 0.30.707i C 0.619j 0.342k,
rBG D 0.2121i C 0.1857j 0.1026k.
rAG D rAB C rBG D 0.5121i C 0.7053j 0.1026k.
W D 160j
m
0m
y
C
60
i
rAG ð W D 0.5121
0
j
0.7053
160
k
0.1026 0
G
D 16.44i C 0j 81.95k D 16.4i C 0j 82k (N m)
B
W
600 mm
600 mm
C
600 mm
G
B
W
A
A
z
x
x
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1
Problem 2.165 The rope CE exerts a 500-N force T
on the hinged door.
E
(0.2, 0.4, ⫺0.1) m
y
(a)
(b)
Express T in terms of components.
Determine the vector component of T parallel to
the line from point A to point B.
T
D
C
(0, 0.2, 0) m
A (0.5, 0, 0) m
x
B
(0.35, 0, 0.2) m
z
Solution:
(a)
F D 500 N
0.2 m 0i C 0.4 m 0.2 mj C 0.1 m 0k
0.2 m2 C 0.2 m2 C 0.1 m2
F D 333i C 333j 166.7k N
(b)
We define the unit vector in the direction of AB and then use the
dot product to find the part of F that is parallel to AB.
eAB D
0.35 m 0.5 mi C 0.2 m 0k
D 0.6i C 0.8k
0.15 m2 C 0.2 m2
Fjj D F Ð eAB eAB D [333i C 333j 166.7k N
Ð 0.6i C 0.8k]0.6i C 0.8k
Working out the algebra we have
Fjj D 200i 267k N
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1
Problem 2.166 In Problem 2.165, let rBC be the position vector from point B to point C. Determine the cross
product rBC ð T.
Solution: The vector from B to C is
rBC D xC xB i C yC yB j C zC zB k
D 0.35i C 0.2j 0.2k m.
The vector T is T D 482i C 60.2j 120k N. The cross product of
these vectors is given by
i
j
k rBC ð T D 0.35 0.2 0.2 D 12.0i 138j 117k N m
482
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1
y
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 .
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
B
TAB
TAC
a
a
C
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
TLA D 0i 1200
0.28 m 0.25 mj D 0i 36j N
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
24 N
jWL j
D
D 2.45 kg
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
36 N
jWB j
D
D 3.67 kg
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 magnitudes of the normal and
friction forces exerted on the crane by the
level ground.
Determine the magnitudes of the normal and
friction forces exerted on the caisson by the
level ground.
45°
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°
Solution:
y


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
x
TL D 1.56 kN,
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
TBC D 0
TAB C
Fx : p
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,
ˇ D 40.6° .
C 3.5
m
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
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 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
dV
W
b2 C h2 C2 D0
D
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
The new angles are given by the cosine law
B
AB2 D 382 C AD2 238AD cos ˛.
β
D
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
L D 25 m,
L
b
A
b D 15 m,
25 m
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
2L C bBC
cos ˇ D
L C b2 C BC2 AB2
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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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
8 in.
y
B
12 in
α
˛ D 33.7°
8 in
A
D
D 0.333
ˇ D 18.4°
β
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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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
W
y
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
A
Strategy: Draw a diagram of the sum of the forces
exerted by the three cables at A.
Solution: Draw the free body diagram of the knot at point A. Then
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° .
β
α
Candidate β
W
Candidate values
for TAC
Fixed direction for
line AB
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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
TBC D 0
FBx : p TAB C p
229
13
TAB
2
15
2
3
2
2
TBC 50 lb D 0
FBy : p TAB C p
229
13
TBC
B
15
15
TCD D 0
TBC C
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
TBC D 0
FBx : p TAB C p
229
13
2
2
2
TBC W1 D 0
FBy : p TAB C p
229
13
15
2
3
TBC
B
15
15
TCD D 0
TBC C
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
0.15 m
0.6 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.
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
β
α
.
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
y
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 ˇ
sin 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 3.53 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
m
Agg 4mg. Thus, the tension, T, in terms of masses and g is T D
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
W
WA
C ,
2
2
WA
3W
WA
7W
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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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 3.56 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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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 3.57 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
m2 D
D 12.5 kg
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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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 3.60 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,
jTj
jLj
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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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
6 12i C 0 4j C 0 2k
rB rA
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
6
4
2
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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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 3.65 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
Equilibrium:
Fy : 3
T 80 lb D 0
5
4
3
TAB D TAC D TAD D 33.3 lb
80 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 3.66 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 rA
.
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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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 3.67 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 rA
.
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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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 3.68 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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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 3.71 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
TAB C p
TAC p
TAD p
TAE D 0
Fx : p
398
374
374
404
18
18
18
18
TAB p
TAC p
TAD p
TAE
Fy : p
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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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 3.72 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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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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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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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
xI xK i C yI yK j C zI zK k
,
eIK 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
(0, 1.4, 1.2) m
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
(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
TEF
BE
(1, 1.2, 0) m
(2, 1, 0) m (3.4, 1, 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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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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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 rA
.
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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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 rA
.
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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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 ,
m Ag
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 ,
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.
and zB D zE C EBeEGz .
Solving the eight equations in the eight unknowns, we obtain
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
F D 36.6 N .
NAx C TAB eABx C FeCDx D 0,
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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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 ˛
y
and ˇ. Solve the equilibrium conditions in terms of distance d. Plot
the result.
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
jWj
1
.
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
jWj
1
.
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
115600 d2
3002 C 1602 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
, LBC D
41 C 40b b2 , LAB D
21
21
21
Equilibrium:
Fx : Fy :
TBC
TAB
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).
0.3 m2 C h2 0.4 m
F D 900 N/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
h
A
0.7 m b2 C h2
B
F D 5200 N/mLBC 0.4 m
Fx : Fy :
b
0.7 m b
TAB C
FD0
0.5 m
LBC
h
h
TAB C
F 490.5 N D 0
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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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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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 3.93* 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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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 3.94* 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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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 3.95 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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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 3.96 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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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 3.97 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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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 3.98 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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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 3.99 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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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 3.100 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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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 3.101 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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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 3.102 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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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 3.103 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
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,
C
(5, 2, –1) m
A
(3, 0, 4) m
x
C
A
The vector locations of the points A, B, C are:
rA D 3i C 0j C 4k,
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
rB rA
jrB rA j
The unit vectors are
eAB D
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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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 3.104 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).
40°
The angle between the normal force and the positive x axis is 90° 75°
60° D 30° .
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
N
75°
T2
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 .
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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 3.106 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 weight is W D 0i mgj.
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.
The angle between the string tension and the positive x axis is (90° 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
y
L
h α
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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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,
TaA D jTaA jeaA ,
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 ,
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.
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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 3.108 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,
x
8 ft
10 ft
B
C
A
rC D 8i C 4j C 0k.
The unit vector eAB is, by definition,
N
W
rB rA
.
eAB 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 .
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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 3.109 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
y
rB rA
.
eAB 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.
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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 3.110* 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.
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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 4.1 The weights W1 D 50 lb and W2 D 20 lb.
Determine the sum of the moments due to the forces
exerted by the suspended weights on the bar AB (a) about
point A; (b) about point B.
14 in
14 in
14 in
A
B
W1
14"
Solution:
(a)
W2
14"
14"
MA D 50 lb14 in 20 lb28 in D 1260 in lb
A
B
MA D 1260 in lb or MA D 1260 in lb CW
(b)
MB D 50 lb28 in C 20 lb14 in D 1680 in lb
50 lb
20 lb
MB D 1680 in lb or MB D 1680 in lb CCW
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1
Problem 4.2 The masses m1 D 20 kg and m2 D 8 kg.
Determine the sum of the moments due to the forces
exerted by the suspended masses on the bar AB (a) about
point A; (b) about point B.
0.35 m
0.35 m
0.35 m
A
B
M1
Solution:
(a)
0.35 m
M2
0.35 m
0.35 m
MA D 20 kg9.81 m/s2 0.35 m
8 kg9.81 m/s2 0.7 m D 123.61 Nm
10°
A
MA D 124 Nm or MA D 124 Nm CW
(b)
B
2
MB D 20 kg9.81 m/s 0.7 m
C 8 kg9.81 m/s2 0.35 m D 164.8 Nm
(20 kg) (9.81 m/s2)
(8 kg) (9.81 m/s2)
MB D 165 Nm or MA D 165 Nm CCW
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1
Problem 4.3 The wheels of the overhead crane exert
downward forces on the horizontal I-beam at B and C.
If the force at B is 40 kip and the force at C is 44 kip,
determine the sum of the moments of the forces on the
beam about (a) point A, (b) point D.
10 ft
(a)
The normal distances from A to the lines of action are DAB D
10 ft, and DAC D 35 ft. The moments are clockwise (negative).
Hence,
(b)
B
A
Solution: Use 2-dimensional moment strategy: determine normal
distance to line of action D; calculate magnitude DF; determine sign.
Add moments.
25 ft
10 ft
15 ft
C
25 ft
D
15 ft
A
D
B
C
MA D 1040 3544 D 1940 ft-kip .
The normal distances from D to the lines of action are DDB D
40 ft, and DDC D 15 ft. The actions are positive; hence
MD D C4040 C 1544 D 2260 ft-kip
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1
Problem 4.4 What force F applied to the pliers is
required to exert a 4 N-m moment about the center of
the bolt at P?
Solution:
MP D 4 N-m D F0.165 m sin 42° ) F D
4 N-m
0.165 m sin 42°
D 36.2 N
P
F
165
mm
42⬚
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1
Problem 4.5 Two forces of equal magnitude F are
applied to the wrench as shown. If a 50 N-m moment is
required to loosen the nut, what is the necessary value
of F?
F
F
F
300 mm
380 mm
30⬚
20⬚
F
Solution:
Mnut center D F cos 30° 0.3 m C F cos 20° 0.38 m D 50 N-m
FD
50 N-m
D 81.1 N
0.3 m cos 30° C 0.38 m cos 20°
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1
Problem 4.6 The sum of the moments of the two
forces about P is zero. What is the magnitude of the
force F?
4 kN
350 mm
450 mm
P
40°
20°
F
Solution:
4 kN
450 mm
350 mm
40°
MP D F cos 20° 0.45 m C 4 kN sin 40° 0.8 m D 0
P
FD
4 kN sin 40° 0.8 m
D 4.86 kN
0.45 m cos 20°
20°
F
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1
Problem 4.7 The gears exert 200-N forces on each
other at their point of contact.
(a)
(b)
Determine the moment about A due to the force
exerted on the left gear.
Determine the moment about B due to the force
exerted on the right gear.
Solution: Use 2-dimensional moment strategy: resolve the forces
into components normal to the radii; calculate magnitude DF, where
F is the normal component; determine sign. The angles between the
forces and the x axis are 270 20 D 250° for the left gear and 90 20 D 70° for the right gear. The forces are FAY D 200j sin 250° j D
187.9 N, and FBY D 200j sin 70° j D 187.9 N. These magnitudes are
normal to the radii. The distances between the points A and B and
their respective action lines are the radii. The radii are RA D 0.120 m,
and RB D 0.080 m. The actions are negative. Thus
MA D 0.120j187.9j D 22.55 N-m,
A
and MB D 0.080j187.9j D 15.0 N-m.
B
20°
200 N
A
B
200 N
20°
120 mm
20°
80 mm
200 N
A
B
80 mm
120 mm
200 N
20°
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1
Problem 4.8 The support at the left end of the beam
will fail if the moment about A of the 15-kN force F
exceeds 18 kN-m. Based on this criterion, what is the
largest allowable length of the beam?
F
30°
B
A
25°
Solution:
MA D L Ð F sin 30° D L
15
2
30°
F = 15 kN
30°
MA D 7.5 L kN Ð m
set
MA D MAmax D 18 kN Ð m D 7.5 Lmax
Lmax D 2.4 m
L
25°
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1
Problem 4.9 The length of the bar AP is 650 mm. The
radius of the pulley is 120 mm. Equal forces T D 50 N
are applied to the ends of the cable. What is the sum of
the moments of the forces (a) about A; (b) about P.
45⬚
A
30⬚
T
T
P
45⬚
Solution:
(a)
MA D 50 N0.12 m 50 N0.12 m D 0
MA D 0
(b)
MP D 50 N0.12 m
50 N cos 30° 0.65 m sin 45° C 0.12 m cos 30° 50 N sin 30° 0.65 m cos 45° C 0.12 m sin 30° MP D 31.4 N-m or MP D 31.4 N-m CW
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1
Problem 4.10 The force F D 12 kN. A structural
engineer determines that the magnitude of the moment
due to F about P should not exceed 5 kN-m. What
is the acceptable range of the angle ˛? Assume that
0 ˛ 90° .
F
α
1m
P
2m
Solution: We have the moment about P
12 kN
MP D 12 kN sin ˛2 m 12 kN cos ˛1 m
α
MP D 122 sin ˛ cos ˛ kN-m
The moment must not exceed 5 kN-m
1m
Thus
5 kN-m ½ j122 sin ˛ cos ˛jkN-m
P
The limits occur when
122 sin ˛ cos ˛ D 5 ) ˛ D 37.3
2m
122 sin ˛ cos ˛ D 5 ) ˛ D 15.83°
So we must have 15.83° ˛ 37.3°
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1
Problem 4.11 The length of bar AB is 350 mm. The
moments exerted about points B and C by the vertical
force F are MB D 1.75 kN-m and MC D 4.20 kN-m.
Determine the force F and the length of bar AC.
B
30°
C
20°
A
F
Solution: We have
1.75 kN-m D F0.35 m sin 30° ) F D 10 kN
4.20 kN-m D FLAC cos 20° ) LAC D 0.447 m
In summary F D 10 kN, LAC D 447 mm
B
C
30°
20°
F
d1
30°
50
0.3
m
0.450
m
20°
d2
600 N
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1
Problem 4.12 Two students attempt to loosen a lug
nut with a lug wrench. One of the students exerts the
two 60-lb forces; the other, having to reach around his
friend, can only exert the two 30-lb forces. What torque
(moment) do they exert on the nut?
Solution: Determine the normal distance from line of action of the
normal force to the lug nut. Calculate moment; determine sign. The
two 60 lb forces act in a positive direction at a distance of 16 in from
the lug nut. The moment due to the 60 lb forces is
M60 D 260 lb16 in
30 lb
60 lb
30°
1 ft
12 in
D 160 ft-lb.
The normal component of the 30 lb force is F30 D 30 cos 30° D 26 lb.
This force acts at a distance of 16 in from the lug nut. The action is
positive. The moment due to the 30 lb forces is
16 in.
M30 D 226 lb16 in
1 ft
12 in
D 69.3 ft-lb.
The total moment is MT D 69.3 C 160 D 229.3 ft-lb
16 in.
60 lb
30° 30 lb
30°
30 lb
60 lb
16 in.
16 in.
30°
30 lb
60 lb
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1
Problem 4.13 Two equal and opposite forces act on
the beam. Determine the sum of the moments of the two
forces (a) about point P; (b) about point Q; (c) about the
point with coordinates x D 7 m, y D 5 m.
y
40 N
P
30⬚
40 N
2m
Solution:
(a)
30⬚
Q
2m
y
40 N
40 N
MP D 40 N cos 30° 2 m C 40 N cos 30° 4 m
D 69.3 N-m CCW
30°
30°
(b)
x
MQ D 40 N cos 30° 2 m D 69.3 N-m CCW
M D 40 N sin 30° 5 m C 40 N cos 30° 5 m
(c)
x
P
2m
2m
Q
40 N sin 30° 5 m 40 N cos 30° 3 m
D 69.3 N-m CCW
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1
Problem 4.14 The moment exerted about point E by
the weight is 299 in-lb. What moment does the weight
exert about point S?
in.
13
S
30°
12
E
40°
in.
Solution: The key is the geometry
From trigonometry,
cos 40° D
Thus
d1 D 12 in cos 30°
d1
and
d2
d1
, cos 30° D
13 in
12 in
D 10.3900
d2 D 13 in cos 40°
d1
S
30°
13 in
12 i
n
W
40°
E
d2
d2 D 9.9600
We are given that
299 in-lb D d2 W D 9.96 W
W D 30.0 lb
Now,
Ms D d1 C d2 W
Ms D 20.3530.0
Ms D 611 in-lb clockwise
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1
Problem 4.15 Three forces act on the square plate. Determine the sum of the moments of the forces (a) about
A, (b) about B, (c) about C.
200 N
C
200 N
3m
Solution: Determine the perpendicular distance between the points
and the lines of action. Determine sign, and calculate moment. (a) The
distances from point A to the lines of action is zero, hence the moment
about A is MA D 0. (b) The perpendicular distances of the lines of
action from B are: 3 m for the force through
p A, with a positive action,
32 C 32 D 2.12 m with a
and for the force through C, DC D 12
negative action. The moment about B is MB D 3200 2.12200 D
175.74 N-m (c) The distance of the force through A from C is 3 m,
with a positive action, and the distance of the force through B from
C is 3 m, with a positive action. The moment about C is MC D
23200 D 1200 N-m.
200 N
B
A
3m
200 N
C
200 N
3m
F
A
3m
200 N
B
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1
Problem 4.16 Three forces act on the piping. Determine the sum of the moments of the three forces about
point P.
2 kN
20⬚
2 kN
4 kN
0.2 m
P
0.2 m
0.2 m
0.2 m
Solution:
20°
4 kN
MP D 4 kN0.2 m C 2 kN0.6 m 2 kN cos 20° 0.2 m
2 kN
0.2 m
2 kN
C 2 kN sin 20° 0.4 m D 10.18 kN-m
P
MP D 0.298 kN-m CCW
0.2 m
0.2 m
0.2 m
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1
800 lb
Problem 4.17 Determine the sum of the moments of
the five forces acting on the Howe truss about point A.
600 lb
600 lb
D
400 lb
400 lb
C
E
8 ft
B
F
A
G
H
4 ft
I
4 ft
J
4 ft
K
4 ft
L
4 ft
4 ft
Solution: All of the moments about A are clockwise (negative).
The equation for the sum of the moments about A in units of ft-lb is
given by:
or
MA D 4400 8600 12800 16600 20400
MA D 33,600 ft-lb.
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1
Problem 4.18 The right support of the truss in
Problem 4.17 exerts an upward force of magnitude G.
(Assume that the force acts at the right end of the truss).
The sum of the moments about A due to the upward
force G and the five downward forces exerted on the
truss is zero. What is the force G?
Solution: Summing moments around A, we get
(ALL UNITS IN lbs)
800 lb
600 lb
600 lb
D
CMA D 4400 8600 12800
400 lb
400 lb
C
E
16600 20400 C 24 G D 0
8 ft
B
F
Solving, we get
A
G
G D 1400 lbs
H
4 ft
I
4 ft
400 lb
4ft
4ft
J
K
L
4 ft
4 ft
4 ft
4 ft
600 lb
800 lb
600 lb
400 lb
4ft
4ft
4ft
G
4ft
A
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G
1
Problem 4.19 The sum of the forces F1 and F2 is
250 N and the sum of the moments of F1 and F2 about
B is 700 N-m. What are F1 and F2 ?
F1
F2
A
B
2m
2m
2m
Solution:
C
F2
F D F1 C F2 D 250 N
F1
MB D 4F1 C 2F2 D 700 N-m
We have two equations in two unknowns. Solving, we have
 −2 m−  −2 m− 
B
F1 D 100 N, F2 D 150 N
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1
Problem 4.20 Consider the beam shown in Problem
4.19. If the two forces exert a 140 kN-m clockwise
moment about A and a 20 kN-m clockwise moment
about B, what are F1 and F2 ?
Solution: Sum of the moments about A:
MptA D 2F1 4F2 D 140 kN-m.
Sum of the moments about B:
MptB D 4F1 C 2F2 D 20 kN-m.
Solving these equations, we obtain
F1 D 30 kN, F2 D 50 kN.
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1
y
Problem 4.21 Three forces act on the car. The sum of
the forces is zero and the sum of the moments of the
forces about point P is zero.
(a)
(b)
3 ft
6 ft
Determine the forces A and B.
Determine the sum of the moments of the forces
about point Q.
x
B
P
Q
2800 lb
A
Solution:
6 ft
(a)
Fy : A C B 2800 lb D 0
MP : 2800 lb6 ft C A9 ft D 0
Solving we find
A D 1867 lb, B D 933 lb
(b)
3 ft
Q
P
2800 lb
MQ D 2800 lb3 ft B9 ft D 0
B
A
MQ D 0
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1
80 lb
Problem 4.22 Five forces act on the piping. The vector
sum of the forces is zero and the sum of the moments
of the forces about point P is zero.
(a)
(b)
45⬚
y
Determine the forces A, B, and C.
Determine the sum of the moments of the forces
about point Q.
2 ft
20 lb
Q
x
A
P
C
2 ft
B
2 ft
2 ft
Solution:
80 lb
The conditions given in the problem are:
45°
y
Fx : A C 80 lb cos 45° D 0
20 lb
2 ft
(a)
P
Fy : B C 20 lb C 80 lb sin 45° D 0
MP : 20 lb2 ft C6 ft 80 lb cos 45° 2 ft
C 80 lb sin 45° 4 ft D 0
Q
x
A
2 ft
B
2 ft
2 ft
C
Solving we have
A D 56.6 lb, B D 24.4 lb, C D 12.19 lb
(b)
MQ : 80 lb cos 45° 2 ft 80 lb sin 45° 2 ft
C20 lb4 ft C B6 ft D 0
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1
Problem 4.23 The weights (in ounces) of fish A, B,
and C are 2.7, 8.1, and 2.1, respectively. The sum of the
moments due to the weights of the fish about the point
where the mobile is attached to the ceiling is zero. What
is the weight of fish D?
12 in
Solution:
Solving
3 in
A
6 in
D
2 in
D D 0.6 oz
12
3
0
2 in
(8.1 + 2.1 +D) = (10.2 + D)
B
7 in
MO D 122.7 310.2 C D D 0
2,7
C
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1
Problem 4.24 The weight W D 1.2 kN. The sum of
the moments about A due to W and the force exerted
at the end of the bar by the rope is zero. What is the
tension in the rope?
Solution:
C
MA D 21.2 C 4T sin 30° D 0
2.4 C 2T D 0
T D 1.2 kN
60°
T
30°
30°
T sin 30°
A
W
2m
2m
A
2m
2m
1.2 kN
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1
Problem 4.25 The 160-N weights of the arms AB and
BC of the robotic manipulator act at their midpoints.
Determine the sum of the moments of the three weights
about A.
150
600
mm
The distance from A to the action line of the weight of the arm AB is:
dAB D 0.300 cos 40° D 0.2298 m
mm
The distance from A to the action line of the weight of the arm BC is
C
20°
B
Solution: The strategy is to find the perpendicular distance from
the points to the line of action of the forces, and determine the sum
of the moments, using the appropriate sign of the action.
40 N
dBC D 0.600cos 40° C 0.300cos 20° D 0.7415 m.
The distance from A to the line of action of the force is
m
0m
0
6
40°
160 N
dF D 0.600cos 40° C 0.600cos 20° C 0.150cos 20° D 1.1644 m.
A
160 N
The sum of the moments about A is
MA D dAB 160 dBC 160 dF 40 D 202 N-m
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1
Problem 4.26 The space shuttle’s attitude thrusters
exert two forces of magnitude F D 7.70 kN. What
moment do the thrusters exert about the center of
mass G?
2.2 m
2.2 m
F
F
G
5°
18 m
Solution: The key to this problem is getting the geometry correct.
The simplest way to do this is to break each force into components
parallel and perpendicular to the axis of the shuttle and then to sum
the moments of the components. (This will become much easier in the
next section)
6°
12 m
F sin 6°
F sin 5°
5˚
6°c
2.2 m
18 m
F cos 5°
FRONT
2.2 m
12 m
F cos 6°
REAR
CMFRONTý D 18F sin 5° 2.2F cos 5°
CMREARý D 2.2F cos 6° 12F sin 6°
CMTOTAL D MFRONT C MREAR
CMTOTAL D 4.80 C 7.19 N-m
CMTOTAL D 2.39 N-m
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1
Problem 4.27 The force F exerts a 200 ft-lb counterclockwise moment about A and a 100 ft-lb
clockwise moment about B. What are F and ?
y
A
(–5, 5) ft
F
θ
(4, 3) ft
x
B
(3, – 4) ft
Solution: The strategy is to resolve F into x- and y-components,
and compute the perpendicular distance to each component from A
and B. The components of F are: F D iFX C jFY . The vector from A
to the point of application is:
rAF D 4 5i C 3 5j D 9i 2j.
The perpendicular distances are dAX D 9 ft, and dAY D 2 ft, and the
actions are positive. The moment about A is MA D 9FY C 2FX D
200 ft-lb. The vector from B to the point of application is rBF D
4 3i C 3 4j D 1i C 7j; the distances dBX D 1 ft and dBY D
7 ft, the action of FY is positive and the action of FX is negative. The moment about B is MB D 1FY 7FX D 100 ft-lb. The
two simultaneous equations have solution: FY D 18.46 lb and FX D
16.92 lb. Take the ratio to find the angle:
D tan1
FY
FX
D tan1
18.46
16.92
y
A
(–5, 5) ft
F
θ
(4, 3) ft
x
B
(3, –4) ft
D tan1 1.091 D 47.5° .
From the Pythagorean theorem
jFj D
F2Y C F2X D
p
18.462 C 16.922 D 25.04 lb
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1
Problem 4.28 Five forces act on a link in the gearshifting mechanism of a lawn mower. The vector sum
of the five forces on the bar is zero. The sum of their
moments about the point where the forces Ax and Ay act
is zero.
(a)
(b)
Determine the forces Ax , Ay , and B.
Determine the sum of the moments of the forces
about the point where the force B acts.
Ay
Solution: The strategy is to resolve the forces into x- and
y-components, determine the perpendicular distances from B to the line
of action, determine the sign of the action, and compute the moments.
The angles are measured counterclockwise from the x axis. The
forces are
F2 D 30i cos 135° C j sin 135° D 21.21i C 21.21j
F1 D 25i cos 20° C j sin 20° D 23.50i C 8.55j.
(a)
The sum of the forces is
Ax
25 kN
F D A C B C F1 C F2 D 0.
20°
Substituting:
650 mm
450 mm
30 kN
45°
and
B
650 mm
350 mm
FX
D AX C BX C 23.5 21.2i D 0,
FY D AY C 21.2 C 8.55j D 0.
Solve the second equation: AY D 29.76 kN. The distances of
the forces from A are: the triangle has equal base and altitude,
hence the angle is 45° , so that the line of action of F1 passes
through A. The distance to the line of action of B is 0.65 m,
with a positive action. The distance to the line of action of the
y-component of F2 is 0.650 C 0.350 D 1 m, and the action is
positive. The distance to the line of action of the x-component
of F2 is 0.650 0.450 D 0.200 m, and the action is positive.
The moment about A is
MA D 8.551 C 23.50.2 C BX 0.65 D 0.
Solve: BX D 20.38 kN. Substitute into the force equation to
obtain AX D 18.09 kN
(b)
The distance from B to the line of action of the y-component of
F1 is 0.350 m, and the action is negative. The distance from B
to the line of action of AX is 0.650 m and the action is negative.
The distance from B to the line of action of AY is 1 m and the
action is positive. The distance from B to the line of action of
the x-component of F2 is 0.450 m and the action is negative. The
sum of the moments about B:
MB D 0.35021.21 0.65018.09
C 129.76 0.45023.5 D 0
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1
Problem 4.29 Five forces act on a model truss built by
a civil engineering student as part of a design project.
The dimensions are b D 300 mm and h D 400 mm; F D
100 N. The sum of the moments of the forces about the
point where Ax and Ay act is zero. If the weight of the
truss is negligible, what is the force B?
F
F
60°
Solution: The x- and y-components of the force F are
F D jFji cos 60° C j sin 60° D jFj0.5i C 0.866j.
The distance from A to the x-component is h and the action is positive.
The distances to the y-component are 3b and 5b. The distance to B is
6b. The sum of the moments about A is
60°
MA D 2jFj0.5h 3bjFj0.866 5bjFj0.866 C 6bB D 0.
Substitute and solve: B D
h
1.6784jFj
D 93.2 N
1.8
Ax
Ay
b
b
b
b
b
b
B
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1
Problem 4.30 Consider the truss shown in Problem
4.29. The dimensions are b D 3 ft and h D 4 ft; F D
300 lb. The vector sum of the forces acting on the truss
is zero, and the sum of the moments of the forces about
the point where Ax and Ay act is zero.
(a)
(b)
Determine the forces Ax , Ay , and B.
Determine the sum of the moments of the forces
about the point where the force B acts.
Solution: The forces are resolved into x- and y-components:
Solve the first: Ax D 300 lb. The distance from point A to the
x-components of the forces is h, and the action is positive. The
distances between the point A and the lines of action of the ycomponents of the forces are 3b and 5b. The actions are negative.
The distance to the line of action of the force B is 6b. The action
is positive. The sum of moments about point A is
F D 300i cos 60° C j sin 60° D 150i 259.8j.
(a)
The sum of the forces:
F D 2F C A C B D 0.
MA D 2150 h 3b259.8 5b259.8 C 6b B D 0.
The x- and y-components:
Substitute and solve: B D 279.7 lb. Substitute this value into the
force equation and solve: Ax D 519.6 279.7 D 239.9 lb
Fx D Ax 300i D 0,
(b)
Fy D 519.6 C Ay C Bj D 0.
The distances from B and the line of action of AY is 6b and the
action is negative. The distance between B and the x-component
of the forces is h and the action is positive. The distance between
B and the y-components of the forces is b and 3b, and the action
is positive. The sum of the moments about B:
MB D 6b239.9 C 2150 h C b259.8 C 3b259.8 D 0
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1
Problem 4.31 The mass m D 70 kg. What is the moment about A due to the force exerted on the beam at B
by the cable?
B
A
45°
30°
3m
m
Solution: The strategy is to resolve the force at B into components
parallel to and normal to the beam, and solve for the moment using
the normal component of the force. The force at B is to be determined
from the equilibrium conditions on the cable juncture O. Angles are
measured from the positive x axis. The forces at the cable juncture are:
FOB D jFOB ji cos 150° C j sin 150° D jFOB j0.866i C 0.5j
FOB
FOC
O
W
FOC D jFOC ji cos 45° C j sin 45° D jFOC j0.707i C 0.707j.
W D 709.810i 1j D 686.7j (N).
The equilibrium conditions are:
Fx D 0.866jFOB j C 0.7070jFOC ji D 0
FY D 0.500jFOB j C .707jFOC j 686.7j D 0.
Solve: jFOB j D 502.70 N. This is used to resolve the cable tension at B:
FB D 502.7i cos 330° C j sin 330° D 435.4i 251.4j. The distance
from A to the action line of the y-component at B is 3 m, and the
action is negative. The x-component at passes through A, so that the
action line distance is zero. The moment at A is MA D 3251.4 D
754.0 N-m
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1
Problem 4.32 The masses m1 D 6 kg and m2 D 12 kg
are suspended by the cable system shown. The cable BC
is horizontal. Determine the angle ˛ and the moment
about point P due to the force exerted on the vertical
post by the cable CD.
A
D
50⬚
B
C
0.8 m
a
m2
m1
P
CD
Solution: We start with the FBD of point C
Fx : CD cos ˛ BC D 0
BC
α
C
2
Fy : CD sin ˛ 12 kg9.81 m/s D 0
This gives two equations, however, we have three unknowns.
12 kg (9.81 m/s2)
We move to point B
AB
Fx : BC AB cos 50° D 0
Fy : AB sin 50° 6 kg9.81 m/s2 D 0
50°
We now have four equations in the four unknowns AB, BC, CD, ˛.
BC
B
Solving we find
AB D 76.8 N, BC D 49.4 N, CD D 127.6 N, ˛ D 67.2°
6 kg (9.81 m/s2)
˛ D 67.2°
D
Now we can find the moment about point P
α
MP D CD cos ˛0.8 m D 39.5 N-m CW
0.8 m
CD
P
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1
Problem 4.33 The bar AB exerts a force at B that helps
support the vertical retaining wall. The force is parallel
to the bar. The civil engineer wants the bar to exert a
38 kN-m moment about O. What is the magnitude of
the force the bar must exert?
B
4m
A
1m
O
1m
3m
Solution: The strategy is to resolve the force at B into components
parallel to and normal to the wall, determine the perpendicular distance
from O to the line of action, and compute the moment about O in terms
of the magnitude of the force exerted by the bar.
By inspection, the bar forms a 3, 4, 5 triangle. The angle the bar makes
with the horizontal is cos D 35 D 0.600, and sin D 45 D 0.800. The
force at B is FB D jFB j0.600i C 0.800j. The perpendicular distance
from O to the line of action of the x-component is 4 C 1 D 5 m, and
the action is positive. The distance from O to the line of action of
the y-component
is 1 m, and the action is positive. The moment about
O is
MO D 50.600jFB j C 10.800jFB j D 3.8jFB j D 38 kN, from
which jFB j D 10 kN
FB
B
4m
θ
O
1m
A
1m
3m
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1
Problem 4.34 A contestant in a fly-casting contest
snags his line in some grass. If the tension in the line is
5 lb, what moment does the force exerted on the rod by
the line exert about point H, where he holds the rod?
H
6 ft
4 ft
7 ft
Solution: The strategy is to resolve the line tension into a component normal to the rod; use the length from H to tip as the perpendicular distance; determine the sign of the action, and compute the
moment.
The line and rod form two right triangles, as shown in the sketch. The
angles are:
˛ D tan1
2
D 15.95°
7
ˇ D tan1
6
15
15 ft
α
β
α
2 ft
7 ft
6 ft
15 ft
β
D 21.8° .
The angle between the perpendicular distance line and the fishing line
is D ˛ C ˇ D 37.7° . The force normal to the
p distance line is F D
5sin 37.7° D 3.061 lb. The distance is d D 22 C 72 D 7.28 ft, and
the action is negative. The moment about H is MH D 7.283.061 D
22.3 ft-lb Check: The tension can be resolved into x and y components,
Fx D F cos ˇ D 4.642 lb, Fy D F sin ˇ D 1.857 lb.
The moment is
M D 2Fx C 7Fy D 22.28 D 22.3 ft-lb. check.
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1
A
Problem 4.35 The cables AB and AC help support the
tower. The tension in cable AB is 5 kN. The points A,
B, C, and O are contained in the same vertical plane.
(a)
(b)
What is the moment about O due to the force exerted
on the tower by cable AB?
If the sum of the moments about O due to the forces
exerted on the tower by the two cables is zero, what
is the tension in cable AC?
Solution: The strategy is to resolve the cable tensions into components normal to the vertical line through OA; use the height of the
tower as the perpendicular distance; determine the sign of the action,
and compute the moments.
(a)
(b)
20 m
60°
45°
C
O
A
B
FN
60°
FN
A
45°
The component normal to the line OA is FBN D 5cos 60° D
2.5 kN. The action is negative. The moment about O is MOA D
2.520 D 50 kN-m
By a similar process, the normal component of the tension in
the cable AC is FCN D jFC j cos 45° D 0.707jFC j. The action is
positive. If the sum of the moments is zero,
MO D 0.70720jFC j 50 D 0,
from which
jFC j D
50 kN m
D 3.54 kN
0.70720 m
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1
Problem 4.36 The cable from B to A (the sailboat’s
forestay) exerts a 230-N force at B. The cable from B to
C (the backstay) exerts a 660-N force at B. The bottom
of the sailboat’s mast is located at x D 4 m, y D 0. What
is the sum of the moments about the bottom of the mast
due to the forces exerted at B by the forestay and backstay?
y
B (4,13) m
A
(0,1.2) m
C
(9,1) m
B (4,13)
Solution: Triangle ABP
tan ˛ D
x
4
, ˛ D 18.73°
11.8
Triangle BCQ
230 N
tan ˇ D
660 N
5
, ˇ D 22.62°
12
CMO D 13230 sin ˛ 13660 sin ˇ
β
α
CMO D 2340 N-m
P
A (0,1.2)
C (9,1)
O (4,0)
Q
660 sin β
230 sin α
α
β
13 m
O
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1
Problem 4.37 The cable AB exerts a 290-kN force on
the building crane’s boom at B. The cable AC exerts a
148-kN force on the boom at C. Determine the sum of
the moments about P due to the forces the cables exert
on the boom.
A
B
8m
C
G
Boom
P
16 m
38 m
56 m
Solution:
A
56
8
290
8
8
290 kN56 m p
148 kN16 m
MP D p
3200
320
B
D 3.36 MNm
kN
40 m
8
8
14
C
kN
16
16 m
P
MP D 3.36 MN-m CW
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1
Problem 4.38 The mass of the building crane’s boom
in Problem 4.37 is 9000 kg. Its weight acts at G. The
sum of the moments about P due to the boom’s weight,
the force exerted at B by the cable AB, and the force
exerted at C by the cable AC is zero. Assume that the
tensions in cables AB and AC are equal. Determine the
tension in the cables.
Solution:
A
56
8
8
8
8
TAB 56 m p
TAC 16 m
MP D p
3200
320
C 9000 kg9.81 m/s2 38 m D 0
using TAB D TAC we solve and find
T AB
B
18 m
C
16
TA
22 m
C
16 m
P
9000 kg (9.81 m/s2)
TAB D TAC D 223 kN
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1
Problem 4.39 The mass of the luggage carrier and the
suitcase combined is 12 kg. Their weight acts at A. The
sum of the moments about the origin of the coordinate
system due to the weight acting at A and the vertical
force F applied to the handle of the luggage carrier
is zero. Determine the force F (a) if ˛ D 30° ; (b) if
˛ D 50° .
F
Solution: O is the origin of the coordinate system
12 kg9.81 m/s2 0.28 cos ˛ 0.14 sin ˛ D 0
Solving we find
x
FD
0.28 m
12 kg9.81 m/s2 0.28 cos ˛ 0.14 sin ˛
1.2 m cos ˛
(a) For ˛ D 30° We find
F D 19.54 N
For ˛ D 50° We find
F D 11.10 N
(b)
y
MO D F1.2 m cos ˛
0.14 m
1.2 m
A
a
C
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1
Problem 4.40 The hydraulic cylinder BC exerts a
300-kN force on the boom of the crane at C. The force
is parallel to the cylinder. What is the moment of the
force about A?
Solution: The strategy is to resolve the force exerted by the hydraulic cylinder into the normal component about the crane; determine the
distance; determine the sign of the action, and compute the moment.
Two right triangles are constructed: The angle formed by the hydraulic
cylinder with the horizontal is
ˇ D tan1
2.4
1.2
D 63.43° .
The angle formed by the crane with the horizontal is
C
˛ D tan1
1.4
3
D 25.02° .
A
2.4 m
1m
B
1.8 m
1.2 m
7m
The angle between the hydraulic cylinder and the crane is D ˇ ˛ D
38.42° . The normal component of the force is: Fp
N D 300sin 38.42° D 186.42 kN. The distance from point A is d D 1.42 C 32 D 3.31 m.
The action is positive. The moment about A is MO D C3.31186.42 D
617.15 kN-m Check: The force exerted by the actuator can be resolved
into x- and y-components, Fx D F cos ˇ D 134.16 kN, Fy D F sin ˇ
D 268.33 kN. The moment about the point A is M D 1.4Fx C 3.0
Fy D 617.15 kN m. check.
β
α
1.2 m
α
3m
2.4 m
β
1.4 m
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1
Problem 4.41 The hydraulic cylinder BC exerts a
2200-lb force on the boom of the crane at C. The force
is parallel to the cylinder. The angle ˛ D 40° . What is
the moment of the force about A?
Solution: Define the positive x direction to the right and the positive y direction as upward. Place a coordinate origin at A. The vector
from A to B is given as rAB D 6i ft. The location of point C in the xy
coordinates is given by
rAC D 9 cos 40° i C 9 sin 40° j D 6.89i C 5.79j ft.
t
6f
The unit vector from B to C is given by
xC xB i C yC yB j
eBC D xC xB 2 C yC yB 2
t
9f
C
Thus, the force along BC is FBC D 2200eBC D 337i C 2174j lb. The
moment of this force about point A is MA D 62174 D 13040 ft-lb
α
A
D 0.153i C 0.988j.
B
6 ft
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1
Problem 4.42 The hydraulic cylinder exerts an 8-kN
force at B that is parallel to the cylinder and points from
C toward B. Determine the moments of the force about
points A and D.
1m
D
C
Hydraulic
cylinder
1m
0.6 m
B
A
0.6 m
Scoop
0.15 m
Solution: Use x, y coords with origin A. We need the unit vector
from C to B, eCB . From the geometry,
eCB D 0.780i 0.625j
5.00 kN
6.25 kN
C (−0.15, + 0.6)
The force FCB is given by
FCB D 0.7808i 0.6258j kN
0.6 m
FCB D 6.25i 5.00j kN
For the moments about A and D, treat the components of FCB as two
separate forces.
0.15 m
CMA D 5, 000.15 0.66.25 kN Ð m
A (0 , 0)
CMA D 3.00 kN Ð m
5.0 kN
m
D
For the moment about D
C
0,4 m
MD D 5 kN1 m C 6.25 kN0.4 m
C
6.25 kN
CMD D 7.5 kN Ð m
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1
Problem 4.43 The structure shown in the diagram is
one of the two identical structures that support the scoop
of the excavator. The bar BC exerts a 700-N force at C
that points from C toward B. What is the moment of this
force about K?
320
mm
C
Shaft
100
mm
Scoop
260
mm
H
B
180
260 mm
mm
J
D
160
mm
L
K
380
mm
1040
mm 1120
mm
Solution:
320
320
700 N0.52 m D 353 Nm
MK D p
108800
80
700 N
520 mm
MK D 353 Nm CW
K
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1
Problem 4.44 In the structure shown in Problem 4.43,
the bar BC exerts a force at C that points from C toward
B. The hydraulic cylinder DH exerts a 1550-N force at D
that points from D toward H. The sum of the moments of
these two forces about K is zero. What is the magnitude
of the force that bar BC exerts at C?
Solution:
320
80
260 mm
MK D p
320
1550 N0.26 m p
F0.52 D 0
1264400
108800
1120
Solving we find
BC
1120
F D 796 N
100
1550 N
260 mm
K
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1
Problem 4.45 Use Eq. (4.2) to determine the moment
of the 50-lb force about the origin O. Compare your
answer with the two-dimensional description of the
moment.
Solution:
M D r ð F D 3j ð 50i
M D 150k ft-lb
y
Two dimensional description
50i (lb)
CMO D dF D 350 D 150 ft-lb
(0, 3, 0) ft
The descriptions match.
O
x
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1
Problem 4.46 Use Eq. (4.2) to determine the moment
of the 80-N force about the origin O letting r be the
vector (a) from O to A; (b) from O to B.
Solution:
(a) MO D rOA ð F
y
D 6i ð 80j D 480k N-m.
80j (N)
B
(b)
(6, 4, 0) m
MO D rOB ð F
D 6i C 4j ð 80j
D 480k N-m.
O
x
A (6, 0, 0) m
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1
Problem 4.47 A bioengineer studying an injury sustained in throwing the javelin estimates that the magnitude of the maximum force exerted was jFj D 360 N and
the perpendicular distance from O to the line of action of
F was 550 mm. The vector F and point O are contained
in the xy plane. Express the moment of F about the
shoulder joint at O as a vector.
Solution: The magnitude of the moment is jFj0.55 m D 360 N
0.55 m D 198 N-m. The moment vector is perpendicular to the xy
plane, and the right-hand rule indicates it points in the positive z direction. Therefore MO D 198k N-m.
y
F
y
550 mm
F
O
O
x
x
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1
y
Problem 4.48 Use Eq. (4.2) to determine the moment
of the 100-kN force (a) about A, (b) about B.
A
100j (kN)
6m
B
x
8m
Solution: (a) The coordinates of A are (0,6,0). The coordinates of
12 m
the point of application of the force are (8,0,0). The position vector
from A to the point of application of the force is rAF D 8 0i C
0 6j D 8i 6j. The force is F D 100j (kN). The cross product is
i
j
rAF ð F D 8 6
0 100
k 0 D 800k (kN-m)
0
(b) The coordinates of B are (12,0,0). The position vector from B to
the point of application of the force is rBF D 8 12 i D 4i. The
cross product is:
i
j
rBF ð F D 4 0
0 100
k 0 D 400k (kN-m)
0
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1
Problem 4.49 The cable AB exerts a 200-N force on
the support at A that points from A toward B. Use
Eq. (4.2) to determine the moment of this force about
point P in two ways: (a) letting r be the vector from P
to A; (b) letting r be the vector from P to B.
y
P (0.9, 0.8) m
(0.3, 0.5) m
A
B
(1, 0.2) m
Solution: First we express the force as a vector. The force points
in the same direction as the position vector AB.
x
AB D 1 0.3 mi C 0.2 0.5 mj D 0.7i 0.3j m
jABj D
p
0.7 m2 C 0.3 m2 D 0.58 m
200 N
0.7i 0.3j
FD p
0.58
(a)
200 N
0.7i 0.3j
MP D PA ð F D 0.6 mi 0.3 mj ð p
0.58
Carrying out the cross product we find
MP D 102.4 N-mk
(b)
200 N
0.7i 0.3j
MP D PB ð F D 0.1 mi 0.6 mj ð p
0.58
Carrying out the cross product we find
MP D 102.4 N-mk
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1
Problem 4.50 The line of action of F is contained in
the xy plane. The moment of F about O is 140k (Nm), and the moment of F about A is 280k (N-m). What
are the components of F?
y
A (0, 7, 0) m
F
(5, 3, 0) m
x
O
Solution: The strategy is to find the moments in terms of the
components of F and solve the resulting simultaneous equations. The
position vector from O to the point of application is rOF D 5i C 3j.
The position vector from A to the point of application is rAF D 5 0i C 3 7j D 5i 4j. The cross products:
i
rOF ð F D 5
FX
j
3
FY
k 0 D 5FY 3FX k D 140k, and
0
i
rAF ð F D 5
FX
j
4
FY
k 0 D 5FY C 4FX k D 280k.
0
y
A
(0,7,0)
F
(5,3,0)
O
x
Take the dot product of both sides with k to eliminate k. The simultaneous equations are:
5FY 3FX D 140, 5FY C 4FX D 280.
Solving: FY D 40, FX D 20, from which F D 20i C 40j (N)
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1
Problem 4.51 Use Eq. (4.2) to determine the sum of
the moments of the three forces (a) about A, (b) about B.
y
6 kN
3 kN
3 kN
B
A
0.2 m
0.2 m
0.2 m
x
0.2 m
Solution:
(a)
MA D 0.2i ð 3j C 0.4i ð 6j C 0.6i ð 3j
D O.
(b)
MB D 0.2i ð 3j C 0.4i ð 6j C 0.6i ð 3j
D O.
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1
Problem 4.52 Three forces are applied to the plate.
Use Eq. (4.2) to determine the sum of the moments of
the three forces about the origin O.
y
200 lb
3 ft
200 lb
3 ft
O
x
6 ft
4 ft
500 lb
Solution: The position vectors from O to the points of application of the forces are: rO1 D 3j, F1 D 200i; rO2 D 10i, F2 D 500j;
rO3 D 6i C 6j, F3 D 200i.
The sum of the moments about O is
i
MO D 0
200
j
j k i
0
3 0 C 10
0 0 0 500
k i
0 C 6
0 200
j k 6 0 lb
0 0
D 600k 5000k 1200k D 5600k ft-lb
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1
Problem 4.53 Three forces act on the plate. Use
Eq. (4.2) to determine the sum of the moments of the
three forces about point P.
y
4 kN
45⬚
3 kN
30⬚
0.18 m
P
0.10 m
20⬚
0.12 m
0.28 m
x
12 kN
Solution:
r1 D 0.12i C 0.08j m, F1 D 4 cos 45° i C 4 sin 45° j kN
r2 D 0.16i m, F2 D 3 cos 30° i C 3 sin 30° j kN
r3 D 0.16i 0.1j m, F3 D 12 cos 20° i 12 sin 20° j kN
MP D r1 ð F1 C r2 ð F2 C r3 ð F3
MP D 0.145 kN-mk D 145 N-mk
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1
Problem 4.54 (a) Determine the magnitude of the
moment of the 150-N force about A by calculating the
perpendicular distance from A to the line of action of
the force.
y
(0, 6, 0) m
150k (N)
(b) Use Eq. (4.2) to determine the magnitude of the
moment of the 150-N force about A.
A
x
(6, 0, 0) m
z
Solution:
(a)
The perpendicular from A to the line of action of the force lies
in the xy plane
p
d
D
62 C 62 D 8.485 m
jMj D dF D 8.485150 D 1270 N-m
(b)
M D 6i C 6j ð 150k D 900j C 900i N-m
p
jMj D
9002 C 9002 D 1270 N-m
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1
Problem 4.55 (a) Determine the magnitude of the
moment of the 600-N force about A by calculating the
perpendicular distance from A to the line of action of
the force.
y
A
(0.6, 0.5, 0.4) m
(b) Use Eq. (4.2) to determine the magnitude of the
moment of the 600-N force about A.
x
0.8 m
600i (N)
z
Solution:
(a)
Choose some point Px, 0, 0.8 m. on the line of action of the
force. The distance from A to P is then
d D x 0.6 m2 C 0 0.5 m2 C 0.8 m 0.4 m2
The perpendicular distance is the shortest distance d which occurs
when x D 0.6 m. We have d D 0.6403 m. Thus the magnitude of
the moment is
M D 600 N0.6403 m D 384 N-m
(b)
Define the point on the end of the rod to be B. Then AB D
0.6i 0.5j C 0.4k m we have
M D AB ð F D 0.6i 0.5j C 0.4k m ð 600 Ni
M D 240j C 300k N-m
Thus the magnitude is
MD
240 Nm2 C 300 Nm2 D 384 N-m
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1
Problem 4.56 The pneumatic support AB exerts a 35N force on the fixture at B that points from A toward
B. Determine the magnitude of the moment of the force
about the hinge at O.
Solution: We identify the force and the position vector.
35 N
FD p
0.42i C 0.14j 0.07k
0.2009
r D OA D 0.48i 0.04j C 0.04k m
Then
M D r ð F D 0.219i C 1.312j C 3.94k N-m ) M D 4.15 N-m
y
B (60, 100, ⫺30) mm
O
z
x
A
(480, ⫺40, 40) mm
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1
Problem 4.57 A force F D 20i 30j C 60k (lb). The
moment of F about a point P is MP D 450i 100j 200k (ft-lb). What is the perpendicular distance from
point P to the line of action of F?
Solution: The magnitude of the moment is
p
jMP j D
4502 C 1002 C 2002 D 502.5 (ft-lb).
The magnitude of the force is
p
jFj D
202 C 302 C 602 D 70 (lb).
The perpendicular distance is
DD
502.5 ft-lb
jMP j
D
D 7.18 ft
jFj
70 lb
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1
Problem 4.58 A force F is applied at the point (8, 6,
13) m. Its magnitude is jFj D 90 N, and the moment of
F about the point (4, 2, 6) is zero. What are the components of F?
Solution:
i
r ð F D 8 4
Fx
j
k 6 2 13 6 D 0.
Fy
Fz From Eq. (3), Fy D Fx , and from Eqs. (1) and (2), Fz D 74 Fx . The
magnitude is
90 N D
F2x C F2y C F2z
Therefore
D
F2x C F2x C
7
4 Fx
2
.
4Fz 7Fy D 0,
1
Solving, we obtain Fx D š40 N. we see that
7Fx 4Fz D 0,
2
F D 40i C 40j C 70k N
4Fy 4Fx D 0.
3
or
F D 40i 40j 70k N.
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1
Problem 4.59
y
The force F D 30i C 20j 10k (N).
(a)
Determine the magnitude of the moment of F
about A.
(b)
Suppose that you can change the direction of F while
keeping its magnitude constant, and you want to
choose a direction that maximizes the moment of
F about A. What is the magnitude of the resulting
maximum moment?
F
A
(8, 2, – 4) m
(4, 3, 3) m
x
z
Solution: The vector from A to the point of application of F is
r D 4i 1j 7k m
and
jrj D
p
42 C 12 C 72 D 8.12 m
(a)
The moment of F about A is
MA
i
D r ð F D 4
30
jMA j D
(b)
j
k 1 7 D 150i 170j C 110k N-m
20 10 p
1502 C 1702 C 1102 D 252 N-m
The maximum moment occurs when r ? F. In this case
jMAmax j D jrjjFj
Hence, we need jFj.
jFj D
p
302 C 202 C 102 D 37.4 N
Thus,
jMAmax j D 8.1237.4 D 304 N-m
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1
Problem 4.60 The direction cosines of the force F are
cos x D 0.818, cos y D 0.182, and cos z D 0.545. The
support of the beam at O will fail if the magnitude of the
moment of F about O exceeds 100 kN-m. Determine the
magnitude of the largest force F that can safely be applied
to the beam.
y
z
O
F
3m
x
Solution: The strategy is to determine the perpendicular distance
from O to the action line of F, and to calculate the largest magnitude of
F from MO D DjFj. The position vector from O to the point of application of F is rOF D 3i (m). Resolve the position vector into components parallel and normal to F. The component parallel to F is rP D
rOF Ð eF eF , where the unit vector eF parallel to F is eF D i cos X C
j cos Y C k cos Z D 0.818i C 0.182j 0.545k. The dot product is
rOF Ð eF D 2.454. The parallel component is rP D 2.007i C 0.4466j 1.3374k. The component normal to F is rN D rOF rP D 3 2i 0.4466j C 1.3374k. The magnitude
p of the normal component is the
perpendicular distance: D D 12 C 0.44662 C 1.3372 D 1.7283 m.
The maximum moment allowed is MO D 1.7283jFj D 100 kN-m,
from which
jFj D
100 kN-m
D 57.86 ¾
D 58 kN
1.7283 m
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1
Problem 4.61 The force F exerted on the grip of the
exercise machine points in the direction of the unit vector
e D 23 i 23 j C 13 k and its magnitude is 120 N. Determine
the magnitude of the moment of F about the origin O.
150 mm
y
F
Solution: The vector from O to the point of application of the
O
200 mm
force is
r D 0.25i C 0.2j 0.15k m
z
250 mm
x
and the force is F D jFje
or
F D 80i 80j C 40k N.
The moment of F about O is
i
MO D r ð F D 0.25
80
j
k 0.2 0.15 N-m
80
40 or
MO D 4i 22j 36k N-m
and
jMO j D
p
42 C 222 C 362 N-m
jMO j D 42.4 N-m
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1
Problem 4.62 The force F in Problem 4.61 points in
the direction of the unit vector e D 23 i 23 j C 13 k. The
support at O will safely support a moment of 560 N-m
magnitude.
(a)
Based on this criterion, what is the largest safe
magnitude of F?
(b)
If the force F may be exerted in any direction, what
is its largest safe magnitude?
Solution: See the figure of Problem 4.61.
The moment in Problem 4.61 can be written as
i
MO D 0.25
2F
3
j
0.2
23 F
k 0.15 where F D jFj
C 13 F MO D 0.0333i 0.1833j 0.3kF
If we set jMO j D 560 N-m, we can solve for jFmax j
560 D 0.353jFmax j
jFmax j D 1586 N
(b)
If F can be in any direction, then the worst case is when r ? F.
The moment in this case is jMO j D jrjjFworst j
p
0.252 C 0.22 C 0.152 D 0.3536 m
And the magnitude of MO is
jrj D
p
jMO j D 0.03332 C 0.18332 C 0.32 F
560 D 0.3536jFWORST j
jMO j D 0.353 F
jFworst j D 1584 N
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1
Problem 4.63 A civil engineer in Boulder, Colorado
estimates that under the severest expected Chinook
winds, the total force on the highway sign will be
F D 2.8i 1.8j (kN). Let MO be the moment due to F
about the base O of the cylindrical column supporting
the sign. The y component of MO is called the torsion
exerted on the cylindrical column at the base, and the
component of MO parallel to the xz plane is called
the bending moment. Determine the magnitudes of the
torsion and bending moment.
y
F
8m
8m
O
x
z
Solution: The total moment is
M D 8j C 8k m ð 2.8i 1.8j kN
D 14.4i C 22.4j 22.4k kN-m
We now identify
Torsion D My D 22.4 kN-m
Bending moment D Mx 2 C Mz 2
D 14.4 kNm2 C 22.4 kNm2 D 26.6 kN-m
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1
Problem 4.64 The weights of the arms OA and AB of
the robotic manipulator act at their midpoints. The direction cosines of the centerline of arm OA are cos x D
0.500, cos y D 0.866, and cos z D 0, and the direction
cosines of the centerline of arm AB are cos x D 0.707,
cos y D 0.619, and cos z D 0.342. What is the sum
of the moments about O due to the two forces?
Solution: By definition, the direction cosines are the scalar components of the unit vectors. Thus the unit vectors are e1 D 0.5i C 0.866j,
and e2 D 0.707i C 0.619j 0.342k. The position vectors of the midpoints of the arms are
r1 D 0.3e1 D 0.30.5i C 0.866j D 0.15i C 0.2598j
r2 D 0.6e1 C 0.3e2 D 0.512i C 0.7053j 0.1026k.
The sum of moments is
y
0
60
mm
B
160 N
M D r1 ð W1 C r2 ð W2
i
j
k i
D 0.15 0.2598 0 C 0.512
0
200 0 0
j
0.7053
160
k
0.1026 0
D 16.42i 111.92k (N-m)
600 mm
A
200 N
O
z
x
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1
Problem 4.65 The tension in cable AB is 100 lb. If
you want the magnitude of the moment about the base
O of the tree due to the forces exerted on the tree by the
two ropes to be 1500 ft-lb, what is the necessary tension
in rope AC ?
Solution: We have the forces
100 lb
TAC
8j C 10k, F2 D p
14i 8j C 14k
F1 D p
164
456
Thus the total moment is
M D 8 ftj ð F1 C F2 D 625 ft lb C 5.24 ft TAC i
y
5.24 ftTAC K
The magnitude squared is then
625 ft lb C 5.24 ft TAC 2 C 5.24 ft TAC 2 D 1500 ft lb2
Solving we find
TAC D 134 lb
(0, 8, 0) ft
A
x
O
B
(0, 0, 10) ft
z
(14, 0, 14) ft
C
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1
Problem 4.66* A force F acts at the top end A of the
pole. Its magnitude is jFj D 6 kN and its x component
is Fx D 4 kN. The coordinates of point A are shown.
Determine the components of F so that the magnitude
of the moment due to F about the base P of the pole is
as large as possible. (There are two answers.)
Solution: The force is given by F D 4 kNi C Fy j C Fz k.
Since the magnitude is constrained we must have
4 kN2 C Fy 2 C Fz 2 D 6 kN2 ) Fz D
20 kN2 Fy 2
Thus we will use (suppressing the units)
y
FD
F
4i C Fy j C
20 Fy 2 k
A
(4, 3, ⫺2) m
The moment is now given by
M D 4i C 3j 2k ð F
M D 2Fy C 3 20 Fy 2 i 8 C 4 20 Fy 2 j C 12 C 4Fy k
The magnitude is
P
x
M2 D 708 5Fy 2 C 64 20 Fy 2 C 12Fy 8 C 20 Fy 2
To maximize this quantity we solve
z
dM2
D 0 for the critical values
dFy
of Fy .
There are three solutions Fy D 4.00, 3.72, 3.38.
The first and third solutions produce the same maximum moment.
The second answer corresponds to a local minimum and is therefore discarded.
So the force that produces the largest moment is
F D 4i 4j C 2k
or
F D 4i 3.38j C 2.92k
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1
y
Problem 4.67 The force F D 5i (kN) acts on the ring
A where the cables AB, AC, and AD are joined. What is
the sum of the moments about point D due to the force
F and the three forces exerted on the ring by the cables?
D (0, 6, 0) m
A
(12, 4, 2) m
Strategy: The ring is in equilibrium. Use what you
know about the four forces acting on it.
C
B
Solution: The vector from D to A is
rDA D 12i 2j C 2k m.
The sum of the moments about point D is given by
F
(6, 0, 0) m
x
(0, 4, 6) m
z
FAD
A
F
MD D rDA ð FAD C rDA ð FAC C rDA ð FAB C rDA ð F
MD D rDA ð FAD C FAC C FAB C F
FAC
FAB
However, we are given that ring A is in equilibrium and this
implies that
FAD C FAC C FAB C F D O D 0
Thus,
MD D rDA ð O D 0
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1
Problem 4.68 In Problem 4.67, determine the moment about point D due to the force exerted on the ring
A by the cable AB.
Solution: We need to write the forces as magnitudes times the
D(0, 6, 0)
appropriate unit vectors, write the equilibrium equations for A in component form, and then solve the resulting three equations for the three
unknown magnitudes. The unit vectors are of the form
FAD
FAC
eAP D
A(12, 4, 2) m
F = 5i (kN)
xP xA i C yP yA j C zP zA k
jrAP j
C(0, 4, 6) m
Where P takes on values B, C, and D
B(6, 0, 0) m
Calculating the unit vectors, we get

e D 0.802i 0.535j 0.267k

 AB
eAC D 0.949i C 0j C 0.316k


eAD D 0.973i C 0.162j 0.162k
From equilibrium, we have
FAB eAB C FAC eAC C FAD eAD C 5i kN D 0
In component form, we get

i: 0.802FAB 0.949FAC 0.973FAD C 5 D 0


j: 0.535FAB C 0FAC C 0.162FAD D 0


k: 0.267FAB C 0.316FAC 0.162FAD D 0
Solving, we get
FAB D 779.5 N, FAC D 1976 N
FAD D 2569 N
The vector from D to A is
rDA D 12i 2j C 2k m
The force FAB is given by
FAB D FAB eAB
FAB D 0.625i 0.417j 0.208k kN
The moment about D is given by
i
MD D rDA ð FAB D 12
0.625
j
k 2
2 0.417 0.208 MD D 1.25i C 1.25j 6.25k kN-m
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1
Problem 4.69 The tower is 70 m tall. The tensions
in cables AB, AC, and AD are 4 kN, 2 kN, and 2 kN,
respectively. Determine the sum of the moments about
the origin O due to the forces exerted by the cables at
point A.
y
A
D
35 m
B
35 m
40 m
C
O
x
40 m
40 m
z
Solution: The coordinates of the points are A (0, 70, 0), B (40, 0,
0), C (40, 0, 40) D(35, 0, 35). The position vectors corresponding
to the cables are:
The sum of the forces acting at A are
TA D 0.2792i 6.6615j C 0.07239k (kN-m)
rAD D 35 0i C 0 70j C 35 0k
The position vector of A is rOA D 70j. The moment about O is M D
rOA ð TA
rAD D 35i 70k 35k
i
M D 0
0.2792
rAC D 40 0i C 0 70j C 40 0k
rAC D 40i 70j C 40k
j
70
6.6615
k
0
0.07239 D 700.07239i j0 k700.2792 D 5.067i 19.54k
rAB D 40 0i C 0 70j C 0 0k
rAB D 40i 70j C 0k
The unit vectors corresponding to these position vectors are:
eAD D
35
70
35
rAD
D
i
j
jrAD j
85.73
85.73
85.73
D 0.4082i 0.8165j 0.4082k
eAC D
40
70
40
rAC
D i
jC
k
jrAC j
90
90
90
D 0.4444i 0.7778j C 0.4444k
eAB D
rAB
40
70
D
i
j C 0k D 0.4962i 0.8682j C 0k
jrAB j
80.6
80.6
The forces at point A are
TAB D 4eAB D 1.9846i 3.4729j C 0k
TAC D 2eAB D 0.8889i 1.5556j C 0.8889k
TAD D 2eAD D 0.8165i 1.6330j 0.8165k.
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1
Problem 4.70 Consider the 70-m tower in Problem 4.69. Suppose that the tension in cable AB is 4 kN,
and you want to adjust the tensions in cables AC and
AD so that the sum of the moments about the origin O
due to the forces exerted by the cables at point A is zero.
Determine the tensions.
Solution: From Varignon’s theorem, the moment is zero only if
the resultant of the forces normal to the vector rOA is zero. From
Problem 4.69 the unit vectors are:
35
70
35
rAD
D
i
j
eAD D
jrAD j
85.73
85.73
85.73
D 0.4082i 0.8165j 0.4082k
eAC D
rAC
40
70
40
D i
jC
k
jrAC j
90
90
90
D 0.4444i 0.7778j C 0.4444k
eAB D
The tensions are TAB D 4eAB , TAC D jTAC jeAC , and TAD D jTAD jeAD .
The components normal to rOA are
FX D 0.4082jTAD j 0.4444jTAC j C 1.9846i D 0
FZ D 0.4082jTAD j C 0.4444jTAC jk D 0.
The HP-28S calculator was used to solve these equations:
jTAC j D 2.23 kN, jTAD j D 2.43 kN
40
70
rAB
D
i
j C 0k D 0.4963i 0.8685j C 0k
jrAB j
80.6
80.6
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1
Problem 4.71 The tension in cable AB is 150 N. The
tension in cable AC is 100 N. Determine the sum of the
moments about D due to the forces exerted on the wall
by the cables.
y
5m
5m
B
C
Solution: The coordinates of the points A, B, C are A (8, 0, 0),
B (0, 4, 5), C (0, 8, 5), D(0, 0, 5). The point A is the intersection of
the lines of action of the forces. The position vector DA is
4m
8m
8m
D
rDA D 8i C 0j 5k.
A
z
x
The position vectors AB and AC are
rAB D 8i C 4j 5k,
rAB D
p
82 C 42 C 52 D 10.247 m.
rAC D 8i C 8j C 5k,
rAC D
p
82 C 82 C 52 D 12.369 m.
The unit vectors parallel to the cables are:
eAB D 0.7807i C 0.3904j 0.4879k,
eAC D 0.6468i C 0.6468j C 0.4042k.
i
MD D 8
181.79
j
0
123.24
k 5 D 123.245i
C32.77 8C32.77 5181.79j C 8123.24k
MD D 616.2i 117.11j 985.9k (N-m)
(Note: An alternate method of solution is to express the moment in
terms of the sum: MD D rDC ð TC C rDB ð TB .
The tensions are
y
TAB D 150eAB D 117.11i C 58.56j 73.19k,
5m
TAC D 100eAC D 64.68i C 64.68j C 40.42k.
The sum of the forces exerted by the wall on A is
5m
B
4m
C
TA D 181.79i C 123.24j 32.77k.
The force exerted on the wall by the cables is TA . The moment about
D is MD D rDA ð TA ,
A
8m
z D
8m
F
x
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1
Problem 4.72 Consider the wall shown in Problem 4.71. The total force exerted by the two cables in
the direction perpendicular to the wall is 2 kN. The
magnitude of the sum of the moments about D due to
the forces exerted on the wall by the cables is 18 kN-m.
What are the tensions in the cables?
Solution: From the solution of Problem 4.71, we have rDA D 8i C
0j 5k. Forces in both cables pass through point A and we can use
this vector to determine moments of both forces about D. The position
vectors AB and AC are
rAB D 8i C 4j 5k,
jrAB j D
p
82 C 42 C 52 D 10.247 m.
rAC D 8i C 8j C 5k,
jrAC j D
p
82 C 82 C 52 D 12.369 m.
The unit vectors parallel to the cables are:
eAB D 0.7807i C 0.3904j 0.4879k,
eAC D 0.6468i C 0.6468j C 0.4042k.
The tensions are
TBA D TBA eAB D TBA 0.7807i C 0.3904j 0.4879k, and
TCA D TCA eAC D TCA 0.6468i C 0.6468j C 0.4042k.
The sum of the forces exerted by the cables perpendicular to the wall
is given by
TPerpendicular D TAB 0.7807 C TAC 0.6468 D 2 kN.
The moments of these two forces about D are given by
MD D rDA ð TCA C rDA ð TBA D rDA ð TCA C TBA .
The sum of the two forces is given by
i
8
MD D TCA C TCB X
j
0
TCA C TCB Y
k
.
5
TCA C TCB Z This expression can be expanded to yield
MD D 5TCA C TCB Y i C [8TCA C TCB Z 5TCA C TCB X ]j
C 8TCA C TCB Y k.
The magnitude of this vector is given as 18 kN-m. Thus, we obtain
the relation
jMD j D
25TCA C TCB 2Y C [8TCA C TCB Z
D 18 kN-m.
5TCA C TCB X ]2 C 64TCA C TCB 2Y
We now have two equations in the two tensions in the cables. Either
algebraic substitution or a numerical solver can be used to give
TBA D 1.596 kN, and TCA D 1.166 kN.
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1
Problem 4.73 The tension in the cable BD is 1 kN. As
a result, cable BD exerts a 1-kN force on the “ball” at
B that points from B toward D. Determine the moment
of this force about point A.
Solution: We have the force and position vectors
FD
1 kN
4i C 2j C 4k, r D AB D 4i C 3j C k m
6
The moment is then
y
M D r ð F D 1.667i 3.33j C 3.33k kN-m
C
(0, 4, ⫺3) m
B
(4, 3, 1) m
D
(0, 5, 5) m
x
A
E
z
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1
Problem 4.74* Suppose that the mass of the suspended
object E in Problem 4.73 is 100 kg and the mass of the
bar AB is 20 kg. Assume that the weight of the bar acts
at its midpoint. By using the fact that the sum of the
moments about point A due to the weight of the bar and
the forces exerted on the “ball” at B by the three cables
BC, BD, and BE is zero, determine the tensions in the
cables BC and BD.
Solution: We have the following forces applied at point B.
F1 D 100 kg9.81 m/s2 j,
F3 D
TBC
F2 D p 4i C j 4k,
33
TBD
4i C 2j C 4k
6
In addition we have the weight of the bar F4 D 20 kg9.81 m/s2 j
The moment around point A is
MA D 4i C 3j C k m ð F1 C F2 C F3 C 2i C 1.5j C 0.5k m ð F4 D 0
Carrying out the cross products and breaking into components we find
Mx D 1079 2.26TBC C 1.667TBD D 0
My D 2.089TBC 3.333TBD D 0
Mz D 4316 C 2.785TBC C 3.333TBD D 0
Only two of these three equations are independent. Solving we find
TBC D 886 N, TBD D 555 N
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1
Problem 4.75 The 200-kg slider at A is held in place
on the smooth vertical bar by the cable AB. Determine
the moment about the bottom of the bar (point C with
coordinates x D 2 m, y D z D 0) due to the force exerted
on the slider by the cable.
y
2m
B
A
5m
2m
2m
C
x
z
Solution: The slider is in equilibrium. The smooth bar exerts no
vertical forces on the slider. Hence, the vertical component of FAB
supports the weight of the slider.
FAB
The unit vector from A to B is determined from the coordinates of
points A and B A2, 2, 0, B0, 5, 2 m
Thus,
H
rAB D 2i C 3j C 2k m
eAB D 0.485i C 0.728j C 0.485k
and
−mg j
FAB D FAB eAB
The horizontal force exerted by the bar on the slider is
H D Hx i C Hz k
Equilibrium requires H C FAB mgj D 0
i: Hx 0.485FAB D 0 m D 200 kg
j: 0.728FAB mg D 0 g D 9.81 m/s2
k: Hz C 0.485FAB D 0
Solving, we get
FAB D 2697N D 2, 70 kN
Hx D 1308N D 1.31 kN
Hz D 1308N D 1.31 kN
rCA D 2j m
FAB D FAB eAB
FAB D 1308i C 1962j C 1308k N
i
Mc D 0
1308
j
2
1962
k 0 1308 Mc D 2616i C 0j C 2616k N-m
Mc D 2.62i C 2.62i kN-m
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1
y
Problem 4.76 To evaluate the adequacy of the design
of the vertical steel post, you must determine the moment
about the bottom of the post due to the force exerted on
the post at B by the cable AB. A calibrated strain gauge
mounted on cable AC indicates that the tension in cable
AC is 22 kN. What is the moment?
5m
5m
C
D
4m
8m
(6, 2, 0) m
B
A
O
z
3m
12 m
x
Solution: To find the moment, we must find the force in cable AB.
In order to do this, we must find the forces in cables AO and AD also.
This requires that we solve the equilibrium problem at A.
Our first task is to write unit vectors eAB , eAO , eAC , and eAD . Each
will be of the form
xi xA i C yi yA j C zi zA k
eAi D xi xA 2 C yi yA 2 C zi zA 2
where i takes on the values B, C, D, and O. We get
eAB D 0.986i C 0.164j C 0k
eAC D 0.609i C 0.609j C 0.508k
eAD D 0.744i C 0.248j 0.620k
D(0, 4, −5) m
C (0, 8, 5) m
TAD
TAC
A (6, 2, 0) m
TAO
B(12, 3, 0) m
TAB
O(0, 0, 0) m
In component form,

T e
C TAC eACx C TAD eADx C TAO eAOx D 0

 AB ABx
TAB eABy C TAC eACy C TAD eADy C TAO eAOy D 0


TAB eABz C TAC eACz C TAD eADz C TAO eAOz D 0
We know TAC D 22 kN. Substituting this in, we have 3 eqns in 3
unknowns. Solving, we get
eAO D 0.949i 0.316j C 0k
We now write the forces as
TAB D 163.05 kN,
TAD D 18.01 kN
TAO D 141.28 kN
We now know that TAB is given as
TAB D TAB eAB
TAB D TAB eAB D 160.8i C 26.8j kN
TAC D TAC eAC
and that the force acting at B is TAB .
TAD D TAD eAD
The moment about the bottom of the post is given by
TAO D TAO eAO
MBOTTOM D r ð TAB D 3j ð TAB We then sum the forces and write the equilibrium equations in component form.
For equilibrium at A,
FA D 0
Solving, we get
MBOTTOM D 482k kN-m
FA D TAB C TAC C TAD C TAO D 0.
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1
Problem 4.77 The force F D 20i C 40j 10k (N).
Use Eqs. (4.5) and (4.6) to determine the moment due
to F about the z axis. (First see if you can write down
the result without using the equations.)
Solution:
Mz axis D [k Ð r ð F]k D [k·8 mi ð [20i C 40j 10k] N]k
0
0
0
Mz axis D 8 m
20 N 40 N
y
0 D 320 N-m
10 N 1
Thus
Mz-axis D 320 N-mk
F
x
(8, 0, 0) m
z
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1
Problem 4.78 Use Eqs. (4.5) and (4.6) to determine
the moment of the 20-N force about (a) the x axis, (b) the
y axis, (c) the z axis. (First see if you can write down
the results without using the equations.)
Solution: The force is parallel to the z axis. The perpendicular
distance from the x axis to the line of action of the force is 4 m. The
perpendicular distance frompthe y axis p
is 7 m and the perpendicular
distance from the z axis is 42 C 72 D 65 m.
By inspection, the moment about the x axis is
y
Mx D 420i (N-m)
(7, 4, 0) m
Mx D 80i N-m
By inspection, the moment about the y axis is My D 720j N-m
20k (N)
x
My D 140j (N-m)
z
By inspection, the moment about the z axis is zero since F is parallel
to the z axis.
Mz D 0 N-m
Now for the calculations using (4.5) and (4.6)
ML D [e Ð r ð F]e
1 0
Mx D 7 4
0 0
0 0 i D 80i N-m
20 0 1
My D 7 4
0 0
0 0 j D 140j N-m
20 0 0
Mz D 7 4
0 0
1 0 k D 0k N-m
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1
y
Problem 4.79 Three forces parallel to the y axis act on
the rectangular plate. Use Eqs. (4.5) and (4.6) to determine the sum of the moments of the forces about the x
axis. (First see if you can write down the result without
using the equations.)
3 kN
x
2 kN
6 kN
600 mm
900 mm
z
Solution: By inspection, the 3 kN force has no moment about the
x axis since it acts through the x axis. The perpendicular distances of
the other two forces from the x axis is 0.6 m. The H 2 kN force has a
positive moment and the 6 kN force has a negative about the x axis.
1 0
M6 kN D 0 0
0 6
Mx D [20.6 60.6]i kN
Mx D 2.4i kN
0 .6 i D 3.6i kN
0
Mx D M3 kN C M2 kN C M6 kN
Mx D 0 C 1.2i 3.6i kN
Mx D 2.4i kN
Calculating the result:
1
0 0 M3 kN D 0
0 0 i D 0i kN
0 3 0 1
0 0 M2 kN D 0
0 .6 i D 1.2i kN
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1
Problem 4.80 Consider the rectangular plate shown in
Problem 4.79. The three forces are parallel to the y
axis. Determine the sum of the moments of the forces
(a) about the y axis, (b) about the z axis.
Solution: (a) The magnitude of the moments about the y axis is
(b) The magnitude of each moment about the z axis is
M D eY Ð r ð F. The position vectors of the three forces are given
in the solution to Problem 4.79. The magnitude for each force is:
0
1 0 eZ Ð r ð F D 0.9 0 0 D 2.7,
0
3 0 0
1 0 0 0 D 0,
eY Ð r ð F D 0.9
0
3 0 0
1 0 eY Ð r ð F D 0.9 0 0.6 D 0,
0
6 0 0
1 0 eY Ð r ð F D 0
0 0.6 D 0
0 2 0 Thus the moment about the y axis is zero, since the magnitude of each
moment is zero.
0
eZ Ð r ð F D 0.9
0 C
0 1 0 0.6 D 5.4,
6 0 0
0 1 eZ Ð r ð F D 0
0 0.6 D 0.
0 2 0 Thus the moment about the z axis is
MZ D 2.7eZ C 5.4eZ D 2.7k (kN-m)
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1
Problem 4.81 The person exerts a force F D 0.2i 0.4j C 1.2k (lb) on the gate at C. Point C lies in the xy
plane. What moment does the person exert about the
gate’s hinge axis, which is coincident with the y axis?
y
A
C
3.5 ft
x
B
2 ft
Solution:
M D [e Ð r ð F]e
e D j,
r D 2i ft,
F is given
0
1
0 0
0 j D 2.4j ft-lb
MY D 2
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1
y
Problem 4.82 Four forces act on the plate. Their
components are
FB
FA D 2i C 4j C 2k (kN),
FA
FB D 3j 3k (kN),
x
FD
FC
FC D 2j C 3k (kN),
FD D 2i C 6j C 4k (kN).
Determine the sum of the moments of the forces
(a) about the x axis; (b) about the z axis.
z
2m
3m
Solution: Note that FA acts at the origin so no moment is generated
about the origin. For the other forces we have
i
j
k j
k i
0
2m
0
0 C 3 m
MO D 3 m
0
0
2 kN 3 kN 3 kN 3 kN
i
C 0
2 kN
j
0
6 kN
k 2m
4 kN MO D 16i C 4j C 15k kN-m
Now we find
Mx D MO Ð i D 16 kN-m, Mz D MO Ð k D 15 kN-m
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1
Problem 4.83
(a)
(b)
y
The force F D 30i C 20j 10k (lb).
What is the moment of F about the y axis?
Suppose that you keep the magnitude of F fixed,
but you change its direction so as to make the
moment of F about the y axis as large as possible.
What is the magnitude of the resulting moment?
F
(4, 2, 2) ft
x
z
Solution:
(a) My D j Ð [4i C 2j C 2k ft ð 30i C 20j 10k lb]
0
1
0 My D 4 ft
2 ft
2 ft D 100 ft lb
30 lb 20 lb 10 lb ) My D 100 ft-lbj
(b)
p
p
Mymax D Fd D 302 C 202 C 102 lb 42 C 22 ft
D 167.3 ft-lb
Note that d is the distance from the y axis, not the distance from
the origin.
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1
Problem 4.84 Suppose that the moment of the force F
shown in Problem 4.83 about the x axis is 80i (ft-lb),
the moment about the y axis is zero, and the moment
about the z axis is 160k (ft-lb). If Fy D 80 lb, what are
Fx and Fz ?
Solution: The magnitudes of the moments:
eX
e ž r ð F D rX
FX
eY
rY
FY
eZ rZ ,
FZ 0
eZ Ð r ð F D 4
FX
0
2
80
1 2 D 320 2FX D 160
FZ Solve: FX D 80 lb, FZ D 40 lb, from which the force vector is F D
80i C 80j C 40k
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1
Problem 4.85 The robotic manipulator is stationary.
The weights of the arms AB and BC act at their
midpoints. The direction cosines of the centerline of arm
AB are cos x D 0.500, cos y D 0.866, cos z D 0, and
the direction cosines of the centerline of arm BC are
cos x D 0.707, cos y D 0.619, cos z D 0.342. What
total moment is exerted about the z axis by the weights
of the arms?
m
0m
y
C
60
160 N
600 mm
B
200 N
A
z
x
Solution: The unit vectors along AB and AC are of the form
e D cos x i C cos y j C cos z k.
The unit vectors are
eAB D 0.500i C 0.866j C 0k and eBC D 0.707i C 0.619j 0.342k.
The vector to point G at the center of arm AB is
rAG D 3000.500i C 0.866j C 0k D 150i C 259.8j C 0k mm,
and the vector from A to the point H at the center of arm BC is
given by
rAH D rAB C rBH D 600eAB C 300eBC
D 512.1i C 705.3j 102.6k mm.
The weight vectors acting at G and H are WG D 200j N, and WH D
160j N. The moment vectors of these forces about the z axis are of
the form
eX
e ž r ð F D rX
FX
ey
rY
FY
ez rZ .
FZ Here, WG and WH take on the role of F, and e D k.
Substituting into the form for the moment of the force at G, we get
0
e ž r ð F D 0.150
0
0
1 0.260 0 D 30 N-m.
200 0 Similarly, for the moment of the force at H, we get
0
e ž r ð F D 0.512
0
0
0.705
160
1 0.103 D 81.9 N-m.
0 The total moment about the z axis is the sum of the two moments.
Hence, Mz axis D 111.9 N-m
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1
Problem 4.86 In Problem 4.85, what total moment is
exerted about the x axis by the weights of the arms?
Solution: The solution is identical to that of Problem 4.85 except
that e D i. Substituting into the form for the moment of the force at
G, we get
1
e Ð r ð F D 0.150
0
0
0 0.260 0 D 0 N-m.
200 0 Similarly, for the moment of the force at H, we get
1
e Ð r ð F D 0.512
0
0
0 0.705 0.103 D 16.4 N-m.
160
0 The total moment about the x axis is the sum of the two moments.
Hence, Mx axis D 16.4 N-m
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1
y
Problem 4.87 Two forces are exerted on the crankshaft
by the connecting rods. The direction cosines of FA are
cos x D 0.182, cos y D 0.818, and cos z D 0.545,
and its magnitude is 4 kN. The direction cosines of
FB are cos x D 0.182, cos y D 0.818, and cos z D
0.545, and its magnitude is 2 kN. What is the sum of
the moments of the two forces about the x axis? (This
is the moment that causes the crankshaft to rotate.)
Solution: The coordinates of the points of action of the two forces
FB
FA
360 mm
O
z
160 mm
80 mm
80 mm x
are A (0.16, 0, 0.08), B (0.36, 0, 0.08). The position vectors are
rOA D 0.16i C 0j C 0.08k (m),
rOB D 0.36i C 0j 0.08k (m).
The unit vectors parallel to the forces are given by the direction
cosines:
eFA D 0.182i C 0.818j C 0.545k,
eFB D 0.182i C 0.818j 0.545k
The forces are
FA D 0.728i C 3.272j C 2.18k (kN)
FB D 0.364i C 1.636j 1.09k (kN)
The magnitude of the moments:
1
eX Ð rA ð FA D 0.16
0.728
1
eX Ð rB ð FB D 0.36
0.364
0
0
3.272
0
0
1.636
0 0.08 D 0.2618,
2.18 0 0.08 D 0.1309
1.09 The sum of the moments about the x axis is
MX D 0.2618eX C 0.1309eX D 0.1309i kN-m.
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1
Problem 4.88 Determine the moment of the 20-N force
about the line AB. Use Eqs. (4.5) and (4.6), letting the unit
vector e point (a) from A toward B, (b) from B toward A.
y
A (0, 5, 0) m
(7, 4, 0) m
20k (N)
B
(– 4, 0, 0) m
x
z
Solution: First, we need the unit vector
Using eAB
xB xA i C yB yA j C zB zA k
eAB D xB xA 2 C yB yA 2 C zB zA 2
0.625
ML D 7
0
0.781 0 1
0 0.625i 0.781j
0
20 eAB D 0.625i 0.781j D eBA
ML D 76.1i 95.1j N-m
Now, the moment of the 20k (N) force about AB is given as
ex
ML D rx
Fx
ey
ry
Fy
ez rz e where e is eAB or eBA
Fz Using eBA
0.625
ML D 7
0
0.781
1
0
0 0 0.625i C 0.781j
20 For this problem, r must go from line AB to the point of application
of the force. Let us use point A.
ML D 76.1i 95.1j N-m
r D 7 0i C 4 5j C 0 0k m
Ł Results are the same
r D 7i 1j C 0k m
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1
Problem 4.90 The force F D 10i C 12j 6k (N).
What is the moment of F about the line OA? Draw a
sketch to indicate the sense of the moment.
y
A
(0, 6, 4) m
F
O
x
(8, 0, 6) m
z
Solution: The strategy is to determine a unit vector parallel to OA
and to use this to determine the moment about OA. The vector parallel
to OA is rOA D 6j C 4k. The magnitude: F. The unit vector parallel
to OA is eOA D 0.8321j C 0.5547k. The vector from O to the point
of application of F is rOF D 8i C 6k. The magnitude of the moment
about OA is
0 0.8321
0
jMO j D eOA Ð rOF ð F D 8
10
12
0.5547 6 6 D 89.8614 C 53.251 D 143.1 N-m.
The moment about OA is MOA D jMOA jeOA D 119.1j C 79.4k (N-m).
The sense of the moment is in the direction of the curled fingers of
the right hand when the thumb is parallel to OA, pointing to A.
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1
Problem 4.91 The tension in the cable AB is 1 kN.
Determine the moment about the x axis due to the force
exerted on the hatch by the cable at point B. Draw a
sketch to indicate the sense of the moment.
y
A
(400, 300, 0) mm
x
600 mm
B
1000 mm
z
Solution: The vector parallel to BA is
rBA D 0.4 1i C 0.3j 0.6k D 0.6i C 0.3j 0.6k.
The unit vector parallel to BA is
eBA D 0.6667i C 0.3333j 0.6667k.
The moment about O is
i
MO D rOB ð T D 1
0.6667
j
0
0.3333
k
0.6
0.66667 MO D 0.2i C 0.2667j C 0.3333k.
The magnitude is
jMX j D eX Ð MO D 0.2 kN-m.
The moment is MX D 0.2i kN-m. The sense is clockwise when
viewed along the x axis toward the origin.
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1
y
Problem 4.92 Determine the moment of the force applied at D about the straight line through the hinges A
and B. (The line through A and B lies in the yz plane.)
6 ft
20i – 60j (lb)
E
A
D
x
4 ft
2 ft
B
z
20°
C
4 ft
Solution: From the figure, we see that the unit vector along the
line from A toward B is given by eAB D sin 20° j C cos 20° k. The
position vector is rAD D 4i ft, and the force vector is as shown in the
figure. The moment vector of a force about an axis is of the form
eX
e ž r ð F D rX
FX
ey
rY
FY
ez rZ .
FZ For this case,
0
e ž r ð F D 4
20
sin 20°
0
60
cos 20° 0 D 240 cos 20° ft-lb
0 D 225.5 ft-lb.
The negative sign is because the moment is opposite in direction to
the unit vector from A to B.
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1
Problem 4.93 In Problem 4.92, the tension in the cable
CE is 160 lb. Determine the moment of the force exerted
by the cable on the hatch at C about the straight line
through the hinges A and B.
Solution: From the figure, we see that the unit vector along the line
from A toward B is given by eAB D sin 20° j C cos 20° k. The position
vector is rBC D 4i ft. The coordinates of point C are (4, 4 sin 20° ,
4 cos 20° ). The unit vector along CE is 0.703i C 0.592j C 0.394k
and the force vector is as shown acting at point D.
The moment vector is a force about an axis is of the form
eX
e ž r ð F D rX
FX
ey
rY
FY
ez rZ .
FZ For this case,
rCE D 4i C 3.368j C 2.242k
TCE D 160eCE D 112.488i C 94.715j C 63.049k
0
4
e ž r ð F D 112.488
sin 20°
0
94.715
cos 20° 0 D 240 cos 20° ft-lb
63.049 D 701 ft-lbs.
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1
Problem 4.94 The coordinates of A are (2.4, 0,
0.6) m, and the coordinates of B are (2.2, 0.7,
1.2) m. The force exerted at B by the sailboat’s main
sheet AB is 130 N. Determine the moment of the force
about the centerline of the mast (the y axis). Draw a
sketch to indicate the sense of the moment.
y
x
B
A
z
Solution: The position vectors:
rOA D 2.4i 0.6k (m), rOB D 2.2i C 0.7j 1.2k (m),
rBA D 2.4 C 2.2i C 0 0.7j C 0.6 C 1.2k (m)
D 0.2i 0.7j C 0.6k (m).
The magnitude is jrBA j D 0.9434 m.
The unit vector parallel to BA is
eBA D 0.2120i 0.7420j C 0.6360k.
The tension is TBA D 130eBA .
The moment of TBA about the origin is
i
MO D rOB ð TBA D 2.2
27.56
or
j
k 0.7
1.2 ,
96.46 82.68 MO D 57.88i C 214.97j C 231.5k.
The magnitude of the moment about the y axis is
jMY j D eY Ð MO D 214.97 N-m.
The moment is MY D eY 214.97 D 214.97j N-m.
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1
Problem 4.95 The tension in cable AB is 200 lb. Determine the moments about each of the coordinate axes
due to the force exerted on point B by the cable. Draw
sketches to indicate the senses of the moments.
y
A
(2, 5, –2) ft
x
z
B (10, –2, 3) ft
y
Solution: The position vector from B to A is
443 ft-lb
rBA D 2 10i C [5 2]j C 2 3k
D 8i C 7j 5k ft,
187 ft-lb
So the force exerted on B is
F D 200
x
rBA
D 136.2i C 119.2j 85.1k lb.
jrBA j
The moment of F about the origin O is
i
rOB ð F D 10
136.2
j
2
119.2
k 3 85.1 z
919 ft-b
D 187i C 443j C 919k ft-lb.
The moments about the x, y, and z axes are
[rOB ð F Ð i]i D 187i ft-lb,
[rOB ð F Ð j]j D 443j ft-lb,
[rOB ð F Ð k]k D 919k ft-lb.
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1
Problem 4.96 The total force exerted on the blades
of the turbine by the steam nozzle is F D 20i 120j C
100k (N), and it effectively acts at the point (100, 80,
300) mm. What moment is exerted about the axis of the
turbine (the x axis)?
y
Fixed
Rotating
x
Solution: The moment about the origin is
i
MO D 0.1
20
j
k 0.08 0.3 120 100 D 44.0i 4.0j 13.6k N-m.
The moment about the x axis is
z
MO Ð ii D 44.0i N-m.
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1
Problem 4.97 The tension in cable AB is 50 N. Determine the moment about the line OC due to the force
exerted by the cable at B. Draw a sketch to indicate the
sense of the moment.
y
A
(0, 7, 0) m
C
(0, 7, 10) m
O
z
x
B
(14, 0, 14) m
Solution: The vector OC is rOC D 7j C 10k. The unit vector
parallel to OC is eOC D 0.573j C 0.819k. The position vectors of A
and B are
rOA D 7j (m), and rOB D 14i C 0j C 14k (m).
The vector parallel to BA is
rBA D 0 14i C 7 0j C 0 14k D 14i C 7j 14k.
The magnitude:
jrBA j D
p
142 C 72 C 142 D 21 m.
The unit vector parallel to BA is
eBA D 0.6667i C 0.3333j 0.6667k.
The tension acting on B is
TBA D 33.335i C 16.665j 33.335k (N).
The moment about the origin is
i
MO D rOB ð TBA D 14
33.335
j
0
16.665
k
14 33.335 D 233.31i C 233.31k.
The magnitude of the moment about the line OC is
jMOC j D eOC Ð MO D 191.2 N-m.
The moment about the line OC is
MOC D 191.1eOC D 109.6j C 156.5k (N-m).
The sense of the moment is in the direction of the curled fingers of
the right hand when the thumb points from O toward C.
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1
Problem 4.98 The tension in cable AB is 80 lb. What
is the moment about the line CD due to the force exerted
by the cable on the wall at B?
y
8 ft
3 ft
Solution: The strategy is to find the moment about the point C
exerted by the force at B, and then to find the component of that
moment acting along the line CD. The coordinates of the points B,
C, D are B (8, 6, 0), C (3, 6, 0), D(3, 0, 0). The position vectors
are: rOB D 8i C 6j, rOC D 3i C 6j, rOD D 3i. The vector parallel to
CD is rCD D rOD rOC D 6j. The unit vector parallel to CD is
eCD D 1j. The vector from point C to B is rCB D rOB rOC D 5i.
B
The position vector of A is rOA D 6i C 10k. The vector parallel to
BA is rBA D rOA rOB D 2i 6j C 10k. The magnitude is jrBA j D
11.832 ft. The unit vector parallel to BA is
C
6 ft
eBA D 0.1690i 0.5071j C 0.8452k.
The tension acting at B is
x
D
TBA D 80eBA D 13.52i 40.57j C 67.62k.
The magnitude of the moment about CD due to the tension acting at
B is
z
A (6, 0, 10) ft
0
jMCD j D eCD Ð rCB ð TBA D 5
13.52
1
0
40.57
0 0 67.62 D 338.1 (ft lb).
The moment about CD is MCD D 338.1eCD D 338.1j (ft lb). The
sense of the moment is along the curled fingers of the right hand when
the thumb is parallel to CD, pointing toward D.
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1
Problem 4.99 The magnitude of the force F is 0.2 N
and its direction cosines are cos x D 0.727, cos y D
0.364, and cos z D 0.582. Determine the magnitude
of the moment of F about the axis AB of the spool.
Solution: We have
rAB D 0.3i 0.1j 0.4k m,
rAB D
0.32 C 0.12 C 0.42 m D
p
0.26 m
y
1
0.3i 0.1j 0.4k
eAB D p
0.26
B
F D 0.2 N0.727i 0.364j C 0.582k
(200, 400, 0) mm
rAP D 0.26i 0.025j 0.11k m
(160, 475, 290) mm
P
A
(⫺100, 500, 400) mm
F
x
Now the magnitude of the moment about the spool axis AB is
0.3
0.1
0.4 0.2 N MAB D p
0.26 m 0.025 m 0.11 m D 0.0146 N-m
0.26 0.727
0.364
0.582 z
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1
Problem 4.100 A motorist applies the two forces
shown to loosen a lug nut. The direction cosines of
4
3
F are cos x D 13
, cos y D 12
, and cos z D 13
. If the
13
magnitude of the moment about the x axis must be 32 ftlb to loosen the nut, what is the magnitude of the forces
the motorist must apply?
y
Solution: The unit vectors for the forces are the direction cosines.
The position vector of the force F is rOF D 1.333k ft. The magnitude
of the moment due to F is
1
0
jMOF j D eX Ð rOF ð F D 0.3077F
0
0
0.9231F
0
1.333 0.2308F jMOF j D 1.230F ft lb.
The magnitude of the moment due to F is
–F
F
jMOF j D eX Ð rOF ð F
1
D 0
.3077F
z
0
0
0.9231F
0
1.333 D 1.230F ft lb.
0.2308F The total moment about the x axis is
16 in
16 in
x
MX D 1.230Fi C 1.230Fi D 2.46Fi,
from which, for a total magnitude of 32 ft lb, the force to be applied is
FD
32
D 13 lb
2.46
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1
Problem 4.101 The tension in cable AB is 2 kN. What
is the magnitude of the moment about the shaft CD due
to the force exerted by the cable at A? Draw a sketch to
indicate the sense of the moment about the shaft.
2m
C
A
Solution: The strategy is to determine the moment about C due
to A, and determine the component parallel to CD. The moment is
determined from the distance CA and the components of the tension,
which is to be found from the magnitude of the tension and the unit
vector parallel to AB. The coordinates of the points A, B, C, and
D are: A (2, 2, 0), B (3, 0, 1), C (0, 2, 0), and D (0,0,0). The unit
vector parallel to CD is by inspection eCD D 1j. The position vectors
parallel to DC, DA, and DB:
rDC D 2j, rDA D 2i C 2j, rDB D 3i C 1k.
2m
The vector parallel to CA is rCA D 2i. The vector parallel to AB is
rAB D rDB rDA D 1i 2j C 1k.
D
B
3m
1m
The magnitude: jrAB j D 2.4495 m. The unit vector parallel to AB is
eAB D 0.4082i 0.8165j C 0.4082k.
The tension is
TAB D 2eAB D 0.8165i 1.633j C 0.8165k.
The magnitude of the moment about CD is
0
jMCD j D eCD Ð rCA ð TAB D 2
0.8164
1
0 0
0 1.633 0.8165 D 1.633 kN-m.
The moment about CD is
MCD D eCD jMCD j D 1.633j (kN-m).
The sense is in the direction of the curled fingers of the right hand
when the thumb is parallel to DC, pointed toward D.
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1
Problem 4.102 The axis of the car’s wheel passes
through the origin of the coordinate system and
its direction cosines are cos x D 0.940, cos y D 0,
cos z D 0.342. The force exerted on the tire by the road
effectively acts at the point x D 0, y D 0.36 m, z D 0
and has components F D 720i C 3660j C 1240k (N).
What is the moment of F about the wheel’s axis?
Solution: We have to determine the moment about the axle where
a unit vector along the axle is
e D cos x i C cos y j C cos z k
e D 0.940i C 0j C 0.342k
The vector from the origin to the point of contact with the road is
r D 0i 0.36j C 0k m
The force exerted at the point of contact is
F D 720i C 3660j C 1240k N
y
The moment of the force F about the axle is
MAXLE D [e Ð r ð F]e
x
z
0.940
MAXLE D 0
720
0
0.36
C3660
0.342 0 0.940i C 0.342k N-m
C1240 MAXLE D 508.260.940i C 0.342k N-m
MAXLE D 478i 174k N-m
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1
Problem 4.103 The direction cosines of the centerline
OA are cos x D 0.500, cos y D 0.866, and cos z D 0,
and the direction cosines of the line AG are cos x D
0.707, cos y D 0.619, and cos z D 0.342. What is the
moment about OA due to the 250-N weight? Draw a
sketch to indicate the sense of the moment about the
shaft.
Solution: By definition, the direction cosines are the scalar components of the unit vectors. Thus the unit vectors are
e1 D 0.5i C 0.866j, and e2 D 0.707i C 0.619j 0.341k.
The force is W D 250j (N). The position vector of the 250 N weight is
rW D 0.600e1 C 0.750e2 D 0.8303i C 0.9839j 0.2565k
The moment about OA is
m
G
m
50
y
7
250 N
MOA D eOA eOA Ð rW ð W
0.5
D 0.8303
0
0.866
0.9839
250
0
0.2565 e1 D 32.06e1
0
D 16i 27.77j (N-m)
A
600 mm
The moment is anti parallel to the unit vector parallel to OA, with the
sense of the moment in the direction of the curled fingers when the
thumb of the right hand is directed oppositely to the direction of the
unit vector.
O
z
x
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1
Problem 4.104 The radius of the steering wheel is
200 mm. The distance from O to C is 1 m. The center C
of the steering wheel lies in the x y plane. The driver
exerts a force F D 10i C 10j 5k (N) on the wheel at A.
If the angle ˛ D 0, what is the magnitude of the moment
about the shaft OC? Draw a sketch to indicate the sense
of the moment about the shaft.
y
Solution: The strategy is to determine the moment about C, and
then determine its component about OC. The radius vectors parallel
to OC and CA are:
rOC D 1i cos 20° C j sin 20° C 0k D 0.9397i C 0.3420j.
The line from C to the x axis is perpendicular to OC since it lies in
the plane of the steering wheel. The unit vector from C to the x axis is
eCX D i cos20 90 C j sin20 90 D 0.3420i 0.9397j,
where the angle is measured positive counterclockwise from the x axis.
The vector parallel to CA is
F
C
A
O
z
rCA D 0.2eCX D C0.0684i 0.1879j (m).
The magnitude of the moment about OC
20°
α
0.9397
jMOC j D eOC Ð rCA ð F D 0.0684
10
0.3420
0.1879
10
0 0 5 x
D 0.9998 D 1 N-m.
The sense of the moment is in the direction of the curled fingers of
the right hand if the thumb is parallel to OC, pointing from O to C.
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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 4.105* The magnitude of the force F is 10 N.
Suppose that you want to choose the direction of the
force F so that the magnitude of its moment about the
line L is a maximum. Determine the components of F
and the magnitude of its moment about L. (There are
two solutions for F.)
y
Solution: The moment of the general force F D Fx i C Fy j C Fz k
about the line is developed by
eBA D
1
3i C 6j 6k
D i C 2j 2k,
9
3
rBP D 12i C 2j 2k m,
MBA D eBA Ð rBP ð F
A (3, 8, 0) m
This expression simplifies to MBA D We also have the constraint that 10 N2 D Fx 2 C Fy 2 C Fz 2
L
F
Since Fx does not contribute to the moment we set it equal to zero.
Solving the constraint equation for Fz and substituting this into the
expression for the moment we find
B
(0, 2, 6) m
P
(12, 4, 4) m
MBA D x
z
22 m
Fy C Fz 3
22
Fy š
3
100 Fy 2 . )
dMBA
D0
dFy
p
p
) Fy D š5 2N ) Fz D š5 2
We thus have two answers:
F D 7.07j C 7.07k N
or F D 7.07j C 7.07k
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1
Problem 4.106 The weight W causes a tension of
100 lb in cable CD. If d D 2 ft, what is the moment
about the z axis due to the force exerted by the cable
CD at point C?
y
(12, 10, 0) ft
(0, 3, 0) ft
Solution: The strategy is to use the unit vector parallel to the bar
to locate point C relative to the origin, and then use this location to
find the unit vector parallel to the cable CD. With the tension resolved
into components about the origin, the moment about the origin can be
resolved into components along the z axis. Denote the top of the bar
by T and the bottom of the bar by B. The position vectors of the ends
of the bar are:
D
W
C
d
x
z
(3, 0, 10) ft
rOB D 3i C 0j C 10k, rOT D 12i C 10j C 0k.
The vector from the bottom to the top of the bar is
rBT D rOT rOB D 9i C 10j 10k.
The magnitude:
jrBT j D
p
92 C 102 C 102 D 16.763 ft.
The unit vector parallel to the bar, pointing toward the top, is
eBT D 0.5369i C 0.5965j 0.5965k.
The position vector of the point C relative to the bottom of the bar is
rBC D 2eBT D 1.074i C 1.193j 1.193k.
The position vector of point C relative to the origin is
rOC D rOB C rBC D 4.074i C 1.193j C 8.807k.
The position vector of point D is
rOD D 0i C 3j C 0k.
The vector parallel to CD is
rCD D rOD rOC D 4.074i C 1.807j 8.807k.
The magnitude is
p
jrCD j D
4.0742 C 1.8072 C 8.8072 D 9.87 ft.
The unit vector parallel to CD is
eCD D 0.4127i C 0.1831j 0.8923k.
The tension is
TCD D 100eCD D 41.27i C 18.31j 89.23k lb.
The magnitude of the moment about the z axis is
0
jMO j D eZ Ð rOC ð TCD D 4.074
41.27
0
1.193
18.31
1 8.807 89.23 D 123.83 ft lb
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1
Problem 4.107* The y axis points upward. The weight
of the 4-kg rectangular plate acts at the midpoint G of
the plate. The sum of the moments about the straight
line through the supports A and B due to the weight of
the plate and the force exerted on the plate by the cable
CD is zero. What is the tension in the cable?
y
Solution: Note that the coordinates of point G are (150, 152.5, 195).
We calculate the moment about the line BA due to the two forces
as follows.
eBA D
0.1i C 0.07j 0.36k
p
0.1445
r1 D 0.2i 0.125j C 0.03k m,
A
(100, 500, 700) mm
F1 D TCD
(100, 250, 0) mm
0.1i C 0.445j C 0.31k
p
0.304125
r2 D 0.15i 0.0275j 0.165k m,
D
F2 D 4 kg9.81 m/s2 j
G
MBA D eBA Ð r1 ð F1 C r2 ð F2 x
B
The moment reduces to
(0, 180, 360) mm
MBA D 3.871 N-m 0.17793 mTCD D 0 ) TCD D 21.8 N
C
(200, 55, 390) mm
z
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1
Problem 4.108 Determine the moment of the couple
(a) about the origin O; (b) about the point with coordinates x D 2 m, y D 4 m.
Solution:
M D 8 N0.4 m D 3.2 N-m
M D 3.2 N-m CW
y
8i (N)
0.2 m
O
x
0.2 m
⫺8i (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 4.109
plane.
The forces are contained in the xy
Solution: The right hand force is
F D [1000 lb]cos 60° i sin 60° j
(a)
(b)
Determine the moment of the couple and represent
it as shown in Fig. 4.28c.
What is the sum of the moments of the two forces
about the point (10, 40, 20) ft?
y
1000 lb
MC D 40i ð 500i 867j ft-lb
60°
x
20 ft
The vector from the x intercept of the left force to that of the right
force is r D 40i ft.
The moment is MC D r ð F
1000 lb
60°
F D C500i 867j lb.
20 ft
MC D 34700 ft-lb k
or MC D 34700 ft-lb) clockwise
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1
Problem 4.110 The moment of the couple is 600k
(N-m). What is the angle ˛?
Solution:
M D 100 N cos ˛4 m C 100 N sin ˛5 m D 600 N-m
y
Solving yields two answers:
˛ D 30.9°
a
or ˛ D 71.8°
(0, 4) m
100 N
100 N
a
(5, 0) m
x
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1
Problem 4.111 Point P is contained in the xy
plane, jFj D 100 N, and the moment of the couple is
500k (N-m). What are the coordinates of P?
Solution: The force is
F D 100i cos30° C j sin30° D 86.6i 50j.
Let r be the distance OP. The vector parallel to OP is
y
P
r D ri cos 70° C j sin 70° D r0.3420i C 0.9397j.
30°
F
The moment is
–F
70°
x
i
M D r ð F D 0.3420r
86.6
From which, r D
j
0.9397r
50.0
k 0 D 98.48rk.
0
500
D 5.077 m. From above,
98.48
r D 5.0770.3420i C 0.9397j.
The coordinates of P are
x D 5.0770.3420 D 1.74 m, y D 5.0770.9397 D 4.77 m
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1
Problem 4.112
plane.
(a)
(b)
(c)
The forces are contained in the xy
y
100 N
Determine the sum of the moments of the two
couples.
What is the sum of the moments of the four forces
about the point (6, 6, 2) m?
Represent the result of (a) as shown in Fig. 4.28c.
4m
100 N
2m
x
2m
100 N
4m
100 N
Solution: The position vectors for the forces on the x axis are
rX1 D 2i, rX2 D 2i. The position vectors for the forces on the y axis
are rY1 D 4j, rY2 D 4j. The force on the positive x axis is FX D
C100j (N). The force on the positive y axis is FY D C100i (N).
(a)
The sum of the moments is
y
400 k
x
M D rX1 rX2 ð FX C rY1 rY2 ð FY
i
j
D 4 0
0 100
(b)
(c) The figure is shown.
k i
0 C 0
0 100
j k 8 0 D 400k (N-m)
0 0
The vector from the point P(6, 6, 2) to the force on the
positive x axis is
rPX1 D rX1 rp D 2 C 6i C 6j 2k.
The vector from the point P to the force on the negative x axis is
rpX2 D rX2 rp D 2 C 6i C 6j 2k.
The vector from point P to the force on the positive y axis is
rpY1 D rY1 rp D C6i C 4 C 6j 2k D 6i C 10j 2k.
The vector from P to the force on the negative y axis is
rPY2 D rY2 rP D 6i C 4 C 6j 2k D 6i C 2j 2k.
The sum of the moments:
M D rpX1 rpX2 ð FX C rpY1 rpY2 ð FY
i
D 4
0
j
0
100
k i
0 C 0
0 100
j k 8 0 D 400k (N-m)
0 0
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1
Problem 4.113 The moment of the couple is 40 kN-m
counterclockwise.
(a)
(b)
Express the moment of the couple as a vector.
Draw a sketch showing two equal and opposite
forces that exert the given moment.
y
40 kN-m
x
Solution:
(a)
The moment is directed along the positive z axis,
M D 40k (kN-m)
A candidate pair of forces is shown in the sketch: 40 kN forces
directed oppositely at 1 m apart.
(b)
y
40 kN
1m
x
40 kN
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1
Problem 4.114 The moments of two couples are shown.
What is the sum of the moments about point P?
y
50 ft-lb
P
x
(– 4, 0, 0) ft
10 ft-lb
Solution: The moment of a couple is the same anywhere in the
plane. Hence the sum about the point P is
M D 50k C 10k D 40k ft lb
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1
Problem 4.115 Determine the sum of the moments
exerted on the plate by the two couples.
y
30 lb
3 ft
30 lb
2 ft
x
20 lb
20 lb
5 ft
4 ft
Solution: The moment due to the 30 lb couple, which acts in a
clockwise direction is
M30 D 330k D 90k ft lb.
The moment due to the 20 lb couple, which acts in a counterclockwise
direction, is
M20 D 920k D 180k ft lb.
The sum of the moments is
M D 90k C 180k D C90k ft lb.
The sum of the moments is the same anywhere on the plate.
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1
Problem 4.116 Determine the sum of the moments
exerted about A by the couple and the two forces.
100 lb
400 lb
900 ft-lb
A
3 ft
B
4 ft
3 ft
4 ft
Solution: Let the x axis point to the right and the y axis point
upward in the plane of the page. The moments of the forces are
M100 D 3i ð 100j D 300k (ft-lb),
and
M400 D 7i ð 400j D 2800k (ft-lb).
The moment of the couple is MC D 900k (ft-lb). Summing the
moments, we get
MTotal D 2200k (ft-lb)
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1
Problem 4.117 Determine the sum of the moments
exerted about A by the couple and the two forces.
100 N
30°
200 N
0.2 m
A
300 N-m
0.2 m
0.2 m
0.2 m
Solution:
MA D 0.2i ð 200j C 0.4i C 0.2j
ð 86.7i C 50j C 300k N-m
MA D 40k C 2.66k C 300k N-m
MA D 262.7k N-m ' 263k N-m
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1
Problem 4.118 The sum of the moments about point
A due to the forces and couples acting on the bar is zero.
(a)
(b)
What is the magnitude of the couple C?
Determine the sum of the moments about point B
due to the forces and couples acting on the bar.
B
4 kN
3m
20 kN-m
C
A
4 kN
2 kN
5 kN
5m
3 kN
3m
Solution:
(a)
MA D 20 kN-m 2 kN5 m 4 kN3 m
3 kN8 C C D 0
C D 26 kN-m
MB D 3 kN3 m 4 kN3 m 5 kN5 m
(b)
C 20 kN-m C 26 kN-m D 0
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1
Problem 4.119 Four forces and a couple act on the
beam. The vector sum of the forces is zero, and the sum
of the moments about the left end of the beam is zero.
What are the forces Ax , Ay , and B?
y
800 N
200 N-m
Ax
B
Ay
x
4m
4m
3m
Solution: The sum of the forces about the y axis is
FY D AY C B 800 D 0.
The sum of the forces about the x axis is
FX D AX D 0.
The sum of the moments about the left end of the beam is
ML D 11B 8800 200 D 0.
From the moments:
BD
6600
D 600 N.
11
Substitute into the forces balance equation to obtain:
AY D 800 600 D 200 N
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1
Problem 4.120 (a) What is the moment of the couple?
(b) Determine the perpendicular distance between the
lines of action of the two forces.
y
(0, 4, 0) m
⫺2i ⫹ 2j ⫹ k (kN)
2i ⫺ 2j ⫺ k (kN)
x
(0, 0, 5) m
z
Solution:
M D 4j 5k m ð 2i 2j k kN
(a)
D 14i 10j 8k kN-m
(b)
MD
FD
142 C 102 C 82 kN-m D 18.97 kN-m
22 C 22 C 12 kN D 3 kN
M D Fd ) d D
18.97 kN-m
M
D
D 6.32 m
F
3 kN
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1
Problem 4.121 Determine the sum of the moments exerted
on the plate by the three couples. (The 80-lb forces are
contained in the xz plane.)
y
3 ft
20 lb
3 ft
20 lb
40 lb
x
8 ft
40 lb
z
Solution: The moments of two of the couples can be determined
from inspection:
M1 D 320k D 60k ft lb.
M2 D 840j D 320j ft lb
The forces in the 3rd couple are resolved:
60°
80 lb
60°
80 lb
The moment is
i
M3 D r3 ð F3 D 6
69.282
j k 0 0 D 240j.
0 40 The sum of the moments due to the couples:
M D 60k C 320j 240j D 80j 60k ft lb
F D 80i sin 60° C k cos 60° D 69.282i C 40k
The two forces in the third couple are separated by the vector
r3 D 6i C 8k 8k D 6i
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1
Problem 4.122 What is the magnitude of the sum of
the moments exerted on the T-shaped structure by the
two couples?
y
3 ft
50i + 20j – 10k (lb)
3 ft
50j (lb)
3 ft
z
–50j (lb)
3 ft
x
–50i – 20j + 10k (lb)
Solution: The moment of the 50 lb couple can be determined by
y
3 ft
inspection:
F
3 ft
50 j (lb)
M1 D 503k D 150k ft lb.
The vector separating the other two force is r D 6k. The moment is
i
M2 D r ð F D 0
50
j
0
20
k 6 D 120i C 300j.
10 3 ft
x
–F
–50 j (lb)
3 ft
z
The sum of the moments is
M D 120i C 300j 150k.
The magnitude is
jMj D
p
1202 C 3002 C 1502 D 356.23 ft lb
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1
Problem 4.123
500 N.
(a)
(b)
The tension in cables AB and CD is
y
A (0, 2, 0) m
Show that the two forces exerted by the cables on
the rectangular hatch at B and C form a couple.
What is the moment exerted on the plate by the
cables?
3m
B
z
x
3m
C
D
Solution: One condition for a couple is that the sum of a pair
of forces vanish; another is for a non-zero moment to be the same
anywhere. The first condition is demonstrated by determining the unit
vectors parallel to the action lines of the forces. The vector position of
point B is rB D 3i m. The vector position of point A is rA D 2j. The
vector parallel to cable AB is
rBA D rA rB D 3i C 2j.
(6, –2, 3) m
The moment about the origin is
MO D rB rC ð TAB D rCB ð TAB ,
which is identical with the above expression for the moment. Let rPC
and rPB be the distances to points C and B from an arbitrary point
P on the plate. Then MP D rPB rPC ð TAB D rCB ð TAB which
is identical to the above expression. Thus the moment is the same
everywhere on the plate, and the forces form a couple.
The magnitude is:
p
jrAB j D
32 C 22 D 3.606 m.
The unit vector:
eAB D
rAB
D 0.8321i C 0.5547j.
jrAB j
The tension is
TAB D jTAB jeAB D 416.05i C 277.35j.
The vector position of points C and D are:
rC D 3i C 3k, rD D 6i 2j C 3k.
The vector parallel to the cable CD is rCD D rD rC D 3i 2j. The
magnitude is jrCD j D 3.606 m, and the unit vector parallel to the cable
CD is eCD D C0.8321i 0.5547j. The magnitude of the tension in
the two cables is the same, and eBA D eCD , hence the sum of the
tensions vanish on the plate. The second condition is demonstrated by
determining the moment at any point on the plate. By inspection, the
distance between the action lines of the forces is
rCB D rB rC D 3i 3i 3k D 3k.
The moment is
i
M D rCB ð TAB D 0
416.05
j
0
277.35
k 3 0 D 832.05i 1248.15j (N-m).
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1
Problem 4.124 The cables AB and CD exert a couple
on the vertical pipe. The tension in each cable is 8 kN.
Determine the magnitude of the moment the cables exert
on the pipe.
(⫺1.6, 2.2, ⫺1.2) m
y
D
C
(0.2, 1.6, ⫺0.2) m
A
(0.2, 0.6, 0.2) m
x
z
B
(1.6, 0, 1.2) m
Solution:
FAB D 8 kN
1.4i 0.6j C 1.0k
p
, rDB D 3.2i 2.2j C 2.4k m
3.32
M D rBD ð FAB D 3.34i C 0.702j C 5.09k kN-m
) M D 6.13 kN-m
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1
Problem 4.125
The bar is loaded by the forces
y
FB
FB D 2i C 6j C 3k (kN),
A
FC D i 2j C 2k (kN),
and the couple
MC D 2i C j 2k (kN-m).
B
MC
C
x
1m
z
1m
FC
Determine the sum of the moments of the two forces
and the couple about A.
Solution: The moments of the two forces about A are given by
MFB D 1i ð 2i C 6j C 3k (kN-m) D 0i 3j C 6k (kN-m) and
MFC D 2i ð 1i 2j C 2k (kN-m) D 0i 4j 4k (kN-m).
Adding these two moments and
MC D 2i C 1j 2k (kN-m),
we get MTOTAL D 2i 6j C 0k (kN-m)
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1
Problem 4.126
In Problem 4.125, the forces
Solution: From the solution to Problem 4.125, the sum of the
moments of the two forces about A is
FB D 2i C 6j C 3k (kN),
MForces D 0i 7j C 2k (kN-m).
FC D i 2j C 2k (kN),
The required moment, MC , must be the negative of this sum.
and the couple
Thus
MCy D 7 (kN-m), and MCz D 2 (kN-m).
MC D MCy j C MCz k (kN-m).
Determine the values for MCy and MCz , so that the sum
of the moments of the two forces and the couple about
A is zero.
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1
y
Problem 4.127 Two wrenches are used to tighten an
elbow fitting. The force F D 10k (lb) on the right wrench
is applied at (6, 5, 3) in, and the force F on the left
wrench is applied at (4, 5, 3) in.
(a)
(b)
(c)
Determine the moment about the x axis due to the
force exerted on the right wrench.
Determine the moment of the couple formed by the
forces exerted on the two wrenches.
Based on the results of (a) and (b), explain why
two wrenches are used.
z
x
F
–F
Solution: The position vector of the force on the right wrench is
rR D 6i 5j 3k. The magnitude of the moment about the x axis is
1 0
0 jMR j D eX Ð rR ð F D 6 5 3 D 50 in lb
0 0
10 (a)
from which MXL D 50i in lb, which is opposite in direction and
equal in magnitude to the moment exerted on the x axis by the
right wrench. The left wrench force is applied 2 in nearer the
origin than the right wrench force, hence the moment must be
absorbed by the space between, where it is wanted.
The moment about the x axis is
MR D jMR jeX D 50i (in lb).
(b)
The moment of the couple is
i
MC D rR rL ð FR D 2
0
(c)
j k 0 6 D 20j in lb
0 10 The objective is to apply a moment to the elbow relative to
connecting pipe, and zero resultant moment to the pipe itself.
A resultant moment about the x axis will affect the joint at the
origin. However the use of two wrenches results in a net zero
moment about the x axis the moment is absorbed at the juncture
of the elbow and the pipe. This is demonstrated by calculating
the moment about the x axis due to the left wrench:
1 0
0 jMX j D eX Ð rL ð FL D 4 5
3 D 50 in lb
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1
Problem 4.128 Two systems of forces act on the beam.
Are they equivalent?
Strategy: Check the two conditions for equivalence.
The sums of the forces must be equal, and the sums of
the moments about an arbitrary point must be equal.
System 1
y
100 N
x
50 N
1m
1m
System 2
y
50 N
x
2m
Solution: The strategy is to check the two conditions for equivalence: (a) the sums of the forces must be equal and (b) the sums of
the moments about an arbitrary point must be equal. The sums of the
forces of the two systems:
FX D 0, (both systems) and
FY1 D 100j C 50j D 50j (N)
FY2 D 50j (N).
The sums of the forces are equal. The sums of the moments about the
left end are:
M1 D 1100k D 100k (N-m)
M2 D 250k D 100k (N-m).
The sums of the moments about the left end are equal. Choose any
point P at the same distance r D xi from the left end on each beam.
The sums of the moments about the point P are
M1 D 50x C 100x 1k D 50x 100k (N-m)
M2 D 502 xk D 50x 100k (N-m).
Thus the sums of the moments about any point on the beam are equal
for the two sets of forces; the systems are equivalent. Yes
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1
Problem 4.129 Two systems of forces and moments
act on the beam. Are they equivalent?
System 1
y
20 lb
50 ft-lb
10 lb
x
2 ft
2 ft
System 2
y
20 lb
30 ft-lb
10 lb
x
2 ft
2 ft
Solution: The sums of the forces are:
FX D 0 (both systems)
FY1 D 10j 20j D 10j (lb)
FY2 D 20j C 10j D 10j (lb)
Thus the sums of the forces are equal. The sums of the moments about
the left end are:
M1 D 204k C 50k D 30k (ft lb)
M2 D C102k 30k D 10k (ft lb)
The sums of the moments are not equal, hence the systems are not
equivalent. No
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1
Problem 4.131 The four systems shown in Problem 4.130
can be made equivalent by adding a couple to one of the
systems. Which system is it, and what couple must be
added?
Solution: From the solution to 4.130, all systems have
F D 10j kN
and systems 1, 2, and 4 have
ML D 80k kN-m
system 3 has
ML D 160k kN-m.
Thus, we need to add a couple M D 80k kN-m to system 3 (clockwise moment).
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1
Problem 4.132 System 1 is a force F acting at a point
O. System 2 is the force F acting at a different point O0
along the same line of action. Explain why these systems
are equivalent. (This simple result is called the principle
of transmissibility.)
System 2
System 1
F
F
O'
O
O
Solution: The sum of forces is obviously equal for both systems.
Let P be any point on the dashed line. The moment about P is the
cross product of the distance from P to the line of action of a force
times the force, that is, M D rPL ð F, where rPL is the distance from
P to the line of action of F. Since both systems have the same line of
action, and the forces are equal, the systems are equivalent.
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1
Problem 4.133 The vector sum of the forces exerted
on the log by the cables is the same in the two cases.
Show that the systems of forces exerted on the log are
equivalent.
A
12 m
B
16 m
C
12 m
D
6m
Solution: The angle formed by the single cable with the positive
E
20 m
Solve:
x axis is
D 180° tan1
12
16
jTL j D 0.3353jT1 j, and
D 143.13° .
jTR j D 0.7160jT1 j.
The single cable tension is
T1 D jTji cos 143.13° C j sin 143.13° D jTj0.8i C 0.6j.
The position vector to the center of the log from the left end is rc D 10i.
The moment about the end of the log is
i
j
M D r ð T1 D jT1 j 10
0
0.8 0.6
k 0 D jTj6k (N-m).
0
The tension in the right hand cable is TR D jT1 j0.71600.9079i C
0.4191j D jT1 j0.6500i C 0.3000. The position vector of the right
end of the log is rR D 20i m relative to the left end. The moments
about the left end of the log for the second system are
i
M2 D rR ð TR D jT1 j 20
0.6500
j
0
0.3000
k 0 D jT1 j6k (N-m).
0
This is equal to the moment about the left end of the log for System
1, hence the systems are equivalent.
For the two cables, the angles relative to the positive x axis are
1 D 180° tan1
2 D 180 tan1
12
6
12
26
D 116.56° , and
D 155.22° .
The two cable vectors are
TL D jTL ji cos 116.56° C j sin 116.56° D jTL j0.4472i C 0.8945j,
TR D jTR ji cos 155.22° C j sin 155.22° D jTR j0.9079i C 0.4191j.
Since the vector sum of the forces in the two systems is equal, two
simultaneous equations are obtained:
0.4472jTL j C 0.9079jTR j D 0.8jT1 j, and
0.8945jTL j C 0.4191jTR j D 0.6jT1 j
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1
Problem 4.134 Systems 1 and 2 each consist of a
couple. If they are equivalent, what is F?
System 1
y
System 2
y
200 N
F
20°
30°
(5, 4, 0) m
5m
200 N
30°
4m
2m
x
20°
F
x
Solution: For couples, the sum of the forces vanish for both systems. For System 1, the two forces are located at r11 D 4i, and r12 D
C5j. The forces are F1 D 200i cos 30° C j sin 30° D 173.21i C 100j.
The moment due to the couple in System 1 is
i
M1 D r11 r12 ð F1 D 4
173.21
j
5
100
k 0 D 1266.05k (N-m).
0
For System 2, the positions of the forces are r21 D 2i, and r22 D
5i C 4j. The forces are
F2 D Fi cos20° C j sin20° D F0.9397i 0.3420j.
The moment of the couple in System 2 is
i
M2 D r21 r22 ð F2 D F 3
0.9397
j
4
0.3420
k 0 D 4.7848Fk,
0
from which, if the systems are to be equivalent,
FD
1266
D 264.6 N
4.7848
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1
Problem 4.135 Two equivalent systems of forces and
moments act on the L-shaped bar. Determine the forces
FA and FB and the couple M.
System 2
System 1
120 N-m
FB
40 N
60 N
3m
FA
M
3m
50 N
3m
Solution: The sums of the forces for System 1 are
FX D 50, and
The sums of the forces for System 2 are
6m
The sum of the moments about the left end for
System 2 is
FY D FA C 60.
3m
M2 D 3FB C M D 150 C M N-m.
Equating the sums of the moments, M D 150 180 D 30 N-m
FX D FB , and
FY D 40.
For equivalent systems: FB D 50 N, and FA D 60 40 D 20 N.
The sum of the moments about the left end for
System 1 is
M1 D 3FA 120 D 180 N-m.
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1
Problem 4.136 Two equivalent systems of forces and
moments act on the plate. Determine the force F and
the couple M.
System 1
30 lb
System 2
10 lb
30 lb
100 in-lb
5 in
5 in
8 in
8 in
M
50 lb
30 lb
F
Solution: The sums of the forces for System 1 are
FX D 30 lb,
FY D 50 10 D 40 lb.
The sums of the forces for System 2 are
FX D 30 lb,
FY D F 30 lb.
For equivalent forces, F D 30 C 40 D 70 lb. The sum of the moments
about the lower left corner for System 1 is
M1 D 530 810 C M D 230 C M in lb.
The sum of the moments about the lower left corner for System 2 is
M2 D 100 in lb.
Equating the sum of moments, M D 230 100 D 130 in lb
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1
Problem 4.137 In system 1, four forces act on the
rectangular flat plate. The forces are perpendicular to
the plate and the 400-kN force acts at its midpoint.
In system 2, no forces or couples act on the plate.
Systems 1 and 2 are equivalent. What are the forces F1 ,
F2 , and F3 .
F1 D
MA1 D
F2
F3
2m
4m
System 2
F2 and
MA2
F2 D 0 and
MA2 D 0
A
8m
400 kN
From system 2,
System 1
F1
Solution: For the two systems to be equivalent
This
F1 D F1 C F2 C F3 400j D 0 or
F1Y D F1 C F2 C F3 400 D 0
1
Summing Moments around A, we get
MA D 4i C 3k ð 400j C 6k ð F1 j
C 8i C 2k ð F3 j
MA D 1600k C 1200i 6F1 i
C 8F3 k 2F3 i kN-m D 0
In component form, we have
x: 6F1 2F3 C 1200 D 0 kN-m
z: 8F3 1600 D 0 kN-m
And the Force equation F1 C F2 C F3 400 D 0 kN
Solving, we get
F3 D 200 kN
F1 D
800
D 133.3 kN
6
F2 D 66.7 kN
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1
Problem 4.138 Three forces and a couple are applied
to a beam (system 1).
(a)
If you represent system 1 by a force applied at A
and a couple (system 2), what are F and M?
If you represent system 1 by the force F (system 3),
what is the distance D?
(b)
System 1
y
30 lb
40 lb
20 lb
30 ft-lb
x
A
2 ft
2 ft
System 2
y
F
M
x
A
System 3
y
F
x
A
D
Solution: The sum of the forces in System 1 is
FX D 0i,
FY D 20 C 40 30j D 10j lb.
The sum of the moments about the left end for System 1 is
M1 D 240 430 C 30k D 10k ft lb.
(a)
For System 2, the force at A is F D 10j lb
The moment at A is M2 D 10k ft lb
(b)
For System 3 the force at D is F D 10j lb. The distance D is
the ratio of the magnitude of the moment to the magnitude of the
force, where the magnitudes are those in System 1:
DD
10
D 1 ft
10
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1
Problem 4.139 Represent the two forces and couple
acting on the beam by a force F. Determine F and determine where its line of action intersects the x axis.
y
60i + 60 j (N)
280 N-m
x
– 40 j (N)
3m
y
Solution: We first represent the system by an equivalent system
consisting of a force F at the origin and a couple M:
This system is equivalent if
3m
F
M
F D 40j C 60i C 60j
x
D 60i C 20j N,
M D 280 C 660
D 80 N-m.
y
F
We then represent this system by an equivalent system consisting of
F alone:
x
For equivalence, M D dFy , so
dD
80
M
D 4 m.
D
Fy
20
d
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1
y
Problem 4.140 The bracket is subjected to three forces
and a couple. If you represent this system by a force F,
what is F, and where does its line of action intersect the
x axis?
400 N
180 N
0.4 m
140 N-m
200 N
0.2 m
x
0.65 m
Solution: We locate a single equivalent force along the x axis a
distance d to the right of the origin. We must satisfy the following
three equations:
Fx D 400 N 200 N D Rx
Fy D 180 N D Ry
MO D 400 N0.6 m C 200 N0.2 m C 180 N0.65 m
C 140 Nm D Ry d
Solving we find
Rx D 200 N, Ry D 180 N, d D 0.317 m
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1
Problem 4.141 The vector sum of the forces acting on
the beam is zero, and the sum of the moments about the
left end of the beam is zero.
Solution: (a) The sum of the forces is
(a)
(b)
(c)
Determine the forces Ax and Ay , and the couple MA .
Determine the sum of the moments about the right
end of the beam.
If you represent the 600-N force, the 200-N force,
and the 30 N-m couple by a force F acting at the
left end of the beam and a couple M, what are F
and M?
y
x
30 N-m
Ay
200 N
380 mm
180 mm
ML D MA 0.38600 30 C 0.560200k D 0,
from which MA D 146 N-m. (b) The sum of the moments about the
right end of the beam is
Ax
FY D AY 600 C 200j D 0,
from which AY D 400 N. The sum of the moments is
600 N
MA
FX D AX i D 0 and
MR D 0.18600 30 C 146 0.56400 D 0.
(c) The sum of the forces for the new system is
FY D AY C Fj D 0,
from F D AY D 400 N, or F D 400j N. The sum of the moments
for the new system is
M D MA C M D 0,
from which M D MA D 146 N-m
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1
Problem 4.142 The vector sum of the forces acting on
the truss is zero, and the sum of the moments about the
origin O is zero.
Solution: (a) The sum of the forces is
(a)
(b)
from which AX D 12 kip
(c)
Determine the forces Ax , Ay , and B.
If you represent the 2-kip, 4-kip, and 6-kip forces
by a force F, what is F, and where does its line of
action intersect the y axis?
If you replace the 2-kip, 4-kip, and 6-kip forces by
the force you determined in (b), what are the vector
sum of the forces acting on the truss and the sum
of the moments about O?
FX D AX 2 4 6i D 0,
FY D AY C Bj D 0.
The sum of the moments about the origin is
MO D 36 C 64 C 92 C 6B D 0,
from which B D 10j kip. (b) Substitute into the force balance equation to obtain AY D B D 10 kip. (b) The force in the new system
will replace the 2, 4, and 6 kip forces, F D 2 4 6i D 12i kip.
The force must match the moment due to these forces: FD D 36 C
60
D 5 ft, or the action
64 C 92 D 60 kip ft, from which D D
12
line intersects the y axis 5 ft above the origin. (c) The new system is
equivalent to the old one, hence the sum of the forces vanish and the
sum of the moments about O are zero.
2 kip
y
3 ft
4 kip
3 ft
6 kip
3 ft
Ax
O
x
B
Ay
6 ft
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1
Problem 4.143 The distributed force exerted on part
of a building foundation by the soil is represented by
five forces. If you represent them by a force F, what is
F, and where does its line of action intersect the x axis?
y
x
80 kN
3m
35 kN
30 kN
40 kN
3m
3m
3m
85 kN
Solution: The equivalent force must equal the sum of the forces
exerted by the soil:
F D 80 C 35 C 30 C 40 C 85j D 270j kN
The sum of the moments about any point must be equal for the two
systems. The sum of the moments are
M D 335 C 630 C 940 C 1285 D 1665 kN-m.
Equating the moments for the two systems FD D 1665 kN-m from
which
DD
1665 kN-m
D 6.167 m.
270 kN
Thus the action line intersects the x axis at a distance D D 6.167 m to
the right of the origin.
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1
Problem 4.144 At a particular instant, aerodynamic
forces distributed over the airplane’s surface exert the
88-kN and 16-kN vertical forces and the 22 kN-m
counterclockwise couple shown. If you represent these
forces and couple by a system consisting of a force F
acting at the center of mass G and a couple M, what are
F and M?
y
88 kN
16 kN
x
G
5m
22 kN-m
5.7 m
9m
Solution:
Fy D 88 kN C 16 kN D Ry
MG D 88 kN0.7 m C 16 kN3.3 m C 22 kN-m D M
Solving we find
Ry D 104 kN, M D 13.2 kN-m
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1
Problem 4.145 If you represent the two forces and
couple acting on the airplane in Problem 4.144 by a
force F, what is F, and where does its line of action
intersect the x axis?
Solution:
Fy D 88 kN C 16 kN D Ry
MOrigin D 88 kN5 m C 16 kN9 m C 22 kN-m D Ry x
Solving we find
F D Ry j D 104 kNj, x D 5.83 m
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1
Problem 4.146 The system is in equilibrium. If you
represent the forces FAB and FAC by a force F acting at
A and a couple M, what are F and M?
y
B
60°
40°
C
FAB
A
A
100 lb
100 lb
FAC
x
Solution: The sum of the forces acting at A is in opposition to the
weight, or F D jWjj D 100j lb.
The moment about point A is zero.
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1
Problem 4.147
(a)
(b)
Three forces act on a beam.
y
Represent the system by a force F acting at the
origin O and a couple M.
Represent the system by a single force. Where does
the line of action of the force intersect the x axis?
30 N
5m
x
O
30 N
6m
4m
50 N
Solution: (a) The sum of the forces is
FX D 30i N, and
FY D 30 C 50j D 80j N.
The equivalent at O is F D 30i C 80j (N). The sum of the moments
about O:
M D 530 C 1050 D 350 N-m
(b) The solution of Part (a) is the single force. The intersection is the
350
D 4.375 m
moment divided by the y-component of force: D D
80
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1
y
Problem 4.148 The tension in cable AB is 400 N, and
the tension in cable CD is 600 N.
(a)
(b)
If you represent the forces exerted on the left post
by the cables by a force F acting at the origin O
and a couple M, what are F and M?
If you represent the forces exerted on the left post
by the cables by the force F alone, where does its
line of action intersect the y axis?
A
400 mm
B
C
300 mm
D
O
Solution: From the right triangle, the angle between the positive
x axis and the cable AB is
D tan1
400
800
D 26.6° .
x
800 mm
Check. (b) The equivalent single force retains the same scalar components, but must act at a point that duplicates the sum of the moments.
The distance on the y axis is the ratio of the sum of the moments to
the x-component of the equivalent force. Thus
419
D 0.456 m
919.6
The tension in AB is
DD
TAB D 400i cos26.6° Cj sin26.6° D 357.77i 178.89j (N).
Check: The moment is
The angle between the positive x axis and the cable CD is
i
M D rF ð F D 0
919.6
˛ D tan1
300
800
D 20.6° .
from which D D
The tension in CD is
300 mm
j
D
389.6
k 0 D 919.6Dk D 419k,
0
419
D 0.456 m, Check.
919.6
TCD D 600i cos20.6° C j sin20.6° D 561.8i 210.67j.
The equivalent force acting at the origin O is the sum of the forces
acting on the left post:
F D 357.77 C 561.8i C 178.89 210.67j
D 919.6i 389.6j (N).
The sum of the moments acting on the left post is the product of the
moment arm and the x-component of the tensions:
M D 0.7357.77k 0.3561.8k D 419k N-m
Check: The position vectors at the point of application are rAB D 0.7j,
and rCD D 0.3j. The sum of the moments is
M D rAB ð TAB C rCD ð TCD i
D 0
357.77
j
0.7
178.89
k i
0 C 0
0 561.8
j
0.3
210.67
k 0 0
D 0.7357.77k 0.3561.8k D 419k
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1
Problem 4.149 Consider the system shown in Problem
4.148. The tension in each of the cables AB and CD is
400 N. If you represent the forces exerted on the right
post by the cables by a force F, what is F, and where
does its line of action intersect the y axis?
Solution: From the solution of Problem 4.148, the tensions are
TAB D 400i cos26.6° Cj sin26.6° D 357.77i C 178.89j,
and
TCD D 400i cos20.6° Cj sin20.6° D 374.42i C 140.74j.
The equivalent force is equal to the sum of these forces:
F D 357.77 374.42i C 178.77 C 140.74j
D 732.19i C 319.5j (N).
The sum of the moments about O is
M D 0.3357.77 C 0.8140.74 C 178.89k D 363k (N-m).
The intersection is D D
363
D 0.496 m on the positive y axis.
732.19
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1
Problem 4.150 If you represent the three forces acting
on the beam cross section by a force F, what is F, and
where does its line of action intersect the x axis?
y
500 lb
800 lb
6 in
x
6 in
z
500 lb
Solution: The sum of the forces is
FX D 500 500i D 0.
FY D 800j.
Thus a force and a couple with moment M D 500k ft lb act on the
cross section. The equivalent force is F D 800j which acts at a positive
500
D 0.625 ft D 7.5 in to the right of the
x axis location of D D
800
origin.
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1
Problem 4.151 The two systems of forces and moments acting on the beam are equivalent. Determine the
force F and the couple M.
System 1
Solution: The sum of the forces on the two systems are equivalent:
the force on System 1 is F1 D 4i C 4j 2k (kN). The moments on the
two systems are equivalent: the moment about the origin for System
1 is the product of the moment arm and the y- and z-components of
the force:
M D 32j C 34k D 6j C 12k. Hence the couple moment on System
2 is M2 D 6j C 12k (kN-m)
y
z
3m
F
x
System 2
M
y
4i + 4j – 2k (kN)
z
3m
x
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1
Problem 4.152
force shown.
(a)
(b)
(c)
The wall bracket is subjected to the
Determine the moment exerted by the force about
the z axis.
Determine the moment exerted by the force about
the y axis.
If you represent the force by a force F acting at O
and a couple M, what are F and M?
y
O
z
10i – 30j + 3k (lb)
12 in
x
Solution:
(a)
The moment about the z axis is negative,
MZ D 130 D 30 ft lb,
(b)
The moment about the y axis is negative,
MY D 13 D 3 ft lb
(c)
The equivalent force at O must be equal to the force at x D 12 in,
thus FEQ D 10i 30j C 3k (lb)
The couple moment must equal the moment exerted by the force at x D
12 in. This moment is the product of the moment arm and the y- and zcomponents of the force: M D 130k 13j D 3j 30k (ft lb).
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1
Problem 4.153 A basketball player executes a “slam
dunk” shot, then hangs momentarily on the rim, exerting
the two 100-lb forces shown. The dimensions are h D
14 12 in, and r D 9 12 in, and the angle ˛ D 120° .
(a)
(b)
If you represent the forces he exerts by a force F
acting at O and a couple M, what are F and M?
The glass backboard will shatter if jMj > 4000 inlb. Does it break?
y
–100j (lb)
O
α
r
–100j (lb)
h
x
z
Solution: The equivalent force at the origin must equal the sum of
the forces applied: FEQ D 200j. The position vectors of the points of
application of the forces are r1 D h C ri, and r2 D ih C r cos ˛ kr sin ˛. The moments about the origin are
M D r1 ð F1 C r2 ð F2 D r1 C r2 ð F
i
j
D 2h C r1 C cos ˛
0
0
100
k
r sin ˛ 0
D 100r sin ˛i 1002h C r1 C cos ˛k.
For the values of h, r, and ˛ given, the moment is M D 822.72i 3375k in lb. This is the p
couple moment required. (b) The magnitude
of the moment is jMj D 822.722 C 33752 D 3473.8 in lb. The backboard does not break.
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1
Problem 4.154
The three forces are parallel to the x axis.
y
(0, 6, 2) ft
(a)
(b)
If you represent the three forces by a force F acting
at the origin O and a couple M, what are F and M?
If you represent the forces by a single force, what is
the force, and where does its line of action intersect
the yz plane?
Strategy: In (b), assume that the force acts at a point
(0, y, z) of the yz plane, and use the conditions for
equivalence to determine the force and the coordinates
y and z. (See Example 4.20.)
300 lb
O
100 lb
(0, 0, 4) ft
z
200 lb
x
y
Solution:
(a) F D 100i C 200i C 300i
D 600i lb.
i
M D 0
200
j k i
0 4 C 0
0 0 300
j k 6 2 0 0
M
0
z
F
D 800j C 600j 1800k
x
D 1400j 1800k ft-lb.
(b)
y
F D 600i lb.
To determine y and z, require that
i
M D 1400j 1800k D 0
600
j
y
0
(0, y, z)
k z 0
F
1400 D 600 z,
1800 D 600 y.
0
z
Solving, y D 3 ft and z D 2.33 ft.
x
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1
Problem 4.155 The normal forces exerted on the car’s
tires by the road are
C
A
0.8 m
NA D 5104j (N),
x
NB D 5027j (N),
NC D 3613j (N),
ND D 3559j (N).
If you represent these forces by a single equivalent force
N, what is N, and where does its line of action intersect
the xz plane?
0.8 m
D
1.4 m
1.4 m
B
z
y
x
Solution: We must satisfy the following three equations
Fy :5104 N C 5027 N C 3613 N C 3559 N D Ry
Mx :5104 N C 3613 N0.8 m
5027 N C 3559 N0.8 m D Ry z
Mz :5104 N C 5027 N1.4 m
3613 N C 3559 N1.4 m D Ry x
Solving we find
Ry D 17303 N, x D 0.239 m, z D 0.00606 m
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1
Problem 4.156 Two forces act on the beam. If you
represent them by a force F acting at C and a couple M,
what are F and M?
Solution: The equivalent force must equal the sum of forces: F D
100j C 80k. The equivalent couple is equal to the moment about C:
M D 380j 3100k D 240j 300k
y
100 N
80 N
z
C
x
3m
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1
Problem 4.157 An axial force of magnitude P acts on
the beam. If you represent it by a force F acting at the
origin O and a couple M, what are F and M?
b
Pi
z
h
O
x
y
Solution: The equivalent force at the origin is equal to the applied
force F D Pi. The position vector of the applied force is r D hj C bk.
The moment is
i
M D r ð P D 0
P
j
h
0
k Cb D bPj C hPk.
0 This is the couple at the origin.
(Note that in the sketch the axis system has been rotated 180 about
the x axis; so that up is negative and right is positive for y and z.)
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1
Problem 4.158
screw.
(a)
(b)
The brace is being used to remove a
If you represent the forces acting on the brace by
a force F acting at the origin O and a couple M,
what are F and M?
If you represent the forces acting on the brace by
a force F0 acting at a point P with coordinates
xP , yP , zP and a couple M0 , what are F0 and M0 ?
y
h
r
h
B
z
O
1
A
2
A
B
x
1
A
2
Solution: (a) Equivalent force at the origin O has the same value
as the sum of forces,
FX D B Bi D 0,
FY D A C 12 A C 12 A j D 0,
thus F D 0. The equivalent couple moment has the same value as the
moment exerted on the brace by the forces,
MO D rAi.
Thus the couple at O has the moment M D rAi. (b) The equivalent
force at xP , yP , zP has the same value as the sum of forces on the
brace, and the equivalent couple at xP , yP , zP has the same moment
as the moment exerted on the brace by the forces: F D 0, M D rAi.
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1
y
Problem 4.159 Two forces and a couple act on the
cube. If you represent them by a force F acting at point
P and a couple M, what are F and M?
P
FB =
2i – j (kN)
FA =
– i + j + k (kN)
x
MC =
4i – 4j + 4k (kN-m)
z
1m
Solution: The equivalent force at P has the value of the sum of
forces,
F
= (2 − 1)i + (1 − 1)j + k, FP = i + k (kN).
The equivalentcouple at P has the moment exerted by the forces and
moment about P. The position vectors of the forces relative to P are:
rA D i j C k, and rB D Ck. The moment of the couple:
M D rA ð FA C rB ð FB C MC
i
j k j k i
D 1 1 1 C 0 0 1 C MC
1 1 1 2 1 0 D 3i 2j C 2k (kN-m).
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1
Problem 4.160 The two shafts are subjected to the
torques (couples) shown.
(a)
(b)
If you represent the two couples by a force F acting
at the origin O and a couple M, what are F and M?
What is the magnitude of the total moment exerted
by the two couples?
y
6 kN-m
4 kN-m
40°
30°
x
z
Solution: The equivalent force at the origin is zero, F D 0 since
there is no resultant force on the system. Represent the couples of
4 kN-m and 6 kN-m magnitudes by the vectors M1 and M2 . The
couple at the origin must equal the sum:
M D M1 C M2 .
The sense of M1 is (see sketch) negative with respect to both y and
z, and the sense of M2 is positive with respect to both x and y.
M1 D 4j sin 30° k cos 30° D 2j 3.464k,
M2 D 6i cos 40° C j sin 40° D 4.5963i C 3.8567j.
Thus the couple at the origin is MO D 4.6i C 1.86j 3.46k (kN-m)
(b) The magnitude
of the total moment exerted by the two couples is
p
jMO j D 4.62 C 1.862 C 3.462 D 6.05 (kN-m)
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1
Problem 4.161 The two systems of forces and
moments acting on the bar are equivalent. If
y
FA D 30i C 30j 20k (kN),
FB D 40i 20j C 25k (kN),
FA
z
A
2m
MB D 10i C 40j 10k (kN-m),
MB
2m
what are F and M?
B
x
FB
System 1
y
F
z
x
M
System 2
Solution:
F D FA C FB D 70i C 10j C 5k kN
M D 2 mi ð FA C 4 mi ð FB C MB
D 10i 20j 30k kNm
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1
Problem 4.162
The forces are
Point G is at the center of the block.
Solution: The equivalent force is the sum of the forces:
FA D 20i C 10j C 20k (lb),
F D 20i C 10 C 10j C 20 10k
D 20i C 20j C 10k (lb).
FB D 10j 10k (lb).
The equivalent couple is the sum of the moments about G. The position
vectors are:
If you represent the two forces by a force F acting at G
and a couple M, what are F and M?
rA D 15i C 5j C 10k (in),
rB D 15i C 5j 10k.
y
The sum of the moments:
MG D rA ð FA C rB ð FB FB
FA
10 in
x
G
i
j
D 15 5
20 10
k i
10 C 15
20 0
j
k 5 10 10 10 D 50i C 250j C 100k (in lb)
20 in
z
30 in
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1
Problem 4.163 The engine above the airplane’s fuselage exerts a thrust T0 D 16 kip, and each of the engines
under the wings exerts a thrust TU D 12 kip. The dimensions are h D 8 ft, c D 12 ft, and b D 16 ft. If you represent the three thrust forces by a force F acting at the
origin O and a couple M, what are F and M?
y
T0
c
O
z
h
2 TU
y
Solution: The equivalent thrust at the point G is equal to the sum
x
O
of the thrusts:
b
T D 16 C 12 C 12 D 40 kip
b
The sum of the moments about the point G is
M D r1U ð TU C r2U ð TU C rO ð TO D r1U C r2U ð TU C rO ð TO .
The position vectors are r1U D Cbi hj, r2U D bi hj, and rO D
Ccj. For h D 8 ft, c D 12 ft, and b D 16 ft, the sum of the moments
is
i
M D 0
0
j
k i
16 0 C 0
0
12 0
j
k 12 0 D 192 C 192i D 0.
0 16 Thus the equivalent couple is M D 0
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1
Problem 4.164 Consider the airplane described in
Problem 4.163 and suppose that the engine under the
wing to the pilot’s right loses thrust.
Solution: The sum of the forces is now
(a)
The sum of the moments is now:
(b)
If you represent the two remaining thrust forces by
a force F acting at the origin O and a couple M,
what are F and M?
If you represent the two remaining thrust forces
by the force F alone, where does its line of action
intersect the xy plane?
F D 12 C 16 D 28k (kip).
M D r2U ð TU C rO ð TO .
For h D 8 ft, c D 12 ft, and b D 16 ft, using the position vectors for
the engines given in Problem 4.163, the equivalent couple is
i
M D 16
0
j
k i j
8 0 C 0 12
0 12 0 0
k 0 D 96i 192j (ft kip)
16 (b) The moment of the single force is
i j
M D x y
0 0
k z D 28yi 28xj D 96i 192j.
28 From which
xD
192
96
D 6.86 ft, and y D
D 3.43 ft.
28
28
As to be expected, z can have any value, corresponding to any point
on the line of action. Arbitrarily choose z D 0, so that the coordinates
of the point of action are (6.86, 3.43, 0).
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1
Problem 4.165 The tension in cable AB is 100 lb, and
the tension in cable CD is 60 lb. Suppose that you want
to replace these two cables by a single cable EF so that
the force exerted on the wall at E is equivalent to the
two forces exerted by cables AB and CD on the walls
at A and C. What is the tension in cable EF, and what
are the coordinates of points E and F?
y
y
C
(4, 6, 0) ft
(0, 6, 6) ft
E
x
A
x
D
(7, 0, 2) ft
B
F
(3, 0, 8) ft
z
z
Solution: The position vectors of the points A, B, C, and D are
For the systems to be equivalent, the moments about the origin must
be the same. The moments about the origin are
rA D 0i C 6j C 6k,
rB D 3i C 0j C 8k,
MO D rA ð FA C rC ð FC i
D 0
42.86
rC D 4i C 6j C 0k, and
j
6
85.71
k i
6 C 4
28.57 25.72
j
k 6
0 51.43 17.14 rD D 7i C 0j C 2k.
D 788.57i C 188.57j 617.14k.
The unit vectors parallel to the cables are obtained as follows:
rAB D rB rA D 3i 6j C 2k,
p
jrAB j D
32 C 62 C 22 D 7,
from which
This result is used to establish the coordinates of the point E. For the
one cable system, the end E is located at x D 0. The moment is
i
M1 D r ð FEF D 0
68.58
j
y
137.14
k z 45.71 eAB D 0.4286i 0.8571j C 0.2857k.
D 45.71y C 137.14zi C 68.58zj 68.58yk
rCD D rD rC D 3i 6j C 2k,
D 788.57i C 188.57j 617.14k,
p
jrCD j D
32 C 62 C 22 D 7,
from which
eCD D 0.4286i 0.8571j C 0.2857k.
Since eAB D eCD , the cables are parallel . To duplicate the force, the
single cable EF must have the same unit vector.
The force on the wall at point A is
FA D 100eAB D 42.86i 85.71j C 28.57k (lb).
The force on the wall at point C is
from above. From which
yD
617.14
D 8.999 . . . D 9 ft
68.58
zD
188.57
D 2.75 ft.
68.58
Thus the coordinates of point E are E (0, 9, 2.75) ft. The coordinates
of the point F are found as follows: Let L be the length of cable
EF. Thus, from the definition of the unit vector, yF yE D Ley with
9
D 10.5 ft. The other coordithe condition that yF D 0, L D
0.8571
nates are xF xE D LeX , from which xF D 0 C 10.50.4286 D 4.5 ft
zF zE D LeZ , from which zF D 2.75 C 10.50.2857 D 5.75 ft The
coordinates of F are F (4.5, 0, 5.75) ft
FC D 60eCD D 25.72i 51.43j C 17.14k (lb).
The total force is
FEF D 68.58i 137.14j C 45.71k (lb),
jFEF j D 160 lb.
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1
Problem 4.166 The distance s D 4 m. If you represent
the force and the 200-N-m couple by a force F acting at
origin O and a couple M, what are F and M?
y
(2, 6, 0) m
s
100i ⫹ 20j ⫺ 20k (N)
O
200 N-m
x
(4, 0, 3) m
z
Solution: The equivalent force at the origin is
F D 100i C 20j 20k.
The strategy is to establish the position vector of the action point of
the force relative to the origin O for the purpose of determining the
moment exerted by the force about the origin. The position of the top
of the bar is
rT D 2i C 6j C 0k. The vector parallel to the bar, pointing toward the
base, is rTB D 2i 6j C 3k, with a magnitude of jrTB j D 7. The unit
vector parallel to the bar is
eTB D 0.2857i 0.8571j C 0.4286k.
The vector from the top of the bar to the action point of the force is
rTF D seTB D 4eTB D 1.1429i 3.4286j C 1.7143k.
The position vector of the action point from the origin is
rF D rT C rTF D 3.1429i C 2.5714j C 1.7143k.
The moment of the force about the origin is
i
MF D r ð F D 3.1429
100
j
2.5714
20
k 1.7143 20 D 85.71i C 234.20j 194.3k.
The couple is obtained from the unit vector and the magnitude. The
sense of the moment is directed positively toward the top of the bar.
MC D 200eTB D 57.14i C 171.42j 85.72k.
The sum of the moments is
M D MF C MC D 142.86i C 405.72j 280k.
This is the moment of the equivalent couple at the origin.
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1
Problem 4.167
1 are
The force F and couple M in system
System 1
System 2
y
F D 12i C 4j 3k (lb),
y
M
M D 4i C 7j C 4k (ft-lb).
F
Suppose you want to represent system 1 by a wrench
(system 2). Determine the couple Mp and the coordinates x and z where the line of action of the force
intersects the xz plane.
O
z
O
x
z
Mp
F
x
(x, 0, z)
Solution: The component of M that is parallel to F is found as
follows: The unit vector parallel to F is
eF D
F
D 0.9231i C 0.3077j 0.2308k.
jFj
The component of M parallel to F is
MP D eF Ð MeF D 4.5444i C 1.5148j 1.1361k (ft-lb).
The component of M normal to F is
MN D M MP D 0.5444i C 5.4858j C 5.1361k (ft-lb).
The moment of F must produce a moment equal to the normal component of M. The moment is
i j k MF D r ð F D x 0 z D 4zi C 3x C 12zj C 4xk,
12 4 3 from which
zD
0.5444
D 0.1361 ft
4
xD
5.1362
D 1.2840 ft
4
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1
Problem 4.168 A system consists of a force F acting
at the origin O and a couple M, where
F D 10i (lb),
M D 20j (ft-lb).
If you represent the system by a wrench consisting of
the force F and a parallel couple Mp , what is Mp , and
where does the line of action F intersect the yz plane?
Solution: The component of M parallel to F is zero, since MP D
eF Ð MeF D 0. The normal component is equal to M. The equivalent
force must produce the same moment as the normal component
i
M D r ð F D 0
10
from which z D
j
y
0
k z D 10zj 10yk D 20j,
0
20
D 2 ft and y D 0
10
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1
Problem 4.169 A system consists of a force F acting
at the origin O and a couple M, where
F D i C 2j C 5k (N),
M D 10i C 8j 4k (N-m).
If you represent it by a wrench consisting of the force F
and a parallel couple Mp , (a) determine Mp , and determine where the line of action of F intersects (b) the xz
plane, (c) the yz plane.
Solution: The unit vector parallel to F is
eF D
F
D 0.1826i C 0.3651j C 0.9129k.
jFj
from which
zD
9.8
5
D 4.9 m, and x D
D 2.5 m
2
2
(a) The component of M parallel to F is
(c) The intersection with the yz plane is
MP D eF Ð MeF D 0.2i C 0.4j C 1.0k (N-m).
i j
MN D r ð F D 0 y
1 2
The normal component is
D 9.8i C 7.6j 5k,
MN D M MP D 9.8i C 7.6j 5k.
The moment of the force about the origin must be equal to the normal
component of the moment. (b) The intersection with the xz plane:
i j k
MN D r ð F D x 0 z D 2zi 5x zj C 2xk
1 2 5
k z D 5y 2zi C zj yk
5
from which
y D 5 m and z D 7.6 m
D 9.8i C 7.6j 5k,
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1
Problem 4.170 Consider the force F acting at the origin
O and the couple M given in Example 4.20. If you represent this system by a wrench, where does the line of action
of the force intersect the xy plane?
Solution: From Example 4.20 the force and moment are F D 3i C
6j C 2k (N), and M D 12i C 4j C 6k (N-m).
The normal component of the moment is
MN D 7.592i 4.816j C 3.061k (N-m).
The moment produced by the force must equal the normal component:
i j
MN D r ð F D x y
3 6
k 0 2
D 2yi 2xj C 6x 3yk D 7.592i 4.816j C 3.061k,
from which
xD
7.592
4.816
D 2.408 m and y D
D 3.796 m
2
2
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1
Problem 4.171 Consider the force F acting at the origin O and the couple M given in Example 4.20. If you
represent this system by a wrench, where does the line
of action of the force intersect the plane y D 3 m?
Solution: From Example 4.20 (see also Problem 4.170) the force
is F D 3i C 6j C 2k, and the normal component of the moment is
MN D 7.592i 4.816j C 3.061k.
The moment produced by the force must be equal to the normal component:
i j
MN D r ð F D x 3
3 6
k z D 6 6zi 2x 3zj C 6x 9k
2
D 7.592i 4.816j C 3.061k,
from which
xD
9 C 3.061
6 7.592
D 2.01 m and z D
D 0.2653 m
6
6
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1
Problem 4.172 A wrench consists of a force of magnitude 100 N acting at the origin O and a couple of magnitude 60 N-m. The force and couple point in the direction
from O to the point (1, 1, 2) m. If you represent the
wrench by a force F acting at point (5, 3, 1) m and a
couple M, what are F and M?
Solution: The vector parallel to the force is rF D i C j C 2k, from
which the unit vector parallel to the force is eF D 0.4082i C 0.4082j C
0.8165k. The force and moment at the origin are
F D jFjeOF D 40.82i C 40.82j C 81.65k (N), and
M D 24.492i C 24.492j C 48.99k (N-m).
The force and moment are parallel. At the point (5, 3, 1) m the equivalent force is equal to the force at the origin, given above. The moment
of this force about the origin is
i
MF D r ð F D 5
40.82
j
3
40.82
k 1 81.65 D 204.13i 367.43j C 81.64k.
For the moments to be equal in the two systems, the added equivalent
couple must be
MC D M MF D 176.94i C 391.92j 32.65k (N-m)
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1
Problem 4.173 System 1 consists of two forces and
a couple. Suppose that you want to represent it by a
wrench (system 2). Determine the force F, the couple
Mp , and the coordinates x and z where the line of action
of F intersects the xz plane.
System 1
y
System 2
y
1000i + 600j (kN-m)
600k (kN)
3m
300j (kN)
Mp
x
x
4m
z
F
z
(x, 0, z)
Solution: The sum of the forces in System 1 is F D 300j C
600k (N). The equivalent force in System 2 must have this value.
The unit vector parallel to the force is eF D 0.4472j C 0.8944k. The
sum of the moments in System 1 is
M D 6003i C 3004k C 1000i C 600j
D 2800i C 600j C 1200k (kN m).
The component parallel to the force is
MP D 599.963j C 1199.93k (kN-m) D 600j C 1200k (kN-m).
The normal component is MN D M MP D 2800i. The moment of
the force
i
MN D x
0
j
0
300
k z D 300zi 600xj C 300xk D 2800i,
600 from which
x D 0, z D
2800
D 9.333 m
300
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1
y
Problem 4.174 A plumber exerts the two forces shown
to loosen a pipe.
(a)
(b)
(c)
What total moment does he exert about the axis of
the pipe?
If you represent the two forces by a force F acting
at O and a couple M, what are F and M?
If you represent the two forces by a wrench
consisting of the force F and a parallel couple Mp ,
what is Mp , and where does the line of action of F
intersect the xy plane?
12 in
6 in
O
z
x
16 in
16 in
50 k (lb)
–70 k (lb)
Solution: The sum of the forces is
(a)
F D 50k 70k D 20k (lb).
The total moment exerted on the pipe is
M D 1620i D 320i (ft lb).
(b)
The equivalent force at O is F D 20k. The sum of the moments
about O is
MO D r1 ð F1 C r2 ð F2 j
k i
16 0 C 18
0
50 0
i
D 12
0
j
k 16
0 0
70 D 320i C 660j.
(c)
The unit vector parallel to the force is eF D k, hence the moment
parallel to the force is MP D eF Ð MeF D 0, and the moment
normal to the force is MN D M MP D 320i C 660j. The force
at the location of the wrench must produce this moment for the
wrench to be equivalent.
i j
MN D x y
0 0
k 0 D 20yi C 20xj D 320i C 660j,
20 from which x D
660
320
D 33 in, y D
D 16 in
20
20
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1
Problem 4.175 The unstretched length of the spring is
1 m, and the spring constant is k D 20 N/m.
(a)
Draw a graph of the moment about A due to the
force exerted by the spring on the circular bar at B
for values of the angle ˛ from zero to 90° .
Use the result of (a) to estimate the angle at
which the maximum moment occurs and the
corresponding value of the maximum moment.
(b)
B
k
4m
3m
a
A
Solution: The unstretched length of spring is 1 m and the spring
N
. Assume that the bar is a quarter circle, with
m
a radius of 4 m. The stretched length of the spring is found from
the Pythagorean Theorem: The height of the attachment point is h D
4 sin ˛ m, and the distance from the center is 4 cos ˛. The stretched
length of the spring is
constant is k D 20
LD
4m
B
k
3m
α
3 h2 C 4 cos ˛2 m.
A
The spring force is F D 20L 1 N. The angle that the spring
makes with a vertical line parallel to A is
ˇ D tan1
3h
4 cos ˛
Moment at B as function of alpha
120
.
The horizontal component of the spring force is FX D F cos ˇ N. The
vertical component of the force is FY D F sin ˇ N. The displacement
of the attachment point to the left of point A is d D 41 cos ˛ m,
hence the action of the vertical component is negative, and the action
of the horizontal component is positive. The moment about A is
MA D dFY C hFX .
M
o
m
e
n
t
,
100
80
60
40
Collecting terms and equations,
N
m
20
h D 4 sin ˛ m,
0
FY D F sin ˇ N,
0
10
20
30
40
50
60
70
80
90
Alpha, deg
FX D F cos ˇ N,
F D 20L 1 N,
LD
3 h2 C 4 cos ˛2 m,
ˇ D tan1
3h
4 cos ˛
.
A programmable calculator or a commercial package such as TK
Solver or Mathcad is almost essential to the solution of this and
the following problems. The commercial package TK Solver PLUS
was used here to plot the graph of M against ˛. Using the graph as a
guide, the following tabular values were taken about the maximum:
˛, deg
Moment, N-m
41.5
42.0
42.5
101.463
101.483
101.472
The maximum value of the moment is estimated at MB D 101.49 N-m,
which occurs at approximately ˛ D 42.2°
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1
Problem 4.176 The exercise equipment shown is used
by resting the elbow on the fixed pad and rotating the
forearm to stretch the elastic cord AB. The cord behaves
like a linear spring, and its unstretched length is 1 ft.
Suppose you want to design the equipment so that the
maximum moment that will be exerted about the elbow
joint E as the forearm is rotated will be 60 ft-lb. What
should the spring constant k of the elastic cord be?
15
in
B
α
E
10 in
A
5 in
Solution: The strategy is to determine the position of maximum
moment as a function of the angle ˛ for a known spring constant and
then from the system linearity use the fact that ratio of the desired
moment to the actual maximum moment is equal to the ratio of the
spring constants, for the same cord elongation. It is convenient to use
k1 D 1 for the initial spring constant. The steps in the algorithm are:
(1)
The position vectors are rEB D 15i cos ˛ C j sin ˛, and rEA D
5i 10j.
(2)
The vector parallel to the cord is rBA D rEA rEB ; its magnitude
is the stretched length of the cord.
B
15 in
α
E
10 in
rBA
.
jrBA j
(3)
The unit vector parallel to the cord is eBA D
(4)
The (unknown) force is F D jFjeBA .
(5)
The moment about the z axis is MZ D k Ð rEB ð F.
(6)
For spring constant k D 1, graph the moment against angle ˛ to
find the maximum moment.
The TK Solver PLUS program was used to graph the moment against
lb
the angle for k D 1 .
ft
The maximum moment occurs at about ˛ D 49° , as shown by the tabulated values at the maximum point, shown to four significant figures
below.
˛, deg
Moment, ft lb
48
49
50
0.4856
0.4856
0.4856
A
Since the system is linear, the ratio of the moments is equal to the
ratio of the applied forces for the same angular position. Thus
5 in
Moment, ft lb vs Alpha, deg
.6
M
o
m
e
n
t
,
.5
.4
.3
f
t
.2
l
b
.1
0
0
10
20
30
40
50
60
70
80
90
Alpha, deg
k2 L
60 ft lb
D
,
0.4856 ft lb
k1 L
from which, since L cancels for the same angular position, and
k1 D 1, k2 D
60
0.4856
k1 D 123.54
lb
.
ft
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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 4.177 The hydraulic cylinder BC exerts a
2200-lb force on the boom of the crane at C. The force
is parallel to the cylinder. Draw a graph of the moment
exerted by the force about A as a function of the angle
˛ for 0 ˛ 90° , and use it to estimate the values of
˛ for which the moment equals 12,000 ft-lb.
t
6f
t
9f
C
α
A
B
6 ft
Solution: Set the coordinate origin at A and the x axis along AB.
The y axis is upward. In these coordinates, with units in feet, B
is at (6,0) and C is at 9 cos ˛, 9 sin ˛. The vector from B to C is
given by
eBC D xC xB i C yC yB j D 9 cos ˛ 6i C 9 sin ˛j.
The force along BC is given by
Moment about A (ft-lb) versus alpha (deg)
14000
12000
10000
M
A
8000
–
f
6000
t
FBC D jFBC jeBC D 2200eBC (lb)
and the moment about A is given by
MA D rAB ð FBC D 6i ð FBC .
l
b
4000
2000
0
0
The magnitude of the moment is given by its magnitude, and since the
problem is planar, the magnitude is the coefficient of the unit vector k.
10
20
30
40
50
Alpha (deg)
60
70
80
90
The TK Solver Plus software package was used to solve this problem
for values of ˛ over the range 0 ˛ 90. Solutions were evaluated
at one degree intervals. The resulting plot of the magnitude of the
moment versus the angle ˛ is shown at the right. From the plot, there
are two values of ˛ where the value of the moment is 12,000 ft-lb.
These values are ˛ ¾
D 88° .
D 29° and ˛ ¾
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1
Problem 4.178 In Problem 4.177, the moment about
A exerted by the 2200-lb force exerted by the hydraulic
cylinder BC depends on the angle ˛. Estimate the maximum value of the moment and the angle ˛ at which it
occurs.
Solution: From the solution plot developed for Problem 4.177, the
maximum value of the moment occurs somewhere near ˛ D 48° . The
maximum value for the moment about A is just over 13,000 ft-lb. We
would expect the maximum to occur at the configuration where AB
and BC are perpendicular. In this case, the maximum value for the
moment would be MMAX D 62200 D 13,200 ft-lb.
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1
Problem 4.179 The support cable extends from the top
of the 3-m column at A to a point B on the line L. The
tension in the cable is 2 kN. The line L intersects the
ground at the point (3, 0, 1) m and is parallel to the unit
vector e D 27 i C 67 j 37 k. The distance along L from the
ground to point B is denoted s. What is the range of
values of s for which the magnitude of the moment about
O due to the force exerted by the cable at A exceeds
5.6 kN-m?
y
(1)
The position vectors are:
rOG D 3i C 0j C 1k, rOA D 0i C 3j C 0k.
(2)
The vector parallel to the line L is
rGB D seGB D
6
2
2
i C j k s.
7
7
7
B
3m
s
O
x
e
(3, 0, 1) m
z
Solution: A programmable calculator or a commercial package
such as TK Solver or Mathcad is almost essential in the solution
to this Problem. The TK Solver Plus package was used here. The
algorithm for computation is outlined as computational steps, which
may be programmed or performed with a calculator:
L
A
Moment vs s
7
M
a 6.5
g
6
n
i
t 5.5
u
d
5
e
M
4.5
4
0
(3)
The vector from O to B is the sum rOB D rOG C rGB .
(4)
The vector parallel to AB is rAB D rOB rOA .
(5)
The unit vector parallel to AB is eAB D
(6)
The tension at A is TAB D 2eAB .
(7)
The moment about O due to the force applied at A is MO D
rOA ð TAB . Completes the computation for a given value of s.
1
2
3
4
5
6
Parameter s
rAB
.
jrAB j
A plot of moment against s obtained using TK Solver 2.0 is shown.
The graph is relatively flat near the values of interest, which leads to
inaccuracies in determining the limits on s. Using the graph as a guide,
the following values were generated by TK Solver.
s, meters
Moment, kN m
1.89
1.90
5.57
5.58
5.596
5.601
5.602
5.5996
Thus the limits occur at about 1.895 < s < 5.575 m.
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1
Problem 4.180 Consider Problem 4.106. Determine
the distance d that causes the moment about the z axis
due to the force exerted by the cable CD at point C to
be a maximum. What is the maximum moment?
y
(12, 10, 0) ft
(0, 3, 0) ft
D
W
C
d
x
z
(3, 0, 10) ft
Solution: A programmable calculator or a commercial package
such as TK Solver or Mathcad is almost essential in the solution
to this Problem. The algorithm for computation is outlined as steps
that may be programmed or performed with a calculator:
(1)
The position vectors for the bottom and top of the bar, and the
pulley D are:
rOB D 3i C 0j C 10k,
rOT D 12i C 10j C 0k
rOD D 0i C 3j C 0k.
(2)
The vector parallel to the bar, pointing to the top, is rBT D rOT rOB .
(3)
The unit vector parallel to the bar is eBT D
(4)
The vector from the bottom of the bar to the point C is rBC D
deBT .
(5)
The vector position of the collar C is rOC D rOB C rBC .
(6)
The vector parallel to CD is rCD D rOD rOC .
(7)
The unit vector parallel to CD is eCD D
(8)
The tension acting at the collar is TCD D 100eCD lb.
(9)
The moment about the z axis is M D k Ð rOC ð TCD .
rBT
.
jrBT j
z axis moment vs d
300
M
o
m
e
n
t
,
280
260
240
220
200
180
M
z 160
140
f
t 120
100
0
2
4
6
8
10
12
14
16
18
Parameter d, ft
rCD
.
jrCD j
A graph of the moment about the z axis against the parameter d
obtained using TK Solver Plus is shown.
Using the graph as a guide, the values from a table in the TK Solver
solution were selected:
d, ft
z axis Moment, ft lb
12.8
13.0
13.2
265.38
265.41
265.38
The maximum appears to occur exactly at d D 13 ft , with a maximum value of MZ D 265.41 ft lb
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1
Problem 4.181 The rod AB must exert a moment of
magnitude 100 ft-lb about the x axis to support the hood
of the car. Draw a graph of the magnitude of the force
the rod must exert on the hood at B as a function of
d for 1 d 4 ft. If you were designing the support
AB, what value of d would you choose, and what is the
magnitude of the force AB must exert on the hood?
B
A
y
(⫺1, 1, 2) ft
B
A
d
x
z
Solution: The steps in an algorithm for computing the magnitude
of the force as a function of the distance d are: (the detailed computations for one value of d are given in the solution to Problem 4.107.)
(1)
y
The position coordinates of A and B are rB D 1i C 1j C 2k, and
rA D 0i C 0j C dk.
(2)
The vector parallel to AB is rAB D rOB rOA .
(3)
The unit vector parallel to AB is eAB D
(4)
(–1, 1, 2) ft
B
rAB
.
jrAB j
The force exerted by the rod AB is F D jFjeAB . The moment
about the x axis is MX D i Ð rOA ð F.
The commercial program TK Solver Plus was used to graph the
magnitude of the force as a function of the distance d. Using the
graph as a guide, the following three values were taken from the table
computed by TK Solver.
d, ft
Force, lb
2.8
3.0
3.2
58.02
57.74
57.96
x
d
z
A
Force vs d
100
F 95
o 90
r 85
c 80
e 75
, 70
65
l 60
b 55
50
1
1.5
2 2.5 3
Distance d, ft
3.5
4
The minimum force for the required moment occurs at d D 3 ft ,
with a value of jFj D 57.74 lb
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1
y
Problem 4.182 Consider the system shown in Problem
4.148. The forces exerted on the left post by cables
AB and CD can be represented by a single force F.
Determine the tensions in the cables so that jFj D 600 N
and the line of action of F intersects the y axis at y D
400 mm.
A
400 mm
B
C
300 mm
D
O
Solution: The strategy is to solve by iteration the simultaneous
conditions on the forces and moments. Equating the horizontal and
vertical components of the forces for the two systems:
TAB cos ˛ C TCD cos ˇ D 600 cos TAB sin ˛ C TCD sin ˇ D 600 sin .
where the angles are
˛ D tan1
400
800
D 26.6°
300 mm
x
800 mm
Error vs angle
300
200
E 100
r
0
r −100
o
−200
r
−300
−400
−500
6
8 10 12 14 16 18 20 22 24 26
Angle, deg
for cable AB, and
ˇ D tan1
300
800
D 20.6°
for cable CD., and the unknown angle applies to the single cable
system. A guess at the unknown angle will yield a solution for the
tensions which may not satisfy the equality of moments condition for
the two systems. This difficulty is resolved as follows: denote the error
in the equality of moments about the origin of the left post as
ε D 0.3TAB cos ˛ C 0.7TCD cos ˇ 0.5600 cos .
Plot the error as a function of the angle over the allowed interval.
For an angle at which ε D 0 the three conditions are satisfied.
The TK Solver software package was used to graph the error as a
function of the angle. The range of allowed angles was determined
by assuming that the single cable could be attached on the right post
at any point between the ground and the 300 mm height.
The zero crossing for the error occurs at about D 22° . A closer look
at tabulated values near this value yields an angle of D 22.11° . At
this angle the tensions in the two cables are TAB D 155.4 N , and
TCD D 445.25 N
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1
Problem 4.183 Suppose you want to represent the
force and the 200-N-m couple in Problem 4.166 by a
force F and a couple M, and choose the distance s so
that the magnitude of M is a minimum. Determine s, F,
and M.
y
(2, 6, 0) m
s
100i ⫹ 20j ⫺ 20k (N)
O
x
200 N-m
(4, 0, 3) m
z
Solution: The steps in an algorithm for entering into a programmable calculator or a commercial software package such as TK
Solver or Mathcad are given: (The details of a computation for
one value of s are given in the solution to Problem 4.150)
(1)
The position vectors for the top and bottom of the bar are:
rT D 2i C 6j C 0k m,
rB D 4i C 0j C 3k m.
(2)
The vector parallel to the bar, pointing top to bottom, is
rTB D rB rT D 2i 6j C 3k m.
(3)
Moment, N-m vs b, m
700
M 650
o
m 600
e
n 550
t
500
,
450
N
M 400
0
1
2
3
4
B, M
5
6
7
The unit vector parallel to the bar, pointing top to bottom is
rTB
.
jrTB j
eTB D
(4)
The distance from the top of the bar to the point of application
of the force is rTF D seTB m.
(5)
The vector from the origin to the point of application of the force
is rOF D rT C rTF m.
(6)
The moment about the origin due to the action of the force is
MOF D rOF ð F N-m.
(7)
The couple moment is MC D MC eBT D MC eTB N-m.
(8)
The total moment is the sum of the moments due to the force
and the couple, MT D MOF C MC .
The force is given: F D 100i C 20j 20k N. The couple moment magnitude is given: MC D 200 N-m.
The strategy is to graph the total moment as a function of s to determine the minimum value of the moment, and the value of s at the
minimum. The TK Solver 2.0 package was used to graph the total
moment as a function of s. Using the graph as a guide, tabulated
values near the minimum were examined. The minimum magnitude of
the moment occurs at s D 4.66 m . The equivalent force at the origin
is F D 100i C 20j 20k N . At s D 4.66 m the equivalent moment
at the origin is
MT D MOF C MC D 137.2i C 437.77j 219.66k N-m
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1
Problem 4.184
force about A.
(a)
(b)
Determine the moment of the 200-N
What is the two-dimensional description of the
moment?
Express the moment as a vector.
y
(– 400, 0, 0) mm
–200 j (N)
x
A
(400, – 200, 0) mm
Solution:
(a)
The two dimensional description. By inspection, the perpendicular distance to the line of action from A is
d D 400 400 mm D 0.8 m.
The moment is positive, since it is counterclockwise. Thus MA D
0.8200 D 160 N-m
(b)
The vector description. The position vectors of A and the point
of action are rOA D 0.4i 0.2j m, and rOF D 0.4i m. The distance from A to the point of action is rAF D rOF rOA D 0.4
0.4i 0.2j D 0.8i C 0.2j m. The moment is
i
j
0.2
MA D rAF ð F D 0.8
0
200
k 0 D 160k N-m
0
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1
Problem 4.185 The Leaning Tower of Pisa is approximately 55 m tall and 7 m in diameter. The horizontal
displacement of the top of the tower from the vertical
is approximately 5 m. Its mass is approximately 3.2 ð
106 kg. If you model the tower as a cylinder and assume
that its weight acts at the center, what is the magnitude
of the moment exerted by the weight about the point at
the center of the tower’s base?
Solution: The position vector of the center of mass, in the coordinates shown, is
5m
y
5m
rCM D 2.5i C 27.5j m
The weight acting at the center of mass is W D mgj
m
W D 9.81
3.2 ð 106 kgj D 31.4j MN
s
The moment is rCM ð W
M D 2.5i C 27.5j ð 31.4j MN-m
M D 78.5k MN-m
x
C
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1
y
Problem 4.186 The cable AB exerts a 300-N force on
the support A that points from A toward B. Determine the
magnitude of the moment the force exerts about point P.
B
(0.3, 0.6) m
A
(⫺0.4, 0.3) m
x
P
(0.5, ⫺0.2) m
Solution:
F D 300 N
0.7i C 0.3j
p
,
0.58
rPA D 0.9i C 0.5j m
MP D rPA ð F D 244 Nmk )
MP D 244 Nm
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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 4.187 Three forces act on the structure. The
sum of the moments due to the forces about A is zero.
Determine the magnitude of the force F.
30⬚
45⬚
2 kN
4 kN
b
A
F
Solution:
b
2b
b
p
MA D 4 kN 2b 2 kN cos 30° 3b
C 2 kN sin 30° b C F4b D 0
Solving we find
F D 2.463 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 4.188 Determine the moment of the 400-N
force (a) about A, (b) about B.
30°
400 N
220 mm
A
260 mm
B
500 mm
Solution: Use the two dimensional description of the moment. The
vertical and horizontal components of the 200 N force are
FY D 400 sin 30° D 200 N,
FX D C400 cos 30° D 346.41 N.
(a)
The moment arm from A to the line of action of the horizontal
component is 0.22 m. The moment arm from A to the vertical
component is zero. The moment about A is negative,
MA D 0.22346.41 D 76.21 N-m
(b)
The perpendicular distances to the lines of action of the vertical
and horizontal components of the force from B are d1 D 0.5 m,
and d2 D 0.48 m. The action of the vertical component is positive, and the action of the horizontal component is negative.
The sum of the moments: MB D C0.5200 0.48346.41 D
66.28 N-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 4.189 Determine the sum of the moments
exerted about A by the three forces and the couple.
A
5 ft
300 lb
800 ft-lb
200 lb
200 lb
6 ft
3 ft
Solution: Establish coordinates with origin at A, x horizontal, and
y vertical with respect to the page. The moment exerted by the couple
is the same about any point. The moment of the 300 lb force about A
is M300 D 6i 5j ð 300j D 1800k ft-lb.
The moment of the downward 200 lb force about A is zero since the
line of action of the force passes through A. The moment of the 200 lb
force which pulls to the right is
M200 D 3i 5j ð 200i D 1000k (ft-lb).
The moment of the couple is MC D 800k (ft-lb). Summing the four
moments, we get
MA D 1800 C 0 C 1000 800k D 1600k (ft-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 4.190 In Problem 4.189, if you represent the
three forces and the couple by an equivalent system
consisting of a force F acting at A and a couple M,
what are the magnitudes of F and M?
Solution: The equivalent force will be equal to the sum of the
forces and the equivalent couple will be equal to the sum of the
moments about A. From the solution to Problem 4.189, the equivalent couple will be C D MA D 1600k (ft-lb). The equivalent force
will be FEQUIV. D 200i 200j C 300j D 200i C 100j (lb)
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1
Problem 4.191 The vector sum of the forces acting on
the beam is zero, and the sum of the moments about A
is zero.
(a)
(b)
30°
220 mm
What are the forces Ax , Ay , and B?
What is the sum of the moments about B?
400 N
Ay
Ax
260 mm
500 mm
B
Solution: The vertical and horizontal components of the 400 N
force are:
FX D 400 cos 30° D 346.41 N,
FY D 400 sin 30° D 200 N.
The sum of the forces is
FX D AX C 346.41 D 0,
from which AX D 346.41 N
FY D AY C B 200 D 0.
The sum of the moments about A is
MA D 0.5B 0.22346.41 D 0,
from which B D 152.42 N. Substitute into the force equation to get
AY D 200 B D 47.58 N
(b) The moments about B are
MB D 0.5AY 0.48346.41 0.26AX C 0.5200 D 0
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1
Problem 4.192 To support the ladder, the force exerted
at B by the hydraulic piston AB must exert a moment
about C equal in magnitude to the moment about C due
to the ladder’s 450-lb weight. What is the magnitude of
the force exerted at B?
6 ft
450 lb
3 ft
A
Solution: The moment about C exerted by the weight is
B
C
MC D 4506 D 2700 ft lb.
6 ft
3 ft
The ladder is at an elevation of 45° from the horizontal. The cylinder
is at an angle
D tan1
3
D 26.56° .
6
6 ft
The vertical and horizontal components of the force at B due to the
cylinder are
FX D F cos 26.57° D 0.8944F lb
450 lb
FY D F sin 26.57° D 0.4472F lb.
B
The moment about C due to these forces is
3 ft
C
A
MC D 30.4472F 30.8944F C 2700 D 0.
6 ft
3 ft
Solving:
FD
2700
D 670.82 lb
4.0249
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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 4.193
(a)
(b)
y
The force F D 60i C 60j (lb).
F
Determine the moment of F about point A.
What is the perpendicular distance from point A to
the line of action of F?
(4, – 4, 2) ft
x
A
(8, 2, 12) ft
z
Solution: The position vector of A and the point of action are
(b)
The magnitude of the moment is
p
6002 C 6002 C 6002 D 1039.3 ft lb.
rA D 8i C 2j C 12k (ft), and rF D 4i 4j C 2k.
jMA j D
The vector from A to F is
p
The magnitude of the force is jFj D 602 C 602 D 84.8528 lb.
The perpendicular distance from A to the line of action is
rAF D rF rA D 4 8i C 4 2j C 2 12k
D 4i 6j 10k.
(a)
DD
1039.3
D 12.25 ft
84.8528
The moment about A is
i
j
k MA D rAF ð F D 4 6 10 60 60
0 D 600i C 600j 600k (ft lb)
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1
y
Problem 4.194 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 moment about the base E due to the
force exerted on the post BE by the cable AB.
C
B
D
A
1m
1m
E
2m
0.3 m
x
z
Solution: The strategy is to develop the simultaneous equations in
the unknown tensions in the cables, and use the tension in AB to find
the moment about E. This strategy requires the unit vectors parallel to
the cables. The position vectors of the points are:
The equilibrium conditions are TAB C TAC C TAD D W. Collect like
terms in i, j, k:
rOA D 1.2j,
FX D 0.2281TAB C 0TAC C 0.9284TAD i D 0
FY D C0.6082 Ð TAB C 0.6247 Ð TAC
rOB D 0.3i C 2j C 1k,
rOC D 2j 1k,
rOD D 2i C 2j,
rOE D 0.3i C 1k.
C 0.3714 Ð TAD 196.2j D 0
FZ D C0.7603 Ð TAB 0.7809 Ð TAC C 0 Ð TAD k D 0
Solve:
TAB D 150.04 N,
The vectors parallel to the cables are:
TAC D 146.08 N,
rAB D rOB rOA D 0.3i C 0.8j C 1k,
TAD D 36.86 N.
rAC D rOC rOA D C0.8j 1k,
rAD D rOD rOA D C2i C 0.8j.
The unit vectors parallel to the cables are:
The moment about E is
ME D rEB ð TAB eAB D TAB rEB ð eAB rAB
D 0.2281i C 0.6082j C 0.7603k :
eAB D
jrAB j
i
D 150 0
0.2281
eAC D 0i C 0.6247j 0.7809k,
D 228i 68.43k (N-m)
j
2
C0.6082
k
0
C0.7603 eAD D C0.9284i C 0.3714j C 0k.
The tensions in the cables are
TAB D TAB eAB ,
TAC D TAC eAC , and
TAD D TAD eAD .
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1
Problem 4.195 Three forces of equal magnitude are
applied parallel to the sides of an equilateral triangle.
(a)
(b)
F
Show that the sum of the moments of the forces is
the same about any point.
Determine the magnitude of the moment.
L
F
Strategy: To do (a), resolve one of the forces into
vector components parallel to the other two forces.
F
Solution: The interior angles of an equilateral triangle are 60° .
The sum of the moments about P is
Assume that the x axis is coincident with the lower side, with the
origin at the lower left corner. Denote the forces by labels 1, 2, 3
counterclockwise beginning with the one coincident with the lower
side, and label the corners 1, 2, 3 beginning with the lower left corner.
The vectors to the lower corners are r1 D 0, and r2 D Li. Let Px, y
be any point in the space. The vector distances from P to the lower
corners are
rP1 D r1 rP D xi yj,
rP2 D r2 rp D L xi yj.
M D rP1 ð F1 C rP1 ð F3 C rP2 ð F2 D F[rP1 ð e1 C e3 C rP2 ð e2 ]
i
i
j k
j
k
y 0 y
0 C FL x p
p
MP D F x
1
1
3
3
0
0
2
2
2
2
p
p
p
1
1
3
3
3
D kF
x C y C kF
L
x y
2
2
2
2
2
The unit vectors parallel to the forces are:
p
e1 D 1i,
p
3
1
j, and
e2 D i C
2
2
D
3
FLk
2
Since the result is independent of the coordinates of P, the moment
about any point is the same.
p
3
1
e3 D i j.
2
2
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1
Problem 4.196 The bar AB supporting the lid of the
grand piano exerts a force F D 6i C 35j 12k (lb) at
B. The coordinates of B are (3, 4, 3) ft. What is the
moment of the force about the hinge line of the lid (the
x axis)?
y
Solution: The position vector of point B is rOB D 3i C 4j C 3k.
The moment about the x axis due to the force is
MX D eX Ð rOB ð F D i Ð rOB ð F
1 0
MX D 3 4
6 35
0 3 D 153 ft lb
12 B
x
A
z
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1
Problem 4.197 Determine the moment of the vertical
800-lb force about point C.
y
800 lb
A (4, 3, 4) ft
B
D (6, 0, 0) ft
x
z
C (5, 0, 6) ft
Solution: The force vector acting at A is F D 800j (lb) and the
position vector from C to A is
rCA D xA xC i C yA yC j C zA zC k
D 4 5i C 3 0j C 4 6k D 1i C 3j 2k (ft).
The moment about C is
i
j
k MC D 1
3
2 D 1600i C 0j C 800k (ft-lb)
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1
Problem 4.198 In Problem 4.197, determine the moment of the vertical 800-lb force about the straight line
through points C and D.
Solution: In Problem 4.197, we found the moment of the 800 lb
force about point C to be given by
MC D 1600i C 0j C 800j (ft-lb).
The vector from C to D is given by
rCD D xD xC i C yD yC j C zD zC k
D 6 5i C 0 0j C 0 6k
D 1i C 0j 6j (ft),
and its magnitude is
jrCD j D
p
p
12 C 62 D 37 (ft).
The unit vector from C to D is given by
6
1
eCD D p i p k.
37
37
The moment of the 800 lb vertical force about line CD is given by
MCD D
D
6
1
p i p k Ð 1600i C 0j C 800j (ft-lb)
37
37
1600 4800
p
37
(ft-lb).
Carrying out the calculations, we get MCD D 1052 (ft-lb)
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1
Problem 4.199 The system of cables and pulleys supports the 300-lb weight of the work platform. If you
represent the upward force exerted at E by cable EF
and the upward force exerted at G by cable GH by a
single equivalent force F, what is F, and where does its
line of action intersect the x axis?
H
F
E
G
B
D
y
A
60°
60°
C
x
8 ft
Solution: The cable-pulley combination does not produce a moment. Hence the equivalent force does not. The equivalent force is
600
j D 300j (lb). The
equal to the total supported weight, or F D C
2
8
force occurs at midpoint of the platform width, x D D 4 ft
2
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1
Problem 4.200
(a)
(b)
Consider the system in Problem 4.199.
What are the tensions in cables AB and CD?
If you represent the forces exerted by the cables at
A and C by a single equivalent force F, what is
F, and where does its line of action intersect the
x axis?
Solution: The vertical component of the tension is each cable must
equal half the weight supported.
TAB sin 60° D 150 lb, from which TAB D
symmetry, the tension TCD D 173.2 lb.
150
D 173.2 lb. By
sin 60°
The single force must equal the sum of the vertical components; since
there is no resultant moment produced by the cables, the force is
F D 300j lb and it acts at the platform width midpoint x D 4 ft.
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1
Problem 4.201 The two systems are equivalent. Determine the forces Ax and Ay , and the couple MA .
System 1
y
20 N
400 mm
Ax
x
30 N
Ay
600 mm
400 mm
System 2
y
8 N-m
400 mm
20 N
MA
10 N
x
80 N
600 mm
400 mm
Solution: The sum of the forces for System 1 is
FX D AX C 20i,
FY D AY C 30j.
The sum of forces for System 2 is
FX D 20i and
FY D 80 10j.
Equating the two systems:
AX C 20 D 20 from which AX D 40 N
AY C 30 D 80 10 from which AY D 40 N
The sum of the moments about the left end for System 1 is
M1 D 0.420 C 301 D 22 N-m.
The sum of moments about the left end for System 2 is
M2 D MA 101 8 D MA 18.
Equating the moments for the two systems:
MA D 18 C 22 D 40 N-m
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1
Problem 4.202 If you represent the equivalent systems
in Problem 4.201 by a force F acting at the origin and
a couple M, what are F and M?
Solution: Summing the forces in System 1, F D AX C 20i C
AY C 30j. Substituting from the solution in Problem 4.201,
F D 20i C 70j. The moment is M D 200.4k C 30k D 22k (N-m)
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1
Problem 4.203 If you represent the equivalent systems
in Problem 4.201 by a force F, what is F, and where does
its line of action intersect the x axis?
Solution: The force is F D 20i C 70j. The moment to be represented is
i
j k M D r ð F D 22k D x
0 0 D 70xk,
20 70 0 from which x D
22
D 0.3143 m
70
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1
Problem 4.204
The two systems are equivalent. If
F D 100i C 40j C 30k (lb),
M0 D 80i C 120j C 40k (in-lb),
determine F0 and M.
System 1
y
System 2
y
4 in
4 in
M
F
F'
M'
6 in
6 in
x
x
6 in
z
6 in
z
Solution: The sum of forces in the two systems must be equal,
thus F0 D F D 100i C 40j C 30k (lb).
The moment for the unprimed system is MT D r ð F C M.
The moment for the primed system is M0T D r0 ð F C M0 .
The position vectors are r D 0i C 6j C 6k, and r0 D 4i C 6j C 6k.
Equating the moments and solving for the unknown moment
i
4
100
M D M0 C r0 r ð F D 80i C 120j C 40k C j
k 0
0 40 30 D 80i C 120j C 40k 120j C 160k
D 80i C 200k (in-lb)
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1
Problem 4.205 The tugboats A and B exert forces
FA D 1 kN and FB D 1.2 kN on the ship. The angle
D 30° . If you represent the two forces by a force F
acting at the origin O and a couple M, what are F
and M?
y
A
FA
Solution: The sums of the forces are:
FX D 1 C 1.2 cos 30° i D 2.0392i (kN)
60 m
O
x
FY D 1.2 sin 30° j D 0.6j (kN).
60 m
FB
The equivalent force at the origin is
FEQ D 2.04i C 0.6j
θ
B
The moment about O is MO D rA ð FA C rB ð FB . The vector positions are
25 m
rA D 25i C 60j (m), and
rB D 25i 60j (m).
The moment:
i
MO D 25
1
j k i
60 0 C 25
0 0 1.0392
j
k 60 0 0.6 0 D 12.648k D 12.6k (kN-m)
Check: Use a two dimensional description: The moment is
MO D 25FB sin 30° C 60FB cos 30° 60FA D 39.46FB 60FA D 12.6 kN-m
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1
Problem 4.206 The tugboats A and B in Problem 4.205
exert forces FA D 600 N and FB D 800 N on the ship.
The angle D 45° . If you represent the two forces by
a force F, what is F, and where does its line of action
intersect the y axis?
Solution: The equivalent force is
Check: Use a two dimensional description:
F D 0.6 C 0.8 cos 45° i C 0.8 sin 45° j D 1.1656i C 0.5656j (kN).
MO D 25FB sin 45° C 60FB cos 45° 60FA
The moment produced by the two forces is
D 24.75FB 60FA D 16.20 kN-m.
The single force must produce this moment.
MO D rA ð FA C rB ð FB .
rA D 25i C 60j (m), and rB D 25i 60j (m).
i
MO D 0
1.1656
The moment:
from which
The vector positions are
i
MO D 25
0.6
j k i
60 0 C 25
0 0 0.5656
j
60
0.5656
k 0 D 16.20k (kN-m)
0
yD
j
y
0.5656
k 0 D 1.1656yk D 16.20k,
0
16.20
D 13.90 m
1.1656
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1
Problem 4.207 The tugboats A and B in Problem 4.205
want to exert two forces on the ship that are equivalent
to a force F acting at the origin O of 2-kN magnitude. If
FA D 800 N, determine the necessary values of FB and
angle .
Solution: The equivalent force at the origin is FA C FB cos 2 C
sin 2 D 20002 . The moment about the origin due to F
FB
must be zero:
A and FB
MO D 60FA C 60FB cos 25FB sin D 0.
Substitute FA D 800 N to obtain 6.76x 2 7616x C 326400 D 0. In
canonical form: x 2 C 2bx C c D 0, where
p b D 563.31, and c D
48284.0, with the solutions x D b š b2 c D 1082.0, D 44.62.
From the second equation, y D 1812.9, D 676.81. The force FB has
two solutions: Solve for FB and : (1)
p
44.62 C 1812.92 D 1813.4 N
These are two equations in two unknowns FB sin and FB cos . For
brevity write x D FB cos , y D FB sin , so that the two equations
become x 2 C 2FA x C F2A C y 2 D 20002 and 60x 25y 60FA D 0.
Eliminate y by solving each equation for y 2 and equating the results:
FB D
60
60 2
x .
y 2 D 20002 x 2 2FA x F2A D FA C
25
25
D tan1
Reduce to obtain the quadratic in x:
1C
60
25
2 C 1C
x 2 C 2FA 1 60
25
60
25
at the angle
p
FB D
1812.9
44.6
D 88.6° , and 2
676.82 C 1082.02 D 1276.2 N,
at the angle
2 x
D tan1
676.8
1082.0
D 32.0°
2 F2A 20002 D 0.
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1
y
Problem 4.208 If you represent the forces exerted by
the floor on the table legs by a force F acting at the
origin O and a couple M, what are F and M?
50 N
Solution: The sum of the forces is the equivalent force at the
origin. F D 50 C 48 C 50 C 42j D 190j (N). The position vectors of
the legs are, numbering the legs counterclockwise from the lower left
in the sketch:
2m
1m
x
z
42 N
48 N
50 N
r1 D C1k,
r2 D 2i C 1k,
r3 D 2i,
r4 D 0.
The sum of the moments about the origin is
i
MO D 0
0
j k i j
0 1 C 2 0
48 0 0 50
k i j
1 C 2 0
0 0 42
k 0 0
D 98i C 184k (N-m).
This is the couple that acts at the origin.
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1
Problem 4.209 If you represent the forces exerted by
the floor on the table legs in Problem 4.208 by a force
F, what is F, and where does its line of action intersect
the xz plane?
Solution: From the solution to Problem 4.208 the equivalent force
is F D 190j. This force must produce the moment M D 98i C 184k
obtained in Problem 4.208.
i
j
M D x
0
0 190
k z D 190zi C 190xk D 98i C 184k,
0
from which
xD
184
D 0.9684 m and
190
zD
98
D 0.5158 m
190
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1
y
Problem 4.210 Two forces are exerted on the crankshaft by the connecting rods. The direction cosines of
FA are cos x D 0.182, cos y D 0.818, and cos z D
0.545, and its magnitude is 4 kN. The direction cosines
of FB are cos x D 0.182, cos y D 0.818, and cos z D
0.545, and its magnitude is 2 kN. If you represent the
two forces by a force F acting at the origin O and a
couple M, what are F and M?
FB
FA
360 mm
O
z
160 mm
80 mm
80 mm x
Solution: The equivalent force is the sum of the forces:
FA D 40.182i C 0.818j C 0.545k
D 0.728i C 3.272j C 2.18k (kN)
FB D 20.182i C 0.818j 0.545k D 0.364iC1.636j1.09k (kN).
The sum: FA C FB D 0.364i C 4.908j C 1.09k (kN)
The equivalent couple is the sum of the moments. M D rA ð FA C
rB ð FB . The position vectors are:
rA D 0.16i C 0.08k,
rB D 0.36i 0.08k.
The sum of the moments:
i
M D 0.16
0.728
j
0
3.272
k i
0.08 C 0.36
2.180 0.364
j
0
1.636
k 0.08 1.090 M D 0.1309i 0.0438j C 1.1125k (kN-m)
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1
Problem 4.211 If you represent the two forces exerted
on the crankshaft in Problem 4.210 by a wrench consisting of a force F and a parallel couple Mp , what are F
and Mp , and where does the line of action of F intersect
the xz plane?
Solution: From the solution to Problem 4.210,
F D 0.364i C 4.908j C 1.09k (kN) and
M D 0.1309i 0.0438j C 1.1125k (kN-m).
The unit vector parallel to F is
eF D
F
D 0.0722i C 0.9737j C 0.2162k.
jFj
The moment parallel to the force is
MP D eF Ð MeF .
Carrying out the operations:
MP D 0.2073eF D 0.01497i C 0.2019j C 0.0448k (kN-m).
This is the equivalent couple parallel to F.
The component of the moment perpendicular to F is
MN D M MP D 0.1159i 0.2457j C 1.0688k.
The force exerts this moment about the origin.
i
MN D x
0.364
j
0
4.908
k z 1.09 D 4.908zi 1.09x C 0.364zj C 4.908xk
D 0.1159i 0.2457j C 1.06884k.
From which
xD
1.0688
D 0.2178 m,
4.908
zD
0.1159
D 0.0236 m
4.908
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1
Problem 5.1 The beam has pin and roller supports and
is subjected to a 4-kN load.
(a)
(b)
Draw the free-body diagram of the beam.
Determine the reactions at the supports.
4 kN
B
A
C
2m
3m
Strategy: (a) Draw a diagram of the beam isolated
from its supports. Complete the free-body diagram of
the beam by adding the 4-kN load and the reactions due
to the pin and roller supports (see Table 5.1). (b) Use the
scalar equilibrium equations (5.4)–(5.6) to determine the
reactions.
Solution:
(a)
(b)
The free-body diagram
The equilibrium equations
Cy
4 kN
Bx
Fx : Bx D 0
Fy : By 4 kN C Cy D 0
By
MB : 4 kN2 m C Cy 3 m D 0
Solving we find
Bx D 0, By D 6.67 kN, Cy D 2.67 kN
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1
Problem 5.2 The beam has a built-in support and is
loaded by a 2-kN force and a 6 kN-m couple.
(a)
(b)
Draw the free-body diagram of the beam.
Determine the reactions at the supports.
2 kN
6 kN-m
A
60°
1m
3m
Solution:
(a)
(b)
2 kN
Ay
The free-body diagram
The equilibrium equations
6 kN-m
60°
Fx : Ax C 2 kN cos 60° D 0
Ax
Fy : Ay C 2 kN sin 60° D 0
MA
MA : MA C 6 kN-m C 2 kN sin 60° 4 m D 0
Solving we find
Ax D 1 kN, Ay D 1.732 kN, MA D 12.93 kN-m
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1
Problem 5.3 The beam is subjected to a load F D
400 N and is supported by the rope and the smooth
surfaces at A and B.
(a)
(b)
Draw the free-body diagram of the beam.
What are the magnitudes of the reactions at A
and B?
F
A
B
30°
45°
1.2 m
y
Solution:
C
FX D 0:
A cos 45° B sin 30° D 0
FY D 0:
A sin 45° C B cos 30° T 400 N D 0
MA D 0:
1.2T 2.7400 C 3.7B cos 30° D 0
1.5 m
1m
A
x
F
45°
B
1.5 m
1.2 m
1m
T
30°
Solving, we get
A D 271 N
B D 383 N
T D 124 N
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1
Problem 5.4
beam.
(a) Draw the free-body diagram of the
(b) Determine the reactions at the supports.
30⬚ 600 lb
B
A
9 ft
5 ft
Solution:
(a)
(b)
The FBD
The equilibrium equations
T
30°
600 lb
MB : 600 lb cos 30° 9 ft T14 ft D 0
Bx
Fx : 600 lb sin 30° C Bx D 0
Fy : T 600 lb cos 30° C By D 0
By
Bx D 300 lb
)
By D 185.6 lb
T D 334 lb
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1
Problem 5.5 (a) Draw the free-body diagram of the
60-lb drill press, assuming that the surfaces at A and B
are smooth.
(b) Determine the reactions at A and B.
60 lb
A
B
10 in
14 in
Solution: The system is in equilibrium.
(a)
(b)
The free body diagram is shown.
The sum of the forces:
FX D 0,
FY D FA C FB 60 D 0
The sum of the moments about point A:
60 lb
MA D 1060 C 24FB D 0,
from which FB D
600
D 25 lb
24
Substitute into the force balance equation:
A
B
10 in
FA D 60 FB D 35 lb
FA
14 in
FB
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1
Problem 5.6 The masses of the person and the diving
board are 54 kg and 36 kg, respectively. Assume that
they are in equilibrium.
(a)
(b)
Draw the free-body diagram of the diving board.
Determine the reactions at the supports A and B.
A
B
WP
WD
1.2 m
2.4 m
4.6 m
Solution:
4.6 m
2.4 m
(a)
(b)
1.2 m
AX
FX D 0:
AX D 0
FY D 0:
AY C BY 549.81 369.81 D 0
MA D 0:
1.2BY 2.4369.81
AY
BY
WD
WP
4.6549.81 D 0
Solving:
AX D 0 N
AY D 1.85 kN
BY D 2.74 kN
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1
y
Problem 5.7 The ironing board has supports at A and
B that can be modeled as roller supports.
(a)
(b)
Draw the free-body diagram of the ironing board.
Determine the reactions at A and B.
A
B
x
3 lb
10 lb
12 in
10 in
20 in
Solution: The system is in equilibrium.
Substitute into the force balance equation:
(a)
(b)
FA D 13 FB D C15.833 lb
The free-body diagram is shown.
The sums of the forces are:
FX D 0,
y
A
B
x
FY D FA C FB 10 3 D 0.
12 in
10 in
10 lb
20 in
FA
FB
10 lb
3 lb
3 lb
The sum of the moments about A is
MA D 12FB 2210 423 D 0,
from which FB D
346
D 28.833 in.
12
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1
Problem 5.8
(a)
(b)
The distance x D 9 m.
Draw the free-body diagram of the beam.
Determine the reactions at the supports.
10 kN
A
B
6m
x
Solution:
(a)
(b)
10 kN
The FBD
The equilibrium equations
x= 9 m
Ax
Fx : Ax D 0
6m
Fy : Ay C By 10 kN D 0
Ay
By
MA : By 6 m 10 kN9 m D 0
Solving we find
Ax D 0, Ay D 5 kN, By D 15 kN
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1
Problem 5.9 An engineer analyzes the beam in
Problem 5.8 and determines that each support will safely
withstand a force of magnitude 20 kN. Based on this
criterion, what is the range of values of the distance x
at which the 10-kN force can safely be applied to the
beam? Assume that 0 x 16 m.
Solution: The equilibrium equations for arbitrary x
10 kN
Fx : Ax D 0
x
Ax
Fy : Ay C By 10 kN D 0
6m
MA : By 6 m 10 kNx D 0
Ay
By
Solving we find
Ax D 0, Ay D
kN
5
x 6 m
, By D
3
m
5
x
3
kN
m
For the reaction at A we must have
5
x 6 m kN 20 kN ) 6 m x 18 m
3
m
For the reaction at B we must have
5
kN 20 kN ) 0 x 12 m
x
3
m
Thus we conclude that we must have
0 x 12 m
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1
Problem 5.10 (a) Draw the free-body diagram of the
beam.
(b) Determine the reactions at the supports.
100 lb
400 lb
900 ft-lb
A
B
3 ft
Solution: (a) Both supports are roller supports. The free body
diagram is shown. (b) The sum of the forces:
and
4 ft
3 ft
100 lb
3 ft
4 ft
400 lb
4 ft
3 ft
4 ft
FX D 0,
A
FY D FA C FB C 100 400 D 0.
The sum of the moments about A is
MA D 3100 C 900 7400 C 11FB D 0.
From which FB D
100 lb
3 ft
900 ft lb
4 ft
3 ft
B
400 lb
4 ft
900 ft lb
FA
FB
2200
D 200 lb
11
Substitute into the force balance equation to obtain
FA D 300 FB D 100 lb
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1
Problem 5.11 The person exerts 20-N forces on the
pliers. The free-body diagram of one part of the pliers
is shown. Notice that the pin at C connecting the two
parts of the pliers behaves like a pin support. Determine
the reactions at C and the force B exerted on the pliers
by the bolt.
25
mm
80
mm
B
C
Cx
Cy
50 mm
45⬚
20 N
C
Solution: The equilibrium equations
MC :B25 mm 20 N cos 45° 80 mm
20 N sin 45° 50 mm D 0
20 N
Fx :Cx 20 N sin 45° D 0
Fy :Cy B 20 N cos 45° D 0
20 N
Solving:
B D 73.5 N, Cx D 14.14 N, Cy D 87.7 N
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1
Problem 5.12
beam.
(a) Draw the free-body diagram of the
8 kN
8 kN
2 kN-m
A
30⬚
B
(b) Determine the reactions at the pin support A.
600
mm
Solution:
(a)
(b)
8 kN
The FBD
The equilibrium equations
500
mm
MA : 8 kN0.6 m C 8 kN1.1 m 2 kNm
600
mm
600
mm
8 kN
2 kN-m
30°
B
Ax
B cos 30° 2.3 m D 0
Fx :Ax B sin 30° D 0
Ay
Fy :Ay 8 kN C 8 kN B cos 30° D 0
Solving
Ax D 0.502 kN, Ay D 0.870 kN, B D 1.004 kN
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1
Problem 5.13
beam.
y
(a) Draw the free-body diagram of the
A
(b) Determine the reactions at the supports.
6m
40 kN
B
x
8m
12 m
Solution:
A
(a)
(b)
The FBD
The equilibrium equations
MB : 40 kN4 m C A6 m D 0
40 kN
Bx
Fx : A C Bx D 0
Fy : 40 kN C By D 0
Solving we find
By
A D Bx D 26.7 kN, By D 40 kN
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1
Problem 5.14
beam.
(a) Draw the free-body diagram of the
A
(b) If F D 4 kN, what are the reactions at A and B?
2 kN-m
F 0.2 m
0.3 m
0.2 m
0.3 m
Solution:
0.4 m
B
2 kN-m
Ax
(a)
(b)
The free-body diagram
The equilibrium equations
F = 4 kN
MA : 2 kN-m 4 kN0.2 m C B1.0 m D 0
Ay
Fx : Ax 4 kN D 0
Fy : Ay C B D 0
Solving:
B
Ax D 4 kN, Ay D 2.8 kN, B D 2.8 kN
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1
Problem 5.15 Consider the beam shown in Problem 5.14. A structural engineer determines that the
support at A can safely be subjected to a force of 12 kN
magnitude and the support at B can safely be subjected
to a force of 15 kN magnitude. Based on this criterion,
what is the largest acceptable magnitude of the force F?
2 kN-m
Solution: The equilibrium equations
Ax
MA : 2 kN-m F0.2 m C B1.0 m D 0
F
Fx : Ax F D 0
Ay
Fy : Ay C B D 0
Solving we find
Ax D F, Ay D 2 kN C F, B D 2 kN C F
AD
Ax 2 C Ay 2 D
B
2 kN2 C 0.8 kNF C 1.04F2
Putting in the limits we have
B D 15 kN D 2 kN C F ) F D 13 kN
A D 12 kN D
2 kN2 C 0.8 kNF C 1.04F2
) F D 11.22 kN
We take the most stringent condition as the answer
Fmax D 11.22 kN
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1
Problem 5.16 The person doing push-ups pauses in
the position shown. His mass is 80 kg. Assume that
his weight W acts at the point shown. The dimensions
shown are a D 250 mm, b D 740 mm, and c D 300 mm.
Determine the normal force exerted by the floor (a) on
each hand, (b) on each foot.
c
W
a
b
Solution: We assume that each hand and each foot carries an
equal load.
FX D 0:
No forces in x-direction
FY D 0:
2FH C 2FF W D 0
MH D 0:
aW C a C b2FF D 0
Solving, we get
W D 784.8 N
FH D 293.3 N
FF D 99.1 N
y
H
F
W
a
2FH
x
b
2FF
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1
Problem 5.17 With each of the devices shown you can
support a load R by applying a force F. They are called
levers of the first, second, and third class.
(a)
(b)
The ratio R/F is called the mechanical advantage.
Determine the mechanical advantage of each lever.
Determine the magnitude of the reaction at A for
each lever. (Express your answer in terms of F.)
F
R
R
A
A
L
L
L
First-class lever
(a)
L
Second-class lever
R
F
Solution: Lever of first kind.
F
A
The sum of the forces is
FY D F C A R D 0.
L
L
Third-class lever
The sum of the moments about A is
MA D FL RL D 0,
L
R
D D1
F
L
from which
(b)
The reaction at A is obtained from the force balance equation:
A D R C F D 2F
Lever of second kind.
(a)
The sum of forces is
FY D A R C F D 0.
The sum of the moments about A is
MA D LR C 2LF D 0,
2L
R
D
D2
F
L
from which
(b)
The reaction at A is obtained from the force balance equation:
A D F C R D F C 2F D F
Lever of third kind.
(a)
The sum of forces is
FY D A R C F D 0.
The sum of moments about A is:
MA D 2LR C LF D 0,
from which:
(b)
L
1
R
D
D
F
2L
2
From the force balance equation
A D F C R D F C
jAj D
F
F
D ,
2
2
F
2
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1
Problem 5.18 A portion of one of the decks of Frank
Lloyd Wright’s Fallingwater is isolated by passing an
imaginary plane A through the deck. The mass of the
isolated part of the deck is 14,700 kg. Treating the plane
A as a fixed support, determine the reactions at A. (These
are internal reactions that the deck’s material, reinforced
concrete, must support at the plane A.)
y
2m
A
x
mg
Solution:
Fx : Ax D 0
Fy : Ay 144.2 kN D 0
MA : MA C 144.2 kN2 m D 0
Solving:
Ax D 0, Ay D 144.2 kN, MA D 288 kN-m
Ay
MA
Ax
144.2 kN
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1
Problem 5.19
beam.
(a) Draw the free-body diagram of the
(b) Determine the tension in the cable and the reactions
at A.
A
B
30°
30 in
30 in
Solution:
(a)
(b)
800 lb
30 in
T
The FBD
The equilibrium equations
C
T
30°
Ax
MA : 800 lb60 in C T30 in
C T sin 30° 90 in D 0
Fx :Ax T cos 30° D 0
Ay
800 lb
Fy :Ay C T C T sin 30° 800 lb D 0
Solving:
Ax D 554 lb, Ay D 160 lb, T D 640 lb
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1
Problem 5.20 The unstretched length of the spring CD
is 350 mm. Suppose that you want the lever ABC to
exert a 120-N normal force on the smooth surface at A.
Determine the necessary value of the spring constant k
and the resulting reactions at B.
C
k
230
mm
D
450
mm
20⬚
180
mm
B
A
Solution: We have
F D k
330
mm
0.23 m2 C 0.3 m2 0.35 m
300
mm
A D 120 N
MB : p
30
1429
F
F0.45 m C A cos 20° 0.18 m
23
30
C A sin 20° 0.33 m D 0
30
FD0
Fx :A cos 20° C Bx C p
1429
23
Fy : A sin 20° C By p
FD0
1429
Bx
A
By
Solving we find:
20°
k D 3380 N/m, Bx D 188 N, By D 98.7 N
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1
Problem 5.21 The mobile is in equilibrium. The fish
B weighs 27 oz. Determine the weights of the fish A, C,
and D. (The weights of the crossbars are negligible.)
12 in
3 in
A
6 in
2 in
B
7 in
2 in
C
D
Solution: Denote the reactions at the supports by FAB , FCD , and
FBCD as shown. Start with the crossbar supporting the weights C and
D. The sum of the forces is
FCD
D
C
7 in
2 in
FY D C D C FCD D 0,
FBCD
from which FCD D C C D.
6 in
For the cross bar supporting the weight B, the sum of the forces is
FY D B C FBCD FCD D 0,
from which, substituting, FBCD D B C C C D.
B
FCD
2 in
FAB
FBCD
A
12 in
3 in
For the crossbar supporting C and D, the sum of the moments about
the support is
MCD D 7D C 2C D 0,
from which D D
2C
.
7
For the crossbar supporting B, the sum of the moments is
MBCD D 6FCD 2B D 0,
from which, substituting from above
FCD D
2C
9C
2B
DCCDDCC
D
,
6
7
7
or C D 7B/27 D 7 oz,
and D D 2C/7 D 2 oz.
The sum of the moments about the crossbar supporting A is
MAB D 12A 3FBCD D 0,
from which, substituting from above,
AD
27 C 7 C 2
3B C C C D
D
D 9 oz
12
4
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1
y
Problem 5.22 The car’s wheelbase (the distance
between the wheels) is 2.82 m. The mass of the car is
1760 kg and its weight acts at the point x D 2.00 m,
y D 0.68 m. If the angle ˛ D 15° , what is the total
normal force exerted on the two rear tires by the sloped
ramp?
x
α
W
Solution: Split W into components:
y
x
α
W cos ˛ acts ? to the incline
W sin ˛ acts parallel to the incline
FX :
R
f W sin ˛ D 0
FY : NR C NF W cos ˛ D 0
0.68
f
MR :
2m
m
2.82
α
m
NF
α = 15°
W = (1760X9.81) N
NR
2W cos ˛ C 0.68W sin ˛ C 2.82NF D 0
Solving: NR D 5930 N, NF D 10750 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 5.23 The car in Problem 5.22 can remain in
equilibrium on the sloped ramp only if the total friction
force exerted on its tires does not exceed 0.8 times the
total normal force exerted on the two rear tires. What is
the largest angle ˛ for which it can remain in equilibrium?
Solution: The solution to Problem 5.22 yielded

 f D W sin ˛
N C NF W cos ˛ D 0
 R
2W cos ˛ C 0.68W sin ˛ C 2.82NF D 0
Our limit is f/NR 0.8, so let us set f D 0.8NR and solve the
resulting relations for ˛max

 0.8NR D W sin ˛max
N C NF W cos ˛max D 0
 R
2W cos ˛ C 6.68 W sin ˛max C 2.82NF D 0
Solving, we get
˛max D 16.1° ,
f D 4788 N,
NR D 5985 N,
NF D 10603 N.
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1
Problem 5.24 The 14.5-lb chain saw is subjected to
the loads at A by the log it cuts. Determine the reactions
R, Bx , and By that must be applied by the person using
the saw to hold it in equilibrium.
y
R
60°
By
1.5 in
7 in
x
A
Bx
5 lb
14.5 lb
10 lb
13 in
6 in
2 in
Solution: The sum of the forces are
FX D 5 C BX R cos 60° D 0.
FY D 10 14.5 C BY R sin 60° D 0.
The sum of the moments about the origin is
MO D 7R cos 60° C 8BY 214.5 1310 51.5 D 0.
From which 7R cos 60° C 8BY 166.5 D 0. Collecting equations and
reducing to 3 equations in 3 unknowns:
BX C 0BY 0.5R D 5
0BX C BY 0.866R D 4.5
0BX C 8BY C 3.5R D 166.5.
Solving:
BX D 11.257 lb,
BY D 15.337 lb,
and R D 12.514 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 5.25 The mass of the trailer is 2.2 Mg (megagrams). The distances a D 2.5 m and b D 5.5 m. The
truck is stationary, and the wheels of the trailer can turn
freely, which means that the road exerts no horizontal
force on them. The hitch at B can be modeled as a pin
support.
(a)
(b)
Draw the free-body diagram of the trailer.
Determine the total normal force exerted on the rear
tires at A and the reactions exerted on the trailer at
the pin support B.
B
W
A
a
b
Solution:
(a)
(b)
The free body diagram is shown.
The sum of forces:
FX D BX D 0.
FY D FA W C FB D 0.
The sum of the moments about A:
MA D aW C a C bFB D 0,
from which
FB D
2.52.2 ð 103 9.81
aW
D
D 6.744 kN
aCb
2.5 C 5.5
Substitute into the force equation:
FA D W FB D 14.838 kN
B
BX
FB
W
A
FA
a
b
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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 5.26 The total weight of the wheelbarrow
and its load is W D 100 lb.
(a)
If F D 0, what are the vertical reactions at A
and B?
What force F is necessary to lift the support at A
off the ground?
(b)
F
W
A
40 in
B
12 in
14 in
Solution: (a) The sum of the forces:
F
FX D AX D 0
FY D AY W C FB D 0.
AX
The sum of the moments about A is
MA D W12 C FB 26 D 0,
A
W
12′′ 14′′
40′′
AY
B
FB
from which
FB D
12W
D 46.1538 lb D 46.2 lb.
26
Substitute into the force equation to obtain:
AY D W FB D 53.8462 lb D 53.8 lb
(b) The sum of the moments about B when the point A is not making
contact with the ground:
MB D 14100 66F D 0,
from which
FD
14100
D 21.2121 D 21.2 lb
66
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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 5.27 The airplane’s weight is W D 2400 lb.
Its brakes keep the rear wheels locked. The front (nose)
wheel can turn freely, and so the ground exerts no horizontal force on it. The force T exerted by the airplane’s
propeller is horizontal.
(a)
(b)
(c)
Draw the free-body diagram of the airplane. Determine the reaction exerted on the nose wheel and
the total normal reaction on the rear wheels
when T D 0,
when T D 250 lb.
T
4 ft
W
A
5 ft
B
2 ft
Solution: (a) The free body diagram is shown. (b) The sum of the
forces:
FX D BX D 0
FY D AY W C BY D 0.
The sum of the moments about A is
MA D 5W C 7BY D 0,
from which BY D
5W
D 1714.3 lb
7
Substitute from the force balance equation:
AY D W BY D 685.7 lb
(c) The sum of the forces:
FX D 250 C BX D 0,
from which BX D 250 lb
FY D AY W C BY D 0.
The sum of the moments about A:
MA D 2504 5W C 7BY D 0,
from which BY D 1571.4 lb. Substitute into the force balance equation
to obtain: AY D 828.6 lb
4 ft
W
A
AY
5 ft
B
2 ft
BX
BY
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1
Problem 5.28 The forklift is stationary. The front
wheels are free to turn, and the rear wheels are locked.
The distances are a D 1.25 m, b D 0.50 m, and c D
1.40 m. The weight of the load is WL D 2 kN, and the
weight of the truck and operator is WF D 8 kN. What
are the reactions at A and B?
WL
WF
A
a
B
b
c
Solution: The sum of the forces:
FX D BX D 0
FY D AY WL WF C BY D 0.
The sum of the moments about A is
WL
MA D CaWL bWF C b C cBY D 0,
from which
BY D
WF
A
AY
a
BX
B
b
c
BY
bWF aWL
D 0.7895 kN.
bCc
Substitute into the force equation to obtain:
AY D WL C WF BY D 9.211 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 5.29 Paleontologists speculate that the stegosaur could stand on its hind limbs for short periods
to feed. Based on the free-body diagram shown and
assuming that m D 2000 kg, determine the magnitudes
of the forces B and C exerted by the ligament — muscle
brace and vertical column, and determine the angle ˛.
580
mm
160
mm
mg
C
B
22°
Solution: Take the origin to be at the point of application of the
force C. The position vectors of the points of application of the forces
B and W are:
α
415
mm
790
mm
rB D 415i C 160j (mm),
rW D 790i C 580j (mm).
The forces are
C D Ci cos90° ˛ C j sin90° ˛
D Ci sin ˛ C j cos ˛.
B D Bi cos270° 22° C j sin270° 22° D B0.3746i 0.9272j.
W D 29.81j D 19.62j (kN).
The moments about C,
i
MC D 415
0.3746B
i
C 790
0
j
160
0.9272B
k 0
0
k 580
0 D 0
19.62 0 j
D 444.72B 15499.8 D 0,
from which
BD
15499.8
D 34.85 kN.
444.72
The sums of the forces:
FX D C sin ˛ 0.3746Bi D 0,
from which C sin ˛ D 13.06 kN.
FY D C cos ˛ 0.9272B 19.62j D 0,
from which C cos ˛ D 51.93 kN.
The angle ˛ is
˛ D tan1
13.06
51.93
D 14.1° .
The magnitude of C,
CD
p
13.062 C 51.932 D 53.55 kN
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1
Problem 5.30 The weight of the fan is W D 20 lb. Its
base has four equally spaced legs of length b D 12 in,
and h D 36 in. What is the largest thrust T exerted by
the fan’s propeller for which the fan will remain in equilibrium?
T
b
W
h
T
Side View
Top View
Solution: Each leg is assumed to be in contact with a rough
surface, with (in two dimensions) two force components each.
T
The four equally spaced legs can be in two positions relative to the
thrust line of action: In the first the distance to the center is b. In the
second, the distance is b sin 45° D 0.707b. Tipping will occur when
the leftmost (or rightmost) leg(s) has zero reaction on the floor.
h
F
W
For each position the sum of the moments about the center is:
MT D bW C Th D 0, and
MT D 0.707bW C Th D 0.
From which the two tipping moment thrusts are:
T1 D
bW
D 6.67 lbs,
h
T2 D
0.707bW
D 4.71 lb
h
b
b
which is the maximum thrust allowed.
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1
Problem 5.31 Consider the fan described in Problem
5.30. As a safety criterion, an engineer decides that the
vertical reaction on any of the fan’s legs should not be
less than 20% of the fan’s weight. If the thrust T is 1 lb
when the fan is set on its highest speed, what is the
maximum safe value of h?
Solution: The total upward reaction of the legs is equal to the
weight of the fan, so that each leg normally bears one quarter of the
weight. Under the condition of maximum tipping moment, with the
legs in the position such that the distance to the center is 0.707b, the
legs in the outer position will each have the reaction of 20 percent of
the weight, so that both will carry 40 percent of the weight. Thus the
legs on the other side must bear 60 percent of the weight. The sum of
the moments at the maximum tipping condition allowed is
MT D 0.707b0.4W C Th 0.707b0.6W D 0,
from which:
hD
0.707b0.2W
D 33.94 in
T
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1
Problem 5.32 To decrease costs, an engineer considers
supporting a fan with three equally spaced legs instead
of the four-leg configuration shown in Problem 5.30. For
the same values of b, h, and W, show that the largest
thrust T for which the fan will remain in equilibrium
with three legs is related to the value with four legs by
b
T
p
Tthree legs D 1/ 2Tfour legs .
Solution: From the solution to Problem 5.30 the maximum thrust is
Tfour legs D
bW
bW sin 45°
D p .
h
2h
b
T
For three legs assume that the legs are in the position shown with
respect to the line of action of the thrust. The distance to the center
b
is b cos 60° D . When the outermost leg has zero reaction, the other
2
legs must bear the weight of the fan. The sum of the moments about
the center when the outer most leg has zero reaction is
MT D from which Tthree legs D
bW
C Th D 0,
2
1
bW
D p Tfour legs .
2h
2
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1
Problem 5.33 A force F D 400 N acts on the bracket.
What are the reactions at A and B?
F
A
80 mm
B
320 mm
Solution: The joint A is a pinned joint; B is a roller joint. The
pinned joint has two reaction forces AX , AY . The roller joint has one
reaction force BX . The sum of the forces is
FX D AX C BX D 0,
F
AY
AX
80
mm
BX
320
mm
FY D AY F D 0,
from which
AY D F D 400 N.
The sum of the moments about A is
MA D 0.08BX 0.320F D 0,
from which
BX D
0.320400
D 1600 N.
0.08
Substitute into the sum of forces equation to obtain:
AX D BX D 1600 N
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1
Problem 5.34 The sign’s weight WS D 32 lb acts at
the point shown. The 10-lb weight of the bar AD acts
at the midpoint of the bar. Determine the tension in the
cable AE and the reactions at D.
11 in
30 in
11 in
E
30°
20°
A
B
C
D
Ws
33 in
Solution:
TAE
MD : 32 lb33 in C 10 lb26 in
20°
TAE cos 20° 52 in sin 30° 30°
TAE sin 20° 52 in cos 30° D 0
Dy
Fx : TAE cos 20° C Dx D 0
Fy : TAE sin 20° 42 lb C Dy D 0
10 lb
Dx
Solving:
TAE D 33.0 lb, Dx D 31.0 lb, Dy D 30.7 lb
32 lb
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1
Problem 5.35 The device shown, called a swape or
shadoof, helps a person lift a heavy load. (Devices of this
kind were used in Egypt at least as early as 1550 B.C.
and are still in use in various parts of the world.) The
dimensions a D 3.6 m and b D 1.2 m. The mass of the
bar and counterweight is 90 kg, and their weight W acts
at the point shown. The mass of the load being lifted
is 45 kg. Determine the vertical force the person must
exert to support the stationary load (a) when the load
is just above the ground (the position shown); (b) when
the load is 1 m above the ground. Assume that the rope
remains vertical.
a
b
25°
W
Solution:
MO : 441 N F3.6 m cos 883 N1.2 m cos D 0
Solving we find
F D 147.2 N
Notice that the angle is not a part of this answer therefore
(a)
F D 147.2 N
(b)
F D 147.2 N
Oy
θ
Ox
F
883 N
441 N
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1
Problem 5.36 This structure, called a truss, has a pin
support at A and a roller support at B and is loaded by
two forces. Determine the reactions at the supports.
Strategy: Draw a free-body diagram, treating the entire
truss as a single object.
30° 2 kN
45°
4 kN
b
B
A
b
b
b
b
Solution:
p
MA : 4 kN 2b 2 kN cos 30° 3 b
C 2 kN sin 30° b C B4 b D 0
Fx : Ax C 4 kN sin 45° 2 kN sin 30° D 0
Fy : Ay 4 kN cos 45° 2 kN cos 30° C B D 0
Solving:
Ax D 1.828 kN, Ay D 2.10 kN, B D 2.46 kN
45°
4 kN
30°
2 kN
Ax
Ay
B
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1
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 a D b D 9 in and c D 13 in.
Treat his shoulder S as a built-in support, and determine
the magnitudes of the reactions at his shoulder. That
is, determine the force and couple his shoulder must
support.
S
8 lb
144 lb
a
b
c
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 D
FH
8 lb
144
C 8 D 80 lb.
2
The sum of the forces on the left arm is the weight of his left arm and
the vertical reaction at the shoulder and hand:
FH
8 lb
144 lb
FH
SX
M
FX D SX D 0.
8 lb
SY
b
FY D FH SY 8 D 0,
c
from which SY D FH 8 D 72 lb. The sum of the moments about the
shoulder is
MS D M C b C cFH b8 D 0,
where M is the couple reaction at the shoulder. Thus
M D b8 b C cFH D 1688 in lb D 1688 (in lb)
1 ft
12 in
D 140.67 ft lb
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1
Problem 5.38
Determine the reactions at A.
A
5 ft
300 lb
800 ft-lb
200 lb
200 lb
6 ft
3 ft
Solution: The built-in support at A is a two-force and couple reaction support. The sum of the forces for the system is
FX D AX C 200 D 0,
from which
AX D 200 lb
FY D AY C 300 200 D 0,
from which AY D 100 lb
The sum of the moments about A:
M D 6300 C 5200 800 C MA D 0,
from which MA D 1600 ft lb which is the couple at A.
AY
MA
300 lb
AX
5 ft
800 ft-lb
200 lb
200 lb
6 ft
3 ft
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1
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
˛ D 15° ; (b) on a down slope with ˛ D 15° .
n
70 i
n
36 i
n
20 i
y
α
3300 lb
Solution: The rear wheels are two force reaction support, and the
front wheels are a one force reaction support. Denote the rear wheels
by A and the front wheels by B, and define the reactions as being
parallel to and normal to the road. The sum of forces:
AX D 854.1 lb.
20 in.
α
FX D AX 3300 sin 15° D 0,
from which
x
BY
3300 lb
AX
AY
36
in.
70
in.
FY D AY 3300 cos 15° C BY D 0.
Since the mass center of the vehicle is displaced above the point A,
a component of the weight (20W sin ˛) produces a positive moment
about A, whereas the other component (36W cos ˛) produces a negative
moment about A. The sum of the moments about A:
MA D 363300 cos 15° C 203300 sin 15° C BY 106 D 0,
from which
BY D
C97669
D 921.4 lb.
106
Substitute into the sum of forces equation to obtain AY D 2266.1 lb
(b) For the car parked down-slope the sum of the forces is
FX D AX C 3300 sin 15° D 0,
from which AX D 854 lb
FY D AY 3300 cos 15° C BY D 0.
The component (20W sin ˛) now produces a negative moment about
A. The sum of the moments about A is
MA D 330036 cos 15° 330020 sin 15° C 106BY D 0,
from which
BY D
131834
D 1243.7 lb.
106
Substitute into the sum of forces equation to obtain AY D 1943.8 lb
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1
Problem 5.40 The weight W of the bar acts at its
center. The surfaces are smooth. What is the tension
in the horizontal string?
L
L
–
2
–L
2
Solution: The surfaces are roller supports, with only one reaction
force, which is normal to the contact surface. Denote the reaction
at the top of the bar by B, the tension in the string by T, and the
reaction at the base of the bar by A. The angle formed by the bar
at the base is ˛ D 45° , since the altitude and base of the triangle are
equal. The reaction at the top of the bar forms the angle 90 ˛ with
the horizontal, and the reaction at the base is vertical. The sum of the
forces is
FX D T C B cos90 ˛ D T C B sin ˛ D 0.
α
90 − α
B
W
T
α
AY
FY D W C AY C B cos ˛ D 0.
The sum of the moments about the lower end is
MA D
WL
2
cos ˛ B
L
p
2
D 0,
from which
BD
W cos ˛
p
.
2
Substitute into the horizontal force equation to obtain the string tension
W
T D p sin ˛ cos ˛
2
W
D p D 0.3536W
2 2
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1
Problem 5.41 The mass of the bar is 36 kg and its
weight acts at its midpoint. The spring is unstretched
when ˛ D 0. The bar is in equilibrium when ˛ D 30° .
Determine the spring constant k.
k
4m
α
2m
Solution:
y
l2 D 42 C 22 242 cos 30°
kδ
Solving, l D 2.48 m
φ
x
lo
4m
The force acting at the top end of the bar is F D kυ where υ D l l0 .
B
4m
α
We also need when ˛ D 30°
sin ˛
sin 30°
sin D
D
z
l
2.48
30°
mg
2m
AY
2m
AX
D 23,78° when ˛ D 30°
Equilibrium equations:
C !
C"
C
FX D 0:
kυ sin C AX D 0
FY D 0:
kυ cos C AY mg D 0
MB D 0:
AY 2 sin ˛ mg1 sin ˛
C AX 2 cos ˛ D 0
Substituting υ, , and ˛ into the equations and solving, we get
AX D 44.1 N
AY D 253.0 N
k D 229 N/m
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1
Problem 5.42 The plate is supported by a pin in a
smooth slot at B. What are the reactions at the supports?
2 kN-m
6 kN-m
A
B
60°
2m
Solution: The pinned support is a two force reaction support. The
smooth pin is a roller support, with a one force reaction. The reaction
at B forms an angle of 90° C 60° D 150° with the positive x axis. The
sum of the forces:
6 kN-m
2 kN-m
FX D AX C B cos 150° D 0
FY D AY C B sin 150° D 0
A
60°
B
The sum of the moments about B is
2m
MB D 2AY C 2 6 D 0,
from which
AY D 4
D 2 kN.
2
2 kN-m
6 kN-m
Substitute into the force equations to obtain
BD
AY
D 4 kN,
sin 150°
and AX D B cos 150° D 3.464 kN.
AX
150°
AY
B
2m
The horizontal and vertical reactions at B are
BX D 4 cos 150° D 3.464 kN,
and BY D 4 sin 150° D 2 kN.
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1
Problem 5.43
Determine the reactions at the supports.
y
40 lb
30⬚
60 lb
1400 in-lb
A
20 in
30 in
15 in
B
x
20 in
Solution:
30°
Fx : Ax C 60 lb sin 30° D 0
Fy : Ay C 40 lb 60 lb cos 30° C B D 0
40 lb
60 lb
1400 in-lb
Ax
MA : 40 lb20 in 60 lb cos 30° 50 in
Ay
B
C 1400 in lb C B85 in D 0
Solving
Ax D 30 lb, Ay D 7.28 lb, B D 4.68 lb
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1
Problem 5.44
(a)
(b)
Consider the beam in Problem 5.43.
If you represent the system of forces and moments
consisting of the 40-lb and 60-lb forces and the
2200 in-lb couple by an equivalent system consisting of a single force F as shown, what is F, and
where does its line of action cross the x axis?
Assume that the beam is subjected only to the force
you determined in part (a) and determine the reactions at the supports. Compare your answers to the
answers to Problem 5.44.
Solution: If these two systems of forces and moments are statically
equivalent then we must have
Fx D 60 lb sin 30°
Fy D 40 lb 60 lb cos 30°
Fy x D 40 lb20 in 60 lb cos 30° 50 in C 1400 in lb
Thus
(a)
(b)
F D 30i 11.96j lb, x D 33.3 in
MA : 11.96 lb33.3 in C B85 in D 0
Fx : Ax C 30 lb D 0
Fy : Ay 11.96 lb C B D 0
B D 4.68 lb
)
Ax D 30 lb
Ay D 7.28 lb
30°
40 lb
60 lb
1400 in-lb
Ax
B
Ay
F
Fy
x
Ax
Fx
85 in
Ay
B
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1
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 B in terms of the
cable tension T.
T
35°
40 mm
B
Brake pad
Wheel rim
45 mm
A
40 mm
Solution: From the force balance equation for the cables: the force
on the brake mechanism TB in terms of the cable tension T is
TB
35°
T 2TB sin 35° D 0,
from which TB D
T
D 0.8717T.
2 sin 35°
40
mm
B
Take the origin of the system to be at A. The position vector of the
point of attachment of B is rB D 45j (mm). The position vector of the
point of attachment of the cable is rC D 40i C 85j (mm).
The force exerted by the brake pad is B D Bi. The force vector due
the cable tension is
AY
45
mm
AX
40
mm
TB D TB i cos 145° C j sin 145° D TB 0.8192i C 0.5736j.
The moment about A is
MA D rB ð B C rC ð TB D 0
i
j
MA D 0 45
B 0
i
k 45 C 40
0 0.8192
j
85
0.5736
k 85 TB D 0
0 MA D 45B C 92.576TB k D 0,
from which B D
92.576TB
D 2.057TB .
45
Substitute the expression for the cable tension:
B D 2.0570.8717T D 1.793T
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1
Problem 5.46 The mass of each of the suspended
boxes is 80 kg. Determine the reactions at the supports
at A and E.
A
B
C
300 mm
D
E
200 mm
Solution: From the free body diagram, the equations of
equilibrium for the rigid body are
and
Fx D AX C EX D 0,
200 mm
y
AY
0.2 m
0.2 m
AX A
x
mg
0.3 m
Fy D AY 2809.81 D 0,
E
mg
EX
MA D 0.3EX 0.2809.81 0.4809.81 D 0.
We have three equations in the three components of the support
reactions. Solving for the unknowns, we get the values
AX D 1570 N,
AY D 1570 N,
and EX D 1570 N.
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1
Problem 5.47 The suspended boxes in Problem 5.46
are each of mass m. The supports at A and E will 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
and
Fx D AX C EX D 0,
Fy D AY 2 m9.81 D 0,
MA D 0.3EX 0.2 m9.81 0.4 m9.81 D 0.
We also need the relation
jAj D
A2X C A2Y D 6000 N.
We have four equations in the three components of the support reactions plus the magnitude of A. This is four equations in four unknowns.
Solving for the unknowns, we get the values
AX D 4243 N,
AY D 4243 N,
EX D 4243 N,
and m D 216.5 kg.
Note: We could have gotten this result by a linear scaling of all of the
numbers in Problem 5.46.
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1
Problem 5.48 The tension in cable BC is 100 lb.
Determine the reactions at the built-in support.
C
6 ft
A
B
300 ft-lb
200 lb
3 ft
3 ft
Solution: The cable does not exert an external force on the system,
and can be ignored in determining reactions. The built-in support is a
two-force and couple reaction support. The sum of forces:
FX D AX D 0.
FY D AY 200 D 0,
MA
6 ft
AY
300 ft-lb
AX
200 lb
3 ft
from which AY D 200 lb.
The sum of the moments about A is
M D MA 3200 300 D 0,
from which MA D 900 ft lb
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1
Problem 5.49 The tension in cable AB is 2 kN. What
are the reactions at C in the two cases?
A
60°
B
2m
1m
C
A
(a)
Solution: First Case: The sum of the forces:
FX D CX T cos 60° D 0,
from which CX D 20.5 D 1 kN
T
T
2m
1m
C
(b)
CY
2m
1m
Case (a)
MC
CX
CY
from which CY D 1.8662 D 3.732 kN.
B
60°
FY D CY C T sin 60° C T D 0,
The sum of the moments about C is
60°
Case (b)
MC
CX
M D MC T sin 60° 3T D 0,
from which MC D 3.8662 D 7.732 kN
Second Case: The weight of the beam is ignored, hence there are no
external forces on the beam, and the reactions at C are zero.
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1
Problem 5.50
Determine the reactions at the supports.
6 in
5 in
50 lb
A
3 in
100 in-lb
Solution: The reaction at A is a two-force reaction. The reaction
at B is one-force, normal to the surface.
3 in
B
The sum of the forces:
30°
FX D AX B cos 60° 50 D 0.
FY D AY C B sin 60° D 0.
AX
50 lb
AY
The sum of the moments about A is
6 in.
100
MA D 100 C 11B sin 60° 6B cos 60° D 0,
from which
11 in.
B
60°
100
D 15.3 lb.
BD
11 sin 60° 6 cos 60° Substitute into the force equations to obtain
AY D B sin 60° D 13.3 lb
and AX D B cos 60° C 50 D 57.7 lb
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1
Problem 5.51 The weight W D 2 kN. Determine the
tension in the cable and the reactions at A.
30°
A
W
0.6 m
AY
Solution: Equilibrium Eqns:
C
FX D 0:
0.6 m
T
AX C T cos 30° D 0
30°
AX
FY D 0:
AY C T C T sin 30° W D 0
MA D 0:
0, 6W C 0.6T sin 30° T
0.6 m
0.6 m
W = 2 kN = 2000 N
C 1, 2T D 0
Solving, we get
AX D 693 N,
AY D 800 N,
T D 800 N
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1
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?
Solution: The equilibrium equations in the solution of problem are
C
FX D 0:
AX C T cos 30° D 0
FY D 0:
AY C T C T sin 30° W D 0
MA D 0:
0, 6W C 0, 6T sin 30° C 1, 2T D 0
We previously had 3 equations in the 3 unknowns AX , AY and T (we
knew W). In the current problem, we know T but don’t know W.
We again have three equations in three unknowns (AX , AY , and W).
Setting T D 6 kN, we solve to get
AX D 5.2 kN
AY D 6.0 kN
W D 15.0 kN
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1
Problem 5.53 The blocks being compressed by the clamp
exert a 200-N force on the pin at D that points from A
toward D. The threaded shaft BE exerts a force on the
pin at E that points from B toward E.
50 mm
(a)
(b)
Draw a free-body diagram of the arm DCE of the
clamp, assuming that the pin at C behaves like a
pin support.
Determine the reactions at C.
E
A
C
50 mm
D
FBE
The free-body diagram
The equilibrium equations
125 mm
B
50 mm
Solution:
(a)
(b)
125 mm
125 mm
Cy
MC : 200 N0.25 m FBE 0.1 m D 0
200 N
Fx : Cx C FBE D 0
Cx
Fy : Cy 200 N D 0
Solving
Cx D 500 N, Cy D 200 N
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1
Problem 5.54 Consider the clamp in Problem 5.53.
The blocks being compressed by the clamp exert a 200N force on the pin at A that points from D toward A.
The threaded shaft BE exerts a force on the pin at B that
points from E toward B.
(a)
(b)
Draw a free-body diagram of the arm ABC of the
clamp, assuming that the pin at C behaves like a
pin support.
Determine the reactions at C.
FBE
Solution:
(a)
(b)
The free-body diagram
The equilibrium equations
Cx
MC : 200 N0.25 m C FBE 0.1 m D 0
Fx : FBE C Cx D 0
200 N
Cy
Fy : 200 N C Cy D 0
Solving we find
Cx D 500 N, Cy D 200 N
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1
Problem 5.55 Suppose that you want to design the
safety valve to open when the difference between the
pressure p in the circular pipe diameter D 150 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
k
A
p
150 mm
Solution: The area of the valve is
aD
0.15
2
150
mm
2
250
mm
D 17.671 ð 103 m2 .
k
The force at opening is
A
F D 10a ð 106 D 1.7671 ð 105 N.
150 mm
The force on the spring is found from the sum of the moments about
A,
MA D 0.15F 0.4kL D 0.
k∆L
A
Solving,
F
0.15F
0.151.7671 ð 105 kD
D
0.4L
0.40.02
D 3.313 ð 106
0.15
m
0.25
m
N
m
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1
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.
C
B
3 ft
A
30⬚
1 ft
Solution: Geometry:
tan D
3 ft 3 ft sin 30°
D 0.4169 ) D 22.63°
1 ft C 3 ft cos 30°
The equilibrium equations
MA : TBC cos 3 ft sin 30° C TBC sin 3 ft cos 30° 10 lb1.5 ft cos 30° D 0
Fx : TBC cos C Ax D 0
Fy : TBC sin 10 lb C Ay D 0
Solving:
Ax D 5.03 lb, Ay D 7.90 lb, T D 5.45 lb
TBC
θ
10 lb
Ax
Ay
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1
Problem 5.57 The crane’s arm has a pin support at A.
The hydraulic cylinder BC exerts a force on the arm at
C in 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?
C
A
2.4 m
1m
B
1.8 m
1.2 m
7m
Solution: The geometry gives
y
tan D 2.4/1.2,
FHX D jFH j cos ,
and FHY D jFH j sin .
FH
2m
mg
θ
C
or D 63.4° .
From the diagram,
FHY
mBg
AX
A
FHX
2.4 m
1 m AY B
1.8 1.7
m
m
7m
x
The force equilibrium equations are
Fx D AX C FHX D 0,
Fy D AY C FHY 200g 800g D 0,
and the moment equation is
MA D 2200g 7800g C 3FHY 2.4FHX D 0.
Solving the five equations simultaneously, we get jFH j D 36.56 kN,
which is the result called for in this problem. Other values obtained in
the solution are
AX D 16.35 kN,
and AY D 22.89 kN.
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1
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?
Solution: The values for the components of A were determined in
the solution to Problem 5.57. The magnitude of the force is
jAj D
A2X C A2Y D 28.13 kN.
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1
Problem 5.59 A speaker system is suspended by the
cables attached at D and 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 A and C.
0.5 m 0.5 m 0.5 m
0.5 m
1m
E
C
A
1m
B
D
G
Solution: The weight of the speaker is W D mg D 1275 N. The
equations of equilibrium for the entire assembly are
1m
AY
1.5 m
CY
E
Fx D CX D 0,
Fy D AY C CY mg D 0
A
C CX
B
D
(where the mass m D 130 kg), and
mg
MC D 1AY 1.5mg D 0.
Solving these equations, we get
1.5 m
CX D 0,
1m
T2
CY D 3188 N,
T1
and AY D 1913 N.
From the free body diagram of the speaker alone, we get
and
mg
Fy D T1 C T2 mg D 0,
Mleft support D 1mg C 1.5T2 D 0.
Solving these equations, we get
T1 D 425. N
and T2 D 850 N
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1
Problem 5.60 The weight W1 D 1000 lb. Neglect the
weight of the bar AB. The cable goes over a pulley at
C. Determine the weight W2 and the reactions at the pin
support A.
B
50°
35°
W1
A
C
W2
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 A to
zero. The angle between the cable and the positive x axis is 35° .
The tension vector in the cable is
T2 D W2 i cos35° C j sin35° .
35°
T2
rB
T1
50°
AY
D W2 0.8192i 0.5736jlb.
AX
Assume a unit length for the bar. The angle between the bar and the
positive x axis is 180° 50° D 130° . The position vector of the tip of
the bar relative to A is
rB D i cos130° C j sin130° , D 0.6428i C 0.7660j.
The tension exerted by W1 is T1 D 1000j. The sum of the moments
about A is:
MA D rB ð T1 C rB ð T2 D rB ð T1 C T2 i
D L 0.6428
0.8191W2
j
0.7660
0.5736W2 1000 MA D 0.2587W2 C 642.8k D 0,
from which W2 D 2483.5 lb
The sum of the forces:
FX D AX C W2 0.8192i D 0,
from which AX D 2034.4 lb
FY D AY W2 0.5736 1000j D 0,
from which AY D 2424.5 lb
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1
Problem 5.61 The dimensions a D 2 m and b D 1 m.
The couple M D 2400 N-m. The spring constant is k D
6000 N/m, and the spring would be unstretched if h D 0.
The system is in equilibrium when h D 2 m and the
beam is horizontal. Determine the force F and the reactions at A.
k
h
A
M
F
a
Solution: We need to know the unstretched length of the spring, l0
b
Unstretched
l0 D a C b D 3 m
(a + b)
We also need the stretched length
l2 D h2 C a C b2
AY
θ
M
l D 3.61 m
AX
a
FS D kl l0 tan D
F
b
h
a C b
D 33.69°
Equilibrium eqns:
C
FX :
AX FS cos D 0
FY :
AY C FS sin F D 0
MA :
M aF C a C bFS sin D 0
a D 2 m,
b D 1 m,
M D 2400 N-m,
h D 2 m,
k D 6000 N/m.
Substituting in and solving, we get
FS D 6000l l0 D 3633 N
and the equilibrium equations yield
AX D 3023 N
AY D 192 N
F D 1823 N
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1
Problem 5.62 The bar is 1 m long, and its weight W
acts at its midpoint. The distance b D 0.75 m, and the
angle ˛ D 30° . The spring constant is k D 100 N/m, and
the spring is unstretched when the bar is vertical. Determine W and the reactions at A.
k
α
W
A
b
p
Solution: The unstretched length of the spring is L D b2 C 12 D
1.25 m. The obtuse angle is 90 C ˛, so the stretched length can be
determined from the cosine law:
β
L22 D 12 C 0.752 20.75 cos90 C ˛ D 2.3125 m2
α
W
from which L2 D 1.5207 m. The force exerted by the spring is
A
T D kL D 1001.5207 1.25 D 27.1 N.
b
The angle between the spring and the bar can be determined from the
sine law:
1.5207
b
D
,
sin ˇ
sin90 C ˛
T
β
from which sin ˇ D 0.4271,
α
ˇ D 25.28° .
W
The angle the spring makes with the horizontal is 180 25.28 90 ˛ D 34.72° . The sum of the forces:
AX
FX D AX T cos 34.72° D 0,
AY
from which AX D 22.25 N.
FY D AY W T sin 34.72° D 0.
The sum of the moments about A is
MA D T sin 25.28° W
2
sin ˛ D 0,
from which
WD
2T sin 25.28°
D 46.25 N.
sin ˛
Substitute into the force equation to obtain: AY D W C T sin 34.72° D
61.66 N
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1
B
Problem 5.63 The boom derrick supports a suspended
15-kip load. The booms BC and DE are each 20 ft long.
The distances are a D 15 ft and b D 2 ft, and the angle
D 30° . Determine the tension in cable AB and the reactions at the pin supports C and D.
E
θ
Solution: Choose a coordinate system with origin at point C, with
the y axis parallel to CB. The position vectors of the labeled points
are:
C
A
D
rD D 2i
a
b
rE D rD C 20i sin 30° C j cos 30° D 12i C 17.3j,
The components:
rB D 20j,
Dx D 0.6jDj D 7.67 kip,
rA D 15i.
The unit vectors are:
eDE D
rE rD
D 0.5i C 0.866j,
jrE rD j
eEB D
rB rE
D 0.976i C 0.2179j.
jrB rE j
eCB D
rB rC
D 1j,
jrB rC j
eAB D
rA rB
D 0.6i 0.8j.
jrA rB j
Dy D 0.866jDj D 13.287 kip,
and Cy D 1jCj D 11.94 kip
B
E
θ
A
C
Fx D 0.5jDj 0.976jTEB j D 0,
TAB
Fy D 0.866jDj C 0.2179jTEB j 15 D 0,
from which
b
a
Isolate the juncture at E: The equilibrium conditions are
D
TEB
TEB
15 kip
C
D
Juncture B Juncture E
jDj D 15.34 kip
and jTEB j D 7.86 kip.
Isolate the juncture at B: The equilibrium conditions are:
and
Fx D 0jCj 0.6jTAB j C 0.976jTEB j,
Fy D 1jCj 0.6jTAB j 0.2179jTEB j D 0,
from which
jTAB j D 12.79 kip,
and jCj D 11.94 kip.
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1
Problem 5.64 The arrangement shown controls the
elevators of an airplane. (The elevators are the horizontal
control surfaces in the airplane’s tail.) The elevators are
attached to member EDG. Aerodynamic pressures on the
elevators exert a clockwise couple of 120 in-lb. Cable
BG is slack, and its tension can be neglected. Determine
the force F and the reactions at pin support A.
E
B
6 in
A
D
2.5 in
C
F
3.5 in
Elevator
120 in-lb
G
2 in
2.5 in
2.5 in
1.5 in
120 in
(Not to scale)
Solution: Begin at the elevator. The moment arms at E and G are
6 in. The angle of the cable EC with the horizontal is
˛ D tan1
12
D 5.734° .
119.5
Denote the horizontal and vertical components of the force on point
E by FX and FY . The sum of the moments about the pinned support
on the member EG is
2 in
FX
α
E
C
A
TEC
FY D
2.5 in
F
3.5 in
TEC
α
C
6 in
120 in-lb
2.5
in
MEG D 2.5FY C 6FX 120 D 0.
This is the tension in the cable EC. Noting that
FX D TEC cos ˛,
and FY D TEC sin ˛,
then TEC D
120
.
2.5 sin ˛ C 6 cos ˛
The sum of the moments about the pinned support BC is
MBC D 2TEC sin ˛ C 6TEC cos ˛ 2.5F D 0.
Substituting:
FD
120
2.5
6 cos ˛ 2 sin ˛
6 cos ˛ C 2.5 sin ˛
D 480.9277 D 44.53 lb.
The sum of the forces about the pinned joint A:
Fx D Ax F C TEC cos ˛ D 0
from which Ax D 25.33 lb,
Fy D Ay C TEC sin ˛ D 0
from which Ay D 1.93 lb
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1
Problem 5.65 In Fig. 5.17, suppose that ˛ D 40° , d D
1 m, a D 200 mm, b D 500 mm, R D 75 mm, and the
mass of the luggage is 40 kg. Determine F and N.
Solution: (See Example 5.5.)
The sum of the moments about the center of the wheel:
MC D dF cos ˛ C aW sin ˛ bW cos ˛ D 0,
from which F D
b a tan ˛W
D 130.35 N.
d
The sum of the forces:
FY D N W C F D 0,
F
from which N D 262.1 N
d
A
d
b
N
h
a
a W
α
R
C
W
b
F
h
R
a
C
N
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1
Problem 5.66 In Fig. 5.17, suppose that ˛ D 35° , d D
46 in, a D 10 in, b D 14 in, R D 3 in, and you don’t
want the user to have to exert a force F larger than
20 lb. What is the largest luggage weight that can be
placed on the carrier?
Solution: (See Example 5.5.) From the solution to Problem 5.65,
the force is
FD
b a tan ˛W
.
d
Solve for W:
WD
Fd
.
b a tan ˛
For F D 20 lb,
W D 131.47 D 131.5 lb
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1
Problem 5.67 One of the difficulties in making design
decisions is that you don’t know how the user will place
the luggage on the carrier in Example 5.5. Suppose you
assume that the point where the weight acts may be
anywhere within the “envelope” R a 0.75c and 0 b 0.75d. If ˛ D 30° , c D 14 in, d D 48 in, R D 3 in,
and W D 80 lb, what is the largest force F the user will
have to exert for any luggage placement?
Solution: (See Example 5.5.) From the solution to Problem 5.65,
the force is
FD
b a tan ˛W
.
d
The force is maximized as
b ! 0.75d,
and a ! R.
Thus
FMAX D
0.75d R tan ˛W
D 57.11 lb
d
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1
Problem 5.68 In our design of the luggage carrier
in Example 5.5, we assumed a user that would hold
the carrier’s handle at h D 36 in above the floor. We
assumed that R D 3 in, a D 6 in, and b D 12 in, and we
chose the dimension d D 4 ft. The resulting ratio of the
force the user must exert to the weight of the luggage
is F/W D 0.132. Suppose that people with a range of
heights use this carrier. Obtain a graph of F/W as a
function of h for 24 h 36 in.
Solution: (See Example 5.5.) From the solution to Problem 5.67,
the force that must be exerted is
FD
from which
b a tan ˛W
,
d
b a tan ˛
F
D
.
W
d
The angle a is given by
˛ D sin1
hR
d
.
F .2
/
W .19
,
d .18
i
m .17
e
n .16
s
i
o .15
n
l .14
e
x .13
e
24
F/W versus height
26
28
30
32
height h, in
34
36
The commercial package TK Solver Plus was used to plot a graph of
F
as a function of h.
W
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1
Problem 5.69 (a) Draw the free-body diagram of the
beam and show that it is statically indeterminate.
(b) Determine as many of the reactions as possible.
20 N-m
A
800 mm
Solution: (a) The free body diagram shows that there are four
unknowns, whereas only three equilibrium equations can be written.
(b) The sum of moments about A is
MA D M C 1.1BY D 0,
from which BY D 20
D 18.18 N.
1.1
The sum of forces in the vertical direction is
300 mm
20 N-m
A
800 mm
B
B
300 mm
BX
AX
AY
800 mm
300 mm
BY
FY D AY C BY D 0,
from which AY D BY D 18.18 N.
The sum of forces in the horizontal direction is
FX D AX C BX D 0,
from which the values of AX and BX are indeterminate.
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1
Problem 5.70 Consider the beam in Problem 5.69.
Choose supports at A and B so that it is not statically
indeterminate. Determine the reactions at the supports.
Solution: One possibility is shown: the pinned support at B is
replaced by a roller support. The equilibrium conditions are:
20 N-m
A
B
FX D AX D 0.
800 mm
The sum of moments about A is
MA D M C 1.1BY D 0,
from which BY D 300 mm
20 N-m
AX
AY
800 mm
300 mm
BY
20
D 18.18 N.
1.1
The sum of forces in the vertical direction is
FY D AY C BY D 0,
from which AY D BY D 18.18 N.
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1
Problem 5.71 (a) Draw the free-body diagram of the
beam and show that it is statically indeterminate. (The
external couple M0 is known.)
M0
A
B
(b) By an analysis of the beam’s deflection, it is determined that the vertical reaction B exerted by the roller
support is related to the couple M0 by B D 2M0 /L. What
are the reactions at A?
Solution:
(a)
C
L
Eqn (3) and Eqn (4) yield
FX :
AX D 0
(1)
MA D MO 2MO
FY :
AY C B D 0
(2)
MA D MO
MA :
MA MO C BL D 0 (3)
Unknowns: MA , AX , AY , B.
3 Eqns in 4 unknowns
MA was assumed counterclockwise
MA D jMO j clockwise
AX D 0
AY D 2MO /L
∴ Statistically indeterminate
(b)
Given B D 2MO /L
(4)
We now have 4 eqns in 4 unknowns and can solve.
Eqn (1) yields AX D 0
AY
MO
MA
AX
L
B
Eqn (2) and Eqn (4) yield
AY D 2MO /L
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1
Problem 5.72 Consider the beam in Problem 5.71.
Choose supports at A and B so that it is not statically
indeterminate. Determine the reactions at the supports.
Solution: This result is not unique. There are several possible
answers
FX :
A
L
AX D 0
FY :
AY C BY D 0
MA :
Mo C BL D 0
O
MO
AX
AX D 0
B D MO /L
MO
L
AY
B
AY D MO /L
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1
Problem 5.73 Draw the free-body diagram of the Lshaped pipe assembly and show that it is statically indeterminate. Determine as many of the reactions as possible.
B
80 N
Strategy: Place the coordinate system so that the x
axis passes through points A and B.
A
300 mm
Solution: The free body diagram shows that there are four reactions, hence the system is statically indeterminate. The sum of the
forces:
and
FX D AX C BX D 0,
FY D AY C BY C F D 0.
A strategy for solving some statically indeterminate problems is to
select a coordinate system such that the indeterminate reactions vanish
from the sum of the moment equations. The choice here is to locate
the x axis on a line passing through both A and B, with the origin at
A. Denote the reactions at A and B by AN , AP , BN , and BP , where the
subscripts indicate the reactions are normal to and parallel to the new
x axis. Denote
F D 80 N,
300
mm
100 N-m
700 mm
The moment about the point A is
MA D LBN 0.3F C M D 0,
from which BN D
M C 0.3F
76
D
D 99.79 N,
L
0.76157
from which
The sum of the forces normal to the new axis is
FN D AN C BN C F cos D 0,
from which
AN D BN F cos D 26.26 lb
The reactions parallel to the new axis are indeterminate.
M D 100 N-m.
The length from A to B is
LD
p
0.32 C 0.72 D 0.76157 m.
The angle between the new axis and the horizontal is
D tan1
0.3
0.7
BN
80 N
AN
D 23.2° .
300
mm
BP
300
mm
AP
100 N-m
700
mm
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1
Problem 5.74 Consider the pipe assembly in Problem
5.73. Choose supports at A and B so that it is not
statically indeterminate. Determine the reactions at the
supports.
Solution: This problem has no unique solution.
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1
Problem 5.75 State whether each of the L-shaped bars
shown is properly or improperly supported. If a bar
is properly supported, determine the reactions at its
supports.
F
C
–12 L
F
L
–1 L
2
A
B
A
L
L
(1)
(2)
–12 L
F
Solution:
is properly constrained. The sum of the forces
FX D F C BX D 0,
45°
C
45°
(1)
B
–12 L
A
B
45°
L
(3)
from which BX D F.
FY D BY C Ay D 0,
from which By D Ay . The sum of the moments about B:
MB D LAY C LF D 0,
from which AY D F, and By D F
(2)
is improperly constrained. The reactions intersect at B, while the
force produces a moment about B.
(3)
is properly constrained. The forces are neither concurrent nor
parallel. The sum of the forces:
FX D C cos 45° B A cos 45° C F D 0.
FY D C sin 45° A sin 45° D 0
from which A D C. The sum of the moments about A:
MA D 12 LF C LC cos 45° C LC sin 45° D 0,
F
F
from which C D p . Substituting and combining: A D p ,
2 2
2 2
F
BD
2
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1
Problem 5.76 State whether each of the L-shaped bars
shown is properly or improperly supported. If a bar
is properly supported, determine the reactions at its
supports.
C
C
–12 L
F
F
–12 L
–12 L
A
B
A
B
L
–12 L
L
45°
(2)
(1)
C
–12 L
F
–12 L
A
B
L
(3)
Solution:
(1)
(2)
(3)
is improperly constrained. The reactions intersect at a point P,
and the force exerts a moment about that point.
is improperly constrained. The reactions intersect at a point P
and the force exerts a moment about that point.
is properly constrained. The sum of the forces:
FX D C F D 0,
from which C D F.
FY D A C B D 0,
from which A D B. The sum of the moments about B: LA C
1
1
L
F LC D 0, from which A D F, and B D F
2
2
2
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1
Problem 5.77 The bar AB has a built-in support at A
and is loaded by the forces
y
A
FB D 2i C 6j C 3k (kN),
FB
z
FC D i 2j C 2k (kN).
(a)
(b)
B
1m
Draw the free-body diagram of the bar.
Determine the reactions at A.
C
1m
Strategy: (a) Draw a diagram of the bar isolated from
its supports. Complete the free-body diagram of the bar
by adding the two external forces and the reactions
due to the built-in support (see Table 5.2). (b) Use the
scalar equilibrium equations (5.16)–(5.21) to determine
the reactions.
FC
Solution:
x
AY
AX
MA = MAX i + MAY j + MAZ K
MA D MAX i C MAY j C MAZ k
FB
(b)
Equilibrium Eqns (Forces)
1m
FX : AX C FBX C FCX D 0
FY :
AY C FBY C FCY D 0
FZ :
AZ C FBZ C FCZ D 0
AZ
B
C
1m
x
FC
Equilibrium Equations (Moments) Sum moments about A
rAB ð FB D 1i ð 2i C 6j C 3k kN-m
rAB ð FB D 3j C 6k (kN-m)
rAC ð FC D 2i ð 1i 2j C 2k kN-m
rAC ð FC D 4j 4k (kN-m)
x:
y:
z:
MA :
MAX D 0
MA :
MAY 3 4 D 0
MA :
MAZ C 6 4 D 0
Solving, we get
AX D 3 kN,
AY D 4 kN,
AZ D 5 kN
MAx D 0,
MAy D 7 kN-m,
MAz D 2 kN-m
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1
y
Problem 5.78 The bar AB has a built-in support at A.
The tension in cable BC is 8 kN. Determine the reactions
at A.
A
z
C
(3,0.5,–0.5)m
2m
B
x
Solution:
AY
MA = MAX i + MAY j + MAZ K
AX
MA D MAx i C MAy j C MAz k
AZ
We need the unit vector eBC
2m
xC xB i C yC yB j C zC zB k
eBC D xC xB 2 C yC yB 2 C zC zB 2
TBC
C (3, 0.5, −0.5)
B (2, 0, 0)
x
eBC D 0.816i C 0.408j 0.408k
TBC D 8 kNeBC
TBC D 6.53i C 3.27j 3.27k (kN)
The moment of TBC about A is
i
MBC D rAB ð TBC D 2
6.53
j
k 0
0 3.27 3.27 MBC D rAB ð TBC D 0i C 6.53j C 6.53k (kN-m)
Equilibrium Eqns.
FX : AX C TBCX D 0
FY : AY C TBCY D 0
FZ : AZ C TBCZ D 0
MX : MAX C MBCX D 0
MY : MAY C MBCY D 0
MZ : MAZ C MBCZ D 0
Solving, we get
AX D 6.53 (kN),
AY D 3.27 (kN),
AZ D 3.27 (kN)
MAx D 0,
MAy D 6.53 (kN-m),
MAz D 6.53 (kN-m)
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1
Problem 5.79 The bar AB has a built-in support at A.
The collar at B is fixed to the bar. The tension in the
cable BC is 10 kN.
(a)
(b)
Draw the free-body diagram of the bar.
Determine the reactions at A.
y
B (5, 6, 1) m
A
x
C (3, 0, 4) m
z
Solution: The position vectors of the ends of the cable are
y
rB D 5i C 6j C k,
A
z
and rC D 3i C 0j C 4k.
z
rBC D rC rB D 2i 6j C 3k,
p
(3,0,4) m
y
The vector parallel to the cable is
jrBC j D
B
(5,6,1) m
x
B
MY
FY
A
T
MZ
FZ
MX
FX
x
22 C 62 C 32 D 7 m.
The unit vector parallel to the cable is
eD
rBC
D 0.2857i 0.8571j C 0.4286k.
jrBC j
The tension vector is
TBC D 10eBC D 2.857i 8.571j C 4.286k.
The force reaction at A is determined by
FR D FA C TBC D 0,
from which FA D 2.857i C 8.571j 4.286k (kN)
The moment reaction at A is given by
MR D MA C rB ð TBC
i
j
D MA C 5
6
2.857 8.571
k 1 D 0
4.286 From which MA D 34.287i C 24.287j C 25.713k (kN-m)
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1
Problem 5.80 Consider the bar in Problem 5.79. The
magnitude of the couple exerted on the bar by the built-in
support is 100 kN-m. What is the tension in the cable?
Solution: From the solution to Problem 5.79, the moment reaction
at A is the solution to the moment equilibrium equation:
MR D MA C rB ð TBC D 0,
from which MA D rB ð TBC .
Noting that TBC D jTBC jeBC , then using the unit vector developed in
the solution to Problem 5.79,
eD
rBC
D 0.2857i 0.8571j C 0.4286k
jrBC j
and the position vector of the end of the bar:
rB D 5i C 6j C k
i
MA D rB ð jTBC jeBC D jTBC j 5
0.2857
j
6
0.8571
k 1 0.4286 D jTBC j3.4287i 2.4287j 2.5713k
Take the magnitude of both sides
p
jMA j D 100 D jTBC j 3.42872 C 2.42872 C 2.57132
D jTBC j4.9261.
Solve: jTBC j D
100
D 20.300 kN
4.9261
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1
Problem 5.81 The total force exerted on the highway
sign by its weight and the most severe anticipated winds
is F D 2.8i 1.8j (kN). Determine the reactions at the
fixed support.
y
F
8m
Solution: The applied load is F D 2.8i 1.8j kN applied at r D
8j C 8k m
The force reaction at the base is
8m
O
x
R D Ox i C O y j C O z k
The moment reaction at the base is
MO D MOx i C MOy j C MOz k
z
For equilibrium we need


Fx : 2.8 kN C Ox D 0



Fy : 1.8 kN C Oy D 0
FDFCRD0)




Fz : 0 C Oz D 0
Ox D 2.8 kN
)
Oy D 1.8 kN
Oz D 0


Mx : 14.4 kN-m C MOx D 0



My : 22.4 kN-m C MOy D 0
M D r ð F C MO D 0 )




Mz : 22.4 kN-m C MOz D 0
MOx D 14.4 kN-m
)
MOy D 22.4 kN-m
MOz D 22.4 kN-m
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1
Problem 5.82 The tension in cable AB is 800 lb.
Determine the reactions at the fixed support C.
y
4 ft
C
5 ft
4 ft
A
x
B
z
(6, 0, 4) ft
Solution: The force in the cable is
F D 800 lb
2i 4j k
p
21
We also have the position vector
rCA D 4i C 5k ft
The force reaction at the base is
R D Cx i C Cy j C Cz k
The moment reaction at the base is
MC D MCx i C MCy j C MCz k
For equilibrium we need


Fx : Cx C 349 lb D 0



FDFCRD0)
Fy : Cy 698 lb D 0




Fz : Cz 175 lb D 0
Cx D 349 lb
)
Cy D 698 lb
Cz D 175 lb


Mx : MCx C 3490 ft-lb D 0



M DrCFCRD0)
My : MCy C 2440 ft-lb D 0




Mz : MCz 2790 ft-lb D 0
MCx D 3490 ft-lb
)
MCy D 2440 ft-lb
MCz D 2790 ft-lb
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1
Problem 5.83 The tension in cable AB is 24 kN.
Determine the reactions in the built-in support D.
2m
C
A
2m
Solution: The force acting on the device is
D
B
F D FX i C FY j C FZ k D 24 kNeAB ,
3m
and the unit vector from A toward B is given by
eAB D
1m
1i 2j C 1k
p
.
6
The force, then, is given by
F D 9.80i 19.60j C 9.80k kN.
The position from D to A is
r D 2i C 2j C 0k m.
The force equations of equilibrium are
DX C FX D 0,
DY C FY D 0,
and DZ C FZ D 0.
The moment equation, in vector form, is
M D MD C r ð F.
Expanded, we get
i
M D MDX i C MDY j C MDZ k C 2
9.80
j
2
19.60
k 0 D 0.
9.80 The corresponding scalar equations are
MDX C 29.80 D 0,
MDY 29.80 D 0,
and MDZ C 219.60 29.80 D 0.
Solving for the support reactions, we get
DX D 9.80 kN,
OY D 19.60 kN,
OZ D 9.80 kN.
MDX D 19.6 kN-m,
MDY D 19.6 kN-m,
and MDZ D 58.8 kN-m.
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1
Problem 5.84 The robotic manipulator is stationary
and the y axis is vertical. The weights of the arms AB and
BC act at their midpoints. The direction cosines of the
centerline of arm AB are cos x D 0.174, cos y D 0.985,
cos z D 0, and the direction cosines of the centerline
of arm BC are cos x D 0.743, cos y D 0.557, cos z D
0.371. The support at A behaves like a built-in support.
y
600 mm
C
160 N
(a)
(b)
B
What is the sum of the moments about A due to
the weights of the two arms?
What are the reactions at A?
600 mm
200 N
A
z
x
Solution: Denote the center of mass of arm AB as D1 and that of
BC as D2 . We need
rAD ,
(a)
rAB ,
and rBD2 .
We now have the geometry determined and are ready to determine the moments of the weights about A.
MW D rAD1 ð W1 C rAD2 ð W2
where
i
rAD1 ð W1 D 0.0522
0
To get these, use the direction cosines to get the unit vectors eAB and
eBC . Use the relation
e D cos X i C cos Y j C cos Z k
j
0.2955
200
k 0 0
rAD1 ð W1 D 10.44k N-m
eAB D 0.174i C 0.985j C 0k
and
i
rAD2 ð W2 D 0.3273
0
eBC D 0.743i C 0.557j 0.371k
rAD1 D 0.3eAB m
Thus,
MW D 17.81i 62.81k (N-m)
rBC D 0.6eBC m
WAB D 200j N
Equilibrium Eqns
FX : AX D 0
WBC D 160j N
Thus rAD1 D 0.0522i C 0.2955j m
rAB D 0.1044i C 0.5910j m
(b)
FY :
AY W1 W2 D 0
FZ :
AZ D 0
rBD2 D 0.2229i C 0.1671j 0.1113k m
Sum Moments about
MW D 0
A : MA C
rBC D 0.4458i C 0.3342j 0.2226k m
and rAD2 D rAB C rBD2
rAD2 D 0.3273i C 0.7581j 0.1113k m
k
0.1113 0
rAD2 ð W2 D 17.81i 52.37k
rAB D 0.6eAB m
rBD2 D 0.3eBC m
j
0.7581
160
MX :
MAx 17.81 D 0 (N-m)
MY : MAy C 0 D 0
MZ :
MZ 62.81 D 0 (N-m)
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1
5.84 (Continued )
Thus:
AX D 0, AY D 360 (N), AZ D 0,
MAx D 17.81 (N-m), MAy D 0, MAz D 62.81 (N-m)
C
W2
W1
D2
B
D1
MA
MA = MAXi + MAYj + MAZk
W1 = 200 N
W2 = 160 N
AX
AZ
AY
2
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Problem 5.85 The force exerted on the grip of the
exercise machine is F D 260i 130j (N). What are the
reactions at the built-in support at O?
150
mm
y
F
O
200
mm
z
250
mm
Solution:
MO D MOx i C MOy j C MOz k
OX
rOP D 0.25i C 0.2j 0.15k
z
Equilibrium (Forces)
0.15 P
m
y
MO
OZ
x
OY
0.2
F = 260 i – 130 j (N)
5m
0.2 m
x
FX :
OX C FX D OX C 260 D 0 (N)
FY :
OY C FY D OY 130 D 0 (N)
FZ :
OZ C FZ D OZ D 0 (N)
Thus, OX D 260 N, OY D 130 N, OZ D 0
Summing Moments about O
MX : MOX C MFX D 0
MY : MOY C MFY D 0
MZ : MOZ C MFZ D 0
where
i
MF D rOP ð F D 0.25
260
j
k 0.2
0.15 130
0 MF D 19.5i 39j 84.5k (N-m)
and from the moment equilibrium eqns,
MOX D 19.5 (N-m)
MOY D 39.0 (N-m)
MOZ D 84.5 (N-m)
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1
Problem 5.86 The designer of the exercise machine
in Problem 5.85 assumes that the force F exerted on the
grip will be parallel to the xy plane and that its magnitude will not exceed 900 N. Based on these criteria, what
reactions must the built-in support at O be designed to
withstand?
Solution: The solution to this problem is similar to that of Problem
5.85. The free-body diagram and the equilibrium equations are similar,
but there are significant differences. For that reason, the solution will
be presented as if Problem 5.35 had not been solved previously.
y
MO
5m
0.1
P
OY
F
θ
OX
θ
O
MO D MOX i C MOY j C MOZ k
0.2 m
F D F cos i C F sin j
OZ
z
0.2
5m
O 360°
jFj D 900 N
x
rOP D 0.25i C 0.2j 0.15k m
Moment (N-m) vs. Theta (deg)
FX :
OX C FX D OX C F cos D 0
FY :
OY C FY D OY C F sin D 0
FZ :
OZ C FZ D OZ C 0 D 0
Moment Magnitude (N-m)
Thus OX D F cos OY D F sin 320
300
280
260
240
220
200
180
160
140
120
0
50
100
OZ D 0
Moment Mag (N-m)
MX : MOX C MFX D 0
MY : MOY C MFY D 0
MZ : MOZ C MFZ D 0
i
where MF D rOP ð F D 0.25
F cos 350
400
300
The moment equilibrium equations are
300
Moment Componente (N-m) vs. Theta (deg)
OX 2 C OY 2 D jFj D 900 N
150 200 250
Theta (deg)
j
0.2
F sin k 0.15 O MOZ
200
MOY
100
0
MOX
−100
−200
−300
0
50
100
150 200 250
Theta (deg)
300
350
400
MF D C0.15 F sin i
0.15 F cos j
C 0.25 F sin 0.2 F cos k
and MO D MF
The first plot on the next page shows jMO j vs for O 360° .
The second plot shows the three components of MO as functions of .
From the analysis and the plot, the support at 0 must be able to exert a
force of 900 N in any direction in the xy plane and it must be able
to exert a moment jMO j ½ 320 (N-m) From the component plots, we
see that the support must provide
135 (N-m) MOX 135 (N-m)
135 (N-m) MOY 135 (N-m)
288 (N-m) MOZ 288 (N-m)
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1
Problem 5.87 The force F acting on the boom ABC
at C points in the direction of the unit vector 0.512i 0.384j C 0.768k and its magnitude is 8 kN. The boom
is supported by a ball and socket at A and the cables BD
and BE. The collar at B is fixed to the boom.
(a)
(b)
y
1.5 m
2m
D
E
Draw the free-body diagram of the boom.
Determine the tensions in the cables and the reactions at A.
1m
2m
A
B
z
2m
C
2m
x
Solution:
(a)
(b)
F
The free-body diagram
We identify the following forces, position vectors, and reactions
rAC D 4 mi, F D 8 kN0.512i 0.384j C 0.768k

2i C 2j C 1.5k


p
T
D
T

BD
 BD
10.25
rAB D 2 mi,

2i
C
j 2k


 TBE D TBE
3
Az
Ax
TBE
B
Ay
TBD
C
R D Ax i C Ay j C Az k
Force equilibrium requires:
F D R C TBD C TBE C F D 0.
F
In component form we have
Fx : Ax C 8 kN0.512 p
Fy : Ay 8 kN0.384 C p
Fz : Az C 8 kN0.768 C p
2
10.25
2
10.25
1.5
10.25
TBD 2
TBE D 0
3
TBD C
1
TBE D 0
3
TBD 2
TBE D 0
3
Moment equilibrium requires:
MA D rAB ð TBD C TBE C rAC ð F D 0.
In components:
Mx : 0 D 0
1.5
TBD 2 m
My : 8 kN0.7684 m p
10.25
C
2
TBE 2 m D 0
3
2
TBD 2 m
Mz : 8 kN0.3844 m C p
10.25
C
1
TBE 2 m D 0
3
Solving five equations for the five unknowns we find
Ax D 8.19 kN, Ay D 3.07 kN, Az D 6.14 kN,
TBD D 0, TBE D 18.43 kN
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1
Problem 5.88 The cables BD and BE in Problem 5.87
will each safely support a tension of 25 kN. Based on
this criterion, what is the largest acceptable magnitude
of the force F?
Solution: We have the force and distances:
rAC D 4 mi, F D F0.512i 0.384j C 0.768k

2i C 2j C 1.5k


p
D
T
T

BD
BD

10.25
rAB D 2 mi,

2i
C
j 2k


 TBE D TBE
3
The moment equations are
1.5
2
TBD 2 m C TBE 2 m D 0
My : F0.7684 m p
3
10.25
2
1
TBD 2 m C TBE 2 m D 0
Mz : F0.3844 m C p
3
10.25
Solving we find
TBE D 2.304F, TBD D 0
Thus:
25 kN D 2.304F ) F D 10.85 kN
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1
Problem 5.89 The suspended load exerts a force F D
600 lb at A, and the weight of the bar OA is negligible.
Determine the tensions in the cables and the reactions at
the ball and socket support O.
y
C
(0, 6, –10) ft
A
(8, 6, 0) ft
B
(0, 10, 4) ft
–Fj
x
O
Solution: From the diagram, the important points in this problem
are A (8, 6, 0), B (0, 10, 4), C (0, 6, 10), and the origin O (0, 0, 0)
with all dimensions in ft. We need unit vectors in the directions A to
B and A to C. Both vectors are of the form
z
eAP D xP xA i C yP yA j C zP zA k,
If we carry through these operations in the sequence described, we get
the following vectors:
where P can be either A or B. The forces in cables AB and AC are
TAB D TAB eAB D TABX i C TABY j C TABZ k,
and TAC D TAC eAB D TACX i C TACY j C TACZ k.
eAB D 0.816i C 0.408j C 0.408k,
eAC D 0.625i C 0j 0.781k,
TAB D 387.1i C 193.5j C 193.5k lb,
The weight force is
jTAB j D 474.1 lb,
F D 0i 600j C 0k,
and the support force at the ball joint is
TAC D 154.8i C 0j 193.5k lb,
S D SX i C SY j C SZ k.
jTAC j D 247.9 lb,
The vector form of the force equilibrium equation (which gives three
scalar equations) for the bar is
MAB D rOA ð TAB D 1161i 1548j C 3871k ft-lb,
MAC D rOA ð TAC D 1161i C 1548j C 929k ft-lb,
TAB C TAC C F C S D 0.
Let us take moments about the origin. The moment equation, in vector
form, is given by
and S D 541.9i C 406.5j C 0k lb
MO D rOA ð TAB C rOA ð TAC
C rOA ð F D 0,
where rOA D 8i C 6j C 0k.
The cross products are evaluated using the form
i
M D r ð H D 8
HX
j
6
HY
k 0 ,
HZ where H can be any of the three forces acting at point A. The vector
moment equation provides another three equations of equilibrium.
Once we have evaluated and applied the unit vectors, we have six
vector equations of equilibrium in the five unknowns TAB , TAC , SX , SY ,
and SZ (there is one redundant equation since all forces pass through
the line OA). Solving these equations yields the required values for
the support reactions at the origin.
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1
Problem 5.90 In Problem 5.89, suppose that the
suspended load exerts a force F D 600 lb at A and bar
OA weighs 200 lb. Assume that the bar’s weight acts at
its midpoint. Determine the tensions in the cables and
the reactions at the ball and socket support O.
Solution: Point G is located at (4, 3, 0) and the position vector of
G with respect to the origin is
rOG D 4i C 3j C 0k ft.
The weight of the bar is
WB D 0i 200j C 0k lb,
and its moment around the origin is
MWB D 0i C 0j 800k ft-lb.
The mathematical representation for all other forces and moments from
Problem 5.89 remain the same (the numbers change!). Each equation
of equilibrium has a new term reflecting the addition of the weight of
the bar. The new force equilibrium equation is
TAB C TAC C F C S C WB D 0.
The new moment equilibrium equation is
MO D rOA ð TAB C rOA ð TAC
C rOA ð F C rOG ð WB D 0.
As in Problem 5.89, the vector equilibrium conditions can be reduced
to six scalar equations of equilibrium. Once we have evaluated and
applied the unit vectors, we have six vector equations of equilibrium
in the five unknowns TAB , TAC , SX , SY , and SZ (As before, there is one
redundant equation since all forces pass through the line OA). Solving
these equations yields the required values for the support reactions at
the origin.
If we carry through these operations in the sequence described, we get
the following vectors:
eAB D 0.816i C 0.408j C 0.408k,
eAC D 0.625i C 0j 0.781k,
TAB D 451.6i C 225.8j C 225.8k lb,
jTAB j D 553.1 lb,
TAC D 180.6i C 0j 225.8k lb,
jTAC j D 289.2 lb,
MAB D rOA ð TAB D 1355i 1806j C 4516k ft-lb,
MAC D rOA ð TAC D 1354i C 1806j C 1084k ft-lb,
and S D 632.3i C 574.2j C 0k lb
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1
Problem 5.91 The 158,000-kg airplane is at rest on
the ground (z D 0 is ground level). The landing gear
carriages are at A, B, and C. The coordinates of the point
G at which the weight of the plane acts are (3, 0.5, 5) m.
What are the magnitudes of the normal reactions exerted
on the landing gear by the ground?
21 m
6m
B
G
A
x
C
6m
y
Solution:
mg
3m
FY D NL C NR C NF W D 0
MR D 3 mg C 21NF D 0
21 m
Side
View
x
Solving,
R
NF D 221.4 kN
F
(NL + NR)
(1)
NF
Z
NL C NR D 1328.6 kN
(2)
0.5 m
W
FY D NR C NL C NF W D 0
(same equation as before)
C
MO D 0.5 W 6NR C 6NL D 0 (3)
Front
View
y
Solving (1), (2), and (3), we get
6
6
NF D 221.4 kN
NR D 728.9 kN
NF
NR
NL
z
NL D 599.7 kN
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1
Problem 5.92 The horizontal triangular plate is
suspended by the three vertical cables A, B, and C.
The tension in each cable is 80 N. Determine the x
and z coordinates of the point where the plate’s weight
effectively acts.
y
A
B
C
0.4 m
0.3 m
(x, 0, z)
x
z
Solution:
80 N
80 N
Mx : 240 Nz 80 N0.4 m D 0
Mz : 80 N0.3 m 240 Nx D 0
80 N
z
x
X
Solving
x D 0.1 m, z D 0.1333 m
240 N
Z
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1
Problem 5.93 The 800-kg horizontal wall section is
supported by the three vertical cables, A, B, and C. What
are the tensions in the cables?
B
7m
C
A
7m
7m
6m
4m
8m
mg
Solution: All dimensions are in m and all forces are in N. Forces
A, B, C, and W act on the wall at (0, 0, 0), (5, 14, 0), (12, 7, 0),
and (4, 6, 0), respectively. All forces are in the z direction. The force
equilibrium equation in the z direction is A C B C C W D 0. The
moments are calculated from
MB D rOB ð Bk,
MC D rOC ð Ck,
and MG D rOG ð Wk.
The moment equilibrium equation is
MO D MB C MC C MG D 0.
Carrying out these operations, we get
A D 3717 N,
B D 2596 N,
C D 1534 N,
and W D 7848 N.
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1
Problem 5.94 The bar AC is supported by the cable
BD and a bearing at A that can rotate about the z axis.
The person exerts a force F D 10j (lb) at C. Determine
the tension in the cable and the reactions at A.
y
A
x
B
C
8 in
14 in
z
(18, ⫺8, 7) in
D
Solution: The force in the cable is
TBD D TBD
10i 8j C 7k
p
213
Ax, MAx
Az
F = 10 lbj
We have the following six equilibrium equations
10
TBD D 0
Fx : Ax C p
213
8
TBD C 10 lb D 0
Fy : Ay p
213
Fz : Az C p
7
213
Ay
MAy
TBD
TBD D 0
Mx : MAx D 0
7
TBD 8 in D 0
My : MAy p
213
8
TBD 8 in C 10 lb22 in D 0
Mz : p
213
Solving we find
Ax D 34.4 lb, Ay D 17.5 lb, Az D 24.1 lb
MAx D 0, MAy D 192.5 lb in, TBD D 50.2 lb
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1
y
Problem 5.95 The L-shaped bar is supported by a
bearing at A and rests on a smooth horizontal surface
at B. The vertical force F D 4 kN and the distance
b D 0.15 m. Determine the reactions at A and B.
F
b
A
x
Solution: Equilibrium Eqns:
B
FX :
0.2 m
OD0
0.3 m
z
FY :
AY C B F D 0
FZ :
AZ D 0
y
Sum moments around A
F
b
x:
Fb 0.3B D 40.15 0.3B D 0
y:
MAY D 0
z: MAZ C 0.2F 0.2B D 0
B
z
3
0.
B
0.2 m
m
MZ
A
AY
AZ
MY
(MAX ≡ 0)
AX ≡ 0
x
b = 0.15 m
F = 4 kN
Solving,
AX D 0,
AY D 2 (kN),
AZ D 0
MAX D 0,
MAY D 0,
MAZ D 0.4 (kN-m)
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1
Problem 5.96 In Problem 5.95, the vertical force F D
4 kN and the distance b D 0.15 m. If you represent the
reactions at A and B by an equivalent system consisting
of a single force, what is the force and where does its
line of action intersect the xz plane?
Solution: We want to represent the forces at A & B by a single
force. From Prob. 5.95
zR 4 D C0.32
zR D C0.15 m
A D C2j (kN),
xR 4 D 0.22
B D C2j (kN)
xR D 0.1 m
MA D 0.4k (kN-m)
y
We want a single equivalent force, R that has the same resultant force
and moment about A as does the set A, B, and MA .
R
F
b
A
R D A C B D 4j (kN)
x
Let R pierce the xz plane at xR , zR B
MX :
zR R D 0.3B
z
MZ :
0.2 m
0.3 m
xR R D 0.2AY
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1
Problem 5.97 In Problem 5.95, the vertical force F D
4 kN. The bearing at A will safely support a force of
2.5-kN magnitude and a couple of 0.5 kN-m magnitude.
Based on these criteria, what is the allowable range of
the distance b?
Solution: The solution to Prob. 5.95 produced the relations
AY C B F D 0 F D 4 kN
Fb 0.3B D 0
MAZ C 0.2F 0.2B D 0
AX D AZ D MAX D MAY D 0
Set the force at A to its limit of 2.5 kN and solve for b. In this case,
MAZ D 0.5 (kN-m) which is at the moment limit. The value for b is
b D 0.1125 m
We make AY unknown, b unknown, and B unknown F D 4 kN,
MAY D C0.5 (kN-m), and solve we get AY D 2.5 at b D 0.4875 m
However, 0.3 is the physical limit of the device.
Thus, 0.1125 m b 0.3 m
y
F
b
A
x
B
z
0.2 m
0.3 m
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1
Problem 5.98 The 1.1-m bar is supported by a ball
and socket support at A and the two smooth walls. The
tension in the vertical cable CD is 1 kN.
(a)
(b)
y
B
Draw the free-body diagram of the bar.
Determine the reactions at A and B.
400 mm
D
x
C
Solution:
(b)
600 mm
700 mm
z
(a)
A
From which,
The ball and socket cannot support a couple reaction, but can
support a three force reaction. The smooth surface supports oneforce normal to the surface. The cable supports one force parallel
to the cable.
The strategy is to determine the moments about A, which will
contain only the unknown reaction at B. This will require the
position vectors of B and D relative to A, which in turn will
require the unit vector parallel to the rod. The angle formed by the
bar with the horizontal is required to determine the coordinates
of B:
p
˛ D cos1
0.62 C 0.72
1.1
BZ D
0.3819
D 0.6365 kN,
0.6
BX D
0.4455
D 0.7425 kN
0.6
.
The reactions at A are determined from the sums of the forces:
D 33.1° .
The coordinates of the points are: A (0.7, 0, 0.6), B 0, 1.1
sin 33.1° , 0 D 0, 0.6, 0, from which the vector parallel to the
bar is
rAB D rB rA D 0.7i C 0.6j 0.6k (m).
FX D BX C AX i D 0, from which AX D 0.7425 kN.
FY D AY 1j D 0, from which AY D 1 kN.
FZ D BZ C AZ k D 0, from which AZ D 0.6365 kN
FB
The unit vector parallel to the bar is
eAB D
rAB
D 0.6364i C 0.5455j 0.5455k.
1.1
T
FY
The vector location of the point D relative to A is
FZ
FX
rAD D 1.1 0.4eAB D 0.7eAB
D 0.4455i C 0.3819j 0.3819k.
The reaction at B is horizontal, with unknown x-component and
z-components. The sum of the moments about A is
i
j
k MA D rAB ð B C rAD ð D D 0 D 0.7 0.6 0.6 BX
0
BZ i
C 0.4455
0
j
0.3819
1
k
0.3819 D 0
0
Expand and collect like terms:
MA D 0.6BZ 0.3819i 0.6BX 0.7BZ j
C0.6BX C 0.4455k D 0.
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1
Problem 5.99 The 8-ft bar is supported by a ball and
socket at A, the cable BD, and a roller support at C. The
collar at B is fixed to the bar at its midpoint. The force
F D 50k (lb). Determine the tension in the cable BD
and the reactions at A and C.
y
A
3 ft
Solution: The strategy is to determine the sum of the moments
B
F
about A, which will involve the unknown reactions at B and C. This
will require the unit vectors parallel to the rod and parallel to the cable.
z
The angle formed by the rod is
4 ft
D
2 ft
3
˛ D sin1
D 22° .
8
C
x
The vector positions are:
rA D 3j,
from which jTj D
rD D 4i C 2k
CY D
and rC D 8 cos 22° i D 7.4162i.
75
D 62.92 lb
1.192
2.036jTj
D 17.27 lb.
7.4162
The reaction at A is determined from the sums of forces:
The vector parallel to the rod is
FX D AX C 0.1160jTji D 0,
rAC D rC rA D 7.4162i 3j.
from which AX D 7.29 lb,
The unit vector parallel to the rod is
eAC D 0.9270i 0.375j.
FY D AY 0.5960jTj C CY j D 0,
from which AY D 20.23 lb
The location of B is
rAB D 4eAC D 3.7081i 1.5j.
FZ D AZ C 0.7946jTj 50k D 0,
from which AZ D 0 lb
The vector parallel to the cable is
rBD D rD rA C rAB D 0.2919i 1.5j C 2k.
y
The unit vector parallel to the cable is
A
eBD D 0.1160i 0.5960j C 0.7946k.
F
B
3 ft
The tension in the cable is T D jTjeBD . The reaction at the roller
support C is normal to the xz plane. The sum of the moments about
A
C
D
z
4 ft
MA D rAB ð F C rAB ð T C rAC ð C D 0
i
D 3.7081
0
2 ft
j
k 1.5
0 0
50 i
C jTj 3.7081
0.1160
j
1.5
0.5960
i
C 7.4162
0
k 0 D 0
0
j
3
CY
x
y
k 0 0.7946 AY
AZ
AX
T
F
z
D
D 75i C 185.4j C jTj1.192i 2.9466j 2.036k
CY
x
C 7.4162CY k D 0,
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1
Problem 5.100 Consider the 8-ft bar in Problem 5.99.
The force F D Fy j 50k (lb). What is the largest value
of Fy for which the roller support at C will remain on
the floor?
Solution: From the solution to Problem 5.99, the sum of the
moments about A is
i
MA D 3.7081
0
j
k 1.5
0 FY 50 i
C jTj 3.7081
0.1160
j
1.5
0.5960
i
C 7.4162
0
k 0 D 0
0
j
3
CY
k 0 0.7946 D 75i C 185.4j C 3.7081FY k
C jTj1.192i 2.9466j 2.036k
C 7.4162CY k D 0,
from which, jTj D
75
D 62.92 lb.
1.192
Collecting terms in k, 3.7081FY C 2.384jTj 7.4162CY D 0.
For CY D 0, FY D
128.11
D 34.54 lb
3.708
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1
Problem 5.101 The tower is 70 m tall. The tension in
each cable is 2 kN. Treat the base of the tower A as a
built-in support. What are the reactions at A?
y
B
Solution: The strategy is to determine moments about A due to
the cables. This requires the unit vectors parallel to the cables.
C
40 m
The coordinates of the points are:
A0, 0, 0, B0, 70, 0, C50, 0, 0,
50 m
E
A
D20, 0, 50, E40, 0, 40.
40 m
D
The unit vectors parallel to the cables, directed from B to the points
E, D, and C
50 m
z
20 m
x
rBE D 40i 70j 40k,
The force reactions at A are determined from the sums of forces. (Note
that the sums of the cable forces have already been calculated and used
above.)
rBD D 20i 70j C 50k,
rBC D 50i 70j.
The unit vectors parallel to the cables, pointing from B, are:
FX D AX C 0.17932i D 0,
from which AX D 0.179 kN,
eBE D 0.4444i 0.7778j 0.4444k,
FY D AY 4.7682j D 0,
eBD D 0.2265i 0.7926j C 0.5661k,
from which AY D 4.768 kN,
eBC D 0.5812i 0.8137j C 0k.
The tensions in the cables are:
TBD D 2eBD D 0.4529i 1.5852j C 1.1323k (kN),
FZ D AZ C 0.2434k D 0,
from which AZ D 0.2434 kN
TBE D 2eBE D 0.8889i 1.5556j 0.8889k (kN),
y
B
TBC D 2eBC D 1.1625i 1.6275j 0k.
TBC
The sum of the moments about A is
TBD
MA D MA C rAB ð TBE
AY ,
C
A
MY
rAB ð TBD C rAB ð TBC D 0
C
D MA C rAB ð TBE C TBC C TBD i
A
MA D M C 0
0.1793
j
70
4.7682
TBE
k 0 D 0
0.2434 AZ ,MA
EA
AX , M A
Z
z
D
X
x
D MAX C 17.038i C MAY C 0j
C MAZ 12.551k D 0
from which
MAX D 17.038 kN-m,
MAY D 0,
MAZ D 12.551 kN-m.
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1
Problem 5.102 Consider the tower in Problem 5.101.
If the tension in cable BC is 2 kN, what must the tensions in cables BD and BE be if you want the couple
exerted on the tower by the built-in support at A to be
zero? What are the resulting reactions at A?
Solution: From the solution to Problem 5.101, the sum of the
moments about A is given by
MA D MA C rAB ð TBE C TBC C TBD D 0.
If the couple MA D 0, then the cross product is zero, which is possible
only if the vector sum of the cable tensions is zero in the x and z
directions. Thus, from Problem 5.101,
ex Ð TBC C jTBE jeBE C jTBD jeBD D 0,
and ez Ð TBC C jTBE jeBE C jTBD jeBD D 0.
Two simultaneous equations in two unknowns result;
0.4444jTBE j C 0.2265jTBD j D 1.1625
0.4444jTBE j C 0.5661jTBD j D 0.
Solve:
jTBE j D 1.868 kN,
jTBD j D 1.467 kN.
The reactions at A oppose the sum of the cable tensions in the x-, y-,
and z-directions.
AX D 0,
AY D 4.243 kN,
AZ D 0.
(These results are to be expected if there is no moment about A.)
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1
Problem 5.103 The space truss has roller supports at
B, C, and D and is subjected to a vertical force F D
20 kN at A. What are the reactions at the roller supports?
y
F
A (4, 3, 4) m
B
Solution: The key to this solution is expressing the forces in terms
of unit vectors and magnitudes-then using the method of joints in three
dimensions. The points A, B, C, and D are located at
D (6, 0, 0) m
x
A4, 3, 4 m, B0, 0, 0 m,
C5, 0, 6 m, D6, 0, 0 m
z
C (5, 0, 6) m
we need eAB , eAC , eAD , eBC , eBD , and eCD . Use the form
ePQ D
xQ xP i C yQ yP j C zQ zP k
[xQ xP 2 C yQ yP 2 C zQ zP 2 ]1/2
F
Joint A :
eAB D 0.625i 0.469j 0.625k
TAD
TAB
eAC D 0.267i 0.802j C 0.535k
eAD D 0.371i 0.557j 0.743k
TAC
D
B
eBC D 0.640i C 0j C 0.768k
C
eBD D 1i C 0j C 0k
eCD D 0.164i C 0j 0.986k
Joint B :
–TAB
We will write each force as a magnitude times the appropriate unit
vector.
TAB D TAB eAB , TAC D TAC eAC
TBD
NBJ
TBC
TAD D TAD eAD , TBC D TBC eBC
Joint C :
TBD D TBD eBD , TCD D TCD eCD
–TAC
Each force will be written in component form, i.e.

TABX D TAB eABX 


TABY D TAB eABY
etc.



TABZ D TAB eABZ
Joint A:
TAB C TAC C TAD C F D 0
TCD
–TBC
NCJ
Joint D :
–TAD
TABX C TACX C TADX D 0
–TBD
TABY C TACY C TADY 20 D 0
–TCD
NDJ
TABZ C TACZ C TADZ D 0
Joint B:
TAB C TBC C TBD C NB j D 0
Joint C:
TAC TBC C TCD C NC j D 0
Joint D:
TAD TBD TCD C ND j D 0
Solving for all the unknowns, we get
NB D 4.44 kN
NC D 2.22 kN
ND D 13.33 kN
Also, TAB D 9.49 kN, TAC D 16.63 kN
TAD D 3.99 kN, TBC D 7.71 kN
TBD D 0.99 kN, TCD D 3.00 kN
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1
Problem 5.104 In Problem 5.103, suppose that you
don’t want the reaction at any of the roller supports to
exceed 15 kN. What is the largest force F the truss can
support?
Solution: The solution to Problem 5.103 is linear in the force
components-hence, it can be scaled. The largest roller reaction is at C,
NC D 13.33 kN. The maximum value is 15 kN. Scaling the force by
the factor 15/13.33 gives
Fmax D
15
13.33
20 kN D 22.5 kN
Fmax D 22.5 kN
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1
Problem 5.105 The 40-lb door is supported by hinges
at A and B. The y axis is vertical. The hinges do not
exert couples on the door, and the hinge at B does not
exert a force parallel to the hinge axis. The weight of
the door acts at its midpoint. What are the reactions at
A and B?
y
4 ft
1 ft
B
Solution: The position vector of the midpoint of the door:
5 ft
rCM D 2 cos 50° i C 3.5j C 2 cos 40° k
A
D 1.2856i C 3.5j C 1.532k.
1 ft
x
The position vectors of the hinges:
40
rA D j,
rB D 6j.
z
The forces are: W D 40j,
y
A D AX i C AY j C AZ k,
BX
B D BX i C BZ k.
BZ
The position vectors relative to A are
AY
rACM D rCM rA D 1.2856i C 2.5j C 1.532k,
rAB D rB rA D 5j.
AZ
The sum of the moments about A
AX
x
z
MA D rACM ð W C rAB ð B
i
D 1.2856
0
W
j
k i
2.5 1.532 C 0
40
0 BX
j k 5 0 D 0
0 BZ MA D 5BZ C 401.532i C 5BX 401.285k D 0,
from which BZ D
401.532
D 12.256 lb
5
and
401.285
D 10.28 lb.
5
BX D
The reactions at A are determined from the sums of forces:
FX D AX C BX i D 0,
from which AX D 10.28 lb,
FY D AY 40j D 0,
from which AY D 40 lb,
FZ D AZ C BZ k D 0,
from which AZ D 12.256 lb
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1
Problem 5.106 The vertical cable is attached at A.
Determine the tension in the cable and the reactions at
the bearing B due to the force F D 10i 30j 10k (N).
y
200 mm
100 mm
100 mm
Solution: The position vector of the point of application of the
force is
B
200 mm
rF D 0.2i 0.2k.
The position vector of the bearing is
F
z
x
A
rB D 0.1i.
The position vector of the cable attachment to the wheel is
rC D 0.1k.
he position vectors relative to B are:
rBC D rC rB D 0.1i C 0.1k,
rBF D rF rB D 0.1i 0.2k.
The sum of the moments about the bearing B is
or
MB D MB C rBF ð F C rBC ð C D 0,
i
j
k j
k i
MB D MB C 0.1
0
0.2 C 0.1 0 0.1 10 30 10 0
T 0 D 6 C 0.1Ti C MBY 1j
C MBZ 3 C 0.1Tk D 0,
from which T D
6
D 60 N,
0.1
MBY D C1 N-m,
MBZ D 0.1T C 3 D 3 N-m.
The force reactions at the bearing are determined from the sums of
forces:
FX D BX C 10i D 0,
from which BX D 10 N.
FY D BY 30 60j D 0,
from which BY D 90 N.
FZ D BZ 10j D 0,
from which BZ D 10 N.
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1
Problem 5.107 In Problem 5.106, suppose that the z
component of the force F is zero, but otherwise F is
unknown. If the couple exerted on the shaft by the
bearing at B is MB D 6j 6k N-m, what are the force
F and the tension in the cable?
Solution: From the diagram of Problem 5.106, the force equilibrium equation components are
and
Fx D BX C FX D 0,
Fy D BY C FY D 0,
Fz D BZ C FZ D 0,
where FZ D 0 is given in the problem statement. The moment
equations can be developed by inspection of the figure also. They are
and
Mx D MBX C MAX C MFX D 0,
MY D MBY C MAY C MFY D 0,
MZ D MBZ C MAZ C MFZ D 0,
where MB D 6j 6k N-m. Note that MBX D 0 can be inferred. The
moments which need to be substituted into the moment equations are
MA D 0.1Ai C 0j C 0.1Ak N-m,
and MF D 0.2FY i 0.2FX j C 0.1FY k N-m.
Substituting these values into the equilibrium equations, we get F D
30i 60j C 0k N, and A D 120 N.
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1
y
Problem 5.108 The device in Problem 5.106 is badly
designed because of the couples that must be supported
by the bearing at B, which would cause the bearing to
“bind”. (Imagine trying to open a door supported by
only one hinge.) In this improved design, the bearings
at B and C support no couples, and the bearing at C
does not exert a force in the x direction. If the force
F D 10i 30j 10k (N), what are the tension in the
vertical cable and the reactions at the bearings B and C?
200 mm
50 mm
100 mm
50 mm
200 mm
B
C
F
z
x
A
Solution: The position vectors relative to the bearing B are: the
position vector of the cable attachment to the wheel is
y
rBT D 0.05i C 0.1k.
BY
The position vector of the bearing C is:
BX CY
BZ
F
CZ
z
rBC D 0.1i.
T
x
The position vector of the point of application of the force is:
rBF D 0.15i 0.2k.
The sum of the moments about B is
MB D rBT ð T C rBC ð C C rBF ð F D 0
i
MB D 0.05
0
i
C 0.15
10
j
j
k i
0 0.1 C 0.1 0
T 0 0 CY
k 0 CZ j
k 0
0.2 D 0
30 10 MB D 0.1T 6i C 0.1CZ C 1.5 2j
C 0.05T C 0.1CY 4.5k D 0.
From which: T D 60 N,
CZ D
0.5
D 5 N,
0.1
CY D
4.5 0.05T
D 15 N.
0.1
The reactions at B are found from the sums of forces:
FX D BX C 10i D 0,
from which BX D 10 N.
FY D BY C CY T 30j D 0,
from which BY D 75 N.
FZ D BZ C CZ 10k D 0,
from which BZ D 15 N
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1
Problem 5.109 The rocket launcher is supported by
the hydraulic jack DE and the bearings A and B. The
bearings lie on the x axis and supports shafts parallel to
the x axis. The hydraulic cylinder DE exerts a force on
the launcher that points along the line from D to E. The
coordinates of D are (7, 0, 7) ft, and the coordinates of
E are (9, 6, 4) ft. The weight W D 30 kip acts at (4.5,
5, 2) ft. What is the magnitude of the reaction on the
launcher at E?
y
W
E
A
B
x
D
3 ft 3 ft
Solution: The position vectors of the points D, E and W are
rD D 7i C 7k,
rE D 9i C 6j C 4k (ft),
rW D 4.5i C 5j C 2k (ft).
The vector parallel to DE is
rDE D rE rD D 2i C 6j 3k.
The unit vector parallel to DE is
eDE D 0.2857i C 0.8571j 0.4286k.
Since the bearings cannot exert a moment about the x axis, the sum of
the moments due to the weight and the jack force must be zero about
the x axis. The sum of the moments about the x axis is:
1
1
0
0 MX D 4.5
5
2 CjFDE j 9
0 30 0 0.2857
0
0
D0
6
4
0.8571 0.4286 D 60 6jFDE j D 0.
From which
jFDE j D
60
D 10 kip
6
y
30 kip
BY
AY
AX
AZ
BZ
x
FDE
z
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1
Problem 5.110 Consider the rocket launcher described
in Problem 5.109. The bearings at A and B do not exert
couples, and the bearing B does not exert a force in the
x direction. Determine the reactions at A and B.
Solution: See the solution of Problem 5.109. The force FDE can
be written
FDE D FDE 0.2857i C 0.8571j 0.4286k.
The equilibrium equations are
FX D AX C 0.2857FDE D 0,
FY D AY C BY C 0.8571FDE 30 D 0,
FZ D AZ C BZ 0.4286FDE D 0,
i
Morigin D 3
AX
j
0
AY
k i
0 C 6
AZ 0
j
0
BY
i
C FDE 7
0.2857
j
0
0.8571
i
j
C 4.5
5
0 30
k 2 D 0
0
k 0 BZ k
7
0.4286 The components of the moment eq. are
5.9997FDE C 60 D 0,
3AZ 6BZ C 5.0001FDE D 0,
3AY C 6BY C 5.9997FDE 135 D 0.
Solving, we obtain
FDE D 10.00 kip, AX D 2.86 kip,
AY D 17.86 kip, AZ D 8.09 kip,
BY D 3.57 kip, BZ D 12.38 kip.
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1
Problem 5.111 The crane’s cable CD is attached to a
stationary object at D. The crane is supported by the
bearings E and F and the horizontal cable AB. The
tension in cable AB is 8 kN. Determine the tension in
the cable CD.
Strategy: Since the reactions exerted on the crane by
the bearings do not exert moments about the z axis, the
sum of the moments about the z axis due to the forces
exerted on the crane by the cables AB and CD equals
zero. (See the discussion at the end of Example 5.10.)
y
Solution: The position vector from C to D is
C
rCD D 3i 6j 3k (m),
A
so we can write the force exerted at C by cable CD as
B
rCD
TCD D TCD
D TCD 0.408i 0.816j 0.408k.
jrCD j
The coordinates of pt. B are x D
4
3 D 2 m, y D 4 m.
6
F
E
z
The moment about the origin due to the forces exerted by the two
cables is
i
j
MO D 2 4
8 0
k i
0 C 3
0 0.408TCD
j
6
0.816TCD
2m
k
0
0.408TCD 2m
D
3m
y
x
C
D 32k 2.448TCD i C 1.224TCD j 4.896TCD k.
A
The moment about the z axis is
B
6m
k Ð MO D 32 4.896TCD D 0,
so TCD D 6.54 kN.
4m
D
3m
x
3m
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1
Problem 5.112 The crane in Problem 5.111 is supported by the horizontal cable AB and the bearings at E and
F. The bearings do not exert couples, and the bearing at
F does not exert a force in the z direction. The tension in
cable AB is 8 kN. Determine the reactions at E and F.
y
Solution: See the solution of Problem 5.111. The force exerted at
C can be written
C
TCD D TCD 0.408i 0.816j 0.408k
8 kN
and the coordinates of pt. B are (2, 4, 0) m.
TCD
B
The equilibrium equations are
EY
FX D EX C FX 8 C 0.408TCD D 0,
FY
z
FY D EY C FY 0.816TCD D 0,
j
0
EY
k i
2 C 0
E Z FX
i
C 3
0.408TCD
O
FX
EX
FZ D EZ 0.408TCD D 0,
i
MO D 0
EX
EZ
k i j k 2 C 2 4 0 0 8 0 0 j
0
FY
j
6
0.816TCD
x
k
D 0.
0
0.408TCD The components of the moment equation are
MX D 2EY C 2FY 2.448TCD D 0,
MY D 2EX 2FX C 1.224TCD D 0,
MZ D 32 4.896TCD D 0.
From the last equation, TCD D 6.54 kN. Solving the other eqs, we
obtain
EX D 667 N,
EY D 1,333 N,
EZ D 2,667 N,
FX D 4,667 N,
FY D 6,667 N.
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1
Problem 5.113 The plate is supported by hinges at A
and B and the cable CE, and it is loaded by the force at
D. The edge of the plate to which the hinges are attached
lies in the yz plane, and the axes of the hinges are
parallel to the line through points A and B. The hinges
do not exert couples on the plate. What is the tension in
cable CE?
y
3m
2i – 6j (kN)
E
A
D
2m
1m
Solution:
B
z
C
20°
2m
F D A C B C FD C TCE D 0
However, we just want tension in CE. This quantity is the only
unknown in the moment equation about the line AB. To get this, we
need the unit vector along CE.
y
3m
Point C is at (2, 2 sin 20° , 2 cos 20° ) Point E is at (0, 1, 3)
AX
A
1m
z
eCE D 0.703i C 0.592j C 0.394k
BZ
BY
AZ
B
20°
FD = 2i – 6j
AY
E
rCE
eCE D
jrCE j
We also need the unit vector eAB . A(0, 0, 0), B0, 2 sin 20° , 2 cos 20° x
D
x
TCE
2m
BX
2m
C
eAB D 0i 0.342j C 0.940k
The moment of FD about A (a point on AB) is
MFD D rAD ð FD1 D 2i ð 2i 6j
MFD D 12k
The moment of TCE about B (another point on line CE) is
MTCE D rBC ð TCE eCE D 2i ð TCE eCE ,
where eCE is given above.
The moment of FD about line AB is
MFDAB D MFD Ð eAB
MFDAB D 11.27 N-m
The moment of TCE about line AB is
MCEAB D TCE 2i ð eCE Ð eAB
MCEAB D TCE 0.788j C 1.184k Ð eAB
MCEAB D 1.382TCE
The sum of the moments about line AB is zero. Hence
MFDAB C MCEAB D 0
11.27 C 1.382TCE D 0
TCE D 8.15 kN
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1
Problem 5.114 In Problem 5.113, the hinge at B does
not exert a force on the plate in the direction of the hinge
axis. What are the magnitudes of the forces exerted on
the plate by the hinges at A and B?
Solution: From the solution to Problem 5.113, TCE D 8.15 kN
y
Also, from that solution,
F = + 2i – 6j (kN)
AY
eAB D 0i 0.342j C 0.940k
AX
We are given that the force at force at hinge B does not exert a force
parallel to AB at B. This implies
θ
z
BY
B Ð eAB D 0.
B Ð eAB D 0.342BY C 0.940BZ D 0
D
AZ
x
TCE
2
BX
C (2, –2sinθ , + 2cosθ)
(1)
BZ
We also had, in the solution to Problem 5.113
eCE D 0.703i C 0.592j C 0.394k
and
TCE D TCE eCE (kN)
For Equilibrium,
F D A C B C TCE C F D 0
FX :
AX C BX C TCE eCEX C 2 D 0 (kN) (2)
FY :
AY C BY C TCE eCEY 6 D 0 (kN)
(3)
FZ :
AZ C BZ C TCE eCEZ D 0 (kN)
(4)
Summing Moments about A, we have
rAD ð F C rAC ð TCE C rAB ð B D 0
rAD ð F D 2i ð 2i 6j D 12k (kN)
rAC ð TCE D 2 sin TZ 2 cos TY i
C 2 cos TX 2TZ j
C 2TY C 2TX sin k
rCE ð B D 2BZ sin 2BY cos i
C 2BX cos j C C2BX sin k
MA D 0,
Hence
x:
2 sin TZ 2 cos TY 2BZ sin 2BY cos D 0
(5)
y:
2 cos TX 2TZ C 2BX cos D 0
(6)
z:
12 C 2TY C 2TX sin C 2BX sin D 0 (7)
Solving Eqns (1)–(7), we get
jAj D 8.53 (kN), jBj D 10.75 (kN)
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1
Problem 5.115 The bar ABC is supported by ball and
socket supports at A and C and the cable BD. The
suspended mass is 1800 kg. Determine the tension in
the cable.
(⫺2, 2, ⫺1) m
y
D
2m
4m
B
A
C
x
4m
z
Solution: We take moments about the line AC to eliminate the
reactions at A and C.
TBD
We have
rAB D 4 mk, TBD D TBD
2i C 2j k
3
Az
rAW D 2i 4k m, W D 1800 kg9.81 m/s2 j
eCA D
1
6i 4k
p
D p 3i 2k
52
13
Ax
Cx
Ay
17.66 kN
The one equilibrium equation we need is
MAC D eCA Ð rAB ð TAB C rAW ð W D 0
Cz
Cy
This equation reduces to the scalar equation
1
p 3i 2k Ð
13
8
8
m TBD i C
m TBD j [4 m][17.66 kN]i
3
3
[2 m][17.66 kN]k
D0
8
1
p 3
m TBD [4 m][17.66 kN] C 2 f[2 m][17.66 kN]g D 0
3
2
Solving we find
TBD D 17.66 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 5.116* In Problem 5.115, assume that the
ball and socket support at A is designed so that it
exerts no force parallel to the straight line from A to
C. Determine the reactions at A and C.
Solution: We have
TBD D TBD
2i C 2j k
3
There are 7 unknowns. We have the following 6 equilibrium equations
Fx : Ax C Cx 2
TBD D 0
3
Fy : Ay C Cy C
2
TBD 17.66 kN D 0
3
Fz : Az C Cz 1
TBD D 0
3
Mx : Cy 4 m D 0
My : Cx 4 m Az 6 m D 0
Mz : Ay 6 m 17.66 kN2 m D 0
The last equation comes from the fact that the ball and socket at A
exerts no force in the direction of the line from A to C
6i C 4k
1
p
D p 6Ax C 4Az D 0
Ax i C Ay j C Az k Ð
52
52
Solving these 7 equations we find
Ax D 3.62 kN, Ay D 5.89 kN, Az D 5.43 kN
Cx D 8.15 kN, Cy D 0, Cz D 0.453 kN
TBD D 17.66 kN
TBD
Az
Ax
Cx
Ay
17.66 kN
Cz
Cy
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1
y
Problem 5.117 The bearings at A, B, and C do not
exert couples on the bar and do not exert forces in the
direction of the axis of the bar. Determine the reactions
at the bearings due to the two forces on the bar.
200 i (N)
300 mm
x
Solution: The strategy is to take the moments about A and solve
C
180 mm
the resulting simultaneous equations. The position vectors of the bearings relative to A are:
B
z
rAB D 0.15i C 0.15j,
A
rAC D 0.15i C 0.33j C 0.3k.
100 k (N)
Denote the lower force by subscript 1, and the upper by subscript 2:
150 mm
rA1 D 0.15i,
rA2 D 0.15i C 0.33j.
y
CY
The sum of the moments about A is:
i
j
j
k i
MA D 0.15 0
0 C 0.15 0.15
0
0 100 BX
0
k 0 BZ j
k i
0 C 0.15 0.33
0 CX
CY
k 0.3 D 0
0 200 N
x
MA D rA1 ð F1 C rAB ð B C rA2 ð F2 C rAC ð C D 0
i
j
C 0.15 0.33
200
0
150 mm
z
CX
BX
BZ
100 N
AY
AZ
MA D 0.15BZ 0.3CY i C 15 C 0.15BZ C 0.3CX j
C 0.15BX 66 0.15CY 0.33CX k D 0.
This results in three equations in four unknowns; an additional equation
is provided by the sum of the forces in the x-direction (which cannot
have a reaction term due to A)
FX D BX C CX C 200i D 0.
The four equations in four unknowns:
0BX C 0.15BZ C 0CX 0.3CZ D 0
0BX C 0.15BZ C 0.3CX C 0CY D 15
0.15BX C 0BZ 0.33CX 0.15CY D 66
BX C 0BZ C CX C 0CZ D 200.
(The HP-28S hand held calculator was used to solve these equations.)
The solution:
BX D 750 N,
BZ D 1800 N,
CX D 950 N,
CY D 900 N.
The reactions at A are determined by the sums of forces:
FY D AY C CY j D 0, from which AY D CY D 900 N
FZ D AZ C BZ C 100k D 0, from which AZ D 1900 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 5.118 The support that attaches the sailboat’s
mast to the deck behaves like a ball and socket support.
The line that attaches the spinnaker (the sail) to the top
of the mast exerts a 200-lb force on the mast. The force
is in the horizontal plane at 15° from the centerline of the
boat. (See the top view.) The spinnaker pole exerts a 50lb force on the mast at P. The force is in the horizontal
plane at 45° from the centerline. (See the top view.)
The mast is supported by two cables, the back stay AB
and the port shroud ACD. (The fore stay AE and the
starboard shroud AFG are slack, and their tensions can
be neglected.) Determine the tensions in the cables AB
and CD and the reactions at the bottom of the mast.
A
A
Spinnaker
50 ft
C
C
F
P
P
6 ft
E
x
B
D
Side View
G
D
15 ft
21 ft
Aft View
x
z (Spinnaker not shown)
Top View
F
200 lb
G
15°
A
B
E
P
C
50
lb
45°
D
Solution: Although the dimensions are not given in the sketch,
assume that the point C is at the midpoint of the mast (25 ft above
the deck). The position vectors for the points A, B, C, D, and P are:
(5) The force due to the spinnaker pole:
rA D 50j,
The sum of the moments about the base of the mast is
FP D 500.707i C 0.707k D 35.35i C 35.35k.
MQ D rA ð FA C rA ð TAB C rA ð TAC C rC ð TCE
rB D 21i,
rP D 6j,
C rP ð FP D 0
MQ D rA ð FA C TAB C TAC C rC ð TCE C rP ð FP D 0.
rC D 25j 7.5k.
The vector parallel to the backstay AB is
From above,
FA C TAB C TAC D FTX i C FTY j C FTZ k
rAB D rB rA D 21i 50j.
D 193.2 0.3872jTAB ji C 0.922jTAB j
The unit vector parallel to backstay AB is
0.9578jTAC jj C 51.76 0.2873jTAC jk
eAB D 0.3872i 0.9220j.
The vector parallel to AC is
i
MQ D 0
FTX
i
C 0
35.35
rAC D rC rA D 25j 7.5k.
The forces acting on the mast are: (1) The force due to the spinnaker
at the top of the mast:
j
k 6
6 D 0
0 35.35 C 50FTX C 212.1k D 0.
(2) The reaction due to the backstay:
(3) The reaction due to the shroud:
k
k i j
7.5
50 C 0 25
FTZ 0 0 0.2873jTAC j D 50FTZ C 250.2873jTAC j C 212.1i
FA D 200i cos 15° C k cos 75° D 193.19i C 51.76k.
TAB D jTAB jeAB
j
50
FTY
Substituting and collecting terms:
2800 7.1829jTAC ji C 9447.9 C 19.36jTAB jk D 0,
from which
TAC D jTAC jeAC
(4) The force acting on the cross spar CE:
TCE D k Ð TAC k D 0.2873jTAC jk.
jTAC j D
2800
D 389.81 lb,
7.1829
jTAB j D 488.0 lb.
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1
5.118 (Continued )
The tension in cable CD is the vertical component of the tension
in AC,
jTCD j D jTAC jj Ð eAC D jTAC j0.9578 D 373.37 lb.
The reaction at the base is found from the sums of the forces:
FX D QX C 193.19 35.35 jTAB j0.3872 D 0,
from which QX D 31.11 lb
FY D QY 0.922jTAB j 0.9578jTAC jj D 0,
from which QY D 823.24 lb
FZ D QZ C 51.76 C 0.2873jTAC j
0.2873jTAC j C 35.35k D 0,
from which QZ D 87.11 lb
Collecting the terms, the reaction is
Q D 31.14i C 823.26j 87.12k (lb)
y
y
FA
TAC
FA
TCD TCE
TAB
z
FP
QX
QY
TCD
FP
QY
QB
z
SIDE VIEW z AFT VIEW
FA
Q
B
x
2
QX
FP TAB
TOP VIEW
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y
Problem 5.119* The bar AC is supported by the cable
BD and a bearing at A that can rotate about the axis AE.
The person exerts a force F D 50j (N) at C. Determine
the tension in the cable.
(0.3, 0.5, 0) m
E
Strategy: Use the fact that the sum of the moments
about the axis AE due to the forces acting on the freebody diagram of the bar must equal zero.
C
(0.82, 0.60, 0.40) m
(0.3, 0.4, 0.3) m
A
B (0.46, 0.46, 0.33) m
x
Solution: We will take moments about the line AE in order to
eliminate all of the reactions at the bearing A. We have:
eAE D
0.1j 0.3k
p
D 0.316j 0.949k
0.1
z
D
(0.7, 0, 0.5) m
rAB D 0.16i C 0.06j C 0.03km,
TBD D TBD
0.24i 0.46j C 0.17k
p
0.2981
rAC D 0.52i C 0.2j C 0.1km,
F D 50jN
Then the equilibrium equation is
MAE D eAE Ð rAB ð TBD C rAC ð F D 0
This reduces to the single scalar equation
TBD D 174.5 N
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1
Problem 5.120* In Problem 5.119, determine the
reactions at the bearing A.
Solution: See the previous problem for setup. We add the reactions
Strategy: Write the couple exerted on the free-body
diagram of the bar by the bearing as MA D MAx i C
MAy j C MAz k. Then, in addition to the equilibrium
equations, obtain an equation by requiring the
component of MA parallel to the axis AE to equal zero.
A D Ax i C Ay j C Az k, MA D MAx i C MAy j C MAz k
(force, moment)
This gives us too many reaction moments. We will add the constrain
that
MA Ð eAE D 0
We have the following 6 equilibrium equations:


Fx : Ax C 0.440TBD D 0



F D A C TBD C F D 0 )
Fy : Ay 0.843TBD C 50 N D 0




Fx : Az C 0.311TBD D 0
MA D MA C rAB ð TBD C rAC ð F D 0


M : MAx 5 N-m C 0.0440 mTBD D 0

 Ax

)
MAy : MAy 0.0366 mTBD D 0




MAz : MAz C 26 N-m 0.161 mTBD D 0
eAE Ð MA D 0
) 0.316MAy 0.949MAz D 0
Solving these 7 equations we find
Ax D 76.7 N, Ay D 97.0 N, Az D 54.3 N
MAx D 2.67 N-m, MAy D 6.39 N-m, MAz D 2.13 N-m
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1
Problem 5.121 The horizontal bar has a mass of 10 kg.
Its weight acts at the midpoint of the bar, and it is
supported by a roller support at A and the cable BC.
Use the fact that the bar is a three-force member to
determine the angle ˛, the tension in the cable BC, and
the magnitude of the reaction at A.
C
A
B
α
30°
2m
Solution: The roller support at A and the cable support at B are
Combining:
one-force supports. The reaction at A is
A D Ai cos 60° C j sin 60° D A0.5i C 0.866j.
The reaction at B is
B D Bi cos ˛ C j sin ˛.
tan ˛ D
49.05
D 1.732,
28.32
from which
˛ D 120°
The sum of the moments about B is
or ˛ D 300° .
Since
MB D CW1 A0.8662 D 0,
from which
AD
9.8110
W
D
D 56.64 N
0.8662
1.732
The sums of the forces:
cos ˛ 0
and sin ˛ ½ 0,
the angle is in the second quadrant, hence ˛ D 120° , and B D
56.64 N
FX D A0.5 C Bcos ˛ D 0,
from which
B cos ˛ D 56.640.5 D 28.32.
49.05
D
sin ˛
C
2m
A
α
B
30°
α
FY D A0.866 W C B sin ˛ D 0,
60°
W
from which
B sin ˛ D 98.1 49.05 D 49.05.
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1
Problem 5.122 The horizontal bar is of negligible
weight. Use the fact that the bar is a three-force member
to determine the angle ˛ necessary for equilibrium.
F
α
30°
4m
9m
Solution: When the action lines of the reactions meet at a point,
and the force does not produce a moment about that point, the system
is in equilibrium. This situation occurs when all three action lines meet
at the point P. Construct the two triangles shown. The hypotenuse of
the left triangle is
hD
A
F
P
30°
9
α
F
4
D 8.
cos 60°
The vertical distance to the point P is D D
angle ˛ is:
90° ˛ D tan1
4
6.9282
9
p
60°
90 − α
82 42 D 6.9282. The
,
from which ˛ D 90° 37.589° D 52.4°
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1
Problem 5.123 The suspended load weighs 1000 lb.
If you neglect its weight, the structure is a three-force
member. Use this fact to determine the magnitudes of
the reactions at A and B.
A
5 ft
B
10 ft
Solution: The pin support at A is a two-force reaction, and the
roller support at B is a one force reaction. The moment about A is
MA D 5B 101000 D 0, from which the magnitude at B is B D
2000 lb. The sums of the forces:
A
5 ft
FX D AX C B D AX C 2000 D 0, from which AX D 2000 lb.
B
10 ft
FY D AY 1000 D 0, from which AY D 1000 lb.
The magnitude at A is A D
p
1000 lb
20002 C 10002 D 2236 lb
AX
AY
5 ft
B
1000 lb
10 ft
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1
Problem 5.124 The weight W D 50 lb acts at the
center of the disk. Use the fact that the disk is a threeforce member to determine the tension in the cable and
the magnitude of the reaction at the pin support.
60°
Solution: Denote the magnitude of the reaction at the pinned joint
by B. The sums of the forces are:
and
W
FX D BX T sin 60° D 0,
FY D BY C T cos 60° W D 0.
The perpendicular distance to the action line of the tension from the
center of the disk is the radius R. The sum of the moments about
the center of the disk is MC D RBY C RT D 0, from which BY D T.
Substitute into the sum of the forces to obtain: T C T0.5 W D 0,
from which
TD
60°
W
2
W D 33.33 lb.
3
Substitute into the sum of forces to obtain
60°
BX D T sin 60° D 28.86 lb.
The magnitude of the reaction at the pinned joint is
BD
p
33.332 C 28.862 D 44.1 lb
T
R
BX
W
BY
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1
Problem 5.125 The weight W D 40 N acts at the
center of the disk. The surfaces are rough. What force
F is necessary to lift the disk off the floor?
F
150 mm
W
50 mm
Solution: The reaction at the obstacle acts through the center of
the disk (see sketch) Denote the contact point by B. When the moment
is zero about the point B, the disk is at the verge of leaving the floor,
hence the force at this condition is the force required to lift the disk.
The perpendicular distance from B to the action line of the weight is
d D R cos ˛, where ˛ is given by (see sketch)
˛ D sin1
Rh
R
D sin1
150 50
150
F
150 mm
50 mm
W
D 41.81° .
α
The perpendicular distance to the action line of the force is
F
D D 2R h D 300 50 D 250 mm.
The sum of the moments about the contact point is
W
MB D R cos ˛W C 2R hF D 0,
from which F D
150 cos 41.81° W
D 0.4472W D 17.88 N
250
h
b
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1
Problem 5.126 Use the fact that the horizontal bar is
a three-force member to determine the angle ˛ and the
magnitudes of the reactions at A and B. Assume that
0 ˛ 90° .
2m
a
3 kN
60⬚ B
A
1m
30⬚
Solution: The forces at A and B are parallel to the respective bars
since these bars are 2-force members. Since the horizontal bar is a
3-force member, all of the forces must intersect at a point. Thus we
have the following picture:
From geometry we see that
d D 1 m cos 30°
d sin 30° D e sin ˛
d cos 30° C e cos ˛ D 3 m
Solving we find
˛ D 10.89°
To find the other forces we look at the force triangle
FB D 3 kN cos 40.89° D 2.27 kN
FA D 3 kN sin 40.89° D 1.964 kN
FA
e
d
α
60°
30°
α
30°
2m
3 kN
1m
FB
3 kN
40.89°
FA
FB
90°
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1
Problem 5.127 The suspended load weighs 600 lb.
Use the fact that ABC is a three-force member to
determine the magnitudes of the reactions at A and B.
3 ft
B
4.5 ft
30⬚
C
45⬚
A
Solution: All of the forces must intersect at a point.
From geometry
tan D
3 ft
D 0.435
3 C 4.5 cos 30° ft
)
D 23.5°
Now using the force triangle we find
FB D 600 lb cot D 1379 lb
FA D 600 lb csc D 1504 lb
FB
θ
FA
600 lb
FB
θ
600 lb
FA
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1
Problem 5.128
member?
(a) Is the L-shaped bar a three-force
2 kN
(b) Determine the magnitudes of the reactions at A and B.
(c) Are the three forces acting on the L-shaped bar
concurrent?
3 kN-m
B
300 mm
150 mm
700 mm
A
250
mm
500
mm
Solution: (a) No. The reaction at B is one-force, and the reaction
at A is two-force. The couple keeps the L-shaped bar from being a three
force member.(b) The angle of the member at B with the horizontal is
˛ D tan1
150
250
D 30.96° .
The sum of the moments about A is
MA D 3 0.52 C 0.7B cos ˛ D 0,
from which B D 6.6637 kN. The sum of forces:
FX D AX C B cos ˛ D 0,
from which AX D 5.7143 kN.
FY D AY B sin ˛ 2 D 0,
from which AY D 5.4281 kN. The magnitude at A:
AD
p
5.712 C 5.432 D 7.88 kN (c) No, by inspection.
α
0.5
m
2 kN
3 kN-m
0.3 m
B
0.7 m
Ay
Ax
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1
Problem 5.129 The bucket of the excavator is
supported by the two-force member AB and the pin
support at C. Its weight is W D 1500 lb. What are the
reactions at C?
14 in.
16 in.
B
A
4 in.
C
W
8 in.
8 in.
Solution: The angle of the member AB relative to the positive x
axis is
˛ D tan1
12
14
D 40.6° .
The moment about the point C is
MC D 4A cos 40.6° C 16A sin 40.6° C 8W D 0,
from which A D 0.5948W D 892.23 lb. The sum of forces:
FX D Cx A cos 40.6° D 0,
from which: CX D 677.4 lb
FY D CY A sin 40.6° W D 0,
from which CY D 919.4 lb
A
α
4 in
CY
CX
W
8 in 8 in
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1
Problem 5.130 The member ACG of the front-end
loader is subjected to a load W D 2 kN and is supported
by a pin support at A and the hydraulic cylinder BC.
Treat the hydraulic cylinder as a two-force member.
A
0.75 m
B
(a)
(b)
Draw a free-body diagrams of the hydraulic
cylinder and the member ACG.
Determine the reactions on the member ACG.
C
1m
G
0.5 m
W
1.5 m
Solution: This is a very simple Problem. The free body diagrams
are shown at the right. From the free body diagram of the hydraulic
cylinder, we get the equation BX C CX D 0. This will enable us to find
BX once the loads on member ACG are known. From the diagram of
ACG, the equilibrium equations are
and
1.5 m
CX
BX
B
AX
Fx D AX C CX D 0,
0.75 m
Fy D AY W D 0,
1m
AY
CX
0.5 m
MA D 0.75CX 3W D 0.
1.5 m
1.5 m
W
Using the given value for W and solving these equations, we get
AX D 8 kN,
AY D 2 kN,
CX D 8 kN,
and BX D 8 kN.
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1
Problem 5.131 In Problem 5.130, determine the reactions of the member ACG by using the fact that it is a
three-force member.
Solution: The easiest way to do this is take advantage of the fact
that for a three force member, the three forces must be concurrent. The
fact that the force at C is horizontal and the weight is vertical make
it very easy to find the point of concurrency. We then use this point
to determine the direction of the force through A. We can even know
which direction this force must take along its line — it must have an
upward component to support the weight — which is down. From the
geometry, we can determine the angle between the force A and the
horizontal.
y
A
A
0.75 m
1m
θ
CX C
1.5 m
x
1.5 m
G
W = 2 kN
tan D 0.75/3,
or D 14.04° .
Using this, we can write force equilibrium equations in the form
Fx D A cos C CX D 0, and
Fy D A sin W D 0.
Solving these equations, we get A D 8.246 kN, and CX D 8 kN. The
components of A are as calculated in Problem 5.130.
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1
Problem 5.132 A rectangular plate is subjected to
two forces A and B (Fig. a). In Fig. b, the two forces
are resolved into components. By writing equilibrium
equations in terms of the components Ax , Ay , Bx , and By ,
show that the two forces A and B are equal in magnitude,
opposite in direction, and directed along the line between
their points of application.
B
B
A
h
A
b
(a)
y
By
Bx
B
h
Ay
Ax
A
x
b
(b)
Solution: The sum of forces:
b
B
FX D AX C BX D 0,
h
A
from which AX D BX
By
y
FY D AY C BY D 0,
Ay
from which AY D By . These last two equations show that A and B
are equal and opposite in direction, (if the components are equal and
opposite, the vectors are equal and opposite). To show that the two
vectors act along the line connecting the two points, determine the
angle of the vectors relative to the positive x axis. The sum of the
moments about A is
Fig a
Ax
Bx
Fig b
x
MA D Bx h bBy D 0,
from which the angle of direction of B is
tan1
BY
BX
D tan1
h
D ˛B .
b
or 180 C ˛B . Similarly, by substituting A:
tan1
AY
AX
D tan1
h
D ˛A ,
b
or 180 C ˛A . But
˛ D tan1
h
b
describes direction of the line from A to B. The two vectors are opposite
in direction, therefore the angles of direction of the vectors is one of
two possibilities: B is directed along the line from A to B, and A is
directed along the same line, oppositely to B.
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1
Problem 5.133 An object in equilibrium is subjected
to three forces whose points of application lie on a
straight line. Prove that the forces are coplanar.
F2
F3
Solution: The strategy is to show that for a system in equilibrium
under the action of forces alone, any two of the forces must lie in
the same plane, hence all three must be in the same plane, since the
choice of the two was arbitrary. Let P be a point in a plane containing
the straight line and one of the forces, say F2 . Let L also be a line,
not parallel to the straight line, lying in the same plane as F2 , passing
through P. Let e be a vector parallel to this line L. First we show
that the sum of the moments about any point in the plane is equal to
the sum of the moments about one of the points of application of the
forces. The sum of the moments about the point P:
F1
F2
F3
F1
P
L
M D r1 ð F1 C r2 ð F2 C r3 ð F3 D 0,
where the vectors are the position vectors of the points of the application of the forces relative to the point P. (The position vectors lie in
the plane.) Define
d12 D r2 r1 ,
and d13 D r3 r1 .
Then the sum of the moments can be rewritten,
M D r1 ð F1 C F2 C F3 C d12 ð F2 C d13 ð F3 D 0.
Since the system is in equilibrium,
F1 C F2 C F3 D 0,
and the sum of moments reduces to
M D d12 ð F2 C d13 ð F3 D 0,
which is the moment about the point of application of F1 . (The vectors
d12 , d13 are parallel to the line L.) The component of the moment
parallel to the line L is
e Ð d12 ð F2 e C e Ð d13 ð F3 e D 0,
or F2 Ð d12 ð ee C F3 Ð d13 ð ee D 0.
But by definition, F2 lies in the same plane as the line L, hence it is
normal to the cross product d12 ð e 6D 0, and the term
F2 Ð d12 ð e D 0.
But this means that
F3 Ð d13 ð ee D 0,
which implies that F3 also lies in the same plane as F2 , since
d13 ð e 6D 0.
Thus the two forces lie in the same plane. Since the choice of the point
about which to sum the moments was arbitrary, this process can be
repeated to show that F1 lies in the same plane as F2 . Thus all forces
lie in the same plane.
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1
Problem 5.134 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 value of the angle ˛ for which the tension
in the string BC is 6 lb. What are the resulting reactions
at A?
C
B
3 ft
A
a
1 ft
Solution: From geometry and equilibrium we have
tan ˇ D
3 ft1 sin ˛
1 C 3 cos ˛ ft
MA : 6 lb cos ˇ3 ft sin ˛ C 6 lb sin ˇ3 ft cos ˛
10 lb1.5 ft cos ˛ D 0
Fx : Ax 6 lb cos ˇ D 0;
Fy : Ay 10 lb C 6 lb sin ˇ D 0
These equations are transcendental and need to be solved using an
equation solver in Mathematica or Matlab or similar program.
˛ D 24.1° ,
Ax D 5.42 lb,
Ay D 7.43 lb
6 lb
β
α
Ax
10 lb
Ay
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1
Problem 5.135 The mass of the bar is 36 kg and its
weight acts at its midpoint. The spring is unstretched
when ˛ D 0, and the spring constant is k D 200 N/m.
Determine the values of ˛ in the range 0 ˛ 90° at
which the bar is in equilibrium.
k
4m
α
2m
C
Solution:
l2BC D 22 C 42 224 cos ˛
T
β
sin ˛/lBC D sin ˇ/2
B
α
FX D AX C T sin ˇ D 0
4m
FY D AY C T cos ˇ W D 0
2m
MA D W sin ˛k C rAB ð T D 0
where
W = mg
AX
rAB D 2 sin ˛i C 2 cos ˛j
AY
T D T sin ˇi C T cos ˇj
k:
MA D W sin ˛ 2T sin ˛ cos ˇ 2T cos ˛ sin ˇ D 0
Solving, we get two roots




˛ D 0°
˛ D 33.0°
 T D 0  and  T D 113.3 N 
υD0
υ D 0.566 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
C
Problem 5.136 The unstretched length of the spring
CD is 350 mm and the spring constant is k D 3400 N/m.
Suppose that you want to choose the dimension h so
that the lever ABC exerts a 280-N normal force on the
smooth surface at A. Determine the dimension h and the
resulting reactions at B.
k
h
D
450
mm
20⬚
180
mm
B
A
330 mm
300 mm
Solution: We have the following equations
F D 3400 N/m
h2 C 0.3 m2 0.35 m
MB : 280 N cos 20° 0.18 m
C 280 N sin 20° 0.33 m
0.3 m
h2 C 0.3 m2
F0.45 m D 0
0.3 m
F C 280 N cos 20° D 0
Fx : Bx C h2 C 0.3 m2
Fx : By h
h2 C 0.3 m2
F 280 N sin 20° D 0
Solving we find
h D 0.298 m, Bx D 439 N, By D 270 N
F
h
0.3 m
Bx
280 N
By
20°
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1
Problem 5.137 Consider the system shown in
Problem 5.62. The bar is 1 m long, and its weight W D
35 N acts at its midpoint. The distance b D 0.75 m. The
spring constant is k D 100 N/m, and the spring is
unstretched when the bar is vertical. Determine the angle
˛ and the reactions at A.
Solution: The unstretched length of the spring is
LD
p
b2 C 12 D 1.25 m.
The obtuse angle is 90 C ˛, so the stretched length can be determined
from the cosine law:
α
W
L22 D 12 C 0.752 20.75 cos90 C ˛
A
D 1.5625 C 1.5 sin ˛.
b
The force exerted by the spring is T D kL D 100L2 1.25 N. The
angle between the spring and the bar can be determined from the sine
law:
L2
b
D
,
sin ˇ
sin90 C ˛
f(A) vs A
.3
.2
.1
b cos ˛
.
from which sin ˇ D
L2
The angle the spring makes with the horizontal is
D 180 ˇ 90 ˛ D 90 ˇ ˛.
The sum of the forces:
f 0
(
−.1
h
) −.2
−.3
42
43
44
45
Angle, deg
46
FX D AX T cos D 0,
from which AX D T cos N.
FY D AY W T sin D 0,
from which AY D W C T sin . The sum of the moments about A is
MA D T sin ˇ W
2
sin ˛ D 0.
The function
f˛ D T sin ˇ W
2
sin ˛
is to be graphed against ˛ to determine the value of ˛ at the zero
crossing. The commercial package TK Solver Plus was used to graph
the function. At the zero crossing ˛ D 44.1° .
AX D 32.6 N,
and AY D 51.2 N
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1
Problem 5.138 The hydraulic actuator BC exerts a
force at C that points along the line from B to C. Treat
A as a pin support. The mass of the suspended load is
4000 kg. If the actuator BC can exert a maximum force
of 90 kN, what is the smallest permissible value of ˛?
2m
3m
C
a
A
B
2m
Solution: Note > 90° , ∴ most adjust from law of sines
FX D AX C FBC cos180° D 0
2m
D
C
mg
FY D AY C FBC sin180° W D 0
3m
AY
y
α
MA D rAC ð FBC C rAD ð W D 0
AX
where rAC D 3 cos ˛i C 3 sin ˛j m
FBC
B
2m
C
rAD D 5 cos ˛i C 5 sin ˛j m
3m
FBC D FBC cos180° i C FBC sin180° j
α
W D 40009.81j N
or k :
MA D 3FBC cos ˛ sin180 x
A
2m
γ
α
γ
γ 180 –
B
(BC)2 = 32 + 22 – 2.2.3 cos α
sin α = sin γ
(BC)
3
3 sin ˛FBC cos180° C 5cos ˛W D 0
Set FBC D 90000 N
Solving-we get
AX D 31.63 kN
AY D 45.02 kN
˛min D 30.8°
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1
Problem 5.139 The beam is in equilibrium in the position shown. Each spring has an unstretched length of
1 m. Determine the distance b and the reactions at A.
200 N/m
1m
Solution: The strategy is to determine the value of b that will
cause the moment about A to equal zero. The angle of the top spring
is
A
1m
400 N/m
˛ D tan1 13 D 18.435° .
b
The angle of the bottom spring is
ˇ D tan1
1
.
b
3m
The stretched length of the top spring is
LT D
p
12 C 32 D 3.1623 m.
α
AX
β
AY
TT
TB
The tension in the top spring is
TT D 2003.1623 1 D 432.456 N.
The tension in the bottom spring is
TB D 400
p
b2 C 1 1 N.
The sum of the moments about the point A:
MA D bTB sin ˇ C 3432.456 sin 18.435°
D bT sin ˇ C 410.26 D 0.
The sum of the forces:
FX D AX TB cos ˇ 432.456 cos 18.435 D 0,
f(b) vs b
20
15
10
5
f 0
(
b −5
) −10
−15
−20
1.8
1.85
1.9
1.95
b, m
2
2.85
from which AX D TB cos ˇ 410.264 N.
FY D AY TB sin ˇ C 432.456 sin 18.435° D 0,
from which AY D TB sin ˇ 136.755 N.
The solution is obtained by graphing the sum of moments equation,
fb D bTB sin ˇ C 410.26,
against b to determine the value of b at the zero crossing.
The commercial package TK Solver Plus was used to graph the function.
At the zero crossing the value of b is b D 1.91 N The reactions at
A are:
AX D 820 N,
and AY D 77.8 N
The values of AX , AY are changing rapidly in the neighborhood of
the zero crossing, so that the results are good only to three significant
figures, at best.
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1
Problem 5.140
(a)
The suspended cable weighs 12 lb.
B
Draw the free-body diagram of the cable. (The
tensions in the cable at A and B are not equal.)
Determine the tensions in the cable at A and B.
What is the tension in the cable at its lowest point?
(b)
(c)
50⬚
A
32⬚
Solution:
(a)
(b)
The FBD
The equilibrium equations
Fx : TA cos 32° C TB cos 50° D 0
Fy : TA sin 32° C TB sin 50 12 lb D 0
Solving we find
TA D 7.79 lb, TB D 10.28 lb
(c)
Consider the FBD where W represents only a portion of the total
weight. We have
Fx : TA cos 32° C T D 0
Solving
T D 6.61 lb
TB
50°
TA
32°
12 lb
TA
32°
T
W
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1
Problem 5.141
support.
Determine the reactions at the fixed
4 kN
3m
A
20 kN-m
2 kN
5m
3 kN
3m
Solution: The equilibrium equations
4 kN
Fx : Ax C 4 kN D 0
Fy : Ay 2 kN 3 kN D 0
Ax
20 kN-m
MA : MA 2 kN5 m 4 kN3 m
3 kN8 m C 20 kN-m D 0
Solving
MA
Ay
2 kN
3 kN
Ax D 4 kN, Ay D 5 kN, MA D 26 kN-m
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1
Problem 5.142 (a) Draw the free-body diagram of the
50-lb plate, and explain why it is statically indeterminate.
y
(b) Determine as many of the reactions at A and B as
possible.
A
12 in
Solution:
(a)
(b)
The pin supports at A and B are two-force supports, thus there
are four unknown reactions AX , AY , BX , and BY , but only three
equilibrium equations can be written, two for the forces, and one
for the moment. Thus there are four unknowns and only three
equations, so the system is indeterminate.
B
x
20 in
50 lb
Sums the forces:
A
FX D AX C BX D 0,
or AX D BX , and
8 in
12 in
8 in
B
20 in
FY D AY C BY 50 D 0.
The sum of the moments about B
50 lb
x
AY
AX
MB D 20AX 5020 D 0,
from which AX D 50 lb,
and from the sum of forces BX D 50 lb.
12 in.
BY
8 in.
BX
50 lb
x
20 in.
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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 5.143 The mass of the truck is 4 Mg. Its
wheels are locked, and the tension in its cable is T D
10 kN.
(003) 676-5942
(a)
(b)
Draw the free-body diagram of the truck.
Determine the normal forces exerted on the truck’s
wheels by the road.
30°
AL's
Tow i n g
2m
T
2.5 m
3m
2.2 m
mg
Solution: The weight is 40009.81 D 39.24 kN. The sum of the
moments about B
MB D 3T sin 30° 2.2T cos 30° C 2.5W 4.5AN D 0
from which
AN D
D
2.5W T3 sin 30° C 2.2 cos 30° 4.5
64.047
D 14.23 N
4.5
The sum of the forces:
FY D AN W C BN T cos 30° D 0,
from which BN D T cos 30° AN C W D 33.67 N
30°
AX
A
AN
W
BX
T
B
BN
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1
Problem 5.144 Assume that the force exerted on the
head of the nail by the hammer is vertical, and neglect
the hammer’s weight.
(a)
(b)
Draw the free-body diagram of the hammer.
If F D 10 lb, what are the magnitudes of the forces
exerted on the nail by the hammer and the normal
and friction forces exerted on the floor by the
hammer?
F
11 in.
65°
2 in.
Solution: Denote the point of contact with the floor by B. The
perpendicular distance from B to the line of action of the force is 11
in. The sum of the moments about B is MB D 11F 2FN D 0, from
11F
D 5.5F. The
which the force exerted by the nail head is FN D
2
sum of the forces:
FX D F cos 25 C Hx D 0,
from which the friction force exerted on the hammer is HX D 0.9063F.
FY D NH FN C F sin 25° D 0,
from which the normal force exerted by the floor on the hammer is
NH D 5.077F
If the force on the handle is
F D 10 lb,
then FN D 55 lb,
HX D 9.063 lb,
and NH D 50.77 lb
F
11 in.
65°
2 in.
F
11 in.
65°
HX
B NH
FN
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1
Problem 5.145 The spring constant is k D 9600 N/m
and the unstretched length of the spring is 30 mm. Treat
the bolt at A as a pin support and assume that the surface
at C is smooth. Determine the reactions at A and the
normal force at C.
A
24 mm
B
15 mm
30 mm
30°
C
k
Solution: The length of the spring is
lD
p
p
302 C 302 mm D 1800 mm
30 mm
AY
l D 42.4 mm D 0.0424 m
The spring force is kυ where υ D l l0 . l0 is give as 30 mm. (We
must be careful because the units for k are given as N/m) We need to
use length units as all mm or all meters). k is given as 9600 N/m. Let
us use l0 D 0.0300 m and l D 0.0424 m
AX
24 mm
B
kδ
Equilibrium equations:
FX D 0:
AX kl l0 sin 45°
NC cos 60° D 0
FY D 0:
50 mm
15 mm
30 mm
45°
60°
50 mm
30 mm
30°
C
NC
AY kl l0 cos 45°
Solving, we get
C NC sin 60° D 0
AX D 126.7 N
MB D 0:
0.024AX C 0.050NC sin 60° AY D 10.5 N
0.015NC cos 60° D 0
NC D 85.1 N
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1
Problem 5.146 The engineer designing the release
mechanism shown in Problem 5.145 wants the normal
force exerted at C to be 120 N. If the unstretched length
of the spring is 30 mm, what is the necessary value of
the spring constant k?
Solution: Refer to the solution of Problem 5.145. The equilibrium
equations derived were
FX D 0:
AX kl l0 sin 45 NC cos 60° D 0
FY D 0:
AY kl l0 cos 45 C NC sin 60° D 0
MB D 0:
0.024AX C 0.050NC sin 60°
0.015NC cos 60° D 0
where l D 0.0424 m, l0 D 0.030 m, NC D 120 N, and AX , AY , and k
are unknowns.
Solving, we get
AX D 179.0 N,
AY D 15.1 N,
k D 13500 N/m
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1
Problem 5.147 The truss supports a 90-kg suspended
object. What are the reactions at the supports A and B?
400 mm
700 mm
300 mm
B
A
Solution: Treat the truss as a single element. The pin support at
A is a two force reaction support; the roller support at B is a single
force reaction. The sum of the moments about A is
MA D B400 W1100 D 0,
from which B D
1100W
D 2.75W
400
B D 2.75909.81 D 2427.975 D 2.43 kN.
The sum of the forces:
FX D AX D 0
FY D AY C B W D 0,
from which AY D W B D 882.9 2427.975 D 1.545 kN
AX
W
B
AY
400 mm
700 mm
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1
Problem 5.148 The trailer is parked on a 15° slope.
Its wheels are free to turn. The hitch H behaves like a
pin support. Determine the reactions at A and H.
y
H
1.4 ft
x
870 lb
1.6 ft
A
8 ft
2.8 ft
15°
Solution: The coordinate system has the x axis parallel to the road.
The wheels are a one force reaction normal to the road, the pin H is
a two force reaction. The position vectors of the points of the center
of mass and H are:
rW D 1.4i C 2.8j ft and
rH D 8i C 1.6j.
The angle of the weight vector realtive to the positive x axis is
˛ D 270° 15° D 255° .
The weight has the components
W D Wi cos 255° C j sin 255° D 8700.2588i 0.9659j
D 225.173i 840.355j (lb).
The sum of the moments about H is
MH D rW rH ð W C rA rH ð A,
i
MH D 6.6
225.355
j
1.2
840.355
j
k k i
0 C 8 1.6 0 D 0
AY
0 0
0
D 5816.55 8AY D 0,
from which AY D
5816.55
D 727.1 lb.
8
The sum of the forces is
FX D HX 225.173i D 0, from which HX D 225.2 lb,
FY D AY C HY 840.355j D 0, from which HY D 113.3 lb
1.4 ft
1.2 ft
W
15°
AY
HX
HY
8 ft
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1
Problem 5.149 To determine the location of the point
where the weight of a car acts (the center of mass),
an engineer places the car on scales and measures the
normal reactions at the wheels for two values of ˛,
obtaining the following results:
y
h
x
B
W
˛
Ay (kN)
B (kN)
10°
10.134
4.357
20°
10.150
3.677
α
Ax
Ay
b
2.7 m
What are the distances b and h?
Solution: The position vectors of the cm and the point B are
These two simultaneous equations in two unknowns were solved using
the HP-28S hand held calculator.
rCM D 2.7 bi C hj,
b D 1.80 m,
rB D 2.7i.
h D 0.50 m
The angle between the weight and the positive x axis is ˇ D 270 ˛.
The weight vector at each of the two angles is
W10 D Wi cos 260° C j sin 260° W10 D W0.1736i 0.9848j
W20 D Wi cos 250° C j sin 250° or
W20 D W0.3420i 0.9397j
The weight W is found from the sum of forces:
FY D AY C BY C W sin ˇ D 0,
from which Wˇ D
AY C BY
.
sin ˇ
Taking the values from the table of measurements:
W10 D 10.134 C 4.357
D 14.714 kN,
sin 260°
[check :W20 D 10.150 C 3.677
D 14.714 kN check ]
sin 250°
The moments about A are
MA D rCM ð W C rB ð B D 0.
Taking the values at the two angles:
i
M10
A D 2.7 b
2.5551
j
h
14.4910
j
k i
0
0 C 2.7
0 0 4.357
k 0 D 0
0
D 14.4903b C 2.5551h 27.3618 D 0
i
M20
A D 2.7 b
5.0327
j
h
13.8272
k i
j
0 C 2.7
0
0 0 3.677
k 0 0
D 013.8272b C 5.0327h 27.4054 D 0
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1
Problem 5.150 The bar is attached by pin supports to
collars that slide on the two fixed bars. Its mass is 10 kg,
it is 1 m in length, and its weight acts at its midpoint.
Neglect friction and the masses of the collars. The spring
is unstretched when the bar is vertical (˛ D 0), and the
spring constant is k D 100 N/m. Determine the values
of ˛ in the range 0 ˛ 60° at which the bar is in
equilibrium.
k
α
Solution: The force exerted by the spring is given by FS D kL L cos ˛. The equations of equilibrium, from the free body diagram, are
and
Spring Constant (K) in N/m vs Alpha (deg)
90000
80000
70000
Fx D NB D 0,
Fy D FS C NA mg D 0,
MB D L sin ˛NA C
L
sin ˛ mg D 0.
2
60000
K
= 50000
N
− 40000
m
30000
20000
10000
These equations can be solved directly with most numerical solvers and
the required plot can be developed. The plot over the given ˛ range is
shown at the left and a zoom-in is given at the right. The solution and
the plot were developed with the TK Solver Plus commercial software
package. From the plot, the required equilibrium value is ˛ ¾
D 59.4° .
0
0
10
20
30
40
50
60
Alpha (deg)
Spring Constant (K) in N/m vs Alpha (deg)
116
114
FS
112
B
y
NB
α
mg
110
K
= 108
N
− 106
m
104
102
A
x
NA
100
98
55
55.5
56
56.5
57
57.5
58
58.5
59
59.5
60
Alpha (deg)
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1
Problem 5.151 The 450-lb ladder is supported by the
hydraulic cylinder AB and the pin support at C. The reaction at B is parallel to the hydraulic cylinder. Determine
the reactions on the ladder.
6 ft
450 lb
3 ft
A
B
C
6 ft
3 ft
Solution: Setting the coordinate origin at C, point A is located at
(9, 0) ft and point B is at (3, 3) ft. The angle ˛ D 45° and point G
is located at (6, 6) ft. The unit vector along the hydraulic cylinder,
AC, is
G
y
6 ft
W = 450 lb
eAB D 0.894i C 0.447j
B
FH
and
FHX D 0.894FH ,
α
A
6 ft
FHY D 0.447FH .
3 ft
x
C
cX
cY
The equations of equilibrium are:
and
Fx D CX C FHX D 0,
Fy D CY C FHY 450 D 0,
MC D 6450 3FHX 3FHY D 0.
All distances are in ft, forces are in lb, and moments in ft-lb. Solving
the equations for the unknown support reactions yields
CX D 600 lb,
CY D 150 lb,
and FH D 671 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 5.152 Consider the crane shown in
Problem 5.138. The hydraulic actuator BC exerts a force
at C that points along the line from B to C. Treat A as a
pin support. The mass of the suspended load is 6000 kg.
If the angle ˛ D 35° , what are the reactions at A?
2m
3m
C
a
A
B
2m
y
D
2m
Solution:
mg
BC2 D 32 C 22 223 cos 35°
C
3m
BC D 1.78 m
AY
sin 35°
sin D
> 90°
BC
3
A
FBC
α = 35°
x
AX
D 104.9°
C
W = mg
= (6000) (9.81)
W = 58860 N
180° D 75.1°
3m
FX D AX C FBC cos180° FY D AY C FBC sin180° W D 0
MA D rAC ð FBC
35°
A
2m
γ
(180° − γ )
B
C rAD ð W D 0
where
rAC D 3 cos35° i C 3 sin35° j
rAD D 5 cos35° i C 5 sin35° j
FBC D FBC cos75.11° i
C FBC sin75.11° j
W D 60009.81j N
Solving for the unknowns, we get
AX D 32.04 kN
AY D 61.68 kN
FBC D 124.73 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 5.153 The horizontal rectangular plate
weighs 800 N and is suspended by three vertical cables.
The weight of the plate acts at its midpoint. What are
the tensions in the cables?
A
C
B
2m
0.5 m
1m
Solution: Choose an origin at A, with the x axis coincident with
y
the line AB, and the y axis normal to the plate, positive upward. The
position vectors of the points B, C, and the cm are
TA
TC
1m
z
rB D 2i,
TB
2m
rC D 1.5i 1k,
0.5 m
800 N
rCM D 1i 0.5k.
x
The sum of the moments about the point A is
MA D rB ð TB C rC ð TC C rCM ð W D 0
i
j
MA D 2 0
0 TB
j
k i
0 C 1.5 0
0 0 TC
k i
j
1 C 1
0
0 0 800
k 0.5 D 0
0 MA D TC 0.5800i C 2TB C 1.5TC 800k D 0
From which TC D 400 N,
TB D
800 1.5TC
D 100 N.
2
The reaction at A is found from the sum of forces:
FY D TA C TB C TC 800 D 0,
from which TA D 800 TB TC D 300 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 5.154 Consider the suspended 800-N plate
in Problem 5.153. The weight of the plate acts at its
midpoint. If you represent the reactions exerted on the
plate by the three cables by a single equivalent force,
what is the force, and where does its line of action intersect the plate?
Solution: The equivalent force must equal the sum of the
reactions: FEQ D TA C TB C TC . From the solution to Problem 5.153,
FEQ D 300 C 100 C 400 D 800 N. The moment due to the action of
the equivalent force must equal the moment due to the reactions: The
moment about A is
i
j
MA D 2 0
0 100
k i
0 C 1.5
0 0
j
0
400
j
k i
0
1 D x
0 0 800
k z 0
MA D 400i C 800k D 800zi C 800xk,
from which z D 0.5 m, and x D 1 m, which corresponds to the
midpoint of the plate. Thus the equivalent force acts upward at the
midpoint of the plate.
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1
y
Problem 5.155 The 20-kg mass is suspended by cables
attached to three vertical 2-m posts. Point A is at (1,
1.2, 0) m. Determine the reactions at the built-in support
at E.
C
B
D
A
Solution: All distances will be in meters, all forces in Newtons,
and all moments in Newton-meters. To solve the three dimensional
point equilibrium problem at A, we will need unit vectors eAB , eAC ,
and eAD . To determine these, we need the coordinates of points A, B,
C, and D. The rest of the problem will require knowing where points
E, G (under C), and H(under D) are located. From the diagram, the
required point locations are A (0, 1.2, 0), B (0.3, 2, 1), C (0, 2, 1),
D (2, 2, 0), E (0.3, 0, 1), G(0, 0, 1), and H(2, 0, 0). The required
unit vectors are calculated from the coordinates of the points of the
ends of the lines defining the vector. These are
1m
2m
0.3 m
x
z
The couple ME is the couple exerted on the post by the built in support.
Solving these equations, we get
E D 34.2i C 91.3j C 114.1k N
eAB D 0.228i C 0.608j C 0.760k,
eAC D 0i C 0.625j 0.781k,
1m
E
and ME D 228.1i C 0j C 68.44k N-m.
jME j D 238.2 N-m.
and eAD D 0.928i C 0.371j C 0k.
Also,
The force TAB in cable AB can be written as
Using a procedure identical to that followed for post EB above, we
can find the built-in support forces and moments for posts CG and
DH. The results for CG are:
TAB D TABX i C TABY j C TABZ k,
where TABX D jTAB jeABX , etc. Similar equations can be written for the
forces in AC and AD. The free body diagram of point A yields the
following three equations of equilibrium.
and
G D 0i C 91.3j 114.1k N
and MG D 228.1i C 0j C 0k N-m.
Fx D TABX C TACX C TADX D 0,
Also,
Fy D TABY C TACY C TADY W D 0,
The results for DH are:
jMG j D 228.1 N-m.
H D 34.2i C 13.7j C 0 kN
Fz D TABZ C TACZ C TADZ D 0,
and MH D 0i C 0j C 68.4k N-m.
where W D mg D 209.81 D 196.2 N. Solving the equations above
after making the substitutions related to the force components yields
the tensions in the cables. They are
jTAB j D 150 N,
Also,
jMG j D 68.4 N-m
B
jTAC j D 146 N, and
jTAD j D 36.9 N.
Now that we know the tensions in the cables, we are ready to tackle
the reactions at E (also G and H). The first step is to draw the free
body diagram of the post EB and to write the equations of equilibrium
for the post. A key point is to note that the force on the post from
cable AB is opposite in direction to the force found in the first part of
the problem. The equations of equilibrium for post EB are
Z
−TAD
C
−TAB
MG
ZM
ME
E
x
EY
D
−TAC
EX
H
GZ
HX
G
HZ
GY GX
HY
EZ
Fx D EX TABX D 0,
Fy D EY TABY D 0,
Fz D EZ TABZ D 0,
and, summing moments around the base point E,
M D ME C 2j ð TAB D 0.
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1
Problem 5.156 In Problem 5.155, the built-in support
of each vertical post will safely support a couple of
800 N-m magnitude. Based on this criterion, what is the
maximum safe value of the suspended mass?
Solution: We have all of the information necessary to solve this
problem in the solution to Problem 5.155 above. All of the force and
moment equations are linear and we know from the solution that a
20 kg mass produces a couple of magnitude 238.2 N-m at support
E and that the magnitudes of the couples at the other two supports
are smaller than this. All we need to do is scale the Problem. The
scale factor is f D 800/238.2 D 3.358 and the maximum value for
the suspended mass is mmax D 20f D 67.16 kg
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1
y
Problem 5.157 The 80-lb bar is supported by a ball
and socket support at A, the smooth wall it leans against,
and the cable BC. The weight of the bar acts at its
midpoint.
(a)
(b)
5 ft
3 ft
Draw the free-body diagram of the bar.
Determine the tension in cable BC and the reactions
at A.
B
C
4 ft
3 ft
3 ft
x
A
z
Solution: (a) The ball and socket is a three reaction force support;
the cable and the smooth wall are each one force reaction supports.
(b) The coordinates of the points A, B and C are A (3, 0, 3), B (5, 4,
0), and C(0, 4, 3).
The vector parallel to the bar is
rAB D rB rA D 2i C 4j 3k.
The length of the bar is
p
jrAB j D
22 C 42 C 32 D 5.3852.
Solve:
jTj D
80
D 23.32 lb
3.43
jBj D
120 2.058jTj
D 18.00 lb.
4
The reactions at A are found from the sums of forces:
The unit vector parallel to the bar is
eAB D 0.3714i C 0.7428j 0.5571k.
FX D AX jTj0.8575 D 0 from which AX D 20 lb
FY D AY 80 D 0, from which AY D 80 lb
FZ D AZ C jTj0.5145 C jBj D 0, from which AZ D 30 lb
The vector parallel to the cable is
rBC D rC rB D 5i C 3k.
B
The unit vector parallel to the cable is
eBC D 0.8575i C 0.5145k.
The cable tension is T D jTjeBC . The point of application of the weight
relative to A is
AY
AX
rAW D 2.6936eAB
AZ
rAW D 1.000i C 2.000j 1.500k.
The reaction at B is B D jBjk, since it is normal to a wall in the yz
plane. The sum of the moments about A is
MA D rAW ð W C rAB ð B C rAB ð T D 0
i
j
k i j k MA D 1
2
1.5 C 2 4 3 0 80
0 0 0 jBj i
C 2
0.8575
j
k 4
3 jTj D 0
0 0.5145 MA D 120 C 4jBj C 2.058jTji 2jBj 1.544jTjj
C 3.43jTj 80k D 0.
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1
Problem 5.158 The horizontal bar of weight W is
supported by a roller support at A and the cable BC. Use
the fact that the bar is a three-force member to determine
the angle ˛, the tension in the cable, and the magnitude
of the reaction at A.
C
A
B
α
W
L/2
L/2
Solution: The sum of the moments about B is
MB D LAY C
L
W D 0,
2
from which AY D
W
. The sum of the forces:
2
FX D T cos ˛ D 0,
from which T D 0 or cos ˛ D 0. The choice is made from the sum of
forces in the y-direction:
FY D AY W C T sin ˛ D 0,
W
. This equation cannot be satisfied
2
W
if T D 0, hence cos ˛ D 0, or ˛ D 90° , and T D
2
from which T sin ˛ D W AY D
T
α
AY
W
L/2
L/2
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1
Problem 6.1 Determine the axial forces in the
members of the truss and indicate whether they are in
tension (T) or compression (C).
A
10 kN
2m
B
C
1m
Solution: In order to avoid solving for reactions we start at joint
A and then work with joint C. Assuming all bars are in tension we
have the FBD for joint A shown.
1
Fx : 10 kN C p FAC D 0 ) FAC D 22.4 kN
5
2
Fy : FAB p FAC D 0 ) FAB D 20 kN
5
10 kN
A
1
2
FAC
FAB
Now work with joint C
1
Fx : p FAC FBC D 0 ) FBC D 10 kN
5
FAC
1
2
C
FBC
Cy
In summary we have
FAC D 22.4 kNC, FAB D 20 kNT, FBC D 10 kNT
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1
Problem 6.2 Determine the axial forces in the
members of the truss and indicate whether they are in
tension (T) or compression (C).
20⬚
800 N
A
0.4 m
C
B
0.7 m
0.7 m
Solution: We start at joint A
Next we move to joint C
7
7
Fx : p FAB C p FAC 800 N sin 20° D 0
65
65
4
4
Fy : p FAB p FAC 800 N cos 20° D 0
65
65
Solving we have
7
Fx : p FAC FBC D 0 ) FBC D 521 N
65
FAC
7
4
FAB D 915 N, FAC D 600 N
C
20°
FCB
800 N
Cy
A
4
7
FAB
In summary we have
4
7
FAB D 915 NC, FAC D 600 NC, FBC D 521 NT
FAC
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1
A
Problem 6.3 Member AB of the truss is subjected to a
1000-lb tensile force. Determine the weight W and the
axial force in member AC.
60 in
W
B
C
60 in
60 in
Solution: Using joint A
1
2
Fx : p 1000 lb p FAC D 0
5
2
1
1000 lb
1
1
Fy : p 1000 lb p FAC W D 0
5
2
Solving we have
FAC D 1265 lb, W D 447 lb
In summary we have
A
2
1
1
FAC
W
W D 447 lb, FAC D 1265 lbC
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1
Problem 6.4 The members of the truss are all of length
L. Determine the axial forces in the members and indicate whether they are in tension (T) or compression (C).
F
B
D
A
C
L
Solution: Start at D
Finally look at C
Fy : F FCD sin 60° D 0 ) FCD D 1.155F
Fx : FAC FBC cos 60° C FCD cos 60° D 0
) FAC D 0.289F
Fx : FBD FCD cos 60° D 0 ) FBD D 0.577F
FBC
FCD
F
FBD
60°
60°
D
FAC
60°
C
FCD
Cy
Next go to B
Fx : FAB cos 60° C FBC cos 60° C FBD D 0
In Summary we have
FAB D 0.577FT
Fy : FAB sin 60° FBC sin 60° D 0
) FAB D 0.577F, FBC D 0.577F
FAB
FBC D 0.577FC
FBD D 0.577FT
B
60°
FAC D 0.289FC
60°
FBD
FCD D 1.155FC
FBC
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1
Problem 6.5 Each suspended weight has mass m D
20 kg. Determine the axial forces in the members of
the truss and indicate whether they are in tension (T) or
compression (C).
A
0.4 m
C
B
D
m
0.32 m
0.16 m 0.16 m
Solution: Assume all bars are in tension. Start with joint D
Finally work with joint A
5
Fy : p TAD 196.2 N D 0
61
6
Fx : p TAD TCD D 0
61
5
5
Fy : p TAB C TAC p TAD D 0
29
61
) TAB D 423 N
T
A
TAD D 306 N, TCD D 235 N
Solving:
m
6
2
TAD
5
5
6
5
5
2
D
TAD
TCD
TAB
TAC
In summary:
TAB D 423 NC
196.2 N
TAC D 211 NT
TAD D 306 NT
Now work with joint C
TBC D 314 NC
5
Fy : p TAC 196.2 N D 0
29
2
Fx : p TAC TBC C TCD D 0
29
Solving:
TCD D 235 NC
TAC D 211 N, TBC D 313 N
TAC
5
2
C
TBC
TCD
196.2 N
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1
Problem 6.6 Determine the largest tensile and compressive forces that occur in the members of the truss,
and indicate the members in which they occur if
(a)
(b)
the dimension h D 0.1 m;
the dimension h D 0.5 m.
Observe how a simple change in design affects the
maximum axial loads.
B
A
h
D
0.7 m
1 kN
0.4 m
C
0.6 m
Solution: To get the force components we use equations of the
form TPQ D TPQ ePQ D TPQX i C TPQY j where P and Q take on the
designations A, B, C, and D as needed.
Equilibrium yields
y
h
B
and
0.4
m
−TAB
−TBD
TBD T
BC
D D
X
At joint A:
1.2 m
DY
−TCD
TCD
−TBC
TAB
TAC
1 kN
0.7 m
−TAC
C
Fx D TABX C TACX D 0,
A
x
CY
0.6 m
1.2 m
Fy D TABY C TACY 1 kN D 0.
At joint B:
and
Fx D TABX C TBCX C TBDX D 0,
Fy D TABY C TBCY C TBDY D 0.
At joint C:
and
and
eAB D 0.986i C 0.164j,
eAC D 0.864i 0.504j,
Fx D TBCX TACX C TCDX D 0,
eBC D 0i 1j,
Fy D TBCY TACY C TCDY C CY D 0.
eBD D 0.768i 0.640j,
At joint D:
(b) For this part of the problem, we set h D 0.5 m. The unit vectors
change because h is involved in the coordinates of point B. The new
unit vectors are
Fx D TCDX TBDX C DX D 0,
and eCD D 0.832i C 0.555j.
We get the force components as above, and the equilibrium forces at
the joints remain the same. Solving the equilibrium equations simultaneously for this situation yields
Fy D TCDY TBDY C DY D 0.
TAB D 1.35 kN,
Solve simultaneously to get
TAC D 1.54 kN,
TAB D TBD D 2.43 kN,
TBC D 1.33,
TAC D 2.78 kN,
TBD D 1.74 kN,
TBC D 0, TCD D 2.88 kN.
and TCD D 1.60 kN.
Note that with appropriate changes in the designation of points, the
forces here are the same as those in Problem 6.4. This can be explained
by noting from the unit vectors that AB and BC are parallel. Also note
that in this configuration, BC carries no load. This geometry is the
same as in Problem 6.4 except for the joint at B and member BC
which carries no load. Remember member BC in this geometry — we
will encounter things like it again, will give it a special name, and will
learn to recognize it on sight.
These numbers differ significantly from (a). Most significantly, member
BD is now carrying a compressive load and this has reduced the loads
in all members except member BD. “Sharing the load” among more
members seems to have worked in this case.
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1
Problem 6.7 This steel truss bridge is in the Gallatin
National Forest south of Bozeman, Montana. Suppose
that one of the tandem trusses supporting the bridge is
loaded as shown. Determine the axial forces in members
AB, BC, BD, and BE.
B
D
F
A
C
17 ft
Solution: We start with the entire structure in order to find the
reaction at A. We have to assume that either A or H is really a roller
instead of a pinned support.
MH : 10 kip17 ft C 10 kip34 ft C 10 kip51 ft
8 ft
H
E
G
10 kip
17 ft
10 kip
17 ft
10 kip
17 ft
Finally work with joint B
17
17
FAB C p
FBE C FBD D 0
Fx : p
353
353
8
8
FAB p
FBE FBC D 0
Fy : p
353
353
A68 ft D 0 ) A D 15 kip
Solving we find FBD D 42.5 kip, FBE D 11.74 kip
B
17 ft
17 ft
8
8
A
10 kip
10 kip
17
10 kip
H
Fy : p
8
353
FBE
FAB
FBC
Now we examine joint A
FBD
17
17 ft
17 ft
FAB C A D 0 ) FAB D 35.2 kip
In Summary we have
17
FAB
FAB D 35.2 kipC, FBC D 10 kipT,
8
FBD D 42.5 kipC, FBE D 11.74 kipT
FAC
A
Now work with joint C
Fy : FBC 10 kip D 0 ) FBC D 10 kip
FBC
FCE
FAC
C
10 kip
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1
Problem 6.8 For the bridge truss in Problem 6.7,
determine the largest tensile and compressive forces that
occur in the members, and indicate the members in
which they occur.
Solution: Continuing the solution to Problem 6.7 will show the
largest tensile and compressive forces that occur in the structure.
Examining joint A we have
17
FAB C FAC D 0 ) FAC D 31.9 kip
Fx : p
353
Examining joint C
Fx : FAC C FCE D 0 ) FCE D 31.9 kip
Examining joint D
Fy : FDE D 0 ) FDE D 0
D
FBD
FDF
FDE
The forces in the rest of the members are found by symmetry. We have
FAB D FFH D 35.2 kipC
FAC D FGH D 31.9 kipT
FBC D FFG D 10 kipT
FBD D FDF D 42.5 kipC
FBE D FEF D 11.74 kipT
FCE D FEG D 31.9 kipT
FDE D 0
The largest tension and compression members are then
FAC D FEG D FCE D FGH D 31.9 kipT
FBD D FDH D 42.5 kipC
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1
Problem 6.9 The trusses supporting the bridge in
Problem 6.8 are called Pratt trusses. Suppose that the
bridge designers had decided to use the truss shown
instead, which is called a Howe truss. Determine the
largest tensile and compressive forces that occur in the
members, and indicate the members in which they occur.
Compare your answers to the answers to Problem 6.8.
Solution: We start with the entire structure in order to find the
reaction at A. We have to assume that either A or H is really a roller
instead of a pinned support.
MH : 10 kip17 ft C 10 kip34 ft C 10 kip51 ft
B
D
F
A
H
C
17 ft
E
8 ft
G
10 kip
17 ft
10 kip
17 ft
10 kip
17 ft
Next work with joint C
8
FCD 10 kip D 0 ) FCD D 11.74 kip
Fy : FBC C p
353
17
FCD FAC D 0 ) FCE D 42.5 kip
Fx : FCE C p
353
A68 ft D 0 ) A D 15 kip
FBC
FCD
17
8
FAC
FCE
C
A
10
kips
10
kips
10
kips
H
10 kip
Now we examine joint A
Finally from joint E we find
Fy : p
8
353
FAB C A D 0 ) FAB D 35.2 kip
17
FAB C FAC D 0 ) FAC D 31.9 kip
Fx : p
353
FAB
17
Fy : FDE 10 kip D 0 ) FDE D 10 kip
FDE
FCE
FEG
E
8
FAC
10 kip
The forces in the rest of the members are found by symmetry. We have
A
FAB D FFH D 35.2 kipC
Now work with joint B
FAC D FGH D 31.9 kipT
17
FAB C FBD D 0 ) FBD D 31.9 kip
Fx : p
353
FBD D FDF D 31.9 kipC
8
FAB FBC D 0 ) FBC D 15 kip
Fy : p
353
B
FBD
17
FBC D FFG D 15 kipT
FCD D FDG D 11.74 kipC
FCE D FEG D 42.5 kipT
8
FDE D 10 kipT
FAB
The largest tension and compression members are then
FBC
FCE D FEG D 42.5 kipT
FAB D FFH D 35.2 kipC
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1
Problem 6.10 The truss shown is part of an airplane’s
internal structure. Determine the axial forces in members
BC, BD, and BE.
8 kN
14 kN
C
A
E
G
H
Solution: First, solve for the support reactions and then use the
method of joints to solve for the reactions in the members.
300 mm
B
8 kN
0.4 m
0.4 m
0.4 m
400 mm
0.4 m
0.3 m
BX
0.8 m
FY
BY
C
F
D
14 kN
400 mm
Fx :
BC D 0
Fy :
AC C CE D 0
Solving, we get
Fx : Bx D 0
400 mm
400 mm
BC D 0,
CE D 10.67 kN T
Fy : By C Fy 8 14 D 0 kN
Joint B :
y
MB : 0.48 C 0.8Fy 1.214 D 0
Solving, we get Bx D 0, By D 5.00 kN Fy D 17.00 kN.
The forces we are seeking are involved at joints B, C, D, and E. The
method of joints allows us to solve for two unknowns at a joint. We
need a joint with only two unknowns. Joints A and H qualify. Joint A
is nearest to the members we want to know about, so let us choose it.
Assume tension in all members.
BC
AB
θ
BE
θ
BD
x
BY
Joint A:
y
We know AB D 13.33 kN BC D 0 By D 5.00 kN.
8 kN
We know 3 of the 5 forces at B Hence, we can solve for the other two.
AC
4
x
θ
AB
3
5
sin D 0.6 cos D 0.8 D 36.87°
Fx :
BD C BE cos AB cos D 0
Fy :
BC C By C BE sin C AB sin D 0
Solving, we get
BE D 5.00 kN T
Fx D AC C AB cos D 0
From Joint C, we had BC D 0
Fy D 8 AB sin D 0
Solving, we get
BD D 14.67 kN C
AC D 10.67 kN T
Thus
BC D 0, BD D 14.67 kN (C)
BE D 5.00 kN T
AB D 13.33 kN C
Joint C : (Again, assume all forces are in tension)
y
AC
CE
x
BC
[AC D 10.67 kN T]
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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 6.11 The loads F1 D F2 D 8 kN. Determine
the axial forces in members BD, BE, and BG.
F1
D
3m
F2
B
E
3m
Solution: First find the external support loads and then use the
method of joints to solve for the required unknown forces. (Assume
all unknown forces in members are tensions).
G
A
C
External loads:
4m
F1 = 8 kN
D
y
3m
B
E
C
8m
DE D 6 kN C
F2 = 8 kN
Joint E :
y
x
C
AY
BD D 10 kN T
3m
G
A
AX
Solving,
4m
DE
GY
F2 = 8 kN
BE
Fx :
Ax C F1 C F2 D 0 (kN)
x
Fy :
Ay C Gy D 0
MA :
8Gy 3F2 6F1 D 0
EG
Solving for the external loads, we get
DE D 6 kN
Ax D 16 kN to the left
Ay D 9 kN downward
Gy D 9 kN upward
Now use the method of joints to determine BD, BE, and BG.
Start with joint D.
Fx D DE EG D 0
Fy D BE C F2 D 0
Solving:
EG D 6 kN C
BE D 8 kN T
Joint D:
y
D
F1 = 8 kN
DE
θ
x
BD
cos D 0.8
sin D 0.6
D 36.87°
Fx : F1 BD cos D 0
Fy :
BD sin DE D 0
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1
6.11 (Continued )
Joint G:
Fx :
CG BG cos D 0
Fy :
BG sin C EG C Gy D 0
y
BG
EG
Solving, we get
BG D 5 kN C
θ
x
CG D 4 kN T
CG
GY
EG D 6 kN C
Thus, we have
BD D 10 kN T
BE D 8 kN T
BG D 5 kN C
Gy D 9 kN
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 6.12 Determine the largest tensile and
compressive forces that occur in the members of the
truss, and indicate the members in which they occur if
(a)
(b)
B
D
h
C
E
the dimension h D 5 in;
the dimension h D 10 in.
20 in
A
20 in
20 in
30⬚
800 lb
Observe how a simple change in design affects the
maximum axial loads.
Solution: Starting at joint A
Finally joint C
20
FAB FAC C 800 lb sin 30° D 0
Fx : p
h2 C 202
Fy : p
h
h2 C 202
FAB 800 lb cos 30° D 0
20
20
FCD C p
FBC FCE C FAC D 0
Fx : p
h2 C 202
h2 C 202
h
h
Fy : p
FCD C p
FBC D 0
h2 C 202
h2 C 202
FBC
FCD
FAB
20
h
20
h
20
h
A
FAC
C
FCE
FAC
(a) Using h D 5 in we find:
FAB D 2860 lbT, FAC D 2370 lbC, FBD D 5540 lbT
800 lb
FBC D 2860 lbC, FCD D 2860 lbT, FCE D 7910 lbC
Next joint B
20
20
FBC C p
FAB D 0
Fx : FBD p
h2 C 202
h2 C 202
FBD D 5540 lbT
)
FCE D 7910 lbC
h
h
FBC p
FAB D 0
Fy : p
h2 C 202
h2 C 202
FBD
B
h
20
FBC
(b) Using h D 10 in we find:
FAB D 1549 lbT, FAC D 986 lbC, FBD D 2770 lbT
FBC D 1549 lbC, FCD D 1549 lbT, FCE D 3760 lbC
h
20
FBD D 2770 lbT
FAB
)
FCE D 3760 lbC
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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 6.13 The truss supports loads at C and E.
If F D 3 kN, what are the axial forces in members BC
and BE?
1m
1m
A
1m
B
D
1m
G
C
E
F
2F
Solution: The moment about A is
AY
AX
MA D 1F 4F C 3G D 0,
1m
from which G D
5
F D 5 kN. The sums of forces:
3
F
1m
FY D AY 3F C G D 0,
1m
DG
from which AY D
45°
4
F D 4 kN.
3
BD
EG
G
Joint G
DG
DE 45°
BE
45°
CE
Joint D
DE
EG
Joint E
FX D AX D 0,
AY
from which AX D 0. The interior angles GDE, EBC are 45° ,
AC
AB
45°
45° BC
AC
CE
F
Joint C
1
from which sin ˛ D cos ˛ D p .
2
Joint A
Denote the axial force in a member joining I, K by IK.
from which
(1)
5
BD D F D 5 kN C.
3
Joint G:
DG
Fy D p C G D 0,
2
from which
p
p
p
5 2
DG D 2G D F D 5 2 kN C.
3
DG
Fx D p EG D 0,
2
from which
DG
5
EG D p D F D 5kN T.
3
2
(2)
G
2F
1m
Joint D:
(3)
Joint E :
BE
Fy D p 2F C DE D 0,
2
p
p
from which BE D 2 2F 2DE D
p
p
2
F D 2 kN T.
3
BE
Fx D CE p C EG D 0,
2
from which
BE
4
CE D EG p D F D 4 kN T.
3
2
DG
Fy D DE p D 0,
2
from which
DE D
5
F D 5 kN T.
3
DG
Fx D BD C p D 0,
2
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1
6.13 (Continued )
(4)
Joint A:
AC
Fy D Ay p D 0,
2
p
p
4 2
F D 4 2 kN T.
from which AC D
3
AC
Fx D AB C p D 0,
2
4
from which AB D F D 4 kN C.
3
(5)
Joint C :
AC
Fy D BC C p F D 0,
2
1
AC
from which BC D F p D F D 1 kN C.
3
2
2
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Problem 6.14 If you don’t want the members of the
truss to be subjected to an axial load (tension or compression) greater than 20 kN, what is the largest acceptable
magnitude of the downward force F?
A
12 m
F
B
Solution: Start with joint A
4m
Fx : FAB cos 36.9° FAC sin 30.5° D 0
C
D
Fy : FAB sin 36.9° FAC cos 30.5° F D 0
3m
A
Finally examine joint D
36.9°
Fy : FBD D 0
30.5°
FBD
FAB
F
FAC
Dx
Now work with joint C
Solving we find
Fx : FCD FBC sin 36.9° C FAC sin 30.5° D 0
Fy : FBC cos 36.9° C FAC cos 30.5° D 0
FBC
FAC
36.9°
FCD
30.5°
FCD
D
FAB D 1.32F, FAC D 2.08F, FCD D 2.4F,
FBC D 2.24F, FBD D 0
The critical member is CD. Thus
2.4F D 20 kN ) F D 8.33 kN
C
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1
Problem 6.15 The truss is a preliminary design for a
structure to attach one end of a stretcher to a rescue
helicopter. Based on dynamic simulations, the design
engineer estimates that the downward forces the stretcher
will exert will be no greater than 1.6 kN at A and at B.
What are the resulting axial forces in members CF, DF,
and FG?
G
300
mm
290
mm
390
mm
150 mm
F
480 mm
C
E
D
200 mm
B
A
Solution: Start with joint C
Fy : p
48
3825
FCF 1.6 kN D 0 ) FCF D 2.06 kN
FCF
39
48
C
FCD
1.6 kN
Now use joint F
59
29
39
FFG p
FDF C p
FCF D 0
Fx : p
3706
3145
3825
3706
48
48
FFG p
FDF p
FCF D 0
3145
3825
Solving we find
FDF D 1.286 kN, FCF D 2.03 kN
Fy : p
15
59
FFG
15
F
39
48
48
29
FDF
FCF
In Summary
FCF D 2.06 kNT, FDF D 1.29 kNC, FCF D 2.03 kNT
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1
Problem 6.16 Upon learning of an upgrade in the helicopter’s engine, the engineer designing the truss does
new simulations and concludes that the downward forces
the stretcher will exert at A and at B may be as large as
1.8 kN. What are the resulting axial forces in members
DE, DF, and DG?
Solution: Assume all bars are in tension.
Next work with joint B
Start at joint C
16
TCF 1.8 kN D 0 ) TCF D 2.32 kN
Fy : p
425
3
Fx : p TBE D 0 ) TBE D 0
13
13
TCF TCD D 0 ) TCD D 1.463 kN
Fx : p
425
2
Fy : p TBE C TBD 1.8 kN D 0 ) TBD D 1.8 kN
13
TBE
TBD
TCF
2
13
B
3
16
C
1.8 kN
TCD
Finally work with joint D
1.8 kN
Next work with joint F
59
29
13
TFG p
TDF C p
TCF D 0
Fx : p
3706
3145
425
Fy : p
15
3706
TFG p
10
29
TDG C p
TDF C TCD D 0
Fx : TDE p
541
3145
21
48
TDG C p
TDF TBD D 0
Fy : p
541
3145
Solving:
TDG D 6.82 kN, TDE D 7.03 kN
TDF
TDG
48
48
TDF p
TCF D 0
3145
425
21
Solving
TDF D 5.09 kN, TFG D 4.23 kN
48
29
10
59
TFG
15
TDE
F
D
TCD
13
48
TBD
16
29
TDF
TCF
In summary:
TDE D 7.03 kNC, TDF D 5.09 kNC, TDG D 6.82 kNT
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1
Problem 6.17 Determine the axial forces in the
members in terms of the weight W.
B
E
1m
A
D
W
1m
C
0.8 m
0.8 m
0.8 m
Solution: Denote the axial force in a member joining two points
I, K by IK. The angle between member DE and the positive x axis
is ˛ D tan1 0.8 D 38.66° . The angle formed by member DB with the
positive x axis is 90° C ˛. The angle formed by member AB with the
positive x axis is ˛.
Joint E :
Fy D DE cos ˛ W D 0,
from which DE D 1.28W C .
Fy D BE DE sin ˛ D 0,
from which BE D 0.8W T
Joint D:
Fx D DE cos ˛ C BD cos ˛ CD cos ˛ D 0,
from which BD CD D DE.
Fy D BD sin ˛ C DE sin ˛ CD sin ˛ D 0,
from which BD C CD D DE.
Solving these two equations in two unknowns:
CD D DE D 1.28W C , BD D 0
Joint B :
Fx D BE AB sin ˛ BD sin ˛ D 0,
from which AB D
BE
D 1.28WT
sin ˛
Fy D AB cos ˛ BC D 0,
from which BC D AB cos ˛ D WC
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1
Problem 6.18 The lengths of the members of the truss
are shown. The mass of the suspended crate is 900 kg.
Determine the axial forces in the members.
A
12 m
B
13 m
5m
C
13 m
12 m
D
40⬚
Solution: Start with joint A
Finally work with joint B
Fx : FAB cos 40° FAC sin 27.4° D 0
Fy : FAB cos 50° FBC sin 50° FBD cos 27.4° D 0
Fy : FAB sin 40° FAC cos 27.4° 900 kg9.81 m/s2 D 0
FAB
50°
A
40°
T
B
50°
27.4°
FAB
27.4°
FBC
FAC
8829 N
Solving we find
Next work with joint C
FBD
Fx : FCD cos 40° FBC cos 50° C FAC sin 27.4° D 0
FAB D 10.56 kN D 10.56 kNT
FAC D 17.58 kN D 17.58 kNC
Fy : FCD sin 40° C FBC sin 50° C FAC cos 27.4° D 0
FBC
FAC
27.4°
FCD D 16.23 kN D 16.23 kNC
FBC D 6.76 kN D 6.76 kNT
FBD D 1.807 kN D 1.807 kNT
50°
C
40°
FCD
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1
Problem 6.19 The loads F1 D 600 lb and F2 D
300 lb. Determine the axial forces in members AE, BD,
and CD.
F1
G
D
F2
B
6 ft
C
3 ft
E
A
4 ft
4 ft
Solution: The reaction at E is determined by the sum of the
moments about G:
F1
GX
MG D C6E 4F1 8F2 D 0,
F2
GY
6 ft
from which
ED
4F1 C 8F2
D 800 lb.
6
E
EG
The interior angle EAG is
˛ D tan1
6
D 36.87° .
8
E
AC
AE
Joint E
From similar triangles this is also the value of the interior angles ACB,
CBD, and CGD. Method of joints: Denote the axial force in a member
joining two points I, K by IK.
from which BD D
Joint E :
4 ft
α
AE
AB
4 ft
BD
α
BC
Joint A
F2
F1
DG
AB
Joint B
CD
α
BD
Joint D
300
F2 C AB
D
D 500 lbC .
0.6
0.6
Fx D BC BD cos ˛ D 0,
Fy D E C AE D 0,
from which BC D BD0.8 D 400 lbT.
from which AE D E D 800 lb C .
Joint D:
Fy D EG D 0,
from which EG D 0.
Fy D BD sin ˛ CD F1 D 0,
from which CD D F1 BD0.6 D 300 lbC
Joint A:
Fy D AE AC cos ˛ D 0,
from which AC D AE
D 1000 lbT.
0.8
Fy D AC sin ˛ C AB D 0,
from which AB D AC0.6 D 600 lbC.
Joint B :
Fy D BD sin ˛ AB F1 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 6.20 Consider the truss in Problem 6.19. The
loads F1 D 450 lb and F2 D 150 lb. Determine the axial
forces in members AB, AC, and BC.
Solution: From the solution to Problem 6.19 the angle ˛ D 36.87°
4F1 C 8F2
D 500 lb. Denote the axial
6
force in a member joining two points I, K by IK.
EG
and the reaction at E is E D
E
Joint E :
AE
Joint E
AC
AB
α
AE
Joint A
BD
α
BC
F2
AB
Joint B
Fy D EG D 0.
Fx D AE C E D 0,
from which AE D E D 500 lbC.
Joint A:
Fx D AE AC cos ˛ D 0,
from which AC D AE
D 625 lbT .
0.8
Fy D AC sin ˛ C AB D 0,
from which AB D AC0.6 D 375 lbC
Joint B:
Fy D BD sin ˛ F2 AB D 0,
from which BD D
F2 C AB
D 375 lbC
0.6
Fx D BC BD cos ˛ D 0,
from which BC D BD0.8 D 300 lbT
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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 6.21 Each member of the truss will safely
support a tensile force of 4 kN and a compressive force
of 1 kN. Determine the largest mass m that can safely
be suspended.
1m
1m
1m
E
F
1m
C
D
m
1m
A
Solution: The common interior angle BAC D DCE D EFD D
EF
CDB is ˛ D tan1 1 D 45° .
W
Joint F
CE
α
CD
AC
DF
Fy D p W D 0,
2
p
from which DF D 2WC.
DF
Fx D EF p D 0,
2
from which EF D WT.
Joint E:
CE
Fx D p C EF D 0
2
from which CE D
p
2WT.
CE
Fy D ED p D 0,
2
α
DF
1
Note cos ˛ D sin ˛ D p . Denote the axial force in a member joining
2
two points I, K by IK.
Joint F :
B
BC
Joint C
ED
CE α
EF
ED
Joint E
BC α BD
AB
CD
α
DF
BD
Joint D
B
Joint B
Joint B:
BD
Fx D AB C p D 0,
2
from which AB D 2WC
This completes the determination of the axial forces in all nine
members. The maximum tensile force occurs in member AC,
p
p
4
AC D 2 2WT, from which the safe load is W D p D 2 D
2 2
1.414 kN. The maximum compression occurs in member BD, BD D
p
1
2 2W C, from which the maximum safe load is W D p D
2 2
0.3536 kN. The largest mass m that can be safely supported is
353.6
D 36.0 kg
mD
9.81
from which ED D WC.
Joint D:
BD
DF
FY D ED C p p D 0,
2
2
p
from which BD D 2 2WC.
BD
DF
FX D p p CD D 0,
2
2
from which CD D WT
Joint C:
CE
AC
Fx D p C p C CD D 0,
2
2
p
from which AC D 2 2WT
AC
CE
Fy D p C p BC D 0,
2
2
from which BC D WC
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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 6.22 The Warren truss supporting the
walkway is designed to support vertical 50-kN loads at
B, D, F, and H. If the truss is subjected to these loads,
what are the resulting axial forces in members BC, CD,
and CE?
B
D
F
H
2m
A
C
6m
Solution: Assume vertical loads at A and I Find the external loads
at A and I, then use the method of joints to work through the structure
to the members needed.
50 kN
50 kN
6m
50 kN
6m
3m
AY
3m
6m
I
6m
D 33.69°
6m
6m
G
AB D 180.3 kN
50 kN
E
Fx :
BC cos C BD AB cos D 0
Fy :
50 AB sin BC sin D 0
x
IY
Solving,
BC D 90.1 kN T
BD D 225 kN C
Fy :
Ay C Iy 450 D 0 (kN)
MA :
350 950 1550 2150 C 24 Iy D 0
Solving
Joint C :
y
Ay D 100 kN
BC
CD
Iy D 100 kN
θ
θ
Joint A:
AC
y
C
CE
x
D 33.69°
AB
AC D 150 kN T
BC D 90.1 kN T
θ
A
x
AC
AY
Fy :
CD sin C BC sin D 0
CE D 300 kN T
D 33.69°
CE AC C CD cos BC cos D 0
Solving,
tan D 23
Fx :
CD D 90.1 kN C
Fx : AB cos C AC D 0
Hence
Fy : AB sin C Ay D 0
Solving,
BC D 90.1 kN T
CD D 90.1 kN C
CE D 300 kN T
AB D 180.3 kN C
AC D 150 kN T
Joint B :
y
50 kN
B
θ
BD
x
θ
BC
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 6.23 For the Warren truss in Problem 6.22,
determine the axial forces in members DF, EF, and FG.
Solution: In the solution to Problem 6.22, we solved for the forces
Solving, we get
in AB, AC, BC, BD, CD, and CE. Let us continue the process. We
ended with Joint C. Let us continue with Joint D.
EF D 0
Joint D:
EG D 300 kN T
y
Note: The results are symmetric to this point!
50 kN
Joint F :
D
BD
DF
y
θ
DF
DE
CD
50 kN
x
θ
BD D 225 kN C
EF
CD D 90.1 kN C
Solving,
50 CD sin DE sin D 0
DF D 300 kN C
EF D 0
DF D 300 kN C
DE D 0
At this point, we have solved half of a symmetric truss with a
symmetric load. We could use symmetry to determine the loads in
the remaining members. We will continue, and use symmetry as a
check.
Joint E :
EF
DE
Fx :
FH DF C FG cos EF cos D 0
Fy :
50 EF sin FG sin D 0
Solving:
FH D 225 kN C
FG D 90.1 kN C
Thus, we have
DF D 300 kN C
EF D 0
FG D 90.1 kN C
y
Note-symmetry holds!
θ
CE
FG
D 33.69°
Fx : DF BD C DE cos CD cos D 0
Fy :
x
θ
θ
D 33.69°
FH
F
θ
E
EG
x
D 33.69°
CE D 300 kN T
DE D 0
Fx : EG CE C EF cos DE cos D 0
Fy : DE sin C EF sin 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 6.24 The Pratt bridge truss supports five
forces (F D 300 kN). The dimension L D 8 m. Determine the axial forces in members BC, BI, and BJ.
L
L
L
L
L
L
B
C
D
E
G
I
J
K
L
M
F
F
L
A
H
F
Solution: Find support reactions at A and H. From the free body
and
Fx D AX D 0,
θ I
AY
L
8
MA D 68HY 3008 C 16 C 24 C 32 C 40 D 0.
L
8
L
K
L
8
L
8
F
L
8
HY
TIJ
x
Joint I
y
TAB
TBI
θ
TAI
H
F
F
F = 300 kN
y
A
I
x
TAI
F
AY
Joint B
y
Fy D TAB sin C AY D 0.
TBC
From these equations,
TAB D 1061 kN
and TAI D 750 kN.
L
M
L
8
Joint A
From the geometry, the angle D 45°
Fx D AX C TAB cos C TAI D 0,
J
F
F
L=8m
From these equations, AY D HY D 750 kN.
G
A
Fy D AY C HY 5300 D 0,
Joint A: From the free body diagram,
F
B
diagram,
F
θ
θ
x
TBJ
TAB
TBI
Joint I: From the free body diagram,
Fx D TIJ TAI D 0,
Fy D TBI 300 D 0.
From these equations,
TBI D 300 kN
and TIJ D 750 kN.
Joint B: From the free body diagram,
Fx D TBC C TBJ cos TAB cos D 0,
Fy D TBI TBJ sin TAB sin D 0.
From these equations,
TBC D 1200 kN
and TBJ D 636 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 6.25 For the roof truss shown, determine the
axial forces in members AD, BD, DE, and DG. Model
the supports at A and I as roller supports.
10 kN
8 kN
E
6 kN
C
8 kN
F
B
6 kN
H
D
3m
3m
Solution: Use the whole structure to find the reaction at A.
Next go to joint C
MI : 6 kN3 m C 8 kN6 m C 10 kN9 m
C 8 kN12 m C 6 kN15 m
C A18 m D 0 ) A D 19 kN
G
3m
3m
3m
3m
Fy : 8 kN FCD C FCE FBC sin 21.8° D 0
Fx : FCE FBC cos 21.8° D 0
Solving:
FCD D 8 kN, FCE D 43.1 kN
10 kN
6 kN
3.6 m
I
A
8 kN
8 kN
8 kN
FCD
6 kN
C
FCD
FBC
A
I
Finally examine joint D
Now work with joint A
Fx : FAD C FDG FBD cos 21.8° C FDE cos 50.19° D 0
Fy : FAB sin 21.8° C A D 0 ) FAB D 51.2 kN
Fx : FAD C FAB cos 21.8° D 0 ) FAD D 47.5 kN
Solving:
Fy : FBD sin 21.8° C FCD C FDE sin 50.19° D 0
FDE D 14.3 kN, FDG D 30.8 kN
FCD
FAB
FDE
FBD
21.8°
A
FAD
50.19°
FAD
A
FDG
In Summary
Next use joint B
D
Fx : FAB C FBC C FBD cos 21.8° D 0
FAD D 47.5 kNT, FBD D 8.08 kNC,
FDE D 14.32 kNT, FDG D 30.8 kNT
Fy : FAB C FBC FBD sin 21.8° 6 kN D 0
Solving:
FBC D 43.1 kN, FBD D 8.08 kN
6 kN
FBC
B
FAB
FBD
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1
Problem 6.26 The Howe truss helps support a roof.
Model the supports at A and G as roller supports. Determine the axial forces in members AB, BC, and CD.
800 lb
600 lb
600 lb
D
400 lb
400 lb
C
E
8 ft
B
F
A
G
H
I
4 ft
4 ft
Solution: The strategy is to proceed from end A, choosing joints
with only one unknown axial force in the x- and/or y-direction, if
possible, and if not, establish simultaneous conditions in the unknowns.
8
12
˛Pitch D tan1
BH D 4
1400 lb
BI
α Pitch
from which the angle
˛HIB D tan1
CI D 8
2.6667
4
D 33.7° .
from which the angle
˛IJC D tan1
5.333
4
CI
AH
BH
400 lb
α Pitch
HI
AB
Joint H
600 lb
CD
α Pitch
BC
α Pitch
BH BI
Joint B
α IJC
CJ
Joint C
BC CI
Joint H :
AH
HI
IJ
Joint I
8
D 5.3333 ft,
12
600 lb
400 lb
G
α Pitch
Joint A
D 2.6667 ft,
4 ft
4 ft 4 ft 4 ft 4 ft 4 ft 4 ft
D 33.7° .
4 ft
A
AB
8
12
4 ft
L
800 lb
The length of the vertical members:
4 ft
K
600 lb
400 lb
The interior angles HIB and HJC differ. The pitch angle is
J
Fy D BH D 0, or, BH D 0.
Fx D AH C HI D 0,
D 53.1° .
from which HI D 2100 lb T
The moment about G:
MG D 4 C 20400 C 8 C 16600 C 12800 24A D 0,
33600
D 1400 lb. Check: The total load is 2800 lb.
24
From left-right symmetry each support A, G supports half the total
load. check.
from which A D
Joint B :
Fx D AB cos ˛Pitch C BC cos ˛Pitch
C BI cos ˛Pitch D 0,
from which BC C BI D AB
The method of joints: Denote the axial force in a member joining two
points I, K by IK.
Joint A:
Fy D AB sin ˛P C 1400 D 0,
from which AB D 1400
D 2523.9 lb C
sin ˛p
Fx D AB cos ˛Pitch C AH D 0,
from which AH D 2523.90.8321 D 2100 lb T
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1
Problem 6.27 The plane truss forms part of the
supports of a crane on an offshore oil platform. The
crane exerts vertical 75-kN forces on the truss at B, C,
and D. You can model the support at A as a pin support
and model the support at E as a roller support that can
exert a force normal to the dashed line but cannot exert
a force parallel to it. The angle ˛ D 45° . Determine the
axial forces in the members of the truss.
C
B
D
1.8 m
2.2 m
F
G
H
A
α
E
3.4 m
3.4 m
3.4 m
3.4 m
Solution: The included angles
D tan1
ˇ D tan1
D tan1
4
3.4
2.2
3.4
1.8
3.4
75 kN 75 kN 75 kN
D 49.64° ,
D 32.91° ,
AY 3.4
m
D 27.9° .
AB
AX
with this relation and the fact that Ex cos 45° C Ey cos 45° D 0, we
obtain Ex D 112.5 kN and Ey D 112.5 kN. From
FAy D Ay 375 C Ey D 0,
The method of joints: Denote the axial force in a member joining two
points I, K by IK.
Joint A:
3.4
m
EY
3.4
m
BF
DE
EX
AF
γ β
CD
θ
DG
EY
Joint E
75 kN
γ
DH
Joint D
DE
β
AF
DH
FG
Joint F
75 kN
BC
CD
GH
β
EH
Joint H
CG
Joint C
FAx D Ax C Ex D 0, AX D EX D 112.5 kN.
from which Ay D 112.5 kN. Thus the reactions at A and E are symmetrical about the truss center, which suggests that symmetrical truss
members have equal axial forces.
β γ
EH
AY
Joint A
75 kN
BC
θ
γ
BF BG
AB
Joint B
MA D 753.41 C 2 C 3 C 43.4Ey D 0.
3.4
m
The complete structure as a free body: The sum of the moments about
A is
EX
AX
Solve:
EH D 44.67 kNC ,
and
DE D 115.8 kNC
Joint F :
Fx D AF cos ˇ C FG D 0,
from which FG D 37.5 kN C
Fx D AB cos C Ax C AF cos ˇ D 0,
Fy D AB sin C Ay C AF sin ˇ D 0,
from which two simultaneous equations are obtained.
Solve:
AF D 44.67 kN C ,
and
AB D 115.8 kN C
Fy D AF sin ˇ C BF D 0,
from which BF D 24.26 kN C
Joint H:
Fx D EH cos ˇ GH D 0,
Joint E:
Fy D DE cos C Ex EH cos ˇ D 0.
Fy D DE sin C Ey C EH sin ˇ D 0,
from which two simultaneous equations are obtained.
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1
6.27 (Continued )
from which GH D 37.5 kN C
from which DG D 80.1 kN T
Fy D EH sin ˇ C DH D 0,
Fx D DE cos CD DG cos D 0,
from which DH D 24.26 kN C
from which CD D 145.8 kN C
Joint B:
Joint C :
Fy D AB sin BF C BG sin 75 D 0,
Fx D CD BC D 0,
from which BG D 80.1 kN T
from which CD D BC Check.
Fx D AB cos C BC C BG cos D 0,
from which BC D 145.8 kN C
Fy D CG 75 D 0,
from which CG D 75 kN C
Joint D:
2
Fy D DE sin DH DG sin 75 D 0,
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Problem 6.28 (a) Design a truss attached to the
supports A and B that supports the loads applied at points
C and D.
(b) Determine the axial forces in the members of the
truss you designed in (a)
1000 lb
C
2000 lb
D
4 ft
A
2 ft
B
5 ft
5 ft
5 ft
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1
Problem 6.29 (a) Design a truss attached to the
supports A and B that goes over the obstacle and
supports the load applied at C.
(b) Determine the axial forces in the members of the
truss you designed in (a).
Obstacle
C
4m
2m
B
A
6m
3.5 m
10 kN
4.5 m
1m
Solution: This is a design problem with many possible solutions
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1
Problem 6.30 Suppose that you want to design a truss
supported at A and B (Fig. a) to support a 3-kN downward load at C. The simplest design (Fig. b) subjects
member AC to 5-kN tensile force. Redesign the truss so
that the largest force is less than 3 kN.
A
A
1.2 m
C
C
B
B
3 kN
3 kN
1.6 m
(a)
Solution: There are many possible designs. To better understand
the problem, let us calculate the support forces in A and B and the
forces in the members in Fig. (b).
(b)
D 36.87°
Ay
Fx :
BC AC cos D 0
Fy :
AC sin 3 kN D 0
Solving: BC D 4 kN C AC D 5 kN C
Ax
A
Thus, AC is beyond the limit, but BC (in compression) is not,
Joint B :
1.2 m
θ
AB
C
B
Bx
x
1.6 m
BX
BC
3 kN
1.2
1.6
D 36.87°
tan D
sin D 0.6
Fx :
Ax C Bx D 0
Fy :
Ay 3 kN D 0
MA :
1.2Bx 1.63 D 0
C
Bx C BC D 0
Fy :
AB D 0
Solving, BC and Bx are both already known. We get AB D 0
cos D 0.8
Fx :
Thus, we need to reduce the load in AC. Consider designs like that
shown below where D is inside triangle ABC. Move D around to adjust
the load.
A
Solving, we get
Ax D 4 kN
Bx D 4 kN
D
Ay D 3 kN
Note: These will be the external reactions for every design that we
produce (the supports and load do not change).
B
C
Reference Solution (Fig. (b))
Joint C :
However, the simplest solution is to place a second member parallel
to AC, reducing the load by half.
AC
θ
BC
3 kN
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1
Problem 6.31 The bridge structure shown in Fig. 6.14
can be given a higher arch by increasing the 15° angles to
20° . If this is done, what are the axial forces in members
AB, BC, CD, and DE? Compare your answers to the
values in Table 6.1.
F
F
b
F
b
F
b
F
b
2b
(1)
F
F
F
F
F
G
H
I
J
K
b
a
b
15⬚
b
15⬚
b
2b
C
D
B
a
A
E
(2)
Solution: Follow the solution method in Example 6.3. F is known
For joint C,
Joint B :
y
F
Fx :
TBC cos 20° C TCD cos 20° D 0
Fy :
F TBC sin 20° TCD sin 20° D 0
TBC
TBC D TCD D 1.46F C
For joint B.
20°
x
α
TAB
Fx :
TBC cos 20 TAB cos ˛ D 0
Fy :
TBC sin 20° F TAB sin ˛ D 0
Solving, we get ˛ D 47.5° and TAB D 2.03F C
Joint C :
For the new truss (using symmetry)
F
C
20°
TBC
20°
TCD
Members
Forces
AG, BH, CI,
DJ, EK
F
AB, DE
2.03F (C)
BC, CD
1.46F (C)
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1
Problem 6.32 Determine the axial forces in the Pratt
truss in Fig. 6.16 and confirm the values in Table 6.2
H
G
I
b
B
C
D
F
F
F
A
F
Solution: The common angle is ˛ D tan1
b
b
b
A F
FY D A 5F C E D 0,
EI
from which A D 2.5F.
The method of joints:
F
Joint E
E
A
F
AG
α
AB
F
b
CD
F
F
b
F E
b
HI
DI
DE
Joint A
GH
GH
α α CG
BG
Joint G
Joint H
BG
Joint E:
Fy D E C EI sin ˛ F D 0,
F
b
α
DE
b
b
D 45°
b
The complete structure as a free body: The moment about A is MA D
bF1 C 2 C 3 C 4 C 4bE D 0, from which E D 2.5F. The reaction
at A:
E
F
Joint D
CI α α
DI
Joint I
EI
HI
AG
BC
F
Joint B
AB
HC
from which EI D 2.12F .
Joint B :
Fx D DE EI cos ˛ D 0,
Fy D BG F D 0,
from which DE D 1.5F .
from which BG D F .
Joint A:
Fy D AG sin ˛ F C A D 0,
from which AG D 2.12F .
Fx D AB C BC D 0,
from which BC D 1.5F
Joint G:
Fx D AG cos ˛ C AB D 0,
Fy D AG cos ˛ CG cos ˛ BG D 0,
from which AB D 1.5F .
from which CG D 0.707F
Joint D:
Fy D DI F D 0,
from which DI D F .
Fx D AG sin ˛ C GH C CG sin ˛ D 0,
from which GH D 2F
Joint H :
Fx D DE CD D 0,
Fy D HC D 0 .
from which CD D 1.5F .
Joint I :
A term by term comparison confirms Table 6.2.
Fy D DI CI sin ˛ EI sin ˛ D 0,
from which CI D 0.707F .
Fx D HI CI cos ˛ C EI cos ˛ D 0,
from which HI D 2F .
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1
Problem 6.33 Determine the axial forces in the
suspension bridge structure in Fig. 6.18, including the
reactions exerted on the towers, and confirm the values
in Table 6.3.
A
E
B
a
L
F
D
C
R
15⬚
15⬚
H
I
J
F
F
F
G
FH
x D HL C HI D 0,
and
Fy D AB sin ˛ BC sin 15° F D 0,
from which.
FJx D IJ C JR D 0,
and
AB D BC
BC sin 15° C F
BC cos 15°
cos 15°
cos ˛
D 38.79°
D 2.39FT .
By symmetry, DE D AB
FIy D CI F D 0,
from which BH D FT , CI D FT , DJ D FT .
b
Fx D BC cos 15° AB cos ˛ D 0,
FJy D DJ F D 0,
b
Joint B: The sum of forces:
˛ D tan1
FH
y D BH F D 0,
b
FIx D HI C IJ D 0,
from which HL D HI D IJ D JR D 0. The sum of the forces in the
y-direction
F
K
b
Solution: The roadway has pinned joints at H, I, and J, and at the
towers. The strategy is to use the method of joints to show that the axial
forces in members BH, CI and DJ are each equal to F. An analysis
of the joints at B and D yields the reaction at the towers. Method of
joints: Denote the axial force in a member joining two points I, K, by
IK. Joints H, I, K. The sum of forces in the x direction,
a
BC
CD
15°
15°
F
Joint C
AB
α
15°
BC
F
Joint B
Joint C:
Fx D BC cos 15° C CD cos 15° D 0,
from which CD D BC
Fy D CD sin 15° C BC sin 15° F D 0,
from which CD D BC D 1.93FT
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1
Problem 6.34 The truss supports a 100-kN load at J.
The horizontal members are each 1 m in length.
(a)
Use the method of joints to determine the axial
force in member DG.
Use the method of sections to determine the axial
force in member DG.
(b)
A
B
C
D
E
F
G
H
1m
J
100 kN
Solution: (a) Start with Joint J
DJ
45°
Fx :
CD DG cos 45° C DJ cos 45° D 0
Fy :
DG sin 45° DJ sin 45° DH D 0
Solving,
CD D 200 kN
J
DG D 141.4 kN C
HJ
(b) Method of Sections
100 kN
CD
D
Fx :
45°
HJ DJ cos 45° D 0
1m
DG
Fy : DJ sin 45° 100 D 0
Solving
J
GH
DJ D 141.4 kN T
H
1m
HJ D 100 kN C
100 kN
Joint H :
DH
CD DG cos 45° GH D 0
Fy :
DG sin 45° 100 D 0
MD :
1GH 1100 D 0
HJ
H
GH
Fx :
Fx : HJ GH D 0
Solving,
Fy : DH D 0
GH D 100 kN C
CD D 200 kN T
DH D 0,
DG D 141.4 kN C
GH D 100 kN C
Joint D
CD
D
x
45°
45°
DJ
DH
DG
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1
Problem 6.35 For the truss in Problem 6.34, use the
method of sections to determine the axial forces in
members BC, CF, and FG.
BC
Solution:
Fx :
C
D
45°
BC CF cos 45 FG D 0
1m
CF
Fy :
CF sin 45° 100 D 0
MC :
1FG 2100 D 0
J
F
FG
G
1m
H
1m
100 kN
Solving
BC D 300 kN T
CF D 141.4 kN C
FG D 200 kN C
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1
Problem 6.36 Use the method of sections to determine
the axial forces in members AB, BC, and CE.
1m
1m
A
B
1m
D
1m
G
C
E
F
2F
Solution: First, determine the forces at the supports
Θ = 45°
D
B
Method of Sections:
y
AY = 1. 33 F
AX
AY
C
1m
F
E
1m
1m
θ
1m
AX = 0
AX = 0
Fx :
Ax D 0
Fy :
Ay C Gy 3F D 0
MA :
1F 22F C 3Gy D 0
C
1m
AY
1m
C
CE
x
F
Fx :
CE C AB D 0
Fy :
BC C Ay F D 0
MB :
1Ay C 1CE D 0
Solving
B
BC
GY
2F
AB
Ax D 0 Gy D 1.67F
Ay D 1.33F
C
Solving, we get
AB D 1.33F C
CE D 1.33F T
BC D 0.33F C
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1
Problem 6.37 The truss supports loads at A and H.
Use the method of sections to determine the axial forces
in members CE, BE, and BD.
18 kN
24 kN
C
A
E
G
H
300 mm
B
400 mm
F
D
400 mm
400 mm
400 mm
Solution: First find the external support loads on the truss
tan D
18 kN
24 kN
C
E
G
H
3
4
D 36.87°
By D 15 kN
0.8 m
BX
FY
BY
1.2 m
CE C BE cos C BD D 0
Fy :
By 18 C BE sin D 0
MB :
C 0.418 0.3CE D 0
0.4 m
Fx :
D
C
Fx :
Bx D 0
Solving,
Fy :
By C Fy 18 24 D 0 (kN)
MB :
0.8Fy 1.224 C 0.418 D 0
Solving:
CE D 24 kN T
BE D 5 kN T
BD D 28 kN C
Bx D 0
By D 15 kN
Fy D 27 kN
Method of sections:
18 kN
C
CE
E
BE
0.3 m
θ
θ
0.4 m
BD
BY
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1
Problem 6.38 The Pratt bridge truss is loaded as
shown. Use the method of sections to determine the axial
forces in members BD, BE, and CE.
B
D
F
A
H
C
17 ft
E
8 ft
G
10 kip
30 kip
20 kip
17 ft
17 ft
17 ft
Solution: Use the whole structure to find the reaction at A.
MH : 20 kip17 ft C 30 kip34 ft
C 10 kip51 ft A68 ft D 0
10
kip
A
30
kip
20
kip
H
) A D 27.5 kip
B
Now cut through BD, BE, CE and use the left section
FBD
8
MB : A17 ft C FCE 8 ft D 0 ) FCE D 58.4 kip
17
FBE
ME : 10 kip17 ft A34 ft FBD 8 ft D 0
C
A
FCE
) FBD D 95.6 kip
8
FBE D 0 ) FBE D 41.1 kip
Fy : A 10 kip p
353
In Summary
10 kip
A
FCE D 58.4 kipT, FBD D 95.6 kipC, FBE D 41.1 kipT
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1
Problem 6.39 The Howe bridge truss is loaded as
shown. Use the method of sections to determine the axial
forces in members BD, CD, and CE.
B
D
F
A
H
C
17 ft
E
10 kip
17 ft
G
30 kip
17 ft
20 kip
17 ft
B
Solution: Use the whole structure to find the reaction at A (same
FBD
as 6.38) A D 27.5 kip
Now cut through BD, CD, and CE and use the left section.
MC : A17 ft FBD 8 ft D 0 ) FBD D 58.4 kip
17
FCD
8
C
A
FCE
MD : A34 ft C 10 kip17 ft C FCE 8 ft D 0
) FCE D 95.6 kip
8 ft
10 kip
8
A
FCD D 0 ) FCD D 41.1 kip
Fy : A 10 kip C p
353
In Summary
FBD D 58.4 kipC, FCE D 95.6 kipT, FCD D 41.1 kipC
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1
Problem 6.40 For the Howe bridge truss in Problem
6.39, use the method of sections to determine the axial
forces in members DF, DG, and EG.
Solution: Same truss as 6.39.
D
Cut through DF, DG, and EG and use left section
8
17
MD : A34 ft C 10 kip17 ft C FEG 8 ft D 0
E
) FEG D 95.6 kip
FDF
FDG
FEG
MG : A51 ft C 10 kip34 ft C 30 kip17 ft FDF 8 ft
A
10 kip
30 kip
D 0 ) FDF D 69.1 kip
Fy : A 10 kip 30 kip p
8
353
FDG D 0 ) FDG D 29.4 kip
In summary
FEG D 95.6 kipT, FDF D 69.1 kipC, FDG D 29.4 kipC
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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 6.41 The Pratt bridge truss supports five
forces F D 340 kN. The dimension L D 8 m. Use the
method of sections to determine the axial force in
member JK.
L
L
L
L
L
L
B
C
D
E
G
I
J
K
L
M
F
F
L
A
H
Solution: First determine the external support forces.
F
AX
L
L
L
F
AY
F
L
F
L
F
HY
F = 340 kN, L = 8 M
C
Solving:
L D 8M
F D 340 kN
Ay D 850 kN
Fx :
Ax D 0
Fy :
Ay 5F C Hy D 0
MA :
6LHy LF 2LF 3LF 4LF 5LF D 0
F
D 45°
L
F
F
Fx :
CD C JK C CK cos D 0
Fy :
Ay 2F CK sin D 0
MC :
LJK C LF 2LAy D 0
Ax D 0,
C
Ay D 850 kN
Solving,
JK D 1360 kN T
Hy D 850 kN
Also, CK D 240.4 kN T
Note the symmetry:
Method of sections to find axial force in member JK.
C
B
θ
I
A
L
CD D 1530 kN C
CD
D
CK
J
K
L
AY
F
F
JK
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1
Problem 6.42 For the Pratt bridge truss in Problem 6.41, use the method of sections to determine the
axial force in member EK.
Solution: From the solution to Problem 6.41, the support forces
are Ax D 0, Ay D Hy D 850 kN.
L
L
L
L
L
L
B
C
D
E
G
I
J
K
L
M
Method of Sections to find axial force in EK.
L
E
DE
A
G
θ
F
EK
Solution:
KL
F
F
F
F
EK D 240.4 kN T
L
F
F
HY
Also, KL D 1360 kN T
DE D 1530 kN C
Fx :
DE EK cos KL D 0
Fy :
Hy 2F EK sin D 0
ME :
LKL LF C 2LHy D 0
H
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1
Problem 6.43 The walkway exerts vertical 50-kN
loads on the Warren truss at B, D, F, and H. Use
the method of sections to determine the axial force in
member CE.
B
D
F
H
2m
A
C
6m
Solution: First, find the external support forces. By symmetry,
Ay D Iy D 100 kN (we solved this problem earlier by the method of
joints).
B
y
50 kN
BD
6m
C
6m
G
6m
I
6m
CE D 300 kN T
Also, BD D 225 kN C
CD D 90.1 kN C
D
2m
A
Solving:
E
CD
θ
CE
x
AY
tan D
2
3
D 33.69°
Fx :
BD C CD cos C CE D 0
Fy :
Ay 50 C CD sin D 0
MC :
6Ay C 350 2BD 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 6.44 The mass m D 120 kg. Use the method
of sections to determine the axial forces in members BD,
CD, and CE.
1m
1m
1m
E
F
1m
C
D
m
1m
A
B
Solution: Cut through BD, CD, and CE and use the top section.
1
MD : p FCE 1 m 1177.2 N1 m D 0
2
1
MC : 1177.2 N2 m p FBD 1 m D 0
2
1
1
Fx : p FCE p FBD FCD D 0
2
2
Solving
FCE D 1665 N, FBD D 3330 N, FCD D 1177 N
E
1m
FCE
1m
(120 kg)(9.81 m/s2)
FCD
D
FBD
In summary
FCE D 1.66 kNT, FBD D 3.33 kNC, FCD D 1.18 kNT
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1
Problem 6.45 For the roof truss shown, use the
method of sections to determine the axial forces in
members AD, BC, and BD. Model the supports at A and
I as roller supports.
8 kN
6 kN
C
6 kN
E
6 kN
F
B
6 kN
H
D
3m
3m
G
3m
3m
3m
3m
Solution: From the entire structure
Now cut through AD, BC, BD and use the left section
MI : 6 kN3 m C 6 kN6 m C 8 kN9 m C 6 kN12 m
C 6 kN15 m A18 m D 0 ) A D 16 kN
3.6 m
I
A
MB : A3 m C FAD 1.2 m D 0
2
MD : A6 m C 6 kN3 m p FBC 6 m D 0
29
2
MA : 6 kN3 m p FBD 6 m D 0
29
8 kN
6 kN
6 kN
6 kN
Solving:
6 kN
FAD D 40.0 kN, FBC D 35.0 kN, FBD D 8.08 kN
6 kN
FBC
5
B
2
2
5
A
I
A
FBD
FAD
A
Summary:
FAD D 40.0 kNT, FBC D 35.0 kNC, FBD D 8.08 kNC
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1
Problem 6.46 For the roof truss in Problem 6.45, use
the method of sections to determine the axial forces in
members CE, DE, and DG.
6 kN
Solution: From the previous problem we know that A D 16 kN
Cut through CE, DE, and DG. Use the left section
6 kN
2
FED
5
MD : A6 m C 6 kN3 m p FCE 2.4 m D 0
29
6
5
ME : A9 m C 6 kN6 m C 6 kN3 m C FDG 3.6 m D 0
Solving:
FCE
5
6
MA : 6 kN3 m 6 kN6 m C p FED 6 m D 0
61
D
FDG
FED D 11.72 kN, FCE D 35.0 kN, FDG D 25.0 kN
Summary:
A
FED D 11.72 kNT, FCE D 35.0 kNC, FDG D 25.0 kNT
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1
Problem 6.47 The Howe truss helps support a roof.
Model the supports at A and G as roller supports.
2 kN
2 kN
2 kN
(a)
(b)
Use the method of joints to determine the axial
force in member BI.
Use the method of sections to determine the axial
force in member BI.
D
2 kN
2 kN
C
E
4m
B
F
A
G
Solution: The pitch of the roof is
˛ D tan1
H
4
D 33.69° .
6
2m
I
2m
J
K
2m
L
2m
2m
2m
F
F
This is also the value of interior angles HAB and HIB. The complete
structure as a free body: The sum of the moments about A is
F
F = 2 kN
F
MA D 221 C 2 C 3 C 4 C 5 C 62G D 0,
G
A
30
from which G D
D 5 kN. The sum of the forces:
6
FY D A 52 C G D 0,
2m 2m 2m 2m 2m 2m
BH
AB
from which A D 10 5 D 5 kN.
The method of joints: Denote the axial force in a member joining I, K
by IK.
(a)
(a)
A
α
AH
Joint A
AH
HI
2 kN
BH
Joint B
Joint H
Joint A:
F
B
Fy D A C AB sin ˛ D 0,
α
(b)
A
5
from which AB D
D
D 9.01 kN (C).
sin ˛
0.5547
Fx D AB cos ˛ C AH D 0,
BC
α
BI
α
AB
A
BC
α
α
BI
HI
2m
from which AH D AB cos ˛ D 7.5 kN (T).
Joint H :
Fy D BH D 0.
Joint B :
Fx D AB cos ˛ C BI cos ˛ C BC cos ˛ D 0,
Fy D 2 AB sin ˛ BI sin ˛ C BC sin ˛ D 0.
Solve: BI D 1.803 kN C , BC D 7.195 kN C
(b)
Make the cut through BC, BI and HI. The section as a free body:
The sum of the moments about B:
MB D A2 C HI2 tan ˛ D 0,
from which HI D
3
A D 7.5 kNT. The sum of the forces:
2
Fx D BC cos ˛ C BI cos ˛ C HI D 0,
Fy D A F C BC sin ˛ BI sin ˛ D 0.
Solve: BI D 1.803 kN C .
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1
Problem 6.48 Consider the truss in Problem 6.47. Use
the method of sections to determine the axial force in
member EJ.
Solution: From the solution to Problem 6.47, the pitch angle is
˛ D 36.69° , and the reaction G D 5 kN. The length of member EK is
LEK D 4 tan ˛ D
16
D 2.6667 m.
6
The interior angle KJE is
ˇ D tan1
LEK
2
DE
β
F
E
F
EJ
α
JK
2m
G
2m
D 53.13° .
Make the cut through ED, EJ, and JK. Denote the axial force in a
member joining I, K by IK. The section as a free body: The sum of
the moments about E is
ME D C4G 2F JK2.6667 D 0,
from which JK D
20 4
D 6 kN T.
2.6667
The sum of the forces:
Fx D DE cos ˛ EJ cos ˇ JK D 0.
Fy D DE sin ˛ EJ sin ˇ 2F C G D 0,
from which the two simultaneous equations:
0.8321DE C 0.6EJ D 6,
0.5547DE 0.8EJ D 1.
Solve: EJ D 2.5 kN C .
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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 6.49 Use the method of sections to determine
the axial force in member EF.
10 kip
A
4 ft
10 kip
C
B
4 ft
E
D
4 ft
G
F
4 ft
I
H
12 ft
Solution: The included angle at the apex BAC is
˛ D tan1
12
16
D 36.87° .
The interior angles BCA, DEC, FGE, HIG are D 90° ˛ D 53.13° .
The length of the member ED is LED D 8 tan ˛ D 6 ft. The interior
angle DEF is
ˇ D tan1
4
LED
F = 10 kip
F = 10 kip
DF
β
EF
E
γ
EG
D 33.69° .
The complete structure as a free body: The moment about H is MH D
280
1012 1016 C I12 D 0, from which I D
D 23.33 kip.
12
The sum of forces:
Fy D Hy C I D 0,
from which Hy D I D 23.33 kip.
Fx D Hx C 20 D 0,
from which Hx D 20 kip. Make the cut through EG, EF, and DE.
Consider the upper section only. Denote the axial force in a member
joining I, K by IK. The section as a free body: The sum of the moments
about E is ME D 104 108 C DFLED D 0, from which DF D
120
D 20 kip.
6
The sum of forces:
Fy D EF sin ˇ EG sin DF D 0,
Fx D EF cos ˇ C EG cos C 20 D 0,
from which the two simultaneous equations: 0.5547EF C 0.8EG D
20, and 0.8320EF 0.6EG D 20. Solve: EF D 4.0 kip (T)
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1
Problem 6.50 For the bridge truss shown, use the
method of sections to determine the axial forces in
members CE, CF, and DF.
200 kN
B
200 kN
200 kN
D
F
200 kN
200 kN
H
J
E
G
C
3m
7m
4m
I
A
5m
5m
5m
5m
Solution: From the entire structure we find the reactions at A
Now we cut through DF, CF, and CE and use the left section.
Fx : Ax D 0
) FDF D 375 kN
MI : 200 kN5 m C 200 kN10 m C 200 kN15 m
C 200 kN20 m Ay 20 m D 0 ) Ay D 500 kN
200 kN
200 kN
200 kN
200 kN
MC : 200 kN5 m Ay 5 m C Ax 3 m FDF 4 m D 0
200 kN
MF : 200 kN10 m C 200 kN5 m Ay 10 m C Ax 7 m
1
5
C p FCE 4 m p FCE 5 m D 0 ) FCE D 680 kN
26
26
5
5
Fx : Ax C FDF C p FCE C p FCF D 0
26
41
) FCF D 374 kN
200 kN
200 kN
Ax
D
I
FDF
Ay
FCF
5
4
FCE
1
C
5
Ax
Ay
Summary:
FDF D 375 kNC, FCE D 680 kNT, FCF D 374 kNC
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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 6.51 The load F D 20 kN and the dimension
L D 2 m. Use the method of sections to determine the
axial force in member HK.
L
L
A
B
D
E
C
F
Strategy: Obtain a section by cutting members HK,
HI, IJ, and JM. You can determine the axial forces in
members HK and JM even though the resulting freebody diagram is statically indeterminate.
L
F
G
L
I
H
J
K
M
L
Solution: The complete structure as a free body: The sum of the
2L
F
moments about K is MK D FL2 C 3 C ML2 D 0, from which
5F
D 50 kN. The sum of forces:
MD
2
FY D KY C M D 0,
L
F
2L
from which KY D M D 50 kN.
FX D KX C 2F D 0,
KX
M
KY
from which KX D 2F D 40 kN.
The section as a free body: Denote the axial force in a member joining
I, K by IK. The sum of the forces:
from which HI IJ D Kx . Sum moments about K to get MK D
ML2 C JML2 IJL C HIL D 0.
HK
KX
Fx D Kx HI C IJ D 0,
Substitute HI IJ D Kx , to obtain JM D M HI
IJ
JM
L
M
KY
2L
Kx
D 30 kN C.
2
Fy D Ky C M C JM C HK D 0,
from which HK D JM D 30 kNT
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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 6.52 The weight of the bucket is W D
1000 lb. The cable passes over pulleys at A and D.
D
A
C
(a)
(b)
Determine the axial forces in member FG and HI.
By drawing free-body diagrams of sections, explain
why the axial forces in members FG and HI are
equal.
F
B
H
3 ft 6 in
J
3 ft
E
3 ft 3 in L
G
I
35°
3 ft
W
K
Solution: The truss is at angle ˛ D 35° relative to the horizontal.
The angles of the members FG and HI relative to the horizontal are
ˇ D 45° C 35° D 80° . (a) Make the cut through FH, FG, and EG,
and consider the upper section. Denote the axial force in a member
joining, ˛, ˇ by ˛ˇ.
W
FH
β
3.25 ft
The section p
as a free body: The perpendicular distance from point F
is LFW D 3 2 sin ˇ C 3.5 D 7.678 ft.
3 ft
The sum of the moments about F is MF D WLFW C W3.25 jEGj3 D 0, from which EG D 1476.1 lb C.
W
α
FG
EG
W
3.5 ft
W
The sum of the forces:
FY D FG sin ˇ FH sin ˛ EG sin ˛ W sin ˛ W D 0,
JH
HI
GI
FX D FG cos ˇ FH cos ˛ EG cos ˛ W cos ˛ D 0,
from which the two simultaneous equations:
0.9848FG 0.5736FH D 726.9, and 0.1736FG 0.8192FH D
389.97.
Solve: FG D 1158.5 lb C , and FH D 721.64 lb T. Make the
cut through JH, HI, and GI, and consider the upper section.
(b) Choose a coordinate system with the y axis parallel to JH. Isolate
a section by making cuts through FH, FG, and EG, and through HJ,
HI, and GI. The free section of the truss is shown. The sum of the
forces in the x- and y-direction are each zero; since the only external
x-components of axial force are those contributed by FG and HI, the
two axial forces must be equal:
The section as a free body: The perpendicular distance from
point
p
H to the line of action of the weight is LHW D 3 cos ˛ C 3 2 sin ˇ C
3.5 D 10.135 ft. The sum of the moments about H is MH D WL jGIj3 C W3.25 D 0, from which jGIj D 2295 lb C.
Fx D HI cos 45° FG cos 45° D 0,
from which HI D FG
FY D HI sin ˇ JH sin ˛ GI sin ˛ W sin ˛ W D 0,
FX D HI cos ˇ JH cos ˛ GI cos ˛ W cos ˛ D 0,
from which the two simultaneous equations:
0.9848HI 0.5736JH D 257.22,
and
0.1736HI 0.8192JH D 1060.8.
Solve:
and
HI D 1158.5 lbC ,
JH D 1540.6 lbT .
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1
Problem 6.53 Consider the truss in Problem 6.52. The
weight of the bucket is W D 1000 lb. The cable passes
over pulleys at A and D. Determine the axial forces in
members IK and JL.
Solution: Make a cut through JL, JK, and IK, and consider the
upper section. Denote the axial force in a member joining, ˛, ˇ by
˛ˇ. The section as a free body: The perpendicular distance p
from point
J to the line of action of the weight is L D 6 cos ˛ C 3 2 sin ˇ C
3.5 D 12.593 ft. The sum of the moments about J is MJ D WL C
W3.25 IK3 D 0, from which IK D 3114.4 lbC.
The sum of the forces:
Fx D JL cos ˛ IK cos ˛
W
W
β
JL
α
3.5 ft
3.25 ft
3 ft
JK
IK
W cos ˛ JK cos ˇ D 0,
and
Fy D JL sin ˛ IK sin ˛
W sin ˛ W JK sin ˇ D 0,
from which two simultaneous equations:
0.8192JL C 0.1736JK D 1732
and
0.5736JL C 0.9848JK D 212.75.
Solve:
and
JL D 2360 lbT ,
JK D 1158.5 lbC .
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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 6.54 The truss supports loads at N, P, and R.
Determine the axial forces in members IL and KM.
2m
2m
2m
2m
2m
K
M
O
Q
I
L
N
P
1 kN
2 kN
1m
J
R
2m
Solution: The strategy is to make a cut through KM, IM, and
IL, and consider only the outer section. Denote the axial force in a
member joining, ˛, ˇ by ˛ˇ.
H
G
1 kN
2m
F
E
2m
The section as a free body: The moment about M is
D
MM D IL 21 42 61 D 0,
from which
C
2m
A
B
IL D 16 kN C .
6m
The angle of member IM is ˛ D tan1 0.5 D 26.57° .
The sums of the forces:
KM
α
Fy D IM sin ˛ 4 D 0,
1m
IM
IL
4
from which IM D D 8.944 kN (C).
sin ˛
1 kN
2 kN
2m
2m
1 kN
2m
Fx D KM IM cos ˛ IL D 0,
from which KM D 24 kNT
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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 6.55 Consider the truss in Problem 6.54.
Determine the axial forces in members HJ and GI.
Solution: The strategy is to make a cut through the four members
AJ, HJ, HI, and GI, and consider the upper section. The axial force
in AJ can be found by taking the moment of the structure about B.
The complete structureas a free body: The angle formed by AJ with the
4
D 26.57° . The moment about B is MB D
vertical is ˛ D tan1
8
6AJ cos ˛ 24 D 0, from which AJ D 4.47 kN (T).
1m
I
AJ HJ
αβ
γ
HI
GI
2m
2m
1 kN
2 kN
1 kN
2m
2m
2m
The section as a free body: The
of members HJ and HI
angles
relative
2
1.5
D 14.0° , and D tan1
D
to the vertical are ˇ D tan1
8
2
°
36.87 respectively. Make a cut through the four members AJ, HJ,
HI, and GI, and consider the upper section. The moment about
the point I is MI D 24 C 2AJ cos ˛ C 2HJ cos ˇ D 0. From which
HJ D 8.25 kN T . The sums of the forces:
Fx D AJ sin ˛ C HJ sin ˇ HI sin D 0,
from which HI D
AJ sin ˛ HJ sin ˇ
22
D
D 0.
sin sin FY D AJ cos ˛ HJ cos ˇ HI cos GI 4 D 0,
from which GI D 16 kN C
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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 6.56 Consider the truss in Problem 6.54. By
drawing free-body diagrams of sections, explain why the
axial forces in members DE, FG, and HI are zero.
Solution: Define ˛, ˇ to be the interior angles BAJ and ABJ
respectively. The sum of the forces in the x-direction at the base yields
AX C BX D 0, from which Ax D Bx . Make a cut through AJ, BD and
BC, from which the sum of forces in the x-direction, Ax BD sin ˇ D
0. Since Ax D AJ sin ˛, then AJ sin ˛ BD sin ˇ D 0. A repeat of the
solution to Problem 6.55 shows that this result holds for each section,
where BD is to be replaced by the member parallel to BD. For example:
make a cut through AJ, FD, DE, and CE. Eliminate the axial force
in member AJ as an unknown by taking the moment about A. Repeat
the solution process in Problem 6.55, obtaining the result that
DE D
AJ sin ˛ DF sin ˇ
D0
cos DE
where DE is the angle of the member DE with the vertical. Similarly,
a cut through AJ, FH, FG, and EG leads to
FG D
AJ sin ˛ FH sin ˇ
D 0,
cos FG
and so on. Thus the explanation is that each member BD, DF, FH and
HJ has equal tension, and that this tension balances the x-component
in member AJ
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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 6.57 The mass of the suspended weight is
2000 kg. Determine the axial forces in the bars AB and
AC.
Strategy:
y
D (4, 7, 0) m
A (6, 5, 3) m
Draw the free-body diagram of joint A.
B
C (10, 0, 0) m
x
z
Solution: The forces acting on point A are
6i 5j 3k
p
70
FAB D FAB
FAD D FAD
2i C 2j 3k
p
17
, FAC D FAC
4i 5j 3k
p
50
, W D 2000 kg9.81 m/s2 j
Adding components we have the following equations
6
4
2
Fx : p FAB C p FAC p FAD D 0
70
50
17
5
5
2
Fy : p FAB p FAC C p FAD 19.62 kN D 0
70
50
17
3
3
3
Fz : p FAB p FAC p FAD D 0
70
50
17
Solving we find
FAB D 14.07 kN, FAC D 7.93 kN, FAD D 11.56 kN
Summary:
FAB D 14.07 kNC, FAC D 7.93 kNC
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1
Problem 6.58 The space truss supports a vertical 10kN load at D. The reactions at the supports at joints A,
B, and C are shown. What are the axial forces in the
members AD, BD, and CD?
y
10 kN
D (4, 3, 1) m
Ay
Ax A
Cy
C (6, 0, 0) m
Az
By
z
Solution: Consider the joint D only. The position vectors parallel
to the members from D are
Cz
B (5, 0, 3) m
x
10 kN
rDA D 4i 3j k,
rDB D i 3j C 2k,
TDA
TDC
TDB
rDC D 2i 3j k.
The unit vectors parallel to the members from D are:
eDA D
rDA
D 0.7845i 0.5883j 0.1961k
jrDA j
eDB D
rDB
D 0.2673i 0.8018j C 0.5345k
jrDB j
eDC D
rDC
D 0.5345i 0.8018j 0.2673k
jrDC j
The equilibrium conditions for the joint D are
F D TDA eDA C TDB eDB C TDC eDC FD D 0,
from which
Fx D 0.7845TDA C 0.2673TDB C 0.5345TDC D 0
Fy D 0.5883TDA 0.8018TDB 0.8108TDC 10 D 0
Fz D 0.1961TDA C 0.5345TDB 0.2673TDC D 0.
Solve:
TDA D 4.721 kN C , TDB D 4.157 kN C
TDC D 4.850 kN C
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1
Problem 6.59 Consider the space truss in Problem 6.58. The reactions at the supports at joints A, B,
and C are shown. What are the axial forces in members
AB, AC, and AD?
Solution: The reactions at A are required for a determination of
Ay
the equilibrium conditions at A.
The complete structure as a free body: The position vectors are rAB D
5i C 3k, rAC D 6i, rAD D 4i C 3j C k. The sum of the forces:
and
Ax
TAC
Az
Fx D Ax D 0,
TAD
TAB
Fy D Ay C Cy C By 10 D 0,
Fz D Az C Cz D 0.
The moments due to the reactions:
M D rAB ð FB C rAC ð FC C rAD ð FD D 0
i
M D 5
0
j
0
By
k i
3 C 6
0 0
j
0
Cy
j
k k i
3
1 D 0
0 C 4
Cz 0 10 0 D 3By C 10i 6Cz j C 5By C 6Cy 40k D 0.
These equations for the forces and moments are to be solved for the
unknown reactions. The solution:
Ax D Cz D 0,
Ay D 2.778 kN,
By D 3.333 kN,
and Cy D 3.889 kN
The method of joints: Joint A: The position vectors are given above.
The unit vectors are:
eAB D 0.8575i C 0.5145k,
eAC D i,
eAD D 0.7845i C 0.5883j C 0.1961k.
The equilibrium conditions are:
F D TAB eAB C TAC C eAC C TAD eAD C A D 0,
from which
Fx D 0.8575TAB C TAC C 0.7845TAD D 0
Fy D 0TAB C 0TAC C 0.5883TAD C 2.778 D 0
Fz D 0.5145jTAB j C 0jTAC j C 0.1961jTAD j D 0.
Solve:
TAB D 1.8 kN T , TAC D 2.16 kN T
TAD D 4.72 kN C
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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 6.60 The space truss supports a vertical load
F at A. Each member is of length L, and the truss rests on
the horizontal surface on roller supports at B, C, and D.
Determine the axial forces in members AB, AC, and AD.
F
A
B
D
C
Solution: By symmetry, the axial forces in members AB, AC, and
AD are equal. We just need to determine the angle between each of
these members and the vertical:
we see that
b
D tan 30°
L
2
F
and
A
TAB
TAD = TAB
bCc
D tan 60° ,
L
2
from which we obtain
TAC = TAB
1
Ltan 60° tan 30° .
2
c
Then D arcsin
L
cD
θ
θ
θ
D 35.26°
F C 3TAB cos D 0,
so TAB D TAC D TAD D F
.
3 cos and
From the top view,
L
TAB D TAC D TAD D F
3 cos 35.26°
D 0.408F.
C
60°
b
30°
L /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.
1
Problem 6.61 For the truss in Problem 6.60, determine
the axial forces in members AB, BC, and BD.
Solution: See the solution of Problem 6.60. The axial force in
member AB is TAB D 0.408F, and the angle between AB and the
vertical is D 35.26° . The free-body diagram of joint B is
From the equilibrium equation
TAB sin C 2TBC cos 30° D 0,
we obtain
TAB
TBC D TBD D 0.136F.
θ
TBD = TBC
30°
TBC
30°
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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 6.62 The space truss has roller supports at B,
C, and D and supports a vertical 800-lb load at A. What
are the axial forces in members AB, AC, and AD?
y
800 lb
A (4, 3, 4) ft
B
D (6, 0, 0) ft
x
z
Solution: The position vectors of the points A, B, C, and D are
rA D 4i C 3j C 4k,
rC D 5i C 6k,
rD D 6i.
C (5, 0, 6) ft
The equilibrium conditions at point A:
The position vectors from joint A to the vertices are:
Fx D 0.6247TAB C 0.2673TAC C 0.3714TAD D 0
Fy D 0.4685TAB 0.8018TAB 0.5570TAD 800 D 0
Fz D 0.6247TAB C 0.5345TAC 0.7428TAD D 0.
rAB D rB rA D 4i 3j 4k,
800 lb
rAC D rC rA D 1i 3j C 2k,
rAD D rD rA D 2i 3j 4k
Joint A: The unit vectors parallel to members AB, AC, and AD are
eAB D
rAB
D 0.6247i 0.4685j 0.6247k,
jrAB j
eAC D
rAC
D 0.2673i 0.8018j C 0.5345k,
jrAC j
and eAD D
rAD
D 0.3714i 0.5570j 0.7428k.
jrAD j
TAB
TAD
TAC
Solve:
and
TAB D 379.4 lb C , TAC D 665.2 lb C ,
TAD D 159.6 lb C
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1
Problem 6.63 The space truss shown models an
airplane’s landing gear. It has ball and socket supports
at C, D, and E. If the force exerted at A by the wheel is
F D 40j (kN), what are the axial forces in members AB,
AC, and AD?
y
E (0, 0.8, 0) m
D
0.4 m
B
x
(1, 0, 0) m
0.6 m
A
(1.1, – 0.4, 0) m
C
z
F
Solution: The important points in this problem are A (1.1, 0.4,
0), B (1, 0, 0), C (0, 0, 6), and D (0, 0, 0.4). We do not need point
E as all of the needed unknowns converge at A and none involve the
location of point E. The unit vectors along AB, AC, and AD are
y
E
(0, 0.8, 0) m
uAB D 0.243i C 0.970j C 0k,
uAC D 0.836i C 0.304j C 0.456k,
and uAD D 0.889i C 0.323j 0.323k.
D
0.4
m
0.6
m
TAD
C
The forces can be written as
z
B
x
(1, 0, 0) m
TAB
TAC
F
A
(1.1, −0.4, 0) m
TRS D TRS uRS D TRSX i C TRSY j C TRSZ k,
where RS takes on the values AB, AC, and AD. We now have three
forces written in terms of unknown magnitudes and known directions.
The equations of equilibrium for point A are
Fx D TAB uABX C TAC uACX C TAD uADX C FX D 0,
and
Fy D TAB uABY C TAC uACY C TAD uADY C FY D 0,
Fz D TAB uABZ C TAC uACZ C TAD uADZ C FZ D 0,
where F D FX i C FY j C FZ k D 40j kN. Solving these equations for
the three unknowns, we obtain TAB D 45.4 kN (compression), TAC D
5.26 kN (tension), and TAD D 7.42 kN (tension).
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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 6.64 If the force exerted at point A of the
truss in Problem 6.63 is F D 10i C 60j C 20k (kN), what
are the axial forces in members BC, BD and BE?
Solution: The important points in this problem are A (1.1, 0.4,
0), B (1, 0, 0), C (0, 0, 0.6), D (0, 0, 0.4), and E (0, 0.8, 0). The
unit vectors along AB, AC, AD, BC, BD, and BE are
y
E
(0, 0.8, 0) m
uAB D 0.243i C 0.970j C 0k,
uAC D 0.836i C 0.304j C 0.456k,
0.4
m
0.6
m
uAD D 0.889i C 0.323j 0.323k,
TBC
C
z
uBC D 0.857i C 0j C 0.514k,
DT TDE
AD
F
B
x
(1, 0, 0) m
TAB
A
(1.1, −0.4, 0) m
uBD D 0.928i C 0j 0.371k,
and uBE D 0.781i C 0.625j C 0k.
The forces can be written as TRS D TRS uRS D TRSX i C TRSY j C
TRSZ k, where RS takes on the values AB, AC, and AD when dealing
with joint A and AB, BC, BD, and BD when dealing with joint B. We
now have three forces written in terms of unknown magnitudes and
known directions.
Joint A: The equations of equilibrium for point A are,
and
Fx D TAB uABX C TAC uACX C TAD uADX C FX D 0,
Fy D TAB uABY C TAC uACY C TAD uADY C FY D 0,
Fz D TAB uABZ C TAC uACZ C TAD uADZ C FZ D 0,
where F D FX i C FY j C FZ k D 10i C 60j C 20k kN. Solving these
equations for the three unknowns at A, we obtain TAB D 72.2 kN
(compression), TAC D 13.2 kN (compression), and TAD D 43.3 kN
(tension).
Joint B: The equations of equilibrium at B are
and
Fx D TAB uABX C TBC uBCX C TBD uBDX C TBE uBEX D 0,
Fy D TAB uABY C TBC uBCY C TBD uBDY C TBE uBEY D 0,
Fz D TAB uABZ C TBC uBCZ C TBD uBDZ C TBE uBEZ D 0.
Since we know the axial force in AB, we have three equations in the
three axial forces in BC, BD, and BE. Solving these, we get TBC D
32.7 kN (tension), TBD D 45.2 kN (tension), and TBE D 112.1 kN
(compression).
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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 6.65 The space truss is supported by roller
supports on the horizontal surface at C and D and a ball
and socket support at E. The y axis points upward. The
mass of the suspended object is 120 kg. The coordinates
of the joints of the truss are A: (1.6, 0.4, 0) m, B: (1.0,
1.0, 0.2) m, C: (0.9, 0, 0.9) m, D: (0.9, 0, 0.6) m,
and E: (0, 0.8, 0) m. Determine the axial forces in
members AB, AC, and AD.
y
B
E
D
A
C
x
z
Solution: The important points in this problem are A: (1.6, 0.4,
y
0) m, B: (1, 1, 0.2) m, C: (0.9, 0, 0.9) m, and D: (0.9, 0, 0.6) m.
We do not need point E as all of the needed unknowns converge at A
and none involve the location of point E. The unit vectors along AB,
AC, and AD are
E
TAB
D
uAB D 0.688i C 0.688j 0.229k,
TAD
uAC D 0.579i 0.331j C 0.745k,
C
and uAD D 0.697i 0.398j 0.597k.
The forces can be written as TRS D TRS uRS D TRSX i C TRSY j C
TRSZ k, where RS takes on the values AB, AC, and AD. We now
have three forces written in terms of unknown magnitudes and known
directions. The equations of equilibrium for point A are
and
B
z
A
mg
x
TAC
L
Fx D TAB uABX C TAC uACX C TAD uADX C FX D 0,
Fy D TAB uABY C TAC uACY C TAD uADY C FY D 0,
Fz D TAB uABZ C TAC uACZ C TAD uADZ C FZ D 0,
where F D FX i C FY j C FZ k D mgj D 1177j N. Solving these
equations for the three unknowns, we obtain TAB D 1088 N (tension),
TAC D 316 N (compression), and TAD D 813 N (compression).
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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 6.66 The free-body diagram of the part of the
construction crane to the left of the plane is shown. The
coordinates (in meters) of the joints A, B, and C are (1.5,
1.5, 0), (0, 0, 1), and (0, 0, 1), respectively. The axial
forces P1 , P2 , and P3 are parallel to the x axis. The axial
forces P4 , P5 , and P6 point in the directions of the unit
vectors
e4 D 0.640i 0.640j 0.426k,
e5 D 0.640i 0.640j 0.426k,
e6 D 0.832i 0.555k.
The total force exerted on the free-body diagram by the
weight of the crane and the load it supports is Fj D
44j (kN) acting at the point (20, 0, 0) m. What is
the axial force P3 ?
Strategy: Use the fact that the moment about the line
that passes through joints A and B equals zero.
y
A
F
z
B
P6
P2
C
P5
P1
P4
P3
x
Solution: The axial force P3 and F are the only forces that exert
moments about the line through A and B. The moment they exert about
pt B is

 
i
j
k
i
0
1  C  0
MB D  20
0
44 0
P3

j k
0 2 
0 0
D 44i 2P3 j C 880k (kN-m).
The position vector from B to A is
rBA D 1.5i C 1.5j k (m),
and the unit vector that points from B toward A is
eBA D
rBA
D 0.640i C 0.640j 0.426k.
jrBA j
From the condition that
eBA Ð MB D 0.64044 C 0.6402P3 0.426880 D 0,
we obtain P3 D 315 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 6.67 In Problem 6.66, what are the axial
forces P1 , P4 , and P5 ?
Strategy: Write the equilibrium equations for the
entire free-body diagram.
Solution: The equilibrium equations are
Fx D P1 C P2 C P3 C 0.64P4 C 0.64P5 C 0.832P6 D 0,
Fy D 0.64P4 0.64P5 44 D 0,
Fz D 0.426P4 C 0.426P5 0.555P6 D 0,

 
i
j
k
i
0
1  C  0
MB D  20
0
44 0
P3

j k
0 2 
0 0


i
j
k

C 1.5 1.5 1 
P1
0
0

i
C  1.5
0.64P4

i
C  1.5
0.64P5
j
1.5
0.64P4

k

1
0.426P4
j
1.5
0.64P5

k
1  D 0.
0.426P5
The components of the moment equation are
MBx D 44 1.279P4 0.001P5 D 0,
MBy D 2P3 P1 0.001P4 1.279P5 D 0,
MBz D 880 1.5P1 1.92P4 1.92P5 D 0.
Solving these equations, we obtain
P1 D 674.7 kN,
P2 D P3 D 315.3 kN,
P4 D P5 D 34.4 kN,
and P6 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 6.68 The mirror housing of the telescope is
supported by a 6-bar space truss. The mass of the housing
is 3 Mg (megagrams), and its weight acts at G. The
distance from the axis of the telescope to points A, B,
and C is 1 m, and the distance from the axis to points
D, E, and F is 2.5 m. If the telescope axis is vertical
(˛ D 90° ), what are the axial forces in the members of
the truss?
A
G
F
C
B
60°
4m
60°
B
60°
60°
C
E
x
60°
y
60°
60°
D
A
F
G
60°
60°
Mirror housing
y
C
B
60°
E
F
A
G
60°
D
α
1m
END VIEW y
D
Solution: A cut through the 6-bar space truss leads to six equations
in the unknowns (see Problem 6.59). However for this problem an
alternate strategy based on reasonable assumptions about the equality
of the tensions is used to get the reactions. Assume that each support
carries one-third of the weight, which is equally divided between the
two bars at the support.
y
Mirror housing
z
E
x
z
60°
A
F
G C
B
The coordinate system has its origin in the upper platform, with the
x axis passing though the point C. The coordinates of the points are:
D
α
1m
A cos 60° , sin 60° , 0 D 0.5, 0.866, 0,
E
4m
B cos 60° , sin 60° , 0 D 0.5, 0.866, 0,
r=1m
A
C1, 0, 0,
y
C
B
x
D2.5, 0, 4,
4m
E2.5 cos 60° , 2.5 sin 60° , 4 D 1.25, 2.165, 4,
R = 2.5 m
F2.5 cos 60° , 2.5 sin 60° , 4 D 1.25, 2.165, 4.
Consider joint B in the upper housing. The position vectors of the
points E and D relative to B are
D
E
rBD D 2i C 0.866j 4k,
rBE D 1.75i 1.299j 4k.
The unit vectors are
eBD D 0.4391i C 0.1901j 0.8781k,
and eBE D 0.3842i 0.2852j 0.8781k.
The weight is balanced by the z components:
Fz D W
0.8781TBD 0.8781TBE D 0.
3
Assume that the magnitude of the axial force is the same in both
members BD and BE, TBE D TBD . The weight is W D 39.81 D
29.43 kN. Thus the result: TBE D TBD D 5.5858 kN C . From
symmetry (and the assumptions made above) the axial force is the
same in all members.
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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 6.69 Consider the telescope described in
Problem 6.68. Determine the axial forces in the members
of the truss if the angle ˛ between the horizontal and the
telescope axis is 20° .
Solution: The coordinates of the points are,
y
F
A cos 60° , sin 60° , 0 D 0.5, 0.866, 0 m,
A
B cos 60° , sin 60° , 0 D 0.5, 0.866, 0 m,
x
C
D
B
C1, 0, 0 m,
E
D2.5, 0, 4 m,
E2.5 cos 60° , 2.5 sin 60° , 4 D 1.25, 2.165, 4 m,
F2.5 cos 60° , 2.5 sin 60° , 4 D 1.25, 2.165, 4 m.
20000
The coordinates of the center of gravity are G (0, 0, 1) (m). Make a
cut through the members just below the upper platform supports, such
that the cut members have the same radial distance from the axis as
the supports. Consider the upper section.
A
x
i
a
l
The section as a free body: The strategy is to sum the forces and
moments to obtain six equations in the six unknown axial forces. The
axial forces and moments are expressed in terms of unit vectors. The
position vectors of the points E, D, and F relative to the points A, B,
and C are required to obtain the unit vectors parallel to the members.
The unit vectors are obtained from these vectors. The vectors and their
associated unit vectors are given in Table I. Note: While numerical
values are shown below to four significant figures, the calculations
were done with the full precision permitted (15 digits for TK Solver
Plus.)
Vector
x
y
z
rAD
rAF
rBD
rBE
rCE
rCF
2
1.75
2
1.75
0.25
0.25
0.866
1.299
0.866
1.299
2.165
2.165
4
4
4
4
4
4
Table I
Unit
Vector
eAD
eAF
eBD
eBE
eCE
eCF
Axial Forces in Bars
25000
|AF| & |CF|
15000
10000
|AD| & |BD|
5000
0
−5000
F
, −10000
N −15000
−20000
|CE| & |BD|
−25000
−100
−50
0
50
100
alpha, deg
x
y
z
0.4391
0.3842
0.4391
0.3842
0.0549
0.0549
0.1901
0.2852
0.1901
0.2852
0.4753
0.4753
0.8781
0.8781
0.8781
0.8781
0.8781
0.8781
The equilibrium condition for the forces is
jTAB jeAD C jTAF jeAF C jTBD jeBD C jTBE jeBE C jTCE jeCE
C jTCF jeCF C W D 0.
This is three equations in six unknowns. The unit vectors are given in
Table I. The weight vector is W D jWjj cos ˛ k sin ˛, where ˛ is
the angle from the horizontal of the telescope housing. The remaining
three equations in six unknowns are obtained from the moments:
rA ð TAD C TAF C rB ð TBD C TBE C rC ð TCE
C TCF C rG ð W 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
6.69 (Continued )
Carry out the indicated operations on the moments to obtain the vectors
defining the moments:
The six equations in six unknowns are:
i
rA ð TAD D jTAD j 0.5
0.4391
jTAD jeADx C jTAF jeAFx C jTBD jeBDx C jTBE jeBEx C jTCE jeCEx
k
0
0.8781 j
0.866
0.1901
C jTCF jeCFx C Wx D 0
jTAD jeADy C jTAF jeAFy C jTBD jeBDy C jTBE jeBEy C jTCE jeCEy
D jTAD j0.7605i 0.4391j C 0.4753
C jTCF jeCFy C Wy D 0
D jTAD jiuADx C juADy C juADz i
rA ð TAF D jTAF j 0.5
0.3842
j
0.866
0.2852
k
0
0.8781 jTAD jeADz C jTAF jeAFz C jTBD jeBDz C jTBE jeBEz C jTCE jeCEz
C jTCF jeCFz C Wz D 0
jTAD juADx C jTAF juAFx C jTBD juBDx C jTBE juBEx C jTCE juCEx
D jTAF j0.7605i 0.4391j 0.4753k
D jTAF jiuAFx C juAFy C kuAFz i
rB ð TBD D jTBD j 0.5
0.4391
j
0.866
0.1901
k
0
0.8781 D jTBD j0.7605i 0.4391j 0.4753k
D jTBD jiuBDx C juBDy C kuBDz i
rB ð TBE D jTBE j 0.5
0.3842
j
0.866
0.2852
k
0
0.8781 C jTCF juCFx C MWx D 0
jTAD juADy C jTAF juAFy C jTBD juBDy C jTBE juBEy C jTCE juCEy
C jTCF juCFy D 0,
jTAD juADz C jTAF juAFz C jTBD juBDz C jTBE juBEz C jTCE juCEz
C jTCF juCFz D 0
This set of equations was solved by iteration using TK Solver 2. For
˛ D 20° the results are:
jTAD j D jTBD j D 1910.5 N C ,
D jTBE j0.7605i 0.4391j 0.4753k
jTAF j D jTCF j D 16272.5 N T ,
D jTBE jiuBEx C juBEy C kuBEz jTBE j D jTCE j D 19707 N C .
i
rC ð TCE D jTCE j 1
0.0549
j
0
0.4753
k
0
0.8781 D jTCE j0i C 0.8781j 0.4753k
D jTCE jiuCEx C juCEy C kuCEz i
rC ð TCF D jTCF j 1
0.0549
j
0
0.4753
k
0
0.8781 D jTCF j0i C 0.8781j C 0.4753k
D jTCF jiuCFx C juCFy C kuCFz i
j
rG ð W D jWj 0
0
0 cos ˛
k 1 sin ˛ Check: For ˛ D 90° , the solution is jTAD j D jTAF j D jTBD j D jTBE j D
jTCE j D jTCF j D 5585.8 N C, which agrees with the solution to
Problem 6.68, obtained by another method. check.
Check: The solution of a six-by-six system by iteration has risks, since
the matrix of coefficients may be ill-conditioned. As a reasonableness
test for the solution process, TK Solver Plus was used to graph the
axial forces in the supporting bars over the range 90° < ˛ < 90° .
The graph is shown. The negative values are compression, and the
positive values are tension. When ˛ D 90° , the telescope platform is
pointing straight down, and the bars are in equal tension, as expected.
When ˛ D 90° the telescope mount is upright and the supporting bars
are in equal compression, as expected. The values of compression
and tension at the two extremes are equal and opposite in value,
and the values agree with those obtained by another method (see
Problem 6.58), as expected. Since the axial forces go from tension
to compression over this range of angles, all axial forces must pass
through zero in the interval. check.
D jWji cos ˛ j0 C k0 D iMWx 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 6.70 Determine the reactions on member AB
at A. (Notice that BC is a two-force member.)
200 N
B
A
400 mm
C
300 mm
300 mm
400 mm
Solution: Since BC is a two force member, the force in BC must
be a long the line between B and C.
y
200 N
AX
0.3 m
0.3 m
B
FBC
x
AY
0.4 m
45°
0.4 m
Fx :
Ax C FBC cos 45° D 0
Fy :
Ay FBC sin 45° 200 D 0
MA :
0.3200 0.6CFBC sin 45° D 0
Solving:
Ax D 100 N,
Ay D 100 N
FBC D 141.2 N (compression)
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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 6.71
ABC.
Determine the reactions on member
A
0.2 m
3 kN
0.2 m
B
D
0.2 m
E
C
0.8 m
Solution: First examine the entire frame to find the reactions at C.
Ay
Fx : 3 kN C Cx D 0 ) Cx D 3 kN
Ax
ME : 3 kN0.4 m Cy 0.8 m D 0 ) Cy D 1.5 kN
3 kN
A
3 kN
FBD
B
D
B
Cx
Cy
Cx
Ey
Solving we find
Cy
Note that BD is a 2-force body. Now isolate body ABC
Ax D 3 kN, Ay D 1.5 kN
Bx D FBD D 3 kN, By D 0
MA : 3 kN0.2 m C FBC 0.4 m C Cx 0.6 m D 0
Cx D 3 kN, Cy D 1.5 kN
Fy : Cy C Ay D 0
Fx : Cx C FBD C Ax C 3 kN D 0
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1
Problem 6.72 For the frame shown, determine the
reactions at the built-in support A and the force exerted
on member AB at B.
A
200 lb
B
6 ft
C
20°
6 ft
3 ft
3 ft
Solution: Element AB: The equilibrium equations are:
and
MA
FX D AX C BX D 0,
AX
AY
BY
A
B
FY D AY C BY D 0,
6 ft
MA D MA C 6BY D 0.
6 ft
Element BC: The equilibrium conditions are
FX D BX C sin20° D 0,
BY
FY D BY 200 C C cos20° D 0,
and, summing moments around B,
BX
BX
B
200 lb
6 ft
C
MB D 3200 6C sin20° C 6C cos20° D 0.
6 ft
We have six equations in six unknowns. Solving simultaneously yields
AX D 57.2 lb, AY D 42.8 lb, BX D 57.2 lb, BY D 42.8 lb, C D
167.3 lb, and MA D 256.6 ft-lb.
3 ft 3 ft
20°
C
20°
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1
Problem 6.73 The force F D 10 kN. Determine the
forces on member ABC, presenting your answers as
shown in Fig. 6.35.
F
E
D
A
B
C
1m
1m
Solution: The complete structure as a free body: The sum of the
MG D C3F 5A D 0,
2m
1m
F
moments about G:
G
GY
GX
A
2m
3F
D 6 kN which is the reaction of the floor. The
5
sum of the forces:
3m
from which A D
F
Fy D Gy F C A D 0,
1m
A
Fx D Gx D 0.
1m
Element DEG: The sum of the moments about D
M D F C 3E C 4Gy D 0,
from which E D
F 4Gy
10 16
D
D 2 kN.
3
3
2m
E
1m
F
from which Gy D F A D 10 6 D 4 kN.
GY
D
A
1m
C = −E
B = −D
8 kN
2 kN
B
C
3m
6 kN
The sum of the forces:
Fy D Gy F C E C D D 0,
from which D D F E Gy D 10 C 2 4 D 8 kN.
Element ABC : Noting that the reactions are equal and opposite:
B D D D 8 kN ,
and
C D E D 2 kN .
The sum of the forces:
Fy D A C B C C D 0,
from which A D 8 2 D 6 kN. Check
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1
Problem 6.74 Consider the frame in Problem 6.73.
The cable CE will safely support a tension of 10 kN.
Based on this criterion, what is the largest downward
force F that can be applied to the frame?
Solution: From the solution to Problem 6.73: E D
F 4Gy
,
3
F
3
F. Back substituting, E D or F D 5E,
5
5
from which, for E D 10 kN, F D 50 kN
Gy D F A, and A D
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1
Problem 6.75 The tension in cable BD is 500 lb.
Determine the reactions at A for cases (1) and (2).
E
G
6 in
D
6 in
A
B
C
300 lb
8 in
8 in
(1)
G
E
6 in
D
6 in
A
B
C
300 lb
8 in
8 in
Solution: Case (a) The complete structure as a free body: The sum
(2)
of the moments about G:
Gy
MG D 16300 C 12Ax D 0,
(a) 12 in
from which Ax D 400 lb . The sum of the forces:
Ey
Gy
Ay
Ax
Gx
16 in
Ex
300 lb
Fx D Ax C Gx D 0,
Ay
from which Gx D 400 lb.
Gx
(b)
B
α
Ax
Fy D Ay 300 C Gy D 0,
8 in
8 in
Cy
Cx
300 lb
from which Ay D 300 Gy . Element GE : The sum of the moments
about E:
ME D 16Gy D 0,
from which Gy D 0, and from above Ay D 300 lb.
Case (b) The complete structure as a free body: The free body diagram,
except for the position of the internal pin, is the same as for case (a).
The sum of the moments about G is
MG D 16300 C 12Ax D 0,
from which Ax D 400 lb .
Element ABC : The tension at the lower end of the cable is up and to
the right, so that the moment exerted by the cable tension about point
C is negative. The sum of the moments about C:
MC D 8B sin ˛ 16Ay D 0,
noting that B D 500 lb and ˛ D tan1
then
6
D 36.87° ,
8
Ay D 150 lb.
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1
Problem 6.76 Determine the reactions on member
ABCD at A, C, and D.
B
A
0.4 m
E
C
600 N
0.4 m
D
0.6 m
0.4 m
0.4 m
Solution: Consider the entire structure first
MA : Dy 0.6 m 600 N1.0 m D 0 ) Dy D 1000 N
Fx : Ax D 0
Fy : Ay C Dy 600 N D 0 ) Ay D 400 N
Ax
Ay
E
C
600 N
Dy
Now examine bar CE. Note that the reactions on ABD are opposite to
those on CE.
ME : 600 N0.4 m C Cy 0.8 m D 0 ) Cy D 300 N
MB : Cx 0.4 m 600 N0.4 m D 0 ) Cx D 600 N
T
Cy
Cx
E
600 N
In Summary we have
Ax D 0, Ay D 400 N
Cx D 600 N, Cy D 300 N
Dx D 0, Dy D 1000 N
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1
Problem 6.77 Determine the reactions on member
ABC at B and C.
4 kN
A
3 kN
0.2 m
D
B
0.2 m
C
E
0.2 m
0.2 m
Solution: Consider the whole structure first
Finally examine bar ABC
Fx : 4 kN C Cx D 0 ) Cx D 4 kN
MA : Bx 0.2 m C Cx 0.4 m D 0 ) Bx D 8 kN
Ay
ME : 4 kN0.4 m Cy 0.4 m D 0 ) Cy D 4 kN
4 kN
A
3 kN
4 kN
Ax
D
B
Bx
By
Cx
Cx
Ey
Cy
Cy
Next examine Bar BD (note that reactions on BD at B are opposite to
those on ABC at B)
MD : By 0.2 m 3 kN0.2 m D 0 ) By D 3 kN
3 kN
In Summary we have
Bx D 8 kN, By D 3 kN, Cx D 4 kN, Cy D 4 kN
Dy
Bx
Dx
By
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1
Problem 6.78 An athlete works out with a squat thrust
machine. To rotate the bar ABD, she must exert a vertical
force at A that causes the magnitude of the axial force in
the two-force member BC to be 1800 N. When the bar
ABD is on the verge of rotating, what are the reactions
on the vertical bar CDE at D and E?
0.6 m
0.6 m
C
A
0.42 m
B
D
1.65 m
E
Solution: Member BC is a two force member. The force in BC is
along the line from B to C.
C
y
FBC
Ay
0.6 m
0.6 m
tan Θ =
Dy 0.42 m
θ
D
0.42
0.6
Dx
x (FBC = 1800 N)
Θ = 34.990
FBC D 1800 N
tan D
0.42
D 34.99° .
0.6
Fx :
Dx FBC cos D 0
Fy :
Ay FBC sin C Dy D 0
MD :
1.2Ay C 0.6FBC sin D 0
C
Solving, we get
Dx D 1475 N
Dy D 516 N
Ay D 516 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 6.79 The frame supports a 6-kN load at C.
Determine the reactions on the frame at A and D.
6 kN
0.4 m
A
B
1.0 m
C
0.5 m
D
E
F
Solution: Note that members BE and CF are two force members.
Consider the 6 kN load as being applied to member ABC.
Ay
Ax
0.4 m
0.4 m
6 kN
1.0 m
B
0.8 m
C
FCF
FBE
φ
θ
tan D
0.5
0.4
D 51.34°
tan D
0.5
0.2
D 68.20°
Member DEF
FBE
θ
Dx
0.8 m
FCF
E
F
φ
0.4 m
Dy
Equations of equilibrium:
Member ABC:
Fx :
Ax C FBE cos FCF cos D 0
Fy :
Ay FBE sin FCF sin 6 D 0
MA :
0.4FBE sin 1.4FCF sin 1.46 D 0
C
Member DEF:
Fx :
Dx FBE cos C FCF cos D 0
Fy :
Dy C FBE sin C FCF sin D 0
MD :
0.8FBE sin C 1.2FCF sin D 0
C
Unknowns Ax , Ay , Dx , Dy , FBE , FCF we have 6 eqns in 6
unknowns.
Solving, we get
Ax D 16.8 kN
Ay D 11.25 kN
Dx D 16.3 kN
Dy D 5.25 kN
Also, FBE D 20.2 kN T
FCF D 11.3 kN C
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1
Problem 6.80 The mass m D 120 kg. Determine the
forces on member ABC, presenting your answers as
shown in Fig. 6.35.
A
B
C
300 m m
D
E
m
200 mm
Solution: The equations of equilibrium for the entire frame are
AX
AY
CX
CX
DY
FY D AY 2mg D 0,
CY
DY
EX
MA D 0.3EX 0.2mg 0.4mg D 0.
Solving yields AX D 2354 N, AY D 2354 N, and EX D 2354 N.
2354 N
Member ABC: The equilibrium equations are
2354 N A
4708 N
and
CY
BY
and summing moments at A,
200 mm
BY
FX D AX C EX D 0,
m
B
4708 N 2354 N
C
2354 N
B
FX D AX C CX D 0,
2354 N
2354 N
4708 N
FY D AY BY C CY D 0,
MA D 0.2BY C 0.4CY D 0.
We have three equations in the three unknowns BY , CX , and CY .
Solving, we get BY D 4708 N, CX D 2354 N, and CY D 2354 N. This
gives all of the forces on member ABC. A similar analysis can be made
for each of the other members in the frame. The results of solving for
all of the forces in the frame is shown in the figure.
4708 N
E
2354 N
C
D
1177 N
1177 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 6.81 The mass of the suspended weight is
12 kg. Determine the reactions on member ABC.
C
160 mm
B
m
D
240 mm
E
A
160
mm
160
mm
160
mm
Solution: Start with the entire structure.
C
Fx : Ax D 0
B
ME : 117.7 N0.16 m Ay 0.32 m D 0 ) Ay D 58.9 N
D
117.7 N
Now take advantage of the fact that CD is a 2-force member. Examine
body ABC
13
FCD 0.32 m
MB : Ax 0.24 m 117.7 N0.48 m p
233
Cp
8
233
Ax
Ay
FCD 0.16 m D 0
Fx : Ax C Bx p
8
233
Ey
FCD D 0
C
13
By
13
FCD D 0
Fy : Ay C By 117.7 N p
233
8
Bx
Solving we find
FCD
117.7 N
FCD D 299.5 N, Bx D 157.0 N, By D 78.5 N
Note that
8
13
FCD , Cy D p
FCD
Cx D p
233
233
Ax
In summary
Ax D 0, Ay D 58.9 N
Ay
Bx D 157.0 N, By D 78.5 N
Cx D 157.0 N, Cy D 225 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 6.82 The weight of the suspended object is
W D 50 lb. Determine the tension in the spring and the
reactions at F. (The slotted member DE is vertical.)
A
B
4 in
E
6 in
W
C
10 in
F
D
8 in
8 in
10 in
Solution: Start with member AB
Finally examine DCE
1
MA : 50 lb8 in C p FB 16 in D 0 ) FB D 35.4 lb
2
10 in
MD : T16 in C FC 10 in D 0 ) T D 62.5 lb
FB
T
1
1
Ax
FC
Ay
50 lb
Now examine BCF
p
MF : FB 20 2 in FC 10 in D 0 ) FC D 100 lb
1
Fx : p FB C FC C Fx D 0 ) Fx D 75 lb
2
1
Fy : p FB C Fy D 0 ) Fy D 25 lb
2
Dx
Dy
Summary
Tension in Spring D 62.5 lb
Fx D 25 lb, Fy D 75 lb
1
1
FB
FC
Fx
Fy
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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 6.83 The mass m D 50 kg. Determine the
forces on member ABCD, presenting your answers as
shown in Fig. 6.35.
1m
1m
D
E
1m
C
m
1m
B
1m
A
Solution: The weight of the mass hanging is W D mg D
509.81 D 490.5 N The complete structure as a free body: The sum
of the moments about A:
MA D 2W C Fy D 0,
F
Fy D Ey C Cy D 0,
from which Cy D 981 N.
Fx D Ex C Cx D 0,
from which Fy D 981 N. The sum of the forces:
from which Cx D 981 N, and
Fy D Ay C Fy W D 0,
from which Ay D 490.5 N,
Element ABCD: All reactions on ABCD have been determined
above. The components at B and C have the magnitudes
p
B D C D 9812 C 9812 D 1387 N , at angles of 45° .
Fx D Ax C Fx D 0,
from which Ax D Fx . Element BF: The sum of the moments
about F:
MF D Bx By D 0,
from which By D Bx . The sum of the forces:
Dy
Dy
Cy
Cx
Cy
Bx
By
Fy D By C Fy D 0,
Ey
Ey
W
Ex
By
Cx
Bx
Fy
Ay
Ax
from which By D 981 N, and Bx D 981 N.
Ex
Dx
Dx
Fx
Fx D Bx C Fx D 0,
490.5 N
from which Fx D 981 N, and from above, Ax D 981 N ,
Element DE: The sum of the moments about D:
MD D Ey 2W D 0,
D
C
981 N
1387 N
45°
B
from which Ey D 981 N. The sum of the forces:
Fy D Dy Ey W D 0,
A
45°
1387 N
981 N
490.5 N
from which Dy D 490.5 N .
Fx D Dx Ex D 0,
from which Dx D Ex . Element CE : The sum of the moments
about C:
MC D Ey Ex D 0,
from which Ex D 981 N, and from above Dx D 981 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 6.84
Determine the forces on member BCD.
400 lb
6 ft
B
A
4 ft
C
4 ft
D
E
8 ft
Solution: The following is based on free body diagrams of the
elements: The complete structure as a free body: The sum of the
moments about D:
MD D 6400 C 8Ey D 0,
from which Ey D 300 lb. The sum of the forces:
Fx D Dx D 0.
MA D 8By 6400 D 0,
from which By D 300 lb. The sum of forces:
The reactions are now known:
By D 300 lb , Bx D 400 lb , Cy D 200 lb ,
Dx D 0 , Dy D 100 lb ,
where negative sign means that the force is reversed from the direction
shown on the free body diagram.
Fy D Ey C Dy 400 D 0,
from which Dy D 100 lb. Element AB: The sum of the moments
about A:
Element BCD:
Fy D By Ay 400 D 0,
400 lb
Ay
Ax
Ax
By
Bx
Cy
Cx E
Ay
Bx
Cy
Dy
Cx
By
Dx
from which Ay D 100 lb.
Fx D Ax Bx D 0,
from which (1) Ax C Bx D 0 Element ACE: The sum of the moments
about E:
ME D 8Ax C 4Cx 8Ay C 4Cy D 0,
from which (2) 2Ax C Cx 2Ay C Cy D 0. The sum of the forces:
Fy D Ay C Ey Cy D 0,
from which Cy D 200 lb .
Fx D Ax Cx D 0,
from which (3) Ax D Cx . The three numbered equations are solved:
Ax D 400 lb, Cx D 400 lb , and Bx D 400 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 6.85
Determine the forces on member ABC.
E
6 kN
1m
D
C
1m
A
B
2m
Solution: The frame as a whole: The equations of equilibrium are
EX
FX D AX C EX D 0,
FY D AY C EY 6000 N D 0,
and, with moments about E,
EY
E DY
DX D
DX D
DY
BX B
B Y BX
A
B
AX
BY
AY
2m
CY
1m
6 kN
C
CY
C
ME D 2AX 56000 D 0.
Solving for the support reactions, we get AX D 15,000 N and EX D
15,000 N. We cannot yet solve for the forces in the y direction at A
and E.
Member ABC: The equations of equilibrium are
FX D AX BX D 0,
FY D AY BY CY D 0,
and summing moments about A,
MA D 2BY 4CY D 0.
Member BDE: The equations of equilibrium are
FX D EX C DX C BX D 0,
FY D EY C DY C BY D 0,
and, summing moments about E,
ME D 1DY C 1DX C 2BY C 2BX D 0.
Member CD: The equations of equilibrium are
FX D DX D 0,
FY D DY C CY 6000 D 0,
and summing moments about D,
MD D 46000 C 3CY D 0.
Solving these equations simultaneously gives values for all of the
forces in the frame. The values are AX D 15,000 N, AY D 8,000 N,
BX D 15,000 N, BY D 16,000 N, CY D 8,000 N, DX D 0, and
DY D 2,000 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 6.86
Determine the forces on member ABD.
8 in
8 in
8 in
A
60 lb
8 in
60 lb
B
E
8 in
C
Solution: The equations of equilibrium for the frame as a whole are
FX D AX C CX D 0,
and
FY D AY 60 60 D 0,
D
AY
AX
BX
DX
BY
BX
60 lb
BY
B
CX
EX
EY
DY
DY
MA D 16CX 1660 2460 D 0.
60 lb
EY
EX
DX
Solving these three equations yields
AX D 150 lb,
AY D 120 lb,
and CX D 150 lb.
Member ABD: The equilibrium equations for this member are:
FX D AX BX DX D 0,
and
FY D AY BY DY D 0,
MA D 8BY 8DY 8BX 16DX D 0.
Member BE: The equilibrium equations for this member are:
FX D BX C EX D 0,
and
FY D BY C EY 60 60 D 0,
MB D 860 1660 C 16EY D 0.
Member CDE: The equilibrium equations for this member are:
and
FX D CX C DX EX D 0,
FY D DY EY D 0,
MD D 8EX 16EY D 0.
Solving these equations, we get BX D 180 lb, BY D 30 lb, DX D
30 lb, DY D 90 lb, EX D 180 lb, and EY D 90 lb. Note that we have
12 equations in 9 unknowns. The extra equations provide a check.
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1
Problem 6.87 The mass m D 12 kg. Determine the
forces on member CDE.
A
200 mm
100 mm
E
B
200 mm
C
D
200 mm
Solution: Start with a free-body diagram of the entire frame.
m
400 mm
Ax
Eq. for entire frame:
C
!
C "
C
∴
Ay
Fx D 0:
Ax C Cx D 0 ) Ax D Cx
1
B
Fy D 0:
Ay 117.7 D 0 ) Ay D 117.7 N
Mc D 0:
Ax 0.4 117.70.7 D 0 Ax D 206 N
Cx
W = (9.81) (12)
W = 117.7
D
Ax D Cx D 206 N.
Now look at free-body diagram ABD.
Ax = 206
Eq. for ABD:
C
MB D 0:
Ay = 117.7
T = 117.7
By
Dx 0.2 117.70.2 C 2060.2 117.70.1 D 0
Bx
Dx D 29.45 N
C
!
Fx D 0:
Dx
206 C 117.7 C Bx 29.45 D 0 Bx D 117.75 N
C "
Dy
Fy D 0:
117.7 By C Dy D 0
Draw free-body diagram of CDE
Ey
Ex
Eq. for CDE:
C
!
Fx D 0:
Dx = 29.45
206 C 29.45 C Ex D 0 Ex D 235.45 or Ex D 235.45
C
Cx = 206
MD D 0:
Ex 0.2 C Ey 0.4 D 0 Ey D
Dy
235.450.2
Ex 0.2
D
0.4
0.4
Ey D 117.7 or Ey D 117.7 N
C "
Fy D 0:
Ey Dy D 0 or Dy D Ey D 117.7 Dy D 117.7 N#
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1
Problem 6.88 The weight W D 80 lb. Determine the
forces on member ABCD.
11 in
5 in
12 in
3 in
B
A
D
C
8 in
W
E
Solution: The complete structure as a free body: The sum of the
moments about A:
Ay
Fx D Ex C Ax D 0,
Cx
By
Cx
Ey
F
Dy
Dx
Cy
Bx
Ax
MA D 31W C 8Ex D 0,
from which Ex D 310 lb. The sum of the forces:
F
Cy
W
Ex
from which Ax D 310 lb .
Fy D Ey C Ay W D 0,
from which (1) Ey C Ay D W.
Element CFE: The sum of the forces parallel to x:
Fx D Ex Cx D 0,
from which Cx D 310 lb . The sum of the moments about E:
ME D 8F 16Cy C 8Cx D 0.
For frictionless pulleys, F D W, and thus Cy D 195 lb . The sum of
forces parallel to y:
Fy D Ey Cy C F D 0,
from which Ey D 115 lb .
Equation (1) above is now solvable: Ay D 35 lb .
Element ABCD: The forces exerted by the pulleys on element ABCD
are, by inspection: Bx D W D 80 lb , By D 80 lb , Dx D 80 lb ,
and Dy D 80 lb , where the negative sign means that the force is
reversed from the direction of the arrows shown on the free body
diagram.
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1
Problem 6.89 The woman using the exercise machine
is holding the 80-lb weight stationary in the position
shown. What are the reactions at the built-in support E
and the pin support F? (A and C are pinned connections.)
2 ft 2 in
B
A
9 in
1 ft 6 in
2 ft
D
C
60⬚
6 ft
80 lb
E
Solution: The complete structure as a free body: The sum of the
26
in
moments about E:
F
42
in
60°
M D 26W 68W sin 60° C 50Fy 81W cos 60° C ME D 0
W
from which (1) 50Fy C ME D 10031. The sum of the forces:
Fx D Fx C W cos 60° C Ex D 0,
ME
from which (2) Fx C Ex D 40.
W
81 in
Ey
Ex
Fx
50 in
Fy D W W sin 60° C Ey C Fy D 0,
from which (3) Ey C Fy D 149.28
Ay
Cy
Element CF: The sum of the moments about F:
Fy
Ax
Cx
M D 72Cx D 0,
from which Cx D 0. The sum of the forces:
ME
Fx D Cx C Fx D 0,
Fx
from which Fx D 0 . From (2) above, Ex D 40 lb
Element AE: The sum of the moments about E:
Fy
Ex
Ey
M D ME 72Ax D 0, .
from which (4) ME D 72Ax . The sum of the forces:
Fy D Ey C Ay D 0,
from which (5) Ey C Ay D 0.
Fx D Ax C Ex D 0;
from which Ax D 40 lb, and from (4) ME D 2880 in lb D 240 ft lb .
From (1) Fy D 143.0 lb , and from (2) Ey D 6.258 lb . This
completes the determination of the 5 reactions on E and F.
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1
Problem 6.90 Determine the reactions on member
ABC at A and B.
80 lb
E
9 in
B
C
8 in
D
A
13 in
Solution: We first examine the entire structure.
Next examine body ABC
MD : Ay 13 in C 80 lb21 in D 0
Solving:
Ay D 129.2 lb
80 lb
4 in
MB : Ax 8 in Ay 13 in C 80 lb 4 in D 0
Fx : Ax C Bx D 0
Fy : Ay C By C 80 lb D 0
80 lb
Bx
By
Dx
Ax
Ax
Ay
Dy
Ay
Solving and summarizing we have
Ax D 170 lb, Ay D 129.2 lb
Bx D 170 lb, By D 209 lb
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1
Problem 6.91 The mass of the suspended object is
m D 50 kg. Determine the reactions on member ABC.
0.2 m
A
B
0.6 m
E
D
0.8 m
C 0.2 m
0.6 m
m
Solution: Begin with an examination of the pulley at B.
Finally look at member ABC
1
Fx : Bx C p 490.5 N D 0 ) Bx D 347 N
2
1
Fy : By 490.5 N p 490.5 N D 0 ) By D 837 N
2
1
By
MC : Ax 0.6 m Ay 1.4 m Bx 0.6 m By 0.6 m D 0
) Ay D 771 N
1
Fx : Ax C Bx C Cx D 0 ) Cx D 961 N
Fy : Ay C By C Cy D 0 ) Cy D 66.6 N
By
Ay
Bx
490.5 N
Ax
Bx
490.5 N
Cy
Now examine the entire structure
MD : 490.5 N1.6 m Ax 0.6 m D 0 ) Ax D 1308 N
Cx
In Summary
Ay
Ax D 1308 N, Ay D 771 N
Ax
Bx D 347 N, By D 837 N
Cx D 961 N, Cy D 66.6 N
Dy
Dx
490.5 N
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1
Problem 6.92 The unstretched length of the string is
LO . Show that when the system is in equilibrium the
angle ˛ satisfies the relation sin ˛ D 2LO 2F/kL.
F
1– L
4
1– L
4
k
1– L
2
α
α
Solution: Since the action lines of the force F and the reaction E
The solution for angle ˛: The spring force is
are co-parallel and coincident, the moment on the system is zero, and
the system is always in equilibrium, for a non-zero force F. The object
is to find an expression for the angle ˛ for any non-zero force F.
Cy D T D k
The complete structure as a free body:
L
sin ˛ LO ,
2
L
sin ˛ LO D 2F.
2
2F
2 LO k
Solve: sin ˛ D
L
from which k
The sum of the moments about A
MA D FL sin ˛ C EL sin ˛ D 0,
from which E D F. The sum of forces:
F
Fx D Ax D 0,
L
from which Ax D 0.
α
Fy D Ay C E F D 0,
from which Ay D 0, which completes a demonstration that F does not
exert a moment on the system. The spring C: The elongation of the
L
spring is s D 2 sin ˛ LO , from which the force in the spring is
4
TDk
L
sin ˛ LO
2
Ax
Ay
E
By
Cy
Bx
L
4
α
E
L
4
Element BE: The strategy is to determine Cy , which is the spring force
on BE. The moment about E is
L
L
L
ME D Cy cos ˛ By cos ˛ Bx cos ˛ D 0,
4
2
2
from which
Cy
C By D Bx . The sum of forces:
2
Fx D Bx D 0,
from which Bx D 0.
Fy D Cy C By C E D 0,
from which Cy C By D E D F. The two simultaneous equations
are solved: Cy D 2F, and By D F.
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1
Problem 6.93 The pin support B will safely support a
force of 24-kN magnitude. Based on this criterion, what
is the largest mass m that the frame will safely support?
C
500 mm
100 mm
E
D
B
300 mm
m
A
300 mm
Solution: The weight is given by W D mg D 9.81 g
Sum the forces in the x-direction:
Fx D Ax D 0,
Cx
Cx
W
By
W
Bx
By Bx
W
Ay
from which Ax D 0
Element ABC: The sum of the moments about A:
400 mm
400 mm
Cy
Cy
The complete structure as a free body:
F
Ey
Ex
Ey
Ex
F
Ax
MA D C0.3Bx C 0.9Cx 0.4W D 0,
from which (1) 0.3Bx C 0.9Cx D 0.4W. The sum of the forces:
Fx D Bx Cx C W C Ax D 0,
from which (2) Bx C Cx D W. Solve the simultaneous equations (1)
5
and (2) to obtain Bx D W
6
Element BE : The sum of the moments about E:
ME D 0.4W 0.7By D 0,
from which By D
4
W. The magnitude of the reaction at B is
7
2
4
5 2
C
D 1.0104W.
6
7
jBj D W
24
D 23.752 kN is the
1.0104
maximum load that can be carried. Thus, the largest mass that can be
supported is m D W/g D 23752 N/9.81 m/s2 D 2421 kg.
For a safe value of jBj D 24 kN, W D
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1
Problem 6.94
Determine the reactions at A and C.
C
A
3 ft
72 ft-lb
36 lb
3 ft
B
18 lb
4 ft
Solution: The complete structure as a free body:
The sum of the moments about A:
8 ft
Cy
Ay
Ax
Cx 3 ft
72 ft-lb
36 lb
MA D 418 C 336 C 12Cy 72 D 0,
18 lb
from which Cy D 3 lb. The sum of the forces:
4 ft
Ay
Fy D Ay C Cy 18 D 0,
8 ft
Ax
from which Ay D 15 lb.
Fx D Ax C Cx C 36 D 0,
72 ft-lb
By 6 ft
Bx
18 lb
from which (1) Cx D Ax 36
Element AB: The sum of the forces:
Fy D Ay By 18 D 0,
from which By D 3 lb. The sum of the moments:
MA D 6Bx 418 4By 72 D 0,
from which Bx D 22 lb. The sum of the forces:
Fx D Ax C Bx D 0,
from which Ax D 22 lb From equation (1) Cx D 14 lb
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1
Problem 6.95
Determine the forces on member AD.
200 N
130 mm
D
400 mm
C
A
B
400 mm
Solution: Denote the reactions of the support by Rx and Ry . The
400 mm
Dy
200 N
complete structure as a free body:
from which Rx D 400 N. The sum of moments:
MA D 800C 400930 C 400530 400200 D 0,
Ay
Ax
Ry
Dy
400 N
Ay
Fx D Rx 400 D 0,
400 N
By
400 N
Ax
Rx
Dx
Dx
Bx
By
Bx
C
from which C D 300 N.
Fy D C C Ry 400 200 D 0,
from which Ry D 300 N. Element ABC : The sum of the moments:
MA D 4By C 8C D 0,
from which By D 600 N. Element BD: The sum of the forces:
Fy D By Dy 400 D 0,
from which Dy D 200 N.
Element AD: The sum of the forces:
Fy D Ay C Dy 200 D 0,
from which Ay D 0: Element AD: The sum of the forces:
and
Fx D Ax C Dx D 0
MA D 400200 C 800Dy 400Dx D 0
Ax D 200 N, and Dx D 200 N.
Element BD: The sum of forces:
Fx D Bx Dx 400 D 0
from which Bx D 600 N. This completes the solution of the nine
equations in nine unknowns, of which Ax , Ay , Dx , and Dy are the
values required by the Problem.
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1
Problem 6.96 The frame shown is used to support
high-tension wires. If b D 3 ft, ˛ D 30° , and W D
200 lb, what is the axial force in member HJ?
A
B
α
C
D
α
E
G
F
W
H
α
I
J
α
W
W
b
Solution: Joints B and E are sliding joints, so that the reactions
are normal to AC and BF, respectively. Member HJ is supported by
pins at each end, so that the reaction is an axial force. The distance
h D b tan ˛ D 1.732 ft
Member ABC. The sum of the forces:
B
b
b
Ay
Ax
B
h
b G
Dy
Dx E
y
W
Fx D Ax C B sin ˛ D 0,
b
Gx
H
W
W
Fy D Ay W B cos ˛ D 0.
The sum of the moments about B:
MB D bAy hAx C bW D 0.
These three equations have the solution: Ax D 173.21 lb, Ay D
100 lb, and B D 346.4 lb.
Member BDEF: The sum of the forces:
Fx D Dx B sin ˛ E sin ˛ D 0,
Fy D Dy W C B cos ˛ E cos ˛ D 0.
The sum of the moments about D:
MD D 2bW bE cos ˛ hE sin ˛ bB cos ˛ C hB sin ˛ D 0.
These three equations have the solution: Dx D 259.8 lb, Dy D
350 lb, E D 173.2 lb.
Member EGHI: The sum of the forces:
Fx D Gx C E sin ˛ H cos ˛ D 0,
Fy D Gy W C E cos ˛ C H sin ˛ D 0.
The sum of the moments about H:
MH D bGy hGx C bW C 2bE cos ˛ 2hE sin ˛ D 0.
These three equations have the solution: Gx D 346.4 lb, Gy D 200 lb,
and H D 300 lb. This is the axial force in HJ.
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1
Problem 6.97 The truck and trailer are parked on a 10°
slope. The 14,000-lb weight of the truck and the 8000-lb
weight of the trailer act at the points shown. The truck’s
brakes prevent its rear wheels at B from turning. The
truck’s front wheels at C and the trailer’s wheels at A
can turn freely, which means they do not exert friction
forces on the road. The trailer hitch at D behaves like a
pin support. Determine the forces exerted on the truck
at B, C, and D.
2 ft
y
9 ft
3 ft
14 ft
D
4 ft
3 ft
8 kip
6 ft
14 kip
B
10⬚
5 ft 6 in
x
C
A
Strategy: Draw the individual free-body diagrams of
the truck and trailer.
Solution: Determine the Reactions at the Supports The reactions
25 ft
y
in this example are the forces exerted on the truck and trailer by the
road. We draw the free-body diagram of the connected truck and trailer
in Fig. a. Because the tires at B are locked, the road can exert both a
normal force and a friction force, but only normal forces are exerted
at A and C. The equilibrium equations are
22 ft
14 ft
4 ft
D
3 ft
8 kip
Fx D Bx 8 sin 10° 14 sin 10° D 0,
Bx
6 ft
By
A
Fy D A C By C C 8 cos 10° 14 cos 10° D 0,
Mpoint A D 14By C 25C C 68 sin 10° 2 ft
Dx
48 cos 10° C 314 sin 10° Dy
8 kip
2214 cos 10° D 0.
11 ft
8 ft
16 ft
4 ft
Dx
Dy
3 ft
D
5 ft 6 in
By
6 ft
C
14 kip
(a)
x
Bx
14 kip
C
A
From the first equation we obtain the reaction Bx D 3.82 kip, but we
can’t solve the other two equations for the three reactions A, By , and C.
(b)
(c)
Analyze the Members We draw the free-body diagrams of the trailer
and truck in Figs. b and c, showing the forces Dx and Dy exerted at
the hitch. Only three unknown forces appear on the free-body diagram
of the trailer. From the equilibrium equations for the trailer,
Fx D Dx 8 sin 10° D 0,
Fy D A C Dy 8 cos 10° D 0,
Mpoint D D 0.58 sin 10° C 128 cos 10° 16A D 0.
we obtain A D 5.95 kip, Dx D 1.39 kip, and Dy D 1.93 kip. (Notice
that by summing moments about D, we obtained an equation containing
only one unknown force.)
The equilibrium equations for the truck are
Fx D Bx Dx 14 sin 10° D 0,
Fy D By C C Dy 14 cos 10° D 0,
Mpoint B D 11C C 5.5Dx 2Dy C 314 sin 10° 814 cos 10° D 0.
Using the known values of Dx and Dy , we can solve these equations,
obtaining Bx D 3.82 kip, By D 6.69 kip, and C D 9.02 kip.
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1
Problem 6.99 Figure a is a diagram of the bones and
biceps muscle of a person’s arm supporting a mass.
Tension in the biceps muscle holds the forearm in the
horizontal position, as illustrated in the simple mechanical model in Fig. b. The weight of the forearm is 9 N,
and the mass m D 2 kg.
(a)
(b)
Determine the tension in the biceps muscle AB.
Determine the magnitude of the force exerted on
the upper arm by the forearm at the elbow joint C.
B
290
mm
(a)
A
50
mm
C
9N
m
150 mm
200 mm
(b)
Solution: Make a cut through AB and BC just above the elbow
joint C. The angle formed
by the biceps muscle with respect to the
290
forearm is ˛ D tan1
D 80.2° . The weight of the mass is W D
50
29.81 D 19.62 N.
T
W
9N
200
mm
α Cy
50
150 mm
mm
Cx
The section as a free body: The sum of the moments about C is
MC D 50T sin ˛ C 1509 C 350W D 0,
from which T D 166.76 N is the tension exerted by the biceps muscle
AB. The sum of the forces on the section is
FX D Cx C T cos ˛ D 0,
from which Cx D 28.33 N.
FY D Cy C T sin ˛ 9 W D 0,
from which Cy D 135.72. The magnitude of the force exerted by the
forearm on the upper arm at joint C is
FD
C2x C C2y D 138.65 N
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1
Problem 6.100 The clamp presses two blocks of wood
together. Determine the magnitude of the force the
members exert on each other at C if the blocks are
pressed together with a force of 200 N.
125 mm
125 mm
B
50 mm
E
A
50 mm
Solution: Consider the upper jaw only.
125 mm
C
50 mm
The section ABC as a free body:
D
The sum of the moments about C is
MC D 100B 250A D 0,
B
from which, for A D 200 N, B D 500 N. The sum of the forces:
Cy
100 mm
Fx D Cx B D 0,
A
Cx
from which Cx D 500 N,
250 mm
Fy D Cy C A D 0,
from which Cy D 200 N. The magnitude of the reaction at C:
CD
C2x C C2y D 538.52 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 6.101 The pressure force exerted on the
piston is 2 kN toward the left. Determine the couple
M necessary to keep the system in equilibrium.
B
300 mm
350 mm
45°
A
C
M
400 mm
Solution: From the diagram, the coordinates of point B are d, d
b can be determined from the
where d D 0.3 cos45° . The distance
Pythagorean Theorem as b D 0.352 d2 . From the diagram, the
angle D 37.3° . From these calculations, the coordinates of points B
and C are B (0.212, 0.212), and C (0.491, 0) with all distances being
measured in meters. All forces will be measured in Newtons.
B
0.3 m
45°
A
0.35 m
d
d
θ
b
C
The unit vector from C toward B is uCB D 0.795i C 0.606j.
y
The equations of force equilibrium at C are
and
FX D FBC cos 2000 D 0,
FBC
FBCY
c
2000 N
x
FBCX
N
FY D N FBC sin D 0.
Solving these equations, we get N D 1524 Newtons(N), and FBC D
2514 N.
The force acting at B due to member BC is FBC uBC D 2000i C
1524j N.
B
y
FBC uCB
M
A
rAB
x
The position vector from A to B is rAB D 0.212i C 0.212j m, and the
moment of the force acting at B about A, calculated from the cross
product, is given by MFBC D 747.6k N-m (counter - clockwise). The
moment M about A which is necessary to hold the system in equilibrium, is equal and opposite to the moment just calculated. Thus,
M D 747.6k N-m (clockwise).
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1
Problem 6.102 In Problem 6.101, determine the forces
on member AB at A and B.
Solution: In the solution of Problem 6.101, we found that the force
acting at point B of member AB was FBC uBC D 2000i C 1524j N,
and that the moment acting on member BC about point A was given by
M D 747.6k N-m (clockwise). Member AB must be in equilibrium,
and we ensured moment equilibrium in solving Problem 6.101.
FBC uCB
y
B
M
From the free body diagram, the equations for force equilibrium are
and
AX
A
x
FX D AX C FBC uBCX D AX 2000 N D 0,
AY
FY D AY C FBC uBCY D AY C 1524 N D 0.
Thus, AX D 2000 N, and AY D 1524 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 6.103 The mechanism is used to weigh mail.
A package placed at A causes the weighted point to rotate
through an angle ˛. Neglect the weights of the members
except for the counterweight at B, which has a mass of
4 kg. If ˛ D 20° , what is the mass of the package at A?
A
100
mm
100
mm
B
30°
α
Solution: Consider the moment about the bearing connecting the
motion of the counter weight to the motion of the weighing platform. The moment arm of the weighing platform about this bearing
is 100 cos30 ˛. The restoring moment of the counter weight is
100 mg sin ˛. Thus the sum of the moments is
M D 100 mB g sin ˛ 100 mA g cos30 ˛ D 0.
Define the ratio of the masses of the counter weight to the mass of the
mB
. The sum of moments equation reduces to
package to be RM D
mA
M D RM sin ˛ cos30 ˛ D 0,
from which RM D
is mA D
cos30 ˛
D 2.8794, and the mass of the package
sin ˛
4
D 1.3892 D 1.39 kg
RM
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1
Problem 6.104 The scoop C of the front-end loader
is supported by two identical arms, one on each side
of the loader. One of the two arms (ABC) is visible in
the figure. It is supported by a pin support at A and
the hydraulic actuator BD. The sum of the other loads
exerted on the arm, including its own weight, is F D
1.6 kN. Determine the axial force in the actuator BD
and the magnitude of the reaction at A.
A
D
Solution: The section ABC as a free body: The sum of the moments
from which BD D 4 kN.
1m
Ax
0.8 m
BD
F
The sum of the forces:
1m
Ay
about A:
MA D 0.8BD 2F D 0,
B
F
0.2 m
C
0.8 m
2m
Fx D Ax C BD D 0,
from which Ax D 4 kN.
Fy D Ay F D 0,
from which Ay D 1.6 kN. The magnitude of the reaction at A is
AD
A2x C A2y D 4.308 kN
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1
Problem 6.105 The mass of the scoop is 220 kg, and
its weight acts at G. Both the scoop and the hydraulic
actuator BC are pinned to the horizontal member at B.
The hydraulic actuator can be treated as a two-force
member. Determine the forces exerted on the scoop at
B and D.
1m
D
C
1m
0.6 m
G
B
A
0.15 m
0.6 m
0.3 m
Scoop
Solution: We need to know the locations of various points in the
Problem . Let us use horizontal and vertical axes and define the coordinates of point A as (0,0). All distances will be in meters (m) and all
forces will be in Newtons (N). From the figure in the text, the coordinates in meters of the points in the problem are A (0, 0), B (0.6, 0),
C (0.15, 0.6), D (0.85, 1), and the x coordinate of point G is 0.9 m.
The unit vector from C toward D is given by uCD D 0.928i C 0.371j,
and the force acting on the scoop at D is given by D D DX i C DY j D
0.928Di C 0.371Dj. From the free body diagram of the scoop, the
equilibrium equations are
and
DY
y
D
DX
C
mg
B
A
x
BX
BY
FX D BX C DX D 0,
D uCD
DY
FY D BY C DY mg D 0,
DX
MB D 0.3 mg C xBD DY yBD DX D 0.
From the geometry, xBD D 0.25 m, and yBD D 1 m. Solving the
equations of equilibrium, we obtain BX D 719.4 N, BY D 2246 N, and
D D 774.8 N (member CD is in tension).
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1
Problem 6.106 The woman exerts 20-N forces on the
handles of the shears. Determine the magnitude of the
forces exerted on the branch at A.
20 N
D
C
B
A
E
36 mm 25 mm 25 mm
65 mm
Solution: Assume that the shears are symmetrical.
Consider the 2 pieces CD and CE
20 N
Now examine CD by itself
MC D 20 N90 mm C Dy 25 mm D 0 ) Dy D 72 N
20 N
Fx D 0 ) D x D E x
Dy
Fy D 0 ) Dy D Ey
Cy
MC D 0 ) D x D E x D 0
Dx = 0
20 N
Dy
Cx
Finally examine DBA
Dx
MB : A36 mm Dy 50 mm D 0
A
C
Dx = 0
By
Ex
Ey
Bx
20 N
Dy
Solving we find
A D 100 N
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1
Problem 6.107 The person exerts 40-N forces on the
handles of the locking wrench. Determine the magnitude
of the forces the wrench exerts on the bolt at A.
A
B
40 N
8 mm 40 mm
E
C
D
50 mm
30 mm
75 mm
40 N
75
FDE
Solution: Recognize that DE is a 2-force member. Examine
part CD
8
75
FDE C Cx D 0
Fx : p
5689
Fy : p
8
5689
D
Cx
Cy
FDE C Cy C 40 N D 0
8
FDE 30 mm C 40 N105 mm D 0
MC : p
5689
40 N
By
Solving we find
Cx D 1312.5 N, Cy D 100 N, FDE D 1320 N
A
Bx
Now examine ABC
MB : A50 mm Cx 40 mm D 0
Fx : Bx Cx D 0
Cx
Fy : By Cy A D 0
Solving:
A D 1050 N, Bx D 1312.5 N
Answer:
A D 1050 N
Cy
By D 1150 N
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1
Problem 6.108 In Problem 6.107, determine the
magnitude of the force the members of the wrench exert
on each other at B and the axial force in the two-force
member CD.
Solution: From the previous problem we have
BD
Bx 2 C By 2 D 1745 N
FDE D 1320 NC
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1
Problem 6.109 The device is designed to exert a large
force on the horizontal bar at A for a stamping operation.
If the hydraulic cylinder DE exerts an axial force of
800 N and ˛ D 80° , what horizontal force is exerted on
the horizontal bar at A?
90°
D
m
α
25
0m
0
25
B
mm
25
0m
m
A
E
C
400 mm
Solution: Define the x-y coordinate system with origin at C. The
projection of the point D on the coordinate system is
Fy
B
Ry D 250 sin ˛ D 246.2 mm,
and Rx D 250 cos ˛ D 43.4 mm.
Py
D
Px
Fx
Cx
Cy
The angle formed
by member
DE with the positive x axis is D
Ry
1
180 tan
D 145.38° . The components of the force
400 Rx
produced by DE are Fx D F cos D 658.3 N, and Fy D F sin D
454.5 N. The angle of the element AB with the positive x axis is ˇ D
180 90 ˛ D 10° , and the components of the force for this member
are Px D P cos ˇ and Py D P sin ˇ, where P is to be determined. The
angle of the arm BC with the positive x axis is D 90 C ˛ D 170° .
The projection of point B is Lx D 250 cos D 246.2 mm, and Ly D
250 sin D 43.4 mm. Sum the moments about C:
MC D Rx Fy Ry Fx C Lx Py Ly Px D 0.
Substitute and solve: P D 2126.36 N, and Px D P cos ˇ D 2094 N is
the horizontal force exerted at A.
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1
Problem 6.110 This device raises a load W by
extending the hydraulic actuator DE. The bars AD and
BC are 4 ft long, and the distances b D 2.5 ft and h D
1.5 ft. If W D 300 lb, what force must the actuator exert
to hold the load in equilibrium?
b
W
A
B
C
D
h
Solution: The angle ADC is ˛ D sin1
h
D 22.02° . The
4
(1)
dCx hCy dB D 0. The sum of the forces:
distance CD is d D 4 cos ˛.
E
Fx D Cx Ex D 0, from which
The complete structure as a free body: The sum of the forces:
(2)
Fy D W C Cy C Dy D 0.
Ex Cx D 0,
Fx D Cx C Dx D 0.
Fy D Cy Ey C B D 0,
from which
The sum of the moments about C:
W
MC D bW C dDy D 0.
A
These have the solution:
B
Ex
B
A
Cy
Cy D 97.7 lb,
Ey
Cx E
Ex
Dy D 202.3 lb,
Dx
and Cx D Dx .
Divide the system into three elements: the platform carrying the
weight, the member AB, and the member BC.
(3)
Dy
Cy Ey C B D 0
Element AD: The sum of the moments about E:
The Platform: (See Free body diagram) The moments about the
point A:
MA D bW dB D 0.
The sum of the forces:
Fy D A C B C W D 0.
ME D
d
h
d
Dy C
Dx A D 0,
2
2
2
from which
(4)
dDy C hDx dA D 0.
These are four equations in the four unknowns: EX , EY , Dx , CX and
DX
These have the solution:
Solving, we obtain Dx D 742 lb.
B D 202.3 lb,
and A D 97.7 lb.
Element BC: The sum of the moments about E is
MC D h
d
d
Cy C
Cx C
B D 0, from which
2
2
2
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1
Problem 6.111 The four-bar linkage operates the forks
of a fork lift truck. The force supported by the forks is
W D 8 kN. Determine the reactions on member CDE.
0.7 m
0.15 m
0.2 m
W
C
0.15 m
B
D E
Forks
0.2 m
0.3 m
A
F
0.2 m
Solution: Consider body BC. Note that AB is a 2-force body.
W = 8 kN
Fx : Cx D 0
MB : Cy 0.2 m 8 kN0.9 m D 0
Cx
) Cx D 0, Cy D 36 kN
Now examine CDE. Note that DF is a 2-force body.
3
ME : Cy 0.15 m Cx 0.15 m C p FDF 0.15 m D 0
13
2
Fx : Cx C Ex C p FDF D 0
13
3
Fy : Cy C Ey p FDF D 0
13
Solving we find
Note that
Cy
FAB
Cy
Cx
FDF D 43.3 kN, Ex D 24 kN, Ey D 0
2
3
Dx D p FDF , Dy D p FDF
13
13
Summary:
Cx D 0, Cy D 36 kN
Dx D 24 kN, Dy D 36 kN
Ex D 24 kN, Ey D 0
Ey
Ex
D
3
2
FDF
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1
Problem 6.112 A load W D 2 kN is supported by the
members ACG and the hydraulic actuator BC. Determine the reactions at A and the compressive axial force
in the actuator BC.
A
0.75 m
B
C
1m
G
0.5 m
W
1.5 m
1.5 m
Solution: The sum of the moments about A is
MA D 0.75BC 32 D 0,
from which BC D 8 kN is the axial force. The sum of the forces
FX D AX C BC D 0,
from which AX D 8 kN.
FY D AY 2 D 0,
from which AY D 2 kN.
AY
A
0.75 m X
BC
W
3m
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1
Problem 6.113 The dimensions are a D 260 mm, b D
300 mm, c D 200 mm, d D 150 mm, e D 300 mm, and
f D 520 mm. The ground exerts a vertical force F D
7000 N on the shovel. The mass of the shovel is 90 kg
and its weight acts at G. The weights of the links AB
and AD are negligible. Determine the horizontal force P
exerted at A by the hydraulic piston and the reactions on
the shovel at C.
b
P
Shovel
A
Solution: The free-body diagram of the shovel is from which we
obtain the equations
Fx D Cx T cos ˇ D 0,
(1)
Fy D Cy C T sin ˇ C F mg D 0,
(2)
D
a
d
B
C
G
c
e
MptC D fF emg C b cT sin ˇ
C dT cos ˇ D 0.
(3)
f
F
The angle ˇ D arctan[a d/b].
From the free-body diagram of joint A,
T
B
β
P
d
b−c
T
CX
we obtain the equation
CY
mg
F D P C T cos ˇ D 0. (4)
Substituting the given information into Eqs. (1)–(4) and solving, we
obtain
T D 19, 260 N,
e
P D 18, 080 N,
Cx D 18, 080 N,
and Cy D 513 N.
f
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F
1
Problem 6.114 The structure shown in the diagram
(one of the two identical structures that support the
scoop of the excavator) supports a downward force F D
1800 N at G. Members BC and DH can be treated as
two-force members. Determine the reactions on member
CDK at K.
320
mm
C
Shaft
100
mm
Scoop
260
mm
H
D
260
mm
B
180
mm
J
160
mm
L
K
1040
mm
380
mm
1120
mm
200
mm
Solution: Start with the scoop
Now examine CDK
56
4
FDH 0.26 m p FBC 0.52 m D 0
MK : p
3161
17
56
4
FDH C p FBC C Kx D 0
Fx : p
3161
17
5
1
FDH p FBC C Ky D 0
Fy : p
3161
17
4
1
MJ : p FBC 0.44 m p FBC 0.06 m
17
17
1800 N0.2 m D 0
) FBC D 873 N
FBC
1
Solving we find
4
G
F
Kx D 847 N, Ky D 363 N
4
1800 N
1
FBC
Jx
56
5
Jy
FDH
Kx
Ky
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1
Problem 6.115 (a) For each member of the truss,
obtain a graph of (axial force)/F as a function of x for
0 x 2 m.
(b) If you were designing this truss, what value of x
would you choose based on your results in (a)?
E
C
1m
A
D
B
x
F
2m
Solution: The
angle of member CD with the positive x axis is
EY
1
, which lies in the interval 90° > ˇ > 26.6° . The angle
x
1
of member AC with the negative x axis is ˛ D tan1
, which
2x
lies in the interval 26.57° ˛ < 90° .
ˇ D tan1
EX
1m
D
The complete structure as a free body: The sum of the moments about
E is
ME D 2F C D D 0,
AB
AC
α
from which D D 2F. The sum of forces
Joint A
FX D EX C D D 0,
FY D EY F D 0,
from which EY D F. These values are to be used as checks on the
joint analysis.
The method of joints: Divide each axial force by the applied force F.
(This is equivalent to adopting the value F D 1)
Joint A: The sums of forces:
and
FY D AC sin ˛ 1 D 0,
FX D AC cos ˛ AB D 0
A 2.5
x
2
i 1.5
a
1
l
.5
0
f
o −.5
r −1
c −1.5
e
−2
/
−2.5
r
BC
AB
BD
1
Joint B
EY
EX
CE
CE
α
β
AC
CD BC
Joint C
from which EX D 2F.
F
2m
DE
Joint E
Axial forces / Force ratio
CE
AC
ED
BC
AB & BD
CD
0 .2 .4 .6 .8 1 1.2 1.4 1.6 1.8 2
X, m
Joint B: The sums of forces:
and
FY D BC D 0,
FX D BD C AB D 0
Joint C: The sums of forces:
and
FY D CD sin ˇ AC sin ˛ D 0,
FX D CE C AC cos ˛ CD cos ˇ D 0.
The TK Solver Plus commercial package was used to plot the axial
force ratios as a function of x.
(b) The minimum compressive axial forces occur at about x D 0.8 m,
so if compressive axial forces are a concern for safety reasons, this
configuration offers advantages. However, note that as lim x ! 2, the
truss approaches a rectangle with CD as a cross brace. While the
compressive force in CD is increased, the compressive force in BD is
reduced (AB doesn’t count, it no longer has any length) and the axial
tension in AC is reduced. None of the axial tensions are increased. Thus
this configuration offers the advantages of material saving, without
safety penalties.
Joint E: The sums of forces:
and
FX D EX C CE D 0
FY D EY DE D 0.
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1
Problem 6.116 The mechanism shown is used to
weigh mail. A package placed at A causes the weighted
pointer to rotate through an angle ˛. Neglect the weights
of the members except for the counterweight at B, which
has a mass of 4 kg.
(a)
A
Obtain a graph of the angle ˛ as a function of the
mass of the mail for values of the mass from 0
to 2 kg.
Use the results of (a) to estimate the value of ˛
when the mass is 1 kg.
(b)
100
mm
100
mm
B
30⬚
a
Solution: First examine the support at A. Notice the 2-force link
mg
Summing moments about B we find that T D 0.
Summing forces in the x direction we find that Bx D 0
Finally summing forces in the y direction we find that By D mg
Now examine the pointer arm.
(a)
MO : mg0.1 m cos30° ˛ C 4 kgg0.1 m sin ˛ D 0
We can rearrange this equation in the form
m D 4 kg
sin ˛
cos30° ˛
T
Bx
From this equation we see that
mD0)˛D0
m D 2 kg ) ˛ D 30°
By
Thus we have the plot
Problem 6.116
mg
Oy
2
1.8
1.6
30°- α
Ox
1.4
mass
1.2
α
1
0.8
0.6
0.4
0.2
5
(b)
(4 kg)g
0
5
10
15
alpha
20
25
30
From the plot we estimate that when m D 1 kg, ˛ ³ 15° . Using
a root solver we find that
m D 1 kg ) ˛ D 13.90°
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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 6.117 A preliminary design for a bridge
structure is shown. The forces F are the loads the
structure must support at G, H, I, J, and K. Plot the
axial forces in members AB and BC as a function of the
angle ˇ. Use your graphs to estimate the value of ˇ for
which the maximum compressive load in any member
of the bridge does not exceed 2F. Draw a sketch of the
resulting design.
F
F
F
F
F
G
H
I
J
K
b
b
β
b
β
b
2b
C
α
B
D
A
α
E
Solution: Follow Example 6.3. Draw a Freebody diagram of
Joint B
Assume a unit load F.
Fx : TAB cos ˛ C TBC cos ˇ D 0
F
TBC
(1)
Fy : F C TBC sin ˇ TAB sin ˛ D 0 (2)
β
B
Joint C :
α
F
TAB
From the plot, jTAB j D 2F at ˇ ¾
D 20.5° (actually 20.7° ). A C ˇ D
20.7° , ˛ D 48.6° and the loads are
β
β
TCD
MEMBER
LOAD
AG, BH, CI, DJ, EK,
AB, DE
BC, CD
F
ZF
1.414F
F
F
F
F
F
TBC
Fx :
TBC cos ˇ C TCD cos ˇ D 0
∴TBC D TCD
β
(obvious from summary)
Fy :
α
F 2TBC sin ˇ D 0 (3)
β
α
Solving the three eqns in 3 unknown F D 1, for 15° ˇ 30° and
plotting the results, we get
Forces in members AB and BC
−8
TBC
Forces (multiples of F)
−1
−1.2
−1.4
−1.6
TAB
−1.8
−2
−2.2
−2.4
14
16
18
20 22 24 26
Angle β (degrees)
28
30
32
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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 6.118 The hydraulic cylinder DE exerts an
axial force of 800 N.
(b)
D
m
0m
25
A
a
m
B
0m
Obtain a graph of the horizontal component of force
exerted on the horizontal bar at A by the rod AB
for values of ˛ from 45° to 85° .
Use the results of (a) to estimate the value of ˛ for
which the horizontal force is 2 kN.
25
(a)
90⬚
25
0m
m
E
C
400 mm
Solution: The algorithm is taken directly from the solution to
Problem 6.109 (which see). The parameters are: F D 800 N, b D
400 mm, LCD D 250 mm, LBC D 250 mm, and LAB D 250 mm.
(1)
Adopt a value of ˛ in the interval, and compute in order:
(2)
Ry D LCD sin ˛, Rx D LCD cos ˛
(3)
D 180° tan1
(4)
Fx D F cos , Fy D F sin (5)
ˇ D 90° ˛
(6)
D 90° C ˛
(7)
Lx D L cos , Ly D L sin (8)
Px D
Ry
b Rx
Rx Fy Ry Fx
Ly cos ˇ Lx sin ˇ
cos ˇ
where Px is the horizontal force.
(b) From an inspection of a Table of values, the force Px D 2 kN
occurs at an angle ˛ D 79.5°
Piston force vs angle
4.5
4
F 3.5
o
3
r
c 2.5
e
2
,
1.5
k
N
1
.5
0
45
50
55
60
65
70
75
80
85
Angle, deg
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1
Problem 6.119 The weight of the suspended object
is 10 kN. The two members have equal cross-sectional
areas A, and each will safely support an axial force of
40A MN, where A is in square meters. Determine the
value of h that minimizes the total volume of material
in the two members.
1m
0.5 m
h
Solution: Assume that the maximum safe axial force is to be
always imposed. The cross sectional area can always be modified to
assure that this is true. However, one member will have a higher axial
force than the other, and that member will determine the area for
maximum safe axial force.
Make a cut near the pin supports.
The sum of the moments about C is
B
C
MC D W C 1.5By D 0,
α
2
from which By D
W.
3
β
D
W
The sum of the forces
Fy D Cy C By W D 0,
from which Cy D
1
W.
3
The vertical components of the forces:
Cy TCD sin ˛ D 0,
from which TCD sin ˛ D
1
W.
3
Axial force vs h
.02
.018
F .016
o .014
r
.012
c
e .01
, .008
.006
M .004
N .002
0
Similarly,
TBD sin ˇ D
TCD
.5
0
1
2
W
3
The algorithm for computing the volume:
(1)
The angle ˇ D tan1 2 h.
(2)
The tension TBD D
(3)
The lengths of the members:
LCD D
and LBD D
1.5
2
h, m
The angle is larger than the angle, and both are in the first quadrant,
hence sin ˇ > sin ˛, so it is not obvious from the expressions for the
axial force which will be the larger. A graph of the tensions in the two
members demonstrates that TBD is the larger. Thus, TBD D 40A MN,
TBD
.
from which A D
40
(4)
TBD
2W
3 sin ˇ
Volume vs h
.05
.045
.04
.035
4 .03
0 .025
V .02
.015
.01
.005
0
0
.2
.4
.6
h, m
.8
1
1.2
p
h2 C 1
p
h2 C 0.25
The volume of the two members:
V D ALCD C LBD D
TBD
LCD C LBD .
40
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1
6.119 (Continued )
TK Solver Plus was used to obtain the graphs.
From the graph of the volume, a minimum volume occurs for h ¾
D 0.6.
A Table of values:
h, m
0.59
0.60
0.61
0.62
40V
0.16904
0.16898
0.16896
0.16898
shows a minimum at h D 0.61 m
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 6.120 The bars AD and BC are 4 ft long, the
distance b D 2.5 ft, and W D 300 lb. If the largest force
the hydraulic actuator DE can exert is 1000 lb, what is
the smallest height h at which the load can be supported?
b
W
A
B
h
C
D
E
Solution: Follow the solution of Problem 6.110, solve the problem
for the specified range of values for h and plot the results. The resulting
plot is shown at the right. From the plot, it is seen that the minimum
value for h is about h D 1.15 ft.
Actuator load (lb) vs platform height (ft)
−600
A
c −800
t
u−1000
a
t −1200
o−1400
r
−1600
L
o−1800
a
d−2000
l −2200
−2400
.5
.6
.7
.8
.9
1
1.1
1.2
Height of platform, h (ft)
1.3
1.4
1.5
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1
Problem 6.121 The linkage is in equilibrium under
the action of the couples MA and MB . When ˛A D 60° ,
˛B D 70° . For the range of 0 ˛A 180° , estimate the
maximum positive and negative values of MA /MB and
the values of ˛A at which they occur.
0.7 m
0.15 m
0.2 m
W
C
0.15 m
D E
B
Forks
0.2 m
0.3 m
A
F
0.2 m
Solution: The algorithm is taken from Problem 6.111, with the
addition of the relationship between the angles. (See Figure with
Problem 6.111 for dimensions.). The algorithm is:
(a)
For a value of ˛A , 0° ˛A 180° , solve for the angle
˛B and the angle of the linkage connection ˇ from the
two simultaneous equations: (1) RA cos ˛A RB cos ˛B C 350 D
L cos ˇ, (2) RA sin ˛A RB sin ˛B D L sin ˇ, where the length of
the linkage connection L is
LD
RA sin60° RB sin70° 2 C RA cos60° RB cos70° C 3502
D 355.367 mm
(b)
Ratio of moments
1
.8
.6
M .4
a .2
/
0
M −.2
b
−.4
−.6
−.8
−1
0
50
100
Angle, deg
150
200
Compute the positions of the crank ends:
xA D RA cos ˛A ,
yA D RA sin ˛A ,
xB D RB cos ˛B ,
yB D RB sin ˛B .
(d)
Get the components of unit vector parallel to the linkage:
eBAX D cos ˇ,
eBAY D sin ˇ.
(e)
Compute the moments:
MB D xB eABY yB eABX ,
and MA D xA eABY yA eABX (f)
MA
. The graph is shown. The minimum
MB
occurs at ˛A D 78° , with a value of
Take the ratio: R D
R D 0.60289 Ð Ð Ð D 0.603 .
The maximum occurs at ˛A D 180° , with a value of R D 0.75
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1
Problem 6.122 A load W D 2 kN is supported by the
member ACG and the hydraulic actuator BC. If the
actuator BC can exert a maximum axial force of 12 kN,
what is the largest height above the ground at which the
center of mass G can be supported?
A
0.75 m
B
C
1m
G
0.5 m
W
1.5 m
1.5 m
Solution: The points G and C describe arcs of circles as the height
above ground changes. The radii of these circles (see sketch of loader
in Problem 6.112 for dimensions) are
p
RAG D 1.752 C 32 D 3.473 m,
A
RAC
RAG
B
C
p
and RAC D 0.752 C 1.52 D 1.677 m.
O G
The angles made by these radii from the horizontal line through point
A when the point G is 0.5 m above the ground are
AC D tan1
and ˇAG D tan1
0.75
1.5
1.75
3
D 26.57° ,
D 30.25° .
The algorithm for computing height as a function of the hydraulic
actuator force is developed from the geometry and the equilibrium
conditions: (1) The vertical distance from A to C as the height is
increased: h D RAC sinAC ˛, where ˛ is the angle of the element
AG, (0 < ˛), measured from the initial height of point G of 0.5 m,
and a negative h means that h is below the point A. (2) The horizontal
distance from A to C: d D RAC cosAC ˛. (3) The height above
the ground of point G: H D 2.25 RAG sinˇAG ˛. (4) The angle
of thehydraulic actuator
BC with a horizontal line through B is D
0.75 C h
. (5) The sum of the moments about point A is
tan1
d
H
13
F 12.5
o 12
r 11.5
c 11
e 10.5
, 10
9.5
9
k 8.5
N
0
Force vs height
.5 1 1.5 2 2.5 3 3.5 4
Height, H
MA D hBC cos C dBC sin WRAG cosˇAG ˛ D 0.
Check: When ˛ D 0 these reduce to the algorithm used in the solution
to Problem 6.112. check. The TK Solver Plus commercial package
was used to graph the force against height. The hydraulic actuator
reaches the limit of BC D 12 kN when the height is H D 3.54 m
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1
Problem 6.123 The crane exerts vertical 75-kN forces
on the truss at B, C, and D. You can model the support
at A as a pin support and model the support at E as a
roller support that can exert a force normal to the dashed
line but cannot exert a force parallel to it. Determine the
value of the angle ˛ for which the largest compressive
force in any of the members is as small as possible. What
are the resulting axial forces in the members?
C
B
D
1.8 m
2.2 m
A
3.4 m
F
G
H
3.4 m
3.4 m
a
E
3.4 m
Solution: The algorithm is taken directly from the solution to
Problem 6.27, with the addition of a MAX function. The algorithm
is as follows:
(1)
Compute the angles of the members: AB and ED relative to the
horizontal,
4
D 49.64° .
D tan1
3.4
AF and EH relative to the horizontal:
2.2
ˇ D tan1
D 32.91° .
3.4
BG and DG relative to the horizontal:
1.8
D tan1
D 27.9° .
3.4
Min-Max force
M 400
a 350
x
300
( 250
c
) 200
F 150
o 100
r
c 50
e
0
0
10
20
30
40
Angle, deg
50
60
(See the sketch with Problem 6.27.)
(2)
For a given value of ˛, get the unit vector components for the
reaction angle at E:
(h)
AB sin BF BG sin 75 D 0,
eE D i cos90 C ˛ C j sin90 C ˛ D i sin ˛ C j cos ˛
(3)
The complete structure as a free body: From the sum of moments:
The x- and y-components:
Ex
and AB cos C BC C BG cos D 0.
(i)
DE sin DH DG sin 75 D 0,
and DE cos CD DG cos D 0
(a constant). The reactions at A:
(j)
Joint C : CD D BC Check. CG D 75 kN C
Ax D Ex .
(k)
Use standard MAX function to get maximum compressive axial
force:
Ay D 375 Ey D 112.5,
from which Ay D 112.5 kN. (a constant)
Two equations are developed for each joint:
Joint A:
AB cos C Ax C AF cos ˇ D 0,
and AB sin C Ay C AF sin ˇ D 0.
(e)
(f)
(g)
Joint D:
D E sin ˛ kN
and Ey D 112.5 kN
(d)
Joint B:
Joint E:
FMAX D MAXAB, AF, FG, BF, BC, BG, CG,
where it is necessary to test only one of a matching value pair.
The maximum axial compressive force for all members is graphed
against the angle. The min-max point occurs at about
˛ D 50.94 . . .° D 51° ,
where AF D EH D 113.9 kN C,
DE cos Ex C EH cos ˇ D 0
AB D DE D 66.5 kN C,
and DE sin C Ey C EH sin ˇ D 0
BF D DH D 61.9 kN C,
Joint F :
BG D DG D 80.2 kN T,
AF cos ˇ C FG D 0,
BC D CD D 113.9 kN C
and AF sin ˇ C BF D 0.
FG D GH D 95.6 kN C,
Joint H:
EH cos ˇ GH D 0,
and EH sin ˇ C DH D 0
and
CG D 75 kN C.
Check. The minimum maximum compressive stress at ˛ D 51°
is R D 113.9 kN which occurs in elements AF, EH and BC, CD.
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1
Problem 6.124 Draw graphs of the magnitudes of the
axial forces in the members BC and BD as functions of
the dimension h for 0.5 m h 1.5 m.
B
A
D
h
1 kN
0.7 m
0.4 m
C
0.6 m
Solution: Joint A
2000
L 1500
o
a 1000
d 500
s
0
-
TAB
β
−1000
1 kN
TBC
−1500
−2000
.5
TBD
N −500
TAC
Loads in members BC and BD (N) vs h (m)
2500
α
1.2 m
.6
.7
.8
.9
1
1.1 1.2
Heigth h of joint B (m)
1.3
1.4
1.5
Fx D TAB cos ˛ TAC cos ˇ D 0,
Joint C
Fy D TAB sin ˛ TAC sin ˇ 1 D 0.
TBC
The angles are
˛ D arctan
ˇ D arctan
TCD
h 0.7
1.2
0.7
1.2
.
Fx D TAC cos ˇ TCD cos υ D 0,
where
γ
TAB
α
υ D arctan
0.4
0.6
.
These equations were solved for 0.5 m h 1.5 m. The graphs of
the results are shown.
TBD
β
Joint B
TAC
δ
,
TBC
Fx D TAB cos ˛ TBD cos D 0,
Fy D TBC TAB sin ˛ TBD sin D 0,
where
D arctan
h 0.4
0.6
.
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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 6.125 For the truss in Problem 6.124, determine the value of the dimension h in the range 0.5 h 1.5 m so that the magnitude of the largest axial
force in any of the members, tensile or compressive,
is a minimum. What are the resulting axial forces in the
members?
Solution: The plots obtained using this procedure are shown. From
the absolute value plot, we see that the minimum occurs where h ¾
D
1.13 m. Solving the original problem for this value of h gives the
following values for the axial forces in the various members:
TAB D 1128 N (tension),
TAC D 1230 N (compression),
TBC D 1672 N (compression),
TBD D 1672 N (tension),
and TCD D 1276 N (compression).
Note that the axial forces in BC and BD are the limiting factors in the
loading of the truss. All other forces are smaller in magnitude.
Loads in all members (N) vs h (m)
L
o
a
d
s
2500
2000
1500
1000
500
i
0
n
−500
M −1000
e
m −1500
b −2000
e −2500
r
s −3000
.5
.6
.7
.8
.9
1
1.1
1.2
1.3
1.4
.1.5
Height h of joint B (m)
Absolute values of loads in all members (N) vs h (m)
A
b
s
o
l
u
t
e
L
o
a
d
s
-
3000
2500
2000
1500
1000
500
0
.5
.6
.7
.8
.9
1
1.1
1.2
1.3
1.4
.1.5
Height of joint B (m)
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1
Problem 6.131 Consider the truss in Problem 6.130.
Determine the axial forces in members CD, GD,
and GH.
Solution: Use the results of the solution of Problem 6.130:
BC D 120 kN C,
BG D 42.43 kN T,
and FG D 90 kN T.
BC
CD
CG
Joint C
BG
α
CG
GD
α
GH
80 kN
Joint G
The angle of the cross-members with the horizontal is ˛ D 45° .
Joint C:
Fx D BC C CD D 0,
from which CD D 120 kN C
FY D CG D 0,
from which CG D 0.
Joint G:
Fy D BG sin ˛ C GD sin ˛ C CG 80 D 0,
from which GD D 70.71 kN T .
Fy D BG cos ˛ C GD cos ˛ FG C GH D 0,
from which GH D 70 kN T
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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 6.132 The truss supports loads at F and H.
Determine the axial forces in members AB, AC, BC, BD,
CD, and CE.
200 lb
F
100 lb
4 in
D
H
B
4 in
E
C
4 in
J
G
A
I
6 in
6 in
6 in
Solution: The complete structure as a free body: The sum of the
200 lb
moments about I:
6 in
100 lb
MA D 1006 C 20012 24AY D 0,
from which AY D 125 lb. The sum of forces:
Ax
Fx D Ax D 0.
The method of joints: The angles of the inclined members with the
horizontal are
12
in
6
in
6
in
CD
AB
BD
α
BC
AC
α
˛ D tan1 0.6667 D 33.69°
Joint A:
I
Ay
Ay
Joint A
AB
Joint B
CE
BC
α
AC
Joint C
Fx D AC cos ˛ D 0,
from which AC D 0.
Fy D Ay C AB C AC sin ˛ D 0,
from which AB D 125 lb C
Joint B :
Fyt D AB C BD sin ˛ D 0,
from which BD D 225.3 lb C .
Fx D BD cos ˛ C BC D 0,
from which BC D 187.5 lb T
Joint C :
Fx D BC AC cos ˛ C CE cos ˛ D 0,
from which CE D 225.3 lb T
Fy D AC sin ˛ C CD C CE sin ˛ D 0,
from which CD D 125 lb C
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1
Problem 6.133 Consider the truss in Problem 6.132.
Determine the axial forces in members EH and FH.
Solution: Use the results from the solution to Problem 6.132:
DF
α
DE
CE D 225.3 lb T,
BD
CD D 125 lb C,
200 lb
α
α
FH
DF
CD
EF
Joint D
Joint F
EF
DE
α
CE
EH
α
EG
Joint E
BD D 225.3 lb C.
The method of joints: The angle of inclined members with the horizontal is ˛ D 33.69° .
Joint D:
Fy D BD sin ˛ CD C DF sin ˛ D 0,
from which DF D 450.7 lb C.
Fx D DF cos ˛ C DE BD cos ˛ D 0,
from which DE D 187.5 lb T
Joint F :
Fx D DF cos ˛ C FH cos ˛ D 0,
from which FH D 450.7 lb C
Fy D 200 DF sin ˛ FH sin ˛ EF D 0,
from which EF D 300 lb T
Joint E :
Fy D CE sin ˛ C EF EG sin ˛ D 0,
from which EG D 315 lb T
Fx D DE C EH CE cos ˛ C EG cos ˛ D 0,
from which EH D 112.5 lb T
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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 6.134 Determine the axial forces in members
BD, CD, and CE.
10 kN
A
2m
14 kN
C
B
2m
D
E
2m
G
F
2m
I
H
6m
Solution: Use the method of sections
y
A
10 kN
2m
Θ
B
1.5 m
Θ
14 kN
FBD
C
x
Θ
FCD
FCE
D
tan D
2
1.5
D 53.13°
Fx :
FCE cos FCD cos C 24 D 0
Fy :
FBD FCD sin FCE sin D 0
MB :
210 1.5FCD sin 1.5FCE sin D 0
3 eqns-3 unknowns.
Solving
FBD D 13.3 kN,
FCD D 11.7 kN,
FCE D 28.3 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 6.135 For the truss in Problem 6.134, determine the axial forces in members DF, EF, and EG.
Solution: Use method of sections
A
10 kN
2m
A
14 kN
10 kN
C
B
2m
2m
14 kN
D
E
2m
2m
D
FDF
G
F
2m
3
E
I
H
φ
Θ
2
FEG
FEF
3
tan D
6m
1.5
2
1.5
D 53.13°
tan D
2
3
D 33.69°
Fx :
24 C FEG cos FEF cos D 0
Fy :
FDF FEF sin FEG sin D 0
ME :
3FDF 214 410 D 0
Solving,
FEG D 32.2 kN C
FDF D 22.67 kN T
FEF D 5.61 kN T
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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 6.136 The truss supports a 400-N load at G.
Determine the axial forces in members AC, CD, and CF
400 N
A
C
E
G
300 mm
600 mm
H
F
D
B
300 mm
Solution: The complete structure as a free body: The sum of the
MA D 900400 C 600B D 0,
600 mm
B
Fx D Ax C B D 0,
AB
B
from which Ax D 600 N.
The method of joints: The angle from the horizontal of element BD is
D tan1
300
900
θ
BD
CD
AD
αAD
DF
θ
BD
AY
AX
AC
αAD
AD
AB
Joint A
Joint B
Fy D Ay 400 D 0,
from which Ay D 400 N.
AC
Joint D
CE
αCF
CF
CD
Joint C
D 18.43° .
The angle from the horizontal of element AD is
˛AD
400 N
Ax
from which B D 600 N. The sum of forces:
300 mm
900 mm
Ay
moments about A:
300 mm
D 90 tan1
300
600 300 tan Joint D:
D 59.04° .
Fx D AD cos ˛AD BD cos C DF cos D 0,
from which DF D 505.96 N C
The angle from the horizontal of element CF is
˛CF D 90 tan1
300
6001 tan Fy D AD sin ˛AD C CD BD sin C DF sin D 0,
D 53.13° .
Joint B:
Fx D B C BD cos D 0,
from which CD D 240 N C
Joint C :
Fy D CD CF sin ˛CF D 0,
from which BD D 632.5 N C
from which CF D 300 N T
Fy D AB C BD sin D 0,
from which AB D 200 N T
Joint A:
Fy D Ay AD sin ˛AD AB D 0,
from which AD D 233.2 N T
Fx D Ax C AC C AD cos ˛AD D 0,
from which AC D 480 N T
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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 6.137 Consider the truss in Problem 6.136.
Determine the axial forces in members CE, EF,
and EH.
Solution: Use the results of the solution of Problem 6.136:
AC
CE
αCF
AC D 480 N T,
CD
CF D 300 N T,
CF
Joint C
CF
αCF
θ
DF
EF
EG
αEH
CE
FH
EF
Joint F
EH
Joint E
DF D 505.96 N C,
Joint F :
D 18.4° ,
˛CF D 53.1° .
The method of joints: The angle from the horizontal of element EH is
˛EH D 90 tan1
300
600 900 tan Fy D CF cos ˛CF DF cos C FH cos D 0,
from which FH D 316.2 N C
D 45°
Fy D EF C CF sin ˛CF DF sin C FH sin D 0,
from which EF D 300 N C
Joint C:
Joint E :
Fx D AC C CE C CF cos ˛CF D 0,
Fy D EH sin ˛EH EF D 0,
from which CE D 300 N T
from which EH D 424.3 N T
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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 6.138 Consider the truss in Problem 6.136.
Which members have the largest tensile and compressive
forces, and what are their values?
Solution: The axial forces for all members have been obtained in
Problems 6.136 and 6.137 except for members EG and GH. These are:
CE
Joint E:
EF
Fx D CE C EG C EH cos ˛EH D 0,
EG
αEH
Joint E
EH
EG
400 N
GH
Joint G
from which EG D 0
Joint G:
Fy D GH 400 D 0,
from which GH D 400 N C.
This completes the determination for all members. A comparison of
tensile forces shows that AC D 480 N T is the largest value, and a
comparison of compressive forces shows that BD D 632.5 N C
is the largest value.
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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 6.139 The Howe truss helps support a roof.
Model the supports at A and G as roller supports. Use
the method of joints to determine the axial forces in
members BC, CD, CI, and CJ.
6 kN
4 kN
4 kN
D
2 kN
2 kN
C
E
4m
B
F
A
G
H
I
2m
Solution: The free body diagrams for the entire truss and the
required joints are shown.
The whole truss: The equations of equilibrium for the entire truss are:
FX D 0,
2m
J
K
2m
L
2m
2m
2m
6 kN
y
4 kN
D E 4 kN
C
2 kN
F 2 kN
4m
B
A
G
x
H I
J K L
AY
12 m
GY
FY D AY C GY 18 kN D 0.
y
Instead of using the moment equation here (it would work), we see
that the loading is symmetric. Thus, AY D GY D 9 kN.
A
TAH
We need unit vectors along AB, BC, CD, (note that these are the same),
and along BI, and CJ. We get
E
uAB D uBC D uCD D 0.832i C 0.555j,
TDF
F
uBI D 0.832i 0.555j,
y
TAB
TFH
TEF
TFG
TBX
THI
x
TAH H THI
y
4 kN
y
TCX
I
x
TBH
x
TIJ
TBC
C
TCD
x
TCJ
TCI
and uCJ D 0.6i 0.8j.
Joint C:
Joint A:
The equations of equilibrium are
and
FX D TAB uABX C TAH D 0
FY D TAB uABY C AY D 0.
Joint H: The equations of equilibrium are
and
FX D TAH C THI D 0,
FY D TBH D 0.
Joint B:
FX D TBC uBCX C TCJ uCJX C TCD uCDX D 0,
FY D TBC uBCY C TCJ uCJY C TCD uCDY TCI 4 D 0.
Solving these equations in sequence (we can solve at each joint before
going to the next), we get
TAB D 16.2 kN, TAH D 13.5 kN, TBH D 0 kN,
THI D 13.5 kN, TBC D 14.4 kN, TBI D 1.80 kN,
TIJ D 12.0 kN, TCI D 1.00 kN, TCJ D 4.17 kN,
and TCD D 11.4 kN.
FX D TAB uABX C TBC uBCX C TBI uBIX D 0,
FY D TAB uABY C TBC uBCY C TBI uBIY TBH 2 D 0,
Joint I:
and
FX D THI C TIJ TBI uBIX D 0,
FY D TCI TBI uBIY D 0,
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1
Problem 6.140 For the roof truss in Problem 6.139,
use the method of sections to determine the axial forces
in members CD, CJ, and IJ.
Solution: The free body diagram of the section is shown at the
right. The support force at A is already known from the solution to
Problem 6.139. The equations of equilibrium for the section are
and
FX D TCD uCDX C TCJ uCJX C TIJ D 0,
FY D TCD uCDY C TCJ uCJY C AY D 0,
MC D yC TIJ 4AY D 0.
Solving, we get
TIJ D 12.0 kN,
TCJ D 4.17 kN,
and TCD D 11.4 kN.
Note that these values check with the values obtained in Problem 6.139.
4 kN
TCD
2 kN
C
I
D
TCJ
TIJ
J
AY
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1
Problem 6.141 A speaker system is suspended from
the truss by cables attached at D and E. The mass of
the speaker system is 130 kg, and its weight acts at G.
Determine the axial forces in members BC and CD.
0.5 m 0.5 m 0.5 m
0.5 m
1m
C
E
A
1m
B
D
G
Solution: The speaker as a free body: The weight of the speaker
is W D 1309.81 D 1275.3 N. Make a cut through the suspension
cables D, E, the sum of the moments about cable D is
1m
Cx
B
0.5 m
The structure as a free body: The sum of the moments about C is
D
D
W
Fy D D C E W D 0,
from which D D 425.1 N.
E
E
MD D 1W C 1.5E D 0,
from which E D 850.2 N. The sum of the forces:
2m
Cy
A
1 m 0.5 m
CY
CE
α
CD
E
DE
BD
MC D C1A 0.5D 2E D 0,
Joint E
β
α
DE
D
Joint D
AC
CE
β
β
BC
CD
Joint C
from which A D 1912.95 N. The sum of the forces:
Joint D:
Fy D A C Cy W D 0,
Fy D CD sin ˇ C DE sin ˛ D D 0,
from which Cy D 637.65 N and
from which CD D 1425.8 N T
Fx D Cx D 0.
Joint C :
The method of joints:
angle of member DE relative to the hori The
1
D 33.69° . The angles of members AB, BC,
zontal is ˛ D tan1
1.5
1
and CD are ˇ D 90 tan 0.5 D 63.43° .
Fy D CD sin ˇ BC sin ˇ C Cy D 0,
from which BC D 2138.7 N T
Joint E :
Fy D E DE sin ˛ D 0,
from which DE D 1532.72 N C.
Fx D CE DE cos ˛ D 0,
from which CE D 1275.3 N T
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1
y
Problem 6.142 The mass of the suspended object is
900 kg. Determine the axial forces in the bars AB
and AC.
Strategy:
D (0, 4, 0) m
Draw the free-body diagram of joint A.
A (3, 4, 4) m
B (0, 0, 3) m
C (4, 0, 0) m
x
z
Solution: The free-body diagram of joint A is.
TAD
TAB
TAC
(900) (9.81) N
The position vectors from pt A to pts B, C, and D are
rAB D 3i 4j k (m),
rAC D i 4j 4k (m),
rAD D 3i 4k (m).
Dividing these vectors by their magnitudes, we obtain the unit vectors
eAB D 0.588i 0.784j 0.196k,
eAC D 0.174i 0.696j 0.696k,
eAD D 0.6i 0.8k.
From the equilibrium equation
TAB eAB C TAC eAC C TAD eAD 9009.81j D O,
We obtain the equations
0.588TAB C 0.174TAC 0.6TAD D 0,
0.784TAB 0.696TAC 9009.81 D 0,
0.196TAB 0.696TAC 0.8TAD D 0.
Solving, we obtain
TAB D 7200 N,
TAC D 4560 N,
TAD D 5740 N.
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1
Problem 6.143 Determine the forces on member ABC,
presenting your answers as shown in Fig. 6.35. Obtain
the answer in two ways:
(a)
(b)
When you draw the free-body diagrams of the
individual members, place the 400-lb load on the
free-body diagram of member ABC.
When you draw the free-body diagrams of the
individual members, place the 400-lb load on the
free-body diagram of member CD.
1 ft
1 ft
200 lb
C
D
400 lb
1 ft
B
1 ft
E
Solution: The angle of element BE relative to the horizontal is
˛ D tan1
1
2
1 ft
D 26.57° .
A
The complete structure as a free body: The sum of the moments
about A:
Cy
MA D 3400 1200 C 2Fy D 0
from which Fy D 700 lb. The sum of forces:
F
400 lb
C
Cx y
B
200 lb
Dx
Dy
Dy
Dx
B
Fy D Ay C Fy 200 D 0,
Ax
from which Ay D 500 lb .
Cx
B
Ay
B
Fx
Fy
Fx D Ax C Fx C 400 D 0.
100 lb
(a)
400 lb
Element CD: The sum of the moments about D:
26.6°
MD D 200 C 2Cy D 0,
1341
lb
from which Cy D 100 lb .
400 lb
400 lb
500 lb
Fy D Dy Cy 200 D 0,
from which Dy D 100.
from which Check:
Fx D Cx Dx D 0,
BD
from which Dx D Cx .
500 C 100
D 1341.6 lb.
sin ˛
Element DEF: The sum of the moments about F:
check. From above: Cx D Dx D 400 lb .
MF D 3Dx C B cos ˛ D 0,
cos ˛ .
from which Dx D B
3
Fy D Fy C B sin ˛ C Dy D 0,
from which
BD
700 C 100
D 1341.6 lb , and Dx D
sin ˛
Fx D 400 C Cx C B cos ˛ C Ax D 0,
from which Ax D 400 lb .
(b)
When the 400 lb load is applied to element CD instead, the
following changes to the equilibrium equations occur: Element
CD:
400 lb
Fx D Cx Dx C 400 D 0,
Element ABC:
from which Cx C Dx D 400. Element ABC:
MA D 2B cos ˛ 3400 3Cx D 0.
The sum of the forces
Fy D Cy B sin ˛ C Ay D 0,
Fx D Cx C Ax B cos ˛ D 0.
Element DEF : No changes. The changes in the solution for
Element ABC Cx D 800 lb when the external load is removed,
instead of Cx D 400 lb when the external load is applied, so that
the total load applied to point C is the same in both cases.
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1
Problem 6.144 The mass m D 120 kg. Determine the
forces on member ABC.
A
B
C
300 mm
D
m
E
200 mm
Solution: The weight of the hanging mass is given by
m
W D mg D 120 kg 9.81 2 D 1177 N.
s
FX D AX C EX D 0,
Cx
B
W
B
Cx
B
Cy
B
Ex
and
Cy
Ay
Ax
The complete structure as a free body: The equilibrium equations are:
200 mm
FY D AY W D 0,
MA D 0.3EX 0.4W D 0.
Solving, we get
AX D 1570 N,
AY D 1177 N,
and EX D 1570 N.
Element ABC: The equilibrium equations are
and:
FX D Ax C CX D 0,
FY D AY C CY BY W D 0,
MA D 0.2BY C 0.4cY 0.4W D 0.
Solution gives BY D 2354 N (member BD is in tension),
CX D 1570 N,
and CY D 2354 N.
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1
Problem 6.145 Determine the forces on member ABC,
presenting your answers as shown in Fig. 6.35.
D
400 lb
2 ft
200 ft-lb
A
1 ft
B
C
100 lb
1 ft
E
2 ft
2 ft
2 ft
Solution: The complete structure as a free body: The sum of the
moments:
By
MA D 1001 4006 200 C 4E D 0,
from which E D 625 lb. The sum of the forces:
Ax
Fy D Ay C E 400 D 0,
Dy
Bx
Ay
Dy
Dx
Bx
By
Cx
Dx
400 lb
Cy
Cx
Cy200 ft-lb
100 lb
from which Ay D 225 lb.
E
Fx D Ax C 100 D 0,
from which Ax D 100 lb. These results are used as a check on the
solution below.
100 lb
Element ECD: (See the free body diagram.) The sum of the moments
about E:
225
lb
ME D 4Dx 2Cx 100 D 0,
from which (1) 4Dx C 2Cx D 100. The sum of the forces:
Fx D Dx C Cx C 100 D 0,
from which (2) Dx C Cx D 100.
Fy D E C Cy C Dy D 0,
675 lb
400 lb
150
lb
200
ft–lb
50 lb
50 lb
The sum of the forces:
Fx D Ax Bx Cx D 0,
from which (8) Ax Bx Cx D 0.
Fy D Ay By Cy 400 D 0,
from which (9) Ay By Cy D 400. These nine equations are solved
for the nine reactions The reactions are DX D 50 lb, DY D 50 lb,:
thus (3) Dy C Cy C E D 0.
Element BD: The sum of the moments about B:
MB D 2Dx 2Dy D 0,
from which (4) Dx Dy D 0. The sum of the forces:
CX D 150 lb, CY D 675 lb, BX D 50 lb,
BY D 50 lb, AX D 100 lb, AY D 225 lb,
and E D 625 lb.
Fx D Bx Dx D 0,
from which (5) Bx Dx D 0.
Fy D By Dy D 0,
from which (6) By Dy D 0
Element ABC : The sum of the moments about A:
MA D 2By 4Cy 200 6400 D 0,
from which (7) By C 2Cy D 1300.
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1
Problem 6.146 Determine the force exerted on the bolt
by the bolt cutters and the magnitude of the force the
members exert on each other at the pin connection A.
90 N
A
80
mm
160
mm
540 mm
100
mm
90 N
Solution: Element AB: The moment about A is
C
90 N
A
MA D 10B 54F D 0,
where F D 90 N. From which B D 486 N. The sum of the forces:
B
Fy D A C B F D 0,
FC
8
cm
B
16
cm
10
cm
54 cm
from which A D 576 N
Element BC: The moment about C:
MC D 16B 8FC D 0,
from which the cutting force is FC D 972 N
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1
Problem 6.147 The 600-lb weight of the scoop acts at
a point 1 ft 6 in to the right of the vertical line CE. The
line ADE is horizontal. The hydraulic actuator AB can
be treated as a two-force member. Determine the axial
force in the hydraulic actuator AB and the forces exerted
on the scoop at C and E.
B
C
2 ft
A
Solution: The free body diagrams are shown at the right. Place the
coordinate origin at A with the x axis horizontal. The coordinates (in
ft) of the points necessary to write the needed unit vectors are A (0,
0), B (6, 2), C (8.5, 1.5), and D (5, 0). The unit vectors needed for
this problem are
uBA D 0.949i 0.316j,
E
D
5 ft
TCB
1 ft 6 in
1 ft
2 ft 6 in
Scoop
C
1.5 ft
1.5 ft
G
EX
E
EY
uBC D 0.981i 0.196j,
600 lb
and uBD D 0.447i 0.894j.
y
The scoop: The equilibrium equations for the scoop are
and
FX D TCB uBCX C EX D 0,
TBA
x
TCB
FY D TCB uBCY C EY 600 D 0,
TBD
MC D 1.5EX 1.5600 lb D 0.
Solving, we get
EX D 600 lb,
EY D 480 lb,
and TCB D 611.9 lb.
Joint B: The equilibrium equations for the scoop are
and
FX D TBA uBAX C TBD uBDX C TCB uBCX D 0,
FY D TBA uBAY C TBD uBDY C TCB uBCY D 0.
Solving, we get
TBA D 835 lb,
and TBD D 429 lb.
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1
Problem 6.148 Determine the force exerted on the bolt
by the bolt cutters.
100 N
A
75
mm
40 mm
C 55 mm
B
D
90 mm
60 mm 65 mm
300 mm
100 N
Solution: The equations of equilibrium for each of the members
will be developed.
AY
AX
F
Member AB: The equations of equilibrium are:
and
100 N
λ
40 mm
B
55 mm
75 mm
A
FX D AX C BX D 0,
BX
BY
FY D AY C BY D 0,
MB D 90F 75AX 425100 D 0
90 mm
60 mm 65 mm
AX
AY
300 mm
Member BD: The equations are
CY
FX D BX C DX D 0,
40 mm
75 mm
and
FY D BY C DY C 100 D 0,
MB D 15DX C 60DY C 425100 D 0.
90 mm
Member AC: The equations are
and
FX D AX C CX D 0,
FY D AY C CY C F D 0,
60 mm 65 mm
300 mm
BY
DX
B
BX
D
DY
MA D 90F C 125CY C 40CX D 0.
Member CD: The equations are:
D
F
C
55 mm X
C
60 mm 65 mm
300 mm
100 N
FX D CX DX D 0,
CY
FY D CY DY D 0.
Solving the equations simultaneously (we have extra (but compatible)
equations, we get F D 1051 N, AX D 695 N, AY D 1586 N, BX D
695 N, BY D 435 N, CX D 695 N, CY D 535 N, DX D 695 N,
and Dy D 535 N
C
CX
DX
D
DY
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1
Problem 6.149 For the bolt cutters in Problem 6.148,
determine the magnitude of the force the members exert
on each other at the pin connection B and the axial force
in the two-force member CD.
Solution: From the solution to 6.148, we know BX D 695 N,
and BY D 435 N. We also know that CX D 695 N, and CY D 535 N,
from which the axial load in
member CD can be calculated. The load
in CD is given by TCD D C2X C C2Y D 877 N
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1
Problem 6.150 The weights W1 D 4 kN and W2 D
10 kN. Determine the forces on member ACDE at points
A and E.
H
G
350 mm
100 mm
600 mm
F
E
D
350 mm
C 100
mm
600 mm
A
600 mm
Solution: The complete structure as a free body: The horizontal
˛ D tan1
1200
1100
850 mm
H
600
mm
G
G
F
F
D
MA D 850W1 C 2800W2 C 1100BC sin ˛ 600BC cos ˛ D 0
Fx D Ax C BC cos ˛ D 0,
β
W1
from which BC D 77.45 kN. The sum of the forces:
250 800 mm
mm
D 47.49° .
The sum of the moments about the point A is
300
mm
B
distance from A to C is 1100 mm. The vertical distance is 600 mm.
The horizontal distances to the action line of the weights W1 , W2 are
850 mm and 2800 mm. The angle of the hydraulic lifter BC relative
to the horizontal is
W1
W2
100
mm
D
Ey
Ay
W2
Ex
Ex
Ey
α
from which Ax D 52.33 kN
C
Ax
Fy D Ay C BC sin ˛ W1 W2 D 0,
from which AY D 43.09 kN. These results are to be used as a check
on the solution below.
The sum of forces:
The angle of member DH relative to the horizontal is
ˇ D tan1
600
100
and
350
600
D 30.25° .
The bucket as a free body: The sum of the moments about point E:
ME D 500G 300W2 F150 cos 100 sin D 0.
The sum of the forces on the bucket:
and
Fy D Ay W1 C C sin ˛ C D sin ˇ C Ey D 0.
D 80.54° .
The angle of the force F relative to the horizontal is
D tan1
Fx D Ax C C cos ˛ D cos ˇ C Ex D 0,
These equilibrium conditions lead to 8 equations in 8 unknowns, which
are solved by iteration using TK Solver Plus The results:
G D 7.53 kN , D D 4.93 kN , F D 9.65 kN ,
Ex D 0.811 kN , Ey D 14.9 kN , Ax D 52.3 kN ,
Ay D 43.1 kN , and C D 77.4 kN :
The last three values check the results obtained from the solution for
the complete structure as a free body.
Fx D Ex C F cos G D 0,
Fy D Ey W2 F sin D 0.
The linkage H as a free body: The sum of the forces:
and
Fx D F cos C G C D cos ˇ D 0,
Fy D F sin D sin ˇ D 0.
The member ACDE as a free body: The sum of the moments about
point A:
MA D 850W1 C 600 cos ˛ 1100 sin ˛
C D 1900 sin ˇ C 950 cos ˇ C 2500Ey 1050Ex D 0.
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1
Problem 7.1 If a D 2, what is the x coordinate of the
centroid of the area?
Strategy: The x coordinate of the centroid is given by
Eq. (7.6). For the element of area dA, use a vertical strip
of width dx. (See Example 7.1).
y
y x2 2
a
x
Solution: Use a vertical strip
2
xD 2
xy dx
x dA
D
0
2
dA
y dx
0
xx 2 C 2 dx
6
D 0 2
D
5
2
x C 2 dx
0
x D 1.2
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1
Problem 7.2 Determine the y coordinate of the
centroid of the area shown in Problem 7.1 if a D 3.
Solution: Use a vertical strip
3
yD
0
1
yy dx
2
3
3
D
0
3
2
x C 2 dx
y dx
0
1 2
x C 22 dx
2
D
161
50
0
y D 3.22
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1
Problem 7.3
of the area?
What is the x coordinate of the centroid
Solution: Use a horizontal strip
1.23
y
xD
0
1
xx dy
2
1.23
x dy
0
1.728
D
0
1 1/3 2
y dy
2
1.728
y 1/3 dy
D
12
25
0
x D 0.48
y ⫽ x3
1.2
x
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1
Problem 7.4 What is the y coordinate of the centroid
of the area in Problem 7.3?
Solution: Use a horizontal strip
1.23
1.728
yx dy
yD
0
1.23
x dy
0
yy 1/3 dy
864
D
D 0 1.728
875
1/3
y dy
0
y D 0.987
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1
Problem 7.5
of the area.
Determine the coordinates of the centroid
y
6
2
3
9
x
Solution: Use a vertical strip - The equation of the line is y D
8
2x
3
9
xy dx
xD
3
9
y dx
3
9
yD
3
9 2
x 8 x dx
11
3
D 3 9 D
2
2
8 x dx
3
3
1
yy dx
2
9
y dx
3
1
2 2
8 x dx
13
2
3
D
D 3 9 6
2
8 x dx
3
3
9
x D 5.5
y D 2.17
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1
Problem 7.6 Determine the x coordinate of the
centroid of the area and compare your answers to the
values given in Appendix B.
y
y = cx n
0
a
x
Solution:
b cxn
AD
dy dx D
0
xD
1
A
0
cbnC1
provided that n > 1
nC1
b cxn
x dy dx D
0
0
bn C 1
nC2
Matches the appendix
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1
Problem 7.7 Determine the y coordinate of the centroid
of the area and compare your answer to the value given
in Appendix B.
Solution: See solution to 7.6
yD
1
A
b cxn
y dy dx D
0
0
bn cn C 1
4n C 2
Matches the appendix
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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 7.8 Suppose that an art student wants to paint
a panel of wood as shown, with the horizontal and
vertical lines passing through the centroid of the painted
area, and asks you to determine the coordinates of the
centroid. What are they?
y
y = x + x3
0
1 ft
x
Solution: The area:
1
AD
x C x 3 dx D
0
2
1
x
3
x4
D .
C
2
4 0
4
The x-coordinate:
1
xx C x 3 dx D
0
3
1
x
8
x5
D
C
.
3
5 0
15
Divide by the area: x D
32
D 0.711
45
The y-coordinate: The element of area is dA D 1 x dy. Note that
dy D 1 C 3x 2 dx, hence dA D 1 x1 C 3x 2 dx. Thus
1
yA D
y dA D
A
x C x 3 1 x1 C 3x 2 dx,
0
from which
1
x x 2 C 4x 3 4x 4 C 3x 5 3x 6 dx
0
D
1
4
4
3
3
1
C C D 0.4381.
2
3
4
5
6
7
Divide by A y D 0.5841
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1
Problem 7.9 The y coordinate of the centroid of the
area is y D 1.063. Determine the value of the constant
c and the x coordinate of the centroid.
y
y = cx 2
0
Solution:
x
4
y = cx2
y
yD 2
y dA
dA
dA D y dx
4
y
y dx
2
y D 2 4
D
1
2
4
24
y dx
2
y/2
C2 x 4 dx
Cx 2 dx
2
1
2
3
4
x
5 4
x 32
1024
c2
C
5 2
5
5
yD
3 4 D
8
64
x 2
2C
3
3
3 2
y D 5.314C
But y D 1.063
∴C D 0.200
Now we have C known and y D Cx 2
4 4
x 4 xD D 24
D 2 4
x 3 dA
Cx 2 dx
2
3 4
x dA
Cx 3 dx
2
256 16
4
D 3.214
xD
64 8
3
x D 3.214
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1
Problem 7.10 Determine the coordinates of the centroid of the metal plate’s cross-sectional area.
y
1
y = 4 – – x 2 ft
4
x
Solution: Let dA be a vertical strip:
The area dA D y dx D
4
1 2
x
4
dx. The curve intersects the x axis
1
where 4 x 2 D 0, or x D š4.
4
Therefore
4
x4
1
2x 2 4x x 3 dx
16 4
4
D 44 D 0.
x D A
D 3 4
1
x
dA
4 x 2 dx
4x
A
4
4
12 4
4
x dA
To determine y, let y in equation (7.7) be the height of the midpoint
of the vertical strip:
4
y dA
y D A
D
1
4 2
dA
A
1
1
4 x 2 dx
4 x2
4
4
4
1 2
4 x
dx
4
4
4
x5
x3
1 4
C
8x x
8 x2 C
dx
3
532 4
32
D 4 4 D
4
3
x2
x
dx
4
4x
4
4
12 4
4
D
34.1
D 1.6 ft.
21.3
y
dA
y
x
x
dx
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1
4
3
y, m
Problem 7.11 An architect wants to build a wall with
the profile shown. To estimate the effects of wind loads,
he must determine the wall’s area and the coordinates
of its centroid. What are they?
y = 2 + 0.02x2
2
1
0
0
2
4
6
8
10
x, m
Solution:
10
Area D
10
y dx D
0
2 C 0.02x 2 dx
0
10
x3
D 26.67 m2
Area D 2x C 0.02
3 0
dA D y dx D 2 C 0.02x 2 dx
Y
X
10
10
x dA
x D 0 10
2x C 0.02x 3 dx
0
D
26.67
dA
0
2
xD
10
x2
x4 C 0.02
2
4 0
m
26.67
100 C 0.02
xD
26.67
104
4
D
150
26.67
x D 5.62 m
10
10 y
2 C 0.02x 2 2 dx
y dx
1 0
2
D
y D 0 10
2
26.67
dA
0
yD
1
226.67
10
4x C 0.08
yD
yD
4 C 0.08x 2 C 0.0004x 4 dx
0
3
5 10
x
x
C 0.0004
3
5
0
226.67
74.67
53.34
y D 1.40 m
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1
Problem 7.12 Determine the coordinates of the centroid of the area.
y
y⫽⫺
1 2
x ⫹ 4x ⫺ 7
4
x
Solution: Use a vertical strip. We first need to find the x intercepts.
1
y D x 2 C 4x 7 D 0 ) x D 2, 14
4
14
xy dx
xD
2
14
y dx
2
14
yD
2
14 1
x x 2 C 4x 7 dx
4
D 2 14 D8
1 2
x C 4x 7 dx
4
2
1
yy dx
2
14
y dx
2
2
1
1
x 2 C 4x 7 dx
18
2
4
D 2 14 D
5
1 2
x C 4x 7 dx
4
2
14
xD8
y D 3.6
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1
Problem 7.13 Determine the coordinates of the centroid of the area.
y
y⫽⫺
1 2
x ⫹ 4x ⫺ 7
4
y⫽5
x
Solution: Use a vertical strip. We first need to find the x intercepts.
1
y D x 2 C 4x 7 D 5 ) x D 4, 12
4
12
xy dx
xD
4
12
y dx
4
12 1
x x 2 C 4x 7 5 dx
4
D 4 12 D8
1 2
x C 4x 7 5 dx
4
4
12
yD
yc y dx
12
y dx
4
4
12
D
4
1
2
1
1
x 2 C 4x 7 C 5
x 2 C 4x 7 5 dx
33
4
4
D
12 5
1 2
x C 4x 7 5 dx
4
4
x D 8, y D 6.6
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1
Problem 7.14 Determine the x coordinate of the centroid of the area.
y
y = x3
y=x
x
Solution: Work this problem like Example 7.2
1
1
x dA
0
xx x 3 dx
y
y = x3
0
xD 1
D 1
dA
0
y=x
x x 3 dx
0
3
1
x
x5
1
1
2
3
5 0
3
5
15
D
xD D 0.533
1 D 1
1
1
x4
x2
4
2
4
2
4 0
dA
x
x D 0.533
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1
Problem 7.15 Determine the y coordinate of the centroid of the area shown in Problem 7.14.
Solution: Solve this problem like example 7.2.
1
y dA
y D A
D
dA
A
0
1
x C x 3 x x 3 dx
2
1
x x 3 dx
0
3
1
x
x7
1
3
7 0
D y D 0 1
1
2
2
x4
x
3
x x dx
2
0
2
4 0
1
x 2 x 6 dx
1
1
4
8
3
7
21
D
D
D 0.381
yD 1
1
2
21
2
2
4
y D 0.381
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1
Problem 7.16 Determine the coordinates of the centroid of the area.
y
(1, 1)
y = x 1/2
y = x2
Solution:
Let dA be a vertical strip: The area dA D
x1/2 x 2 dx, so
y
y = x1/2
5/2
1
x4
x
x 3/2 x 3 dx
x dA
5/2
4 0
D 0 1 x D A
D 0.45.
D 3/2
3 1
x
x
1/2
2
dA
x x dx
A
0
3/2
3 0
1
x
(1, 1)
If we use a horizontal strip to obtain y, we obtain
y = x2
dA
x
x
dx
1
y 3/2 y 3 dy
y D A
D 0 1 D 0.45
dA
y 1/2 y 2 dy
y dA
A
0
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1
y
Problem 7.17 Determine the x coordinate of the centroid of the area.
y = x 2 – 20
y=x
x
Solution: The intercept of the straight line with the parabola
occurs at the roots of the simultaneous equations: y D x, and y D x 2 20. This is equivalent to the solution of the quadratic x 2 x 20 D 0,
x1 D 4, and x2 D 5. These establish the limits on the integration.
The area: Choose a vertical strip dx wide. The length of the strip
is x x 2 C 20, which is the distance between the straight line
y D x and the parabola y D x 2 20. Thus the element of area is
dA D x x 2 C 20 dx and
C5
AD
x x 2 C 20 dx D
4
C5
2
x3
x
C 20x
D 121.5.
2
3
4
The x-coordinate:
C5
xA D
x dA D
A
D
xD
x 2 x 3 C 20x dx
4
3
C5
x
x4
C 10x 2
D 60.75.
3
4
4
60.75
D 0.5
121.5
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1
Problem 7.18 Determine the y coordinate of the centroid of the area in Problem 7.17.
Solution: Use the results of the solution to Problem 7.17 in the
following.
The y-coordinate: The centroid of the area element occurs at the
midpoint of the strip enclosed by the parabola and the straight line,
and the y-coordinate is:
yDx
1
1
x x 2 C 20 D
x C x 2 20.
2
2
yA D
y dA D
A
yD
5
1
x C x 2 20x x 2 C 20 dx
2
4
D
C5
1
x 4 C 41x 2 400 dx
2
4
D
5
5
1
x
41x 3
C
400x
D 923.4.
2
5
3
4
923.4
D 7.6
121.5
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1
Problem 7.19
of the area?
What is the x coordinate of the centroid
y
y⫽⫺
1 2
x ⫹ 2x
6
2
Solution: Use vertical strips, do an integral for the parabola then
subtract the square
x
6
2
First find the intercepts
1
y D x 2 C 2x D 0 ) x D 0, 12
6
12 1
x x 2 C 2x dx 722
65
6
D
x D 0 12 11
1 2
x C 2x dx 22
6
0
x D 5.91
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1
Problem 7.20 What is the y coordinate of the centroid
of the area in Problem 7.19?
Solution: Use vertical strips, do an integral for the parabola then
subtract the square
First find the intercepts
1
y D x 2 C 2x D 0 ) x D 0, 12
6
2
1
x 2 C 2x dx 122
139
6
y D 0 12 D
55
1 2
x C 2x dx 22
6
0
12
1
2
y D 2.53
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1
y
Problem 7.21 An agronomist wants to measure the
rainfall at the centroid of a plowed field between two
roads. What are the coordinates of the point where the
rain gauge should be placed?
0.5 mi
Solution: The area: The element of area is the vertical strip yt yb long and dx wide, where yt D mt x C bt and yb D mb x C bb are
the two straight lines bounding the area, where
0.8 0.3
D 0.3846,
mt D
1.3 0
0.3 mi
0.3 mi
x
0.5 mi
0.6 mi
0.2 mi
and bt D 0.8 1.3 mt D 0.3.
Similarly:
0.3 0
D 0.2308,
1.3 0
mb D
and bb D 0.
The element of area is
dA D yt yb dx D mt mb x C bt bb dx
D 0.1538x C 0.3 dx,
from which
1.1
0.1538x C 0.3 dx
AD
0.5
1.1
x2
D 0.2538 sq mile.
D 0.1538 C 0.3x
2
0.5
The x-coordinate:
1.1
0.1538x C 0.3x dx
x dA D
A
0.5
1.1
x2
x3
D 0.2058.
D 0.1538 C 0.3
3
2 0.5
x D 0.8109 mi
The y-coordinate: The y-coordinate of the centroid of the elemental
area is
y D yb C 12 yt yb D 12 yt C yb D 0.3077x C 0.15.
Thus,
yA D
y dA
A
1.1
0.3077x C 0.150.1538x C 0.3 dx
D
0.5
1.1
D
0.0473x 2 C 0.1154x C 0.045 dx
0.5
1.1
x3
x2
D 0.0471 C 0.1153 C 0.045x
D 0.1014.
3
2
0.5
Divide by the area: y D
0.1014
D 0.3995 mi
0.2538
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1
Problem 7.22 The cross section of an earth-fill dam is
shown. Determine the coefficients a and b so that the y
coordinate of the centroid of the cross section is 10 m.
y
y = ax – bx3
x
100 m
Solution: The area: The elemental area is a vertical strip of length
y and width dx, where y D ax bx 3 . Note that y D 0 at x D 100, thus
b D a ð 104 . Thus
100
dA D a
AD
x 104 x 3 dx
0
A
D 0.5a[x 2 0.5 ð 104 x 4 ]100
0
D 0.5a ð 104 0.25b ð 108 ,
and the area is A D 0.25a ð 104 . The y-coordinate: The y-coordinate
of the centroid of the elemental area is
y D 0.5ax bx3 D 0.5ax 104 x 3 ,
from which
yA D
y dA
A
100
D 0.5a2
x 104 x 3 2 dx
0
100
D 0.5a2
x 2 2104 x 4 C 108 x 6 dx
0
D 0.5a2 100
3
x
2x 5
x7
104 C 108 3
5
7 0
D 3.81a2 ð 104 .
Divide by the area:
yD
3.810a2 ð 104
D 15.2381a.
0.25a ð 104
For y D 10, a D 0.6562 , and b D 6.562 ð 105 m2
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1
Problem 7.23 The Supermarine Spitfire used by Great
Britain in World War II had a wing with an elliptical
profile. Determine the coordinates of its centroid.
y
2
y2
a2
b2
x + — =1
—
Solution:
2
2
y
a2
b2
x
2b
y =1
x + —
—
b
a
x
b
a
By symmetry, y D 0.
From the equation of the ellipse,
yD
bp 2
a x2
a
By symmetry, the x centroid of the wing is the same as the x centroid
of the upper half of the wing. Thus, we can avoid dealing with š
values for y.
y = ab a2 – x2
y
b
dA = y dx
0
a
x
dx
b a 2
x a x 2 dx
a 0
a
xD D
b
dA
a2 x 2 dx
a 0
x dA
Using integral tables
x
a2 x 2 dx D a2 x 2 3/2
3
p
x
a2
x a2 x 2
C
sin1
a2 x 2 dx D
2
2
a
Substituting, we get
3/2 a
/3
a2 x 2
xD
0
p
x a
a2
x a2 x 2
C
sin1
2
2
a
0
xD
xD
0 C a3 /3
a3 /3
D 2
2
a /4
a
00
0C
2 2
4a
3
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1
Problem 7.24 Determine the coordinates of the
centroid of the area.
y
Strategy: Write the equation for the circular boundary
in the form y D R2 x 2 1/2 and use a vertical “strip”
of width dx as the element of area dA.
R
x
Solution: The area: The equation of the circle is x2 Cpy 2 D R2 .
2
2
Take the elemental area to be a vertical strip of height
p y D R x
and width dx, hence the element of area is dA D R2 x 2 dx. The
Acircle
R2
area is A D
D
. The x-coordinate:
4
4
xA D
x dA D
A
xD
R
R R2 x 2 3/2
R3
x R2 x 2 dx D D
:
3
3
0
0
4R
3
The y-coordinate: The y-coordinate of the centroid of the element of
area is at the midpoint:
p
y D 12 R2 x 2 ,
hence yA D
y dA D
A
D
yD
R
1
R2 x 2 dx
2
0
R
x3
R3
1
R2 x D
2
3 0
3
4R
3
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1
Problem 7.25* If R D 6 and b D 3, what is the y
coordinate of the centroid of the area?
y
R
x
b
Solution: We will use polar coordinates. First find the angle ˛
˛ D cos1
b
3
D cos1
D 60° D
R
6
3
˛ R
AD
/3 6
rdrd D 6
rdrd D
0
1
yD
A
0
0
/3 6
0
R
0
α
b
6
r 2 sin drd D D 1.910
0
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1
Problem 7.26* What is the x coordinate of the centroid of the area in Problem 7.25?
Solution: See the solution to 7.25
xD
1
A
/3 6
0
0
r 2 cos drd D
p
6 3
D 3.31
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1
Problem 7.27
centroids.
Determine the coordinates of the
y
40 mm
x
40 mm
80 mm
80 mm
Solution: Break into 2 triangles
xD
80
1
1
2 16040 C 3 160
2 16040
1
1
2 16040 C 2 16040
1
yD
1
1
3 40
y
80, 40
200
D
3
1
1
2 16040 C 3 40
2 16040
1
1
16040
C
16040
2
2
I
D0
II
x
160, 0
x D 66.7 mm
yD0
0, −40
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1
Problem 7.28
centroids.
y
Determine the coordinates of the
20 mm
60 mm
Solution: Let us solve this problem by using symmetry and by
breaking the composite shape into parts.
70 mm
l1
20 mm
A1
y
x
30 mm
h1
A2
h2
60 mm
l2
x
l1 D 70 mm
h1 D 20 mm
l2 D 30 mm
h2 D 60 mm
A1 D l1 h1 D 1400 mm2
A2 D l2 h2 D 1800 mm2
By symmetry,
x1 D 0
x2 D 0
y1 D 70 mm
y2 D 30 mm
For the composite,
xD
0C0
x1 A1 C x2 A2
D
D0
A1 C A2
320 mm2
yD
y1 A1 C y2 A2
A1 C A2
yD
152000
701400 C 301800
D
3200
3200
y D 47.5 mm
xD0
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1
Problem 7.29
centroids.
Determine the coordinates of the
Solution: Break into a rectangle, a triangle and a circular hole
xD
y
yD
1
2
2 86 4[2 ]
D 6.97 in
1
108 C 2 86 22
5[108] C 12
4[108] C 13 8
1
2 86
3[22 ]
108 C 12 86 22
D 3.79 in
2 in
8 in
x D 6.97 in
y D 3.79 in
3 in
x
4 in
6 in
10 in
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1
Problem 7.30
centroids.
Determine the coordinates of the
y
10 in
x
20 in
Solution: The strategy is to find the centroid for the half circle
area, and use the result in the composite algorithm. The area: The
element of area is a vertical
strip y high and dx wide. From the equation
p
of the circle, y D š R2 x 2 . The
p height of the strip will be twice
the positive value, so that dA D 2 R2 x 2 dx, from which
R
dA D 2
AD
R2 x 2 1/2 dx
0
A
p
R
x R2 x 2
R2
R2
1 x
D2
D
C
sin
2
2
R
2
0
The x-coordinate:
x dA D 2
A
R x R2 x 2 dx
0
R
2R3
R2 x 2 3/2
.
D
D2 3
3
0
Divide by A: x D
4R
3
The y-coordinate: From symmetry, the y-coordinate is zero.
420
D 8.488 in. For
3
the inner half circle x2 D 4.244 in. The areas are
The composite: For a complete half circle x1 D
A1 D 628.32 in2
and A2 D 157.08 in2 .
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1
Problem 7.31
centroids.
Determine the coordinates of the
y
0.8 m
x
0.6 m
1.0 m
Solution: Use a big triangle and a triangular hole
2
xD 3
1
yD 3
1.0
1
0.8
1
2
1
2 1.00.8 0.6 C 3 0.4
2 0.40.8
1
1
2 1.00.8 2 0.40.8
1
1
2 1.00.8 3 0.8
2 0.40.8
1
1
2 1.00.8 2 0.40.8
D 0.533
D 0.267
x D 0.533 m
y D 0.267 m
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1
Problem 7.32
centroids.
y
Determine the coordinates of the
0.5 m
x
1m
Solution: Use a full circle, a quarter circular hole, and a triangle
1
0.52
1
C 1.0 1.00.5
4
3
2
2
1
0.5
C 1.00.5
0.52 4
2
0[0.52 ] xD
40.5
3
1
40.5
0.52
1
0[0.52 ] 0.5 1.00.5
3
4
3
2
yD
1
0.52
2
C 1.00.5
0.5 4
2
x D 0.0497 m
yD0
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1
Problem 7.33
centroids.
Determine the coordinates of the
y
400
mm
300 mm
x
300
mm
300
mm
Solution: Break into 4 pieces (2 rectangles, a quarter circle, and
a triangle)
4[0.4]
[0.4]2
0.2[0.40.3] C 0.4 3
4
1
C 0.550.30.7 C 0.8 0.30.7
2
xD
[0.4]2
1
0.40.3 C
C 0.30.7 C 0.30.7
4
2
[0.4]2
4[0.4]
0.15[0.40.3] C 0.3 C
3
4 1
1
0.7
0.30.7
C 0.350.30.7 C
3
2
yD
2
1
[0.4]
C 0.30.7 C 0.30.7
0.40.3 C
4
2
x D 0.450 m, y D 0.312 m
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1
Problem 7.34
centroids.
Determine the coordinates of the
Solution: Divide the object into four areas: (1) The rectangle 18 in
by 18 in, (2) The triangle of altitude 18 in and base 6 in, and (3) the
semi circle with radius 9 in and (4) The object itself.
y
The areas and their centroids are determined by inspection:
(1)
(2)
(3)
A1 D 182 D 324 in2 , x1 D 9 in, y1 D 9 in.
A2 D 12 186 D 54 in2 , x2 D 9 in, y2 D 6 in.
92
49
A3 D
D 127.2 in2 , x3 D 9 in, y3 D 18 C
D 21.8 in.
2
3
The composite area: A D A1 A2 C A3 D 397.2 in2 .
The composite centroid:
18 in
xD
A1 x1 A2 x2 C A3 x3
D 9 in
A
yD
A1 y1 A2 y2 C A3 y3
D 13.51 in
A
x
6 in
6 in
6 in
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1
Problem 7.35
centroids.
y
Determine the coordinates of the
20 mm
30 mm
20 mm
10
mm
30 mm
x
90 mm
Solution: Determine this result by breaking the compound object
into parts
For the composite:
y
m
20 m
50 m
m
m
10 m
30 mm
A1
=
A2
+
+
–
x1 A1 C x2 A2 C x3 A3 x4 A4
A1 C A2 C A3 A4 xD
155782
D 35.3 mm
4414.2
yD
y1 A1 C y2 A2 C y3 A3 y4 A4
A1 C A2 C A3 A4
yD
146675
D 33.2 mm
4414.2
A4
40 mm
20 mm
30 mm
90 mm
A1 :
A3
xD
x
A1 D 3090 D 2700 mm2
The value for y is not the same as in the new problem statement. This
value seems correct. (The x value checks).
x1 D 45 mm
y1 D 15 mm
A2 : (sits on top of A1 )
A2 D 4050 D 2000 mm2
x2 D 20 mm
y2 D 30 C 25 D 55 mm
A3 :
A3 D
1 2
r D 202 D 628.3 mm2
2 0
2
x3 D 20 mm
y3 D 80 mm C
A4 :
4r0
D 88.49 mm
3
A4 D 3020 C ri2
A4 D 600 C 102 D 914.2 mm2
x4 D 20 mm
y4 D 50 C 15 D 65 mm
Area (composite)
D A1 C A2 C A3 A4
D 4414.2 mm2
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1
Problem 7.36
centroids.
Determine the coordinates of the
y
5 mm
15 mm
50 mm
5 mm
5 mm
15 mm
x
15 mm
10 15 15 10
mm mm mm mm
Solution: Comparison of the solution to Problem 7.29 and our
areas 1, 2, and 3, we see that in order to use the solution of
Problem 7.29, we must set a D 25 mm, R D 15 mm, and r D 5 mm.
If we do this, we find that for this shape, measuring from the y axis,
x D 18.04 mm. The corresponding areas for regions 1, 2, and 3 is
1025 mm2 . The centroids of the rectangular areas are at their geometric
centers. By inspection, we how have the following information for the
five areas
Area 1: Area1 D 1025 mm2 , x1 D 18.04 mm, and y1 D 50 mm.
Area 2: Area2 D 1025 mm2 , x2 D 18.04 mm, and y2 D 0 mm.
Area 3: Area3 D 1025 mm2 , x3 D 18.04 mm, and y3 D 0 mm.
y
1
5 mm
5
15
15 mm
50 mm
y 4
5 mm
3
2
15
15 mm
5 mm
x
15 mm
10 15 15 10
mm mm mm mm
Area 4: Area4 D 600 mm2 , x4 D 0 mm, and y4 D 25 mm.
Area 5: Area5 D 450 mm2 , x5 D 7.5 mm, and y5 D 50 mm.
Combining the properties of the five areas, we can calculate the
centroid of the composite area made up of the five regions shown.
AreaTOTAL D Area1 C Area2 C Area3 C Area4 C Area5
D 4125 mm2 .
Then, x D x1 Area1 C x2 Area2 C x3 Area3 C x4 Area4
C x5 Area5 /AreaTOTAL D 3.67 mm,
and y D y1 Area1 C y2 Area2 C y3 Area3 C y4 Area4
C y5 Area5 /AreaTOTAL D 21.52 mm.
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1
y
Problem 7.37 The dimensions b D 42 mm and h D
22 mm. Determine the y coordinate of the centroid of
the beam’s cross section.
200 mm
h
120 mm
x
b
Solution: Work as a composite shape
y
100 mm
100 mm
A2
A
1
h
120 mm
x
b
b D 42 mm
h D 22 mm
A1 D 120 b mm2 D 5040 mm2
x1 D 0
y1 D 60 mm
by symmetry
A2 D 200 h D 4400 mm2
x2 D 0
y2 D 120 C
xD
h
D 131 mm
2
0C0
A1 x1 C A2 x2
D
A1 C A2
9440
xD0
yD
A1 y1 C A2 y2
D 93.1 mm
A1 C A2
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1
Problem 7.38 If the cross-sectional area of the beam
shown in Problem 7.37 is 8400 mm2 and the y coordinate of the centroid of the area is y D 90 mm, what are
the dimensions b and h?
Solution: From the solution to Problem 7.37
A1 D 120 b, A2 D 200 h
and y D
y1 A1 C y2 A2
A1 C A2
h
200 h
60120 b C 120 C
2
yD
120 b C 200 h
where
y1 D 60 mm
y D 90 mm
A1 C A2 D 8400 mm2
Also, y2 D 120 C h/2
Solving these equations simultaneously we get
h D 18.2 mm
b D 39.7 mm
200 mm
h
A2
A1
120 mm
b
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1
Problem 7.39 Determine the y coordinate of the centroid of the beam’s cross section.
y
5 in
2 in
8 in
x
3 in
5 in
5 in
3 in
Solution: Take advantage of the symmetry. Work with only one
half of the structure. Break into 2 rectangles, a quarter circle, and a
quarter circular hole.
4[5]
[5]2
438 C 11.533 C 8 C
3
4
4[2]
[2]2
8C
3
4
yD
[2]2
[5]2
38 C 33 C
4
4
y D 7.48 in
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1
y
Problem 7.40 Determine the coordinates of the centroid of the airplane’s vertical stabilizer.
11 m
48°
x
70°
12.5 m
Solution: We work with a rectangle and two triangular holes
y
e
We have
d D 12.5 m C 11 m cot 70° D 16.50 m
11 m
e D 11 m tan 48° D 12.22 m
48°
In the x direction
xD
x
1
1
1
2
2 d[d11] 3 e 2 e11 12.5 C 3 d 12.5
1
2 d 12.511
70°
12.5 m
d
d11 12 e11 12 d 12.511
1
yD 2
11[d11] 23 11
1
1
1
2 e11 3 11
2 d 12.5 m11 m
1
1
d11 2 e11 2 d 12.511
Solving we find
x D 9.64 m, y D 4.60 m
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1
Problem 7.41 The area has elliptical boundaries. If
a D 30 mm, b D 15 mm, and ε D 6 mm, what is the x
coordinate of the centroid of the area?
y
Solution: The equation of the outer ellipse is
y2
x2
C
D1
a C ε2
b C ε2
b
and for the inner ellipse
y2
x2
C 2 D1
2
a
b
x
a
We will handle the problem by considering two solid ellipses
For any ellipse
ˇ ˛ 2
x ˛ x 2 dx
x dA
˛ 0
D
xD ˇ
dA
˛2 x 2 dx
˛
y
ε
From integral tables
b
˛2 x 2 3/2
x
3
p
x
x ˛2 x 2
˛2
˛2 x 2 dx D
C
sin1
2
2
˛
˛
3/2
˛2 x 2
0
Substituting x D
p
x ˛
˛2
x ˛2 x 2
C
sin
3
2
˛
˛2 x 2 dx D ε
a
A1
=
–
x
A2
0
xD
0C
xD
dA D
ˇ
˛
2
β
2
˛3 /3
D 2
˛ /4
00
β
y = α √α 2 – x2
dA = y dx
α
ˇ
˛
˛
0
0
˛2 2
2
x
For the composite
˛2 C x 2 dx
p
x ˛
˛2
x ˛2 x 2
C
sin1
2
2
˛
ˇ
D
˛
Area D
y
˛2 4˛
3
Also Area D
0 C ˛3 /3
D ˛ˇ/4
(The area of a full ellipse is ˛ˇ so this checks.
Now for the composite area.
xD
x1 A1 x2 A2
A1 A2
Substituting, we get
x1 D 15.28 mm
x2 D 12.73 mm
A1 D 2375 mm2
A2 D 1414 mm2
and x D 19.0 mm
For the outer ellipse, ˛ D a C ε ˇ D b C ε and for the inner ellipse
˛Da ˇDb
Outer ellipse
x1 D
4a C ε
3
A1 D
a C εb C ε
4
Inner Ellipse
x2 D
4a
3
A2 D
ab
4
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1
Problem 7.42 By determining the x coordinate of the
centroid of the area shown in Problem 7.41 in terms of
a, b, and ε, and evaluating its limit as ε ! 0, show that
the x coordinate of the centroid of a quarter-elliptical
line is
xD
4aa C 2b
.
3a C b
Solution: From the solution to 7.41, we have
x1 D
4a C ε
3
4a
x2 D
3
so x1 A1 D
x2 A2 D
A1 D
x1 A1 x2 A2 D 13 a2 b C 2abε C bε2
a C εb C ε
4
ab
A2 D
4
C a2 ε C 2aε2 C ε3 a2 b
x1 A1 x2 A2 D 13 2ab C a2 ε
C 2a C bε2 C ε3 a C ε2 b C ε
3
Finally x D
a2 b
3
A1 A2 D
ab C aε C bε C ε2 ab
4
A1 A2 D
aε C bε C ε2 4
x1 A1 x2 A2
A1 A2
1
2ab C a2 C 2a C bε C ε2 ε
3
xD
[a C b C ε] ε
4
xD
42a C bε
4 2
4aa C 2b
C
C
ε
3a C b
3
3
Taking the limit as ε ! 0
xD
4aa C 2b
3a C b
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1
Problem 7.43 Three sails of a New York pilot
schooner are shown. The coordinates of the points are
in feet. Determine the centroid of sail 1.
Solution: Divide the object into three areas: (1) The triangle with
altitude 21 ft and base 20 ft. (2) The triangle with altitude 21 ft
and base 20 16 D 4ft, and (3) the composite sail. The areas and
coordinates are:
(1)
A1 D 210 ft2 ,
x1 D
y1 D
1
2
3
(2)
2
20 D 13.33 ft,
3
1
21 D 7 ft.
3
A2 D 42 ft2 ,
(a)
x2 D 16 C
y
y
y
y2 D 7 ft.
(14, 29)
(12.5, 23)
(20, 21)
(3, 20)
1
(16, 0)
(3)
(3.5, 21)
2
The composite area: A D A1 A2 D 168 ft2 .
The composite centroid:
3
x
x
x
(10, 0)
(23, 0)
2
4 D 18.67 ft,
3
xD
A1 x1 A2 x2
D 12 ft ,
A
yD
A1 y1 A2 y2
D 7 ft
A
(b)
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1
Problem 7.44
Problem 7.43.
Determine the centroid of sail 2 in
Solution: Divide the object into five areas: (1) a triangle on the
(3)
left with altitude 20 ft and base 3 ft, (2) a rectangle in the middle 23 ft
by 9.5 ft, (3) a triangle at the top with base of 9.5 ft and altitude of
3 ft. (4) a triangle on the right with altitude of 23 ft and base of 2.5 ft.
(5) the composite sail. The areas and centroids are:
(1)
A1 D
x1 D
y1 D
(2)
(4)
y2 D
9.5
2
y4 D
D 7.75 ft,
A4 D
(5)
2
3 D 22 ft
3
1
2.523 D 28.75 ft2 ,
2
x4 D 10 C
A2 D 239.5 D 218.5 ft2 ,
1
9.5 D 6.167 ft,
3
y3 D 20 C
1
20 D 6.67 ft.
3
x2 D 3 C
1
39.5 D 14.25 ft2 ,
2
x3 D 3 C
320
D 30 ft2 ,
2
2
3 D 2 ft,
3
A3 D
2
2.5 D 11.67 ft,
3
1
23 D 7.66 ft
3
The composite area: A D A1 C A2 A3 A4 D 205.5 ft2 .
The composite centroid:
23
D 11.5 ft
2
xD
A1 x1 C A2 x2 A3 x3 A4 x4
D 6.472 ft ,
A
yD
A1 y1 C A2 y2 A3 y3 A4 y4
D 10.603 ft
A
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1
Problem 7.45
Problem 7.43.
Determine the centroid of sail 3 in
Solution: Divide the object into six areas: (1) The triangle Oef,
with base 3.5 ft and altitude 21 ft. (2) The rectangle Oabc, 14 ft by
29 ft. (3) The triangle beg, with base 10.5 ft and altitude 8 ft. (4) The
triangle bcd, with base 9 ft and altitude 29 ft. (5) The rectangle agef
3.5 ft by 8 ft. (6) The composite, Oebd. The areas and centroids are:
(1)
a
f
g
b
e
A1 D 36.75 ft2 ,
x1 D 1.167 ft,
o
c
d
y1 D 14 ft.
(2)
A2 D 406 ft2 ,
(5)
A5 D 28 ft2 ,
x5 D 1.75 ft,
x2 D 7 ft,
y5 D 25 ft.
y2 D 14.5 ft.
(6)
(3)
The composite area:
A3 D 42 ft2 ,
A D A1 C A2 A3 C A4 A5 D 429.75 ft2 .
x3 D 7 ft,
The composite centroid:
y3 D 26.33 ft
(4)
A4 D 130.5 ft2 ,
xD
A1 x1 C A2 x2 A3 x3 C A4 x4 A5 x5
D 10.877 ft
A
yD
A1 y1 C A2 y2 A3 y3 C A4 y4 A5 y5
D 11.23 ft
A
x4 D 17 ft,
y4 D 9.67 ft.
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1
Problem 7.46 The value of the distributed load w at
x D 6 m is 240 N/m.
(a)
(b)
(c)
The equation for the loading curve is w D 80x 240N/m. Use Eq. (7.10) to determine the magnitude of the total force exerted on the beam by the
distributed load.
If you use the area analogy to represent the
distributed load by an equivalent force, what is the
magnitude of the force and where does it act?
Determine the reactions on the beam at A and B.
y
w
240 N/m
B
A
3m
3m
Solution:
360 N
(a)
RD
x
6
80x 240 dx D 360 N
w dx D
3
(b)
R D 12 240 N/m3 m D 360 N
5m
Ax
x D 3 m C 23 3 m D 5 m
(c) Equilibrium equations
6m
Ay
B
MA : B6 m 360 N5 m D 0
Fx : Ax D 0
Fy : Ay C B 360 N D 0
)
Ax D 0
Ay D 60 N
B D 300 N
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1
Problem 7.47
support A.
Determine the reactions at the fixed
y
200 N/m
A
x
100 N/m
2m
2m
2m
2m
Solution: Replace the distributed loads by equivalent single forces
200 N
Fx : Ax D 0
MA
Fy : Ay C 200 N 200 N D 0
2m
2m
MA : MA C 200 N3 m 200 N 6 m C 13 2 m D 0
2m
Ax
Solving we find:
Ax D 0, Ay D 0, MA D 733 N-m
2m
Ay
200 N
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1
Problem 7.48
support A.
Determine the reactions at the built-in
y
200 N/m
100 N/m
x
A
3m
3m
Solution: Replace the distributed load by two equivalent forces
Equilibrium Equations
Fx : Ax D 0
Fy : Ay 300 N 150 N D 0
MA : MA 300 N4.5 m 150 N5 m D 0
Solving:
Ax D 0, Ay D 450 N, MA D 2100 N-m
150 N
300 N
Ax
MA
Ay
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1
Problem 7.49
Determine the reactions at A and B.
y
x
A
B
L/2
L/2
Solution: Let us break the load into two parts and use the area
analogy.
Now we can find the support reactions
y
3L
4
L
2
L
2
B
L
3
L
2
L
xL
2
Load 1
L1 D
0
L1 D
By
Ay
For Load L2
L/2
2ω0 2ω0
x dx D
L
L 2
Fx :
Ax D 0
Fy :
Ay C By MA :
By
L/2
x2 0
Lω0
ω0 L 2
D
L 4
4
Solving the third eqn.
By D
x1 D L/3
From the second eqn,
3Lω0
11
Lω0
C
D
Lω0
6
4
12
Load 2
L2 D
L/2
Lω0
Lω0
D0
4
2
Lω0
L
Lω0
3L
L
D0
2
4
3
2
4
using the area analogy, load L1 acts 2/3 of the distance from the origin
to L/2. Thus
L
4
Ax
2ω0
x for 0 x L/2
L
ωx D ω0 for
2
L
x
For Load L1
ωx D
L
L2
L1
A
L
ω0 dx D ω0 x L/2
Ay C By D
D
ω0 L
2
3
Lω0
4
Hence Ay D
3
Lω0
lω0 By D 4
6
And from the area analogy, L2 acts half way between L/2 and L.
3L
.
x2 D
4
Ax D 0 Ay D Lω0 /6
By D 11 Lω0 /12
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1
Problem 7.50
support A.
Determine the reactions at the fixed
y
= 3(1 – x 2/25) kN/m
Solution: The free-body diagram of the beam is: The downward
force exerted by the distributed load is
w dx D
L
= 3(1 – x 2/25) kN/m
5 x2
3 1
dx
25
0
5
x3
D 10 kN.
D3 x
75 0
x
A
5m
Ma
5m
Ax
x
Ay
The clockwise moment about the left end of the beam due to the
distributed load is
xw dx D
L
5 x3
3 x
dx
25
0
D3
2
5
x4
x
D 18.75 kN-m.
2
100 0
From the equilibrium equations
Fx D Ax D 0,
Fy D Ay 10 D 0,
mleftend D Ma C 5Ay 18.75 D 0,
we obtain
Ax D 0,
Ay D 10 kN,
and Ma D 31.25 kN-m.
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1
Problem 7.51 An engineer measures the forces
exerted by the soil on a 10-m section of a building
foundation and finds that they are described by the
distributed load w D 10x x 2 C 0.2x 3 kN[/]m.
(a)
(b)
Determine the magnitude of the total force exerted
on the foundation by the distributed load.
Determine the magnitude of the moment about A
due to the distributed load.
y
2m
10 m
A
x
Solution:
(a)
The total force is
12
FD
10x C x 2 0.2x3 dx
0
10
x3
0.2 4
D 5x 2 C
x
3
4
0
jFj D 333.3 kN
(b)
The moment about the origin is
10
MD
10x C x 2 0.2x3 x dx
0
1
0.2 5 10
10
x
,
D x3 x4 C
3
4
5
0
jMj D 1833.33 kN.
The distance from the origin to the equivalent force is
dD
jMj
D 5.5 m,
F
from which
jMA j D d C 2F D 2500 kN 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 7.52
A and B.
Determine the reactions on the beam at
3 kN/m
2 kN/m
A
B
4m
2m
Solution: Replace the distributed load with three equivalent single
forces.
The equilibrium equations
Fx : Ax D 0
Fy : Ay C B 8 kN 2 kN 3 kN D 0
MA : B4 m 8 kN2 m 2 kN
2
34 m
3 kN 4 m C 13 2 m D 0
Ax D 0, Ay D 4.17 kN B D 8.83 kN
2 kN
8 kN
3 kN
Ax
Ay
B
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1
Problem 7.53 The aerodynamic lift of the wing is described
by the distributed load
y
p
w D 300 1 0.04x 2 N/m.
The mass of the wing is 27 kg, and its center of mass is
located 2 m from the wing root R.
x
R
2m
(a)
(b)
Determine the magnitudes of the force and the moment
about R exerted by the lift of the wing.
Determine the reactions on the wing at R.
5m
Solution:
(a)
w
The force due to the lift is
M
5
2 1/2
3001 0.04x F D w D
dx,
mg
0
FD
300
5
F D 60
5
R
FR
2m
25 x 2 1/2 dx
3m
0
5
p
25 1 x x 25 x 2
C
sin
D 375 N,
2
2
5
0
jFj D 1178.1 N.
The moment about the root due to the lift is
5
M D 300
1 0.04x 2 1/2 x dx,
0
M D 60
25 x 2 3/2
3
5
D
0
60253/2
D 2500
3
jMj D 2500 Nm.
(b)
The sum of the moments about the root:
M D MR C 2500 27g2 D 0,
from which MR D 1970 N-m. The sum of forces
Fy D FR C 1178.1 27g D 0,
from which FR D 1178.1 C 27g D 913.2 N
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1
Problem 7.54
and B.
Determine the reactions on the bar at A
400 lb/ft
B
y
2 ft
600 lb/ft
400 lb/ft
2 ft
x
4 ft
A
4 ft
Solution: First replace the distributed loads with three equivalent forces.
The equilibrium equations
Fx : Bx C 800 lb D 0
MB : 800 lb1 ft A4 ft C 1600 lb6 ft
C 400 lb6.67 ft D 0
Fy : A C By 1600 lb 400 lb D 0
Solving:
A D 3267 lb, Bx D 800 lb, By D 1267 lb
Bx
800 lb
By
400 lb
1600 lb
A
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1
Problem 7.55
y
Determine the reactions at A and B.
20 kN-m
4 kN/m
A
6 kN
6m
x
B
6m
6m
Solution: Break the load into two parts and find the equivalent
concentrated load for each part. Then find the reactions at A and B
y
w1 (x)
w2 (x)
4 kN/m
x
6
w1 x D 4 kN/m
w2 x D
12
6 m ð 12 m
2
x C 12 kN/m
3
12
F1 D
18
12 m ð 18 m
12
4 dx
w1 x dx D
6
6
12
F1 D 4x D 24 kN
6
By symmetry, F1 is applied at x D 9 m
18
F2 D
w2 x dx D
12
18 2
x C 12 dx
3
12
2
18
x
F2 D C 12x
D 108 96 kN
3
12
F2 D 12 kN
By the area analogy, this load is applied at x D 14 m ( 13 of the way
from 12 to 18).
y
24 kN
3m
6m
5m
12 kN
20 kN-m
x
AX
6 kN
6m
AY
BY
Fx :
Ax D 0
Fy :
Ay C By 24 12 6 D 0
MA :
66 C 20 324 C 6By 812 D 0
Solving
Ax D 0, Ay D 23.3 kN, By D 18.7 kN
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1
Problem 7.56 Determine the axial forces in members
BD, CD, and CE of the truss and indicate whether they
are in tension (T) or compression (C).
2m
2m
B
A
2m
2m
H
F
D
2m
C
G
E
4 kN/m
8 kN/m
Solution: We start by analyzing the horizontal bar with the distributed load
MG : 16 kN0.667 m C 32 kN2 m
FC
FG
FC 4 m D 0 ) FC D 18.67 kN
Fy D FC C FG 32 kN 16 kN D 0
32 kN
) FG D 29.33 kN
Now work with the whole structure in order to find the reactions at A
Fx : Ax D 0 ) Ax D 0
16 kN
Ax
MH : FG 2 m C FC 6 m
Ay
Ay 8 m D 0 ) Ay D 21.3 kN
H
Finally, cut through the truss and look at the left section
MC : Ax 2 m Ay 2 m BD2 m D 0
FC
MD : Ay 4 m C FC 2 m C CE2 m D 0
Ax
FG
B
BD
1
Fy : Ay FC C p CD D 0
2
CD
Solving we find
BD D 21.3 kN, CD D 3.77 kN, CE D 24 kN
Ay
C
CE
In summary:
FC
BD D 21.3 kNC, CD D 3.77 kNC, CE D 24 kNT
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1
Problem 7.57 Determine the reactions on member
ABC at A and B.
400 N/m
200 N/m
C
160 mm
B
D
240 mm
E
A
400 N/m
160
mm
Solution: Work on the entire structure first to find the reactions at
A. Replace the distributed forces with equivalent concentrated forces
160
mm
160
mm
48 N
96 N
Fx : Ax C 160 N D 0
ME : 96 N0.08 m C 48 N0.16 m
160 N0.2 m Ay 0.32 m D 0
Solving:
Ax D 160 N, Ay D 52 N
Now look at body ABC. Take advantage of the two-force body CD.
160 N
MB : Ax 0.24 m C 160 N0.04 m
48 N0.16 m 96 N0.24 m
7
4
p CD0.32 m C p CD0.16 m D 0
65
65
Ax
E
Ay
4
Fx : Ax C Bx C 160 N p CD D 0
65
48 N
7
Fy : Ay C By 48 N 96 N p CD D 0
65
96 N
Solving:
Ax D 160 N, Ay D 52 N
Bx D 157 N, By D 78.4 N
C
7
By
4
Bx
CD
160 N
Ax
Ay
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1
Problem 7.58
of the frame.
Determine the forces on member ABC
A
1m
3 kN/m
B
1m
C
2m
Solution: The free body diagram of the member on which the
distributed load acts is
From the equilibrium equations
Fx D Bx D 0,
Fy D By C E 12 D 0,
2m
1m
(4 m)(3 kN/m) = 12 kN
BX
2m
BY
1m
E
AX
4 kN
mleftend D 3E 212 D 0,
AY
CX
we find that Bx D 0, By D 4 kN, and E D 8 kN. From the lower fbd,
writing the equilibrium equation
mleftend D 2Cy 48 D 0,
we obtain Cy D 16 kN. Then from the middle free body diagram,
we write the equilibrium equations
8 kN
CY
CX
DX
DY
2m
CY
2m
Fx D Ax C Cx D 0,
Fy D Ay 4 16 D 0,
mrightend D 2Ax 2Ay C 14 D 0
obtaining Ax D 18 kN, Ay D 20 kN, Cx D 18 kN.
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1
y
Problem 7.59 Determine the coordinates of the
centroid of the truncated conical volume.
Strategy:
Use the method described in Example 7.8.
z
R
x
h–
2
h–
2
Solution: Refer to Example 7.8.
y
dV
z
x
x
dx
(a) An element dV in the form of a disk.
y
h
R
–x
h
h–
2
R
x
x
dx
(b) The radius of the element is (R/h)x.
h
xdV
x D V
R2 3
x dx
h2
R2 2
x dx
h2
h/2
D h
dV
V
h/2
4 h
4
x /4 h/2
h /4 h4 /64
xD h D 3
h /3 h3 /24
x 3 /3 h/2
xD
45
h
56
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1
Problem 7.60 A grain storage tank has the form of a
surface of revolution with the profile shown. The height
of the tank is 7 m and its diameter at ground level is
10 m. Determine the volume of the tank and the height
above ground level of the centroid of its volume.
y
y = ax1/2
7m
10 m
x
Solution:
O
y
y = ax1/2
y
dx
dV = π y2dx
x
dV D y 2 dx
7
7
y 2 dx
x D 0 7
a2 x dx
D 0 7
y 2 dx
a2 x dx
0
0
3 7
x /3 0
xD 7 D 4.67 m
x 2 /2 0
The height of the centroid above the ground is 7 m x
h D 2.33 m
The volume is
7
VD
a2 x dx D a2
0
49
2
m3
To determine a,
y D 5, m when x D 7 m.
p
y D ax 1/2 , 5 D a 7
p
a D 5/ 7a2 D 25/7
VD
25
7
49
2
D 275 m3
V D 275 m3
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1
y
Problem 7.61 The object shown, designed to serve
as a pedestal for a speaker, has a profile obtained by
revolving the curve y D 0.167x 2 about the x axis. What
is the x coordinate of the centroid of the object?
z
x
0.75 m
0.75 m
Solution:
y = 0.167 x2
dV = π y2dx
x
dv
1.50
xdV
x D V
D 0.75
1.50
dV
V
x0.167x2 2 dx
0.167x 2 2 dx
0.75
1.5
0.1672
Ð 0.75
xD
1.5
0.1672
x 5 dx
x 4 dx
6 1.5
x /6 0.75
D 1.5
x 5 /5 0.75
0.75
x D 1.27 m
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1
y
Problem 7.62 The volume of a nose cone is generated
by rotating the function y D x 0.2x 2 about the x axis.
(a)
(b)
What is the volume of the nose cone?
What is the x coordinate of the centroid of
the volume?
z
x
2m
Solution:
2m
(a)
VD
y 2 dx D
0
2m
(b)
xD
2m
x 0.2x 2 2 dx D 4.16 m3
0
2
xy 2 dx
0
V
D
xx 0.2x 2 2 dx
0
4.155 m3
D 1.411 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 7.63
rical volume.
Determine the centroid of the hemisphey
R
z
x
Solution: The equation of the surface of a sphere is x2 C y 2 C
z 2 D R2 .
The volume: The element of volume is a disk of radius and thickness
dx. The radius of the disk at any point within the hemisphere is 2 D
y 2 C z2 . From the equation of the surface of the sphere, 2 D R2 x 2 . The area is 2 , and the element of volume is dV D R2 x 2 dx, from which
Vsphere
2 3
D
R .
2
3
VD
The x-coordinate is:
R
x dV D R2 x 2 x dx
0
V
D
D
x4
R2 x 2
2
4
R
0
4
R .
4
Divide by the volume:
xD
R4
4
3
2R3
D
3
R.
8
By symmetry, the y- and z-coordinates of the centroid are zero.
y
x
R
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1
Problem 7.64 The volume consists of a segment of a
sphere of radius R. Determine its centroid.
y
x
R
R
2
z
Solution: The volume: The element of volume is a disk of radius
and thickness dx. The area of the disk is 2 , and the element of
volume is 2 dx. From the equation of the surface of a sphere (see
solution to Problem 7.63) 2 D R2 x 2 , from which the element of
volume is dV D R2 x 2 dx. Thus
R
dV D VD
V
R2 x 2 dx
R/2
R
x3
5
D
R3 .
D R2 x 3 R/2
24
The x-coordinate:
R
x dV D V
R2 x 2 x dx
R/2
D
x4
R2 x 2
2
4
R
D
R/2
9 4
R .
64
Divide by the volume:
xD
9R4
64
24
5R3
D
27
R D 0.675R.
40
By symmetry the y- and z-coordinates are zero.
y
R
–
2
x
R
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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 7.65
A volume of revolution is obtained
x2
y2
by revolving the curve 2 C 2 D 1 about the x axis.
a
b
Determine its centroid.
y
y2
x 2 + ––
––
=1
a 2 b2
z
x
Solution: The volume: The element of volume is a disk of radius
y and thickness dx. The area of the disk is y 2 . From the equation for
the surface of the ellipse,
y
x2 + y 2 = 1
a2 b2
x2
y 2 D b2 1 2
a
and dV D y 2 dx D b2
x2
1 2 dx,
a
x
from which
dV D b2
VD
a
1
0
V
x2
a2
dx
a
2b2 a
x3
D b2 x 2 D
.
3a 0
3
The x-coordinate:
a
x dV D b2
1
0
V
D b2
x2
a2
x dx
2
a
x4
b2 a2
x
2 D
.
2
4a 0
4
Divide by volume:
xD
b2 a2
4
3
2b2 a
D
3
a.
8
By symmetry, the y- and z-coordinates of the centroid are zero.
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1
Problem 7.66 The volume of revolution has a cylindrical hole of radius R. Determine its centroid.
y
z
h
R
R+a
x
Solution: The volume: The element of volume is a disk of radius
y and thickness
dx. The area of the disk is y 2 R2 . The radius is
a
x C R. The volume element is
yD
h
dV D a
h
2
xCR
y
R
R+a
dx R2 dx.
h
Denote
mD
a
h
, dV D m2 x 2 C 2mRx dx,
from which
VD
h
dV D m
mx 2 C 2Rx dx
0
V
h
3
mh
x
CR .
D m m C Rx 2 D mh2
3
3
0
The x-coordinate:
h
x dV D m
mx 3 C 2Rx 2 dx
0
V
h
4
2Rx 3
x
D m m C
4
3 0
2R
h
D mh3 m C
.
4
3
Divide by the volume:
2R
2R
h
a
C
m C
4
3
4
3
.
D h a
xDh h
CR
m CR
3
3
By symmetry, the y- and z-coordinates of the centroid are zero.
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1
y
Problem 7.67 Determine the coordinates of the centroid of the line. (See Example 7.9.)
y ⫽x2
⫺1
2
x
Solution:
2
2 dy
x
1C
dx
xds
x 1 C 2x2 dx
dx
1
1
D
x D 12
D
2 2
2
dy
ds
1 C 2x2 dx
1
C
dx
1
1
dx
1
2
2
2
2 dy
y 1C
dx
x 2 1 C 2x2 dx
dx
1
1
D
D
y D 12
2 2
2
dy
ds
1 C 2x2 dx
1
C
dx
1
1
dx
1
2
2
yds
x D 0.801
y D 1.482
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1
Problem 7.68 Determine the x coordinate of the
centroid of the line.
Solution: The length: Noting that
of length is
y
1C
dL D
dy
dx
2
dx D
p
dy
D x 11/2 , the element
dx
x dx
from which
2
y = – (x – 1)3/2
3
5
dL D
LD
x1/2 dx D
1
L
2 3/2 5
D 6.7869.
x
3
1
The x-coordinate:
5
x dL D
0
5
x
L
x 3/2 dx D
0
Divide by the length: x D
2 5/2 5
x
D 21.961.
5
1
21.961
D 3.2357
6.7869
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1
Problem 7.69 Determine the x coordinate of the
centroid of the line.
y
2
y = – x 3/2
3
0
x
2
Solution: The length: Noting that
length is
1C
dL D
dy
dx
2
dx D
dy
D x1/2 the element of
dx
p
1 C x dx
from which
2
dL D
LD
1 C x1/2 dx D
0
L
2
2
1 C x3/2 D 2.7974
3
0
The x-coordinate:
2
x dL D
x1 C x1/2 dx D 2
0
L
D2
1 C x5/2
1 C x3/2
5
3
2
0
5/2
33/2
1
3
1
C
D 3.0379.
5
3
5
3
Divide by the length: x D 1.086
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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 7.70
arc.
Determine the centroid of the circular
y
α
x
R
Solution: The length: From the equation for the circle,
yD
p
dy
R2 x 2 and
D R2 x 2 1/2 x.
dx
The element of length
1C
dL D
dy
dx
2
dx D RR2 x 2 1/2 dx,
from which
R
RR2 x 2 1/2 dx
dL D
LD
R cos ˛
L
x R
D R sin1
R R cos ˛
DR
DR
2
sin1 cos ˛
sin1 sin
˛
D R˛
2
2
Check: L D R˛ from the definition of ˛. check.
The x-coordinate:
R
x dL D R
xR2 x 2 1/2 dx
R cos ˛
L
R
D R R2 x 2 1/2 R cos ˛ D R2 sin ˛
Divide by the length: x D
R
sin ˛.
˛
The y-coordinate:
The y-coordinate of the centroid of each element is
p
y D y D R2 x 2 . Hence
R
y dL D R
R2 x 2 1/2 R2 x 2 1/2 dx
R cos ˛
L
Rc
DR
dx
R cos ˛
D R2 1 cos ˛.
Divide by the length:
yD
R
1 cos ˛
˛
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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 7.71
y
Determine the centroids of the volumes.
15 in
20 in
15 in
x
15 in
35 in
z
Solution: The Rectangle:
y
15 in
AreaR D 30 ð 35 D 1050 in2 ,
20 in
15 in
xR D 35/2 D 17.5 in,
30 in
yR D 30/2 D 15 in
15 in
The Triangle:
35 in
x
AreaT D 2015/2 D 150 in2 ,
xT D 15 C
2
20 D 28.33 in,
3
yT D 30 15/3 D 25 in
The Solid:
x D xR AreaR xT AreaT /AreaR AreaT D 15.7 in,
y D y D yR AreaR yT AreaT /AreaR AreaT D 13.3 in,
and from symmetry, z 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 7.72
Determine the centroids of the volumes.
y
600
mm
300
mm
x
400
mm
z
800
mm
1400
mm
Solution: Divide the figure into a rectangular and 2 triangular parts
V D 0.40.60.8 C 12 0.80.80.4 C 12 0.60.80.3
D 0.192 C 0.128 C 0.072 D 0.392 m3
xD
yD
zD
0.1920.3 C 0.128 0.6 C 13 0.8 C 0.0720.3
D 0.485 m
0.392
0.1920.2 C 0.128
1
3 0.4
C 0.072 0.4 C 13 0.3
0.392
0.1920.4 C 0.1280.4 C 0.072
0.392
1
D 0.233 m
3 0.8
D 0.376 m
x D 485 mm, y D 233 mm, z D 376 mm
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1
Problem 7.73
Determine the centroids of the volumes.
y
z
R
x
4R
Solution: The object will be divided into a cone and a hemisphere.
From symmetry y D z D 0
Using tables we have in the x direction
xD
3
4R
4
1 2
3R
2R3
R [4R] C 4R C
83R
3
8
3
D
24
2R3
1 2
R [4R] C
3
3
In summary
xD
83R
, y D 0, z D 0
24
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1
Problem 7.74
Determine the centroids of the volumes.
y
200 mm
300 mm
Solution: We have a hemisphere and a hemispherical hole. From
symmetry y D z D 0
z
x
In the x direction we have
3[300 mm]
2[300 mm]3
8
3
3[200 mm]
2[200 mm]3
8
3
xD
2[300 mm]3
2[200 mm]3
3
3
We have
x D 128 mm, y D 0, z D 0
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1
Problem 7.75
y
Determine the centroids of the volumes.
z
60 mm
90 mm
360 mm
x
460 mm
Solution: This is a composite shape. Let us consider a solid
cylinder and then subtract the cone. Use information from the appendix
Cylinder
Cone
Volume
Volume (mm3 )
x
x (mm)
R2 L
1.1706 ð 107
1.3572 ð 106
L/2
L-h/4
230
370
1
2
3 r h
R D 90 mm
L D 460 mm
r D 60 mm
h D 360 mm
xD
XCyL VCyL XCONE VCONE
VCyL VCONE
x D 211.6 mm
y D z D 0 mm
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1
Problem 7.76
Determine the centroids of the volumes.
y
20 mm
25 mm
75 mm
x
120 mm
25 mm
100 mm
z
Solution: Break the composite object into simple shapes, find the
volumes and centroids of each, and combine to find the required
centroid.
Object
Volume (V)
x
y
z
1
2
LWH
hWD
0
0
L/2
D/2
3
R2 D/2
0
4
r 2 D
0
H/2
H
C h/2 4R
HChC
3
H C h
y
x
(H)
25 mm
+
D/2
D/2
12
(L)
0m
m
100
mm (W)
y
where R D W/2. For the composite,
z
x1 V1 C x2 V2 C x3 V3 x4 V4
V1 C V2 C V3 V4
+
with similar eqns for y and z
–
50
xD
mm
1
3
x
The dimensions, from the figure, are
L D 120 mm
W D 100 mm
H D 25 mm
r D 20 mm
h D 75 mm
D D 25 mm
R D 50 mm
Object
V mm3
x (mm)
y (mm)
z (mm)
C1
C2
C3
4
300000
187500
98175
31416
0
0
0
0
12.5
62.5
121.2
100
60
12.5
12.5
12.5
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1
7.76 (Continued )
y
Substituting into the formulas for the composite, we get
D)
xD0
mm
100
(
mm
25
y D 43.7 mm
)
(h
mm
75
z D 38.2 mm
x
2
H
y
z
r = 20 mm
x
4
z
2
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Problem 7.77
Determine the centroids of the volumes.
y
1.75 in
1 in
5 in
z
4 in
1 in
x
Solution: Divide the object into six volumes: (1) A cylinder 5 in
long of radius 1.75 in, (2) a cylinder 5 in long of radius 1 in, (3) a
block 4 in long, 1 in thick, and 21.75 D 3.5 in wide. (4) Semicylinder 1 in long with a radius of 1.75 in, (5) a semi-cylinder 1 in
long with a radius of 1.75 in. (6) The composite object. The volumes
and centroids are:
1 in
x
1.75 in
z
z
x 5 in
Volume
V1
V2
V3
V4
V5
Vol, cu in
48.1
15.7
14
4.81
4.81
x, in
0
0
2
0.743
0
y, in
2.5
2.5
0.5
0.5
4.743
z, in
0
0
0
0
0
4 in
y
1 in
x
The composite volume is V D V1 V2 C V3 V4 C V5 D 46.4 in3 .
The composite centroid:
xD
V1 x1 V2 x2 C V3 x3 V4 x4 C V5 x5
D 1.02 in,
V
yD
V1 y1 V2 y2 C V3 y3 V4 y4 C V5 y5
D 1.9 in,
V
zD0
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1
Problem 7.78
y
Determine the centroids of the volumes.
30 mm
60 mm
z
x
180
mm
Solution: Consider the composite volume as being made up of
three volumes, a cylinder, a large cone, and a smaller cone which is
removed
V
x
Cylinder
r 2 L/2
1 2
R L
3
1 2 L
r
3
2
L/4
Cone 1
Cone 2
Cylinder
Cone 1
Cone 2
y
2R
Object
180
mm
2r
3L/4
L/2
3(L/2)/4
(mm3 )
(mm)
5.089 ð 105
1.357 ð 106
1.696 ð 105
90
270
135
L/2
y
60 mm
Cylinder
L D 360 mm
1
x
r D 30 mm
R D 60 mm
y
For the composite shape
xCyl VCyL C x1 V1 x2 V2
VCyL C V1 V2
120 mm
cone
+
xD
360 mm
2
x
x D 229.5 mm
y
cone
–
60 mm
3
x
180 mm
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1
Problem 7.79 The dimensions of the Gemini spacecraft (in meters) are a D 0.70, b D 0.88, c D 0.74,
d D 0.98, e D 1.82, f D 2.20, g D 2.24, and h D 2.98.
Determine the centroid of its volume.
y
g
e
b
a
Solution: The spacecraft volume consists of three truncated cones
and a cylinder. Consider the truncated cone of length L with radii at
the ends R1 and R2 , where R2 > R1 . Choose the origin of the x –y
coordinate system at smaller end. The radius of the cone is a linear
function of the length; from geometry, the length of the cone before
truncations was
R2 L
with volume
R2 R1 (1)
HD
(2)
R22 H
. The length of the truncated portion is
3
(3)
D
(4)
R1 L
with volume
R2 R1 R12 . The volume of the truncated cone is the difference of the
3
two volumes,
(5)
L
VD
3
cone is
(6)
x D
(7)
xh D
(8)
R23 R13
R2 R1
. The centroid of the removed part of the
3
, and the centroid of the complete cone is
4
c
f
d
h
x
Beginning from the left, the volumes are (1) a truncated cone, (2) a
cylinder, (3) a truncated cone, and (4) a truncated cone. The algorithm
and the data for these volumes were entered into TK Solver Plus and
the volumes and centroids determined. The volumes and x-coordinates
of the centroids are:
Volume
V1
V2
V3
V4
Composite
Vol, cu m
0.4922
0.5582
3.7910
11.8907
16.732
x, m
0.4884
1.25
2.752
4.8716
3.999
The last row is the composite volume and x-coordinate of the centroid
of the composite volume.
The total length of the spacecraft is 5.68 m, so the centroid of the
volume lies at about 69% of the length as measured from the left
end of the spacecraft. Discussion: The algorithm for determining the
centroid of a system of truncated cones may be readily understood
if it is implemented for a cone of known dimensions divided into
sections, and the results compared with the known answer. Alternate
algorithms (e.g. a Pappus-Guldinus algorithm) are useful for checking
but arguably do not simplify the computations End discussion.
3
H, measured from the pointed end. From the
4
composite theorem, the centroid of the truncated cone is
Vh xh V x
C x, where x is the x-coordinate of the left
V
hand edge of the truncated cone in the specific coordinate system.
These eight equations are the algorithm for the determination of
the volumes and centroids of the truncated cones forming the
spacecraft.
xD
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1
y
Problem 7.80 Two views of a machine element are
shown. Determine the centroid of its volume.
y
24 mm
Solution: We divide the volume into six parts as shown. Parts 3
and 6 are the “holes”, which each have a radius of 8 mm. The volumes
are
8 mm
18 mm
60 mm
V1 D 604850 D 144,000 mm3 ,
V2 D 12 242 50 D 45, 239 mm3 ,
x
V3 D 82 50 D 10, 053 mm3 ,
V4
8 mm
z
20
mm
D 163620 D 11, 520 mm3 ,
16
mm
50 mm
V5 D 12 182 20 D 10, 179 mm3 ,
y
V6 D 82 20 D 4021 mm3 .
The coordinates of the centroids are
2
x1 D 25 mm,
3
y1 D 30 mm,
6
z1 D 0,
x2 D 25 mm,
y2 D 60 C
424
D 70.2 mm,
3
z2 D 0,
5
z5 D 24 C 16 C
x6 D 10 mm,
y3 D 60 mm,
y6 D 18 mm,
x4 D 10 mm,
x5 D 10 mm,
y5 D 18 mm,
418
D 47.6 mm,
3
z6 D 24 C 16 D 40 mm.
The x coordinate of the centroid is
y4 D 18 mm,
z4 D 24 C 8 D 32 mm,
4
z
x3 D 25 mm,
z3 D 0,
1
xD
x1 V1 C x2 V2 x3 V3 C x4 V4 C x5 V5 x6 V6
D 23.65 mm.
V1 C V2 V3 C V4 C V5 V6
Calculating the y and z coordinates in the same way, we obtain y D
36.63 mm and z D 3.52 mm
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1
Problem 7.81
Determine the centroid of the line.
y
4 in
Solution: Break into 3 line segments
xD
3 in
yD
2 in
52 in2 C 9 in6 in C 12 in4 in
D 7.18 in
52 in2 C 6 in C 4 in
x
6 in
6 in
52 in2 C 4 in6 in C 2 in4 in
D 2.70 in
52 in2 C 6 in C 4 in
x D 7.18 in
y D 2.70 in
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1
Problem 7.82
Determine the centroids of the lines.
y
3m
x
6m
Solution: The object is divided into two lines and a composite.
(1)
(2)
(3)
L1 D 6 m, x1 D 3 m, y1 D 0.
6
L2 D 3 m, x2 D 6 C m (Note: See Example 7.13) y2 D 3.
The composite length: L D 6 C 3 m. The composite centroid:
xD
L1 x1 C L2 x2
D 6 m,
L
yD
3
D 1.83 m
2C
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1
Problem 7.83
y
Determine the centroids of the lines.
2m
2m
x
2m
2m
Solution: Break the composite line into three parts (the quarter
circle and two straight line segments) (see Appendix B).
Part 1
Part 2
Part 3
xi
yi
Li
2R/
3 m
0
2R/
0
3 m
R/2
2 m
2 m
xD
x1 L 1 C x 2 L 2 C x 3 L 3
D 1.4 m
L1 C L2 C L3
yD
y 1 L 1 C y2 L 2 C y3 L 3
D 1.4 m
L1 C L2 C L3
(R D 2 m)
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1
Problem 7.84 The semicircular part of the line lies in
the x –z plane. Determine the centroid of the line.
y
100 mm
160 mm
x
120 mm
z
Solution: The bar is divided into three segments plus the
composite. The lengths and the centroids are given in the table: The
composite length is:
3
LD
y
100 mm
2
3
160 mm
Li .
iD1
z
The composite coordinates are:
3
xD
L
and y D iD1
120 mm
x
Li xi
iD1
3
1
,
Li yi
L
Segment
Length, mm
L1
120
L2
100
x, mm
240
0
y, mm
z, mm
0
120
50
0
L3
188.7
80
50
0
Composite
665.7
65.9
21.7
68.0
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1
Problem 7.85
Determine the centroid of the line.
y
200 mm
60⬚
x
Solution: Break into a straight line and an arc.
1
° 2
°
2 200 mm tan 60 cos 30 C
2/3
0
1
C cos d
cos 60°
D 332 mm
xD
2/3
200 mm tan 60° C
200 mmd
200 mm2
0
1
°
°
2 200 mm tan 60 200 mm sin 60 2/3
2
200 mm sin d
D 118 mm
2/3
200 mm tan 60° C
200 mmd
C
yD
0
0
x D 332 mm, y D 118 mm
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1
Problem 7.86 Use the first Pappus–Guldinus theorem
to determine the area of the curved part of the surface
of the truncated cone.
y
z
R
x
h
2
h
2
Solution: Work with the solid line shown.
The surface area is given by
A D 2yL D 2
3R
4
2 2
h
R
C
2
2
3R/4
R
R/2
h/2
h/4
h/4
3R p 2
AD
h C R2
4
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1
Problem 7.87 Use the second Pappus–Guldinus theorem to determine the volume of the truncated cone.
Solution: Work with the trapezoidal area
A D R/2h/2 C 1/2R/2h/2 D
3Rh
8
R
R/2
yD
7R
R/2h/2R/4 C 1/2R/2h/2[1/3R/2 C R/2]
D
A
18
V D 2yA D 2
VD
7R
18
3Rh
8
D
h/2
h/4
h/4
7R2 h
24
7R2 h
24
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1
Problem 7.88 Use the second Pappus–Guldinus
theorem to determine the volume generated by revolving
the curve about the x axis.
Solution: The area: The element of area is the vertical strip of
height y and width dx. Thus
1
AD
y
1
y dx D
0
x 2 dx.
0
Integrating,
(1, 1)
AD
3 1
1
x
D .
3 0
3
The y-coordinate:
y = x2
1
y dA D
y1 x dy
0
A
x
1
D
y y 3/2 dy D
0
Divide by the area: y D
y
2
1
y
1
2y 5/2
D
.
2
5
10
0
3
. The Volume: V D 2yA D
10
5
(1, 1)
y = x2
x
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1
Problem 7.89 Use the second Pappus–Guldinus
theorem to determine the volume generated by revolving
the curve about the y axis.
Solution: The x coordinate of the centroid:. The element of area
is the vertical strip of height 1 y and width dx. Thus
1
AD
1
1 y dx D
0
1 x 2 dx.
0
Integrating,
1
2
x3
D .
AD x
3 0
3
1
x dA D
A
x x 3 dx D
0
divide by the area: x D
2
1
x4
x
1
D ,
2
4 0
4
3
. The volume is V D 2xA D
8
2
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1
Problem 7.90 The length of the curve is L D 1.479,
and the area generated by rotating it about the x axis
is A D 3.810. Use the first Pappus–Guldinus theorem to
determine the y coordinate of the centroid of the curve.
Solution: The surface area is A D 2yL, from which
yD
A
D 0.41
2L
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1
Problem 7.91 Use the first Pappus–Guldinus theorem
to determine the area of the surface generated by
revolving the curve about the y axis.
Solution: The length of the line is given in Problem 7.90. L D
1.479. The elementary length of the curve is
dy
dx
2
dL D
1C
Noting
dy
D 2x, the element of line is dL D 1 C 4x 2 1/2 .
dx
dx.
The x-coordinate:
1
x dL D
x1 C 4x 2 1/2 dx
0
L
D
1
1 53/2 1
1 C 4x 2 3/2 0 D
D 0.8484.
12
12
Divide by the length to obtain x D 0.5736. The surface area is A D
2xL D 5.33
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1
Problem 7.92 A nozzle for a large rocket engine is
designed by revolving the function y D 23 x 13/2 about
the y axis. Use the first Pappus–Guldinus theorem to
determine the surface area of the nozzle.
y
y = _2 (x – 1)3/2
3
x
5 ft
Solution: The length: Noting that
of length is
1C
dL D
dy
dx
2
dx D
dy
D x 11/2 , the element
dx
p
x dx
from which
LD
5
dL D
x1/2 dx D
1
L
2 3/2 5
x
D 6.7869 ft
3
1
The x-coordinate:
5
x dL D
L
x 3/2 dx D
1
2 5/2 5
x
D 21.961.
5
1
Divide by the length: x D 3.2357. The area
A D 2xL D 138 ft2
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1
Problem 7.93 The coordinates of the centroid of the
line are x D 332 mm and y D 118 mm.
Use the first Pappus-Guldinus theorem to determine the
area of the surface of revolution obtained by revolving
the line about the x axis.
y
200 mm
60⬚
x
Solution:
L D 200 mm tan 60° C 200 mm
120°
180°
D 765 mm
A D 2yL D 20.118 m0.765 m D 0.567 m2
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1
Problem 7.94 The coordinates of the centroid of the
area between the x axis and the line in Problem 7.93 are
x D 355 mm and y D 78.4 mm. Use the second PappusGuldinus theorem to determine the volume obtained by
revolving the area about the x axis.
Solution: The area is
AD
1
0.2 m0.2 m tan 60° C
2
120°
0.2 m2 D .0765 m2
360°
V D 2yA D 20.0784 m0.0765 m2 D 0.0377 m3
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1
Problem 7.95 The volume of revolution contains a
hole of radius R.
(a)
(b)
R+a
Use integration to determine its volume.
Use the second Pappus–Guldinus theorem to determine its volume.
R
h
Solution:
(a)
The element of volume is a disk of radius y and thickness dx.
The area of the disk is y 2 R2 . The radius is
yD
a
h
from which dV D Denote m D
a
h
x C R,
2
xCR
dx R2 dx.
a
, dV D m2 x 2 C 2mRx dx,
h
from which
h
dV D m
VD
mx 2 C 2Rx dx
0
V
3
h
x
mh
CR
D m m C Rx 2 D mh2
3
3
0
D ah
(b)
a
3
CR .
The area of the triangle is A D 12 ah. The y-coordinate of the
centroid is y D R C 13 a. The volume is
V D 2yA D ahR C 13 a
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1
Problem 7.96
revolution.
Determine the volume of the volume of
Solution: The area of the semicircle is A D
y DRC
4r
. The volume is
3
V D 2
r 2
2
RC
4r
3
r 2
. The centroid is
2
4r
D 2 r 2 R C
.
3
For r D 40 mm and R D 140 mm, V D 2.48 ð 103 m3
140
mm
80
mm
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1
Problem 7.97 Determine the surface area of the volume
of revolution in Problem 7.96.
Solution: The length and centroid of the semicircle is Lo D r,
yDRC
y D R.
2r
. The length and centroid of the inner line is Li D 2r, and
2r
C 22rR D 2rR C 2r C 2R.
A D 2r R C
For r D 40 mm and R D 140 mm, A D 0.201 m2
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1
Problem 7.98 The volume of revolution has an
elliptical cross section. Determine its volume.
230 mm
130 mm
180 mm
Solution: Use the second theorem of Pappus-Guldinus. The
centroid of the ellipse is 180 mm from the axis of rotation. The area
of the ellipse is ab where a D 115 mm, b D 65 mm.
2b
The centroid moves through a distance jdj D 2R D 2 (180 mm) as
the ellipse is rotated about the axis.
2a
V D Ad D abd D 2.66 ð 107 mm3
v D 0.0266 m3
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1
Problem 7.99 The mass of the homogeneous flat plate
is 80 kg. What are the reactions at A and B?
400
mm
Strategy: The center of mass of the plate is coincident
with the centroid of its area. Determine the horizontal
coordinate of the centroid and assume that the plate’s
weight acts there.
300
mm
B
A
300
mm
300
mm
Solution: The weight is located at the center of mass which was
found in problem 7.33
x D 0.450 m, y D 0.312 m
784.8 N
The equilibrium equations
MA : B1.0 m 784.8 N0.450 m D 0
Fx : Ax D 0
Ax
Fy : Ay C B 784.8 N D 0
Solving:
Ay
B
Ax D 0, Ay D 432 N, B D 353 N
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1
Problem 7.100 The mass of the homogeneous flat
plate is 50 kg. Determine the reactions at the supports A
and B.
100 mm
400 mm
200 mm
A
B
600 mm
800 mm
Solution: Divide the object into three areas and the composite.
Since the distance to the action line of the weight is the only item of
importance, and since there is no horizontal component of the weight,
it is unnecessary to determine any centroid coordinate other than the xcoordinate. The areas and the x-coordinate of the centroid are tabulated.
The last row is the composite area and x-coordinate of the centroid.
500 N
AX
X
B
AY
Area
A, sq mm
x
Rectangle
3.2 ð 105
400
Circle
3.14 ð 104
600
Triangle
1.2 ð 105
1000
Composite
4.09 ð 105
561
600 mm
1400 mm
The composite area is A D Arect Acirc C Atriang . The composite xcoordinate of the centroid is
xD
Arect xrect Acirc xcirc C Atriang xtriang
.
A
The sum of the moments about A:
MA D 500561 C 1400B D 0,
from which B D 200 N. The sum of the forces:
Fy D Ay C B 500 D 0,
from which Ay D 300 N.
Fx D Ax D 0
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1
Problem 7.101 The suspended sign is a homogeneous
flat plate that has a mass of 130 kg. Determine the axial
forces in members AD and CE. (Notice that the y axis
is positive downward.)
A
2m
4m
C
1m
E
B
x
D
y
y = 1 + 0.0625x2
Solution: The strategy is to determine the distance to the action
line of the weight (x-coordinate of the centroid) from which to apply
the equilibrium conditions to the method of sections.
The area: The element of area is the vertical strip of length y and width
dx. The element of area dA D y dx D 1 C ax 2 dx, where a D 0.0625.
Thus
4
ax 3
1 C ax 2 dx D x C
D 5.3333 sq ft.
3 0
0
4
dA D
AD
A
The x-coordinate:
4
x dA D
A
x1 C ax 2 dx D
0
Divide A: x D
4
2
ax 4
x
C
D 12.
2
4 0
12
D 2.25 ft.
5.3333
The equilibrium conditions: The angle of the member CE is
˛ D tan1 14 D 14.04° .
The weight of the sign is W D 1309.81 D 1275.3 N. The sum of the
moments about D is
MD D 2.25W C 4CE sin ˛ D 0,
from which CE D 2957.7 N T .
Method of sections: Make a cut through members AC, AD and BD
and consider the section to the right. The angle of member AD is
ˇ D tan1 12 D 26.57° .
The section as a free body: The sum of the vertical forces:
FY D AD sin ˇ W D 0
from which AD D 2851.7 N T
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1
Problem 7.102 The bar has a mass of 80 kg. What are
the reactions at A and B?
A
2m
2m
B
y
Solution: Break the bar into two parts and find the masses and
centers of masses of the two parts. The length of the bar is
X1
m1g
m2g
AX
L D L1 C L2 D 2 m C 2R/4R D 2 m
X2
L D2C m
AY
Lengthi (m)
Part
1
2
2
m1 D 31.12 kg
x1 D 1 m
m2 D 48.88 kg
x2 D 3.27 m
Fx :
Ax D 0
Fy :
Ay C By m1 g m2 g D 0
MA :
x1 m1 g x2 m2 g C 4By D 0
Massi (kg)
2
80
2C
80
2C
xi (m)
x
4m
BY
1
2C
2R
Solving
Ax D 0, Ay D 316 N, B D 469 N
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1
Problem 7.103 The mass of the bar per unit length is
2 kg/m. Choose the dimension b so that part BC of the
suspended bar is horizontal. What is the dimension b,
and what are the resulting reactions on the bar at A?
A
1m
30⬚
B
b
Solution: We must have
C
Ay
MA : g1.0 m0.5 m cos 30° gb
b
1.0 m cos 30°
2
Ax
D0
) b D 2.14 m
Then
Fx : Ax D 0
ρg(1.0 m)
ρ gb
Fy : Ay g1.0 m gb D 0
)
Ax D 0, Ay D 61.6 N, b D 2.14 m
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1
Problem 7.104 The semicircular part of the homogeneous slender bar lies in the x –z plane. Determine the
center of mass of the bar.
Solution: The bar is divided into three segments plus the composite. The lengths and the centroids are given in the table: The
composite length is:
y
3
LD
Li .
iD1
The composite coordinates are:
10 in
3
16 in
12 in
xD
Li xi
iD1
L
,
z
x
3
and y D iD1
Li yi
L
Segment
Length, in
10
x, in
24
0
L1
12
L2
L3
Composite
y, in
5
0
18.868
8
5
0
66.567
6.594
2.168
6.796
0
z, in
12
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1
y
Problem 7.105 The 10-ft horizontal cylinder with 1ft radius is supported at A and B. Its weight density is
D 1001 0.002x 2 lb/ft3 . What are the reactions at
A and B?
10 ft
A
1 ft
z
x
B
Solution: The weight: Denote a D 0.002. The element of volume
y
is a disk of radius R D 1 ft, thickness dx, and weight
1 ft
x
dW D R2 dx,
A
B
from which
WD
dW D 100R2
L
10 ft
1 ax 2 dx
0
L
ax 3
D 100R2 x D 2932.15 lb
3 0
The x-coordinate of the mass center:
L
x dW D 100R2
1 ax 2 x dx
0
W
D
L
25R2 1 ax 2 2 0 D 14137.17.
a
Divide by W: x D 4.8214 ft.
The equilibrium conditions: The sum of the moments about A is
MA D Wx C 10B D 0,
from which
BD
Wx
2932.154.8214
D
L
10
D 1413.7 lb .
The sum of the vertical forces:
FY D A C B W D 0,
from which A D 1518.4 lb . The horizontal components of the reactions are zero:
FX D 0
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1
Problem 7.106 A horizontal cone with 800-mm length
and 200-mm radius has a built-in support at A. Its mass
density is D 60001 C 0.4x 2 kg/m3 , where x is in
meters. What are the reactions at A?
y
200 mm
A
x
800 mm
Solution: The strategy is to determine the distance to the line of
action of the weight, from which to apply the equilibrium conditions.
The mass: The element of volume is a disk of radius y and thickness
dx. y varies linearly with x: y D 0.25x. Denote a D 0.4. The mass of
the disk is
dm D y 2 dx D 60001 C ax 2 0.25x2 dx
y
200
mm
x
A
800 mm
D 3751 C ax 2 x 2 dx,
from which
0.8
m D 375
1 C ax 2 x 2 dx D 375
0
0.8
3
x5
x
Ca
3
5 0
D 231.95 kg
The x-coordinate of the mass center:
0.8
x dm D 375
1 C ax 2 x 3 dx D 375
0
m
0.8
4
x6
x
Ca
4
6 0
D 141.23.
Divide by the mass: x D 0.6089 m
The equilibrium conditions: The sum of the moments about A:
M D MA mgx D 0,
from which
MA D mgx D 231.949.810.6089
D 1385.4 N-m .
The sum of the vertical forces:
FY D AY mg D 0
from which AY D 2275.4 N . The horizontal component of the reaction is zero,
FX D 0.
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1
Problem 7.107 The circular cylinder is made of
aluminum (Al) with mass density 2700 kg/m3 and iron
(Fe) with mass density 7860 kg/m3 .
(a)
(b)
y
Determine the centroid of the volume of the
cylinder.
Determine the center of mass of the cylinder.
Al
z
Fe
200 mm
600 mm
x
600 mm
Solution:
y
AI
(a)
Fe
The volume of the cylinder is
V D 0.12 1.2 D 0.0377 m3 .
200 mm
x
600 mm
y
z
600 mm
The volume of the parts:
VAl D VFe D
V
D 0.0188 m3 .
2
The centroid of the first part is xAl D 0.3 m, yAl D zAl D 0. The
centroid of the iron part is
xFe D 0.6 C 0.3 D 0.9 m,
yFe D zFe D 0.
The composite centroid
xD
1.2
VAl 0.3 C VFe 0.9
D
D 0.6 m,
V
2
y D z D 0.
(b)
The mass center: The mass of the aluminum part is mAl D
VAl 2700 D 50.89 kg. The mass of the iron part is mFe D
VFe 7860 D 148.16 kg. The composite mass is m D mAl C
mFe D 199.05 kg. The composite center of mass is
xm D
50.890.3 C 148.160.9
D 0.7466 m
199.05
ym D zm D 0
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1
Problem 7.108 The cylindrical tube is made of
aluminum with mass density 2700 kg/m3 . The cylindrical plug is made of steel with mass density
7800 kg/m3 . Determine the coordinates of the center of
mass of the composite object.
y
z
x
y
y
A
20 mm
x
z
35 mm
100
mm
100
mm
A
Section A-A
Solution: The volume of the aluminum tube is
VAl D 0.0352 0.022 0.2 D 5.18 ð 104 m3 .
The mass of the aluminum tube is mAl D 2700VAl D 1.4 kg. The
centroid of the aluminum tube is xAL D 0.1 m, yAl D zAl D 0.
The volume of the steel plug is VFe D 0.022 0.1 D 1.26 ð
104 m3 . The mass of the steel plug is mFe D 7800VFe D
0.9802 kg. The centroid of the steel plug is xFe D 0.15 m, yFe D
zFe D 0.
The composite mass is m D 2.38 kg. The composite centroid is
xD
mAl 0.1 C mFe 0.15
D 0.121 m
m
yDzD0
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1
Problem 7.109 The truncated conical bar is made of
bronze with weight density 0.28 lb/in3 and titanium with
weight density 0.16 lb/in3 . Determine the coordinates of
the center of mass of the bar.
y
z
x
Bronze
y
Titanium
8 in
4 in
12 in
Solution: From the table in appendix C we know that
VD
x
8 in
x
R2 h
3h
, xD
3
4
2R
We have
B D 0.28 lb/in3 , T D 0.16 lb/in3
For each part we will use a full cone and subtract a smaller cone.
h
C T
xW D B
[3.2]2 [32]
3
B
C T
T
xD
[4]2 [40]
3
[2]2 [20]
3
[4]2 [40]
3
B
[2]2 [20]
3
T
[3.2]2 [32]
3
3[32]
20
4
3[20]
20
4
3[40]
20
4
[3.2]2 [32]
3
8 in
[3.2]2 [32]
3
4 in
W D B
6.4 in
From symmetry we know that y D z D 0
20 in
12 in
8 in
3[32]
20
4
xW
D 10.84 in, y D z D 0
W
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1
Problem 7.110 A machine consists of three parts. The
masses and the locations of the centers of mass of two
of the parts are:
Part
1
2
Mass (kg)
2.0
4.5
x (mm)
100
150
y (mm)
50
70
z (mm)
20
0
The mass of part 3 is 2.5 kg. The design engineer wants
to position part 3 so that the center of mass of location
of the machine is x D 120 mm, y D 80 mm, and z D 0.
Determine the necessary position of the center of mass
of part 3.
Solution: The composite mass is m D 2.0 C 4.5 C 2.5 D 9 kg.
The location of the third part is
x3 D
1209 2100 4.5150
D 82 mm
2.5
y3 D
809 250 4.570
D 122 mm
2.5
z3 D
220
D 16 mm
2.5
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1
Problem 7.111 Two views of a machine element
are shown. Part 1 is aluminum alloy with density
2800 kg/m3 , and part 2 is steel with density 7800 kg/m3 .
Determine the coordinates of its center of mass.
y
y
1
24 mm
2
8 mm
18 mm
x
20
mm
60 mm
8 mm
z
16
mm
50 mm
Solution: The volumes of the parts are
V1 D 6048 C 12 242 82 50
D 179, 186 mm3 D 17.92 ð 105 m3 ,
V2 D 1636 C 12 182 82 20 D 17, 678 mm3
D 1.77 ð 105 m3 ,
so their masses are
m1 D S1 V1 D 280017.92 ð 105 D 0.502 kg,
m2 D S2 V2 D 78001.77 ð 105 D 0.138 kg.
The x coordinates of the centers of mass of the parts are x1 D 25 mm,
x2 D 10 mm, so
xD
x1 m1 C x2 m2
D 21.8 mm
m1 C m 2
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1
Problem 7.112 The airplane’s total weight is W D
2400 lb. The weight of the engine and propeller is
525 lb and their combined center of mass is 6.5 ft to
the left of point B. If these items are to be removed
for maintenance, will it be necessary to place a support
under the airplane’s tail to prevent it from falling?
W
A
5 ft
B
2 ft
Solution: The problem asks us to locate the center of mass of the
airplane without the engine and propeller
We have
525 lb
1875 lb
525 lb6.5 ft C 1875 lbd D 2400 lb2 ft
Solving:
d D 0.74 ft ) No support is necessary
d
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1
Problem 7.113 With its engine removed, the mass of
the car is 1100 kg and its center of mass is at C. The
mass of the engine is 220 kg.
E
(a)
(b)
Suppose that you want to place the center of mass
E of the engine so that the center of mass of the
car is midway between the front wheels A and the
rear wheels B. What is the distance b?
If the car is parked on a 15° slope facing up the
slope, what total normal force is exerted by the
road on the rear wheels B?
C
0.6 m
0.45 m
A
B
1.14 m
b
2.60 m
Solution:
(a)
The composite mass is m D mC C mE D 1320 kg. The xcoordinate of the composite center of mass is given:
xD
2.6
D 1.3 m,
2
from which the x-coordinate of the center of mass of the engine is
xE D b D
1.3 m 1.14 mC D 2.1 m.
mE
The y-coordinate of the composite center of mass is
yD
(b)
0.45 mC C 0.6 mE
D 0.475 m.
m
Assume that the engine has been placed in the new position, as
given in Part (a). The sum of the moments about B is
MA D 2.6A C ymg sin15° 2.6 xmg cos15° D 0,
from which A D 5641.7 N. This is the normal force exerted by
the road on A. The normal force exerted on B is obtained from;
FN D A mg cos15° C B D 0,
from which B D 6866 N
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1
Problem 7.114 The airplane is parked with its landing
gear resting on scales. The weights measured at A, B, and
C are 30 kN, 140 kN, and 146 kN, respectively. After a
crate is loaded onto the plane, the weights measured at A,
B, and C are 31 kN, 142 kN, and 147 kN, respectively.
Determine the mass and the x and y coordinates of the
center of mass of the crate.
B
6m
A
x
C
6m
10 m
y
Solution: The weight of the airplane is WA D 30 C 140 C 146 D
316 kN. The center of mass of the airplane:
Myaxis D 3010 xA WA D 0,
from which xA D 0.949 m.
Mxaxis D 140 1466 C yA WA D 0,
from which yA D 0.114 m. The weight of the loaded plane:
W D 31 C 142 C 147 D 320 kN.
The center of mass of the loaded plane:
Myaxis D 3110 xW D 0,
from which x D 0.969 m.
Mxaxis D 142 1476 C yW D 0,
from which y D 0.0938 m. The weight of the crate is Wc D W WA D 4 kN. The center of mass of the crate:
xc D
Wx WA xA
D 2.5 m,
Wc
yc D
Wy WA yA
D 1.5 m.
Wc
The mass of the crate:
mc D
Wc ð 103
D 407.75 kg
9.81
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1
Problem 7.115 A suitcase with a mass of 90 kg is
placed in the trunk of the car described in Example 7.20.
The position of the center of mass of the suitcase is
xs D 0.533 m, ys D 0.762 m, and zs D 0.305 m. If
the suitcase is regarded as part of the car, what is the
new position of the car’s center of mass?
Solution: In Example 7.20, the following results were obtained
for the car without the suitcase
The new center of mass is at
xN D
Wc D 17303 N
xc Wc C xs Ws
Wc C Ws with similar eqns for yN and zN
xc D 1.651 m
Solving, we get
yc D 0.584 m
xN D 1.545 m, yN D 0.593 m, zN D 0.717 m
zc D 0.769 m
For the suitcase
Ws D 90 g,
y D 0.762 m,
xs D 0.533 m,
z D 0.305 m.
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1
Problem 7.116 A group of engineering students
constructs a miniature device of the kind described in
Example 7.20 and uses it to determine the center of mass
of a miniature vehicle. The data they obtain are shown
in the following table:
Wheelbase = 36 in
Track = 30 in
Left front wheel, NLF
Right front wheel, NRF
Left rear wheel, NLR
Right rear wheel, NRR
Measured Loads (lb)
˛D0
˛ D 10°
35
32
36
33
27
34
29
30
Determine the center of mass of the vehicle. Use the
same coordinate system as in Example 7.20.
Solution: The weight of the go-cart: W D 35 C 36 C 27 C 29 D
127 lb. The sum of the moments about the z axis
With the go-cart in the tilted position, the sum of the moments about
the z axis
Mzaxis D WheelbaseNLF C NRF xW D 0,
Mzaxis D WheelbaseNLF C NRF C yW sin10° xW cos10° D 0,
from which
xD
3635 C 36
D 20.125 in.
W
The sum of the moments about the x axis:
Mxaxis D zW TrackNRF C NRR D 0,
from which
yD
xW cos10° 3632 C 33
W sin10° D 8.034 in.
from which
zD
3036 C 29
D 15.354 in.
W
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1
Problem 7.117 Determine the centroid of the area by
letting dA be a vertical strip of width dx.
y
(1, 1)
y = x2
x
Solution: The area: The length of the vertical strip is 1 y, so
that the elemental area is dA D 1 y dx D 1 x 2 dx. The area:
1
x3
1
2
1 x 2 dx D x D1 D .
3 0
3
3
0
1
dA D
A
The x-coordinate:
1
xA D
x dA D
x1 x 2 dx D
0
A
2
1
x
1
x4
3
D : xD
2
4 0
4
8
The y-coordinate: The y-coordinate of the centroid of each element of
area is located at the midpoint of the vertical dimension of the area
element.
y D y C 12 1 x 2 .
Thus
1
y dA D
0
A
D
yD
x2 C
1
1 x 2 1 x 2 dx
2
1
1
2
x5
D .
x
2
5 0
5
3
5
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1
Problem 7.118 Determine the centroid of the area in
Problem 7.117 by letting dA be a horizontal strip of
height dy.
Solution: The area: The length of the horizontal strip is x, hence
Divide by the area: x D
the element of area is
3
8
The y-coordinate:
dA D x dy D y 1/2 dy.
yA D
Thus
y dA D
A
1
AD
0
y 1/2 dy D
1
2y 3/2
3
0
D
2
3
1
D
0
1 y y 1/2 dy
0
y 3/2 dy D
2y 5/2
5
Check:
The x-coordinate: The x-coordinate of the centroid of each element of
area is x D 12 x D 12 y 1/2 . Thus
Divide by the area: y D
1
D
0
2
.
5
3
5
1
2 1
1
y
1
1
1
y 1/2 dA D
y dy D
D .
2
2
2
2
4
A
0
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 7.119
y
Determine the centroid of the area.
60 cm
x
80 cm
Solution: The strategy is to develop useful general results for the
triangle and the rectangle.
Divide by the area: y D
The rectangle: The area of the rectangle of height h and width w is
w
AD
xD
60 cm
a
20 cm. The composite:
3
404800 C 1001800
xR AR C xT AT
D
AR C AT
4800 C 1800
h dx D hw D 4800 cm2 .
0
D 56.36 cm
The x-coordinate:
w
0
2 w x
1
hw2 .
hx dx D h
D
2 0
2
yD
304800 C 201800
4800 C 1800
D 27.27 cm
Divide by the area: x D
w
D 40 cm
2
The y-coordinate:
w
1
1
h2 dx D
h2 w.
2
2
0
Divide by the area: y D
1
h D 30 cm
2
The triangle: The area of the triangle of altitude a and base b is
(assuming that the two sides a and b meet at the origin)
b
yx dx D
AD
0
b
b
ax 2
a
ax
x C a dx D b
2b
0
0
ab
ab
C ab D
D 1800 cm2
D 2
2
Check: This is the familiar result. check.
The x-coordinate:
b
b
ax 3
ax 2
a
ab2
C
.
x C a x dx D D
b
3b
2 0
6
0
Divide by the area: x D
b
D 20 cm
3
The y-coordinate:
y dA D
A
b 2
1
a
x C a dx
2
b
0
D
3 b
b a
ba2
xCa
.
D
6a
b
6
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 7.120
Determine the centroid of the area.
y
40 mm
20 mm
Solution: Divide the object into five areas:
40 mm
(1)
(2)
(3)
(4)
(5)
The rectangle 80 mm by 80 mm,
The rectangle 120 mm by 80 mm,
the semicircle of radius 40 mm,
The circle of 20 mm radius, and
the composite object. The areas and centroids:
80 mm
(1)
A1 D 6400 mm2 ,
x1 D 40 mm, y1 D 40 mm,
(2)
A2 D 9600 mm2 ,
x2 D 120 mm, y2 D 60 mm,
(3)
A3 D 2513.3 mm2 ,
x3 D 120 mm, y3 D 136.98 mm,
(4)
A4 D 1256.6 mm2 ,
x4 D 120 mm, y4 D 120 mm.
(5)
The composite area: A D A1 C A2 C A3 A4 D 17256.6 mm2 . The
composite centroid:
x
xD
A1 x1 C A2 x2 C A3 x3 A4 x4
D 90.3 mm .
A
yD
A1 y1 C A2 y2 C A3 y3 A4 y4
D 59.4 mm
A
120 mm
160 mm
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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 7.121 The cantilever beam is subjected to a
triangular distributed load. What are the reactions at A?
Solution: The load distribution is a straight
line with intercept
w D 200 N/m at x D 0, and slope 200
10
D 20 N/m2 . The sum
200 N/m
of the moments is
x
10
A
20x C 200x dx D 0,
M D MA 10 m
0
from which
10
20
D 3333.3 N-m.
MA D x 3 C 100x 2
3
0
The sum of the forces:
10
200 N/m
AX
MA
AY
10 m
20x C 200 dx D 0,
Fy D Ay 0
from which
10
Ay D 10x2 C 200x 0 D 1000 N,
and
Fx D Ax 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 7.122
of the frame?
What is the axial load in member BD
C 100 N/m
5m
B
D
5m
A
E
10 m
Solution: The distributed load is two straight lines: Over the
interval 0 y 5 the intercept is w D 0 at y D 0 and the slope is
100
D 20.
C
5
Over the interval 5 y 10, the load is a constant w D 100 N/m.
The moment about the origin E due to the load is
5
ME D
10
100y dy,
20yy dy C
0
Cy
Cx
Cx
By
Bx
Ay
Cy
Bx
By
Dy
Dx
Dx Dy
Ey
Ex
5
from which
ME D
20 3 5
100 2 10
y
y
C
D 4583.33 N-m.
3
2
0
5
Check: The area of the triangle is
F1 D 12 5100 D 250 N.
The area of the rectangle: F2 D 500 N. The centroid distance for the
triangle is
d1 D 23 5 D 3.333 m.
The centroid distance of the rectangle is d2 D 7.5 m. The moment
about E is
ME D d1 F1 C d2 F2 D 4583.33 Nm check.
The Complete Structure: The sum of the moments about E is
M D 10AR C ME D 0,
where AR is the reaction at A, from which AR D 458.33 N.
The element ABC : Element BD is a two force member, hence By D 0.
The sum of the moments about C:
MC D 5Bx 10Ay D 0,
where Ay is equal and opposite to the reaction of the support, from
which
Bx D 2Ay D 2AR D 916.67 N.
Since the reaction in element BD is equal and opposite, Bx D
916.67 N, which is a tension in BD.
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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 7.123 An engineer estimates that the maximum wind load on the 40-m tower in Fig. a is described
by the distributed load in Fig. b. The tower is supported
by three cables A, B, and C from the top of the tower to
equally spaced points 15 m from the bottom of the tower
(Fig. c). If the wind blows from the west and cables B
and C are slack, what is the tension in cable A? (Model
the base of the tower as a ball and socket support.)
200 N/m
B
N
A
40 m
15 m
C
400 N/m
(a)
Solution: The load distribution is a straight line with the intercept
w D 400 N/m, and slope 5. The moment about the base of the tower
due to the wind load is
40
(c)
200 N/m
θ
40 m
5y C 400y dy,
TA
40
5
D 213.33 kN-m,
MW D y 3 C 200y 2
3
0
Fx
MW D
(b)
0
clockwise about the base, looking North. The angle formed by the
cable with the horizontal at the top of t