Problem 1.1 Express the fractions
significant digits.
1
3
and
2
3
to three
Solution:
1/3 = 0.3333. . = 0.333
2/3 = 0.6666. . = 0.667
Problem 1.2 What is the value of e (the base of the
natural logarithms) to five significant digits?
Solution:
The value of e is 2.718281828. . . . Record the first five digits, and
round the last digit to the nearest integer. The result is e = 2.7183 to
five significant digits.
Problem 1.3 A machinist drills a circular hole in a
panel with radius r = 5 mm. Determine the circumference C and area A of the hole to four significant digits.
Solution:
C = 2πr = 10π = 31.42 mm
A = πr 2 = 25π = 78.54 mm2
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 = 3.281 ft. The goal width,
1 m
w = 24 ft
= 7.3148 m = 7.31 m.
3.281 ft
The goal height is given by
1 m
h = 8 ft
= 2.438 m = 2.44 m.
3.281 ft
Problem 1.5 The central span of the Golden Gate
Bridge is 1280 m long. What is its length in miles
to three significant digits?
Solution:
(1280 m)
39.37 in
1 m
1 ft
12 in
= 0.7953. . mi = 0.795 mi
1 mi
5280 ft
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 = 1/4 in., 1/2 in., 3/4 in., and 1 in., and the car has nuts
with dimensions n = 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:
1 inch
0.19685 − 0.25
5 mm
= 0.19685.. in,
100
25.4 mm
0.19685
10 mm
15 mm
1 inch
25.4 mm
1 inch
25.4 mm
= 0.3937.. in,
= 0.5905.. in,
1 inch
25.4 mm
0.5905 − 0.5
0.5905
100 = +15.3%
0.7874 − 0.75
100 = +4.7%
0.7874
1 inch
0.9843 − 1.0
25 mm
= 0.9843.. in,
100 = −1.6%
25.4 mm
0.9843
20 mm
= −27.0%
100 = −27.0%
0.3937 − 0.5
0.3937
= 0.7874.. in,
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.
Problem 1.7 The orbital velocity of the International
Space Station is 7690 m/s. Determine its velocity in
km/hr and in mi/hr to three significant digits.
Solution:
m
1m
3600 s
m
= 27,684 = 27,700
s
1000 m
1 hr
hr
39.37 in 1 ft 1 mi 3600 s 7690 m
s
1 m
12 in
5280 ft
1 hr
7690
= 17,202. . = 17,200 mi/hr
Problem 1.8 High-speed “bullet trains” began running
between Tokyo and Osaka, Japan, in 1964. If a bullet
train travels at 240 km/hr, what is its velocity in mi/hr to
three significant digits?
Solution:
Convert the units using Table 1.2. The results are:
m 1 mile 1 ft
1000 m
240
hr
5280 ft
0.3048 m
1m
mile
mile
= 149.12908 . . .
= 149
hr
hr
n
Problem 1.9 In December 1986, Dick Rutan and
Jeana Yeager flew the Voyager aircraft around the world
nonstop. They flew a distance of 40,212 km in 9 days,
3 minutes, and 44 seconds.
(a) Determine the distance they flew in miles to three
significant digits.
(b) Determine their average speed (the distance flown
divided by the time required) in kilometers per hour,
miles per hour, and knots (nautical miles per hour)
to three significant digits.
Solution:
Convert the units using Table 1.2.
1000 m
1 ft
1 mile
(a) 40,212 m
1m
0.3048 m
5280 ft
= 24,987 mi = 25,000 mi
(b)
The time of flight is
3
44
9 days 3 min 44 sec = (9)(24)+
+
hours
60
3600
= 216.062 hours.
The average speed is
40,212 m
m
= 186.11 . Converting,
216.062 hours
hr
m
1 mile
mi
mi
186.11
= 115.7
= 116
, or
hr
1.609 m
hr
hr
m 1 nautical mile
186.11
= 100.49 knots
hr
1.852 m
= 100 knots to three significant digits.
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:
6 mm
1 m
10 µs
m
(a) 5
= 5000
.
µs
1000 mm
1s
s
Next, use this result to get (b):
m
1 ft
1 mi
(b) 5000
s
0.3048 m
5280 ft
= 3.10685 . . .
= 3.11
mi
s
mi
s
Problem 1.11 The kinetic energy of a particle of mass
m is defined to be 12 mv 2 , 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
= 147.46 = 147
1 ft
0.3048 m
2
slug-ft2
s2
Problem 1.12 The acceleration due to gravity at sea
level in SI units is g = 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:
m
1 ft
ft
ft
g = 9.81
=
32.185
.
.
.
=
32.2
s2
0.3048 m
s2
s2
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:
ft
1 furlong
3600 s
24 hr
14 day
5
s
660 ft
1 hr
1 day
1 fortnight
furlongs
= 9160
fortnight
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:
1 ft 2 0.3048 m 2
480 in2
= 0.30967 . . . m2 = 0.310 m2
12 in
1 ft
Problem 1.15 At sea level, the weight density (weight
per unit volume) of water is approximately 62.4 lb/ft3 .
1 lb = 4.448 N, 1 ft = 0.3048 m, and g = 9.81 m/s2 .
Using only this information, determine the mass density
for water in kg/m3 .
Solution:
Get wt. density in N/m3 first.
3
lb
4.448 N
1 ft
N
62.4 3
= 9801.77
1 lb
0.3048 m
m3
ft
(carry extra significant figures till end—then round)
weight = mass · g
mass =
weight
g
9801.77
N
m3
s2
9.81 m
= 999
N-s2
m
1
m3
= 999 kg/m3
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:
2
lb
4.448 N
12 in 2
1 ft
300
2
1 lb
1 ft
0.3048 m
in
N
= 2.0683 . . . (106 )
= 2.07(106 ) Pa
m2
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
12-horsepower 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
746 watt
(a) 12 hp
= 8950 watt
1 hp
746 watt
(b) 105 hp
= 7.46(107 ) watt
1 hp
Problem 1.18 In SI units, the universal gravitational
constant G = 6.67 × 10−11 N-m2 /kg2 . Determine the
value of G in U.S. Customary units.
Solution:
Convert units using Table 1.2. The result:
2 N-m2
1 lb
1 ft
14.59 kg 2
6.67(10−11 )
2
4.448 N
0.3048 m
1 slug
kg
2
2
lb-ft
lb-ft
= 3.43590 . . . (10−8 )
= 3.44(10−8 )
slug2
slug2
Problem 1.19 If the earth is modeled as a homogenous
sphere, the velocity of a satellite in a circular orbit is
2
gRE
,
v=
r
where RE is the radius of the earth and r is the radius of
the orbit.
(a) If g is in m/s2 and RE and r are in meters, what are
the units of v?
(b) If RE = 6370 km and r = 6670 km, what is the
value of v to three significant digits?
(c) For the orbit described in Part (b), what is the value
of v in mi/s to three significant digits?
Solution:
For (a), substitute the units into the expression and reduce:
2
g m
2 gRE
m3
m
s2 (RE m)
(a)
=
=v
2
(rm)
r
ms
s
Hence, the units are m/s
For (b), substitute
the numerical values into the expression, using
g = 9.81 sm
2 .
v
=
=
9.81
m
m 2
(6370 m) 103 m
s2
m
(6670 m) 103 m
m
m
59.679 . . . (106 )
= 7.7252 . . . (103 )
s
s
(b) v = 7730
m
s
For (c), convert units using Table 1.2. The result:
(c) v
m
1 ft
1 mile
s
0.3048 m
5280 ft
mile
mile
= 4.803 . . .
= 4.80
s
s
= 7730
Problem 1.20
T =
In the equation
1 2
Iω ,
2
the term I is in kg-m2 and ω is in s−1 .
(a) What are the SI units of T ?
(b) If the value of T is 100 when I is in kg-m2 and ω is
in s−1 ,
what is the value of T when it is expressed in
U.S. Customary base units?
Solution:
For (a), substitute the units into the expression for T :
1
kg-m2
(a) T =
(I kg-m2 )(ωs−1 )2 =
2
s2
For (b), convert units using Table 1.2. The result:
2
1 slug
1 ft
14.59 kg
0.3048 m
2
slug-ft
slug-ft2
= 73.7759 . . .
= 73.8
2
2
s
s
(b) 100
kg-m2
s2
Problem 1.21 The aerodynamic drag force D exerted
on a moving object by a gas is given by the expression
1
D = CD S ρv 2 ,
2
where the drag coefficient CD is dimensionless, S is
reference area, ρ is the mass per unit volume of the gas,
and v is the velocity of the object relative to the gas.
(a) Suppose that the value of D is 800 when S, ρ, and
v are expressed in SI base units. By converting
units, determine the value of D when S, ρ, and v
are expressed in U.S. Customary units.
(b) The drag force D is in newtons when the expression
is evaluated in SI base units and is in pounds when
the expression is evaluated in U.S. Customary base
units. Using your result from part (a), determine the
conversion factor from newtons to pounds.
Solution:
For (a), we just carry out the conversion unit by unit. We get:
2
kg
m
kg m
(a) 800( m2 )
=
800
m3
s
s2
0.0685 slug
3.281 ft
1
= 800
1 kg
1 m
s2
slug ft
= 180
s2
(b) From (a), 800 N = 180 lb. Hence, 1 N = 0.225 lb.
Problem 1.22 The pressure p at a depth h below the
surface of a stationary liquid is given by
p = ps + γh,
where ps is pressure at the surface and γ is a constant.
(a) If p is in newtons per meter squared and h is in
meters, what are the units of γ?
(b) For a particular liquid, the value of γ is 9810 when
p is in newtons per meter squared and h is in meters.
What is the value for γ when p is in pounds per foot
squared and h is in feet?
Solution:
The units of γh are the same as the units of P . Thus, in units
N
= γ( m)
m2
units of γ ∼ N/m3
N to lb
We must convert 9810 m
3
ft3
N
1 lb
0.3048 m 3
9810
= 62.4 lb/ft3
3
m
4.448 N
1 ft
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.
m=
w
180 N-s2
=
= 111.1 kg
g
1.62 m
This is the total allowable mass. Thus, the suit & equipment can have
mass of
mS/E = 111.1 kg − 57 kg = 54.1 kg
Problem 1.24 A person has a mass of 50 kg.
(a) The acceleration due to gravity at sea level is g =
9.81 m/s2 . What is the person’s weight at sea level?
(b) The acceleration due to gravity on the moon is g =
1.62 m/s2 . What would the person weigh on the
moon?
Solution:
Use Eq (1.6).
m
(a) We = 50 kg 9.81 2 = 490.5 N = 491 N, and
s
m
(b) Wmoon = 50 kg 1.62 2 = 81 N.
s
Problem 1.25 The acceleration due to gravity at sea
level is g = 9.81 m/s2 . The radius of the earth is
6370 km. The universal gravitation constant is G =
6.67 × 10−11 N-m2 /kg2 . Use this information to determine the mass of the earth.
Solution:
Use Eq (1.3) a =
mE =
GmE
R2
. Solve for the mass,
m 2
(9.81 m/s2 )(6370 m)2 103 m
gR2
=
2
G
6.67(10−11 ) N-m2
kg
= 5.9679 . . . (1024 ) kg = 5.97(1024 ) kg
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).
2
2
RE 2
WE
RE
3960
W = mg
=
g
= WE
r
g
RE + H
3960 + 200
= (180)(0.90616) = 163 lb
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 = 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.
Solution:
Use Eq (1.4), rewritten to apply to the Moon. . . a = gM
a = (1.62 m/s2 )
RM
RM +RM
2
= (1.62 m/s2 )
RM
r
2
2
1
= 0.405 m/s2
2
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 = 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 decrease to 0.99 mg?
Solution:
Use a variation of Eq (1.5).
2
RE
W = mg
= 0.99 mg
RE + h
Solve for the radial height,
1
h = RE √
− 1 = (6370)(1.0050378 − 1.0)
0.99
= 32.09 . . . m = 32,100 m = 32.1 m
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 = 6.67 × 10−11 N-m2 /kg2 .)
Solution:
Use Eq (1.1) F =
F =
Gm1 m2
.
r2
Substitute:
(6.67)(10−11 )(0.2)(0.2)
= 2.668(10−12 ) N
12
Problem 1.30 At a point between the earth and the
moon, the magnitude of the earth’s gravitational acceleration equals the magnitude of the moon’s gravitational
acceleration. 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 of 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 + rMp = rEM = 383,000 m. The
fact that the gravitational attractions by the Earth and the Moon at this
point are equal leads to the equation
RE 2
RM 2
gE
= gM
,
rEp
rMp
where rEM = 383,000 m. Substituting the correct numerical values
leads to the equation
2
m
6370 m 2
m
1738 m
9.81
=
1.62
,
s2
rEp
s2
rEM − rEp
where rEp is the only unknown. Solving, we get rEp = 344,770 m =
345,000 m.
Problem 2.1 The magnitudes |FA | = 60 N and
|FB | = 80 N. The angle α = 45◦ . Graphically determine the magnitude of the sum of the forces F =
FA + FB and the angle between FB and F.
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 α, as we did in Example 2.1. Then you
can measure the magnitude of their sum and the angle
between their sum and FB .
Solution: The graphical construction is shown:
The angle β is graphically determined to be about 26◦ and the angle
θ is about θ = 19◦ . The magnitude of the sum |F| = |FA + FB | is
about |F| = 130 N. (These values check with a determination using
trigonometry, β = 25.9◦ ,
FA
FB
F
FA
45˚
θ = 19.1◦ , and |F| = 129.6 N .)
FB
Problem 2.2 The magnitudes |FA | = 40 N and
|FA + FB | = 80 N. The angle α = 60◦ . Graphically
determine the magnitude of FB .
Solution: Measuring, FB ∼
= 52 N
FA
FB
FA + FB
FB
FA
60°
0
40
50
80
100
N
Problem 2.3 The magnitudes |FA | = 100 lb and
|FB | = 140 lb. The angle α = 40◦ . Use trigonometry to determine the magnitude of the sum of the forces
F = FA + FB and the angle between FB and F.
Strategy: Use the laws of sines and cosines to analyze
the triangles formed by the parallelogram rule for the
sum of the forces as we did in Example 2.1. The laws of
sines and cosines are given in Section A.2 of Appendix A.
Solution: The construction is shown. Use the cosine law to determine the magnitude |F|.
|F|2 = |FA |2 + |FB |2 − 2|FA ||FB | cos 140◦
2
2
2
|F| = (100) + (140) − 2 (100) (140) cos 140
or
|F| =
F
FA
140
◦
FB
5.1049 . . . (104 ) = 225.94 . . . = 225.9 lb
Use the law of sines to determine the angle θ., i.e.,
|FB |
|F|
|F|
sin 140◦
=
40°
|FB |
.
sin θ
From which we get sin θ =
sin 140◦ = 0.398, and θ =
23.47◦ . The angle between FB and F is
Thus β = 40 − 23.47 = 16.53◦
Problem 2.4 The magnitudes |FA | = 40 N and |FA +
FB | = 80 N. The angle α = 60◦ . Use trigonometry to
determine the magnitude of FB .
Solution: Draw the force triangle.
From the law of sines
FA
|FA + FB |
|FA |
=
sin 120◦
sin θ
FB
|FA |
sin 120◦
|FA + FB |
40
sin θ =
(0.866)
80
sin θ =
θ = 25.66◦
θ + γ + 120
◦
= 180◦ ⇒ γ = 34.34◦
From the law of sines,
|FB |
|FA + FB |
=
sin γ
sin 120◦
sin γ
|FB | =
(80 N)
sin 120◦
FB
FA
FA
120
|FB | = 52.1 N
FB
°
60°
Problem 2.5 The magnitudes |FA | = 100 lb and
|FB | = 140 lb. If α can have any value, what are the
minimum and maximum possible values of the magnitude of the sum of the forces F = FA + FB , and what
are the corresponding values of α?
Solution: A graphical construction shows that the magnitude is
a minimum when the two force vectors are opposed, and a maximum
when both act in the same direction. The corresponding values are
|F|max = |FA + FB | = |100 + 140| = 240 lb, and α = 0◦ .
|F|min = |FA + FB | = |100 − 140| = 40 lb, and α = 180◦ .
Problem 2.6 The angle θ = 30◦ . What is the magnitude of the vector rAC ?
150 mm
60 mm
B
rAB
rBC
C
A
rAC
Solution:
B
150
mm
60
30°
A
mm
rAC
C
From the law of sines
BC
AB
AC
=
=
sin 30◦
sin α
sin γ
We know BC and AB. Thus
150
60
=
⇒ α = 11.54◦
sin 30◦
sin α
Also 30◦ + α + γ = 180◦ ⇒ γ = 138.46◦
Now, from the law of sines
150
AC
=
sin 30◦
sin 138.46◦
AC = |rAC | = 199 mm
Problem 2.7 The vectors FA and FB represent the
forces exerted on the pulley by the belt. Their magnitudes are |FA | = 80 N and |FB | = 60 N. What is the
magnitude |FA + FB | of the total force the belt exerts
on the pulley?
FB
45°
FA
10°
Solution:
FB
FB
+F
FA
45°
FA
35
10°
10°
FA
B
5°
14
45°
35°
Law of cosines
|FA + FB | = (80)2 + (60)2 − 2(80)(60) cos 145◦
|FA + FB | = 133.66 ≈ 134 N
Law of sines
|FB |
|FA + FB |
60
133.66
=
⇒
=
β = 14.92◦
sin β
sin 145
sin β
sin 145
∴ |FA + FB | = 134 N
Problem 2.8 The magnitude of the vertical force F is
80 N. If you resolve it into components FAB and FAC
that are parallel to the bars AB and AC, what are the
magnitudes of the components?
B
30°
C
20°
A
F
Solution: F is made up of components in the two known directions. Since we also know the direction of F, we can draw a force
triangle (FAB + FAC = F).
Thus, we have a triangle DEF as shown
γ + 110◦ + 30◦ = 180◦
γ = 40
FAC
20°
FAB
30° 80 N
◦
From the law of sines,
80
|FAC |
|FAB |
=
=
sin 40◦
sin 30◦
sin 110◦
|FAB | = 117 N |FAC | = 62.2 N
F
FAC
D
110°
60°
80 N
FAB
30°
E
Problem 2.9 The rocket engine exerts an upward force
of 4 MN (meganewtons) magnitude on the test stand. If
you resolve the force into vector components parallel to
the bars AB and CD, what are the magnitudes of the
components?
Solution:
The vector diagram construction is shown. From the
law of sines,
|FAB |
|F|
=
,
sin 45◦
sin 75◦
from which
sin 45◦
= (4)(0.732) = 2.928 . . . = 2.93 MN
◦
sin 75
sin 60
|FBC | = |F|
= (4)(0.8966) = 3.586 . . . = 3.59 MN
sin 75
|FAB | = |F|
B
30°
A
FBA
A
D
38˚
45˚
45˚
45°
C
B
FBC
68˚
45˚
45˚ F
D
C
B
A
30°
D
45°
C
Problem 2.10 If F is resolved into components parallel to the bars AB and BC, the magnitude of the component parallel to bar AB is 4 kN. What is the magnitude
of F?
F
B
150 mm
Solution:
A
F
400 mm
B
150 mm
A
C
100 mm
400 mm
Call the force in AB FAB and the force in BC ∼ FBC . Then
FAB + FBC = F. We know the geometry and that |FAB | = 4 kN.
Draw a diagram of the geometry, get the angles, then draw the force
triangle.
100 mm
B
150 mm
φ
θ
A
C
500 mm
tan θ =
150
400
θ = 20.6◦
tan φ =
150
100
φ = 56.3◦
θ
θ
FAB
γ
δ
FBC
FAB = 4 kN
F
α
φ
γ + θ = 90◦
γ = 69.4◦
α + φ = 90◦
α = 33.7◦
δ + α + γ = 180◦
δ = 76.9◦
From the law of sines
|F|
|FAB |
=
sin δ
sin α
sin(76.9◦ )
|F| =
(4) = 7.02 kN
sin(33.7◦ )
C
100 mm
Problem 2.11 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 + L + 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
|W| = 3500 N
|L| = 3500 cos 25◦
|D| = 3500 sin 25◦
L
|L| = 3170 N
|D| = 1480 N
25°
D
L
25°
W
W
65°
cos 25◦ =
sin 25
◦
|L|
|W|
25°
D
|D|
=
|W|
Problem 2.12 The suspended weight exerts a downward 2000-lb force F at A. If you resolve F into vector
components parallel to the wires AB, AC, and AD, the
magnitude of the component parallel to AC is 600 lb.
What are the magnitudes of the components parallel to
AB and AD?
B
60°
70°
C
45°
D
A
Solution: We resolve the force exerted by the weight into components parallel to the wires:
45°
Setting |W| = 2000 lb and |FAC | = 600 lb and solving, we obtain FAB =
1202 lb, FAD = 559 lb.
70°
W
60°
FAB
|FAD | sin 45◦ + |FAC | sin 70◦ + |FAB | sin 60◦ = |W|,
|FAD | cos 45◦ + |FAC | cos 70◦ − |FAB | cos 60◦ = 0.
FAD
FAC
We see that
Problem 2.13 The wires in Problem 2.12 will safely
support the weight if the magnitude of the vector component of F parallel to each wire does not exceed 2000 lb.
Based on this criterion, how large can the magnitude of
F be? What are the corresponding magnitudes of the
vector components of F parallel to the three wires?
From Problem 2.12 above, we have FAC = 600 lb., FAB =
1202 lb. The largest force is FAB = 1202 lb. We want this value to be 2000 lb
and to have all other values scaled accordingly. Hence, we multiply all forces
by k = 2000
= 1.664. Multiplying all of the forces in Problem 2.12 by this
1202
factor, we get
Solution:
FAB = 2000 lb, FAC = 999 lb,
FAD = 931 lb, and F = 3329 lb.
Problem 2.14 Two vectors rA and rB have magnitudes |rA | = 30 m and |rB | = 40 m. Determine the
magnitude of their sum rA + rB
(a) if rA and rB have the same direction.
(b) if rA and rB are perpendicular.
Solution: The vector constructions are shown.
(a) The magnitude of the sum is the sum of the magnitudes for vectors in the
same direction:
|rA + rB | = 30 + 40 = 70 m
From the cosine law (which reduces to the Pythagorean Theorem for a right
triangle)
|rA + rB |2 = |r2A + |r2B = (30)2 + (40)2 = 2500 |rA + rB | = 50 m .
rB
rA
(a)
rA + r B
rB
(b)
rA
Problem 2.15
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
|W| = 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:
|W|
|FA |
|FB |
=
=
sin 140◦
sin 20◦
sin 20◦
Solving:
|FA | = |FB | = |W|
= 319.25 . . . = 319.3 lb
FB
FA
40°
sin 20
sin 140
20°
FB
20°
20
FA
40°
20
FB
20°
20°
140
W
20
20
FA
W
W
Problem 2.16 The rope ABC exerts the forces FBA
and FBC on the block at B. Their magnitudes are
|FBA | = |FBC | = 800 N. Determine |FBA + FBC |
(a) graphically and (b) by using trigonometry.
FBC
C
B
20°
B
FBA
A
Solution:
(a)
(b)
The vector graphical construction is shown.
The angles are derived from the rule that for equal legs in a triangle the opposite interior angles are equal, and the rule that the
sum of the interior angles is 180 deg. Thus from the problem
statement the 70◦ angle is determined; from the equality of angles and the sum of interior angles the other two 55◦ angles are
derived. The magnitude of the sum of the two vectors is then
measured from the graph.
The trigonometric solution follows from the law of sines:
|FAB + FBC |
|FAB |
|FBC |
=
=
.
sin 70◦
sin 55◦
sin 55◦
Solve:
|FAB + FBC |
= |FAB |
sin 70◦
sin 55◦
= 800(1.1471 . . .)
= 917.72 . . . |FA + FB | = 917.7 N
FBC
20°
B
20°
55°
FBA + FBC
55°
70°
20°
FBA
Problem 2.17 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 |FA | = 1000 lb. Determine |FB | and
|FA +FB | (a) graphically and (b) by using trigonometry.
L
50°
FA
30°
FB
TOP VIEW
Solution: The graphical construction is shown. The sum of
the interior angles must be 180◦ . (a) The magnitudes of |FB | and
|FA + FB | are determined from measurements. (b) The trigonometric solution is obtained from the law of sines:
L
50°
|FA + FB |
|FA |
|FB |
=
=
sin 100
sin 30
sin 50
sin 50
from which |FB | = |FA |
= 1000(1.532) = 1532 lb
sin 30
sin 100
|FA + FB | = |FA |
= 1000(1.9696) = 1970 lb
sin 30
30°
FA
TOP VIEW
FB
FA + FB
38°
38°
FB
156°
38°
50°
50°
FA
Problem 2.18 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
α (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 |rBC | is obtained by the cosine law:
2
2
2
|rBC | = (400) + (600) − 2(400)(600) cos 40
or
B
F
◦
C
|rBC | = 390.25 = 390.3 m
The three right triangles are shown. The distance BD is BD =
(400) sin 60◦ = 346.41 m. The distance CE is CE =
600 sin 20◦ = 205.2 m. The distance F C is F C = (346.4 −
205.2) = 141.2 m.
The angle α is sin α =
141.2
390.3
= 0.36177 . . ., or α = 21.2◦
60°
20°
A
D
E
FB
Problem 2.19 The vector r extends from point A to
the midpoint between points B and C. Prove that
r=
C
1
(rAB + rAC ).
2
rAC
r
rAB
B
A
Solution:
The proof is straightforward:
C
r = rAB + rBM , and r = rAC + rCM .
Add the two equations and note that rBM + rCM = 0, since the two
vectors are equal and opposite in direction.
1
Thus 2r = rAC + rAB , or r =
(rAC + rAB )
2
rAC
r
M
A
rAB
Problem 2.20
explain why
By drawing sketches of the vectors,
U + (V + W) = (U + V) + 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 + W, and then the addition
of U. The result is the vector U + (V + W).
(b) shows the addition of U + V, and then the addition of W, leading
to the result (U + V) + W.
The final vector in the two sketches is the same vector, illustrating that
associativity of vector addition is intuitively reasonable.
V
U
V+W
(a)
W
U+[V+W]
V
U
U+V
(b)
W
[U+V]+W
Problem 2.21 A force F = 40 i − 20 j (N). What is Solution:
its magnitude |F|?
Strategy: The magnitude of a vector in terms of its
components is given by Eq. (2.8).
|F| =
√
402 + 202 = 44.7 N
Problem 2.22 An engineer estimating the components Solution:
of a force F = Fx i + Fy j acting on a bridge abutment
|F| = |Fx |2 + |Fy |2
has determined that Fx = 130 MN, |F| = 165 MN, and
Fy is negative. What is Fy ?
Thus |Fy | = |F|2 − |Fx |2 (mN)
|Fy | =
1652 − 1302 (mN)
|Fy | = 101.6 mN
Fy = −102 mN
Problem 2.23 A support is subjected to a force F =
Fx i + 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 ≥ (Fx )2 + (80)2 , from which (100)2 − (80)2 ≥ (Fx )2 .
√
Thus |Fx | ≤ 3600, or −60 ≤ (Fx ) ≤ +60 (N)
Problem 2.24 If FA = 600i − 800j (kip) and FB = Solution: Take the scalar multiple of FB , add the components of the two
200i − 200j (kip), what is the magnitude of the force forces as in Eq. (2.9), and use the definition of the magnitude. F = (600 −
2(200))i + (−800 − 2(−200))j = 200i − 400j
F = FA − 2FB ?
|F| =
(200)2 + (−400)2 = 447.2 kip
Problem 2.25 If FA = i − 4.5j (kN) and FB = Solution: Take the scalar multiples and add the components.
−2i − 2j (kN), what is the magnitude of the force F = F = (6 + 4(−2))i + (6(−4.5) + 4(−2))j = −2i − 35j, and
6FA + 4FB ?
|F| =
(−2)2 + (−35)2 = 35.1 kN
Problem 2.26 Two perpendicular vectors U and V lie
in the x-y plane. The vector U = 6i − 8j and |V| = 20.
What are the components of V?
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 ±90◦ 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 = 6i
62 + (−8)2
−
8j
62 + (−8)2
y
V2
6
V1
Expressed in terms of an angle,
U
◦
◦
eU = i cos α − j sin α = i cos(53.1 ) − j sin(53.1 )
Add ±90◦ to find the two unit vectors that are perpendicular to this
unit vector:
ep1 = i cos(143.1◦ ) − j sin(143.1◦ ) = −0.8i − 0.6j
ep2 = i cos(−36.9◦ ) − j sin(−36.9◦ ) = 0.8i + 0.6j
Take the scalar multiple of these unit vectors to find the two vectors
perpendicular to U.
V1 = |V|(−0.8i − 0.6j) = −16i − 12j.
The components are Vx = −16, Vy = −12
V2 = |V|(0.8i + 0.6j) = 16i + 12j.
The components are Vx = 16, Vy = 12
x
= 0.6i − 0.8j
8
Problem 2.27 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 = |F| cos 60◦ = (40)(0.5) = 20 (N)
Fy = −|F | sin 60◦ = −(40)(0.866) = −34.6 (N)
y
F = 20i − 34.6j (N)
y
60°
F
60°
F
x
x
Problem 2.28 A person exerts a 60-lb force F to push Solution: The strategy is to express the force F in terms of the angle. Thus
a crate onto a truck. Express F in terms of components. F = (i|F| cos(20◦ ) + j|F| sin(20◦ ))
F = (60)(0.9397i + 0.342j) or F = 56.4i + 20.5j (lb)
y
y
F
20°
F
20°
x
x
Problem 2.29 The missile’s engine exerts a 260-kN
force F. Express F in terms of components using the
coordinate system shown.
y
F
40°
Solution:
Fx = |F| cos 40◦
Fx = 199 N
x
Fy = |F| sin 40◦
Fy = 167 N
F = 199i + 167j (N)
y
F
40°
x
Problem 2.30 The coordinates of two points A and B
of a truss are shown. Express the position vector from
point A to point B in terms of components.
y
A (6, 4) m
B
(2, 1) m
Solution: The strategy is find the distance along each axis by
taking the difference between the coordinates.
x
A(6, 4)
rAB = (2 − 6)i + (1 − 4)j = −4i − 3j (m)
B(2, 1)
Problem 2.31 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 .
y
E
D
2m
F
C
A
x
B
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
y
rAC = rAB + rBC = (xB − xA )i + (yB − yA )j
+ (xC − xB )i + (yC − yB )j
E
D
2m
rAC = (2m − 0)i + (0 − 0)j + (2m cos 60◦ − 0)i
or
+ (2m cos 60◦ − 0)j,
F
giving
◦
◦
rAC = (2m + 2m cos 60 )i + (2m sin 60 )j. For rAF , we have
rAF
C
rAC
rAF = (xF − xA )i + (yF − yA )j
= (−2m cos 60◦ xF − 0)i + (2m sin 60◦ − 0)j.
A
B
x
Problem 2.32 For the hexagonal structural element in
Problem 2.31, determine the components of the vector
rAB − rBC .
rAB − rBC .
The angle between BC and the x-axis is 60◦ .
Solution:
y
E
D
F
2m
C
A
B
x
rBC = 2 cos(60◦ )i + 2(sin(60◦ )j (m)
rBC = 1i + 1.73j (m)
rAB − rBC = 2i − 1i − 1.73j (m)
rAB − rBC = 1i − 1.73j (m)
Problem 2.33 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
Let the x-component of point B be xB . The vector
from A to B can be written as
Solution:
rAB = (xB − xA )i + (yB − yA )j (m)
or rAB = (xB − 1.8)i + (0.6 − 3.0)j (m)
rAB = (xB − 1.8)i − 2.4j (m)
We also know |rAB | = 3.0 m. Thus
32 = (xB − 1.80)2 + (−2.4)2
Solving, xB = 3.60. Thus
rAB = 1.80i − 2.40j (m)
Problem 2.34 (a) Express the position vector from
point A of the front-end loader to point B in terms of
components.
(b) Express the position vector from point B to point C
in terms of components.
(c) Use the results of (a) and (b) to determine the distance
from point A to point C.
y
98 in.
45 in.
C
B
A
55 in.
50 in.
35 in.
x
50 in.
The coordinates are A(50, 35); B(98, 50); C(45, 55).
The vector from point A to B:
Solution:
(a)
y
rAB = (98 − 50)i + (50 − 35)j = 48i + 15j (in.)
(b)
The vector from point B to C is
rBC = (45 − 98)i + (55 − 50)j = −53i + 5j (in.).
(c)
98 in.
45 in.
The distance from A to C is the magnitude of the sum of the
vectors,
C
55 in.
B
A
50 in.
35 in.
x
rAC = rAB + rBC = (48 − 53)i + (15 + 5)j = −5i + 20j.
The distance from A to C is
|rAC | = (−5)2 + (20)2 = 20.62 in.
50 in.
Problem 2.35 Consider the front-end loader in Problem 2.34. To raise the bucket, the operator increases the
length of the hydraulic cylinder AB. The distance between points B and C remains constant. If the length
of the cylinder AB is 65 in., what is the position vector
from point A to point B?
Solution: Assume that the two points A and C are fixed. The
strategy is to determine the unknown angle θ from the geometry.
From
Problem 2.34 |rAC | = 20.6 and the angle β is tan β = −20
=
5
√
◦
2
2
−4, β = 76 . Similarly, |rCB | = 53 + 5 = 53.2. The angle a
is found from the cosine law:
cos α =
(20.6)2 + (65)2 − (53.2)2
= 0.6776,
2(20.6)(65)
α = 47.3◦ . Thus the angle θ is
θ = 180◦ − 47.34◦ − 75.96◦ = 56.69 . . . = 56.7◦ . The vector
rAB = 65(i cos θ + j sin θ) = 35.69 . . . i + 54.32 . . . j
= 35.7i + 54.3j (in.)
B
53.2
C
20.6
α
β
A
65
θ
Problem 2.36 Determine the position vector rAB in
terms of its components if: (a) θ = 30◦ , (b) θ = 225◦ .
y
150 mm
60 mm
B
rAB
rBC
C
x
A
Solution:
(a) rAB = (60) cos(30◦ )i + (60) sin(30◦ )j, or
rAB = 51.96i + 30j mm. And
(b) rAB = (60) cos(225◦ )i + (60) sin(225◦ )j or
y
150
mm
60
mm
B
rAB = −42.4i − 42.4j mm.
FAB
A
Problem 2.37 In problem 2.36 determine the position
vector rBC in terms of its components if: (a) θ = 30◦ ,
(b) θ = 225◦ .
Solution:
(a)
From Problem 2.36, rAB = 51.96i + 30j mm. Thus, the
coordinates of point B are (51.96, 30) mm. The vector rBC is
given by rBC = (xC − xB )i + (yC − yB )j, whereyC = 0.
The magnitude of the vector rBC is 150 mm. Using these facts,
we find that yBC = −30 mm, and xBC = 146.97 mm.
(b) rAB = (60) cos(225◦ )i + (60) sin(225◦ )j or
rAB = −42.4i − 42.4j mm.
From Problem 2.36, rAB = −42.4i − 42.4j mm. Thus, the
coordinates of point B are (−42.4, −42.4) mm. The vector rBC
is given by rBC = (xC − xB )i + (yC − yB )j, where yC = 0.
The magnitude of the vector rBC is 150 mm. Using these facts,
we find that yBC = 42.4 mm, and xBC = 143.9 mm.
β
FBC
C
F
x
Problem 2.38 A surveyor measures the location of
point A and determines that rOA = 400i + 800j (m).
He wants to determine the location of a point B so that
|rAB | = 400 m and |rOA + rAB | = 1200 m. What are
the cartesian coordinates of point B?
y
B
A
N
rAB
rOA
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 α, β, and θ. The magnitude of OA is
|rOA | = (400)2 + (800)2 = 894.4.
Proposed
roadway
x
O
The angle β is determined by
tan β =
800
= 2, β = 63.4◦ .
400
B
The angle α is determined from the cosine law:
cos α =
(894.4)2
(1200)2
+
−
2(894.4)(1200)
(400)2
A
B
y
= 0.9689.
α
α = 14.3◦ . The angle θ is θ = β ± α = 49.12◦ , 77.74◦ .
The two possible sets of coordinates of point B are
α
θ
β
rOB = 1200(i cos 77.7 + j sin 77.7) = 254.67i + 1172.66j (m)
rOB = 1200(i cos 49.1 + j sin 49.1) = 785.33i + 907.34j (m)
x
0
The two possibilities lead to B(254.7 m, 1172.7 m) or B(785.3 m,
907.3 m)
Problem 2.39 Bar AB is 8.5 m long and bar AC is
6 m long. Determine the components of the position
vector rAB from point A to point B.
y
3m
B
x
C
A
Solution: The key to this solution is to find the coordinates of
point A. We know the lengths of all three sides of the triangle. The
law of cosines can be used to give us the angle θ.
y
|rAC |2 = |rAB |2 + |rBC |2 − 2|rAB ||rBC | cos θ
62 = 8.52 + 32 − 2(8.5)(3) cos θ
cos θ = 0.887
3m
θ = 27.5◦
rAB = |rAB | cos θi − |rAB | sin θj
rAB = (8.5) cos 27.5◦ i − (8.5) sin 27.5◦ j (m)
x
B
θ
C
rAB = 7.54i − 3.92j m
6m
8.5 m
A
Problem 2.40 For the truss in Problem 2.39, determine
the components of a unit vector eAC that points from
point A toward point C.
Strategy: Determine the components of the position
vector from point A to point C and divide the position
vector by its magnitude.
Solution: From the solution of Problem 2.39, point A is located
at (7.54, −3.92). From the diagram, Point C is located at (3.0, 0).
The vector from A to C is
rAC = (xC − xA )i + (yC − yA )j (m)
rAC = (3 − 7.54)i + (0 − (−3.92))j (m)
rAC = −4.54i + 3.92j (m)
and |rAC | = 6 m
Thus eAC =
rAC
4.54
3.92
=−
i+
j
|rAC |
6
6
eAC = −0.757i + 0.653j
Problem 2.41 The x and y coordinates of points A,
B, and C of the sailboat are shown.
(a) Determine the components of a unit vector that is
parallel to the forestay AB and points from A toward B.
(b) Determine the components of a unit vector that is
parallel to the backstay BC and points from C toward B.
y
B (4,13) m
C
(9,1) m
A
(0,1.2) m
Solution:
x
y
rAB = (xB − xA )i + (yB − yA )j
rCB = (xB − xC )i + (yC − yB )j
B (4, 13) m
Points are: A (0, 1.2), B (4, 13) and C (9, 1)
Substituting, we get
rAB = 4i + 11.8j (m), |rAB | = 12.46 (m)
rCB = −5i + 12j (m), |rCB | = 13 (m)
The unit vectors are given by
eAB =
rAB
rCB
and eCB =
|rAB |
|rCB |
Substituting, we get
eAB = 0.321i + 0.947j
eCB = −0.385i + 0.923j
A
(0, 1.2) m
C
(9, 1) m
x
Problem 2.42 Consider the force vector F = 3i − 4j Solution: The magnitude of the force vector is
(kN). Determine the components of a unit vector e that
|F| = 32 + 42 = 5.
has the same direction as F.
The unit vector is
F
3
4
= i − j = 0.6i − 0.8j
|F|
5
5
|e| = 0a.62 + 0.82 = 1
e =
Problem 2.43 Determine the components of a unit
vector that is parallel to the hydraulic actuator BC and
points from B toward C.
As a check, the magnitude:
y
1m
D
C
1m
0.6 m
B
A
0.15 m
Solution:
x
0.6 m
Scoop
Point B is at (0.75, 0) and point C is at (0, 0.6). The
y
vector
1m
rBC = (xC − xB )i + (yC − yB )j
D
rBC = (0 − 0.75)i + (0.6 − 0)j (m)
rBC = −0.75i + 0.6j (m)
|rBC | = (0.75)2 + (0.6)2 = 0.960 (m)
eBC =
C
1m
0.6 m
rBC
−0.75
0.6
=
i+
j
|rBC |
0.96
0.96
eBC = −0.781i + 0.625j
Problem 2.44 The hydraulic actuator BC in Problem
2.43 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 2.43,
eBC = −0.781i + 0.625j
The vector F is given by F = |F|eBC
F = (1.2)(−0.781i + 0.625j) (k · N)
F = −937i + 750j (N)
B
A
0.15 m
0.6 m
x
Scoop
Problem 2.45 A surveyor finds that the length of the
line OA is 1500 m and the length of line OB is 2000 m.
(a) Determine the components of the position vector
from point A to point B.
(b) Determine the components of a unit vector that
points from point A toward point B.
y
N
A
Proposed bridge
B
60°
30°
O
Solution:
We need to find the coordinates of points A and B
rOA = 1500 cos 60◦ i + 1500 sin 60◦ j
rOA = 750i + 1299j (m)
Point A is at (750, 1299) (m)
rOB = 2000 cos 30◦ i + 2000 sin 30◦ j (m)
rOB = 1723i + 1000j (m)
Point B is at (1732, 1000) (m)
(a) The vector from A to B is
rAB = (xB − xA )i + (yB − yA )j
rAB = 982i − 299j (m)
(b)
The unit vector eAB is
eAB =
rAB
982i − 299j
=
|rAB |
1026.6
eAB = 0.957i − 0.291j
y
N
A
Proposed
bridge
B
60°
30°
O
River
x
River
x
Problem 2.46 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 × 106 km, the distance from
the Sun to Venus is 108 × 106 km, and the distance from
the Sun to the Earth is 150 × 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 = |rE |(− sin 20◦ i) + |rE |(cos 20◦ )j (km)
rM = |rM | cos 0◦ i (km)
E
6
6
rE = (−51.3 × 10 )i + (141 × 10 )j (km)
y
rM = 57 × 106 i (km)
rEM = (xM − xE )i + (yM − yE )j (km)
20°
rEM = (108.3 × 106 )i − (141 × 106 j) (km)
|rEM | = 177.8 × 106 (km)
eEM =
rEM
= +0.609i − 0.793j
|rEM |
S
M
x
40°
V
Problem 2.47 For the positions described in Problem 2.46, determine the components of a unit vector that
points from Earth toward Venus.
Solution:
From the solution to Problem 2.47,
6
6
rE = (−51.3 × 10 )i + (141 × 10 )j (km)
The position of Venus is
rV = −|rV | cos 40◦ i − |rV | sin 40◦ j (km)
rV = (−82.7 × 106 )i − (69.4 × 106 )j (km)
rEV = (xV − xE )i + (yV − yE )j (km)
rEV = (−31.4 × 106 )i − (210.4 × 106 )j (km)
|rEV | = 212.7 × 106 (km)
eEV =
rEV
|rEV |
eEV = −0.148i − 0.989j
Problem 2.48 The rope ABC exerts forces FBA
and FBC on the block at B. Their magnitudes are
|FBA | = |FBC | = 800 N. Determine the magnitude
of the vector sum of the forces by resolving the forces
into components, and compare your answer with that of
Problem 2.16.
FBC
C
20°
B
B
FBA
A
Solution: The strategy is to use the magnitudes and the angles to
determine the force vectors, and then to determine the magnitude of
their sum. The force vectors are:
FBC
FBA = 0i − 800j, and
◦
20°
◦
FBC = 800(i cos 20 + j sin 20 ) = 751.75i + 273.6j
y
B
The sum is given by: FBA + FBC = 751.75i − 526.4j
x
The magnitude is given by
|FBA + FBC | = (751.75)2 + (526.4)2 = 917.7 N
Problem 2.49 The magnitudes of the forces are |F1 | =
|F2 | = |F3 | = 5 kN. What is the magnitude of the vector
sum of the three forces?
FBA
y
30°
F1
F3
45°
F2
x
Solution: The strategy is to use the magnitudes and the angles to
determine the force vectors, and then to take the magnitude of their
sum. The force vectors are:
F1 = 5i + 0j (kN),
◦
y
F1
30°
F3
◦
F2 = 5(i cos(−45 ) + j sin(−45 )) = 3.54i − 3.54j
F3 = 5(i cos 210◦ + j sin 210◦ ) = −4.33i − 2.50j
The sum is given by F1 + F2 + F3 = 4.21i − 6.04j and the magnitude is
|F1 + F2 + F3 | = (4.21)2 + (6.04)2 = 7.36 kN
45°
F2
x
Problem 2.50 Four groups engage in a tug-of-war.
The magnitudes of the forces exerted by groups B, C,
and D are |FB | = 800 lb, |FC | = 1000 lb, |FD | =
900 lb. If the vector sum of the four forces equals zero,
what are the magnitude of FA and the angle α?
y
FB
FC
70°
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
30°
20°
α
FB = 800(i cos 110◦ + j sin 110◦ ) = −273.6i + 751.75j
FD
FC = 1000(i cos 30◦ + j sin 30◦ ) = 866i + 500j
FA
FD = 900(i cos(−20◦ ) + j sin(−20◦ )) = 845.72i − 307.8j
FA = |FA |(i cos(180 + α) + j sin(180 + α))
x
= |FA |(−i cos α − j sin α)
The sum vanishes:
FA + FB + FC + FD = i(1438.1 − |FA | cos α)
y
+ j(944 − |FA | sin α) = 0
|FA | =
(1438)2
+
The angle is: tan α =
(944)2
944
1438
FC
70°
From which FA = 1438.1i + 944j. The magnitude is
FB
30°
20°
FD
α
= 1720 lb
FA
x
= 0.6565, or α = 33.3◦
Problem 2.51 The total thrust exerted on the launch
vehicle by its main engines is 200,000 lb parallel to the
y axis. Each of the two small vernier engines exert a
thrust of 5000 lb in the directions shown. Determine the
magnitude and direction of the total force exerted on the
booster by the main and vernier engines.
y
Solution: The strategy is to use the magnitudes and angles to
determine the force vectors. The force vectors are:
FM E = 0i − 200j (kip)
x
Vernier
engines
FLV = 5(i cos 240◦ + j sin 240◦ ) = −2.5i − 4.33j (kip)
FRV = 5(i cos 285◦ + j sin 285◦ ) = 1.29i − 4.83j (kip)
The sum of the forces:
FM E + FRV + FLV = −1.21i − 209.2j (kip).
The magnitude of the sum is
|FR | = (1.21)2 + (209.2)2 = 209.2 (kip)
30°
15°
The direction relative to the y-axis is
tan α =
1.21
= 0.005784, or α = 0.3314◦
209.2
measured clockwise from the negative y-axis.
5k
5k
30°
200k
15°
Problem 2.52 The magnitudes of the forces acting on
the bracket are |F1 | = |F2 | = 2 kN. If |F1 + F2 | =
3.8 kN, what is the angle α? (Assume 0 ≤ α ≤ 90◦ )
F2
α
F1
Solution:
Let |F1 | = |F2 | = a = 2 kn
and |F1 + F2 | = b
F2
α
F1
F2
F1
β
α
F1 + F2
Angle β is given by
α
α
+β+
= 180◦
2
2
β = 180◦ − α
b = 3.8 kN
a = 2 kN
From the law of cosines
b2 = a2 + a2 − 2a2 cos β
β = 143.6◦
α = 180◦ − β
α
α = 36.4◦
Problem 2.53 The figure shows three forces acting on
a joint of a structure. The magnitude of Fc is 60 kN, and
FA + FB + FC = 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 = |FA | cos 40i + |FA | sin 40j (kN)
FA
195°
FB = |FB | cos 195◦ i + |FB | sin 195j (kN)
40°
FC = |FC | cos 270◦ i + |FC | sin 270◦ j (kN)
Thus FC = −60j kN
Since FA + FB + FC = 0, their components in each direction must
also sum to zero.
FAx + FBx + FCx
FAy + FBy + FCy
x
FB
=0
=0
270°
FC
Thus,
y
|FA | cos 40◦ + |FB | cos 195◦ + 0 = 0
|FA | sin 40◦ + |FB | sin 195◦ − 60 (kN) = 0
FC
Solving for |FA | and |FB |, we get
|FA | = 137 kN, |FB | = 109 kN
FB
15°
x
40°
FA
Problem 2.54 Four forces act on a beam. The vector
sum of the forces is zero. The magnitudes |FB | = 10 kN
and |FC | = 5 kN. Determine the magnitudes of FA
and FD .
Solution: Use the angles and magnitudes to determine the vectors,
and then solve for the unknowns. The vectors are:
FA = |FA |(i cos 30◦ + j sin 30◦ ) = 0.866|FA |i + 0.5|FA |j
FD
30°
FA
FC
FB
From the second equation we get |FA | = 10 kN . Using this value in the first
equation, we get |FD | = 8.7 kN
FB = 0i − 10j, FC = 0i + 5j, FD = −|FD |i + 0j.
Take the sum of each component in the x- and y-directions:
Fx = (0.866|FA | − |FD |)i = 0
and
Fy = (0.5|FA | − (10 − 5))j = 0.
FD
30°ν
FA
FB
FC
Problem 2.55 Six forces act on a beam that forms part
of a building’s frame. The vector sum of the forces is
zero. The magnitudes |FB | = |FE | = 20 kN, |FC | =
16 kN, and |FD | = 9 kN. Determine the magnitudes of
FA and FG .
FA
FC
70°
40°
50°
40°
FB
Solution:
FG
FD
FE
Write each force in terms of its magnitude and direction
y
as
F = |F| cos θi + |F| sin θj
where θ is measured counterclockwise from the +x-axis.
Thus, (all forces in kN)
FA = |FA | cos 110◦ i + |FA | sin 110◦ j (kN)
FB = 20 cos 270◦ i + 20 sin 270◦ j (kN)
θ
FC = 16 cos 140◦ i + 16 sin 140◦ j (kN)
FD = 9 cos 40◦ i + 9 sin 40◦ j (kN)
x
FE = 20 cos 270◦ i + 20 sin 270◦ j (kN)
FG = |FG | cos 50◦ i + |FG | sin 50◦ j (kN)
We know that the x components and y components of the forces must
add separately to zero.
Thus
FAx + FBx + FCx + FDx + FEx + FGx
FAy + FBy + FCy + FDy + FEy + FGy
=0
=0
|FA | cos 110◦ + 0 − 12.26 + 6.89 + 0 + |FG | cos 50◦ = 0
|FA | sin 110◦ − 20 + 10.28 + 5.79 − 20 + |FG | sin 50◦ = 0
Solving, we get
|FA | = 13.0 kN
FA
|FG | = 15.3 kN
70°
FC
FG
FD
40°
FB
50°
40°
FE
Problem 2.56 The total weight of the man and parasail Solution: Three forces in equilibrium form a closed triangle. In this inis |W| = 230 lb. The drag force D is perpendicular to stance it is a right triangle. The law of sines is
the lift force L. If the vector sum of the three forces is
|W|
|L|
|D|
=
=
sin 90◦
sin 70◦
sin 20◦
zero, what are the magnitudes of L and D?
From which:
y
|L| = |W| sin 70◦ = (230)(0.9397) = 216.1 lb
L
20°
|D| = |W| sin 20◦ = (230)(0.3420) = 78.66 lb
y
D
20°
L
D
x
W
x
D
W
W
20°
L
Problem 2.57 Two cables AB and CD extend from
the rocket gantry to the ground. Cable AB exerts a force
of magnitude 10,000 lb on the gantry, and cable CD
exerts a force of magnitude 5000 lb.
(a) Using the coordinate system shown, express each of
the two forces exerted on the gantry by the cables in
terms of scalar components.
(b) What is the magnitude of the total force exerted on
the gantry by the two cables?
y
A
40°
C
30°
D
Solution: Use the angles and magnitudes to determine the components of the two forces, and then determine the magnitude of their
sum. The forces:
(a) FAB
FCD
◦
A
◦
= 10(i cos(−50 ) + j sin(−50 ))
50 °
= 6.428i − 7.660j kip
40°
= 5(i cos(−60◦ ) + j sin(−60◦ )) = 2.50i − 4.330j kip
C
60°
The sum: FAB + FCD = 8.928i − 11.990j,
The magnitude is:
(b) |FAB + FCD | =
B
30°
(8.928)2 + (11.99)2 = 14.95 kip
D
B
x
Problem 2.58 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:
|FA | = |FB | = |FC |. The magnitude of the vector sum
of the three forces is 200 kN. What is |FA |?
FC
FA
6m
A
4m
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:
4m
θA = tan−1
= 33.7◦ ,
6m
8
θB = tan−1
= 53.1◦
6
12
θC = tan−1
= 63.4◦ .
6
Measure the angles counterclockwise form the x-axis. The force vectors acting along the cables are:
FA = |FA |(i cos 303.7◦ + j sin 303.7◦ ) = 0.5548|FA |i − 0.8319|FA |j
FB = |FB |(i cos 323.1◦ + j sin 323.1◦ ) = 0.7997|FB |i − 0.6004|FB |j
FC = |FC |(i cos 333.4◦ + j sin 333.4◦ ) = 0.8944|FC |i−0.4472|FC |j
The sum of the forces are, noting that each is equal in magnitude, is
F = (2.2489|FA |i − 1.8795|FA |j).
The magnitude of the sum is given by the problem:
200 = |FA | (2.2489)2 + (1.8795)2 = 2.931|FA |,
from which |FA | = 68.24 kN
6m
A
4m
B
C
4m
4m
B
4m
C
4m
FB
Problem 2.59 The cable from B to A on the sailboat
shown in Problem 2.41 exerts a 230-N force at B. The
cable from B to C exerts a 660-N force at B. What is
the magnitude of the total force exerted at B by the two
cables? What is the magnitude of the downward force
(parallel to the y axis) exerted by the two cables on the
boat’s mast?
Solution: Find unit vectors in the directions of the two forcesexpress the forces in terms of magnitudes times unit vectors-add the
forces.
Unit vectors:
eBA =
=
y
B (4, 13) m
rBA
(xA − xB )i + (yA − yB )j
= |rBA |
(xA − xB )2 + (yA − yB )2
(0 − 4)i + (1.2 − 13)j
√
42 + 11.82
eBA = −0.321i − 0.947j
Similarly,
eBC = 0.385i − 0.923j
FBA = |FBA |eBA = −73.8i − 217.8j kN
FBC = |FBC |eBC = 254.1i − 609.2j kN
Adding
F = FBA + FBC = 180.3i − 827j kN
|F| =
C
(9, 1) m
A
(0, 1.2) m
Fx2 + Fy2 = 846 kN (Total force)
Fy = −827 kN (downward force)
Problem 2.60 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 |FAB | = 4 kN. If the vector sum of
the three forces equals zero, what are the magnitudes of
FAC and FAD ?
Solution:
eAD = √
eAC = √
eAB = √
y
B
(– 4, 1) m
FAC
C
22 + 3 2
−4
42 + 1 2
4
42 + 2 2
i+ √
i+ √
i+ √
−3
22 + 3 2
1
42 + 1 2
2
42 + 2 2
D
j = −0.5547i − 0.8320j
(–2, – 3) m
j = −0.9701i + 0.2425j
j = 0.89443i + 0.4472j
The forces are FAD = |FAD |eAD , FAC = |FAC |eAC ,
FAB = |FAB |eAB = 3.578i + 1.789j. Since the vector sum of
the forces vanishes, the x- and y-components vanish separately:
B
C
Fx = (−0.5547|FAD | − 0.9701|FAC | + 3.578)i = 0, and
Fy = (−0.8320|FAD | + 0.2425|FAC | + 1.789)j = 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: |FAC | = 2.108 kN , |FAD | = 2.764 kN
A
D
(4, 2) m
x
A
FAD
Determine the unit vectors parallel to each force:
−2
FAB
x
Problem 2.61 The distance s = 45 in.
(a) Determine the unit vector eBA that points from B
toward A.
(b) Use the unit vector you obtained in (a) to determine
the coordinates of the collar C.
y
The unit vector from B to A is the vector from B to A divided
by its magnitude. The vector from B to A is given by
Solution:
rBA = (xA − xB )i + (yA − yB )j or rBA = (14 − 75)i + (45 − 12)j in.
Hence, vector from B to A is given by rBA = (−61)i + (33)j in. The
magnitude of the vector from B to A is 69.4 in. and the unit vector from B
toward A is eBA = −0.880i + 0.476j.
y
A
A
(14, 45) in
(14, 45) in
s
C
C
B
(75, 12) in
x
s
B
(75, 12) in
x
Problem 2.62 In Problem 2.61, determine the x and y Solution: The coordinates of the point C are given by
coordinates of the collar C as functions of the distance s. xC = xB + s(−0.880) and yC = yB + s(0.476).
Thus, the coordinates of point C are xC = 75 − 0.880s in. and yC =
12+0.476s in. Note from the solution of Problem 2.61 above, 0 ≤ s ≤ 69.4 in.
Problem 2.63 The position vector r goes from point
A to a point on the straight line between B and C. Its
magnitude is |r| = 6 ft. Express r in terms of scalar
components.
y
B
(7, 9) ft
r
A (3, 5) ft
C
(12, 3) ft
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
rAB = (7 − 3)i + (9 − 5)j = 4i + 4j,
|rAB | = 5.6568
rAC = (12 − 3)i + (3 − 5)j = 9i − 2j,
|rAC | = 9.2195
rBC = (12 − 7)i + (3 − 9)j = 5i − 6j,
|rBC | = 7.8102
The unit vector parallel to BC is
eBC =
rBC
= 0.6402i − 0.7682j = i cos 50.19◦ − j sin 50.19◦ .
|rBC |
Add ±90◦ to the angle to find the two possible perpendicular vectors:
x
where s is the semiperimeter, s = 12 (|rAC | + |rAB | + |rBC |). Substituting
values, s = 11.343, and area = 22.0 and the magnitude of the perpendicular
2(22)
is |rAP | = 7.8102 = 5.6333. The angle between the vector r and the perpendicular rAP is β = cos−1 5.6333
= 20.1◦ . Thus the angle between the
6
vector r and the x-axis is α = 39.8 ± 20.1 = 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 = 6(i cos 19.7◦ + j sin 19.7◦ ) = 5.65i + 2.02j
eAP 1 = i cos 140.19◦ − j sin 140.19◦ , or
eAP 2 = i cos 39.8◦ + 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
2(area)
BC. The altitude is given by h =
. From geometry, the area
base
of a triangle with known sides is given by
area = s(s − |rBC |)(s − |rAC |)(s − |rAB |),
y
B[7,9]
P
r
A[3,5]
C[12,3]
x
Problem 2.64 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:
rAB = (10 − 3)i + (9 − 4)j = 7i + 5j,
|rAB | = 8.602
Check: An alternate solution: Find the angle of the line AB:
5
θ = tan−1
= 35.54◦ .
7
rCA = (3 − 9)i + (4 − 3) = −6i + 1j,
|rCA | = 6.0828
The components of s,
rCB = (10 − 9)i + (9 − 3)j = 1i + 6j,
|rCB | = 6.0828
The ratio of the magnitudes of the two parts of the line is
|rBP |
s
=R=
|rP A |
|rBC | − s
Since the ratio is a scalar, then rBP = RrP A , from which (r −
rCA ) = R(rCB − r).
+rCA
. Substitute the values of the
Solve for the vector r, r = RrCB
1+R
s
vectors, note that R = 8.602−s , and reduce algebraically:
s = |s|(i cos θ + j sin θ) = |s|(0.8138i + 0.5812j).
The coordinates of point P (3+0.8138|s|, 4+0.5812|s|). Subtract coordinates
of point C to get
r = (0.8135|s| − 6)i + (0.5812|s| + 1)j .
check .
B[10,9] m
y
r = (0.8138s − 6)i + (0.5813s + 1)j (m) :
P
8
r
C[9,3] m
A[3,4] m
Problem 2.65 A vector U = 3i − 4j − 12k. What is
its magnitude?
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 magni-
tude is
|U| =
32 + (−4)2 + (−12)2 = 13
Problem 2.66 A force vector F = 20i + 60j − 90k
(N). Determine its magnitude.
Solution: Use definition given in Eq. (14). The magnitude of the
vector is
|F| = (20)2 + (60)2 + (−90)2 = 110 N
x
Problem 2.67 An engineer determines that an attachment point will be subjected to a force F = 20i + 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 ?
Fy2LIMIT = 802 − 202 − (45)2
Solution:
80
2
≥
Fx2
+
Fy2
+
Fz2
Fy2LIMIT = 3975
802 ≥ 202 + Fy2 + (45)2
Fy LIMIT = +63.0, −63.0 (kN)
To find limits, use equality.
|Fy LIMIT | ≤ 63.0 kN − 63.0 kN ≤ Fy ≤ 63.0 kN
Problem 2.68 A vector U = Ux i + Uy j + Uz k. Its
magnitude is |U| = 30. Its components are related by
the equations Uy = −2Ux and Uz = 4Uy . Determine
the components.
Solution: Substitute the relations between the components, determine the magnitude, and solve for the unknowns. Thus
U = +3.61i + (−2(3.61))j + (4(−2)(3.61))k
= 3.61i − 7.22j − 28.9k
U = Ux i + (−2Ux )j + (4(−2Ux ))k = Ux (1i − 2j − 8k)
where Ux can be factored out since it is a scalar. Take the magnitude,
noting that the absolute value of |Ux | must be taken:
30 = |Ux | 12 + 22 + 82 = |Ux |(8.31).
U = −3.61i + (−2(−3.61))j
+ 4(−2)(−3.61)k = −3.61i + 7.22j + 28.9k
Solving, we get |Ux | = 3.612, or Ux = ±3.61. The two possible
vectors are
Problem 2.69 A vector U = 100i + 200j − 600k, Solution: The resultant is
and a vector V = −200i + 450j + 100k. Determine the −2U + 3V = (−2(100) + 3(−200))i + (−2(200) + 3(450))j
magnitude of the vector −2U + 3V.
+ (−2(−600) + 3(100))k
−2U + 3V = −800i + 950j + 1500k
The magnitude is:
| − 2U + 3V| =
(−800)2 + (950)2 + (1500)2 = 1947.4
Problem 2.70 Two vectors U = 3i − 2j + 6k and Solution: The magnitudes:
√
√
V = 4i + 12j − 3k.
|U| = 32 + 22 + 62 = 7 and |V| = 42 + 122 + 32 = 13
(a)
(a) Determine the magnitudes of U and V.
The resultant vector
(b) Determine the magnitude of the vector 3U + 2V.
3U + 2V = (9 + 8)i + (−6 + 24)j + (18 − 6)k
= 17i + 18j + 12k
(b)
The magnitude |3U + 2V| =
√
172 + 182 + 122 = 27.51
Problem 2.71 A vector U = 40i − 70j − 40k.
Solution: √The magnitude:
(a) |U| = 402 + 702 + 402 = 90
(a) What is its magnitude?
cosines:
(b) What are the angles θx , θy , and θz between U and (b) The direction
40
70
40
the positive coordinate axes?
U = 90
i−
j−
90
90
90
Strategy: Since you know the components of U,
= 90(0.4444i − 0.7777j − 0.4444k)
you can determine the angles θx , θy , and θz from
Eqs. (2.15).
◦
◦
U = 90(i cos 63.6 + j cos 141.1 + k cos 116.4◦ )
Problem 2.72 A force F = 600i − 700j + 600k (lb).
What are the angles θx , θy , and θz between the vector F
and the positive coordinate axes?
Solution: The
√
6002
|F| =
+
The unit vector is:
e=
7002
magnitude:
+
6002
The angles are
= 1100
θx = cos−1 (0.5455) = 56.9◦ , θy = cos−1 (−0.6364) = 129.5◦ ,
F
600
700
600
=
i−
j+
k = 0.5455i − 0.6364j + 0.5455k
|F|
1100
1100
1100
Problem 2.73 The cable exerts a 50-lb force F on the
metal hook at O. The angle between F and the x axis is
40◦ , and the angle between F and the y axis is 70◦ . The
z component of F is positive.
(a) Express F in terms of components.
(b) What are the direction cosines of F?
Strategy: Since you are given only two of the angles between F and the coordinate axes, you must
first determine the third one. Then you can obtain
the components of F from Eqs. (2.15).
and θz = cos−1 (0.5455) = 56.9◦
y
70
F
40
x
O
z
Solution: Use Eqs. (2.15) and (2.16).
The force F = 50(i cos 40◦ + j cos 70◦ + k cos θz ). Since
12 = cos2 40◦ + cos2 70◦ + cos2 θz , by definition (see
Eq. (2.16)) then
√
cos θz = ± 1 − 0.5868 − 0.1170 = ±0.5442.
Thus the components of F are
(a)
(b)
F = 50(0.7660i + 0.3420j + 0.5442k),
= 38.3i + 17.1j + 27.2k (lb)
the direction cosines are
cos θx = 0.7660, cos θy = 0.3420, cos θz = 0.5442
Problem 2.74 A unit vector has direction cosines Solution: Use Eq. (2.15) and (2.16). The third direction cosine is
cos θx = −0.5 and cos θy = 0.2. Its z component is
cos θz = ± 1 − (0.5)2 − (0.2)2 = +0.8426.
positive. Express it in terms of components.
The unit vector is
u = −0.5i + 0.2j + 0.8426k
Problem 2.75 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) What is the angle between T and the z axis?
(b) Express T in terms of components.
◦
l = cos 120 = −0.5, m = cos 130◦ = −0.6428
from which the z-direction cosine is
n = cosθz = ± 1 − (0.5)2 − (0.6428)2 = +0.5804.
Thus the angle between T and the z-axis is
(a)
y
y
130°
x
x
120°
z
z
θz = cos−1 (0.5804) = 54.5◦ , and the thrust is
T = 200(−0.5i − 0.6428j + 0.5804k), or:
(b)
T
The x- and y-direction cosines are
Solution:
T = −100i − 128.6j + 116.1k (kN)
Problem 2.76 The position vector from a point A to
a point B is 3i + 4j − 4k (ft). The position vector from
point A to point C is −3i + 13j − 2k. ft
(a) What is the distance from point B to point C?
(b) What are the direction cosines of the position vector
from point B to point C?
Solution:
The vector from point B to point C is rBC = rAC − rAB .
Thus
rBC = (−3 − 3)i + (13 − 4)j + (−2 − (−4))k = −6i + 9j + 2k.
The distance between
points B and C is
√
(a) |rBC | = 62 + 92 + 22 = 11 (ft). The direction cosines are
cos θx = −6
= −0.5454,
11
2
cos θz = 11
= 0.1818
(b)
cos θy =
9
11
= 0.8182,
Problem 2.77 A vector U = 3i − 2j + 6k. Deter- Solution: By definition, the unit vector is the vector whose components
mine the components of the unit vector that has the same are the direction cosines
√ of U. (See discussion following Eq. (2.15)). The
magnitude is |U| = 32 + 22 + 62 = 7. Thus the unit vector is
direction as U.
u=
U
|U|
=
3
i
7
− 27 j + 67 k
Problem 2.78 A force vector F = 3i − 4j − 2k (N). Solution: By definition, the unit vector is the vector whose components are
the direction cosines of F. The magnitude is
(a) What is the magnitude of F?
(b) Determine the components of the unit vector that (a) |F| = 32 + 42 + 22 = 5.385 (N) The unit vector is
has the same direction as F.
(b) e =
3
4
2
i−
j−
k
5.385
5.385
5.385
= 0.5571i − 0.7428j − 0.3714k
Problem 2.79 A force vector F points in the same Solution: By definition, F = |F|e, where e is a unit vector in the direction
direction as the unit vector e = 27 i − 67 j − 37 k. The of F. (See discussion following Eq. (2.16).) Thus
magnitude of F is 700 lb. Express F in terms of com- F = 700 2 i − 6 j − 3 k = 200i − 600j − 300k
7
7
7
ponents.
Problem 2.80 A force vector F points in the same
direction as the position vector r = 4i + 4j − 7k (m).
The magnitude of F is 90 kN. Express F in terms of
components.
Solution: By definition, F = |F|e, where e is a unit vector in the direction
of
The magnitude is |r| =
√ F. Find the unit vector from the position4vector.
42 + 42 + 72 = 9; the unit vector is e = 9 i + 49 j − 79 k. The components
are
4
4
7
F = 90
i + j − k = 40i + 40j − 70k (kN)
9
9
9
Problem 2.81 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 θx = 0.768, cos θy =
0.384, cos θz = 0.512. The vector rB from the shuttle
to satellite B has magnitude 4 km and direction cosines
cos θx = 0.743, cos θy = 0.557, cos θz = −0.371.
What is the distance between the satellites?
B
rB
x
y
rA
A
z
Solution:
The two position vectors are:
rA = 2(0.768i+ 0.384j+ 0.512k) = 1.536i + 0.768j + 1.024k (km)
B
rB = 4(0.743i+ 0.557j− 0.371k) = 2.972i + 2.228j − 1.484k (km)
rB
The distance is the magnitude of the difference:
x
|rA − rB |
= (1.536−2.927)2 + (0.768−2.228)2 + (1.024−(−1.484))2
y
rA
A
z
= 3.24 (km)
Problem 2.82 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.
y
4m
10 m
4m
A
10 m
8m
B
b
8m
z
C
Solution: The coordinates are A(0, 16, 14), and B(10, 8, 4). The
vector from A to B is
10 m
4m
rAB = (10 − 0)i + (8 − 16)j + (4 − 14)k = 10i − 8j − 10k.
and
10
16.2
= 0.6155,
cos θy =
−8
16.2
= −0.4938,
cos θz =
−10
16.2
= −0.6155 .
y
4m
A
10 m
The magnitude is
√
(a)
|rAB | = 102 + 82 + 102 = 16.2 m , and
(b) The direction cosines are
cos θx =
x
B
8m
b
z
8m
C
x
Problem 2.83 Consider the structure described in Problem 2.82. After returning to the United States, an archaeologist discovers that he lost the notes containing the
dimension b, but other notes indicate that the distance
from point B to point C is 16.4 m. What are the direction
cosines of the vector from B to C?
The coordinates of B and C are B(10, 8, 4) and C(10 +b, 0,
18). The vector from B to C is
Solution:
rBC = (10 + b − 10)i + (0 − 8)j + (18 − 4)k = bi − 8j + 14k.
The magnitude of this vector is known:
16.4 = b2 + 82 + 142 = b2 + 260, from which
b2 = (16.4)2 − 260 = 8.96, or b = ±3 = +3 m.
The direction cosines are
cos θx =
3
16.4
= 0.1829,
cos θz =
14
16.4
= 0.8537
Problem 2.84 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.
cos θy =
−8
16.4
= −0.4878,
y
rAR
rBR
A
x
B (5,0,2) km
z
Solution:
The vector rAR is given by
rAR = 4i + 4j + 2k km
y
and the magnitude of rAR is given by
|rAR | = (4)2 + (4)2 + (2)2 km = 6 km.
rAR
The unit vector along AR is given by
A
uAR = rAR /|rAR |.
Thus, uAR = 0.667i + 0.667j + 0.333k
and the direction cosines are
cos θx = 0.667, cos θy = 0.667, and cos θz = 0.333.
The vector rBR is given by
rBR = (xR − xB )i + (yR − yB )j + (zR − zB )k km
= (4 − 5)i + (4 − 0)j + (2 − 2)k km
and the magnitude of rBR is given by
|rBR | = (1)2 + (4)2 + (0)2 km = 4.12 km.
The unit vector along BR is given by
eBR = rBR /|rBR |.
Thus, uBR = −0.242i + 0.970j + 0k
and the direction cosines are
cos θx = −0.242, cos θy = 0.970, and cos θz = 0.0.
z
rBR
x
B (5,0,2) km
Problem 2.85 In Problem 2.84, 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 θx = 0.535, cos θy = 0.802,
and cos θz = 0.267, and the person at B determines
that the direction cosines of rBR are cos θx = −0.576,
cos θy = 0.798, and cos θz = −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 = (xB − xA )i + (yB − yA )j + (zB − zA )k or
rAB = (5 − 0)i + (0 − 0)j + (2 − 0)k = 5i + 2k km.
The magnitude of rAB is given by |rAB | =
(5)2 + (2)2 =
5.39 km. The unit vector along AB, uAB , is given by
uAB = rAB /|rAB | = 0.928i + 0j + 0.371k km.
The unit vector along the line AR,
uAR = cos θx i + cos θy j + cos θz k = 0.535i + 0.802j + 0.267k.
Similarly, the vector along BR, uBR = −0.576i + 0.798 − 0.177k.
From the diagram in the problem statement, we see that rAR = rAB +
rBR . Using the unit vectors, the vectors rAR and rBR can be written
as
rAR = 0.535rAR i + 0.802rAR j + 0.267rAR k, and
rBR = −0.576rBR i + 0.798rBR j − 0.177rBR k.
Substituting into the vector addition rAR = rAB +rBR and equating
components, we get, in the x direction, 0.535rAR = −0.576rBR ,
and in the y direction, 0.802rAR = 0.798rBR . Solving, we get that
rAR = 4.489 km. Calculating the components, we get
rAR = rAR eAR = 0.535(4.489)i + 0.802(4.489)j + 0.267(4.489)k.
Hence, the coordinates of the rocket, R, are (2.40, 3.60, 1.20) km.
Problem 2.86 The height of Mount Everest was originally measured by a surveyor using the following procedure. He first measured the distance between two points
A and B of equal altitude. Suppose that they are 10,000 ft
above sea level and are 32,000 ft apart. He then used a
theodolite to measure the direction cosines of the vectors
from point A to the top of the mountain P and from point
B to P . Suppose that for rAP , the direction cosines are
cos θx = 0.509, cos θy = 0.509, cos θz = 0.694, and
for rBP they are cos θx = −0.605, cos θy = 0.471,
cos θz = 0.642. The z axis of the coordinate system is
vertical. What is the height of Mount Everest above sea
level?
Solution: Construct the two triangles: (a) Triangle ABP , which
has one known side, AB, and two known adjacent interior angles θA
and θB (b) Triangle AOP , which is a right triangle with a derived
known interior angle θAO . From triangle ABP , determine the length
of AP , and from triangle AP O and the derived interior angle, determine the height above the base, OP . The interior angles of the triangle
ABP are
θA = θAX = cos−1 (0.509) = 59.4◦ ,
z
P
y
B
A
P
z
y
x
B
θB = 180 − θBX = 180 − cos−1 (−0.605)
= 180◦ − 127.2◦ = 52.77◦ and
A
β = 180 − 59.4 − 52.77 = 67.83◦ .
|rAB |
|rBP |
|rAP |
=
=
.
sin 67.83◦
sin 59.4◦
sin 52.77◦
Therefore the length of side AP is
sin 52.77
|rAP | = |rAB |
= 32000(0.8598) = 27512.9 ft.
sin 67.83
The interior angle of the triangle AP O is θAO = 90 − θAZ =
90 − cos−1 (0.694) = 90 − 46.05 = 43.95◦ . Therefore the length
of the side OP is hOP = |rAP | sin 43.95◦ = 27512.9(0.6940) =
19093.9 ft. Check: The z-component of |rAP | is hop =
|rAP | sin θAZ = 19093.9 ft. check.
The base is 10000 ft above sea level, hence the height of P above sea
level is
P = 19093.9 + 10000 = 29094 ft
P
z
θAZ
From the law of sines:
β
y
O
θA
A
θB
x
B
x
Problem 2.87 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.
Strategy: You can resolve rOA into a vector from O
to B and a vecotr 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.
y
A
rOA
O
x
30°
60°
Solution:
See Example 2.10. The length BA is, from the right
triangle OAB,
|rAB | = |rOA | sin 30◦ = 20(0.5) = 10 ft.
B
z
Similarly, the length OB is
|rOB | = |rOA | cos 30◦ = 20(0.866) = 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,
The vector rOA is given by rOA = rOB + rBA , from which
rOA = 15i + 10j + 8.66k (ft)
A
rOB = |rOB |(i cos 30◦ + j cos 90◦ + k cos 60◦ ) or
rOA
y
rOB = 15i + 0j + 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 = |rBA |(i cos 90◦ + j cos 0◦ + k cos 90◦ ) = 0i + 10j + 0k.
30°
O
x
Q
z
P
60°
Problem 2.88 The magnitude of r is 100 in. The straight line from the head of r to point A is parallel to the x
axis, and point A is contained in the y-z plane. Express
r in terms of scalar components.
B
y
A
Bedford
Falls
r
45°
60°
The vector r can be expressed as the sum of the two
vectors, r = rOA +rAP , both of which can be resolved into direction
cosine components. The magnitudes can be determined from the law
of sines for the triangle OAP .
Solution:
O
rOA = |rOA |(i cos 90◦ + j cos 30◦ + k cos 60◦ )
rOA = |rOA |(0i + 0.866j + 0.5k). Similarly,
rAP = |rAP |(i cos 0 + j cos 90 + k cos 90) = |rAP |(1i + 0j + 0k)
z
Since rAP is parallel to the x-axis, it makes an angle of 90◦ with the
y-z plane, and the triangle OAP is a right triangle. From the law of
sines
|r|
|rAP |
|rOA |
=
=
,
sin 90◦
sin 45◦
sin 45◦
y
from which |rAP | = |rOA | = 100(0.707) = 70.7. Substituting
these values into the vectors
r = rOA + rAP = 70.7(1i + 0.866j + 0.5k)
= 70.7i + 61.2j + 35.4k (in.)
P
A
r
45°
60°
z
x
x
Problem 2.89 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 = 100 N.
(a) What is the magnitude of F?.
(b) Determine the angles θx , θy , and θz between F and
the positive coordinate axes.
y
F
x
20°
O
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:
Solution:
60°
A
|rAp |
|F|
|rOA |
=
=
,
sin 20◦
sin 90◦
sin 70◦
thus |rAp | = |F|(0.342) and |rOA | = |F|(0.9397). The components of the two vectors are from the geometry
z
rOA = |rOA |(i cos 30◦ + j cos 90◦ + k cos 60◦ )
(b) θx = cos−1 (0.8138) = 35.5◦ ,
= |rOA |(0.866i + 0j + 0.5k) and
rAp = |rAp |(i cos 90◦ + j cos 0◦ + k cos 90◦ ) = |rAp |(0i + 1j + 0k)
θy = cos−1 (0.342) = 70◦ and θz = cos−1 (0.4699) = 62◦
Noting F = rOA + rAp , then from above
y
F = |F|(0.3420)(0i + 1j + 0k) + |F|(0.9397)(0.866i + 0j + 0.5k)
P
F = |F|(0.8138i + 0.342j + 0.4699k)
F
The x-component is given to be 100 N. Thus,
Problem 2.90 The position of a point P on the surface of the earth is specified by the longitude λ, 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 .)
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 = rOB + rBP . The vector rOB =
|rOB |(i cos λ + 0j + k sin λ). From geometry, the magnitude is
|rOB | = RE cos θ.
The vector rBP = |rBP |(0i + 1j + 0k). From geometry, the
magnitude is |rBP | = RE sin θP . Substitute: P = rOB +
rBP = RE (i cos λ cos θ + j sin θ + k sin λ cos θ). Substitute
from the problem statement: λ = +30◦ , θ = 45◦ . Hence
P = RE (0.6124i + 0.707j + 0.3536k)
20°
O
100
(a) |F| =
= 122.9 N The angles are given by
0.8138
z
q
x
60°
A
y
N
P
L
z
O
λ
G
Equator
x
y
P
O
z
B
λ
x
Problem 2.91 An engineer calculates that the magni- Solution: The components of the position vector from B to A are
tude of the axial force in one of the beams of a geodesic rBA = (Ax − Bx )i + (Ay − By )j + (Az − Bz )k
dome is |P = 7.65 kN. The cartesian coordinates of
= (−12.4 + 9.2)i + (22.0 − 24.4)j
the endpoints A and B of the straight beam are (−12.4,
+(−18.4 + 15.6)k
22.0, −18.4) m and (−9.2, 24.4, −15.6) m, respectively.
Express the force P in terms of scalar components.
= −3.2i − 2.4j − 2.8k (m).
Dividing this vector by its magnitude, we obtain a unit vector that points from
B toward A:
B
eBA = −0.655i − 0.492j − 0.573k.
Therefore
P
P = |P|eBA
= 7.65 eBA
A
= −5.01i − 3.76j − 4.39k (kN).
Problem 2.92 The cable BC exerts an 8-kN force F
on the bar AB at B.
(a) Determine the components of a unit vector that
points from B toward point C.
(b) Express F in terms of components.
y
B (5, 6, 1) m
F
A
x
C (3, 0, 4) m
z
Solution:
(a) eBC
=
eBC
−2i − 6j + 3k
2
6
3
= √
=− i− j+ k
7
7
7
22 + 6 2 + 3 2
eBC
= −0.286i − 0.857j + 0.429k
(b) F
y
rBC
(xC − xB )i + (yC − yB )j + (zC − zB )k
= |rBC |
(xC − xB )2 + (yC − yB )2 + (zC − zB )2
B (5, 6, 1) m
F
= |F|eBC = 8eBC = −2.29i − 6.86j + 3.43k (kN)
A
x
C (3, 0, 4) m
z
Problem 2.93 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
20°
B
C
4 ft
Find the unit vector eCE and multiply it times the magnitude of the force to get the vector in component form,
Solution:
eCE =
rCE
(xE − xC )i + (yE − yC )j + (zE − zC )k
= |rCE |
(xE − xC )2 + (yE − yC )2 + (zE − zC )2
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)
eCE =
(0 − 4)i + (2 − (−1.37))j + (6 − 3.76)k
√
42 + 3.372 + 2.242
eCE = −0.703i + 0.592j + 0.394k
T = 50eCE (lb)
T = −35.2i + 29.6j + 19.7k (lb)
y
6 ft
A
E
D
T
2 ft
x
4 ft
T
z
B
C
20°
4 ft
Problem 2.94 What are the direction cosines of the
force T in Problem 2.93?
Solution:
From the solution to Problem 2.93,
eCE = −0.703i + 0.592j + 0.394k
However
eCE = cos θx i + cos θy j + cos θz k
Hence,
cos θx = −0.703
cos θy = 0.592
cos θz = 0.394
4 ft
T
2 ft
x
Problem 2.95 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.
y
8 ft
C
8 ft
6 ft
B
x
FAB
z
Solution:
A (6, 0, 10) ft
The coordinates of B are B(0,6,8). The position vector
from A to B is
rAB = (0 − 6)i + (6 − 0)j + (8 − 10)k = −6i + 6j − 2k
√
The magnitude is |rAB | = 62 + 62 + 22 = 8.718 ft.
The unit vector is
uAB
FAC
y
8 ft
C
8 ft
B
6 ft
x
−6
6
2
=
i+
j−
k
8.718
8.718
8.718
FAB = |FAB |uAB = 200(−0.6882i + 0.6882j − 0.2294k) or
uAB = −0.6882i + 0.6882j − 0.2294k.
z
A(6, 0, 10) ft
The components of the force are
FAB = |FAB |uAB = 200(−0.6882i + 0.6882j − 0.2294k) or
FAB = −137.6i + 137.6j − 45.9k
Problem 2.96 Consider the cables and wall described
in Problem 2.95. 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 2.81. From Problem 2.81
the force FAB is
FAB = −137.6i + 137.6j − 45.9k
The coordinates of C are C(8,6,0). The position vector from A to C is
rAC = (8 − 6)i + (6 − 0)j + (0 − 10)k = 2i + 6j − 10k.
√
The magnitude is |rAC | = 22 + 62 + 102 = 11.83 ft.
The unit vector is
uAC =
2
6
10
i+
j−
k = 0.1691i + 0.5072j − 0.8453k.
11.83
11.83
11.83
The force is
FAC = |FAC |uAC = 100uAC = 16.9i + 50.7j − 84.5k.
The resultant of the two forces is
FR = FAB + FAC = (−137.6 + 16.9)i + (137.6 + 50.7)j
+ (−84.5 − 45.9)k.
FR = −120.7i + 188.3j − 130.4k.
The magnitude is
|FR | =
√
120.72 + 188.32 + 130.42 = 258.9 lb
Problem 2.97 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
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:
40 m
z
The resultant force exerted on the tower by the cables is:
FR = FAB + FAC + FAD = −0.9872i − 4.5654j − 0.2022k kN
rAD = (−60 − 0)i + (0 − 70)j + (−60 − 0)k
A
rAD = −60i − 70k − 60k
rAC = (−40 − 0)i + (0 − 70)j + (40 − 0)k
A
rAC = −40i − 70j + 40k
rAB = (40 − 0)i + (0 − 70)j + (0 − 0)k
D
rAB = 40i − 70j + 0k
60 m
60 m
The unit vectors corresponding to these position vectors are:
uAD =
rAD
−60
70
60
=
i−
j−
k
|rAD |
110
110
110
B
40 m
= −0.5455i − 0.6364j − 0.5455k
uAC =
rAC
40
70
40
=− i−
j+
k
|rAC |
90
90
90
E
A
rAB
40
70
=
i−
j + 0k = 0.4963i − 0.8685j + 0k
|rAB |
80.6
80.6
The forces are:
40 m
40 m
= −0.4444i − 0.7778j + 0.4444k
uAB =
C
FAD
FAB = |FAB |uAB = 0.9926i − 1.737j + 0k
FAC = |FAC |uAC = −0.8888i − 1.5556j + 0.8888
FAD = |FAD |uAD = −1.0910i − 1.2728j − 1.0910k
FAC
FAB
x
Problem 2.98 Consider the tower described in Problem 2.97. 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:
uAC =
From the solution of Problem 2.83, the unit vectors are:
rAC
40
70
40
=− i−
j+
k
|rAC |
90
90
90
Taking the sum of the forces:
FR = FAB + FAC + FAD = (0.9926 − 0.4444|FAC | − 0.5455|FAD |)i
+(−1.737 − 0.7778|FAC | − 0.6364|FAD |)j
= −0.4444i − 0.7778j + 0.4444k
uAD =
rAD
−60
70
60
=
i−
j−
|rAD |
110
110
110
= −0.5455i − 0.6364j − 0.5455k
+(0.4444|FAC | − 0.5455|FAD |)k
The sum of the x- and z-components vanishes, hence the set of simultaneous
equations:
0.4444|FAC | + 0.5455|FAD | = 0.9926 and
From the solution of Problem 2.83 the force FAB is
FAB = |FAB |uAB = 0.9926i − 1.737j + 0k
The forces FAC and FAD are:
FAC = |FAC |uAC = |FAC |(−0.4444i − 0.7778j + 0.4444k)
FAD = |FAD |uAD = |FAD |(−0.5455i − 0.6364j − 0.5455k)
0.4444|FAC | − 0.5455|FAD | = 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:
|FAC | = 1.1168 kN
Problem 2.99 Express the position vector from point
O to the collar at A in terms of scalar components.
|FAD | = 0.9098 kN
y
6 ft
7 ft
A
x
O
4 ft
4 ft
z
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 rT A 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 = 0i + 7j + 0k. The position vector from the
top of the bar to the base is:
rT B = (4 − 0)i + (0 − 7)j + (4 − 0)k. or
rT B = 4i − 7j + 4k. The unit vector pointing from the top of the
bar to the base is
uT B =
6 ft
7 ft
A
O
rT B
4
7
4
= i − j + k = 0.4444i − 0.7778j + 0.4444k.
|rT B |
9
9
9
The collar position is
rT A = |rT A |uT B = 6(0.4444i − 0.7778j + 0.4444k)
= 2.6667i − 4.6667j + 2.6667,
measured along the bar. The sum of the two vectors is the position
vector of A from origin O:
rOA = (2.6667 + 0)i + (−4.6667 + 7)j + (2.6667 + 0)k
= 2.67i + 2.33j + 2.67k ft
4 ft
4 ft
Problem 2.100 The cable AB exerts a 32-lb force T
on the collar at A. Express T in terms of scalar components.
y
4 ft
B
T
6 ft
7 ft
A
x
4 ft
4 ft
z
Solution: The coordinates of point B are B (0, 7, 4). The vector
position of B is rOB = 0i + 7j + 4k.
The vector from point A to point B is given by
rAB = rOB − rOA .
From Problem 2.86, rOA = 2.67i + 2.33j + 2.67k. Thus
rAB = (0 − 2.67)i + (7 − 2.33)j + (4 − 2.67)j
rAB = −2.67i + 4.67j + 1.33k.
The magnitude is
|rAB | = 2.672 + 4.672 + 1.332 = 5.54 ft.
The unit vector pointing from A to B is
uAB =
rAB
= −0.4819i + 0.8429j + 0.2401k
|rAB |
The force T is given by
TAB = |TAB |uAB = 32uAB = −15.4i + 27.0j + 7.7k (lb)
y
4 ft
6 ft
B
A
7 ft
x
4 ft
z
4 ft
Problem 2.101 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 = 4 cos(20◦ )i+4 sin(20◦ )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 = (xA − xB )i + (yA − yB )j + (zA − zB )k =
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
y
3 ft
B
B
A
4m
A
4 ft
20°
20°
4ft
4m
x
x
z
z
Problem 2.102 The cable AB in Problem 2.101 exerts Solution: We know rBA = 3.76i − 2.63j − 3k m from Problem 2.101.
a 60-N force T on the collar at A that is directed along The unit vector uAB = −rBA /|rBA |. The unit vector is uAB = −0.686i +
the line from A toward B. Express T in terms of scalar 0.480j + 0.547k. Hence, the force vector T is given by
components.
T = |T|(−0.686i+ 0.480j+ 0.547k) N = −41.1i + 28.8j + 32.8k N
Problem 2.103
tors
Determine the dot product of the vec-
U = 8i − 6j + 4k
and V = 3i + 7j + 9k.
Solution:
U · V = Ux Vx + Uy Vy + Uz Vz
= (8)(3) + (−6)(7) + (4)(9)
U · V = 18
Problem 2.104 Determine the dot product U · V of Solution:
the vectors U = 40i + 20j + 60k and V = −30i + 15k.
Use Eq. 2.23.
Problem 2.105 What is the dot product of the position Solution:
vector r = −10i + 25j (m) and the force
Use Eq. (2.23).
U · V = (40)(−30) + (20)(0) + (15)(60) = −300
F = 300i + 250j + 300k (N)?
F · r = (300)(−10) + (250)(25) + (300)(0) = 3250 N-m
Problem 2.106 What is the dot product of the position Solution: Use Eq. (2.23).
vector r = 4i − 12j − 3k (ft) and the force F = 20i +
r · F = 4(20) + 30(−12) − 10(−3) = −250 ft lb
30j − 10k (lb)?
When the vectors are perpendicular, U · V ≡ 0.
Problem 2.107 Two perpendicular vectors are given Solution:
Thus
in terms of their components by
U · V = Ux Vx + Uy Vy + Uz Vz = 0
U = Ux i − 4j + 6k
= 3Ux + (−4)(2) + (6)(−3) = 0
and V = 3i + 2j − 3k.
3Ux = 26
Use the dot product to determine the component Ux .
Problem 2.108
Three vectors
Ux = 8.67
Solution:
For mutually perpendicular vectors, we have three equations,
i.e.,
U = Ux i + 3j + 2k
U·V =0
V = −3i + Vy j + 3k
U·W =0
W = −2i + 4j + Wz k
V·W =0
are mutually perpendicular. Use the dot product to determine the components Ux , Vy , and Wz
Thus

= 0
3 Eqns
=0
 3 Unknowns
=0
−3Ux + 3Vy + 6
−2Ux + 12 + 2Wz
+6 + 4Vy + 3Wz
Solving, we get
Ux
Vy
Wz
Problem 2.109 The magnitudes |U| = 10 and |V| =
20.
(a) Use the definition of the dot product to determine
U · V.
(b) Use Eq. (2.23) to obtain U · V.
= 2.857
= 0.857
= −3.143
y
V
U
45°
30°
x
Solution:
(a)
The definition of the dot product (Eq. (2.18)) is
U · V = |U||V| cos θ. Thus
◦
y
◦
U · V = (10)(20) cos(45 − 30 ) = 193.2
(b)
U
V
The components of U and V are
U = 10(i cos 45◦ + j sin 45◦ ) = 7.07i + 7.07j
45°
V = 20(i cos 30◦ + j sin 30◦ ) = 17.32i + 10j
From Eq. (2.23)
U · V = (7.07)(17.32) + (7.07)(10) = 193.2
30°
x
Problem 2.110 By evaluating the dot product U · V,
prove the identity cos(θ1 − θ2 ) = cos θ1 cos θ2 +
sin θ1 sin θ2 .
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 · V = |U||V| cos(θ1 − θ2 ). From Eq. (2.23) and the figure,
y
U = |U|(i cos θ1 + j sin θ2 ), V = |V|(i cos θ2 + j sin θ2 ),
U
θ1
and the dot product is U·V = |U||V|(cos θ1 cos θ2 +sin θ1 sin θ2 ).
Equating the two results:
U · V = |U||V| cos(θ1 − θ2 ) = |U||V|(cos θ1 cos θ2 + sin θ1 sin θ2 ),
θ2
from which if |U| = 0 and |V| = 0, it follows that
cos(θ1 − θ2 ) = cos θ1 cos θ2 + sin θ1 sin θ2 ,
V
x
Q.E.D.
Problem 2.111 Use the dot product to determine the
angle between the forestay (cable AB) and the backstay
(cable BC) of the sailboat in Problem 2.41.
The unit vector from B to A is
rBA
=
= −0.321i − 0.947j
|rBA |
Solution:
eBA
y
The unit vector from B to C is
rBC
eBC =
= 0.385i − 0.923j
|rBC |
B (4,13) m
From the definition of the dot product, eBA · eBC = 1 · 1 · cos θ,
where θ is the angle between BA and BC. Thus
cos θ = (−0.321)(0.385) + (−0.947)(−0.923)
cos θ = 0.750496
θ = 41.3◦
A
(0,1.2) m
C
(9,1) m
x
Problem 2.112 What is the angle θ between the straight lines AB and AC?
y
B
(–4, 5, –4) ft
Solution:
From the given coordinates, the position vectors are:
A
(8, 6, 4) ft
θ
rOB = −4i + 5j − 4k, rOA = 8i + 6j + 4k, and
x
rOC = 6i + 0j + 6k.
The straight lines correspond to the vectors:
rAB = rOB − rOA = −12i − j − 8k,
z
C (6, 0, 6) ft
rAC = rOC − rAC = −2i − 6j + 2k
The dot product is given by
y
rAB · rAC = (−2)(−12) + (−1)(−6) + (+2)(−8) = 14.
B (–4, 5, –4)
The magnitudes of the vectors are:
|rAC = 22 + 62 + 22 | = 6.6333, and
|rAB = 122 + 12 + 82 | = 14.456.
A (8, 6, 4)
θ
From the definition of the dot product, the angle is
x
rAC · rAB
14
cos θ =
=
= 0.1460.
|rAC ||rAB |
(14.456)(6.633)
z
C (6, 0, 6)
Take the principal value: θ = 81.6◦
Problem 2.113 The ship O measures the positions of
the ship A and the airplane B and obtains the coordinates
shown. What is the angle θ between the lines of sight
OA and OB?
y
B
(4, 4, –4) km
θ
x
O
A
(6, 0, 3) km
z
Solution:
From the coordinates, the position vectors are:
rOA = 6i + 0j + 3k and rOB = 4i + 4j − 4k
The dot product: rOA · rOB = (6)(4) + (0)(4) + (3)(−4) = 12
The magnitudes: |rOA | = 62 + 02 + 32 = 6.71 km and
|rOA | = 42 + 42 + 42 = 6.93 km.
·rOB
From Eq. (2.24) cos θ = |rrOA||r
= 0.2581, from which θ =
OA
OB |
◦
±75 . From the problem and the construction, only the positive angle
makes sense, hence θ = 75◦
B (4, 4, –4) km
y
O
z
θ
x
A (6, 0, 3) km
Problem 2.114 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 θx = 0.768, cos θy =
0.384, cos θz = 0.512. The vector rB from the shuttle
to satellite B has magnitude 4 km and direction cosines
cos θx = 0.743, cos θy = 0.557, cos θz = −0.371.
What is the angle θ between the vectors rA and rB ?
B
rB
x
θ
y
rA
A
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 =
cos θx i + cos θy j + cos θz 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 .
B
cos θ = cos θxA cos θxB + cos θyA cos θyB + cos θzA cos θzB .
rB
x
Evaluation of the relation yields
θ
cos θ = 0.594 ⇒ θ = 53.5◦ .
A
y
rA
z
Problem 2.115 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.
y
B (5, 6, 1) m
F
Solution: Eqn. 2.26 is UP = (e · U)e where U is the vector
for which you want the component parallel to the direction indicated
by the unit vector e.
For the problem at hand, we must find two unit vectors. We need
eBC to be able to write the force F(F = |F|eBC ) and eBA ∼ the
direction parallel to the bar.
eBC
rBC
(xC − xB )i + (yC − yB )j + (zC − zB )k
=
= |rBC |
(xC − xB )2 + (yC − yB )2 + (zC − zB )2
eBC =
eBC
A
x
C (3, 0, 4) m
z
(3 − 5)i + (0 − 6)j + (4 − 1)k
√
22 + 6 2 + 3 2
B (5, 6, 1) m
2
6
3
=− i− j+ k
7
7
7
Similarly
F
−5i − 6j − 1k
eBA = √
52 + 6 2 + 1 2
A
eBA = −0.635i − 0.762j − 0.127k
Now F = |F|eBC = 800 eBC
x
F = −228.6i − 685.7j + 342.9k N
FP = (F · eBA )eBA
FP = (624.1)eBA
FP = −396.3i − 475.6j − 79.3k N
C (3, 0, 4) m
z
Problem 2.116 The force F = 21i + 14j (kN). Resolve it into vector components parallel and normal to the
line OA.
y
F
O
x
z
Solution:
A (6, – 2, 3) m
The position vector of point A is
y
rA = 6i − 2j + 3k
√
The magnitude is |rA | = 62 + 22 + 32 = 7. The unit vector
rA
parallel to OA is eOA = |r | = 67 i − 27 j + 37 k
A
(a) The component of F parallel to OA is
1
(F · eOA ) eOA = ((3)(6) + (−2)(2))
(6i − 2j + 3k)
7
F
x
FP = 12i − 4j + 6k (kN)
(b)
The component of F normal to OA is
FN
z
= F − Fp = (21 − 12)i + (14 − (−4))j + (0 − 6)k
A
(6, – 2, 3) m
= 9i + 18j − 6k (kN)
Problem 2.117 At the instant shown, the Harrier’s
thrust vector is T = 3800i + 15,300j − 1800k (lb), and
its velocity vector is v = 24i + 6j − 2k (ft/s). Resolve T
into vector components parallel and normal to v. (These
are the components of the airplane’s thrust parallel and
normal to the direction of its motion.)
y
v
T
x
Solution:
|v| =
The magnitude of the velocity vector is given by
vx2 + vy2 + vz2 = 242 + 62 + (−2)2 .
y
Thus, |v| = 24.8 ft/s. The components of the unit vector in the
direction of the velocity vector are given by
v
vx
vy
vz
ex =
, ey =
, and ez =
.
|v|
|v|
|v|
Substituting numerical values, we get ex = 0.967, ey = 0.242, and
ez = −0.0806. The dot product of T and this unit vector gives the
component of T parallel to the velocity. The resulting equation is
Tparallel = Tx ex + Ty ey + Tz ez . Substituting numerical values,
we get Tparallel = 7232.12 lb. The magnitude of the vector T is
15870 lb. Using the Pythagorean Theorem, we get
Tnormal =
|T |2 − (Tparallel )2 = 14130 lb.
T
x
Problem 2.118 Cables extend from A to B and from
A to C. The cable AC exerts a 1000-lb force F at A.
(a) What is the angle between the cables AB and AC?
(b) Determine the vector component of F parallel to the
cable AB.
y
(0, 7, 0) ft
A
F
x
B
C
(14, 0, 14) ft
(0, 0, 10) ft
z
Solution:
(a)
Use Eq. (2.24) to solve.
From the coordinates of the points, the position vectors are:
y
rAB = (0 − 0)i + (0 − 7)j + (10 − 0)k
rAB = 0i − 7j + 10k
A
(0, 7, 0) ft
rAC = (14 − 0)i + (0 − 7)j + (14 − 0)k
rAC = 14i − 7j + 14k
The magnitudes are:
|rAB | = 72 + 102 = 12.2 (ft) and
|rAB | = 142 + 72 + 142 = 21.
x
B
The dot product is given by
C
rAB · rAC = (14)(0) + (−7)(−7) + (10)(14) = 189.
The angle is given by
cos θ =
(b)
(0, 0, 10) ft
z
(14, 0, 14) ft
189
= 0.7377,
(12.2)(21)
from which θ = ±42.5◦ . From the construction: θ = +42.5◦
The unit vector associated with AB is
eAB =
rAB
= 0i − 0.5738j + 0.8197k.
|rAB |
The unit vector associated with AC is
rAC
eAC =
= 0.6667i − 0.3333j + 0.6667k.
|rAC |
Thus the force vector along AC is
FAC = |F|eAC = 666.7i − 333.3j + 666.7k.
The component of this force parallel to AB is
(FAC · eAB )eAB = (737.5)eAB = 0i − 423.2j + 604.5k (lb)
Problem 2.119 Consider the cables AB and AC Solution: From Problem 2.100, rAB = 0i − 7j + 10k, and eAC =
shown in Problem 2.118. Let rAB be the position vector 0.6667i−0.3333j+0.6667k. Thus rAB ·eAC = 9, and (rAB ·eAC )eAC =
from point A to point B. Determine the vector compo- 6i − 3j + 6k
nent of rAB parallel to the cable AC.
Problem 2.120 The force F = 10i + 12j − 6k (N).
Determine the vector components of F parallel and normal to line OA.
y
A
(0, 6, 4) m
F
O
Solution:
Find eOA =
Then
x
rOA
|rOA |
z
FP = (F · eOA )eOA
y
and FN = F − FP
eOA =
0i + 6j + 4k
6j + 4k
√
= √
52
62 + 4 2
eOA =
6
4
j+
k = 0.832j + 0.555k
7.21
7.21
A
(0, 6, 4) m
FP = [(10i + 12j − 6k) · (0.832j + 0.555k)]eOA
F
O
FP = [6.656]eOA = 0i + 5.54j + 3.69k (N)
FN = F − FP
x
FN = 10i + (12 − 5.54)j + (−6 − 3.69k)
FN = 10i + 6.46j − 9.69k N
z
Problem 2.121 The rope AB exerts a 50-N force T on
collar A. Determine the vector component of T parallel
to bar CD.
y
0.15 m
0.4 m
B
C
T
0.2 m
0.3 m
A
0.5 m
Solution: The vector from C to D is rCD = (xD − xC )i +
(yD − yC )j + (zD − zC )k. The magnitude of the vector
|rCD | = (xD − xC )2 + (yD − yC )2 + (zD − zC )2 .
O
D
0.2 m
z
The components of the unit vector along CD are given by uCDx =
(xD − xC )/|rCD |, uCDy = (yD − yC )/|rCD |, etc. Numerical
values are |rCD | = 0.439 m, uCDx = −0.456, uCDy = −0.684,
and uCDz = 0.570. The coordinates of point A are given by xA =
xC + |rCA |eCDx , yA = yC + |rCA |uCDy , 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 |rAB | = 0.458 m, uABx = −0.674,
uABy = 0.735, and uABz = 0.079. The force T is given by
T = |T|uAB . The result is T = −33.7i + 36.7j + 3.93k N.
The component of T parallel to CD is given
y
0.15 m
B
0.4 m
C
T
A
0.5 m
0.2 m
O
Tparallel = T • uCD = −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).
x
0.25 m
x
D
z
0.2 m
0.3 m
0.25 m
Problem 2.122 In Problem 2.121, determine the vector component of T normal to the bar CD.
Solution: From the solution of Problem 2.121, |T | = 50 N, and the component of T parallel to bar CD is Tparallel = −7.52 N. The component of T
normal to bar CD is given by
Tnormal =
Problem 2.123 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?
|T|2 − (Tparallel )2 = 49.4 N.
y
B
(0, 6, 0) ft
F
2 ft
A
x
8 ft
10 ft
z
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 = 8j + 2k. Let us find the component
of F parallel to this line.
The unit vector in the direction normal to the surface is
eN =
2
y
8
rN
8j + 2k
= √
= 0.970j + 0.243k
|rN |
82 + 2 2
The unit vector eAB can be found by
(xB − xA )i + (yB − yA )j + (zB − zA )h
eAB = (xB − xA )2 + (yB − yA )2 + (zB − zA )2
Point B is at (0, 6, 0) (ft) and A is at (5, 1, 4) (ft).
Substituting, we get
2
z
8
y
eAB = −0.615i + 0.615j − 0.492k
Now F = |F|eAB = (0.2)eAB
B
(0, 6, 0) ft
F = −0.123i + 0.123j − 0.0984k (lb)
F
The component of F normal to the surface is the component parallel
to the unit vector eN .
C
2 ft
FNORMAL = (F · eN )eN = (0.955)eN
A
FNORMAL = 0i + 0.0927j + 0.0232k lb
x
8 ft
10 ft
z
Problem 2.124 In Problem 2.123, what is the vector
component of F parallel to the surface?
Solution:
From the solution to Problem 2.123,
Thus
F = −0.123i + 0.123j − 0.0984k (lb) and
Fparallel = F − FNORMAL
FNORMAL = 0i + 0.0927j + 0.0232k (lb)
Substituting, we get
The component parallel to the surface and the component normal to
the surface add to give F(F = FNORMAL + Fparallel ).
Fparallel = −0.1231i + 0.0304j − 0.1216k lb
Problem 2.125 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 = 50i + 80j +
180k (m) and his velocity is v = −2.2j − 3.6k (m/s).
The position of the airlock is rA = −12i + 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 = (xA − xG )i + (yA −
yG )j + (zA − zG )k = (−12 − 50)i + (0 − 80)j + (20 − 180)k m.
The dot product between v and rGA is v • rGA = |v||rGA | cos θ =
vx xGA +vy yGA +vz zGA , where θ is the angle between v and rGA .
Substituting in the numerical values, we get θ = 19.7◦ .
y
G
A
z
x
Problem 2.126 In Problem 2.125, determine the vector component of the astronaut’s velocity parallel to the
line from his position to the airlock’s position.
The dot product v • rGA = vx xGA + vy yGA +
vz zGA = 752 (m/s)2 and the component of v parallel to GA is
vparallel = |v| cos θ where θ is defined as in Problem 2.125 above.
Solution:
vparallel = (4.22)(0.941) = 3.96 m/s
Problem 2.127 Point P is at longitude 30◦ W and latitude 45◦ N on the Atlantic Ocean between Nova Scotia
and France. (See Problem 2.90.) 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 = RE θ, where θ is in radian measure. From Eqs. (2.18)
and (2.24), the angular separation of P and Q is given by
P·Q
cos θ =
.
|P||Q|
The strategy is to determine the angle θ 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 = rOB + rBP . The vector rOB =
|rOB |(i cos λP + 0j + k sin λP ). From geometry, the magnitude
is |rOB | = RE cos θP . The vector rBP = |rBP |(0i + 1j + 0k).
From geometry, the magnitude is |rBP | = RE sin θP . Substitute
and reduce to obtain:
The dot product is
2
P · Q = RE
(cos(λP − λQ ) cos θP cos θQ + sin θP sin θQ )
Substitute:
cos θ =
P·Q
= cos(λP − λQ ) cos θP cos θQ + sin θP sin θQ
|P||Q|
Substitute λP = +30◦ , λQ = −60◦ , θp = +45◦ ,
θQ = +20◦ , to obtain cos θ = 0.2418, or θ = 1.326 radians.
Thus the distance is d = 1.326RE
y
P
P = rOB + rBP = RE (i cos λP cos θP + j sin θP + k sin λP cos θP ).
A similar argument for the point Q yields
Q = rOC + rCQ = RE (i cos λQ cos θQ + j sin θQ + k sin λQ cos θQ )
Using the identity cos2 β + sin2 β = 1, the magnitudes are
Problem 2.128 Determine the cross product U × V
of the vectors U = 8i − 6j + 4k and V = 3i + 7j + 9k.
Strategy: Sine the vectors are expressed in terms of
their components, you can use Eq. (2.34) to determine
their cross product.
Solution:
i
U × V = 8
3
θ
j
−6
7
k
4
9
= (−54 − 28)i + (12 − 72)j + (56 + 18)k
U × V = −82i − 60j + 74k
Q
RE
45°
b
30°
60°
x
|P| = |Q| = RE
N
G
20°
c
Problem 2.129 Two vectors U = 3i + 2j and V = Solution:
2i + 4j.
i
(a) What is the cross product U × V?
U × V = 3
2
(b) What is the cross product V × U?
Use Eq. (2.34) and expand into 2 by 2 determinants.
j k
2 0 = i((2)(0) − (4)(0)) − j((3)(0) − (2)(0))
4 0
i
V × U = 2
3
+ k((3)(4) − (2)(2)) = 8k
k
0 = i((4)(0) − (2)(0)) − j((2)(0) − (3)(0))
0
j
4
2
+ k((2)(2) − (3)(4)) = −8k
Problem 2.130 What is the cross product r × F of Solution: Use Eq. (2.34) and expand into 2 by 2 determinants.
the position vector r = 2i + 2j + 2k (m) and the force
i j
k F = 20i − 40k (N)?
r×F= 2 2
2 = i((2)(−40) − (0)(2)) − j((2)(−40)
20
0
−40 − (20)(2)) + k((2)(0) − (2)(20))
r × F = −80i + 120j − 40k (N-m)
Problem 2.131
Determine the cross product r × F of Solution:
the position vector r = 4i − 12j + 3k (m) and the force r × F = 4i
16
F = 16i − 22j − 10k (N).
−12
−22
j
k
3
−10 r × F = (120 − (−66))i + (48 − (−40))j
+ (−88 − (−192))k (N-m)
r × F = 186i + 88j + 104k (N-m)
Problem 2.132 Consider the vectors U = 6i−2j−3k
and V = −12i + 4j + 6k.
(a) Determine the cross product U × V.
(b) What can you conclude about U and V from the
result of (a)?
Solution:
For (a) Use Eq. (2.34) and expand into 2 by 2 determi-
nants.
i
U×V =
6
−12
j
−2
4
k
−3 6
= i((−2)(6) − (4)(−3)) + j((6)(6)
− (−12)(−2)) + k((6)(4) − (−12)(−2))
U × V = 0i + 0j + 0k
(b) From the definition of the cross product (see Eq. (2.28)) U × V =
|U||V| sin θe, where θ is the angle between the two vectors, and e is a unit
vector perpendicular to both U and V. If U × V = 0 and if |U| = 0 and
|V| = 0 then since by definition e = 0, sin θ must be zero: sin θ = 0, and
θ = 0◦ or θ = 180◦ , and the two vectors are said to be parallel. (A graphical
construction confirms this interpretation.)
Problem 2.133 The cross product of two vectors U
and V is U × V = −30i + 40k. The vector V =
4i − 2j + 3k. Determine the components of U.
Solution:
We know
i
j
Uy
U × V = Ux
4
−2
k
Uz 3
U × V = (3Uy + 2Uz )i + (4Uz − 3Ux )j + (−2Ux − 4Uy )k (1)
We also know
U × V = −30i + 0j + 40k (2)
Equating components of (1) and (2), we get
3Uy + 2Uz = −30
4Uz − 3Ux = 0
−2Ux − 4Uy = 40
Setting Ux = 4 and solving, we get
U = 4i − 12j + 3k
Problem 2.134 The magnitudes |U| = 10 and |V| =
20.
(a) Use the definition of the cross product to determine
U × V.
(b) Use the definition of the cross product to determine
V × U.
(c) Use Eq. (2.34) to determine U × V.
(d) Use Eq. (2.34) to determine V × U.
y
V
U
30°
45°
x
Solution: From Eq. (228) U × V = |U||V| sin θe. From the
sketch, the positive z-axis is out of the paper. For U × V, e = −1k
(points into the paper); for V × U, e = +1k (points out of the paper).
The angle θ = 15◦ , hence (a) U × V = (10)(20)(0.2588)(e) =
51.8e = −51.8k. Similarly, (b) V × U = 51.8e = 51.8k (c) The
two vectors are:
U = 10(i cos 45◦ + j sin 45) = 7.07i + 0.707j,
V = 20(i cos 30◦ + j sin 30◦ ) = 17.32i + 10j
i
U × V = 7.07
17.32
j
k
7.07 0 = i(0) − j(0) + k(70.7 − 122.45)
10 0 = −k51.8
i
(d) V × U = 17.32
7.07
j
10
7.07
k
0 = i(0) − j(0) + k(122.45 − 70.7)
0
= 51.8k
y
V
U
45°
30°
x
Problem 2.135 The force F = 10i − 4j (N). Determine the cross product rAB × F.
y
(6, 3, 0) m
A
rA B
x
z
(6, 0, 4) m
B
F
Solution:
The position vector is
y
rAB = (6 − 6)i + (0 − 3)j + (4 − 0)k = 0i − 3j + 4k
A (6, 3, 0)
The cross product:
i
j k
rAB × F = 0 −3 4 = i(16) − j(−40) + k(30)
10 −4 0 rA B
x
= 16i + 40j + 30k (N-m)
z
Problem 2.136 By evaluating the cross product U ×
V, prove the identity sin(θ1 − θ2 ) = sin θ1 cos θ2 −
cos θ1 sin θ2 .
F
B (6, 0, 4)
y
U
V
θ1
θ2
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 × V =
|U||V| sin(θ1 − θ2 )e. Since the positive z-axis is out of the paper, and e points into the paper, then e = −k. Take the dot product
of both sides with e, and note that k · k = 1. Thus
(U × V) · k
sin(θ1 − θ2 ) = −
|U||V|
x
Solution:
The vectors are:
U = |U|(i cos θ1 + j sin θ2 ), and V = |V|(i cos θ2 + j sin θ2 ).
The cross product is
i
U × V = |U| cos θ1
|V| cos θ
2
j
|U| sin θ1
|V| sin θ2
k
0
0
= i(0) − j(0) + k(|U||V|)(cos θ1 sin θ2 − cos θ2 sin θ1 )
Substitute into the definition to obtain: sin(θ1 −θ2 ) = sin θ1 cos θ2 −
cos θ1 sin θ2 . Q.E.D.
y
U
V
θ1
θ2
x
Problem 2.137 Use the cross product to determine the
components of a unit vector e that is normal to both of
the vectors U = 8i − 6j + 4k and V = 3i + 7j + 9k.
First, find U × V = R
i
j k
R = U × V = 8
−6
4
3
7 9
Solution:
R = (−54 − 28)i + (12 − 72)j + (56 − (−18)) k
R = −82i − 60j + 74k
R
−82i − 60j + 74k
eR = ±
=±
|R|
125.7
er = ±(−0.652i − 0.477j + 0.589k)
Problem 2.138 (a) What is the cross product rOA ×
rOB ? (b) Determine a unit vector e that is perpendicular
to rOA and rOB .
y
B ( 4, 4, –4) m
rOB
O
x
rOA
A (6, –2, 3) m
z
Solution:
The two radius vectors are
rOB = 4i + 4j − 4k, rOA = 6i − 2j + 3k
The cross product is
i
j
rOA × rOB = 6 −2
4 4
y
B ( 4, 4, –4)
(a)
k 3 −4 rOB
= i(8 − 12) − j(−24 − 12)
O
+ k(24 + 8)
= −4i + 36j + 32k (m2 )
The magnitude is
|rOA × rOB | = 42 + 362 + 322 = 48.33 m2
(b)
The unit vector is
rOA × rOB
e=±
= ±(−0.0828i + 0.7448j + 0.6621k)
|rOA × rOB |
(Two vectors.)
z
x
rOA
A(6, –2, 3)
Problem 2.139 For the points O, A, and B in Problem 2.138, 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:
rOA = 6i − 2j + 3k (m)
rOB = 4i + 4j − 4k (m)
rOA × rOB = C
(C is ⊥ to both rOA and rOB )
i
j
k
(+8 − 12)i
C = 6
−2
3 = +(12 + 24)j
4
4
−4 +(24 + 8)k
(The magnitude of C is 338.3)
We now want to find the length of the projection, P , of line OB in direction ec .
P = rOB · eC
= (4i + 4j − 4k) · eC
P = 6.90 m
y
B ( 4, 4, –4) m
C = −4i + 36j + 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 × rOA
is such a line. We then need to find the component of rOB in this
direction and compute its magnitude.
i
j
k
C × rOA = −4
+36
32 6
−2
3
rOB
O
x
rOA
C = 172i + 204j − 208k
The unit vector in the direction of C is
eC
P
A(6, –2, 3) m
z
C
=
= 0.508i + 0.603j − 0.614k
|C|
Problem 2.140 The cable BC exerts a 1000-lb force
F on the hook at B. Determine rAB × F.
y
B
F
6 ft
rAB
x
8 ft
C
rAC
4 ft
4 ft
A
12 ft
z
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
y
rOA = 16i + 0j + 12k, rOB = 4i + 6j + 0k, rOC = 4i + 0j + 8k.
B
The force F acts along the unit vector
eBC =
rBC
rOC − rOB
rAB
=
=
|rBC |
|rOC − rOB |
|rAB |
6 ft
x
Noting rOC −rOB
√ = (4−4)i+(0−6)j+(8−0)k = 0i−6j+8k
|rOC − rOB | = 62 + 82 = 10. Thus
eBC = 0i − 0.6j + 0.8k, and F = |F|eBC = 0i − 600j + 800k (lb).
rAB = (4 − 16)i + (6 − 0)j + (0 − 12)k = −12i + 6j − 12k
Thus the cross product is
i
j
rAB × F = −12
6
0
−600
8 ft
C
4 ft
4 ft
The vector
k −12 = −2400i + 9600j + 7200k (ft-lb)
800 r
12 ft
A
Problem 2.141
The cable BC shown in Problem 2.140 exerts a 300-lb force F on the hook at B.
(a) Determine rAB × F and rAC × F.
(b) Use the definition of the cross product to explain
why the result of (a) are equal.
Solution:
(a) From Problem 2.140, the unit vector
eBC = 0i − 0.6j + 0.8k, and rAB = −12i + 6j − 12k
Thus F = |F|eBC = 0i − 180j + 240k, and the cross product is
i
j
k rAB × F = −12
6
−12 = −720i + 2880j + 2160k (ft-lb)
0
−180 240 The vector rAC = (4 − 16)i + 0j + (8 − 12)k = −12i + 0j − 4k.
Thus the cross product is
i
j
k rAC × F = −12
0
−4 = −720i + 2880j + 2160k (ft-lb)
0
−180 240 (b) The definition of the cross product is r × F = |r||F| sin θe.
Since the two cross products above are equal, |rAB ||F| sin θ1 e =
|rAC ||F| sin θ2 e. Note that rAC = rAB + rBC from Problem 2.116, hence rAC × F = rAB × F + rBC × F =
|rAB ||F| sin θ1 e + |rBC ||F| sin 0e = |rAB ||F| sin θ1 e, since
rBC and F are parallel. Thus the two results are equal.
Problem 2.142 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 × T.
y
0.15 m
0.4 m
B
C
T
0.3 m
0.2 m
A
0.5 m
O
x
0.25 m
D
0.2 m
z
Solution: The vector from C to D is rCD = (xD − xC )i +
(yD − yC )j + (zD − zC )k. The magnitude of the vector
|rCD | = (xD − xC )2 + (yD − yC )2 + (zD − zC )2 .
The components of the unit vector along CD are given by uCDx =
(xD − xC )/|rCD |, uCDy = (yD − yC )/|rCD |, etc. Numerical
values are |rCD | = 0.439 m, uCDx = −0.456, uCDy = −0.684,
and uCDz = 0.570. The coordinates of point A are given by xA =
xC + |rCA |uCDx , yA = yC + |rCA |uCDy , etc. The coordinates
of point A are (0.309, 0.162, 0.114) m. The vector rCA is given by
rCA = (xA −xC )i+(yA −yC )j+(zA −zC )k. The vector rCA is
rCA = (−0.091)i + (−0.137)j + (0.114)k 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 |rAB | = 0.458 m, uABx =
−0.674, uABy = 0.735, and uABz = 0.079. The force T is given
by T = |T|uAB . The result is T = −33.7i + 36.7j + 3.93k N.
The cross product rCA × T can now be calculated.
i
j
k rCA × T = −0.091 −0.138 0.114 −33.7
36.7
3.93 = (−4.65)i + (−3.53)j + (−7.98)k N-m
y
0.15 m
0.4 m
B
C
T
0.2 m
A
0.5 m
O
D
z
0.2 m
0.3 m
x
0.25 m
Problem 2.143 In Problem 2.142, let rCB be the position vector from point C to point B. Determine the
cross product rCB × T and compare your answer to the
answer to Problem 2.142.
Solution:
We need rCB and T in component form.
y
rCB = (xB − xC )i + (yB − yC )j + (zB − zC )k
0.15 m
where B is at (0, 0.5, 0.15) (m) and C is at (0.4, 0.3, 0) (m)
rCB = −0.4i + 0.2j + 0.15k (m)
0.4 m
We now need to find T . From Problem 2.142, its magnitude is 50 N.
We need a unit vector eAB to be able to write T as T = 50 eAB
and then perform the required cross product. We need the coordinates
of point A. Let us find eCA = eCD and use this plus the known
location of C to get the location of A. Point D is located at (0.2, 0,
0.25)
eCD = eCA =
B
T
A
O
z
xA = xC + dAC (eCDx )
yA = yC + dAC (eCDy )
zA = zC + dAC (eCDz )
Recall C is at (0.4, 0.3, 0)
Substituting, we find A is at (0.309, 0.163, 0.114).
We now need the unit vector from A to B.
rAB
(xB − xA )i + (yB − yA )j + (zB − zA )k
=
|rAB |
|rAB |
or
eAB = −0.674i + 0.735j + 0.078k
We now want T = |T|eAB = 50 eAB we get
T = 33.69i + 36.74j + 3.93k (N)
k
+0.15 3.93 rCB × T = −4.72i + 6.626j − 21.434k (N-m)
0.3 m
x
D
0.2 m
From the diagram, dAC = 0.2 m
we can now form rCB × T
i
j
+0.2
rCB × T = −0.4
33.69
36.74
0.2 m
0.5 m
rCD
=0
|rCD |
eCD = −0.456i − 0.684j + 0.570k
eAB =
C
0.25 m
Problem 2.144 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
(0, 3, 0) m
A
C
D
(0, 0, 3) m
(4, 0, 0) m
z
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:
y
rOA = 0i + 3j + 0k, rOD = 0i + 0j + 3k, and
rAD = (0 − 0)i + (0 − 3)j + (3 − 0)k = 0i − 3j + 3k, rAC
[4,0,0]
= (4 − 0)i + (0 − 3)j + (0 − 0)k = 4i − 3j + 0k.
The magnitude |R| = 19.21 (m). The unit vector is
eAB =
R
= 0.4685i + 0.6247j + 0.6247k.
|R|
Thus the vector collinear with AB is
rAB = 6eAB = +2.811i + 3.75j + 3.75k.
Using the coordinates of point A:
xB = 2.81 + 0 = 2.81 (m)
yB = 3.75 + 3 = 6.75 (m)
zB = 3.75 + 0 = 3.75 (m)
B
A
[0,3,0]
rOC = 4i + 0j + 0k. The vectors collinear with the bars are:
The vector collinear with rAB is
i
j k
R = rAD × rAC = 0 −3 3 = 9i + 12j + 12k
4 −3 0 (xB, yB, zB)
C
D
z
[0,0,3]
x
x
Problem 2.145
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
A
(3, 0, 0) m
C
x
(0, 0, 4) m
z
Solution: The strategy is to find the unit vector perpendicular to
the plane. The projection of this unit vector on the vector OP : rOP ·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 · e, rOB · e, or
rOC · e) is the distance from the origin to the plane along this same
perpendicular. Thus the distance of P from the plane is
y
P[9,6,5]
B[0,5,0]
d = rOP · e − rOA · e.
The position vectors are: rOA = 3i, rOB = 5j, rOC = 4k and
rOP = 9i + 6j + 5k. The unit vector perpendicular to the plane is
found from the cross product of any two vectors lying in the plane.
Noting: rBC = rOC − rOB = −5j + 4k, and rBA = rOA −
rOB = 3i − 5j. The cross product:
i
j k
rBC × rBA = 0 −5 4 = 20i + 12j + 15k.
3 −5 0 The magnitude is |rBC × rBA | = 27.73, thus the unit vector is
e = 0.7212i + 0.4327j + 0.5409k. The distance of point P from
the plane is d = rOP ·e−rOA ·e = 11.792−2.164 = 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 .
Problem 2.146 Consider vectors U = 3i − 10j, V =
−6j + 2k, and W = 2i + 6j − 4k.
(a) Determine the value of the mixed triple product U ·
(V × W) by first evaluating the cross product V ×
W and then taking the dot product of the result with
the vector U.
(b) Determine the value of the mixed triple product U ·
(V × W) by using Eq. (2.36).
Solution:
i
V × W = 0
2
(a) The cross product
j
k
−6
2 = (+24 − 12)i − (0 − 4)j + (0 + 12)k
6
−4 = 12i + 4j + 12k
Take the dot product: U·(V×W) = (3)(12)+(4)(−10)+0 = −4
(b) Eq. (2.36) expresses the mixed triple product as a 3X3 determinant.
3 −10 0 U · (V × W) = 0 −6
2 = (3)(24 − 12) − (−10)(−4) + (0)
2
6
−4 = 36 − 40 = −4
x
O
A[3,0,0]
z
C[0,0,4]
Problem 2.147 For the vectors U = 6i + 2j − 4k,
V = 2i + 7j, and W = 3i + 2k, evaluate the following
mixed triple products: (a) U·(V×W); (b) W·(V×U);
(c) V · (W × U).
Solution:
Use Eq. (2.36).
6 2
−4 (a) U · (V × W) = 2 7
0
3 0
2
= 6(14) − 2(4) + (−4)(−21) = 160
3
(b) W · (V × U) = 2
6
0
7
2
2
0
−4 = 3(−28) − (0) + 2(4 − 42) = −160
2 7
0
(c) V · (W × U) = 3 0
2
6 2
−4 = 2(−4) − 7(−12 − 12) + (0) = 160
Problem 2.148 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
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 = 200i mm, rOA = −60i + 90j + 30k mm, and
rOC = −40i + 0j + 100k mm. The mixed triple product of the three
vectors is the volume of the parallelepiped.
The volume is
−60 90 30 rOA · (rOC × rOD ) = −40 0 100 200 0
0 Solution:
= −60(0) + 90(200)(100) + (30)(0) mm3
= 1,800,000 mm3
y
(140, 90, 30)
mm
E
A
B
F
O
D
G
C
z
(160, 0, 100)
mm
(200, 0, 0)
mm
x
Problem 2.149
that
By using Eqs. (2.23) and (2.34), show
Ux
U · (V × W) = Vx
Wx
.
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 · V = UX VX + UY VY + UZ VZ . Eq. (2.34) is the determinant representation of the cross product:
i
j
k Eq. (2.34) U × V = UX UY UZ V
V
V
X
Y
Z
For notational convenience, write P = (U × V). Expand the determinant about its first row:
U
UX UZ UZ + k UX UZ P = i Y
−
j
VX VZ VX VZ VY VZ Since the two-by-two determinants are scalars, this can be written in the form:
P = iPX + jPY + kPZ where the scalars PX , PY , and PZ are the two-bytwo determinants. Apply Eq. (2.23) to the dot product of a vector Q with P.
Thus Q · P = QX PX + QY PY + QZ PZ . Substitute PX , PY , and PZ into
this dot product
U
U
U
UZ UZ UZ Q · P = QX Y
− QY X
+ Qz X
VY VZ
VX VZ
VX VZ But this expression can
rectly, thus:
QX
Q · (U × V) = UX
V
X
be collapsed into a three-by-three determinant diQY
UY
VY
QZ UZ . This completes the demonstration.
VZ Problem 2.150 The vectors U = i + UY j + 4k, V =
2i + j − 2k, and W = −3i + 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 · (V × W) = 0, for example.
1 UY
4 U · (V × W) = 2
1 −2 −3 1 −2 = 1(−2 + 2) − (UY )(−4 − 6) + (4)(2 + 3)
= +10UY + 20 = 0
Thus UY = −2
Problem 2.151 The magnitude of F is 8 kN. Express Solution: The unit vector collinear with the force F is developed as follows:
The collinear vector is √
r = (7 − 3)i + (2 − 7)j = 4i − 5j
F in terms of scalar components.
2
2
The magnitude: |r| = 4 + 5 = 6.403 m. The unit vector is
r
e = |r|
= 0.6247i − 0.7809j. The force vector is
y
(3, 7) m
F = |F|e = 4.997i − 6.247j = 5i − 6.25j (kN)
y
F
(3,7)m
F
(7, 2) m
x
(7,2)m
x
Problem 2.152 The magnitude of the vertical force W
is 600 lb, and the magnitude of the force B is 1500 lb.
Given that A + B + W = 0, determine the magnitude
of the force A and the angle α.
W
B
50°
α
A
Solution: The strategy is to use the condition of force balance to
determine the unknowns. The weight vector is W = −600j. The
vector B is
B = 1500(i cos 50◦ + j sin 50◦ ) = 964.2i + 1149.1j
The vector A is A = |A|(i cos(180 + α) + j sin(180 + α))
A = |A|(−i cos α − j sin α). The forces balance, hence A +
B + W = 0, or (964.2 − |A| cos α)i = 0, and (1149.1 − 600 −
|A| sin α)j = 0. Thus |A| cos α = 964.2, and |A| sin α = 549.1.
Take the ratio of the two equations to obtain tan α = 0.5695, or
B
50°
α
α = 29.7◦ . Substitute this angle to solve: |A| = 1110 lb
Problem 2.153
W
A
What are the direction cosines of F?
y
F = 20i + 10j – 10k (lb)
A
θ
(4, 4, 2) ft
B (8, 1, – 2) ft
x
z
Solution: Use the definition of the direction cosines and the ensuing discussion.
√
The magnitude of F: |F| = 202 + 102 + 102 = 24.5.
Fx
20
The direction cosines are cos θx = |F|
= 24.5
= 0.8165,
cos θy
y
F = 20i + 10j – 10k (lb)
A
Fy
10
=
=
= 0.4082
|F|
24.5
cos θz =
θ
(4, 4, 2) ft
Fz
−10
=
= −0.4082
|F|
24.5
B (8, 1, – 2) ft
x
z
Problem 2.154 Determine the scalar components of Solution: Use the definition of the unit vector, we get
a unit vector parallel to line AB that points from A to- The position vectors are: rA = 4i + 4j + 2k, rB = 8i + 1j − 2k. The vector
from A to B is rAB = (8√
− 4)i + (1 − 4)j + (−2 − 2)k = 4i − 3j − 4k.
ward B.
The magnitude: |rAB | = 42 + 32 + 42 = 6.4. The unit vector is
eAB =
rAB
3
4
4
i−
j−
k = 0.6247i − 0.4688j − 0.6247k
=
|rAB |
6.4
6.4
6.4
Problem 2.155 What is the angle θ between the line
AB and the force F?
Solution:
Use the definition of the dot product Eq. (2.18), and
Eq. (2.24):
cos θ =
rAB · F
.
|rAB ||F|
From the solution to Problem 2.130, the vector parallel to AB is
rAB = 4i − 3j − 4k, with a magnitude |rAB | = 6.4. From Problem 2.129, the force is F = 20i + 10j − 10k, with a magnitude of
|F| = 24.5. The dot product is rAB · F = (4)(20) + (−3)(10) +
90
(−4)(−10) = 90. Substituting, cos θ = (6.4)(24.5)
= 0.574,
θ = 55◦
Problem 2.156 Determine the vector component of F Solution: Use the definition in Eq. (2.26): UP = (e · U)e, where e is
parallel to a line L. From Problem 2.130 the unit vector parallel to line AB is
that is parallel to the line AB.
eAB = 0.6247i − 0.4688j − 0.6247k. The dot product is
e · F = (0.6247)(20) + (−0.4688)(10) + (−0.6247)(−10) = 14.053.
The parallel vector is
(e · F)e = (14.053)e = 8.78i − 6.59j − 8.78k (lb)
Problem 2.157 The magnitude of FB is 400 N and
|FA + FB | = 900 N. Determine the components of FA .
y
FB
FA
60°
40°
30°
x
50°
z
Solution:
Setting
|FB | = 400 N
900 N = |FA + FB |
= [(0.587FA − 100)2 + (0.643FA + 346)2
We need to write each vector in terms of its known or unknown components. From the diagram
+(0.492FA + 173)2 ]1/2
FAx = (|FA | cos 40◦ ) cos 40◦ = 0.587
and solving, we obtain FA = 595 N. Substituting this result into Eq. (1),
FAz = (|FA | cos 40◦ ) cos 50◦ = 0.492
FA = 349i + 382j + 293k (N).
FAy = |FA | sin 40◦ = 0.642
FBx = −(400 cos 60◦ ) cos 60◦
y
FBz = (400 cos 60◦ ) cos 30◦
FB
FBy = 400 sin 60◦
FA
Let FA = |FA | and FB = |FB | = 400 N.
The components of the vectors are
FA = FA cos 40◦ sin 50◦ i + FA sin 40◦ j + FA cos 40◦ cos 50◦ k
= FA (0.587i + 0.643j + 0.492k),
60°
(1)
30°
FB = −FB cos 60◦ sin 30◦ i + FB sin 60◦ j + FB cos 60◦ cos 30◦ k
= −100i + 346j + 173k (N).
40°
20°
z
50°
x
40°
Problem 2.158 Suppose that the forces FA and FB
shown in Problem 2.163 have the same magnitude and
FA · FB = 600 N2 . What are FA and FB ?
Solution:
From Problem 2.163, the forces are:
FA = |FA |(i cos 40◦ sin 50◦ + j sin 40◦ + k cos 40◦ cos 50◦ )
= |FA |(0.5868i + 0.6428j + 0.4924k)
FB = |FB |(−i cos 60◦ sin 30◦ + j sin 60◦ + k cos 60◦ cos 30◦ )
= |FB |(−0.25i + 0.866j + 0.433k)
The dot product: FA · FB = |FA ||FB |(0.6233) = 600 N2 , from
600
|FA | = |FB | =
= 31.03 N,
0.6233
and
FA = 18.21i + 19.95j + 15.28k (N)
FB = −7.76i + 26.87j + 13.44k (N)
Problem 2.159 The rope CE exerts a 500-N force T
on the door ABCD. Determine the vector component
of T in the direction parallel to the line from point A to
point B.
y
D
C
(0,0.2,0) m
E
(0.4,0.25,–0.1) m
A (0.5,0,0) m
x
T
B
(0.35,0,0.2) m
z
Two vectors are needed, rCE and rAB . The end
points of these vectors are given in the figure. Thus, rCE =
(xE − xC )i + (yE − yC )j + (zE − zC )k and a similar form holds
for rAB . Calculating these vectors, we get
Solution:
y
D
rCE = 0.4i + 0.05j − 0.1k m and rAB = −0.15i + 0j + 0.2k m.
The unit vector along CE is eCE = 0.963i + 0.120j − 0.241k
and the force T, is T = |T|eCE . Hence, T = 500(0.963i +
0.120j − 0.241k) = 482i + 60.2j − 120k N. The unit vector along
AB is given by eAB = −0.6i + 0j + 0.8k and the component of
T parallel to AB is given by TAB = T • eAB . Thus, TAB =
(482)(−0.6) + (60.2)(0) + (−120)(0.8) = −385.2 N
Problem 2.160 In Problem 2.169, let rBC be the position vector from point B to point C. Determine the
cross product rBC × T.
Solution:
The vector from B to C is
rBC = (xC − xB )i + (yC − yB )j + (zC − zB )k
= −0.35i + 0.2j − 0.2k m.
The vector T is T = 482i + 60.2j − 120k N. The cross product of
these vectors is given by
i
j
k rBC × T = −0.35 0.2 −0.2 = −12.0i − 138j − 117k N m
482
60.2 −120 C
(0,0.2,0) m
z
E (0.4,0.25,–0.1) m
T
A (0.5,0,0) m
x
B
(0.35,0,0.2) m
Problem 3.1 The figure shows the external forces acting on an object in equilibrium. The forces F1 = 32 N
and F3 = 50 N. Determine F2 and the angle α.
y
F1
30°
12°
Solution:
x
α
F3
Write the forces in component form.
F2
F1 = 32 sin 30◦ i + 32 cos 30◦ j
F1 = 16i + 27.7j N
F2 = −50 cos 12◦ i − 50 sin 12◦ j
y
F2 = −48.9i − 10.4j (N)
F1
F2 = F2 cos αi − F2 sin αj
30°
Sum components in x and y directions
F = 16 − 48.9 + F2 cos α = 0
x
Fy = 27.7 − 10.4 − F2 sin α = 0
12°
x
α
F3
Solving, we get
Fz = 37.2 N
α = 27.73◦
Problem 3.2 The force F1 = 100 N and the angle α =
60◦ . The weight of the ring is negligible. Determine the
forces F2 and F3 .
F2
y
F2
30°
x
F1
α
F3
Solution:
Write the forces in component form.
F1 = F1 i + 0j
F2 = −F2 cos 30◦ i + F2 sin 30◦ j
F3 = −F3 cos αi − F3 sin αj
We know
F = 0, thus
Fx = 0 and
Fy = 0. Writing the
equilibrium equations, we have
F
= F1 − F2 cos 30◦ − F3 cos α = 0
x
Fy = F2 sin 30◦ − F3 sin α = 0
F1 = 100 N, α = 60◦
y
F2
30°
x
F1
α
Solving, we get
F2 = 86.6 N, F3 = 50 N
F3
Problem 3.3 Consider the forces shown in Problem 3.2. Suppose that F2 = 100 N and you want to
choose the angle α so that the magnitude of F3 is a minimum. What is the resulting magnitude of F3 ?
Strategy: Draw a vector diagram of the sum of the
three forces.
Solution: |F2 | = 100 N, F1 is horizontal, and
From the diagram, α = 90◦ and |F3 | = 50 N
F = 0.
F1
α
F2
F3 min
60°
30°
x
Problem 3.4 The beam is in equilibrium. If Ax =
77 kN, B = 400 kN, and the beam’s weight is negligible,
what are the forces Ay and C?
Ax
Ay
B
30°
2m
C
4m
Solution:
 +

→
F
x
+
↑
Fy


Ax
Solving, we get
= Ax − C sin 30◦ = 0
Ax
cos 30◦
= Ay − B + C
=0
= 77 kN, B = 400 kN
B
Ay
30°
2m
Ay = 267 kN
C = 154 kN
4m
Problem 3.5 Suppose that the mass of the beam shown
in Problem 3.4 is 20 kg and it is in equilibrium. The force
Ay points upward. If Ay = 258 kN and B = 240 kN,
what are the forces Ax and C?
Solution:
 +

→
F
x
+
↑
Fy


Ay
= 0 = Ax − C sin 30◦ = 0
= 0 = Ay − B − (20)(9.81) + C cos 30◦ = 0
= 258 kN, B = 240 kN
Ax
C
Ay
30°
B
(20 kg) (9.81) m/s2
Solving, we get
Ax = 103 kN
C = 206 kN
C
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 = |T|(i cos 22◦ + j sin 22◦ ) = |T|(0.927i + 0.375j)
P = 800(i cos 270◦ + j sin 270◦ ) = 0i − 800j
M = |M|(i cos 144◦ + j sin 144◦ ) = |M|(−0.809i + 0.588j)
Apply the equilibrium conditions,
F=0=T+M+P=0
Collect like terms:
Fx = (0.927|T| − 0.809|M|)i = 0
Fy = (0.375|T| − 0.588|M| − 800)j = 0
Solve the first equation, |T| = 0.809
|M| = 0.873|M|
0.927
Substitute this value into the second equation, reduce algebraically,
and solve: |M| = 874 N, |T| = 763.3 N
22°
T
P
M
36°
Problem 3.7 The two springs are identical, with unstretched lengths 250 mm and spring constants k =
1200 N/m.
(a) Draw the free-body diagram of block A.
(b) Draw the free-body diagram of block B.
(c) 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
TU A = 0i +
N
1200
m
(0.3 m − 0.25 m)j = 0i + 60j N
A
Similarly, the tension in the lower spring acts on block A in the negative
Y direction
280 mm
TLA = 0i −
N
1200
m
(0.28 m − 0.25 m)j = 0i − 36j N
B
The weight is WA = 0i − |WA |j
The equilibrium conditions are
F=
Fx +
Fy = 0,
F = WA + TU A + TLA = 0
Tension,
upper spring
Collect and combine like terms in i, j
Fy = (−|WA | + 60 − 36)j = 0
A
Solve |WA | = (60 − 36) = 24 N
Tension,
lower
spring
The mass of A is
mA =
Weight,
mass A
|WL |
24 N
=
= 2.45 kg
|g|
9.81 m/s2
The free body diagram for block B is shown.
The tension in the lower spring TLB = 0i + 36j
y
The weight: WB = 0i − |WB |j
Apply the equilibrium conditions to block B.
B
x
F = WB + TLB = 0
Collect and combine like terms in i, j:
Fy = (−|WB | + 36)j = 0
Solve: |WB | = 36 N
The mass of B is given by mB =
|WB |
|g|
=
Tension,
lower spring
36 N
9.81 m/s2
= 3.67 kg
Weight,
mass B
Problem 3.8 The two springs in Problem 3.7 are identical, with unstretched lengths 250 mm and spring constants k. 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.
Problem 3.9 The 200-kg horizontal steel bar is suspended by the three springs. The stretch of each spring
is 0.1 m. The constant of spring B is kB = 8000 N/m.
Determine the constants kA = kC of springs A and C.
Solution:
δ = 0.1 m
KA = K C
+↑
Fy = KA δ + KB δ + KC δ − (200)(9.81) = 0
2KA (0.1) + (8000)(0.1) = 1962 N
Solving
KA = 5810 N/m = KC
A
KA δ
B
C
KB δ
KC δ
(200) (9.81) N
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 TU A − (mA + mB )g = 0,
where TU A = k(U − 0.25) N. The equation of equilibrium for block
B is TU B − mB g = 0, where TU B = k(L − 0.25) N. The equation of equilibrium for block A alone is TU A + TLA − mA g = 0 where
TLA = −TU B . Using g = 9.81 m/s2 , and solving simultaneously, we get
k = 1962 N/m, mA = 4 kg, and mB = 6 kg .
A
B
C
Problem 3.10 The mass of the crane is 20 Mg (megagrams), and the tension in its cable is 1 kN. The crane’s
cable is attached to a caisson whose mass is 400 kg. Determine the magnitudes of the normal and friction forces
exerted on the crane by the level ground.
Strategy: Draw the free-body diagram of the crane and
the part of its cable within the dashed line.
45°
Solution: Resolve the forces into scalar components, and solve
the equilibrium equations.
The external forces are the weight, the friction force, the normal force,
and the tension in the cable. The weight vector is
W = 0i − mc |g|j = 0i − (20000 kg)(9.81 m/s2 )j W = 0i − 196,200j
45°
The normal force vector is N = (0i + Ny + j). The friction force
by definition acts at right angles to the normal force, in a direction that
holds the crane in place.
y
Fx = −|Fx |i + 0j
F
45°
T
The angle between the tension vector and the positive x axis is −45◦ =
315◦ , hence the tension vector projection is
W
N
x
T = |T|(i cos 315◦ + j sin 315◦ ) = 707i − 707j N
Solve:
The equilibrium conditions are,
Fx = (−|Fx | + 707)i = 0,
|Fx | = 707 N
Fy = (|Ny | − 707 − 196200)j = 0
|Ny | = 196200 + 707 = 196,907 N.
Thus the friction force is directed toward the left, and the normal force
acts upward.
Problem 3.11 What is the tension in the horizontal
cable AB in Example 3.1 if the 20◦ angle is increased
to 25◦ ?
Solution:
A
m = 1440 kg
mg = (1440)9.81
Fx = T − N sin 25◦ = 0
Fy = N cos 25◦ − mg = 0
or
T − N sin 25◦ = 0
N cos 25◦ − 14.126 kN = 0
25°
y
Solving, we get
T
T = 6.59 kN, N = 15.59 kN
x
25°
mg
N
B
Problem 3.12 The 2400-lb car will remain in equilibrium on the sloping road only if the friction force exerted
on the car by the road is not greater than 0.6 times the
normal force. What is the largest angle α for which the
car will remain in equilibrium?
α
Solution:
y
Fx = W sin α − f = 0,
Fy = N − W cos α = 0.
Set f = 0.6 N and write the equilibrium equations as
α
W sin α = 0.6 N, (1)
W cos α = N.
W
(2)
f
Divide Eq. (1) by Eq. (2):
sin α
= tan α = 0.6.
cos α
Solving,
N
α = 31.0◦
x
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 =
2500 N/m and the stretch of the spring is 0.055 m. What
is the mass of the crate?
20°
20
Solution:
K = 2500 N/m
δ = 0.055 m
+
Fx = −Kδ + m(9.81) sin 20◦ = 0
+
Fy = N-m(9.81) cos 20◦ = 0
20°
−(2500)(0.055) + 3.355 m = 0
N − 9.218 m = 0
y
Solving, m = 41.0 kg, N = 378 (N)
Kδ
x
N
20°
mg = 9.81 m
Problem 3.14 A 600-lb box is held in place on the
smooth bed of the dump truck by the rope AB.
(a) If α = 25◦ , what is the tension in the rope?
(b) If the rope will safely support a tension of 400 lb,
what is the maximum allowable value of α?
B
A
α
Solution: Isolate the box. Resolve the forces into scalar components, and solve the equilibrium equations.
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 − α) (or 270 + α). The weight vector is
W = |W|(i sin α − j cos α) = (600)(i sin α − j cos α)
The projections of the rope tension and the normal force are
T = −|Tx |i + 0j
N = 0i + |Ny |j
The equilibrium conditions are
F=W+N+T=0
Substitute, and collect like terms
Fx = (600 sin α − |Tx |)i = 0
Fy = (−600 cos α + |Ny |)j = 0
Solve for the unknown tension when
α = 25◦ |Tx | = 600 sin α = 253.6 lb.
For a tension of 400 lb, (600 sin α−400) = 0. Solve for the unknown
angle
sin α =
400
= 0.667 or α = 41.84◦
600
A
B
α
y
T
x
N
W
α
Problem 3.15 Three forces act on the free-body diagram of a joint of a structure. If the structure is in equilibrium and FA = 4.20 kN, what are FB and FC ?
FC
FB
15°
40°
FA
Solution:

F

x
Fy

FA
= FA cos 40◦ − FB cos 15◦ = 0
= FA sin 40◦ − FB sin 15◦ − FC = 0
= 4.20 kN
Substitute in the value for FA and solve the resulting two equations in
two unknowns. We get
FB = 3.33 kN, FC = 1.84 kN
FC
FB
15°
40°
FA
y
FA
40°
x
15°
FB
FC
Problem 3.16 The weights of the two blocks are W1 =
200 lb and W2 = 50 lb. Neglecting friction, determine
the force the man must exert to hold the blocks in place?
30°
W1
30°
W2
Solution: Isolate block W2 and apply equilibrium conditions.
Repeat for block W1 .
T2
y
For W2 : The weight vector: W2 = 0i − 50j
The rope tension: T2 = 0i + |T2 |j
The equilibrium conditions are
Fx = 0,
x
W2
Fy = (−50 + |T2 |)j = 0, or |T2 = 50
For W1 : The magnitude of the rope tension |T2 | is unchanged by
passage over the frictionless lower pulley, hence,
y
T1
x
T2 = |T2 |i + 0j = 50i + 0j.
The rope tension T1 : T1 = −|T1 |i + 0j. The normal force is
N = 0i + |N|j. The angle between the x axis and the weight vector
is −(90 − α) (or 270 + α). The projection of the weight vector is
W = |W|(i sin α − j cos α) = 100i − 173.2j.
The equilibrium conditions are
F = T1 + T2 + N + W1 = 0
Substitute and collect like terms,
Fx = (−|T1 + 50 + 100)i = 0,
Fy = (|N − 173.2)j = 0
Solve: |T1 | = 150 lb. Since the frictionless pulley does not change
the magnitude of the rope tension, then the tension at the man’s hands is
|T1 | = 150 lb.
30°
W1
30°
W2
N
α
W1
T2
Problem 3.17 The two springs have the same unstretched length, and the inclined surface is smooth. Show
that the magnitudes of the forces exerted by the two
springs are
W sin α
,
F1 = 1 + kk21
W sin α
F2 = 1 + kk12
Solution: Isolate the block. Apply the linear spring forcedeflection relations to find the ratios of the spring forces. The spring
forces are,
F1 = −|F1 |i + 0j,
F2 = −|F2 |i + 0j.
The normal force is, N = 0i + |N|j. The angle between the x axis
and the weight vector is −(90 − α) (or 270 + α). The weight vector
is
W = |W|(i sin α − j cos α).
The equilibrium conditions are
F = W + N + F1 + F2 = 0.
Substitute and collect like terms,
Fx = (|W| sin α − |F1 | − |F2 |)i = 0,
Fy = (|N| − W cos α)j = 0.
For equal extensions, ∆L1 = ∆L2 = ∆L, the forces are |F1 | =
k1 ∆L, and |F2 | = k2 ∆L.
|F |
The ratio is, |F1 | = kk1 .
2
2
Substitute to eliminate the unknowns |F2 |, and |F1 |. Solve,
|F2 | =
|W| sin α
,
1 + kk1
|F1 | =
2
|W| sin α
1 + kk2
1
k1
k2
W
α
y
F1
F2
x
N
W
α
k1
k2
W
α
Problem 3.18 A 10-kg painting is suspended by a
wire. If α = 25◦ , what is the tension in the wire?
α
α
Solution: Isolate support pin fixed to the wall or other support.
The angle of the right hand wire with the positive x axis is −α, hence
the tension is
α
α
F2 = |F2 |(i cos α − j sin α)
The angle of the left hand wire is (180◦ + α) hence
F1 = |F1 |(−i cos α − j sin α).
The weight is W = 0i + |W|j
The equilibrium conditions are
y
W
F = W + F1 + F2 = 0
x
Substitute the vector forces, and collect like terms,
Fx = (|F2 | cos α − |F1 | cos α)i = 0,
Fy = (|W| − |F2 | sin α − |F1 | sin α)j = 0.
1
2
α
α
F2
With α = 25◦ and |W| = (10 kg) 9.81 sm2
Thus |F1 | = |F2 |, and
|F1 | = |F2 | =
F1
|W|
sin α
|F1 | = |F2 | =
.
1
2
98.1
0.423
= 116.06 N
Problem 3.19 If the wire supporting the suspended
painting in Problem 3.18 breaks when the tension exceeds 150 N and you want a 100 percent safety factor
(that is, you want the wire to be able to support twice the
actual weight of the painting), what is the smallest value
of α you can use?
Solution:
From Problem 3.18
|F1 | = |F2 | =
1
2
W
sin α
and |W| = (10 kg) 9.81
m
s2
y
W
= 98.1 N.
x
Thus
sin α =
1
2
98.1
|F|
For a tension |F| =
150
= 75,
2
sin α =
1
2
98.1
75
= 0.654 or α = 40.8◦
F1
α
α
F2
= 98.1 N
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?
14°
Solution:
Fx = TR cos(15◦ ) − TL cos(14◦ ) = 0
Fy = TR sin(15◦ ) + TL sin(14◦ ) − 150 = 0
Solving, we get TL = 299 lb, TR = 300 lb
14°
15°
y
14°
TR
TL
15°
x
150 lb
15°
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?
Solution:
Fx
= TR cos(15◦ ) − TR cos(14◦ ) = 0
Fy
= TR sin(15◦ ) + TR sin(14◦ ) − mg = 0
y
Solving, we get
TL = 1.56 kN,
TR = 1.57 kN
TR
TL
14°
15°
x
mg = (80) (9.81) N
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:

 Fx
Fy

mg
Eqns. of Equilibrium:
= T2 cos 30◦ − T1 sin 5◦ = 0
= T1 cos 5◦ − T2 sin 30◦ − mg = 0
= (180)(9.81) N
Solving, we get
T1 = 1867 N
T2 = 188 N
5°
5°
y
T1
30°
x
30°
T2
mg = (180) (9.81) N
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°
Solution Eqns. of Equilibrium

 Fx
Fy

mg
= T2 cos 30◦ − T1 sin 5◦ = 0
= T1 cos 5◦ − T2 sin 30◦ − mg = 0
= (180)(1.62) N
Solving, we get
T1 = 308 N T2 = 31.0 N
5°
30°
5°
y
T1
x
30°
T2
mg = (180) (1.62) N
Problem 3.24 A student on his summer job needs to
pull a crate across the floor. Pulling as shown in Fig. a, he
can exert a tension of 60 lb. He finds that the crate doesn’t
move, so he tries the arrangement in Fig. b, exerting a
vertical force of 60 lb on the rope. What is the magnitude
of the horizontal force he exerts on the crate in each case?
20°
(a)
10°
(b)
Solution:
(a)
The force diagram for part (a) is as shown. The horizontal component of the 60 lb is
20°
Fhoriz = (60) cos(20◦ ) = 56.4 lb .
(b)
(a)
The free body diagram for the point where the student’s hands
grasp the rope is shown to the right. The equations of equilibrium are
and
10°
Fx = Ffloor cos(10◦ ) − Fbox = 0,
(b)
Fy = 60 lb − Ffloor sin(10◦ ) = 0.
Solving these two equations simultaneously, we find that
y
Ffloor = 345.5 lb, and Fbox = 340.3 lb .
60 lb
Note: We should keep this problem in mind when we try to exert
a large force on an object. Here, the floor did most of the pulling
and the arrangement amplified the student’s effort by a factor
of almost six. Note that the angles are critical in this Problem.
Small changes can make big differences.
20°
x
y
60 lb
10°
Fbox
x
Ffloor
Problem 3.25 The 140-kg traffic light is suspended
above the street by two cables. What is the tension in
the cables?
20 m
20 m
B
C
12 m
A
Solution: Isolate the traffic light. From symmetry, the angles α
formed by the suspension cables are equal.
tan α =
12 m
20 m
α = 30.964◦ ∼
= 31◦
= 0.6,
The angle formed by cable C and the +x axis is α. The tension is
C = |C|(i cos α + j sin α).
The angle formed by cable B and the +x axis is (180◦ − α). The
tension is
B = |B|(i cos(180 − α) + j sin(180 − α)).
The weight is W = 0i − j|W|. The equilibrium conditions are
F = W + B + C = 0.
Substitute, and collect like terms. From the first equation, |B| = |C|.
Substitute this into the second equation
|B| = |C| =
1
2
|W|
sin α
.
For values of
m
s2
|W| = (140 kg) 9.81
= 1373.4 N
and α = 30.96 ∼
= 31◦ ,
1
2
|B| = |C| =
1373.4
sin α
20 m
20 m
12 m
C
B
A
y
B
= 1334.7 N.
α
α
W
x
C
Problem 3.26 Consider the suspended traffic light in
Problem 3.25. To raise the light temporarily during a
parade, an engineer wants to connect the 17-m length
of cable DE to the midpoints of cables AB and AC as
shown. However, for safety considerations, he doesn’t
want to subject any of the cables to a tension larger than
4 kN. Can he do it?
17 m
D
E
B
C
A
Determine the length of AC and AB from Problem 3.26: The distance between support poles is 40 m. The vertical
drop distance of the light is 12 m. Each triangle is a right triangle so
that the length of the cables is
Solution:
DAC = DAB =
(20)2 + (12)2 = 23.32 m.
17 m
Since the cable DE is attached to the midpoint of the cables AC and
AB, AD and AE are each half of this distance, or 11.66 m. The cable
DE is given to be 17 m. From this, the angle α is found:
cos α =
D
E
B
C
A
8.5 m
11.66 m
α = 43.2◦ .
= 0.729
Isolate the traffic light as shown. The angle formed by cable AD
and the positive x axis is (180◦ − α). The tension is: TAD =
|TAD |(i cos(180 − α) + j sin(180 − α)).
The angle formed by cable AE and the positive x axis is α, hence the
tension is TAE = |TAE |(i cos α + j sin α).
y
x
The weight is W = 0i − j|W|. The equilibrium conditions are
D
α
F = W + TAD + TAE = 0.
α
y
x
E
A
D
α
Substitute and collect like terms
W
Fx = (−|TAD | cos α + |TAE | cos α)i = 0,
Fy = (|TAD | sin α + |TAE | sin α − |W|)j = 0.
Solve, |TAD | = |TAE |. . , |TAD | = |TAE | =
For
|W| = (140 kg) 9.81
m
s2
and α = 43.2◦ , |TAD | = |TAE | =
1
2
2
1373.4
sin 43.2
|W|
sin α
= 1003 N.
Isolate the cable juncture E as shown. The angle θ is found as follows:
The cable EC is 11.66 m. The distance between poles 40 m. The cable
DE is 17 m, and cable DE is horizontal. Thus EC projects onto the
x-axis
HEC =
1
2
cos θ =
.
TAE = |TAE |(i cos(180 + α) + j sin(180 + α))
TAE = |TAE |(−i cos α − j sin α).
The tension in ED is: TED = −|TED |i + 0j.
The angle between CE and the positive x axis is θ.
The tension in CE is: TCE = |TCE |(i cos θ + j sin θ).
The equilibrium conditions are
(40 − 17) = 11.5 m.
11.5
11.66
A
Check: Use components: The angle between AE and the positive x axis is
(180◦ + α). The tension in AE is
F = TAE + TED + TCE = 0.
Substitute and collect like terms:
The ratio is the cosine of the angle,
= 0.9863,
◦
or θ = 9.5 .
From the law of sines:
|TAD |
|TCE |
=
sin θ
sin(180 − α)
from which |TCE | = 4159.6 N.
C
Thus the tension in the cable CE exceeds the allowable limit of 4 kN.
= 1373.4 N
1 θ
E
Fx = (−|TAE | cos α − |TED | + |TCE | cos θ)i = 0,
Fy = (−|TAE | sin α + |TCE | sin θ)j = 0.
From the second equation, |TCE | =
sin α
sin θ
|TAE |.
For α = 43.2◦ , θ = 9.5◦ , and |TAE | = 1003.2 N,
|TCE | =
0.6845
0.1651
(1003.2) = 4159.6 N
check .
Problem 3.27 The mass of the suspended crate is 5 kg.
What are the tensions in the cables AB and AC?
10 m
B
C
5m
7m
Solution:
Find the interior angles in the figure, then apply the
equilibrium conditions to the isolated crate. Given the triangle shown,
with known sides A, B, and C, find the unknown interior angles α, β,
and γ using the. law of cosines
A
B 2 = A2 + C 2 − 2AC cos β
Solve: cos β =
cos γ =
A2 + C 2 − B 2
. Similarly,
2AC
A2 + B 2 − C 2
.
2AB
10 m
27.66◦
For A = 10, B = 7, C = 5, γ =
third angle is
and β =
40.54◦ .
The
B
C
α = (180 − 27.66 − 40.64) = 111.8◦
5m
7m
A
Isolate the cable juncture at A. The angle between the positive x axis
and the tension TAC is γ. The tension is
TAC = |TAC |(i cos γ + j sin γ).
The angle between the positive x axis and the tension TAB is
(180◦ − β),
A
TAB = |TAB |(i cos(180 − β) + j sin(180 − β))
β
TAB = |TAB |(−i cos β + j sin β).
C
◦
◦
γ
α
B
The weight is W = 0i − |W|j.
The equilibrium conditions are
y
F = W + TAB + TAC = 0.
Fx = (|TAC | cos γ − |TAB | cos β)i = 0
Fy = (|TAC | sin γ + |TAB | sin β − |W|) = 0
Solve: |TAC | =
For
and
cos β
cos γ
|TAB |,
|TAB | =
|W| cos γ
sin(β + γ)
,
|TAC | =
|W| cos β
sin(β + γ)
.
|W| = (5 kg) 9.81
β = 40.54◦ ,
|TAB | = 46.79 N,
|TAC | = 40.15 N
C
γ
β
Substitute and collect like terms,
x
B
m
s2
= 49.05 N,
γ = 27.66◦
A
W
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 α, hence
TU R = |TU |(i cos α + j sin α).
45°
30°
The angle between the positive x and the left hand upper pulley is
(180◦ − β), hence
TU L = |TU |(i cos(180 − β) + j sin(180 − β))
W
= |TU |(−i cos β + j sin β).
The lower cable exerts a force: TL = −|TL |i + 0j
The weight:
W = 0i − |W|j
TU
The equilibrium conditions are
F = W + T U L + T U R + TL = 0
y
TL
W
Fx = (−|TU | cos β + |TU | cos α − |TL |)i = 0
Fy = (|TU | sin α + |TU | sin β − |W|)j = 0.
Solve:
|TU | =
|W|
(sin α + sin β)
,
|TL | = |TU |(cos α − cos β).
From which
For
α
β
Substitute and collect like terms,
TU
|TL | = |W|
α = 30◦ and β = 45◦
|TU | = 0.828|W|,
|TL | = 0.132|W|
cos α − cos β
sin α + sin β
.
x
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?
y
(10, 9) m
(3, 8) m
C
B
(6, 4.5) m
A
x
We need to find unit vectors eAB and eAC . Then write
TAB = TAB eAB and TAC = TAC eAC . Finally, write and solve
the equations of equilibrium.
Solution:
For the ring at A.
(10, 9) m
(3, 8) m
y
C
B
From the known locations of points A, B, and C,
eAB =
rAB
|rAB |
eAC =
rAB = −3i + 3.5j m
rAC
|rAC |
A
|rAB | = 4.61 m
(6, 4.5) m
rAC = 4i + 4.5j m |rAC | = 6.02 m
eAB = −0.651i + 0.759j
eAC = 0.664i + 0.747j
x
TAB = −0.651TAB i + 0.759TAB j
TAC = 0.664TAC i + 0.747TAC j
W = −mgj = −(100)(9.81)j N
y
TAB
For equilibrium,
TAC
TAB + TAC + W = 0
In component form, we have
Fx
= −0.651TAB + 0.664TAC = 0
Fy
= +0.759TAB + 0.747TAC − 981 = 0
Solving, we get
TAB = 658 N, TAC = 645 N
A (6, 4, 5)
B (3, 8)
C (10, 9)
x
A
mg = (100) (9.81) N
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:
1.5 m
Isolate the platform. The angles α and β are
B
Also,
tan α =
1.5
3.5
= 0.429,
α = 23.2◦ .
3.0 m
tan β =
3.0
3.5
= 0.857,
β = 40.6◦ .
1.5 m
A
D
C 3.5
m
The angle between the tether AB and the positive x axis is (180◦ −β),
4.0
m
hence
TAB = |TAB |(i cos(180◦ − β) + j sin(180◦ − β))
TAB = |TAB |(−i cos β + j sin β).
The angle between the tether AC and the positive x axis is (180◦ +α).
The tension is
TAC = |TAC |(i cos(180◦ + α) + j sin(180◦ + α))
y
B
x
β
α
C
= |TAC |(−i cos α − j sin α).
The tether AD is aligned with the positive x axis, TAD = |TAD |i +
0j.
The equilibrium condition:
F = TAD + TAB + TAC = 0.
Substitute and collect like terms,
Fx = (−|TAB | cos β − |TAC | cos α + |TAD |)i = 0,
Fy = (|TAB | sin β − |TAC | sin α)j = 0.
D
A
Solve:
|TAB | =
sin α
sin β
|TAD | =
|TAC | sin(α + β)
sin β
|TAC |,
.
For |TAC | = 2 N, α = 23.2◦ and β = 40.6◦ ,
|TAB | = 1.21 N, |TAD | = 2.76 N
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°
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
Fy

mg
= FFEET cos 4◦ − FBACK cos 3◦ − TROPE sin 10◦ = 0
= FFEET sin 4◦ +FBACK sin 3◦ +TROPE cos 10◦ − mg = 0
= (72)9.81 N, TROPE = 640 N
Solving, we get
FBACK = 559 N, FFEET = 671 N
10°
TROPE
10° y
FFEET
FBACK
3°
x
4°
mg = (72) (9.81) N
3°
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.
Eqns. of Equilibrium:
Fx
= T sin 20◦ + N cos 45◦ = 0
Fy
= N sin 45◦ + T cos 20◦ − mg = 0
(T = 200 N)
g = 9.81 m/s2
Substituting for T and g, we have two eqns in two unknowns
(N and m).
Solving, we get N = −96.7 N, m = 12.2 kg.
20°
200 Ν
A
45°
y
T
200 N
20°
x
N
45°
mg = (9.81) g
Problem 3.33 The unstretched length of the spring
AB is 660 mm, and the spring constant k = 1000 N/m.
What is the mass of the suspended object?
400 mm
600 mm
B
350 mm
A
Solution: Use the linear spring force-extension relation to find
the magnitude of the tension in spring AB. Isolate juncture A. The
forces are the weight and the tensions in the cables. The angles are
tan α =
350
600
= 0.5833, α = 30.26◦ .
tan β =
350
400
= 0.875, β = 41.2◦ .
400 mm
600 mm
B
350 mm
A
The angle between the x axis and the spring is α. The tension is
TAB = |TAB |(i cos α + j sin α).
C
B
β
The angle between the x axis and AC is (180 − β). The tension is
α
y
TAC = |TAC |(i cos(180 − β) + j sin(180 − β))
A
TAC = |TAC |(−i cos β + j sin β).
The weight is: W = 0i − |W|j.
The equilibrium conditions:
W
F = W + TAB + TAC = 0.
x
Substitute and collect like terms
The tension is |TAB | = k∆L = (1000)(0.03462) = 34.6 N.
Fx = (|TAB | cos α − |TAC | cos β)i = 0,
For α = 30.26◦ and β = 41.2◦ ; the weight is
Fy = (|TAC | sin α + |TAB | sin β − |W|)j = 0.
Solve: |TAC | =
cos α
cos β
|TAB | and |W| =
sin(α+β)
cos β
|TAB |.
The tension |TAB | is found from the linear spring force-deflection
relation. The spring extension is
∆L =
(350)2 + (600)2 − 660 = 694.62 − 660 = 34.62 mm
|W| =
0.948
0.752
The mass is m =
(34.6) = 43.62 N;
|W|
|g|
=
43.62
9.81
= 4.447 kg
Problem 3.34 The unstretched length of the spring in
Problem 3.33 is 660 mm. If the mass of the suspended
object is 10 kg and the system is in equilibrium in the
position shown, what is the spring constant?
First, find the distance AB to determine the stretched
length of the spring. Write unit vectors from A toward B and from A
toward C. Write the forces, in terms of these unit vectors. Then write
the equations of equilibrium and solve for the unknowns.
Solution:
From the diagram, A is at (0, 0), B is at (0.6, 0.35) m, and C is at
(−0.4, 0.35) m.
400 mm
600 mm
B
C
350 mm
eAB
=
eAC
=
T
AB
TAC
W
rAB
|rAB |
rAC
|rAC |
= 0.864i + 0.504j
= −0.753i + 0.659j
A
= 0.864TAB i + 0.504TAB j
= −0.753TAC i + 0.659TAC j
= 0i − 98.1j (N)
From our calculations
|rAB | = 0.695 m
y
C
B
••• the stretched length of the spring.
Thus, the stretch in the spring is given by
δ = |rAB | − unstretched
TAB
TAC
A
(0,0)
δ = 0.6946 − 0.6600 = 0.0346 (m)
mg = (10) (9.81) N
We know that
TAB = Kδ = 0.0346
The equilibrium equations are
Fx
= 0.864TAB − 0.753TAC = 0
Fy
= 0.504TAB + 0.659TAC − 98.1 = 0
Solving, we get
TAB = 77.88 N
TAC = 89.38 N
Finally solving for K, we get
K = 2250 N/m
x
Problem 3.35 The collar A slides on the smooth vertical bar. The masses mA = 20 kg and mB = 10 kg.
When h = 0.1 m, the spring is unstretched. When the
system is in equilibrium, h = 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 =
(0.25)2 + (0.1)2 and Ls =
Lu
0.1 m
0.25 m
(0.25)2 + (0.3)2 .
Ls
The stretch in the spring when in equilibrium is given by δ = Ls −Lu .
Carrying out the calculations, we get Lu = 0.269 m, Ls = 0.391 m,
and δ = 0.121 m. The angle, θ, between the rope at A and the
horizontal when the system is in equilibrium is given by tan θ =
0.3/0.25, or θ = 50.2◦ . From the free body diagram for mass A, we
get two equilibrium equations. They are
and
Lu
0.3 m
0.25 m
Fx = −NA + T cos θ = 0
T
Fy = T sin θ − mA g = 0.
NA
A
We have two equations in two unknowns and can solve. We get NA =
163.5 N and T = 255.4 N. Now we go to the free body diagram for
B, where the equation of equilibrium is T − mB g − kδ = 0. This
equation has only one unknown. Solving, we get k = 1297 N/m
mA g
0.25 m
T
h
B
A
B
k
mBg
Kδ
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 = σ. The “cost”
of your design is
√ the total volume of material in the two
wires, V = 2A b2 + h2 . Determine the value of h that
minimizes the cost.
b
b
h
W
Solution:
From the equation
Fy = 2T sin θ − W = 0,
W
2 sin θ
=
W
Since T /A = σ, A =
T
σ
=
we obtain T =
√
b2 +h2
.
2h
W
√
b2 +h2
2σh
√
and the “cost” is V = 2A b2 + h2 =
W (b2 +h2 )
.
σh
To determine the value of h that minimizes V , we set
dV
W
=
dh
σ
−
(b2 + h2 )
+2 =0
2
h
and solve for h, obtaining h = b.
T
T
θ
θ
W
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
C
E
45°
Solution: Isolate juncture A, and solve the equilibrium equations.
Repeat for the cable juncture C.
30°
A
The angle between the cable AC and the positive x axis is α. The
tension in AC is TAC = |TAC |(i cos α + j sin α)
The angle between the x axis and AB is (180◦ − β). The tension is
TAB = |TAB |(i cos(180 − β) + j sin(180 − β))
TAB = (−i cos β + j sin β).
The weight is W = 0i − |W|j.
Solve: |TCE | = |TCA | cos α,
The equilibrium conditions are
|TCD | = |TCA | sin α;
F = 0 = W + TAB + TAC = 0.
for |TCA | = 732 lb and α = 30◦ ,
Substitute and collect like terms,
|TAB | = 896.6 lb,
Fx = (|TAC | cos α − |TAB | cos β)i = 0
|TCE | = 634 lb,
Fy = (|TAB | sin β + |TAC | sin α − |W|)j = 0.
|TCD | = 366 lb
Solving, we get
|TAB | =
cos α
cos β
|TAC |
and
|TAC | =
|W| cos β
sin(α + β)
20 ft
,
B
|W| = 1000 lb, and α = 30◦ , β = 45◦
|TAC | = (1000)
|TAB | = (732)
D
C
0.7071
0.9659
0.866
0.7071
18 ft
= 732.05 lb
E
45°
30°
= 896.5 lb
A
Isolate juncture C. The angle between the positive x axis and the cable
CA is (180◦ − α). The tension is
TCA = |TCA |(i cos(180◦ + α) + j sin(180◦ + α)),
or TCA = |TCA |(−i cos α − j sin α).
The tension in the cable CE is
B
C
A
β
α
y
TCE = i|TCE | + 0j.
x
W
The tension in the cable CD is TCD = 0i + j|TCD |.
The equilibrium conditions are
F = 0 = TCA + TCE + TCD = 0
D
Substitute t and collect like terms,
Fx = (|TCE | − |TCA | cos α)i = 0,
C
Fy = (|TCD | − |TCA | sin α)j = 0.
α
90°
E
y
A
x
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.
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
AB =
18 ft
20 ft
D
B
C
α
β
20 ft
= 28.284 ft.
cos 45◦
A
The length AC is given by
AC =
18 ft
= 20.785 ft.
cos 30◦
The altitude of the triangle for which AC is the hypotenuse is
18 tan 30◦ = 10.392 ft. The distance CD is given by 20−10.392 =
9.608 ft.
38
B
D
β
α
β
20.784+9.608
= 30.392
α
28.284
The distance AD is given by
AD = AC + CD = 20.784 + 9.608 = 30.392
A
The new angles are given by the cosine law
B
AB 2 = 382 + AD2 − 2(38)(AD) cos α.
D
β
Reduce and solve:
cos α =
382 + (30.392)2 − (28.284)2
2(38)(30.392)
cos β =
(28.284)2 + (38)2 − (30.392)2
2(28.284)(38)
A
= 0.6787, α = 47.23◦ .
y
= 0.6142, β = 52.1◦ .
x
W
Isolate the juncture A. The angle between the cable AD and the
positive x axis is α. The tension is:
TAD = |TAD |(i cos α + j sin α).
α
Solve: |TAB | =
|TAD | =
cos α
cos β
|TAD |,
|W| cos β
sin(α + β)
.
The angle between x and the cable AB is (180◦ − β). The tension is
and
TAB = |TAB |(−i cos β + j sin β).
For |W| = 1000 lb, and α = 51.2◦ , β = 47.2◦
The weight is W = 0i − |W|j
The equilibrium conditions are
F = 0 = W + TAB + TAD = 0.
Substitute and collect like terms,
Fx = (|TAD | cos α − |TAB | cos β)i = 0,
Fy = (|TAB | sin β + |TAD | sin α − |W|)j = 0.
|TAD | = (1000)
0.6142
0.989
= 621.03 lb,
|TAB | = (622.3)
0.6787
0.6142
= 687.9 lb
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
30°
C
50°
B
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 α = 70◦ ; the tension in cable
AB is TAB = |TAB |(i cos α + j sin α). The weight is W =
0i − |W|j. The tension in cable AT is T = −|T|i + 0j. The
equilibrium conditions are
Solution:
A
70°
F = W + T + TAB = 0.
Substitute and collect like terms
Fx (|TAB | cos α − |T|)i = 0,
D
Fy = (|TAB | sin α − |W|)j = 0.
C
Solve: the tension in cable AB is |TAB | =
For |W| = (1250 kg) 9.81
|TAB | =
12262.5
0.94
m
s2
|W|
.
sin α
= 12262.5 N and α = 70◦
70°
A
T
= 13049.5 N
Isolate juncture B. The angles are α = 50◦ , β = 70◦ , and the tension
cable BC is TBC = |TBC |(i cos α + j sin α). The angle between
the cable BA and the positive x axis is (180 + β); the tension is
TBA = |TBA |(i cos(180 + β) + j sin(180 + β))
y
= |TBA |(−i cos β − j sin β)
B
x
The tension in the left horizontal cable is T = −|T|i + 0j. The
equilibrium conditions are
α
A
T
F = TBA + TBC + T = 0.
Substitute and collect like terms
W
Fx = (|TBC | cos α − |TBA | cos β − |T|)i = 0
y
Fy = (|TBC | sin α − |TBA | sin β)j = 0.
Solve: |TBC | =
sin β
sin α
|TBA |.
For |TBA | = 13049.5 N, and α =
|TBC | = (13049.5)
0.9397
0.7660
β=
70◦ ,
sin β
sin α
β
= 16007.6 N
A
y
D
x
|TCB |.
Substitute: |TCD | = (16007.6)
α
B
T
Isolate the cable juncture C. The angles are α = 30◦ , β = 50◦ . By
symmetry with the cable juncture B above, the tension in cable CD is
|TCD | =
C
x
50◦ ,
0.7660
0.5
= 24525.0 N.
50°
B
T
α
C
β
This completes the problem solution.
B
30°
Problem 3.40 A truck dealer wants to suspend a 4Mg (megagram) truck as shown for advertising. The
distance b = 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:
b = 15 m,
L = 25 m,
40 m
b
C
A
AB + BC = S = 42 m.
B
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 + d2 = AB 2 , L2 + d2 = BC 2 .
Subtract the first equation from the second to eliminate the unknown d,
L2 − b2 = BC 2 − AB 2 .
Note that BC 2 − AB 2 = (BC − AB)(BC + AB).
15 m
Substitute and reduce to the pair of simultaneous equations in the unknowns
BC − AB =
L2 − b2
S
,
25 m
b
A
L
β
C
α
BC + AB = S
B
Solve: BC =
1
2
L2 − b2
+S
S
=
1
2
252 − 152
+ 42
42
y
A
= 25.762 m
α
B
β
and AB = S − BC = 42 − 25.762 = 16.238 m.
C
W
The interior angles are found from the cosine law:
cos α =
(L + b)2 + BC 2 − AB 2
2(L + b)(BC)
cos β =
(L + b)2 + AB 2 − BC 2
2(L + b)(AB)
x
= 0.9704
α = 13.97◦
= 0.9238
◦
Substitute and collect like terms
β = 22.52
Isolate cable juncture B. The angle between BC and the positive x
axis is α; the tension is
TBC = |TBC |(i cos α + j sin α)
The angle between BA and the positive x axis is (180◦ − β); the
tension is
TBA = |TBA |(i cos(180 − β) + j sin(180 − β))
= |TBA |(−i cos β + j sin β).
Fx = (|TBC | cos α − |TBA | cos β)i = 0,
Fy = (|TBC | sin α + |TBA | sin β − |W|)j = 0
Solve:
|TBC | =
and |TBA | =
cos β
cos α
|W| cos α
sin(α + β)
and α = 13.97◦ , β = 22.52◦ ,
|TBA | = 64033 = 64 kN,
The equilibrium conditions are
|TBC | = 60953 = 61 kN
F = W + TBA + TBC = 0.
.
For |W| = (4000)(9.81) = 39240 N,
The weight is W = 0i − |W|j.
|TBA |,
Problem 3.41 The distance h = 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 α =
tan β =
8 in.
8
12
4
12
B
= 0.667
α = 33.7◦
12 in.
A
D
C
= 0.333
β = 18.4◦
12 in.
h
8 in.
The angle between the cable AB and the positive x axis is (180◦ −α),
the tension in AB is:
8 in.
12 in.
TAB = |TAB |(i cos(180 − α) + j sin(180 − α))
TAB = |TAB |(−i cos α + j sin α).
The angle between AC and the positive x axis is (180 + β). The
tension is
y
B
12 in
α
8 in
A
D
TAC = |TAC |(i cos(180 + β) + j sin(180 + β))
TAC = |TAC |(−i cos β − j sin β).
x
TAD = |TAD |i + 0j.
The equilibrium conditions are
F = TAC + TAB + TAD = 0.
Substitute and collect like terms,
Fx = (−|TAB | cos α − |TAC | cos β + |TAD |)i = 0
Fy = (|TAB | sin α − |TAC | sin β)j = 0.
Solve: |TAB | =
and |TAC | =
sin β
sin α
|TAC |,
sin α
sin(α + β)
|TAD |.
For |TAD | = 200 lb, α = 33.7◦ , β = 18.4◦
|TAC | = 140.6 lb,
4 in
C
The tension in the cable AD is
β
|TAB | = 80.1 lb
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 α, but you can choose
the angle β by placing point C wherever you wish. Show
that to minimize the tensions in cables AB and BC, you
must choose β = α if the angle α ≥ 45◦ .
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 α 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 β. By inspection, we see that the
minimum simultaneous values for TAC and TAB occur when the two
forces are equal. This occurs when α = β. Note: this does not happen
when α < 45◦ .
In this case, we solved the problem without writing the equations of
equilibrium. For reference, these equations are:
and
Fx = −TAB cos α + TAC cos β = 0
Fy = TAB sin α + TAC sin β − W = 0.
α
B
β
C
A
W
y
TAC
TAB
B
α
A
x
W
Possible locations
for C lie on line
C?
C?
B
α
TAB
Candidate β
W
Candidate values
for TAC
Fixed direction for
line AB
B
β
α
A
W
C
Problem 3.43 In Problem 3.42, suppose that you have
no control over the angle α and you want to design the
cable system so that the tension in cable AC is minimum.
What is the required angle β?
Solution: From Problem 3.32 above, the angle required to minimize the tension in cable AC, for large values of α is β = α. However,
for small values of ∝, the situation is different. In this situation, the
force triangle is as shown in the figure. It is obvious from the figure
that the minimum value for tension in cable AC is obtained when the
TAC is perpendicular to TAB .
Possible locations
for C lie on line
C?
B
C?
α
TAB
Candidate β
W
Fixed direction for
line AB
Candidate values
for TAC
Problem 3.44 The masses of the boxes on the left and
right are 25 kg and 40 kg, respectively. The surfaces are
smooth and the boxes are in equilibrium. Determine the
tension in the cable and the angle α.
Solution:
α
30°
α
30°
We now need to write the equilibrium equations for
each box.
For the left box,
Fx
= T − mL g sin α = 0
Fy
= NL − mL g cos α = 0
For the right box,
Fx
Fy mLg = (25) (9.81) N
= −T + mR g sin 30◦ = 0
= NR − mR g cos 30◦ = 0
We have four equations in the four unknowns T , NL , NR , and α.
(mL = 25 kg, mR = 40 kg). Solving, we get
NL = 147 N, NR = 340 N
T
y
α
mRg = 40 (9.81) N
x
30°
T
y′
T = 196.2 N, α = 53.1◦
NL
α
NR
30°
x′
Problem 3.45 Consider the system shown in Prob- Solution: Use the free body diagrams of Problem 3.44. The equations of
lem 3.44. The angle α = 45◦ , the surfaces are smooth, equilibrium are the same as for Problem 3.44.
and the boxes are in equilibrium. Determine the ratio of

T − mL g sin α = 0
the mass of the right box to the mass of the left box.




NL − mL g cos α = 0
−T + mR g sin 30◦ = 0
NR − mR g cos 30◦ = 0
where g = 9.81 m/s2 , mL = 1, α = 45◦ .
Solving, we get mR = 1.41.
∴ mR/mL = 1.41/1 = 1.41
Problem 3.46 The 3000-lb car and the 4600-lb tow
truck are stationary. The muddy surface on which the car
rests exerts a negligible friction force on the car. What
is the tension in the tow cable?
10°
18°
26°
Solution: From the geometry, the angle between the cable and
the x axis is 8◦ . From the free body diagram, the equations of equilibrium are
and
Fx = −T cos(8◦ ) + 3000 sin(26◦ ) = 0
Fy = N − 3000 cos(26◦ ) = 0.
The first equation can be solved for the tension in the cable. The tension
is T = 3000 sin(26◦ )/ cos(8◦ ) = 1328 lb.
18°
10°
26°
N
18°
T
y
3000 lb
25°
x
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) Draw the free-body diagram of the cylinder. (The
cylinder’s weight is negligible).
(b) Determine the magnitudes of the forces exerted by
the links AC and CD.
1m
D
C
Hydraulic
cylinder
1m
0.6 m
B
A
0.15 m
0.6 m
From the figure, if C is at the origin, then points A, B,
and D are located at
Scoop
Solution:
1m
D
A(0.15, −0.6)
B(0.75, −0.6)
D(1.00, 0.4)
C
Hydraulic
cylinder
1m
0.6 m
and forces FCA , FBC , and FCD are parallel to CA, BC, and CD,
respectively.
A
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
0.15 m
0.6 m
B
Scoop
y
eCA
rCA
=
= 0.243i − 0.970j
|rCA |
eCB =
eCD =
rCB
= 0.781i − 0.625j
|rCB |
rCD
= 0.928i + 0.371j
|rCD |
Now we write the forces in terms of magnitudes and unit vectors. We
can write FBC as FCB = −8eCB kN or as FCB = 8(−eCB )
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 = −6.25i + 5.00j kN
Similarly,
FCA = 0.243FCA i − 0.970FCA j
FCD = 0.928FCD i + 0.371FCD j
For equilibrium, FCA + FCB + FCD = 0
In component form, this gives
D
FCD
Fx
= 0.243FCA + 0.928FCD − 6.25 (kN) = 0
Fy
= −0.970FCA + 0.371FCD + 5.00 (kN) = 0
Solving, we get
FCA = 7.02 kN, FCD = 4.89 kN
C
x
FBC
FCA
A
B
Problem 3.48 The 50-lb cylinder rests on two smooth
surfaces.
(a) Draw the free-body diagram of the cylinder.
(b) If α = 30◦ , what are the magnitudes of the forces
exerted on the cylinder by the left and right surfaces?
α
45°
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 + β). The normal force is
45°
α
NR = |NR |(i cos(90 + β) + j sin(90 + β))
NR = |NR |(−i sin β + j cos β).
The angle between the positive x axis and the left hand force is normal
(90 − α); the normal force is NL = |NL |(i sin α + j cos α). The
weight is W = 0i − |W|j. The equilibrium conditions are
y
F = W + NR + NL = 0.
NL
Substitute and collect like terms,
β
α
W
NR
x
Fx = (−|NR | sin β + |NL | sin α)i = 0,
Fy = (|NR | cos β + |NL | cos α − |W|)j = 0.
Solve:
|NR | =
and |NL | =
sin α
sin β
|NL |,
|W| sin β
sin(α + β)
.
For |W| = 50 lb, and α = 30◦ , β = 45◦ , the normal forces are
|NL | = 36.6 lb,
|NR | = 25.9 lb
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 α 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).
|NR | =
sin α
sin β
|NL |
|NL | =
|W| sin β
sin(α + β)
y
β
α
.
NR
NL
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 = −|NR |i + 0j. The angle between
the left hand normal force and the positive x is 180 − (α + β). The
normal force is NL = |NL |(−i cos(α + β) + j sin(α + β)).
W
x
The angle between the weight vector and the positive x axis is −β.
The weight vector is W = |W|(i cos β − j sin β).
Substitute and collect like terms,
The equilibrium conditions are
Fx = (−|NR | − |NL | cos(α + β) + |W| cos β)i = 0,
Fy = (|NL | sin(α + β) − |W| sin β)j = 0.
F = W + NR + NL = 0.
Solve: |NL | =
|W| sin β
sin(α + β)
Problem 3.50 The 50-kg sphere is at rest on the
smooth horizontal surface. The horizontal force F =
500 N. What is the normal force exerted on the sphere
by the surface?
30°
F
Solution: Isolate the sphere and solve the equilibrium equations.
The angle between the cable and the positive x is (180 − α). The
tension:
30°
T = |T|(−i cos α + j sin α).
The other forces are
F = |F|i + 0j,
F
N = 0i + |N|j,
W = 0i − |W |j.
The equilibrium conditions are
F = T + F + N + W = 0.
T
α
Substitute and collect like terms,
Fx = (−|T| cos α + |F|)i = 0,
F
Fy = (|N| − |W| + |T| sin α)j = 0.
Solve: |T| =
|F|
cos α
, and |N| = |W| − |F| tan α.
W
N
For |W| = (50)(9.81) = 490.5 N, |F| = 500 N, and α = 30◦
|N| = 490.5 − (500)(0.577) = 201.8 N
Problem 3.51
Consider the stationary sphere in
Problem 3.50.
(a) Draw a graph of the normal force exerted on the
sphere by the surface as a function of the force F
from F = 0 to F = 1 kN.
(b) In the result of (a), notice that the normal force decreases to zero and becomes negative as F increases.
What does that mean?
Solution:
From the solution of Problem 3.50,
|N| = |W| − |F| tan α.
(a)
(b)
The commercial package TK Solver Plus was used to produce
the graph of the normal force vs. the applied force, for |W| =
(50)(9.81) = 490.5 N and α = 30◦ , as shown.
The normal force becomes negative when the cylinder is lifted
from the surface (it would take a negative force to keep it in
contact with the surface).
N
o
r
m
a
l
Normal Force vs Force
500
400
300
200
F 100
o
r
0
c
e −100
0
200
400
600
Force
800
1000
Problem 3.52 The 1440-kg car is moving at constant
speed on a road with the slope shown. The aerodynamic
forces on the car are the drag D = 530 N, which is
parallel to the road, and the lift L = 360 N, which is
perpendicular to the road. Determine the magnitudes of
the total normal and friction forces exerted on the car by
the road.
Solution:
From the free body diagram, the equations of equilib-
rium are
Fx = f − D − W sin 15◦ = 0
Fy = L + N − W cos 15◦ = 0
W = mg = (1440)(9.81) N
L = 360 N, D = 530 N
m = 1440 kg, g = 9.81 m/s2
Solving, we get
f = 4.19 kN, N = 13.29 kN
L
D
15°
y
L
x
F
D
15°
W
N
15°
L
D
15°
Problem 3.53 The device shown is towed beneath a
ship to measure water temperature and salinity. The mass
of the device is 130 kg. The angle α = 20◦ . The motion
of the water relative to the device causes a horizontal
drag force D. The hydrostatic pressure distribution in
the water exerts a vertical “buoyancy” force B. The
magnitude of the buoyancy force is equal to the product
of the volume of the device, V = 0.075 m3 , and the
weight density of the water, γ = 9500 N/m3 . Determine
the drag force D and the tension in the cable.
α
B
D
Solution: Calculate the magnitude of the buoyancy force. Draw
a free body diagram of the device. The drag, buoyancy and drag
forces are
y
T
D = |D|i + 0j,
B = 0i + |B|j,
W = 0i − j|W|.
The angle between the tow cable and the positive x axis is (90◦ + α);
the cable tension is
T = |T|(i cos(90 + α) + j sin(90 + α))
T = |T|(−i sin α + j cos α).
The equilibrium conditions are
F = W + B + T + D = 0.
Substitute and collect terms
Fx = (|D| − |T| sin α)i = 0
Fy = (|T| cos α + |B| − |W|)j = 0.
The magnitude of the buoyancy force is
B = ρV = (970)(0.15) = 145.5 N.
Solve: |D| = |T| sin α, and |T| =
|W|−|B|
cos α
.
For |W| = (130)(9.81) = 1275.3 N, and α = 20◦ , the tension in
the cable and the drag are |T| = 1202 N, |D| = 411.2 N
α
B
α
D
B
W
x
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 = mg and WA = mA g. The
equilibrium equations for the lower pulley, middle pulley, and upper
pulley are, respectively, A − 2T − W = 0, B − 2A − W = 0,
and C − 2B − W = 0. The equilibrium equation for the weight is
T + A + B − WA = 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 = 2T + W and B = 4T + 3W .
Substituting for A and B in the equilibrium equation for the weight,
we get 7T = WA − 4W = m
A g − 4mg. Thus, the tension, T, in
terms of masses and g is T = g7 (mA − 4m)
T
A
C
B
B
B
W
A
A
T
A
T
W
W
T
WA
B
C
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 = mg and WA = mA g. The
equilibrium equations for the weight A, the lower pulley, second pulley,
third pulley, and the top pulley are, respectively, B − WA = 0,
2C − B − W = 0, 2D − C − W = 0, 2T − D − W = 0, and
FS − 2T − W = 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
B = WA ,
D =
C=
WA
W
+
,
2
2
3W
WA
7W
WA
+
, and T =
+
,
4
4
8
8
or in terms of the masses,
g
(mA + 7m).
8
T =
T
A
Fs
W
T
W
D
D
D
C
W
C
C
B
B
WA
T
T
T
W
T
A
Problem 3.56 The system is in equilibrium. What are
the coordinates of A?
y
b
Solution: Determine from geometry the coordinates x, y. Isolate
the cable juncture A. Since the frictionless pulleys do not change the
magnitude of cable tension, and since each cable is loaded with the
same weight, arbitrarily set this weight to unity, |W| = 1. The angle
between the cable AB and the positive x axis is α; the tension in AB is
h
x
W
|TAB | = i cos α + j sin α.
(x, y)
A
The angle between AC and the positive x axis is (180◦ − β); the
tension is
W
W
TAC = |TAC |(−i cos β + j sin β).
The weight is |W| = 0i − j1. The equilibrium conditions are
y
F = TAB + TAC + W = 0.
b
Substitute and collect like terms,
Fx = (cos α − cos β)i = 0,
h
From the first equation cos α = cos β. On the realistic assumption
that both angles are in the same quadrant, then α = β. From the
second equation sin α = 12 or α = 30◦ . With the angles known,
geometry can be used to determine the coordinates x, y. The origin
of the x, y coordinate system is at the pulley B, so that the coordinate
x of the point A is positive. Define the positive distance ε as shown,
so that
ε
x
h+ε
b−x
Similarly,
x
W
Fy = (sin α + sin β − 1)j = 0.
A
W
W
y
B
= tan α.
C
= tan α.
A
β
α
Reduce to obtain
x = b − h cot α − ε cot α.
W
Substitute into the first equation to obtain
x=
1
2
x
(b − h cot α).
y
Multiply this equation by tan α and use ε = x tan α to obtain
ε=
tan α
2
b
1
2
(b − h cot α),
α
h
The sign of the coordinate y is determined as follows: Since the coordinate x is positive, the condition (b − h cot α) > 0 is required;
with this inequality satisfied (as it must be, or the problem is invalid),
ε is also positive, as required. But the angle α is in the first quadrant,
so that the point A is below the pulley B. Thus y = −ε and the
coordinates of the point A are:
x=
C
(b − h cot α).
1
y = − (b tan α − h),
2
α = 30◦
B
0
x
α
0
x
α
α
A
Problem 3.57 The light fixture of weight W is suspended from a circular arch by a large number N of equally
spaced cables. The tension T in each cable is the same.
Show that
πW
T =
.
2N
dθ
θ
Strategy: Consider an element of the arch defined by
an angle dθ measured from the point where the cables
join:
Since the total angle described by the arch is π radians,
the number of cables attached to the element is (N/pi)dθ.
You can use this result to write the equilibrium equations
for the part of the cable system where the cables join.
Solution: The angle between any cable and
πthe positive x axis is
kδθ, where k = 0, 1, 2, 3 . . . K, where K = δθ
is the number of
intervals, one less than the number of cables. The tension in the kth
cable is Tk = |Tk |(i cos kδθ + j sin kδθ). The weight is W =
0i − |W|j.
The equilibrium conditions are
Substitute into the solution for the tension



 

|W| sin 12 δθ
|W|
=
= |W| tan
|T| = 


K
cos 12 δθ


sin kδθ
k=0
F=W+
K
k=0
Tk = 0
If δθ =
where N is the number of cables.
|T| =
Substitute and collect like terms:
Fx =
K
k=0
Fy =
(|Tk | cos kδθ)i = 0
K
k=0
(|Tk | sin kδθ) − |W|
j=0
Since the tension in each cable is the same, |Tk | = |T|, the tension
can be removed from the sum, and the second equation solved for the
tension:






|W|
.
|T| = 


K


sin kδθ
k=0
The trigonometric sum can be found in handbooks1 :
K
sin kδθ =
sin
1
2
k=0
(N + 1)δθ sin
sin 12 δθ
1
2
N δθ
.
The angle is divided into K intervals over the arc, K =
Substitute into the sum to obtain
K
k=0
sin kδθ =
cos
1
δθ
2
sin 12 δθ
.
π
.
δθ
π
K
∼
=
|W|π
2N
π
N
1, tan
δθ
2
∼
=
δθ
,
2
therefore:
1
δθ .
2
Problem 3.58 The solution to Problem 3.57 is an
“asymptotic” result whose accuracy increases as N increases. Determine the exact tension Texact for N = 3,
5, 9, and 17, and confirm the numbers in the following
table. (For example, for N = 3, the cables are attached
at θ = 0, θ = 90◦ , and θ = 180◦ ).
N
3
5
9
17
Texact
πW/2N
1.91
1.32
1.14
1.07
Solution:
where N − 1 is the number of intervals in an arc of π radians. If the angle
increment δθ is sufficiently small,
From Problem 3.57, the tension is






|W|

|T| = 
 
K


sin kδθ
tan
k=01
sin(kδθ) =
sin
1
2
k=0
(K + 1)δθ sin
sin 12 δθ
1
2
Kδθ
N
where N is the number of cables. The angle is divided into K segments
over the interval, thus
K=
π
+ 1,
δθ
π
and N =
δθ
since the number of cables is one more than the number of intervals
Substitute this into the sum to obtain
K
sin(kδθ) =
k=0
cos
1
δθ
2
sin 12 δθ
.
Substitute this into the expression for the tension:



 

|W| sin 12 δθ
|W|
=
|T| = 
= |W| tan


N
cos 12 δθ


sin kδθ
k=0
Since δθ =
π
,
N −1
the exact solution for the tension in a cable is given by
|T| = |W| tan
δθ
|W|π ∼ |W|π
∼
, and |T| =
=
=
2
2(N − 1)
2N
is the asymptotic solution. The asymptotic solution, the exact solution, and the
ratio of the exact solution to the asymptotic solution for the two configurations
are given in the table below for 3, 5, 9, and 17 cables for the two configurations.
where the denominator is
K
δθ
2
π
)
2(N − 1
1
δθ .
2
3
5
7
9
17
π
2N
0.5235
0.3142
0.2244
0.1745
0.0924
tan
π
2(N −1)
1
0.4142
0.2679
0.1989
0.0985
Ratio
1.909
1.318
1.194
1.140
1.066
Problem 3.59 If the coordinates of point A in Example 3.5 are changed to (0, −2, 0) m, what are the tensions
in cables AB, AC, and AD?
Solution:
We need to write unit vectors eAB , eAC , and eAD .
eAB = 0.816i + 0.408j + 0.408k
eAC = −0.577i + 0.577j − 0.577k
eAD = −0.640i + 0.426j + 0.640k
We now need to write the four forces acting at point A.

T

 AB
TAC

 TAD
W
= 0.816TAB i + 0.408TAB j + 0.408TAB k
= −0.577TAC i + 0.577TAC j − 0.577TAC k
= −0.640TAD i + 0.426TAD j + 0.640TAD k
= −981j (N)
Equilibrium: TAB + TAC + TAD + W = 0


 Fx
Fy


Fz
= 0.816TAB − 0.577TAC − 0.640TAD = 0
= 0.408TAB + 0.577TAC + 0.426TAD − 981 = 0
= +0.408TAB − 0.577TAC + 0.640TAD = 0
Solving, we get
TAB = 848 N
TAC = 900 N
TAD = 271 N
C
B
TAC
TAB
D
TAD
A
W = mg j
= (100) (9.81) Nj
y
C
(Ð2, 0, Ð2) m
B
(Ð3, 0, 3) m
D
x
(4, 0, 2) m
z
A
(0, Ð2, 0) m
100 kg
Problem 3.60 The force F = 5i (kN) acts on point
A where the cables AB, AC, and AD are joined. What
are the tensions in the three cables?
Strategy: Isolate part of the cable system near point
A. See Example 3.5.
y
D (0, 6, 0) m
A
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 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:
y
D
(0, 6, 0) m
A
A(12, 4, 2),
B(6, 0, 0),
C(0, 4, 6),
D(0, 6, 0).
C
(0, 4, 6) m
The unit vector eAB is, by definition,
eAB =
=
rB − rA
(6 − 12)i + (0 − 4)j + (0 − 2)k
=
|rB − rA |
(6 − 12)2 + (4)2 + (2)2
−6
4
2
i−
j−
k
7.483
7.483
7.483
eAB = 0.8018i − 0.5345j − 0.267k.
Similarly, the other unit vectors are
eAB = −0.9487i + 0j + 0.3163k,
eAD = −0.9733i + 0.1622j − 0.1622k.
The tensions in the cables are expressed in terms of the unit vectors,
TAB = |TAB |eAB ,
TAC = |TAC |eAC ,
TAD = |TAD |eAD .
The external force acting on the juncture is, F = 5i + 0j + 0k. The
equilibrium conditions are
F = 0 = TAB + TAC + TAD + F = 0.
Substitute and collect like terms,
Fx = (−0.8018|TAB | − 0.9487|TAC | − 0.9733|TAD | + 5)i = 0
Fy = (−0.5345|TAB | + 0|TAC | − 0.1622|TAD |)j = 0
Fz = (−0.2673|TAB | − 0.3163|TAC | − 0.1622|TAD |)k = 0.
A hand held calculator was used to solve these simultaneous equations.
The results are:
|TAB | = 0.7795 kN,
|TAC | = 1.9765 kN,
|TAD | = 2.5688 kN.
F
(12, 4, 2) m
z
B
(6, 0, 0) m
x
Problem 3.61 The cables in Problem 3.60 will safely
support a tension of 25 kN. Based on this criterion, what
is the largest safe magnitude of the force F = F i?
Solution: This problem offers a new challenge. We need to be
able to solve the problem with one of the forces FAB , FAC , or FAD
equal to 25 kN and the other two forces must be smaller. Note that in
all of our earlier work, forces have appeared linearly in our equations
of equilibrium. This means that if we increase F by some factor, all
other forces increase by the same factor.
y
D (0, 6, 0) m
A
Plan of Attack: Assume F has a value of 1 kN and solve for all forces.
Find the largest force in the three cables and scale it up to 25 kN—
increasing all forces by the same scale factor.
We must write our forces in terms of unit vectors.
eAB =
rAB
,
|rAB |
eAC =
rAC
,
|rAC |
eAD =
rAD
|rAD |
where the points A, B, C, and D are
A: (12, 4, 2) m, B: (6, 0, 0) m
C: (0, 4, 6) m, D: (0, 6, 0) m
The unit vectors are
eAB
eAC
eAD
= −0.802i − 0.535j − 0.267k
= −0.949i + 0j + 0.316k
= −0.973i + 0.162j − 0.162k
The forces are

T

 AB
TAC

 TAD
F
= −0.802TAB i − 0.535TAB j − 0.267TAB k
= −0.949TAC i + 0j + 0.316TAC k
= −0.973TAD i + 0.162TAD j − 0.162TAD k
= Fi
Summing forces in the three coord. directions, we get


 Fx = −0.802TAB − 0.949TAC − 0.973TAD + F = 0
Fy = −0.535TAB + 0.162TAD = 0


Fz = −0.267TAB + 0.316TAC − 0.162TAD = 0
We set F = 1 and solve the three eqns in 3 unknowns.
Solving, we get
FAB = 0.155 kN, FAC = 0.395 kN
and FAD = 0.513 for F = 1 kN
Scaling, we want FAD → 25 kN we get F = 48.7 kNi. When
|FAD | = 25 kN
(12, 4, 2) m
C
B
(0, 4, 6) m
z
F
(6, 0, 0) m
x
Problem 3.62 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?
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, E are:
y
(0, 5, 0) ft
C
(0, 6, 6) ft
D
(5, 4, 3) ft (8, 4, 3) ft
A
B
x
A(5, 4, 3),
B(8, 4, 3),
C(0, 5, 0),
D(0, 6, 6),
E(3, 0, 3).
E
(3, 0, 3) ft
The vector locations of these points are,
z
rA = 5i + 4j + 3k,
rB = 8i + 4j + 3k,
rC = 0i + 5j + 0k,
rD = 0i + 6j + 6k,
y
rE = 3i + 0j + 3k.
The unit vector parallel to the tension acting between the points A, B
in the direction of B is by definition
eAB =
(0, 5, 0) ft
C
(0, 6, 6) ft
D
rB − rA
.
|rB − rA |
E
(3, 0, 3) ft
z
eAB = 1i + 0j + 0k
C
eAC = −0.8452i + 0.1690j − 0.5071k
A
B
eAD = −0.8111i + 0.3244j + 0.4867k
D
eAE = −0.4472i − 0.8944j + 0k
E
The tensions in the cables are,
TAB = |TAB |eAB = 40eAB ,
TAD = |TAD |eAD ,
TAC = |TAC |eAC ,
TAE = |TAE |eAE .
The equilibrium conditions are
F = 0 = TAB + TAC + TAD + TAE = 0.
Substitute the tensions,
Fx = (40 − 0.8452|TAC | − 0.8111|TAD | − 0.4472|TAE |)i = 0
Fy = (+0.1690|TAC | − 0.3244|TAD | − 0.8944|TAE |)j = 0
Fz = (−0.5071|TAC | − 0.4867|TAD |)k = 0.
This set of simultaneous equations in the unknown forces may be
solved using any of several standard algorithms.: The results are:
|TAE | = 11.7 lb,
A
B
x
Perform this operation for each unit vector. We get
(5, 4, 3) ft (8, 4, 3) ft
|TAC | = 20.6 lb,
|TAD | = 21.4 lb.
Problem 3.63 The bulldozer exerts a force F = 2i (kip)
at A. What are the tensions in cables AB, AC, and AD?
y
6 ft
C
8 ft
2 ft
B
A
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:
3 ft
D
z
4 ft
8 ft
x
A(8, 0, 0),
B(0, 3, 8),
C(0, 2, −6),
D(0, −4, 0).
The radius vectors for these points are
rA = 8i + 0j + 0k,
rB = 0i + 3j + 8k,
rC = 0i + 2j + 6k,
rD = 0i + 4j + 0k.
y
By definition, the unit vector parallel to the tension in cable AB is
6 ft
eAB
rB − rA
=
.
|rB − rA |
C
8 ft
B
Carrying out the operations for each of the cables, the results are:
z
2 ft
A
3 ft
D
4 ft
8 ft
eAB = −0.6835i + 0.2563j − 0.6835k,
eAC = −0.7845i + 0.1961j − 0.5883k,
eAD = −0.8944i + 0.4472j + 0k.
The tensions in the cables are expressed in terms of the unit vectors,
TAB = |TAB |eAB ,
TAC = |TAC |eAC ,
TAD = |TAD |eAD .
The external force acting on the juncture is F = 2000i + 0j + 0k.
The equilibrium conditions are
F = 0 = TAB + TAC + TAD + F = 0.
Substitute the vectors into the equilibrium conditions:
Fx = (−0.6835|TAB |− 0.7845|TAC |− 0.8944|TAD |+2000)i = 0
Fy = (0.2563|TAB | + 0.1961|TAC | − 0.4472|TAD |)j = 0
Fz = (0.6835|TAB | − 0.5883|TAC | + 0|TAD |)k = 0
The commercial program TK Solver Plus was used to solve these
equations. The results are
|TAB | = 780.31 lb ,
|TAC | = 906.9 lb ,
|TAD | = 844.74 lb .
x
Problem 3.64 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?
y
A (0, 8, 0) m
C (10,0, –12) m
Solution:
Fy = 1000 − (90)(9.81) − T = 0
D
(–16, 0, 4) m
x
B (16, 0, 16) m
z
T = 117.1 N
y
A(0, 8, 0)
B(16, 0, 16)
C(10, 0, −12)
D(−16, 0, 4)
We need to write unit vectors eAB , eAC , and eAD .
eAB = 0.667i − 0.333j + 0.667k
A (0, 8, 0) m
eAC = 0.570i − 0.456j − 0.684k
eAD = −0.873i − 0.436j + 0.218k
We now write the forces in terms of magnitudes and unit vectors

F

 AB
FAC

 FAD
T
C (10, 0, Ð12) m
D
(Ð16, 0, 4) m
z
x
B (16, 0, 16) m
= 0.667FAB i − 0.333FAB j + 0.667FAB k
= 0.570FAC i − 0.456FAC j − 0.684FAC k
= −0.873FAD i − 0.436FAC j + 0.218FAC k
= 117.1j (N)
1000 N
(90) g
The equations of equilibrium are
Fx = 0.667FAB + 0.570FAC − 0.873FAD = 0
Fy = −0.333FAB − 0.456FAC − 0.436FAC + 117.1 = 0
T
Fz = 0.667FAB − 0.684FAC + 0.218FAC = 0
y
Solving, we get
T
(0, 8, 0)
FAB = 64.8 N ∼ 2 students
A
FAC = 99.8 N ∼ 3 students
FAC
FAD = 114.6 N ∼ 3 students
FAD
C (10, 0, −12) m
D
x
(−16, 0, 4)
z
B (16, 0, 16) m
Problem 3.65 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
Points A, B, C, and D are located at
Solution:
y
A(0, 1.2, 0),
C(0, 2, −1),
B(−0.3, 2, 1),
D(2, 2, 0)
C
FAC
B
FAB
FAD
Write the unit vectors eAB , eAC , eAD
D
A
eAB = −0.228i + 0.608j + 0.760k
eAC = 0i + 0.625j − 0.781k
eAD = 0.928i + 0.371j + 0k
W
x
(20) (9.81) N
z
The forces are
y
C
FAB = −0.228FAB i + 0.608FAB j + 0.760FAB k
B
FAC = 0FAC i + 0.625FAC j − 0.781FAC k
D
FAD = 0.928FAD i + 0.371FAD j + 0k
A
W = −(20)(9.81)j
The equations of equilibrium are


 Fx
Fy


Fz
1m
1m
= −0.228FAB + 0 + 0.928FAD = 0
= 0.608FAB + 0.625FAC + 0.371FAD − 20(9.81) = 0
= 0.760FAB − 0.781FAC + 0 = 0
We have 3 eqns in 3 unknowns solving, we get
FAB = 150.0 N
FAC = 146.1 N
FAD = 36.9 N
0.3 m
z
2m
x
Problem 3.66 The weight of the horizontal wall section is W = 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 = W
k. From the figure, the coordinates of the points in feet are
A(4, 6, 10),
B(0, 0, 0),
C(12, 0, 0),
and D(4, 14, 0).
The three unit vectors are of the form
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:
eAB = −0.324i − 0.487j − 0.811k,
eAC = 0.566i − 0.424j − 0.707k,
and eAD = 0i + 0.625j − 0.781k.
The forces are
TAC = TAC eAC ,
and TAD = TAD eAD .
The equilibrium equation for the knot at point A is
T + TAB + TAC + TAD = 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 = 9393 lb, TAC = 5387 lb, and
TAD = 10,977 lb
A
6 ft
7 ft
D
10 ft
14 ft
C
B
4 ft
8 ft
W
T
z
A
y
TD
10 ft
TB
D
7 ft
TC
6 ft
14 ft
C
B
4 ft
X
8 ft
W
D
10 ft
6 ft
((xI xA )i + (yI yA )j + (zI − ZA )k)
eAI = ,
(xI xA )2 + (yI yA )2 + (zI − ZA )2
TAB = TAB eAB ,
A
C
B
4 ft
7 ft
8 ft
W
14 ft
Problem 3.67 In Problem 3.66, each cable will safely
support a tension of 40,000 lb. Based on this criterion,
what is the largest safe value of the weight W ?
Solution: There are two possible solutions to this problem, depending on how we interpret the problem. One solution considers the cable
extending upward from A as one of the cables subject to the 40,000 lb
limit and the other does not.
(a)
Assume that the cable upward from A is subject to the limit. From
the solution to Problem 3.66, we see that the largest tension in
the cables is the tension in the cable extending upward from A.
If we double the weight, we increase the tension in this cable to
40,000 lb. For this case, WMAX = 40,000 lb.
(b)
Assume that the cable upward from A is not subject to the limit.
From the solution to Problem 3.66, the largest force in the three
supporting cables is TAD = 10977 lb. The scale factor must
increase this force to 40,000 lb. The scale factor, f , is given by
f = 40,000/10,977 = 3.644.
The maximum allowable weight is
WMAX = 20,000f = (20,000)(3.644) = 72,880 lb.
Problem 3.68 The 680-kg load suspended from the Solution:
helicopter is in equilibrium. The aerodynamic drag force
on the load is horizontal. The y axis is vertical, and cable Fx = TOA sin 10◦ − D = 0,
OA lies in the x-y plane. Determine the magnitude of Fy = TOA cos 10◦ − (680)(9.81) = 0.
the drag force and the tension in cable OA.
Solving, we obtain D = 1176 N, TOA = 6774 N.
y
B
(2, 2, 0)
m
y
C
(5, 2, Ð1) m
A
x
10°
A
z
O
(3, 0, 4) m
x
y
TOA
B
C
D
10°
D
x
(680) (9.81) N
Problem 3.69 In Problem 3.68, 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 = −3.3i − 4.5j (m),
rOC = 1.1i − 5.3j + k (m),
rOD = 1.6i − 5.4j − k (m).
Dividing by the magnitudes, we obtain the unit vectors
eOB = −0.591i − 0.806j,
eOC = 0.200i − 0.963j + 0.182k,
eOD = 0.280i − 0.944j − 0.175k.
Using these unit vectors, we obtain the equilibrium equations
Fx = TOA sin 10◦ − 0.591TOB + 0.200TOC + 0.280TOD = 0,
Fy = TOA cos 10◦ − 0.806TOB − 0.963TOC − 0.944TOD = 0,
Fz = 0.182TOC − 0.175TOD = 0.
From the solution of Problem 3.68, TOA = 6774 N. Solving these
equations, we obtain
TOB = 3.60 kN,
TOC = 1.94 kN,
TOD = 2.02 kN.
y
TOA
10°
x
TOB
TOC
TOD
Problem 3.70 The small sphere A weighs 20 lb, and
its coordinates are (4, 0, 6) ft. It is supported by two
smooth flat plates labeled 1 and 2 and the cable AB.
The unit vector e1 = 49 i + 79 j + 49 k is perpendicular to
9
2
6
plate 1, and the unit vector e2 = − 11
i + 11
j + 11
k
is perpendicular to plate 2. What is the tension in the
cable?
y
B
(0, 4, 0) ft
2
1
e2
e1
x
A
Solution:
A and B are located at A (4, 0, 6), B (0, 4, 0) feet.
z
The vector locations of the points A and B are:
rA = 4i + 0j + 6k,
rB = 0i + 4j + 0k.
The unit vector parallel to the tension acting from A toward B is
A
eAB = |rrB −r
.
−r |
B
A
The weight is W = 0i − |W|j0k = 0i − 20j + 0k.
The unit vectors are
eAB = −0.4851i + 0.4851j − 0.7276k
e1 = 0.4444i + 0.7778j + 0.4444k
e2 = −0.8182i + 0.1818j + 0.5455k
where the values of the last two were given by the problem statement.
The forces are expressed in terms of the unit vectors,
TAB = |TAB |eAB ,
N1 = |N1 |e1 ,
N2 = |N2 |e2 .
The equilibrium conditions are
F = 0 = TAB + N1 + N2 + W = 0
Substitute the force vectors and collect like terms,
Fx = (−0.4851|TAB | + 0.4444|N1 | − 0.8182|N2 |)i = 0
Fy = (0.4851|TAB | + 0.7778|N1 | − 0.1818|N2 | − 20)j = 0
Fz = (−0.7276|TAB | + 0.4444|N1 | − 0.5455|N2 |)k = 0
An electronic calculator was used to solve these equations. The solution is:
|TAB | = 12.34 lb ,
|N1 | = 17.51 lb ,
|N2 | = 2.19 lb .
y
(0, 4, 0) ft
B
x
2
1
e2
e1
A
z
Problem 3.71 The 1350-kg car is at rest on a plane
surface. The unit vector en = 0.231i + 0.923j + 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.
y
en
x
z
Solution:
The weight force is
W = −mgj = −(1350)(9.81)j = −13240j N.
The component of W normal to the surface is
F N = W · e = W x ex + W y e y + W z e z = W y ey
= (−13240)(0.923) = −12220 N.
The component of W tangent to the surface (the friction force) can be
calculated from
FT =
2 =
W 2 − FN
(13240)2 − (12220)2 = 5096 N.
Thus, FN = 12220 N and FT = 5096 N.
y
en
x
z
Problem 3.72 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?
y
G
(0, 1.4, –1.2) m
Solution:
From the figure, the coordinates of the points (in me-
E
F
ters) are
A(3.4, 1, 0),
E(0.9, 1.2, 0),
B(1.8, 1, 0),
C(2, 0, 1),
F (0, 1.4, 1.2),
D(2, 0, −1),
(3.4, 1, 0) m
(2, 1, 0) m
(1, 1.2, 0) m
B
(0, 1.4, 1.2) m
A
and G(0, 1.4, −1.2).
(2.2, 0, –1) m
D
The unit vectors are of the form
((xI − xK )i + (yI − yK )j + (zI − zK )k)
eIK = ,
(xI − xK )2 + (yI − yK )2 + (zI − zK )2
z
x
(2.2, 0, 1) m
C
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 .
y
(0, 1.4, −1.2) m
G
Substitution of the coordinates of the points yields the following six
unit vectors:
E
F
(1, 1.2, 0) m
eBC = 0.140i − 0.707j + 0.707k,
(3.4, 1, 0) m
A
(2, 1, 0) m
eBA = 1i + 0j + 0k,
B
(0, 1.4, 1.2) m
eBD = 0.140i − 0.707j − 0.707k,
(2, 0, −1) m
D
eBE = −0.981i + 0.196j + 0k,
x
C
eEF = −0.635i + 0.127j + 0.762k,
(2, 0, 1) m
z
and eEG = −0.635i + 0.127j − 0.762k.
The forces are of the form TIK = TIK eIK where IK takes on the
same values as above. The known force magnitude |TBA | = 900 kN.
Thus,
y
E
TBA = TBA eBA = 900(1i + 0j + 0k) kN = 900i kN.
The vector equation of equilibrium at point B (see the first free body
diagram) is
(0, 1.4, −1.2) m
G
F
(3.4, 1, 0) m
TBE (2, 1, 0) m
(1, 1.2, 0) m
B
(0, 1.4, 1.2) m
A
TBD
TBC
TBA + TBC + TBD + TBE = 0.
(2, 0, 1) m
y
TBD = 127.3 kN,
(0, 1.4, −1.2) m
G
TEG
and TBE = 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 + TEF + TEG = 0. Using the unit
vectors as above and solving for TEF and TEG , we get TEF =
TEG = 708.7 kN.
(2, 0, −1) m
x
z
The result is
TBC = 127.3 kN,
D
C
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 .
TBA
F
E −T (2, 1, 0) m (3.4, 1, 0) m
BE
TEF
(1, 1.2, 0) m
B
A
(0, 1.4, 1.2) m
D
C
z
(2, 0, −1) m
x
(2, 0, 1) m
Problem 3.73 The cables of the system in Prob- Solution: The largest load found in the solution of Problem 3.72 is
lem 3.72 will each safely support a tension of 1500 kN. TBE = 917.8 kN. The scale factor, scaling this force up to 1500 kN is
= 1.634. The largest safe value for the load in cable
Based on this criterion, what is the largest safe value of fAB=is(1500/917.8)
TAB max = TBA f = (900)(1.634) = 1471 kN.
the tension in cable AB?
Problem 3.74 The 200-kg slider at A is held in place
on the smooth vertical bar by the cable AB.
(a) Determine the tension in the cable.
(b) Determine the force exerted on the slider by the bar.
y
2m
B
A
5m
2m
x
2m
z
The coordinates of the points A, B are A(2, 2, 0),
B(0, 5, 2). The vector positions
Solution:
rA = 2i + 2j + 0k,
y
2m
rB = 0i + 5j + 2k
The equilibrium conditions are:
B
F = T + N + W = 0.
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 = 0i+1j+0k.
The dot product of the unit vector and the normal force vanishes: eB ·
N = 0. Take the dot product of eB with the equilibrium conditions:
eB · N = 0.
eB · F = eB · T + eB · W = 0.
A
2m
5m
x
2m
z
The weight is
eB · W = 1j · (−j|W|) = −|W| = −(200)(9.81) = −1962 N.
T
The unit vector parallel to the cable is by definition,
eAB =
rB − rA
.
|rB − rA |
N
Substitute the vectors and carry out the operation:
W
eAB = −0.4851i + 0.7278j + 0.4851k.
(a)
The tension in the cable is T = |T|eAB . Substitute into the
modified equilibrium condition
eB F = (0.7276|T| − 1962) = 0.
Solve: |T| = 2696.5 N from which the tension vector is
T = |T|eAB = −1308i + 1962j + 1308k.
(b)
The equilibrium conditions are
F = 0 = T + N + W = −1308i + 1308k + N = 0.
Solve for the normal force: N = 1308i − 1308k. The magnitude is |N| = 1850 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 3.75 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.
Problem 3.75 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) Determine the tension in the cable.
(b) Determine the normal force exerted on the slider by
the bar.
y
3 ft
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.
A
4 ft
20°
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
x
4 ft
z
A(4 cos α, 4 sin α, 0) = A(3.76, 1.37, 0), B(0, 4, 3).
The vector positions are
rA = 3.76i + 1.37j + 0k,
y
rB = 0i + 4j + 3k
The equilibrium conditions are:
3 ft
F = T + N + W = 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 |T| = 137.1 lb.
The dot product with the normal force is zero, eB N = 0. Take the
dot product of the unit vector and the equilibrium condition:
eB F = eB T + eB W = 0.
A
B
20°
4 ft
4 ft
z
The weight is
eB W = eB (−j|W|) = −0.9397|W| = −(0.9397)(100) = −94 lb.
T
N
The unit vector parallel to the cable is by definition,
eAB =
rB − rA
.
|rB − rA |
Substitute the vectors and carry out the operation
eAB = −0.6856i + 0.4801j + 0.5472k.
(a)
The tension in the cable is T = |T|eAB . Substitute into the
modified equilibrium condition
eB F = (0.6854|T| − 94) = 0.
Solve: |T| = 137.1 lb, from which the tension vector is
T = |T|eAB = −94i + 65.8j + 75k
(b)
Substitute T into the original equilibrium conditions,
F
= 0 = T + N + W = −94i + 65.8j
+75k + N − 100j = 0.
Solve for the normal force exerted by the bar on the slider
N = 94i + 34.2j − 75k (lb)
W
x
Problem 3.76 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?
y
0.15 m
0.4 m
B
C
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
eCD = eCDx i + eCDy j + eCDz k = −0.456i − 0.684j + 0.570k.
The location of point A is given by xA = xC + dCA eCDx , with
similar equations for yA and zA . From the figure, dCA = 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
0.2 m
A
O
x
0.25 m
D
0.2 m
z
eAB = eABx i + eABy j + eABz k = −0.674i + 0.735j + 0.079k.
y
The tension force in the cable can now be written as
0.15 m
TAB = −0.674TAB i + 0.735TAB j + 0.079TAB k.
B
0.4 m
C
From the free body diagram, the equilibrium equations are:
FN x + TAB eABx = 0,
and FN z + TAB eABz = 0.
z
0.2 m
0.4 m
W C
B
TAB
0.5 m
FN x = 38.9 N,
FN y = 36.1 N, and FN z = −4.53 N.
z
Solution: The solution to Problem 3.76 above provides the magnitudes of the components of the normal force exerted on the collar
at A.
|FN | =
(FN x )2 + (FN y )2 + (FN z )2 .
Substituting in the values found in Problem 3.77, we get
|FN | = 53.2 N.
x
y
We now have four equations in our four unknowns. Substituting in the
numbers and solving, we get
Problem 3.77 In Problem 3.76, determine the magnitude of the normal force exerted on the collar A by the
smooth bar.
0.25 m
D
0.15 m
FN x eCDx + FN y eCDy + FN z eCDz = 0.
TAB = 57.7 N,
0.2 m 0.3 m
A
0.5 m
FN y + TAB eABy − mg = 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 · eCD = 0. Expanding this, we get
0.3 m
0.5 m
0.2 m
FN
D
A
0.2 m 0.3 m
0.25 m
x
Problem 3.78 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 .
y
(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
eCD = erCDx i + eCDy j + eCDz k,
F
B
(0, 0, 4) m
z
y
Using the coordinates of points C, D, E, and G from the picture, the
unit vectors are
(0, 5, 0) m
eCD = −0.625i + 0.781j + 0k,
F
(0, 3, 0) m
and eEG = 0i + 0.6j + 0.8k.
3m
A
The location of point A is given by
B
yA = yC + CAeCDy ,
z
and zA = zC + CAeCDz ,
where CA = 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:
xB = xA + ABeABx ,
(0, 0, 4) m
y
D (0, 5, 0)
m
F
G (0, 3, 0)
m
yB = yA + ABeABy ,
NA
TAB
The equations based on line EG are:
B
mBg
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
z
3m
A
TAB
NB
yB = yE + EBeEGy ,
and zB = zE + EBeEGz .
x
(4, 0, 0) m
and zB = zA + ABeABz .
xB = xE + EBeEGx ,
x
(4, 0, 0) m
and eEG = erEGx i + eEGy j + eEGz k.
xA = xC + CAeCDx ,
3m
A
mAg
C (4, 0, 0) m
x
E (0, 0, 4) 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
NAx + TAB eABx + F eCDx = 0,
NA · eCD = NAx eCDx + NAy eCDy + NAz eCDz = 0
NAy + TAB eABy + F eCDy − mA g = 0,
for slider A and
and NAz + TAB eABz + F eCDz = 0.
From the free body diagram for B, we get
NB · eEG = NBx eEGx + NBy eEGy + NBz eEGz = 0.
Solving the eight equations in the eight unknowns, we obtain
NBx − TAB eABx = 0,
Nby − TAB eABy − mB g = 0,
and NBz − TAB eABz = 0.
F = 36.6 N .
Other values obtained in the solution are EB = 2.56 m,
NAx = 145 N,
NBx = −122 N,
NAy = 116 N,
NBy = 150 N,
NAz = −112 N,
and
NBz = 112 N.
Problem 3.79 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
Solution: Isolate the crate, and solve the equilibrium conditions.
The weight is W = 0i − 100j. The angle between the normal force
and the positive x axis is (90 − 30) = 60◦ . The normal force is
100 lb
N = |N|(i cos 60 + j sin 60) = |N|(0.5i + 0.866j).
30°
The angle between the string tension and the positive x axis is (180◦ −
45◦ ) = 135◦ , hence the tension is
T = |T|(i cos 135◦ + j sin 135◦ ) = |T|(−0.7071i + 0.7071j.
The equilibrium conditions are
y
T
F = W + N + T = 0.
Substituting, and collecting like terms
Fx = (0.5|N| − 0.7071|T|)i = 0
Fy = (0.866|N| + 0.7071|T| − 100)j = 0
Solve: |T| = 51.8 lb, |N| = 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 |W| sin 30◦ − |T| cos 15◦ = 0
from which |T| =
sin 30◦
cos 15◦
100 = 51.76 = 51.8 lb.
The equilibrium equation for the y-coordinate is
Fy = |N| − W cos 30◦ + |T| sin 15◦ − 0,
from which |N| = 73.2 lb. check.
A
45°
B
100 lb
30°
N
W
β
x
Problem 3.80 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 |W|. The
force at B is
60°
20°
FB = |W|(i cos 60◦ + j sin 60◦ ).
25°
B
FB = |W|(0.5i + 0.866j).
A
The angles at A relative to the positive x axis are: 180◦ and 180◦ −
25◦ = 155◦ . The force at A is the sum of the two forces:
W
FA = |W|(i cos 180◦ + j sin 180◦ ) + |W|(i cos 155◦ + j sin 155◦ )
x
FA = |W|(−1.906i + 0.4226j).
The total force exerted by the patient’s leg is FP = FH i − 32.2j,
where FH is an unknown component. The equilibrium conditions are
F = FA + FtB + FP = 0,
60°
20°
from which:
and
25°
FX = (0.5|W| − 1.906|W| + FH )i = 0
B
A
FY = (0.866|W| + 0.4226|W| − 32.2)j = 0.
W
Solve for the weight: |W| =
32.2
1.2886
= 25 lb .
Problem 3.81 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°
Resolve the tensions at A and B into scalar components. Solve the equilibrium equations. The tension at B is
Solution:
TB = |TB |(i cos 40◦ + j sin 40◦ )
A
55°
TB = |TB |(0.7660i + 0.6428j).
B
40°
The angle at A relative to the positive x axis is 180◦ − 55◦ = 125◦ .
The tension at A:
TA = |TA |(i cos 125◦ + j sin 125◦ ) = |TA |(−0.5736i + 0.8192j).
from which
The weight is: W = 0i − 200j. The equilibrium conditions are
F = TA + TB + W = 0,
Solve:
Fx = (0.766|TB | − 0.5736|TA |)i = 0
Fy = (0.6428|TB | − 0.8192|TA | − 200)i = 0.
|TB | = 115.1 lb,
|TA | = 153.8 lb.
Problem 3.82 The cable AB is horizontal, and the box
on the right weighs 100 lb. The surfaces are smooth.
(a) What is the tension in the cable?
(b) What is the weight of the box on the left?
A
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
(a)
For right hand box. The weight is W = 0i − 100j. The angle
between the normal force and the positive x axis is (90◦ −40◦ ) =
50◦ . The force:
20°
40°
N = |N|(i cos 50◦ + j sin 50◦ ) = |N|(0.6428i + 0.7660j).
The cable tension is T = −|T|i + 0j. The equilibrium conditions are
F
from which
Fx
and
Fy
= T + N + W = 0,
T
= (0.6428|N| − |T|)i = 0
= (0.7660|N| − 100)j = 0
N
40°
W
Solve: |T| = 83.9 lb
(b)
For left hand box: The weight W = 0i − |W|j. The angle
between the normal force and the positive x axis is (90◦ +20◦ =
110◦ . The normal force:
y
T
N = |N|(−0.3420i + 0.9397j).
The cable tension is: T = |T|i + 0j. The equilibrium conditions are:
F
= W + N + T = 0,
from which:
Fx
and
Fy
= (−0.342|N| + 83.9)i = 0
= (−0.940|N| − |W|)j = 0.
Solving for the weight of the box, we get |W| = 230.6 lb.
B
20°
W
N
x
Problem 3.83 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
eAC
= −0.243i + 0.970j
= 0.316i + 0.949j
The forces are
TAB
TAC
W
A
= −0.243TAB i + 0.970TAB j
= 0.316TAC i + 0.949TAC j
= −5886j N
= −0.243TAB + 0.316TAC = 0
Fy
= 0.970TAB + 0.949TAC − 5886 = 0
Solving, TAB = 3.47 kN, TAC = 2.66 kN
B
C
(5, 14) m
y
A
(3,8)
W = –mg j
= (600)(9.81)N
Fx
(1.5, 14) m
TAC
TAB
(3, 8) m
x
Problem 3.84 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 = 0,
R − 3S = 0,
F
F − 3R = 0,
R
and 2T + 2S + 2R − mA g = 0.
R
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 = 3T , and
R = 3S = 9T . Substituting these into the last equilibrium equation
results in 2T + 2(3T ) + 2(9T ) = mA g.
R
R
S
S
S
Solving, we get T = mA g/26 .
S
T
T
S
S
R
R
T
T
T
A
mAg
T
A
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 = 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 = 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.
Problem 3.85 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 = 0, or F = 20 lb
F
A
F
F
F F
F
F
F
60 lb.
Problem 3.86 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
B
45°
F
A
F = N1 + N2 + W = 0
Fx = (0.7071|N1 | − |N2 |)i = 0
20°
Fy = (0.7071|N1 | − 490.5)j = 0.
y
Solve: N1 = 693.7 N,
|N2 | = 490.5 N
B
Isolate the bottom block. The weight is
N2
W = 0i − |W|j = 0i − (42)(9.81)j = 0i − 412.02j (N).
N1
W
α
x
The angle between the normal force N1 and the positive x axis is
(270◦ − 45◦ ) = 225◦ .
y
The normal force:
N1
N1 = |N1 |(i cos 225◦ + j sin 225◦ ) = |N1 |(−0.7071i − 0.7071j).
F
β
α
A
The angle between the normal force N3 and the positive x-axis is
(90◦ − 20◦ ) = 70◦ .
x
N3
The normal force is
N1 = |N3 |(i cos 70◦ + j sin 70◦ ) = |N3 |(0.3420i − 0.9397j).
The force is . . . F = |F|i + 0j. The equilibrium conditions are
F = W + N1 + N3 + F = 0,
from which:
Fx = (−0.7071|N1 | + 0.3420|N3 | + |F|)i = 0
Fy = (−0.7071|N1 | + 0.9397|N3 | − 412)j = 0
For |N1 | = 693.7 N from above:
|F| = 162 N
W
Problem 3.87 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?
y
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 A(6, 2, 0), B(12, 3, 0), C(0, 8, 5),
D(0, 4, −5), O(0, 0, 0).
4m
8m
(6, 2, 0) m
O
The vector locations of these points are:
z
rA = 6i + 2j + 0k,
rB = 12i + 3j + 0k,
rD = 0i + 4j − 5k,
rO = 0i + 0j + 0k.
3m
rC = 0i + 8j + 5k,
12 m
x
The unit vector parallel to the tension acting between the points A, B
in the direction of B is by definition
eAB =
y
rB − rA
.
|rB − rA |
5m
5m
Perform this for each of the unit vectors
D
4m
C
eAB = +0.9864i + 0.1644j + 0k
8m
eAC = −0.6092i + 0.6092j + 0.5077k
A
eAO = −0.9487i − 0.3162j + 0k
The tensions in the cables are expressed in terms of the unit vectors,
TAB = |TAB |eAB = 50eAB ,
TAD = |TAD |eAD ,
TAC = |TAC |eAC ,
TAO = |TAO |eAO .
The equilibrium conditions are
F = 0 = TAB + TAC + TAD + TAO = 0.
Substitute and collect like terms,
Fx = (0.9864(50) − 0.6092|TAC | − 0.7422|TAD |
−0.9487|TAO |)i = 0
Fy = (0.1644(50) + 0.6092|TAC | + 0.2481|TAD |
−0.3162|TAO |)j = 0
Fz = (+0.5077|TAC | − 0.6202|TAD |)k = 0.
This set of simultaneous equations in the unknown forces may be
solved using any of several standard algorithms. The results are:
|TAO | = 43.3 kN,
|TAC | = 6.8 kN,
O
(6, 2, 0) m
eAD = −0.7442i + 0.2481j − 0.6202k
|TAD | = 5.5 kN.
B
A
Problem 3.88 The 1350-kg car is at rest on a plane
surface with its brakes locked. The unit vector en =
0.231i+0.923j+0.308k is perpendicular to the surface.
The y axis points upward. The direction cosines of the
cable from A to B are cos θx = −0.816, cos θy = 0.408,
cos θz = −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
B
FS + TAB + W = 0.
ep
Writing these forces in terms of components, we have
A
W = −mgj = −(1350)(9.81) = −13240j N,
FS = FSx i + FSy j + FSz k,
x
and TAB = TAB eAB ,
z
where
y
eAB = cos θx i + cos θy j + cos θz k = −0.816i + 0.408j − 0.408k.
en
"car"
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 = 0,
FSx = 979.2 N,
FSy = 12, 754 N,
and FSz = 489.6 N.
The next step is to find the component of FS normal to the surface.
This component is given by
FN = FN · en = FSx eny + FSx eny + FSz enz .
Substitution yields
FN = 12149 N .
From its components, the magnitude of FS is FS = 12800 N. Using
the Pythagorean theorem, the friction force is
f =
x
z
Substituting in the numbers and solving, we get
2 = 4033 N.
FS2 − FN
FS
F
FSy − TABy − W = 0,
and FSz − TABz = 0.
FN
TAB
W
Problem 4.1 Determine the moment of the 50-N force
about (a) point A, (b) point B.
5m
50 N
B
P
3m
A
Solution: Use 2-dimensional moment strategy: determine normal distance to line of action D; calculate magnitude DF ; determine sign.
(a)
(b)
The perpendicular distance from A to line of action is D = 0,
hence the moment MA = DF = 0.
50 N
5m
F
The perpendicular distance from B to line of action is 3 m (the
triangle is a 3-4-5 right triangle), and the action is counter clockwise, hence MB = +(3)(50) = +150 N-m.
Problem 4.2 The radius of the pulley is r = 0.2 m
and it is not free to rotate. The magnitudes of the forces
are |FA | = 140 N and |FB | = 180 N.
(a) What is the moment about the center of the pulley
due to the force FA ?
(b) What is the sum of the moments about the center of
the pulley due to the forces FA and FB ?
P
B
3m
A
FB
45
45°
r
FA
10°
10
Solution:
FB
+MA = r|FA | = (0.2)140 N-m
MA = 28 N-m
45°
+MB = −r|FB | = −(0.2)180
+MB = −36 N-m
+MA + MB = 28 − 36 = −8 Nm
+(MA + MB ) = −8 N-m
r
FA
FB
r
10°
45°
FA
r
10°
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
25 ft
B
A
Solution: Use 2-dimensional moment strategy: determine normal distance to line of action D; calculate magnitude DF ; determine
sign. Add moments.
(a)
The normal distances from A to the lines of action are DAB =
10 ft, and DAC = 35 ft. The moments are clockwise (negative).
Hence,
10 ft
C
25 ft
(b)
D
15 ft
A
D
B
15 ft
C
MA = −10(40) − 35(44) = −1940 ft-kip .
The normal distances from D to the lines of action are DDB =
40 ft, and DDC = 15 ft. The actions are positive; hence
MD = +(40)(40) + (15)(44) = 2260 ft-kip
Problem 4.4 If you exert a 90-N force on the wrench in
the direction shown, what moment do you exert about the
center of the nut? Compare your answer to the moment
exerted if you exert the 90-N force perpendicular to the
shaft of the wrench.
500 mm
90 N
m
0m
45
Solution:
M = d1 · F = (.45)
MP = d2 · F = (0.5)
90 = 40.5 N-m
clockwise
for direction shown
90 = 45 N-m
clockwise
for perpendicular force
500 mm
90 N
m
0m
45
Problem 4.5 If you exert a force F on the wrench in
the direction shown and a 50 N-m moment is required to
loosen the nut, what force F must you apply?
Nut
265 mm
300 mm
F
Solution: Use 2-dimensional moment strategy: determine normal distance to line of action D; calculate magnitude DF ; determine
sign. Solve for unknown force.
F
Nut
The normal distance from the nut center to the line of action is D =
0.265 m.
265 mm
Thus to loosen the nut, 50 = D|F| = 0.265|F|. Solve: |F| =
50
= 188.7 N in the direction shown.
0.265
300 mm
F
Problem 4.6 The support at the left end of the beam
will fail if the moment about P due to the 20-kN force
exceeds 35 kN-m. Based on this criterion, what is the
maximum safe value of the angle α in the range 0 ≤ α ≤
90◦ ?
20 kN
P
α
2m
Solution:
20 kN
MP = dF sin α
P
MP = (2)(20 kN) sin α kN-m
α
Set MP = 35 kN-m and solve
for
α sin α =
35
40
2m
αmax = 61.0◦
20 kN
α
P
2m
Problem 4.7 The gears exert 200-N forces on each
other at their point of contact.
(a) Determine the moment about A due to the force
exerted on the left gear.
(b) 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) = 250◦ for the left gear and (90 − 20 = 70◦
for the right gear. The forces are FAY = 200| sin 250| = 187.9 N, and
FBY = 200| sin 70◦ | = 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 = 0.120 m, and RB = 0.080 m. The actions are
negative. Thus
MA = −0.120(|187.9|) = −22.55 N-m,
and MB = −0.080(|187.9|) = −15.0 N-m.
A
B
20°
200 N
A
B
200 N
120 mm
20°
80 mm
20°
200 N
A
B
80 mm
120 mm
200 N
20°
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
Solution:
◦
MA = L · F sin 30 = L
15
2
25°
MA = 7.5 L kN · m
set MA = MAmax = 18 kN · m = 7.5 Lmax
Lmax = 2.4 m
F
30°
30°
30°
B
L
A
25°
F = 15 kN
25°
Problem 4.9
about P .
Determine the moment of the 80-lb force
80 lb
20°
40°
P
Solution: Use 2-dimensional moment strategy: resolve the force
into a component normal to the beam; calculate magnitude DF , where
F is the component normal to the beam; determine sign.
3 ft
F
20°
The angle between the beam and the force is (180◦ − 40◦ − 20◦ ) =
120◦ . The component of the force normal to the beam is
FN = |F| sin 120◦ = (80)(0.866) = 69.3 lb.
40°
3 ft
P
The normal distance from P to the action line is the length of the beam,
and the action is positive. Thus
M = (3)(69.3) = 207.8 ft-lb
Problem 4.10 The 20-N force F exerts a 20 N-m counterclockwise moment about P .
(a) What is the perpendicular distance from P to the
line of action of F ?
(b) What is the angle α?
F
α
1m
P
2m
Solution: Use 2-dimensional moment strategy: determine normal distance to line of action D; calculate magnitude DF ; determine sign.
α
(a)
(b)
The moment is |M| = 20 = 20 D, from which the perpendicular
distance is D = 2020N-m
N =1m
1m
The angle between the force and the line from P is (α − β),
where β = tan−1 12 = 26.6◦ . The component of the force
normal√to the line from P is FN = √
|F| sin(α − β), thus M =
20 = 12 + 22 |F| sin(α − β) = 5(20) sin(α − β). Solve:
α = β + sin−1 (0.4472). Thus α = 26.6◦ + 26.6◦ = 53.1◦ ,
α = 26.6◦ + 153.4◦ = 180◦ .
P
2m
α
P
β
2
1
Problem 4.11 The lengths of bars AB and AC are
350 mm and 450 mm respectively. The magnitude of
the vertical force at A is |F| = 600 N. Determine the
moment of F about B and about C.
B
30°
C
20°
A
F
Solution:
sin 30◦ =
The geometry is the key to this problem
d1
0.35 m
d1 = 0.175 m
cos 20◦ =
d2
0.45 m
d2 = 0.423 m
+MB = −d1 F = −(0.175)(600)−
+MB = −105 N-m = 105 N-m clockwise
+MC = −d2 F = −(0.423)(600) N-m
+MC = −254 N-m
= 254 N-m clockwise
B
30°
C
20°
A
F
d1
30°
0m
5
0.3
0.450
m
20°
d2
600 N
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 = 2(60 lb)(16 in)
30 lb
= 160 ft-lb.
The normal component of the 30 lb force is F30 = 30 cos 30◦ = 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
30°
60 lb
1 ft
12 in
16 in.
M30 = 2(26 lb)(16 in)
1 ft
12 in
= 69.3 ft-lb.
The total moment is MT = 69.3 + 160 = 229.3 ft-lb
16 in.
60 lb
30° 30 lb
30°
16 in.
60 lb
30 lb
16 in.
30°
30 lb
Problem 4.13 The two students described in Problem 4.12, having failed to loosen the lug nut, try a different tactic. One of them stands on the lug wrench, exerting
a 150-lb force on it. The other pulls on the wrench with
the force F . If a torque of 245 ft-lb is required to loosen
the lug nut, what force F must the student exert?
150 lb
F
20°
16 in.
Solution:
60 lb
16 in.
The normal component of the force is
FN = |F| cos 20◦ = 0.9397|F|
150 lb
F
The total moment is
M = (150)(16 in)
1 ft
12 in
+ (|F|)(0.9397)(16 in)
1 ft
12 in
20°
= 200 + 1.253|F| ft-lb.
If the moment required is 245 ft-lb, then
|F| =
1
1.253
16 in.
(245 − 200) = 35.9 lb
16 in.
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?
S
13
in.
30°
12
Solution:
E
40°
in.
The key is the geometry
From trigonometry,
cos 40◦ =
d2
d1
, cos 30◦ =
13 in
12 in
S
13
in.
30°
Thus d1 = (12 in) cos 30◦
d1 = 10.39
12
and d2 = (13 in) cos 40◦
E
40°
in.
d2 = 9.96
We are given that
299 in-lb = d2 W = 9.96 W
d1
W = 30.0 lb
S
30°
Now,
13 in
12 i
n
Ms = (d1 + d2 )W
W
40°
Ms = (20.35)(30.0)
E
d2
Ms = 611 in-lb clockwise
Problem 4.15 Three forces act on the square plate. Solution: Determine the perpendicular distance between the points and the
Determine the sum of the moments of the forces (a) about 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 = 0.
A, (b) about B, (c) about C.
(b) The perpendicular distances of the lines of action from B are: 3 m for the
200 N
C
200 N
force
√through A, with a positive action, and for the force through C, DC =
1
32 + 32 = 2.12 m with a negative action. The moment about B is
2
MB = (3)(200) − 2.12(200) = 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 = 2(3)(200) = 1200 N-m.
3m
200 N
B
A
3m
C
200 N
200 N
3m
F
A
3m
200 N
B
Problem 4.16 Determine the sum of moments of the
three forces about (a) point A, (b) point B, (c) point C.
100 lb
200 lb
A
100 lb
B
2 ft
C
2 ft
2 ft
2 ft
Solution:
200 lb
The sum of the moments about A:
(a)
100 lb
MA = −(2)(100) + (4)(200) − (6)(100) = 0.
100 lb
A
C
B
The sum of the moments about B:
(b)
2 ft
2 ft
2 ft
2 ft
MB = +(2)(100) − (2)(100) = 0
The sum of the moments about C:
(c)
MC = +(6)(100) − (4)(200) + (2)(100) = 0.
Problem 4.17 Determine the sum of the moments of
the five forces acting on the Howe truss about point A.
800 lb
600 lb
600 lb
D
400 lb
400 lb
E
C
8 ft
F
B
G
A
H
4 ft
I
4 ft
J
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:
4 ft
4 ft
4 ft
600 lb
600 lb
MA = −4(400) − 8(600) − 12(800) − 16(600) − 20(400)
L
800 lb
D
400 lb
or
K
400 lb
E
C
MA = −33,600 ft-lb.
8 ft
F
B
G
A
H
4 ft
I
4 ft
J
4 ft
K
4 ft
L
4 ft
4 ft
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?
800 lb
D
400 lb
400 lb
C
E
8 ft
F
B
Summing moments around A, we get
Solution:
600 lb
600 lb
A
G
(ALL UNITS IN lbs)
H
4 ft
+MA = −(4)(400) − (8)(600) − (12)(800)
I
4 ft
J
4 ft
K
4 ft
L
4 ft
4 ft
−(16)(600) − (20)(400) + 24 G = 0
400 lb
600 lb
800 lb
600 lb
400 lb
G
Solving, we get
G = 1400 lbs
4ft
4ft
4ft
4ft
F1
F2
A
B
2m
Solution:
+
F = F1 + F2 = 250 N
MB = 4F1 + 2F2 = 700 N-m
We have two equations in two unknowns. Solving, we have
F1 = 100 N, F2 = 150 N
F1
F2
A
B
2m
2m
2m
F2
F1
B
2m
2m
2m
4ft
G
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 ?
4ft
A
2m
2m
Problem 4.20 Consider the beam shown in Prob- Solution: Sum of the moments about A:
lem 4.19. If the two forces exert a 140 kN-m clockwise moment about A and a 20 kN-m clockwise moment M
(ptA) = −2F1 − 4F2 = −140 kN-m.
about B, what are F1 and F2 ?
Sum of the moments about B:
M(ptB) = 4F1 + 2F2 = −20 kN-m.
Solving these equations, we obtain
F1 = −30 kN, F2 = 50 kN.
Problem 4.21 The force |F| = 140 lb. 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.
(a) What are the forces Ax , Ay , and B?
(b) What is the sum of the moments about the right end
of the beam?
Solution:
The forces are:
AX = |AX |(1i + 0j), AY = |AY |(0i + 1j)
B = |B|(0i + 1j), F = 140(0i − 1j).
(a)
The sum of the forces:
F = AX + AY + B + F = 0.
Substitute and collect like terms:
FX = (|AX |)i = 0.
FY = (AY + B − 140)j = 0.
It follows that |AX | = 0.
The sum of the moments about the left end is
M = (+0|AY | − 140(8) + |B|14) = 0.
Solve: |B| = 80 lb, from which the sum of forces equation
yields |AY | = 60 lb
(b)
The moments about the right end are:
M = (+6)((140)) − (14)(60) = 0
F
Ax
B
Ay
8 ft
6 ft
F
Ax
B
Ay
8 ft
6 ft
Problem 4.22 The vector sum of the three forces is
zero, and the sum of the moments of the three forces
about A is zero.
(a) What are FA and FB ?
(b) What is the sum of the moments of the three forces
about B?
Solution:
80 N
A
B
FA
FB
900 mm
The forces are:
400 mm
FA = |FA |(0i + 1j), FB = |FB |(0i + 1j),
and F = 80(0i − 1j).
The sum of the forces is:
F = FA + FB + F = 0,
from which
FY = (|FA | + |FB | − 80)j = 0.
The sum of the moments:
MA = −(0.9)(80) + (1.3)(|FB |) = 0.
(a) Solve these two equations to obtain: |FB | = 55.4 N, and |FA | =
24.6 N (b) The moments about B:
MB = (80)(0.4) − (1.3)|FA | = 0
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
3 in
A
6 in
2 in
B
7 in
2 in
C
D
Solution:
Solving
MO = (12)(2.7) − 3(10.2 + D)
12 in
D = 0.6 oz
3 in
A
6 in
2 in
B
7 in
2 in
C
D
12
3
0
(8.1 + 2.1 +D) = (10.2 + D)
2,7
Problem 4.24 The weight W = 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?
60°
Solution:
+
A
MA = −(2)(1.2) + 4(T sin 30◦ ) = 0
W
2m
−2.4 + 2T = 0
2m
T = 1.2 kN
T
30°
60°
30°
T sin 30°
2m
A
2m
A
W
2m
1.2 kN
2m
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
C
20°
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.
B
m
0m
40°
60
The distance from A to the action line of the weight of the arm AB is:
dAB = (0.300) cos 40◦ = 0.2298 m
mm
A
160 N
160 N
The distance from A to the action line of the weight of the arm BC is
dBC = (0.600)(cos 40◦ ) + (0.300)(cos 20◦ ) = 0.7415 m.
m
150 m
0 mm
The distance from A to the line of action of the force is
◦
◦
60
20°
◦
dF = (0.600)(cos 40 ) + (0.600)(cos 20 ) + (0.150)(cos 20 )
A
MA = −dAB (160) − dBC (160) − dF (40) = −202 N-m
40°
160 N
The sum of the moments about A is
B
m
0m
60
= 1.1644 m.
160 N
C
40 N
40 N
Problem 4.26 The space shuttle’s attitude thrusters
exert two forces of magnitude F = 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°
6°
12 m
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)
+MFRONT⊕ = (18)Fsin 5◦ − (2.2)Fcos 5◦
2.2 m
2.2 m
F
F
G
5°
12 m
18 m
+MREAR⊕ = (2.2)Fcos 6◦ − (12)Fsin 6◦
6°
+MTOTAL = MFRONT + MREAR
F sin 6°
F sin 5°
+MTOTAL = −4.80 + 7.19 N-m
5û
6°c
+MTOTAL = 2.39 N-m
2.2 m
18 m
F cos 5°
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 θ?
2.2 m
12 m
F cos 6°
REAR
FRONT
y
A
(−5, 5) ft
F
θ
(4, 3) ft
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 = iFX + jFY . The vector from
A to the point of application is:
Solution:
x
rAF = (4 − (−5))i + (3 − 5)j = 9i − 2j.
The perpendicular distances are dAX = 9 ft, and dAY = 2 ft, and
the actions are positive. The moment about A is MA = (9)FY +
(2)FX = 200 ft-lb. The vector from B to the point of application is
rBF = (4 − 3)i + (3 − (−4))j = 1i + 7j; the distances dBX = 1 ft
and dBY = 7 ft, the action of FY is positive and the action of FX
is negative. The moment about B is MB = (1)FY − (7)FX =
−100 ft-lb. The two simultaneous equations have solution: FY =
18.46 lb and FX = 16.92 lb. Take the ratio to find the angle:
θ = tan−1
FY
FX
= tan−1
18.46
16.92
= tan−1 (1.091) = 47.5◦ .
B
(3, −4) ft
y
A
(−5, 5) ft
F
θ
(4, 3) ft
x
From the Pythagorean theorem
|F| =
2 =
FY2 + FX
18.462 + 16.922 = 25.04 lb
B
(3, − 4) ft
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) Determine the forces Ax , Ay , and B.
(b) Determine the sum of the moments of the forces
about the point where the force B acts.
Ay
Ax
25 kN
20°
650 mm
450 mm
30 kN
45°
B
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.
Solution:
650 mm
The angles are measured counterclockwise from the x-axis. The
forces are
350 mm
Ay
Ax
F2 = 30(i cos 135◦ + j sin 135◦ ) = −21.21i + 21.21j
F1 = 25(i cos 20◦ + j sin 20◦ ) = 23.50i + 8.55j.
(a)
F1 = 25 kN
The sum of the forces is
20°
F2 = 30 kN
650 mm
450 mm
45°
F = A + B + F1 + F2 = 0.
B
650 mm
Substituting:
and
FX
= (AX + BX + 23.5 − 21.2)i = 0,
FY
Solve: BX = −20.38 kN. Substitute into the force equation to obtain
AX = 18.09 kN
= (AY + 21.2 + 8.55)j = 0.
Solve the second equation: AY = −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 + 0.350) = 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) = 0.200 m, and the action is positive.
The moment about A is
350 mm
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:
(b)
MB = −(0.350)(21.21) − (0.650)(18.09)
+ (1)(29.76) − (0.450)(23.5) = 0
MA = (8.55)(1) + (23.5)(0.2) + (BX )(0.65) = 0.
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 = 300 mm and h = 400 mm;
F = 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°
60°
h
Ax
Ay
Solution:
b
b
b
b
b
b
B
The x- and y-components of the force F are
F
F = −|F|(i cos 60◦ + j sin 60◦ ) = −|F|(0.5i + 0.866j).
F 60°
60°
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
MA = 2|F|(0.5)(h) − 3b|F|(0.866) − 5b|F|(0.866) + 6bB = 0.
Substitute and solve: B =
1.6784|F|
1.8
= 93.2 N
h
Ax
Ay
b
b
b
b
b
b
B
Problem 4.30 Consider the truss shown in Problem 4.29. The dimensions are b = 3 ft and h = 4 ft;
F = 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) Determine the forces Ax , Ay , and B.
(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 = 300 lb. The distance from point A to the xcomponents of the forces is h, and the action is positive. The distances
between the point A and the lines of action of the y-components 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 = −300(i cos 60◦ + j sin 60◦ ) = −150i − 259.8j.
(a)
The sum of the forces:
F = 2F + A + B = 0.
The x- and y-components:
Fx = (Ax − 300)i = 0,
MA = 2(150) h − 3b(259.8) − 5b(259.8) + 6b B = 0.
Substitute and solve: B = 279.7 lb. Substitute this value into the force
equation and solve: Ax = 519.6 − 279.7 = 239.9 lb
(b)
Fy = (−519.6 + Ay + B)j = 0.
Problem 4.31 The mass m = 70 kg. What is the
moment about A due to the force exerted on the beam at
B by the cable?
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 = −6b(239.9) + 2(150) h + b(259.8) + 3b(259.8) = 0
B
A
45°
30°
3m
m
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:
Solution:
A
C
B
30°
3m
30°
0
FOB = |FOB |(i cos 150◦ + j sin 150◦ ) = |FOB |(−0.866i + 0.5j)
FOC = |FOC |(i cos 45◦ + j sin 45◦ ) = |FOC |(0.707i + 0.707j).
m
W = (70)(9.81)(0i − 1j) = −686.7j (N).
The equilibrium conditions are:
Fx = (−0.866|FOB | + 0.7070|FOC |)i = 0
FY = (0.500|FOB | + (.707|FOC |) − 686.7)j = 0.
Solve:
|FOB | = 502.70 N. This is used to resolve
the cable tension at B:
FB
=
502.7(i cos 330◦ +
j sin 330◦ )
=
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 = −3(251.4) = −754.0 N-m
FOB
FOC
O
W
45°
Problem 4.32 Consider the system shown in Problem 4.31. The beam will collapse at A if the magnitude
of the moment about A due to the force exerted on the
beam at B by the cable exceeds 2 kN-m. What is the
largest mass m that can be suspended?
Solution: The strategy is to determine the tension in the cable at
B that corresponds to the 2 kN-m moment at A, and then determine the
mass that will exert this tension. From the solution to Problem 4.31,
the y-component of the cable tension at B that corresponds to the 2 kN
moment is
−0.5|FOB | =
−2000 N-m
= 666.67 N,
3m
from which |FOB | = 1333 N. The two equilibrium equations for the
cable juncture O in the solution for Problem 4.31 are:
FX = (−0.866|FOB | + 0.7070|FOC |)i = 0
FY = (0.500|FOB | + (.707|FOC |) − |W|)j = 0.
Substitute the value of |FOB |, and rewrite these in the form of simultaneous equations in two unknowns:
(0.707|FOC | + 0|W|) = 1154.3
(0.707|FOC | − |W|) = −666.7.
Solve: |W| = 1821 N, from which: m =
1821
9.81
= 185.66 kg
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
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.
Solution:
FB
B
By inspection, the bar forms a 3, 4, 5 triangle. The angle the bar makes
with the horizontal is cos θ = 35 = 0.600, and sin θ = 45 = 0.800.
The force at B is FB = |FB |(−0.600i+0.800j). The perpendicular
distance from O to the line of action of the x-component is (4 + 1) =
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 = 5(0.600)|FB | + 1(0.800)|FB | = 3.8|FB | =
38 kN, from which |FB | = 10 kN
4m
θ
O
1m
3m
A
1m
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
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.
7 ft
15 ft
The line and rod form two right triangles, as shown in the sketch. The
angles are:
2
= 15.95◦
7
6
β = tan−1
= 21.8◦ .
15
H
α = tan−1
6 ft
4 ft
15 ft
7 ft
The angle between the perpendicular distance line and the fishing line
is θ = α + β = 37.7◦ . The force normal to the distance
line is
√
F = 5(sin 37.7◦ ) = 3.061 lb. The distance is d = 22 + 72 =
7.28 ft, and the action is negative. The moment about H is MH =
−7.28(3.061) = −22.3 ft-lb Check: The tension can be resolved
into x and y components,
α
Fx = F cos β = 4.642 lb, Fy = −F sin β = −1.857 lb.
β
α
The moment is
2 ft
7 ft
6 ft
M = −2Fx + 7Fy = −22.28 = −22.3 ft-lb. check.
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) What is the moment about O due to the force exerted
on the tower by cable AB?
(b) 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?
A
20 m
60°
45°
C
O
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)
The component normal to the line OA is FBN = 5(cos 60◦ ) =
2.5 kN. The action is negative. The moment about O is MOA =
−2.5(20) = −50 kN-m
(b)
By a similar process, the normal component of the tension in the
cable AC is FCN = |FC | cos 45◦ = 0.707|FC |. The action
is positive. If the sum of the moments is zero,
MO = (0.707(20)|FC | − 50) = 0,
β
15 ft
B
A
20 m
45°
60°
C
B
from which
|FC | =
50 kN m
= 3.54 kN
(0.707)(20 m)
A
FN
60°
FN
45°
A
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 = 4 m, y = 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
C
(9,1) m
A
(0,1.2) m
Solution:
tan α =
x
Triangle ABP
y
4
, α = 18.73◦
11.8
B (4,13) m
Triangle BCQ
tan β =
5
, β = 22.62◦
12
+MO = (13)(230) sin α − (13)(660) sin β
+MO = −2340 N-m
B (4,13)
C
(9,1) m
A
(0,1.2) m
230 N
660 N
660 sin β
230 sin α
β
α
α
β
P
A (0,1.2)
13 m
C (9,1)
Q
O (4,0)
O
x
Problem 4.37 The tension in each cable is the same.
The forces exerted on the beam by the three cables exert
a 1.2 kN-m counterclockwise moment about O. What is
the tension in the cables?
Solution: The strategy is to resolve cable tensions into compo-
nents normal to the beam; use the distances from O to attachment
point; determine action, and compute the moment in terms of cable
tensions. From the known moment, solve for the tensions.
1m
O
Denote the cables as 1, 2, and 3, starting near the root.
1m
The angle formed by the first cable with the beam is
θ1 = tan−1
1m
1m
1
= 45◦ .
1
The component normal to the beam is: F1 = |T| sin 45◦ =
0.707|T|. Similarly,
1
= 26.57◦ ,
2
θ2 = tan−1
1m
F2 = |T| sin 26.57◦ = 0.4472|T|,
1
θ3 = tan−1
= 18.43◦ ,
3
O
1m
1m
1m
and F3 = |T| sin 18.43◦ = 0.3162|T|.
The actions are positive. The sum of the moments about O
MO = (1)(0.707|T|) + 2(0.4472|T|) + 3(0.3162|T|)
Solving:
|T| =
1.2
= 0.4706 kN
2.55
= 1.2 kN m.
Problem 4.38 The tension in cable AB is 300 lb. The
sum of the moments about O due to the forces exerted
on the beam by the two cables is zero. What is the
magnitude of the sum of the forces exerted on the beam
by the two cables?
Solution: The strategy is to resolve the cable tensions into a components parallel to and normal to the beam; use the length of the beam
as the distance; determine the sign of the action, and compute the
moments about O. The moment balance is solved for the unknown
force in CA. The parallel components are used to find the magnitude
of the force.
B
6 ft
A
O
4 ft
C
12 ft
The angle formed by cable AB is
θAB = tan−1
6
12
= 26.6◦ .
B
The component normal to the beam is
6 ft
FN B = |TAB | sin 26.6◦ = (300)(0.4472) = 134.16 lb.
A
The parallel component is
O
◦
FP B = |TAB | cos 26.6 = (300)(0.8944) = 268.32 lb.
Similarly for the cable AC,
θCA = tan
−1
4
12
4 ft
C
12 ft
◦
= 18.43 ,
◦
FN C = |TAC | sin 18.43 = 0.3162|TAC |,
FP C = |TAC | cos 18.43◦ = 0.9487|TAC |
The action of AB is positive, the action of AC is negative. The moments about
O
MO = 12(134.16) − 12(0.3162)(|TAC |) = 0.
Solve: |TAC | = 424.3 lb. The sum of the forces acting on the beam is
FP = 268.32 + (0.9487)(424.3) = 670.8 lb
Problem 4.39 The beam shown in Problem 4.38 will MO = rOA × FAC + rOA × FAB = rOA × (FAC + FAB ) = 0, or
safely support the force exerted by the two cables at A if MO = 12i × ((−FAC cos 18.43◦ i − FAC sin 18.43◦ j
the magnitude of the horizontal component of the total
force exerted at A does not exceed 1000 lb and the sum
− FAB cos 26.57◦ i + FAB sin 26.57◦ j)) = 0.
of the moments about O due to the forces exerted by the
cables equals zero. Based on these criteria, what are the Carrying out the vector operations, we get
maximum permissible tensions in the two cables?
Solution:
From Problem 4.38, the forces are
gives
◦
MO = 12(−FAC sin 18.43◦ + FAB sin 26.57◦ )k = 0.
MO = 12(−FAC sin 18.43◦ + FAB sin 26.57◦ )k = 0
◦
FAC = −FAC cos 18.43 i − FAC sin 18.43 j, and
FAB = −FAB cos 26.57◦ i + FAB sin 26.57◦ j.
Solving the force equation and the moment equation simultaneously, we obtain
FAB = 447 lb and FAC = 632 lb.
The vector from O to A is given by rOA = 12i ft. The equation for the
sum of the horizontal components is −FAC cos 18.43◦ − FAB cos
26.57◦ = −1000 lb. The moment equation is
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
β = tan
−1
2.4
1.2
C
A
2.4 m
1m
= 63.43 .
1.8 m
The angle formed by the crane with the horizontal is
α = tan−1
1.4
3
B
◦
1.2 m
7m
= 25.02◦ .
The angle between the hydraulic cylinder and the
crane is θ = β − α = 38.42◦ .
The normal component of the force is:
FN
=
(300)(sin
38.42◦ )
√
= 186.42 kN. The distance from point A is d = 1.42 + 32 =
3.31 m.
The action is positive.
The moment about A is
MO = +3.31(186.42) = 617.15 kN-m Check: The
force exerted by the actuator can be resolved into x- and ycomponents, Fx = F cos β = 134.16 kN, Fy = F sin β
= 268.33 kN. The moment about the point A is M = −1.4Fx + 3.0
Fy = 617.15 kN m. check.
β
α
1.2 m
C
1m
2.4 m
β
1.4 m
A
2.4 m
α
3m
B
1.8 m
1.2 m
7m
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 α = 40◦ . What is
the moment of the force about A?
t
6f
t
9f
C
α
A
B
6 ft
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 = 6i ft. The location of point C in the
xy coordinates is given by
t
6f
rAC = 9 cos 40◦ i + 9 sin 40◦ j = 6.89i + 5.79j ft.
t
9f
The unit vector from B to C is given by
(xC − xB )i + (yC − yB )j
eBC = (xC − xB )2 + (yC − yB )2
A
= 0.153i + 0.988j.
α
C
B
6 ft
Thus, the force along BC is FBC = 2200eBC = 337i + 2174j lb.
The moment of this force about point A is MA = 6(2174) =
13040 ft-lb
Problem 4.42 The hydraulic cylinder BC in Problem 4.41 exerts a 2200-lb force on the boom of the
crane at C. The force is parallel to the cylinder. The
cable supporting the suspended crate exerts a downward
force at the end of the boom equal to the weight of the
crate. The angle α = 35◦ . If the sum of the moments
about A due to the two forces exerted on the boom is
zero, what is the weight of the crate?
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 = 6i ft. The location of point C in the xy coordinates is given
by rAC = 9 cos 35◦ i + 9 sin 35◦ j = 7.37i + 5.16j ft. The unit vector from
B to C is given by
Solution:
(xC − xB )i + (yC − yB )j
eBC = (xC − xB )2 + (yC − yB )2
= 0.257i + 0.966j.
Thus, the force along BC is
FBC = 2200
eBC = 565i + 2126j lb.
The moment of the force in BC about point A is MA = 6(2126) =
12756 ft-lb. The moment of the weight of the crate about A is given by
MAW = (15 cos 35◦ )(−W ) = −12.29W ft-lb. Summing the two moments
and setting the sum to zero, we get
M = (−12.29W + 12756)k ft-lb = 0.
Solving, we get W = 1038 lb.
Problem 4.43 The unstretched length of the spring is
1 m, and the spring constant is k = 20 N/m. If α = 30◦ ,
what is the moment about A due to the force exerted by
the spring on the circular bar at B?
B
k
4m
3m
α
A
Solution: Assume that the bar is a quarter circle, with a radius of
4 m. The stretched length of the spring is found from the Pythagorean
Theorem: The vertical height from the floor to the attachment point on
the bar is h = 4 sin α = 2 m, and the distance
from the wall is 4 cos α.
The stretched length of the spring is L = (3 − h)2 + (4 cos α)2 =
3.6055 m. The spring force is F = (20)(3.6055 − 1) = 52.1 N.
The angle that the spring makes with the horizontal is
β = tan−1
3−h
4 cos α
= tan−1 (0.2887) = 16.1◦ .
The horizontal component of the spring force is FX = F cos 16.1◦ =
(52.1)(0.9608) = 50.07 N. The vertical component of the force is
FY = F sin 16.1◦ = (52.1)(0.2773) = 14.45 N.
The distance from A to the spring attachment point to the left of A is
d = 4(1 − cos α) = 0.536 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 = −0.536(14.45) + 2(50.07) =
92.4 N-m.
B
k
4m
3m
α
A
Problem 4.44 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
1m
0.6 m
B
A
0.15 m
0.6 m
Scoop
Solution: Use x, y coords with origin A. We need the unit vector
from C to B, eCB . From the geometry,
1m
D
eCB = 0.780i − 0.625j
C
1m
The force FCB is given by
0.6 m
B
FCB = (0.780)8i − (0.625)8j kN
x
FCB = 6.25i − 5.00j kN
0.15 m
0.6 m
Scoop
For the moments about A and D, treat the components of FCB as two
separate forces.
5.00 kN
+MA = (5, 00)(0.15) − (0.6)(6.25) kN · m
+MA = −3.00 kN · m
6.25 kN
C (−0.15, − 0.6)
For the moment about D
+
0.6 m
MD = (5 kN)(1 m) + (6.25 kN)(0.4 m)
+MD = 7.5 kN · m
0.15 m
A (0 , 0)
5.0 kN
m
D
0,4 m
C
6.25 kN
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.
y
50 i (lb)
(0, 3, 0) ft
x
O
Solution:
y
M = r × F = 3j × 50i
M = −150k ft-lb
50 i (lb)
(0, 3, 0) ft
Two dimensional description
+MO = −dF = −(3)(50) = −150 ft-lb
x
The descriptions match.
O
Problem 4.46 Use Eq. (4.2) to determine the moment Solution:
of the 80-N force about the origin O letting r be the (a) MO = rOA × F
vector (a) from O to A; (b) from O to B.
= 6i × 80j = 480k (N-m).
y
(b) MO
B
O
= rOB × F
80j (N)
= (6i + 4j) × 80j
(6, 4, 0) m
= 480k (N-m).
x
A (6, 0, 0) m
Problem 4.47 A bioengineer studying an injury sustained in throwing the javelin estimates that the magnitude of the maximum force exerted was |F| = 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 x − y plane. Express the moment of F about the
shoulder joint at O as a vector.
Solution: The magnitude of the moment is |F|(0.55 m) = (360 N)
(0.55 m) = 198 N-m. The moment vector is perpendicular to the x − y plane,
and the right-hand rule indicates it points in the positive z direction. Therefore
MO = 198k (N-m).
y
F
y
550 mm
F
O
O
x
x
Problem 4.48 Use Eq. (4.2) to determine the moment
of the 100-kN force (a) about A, (b) about B.
y
A
100j (kN)
6m
B
8m
12 m
(a) The coordinates of A are (0,6,0). The coordinates
of 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 = (8 − 0)i +
(0 − 6)j = 8i − 6j. The force is F = 100j (kN). The cross product is
Solution:
i
rAF × F = 8
0
j
−6
100
y
A
k
0 = 800k (kN-m)
0
B
8m
(b) The coordinates of B are (12,0,0). The position vector from B to
the point of application of the force is rBF = (8 − 12) i = −4i. The
cross product is:
i
rBF × F = −4
0
j
0
100
100j (kN)
6m
12 m
k
0 = −400k (kN-m)
0
Problem 4.49 The line of action of the 100-lb force
is contained in the x − y plane.
(a) Use Eq. (4.2) to determine the moment of the force
about the origin O.
(b) Use the result of (a) to determine the perpendicular
distance from O to the line of action of the force.
y
100 lb
30°
(10, 5, 0) ft
x
O
The resolved force is F = 100(i cos 150◦ +
j sin 150◦ ) = −86.6i + 50j. The position vector to the point of
application: r = 10i + 5j.
Solution:
(a)
y
100 lb
The cross product:
30°
(10, 5, 0)
r×F=
(b)
i
10
−86.6
j
5
50
k
0 = (500 + 433)k. = 933k ft-lb
0
The magnitude of the moment is |M| = 933 ft-lb. The magnitude of the force is |F| = 100 lb. The distance is
D=
933 ft-lb
= 9.33 ft.
100 lb
x
O
x
Problem 4.50 The line of action of F is contained in
the x − y 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 = 5i + 3j.
The position vector from A to the point of application is rAF =
(5 − 0)i + (3 − 7)j = 5i − 4j. The cross products:
rOF
i
×F = 5
FX
j
3
FY
k
0 = (5FY − 3FX )k = 140k, and
0
i
5
FX
j
−4
FY
k
0 = (5FY + 4FX )k = 280k.
0
rAF × F =
y
A
(0,7,0)
F
(5,3,0)
x
O
Take the dot product of both sides with k to eliminate k. The simultaneous equations are:
5FY − 3FX = 140, 5FY + 4FX = 280.
Solving: FY = 40, FX = 20, from which F = 20i + 40j (N)
Problem 4.51 To test the bending stiffness of a light
composite beam, engineering students subject it to the
vertical forces shown. Use Eq. (4.2) to determine the
moment of the 6-kN force about A.
y
3 kN
3 kN
6 kN
A
B
0.2 m
0.2 m
x
0.2 m
0.2 m
Solution:
MA = r × F
= 0.4i × 6j
= 2.4k (kN-m).
Problem 4.52 Consider the beam and forces shown in Solution:
Problem 4.51. Use Eq. (4.2) to determine the sum of the (a) MA = 0.2i × (−3j) + 0.4i × 6j + 0.6i × (−3j)
moments of the three forces (a) about A, (b) about B.
= O.
(b) MB
= (−0.2i) × (−3j) + (−0.4i) × 6j + (−0.6i) × (−3j)
= O.
Problem 4.53 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 = 3j, F1 = −200i; rO2 = 10i, F2 =
−500j; rO3 = 6i + 6j, F3 = 200i.
y
3 ft
The sum of the moments about O is
MO
i
=
0
−200
j k
i
3 0 + 10
0 0
0
200 lb
200 lb
j
0
−500
k
i
0 + 6
0
200
3 ft
j k
6 0 lb
0 0
O
6 ft
4 ft
500 lb
= 600k − 5000k − 1200k = −5600k ft-lb
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.
(b) Use Eq. (4.2) to determine the magnitude of the moment of the 150-N force about A.
y
(0, 6, 0) m
150k (N)
A
x
(6, 0, 0) m
z
Solution:
The perpendicular from A to the line of action of the force lies
in the x − y plane
(a)
y
(0, 6, 0) m
62 + 62 = 8.485 m
d
=
|M|
= dF = (8.485)(150) = 1270 N-m
150k (N)
A
(b) M
|M|
= (−6i + 6j) × (150k) = −900j + 900i N-m
= 9002 + 9002 = 1270 N-m
(6, 0, 0) m
z
x
Problem 4.55 A force F = −4i + 6j − 2k (kN) is Solution: The vector from P (2, 2, 2) m, to the point of application of the
applied at the point (8, 4, −4) m. What is the magnitude force (8, 4, −4) m, is
of the moment of F about the point P with coordinates
(2, 2, 2) m? What is the perpendicular distance D from r = 6i + 2j − 6k m
P to the line of action of F?
The moment of the force F about P is
i j k
MP = r × F = 6 2 −6 = 32i + 36j + 44k kN-m
−4 6 −2
|MP | = 322 + 362 + 442 = 65.2 kN-m
|F| = 42 + 62 + 22 = 7.48 kN
|MP | = |F|D
D=
|MP |
= 8.72 m
|F|
Problem 4.56 A force F = 20i − 30j + 60k (N) is Solution: The vector from P (2, 3, 6) m. to the point of application of the
applied at the point (2, 3, 6) m. What is the magnitude force (−2, −1, −1) m is
of the moment of F about the point P with coordinates
(−2, −1, −1) m? What is the perpendicular distance D r = −4i − 4j − 7k m
from P to the line of action of F?
The moment of the force about P is
i
j
k
MP = r × F = −4 −4 −7 = −450i + 100j + 200k N-m
20 −30 60
|MP | = 4502 + 1002 + 2002 = 502 N-m
|F| = 202 + 302 + 602 = 70 N
|MP | = |F|D
Problem 4.57 A force F = 20i − 30j + 60k (lb). The
moment of F about a point P is MP = 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
|MP | =
4502 + 1002 + 2002 = 502.5 (ft-lb).
The magnitude of the force is
|F| =
202 + 302 + 602 = 70 (lb).
The perpendicular distance is
D=
|MP |
502.5 ft-lb
=
= 7.18 ft
|F|
70 lb
D = |MP |/|F| = 7.18 m
Problem 4.58 A force F is applied at the point (8, 6,
13) m. Its magnitude is |F| = 90 N, and the moment of F
about the point (4, 2, 6) is zero. What are the components
of F?
Solution:
i
r×F = 8−4
Fx
j
6−2
Fy
k
13 − 6 = 0.
Fz
From Eq. (3), Fy = Fx , and from Eqs. (1) and (2), Fz =
tude is
90 N =
Fx2 + Fy2 + Fz2
Therefore
=
Fx2 + Fx2 +
4Fz − 7Fy = 0, (1)
7
Fx
4
2
.
7Fx − 4Fz = 0, (2)
Solving, we obtain Fx = ±40 N. we see that
4Fy − 4Fx = 0. (3)
F = 40i + 40j + 70k (N)
or
F = −40i − 40j − 70k (N).
Problem 4.59 The force F = 30i + 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?
y
F
(8, 2, −4) m
A
(4, 3, 3) m
x
z
Solution:
The vector from A to the point of application of F is
y
r = 4i − 1j − 7k m
F
and
|r| =
(a)
|MA |
=r×F=
=
1502
+
i
4
30
1702
j
−1
20
+
x
k
−7 = 150i − 170j + 110k N-m
−10
1102
= 252 N-m
The maximum moment occurs when r ⊥ F. In this case
|MAmax | = |r||F|
Hence, we need |F|.
|F| =
(8, 2, −4) m
(4, 3, 3) m
The moment of F about A is
MA
(b)
A
42 + 12 + 72 = 8.12 m
302 + 202 + 102 = 37.4 (N)
Thus,
|MAmax | = (8.12)(37.4) = 304 N-m
z
7
F .
4 x
The magni-
Problem 4.60 The direction cosines of the force F are
cos θx = 0.818, cos θy = 0.182, and cos θz = −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
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|F|. The position vector from
O to the point of application of F is rOF = 3i (m). Resolve
the position vector into components parallel and normal to F. The
component parallel to F is rP = (rOF · eF )eF , where the
unit vector eF parallel to F is eF = i cos θX + j cos θY +
k cos θZ = 0.818i + 0.182j − 0.545k. The dot product is
rOF · eF = 2.454.
The parallel component is rP =
2.007i + 0.4466j − 1.3374k. The component normal to F
is rN = rOF − rP = (3 − 2)i − 0.4466j + 1.3374k.
The magnitude of the
√ normal component is the perpendicular
distance: D =
12 + 0.44662 + 1.3372 = 1.7283 m.
The maximum moment allowed is MO = 1.7283|F| = 100 kN-m,
from which
|F| =
O
z
F
3m
x
y
F
O
x
3m
z
100 kN-m
= 57.86 ∼
= 58 kN
1.7283 m
Problem 4.61 The force F exerted on the grip of the
exercise machine points in the direction of the unit vector
e = 23 i− 23 j+ 13 k and its magnitude is 120 N. Determine
the magnitude of the moment of F about the origin O.
Solution:
150 mm
y
F
The vector from O to the point of application of the
O
force is
200 mm
z
r = 0.25i + 0.2j − 0.15k m
250 mm
x
and the force is F = |F|e
or
F = 80i − 80j + 40k N.
150 mm
y
The moment of F about O is
i
MO = r × F = 0.25
80
j
0.2
−80
k
−0.15 N-m
40
O
200 mm
z
or
MO = −4i − 22j − 36k N-m
and
|MO | =
F
42 + 222 + 362 N-m
|MO | = 42.4 N-m
250 mm
x
Problem 4.62 The force F in Problem 4.61 points in
the direction of the unit vector e = 23 i − 23 j + 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.
If we set |MO | = 560 N-m, we can solve for |Fmax |
The moment in Problem 4.61 can be written as
MO
i
= 0.25
2
F
3
j
0.2
− 23 F
k
−0.15 where F = |F|
+ 13 F
560 = 0.353|Fmax |
|Fmax | = 1586 N
MO = (−0.0333i − 0.1833j − 0.3k)F
If F can be in any direction, then the worst case is when r ⊥ F. The
moment in this case is |MO | = |r||Fworst |
And the magnitude of MO is
|r| =
|MO | = (
(b)
0.252 + 0.22 + 0.152 = 0.3536 m
560 = (0.3536)|FWORST |
0.03332 + 0.18332 + 0.32 )F
|Fworst | = 1584 N
|MO | = 0.353 F
Problem 4.63 An engineer estimates that under the
most adverse expected weather conditions, the total force
on the highway sign will be F = ±1.4i−2.0j (kN). What
moment does this force exert about the base O?
y
F
N
O
C
C
TU
8m
O
8m
x
z
Solution: The coordinates of the point of application of the force
are: (0, 8, 8). The position vector is rOF = 8j + 8k. The cross
product is
rOF × F =
i
0
±1.4
F
j k
8 8 = 16i − (∓1.4)(8)j + (∓1.4)(8)k
−2 0
MO = 16i ± 11.2j ∓ 11.2k (N-m)
8m
8m
O
x
Check: Use perpendicular distances to forces:
MX = 8(2) = 16,
MY = 8(±1.4) = ±11.2,
MZ = −8(±1.4) = ∓11.2 .
z
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 θx = 0.500, cos θy = 0.866, and cos θz = 0, and
the direction cosines of the centerline of arm AB are
cos θx = 0.707, cos θy = 0.619, and cos θz = −0.342.
What is the sum of the moments about O due to the two
forces?
m
y
0m
Solution: By definition, the direction cosines are the scalar components
of the unit vectors. Thus the unit vectors are e1 = 0.5i + 0.866j, and e2 =
0.707i+0.619j−0.342k. The position vectors of the midpoints of the arms are
r1 = 0.3e1 = 0.3(0.5i + 0.866j) = 0.15i + 0.2598j
r2 = 0.6e1 = 0.3e2 = 0.512i + 0.7053j − 0.0943k.
The sum of moments is
M = r1 × W1 r2 × W2
i
= 0.15
0
B
60
j
0.2598
−200
k
i
0 + 0.512
0
0
j
0.7053
−160
= −16.42i − 111.92k (N-m)
y
160 N
600
600 mm
A
mm
C
160 N
600 mm
B
200 N
200 N
O
A
z
x
Problem 4.65 The tension in cable AC is 100 lb. Determine the moment about the origin O due to the force
exerted at A by cable AC. Use the cross product, letting
r be the vector (a) from O to A, (b) from O to C
Solution:
y
The position vectors of the point A and C are:
rOA = 8j, rOC = 14i + 14k. The vector parallel to cable AC is
rAC = rOC − rOA = 14i − 8j + 14k, with magnitude |rAC | =
21.3542 ft. The unit vector parallel to AC is
eAC =
(0, 8, 0) ft
A
rAC
= 0.6556i − 0.3746j + 0.6556k
|rAC |
The tension in cable AC is TAC = 100eAC = 65.56i − 37.46j +
65.56k.
(a)
MO = rOA × TAC
=
i
0
65.56
j
8
−37.46
k
0
65.56
= 524.4i − 524.4k (ft-lb)
(b)
x
The moment using the vector from O to A is
O
B
C
z
y
A(0,8,0)
The moment using the vector from O to C is
MO = rOC × TAC
=
i
14
65.56
j
0
−37.46
(14, 0, 14) ft
(0, 0, 10) ft
k
14
65.56
= 524.4i − 524.4k (ft-lb)
O
z
B(0,0,10)
x
C(14,0,14)
k
−0.1026
0
Problem 4.66 Consider the tree in Problem 4.65. The
tension in cable AB is 100 lb, and the tension in cable
AC is 140 lb. Determine the magnitude of the sum of
the moments about O due to the forces exerted at A by
the two cables.
Solution: From the solution to Problem 4.65 the unit vector parallel to AC is eAC = 0.6556i − 0.3746j + 0.6556k. The tension in
cable AC is TAC = 140eAC = 91.784i − 52.444j + 91.784k.
The position vector parallel to cable AB is rAB = −8j + 10k.
The magnitude is |rAB | = 12.8062 ft. The unit vector is eAB =
−0.6247j + 0.7809k. The tension in cable AB is TAB =
100eAB = −62.47j + 78.09k (lb). The sum of the moments is
MO = (rOA × TAB ) + (rOA × TAC ) = (rOA × (TAB + TAC ))
=
i
0
91.78
j
8
−114.91
k
0
= 1359i − 734.2k
169.87
The magnitude of the sum of moments: |MO | = 1544.65 ft lb
Problem 4.67 The force F = 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?
Strategy: The ring is in equilibrium. Use what you
know about the four forces acting on it.
y
D (0, 6, 0) m
A
F
(12, 4, 2) m
C
B
(6, 0, 0) m
x
(0, 4, 6) m
z
Solution:
The vector from D to A is
y
D (0, 6, 0) m
rDA = 12i − 2j + 2k m.
A
The sum of the moments about point D is given by
F
(12, 4, 2) m
C
MD = rDA × FAD + rDA × FAC + rDA × FAB + rDA × F
MD = rDA × (FAD + FAC + FAB + F)
However, we are given that ring A is in equilibrium and this implies that
(FAD + FAC + FAB + F) = O = 0
B
z
FAD
A
Thus,
MD = rDA × (O) = 0
FAC
(6, 0, 0) m
x
(0, 4, 6) m
FAB
F
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
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
eAP
D(0, 6, 0)
FAD
FAC
(xP − xA )i + (yP − yA )j + (zP − zA )k
=
|rAP |
A(12, 4, 2) m
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
eAB
eAC
eAD
= −0.802i − 0.535j − 0.267k
= −0.949i + 0j + 0.316k
= −0.973i + 0.162j − 0.162k
From equilibrium, we have
FAB eAB + FAC eAC + FAD eAD + 5i (kN) = 0
In component form, we get
i:
j:
k:
−0.802FAB − 0.949FAC − 0.973FAD + 5 = 0
−0.535FAB + (0)FAC + 0.162FAD = 0
−0.267FAB + 0.316FAC − 0.162FAD = 0
Solving, we get
FAB = 779.5 N, FAC = 1976 N
FAD = 2569 N
The vector from D to A is
rDA = 12i − 2j + 2k m
The force FAB is given by
FAB = FAB eAB
FAB = −0.625i − 0.417j − 0.208k (kN)
The moment about D is given by
MD = rDA × FAB =
i
12
−0.625
j
−2
−0.417
MD = 1.25i + 1.25j − 6.25k (kN-m)
k
2
−0.208
F = 5i (kN)
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:
y
A
rAD = (−35 − 0)i + (0 − 70)j + (−35 − 0)k
rAD = −35i − 70k − 35k
rAC = (−40 − 0)i + (0 − 70)j + (40 − 0)k
D
rAC = −40i − 70j + 40k
rAB = (40 − 0)i + (0 − 70)j + (0 − 0)k
35 m
40 m
rAB = 40i − 70j + 0k
B
35 m
C
40 m
x
O
40 m
z
The unit vectors corresponding to these position vectors are:
eAD
rAD
−35
70
35
=
=
i−
j−
|rAD |
85.73
85.73
85.73
= −0.4082i − 0.8165j − 0.4082k
eAC =
rAC
40
70
40
=− i−
j+
k
|rAC |
90
90
90
= −0.4444i − 0.7778j + 0.4444k
eAB =
rAB
40
70
=
i−
j + 0k = 0.4962i − 0.8682j + 0k
|rAB |
80.6
80.6
The forces at point A are
TAC = 2eAB = −0.8889i − 1.5556j + 0.8889k
TAD = 2eAD = −0.8165i − 1.6330j − 0.8165k.
The sum of the forces acting at A are
TA = 0.2792i − 6.6615j + 0.07239k (kN-m)
The position vector of A is rOA = 70j. The moment about O is M =
rOA × TA
M =
i
0
0.2792
j
70
−6.6615
k
0
0.07239
= (70)(0.07239)i − j0 − k(70)(0.2792) = 5.067i − 19.54k
TAB = 4eAB = 1.9846i − 3.4729j + 0k
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.
The tensions are TAB = 4eAB , TAC = |TAC |eAC , and TAD =
|TAD |eAD . The components normal to rOA are
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:
eAD =
rAD
−35
70
35
=
i−
j−
|rAD |
85.73
85.73
85.73
FX = (−0.4082|TAD | − 0.4444|TAC | + 1.9846)i = 0
FZ = (−0.4082|TAD | + 0.4444|TAC |)k = 0.
= −0.4082i − 0.8165j − 0.4082k
eAC =
rAC
40
70
40
=− i−
j+
k
|rAC |
90
90
90
The HP-28S calculator was used to solve these equations:
= −0.4444i − 0.7778j + 0.4444k
eAB
|TAC | = 2.23 kN, |TAD | = 2.43 kN
rAB
40
70
=
=
i−
j + 0k = 0.4963i − 0.8685j + 0k
|rAB |
80.6
80.6
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
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
B
Solution:
C
4m
rDA = 8i + 0j − 5k.
8m
The position vectors AB and AC are
rAB = −8i + 4j − 5k,
rAC = −8i + 8j + 5k,
rAB =
rAC =
82 + 82 + 52 = 12.369 m.
A
z
x
The unit vectors parallel to the cables are:
MD =
eAB = −0.7807i + 0.3904j − 0.4879k,
eAC = −0.6468i + 0.6468j − 0.4042k.
8m
D
82 + 42 + 52 = 10.247 m.
i
8
181.79
j
0
−123.24
k
−5
= (−123.24)(5)i
+32.77
−((8)(+32.77) − (−5)(181.79))j + (8)(−123.24)k
MD = −616.2i − 117.11j − 985.9k (N-m)
The tensions are
TAB = 150eAB = −117.11i + 58.56j − 73.19k,
(Note: An alternate method of solution is to express the moment in terms of the
sum: MD = (rDC × TC + (rDB × TB ).)
TAC = 100eAC = −64.68i + 64.68j − 40.42k.
y
5m
The sum of the forces exerted by the wall on A is
TA = −181.79i + 123.24j − 32.77k.
The force exerted on the wall by the cables is −TA . The moment
about D is MD = −rDA × TA ,
5m
B
4m
C
A
8m
z D
8m
F
x
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?
From the solution of Problem 4.71, we have rDA =
8i + 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
Solution:
rAB = −8i + 4j − 5k,
|rAB | =
rAC = −8i + 8j + 5k,
|rAC | =
82 + 42 + 52 = 10.247 m.
82 + 82 + 52 = 12.369 m.
The unit vectors parallel to the cables are:
eAB = −0.7807i + 0.3904j − 0.4879k,
eAC = −0.6468i + 0.6468j + 0.4042k.
The tensions are
TBA = −TBA eAB = −TBA (−0.7807i + 0.3904j − 0.4879k), and
TCA = −TCA eAC = −TCA (−0.6468i + 0.6468j + 0.4042k).
The sum of the forces exerted by the cables perpendicular to the wall
is given by
TPerpendicular = TAB (0.7807) + TAC (0.6468) = 2 kN.
The moments of these two forces about D are given by
MD = (rDA × TCA ) + (rDA × TBA ) = rDA × (TCA + TBA ).
The sum of the two forces is given by
MD =
i
8
(TCA + TCB )X
j
0
(TCA + TCB )Y
k
−5
.
(TCA + TCB )Z
This expression can be expanded to yield
MD = 5(TCA + TCB )Y i + [−8(TCA + TCB )Z − 5(TCA + TCB )X ]j
+8(TCA + TCB )Y k.
The magnitude of this vector is given as 18 kN-m. Thus, we obtain
the relation
|MD | =
25(TCA + TCB )2Y + [−8(TCA + TCB )Z
= 18 kN-m.
−5(TCA + TCB )X ]2 + 64(TCA + 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 = 1.596 kN, and TCA = 1.166 kN.
Problem 4.73 The force F = 800 lb. The sum of
the moments about O due to the force F and the forces
exerted at A by the cables AB and AC is zero. What
are the tensions in the cables?
y
C
(0, 6, Ð10) ft
A
(8, 6, 0) ft
B
(0, 10, 4) ft
–Fj
x
O
z
Solution: The coordinates of the points O, A, B, C are O (0, 0,
0), A (8, 6, 0), B (0, 10, 4), C (0, 6, −10). The position vectors are:
y
C
(0, 6, Ð10) ft
rOA = 8i + 6j + 0k.|rOA | = 10 ft
B
rAB = (0 − 8)i + (10 − 6)j + (4 − 0)k
(0, 10, 4) ft
A
(8, 6, 0) ft
= −8i + 4j + 4k|rAB | = 9.798 ft
–Fj
rAC = (0 − 8)i + (6 − 6)j + (−10 − 0)k
= −8i + 0j − 10k.|rAC | = 12.806 ft.
O
The line of action of the forces intersect at point A. By Varignon’s
theorem, the forces at point A are in equilibrium if the moments vanish
about O. The unit vectors are
eOA = 0.8i + 0.6j + 0k,
eAB = −0.8165i + 0.4082j + 0.4082k,
eAC = −0.6247i + 0j − 0.7809k.
The forces are:
FOA = |FOA |eOA ,
FAB = |FAB |eAB ,
FAC = |FAC |eAC .
The equilibrium conditions are
F = F + FOA + FAB + FAC = 0.
Combining and collecting terms
FX = (0.8|FOA | − 0.8165|FAB | − 0.6247|FAC |)i = 0
FY = (0.6|FOA | + 0.4082|FAB | − 800)j = 0
FZ = (0.4082|FAB | − 0.7809|FAC |)k = 0.
These equations were solved using the TK Solver Plus commercial
software package. The result:
|FOA | = 903.23 lb,
|FAB | = 632.13 lb,
|FAC | = 330.48 lb.
z
x
Problem 4.74 In Problem 4.73, the sum of the moments about O due to the force F and the forces exerted
at A by the cables AB and AC is zero. Each cable
will safely support a tension of 2000 lb. Based on this
criterion, what is the largest safe value of the force F ?
Solution: For Problem 4.73, the loads, for a value of |F| = 800 lb, we get
|FOA | = 903.23 lb. |FAB | = 632.13 lb, and |FAC | = 330.48 lb. To scale
the problem, we must make |FAB | = 2000 lb. The problem is linear, so we
may scale everything up. The scaling factor is k = 2000/632.13 = 3.164.
Scaling the 800 lb load by this factor gives FMAX = 2531 lb.
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 = 2 m, y = z = 0) due to the force
exerted on the slider by the cable.
y
2m
B
A
5m
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.
The unit vector from A to B is determined from the coordinates of
points A and B A(2, 2, 0), B(0, 5, 2) m
2m
2m
x
C
z
Thus, rAB = −2i + 3j + 2k m
eAB = −0.485i + 0.728j + 0.485k
and
y
FAB = FAB eAB
2m
B
The horizontal force exerted by the bar on the slider is
H = Hx i + Hz k
Equilibrium requires H + FAB − mgj = 0
5m
i: Hx − 0.485FAB = 0
m = 200 kg
j: 0.728FAB − mg = 0
g = 9.81 m/s2
A
2m
k: Hz + 0.485FAB = 0
2m
Solving, we get
C
z
FAB = 2697N = 2, 70 kN
Hx = 1308N = 1.31 kN
FAB
Hz = −1308N = −1.31 kN
rCA = 2j m
FAB = FAB eAB
H
FAB = −1308i + 1962j + 1308k N
Mc =
i
0
−1308
j
2
1962
k
0
1308
Mc = 2616i + 0j + 2616k N-m
Mc = 2.62i + 2.62i kN-m
−mg j
x
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?
y
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.
y
5m
5m
Our first task is to write unit vectors eAB , eAO , eAC , and eAD . Each
will be of the form
C
(xi − xA )i + (yi − yA )j + (zi − zA )k
eAi = (xi − xA )2 + (yi − yA )2 + (zi − zA )2
D
4m
where i takes on the values B, C, D, and O. We get
8m
eAB = 0.986i + 0.164j + 0k
(6, 2, 0) m
eAC = −0.609i + 0.609j + 0.508k
O
eAD = −0.744i + 0.248j − 0.620k
eAO = −0.949i − 0.316j + 0k
B
A
z
3m
12 m
We now write the forces as
x
TAB = TAB eAB
TAC = TAC eAC
D (0, 4, −5) m
TAD = TAD eAD
C (0, 8, 5) m
TAD
TAO = TAO eAO
TAC
We then sum the forces and write the equilibrium equations in component form.
For equilibrium at A,
FA = 0
FA = TAB + TAC + TAD + TAO = 0.
In component form,
A(6, 2, 0) m
TAB
O(0, 0, 0) m
We now know that TAB is given as
TAB = TAB eAB = 160.8i + 26.8j (kN)
TAB eABx + TAC eACx + TAD eADx + TAO eAOx = 0
TAB eABy + TAC eACy + TAD eADy + TAO eAOy = 0
TAB eABz + TAC eACz + TAD eADz + TAO eAOz = 0
We know TAC = 22 kN. Substituting this in, we have 3 eqns in 3
unknowns. Solving, we get
TAB = 163.05 kN,
TAO
TAD = 18.01 kN
TAO = 141.28 kN
and that the force acting at B is (−TAB ).
The moment about the bottom of the post is given by
MBOTTOM = r × (−TAB ) = 3j × (−TAB )
Solving, we get
MBOTTOM = 482k (kN-m)
B(12, 3, 0) m
Problem 4.77 Use Eqs. (4.5) and (4.6) to determine
the moment of the 40-N force about the z axis. (First
see if you can write down the result without using the
equations.)
y
40j (N)
x
(8, 0, 0) m
z
Solution: By inspection, the moment should be 8 m × 40 N in
the negative z direction (using the right hand rule). Thus, we expect
the result to be
Mzaxis = −320k (N-m)
Equation (4.5) is
ML = [e · (r × F)]e
where Eq. (4.6) can be used to evaluate e · (r × F). Eqn. (4.6) is
ex
e · (r × F) = rx
Fx
ey
ry
Fy
ez
rz
Fz
For our problem, e = kr = 8i and F = −40j
Thus, from Eq. (4.6)
0
e · (r × F) = 8
0
0
0
−40
1
0 = −320 (N-m)
0
and from Eq. (4.5)
ML = (e · (r × F))e = −320k
y
40j(N)
x
(8,0,0) m
z
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.)
y
(7, 4, 0) m
20 k (N)
x
z
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 from
√ the y axis is√7 m and the perpendicular
distance from the z axis is 42 + 72 = 65 m.
By inspection, the moment about the x axis is
Mx = (4)(20)i (N-m)
Mx = 80i (N-m)
By inspection, the moment about the y axis is My = (7)(20)(−j) Nm
My = −140j (N-m)
By inspection, the moment about the z-axis is zero since F is parallel
to the z-axis.
Mz = 0 (N-m)
Now for the calculations using (4.5) and (4.6)
ML = [e · (r × F)]e
1
Mx = 7
0
0
4
0
0
0 i = 80i (N-m)
20
0
My = 7
0
1
4
0
0
0 j = −140j (N-m)
20
0
Mz = 7
0
0
4
0
1
0 k = 0k (N-m)
20
y
(7, 4, 0) m
20k (N)
x
z
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.)
y
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.
y
3 kN
Mx = [(2)(0.6) − (6)(0.6)]i kN
Mx = −2.4i kN
x
2 kN
6 kN
900 mm
Calculating the result:
z
1
M3 kN = 0
0
0
0
−3
0
0 i = 0i kN
0
1
M2 kN = 0
0
0
0
−2
0
.6 i = 1.2i kN
0
1 0 0
M6 kN = 0 0 .6 i = −3.6i kN
0 6 0
Mx = M3 kN + M2 kN + M6 kN
Mx = 0 + 1.2i − 3.6i (kN)
Mx = −2.4i (kN)
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
M = eY · (r × F). The position vectors of the three forces are given
in the solution to Problem 4.79. The magnitude for each force is:
eY · (r × F) =
0
0.9
0
eY
0
· (r × F) = 0.9
0
eY
0
· (r × F) = 0
0
1
0
−3
0
0 = 0,
0
1
0
6
0
0.6
0
= 0,
1
0
−2
0
0.6
0
=0
Thus the moment about the y-axis is zero, since the magnitude of each
moment is zero.
(b) The magnitude of each moment about the z-axis is
eZ · (r × F) =
0
0.9
0
eZ · (r × F) =
0
0.9
0
+
0
eZ · (r × F) = 0
0
1 0
0 0 = −2.7,
−3 0
0 1
0
0.6
6 0
0 1
0 0.6
−2
0
= 5.4,
= 0.
Thus the moment about the z-axis is
MZ = −2.7eZ + 5.4eZ = 2.7k (kN-m)
600 mm
Problem 4.81 The person exerts a force F = 0.2i −
0.4j + 1.2k (lb) on the gate at C. Point C lies in the
x − y 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:
y
M = [e · (r × F)]e
e = j,
MY
0
= 2
.2
r = 2i ft,
1
0
−.4
F is given
0
0 j = −2.4j (ft-lb)
1.2
A
C
3.5 ft
x
B
Problem 4.82 Four forces parallel to the y axis act
on the rectangular plate. The sum of the forces in the
positive y direction is 200 lb. The sum of the moments
of the forces about the x axis is −300i (ft-lb) and the sum
of the moments about the z axis is 400k (ft-lb). What
are the magnitudes of the forces?
2 ft
y
F1
100 lb
F3
x
O
3 ft
F2
4 ft
z
Solution:
Sum of forces in y-direction:
y
F1 − F2 + F3 − 100 = 200 lb (1)
100 lb
Sum of Moments about x-axis:
or
Mx
Mx
1
= 0
0
0
0
−F2
F1
0
1
3 i+ 0
0
0
0
0
F3
0
3 i+0+0
0
x
F2
3 ft
= 3F2 i − 3F3 i = −300i
3F2 − 3F3 = −300 (ft-lb)
(2)
4 ft
Sum of moments about the z-axis:
or
0 1
0
0 0 k+ 4
−100
0
0
z
Mz
0
= 4
0
Mz
= −400k + 4F3 k = 400k (ft-lb)
− 400 + 4F3 = 400 (ft-lb)
F3
0
0
F3
1
0 k+0+0
0
Solving (1), (2), and (3) simultaneously, we get
F1 = 200 lb, F2 = 100 lb, F3 = 200 lb.
(3)
Problem 4.83 The force F = 100i + 60j − 40k (lb).
What is the moment of F about the y axis? Draw a sketch
to indicate the sense of the moment.
y
F
(4, 2, 2) ft
x
z
Solution: The radius vector to the point of application of the force
is r = 4i + 2j − 2k. The magnitude of the moment is
eY · (r × F) =
0
4
100
1
2
60
0
− 2 = −40 ft lb
−40
The moment is MY = −40eY = −40j. The sense of the moment is
the direction of the curled fingers of the right hand if the thumb is held
parallel to the negative y-axis (pointing thumb down in the sketch).
y
F
(4,2,−2) ft
z
x
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 = 80 lb, what are
Fx and Fz ?
Solution:
The magnitudes of the moments:
eX
e • (r × F) = rX
FX
eY
rY
FY
0
4
FX
0
2
80
eZ · (r × F) =
eZ
rZ ,
FZ
1
−2 = 320 − 2FX = 160
FZ
Solve: FX = 80 lb, FZ = 40 lb, from which the force vector is
F = 80i + 80j + 40k
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 θx = 0.500, cos θy = 0.866, cos θz = 0,
and the direction cosines of the centerline of arm BC
are cos θx = 0.707, cos θy = 0.619, cos θz = −0.342.
What total moment is exerted about the z axis by the
weights of the arms?
m
y
0m
C
60
160 N
600 mm
B
200 N
A
z
Solution:
x
The unit vectors along AB and AC are of the form
e = cos θx i + cos θy j + cos θz k.
m
y
0m
C
60
The unit vectors are
eAB = 0.500i + 0.866j + 0k and eBC = 0.707i + 0.619j − 0.342k.
160 N
B
600 mm
The vector to point G at the center of arm AB is
rAG = 300(0.500i + 0.866j + 0k) = 150i + 259.8j + 0k mm,
and the vector from A to the point H at the center of arm BC is
given by
200 N
A
rAH = rAB + rBH = 600eAB + 300eBC
= 512.1i + 705.3j − 102.6k mm.
z
x
The weight vectors acting at G and H are WG = −200j N, and
WH = −160j N. The moment vectors of these forces about the z
axis are of the form
eX
e • (r × F) = rX
FX
ey
rY
FY
ez
rZ .
FZ
Here, WG and WH take on the role of F, and e = k.
Substituting into the form for the moment of the force at G, we get
0
e • (r × F ) = 0.150
0
0
0.260
−200
1
0 = −30 N-m.
0
Similarly, for the moment of the force at H, we get
0
e • (r × F) = 0.512
0
0
0.705
−160
1
−0.103 = −81.9 N-m.
0
The total moment about the z axis is the sum of the two moments. Hence,
Mz−axis = −111.9 N-m
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 = i. Substituting into the form for the moment of the force at
G, we get
1
e · (r × F) = 0.150
0
0
0.260
−200
0
0 = 0 N-m.
0
Similarly, for the moment of the force at H, we get
1
e · (r × F) = 0.512
0
0
0.705
−160
0
−0.103 = −16.4 N-m.
0
The total moment about the x axis is the sum of the two moments.
Hence, Mx−axis = −16.4 N-m
Problem 4.87
Two forces are exerted on the
crankshaft by the connecting rods. The direction cosines
of FA are cos θx = −0.182, cos θy = 0.818, and
cos θz = 0.545, and its magnitude is 4 kN. The direction
cosines of FB are cos θx = 0.182, cos θy = 0.818, and
cos θz = −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.)
y
FB
FA
360 mm
O
160 mm
z
80 mm
80 mm x
Solution: The coordinates of the points of action of the two forces
are A (0.16, 0, 0.08), B (0.36, 0, −0.08). The position vectors are
y
FA
rOA = 0.16i + 0j + 0.08k (m),
360 mm
rOB = 0.36i + 0j − 0.08k (m).
FB
O
The unit vectors parallel to the forces are given by the direction cosines:
eF A = −0.182i + 0.818j + 0.545k,
z
80 mm
160 mm
eF B = 0.182i + 0.818j − 0.545k
x
80 mm
The forces are
The sum of the moments about the x-axis is
FA = −0.728i + 3.272j + 2.18k (kN)
FB = 0.364i + 1.636j − 1.09k (kN)
MX = −0.2618eX + 0.1309eX = −0.1309i kN-m.
The magnitude of the moments:
eX · (rA × FA ) =
1
0.16
−0.728
eX · (rB × FB ) =
1
0.36
0.364
0
0
3.272
0
0
1.636
0
0.08 = −0.2618,
2.18
0
−0.08 = 0.1309
−1.09
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)
Solution:
B
(– 4, 0, 0) m
First, we need the unit vector
(xB − xA )i + (yB − yA )j + (zB − zA )k
eAB = (xB − xA )2 + (yB − yA )2 + (zB − zA )2
x
z
eAB = −0.625i − 0.781j = −eBA
Now, the moment of the 20k (N) force about AB is given as
ex
ML = rx
Fx
ey
ry
Fy
ez
rz e
Fz
y
A (0, 5, 0) m
where e is eAB or eBA
For this problem, r must go from line AB to the point of application
of the force. Let us use point A.
20 k (N)
B
r = 7i − 1j + 0k m
z
Using eAB
Using eBA
−0.625
=
7
0
−0.781
−1
0
0
0 (−0.625i − 0.781j)
20
ML = −76.1i − 95.1j (N-m)
ML =
Problem 4.89 The force F = −10i + 5j − 5k (kip).
Determine the moment of F about the line AB. Draw a
sketch to indicate the sense of the moment.
0.781
−1
0
0
0 (0.625i + 0.781j)
20
are the same
y
B
(6, 6, 0) ft
The moment of F about pt. A is
MA = −6i × F
i
= −6
−10
0.625
7
0
ML = −76.1i − 95.1j (N-m)
∗ Results
Solution:
x
(– 4, 0, 0) m
r = (7 − 0)i + (4 − 5)j + (0 − 0)k m
ML
(7, 4, 0) m
j
0
5
F
k
0
−5
A
x
(6, 0, 0) ft
= −30j − 30k (ft-kip).
z
The unit vector j is parallel to line AB, so the moment about AB is
MAB = (j · MA )j
y
= −30j (ft-kip).
B
−30j (ft-kip)
y
(6, 6, 0) ft
B
Direction of moment
x
F
z
A
x
A (6, 0, 0) ft
z
Problem 4.90 The force F = 10i + 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
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 = 6j+4k. The magnitude: F. The unit vector
parallel to OA is eOA = 0.8321j + 0.5547k. The vector from O to
the point of application of F is rOF = 8i + 6k. The magnitude of
the moment about OA is
x
(8, 0, 6) m
z
y
|MO | = eOA · (rOF × F) =
0
8
10
0.8321
0
12
0.5547
6
−6
(0, 6, 4)
F
A
= 89.8614 + 53.251 = 143.1 N-m.
x
O
The moment about OA is MOA = |MOA |eOA = 119.1j +
79.4k (N-m).
z
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.
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.
Solution:
(8, 0, 6)
y
A
(400, 300, 0) mm
The vector parallel to BA is
x
600 mm
rBA = (0.4 − 1)i + 0.3j − 0.6k = −0.6i + 0.3j − 0.6k.
B
1000 mm
The unit vector parallel to BA is
z
eBA = −0.6667i + 0.3333j − 0.6667k.
y
The moment about O is
MO = rOB × T =
(400, 300, 0) mm
A
i
1
−0.6667
j
0
0.3333
k
0.6
−0.66667
600 mm
MO = −0.2i + 0.2667j + 0.3333k.
The magnitude is
|MX | = eX · MO = −0.2 kN-m.
The moment is MX = −0.2i kN-m. The sense is clockwise when
viewed along the x-axis toward the origin.
B
z
1000 mm
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 y − z
plane.)
y
6 ft
20i – 60j (lb)
E
A
D
4 ft
2 ft
Solution:
From the figure, we see that the unit vector along the
line from A toward B is given by eAB = − sin 20◦ j + cos 20◦ k.
The position vector is rAD = 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 • (r × F) = rX
FX
ey
rY
FY
x
B
z
C
20°
4 ft
ez
rZ .
FZ
y
6 ft
20i Ð60j (lb)
For this case,
0
e • (r × F) = 4
20
− sin 20◦
0
−60
E
cos 20◦
0
= −240 cos 20◦ ft-lb
0
z
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 = − sin 20◦ j + cos 20◦ k.
The position vector is rBC = 4i ft. The coordinates of point C are
(4, −4 sin 20◦ , 4 cos 20◦ ). The unit vector along CE is and the force
vector is as shown in the figure.
The moment vector is a force about an axis is of the form
eX
e • (r × F) = rX
FX
ey
rY
FY
ez
rZ .
FZ
For this case,
e • (r × F) =
0
4
20
− sin 20◦
0
−60
cos 20◦
0
= −240 cos 20◦ ft-lb
0
= −225.5 ft-lb.
The negative sign is because the moment is opposite in direction to the
unit vector from A to B.
D
4 ft
2 ft
= −225.5 ft-lb.
The negative sign is because the moment is opposite in direction to
the unit vector from A to B.
A
B
20°
C
4 ft
x
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 = −2.4i − 0.6k (m), rOB = −2.2i + 0.7j − 1.2k (m),
rBA = (−2.4 + 2.2)i + (0 − 0.7)j + (−0.6 + 1.2)k (m)
= −0.2i − 0.7j + 0.6k (m).
The magnitude is |rBA | = 0.9434 m.
The unit vector parallel to BA is
eBA = −0.2120i − 0.7420j + 0.6360k.
The tension is TBA = 130eBA .
The moment of TBA about the origin is
MO = rOB × TBA =
or
i
−2.2
−27.56
j
0.7
−96.46
k
−1.2 ,
82.68
MO = −57.88i + 214.97j + 231.5k.
The magnitude of the moment about the y-axis is
|MY | = eY · MO = 214.97 N-m.
The moment is MY = eY (214.97) = 214.97j N-m.
y
x
B
A
z
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
Solution:
B (10, –2, 3) ft
The position vector from B to A is
y
443 ft-lb
rBA = (2 − 10)i + [5 − (−2)]j + (−2 − 3)k
= −8i + 7j − 5k (ft),
187 ft-lb
So the force exerted on B is
F = 200
x
rBA
= −136.2i + 119.2j − 85.1k (lb).
|rBA |
The moment of F about the origin O is
rOB × F =
i
10
−136.2
j
−2
119.2
919 ft-b
z
k
3
−85.1
= −187i + 443j + 919k (ft-lb).
The moments about the x, y, and z axes are
[(rOB × F) · i]i = −187i (ft-lb),
[(rOB × F) · j]j = 443j (ft-lb),
[(rOB × F) · k]k = 919k (ft-lb).
Problem 4.96 The total force exerted on the blades
of the turbine by the steam nozzle is F = 20i − 120j +
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)?
Solution:
i
MO = 0.1
20
y
Fixed
Rotating
The moment about the origin is
j
0.08
−120
x
k
0.3
100
= 44.0i − 4.0j − 13.6k (N-m).
The moment about the x axis is
(MO · i)i = 44.0i (N-m).
z
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
x
B
(14, 0, 14) m
z
The vector OC is rOC = 7j + 10k. The unit vector
parallel to OC is eOC = 0.573j + 0.819k. The position vectors of
A and B are
Solution:
y
A
(0, 7, 0) m
rOA = 7j (m), and rOB = 14i + 0j + 14k (m).
The vector parallel to BA is
C
(0, 7, 10) m
x
O
rBA = (0 − 14)i + (7 − 0)j + (0 − 14)k = −14i + 7j − 14k.
B
The magnitude:
|rBA | =
z
142 + 72 + 142 = 21 m.
The unit vector parallel to BA is
eBA = −0.6667i + 0.3333j − 0.6667k.
The tension acting on B is
TBA = −33.335i + 16.665j − 33.335k (N).
The moment about the origin is
MO = rOB × TBA =
i
14
−33.335
j
0
16.665
k
14
−33.335
= −233.31i + 233.31k.
The magnitude of the moment about the line OC is
|MOC | = eOC · MO = 191.2 N-m.
The moment about the line OC is
MOC = 191.1eOC = 109.6j + 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.
(14, 0, 14) m
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
B
C
6 ft
x
D
A (6, 0, 10) ft
z
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 = 8i + 6j, rOC = 3i + 6j, rOD = 3i. The vector
parallel to CD is rCD = rOD − rOC = −6j. The unit vector
parallel to CD is eCD = −1j. The vector from point C to B is
rCB = rOB − rOC = 5i.
The position vector of A is rOA = 6i + 10k. The vector parallel to
BA is rBA = rOA − rOB = −2i − 6j + 10k. The magnitude is
|rBA | = 11.832 ft. The unit vector parallel to BA is
eBA = −0.1690i − 0.5071j + 0.8452k.
The tension acting at B is
TBA = 80eBA = −13.52i − 40.57j + 67.62k.
The magnitude of the moment about CD due to the tension acting at
B is
|MCD | = eCD · (rCB × TBA ) =
0
5
−13.52
−1
0
−40.57
0
0
67.62
= 338.1 (ft lb).
The moment about CD is MCD = 338.1eCD = −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.
y
8 ft
3 ft
C
B
6 ft
D
z
x
A
(6, 0, 10)
Problem 4.99 The universal joint is connected to the
drive shaft at A and A . The coordinates of A are (0, 40,
0) mm, and the coordinates of A are (0, −40, 0) mm.
The forces exerted on the drive shaft by the universal
joint are −30j + 400k (N) at A and 30j − 400k (N)
at A . What is the magnitude of the torque (moment)
exerted by the universal joint on the drive shaft about the
shaft axis O-O ?
y
Universal joint
A
O
Solution: The position vectors of A and A are rOA =
40j (mm), and rOA = −40j (mm). The magnitudes of the moments about the origin are:
1
|MOO | = eX · (rOA × F) = 0
0
0
40
−30
0
0 = 16000 (N-mm),
400
MOO = 16000eX = 16000i
1
× F) = 0
0
|MOO | = eX · (rOA
0
−40
30
Drive shaft
0
0
= 16000 (N-mm)
−400
O'
A'
Universal joint
y
Drive shaft
A
x
O′
0
A′
MOO = 16000eX = 16000i
The sum of the moments is
MOO = 32000i (N-mm) = 32i (N-m)
Problem 4.100 A motorist applies the two forces
shown to loosen a lug nut. The direction cosines of
4
3
F are cos θx = 13
, cos θy = 12
13 , and cos θz = 13 . If
the magnitude of the moment about the x axis must be
32 ft-lb to loosen the nut, what is the magnitude of the
forces the motorist must apply?
y
–F
F
z
Solution: The unit vectors for the forces are the direction cosines.
The position vector of the force F is rOF = −1.333k ft. The magnitude of the moment due to F is
|MOF | = eX · (rOF
1
× F) =
0
0.3077F
0
0
0.9231F
16 in
16 in
0
−1.333
0.2308F
x
|MOF | = 1.230F ft lb.
y
The magnitude of the moment due to −F is
|M−OF | = eX · (r−OF × −F)
=
1
0
−.3077F
0
0
−0.9231F
0
1.333
= 1.230F ft lb.
−0.2308F
The total moment about the x-axis is
F
z
MX = 1.230F i + 1.230F i = 2.46F i,
from which, for a total magnitude of 32 ft lb, the force to be applied is
16 in
32
= 13 lb
F =
2.46
16 in
x
x
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
2m
D
B
3m
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 = −1j. The position vectors
parallel to DC, DA, and DB:
rDC = 2j, rDA = 2i + 2j, rDB = 3i + 1k.
The vector parallel to CA is rCA = 2i. The vector parallel to AB is
rAB = rDB − rDA = 1i − 2j + 1k.
The magnitude: |rAB | = 2.4495 m. The unit vector parallel to AB
is
eAB = 0.4082i − 0.8165j + 0.4082k.
The tension is
TAB = 2eAB = 0.8165i − 1.633j + 0.8165k.
The magnitude of the moment about CD is
|MCD | = eCD · (rCA × TAB ) =
0
2
0.8164
−1
0
−1.633
0
0
0.8165
= 1.633 kN-m.
The moment about CD is
MCD = eCD |MCD | = −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.
2m
C
A
2m
F
D
B
3m
1m
1m
Problem 4.102 The axis of the car’s wheel passes
through the origin of the coordinate system and its direction cosines are cos θx = 0.940, cos θy = 0, cos θz =
0.342. The force exerted on the tire by the road effectively acts at the point x = 0, y = −0.36 m, z = 0 and
has components F = −720i+3660j+1240k (N). What
is the moment of F about the wheel’s axis?
y
x
z
Solution: We have to determine the moment about the axle where
a unit vector along the axle is
e = cos θx i + cos θy j + cos θz k
e = 0.940i + 0j + 0.342k
The vector from the origin to the point of contact with the road is
r = 0i − 0.36j + 0k m
The force exerted at the point of contact is
F = −720i + 3660j + 1240k N
The moment of the force F about the axle is
MAXLE = [e · (r × F)]e
MAXLE =
0.940
0
−720
0
−0.36
+3660
0.342
0
(0.940i + 0.342k) (N-m)
+1240
MAXLE = (−508.26)(0.940i + 0.342k) (N-m)
MAXLE = −478i − 174k (N-m)
y
x
z
Problem 4.103 The direction cosines of the centerline
OA are cos θx = 0.500, cos θy = 0.866, and cos θz = 0,
and the direction cosines of the line AG are cos θx =
0.707, cos θy = 0.619, and cos θz = −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.
m
G
m
50
y
7
250 N
600 mm
A
O
z
x
Solution: By definition, the direction cosines are the scalar components of the unit vectors. Thus the unit vectors are
y
750 mm
e1 = 0.5i + 0.866j, and e2 = 0.707i + 0.619j − 0.341k.
G
The force is W = 250j (N). The position vector of the 250 N weight is
250 N
rW = 0.600e1 + 0.750e2 = 0.8303i + 0.9839j − 0.2565k
600 mm
The moment about OA is
A
MOA = eOA (eOA · (rW × W))
0.5
= 0.8303
0
0.866
0.9839
−250
0
−0.2565 e1 = −32.06e1
0
= −16i − 27.77j (N-m)
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
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 = 10i + 10j − 5k (N) on the wheel at A.
If the angle α = 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
F
C
A
20°
O
α
z
x
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:
◦
y
C
◦
rOC = 1(i cos 20 + j sin 20 + 0k) = 0.9397i + 0.3420j.
A
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
O
20°
z
α
eCX = i cos(20 − 90) + j sin(20 − 90) = 0.3420i − 0.9397j,
x
where the angle is measured positive counterclockwise from the x-axis.
The vector parallel to CA is
rCA = 0.2eCX = +0.0684i − 0.1879j (m).
The magnitude of the moment about OC
0.9397
|MOC | = eOC · (rCA × F) = 0.0684
10
0.3420
−0.1879
10
0
0
−5
= 0.9998 = 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.
Problem 4.105 Consider the steering wheel in Problem 4.104. Determine the moment of F about the shaft
OC of the steering wheel if α = 30◦ . Draw a sketch to
indicate the sense of the moment about the shaft.
Solution:
The position vector of A relative to C is
The sense of the moment is in the direction of the curled fingers of the right
hand with the thumb pointing from 0 to C.
rCA = R cos α sin 20◦ i − R cos α cos 20◦ j − R sin αk
y
= 0.0592i − 0.1628j − 0.1k (m).
The moment about pt. C is
C
R cos α
R = 0.2 m.
1m
MC
i
= 0.0592
10
j
−0.1628
10
k
−0.1
−5
= 1.814i − 0.704j + 2.220k (N-m).
A unit vector parallel to the shaft is
eOC = cos 20◦ i + sin 20◦ j,
and eOC · MC = 1.464 N-m
20°
O
A
x
Problem 4.106 The weight W causes a tension of
100 lb in cable CD. If 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
D
(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:
W
C
d
z
x
(3, 0, 10) ft
rOB = 3i + 0j + 10k, rOT = 12i + 10j + 0k.
The vector from the bottom to the top of the bar is
y
(12, 10, 0)
rBT = rOT − rOB = 9i + 10j − 10k.
The magnitude:
|rBT | =
(0, 3, 0)
92 + 102 + 102 = 16.763 ft.
The unit vector parallel to the bar, pointing toward the top, is
D
W
C
eBT = 0.5369i + 0.5965j − 0.5965k.
x
The position vector of the point C relative to the bottom of the bar is
rBC = 2eBT = 1.074i + 1.193j − 1.193k.
The position vector of point C relative to the origin is
rOC = rOB + rBC = 4.074i + 1.193j + 8.807k.
The position vector of point D is
rOD = 0i + 3j + 0k.
The vector parallel to CD is
rCD = rOD − rOC = −4.074i + 1.807j − 8.807k.
The magnitude is
|rCD | =
4.0742 + 1.8072 + 8.8072 = 9.87 ft.
The unit vector parallel to CD is
eCD = −0.4127i + 0.1831j − 0.8923k.
The tension is
TCD = 100eCD = −41.27i + 18.31j − 89.23k lb.
The magnitude of the moment about the z-axis is
|MO | = eZ · (rOC × TCD ) =
= 123.83 ft lb
0
0
4.074 1.193
−41.27 18.31
1
8.807
−89.23
z
(3, 0, 10) ft
d
Problem 4.107 The rod AB supports the open hood
of the car. The force exerted by the rod on the hood at B
is parallel to the rod. If the rod must exert a moment of
100 ft-lb magnitude about the x axis to support the hood
and the distance d = 2 ft, what is the magnitude of the
force the rod must exert on the hood?
B
A
y
(–1, 1, 2) ft
Solution: The coordinates of B are B (− 1, 1, 2). The position
vectors of A, B are
A
z
rOA = 2k, rOB = −1i + 1j + 2k.
The vector parallel to AB is
rAB = rOB − rOA = −1i + 1j.
The unit vector is eAB = −0.7071i + 0.7071j. The force is FAB =
F eAB . The moment about the origin is
|MX | = eX · (rOA × FAB ) =
1
0
−0.7071F
from which, for
|MX | = 100 ft lb, F =
100
= 70.71 lb
1.414
y
y
(−1, 1, 2) ft
B
x
d
z
A
0
0
0.7071F
0
2 = −1.414F,
0
B
d
x
Problem 4.108 Determine the moment of the couple
and represent it as shown in Fig. 4.28(c).
y
10 j (N)
x
(4, 0, 0) m
–10 j (N)
Solution:
The moment of the couple is
y
M = r × F = 4i × 10j = 40k (N-m)
10j (N)
r
(4, 0, 0)
x
Ð10j (N)
y
40 (N-m)
x
Problem 4.109 The forces are contained in the x − y
plane.
(a) Determine the moment of the couple and represent
it as shown in Fig. 4.28c.
(b) What is the sum of the moments of the two forces
about the point (10, −40, 20) ft?
y
1000 lb
1000 lb
60°
60°
x
20 ft
Solution:
20 ft
The right hand force is
y
F = [1000 (lb)](cos 60◦ i − sin 60◦ j)
1000 lb
1000 lb
F = +500i − 867j lb.
60°
60°
The vector from the x intercept of the left force to that of the right
force is r = 40i ft.
The moment is MC = r × F
MC = 40i × (500i − 867j) (ft-lb)
MC = −34700 (ft-lb) k
or
MC = −34700 (ft-lb) clockwise
x
20 ft
20 ft
Problem 4.110 The forces are contained in the x − y
plane and the moment of the couple is −110k (N-m).
(a) What is the distance b?
(b) What is the sum of the moments of the two forces
about the point (3, −3, 2) m?
y
50 N
35°
x
b
35°
50 N
Solution:
The right hand-force is
FR = 50(cos 35◦ i − sin 35◦ j)
FR = 40.96i − 28.68j N
The moment of the couple is −110 (N-m). Also,
MC = bi × (40.96i − 28.6j) :
−110k = −28.68 bk,
b = 3.84 m
The moment of the couple about any point is the same, −110k (N-m)
To show this, we can find the moments about a point (3, −3, 2) m. Use
the following as the points of application of the two forces.
Left force (0, 0, 0) m
Right force (3.48, 0, 0) m
Moment of Left force
ML = rP O × FL
ML = (−3i + 3j − 2k) × (−40.96i + 28.68j)
ML = 57.36i + 81.92j + 36.9k (N-m)
Moment of Right Force
MR = (0.84i + 3j − 2k) × (40.96i − 28.68j)
MR = −57.36i − 81.92j − 146.9k (N-m)
M = −110k (N-m)
y
50 N
35°
b
35°
50 N
x
Problem 4.111 Point P is contained in the x−y plane,
|F| = 100 N, and the moment of the couple is −500k (Nm). What are the coordinates of P ?
y
P
30°
F
–F
70°
x
Solution:
The force is
F = 100(i cos(−30◦ ) + j sin(−30◦ )) = 86.6i − 50j.
Let r be the distance OP . The vector parallel to OP is
r = r(i cos 70◦ + j sin 70◦ ) = r(0.3420i + 0.9397j).
The moment is
i
M = r × F = 0.3420r
86.6
From which, r =
500
98.48
j
0.9397r
−50.0
k
0 = −98.48rk.
0
= 5.077 m. From above,
r = 5.077(0.3420i + 0.9397j).
The coordinates of P are
x = 5.077(0.3420) = 1.74 m, y = 5.077(0.9397) = 4.77 m
y
P
30°
−F
70°
F
x
Problem 4.112 The forces are contained in the x − y
plane.
(a) Determine the sum of the moments of the two couples.
(b) What is the sum of the moments of the four forces
about the point (−6, −6, 2) m?
(c) Represent the result of (a) as shown in Fig. 4.28c.
y
100 N
4m
100 N
2m
x
2m
100 N
4m
100 N
The position vectors for the forces on the x-axis are
rX1 = 2i, rX2 = −2i. The position vectors for the forces on the
y-axis are rY 1 = 4j, rY 2 = −4j. The force on the positive xaxis is FX = +100j (N). The force on the positive y-axis is FY =
+100i (N).
Solution:
(a)
(c)
The figure is shown.
y
100 N
The sum of the moments is
4m
M = (rX1 − rX2 ) × FX + (rY 1 − rY 2 ) × FY
i
= 4
0
j
0
100
k
i
0 + 0
0
100
100 N
j k
8 0 = −400k (N-m)
0 0
x
100 N
(b)
2m
The vector from the point P (−6, −6, 2) to the force on the
positive x-axis is
rP X1 = rX1 − rp = (2 + 6)i + 6j − 2k.
100 N
The vector from the point P to the force on the negative x-axis is
rpX2 = rX2 − rp = (−2 + 6)i + 6j − 2k.
The vector from point P to the force on the positive y-axis is
rpY 1 = rY 1 − rp = +6i + (4 + 6)j − 2k = 6i + 10j − 2k.
The vector from P to the force on the negative y-axis is
rP Y 2 = rY 2 − rP = 6i + (−4 + 6)j − 2k = 6i + 2j − 2k.
The sum of the moments:
M = (rpX1 − rpX2 ) × FX + (rpY 1 − rpY 2 ) × FY
i
j
= 4
0
0 100
k
i
0 + 0
0
100
2m
j k
8 0 = −400k (N-m)
0 0
y
400 k
x
4m
Problem 4.113 The moment of the couple is 40 kN-m
counterclockwise.
(a) Express the moment of the couple as a vector.
(b) Draw a sketch showing two equal and opposite
forces that exert the given moment.
y
40 kN-m
x
Solution:
directed
The moment is
M = 40k (kN-m)
(b)
A candidate pair of forces is shown in the sketch: 40 kN forces
directed oppositely at 1 m apart.
y
along
the
positive
z-axis,
(a)
y
40 kN
40 kN-m
1m
x
x
40 kN
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 = −50k + 10k = −40k ft lb
y
50 ft lb
P
(−4,0,0)ft
x
10 ft lb
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
y
30 lb
M30 = −3(30)k = −90k ft lb.
3 ft
30 lb
The moment due to the 20 lb couple, which acts in a counterclockwise
direction, is
2 ft
x
20 lb
20 lb
M20 = 9(20)k = 180k ft lb.
5 ft
4 ft
The sum of the moments is
M = −90k + 180k = +90k ft lb.
The sum of the moments is the same anywhere on the plate.
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
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
B
4 ft
3 ft
4 ft
Solution:
400 lb
100 lb
900 ft-lb
M100 = (−3i) × (100j) = −300k (ft-lb),
The moment of the couple is MC = 900k (ft-lb). Summing the
moments, we get
MTotal = −2200k (ft-lb)
B
A
and M400 = (7i) × (−400j) = −2800k (ft-lb).
3 ft
4 ft
3 ft
4 ft
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:
y
MA = (0.2i) × (−200j) + (0.4i + 0.2j)
100 N
30°
×(86.7i + 50j) + 300k (N-m)
MA = −40k + 2.66k + 300k (N-m)
200 N
0.2 m
A
MA = 262.7k (N-m) 263k (N-m)
300 N-m
0.2
m
Problem 4.118
on the object?
0.2
m
0.2
m
What is the sum of the moment exerted
y
40 N
4m
100 N-m
40 N
x
30 N
30 N
6m
3m
Solution: Three couples act on the object. The moments due to
the couples are:
y
40 N
M1 = −100k N-m.
M2 = 9(30)k = 270k N-m,
40 N
M3 = 4(40)k = 160k N-m
30 N
30 N
The sum of the moments:
M = −100k + 270k + 160k = 330k N-m
4m
100 N-m
6m
3m
x
x
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?
Solution:
y
800 N
200 N-m
Ax
B
The sum of the forces about the y-axis is
Ay
FX = AY + B − 800 = 0.
x
4m
The sum of the forces about the x-axis is
3m
FX = AX = 0.
y
The sum of the moments about the left end of the beam is
4m
ML = 11B − 8(800) − 200 = 0.
800 N
B =
200 N-m
Ax
From the moments:
6600
= 600 N.
11
Ay
B
x
Substitute into the forces balance equation to obtain:
4m
AY = 800 − 600 = 200 N
Problem 4.120 The force F = 40i + 24j + 12k (N).
(a) What is the moment of the couple?
(b) Determine the perpendicular distance between the
lines of action of the two forces.
4m
3m
y
F
(6, 3, 2) m
x
(10, 0, 1) m
–F
z
Solution:
(a)
y
The moment of the couple is given
MC = rAB × F
F
MC = (−4i + 3j + 1k) × (40i + 24j + 12k)
MC = 12i + 88j − 216k (N-m)
B
(6, 3, 2) m
(b) |MC | = |d||F| sin 90◦
|F| = Fx2 + Fy2 + Fz2 = 48.2 N
|MC | =
Mx2
+
My2
+
Mz2
x
= 233.5 N
(10, 0, 1) m
|d| = perpendicular distance
|d| = |MC |/|F|
|d| = 4.85 m
A
z
−F
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 x-z plane.)
y
3 ft
20 lb
3 ft
20 lb
40 lb
x
8 ft
40 lb
Solution: The moments of two of the couples can be determined
from inspection:
60°
z
60°
80 lb
M1 = −(3)(20)k = −60k ft lb.
y
3 ft
3 ft
M2 = (8)(40)j = 320j ft lb
80 lb
20 lb
The forces in the 3rd couple are resolved:
20 lb
40 lb
x
F = (80)(i sin 60◦ + k cos 60◦ ) = 69.282i + 40k
8 ft
The two forces in the third couple are separated by the vector
40 lb
80 lb
z
r3 = (6i + 8k) − (8k) = 6i
80 lb
60°
60°
The moment is
The sum of the moments due to the couples:
M3
i
= r3 × F3 =
6
69.282
j k
0 0 = −240j.
0 40
M = −60k + 320j − 240j = 80j − 60k ft lb
Problem 4.122 What is the magnitude of the sum of
the moments exerted on the T -shaped structure by the
two couples?
Solution:
3 ft
y
50i + 20j – 10k (lb)
3 ft
50j (lb)
The moment of the 50 lb couple can be determined by
inspection:
3 ft
z
–50j (lb)
3 ft
M1 = −(50)(3)k = −150k ft lb.
The vector separating the other two force is r = 6k. The moment is
M2 = r × F =
i
0
50
j
0
20
k
6
= −120i + 300j.
−10
–50i – 20j + 10k (lb)
y
3 ft
F
3 ft
50 j (lb)
The sum of the moments is
3 ft
x
M = −120i + 300j − 150k.
–F
–50j (lb)
z
The magnitude is
|M| =
1202 + 3002 + 1502 = 356.23 ft lb
3 ft
x
Problem 4.123 The tension in cables AB and CD is
500 N.
(a) Show that the two forces exerted by the cables on
the rectangular hatch at B and C form a couple.
(b) What is the moment exerted on the plate by the cables?
y
A (0, 2, 0) m
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 = 3i m. The vector position of point A is rA = 2j.
The vector parallel to cable AB is
(6, –2, 3) m
y
A
(0, 2, 0) m
B
x
3m
rBA = rA − rB = −3i + 2j.
C
z
3m
D
(6, –2, 3) m
The magnitude is:
|rAB | =
32 + 22 = 3.606 m.
The moment about the origin is
MO = (rB − rC ) × TAB = rCB × TAB ,
The unit vector:
eAB =
rAB
= −0.8321i + 0.5547j.
|rAB |
which is identical with the above expression for the moment. Let rP C and rP B
be the distances to points C and B from an arbitrary point P on the plate. Then
MP = (rP B − rP C ) × TAB = rCB × 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 tension is
TAB = |TAB |eAB = −416.05i + 277.35j.
The vector position of points C and D are:
rC = 3i + 3k, rD = 6i − 2j + 3k.
The vector parallel to the cable CD is rCD = rD − rC = 3i − 2j.
The magnitude is |rCD | = 3.606 m, and the unit vector parallel to
the cable CD is eCD = +0.8321i − 0.5547j. The magnitude of
the tension in the two cables is the same, and eBA = −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 = rB − rC = 3i − 3i − 3k = −3k.
The moment is
M = rCB × TAB =
i
0
−416.05
= 832.05i − 1248.15j (N-m).
j
0
277.35
k
−3
0
Problem 4.124 Determine the sum of the moments
exerted about P by the couple and the two forces.
y
P
FB =
2i – j (kN)
FA =
– i + j + k (kN)
x
MC =
4i – 4j + 4k (kN-m)
1m
z
Solution:
y
MP = rP A × FA + rP B × FB + MC
P
MP = (−1i − 1j + 1k) × (−1i + 1j + 1k) (kN-m)
+(0i + 0j + 1k) × (2i − 1j + 0k) (kN-m)
B
+4i − 4j + 4k (kN-m)
MP = (−2i + 0j − 2k)
FA =
– i + j + k (kN)
+(1i + 2j + 0k)
x
A
+(4i − 4j + 4k) (kN-m)
z
The bar is loaded by the forces
y
FB = 2i + 6j + 3k (kN),
FB
FC = i − 2j + 2k (kN),
A
MC = 2i + j − 2k (kN-m).
Determine the sum of the moments of the two forces and
the couple about A.
MC
B
and the couple
Solution:
MC =
4i – 4j + 4k (kN-m)
1m
MP = 3i − 2j + 2k (kN-m)
Problem 4.125
FB =
2i – j (kN)
C
x
1m
1m
z
FC
The moments of the two forces about A are given by
y
MF B = (1i) × (2i + 6j + 3k) (kN-m) = 0i − 3j + 6k (kN-m) and
FB
MF C = (2i) × (1i − 2j + 2k) (kN-m) = 0i − 4j − 4k (kN-m).
A
Adding these two moments and
z
MC = 2i + 1j − 2k (kN-m),
we get MTOTAL = 2i − 6j + 0k (kN-m)
1m
B
C
MC
x
1m
FC
Problem 4.126
In Problem 4.125, the forces
FB = 2i + 6j + 3k (kN),
FC = i − 2j + 2k (kN),
Solution: From the solution to Problem 4.125, the sum of the moments of
the two forces about A is
MForces = 0i − 7j + 2k (kN-m).
and the couple
The required moment, MC , must be the negative of this sum.
MC = MCy j + MCz k (kN-m).
Determine the values for MCy and MCz , so that the sum Thus
of the moments of the two forces and the couple about
A is zero.
Problem 4.127 Two wrenches are used to tighten an
elbow fitting. The force F = 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) Determine the moment about the x axis due to the
force exerted on the right wrench.
(b) Determine the moment of the couple formed by the
forces exerted on the two wrenches.
(c) Based on the results of (a) and (b), explain why two
wrenches are used.
MCy = 7 (kN-m), and MCz = −2 (kN-m).
y
z
x
F
–F
from which MXL = 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.
Solution: The position vector of the force on the right wrench is
rR = 6i − 5j − 3k. The magnitude of the moment about the x-axis
is
1
|MR | = eX · (rR × F) = 6
0
(a)
0
−5
0
0
−3 = −50 in lb
10
The moment about the x-axis is
y
z
x
MR = |MR |eX = −50i (in lb).
(b)
The moment of the couple is
–F
F
MC = (rR − rL ) × FR
(c)
i
= 2
0
j
0
0
k
−6 = −20j in lb
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
|MX | = eX · (rL × FL ) = 4
0
0
−5
0
0
3
= 50 in lb
−10
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 = 0, (both systems) and
FY 1 = −100j + 50j = −50j (N)
y
100 N
x
50 N
FY 2 = −50j (N).
1m
The sums of the forces are equal. The sums of the moments about the
left end are:
y
y
M1 = −(1)(100)k = −100k (N-m)
M2 = −(2)(50)k = −100k (N-m).
The sums of the moments about the left end are equal. Choose any
point P at the same distance r = xi from the left end on each beam.
The sums of the moments about the point P are
M1 = (−50x + 100(x − 1))k = (50x − 100)k (N-m)
M2 = (−50(2 − x))k = (50x − 100)k (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
System 1
1m
System 2
50 N
x
2m
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
Solution:
The sums of the forces are:
FX = 0 (both systems)
50 ft-lb
10 lb
20 lb
x
FY 1 = 10j − 20j = −10j (lb)
2 ft
FY 2 = −20j + 10j = −10j (lb)
2 ft
y
20 lb
System 1
y
Thus the sums of the forces are equal. The sums of the moments about
the left end are:
2 ft
10 lb
M1 = (−20)(4)k + 50k = −30k (ft lb)
2 ft
System 2
30 ft-lb
x
2 ft
M2 = (+10(2))k − 30k = −10k (ft lb)
The sums of the moments are not equal, hence the systems are not
equivalent. No
Problem 4.130 Four systems of forces and moments
act on an 8-m beam. Which systems are equivalent?
System 1
System 2
10 kN
8m
10 kN
80 kN-m
8m
System 3
System 4
20 kN
10 kN
20 kN
4m
Solution: For equivalence, the sum of the forces and the sum of
the moments about some point (the left end will be used) must be
the same.
"
"System 1 "System 2 "System 3 "System 4 "
" F (kN) "10j "10j "10j "10j "
" ML (kN-m) "80k "80k "160k "80k "
Systems 1, 2, and 4 are equivalent.
10 kN
8m
x
10 kN
80 kN-m
System 3
20 kN
8m
System 4
20 kN
10 kN
8m
4m
System 2
System 1
y
10 kN
80 kN-m
8m
20 kN
20 kN
10 kN
80 kN-m
4m
4m
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 = 10j kN
and systems 1, 2, and 4 have
ML = 80k (kN-m)
system 3 has
ML = 160k (kN-m)
Thus, we need to add a couple M = −80k (kN-m) to system 3.
(clockwise moment)
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
O 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 beam. 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 = rP L × F, where rP L 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.
System 1
System 2
F
F
O'
O
O
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
E
D
20 m
6m
Solution:
The angle formed by the single cable with the positive
x-axis is
θ = 180◦ − tan−1
A
12
16
= 143.13◦ .
12 m
B
The single cable tension is
16 m
C
T1 = |T|(i cos 143.13◦ + j sin 143.13◦ )
= |T|(−0.8i + 0.6j).
12 m
D
The position vector to the center of the log from the left end is rc = 10i.
The moment about the end of the log is
M = r × T1 = |T1 |
i
10
−0.8
j k
0 0 = |T|(6)k (N-m).
0.6 0
For the two cables, the angles relative to the positive x-axis are
12
= 116.56◦ , and
6
12
θ2 = 180 − tan−1
= 155.22◦ .
26
θ1 = 180◦ − tan−1
6m
E
20 m
Solve:
|TL | = 0.3353|T1 |, and
|TR | = 0.7160|T1 |.
The two cable vectors are
The tension in the right hand cable is TR = |T1 |(0.7160)(−0.9079i +
0.4191j) = |T1 |(−0.6500i + 0.3000). The position vector of the right end
of the log is rR = 20i m relative to the left end. The moments about the left
end of the log for the second system are
M2 = rR × TR = |T1 |
TL = |TL |(i cos 116.56◦ + j sin 116.56◦ )
= |TL |(−0.4472i + 0.8945j),
◦
◦
TR = |TR |(i cos 155.22 + j sin 155.22 )
= |TR |(−0.9079i + 0.4191j).
Since the vector sum of the forces in the two systems is equal, two
simultaneous equations are obtained:
0.4472|TL | + 0.9079|TR | = 0.8|T1 |, and
0.8945|TL | + 0.4191|TR | = 0.6|T1 |
i
20
−0.6500
j
0
0.3000
k
0 = |T1 |(6)k (N-m).
0
This is equal to the moment about the left end of the log for System 1, hence the
systems are equivalent.
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°
Solution:
For couples, the sum of the forces vanish for both systems. For System 1, the two forces are located at r11 = 4i, and
r12 = +5j. The forces are F1 = 200(i cos 30◦ + j sin 30◦ ) =
173.21i + 100j. The moment due to the couple in System 1 is
i
M1 = (r11 − r12 ) × F1 =
4
173.21
j
−5
100
(5, 4, 0) m
5m
200 N
30°
k
0 = 1266.05k (N-m).
0
x
20°
F
System 1
y
F2 = F (i cos(−20◦ ) + j sin(−20◦ )) = F (0.9397i − 0.3420j).
System 2
y
200 N
30°
F
20°
The moment of the couple in System 2 is
i
−3
0.9397
x
4m
For System 2, the positions of the forces are r21 = 2i, and r22 =
5i + 4j. The forces are
M2 = (r21 − r22 ) × F2 = F
2m
5m
j
−4
−0.3420
(5, 4, 0) m
200 N
k
0 = 4.7848F k,
0
30°
2m
x
20°
4m
F
from which, if the systems are to be equivalent,
F =
1266
= 264.6 N
4.7848
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 1
System 2
120 N-m
FB
40 N
60 N
3m
FA
3m
M
50 N
3m
Solution:
6m
3m
The sums of the forces for System 1 are
FX = 50, and
A
FY = −FA + 60.
The sums of the forces for System 2 are
20 m
FX = FB , and
FY = 40.
For equivalent systems: FB = 50 N, and FA = 60 − 40 = 20 N.
45°
C
60°
B
The sum of the moments about the left end for
System 1 is
The sum of the moments about the left end for
System 2 is
M1 = −(3)FA − 120 = −180 N-m.
M2 = −(3)FB + M = −150 + M N-m.
Equating the sums of the moments, M = 150 − 180 = −30 N-m
x
Problem 4.136 Two equivalent systems of forces and
moments act on the plate. Determine the force F and
the couple M .
System 2
System 1
30 lb
30 lb
10 lb
100 in-lb
5 in
5 in
8 in
50 lb
Solution:
30 lb
8 in
M
F
The sums of the forces for System 1 are
FX = 30 lb,
FY = 50 − 10 = 40 lb.
The sums of the forces for System 2 are
FX = 30 lb,
FY = F − 30 lb.
For equivalent forces, F = 30+40 = 70 lb. The sum of the moments
about the lower left corner for System 1 is
M1 = −(5)(30) − (8)(10) + M = −230 + M in lb.
The sum of the moments about the lower left corner for System 2 is
M2 = −100 in lb.
Equating the sum of moments, M = 230 − 100 = 130 in lb
System 1
30 lb
System 2
10 lb
30 lb
100 in-lb
8 in
8 in
M
50 lb
F
30 lb
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 .
System 1
F2
F3
F1
2m
4m
8m
400 kN
Solution:
For the two systems to be equivalent
System 2
F1 =
M A1
F2 and
=
MA2
From system 2,
System 1
y
F2
F2 = 0 and
MA2 = 0
F1
F3
A
2m
This
4m
F1
= F1 + F2 + F3 − 400j = 0 or
F1Y = F1 + F2 + F3 − 400 = 0
8m
400 kN
z
(1)
y
Summing Moments around A, we get
A
x
MA = (4i + 3k) × (−400j) + (6k × F1 j)
+(8i + 2k) × F2 j
z
MA = (−1600k + 1200i) − 6F1 i
+8F3 k − 2F3 i (kN-m) = 0
In component form, we have
x: − 6F1 − 2F3 + 1200 = 0
z: 8F3 − 1600 = 0
(kN-m)
(kN-m)
And the Force equation F1 + F2 + F3 − 400 = 0 kN
Solving, we get
F3 = 200 kN
F1 =
800
= 133.3 kN
6
F2 = 66.7 kN
System 2
x
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 ?
(b) If you represent system 1 by the force F (system 3),
what is the distance D?
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
System 1
y
FX = 0i,
20 lb
30 lb
40 lb
30 ft-lb
x
A
FY = (−20 + 40 − 30)j = −10j lb.
2 ft
2 ft
The sum of the moments about the left end for System 1 is
System 2
y
M1 = (2(40) − 4(30) + 30)k = −10k ft lb.
F
M
x
A
(a)
For System 2, the force at A is F = −10j lb
System 3
y
The moment at A is M2 = −10k ft lb
(b)
For System 3 the force at D is F = −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:
D=
10
= 1 ft
10
F
x
A
D
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
Solution: We first represent the system by an equivalent system
consisting of a force F at the origin and a couple M :
3m
y
This system is equivalent if
F
M
F = −40j + 60i + 60j
x
= 60i + 20j (N),
M = −280 + (6)(60)
= 80 N-m.
y
We then represent this system by an equivalent system consisting of F
alone:
F
For equivalence, M = d(Fy ), so
x
M
80
d =
=
= 4 m.
Fy
20
Problem 4.140 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.
(a) Determine the forces Ax , Ay , and B.
(b) If you represent the forces Ax , Ay , and B by a force
F acting at the right end of the beam and a couple
M , what are F and M ?
d
100 lb
y
Ax
x
1120 in-lb
Ay
B
16 in
Solution:
(a) The sum of the forces
y
FX = AX = 0
FY = (AY + B − 100) = 0.
The sum of the moments:
8 in
M = ((16)B − 24(100) + 1120)k = (16B − 1280)k = 0,
from which B = 80 lb. From the force balance equation: AY =
−80 + 100 = 20 lb. (b) The force at the right end of the beam must
balance the 100 lb force, F = 100j lb. The couple must balance the
existing couple M = −1120k in lb
100 lb
Ax
x
1120 in-lb
Ay
B
16 in
8 in
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.
(a) Determine the forces Ax and Ay , and the couple MA .
(b) Determine the sum of the moments about the right
end of the beam.
(c) 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
600 N
MA
x
Ax
30 N-m
Ay
200 N
380 mm
Solution:
180 mm
(a) The sum of the forces is
FX = AX i = 0 and
FY = (AY − 600 + 200)j = 0,
from which AY = 400 N. The sum of the moments is
ML = (MA − 0.38(600) − 30 + 0.560(200))k = 0,
from which MA = 146 N-m. (b) The sum of the moments about the
right end of the beam is
ML = 0.18(600) − 30 + 146 − 0.56(400) = 0.
(c) The sum of the forces for the new system is
FY = (AY + F )j = 0,
from F = −AY = −400 N, or F = −400j N. The sum of the
moments for the new system is
M = (MA + M ) = 0,
from which M = −MA = −146 N-m
y
600 N
MA
Ax
30 N-m
30 N-m
Ay
380 mm
180 mm
x
200 N
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.
(a) Determine the forces Ax , Ay , and B.
(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?
(c) 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?
2 kip
y
3 ft
4 kip
3 ft
6 kip
3 ft
Ax
O
x
Ay
B
6 ft
Solution:
(a) The sum of the forces is
2 kip
y
FX = (AX − 2 − 4 − 6)i = 0,
3 ft
from which AX = 12 kip
4 kip
FY = (AY + B)j = 0.
3 ft
6 kip
The sum of the moments about the origin is
3 ft
MO = (3)(6) + (6)(4) + (9)(2) + 6(B) = 0,
from which B = −10j kip. (b) Substitute into the force balance equation to obtain AY = −B = 10 kip. (b) The force in the new system
will replace the 2, 4, and 6 kip forces, F = (−2−4−6)i = −12i kip.
The force must match the moment due to these forces: F D = 3(6) +
(6)(4) + (9)(2) = 60 kip ft, from which D = 60
= 5 ft, or the
12
action 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.
Ax O
x
Ay
B
6 ft
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
Solution: The equivalent force must equal the sum of the forces
exerted by the soil:
F = (80 + 35 + 30 + 40 + 85)j = 270j kN
The sum of the moments about any point must be equal for the two
systems. The sum of the moments are
35 kN
30 kN
40 kN
3m
3m
3m
80 kN
3m
85 kN
y
M = 3(35) + 6(30) + 9(40) + 12(85) = 1665 kN-m.
Equating the moments for the two systems F D = 1665 kN-m from
which
x
35 kN
D =
1665 kN-m
= 6.167 m.
270 kN
30 kN
40 kN x
85 kN
80 kN
3m
Thus the action line intersects the x axis at a distance D = 6.167 m
to the right of the origin.
Problem 4.144 After landing, the pilot engages the
airplane’s thrust reversers and engines 1, 2, 3, and 4
exert forces toward the right of magnitudes 39 kN, 40 kN,
42 kN, and 40 kN, respectively. If you represent the four
forces by an equivalent force F, what is F, and what is
the y coordinate of its line of action?
3m
3m
3m
y
1
3m
2
4m
x
4m
Solution: We must find the sum of the forces and the sum of the
moments around the center of mass. We must then find the moment
arm at which the sum of the forces would create the same moment as
the four individual forces.
3
3m
4
F = 39i + 40i + 42i + 40i (kN)
F = 161i (kN)
y
M⊕ = +7j × 39i + 4j × 40i − 4j × 42i − 7j × 40i
F1
1
M⊕ = −273k − 160k + 168k + 280k
F2
2
M⊕ = 15k (kN-m)
3m
4m
We Now need to find the moment arm for the sum of the forces. We
require
z
x
F3
3
4j × 161i = 15k
−161y = 15
y = −0.0932 m
4m
4
F4
3m
Problem 4.145 The pilot of the airplane in Problem 4.144 wants to adjust engine 2 so that the forces
exerted by the engines can be represented by an equivalent force whose line of action intersects the z axis.
When this is done, what force is exerted by engine 2?
Solution:
Here we have F2 unknown and know that
M⊕ = 0.
All else is unchanged from the solution to Problem 4.144. Hence,
we have
and
or
F = 39i + F2 i + 42i + 40i (kN)
F = 121i + F2 i (kN)
M⊕ = 7j × 39i + 4j × F2 i − 4j × 42i − 7j × 40i = 0
M⊕ = −273k − 4F2 k + 168k + 280k = 0
M⊕ = (175 − 4F2 )k = 0
Solving:
F2 = 43.75 (kN-m)
F2 = 43.75i (kN-m)
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°
FAC
FAB
C
A
A
100 lb
100 lb
x
Solution: The sum of the forces acting at A is in opposition to
the weight, or F = |W|j = 100j lb.
The moment about point A is zero.
y
B
60° 40°
A
100 lb
C
FAB
FAC
A
100 lb
x
Problem 4.147 Three forces act on a beam.
(a) Represent the system by a force F acting at the origin
O and a couple M .
(b) Represent the system by a single force. Where does
the line of action of the force intersect the x axis?
y
30 N
5m
x
O
30 N
6m
Solution:
4m
50 N
(a) The sum of the forces is
FX = 30i N, and
FY = (30 + 50)j = 80j N.
The equivalent at O is F = 30i + 80j (N). The sum of the moments
about O:
M = (−5(30) + 10(50)) = 350 N-m
(b) The solution of Part (a) is the single force. The intersection is the
moment divided by the y-component of force: D = 350
= 4.375 m
80
y
30 N
5m
x
O
30 N
6m
4m
50 N
Problem 4.148 The tension in cable AB is 400 N, and
the tension in cable CD is 600 N.
(a) 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 ?
(b) 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?
y
A
400 mm
B
C
300 mm
D
x
O
800 mm
Solution: From the right triangle, the angle between the positive
x-axis and the cable AB is
θ = − tan−1
400
800
300 mm
y
= −26.6◦ .
A
400 mm
The tension in AB is
B
C
TAB = 400(i cos(−26.6◦ )+j sin(−26.6◦ )) = 357.77i − 178.89j (N).
300 mm
D
300 mm
x
800 mm
The angle between the positive x-axis and the cable CD is
α = − tan−1
300
800
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
= −20.6◦ .
The tension in CD is
TCD = 600(i cos(−20.6◦ ) + j sin(−20.6◦ )) = 561.8i − 210.67j.
The equivalent force acting at the origin O is the sum of the forces
acting on the left post:
F = (357.77 + 561.8)i + (−178.89 − 210.67)j
D =
419
= 0.456 m
919.6
Check: The moment is
M = rF × F =
i
0
919.6
j
k
D
0 = −919.6Dk = −419k,
−389.6 0
= 919.6i − 389.6j (N).
from which D =
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 = −0.7(357.77)k − 0.3(561.8)k = −419k N-m
Check: The position vectors at the point of application are rAB =
0.7j, and rCD = 0.3j. The sum of the moments is
M = (rAB × TAB ) + (rCD × TCD )
=
i
0
357.77
j
k
i
0.7
0 +
0
−178.89 0
561.8
j
k
0.3
0
−210.67 0
= −0.7(357.77)k − 0.3(561.8)k = −419k
419
919.6
= 0.456 m, Check.
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 = −400(i cos(−26.6◦ )+j sin(−26.6◦ )) = −357.77i + 178.89j,
and
TCD = −400(i cos(−20.6◦ )+j sin(−20.6◦ )) = −374.42i + 140.74j.
The equivalent force is equal to the sum of these forces:
F = (−357.77 − 374.42)i + (178.77 + 140.74)j
= −732.19i + 319.5j (N).
The sum of the moments about O is
M = 0.3(357.77) + 0.8(140.74 + 178.89)k = 363k (N-m).
The intersection is D =
363
732.19
= 0.496 m on the positive y-axis.
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
y
FX = (500 − 500)i = 0.
500 lb
6 in
800 lb
FY = 800j.
x
6 in
Thus a force and a couple with moment M = 500k ft lb act on
the cross section. The equivalent force is F = 800j which acts at a
positive x-axis location of D = 500
= 0.625 ft = 7.5 in to the right
800
of the origin.
500 lb
z
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 = 4i + 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 = 3(2)j + 3(4)k = 6j + 12k. Hence the couple moment on System 2 is
M2 = 6j + 12k (kN-m)
y
System 1
y
z
3m
F
z
3m
x
F
x
System 2
M
M
y
System 2
y
4i + 4j − 2k (kN)
z
4i + 4j – 2k (kN)
3m
z
x
3m
x
Problem 4.152 The wall bracket is subjected to the
force shown.
(a) Determine the moment exerted by the force about
the z axis.
(b) Determine the moment exerted by the force about
the y axis.
(c) If you represent the force by a force F acting at O
and a couple M, what are F and M?
y
O
10i – 30j + 3k (lb)
12 in
z
x
Solution:
(a)
The moment about the z-axis is negative,
y
O
MZ = −1(30) = −30 ft lb,
(b)
The moment about the y-axis is negative,
MY = −1(3) = −3 ft lb
(c)
The equivalent force at O must be equal to the force at x = 12 in,
thus FEQ = 10i − 30j + 3k (lb)
The couple moment must equal the moment exerted by the force
at x = 12 in. This moment is the product of the moment arm and
the y- and z-components of the force: M = −1(30)k−1(3)j =
−3j − 30k (ft lb).
z
10i Ð30j + 3k (lb)
12 in
x
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 =
14 12 in., and r = 9 12 in., and the angle α = 120◦ .
(a) If you represent the forces he exerts by a force F
acting at O and a couple M, what are F and M?
(b) The glass backboard will shatter if |M| > 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 = −200j. The position vectors of the
points of application of the forces are r1 = (h + r)i, and r2 =
i(h + r cos α) − kr sin α. The moments about the origin are
M = (r1 × F1 ) + (r2 × F2 ) = (r1 + r2 ) × F
i
= 2h + r(1 + cos α)
0
j
0
−100
k
−r sin α
0
= −100(r sin α)i − 100(2h + r(1 + cos α))k.
For the values of h, r, and α given, the moment is M = −822.72i −
3375k in lb. This is the couple
√ moment required. (b) The magnitude
of the moment is |M| = 822.722 + 33752 = 3473.8 in lb. The
backboard does not break.
y
−100j (lb)
α
O
−100j (lb)
r
h
z
x
Problem 4.154 The three forces are parallel to the
x axis.
(a) If you represent the three forces by a force F acting
at the origin O and a couple M, what are F and M?
(b) If you represent the forces by a single force, what is
the force, and where does its line of action intersect
the y − z plane?
Strategy: In (b), assume that the force acts at a
point (0, y, z) of the y − z plane, and use the conditions for equivalence to determine the force and the
coordinates y and z. (See Example 4.20.)
y
(0, 6, 2) ft
300 lb
O
100 lb
(0, 0, 4) ft
z
200 lb
x
Solution:
(a) F
y
= 100i + 200i + 300i
= 600i (lb).
M
=
i
0
200
j k
i
0 4 + 0
0 0
300
j k
6 2
0 0
M
= 800j + 600j − 1800k
0
= 1400j − 1800k (ft-lb).
z
(b) F
F
= 600i (lb).
x
To determine y and z, require that
y
M = 1400j − 1800k =
i
0
600
j
y
0
k
z
0
(0, y, z)
1400 = 600 z,
−1800 = −600 y.
F
Solving, y = 3 ft and z = 2.33 ft.
0
z
x
Problem 4.155 The positions and weights of three
particles are shown. If you represent the weights by a
single force F, determine F and show that its line of
action intersects the x-z plane at
3
x=
3
xi Wi
i=1
3
,
z=
Wi
(x2, y2, z2)
(x1, y1, z1)
zi W i
i=1
3
i=1
y
(x3, y3, z3)
–W2 j
.
–W1 j
x
–W3 j
Wi
i=1
z
Solution:
The single equivalent force must be equal to the sum
of the forces:
y
F = (−W1 − W2 − W3 )j = −
3
i=1
(x2, y2, z2)
Wi
(x1, y1, z1)
(x3, y3, z3)
ÐW2 j
The moment about the x-axis is
x
ÐW1 j
ÐW3 j
MX = eX · (r1 × W1 ) + eX · (r2 × W2 ) + eX · (r3 × W3 )
1
= x1
0
=
3
i=1
0
y1
−W1
0
1
z 1 + x2
0
0
0
y2
−W2
0
1
z 2 + x3
0
0
0
y3
−W3
0
z3
0
z
The moment arm is the ratio of the magnitude of the moment to the magnitude
of the force,
W i zi .
3
Similarly, the moment about the z-axis is
DX =
MZ = −
3
i=1
W i xi .
3
W i zi
i=1
3
i=1
.
Wi
DZ =
W i xi
i=1
3
i=1
Wi
Problem 4.156 Two forces act on the beam. If you Solution: The equivalent force must equal the sum of forces: F = 100j +
represent them by a force F acting at C and a couple M, 80k. The equivalent couple is equal to the moment about C:
what are F and M?
y
M = (3)(80)j − (3)(100)k = 240j − 300k
y
100 N
100 N
80 N
80 N
z
z
C
C
x
3m
3m
x
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 = P i. The position vector of the applied force is r = −hj +
bk. The moment is
i
M = (r × P) = 0
P
j
−h
0
b
Pi
k
+b = bP j + hP k.
0
z
h
O
x
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.)
Problem 4.158 The brace is being used to remove a
screw.
(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?
(b) If you represent the forces acting on the brace by
a force F acting at a point P with coordinates
(xP , yP , zP ) and a couple M , what are F and M ?
y
y
h
h
r
B
O
z
1–
2A
A
B
1–
2A
(a) Equivalent force at the origin O has the same value
as the sum of forces,
Solution:
y
h
FX = (B − B)i = 0,
1
1
FY = −A + A + A j = 0,
2
2
r
O
z
thus F = 0. The equivalent couple moment has the same value as the
moment exerted on the brace by the forces,
MO = (rA)i.
Thus the couple at O has the moment M = 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 = 0, M = rAi.
h
B
1−
2A
A
B
1−
2A
x
x
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?
y
P
FB =
2i – j (kN)
FA =
– i + j + k (kN)
x
MC =
4i – 4j + 4k (kN-m)
1m
z
Solution:
The equivalent force at P has the value of the sum of
forces,
y
F = (2 − 1)i + (1 − 1)j + k, FP = i + k (kN).
P
FB =
2i − j (kN)
FA =
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:
−i + j + k (kN)
x
MC =
4i − 4j + 4k (kN-m)
rA = −i − j + k, and rB = +k. The moment of the couple:
z
M = (rA × FA ) + (rB × FB ) + MC
i
= −1
−1
j k
i
−1 1 + 0
1 1
2
1m
j k
0 1 + MC
−1 0
= 3i − 2j + 2k (kN-m).
Problem 4.160 The two shafts are subjected to the
torques (couples) shown.
(a) If you represent the two couples by a force F acting
at the origin O and a couple M, what are F and M?
(b) 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 = 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:
y
M = M1 + 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.
6 kN-m
4 kN-m
M1 = 4(−j sin 30◦ − k cos 30◦ ) = −2j − 3.464k,
40°
M2 = 6(i cos 40◦ + j sin 40◦ ) = 4.5963i + 3.8567j.
Thus the couple at the origin is MO = 4.6i + 1.86j − 3.46k (kN-m)
(b) The magnitude
of the total moment exerted by the two couples is
√
|MO | = 4.62 + 1.862 + 3.462 = 6.05 (kN-m)
30°
x
z
Problem 4.161 The persons A and B support a bar to
which three dogs are tethered. The forces and couples
they exert are
A
FA = −5i + 15j − 10k (lb),
B
MA = 15j + 10k (ft-lb),
FB = 5i + 10j − 10k (lb),
MB = −10j − 15k (ft-lb).
If person B let go, person A would have to exert a force
F and couple M equivalent to the system both of them
were exerting together. What are F and M?
y
FA
MA
FB
z
6 ft
x
MB
Solution:
The equivalent force at B is the sum of the forces:
A
F = (−5 + 5)i + (15 + 10)j + (−10 − 10)k = 25j − 20k (lb).
B
The equivalent couple at A is the sum of the moments at A
M = (rB × FB ) + MA + MB .
The position vector of B relative to A is rB = 6i. Thus:
y
i
MB = 6
5
j
0
10
k
0 + M A + MB
−10
MB = (60j + 60k) + (15j + 10k) + (−10j − 15k)
= 65j + 55k (ft-lb)
MA
z
FA
FB
6 ft
MB
x
Point G is at the center of the block.
Problem 4.162
The forces are
FA = −20i + 10j + 20k (lb),
FB = 10j − 10k (lb).
If you represent the two forces by a force F acting at G
and a couple M, what are F and M?
y
FB
FA
10 in
x
G
20 in
30 in
z
Solution:
The equivalent force is the sum of the forces:
F = (−20)i + (10 + 10)j + (20 − 10)k
= −20i + 20j + 10k (lb).
The equivalent couple is the sum of the moments about G. The position
vectors are:
rA = −15i + 5j + 10k (in),
rB = 15i + 5j − 10k.
The sum of the moments:
MG = (rA × FA ) + (rB × FB )
i
= −15
−20
j
5
10
k
i
10 + 15
20
0
j
5
10
k
−10
−10
= 50i + 250j + 100k (in lb)
y
FB
FA
10 in
x
20 in
z
30 in
Problem 4.163 The engine above the airplane’s fuselage exerts a thrust T0 = 16 kip, and each of the engines
under the wings exerts a thrust TU = 12 kip. The dimensions are h = 8 ft, c = 12 ft, and b = 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?
Solution:
y
T0
c
O
z
h
2 TU
y
The equivalent thrust at the point G is equal to the sum
of the thrusts:
T = 16 + 12 + 12 = 40 kip
x
O
The sum of the moments about the point G is
b
b
M = (r1U × TU ) + (r2U × TU ) + (rO × TO )
= (r1U + r2U ) × TU + (rO × TO ).
The position vectors are r1U = +bi − hj, r2U = −bi − hj, and
rO = +cj. For h = 8 ft, c = 12 ft, and b = 16 ft, the sum of the
moments is
i
M= 0
0
j
k
i
−16 0 + 0
0
12
0
j
12
0
y
T0
c
z
2 TU
k
0 = (−192 + 192)i = 0.
16
y
Thus the equivalent couple is M = 0
x
b
b
Problem 4.164 Consider the airplane described in For h = 8 ft, c = 12 ft, and b = 16 ft, using the position vectors
Problem 4.163 and suppose that the engine under the for the engines given in Problem 4.147, the equivalent couple is
wing to the pilot’s right loses thrust.
i
j
k i j
k (a) If you represent the two remaining thrust forces by
M = 16 −8 0 + 0 12 0 = 96i − 192j (ft kip)
a force F acting at the origin O and a couple M,
0
0 12 0 0 16 what are F and M?
(b) If you represent the two remaining thrust forces by
the force F alone, where does its line of action intersect the x − y plane?
Solution:
The sum of the forces is now
F = 12 + 16 = 28k (kip).
(b) The moment of the single force is
i j
M = x y
0 0
From which
The sum of the moments is now:
x =
M = (r2U × TU ) + (rO × TO ).
k z = 28yi − 28xj = 96i − 192j.
28 192
96
= 6.86 ft, and y =
= 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 = 0, so that the
coordinates of the point of action are (6.86, 3.43, 0).
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
Solution:
z
The position vectors of the points A, B, C, and D are
y
rA = 0i + 6j + 6k,
y
rB = 3i + 0j + 8k,
C (4, 6, 0) ft
(0, 6, 6) ft
rC = 4i + 6j + 0k, and
A
rD = 7i + 0j + 2k.
B
The unit vectors parallel to the cables are obtained as follows:
F
x
D
(7, 0, 2) ft
x
E
x
(3, 0, 8) ft
z
z
rAB = rB − rA = 3i − 6j + 2k,
|rAB | = 32 + 62 + 22 = 7,
For the systems to be equivalent, the moments about the origin must be the same.
The moments about the origin are
from which
MO = (rA × FA ) + (rC × FC )
eAB = 0.4286i − 0.8571j + 0.2857k.
=
rCD = rD − rC = 3i − 6j + 2k,
|rCD | = 32 + 62 + 22 = 7,
from which
i
0
42.86
Since eAB = eCD , the cables are parallel. To duplicate the force,
the single cable EF must have the same unit vector.
The total force is
FEF = 68.58i − 137.14j + 45.71k (lb),
|FEF | = 160 lb.
k
0
17.14
i
0
68.58
j
y
−137.14
k
z
45.71
= (45.71y + 137.14z)i + (68.58z)j − (68.58y)k
= 788.57i + 188.57j − 617.14k,
The force on the wall at point A is
FC = 60eCD = 25.72i − 51.43j + 17.14k (lb).
j
6
−51.43
This result is used to establish the coordinates of the point E. For the one cable
system, the end E is located at x = 0. The moment is
M1 = r × FEF =
The force on the wall at point C is
k
i
6
+
4
28.57
25.72
= 788.57i + 188.57j − 617.14k.
eCD = 0.4286i − 0.8571j + 0.2857k.
FA = 100eAB = 42.86i − 85.71j + 28.57k (lb).
j
6
−85.71
from above. From which
y =
617.14
= 8.999 . . . = 9 ft
68.58
z =
188.57
= 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 = Ley with the condition that yF = 0,
9
L = 0.8571
= 10.5 ft. The other coordinates are xF − xE = LeX , from
which xF = 0 + 10.5(0.4286) = 4.5 ft zF − zE = LeZ , from which
zF = 2.75 + 10.5(0.2857) = 5.75 ft The coordinates of F are F (4.5, 0,
5.75) ft
Problem 4.166 The distance s = 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
100 i + 20 j – 20 k (N)
O
x
200 N-m
(4, 0, 3) m
z
Solution:
The equivalent force at the origin is
y
(2, 6, 0) m
F = 100i + 20j − 20k.
s
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 = 2i + 6j + 0k. The vector parallel to the bar, pointing toward
the base, is rT B = 2i − 6j + 3k, with a magnitude of |rT B | = 7.
The unit vector parallel to the bar is
eT B = 0.2857i − 0.8571j + 0.4286k.
The vector from the top of the bar to the action point of the force is
rT F = seT B = 4eT B = 1.1429i − 3.4286j + 1.7143k.
The position vector of the action point from the origin is
rF = rT + rT F = 3.1429i + 2.5714j + 1.7143k.
The moment of the force about the origin is
i
MF = r × F = 3.1429
100
j
2.5714
20
k
1.7143
−20
= −85.71i + 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 = −200eT B = −57.14i + 171.42j − 85.72k.
The sum of the moments is
M = MF + MC = −142.86i + 405.72j − 280k.
This is the moment of the equivalent couple at the origin.
O
100 i + 20 j – 20 k (N)
200 N-m
(4, 0, 3) m
z
x
The force F and couple M in system
Problem 4.167
1 are
System 1
System 2
y
F = 12i + 4j − 3k (lb),
y
M = 4i + 7j + 4k (ft-lb).
M
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 x − z plane.
F
O
x
z
eF =
System 1
F
MP = (eF · M)eF = 4.5444i + 1.5148j − 1.1361k (ft-lb).
MN = M − MP = −0.5444i + 5.4858j + 5.1361k (ft-lb).
The moment of F must produce a moment equal to the normal component of M. The moment is
MF = r × F =
i
x
12
j
0
4
k
z = −(4z)i + (3x + 12z)j + (4x)k,
−3
from which
z =
−0.5444
= 0.1361 ft
−4
x =
5.1362
= 1.2840 ft
4
Problem 4.168 A system consists of a force F acting
at the origin O and a couple M, where
F = 10i (lb),
M = 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 y-z plane?
Solution: The component of M parallel to F is zero, since
MP = (eF · M)eF = 0. The normal component is equal to M.
The equivalent force must produce the same moment as the normal
component
M=r×F=
i
0
10
from which z =
20
10
j
y
0
k
z = (10z)j − (10y)k = 20j,
0
= 2 ft and y = 0
y
M
The component of M parallel to F is
The component of M normal to F is
System 2
y
F
= 0.9231i + 0.3077j − 0.2308k.
|F|
x
(x, 0, z)
z
Solution: The component of M that is parallel to F is found as
follows: The unit vector parallel to F is
F
Mp
O
O
z
x
O
z
(x, 0, z)
MP
F
x
Problem 4.169 A system consists of a force F acting
at the origin O and a couple M, where
F = i + 2j + 5k (N),
M = 10i + 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
x − z plane, (c) the y-z plane.
Solution: The unit vector parallel to F is
eF =
F
= 0.1826i + 0.3651j + 0.9129k.
|F|
from which
z =
9.8
5
= −4.9 m, and x =
= −2.5 m
−2
−2
(a) The parallel component of Mt is
(c) The intersection with the y-z plane is
MP = (eF · M)eF = 0.2i + 0.4j + 1.0k (N-m).
i
MN = r × F = 0
1
j
y
2
The normal component is
from which
The moment of the force about the origin must be equal to the normal
component of the moment. (b) The intersection with the x − z plane:
y = 5 m and z = 7.6 m
i j k
MN = r × F = x 0 z
1 2 5
= −(2z)i − (5x − z)j + (2x)k
= 9.8i + 7.6j − 5k,
Problem 4.170 Consider the force F acting at the origin O and the couple M given in Example 4.21. If you
represent this system by a wrench, where does the line
of action of the force intersect the x − y plane?
Solution: From Example 4.21 the force and moment are F =
3i + 6j + 2k (N), and M = 12i + 4j + 6k (N-m).
The normal component of the moment is
MN = 7.592i − 4.816j + 3.061k (N-m).
The moment produced by the force must equal the normal component:
k
0
2
= (2y)i − (2x)j + (6x − 3y)k = 7.592i − 4.816j + 3.061k,
from which
x =
= (5y − 2z)i + (z)j − (y)k
= 9.8i + 7.6j − 5k,
MN = M − MP = 9.8i + 7.6j − 5k.
i j
MN = r × F = x y
3 6
k
z
5
4.816
7.592
= 2.408 m and y =
= 3.796 m
2
2
Problem 4.171 Consider the force F acting at the origin O and the couple M given in Example 4.21. If you
represent this system by a wrench, where does the line
of action of the force intersect the plane y = 3 m?
Solution: From Example 4.21 (see also Problem 4.170) the force
is F = 3i + 6j + 2k, and the normal component of the moment is
MN = 7.592i − 4.816j + 3.061k.
The moment produced by the force must be equal to the normal component:
i j k
MN = r × F = x 3 z = (6 − 6z)i − (2x − 3z)j + (6x − 9)k
3 6 2
= 7.592i − 4.816j + 3.061k,
from which
x =
9 + 3.061
6 − 7.592
= 2.01 m and z =
= −0.2653 m
6
6
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?
The vector parallel to the force is rF = i + j + 2k,
from which the unit vector parallel to the force is eF = 0.4082i +
0.4082j + 0.8165k. The force and moment at the origin are
Solution:
F = |F|eOF = 40.82i + 40.82j + 81.65k (N), and
M = 24.492i + 24.492j + 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
MF = r × F =
i
5
40.82
j
3
40.82
k
1
81.65
= 204.13i − 367.43j + 81.64k.
For the moments to be equal in the two systems, the added equivalent
couple must be
MC = M − MF = −176.94i + 391.92j − 32.65k (N-m)
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 x − z plane.
System 1
y
System 2
y
1000i + 600j (kN-m)
600k (kN)
300j (kN)
3m
F
Mp
x
x
4m
z
(x, 0, z)
z
Solution: The sum of the forces in System 1 is F = 300j +
600k (N). The equivalent force in System 2 must have this value. The
unit vector parallel to the force is eF = 0.4472j + 0.8944k. The
sum of the moments in System 1 is
M = 600(3)i + 300(4)k + 1000i + 600j
= 2800i + 600j + 1200k (kN m).
The component parallel to the force is
MP = 599.963j + 1199.93k (kN-m) = 600j + 1200k (kN-m).
The normal component is MN = M − MP = 2800i. The moment
of the force
i
j
MN = x
0
0 300
k
z = −300zi − 600xj + 300xk = 2800i,
600
from which
x = 0, z =
2800
= −9.333 m
−300
System 2
System 1
y
y
1000i + 600j (kN-m)
600k (kN)
3m
300j (kN)
Mp
x
x
4m
z
F
z
(x, 0, z)
Problem 4.174 A plumber exerts the two forces shown
to loosen a pipe.
(a) What total moment does he exert about the axis of
the pipe?
(b) If you represent the two forces by a force F acting
at O and a couple M, what are F and M?
(c) 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 x − y plane?
y
12 in
6 in
O
z
x
16 in
16 in
50 k (lb)
–70 k (lb)
Solution:
(a)
The sum of the forces is
F = 50k − 70k = −20k (lb).
The total moment exerted on the pipe is
M = 16(20)i = 320i (ft lb).
The equivalent force at O is F = −20k. The sum of the moments about O is
(b)
MO
= (r1 × F1 ) + (r2 × F2 )
i
= 12
0
j
k
i
−16 0 + 18
0
50
0
j
−16
0
k
0
−70
= 320i + 660j.
The unit vector parallel to the force is eF = k, hence the moment
parallel to the force is MP = (eF ·M)eF = 0, and the moment
normal to the force is MN = M − MP = 320i + 660j. The
force at the location of the wrench must produce this moment for
the wrench to be equivalent.
(c)
i
MN = x
0
j
y
0
k
0
= −20yi + 20xj = 320i + 660j,
−20
from which x =
660
20
= 33 in, y =
320
−20
= −16 in
y
12 in
O
6 in
x
z
16 in
16 in
50 k (lb)
−70 k (lb)
Problem 4.175 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
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 = (−6i − 5j) × (300j) = −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
3 ft
A
300 lb
5 ft
800 ft-lb
M200 = (3i − 5j) × (200i) = 1000k (ft-lb).
200 lb
The moment of the couple is MC = −800k (ft-lb). Summing the
four moments, we get
200 lb
6 ft
3 ft
MA = (−1800 + 0 + 1000 − 800)k = −1600k (ft-lb)
Problem 4.176 In Problem 4.175, 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?
Problem 4.177 The vector sum of the forces acting
on the beam is zero, and the sum of the moments about
A is zero.
(a) What are the forces Ax , Ay , and B?
(b) What is the sum of the moments about B?
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 4.175, the equivalent couple will be C = MA =
−1600k (ft-lb). The equivalent force will be FEQUIV. = 200i − 200j +
300j = 200i + 100j (lb)
30°
220 mm
400 N
Ay
Ax
Solution:
The vertical and horizontal components of the 400 N
force are:
260 mm
FX = 400 cos 30◦ = 346.41 N,
FY = 400 sin 30◦ = 200 N.
500 mm
B
The sum of the forces is
FX = AX + 346.41 = 0,
from which AX = −346.41 N
30°
220 mm
400 N
Ax
The sum of the moments about A is
Ay
FY = AY + B − 200 = 0.
MA = 0.5B − 0.22(346.41) = 0,
from which B = 152.42 N. Substitute into the force equation to get
AY = 200 − B = 47.58 N
(b) The moments about B are
MB = −0.5AY − 0.48(346.41) − 0.26AX + 0.5(200) = 0
260 mm
500 mm
B
Problem 4.178 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
The moment about C exerted by the weight is
Solution:
3 ft
A
MC = 450(6) = 2700 ft lb.
C
The ladder is at an elevation of 45◦ from the horizontal. The cylinder
is at an angle
θ = tan−1
B
6 ft
3 ft
3
= 26.56◦ .
6
The vertical and horizontal components of the force at B due to the
cylinder are
FX = F cos 26.57◦ = 0.8944F lb
6 ft
◦
FY = F sin 26.57 = 0.4472F lb.
The moment about C due to these forces is
450 lb
MC = −3(0.4472)F − 3(0.8944)F + 2700 = 0.
B
3 ft
Solving:
F =
C
A
2700
= 670.82 lb
4.0249
6 ft
Problem 4.179 The force F = −60i + 60j (lb).
(a) Determine the moment of F about point A.
(b) What is the perpendicular distance from point A to
the line of action of F?
3 ft
y
F
(4, – 4, 2) ft
x
A
The position vector of A and the point of action are
Solution:
rA = 8i + 2j + 12k (ft), and rF = 4i − 4j + 2k.
(8, 2, 12) ft
z
The vector from A to F is
y
rAF = rF − rOA = (4 − 8)i + (−4 − 2)j + (2 − 12)k
F
= −4i − 6j − 10k.
(a)
The moment about A is
MA = rAF
i
× F = −4
−60
(4, – 4, 2) ft
j
−6
60
= 600i + 600j − 600k (ft lb)
(b)
x
k
−10
0
A
(8, 2, 12) ft
z
The magnitude of the moment is
|MA | =
6002 + 6002 + 6002 = 1039.3 ft lb.
√
The magnitude of the force is |F| = 602 + 602 = 84.8528 lb. The
perpendicular distance from A to the line of action is
D=
1039.3
= 12.25 ft
84.8528
Problem 4.180 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.
y
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:
y
C
B
rOA = 1.2j,
D
A
rOB = −0.3i + 2j + 1k,
rOC = 2j − 1k,
rOD = 2i + 2j,
1m
rOE = −0.3i + 1k.
1m
E
2m
0.3 m
The vectors parallel to the cables are:
x
z
rAB = rOB − rOA = −0.3i + 0.8j + 1k,
rAC = rOC − rOA = +0.8j − 1k,
Solve:
rAD = rOD − rOA = +2i + 0.8j.
TAB = 150.04 N,
The unit vectors parallel to the cables are:
eAB =
rAB
= −0.2281i + 0.6082j + 0.7603k :
|rAB |
eAC = 0i + 0.6247j − 0.7809k,
eAD = +0.9284i + 0.3714j + 0k.
The tensions in the cables are
TAB = TAB eAB ,
TAC = TAC eAC , and
TAD = TAD eAD .
The equilibrium conditions are TAB + TAC + TAD = W. Collect
like terms in i, j, k:
FX = (−0.2281TAB + 0TAC + 0.9284TAD )i = 0
FY = (+0.6082 · TAB + 0.6247 · TAC
+ 0.3714 · TAD − 196.2)j = 0
FZ = (+0.7603 · TAB − 0.7809 · TAC + 0 · TAD )k = 0
TAC = 146.08 N,
TAD = 36.86 N.
The moment about E is
ME = rEB × (−TAB eAB ) = −TAB (rEB × eAB )
= −150
i
0
−0.2281
j
2
+0.6082
= −228i − 68.43k (N-m)
k
0
+0.7603
Problem 4.181 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
The force vector acting at A is F = −800j (lb) and
the position vector from C to A is
Solution:
z
C (5, 0, 6) ft
rCA = (xA − xC )i + (yA − yC )j + (zA − zC )k
= (4 − 5)i + (3 − 0)j + (4 − 6)k = −1i + 3j − 2k (ft).
y
800 lb
A (4, 3, 4) ft
The moment about C is
MC
i
= −1
0
j
3
−800
k
−2 = −1600i + 0j + 800k (ft-lb)
0
Problem 4.182 In Problem 4.181, determine the moment of the vertical 800-lb force about the straight line
through points C and D.
Solution: In Problem 4.181, we found the moment of the 800 lb
force about point C to be given by
MC = −1600i + 0j + 800j (ft-lb).
The vector from C to D is given by
rCD = (xD − xC )i + (yD − yC )j + (zD − zC )k
= (6 − 5)i + (0 − 0)j + (0 − 6)k
= 1i + 0j − 6j (ft),
and its magnitude is
|rCD | =
12 + 6 2 =
√
37 (ft).
The unit vector from C to D is given by
1
6
eCD = √ i − √ k.
37
37
The moment of the 800 lb vertical force about line CD is given by
6
1
√ i − √ k · (−1600i + 0j + 800j (ft-lb))
37
37
−1600 − 4800
√
=
(ft-lb).
37
MCD =
Carrying out the calculations, we get MCD = −1052 (ft-lb)
B
z
D (6, 0, 0) ft
x
C (5, 0, 6) ft
Problem 4.183 The tugboats A and B exert forces
FA = 1 kN and FB = 1.2 kN on the ship. The angle
θ = 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
60 m
O
Solution:
The sums of the forces are:
60 m
x
FB
◦
FX = (1 + 1.2 cos 30 )i = 2.0392i (kN)
FY = (1.2 sin 30◦ )j = 0.6j (kN).
B
θ
The equivalent force at the origin is
25 m
FEQ = 2.04i + 0.6j
The moment about O is MO = rA × FA + rB × FB . The vector
positions are
rA = −25i + 60j (m), and
rB = −25i − 60j (m).
FA
The moment:
MO
i
= −25
1
j
60
0
k
i
0 + −25
0
1.0392
j
k
−60 0
0.6 0
= −12.648k = −12.6k (kN-m)
Check: Use a two dimensional description: The moment is
60 m
B
O
FB
60 m
A
θ
MO = −(25)FB sin 30◦ + (60)(FB cos 30◦ ) − (60)(FA )
25 m
= 39.46FB − 60FA = −12.6 kN-m
Problem 4.184
The tugboats A and B in Problem 4.183 exert forces FA = 600 N and FB = 800 N on
the ship. The angle θ = 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?
The moment:
i
MO = −25
0.6
j
60
0
k
i
0 + −25
0
0.5656
j
−60
0.5656
k
0 = −16.20k (kN-m)
0
Check: Use a two dimensional description:
Solution:
The equivalent force is
◦
◦
F = (0.6 + 0.8 cos 45 )i + 0.8 sin 45 j = 1.1656i + 0.5656j (kN).
The moment produced by the two forces is
MO = −(25)FB sin 45◦ + (60)FB cos 45◦ − 60FA
= 24.75FB − 60FA = −16.20 kN-m.
The single force must produce this moment.
MO = rA × FA + rB × FB .
The vector positions are
rA = −25i + 60j (m), and rB = −25i − 60j (m).
MO =
i
0
1.1656
j
y
0.5656
from which
y =
16.20
= 13.90 m
1.1656
k
0 = −1.1656yk = −16.20k,
0
Problem 5.1 The beam has pin and roller supports and
is subjected to a 4-kN load.
(a) Draw the free-body diagram of the beam.
(b) Determine the reactions at the supports.
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.
4 kN
A
B
2m
3m
Solution:
FX = 0:
AX = 0
FY = 0:
AY + BY − 4 kN = 0
MA = 0:
−2(4 kN) + 3BY = 0
4 kN
A
B
BY = 8/3 kN
2m
AY = 4/3 kN
3m
AX = 0, AY = 1.33 kN, BY = 2.67 kN
4 kN
2m
BX
AX
AY
3m
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) Draw the free-body diagram of the beam.
(b) Determine the reactions at the supports.
BY
2 kN
6 kN-m
A
2m
3m
Solution:
(a)
(b)
2 kN
FX = 0:
AX = 0
FY = 0:
AY − 2 kN = 0
MA = 0:
MA − (2)(2 kN) + 6 kN-m = 0
6 kN-m
A
2m
MA = −2 kNm
AX = 0
3m
AY = 2 kN
y
x
MA
2m
AX
3m
AY
2 kN
6 kN-m
Problem 5.3 The beam is subjected to a load F =
400 N and is supported by the rope and the smooth surfaces at A and B.
(a) Draw the free-body diagram of the beam.
(b) What are
the magnitudes of the reactions at A and B?
F
A
B
30°
45°
1.2 m
1.5 m
1m
Solution:
+
FX = 0:
A cos 45◦ − B sin 30◦ = 0
FY = 0:
A sin 45◦ + B cos 30◦ − T − 400 N = 0
MA = 0:
F
A
B
30°
45°
−1.2T − 2.7(400) + 3.7B cos 30◦ = 0
Solving, we get
A = 271 N
B = 383 N
T = 124 N
1.2 m
1.5 m
y
1m
A
F
45°
x
B
1.5 m
1.2 m
1m
30°
T
Problem 5.4 (a) Draw the free-body diagram of the
beam.
(b) Determine the reactions at the supports.
5 kN
B
A
3m
3m
Solution:
(a)
(b)
5 kN
FX = 0:
AX = 0
Solving:
AX
B
A
FY = 0: AY + BY − 5 kN = 0
+
MA = 0: 3BY − 6(5 kN) = 0
BY
= 10 kN
AY
= −5 kN
3m
3m
=0
AY
BY
5 kN
y
AX
3m
3m
x
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) The free body diagram is shown.
(b) The sum of the forces:
FX = 0,
FY = FA + FB − 60 = 0
The sum of the moments about point A:
MA = −10(60) + 24(FB ) = 0,
from which FB =
600
= 25 lb
24
Substitute into the force balance equation:
FA = 60 − FB = 35 lb
60 lb
A
B
10 in
FA
14 in
FB
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) Draw the free-body diagram of the diving board.
(b) Determine the reactions at the supports A and B.
A
B
WP
WD
1.2 m
2.4 m
4.6 m
Solution:
(a)
(b) (N)
(N)
FX = 0:
AX = 0
FY = 0: AY + BY − (54)(9.81) − 36(9.81) = 0
MA = 0: 1.2BY − (2.4)(36)(9.81)
A
B
− (4.6)(54)(9.81) = 0
Solving:
AX
1.2 m
=0N
AY
= −1.85 kN
BY
= 2.74 kN
WP
WD
2.4 m
4.6 m
4.6 m
2.4 m
1.2 m
AX
Problem 5.7 The ironing board has supports at A and
B that can be modeled as roller supports.
(a) Draw the free-body diagram of the ironing board.
(b) Determine the reactions at A and B.
WD
BY
AY
y
A
B
x
3 lb
10 lb
12 in
Solution: The system is in equilibrium.
(a) The free-body diagram is shown.
(b) The sums of the forces are:
FX = 0,
FY = FA + FB − 10 − 3 = 0.
The sum of the moments about A is
MA = 12FB − 22(10) − 42(3) = 0,
from which FB =
WP
10 in
20 in
Substitute into the force balance equation:
FA = 13 − FB = −15.833 lb
y
A
B
12 in
x
10 in
10 lb
20 in
3 lb
FB
10 lb
3 lb
346
= 28.833 in.
12
FA
Problem 5.8 The distance x = 2 m.
(a) Draw the free-body diagram of the beam.
(b) Determine the reactions at the supports.
10 kN
A
B
x
4m
The system is in equilibrium. The point A is a pinned
support. The point B is a roller support. (a) The free body diagram is
shown. (b) The sums of the forces:
FX = AX = 0,
FY = AY − 10 + FB = 0.
Solution:
The sum of the moments about A is
MA = −2(10) + 4FB = 0,
from which FB =
4m
X
A
B
10 kN
AX
AY
20
= 5 kN.
4
X
4m
Substitute into the force balance equation to obtain
AY = 10 − FB = 5 kN
Problem 5.9 Consider the beam in Problem 5.8. An
engineer determines that each support will safely support a force of 7.5 kN. What is the range of values of
the distance x at which the 10-kN force can safely be
applied?
Solution: From the solution to Problem 5.8 the equations for FB
and AY are:
MA = 4FB − 10x = 0,
from which FB
10x
=
= 2.5x.
4
FY = AY − 10 + FB = 0,
from which AY = 10 − FB = 10 − 2.5x
Solve for the value of x:
x =
and x =
FB
,
2.5
10 − AY
.
2.5
10 kN
Let FB = 7.5 kN,
then AY = 2.5 kN
and x =
7.5
= 3 m.
2.5
Let AY = 7.5 kN,
then FB = 2.5 kN,
and x =
Thus
10 − 7.5
= 1 m.
2.5
1≤x≤3m
FB
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:
FX = 0,
and
FY = FA + FB + 100 − 400 = 0.
The sum of the moments about A is
MA = −3(100) + 900 − 7(400) + 11FB = 0.
From which FB =
4 ft
3 ft
100 lb
4 ft
400 lb
3 ft
4 ft
A
3 ft
4 ft
900 ft lb
100 lb
3 ft
4 ft
2200
= 200 lb
11
3 ft
B
400 lb
4 ft
900 ft lb
FA
Substitute into the force balance equation to obtain
FB
FA = 300 − FB = 100 lb
Problem 5.11 Consider the beam in Problem 5.10.
First represent the loads (the 100-lb force, the 400-lb
force, and the 900 ft-lb couple) by a single equivalent
force; and then determine the reactions at the supports.
Solution:
The equivalent force is the sum of the applied force:
FY = 100 − 400 = −300 lb.
The sum of moments about A is
MA = −(7.33)(300) + 11FB = 0,
The applied moment about the point A is
MA = −3(100) + 900 − 7(400) = −2200 ft lb.
from which FB =
The equivalent force must be applied at a point that produces this
moment about A.
Let x be the distance to the line of action from A: then
FA = 300 − FB = 100 lb.
2200
= 200 lb.
11
Substitute into the force balance equation to obtain:
−300x = −2200 ft lb from which x
300 lb
2200
=
= 7.33 ft
300
to the right of A. The equivalent system is shown. The sum of the
forces:
FX = 0,
FY = FA + FB − 300 = 0.
7.33 ft
11 ft
FA
FB
Problem 5.12 (a) Draw the free-body diagram of the
beam.
(b) Determine the reactions at the support.
2 kN
2.4 kN-m
400 mm
Solution:
B
A
800 mm
400 mm
The equilibrium equations are
FX = AX − FB cos 60◦ = 0,
2 kN
2.4 kN-m
B
A
FY = AY − 2 + FB sin 60◦ = 0,
30°
400 mm 800 mm 400 mm
◦
M(pt A) = −2.4 − (0.4)(2) + (1.6)FB sin 60 = 0.
Solving, we obtain
2 kN
AX = 1.15 kN,
AX
2.4 kN-m
60°
AY 400 mm 800 mm 400 mm F
B
AY = 0,
FB = 2.31 kN.
Problem 5.13 Consider the beam in Problem 5.12.
First represent the loads (the 2-kN force and the 24-kNm couple) by a single equivalent force; then determine
the reactions at the supports.
Solution:
The single force is equivalent to the force and couple if
2 kN
d(2) = (0.4)(2) + 2.4,
so
2.4 kN-m
x
d = 1.6 m.
From the equilibrium equations
FX = AX − FB cos 60◦ = 0,
FY = AY − 2 + FB sin 60◦ = 0,
M(pt B) = −1.6AY = 0,
0.4 m
2 kN
x
d
2 kN
we obtain
AX = 1.15 kN,
AY = 0,
FB = 2.31 kN.
AX
60°
AY
1.6 m
FB
30°
Problem 5.14 If the force F = 40 kN, what are the
reactions at A and B?
y
A
F
6m
B
x
8m
12 m
Solution: There are two force equilibrium equations and one moment equilibrium equation for this Problem. The force equilibrium
equations are
Fx = FAX + FBX = 0,
and
Fy = FBY + F kN = 0.
The moment equilibrium equation around point B is
MB = −(4 m)(F kN)
y
FAX
A
6m
40 kN
B
−(6 m)FAX = 0.
Since we know that F = 40 kN, We have three scalar equations in the
three unknowns. The moment equation can be solved alone, giving
FAX = −26.7 kN. Substituting this value into the force equilibrium
equation in the x direction yields FBX = 26.7 kN. Finally, the y
direction force equation yields FBY = −40 kN.
FBX
x
8m
12 m
FBY
Problem 5.15 In Problem 5.14, the structural designer
determines that the magnitude of the force exerted on the
support A by the beam must not exceed 80 kN, and the
magnitude of the force exerted on the support B must
not exceed 140 kN. Based on these criteria, what is the
largest allowable value of the upward load F ?
Solution:
and
From Problem 5.14, we have
Fx = FAX + FBX = 0,
Fy = FBY + F kN = 0,
MB = −(4 m)(F kN) − (6 m)FAX = 0.
We also know that
|FA | = FAX
2
2 .
and |FB | = FBX
+ FBY
Thus we have five equations relating the six variables F , FAX , FBX ,
FBY , |FA |, and |FB |.
Case 1: If we set |FA | = FAX = 80 kN and solve the equations,
we get answers for all of the forces in the system. The one answer that
concerns us is that we get |FB | = 144.2 kN, which violates the force
condition at B. Thus, the loading in Case 1 cannot be allowed for this
system.Case 2: If we set
2
2
|FB | = FBX
+ FBY
= 140 kN
and solve the equations, we again get answers for all of the forces in the system.
In this case, we get
|FA | = |FAX | = −77.65 kN,
which is under the 80 kN limit for this force. This is the situation we wanted—
one force at its limit and the other under its limit. In this case, the solution of
the equation set yields F = 116.4 kN. Note: The actual answer for F can be
rounded up to F = 116.5 kN, but if we round up, we exceed the load limits
set in our Problem. Be very careful in such cases. (In the real world, we have
safety factors that eliminate the possibility of such situations)
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 = 250 mm, b = 740 mm, and c = 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 = 0: No forces in x-direction
FY = 0: 2FH + 2FF − W = 0
MH = 0: − aW + (a + b)(2FF ) = 0
Solving, we get
W = 784.8 N
FH = 293.3 N
FF = 99.1 N
c
W
a
b
y
H
F
W
a
2FH
b
2FF
x
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) The ratio R/F is called the mechanical advantage.
Determine the mechanical advantage of each lever.
(b) Determine the magnitude of the reaction at A for
each lever. (Express your answer in terms of F .)
R
F
R
A
A
L
L
L
First-class lever
R
F
L
L
Third-class lever
F
R
from which
F
(b)
L
R
The reaction at A is obtained from the force balance equation:
The sum of the moments about A is
MA
= −LR + 2LF = 0,
R
from which
F
A = −F + R = −F + 2F = F
Lever of third kind.
(a) The sum of forces is
FY = A − R + F = 0.
The sum of moments about A is:
MA
= −2LR + LF = 0,
R
F
=
L
1
=
2L
2
From the force balance equation
A = −F + R = −F +
|A| =
F
2
L
L
F
R
A
2L
=
=2
L
The reaction at A is obtained from the force balance equation:
from which:
F
A
Lever of second kind.
(a) The sum of forces is
FY = A − R + F = 0.
(b)
L
L
=
=1
L
A = R + F = 2F
(b)
R
A
The sum of the moments about A is
MA
= F L − RL = 0,
F
F
=− ,
2
2
L
L
Second-class lever
A
Solution: Lever of first kind.
(a) The sum of the forces is
FY = −F + A − R = 0.
F
L
Problem 5.18 (a) Draw the free-body diagram of the
beam. (b) Determine the reactions at the support.
400 lb
A
1400 lb
3 ft
7 ft
Solution:
FX = 0:
AX + 200 N = 0
FY = 0:
AY + 300 N − 200 N = 0
MA = 0:
Solving:
MA − 180 − (0.6)(300) + (0.5)(200) = 0
A
0.5 m
300 N
180 N-m
AX = −200 N
200 N
AY = −100 N
200 N
0.6 m
MA = 260 N-m
0.3 m
AY
AX
MA
A
300 N
0.5 m
180 N-m
200 N
0.6 m
0.3 m
200 N
Problem 5.19
reactions at A.
The force F = 12 kN. Determine the
y
F
A
45°
x
5m
Solution:
FX :
AX − 12 cos(45◦ ) = 0
FY :
AY + 12 sin(45◦ ) = 0
MA :
Solving:
y
F
A
45°
x
MA − (5)(12)(sin 45◦ ) = 0
AX = 8.49 kN
5m
AY = −8.49 kN
MA = 42.4 kN-m
AY
F
45°
AX
5m
MA
Problem 5.20 The built-in support of the beam shown
in Problem 5.19 will fail if the magnitude of the total
force exerted on the beam by the support exceeds 20 kN
or if the magnitude of the couple exerted by the support
exceeds 65 kN-m. Based on these criteria, what is the
maximum force F that can be applied?
Solution:
(a) Assume the limit is a force limit.
F=0 A+F=0
Thus, |A| = |F| and if |A| = 20 kN, the force limit for |F| is
20 kN.
(b) Assume the limit is a moment limit
M
= MA − 5(F sin 45◦ ) = 0
and MA
= 65 kN-m
Here,
F
F
=
65
5
sin 45◦
= 18.38 kN
The limit is |F| = 18.38 kN
y
F
A
45°
x
5m
F
45°
AY
MA
AX
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
FY = −C − D + FCD = 0,
FCD
D
7 in
from which FCD = C + D.
2C
from which D =
.
7
For the crossbar supporting B, the sum of the moments is
MBCD = 6FCD − 2B = 0,
from which, substituting from above
FCD =
2B
2C
9C
=C+D =C+
=
,
6
7
7
or C = 7B/27 = 7 oz,
and D = 2C/7 = 2 oz.
The sum of the moments about the crossbar supporting A is
MAB = 12A − 3FBCD = 0,
from which, substituting from above,
A =
3(B + C + D)
27 + 7 + 2
=
= 9 oz
12
4
12 in
3 in
6 in 2 in
A
7 in
D
B
2 in
C
2 in
FBCD
For the cross bar supporting the weight B, the sum of the forces is
FY = −B + FBCD − FCD = 0,
from which, substituting, FBCD = B + C + D.
For the crossbar supporting C and D, the sum of the moments about
the support is
MCD = 7D + 2C = 0,
C
B
FCD
6 in
2 in
FAB
FBCD
A
12 in
3 in
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 = 2.00 m,
y = 0.68 m. If the angle α = 15◦ , what is the total
normal force exerted on the two rear tires by the sloped
ramp?
y
x
α
y
Solution:
Split W into components:
y
W cos α acts ⊥ to the incline
W sin α acts parallel to the incline
FX : f − W sin α = 0
FY : NR + NF − W cos α = 0
MR : (−2)(W cos α) + (0.68)W sin α + 2.82NF = 0
x
α
Solving: NR = 5930 N, NF = 10750 N
W
y
x
α
2m
m
2.82
R
0.68
α
m
f
NR
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 α for which it can remain in equilibrium?
Solution:
The solution to Problem 5.22 yielded
f = W sin α
NR + NF − W cos α = 0
−2W cos α + 0.68W sin α + 2.82NF = 0
Our limit is f /NR ≤ 0.8, so let us set f = 0.8NR and solve the
resulting relations for αmax
0.8NR = W sin αmax
NR + NF − W cos αmax = 0
−2W cos α + 6.68 W sin αmax + 2.82NF = 0
Solving, we get
αmax = 16.1◦ ,
f = 4788 N,
NR = 5985 N,
NF = 10603 N.
NF
α = 15°
W = (1760X9.81) N
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 = −5 + BX − R cos 60◦ = 0.
FY = 10 − 14.5 + BY − R sin 60◦ = 0.
The sum of the moments about the origin is
MO = 7R cos 60◦ + 8BY − 2(14.5) − 13(10) − 5(1.5) = 0.
From which 7R cos 60◦ + 8BY − 166.5 = 0. Collecting equations
and reducing to 3 equations in 3 unknowns:
BX + 0BY − 0.5R = 5
0BX + BY − 0.866R = 4.5
0BX + 8BY + 3.5R = 166.5.
Solving:
BX = 11.257 lb,
BY = 15.337 lb,
and R = 12.514 lb
y
R
60°
Bx
A
1.5 in
5 lb
By
14.5 lb
10 lb
13 in
6 in
2 in
7 in
x
Problem 5.25 The mass of the trailer is 2.2 Mg (megagrams). The distances a = 2.5 m and b = 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) Draw the free-body diagram of the trailer.
(b) Determine the total normal force exerted on the rear
tires at A and the reactions exerted on the trailer at
the pin support B.
Solution:
(a) The free body diagram is shown.
(b) The sum of forces:
FX = BX = 0.
FY = FA − W + FB = 0.
The sum of the moments about A:
MA = −aW + (a + b)FB = 0,
from which
FB =
aW
2.5(2.2 × 103 )(9.81)
=
= 6.744 kN
a+b
(2.5 + 5.5)
Substitute into the force equation:
FA = W − FB = 14.838 kN
B
W
a
A
b
B
BX
FB
W
A
FA
a
b
B
W
A
a
b
Problem 5.26 The total weight of the wheelbarrow
and its load is W = 100 lb.
(a) If F = 0, what are the vertical reactions at A and B?
(b) What force F is necessary to lift the support at A
off the ground?
F
W
A
40 in
Solution:
B
12 in
14 in
(a) The sum of the forces:
FX = AX = 0
FY = AY − W + FB = 0.
The sum of the moments about A is
MA = −W (12) + FB (26) = 0,
from which
FB =
12W
= 46.1538 lb = 46.2 lb.
26
Substitute into the force equation to obtain:
AY = W − FB = 53.8462 lb = 53.8 lb
(b) The sum of the moments about B when the point A is not making
contact with the ground:
MB = (14)(100) − (66)F = 0,
from which
F =
(14)(100)
= 21.2121 = 21.2 lb
66
F
AX
A
W
12
40
AY
B
14
FB
Problem 5.27 The airplane’s weight is W = 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) 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
(b) when T = 0,
(c) when T = 250 lb.
Solution: (a) The free body diagram is shown. (b) The sum of
the forces:
FX = BX = 0
FY = AY − W + BY = 0.
The sum of the moments about A is
MA = −5W + 7BY = 0,
from which BY = 5W
= 1714.3 lb
7
Substitute from the force balance equation:
AY = W − BY = 685.7 lb
(c) The sum of the forces:
FX = −250 + BX = 0,
from which BX = 250 lb
FY = AY − W + BY = 0.
The sum of the moments about A:
MA = (250)(4) − 5W + 7BY = 0,
from which BY = 1571.4 lb. Substitute into the force balance equation to obtain: AY = 828.6 lb
T
4 ft
W
5 ft
A
4 ft
2 ft B
W
A
AY
5 ft
B
2 ft
BY
BX
T
4 ft
W
A
5 ft
B
2 ft
Problem 5.28 The forklift is stationary. The front
wheels are free to turn, and the rear wheels are locked.
The distances are a = 1.25 m, b = 0.50 m, and c =
1.40 m. The weight of the load is WL = 2 kN, and the
weight of the truck and operator is WF = 8 kN. What
are the reactions at A and B?
Solution:
The sum of the forces:
FX = BX = 0
FY = AY − WL − WF + BY = 0.
The sum of the moments about A is
MA = +aWL − bWF + (b + c)BY = 0,
from which
BY =
bWF − aWL
= 0.7895 kN.
b+c
Substitute into the force equation to obtain:
AY = WL + WF − BY = 9.211 kN
WL
WF
A
B
WL
a
b
c
Problem 5.29 Consider the stationary forklift shown
in Problem 5.28. The front wheels are free to turn, and
the rear wheels are locked. The distances are a = 45 in.,
b = 20 in., and c = 50 in. The weight of the truck and
operator is WF = 3000 lb. For safety reasons, a rule is
established that the reaction at the rear wheels must be
at least 400 lb. If the weight WL of the load acts at the
position shown, what is the maximum safe load?
From the solution to Problem 5.26, BX = 0, and the
sum of the moments:
MA = +aWL − bWF + (b + c)BY = 0,
Solution:
from which
WL =
bWF − (b + c)BY
= 1333.33 − 1.5555BY .
a
Since BY is positive, the maximum load for BY = 400 lb is WL =
711.1 lb
WF
A
AY
a
BX
B
b
c
BY
Problem 5.30 The weight of the fan is W = 20 lb. Its
base has four equally spaced legs of length b = 12 in.,
and h = 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
Solution: Each leg is assumed to be in contact with a rough surface, with (in two dimensions) two force components each.
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◦ = 0.707b. Tipping will occur when
the leftmost (or rightmost) leg(s) has zero reaction on the floor.
For each position the sum of the moments about the center is:
MT = −bW + T h = 0, and
MT = −0.707bW + T h = 0.
Top View
T
h
F
W
From which the two tipping moment thrusts are:
T1 =
bW
= 6.67 lbs,
h
T2 =
0.707bW
= 4.71 lb
h
which is the maximum thrust allowed.
b
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 = (0.707b)(0.4W ) + T h − (0.707b)(0.6W ) = 0,
from which:
h =
0.707b(0.2W )
= 33.94 in
T
b
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
√
T( three legs) = (1/ 2)T( four legs).
Solution:
b
T
From the solution to Problem 5.30 the maximum
thrust is
Tfour legs =
bW sin 45◦
bW
= √ .
h
2h
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 is
b cos 60◦ = 2b . When the outermost leg has zero reaction, the other
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 = −
from which Tthree legs =
T
b
bW
+ T h = 0,
2
bW
1
= √ Tfour legs .
2h
2
Problem 5.33 A force F = 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 = AX + BX = 0,
FY = AY − F = 0,
from which
F
A
80 mm
AY = F = 400 N.
The sum of the moments about A is
MA = 0.08BX − 0.320F = 0,
B
320 mm
from which
BX =
0.320(400)
= 1600 N.
0.08
F
Substitute into the sum of forces equation to obtain:
A
AX = −BX = −1600 N
80 mm
B
320 mm
Problem 5.34 The hanging sign exerts vertical 25-lb
forces at A and B. Determine the tension in the cable
and the reactions at the support at C.
30°
A
C
B
8 ft
1 ft
Solution: The joint C is pinned joint, with two forces, CX , CY .
The sum of the vertical forces is
FX = TC cos 30◦ + CX = 0.
FY = TC sin 30◦ − AY − BY + CY = 0.
The sum of the moments about C is
MC = 9AY + 1BY − 10TC sin 30◦ = 0
from which
TC =
9AY + BY
250
=
= 50 lb.
10 sin 30◦
5
Substitute into the vertical forces sum to obtain:
CY = AY + BY − TC sin 30◦ = 25 lb.
Substitute into the horizontal forces sum to obtain
CX = −TC cos 30◦ = −43.3 lb
30
C
A
B
1 ft
8 ft
1 ft
CY
TC
130°
CX
BY
AY
1 ft
8 ft
1 ft
1 ft
Problem 5.35 This device, called a swape or shadoof,
is used to help a person lift a heavy load. (It was used
in Egypt at least as early as 1550 B.C. and is still in use
in various parts of the world today.) The distances are
a = 12 ft and b = 4 ft. If the load being lifted weighs
100 lb and W = 200 lb, 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 3 ft above the ground. (Assume that
the rope remains vertical.)
a
b
25°
W
Solution: Denote the vertical force exerted by the laborer by F .
(a) The sum of the moments about the fulcrum point is
MF = −aF cos 25◦ + aWL cos 25◦ − bW cos 25◦ = 0
from which
F =
or F =
(b)
(aWL − bW )
,
a
1200 − 800
= 33.33 lb.
12
the force exerted by the laborer is independent of the angle.
When the load is three feet above the ground, the swape is at
a new angle. But since the force is independent of the angle,
F = 33.33 lb
a
b
25°
W
a
25°
b
F
W
WL
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.
3F
F
b
A
B
b
b
b
b
Solution: The truss can be treated as a single member. The pinned
joint at A is a two force support; the roller support at B is a single force
support. The sum of the forces:
FX = AX = 0.
FY = AY + BY − F − 3F = 0.
The sum of moments about A is
MA = −bF − 3b(3F ) + 4bBY = 0,
from which BY = 2.5F.
From the vertical force sum,
AY = 4F − 2.5F = 1.5F
3F
F
b
A
b
b
b
B
b
3F
F
AX
AY
BY
b
2b
b
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 = b = 9 in. and c =
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 =
144
+ 8 = 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:
FX = SX = 0.
FY = FH − SY − 8 = 0,
144 lb
8 lb
from which SY = FH − 8 = 72 lb. The sum of the moments about
the shoulder is
MS = M + (b + c)FH − b8 = 0,
where M is the couple reaction at the shoulder. Thus
M = b8 − (b + c)FH = −1688 in lb = 1688 (in lb)
= 140.67 ft lb
1 ft
12 in
a
b
FH
c
FH
8 lb
8 lb
144 lb
FH
SX
M
8 lb
SY
b
c
Determine the reactions at A.
Problem 5.38
A
5 ft
300 lb
800 ft-lb
200 lb
200 lb
6 ft
3 ft
The built-in support at A is a two-force and couple
reaction support. The sum of the forces for the system is
FX = AX + 200 = 0,
Solution:
from which
AX = −200 lb
FY = AY + 300 − 200 = 0,
from which AY = −100 lb
The sum of the moments about A:
M = −6(300) + 5(200) − 800 + MA = 0,
from which MA = 1600 ft lb which is the couple at A.
A
5 ft
300 lb
800 ft-lb
200 lb
3 ft
200 lb
6 ft
AY
MA
300 lb
AX
5 ft
800 ft-lb
200 lb
200 lb
6 ft
3 ft
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 α = 15◦ ;
(b) on a down slope with α = −15◦ .
n
70 i
n
36 i
n
20 i
y
3300 lb
α
Solution: The rear wheels are two force reaction support, and the
front wheels are a one force reaction support. Denote the rear wheels
by A and the front wheels by B, and define the reactions as being
parallel to and normal to the road. The sum of forces:
FX = AX − 3300 sin 15◦ = 0,
n
70 i
n
36 i
n
20 i
from which
AX = 854.1 lb.
x
y
FY = AY − 3300 cos 15◦ + BY = 0.
3300 lb
α
Since the mass center of the vehicle is displaced above the point A,
a component of the weight (20W sin α) produces a positive moment
about A, whereas the other component (36W cos α) produces a negative moment about A. The sum of the moments about A:
MA = −36(3300 cos 15◦ ) + 20(3300 sin 15◦ ) + BY (106) = 0,
20 in.
α
from which
BY =
+97669
= 921.4 lb.
106
Substitute into the sum of forces equation to obtain AY = 2266.1 lb
(b) For the car parked down-slope the sum of the forces is
FX = AX + 3300 sin 15◦ = 0,
from which AX = −854 lb
FY = AY − 3300 cos 15◦ + BY = 0.
The component (20W sin α) now produces a negative moment about
A. The sum of the moments about A is
MA = −3300(36) cos 15◦ − 3300(20) sin 15◦ + 106BY = 0,
from which
BY =
131834
= 1243.7 lb.
106
Substitute into the sum of forces equation to obtain AY = 1943.8 lb
x
BY
3300 lb
AX
AY
36
in.
70
in.
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 α = 45◦ , since the altitude and base of the triangle are equal.
The reaction at the top of the bar forms the angle (90 − α) with the
horizontal, and the reaction at the base is vertical. The sum of the
forces is
FX = −T + B cos(90 − α) = −T + B sin α = 0.
FY = −W + AY + B cos α = 0.
The sum of the moments about the lower end is
WL
L
MA =
cos α − B √
= 0,
2
2
from which
B =
W cos α
√
.
2
Substitute into the horizontal force equation to obtain the string tension
W
T = √ sin α cos α
2
W
= √ = 0.3536W
2 2
L
L
−
2
L
−
2
Problem 5.41 The mass of the bar is 36 kg and its
weight acts at its midpoint. The spring is unstretched
when α = 0. The bar is in equilibrium when α = 30◦ .
Determine the spring constant k.
k
4m
α
2m
Solution:
l2 = 42 + 22 − 2(4)(2) cos 30◦
Solving, l = 2.48 m
The force acting at the top end of the bar is F = k(δ) where δ = l−l0 .
We also need φ when α = 30◦
k
sin φ
sin α
sin 30◦
=
=
z
l
2.48
φ = 23,78◦ when α = 30◦
Equilibrium equations:
+ →
FX = 0:
kδ sin φ + AX = 0
kδ cos φ + AY − mg = 0
+↑
FY = 0:
+
MB = 0:
AY (2 sin α) − mg(1 sin α)
4m
α
2m
+AX (2 cos α) = 0
Substituting δ, φ, and α into the equations and solving, we get
AX = −44.1 N
A
AY = 253.0 N
k = 229 N/m
y
kδ
x
lo
4m
B
4m
α
30°
mg
2m
AY
AX
2m
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◦ + 60◦ = 150◦ with the positive x-axis.
The sum of the forces:
FX = AX + B cos 150◦ = 0
FY = AY + B sin 150◦ = 0
6 kN-m
2 kN-m
A
The sum of the moments about B is
MB = −2AY + 2 − 6 = 0,
60°
B
2m
from which
AY = −
4
= −2 kN.
2
Substitute into the force equations to obtain
B =
2 kN-m
6 kN-m
AY
= 4 kN,
sin 150◦
and AX = −B cos 150◦ = 3.464 kN.
The horizontal and vertical reactions at B are
AX
150°
AY
B
BX = 4 cos 150◦ = −3.464 kN,
2m
and BY = 4 sin 150◦ = 2 kN.
Problem 5.43 The force F = 800 N, and the couple
M = 200 N-m. The distance L = 2 m. What are the
reactions at A and B?
A
F
L
M
B
L
Solution:
L
A
The sum of the forces:
FX = AX + BX = 0
FY = BY + F = 0,
L
from which AX
F
M
from which BY = −F = −800 N.
The sum of the moments about B is
MB = −LAX + LF − M = 0,
B
L
L
LF − M
=
= 700 N.
L
Substitute into the force equations to obtain BX = −700 N.
AX
L
F
M
BX
BY
L
L
Problem 5.44 The mass of the bar is 40 kg and its
weight acts at its midpoint. Determine the tension in the
cable and the reactions at A.
60°
A
W
2m
Solution:
Equations of equilibrium:
FX = 0:
AX − T cos 30◦ = 0
FY = 0:
AY + T sin 30◦ − mg = 0
MA = 0:
−2 mg + 4(T sin 30◦ ) = 0
2m
Solving, we get
AX = 340 N,
AY = 196 N,
T = 392 N.
60°
A
W
2m
2m
T
AY
y
30°
AX
2m
2m
W
x
W = mg
Problem 5.45 If the length of the cable in Problem
5.44 is increased by 1 m, what are the tension in the
cable and the reactions at A?
Solution:
cos 30◦ =
l1 =
We first need the cable length in Problem 5.44.
4
l1
4
= 4.62 m
cos 30◦
l1
2.31 m
30°
In the new problem,
l2 = l1 + 1 m
4m
l2 = 5.62
From the law of cosines,
42 = 5.622 + 2.312 − 2(5.62)(2.31) cos α
α = 36.3◦
h = 2.31 m
α
From the law of sines
sin α
sin γ
=
4
h
5.62 = l2
β
γ = 20.0◦
γ
Now α + β + γ = 180◦
4m
β = 123.7◦
We now have the geometry defined and can draw a free body diagram
and write the equilibrium eqns.
AY
β = β − 90◦ = 33.7◦
FX = AX − T cos(γ + β ) = 0
AX
β
γ
FY = AY + T sin(γ + β ) − mg = 0
MA = −mg(2 cos β ) + T sin(γ + β )(4 cos β )
−T cos(γ + β )(4 sin β ) = 0
Solving, we get
AX = 282.88 N(282.55)
AY = 7.89 N(7.76)
T = 477.36 N(477.28)
β′
β′
mg
T
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
Fx = AX + EX = 0,
Fy = AY − 2(80)(9.81) = 0,
and
MA = 0.3EX − 0.2(80)(9.81) − 0.4(80)(9.81) = 0.
A
200 mm
B
C
300 mm
D
We have three equations in the three components of the support reactions. Solving for the unknowns, we get the values
E
AX = −1570 N,
200 mm
AY = 1570 N,
200 mm
and EX = 1570 N.
y
AY
0.2 m
0.2 m
AX A
x
mg
0.3 m
E
mg
EX
Problem 5.47 The suspended boxes in Problem 5.46 Solution: Written with the mass value of 80 kg replaced by the symbol m,
are each of mass m. The supports at A and E will each the equations of equilibrium from Problem 5.46 are
safely support a force of 6 kN magnitude. Based on this
Fx = AX + EX = 0,
criterion, what is the largest safe value of m?
and
Fy = AY − 2 m(9.81) = 0,
MA = 0.3EX − 0.2 m(9.81) − 0.4 m(9.81) = 0.
We also need the relation
|A| = A2X + A2Y = 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 = −4243 N,
AY = 4243 N,
EX = 4243 N,
and m = 216.5 kg.
Note: We could have gotten this result by a linear scaling of all of the numbers
in Problem 5.46.
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 = AX = 0.
FY = AY − 200 = 0,
6 ft
C
6 ft
A
B
3 ft
3 ft
from which AY = 200 lb.
300
ft-lb
6 ft
200 lb
The sum of the moments about A is
M = MA − (3)(200) − 300 = 0,
from which MA = 900 ft lb
MA
AY
300 ft-lb
AX
200 lb
3 ft
Problem 5.49 The tension in cable AB is 2 kN. What
are the reactions at C in the two cases?
60°
A
2m
B C
A
1m
2m
(a)
Solution:
FX = CX − T cos 60◦ = 0,
60°
A
B
2m
C
1m
from which CY = −1.866(2) = −3.732 kN.
The sum of the moments about C is
M = MC − T sin 60◦ − 3T = 0,
60°
A
B
2m
C
1m
from which MC = 3.866(2) = 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.
T
T
CY
60°
2m
1m
Case (a)
MC
CX
CY
Case (b)
MC
CX
B C
1m
(b)
First Case: The sum of the forces:
from which CX = 2(0.5) = 1 kN
FY = CY + T sin 60◦ + T = 0,
60°
Problem 5.50
Determine the reactions at the supports.
6 in
5 in
50 lb
A
3 in
100 in-lb
3 in
B
30°
Solution: The reaction at A is a two-force reaction. The reaction
at B is one-force, normal to the surface.
The sum of the forces:
FX = AX − B cos 60◦ − 50 = 0.
FY = AY + B sin 60◦ = 0.
The sum of the moments about A is
MA = −100 + 11B sin 60◦ − 6B cos 60◦ = 0,
from which
B =
6 in
5 in
50 lb
A
100 in lb
3 in
3 in
B
30°
AX
100
= 15.3 lb.
(11 sin 60◦ − 6 cos 60◦ )
50 lb
AY
6 in.
Substitute into the force equations to obtain
100
AY = −B sin 60◦ = −13.3 lb
and AX = B cos 60◦ + 50 = 57.7 lb
B
11 in.
60°
Problem 5.51 The weight W = 2 kN. Determine the
tension in the cable and the reactions at A.
30°
A
W
0.6 m
Solution:
+
0.6 m
Equilibrium Eqns:
FX = 0:
AX + T cos 30◦ = 0
FY = 0:
AY + T + T sin 30◦ − W = 0
MA = 0:
A
30°
◦
(−0, 6)(W ) + (0.6)(T sin 30 )
+ (1, 2)(T ) = 0
W
Solving, we get
0.6 m
0.6 m
AX = −693 N,
AY
AY = 800 N,
T
T = 800 N
30°
AX
0.6 m
0.6 m
W = 2 kN = 2000 N
T
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:
lem are
FX = 0:
AX + T cos 30◦ = 0
FY = 0:
AY + T + T sin 30◦ − W = 0
MA = 0:
◦
+
The equilibrium equations in the solution of prob-
(−0, 6)(W ) + (0, 6)(T sin 30 )
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 = 6 kN, we
solve to get
AX = −5.2 kN
AY = 6.0 kN
W = 15.0 kN
+(1, 2)(T ) = 0
Problem 5.53 The spring constant is k = 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 mm
Solution:
l =
302
A
The spring force is kδ where δ = 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 = 0.0300 m and l = 0.0424 m
Equilibrium equations:
FX = 0:
AX − k(l − l0 ) sin 45◦
50 mm
The length of the spring is
√
+ 302 mm = 1800 mm
l = 42.4 mm = 0.0424 m
30°
C
k
24 mm
B
15 mm
30°
C
k
30 mm
−NC cos 60◦ = 0
FY = 0:
AY − k(l − l0 ) cos 45◦
MB = 0:
50 mm
30 mm
+NC sin 60◦ = 0
(−0.024)AX + (0.050)(NC sin 60◦ )
AY
◦
−(0.015)(NC cos 60 ) = 0
AX
Solving, we get
AX = 126.7 N
24 mm
B
AY = 10.5 N
kδ
NC = 85.1 N
15 mm
30 mm
45°
60
50 mm
30 mm
30°
C
NC
Problem 5.54 The engineer designing the release
mechanism shown in Problem 5.53 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 5.53. The equilibrium
equations derived were
FX = 0:
AX − k(l − l0 ) sin 45 − NC cos 60◦ = 0
FY = 0:
AY − k(l − l0 ) cos 45 + NC sin 60◦ = 0
MB = 0:
−0.024AX + 0.050NC sin 60◦
A
24 mm
B
◦
−0.015NC cos 60 = 0
where l = 0.0424 m, l0 = 0.030 m, NC = 120 N, and AX , AY ,
and k are unknowns.
Solving, we get
15 mm
k
30 mm
AX = 179.0 N,
AY = 15.1 N,
50 mm
30 mm
k = 13500 N/m
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 = 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:
a =π
The area of the valve is
2
0.15
= 17.671 × 10−3 m2 .
2
150
mm
250
mm
The force at opening is
F = 10a × 106 = 1.7671 × 105 N.
k
A
The force on the spring is found from the sum of the moments about
A,
MA = 0.15F − (0.4)k∆L = 0.
150 mm
Solving,
k =
0.15F
0.15(1.7671 × 105 )
=
(0.4)∆L
(0.4)(0.02)
k∆L
A
N
= 3.313 × 106
m
F
0.15
m
0.25
m
C
30°
Problem 5.56 The bar AB is of length L and weight
W , and the weight acts at its midpoint. The angle α =
30◦ . What is the tension in the string?
C
B
L
α
A
L
Solution: The strategy is to determine the angle formed by the
string on the end of the rod (see sketch). The vertical distance from
the pinned joint to the end of the rod is D = L sin α. The horizontal
distance is H = L cos α. The sides of the small triangle formed by
the string is X = L − L cos α and Y = L − L sin α. The angle
formed by the string with the horizontal is
1 − sin α
β = tan−1
= 75◦ .
1 − cos α
The angle relative to the rod is θ = β − α = 45◦ . The moment about
the pinned joint is
L
M =−
W cos α + T L sin θ = 0,
2
from which
T =
√
W (0.866)
3W
W cos α
=
= √ = 0.61237W
2 sin θ
2(0.707)
2 2
C
B
A
L
F
α
L
T
S
β
α
L
L
L
W
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
tan θ = 2.4/1.2,
or θ = 63.4◦ .
From the diagram,
C
FHX = |FH | cos θ,
and FHY = |FH | sin θ.
The force equilibrium equations are
Fx = AX + FHX = 0,
Fy = AY + FHY − (200)g − (800)g = 0,
A
2.4 m
1m
B
1.8 m 1.2 m
7m
and the moment equation is
MA = −(2)(200)g − (7)(800)g + (3)FHY − (2.4)FHX = 0.
y
Solving the five equations simultaneously, we get |FH | = 36.56 kN,
which is the result called for in this problem. Other values obtained in
the solution are
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
|A| = A2X + A2Y = 28.13 kN.
FH
2m
mg
θ
C
AX = −16.35 kN,
and AY = −22.89 kN.
FHY
mBg
P
AX
A
FHX
2.4 m
1 m AY B
1.8 1.7
m
m
7m
x
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 = mg = 1275 N.
The equations of equilibrium for the entire assembly are
Fx = CX = 0,
Fy = AY + CY − mg = 0
(where the mass m = 130 kg), and
MC = −(1)AY − (1.5)mg = 0.
0.5 m 0.5 m0.5 m
C
E
A
1m
Solving these equations, we get
B
CX = 0,
D
CY = 3188 N,
G
and AY = −1913 N.
From the free body diagram of the speaker alone, we get
Fy = T1 + T2 − mg = 0,
and
Mleft support = −(1)mg + (1.5)T2 = 0.
Solving these equations, we get
0.5 m
1m
1m
AY
1.5 m
CY
E
A
C CX
T1 = 425. N
B
and T2 = 850 N
D
mg
1.5 m
1m
T2
T1
mg
Problem 5.60 The weight W1 = 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
B
T2 = W2 (i cos(−35◦ ) + j sin(−35◦ )).
= W2 (0.8192i − 0.5736j)(lb).
Assume a unit length for the bar. The angle between the bar and the
positive x-axis is 180◦ − 50◦ = 130◦ . The position vector of the tip
of the bar relative to A is
W1
50°
A
35°
rB = i cos(130◦ ) + j sin(130◦ ), = −0.6428i + 0.7660j.
The tension exerted by W1 is T1 = −1000j. The sum of the moments
about A is:
MA = (rB × T1 ) + (rB × T2 ) = rB × (T1 + T2 )
i
j
= L −0.6428
0.7660
0.8191W
−0.5736W − 1000 2
2
MA = (−0.2587W2 + 642.8)k = 0,
W2
35°
T2
rB
T1
50°
from which W2 = 2483.5 lb
The sum of the forces:
FX = (AX + W2 (0.8192))i = 0,
from which AX = −2034.4 lb
FY = (AY − W2 (0.5736) − 1000)j = 0,
from which AY = 2424.5 lb
C
AY
AX
Problem 5.61 The dimensions a = 2 m and b = 1 m.
The couple M = 2400 N-m. The spring constant is
k = 6000 N/m, and the spring would be unstretched if
h = 0. The system is in equilibrium when h = 2 m and
the beam is horizontal. Determine the force F and the
reactions at A.
k
h
A
M
F
a
Solution:
b
We need to know the unstretched length of the spring, l0
l0 = a + b = 3 m
k
We also need the stretched length
l2 = h2 + (a + b)2
h
l = 3.61 m
FS = k(l − l0 )
tan θ =
A
M
h
(a + b)
F
a
θ = 33.69◦
Equilibrium eqns:
FX :
AX − FS cos θ = 0
FY :
AY + FS sin θ − F = 0
+
MA :
−M − aF + (a + b)FS sin θ = 0
a = 2 m,
h = 2 m,
b = 1 m,
Substituting in and solving, we get
FS = 6000(l − l0 ) = 3633 N
and the equilibrium equations yield
AX = 3023 N
AY = −192 N
F = 1823 N
Unstretched
(a + b)
AY
M = 2400 N-m,
k = 6000 N/m.
b
θ
M
AX
a
F
b
Problem 5.62 The bar is 1 m long, and its weight W
acts at its midpoint. The distance b = 0.75 m, and the
angle α = 30◦ . The spring constant is k = 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
Solution: The unstretched length of the spring is L =
√
b2 + 12 = 1.25 m. The obtuse angle is 90 + α, so the stretched
length can be determined from the cosine law:
L22 = 12 + 0.752 − 2(0.75) cos(90 + α) = 2.3125 m2
β
from which L2 = 1.5207 m. The force exerted by the spring is
α
T = k∆L = 100(1.5207 − 1.25) = 27.1 N.
The angle between the spring and the bar can be determined from the
sine law:
A
b
1.5207
=
,
sin β
sin(90 + α)
b
from which sin β = 0.4271,
β = 25.28◦ .
T
The angle the spring makes with the horizontal is 180 − 25.28 − 90 −
α = 34.72◦ . The sum of the forces:
FX = AX − T cos 34.72◦ = 0,
β
α
from which AX = 22.25 N.
FY = AY − W − T sin 34.72◦ = 0.
W
AX
The sum of the moments about A is
W
MA = T sin 25.28◦ −
sin α = 0,
2
AY
from which
W =
2T sin 25.28◦
= 46.25 N.
sin α
Substitute into the force equation to obtain:
T sin 34.72◦ = 61.66 N
AY
=
W +
W
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 = 15 ft and b = 2 ft, and the angle
θ = 30◦ . Determine the tension in cable AB and the
reactions at the pin supports C and D.
B
E
θ
C
A
D
a
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:
b
The components:
Dx = 0.6|D| = 7.67 kip,
Dy = 0.866|D| = 13.287 kip,
rD = 2i
rE = rD + 20(i sin 30◦ + j cos 30◦ )
and Cy = 1|C| = 11.94 kip
= 12i + 17.3j,
rB = 20j,
B
rA = −15i.
The unit vectors are:
rE − rD
eDE =
= 0.5i + 0.866j,
|rE − rD |
eEB =
eCB
θ
A
C
rB − r E
= −0.976i + 0.2179j.
|rB − rE |
rB − r C
=
= 1j,
|rB − rC |
eAB =
E
rA − rB
= −0.6i − 0.8j.
|rA − rB |
Isolate the juncture at E: The equilibrium conditions are
Fx = 0.5|D| − 0.976|TEB | = 0,
Fy = 0.866|D| + 0.2179|TEB | − 15 = 0,
from which
|D| = 15.34 kip
and |TEB | = 7.86 kip.
Isolate the juncture at B: The equilibrium conditions are:
Fx = 0|C| − 0.6|TAB | + 0.976|TEB |,
and
Fy = 1|C| − 0.6|TAB | − 0.2179|TEB | = 0,
from which
|TAB | = 12.79 kip,
and |C| = 11.94 kip.
D
b
a
TAB
C
TEB
TEB
15 kip
D
Juncture B Juncture E
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.
Elevator
E
B
6 in
A
D
2.5 in
C
F
3.5 in
2 in
2.5 in
120 in-lb
G
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
α = tan−1
12
= 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
MEG = 2.5FY + 6FX − 120 = 0.
The sum of the forces about the pinned joint A:
Fx = Ax − F + TEC cos α = 0
from which Ax = 25.33 lb,
Fy = Ay + TEC sin α = 0
from which Ay = −1.93 lb
This is the tension in the cable EC. Noting that
FX = TEC cos α,
and FY = TEC sin α,
then TEC
120
=
.
2.5 sin α + 6 cos α
The sum of the moments about the pinned support BC is
MBC = −2TEC sin α + 6TEC cos α − 2.5F = 0.
6 in
D
A
120
in-lb
2.5 in
F
C
3.5 in
Substituting:
120
6 cos α − 2 sin α
F =
2.5
6 cos α + 2.5 sin α
2 in
2.5 in
120 in
(Not to scale)
FX
α
C
A
TEC
2.5 in
F
3.5 in
120 in-lb
20 N-m
A
The sum of forces in the vertical direction is
FY = AY + BY = 0,
from which AY = −BY = 18.18 N.
The sum of forces in the horizontal direction is
FX = AX + BX = 0,
from which the values of AX and BX are indeterminate.
800 mm
B
300 mm
20 N-m
A
20
= −18.18 N.
1.1
B
300 mm
BX
AX
AY
6 in
FY D
2.5
in
800 mm
from which BY = −
E
TEC
α
C
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 = M + 1.1BY = 0,
G
1.5 in
2.5 in
2 in
= (48)(0.9277) = 44.53 lb.
Problem 5.65 (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.
Elevator
E
B
800 mm
300 mm
BY
Problem 5.66 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:
FX = AX = 0.
B
The sum of moments about A is
MA = M + 1.1BY = 0,
from which BY = −
20 N-m
A
800 mm
AX
20
= −18.18 N.
1.1
AY
The sum of forces in the vertical direction is
FY = AY + BY = 0,
300 mm
20 N-m
800 mm
300 mm
BY
from which AY = −BY = 18.18 N.
Problem 5.67 (a) Draw the free-body diagram of the
beam and show that it is statically indeterminate. (The
external couple M0 is known.)
(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 = 2M0 /L.
What are the reactions at A?
M0
A
B
L
Solution:
(a)
+
FX :
AX = 0
(1)
FY :
AY + B = 0
(2)
MA :
M0
A
B
MA − MO + BL = 0 (3)
L
Unknowns: MA , AX , AY , B.
3 Eqns in 4 unknowns
∴ Statistically indeterminate
(b)
AY
Given B = 2MO /L (4)
We now have 4 eqns in 4 unknowns and can solve.
Eqn (1) yields AX = 0
Eqn (2) and Eqn (4) yield
AY = −2MO /L
Eqn (3) and Eqn (4) yield
MA = MO − 2MO
MA = −MO
MA was assumed counterclockwise
MA = |MO | clockwise
AX = 0
AY = −2MO /L
MO
MA
AX
L
B
Problem 5.68 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
A
answers
FX :
FY :
MA :
MO
AX = 0
L
AY + BY = 0
O
−Mo + BL = 0
AX = 0
MO
AX
B = MO /L
L
AY = −MO /L
Problem 5.69 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.
Strategy: Place the coordinate system so that the x
axis passes through points A and B.
Solution: The free body diagram shows that there are four reactions, hence the system is statically indeterminate. The sum of the
forces:
FX = (AX + BX ) = 0,
and
FY = AY + BY + F = 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
B
AY
B
80 N
300
mm
100 N-m
A
300 mm
700 mm
B
80 N
A
300
mm
100 N-m
F = 80 N,
300
mm
M = 100 N-m.
700
mm
The length from A to B is
L = 0.32 + 0.72 = 0.76157 m.
The angle between the new axis and the horizontal is
0.3
θ = tan−1
= 23.2◦ .
0.7
The moment about the point A is
BN
80 N
AN
300
mm
AP
MA = LBN − 0.3F + M = 0,
from which BN =
−M + 0.3F
−76
=
= −99.79 N,
L
0.76157
from which
The sum of the forces normal to the new axis is
FN = AN + BN + F cos θ = 0,
from which
AN = −BN − F cos θ = 26.26 lb
The reactions parallel to the new axis are indeterminate.
300
mm
BP
100 N-m
700
mm
Problem 5.70 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. Please just leave
it out.
Problem 5.71 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
Solution:
B
(1) is properly constrained. The sum of the forces
FX = −F + BX = 0,
from which BX = F .
A
L
L
(1)
(2)
FY = BY + Ay = 0,
MB = −LAY + LF = 0,
(3)
C
45°
from which By = −Ay . The sum of the moments about B:
(2)
–12 L
F
from which AY = F , and By = −F
is improperly constrained. The reactions intersect at B, while
the force produces a moment about B.
is properly constrained. The forces are neither concurrent nor
parallel. The sum of the forces:
–12 L
A
B
L
45°
FX = −C cos 45◦ − B − A cos 45◦ + F = 0.
(3)
FY = C sin 45◦ − A sin 45◦ = 0
from which A = C. The sum of the moments about A:
1
MA = − LF + LC cos 45◦ + LC sin 45◦ = 0,
2
from which C =
B=
F
2
2
F
√
2
. Substituting and combining: A =
2
F
√
2
,
F
C
1–
2
F
L
A
45°
L
(1)
1–
2
B
A
B
(2)
C
45°
–12 L
F
B
A
45°
B
L
(3)
–12 L
45°
Problem 5.72 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
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:
C
FX = C − F = 0,
–1 L
2
F
from which C = F .
FY = −A + B = 0,
A
–1 L
2
B
from which A = B. The sum of the moments about B: LA +
L
F − LC = 0, from which A = 12 F , and B = 12 F
2
L
(2)
P
C
–1 L
2
F
A
C
–1 L
2
B
F
–1 L
2
–1 L
2
L
45°
B
A
B
L
B
A
(1)
L
(3)
–12 L
Problem 5.73 The bar AB has a built-in support at A
and is loaded by the forces
y
FB = 2i + 6j + 3k (kN),
A
FC = i − 2j + 2k (kN).
(a) Draw the free-body diagram of the bar.
(b) Determine the reactions at A.
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.
C
1m
FC
A
=0
FB
z
1m
B
=0
C
1m
Equilibrium Equations (Moments) Sum moments about A
rAB × FB = 1i × (2i + 6j + 3k) kN-m
FC
rAB × FB = −3j + 6k (kN-m)
rAC × FC = 2i × (1i − 2j + 2k) kN-m
rAC × FC = −4j − 4k (kN-m)
x:
y:
z:
MA :
M AX = 0
MA :
M AY − 3 − 4 = 0
MA :
M AZ + 6 − 4 = 0
Solving, we get
AX = −3 kN,
AY = −4 kN,
AZ = −5 kN
MAx = 0,
MAy = 7 kN-m,
MAz = −2 kN-m
x
y
MA = M A X i + M A Y j + M A Z k
=0
B
1m
Solution:
(b) Equilibrium Eqns (Forces)
FX : AX + FBX + FCX
FY : AY + FBY + FCY
FZ : AZ + FBZ + FCZ
FB
z
x
AY
AX
MA = MAX i + MAY j + MAZ K
FB
1m
AZ
B
C
1m
x
FC
Problem 5.74 The bar AB has a built-in support at
A. The tension in cable BC is 8 kN. Determine the
reactions at A.
y
A
z
C
(3,0.5,–0.5)m
2m
B
x
Solution:
y
MA = MAx i + MAy j + MAz k
We need the unit vector eBC
(xC − xB )i + (yC − yB )j + (zC − zB )k
eBC = (xC − xB )2 + (yC − yB )2 + (zC − zB )2
A
z
eBC = 0.816i + 0.408j − 0.408k
C
2m
TBC = (8 kN)eBC
B
TBC = 6.53i + 3.27j − 3.27k (kN)
The moment of TBC about A is
i
MBC = rAB × TBC = 2
6.53
j
0
3.27
k 0 −3.27 MBC = rAB × TBC = 0i + 6.53j + 6.53k (kN-m)
Equilibrium Eqns.
FX :
AX + TBCX = 0
FY :
AY + TBCY = 0
FZ :
AZ + TBCZ = 0
MX :
MAX + MBCX = 0
MY :
MAY + MBCY = 0
MZ :
MAZ + MBCZ = 0
Solving, we get
AX = −6.53 (kN),
AY = −3.27 (kN),
AZ = 3.27 (kN)
MAx = 0,
MAy = −6.53 (kN-m),
MAz = −6.53 (kN-m)
(3,0.5,–0.5) m
x
AY
MA = MAX i + MAY j + MAZ K
AX
AZ
2m
TBC
C (3, 0.5, −0.5)
B (2, 0, 0)
x
Problem 5.75 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) Draw the free-body diagram of the bar.
(b) 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 = 5i + 6j + k,
and rC = 3i + 0j + 4k.
A
The vector parallel to the cable is
z
y
rBC = rC − rB = −2i − 6j + 3k,
|rBC | = 22 + 62 + 32 = 7 m.
(3,0,4) m
B
MY
FY
A
z
The unit vector parallel to the cable is
e =
B
(5,6,1) m
x
rBC
= −0.2857i − 0.8571j + 0.4286k.
|rBC |
T
x
MX
FX
MZ
FZ
The tension vector is
TBC = 10eBC = −2.857i − 8.571j + 4.286k.
The force reaction at A is determined by
FR = FA + TBC = 0,
from which FA = 2.857i + 8.571j − 4.286k (kN)
The moment reaction at A is given by
MR = MA + rB × TBC
i
j
= MA + 5
6
−2.857 −8.571
k 6 =0
4.286 From which MA = −34.287i + 24.287j + 25.713k (kN-m)
Problem 5.76 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 = MA + rB × TBC = 0,
from which MA = −rB × TBC .
Noting that TBC = |TBC |eBC , then using the unit vector developed
in the solution to Problem 5.79,
rBC
= −0.2857i − 0.8571j + 0.4286k
e =
|rBC |
and the position vector of the end of the bar:
rB = 5i + 6j + k
i
MA = −rB × |TBC |eBC = −|TBC | 5
−0.2857
j
6
−0.8571
= −|TBC |(3.4287i − 2.4287j − 2.5713k)
Take the magnitude of both sides
|MA | = 100 = |TBC | 3.42872 + 2.42872 + 2.57132
= |TBC |(4.2961).
Solve: |TBC | =
100
4.2961
= 20.300 kN
k 6 0.4286 Problem 5.77 The force exerted on the highway sign
by wind and the sign’s weight is F = 800i − 600j (N).
Determine the reactions at the built-in support at O.
Solution: The force acting on the sign is
y
F
N
O
C
C
TU
F = FX i + FY j + FZ k = 800i − 600j + 0k N,
and the position from O to the point on the sign where F acts is
r = 0i + 8j + 8k m.
8m
The force equations of equilibrium for the sign are
OX + FX = 0,
O
8m
OY + FY = 0,
x
and OZ + FZ = 0.
Note that the weight of the sign is included in the components of F.
The moment equation, in vector form, is
M = MO + r × F.
z
Expanded, we get
i
M = MOX i + MOY j + MOZ k + 0
800
j
8
−600
k
8 = 0.
0
y
F
The corresponding scalar equations are
MOX − (8)(−600) = 0,
MOY + (8)(800) = 0,
8m
OY
and MOZ − (8)(800) = 0.
MO
Solving for the support reactions, we get
OX = −800 N,
8m
x
OY = 600 N,
OZ
OZ = 0,
MOX = −4800 N-m,
OX
z
MOY = −6400 N-m,
and MOZ = 6400 N-m.
Problem 5.78 In Problem 5.81, the force exerted on
the sign by wind and the sign’s weight is F = ±4.4v 2 i−
600j (N), where v is the component of the wind’s velocity Expanded, we get
perpendicular to the sign in meters per second (m/s). i
j
k
M = MOX i + MOY j + MOZ k + 0
8
8 = 0.
If you want to design the sign to remain standing in
±21560 −600 0 hurricane winds with velocities v as high as 70 m/s, what
reactions must the built-in support at O be designed to The corresponding scalar equations are
withstand?
MOX − (8)(−600) = 0,
Solution: The magnitude of the wind component of the force on
MOY + (8)(±21560) = 0,
the sign is (4.4)(70)2 N = 21.56 kN. The force acting on the sign is
F = FX i + FY j + FZ k = ±21560i − 600j + 0k N,
and the position from O to the point on the sign where F acts is r =
0i + 8j + 8k m. The force equations of equilibrium for the sign are
and MOZ − (8)(±21560) = 0.
Solving for the support force reactions, we get
OX = ±21560 N,
OX + FX = 0,
OY = 600 N,
OY + FY = 0,
OZ = 0,
and OZ + FZ = 0.
Note that the weight of the sign is included in the components of F.
The moment equation, in vector form, is
M = MO + r × F.
and for moments,
MOX = −4800 N-m,
MOY = ±172,500 N-m,
and MOZ = ±172,500 N-m.
Problem 5.79 The tension in cable AB is 24 kN. Determine the reactions in the built-in support D.
2m
C
A
2m
D
B
1m
3m
Solution:
The force acting on the device is
F = FX i + FY j + FZ k = (24 kN)eAB ,
and the unit vector from A toward B is given by
eAB =
1i − 2j + 1k
√
.
6
The force, then, is given by
F = 9.80i − 19.60j + 9.80k kN.
The position from D to A is
r = 2i + 2j + 0k m.
The force equations of equilibrium are
DX + FX = 0,
DY + FY = 0,
and DZ + FZ = 0.
The moment equation, in vector form, is
M = MD + r × F.
Expanded, we get
i
M = MDX i + MDY j + MDZ k + 2
9.80
The corresponding scalar equations are
MDX + (2)(9.80) = 0,
MDY − (2)(9.80) = 0,
and MDZ + (2)(−19.60) − (2)(9.80) = 0.
Solving for the support reactions, we get
DX = −9.80 kN,
OY = 19.60 kN,
OZ = −9.80 kN.
MDX = −19.6 kN-m,
MDY = 19.6 kN-m,
and MDZ = 58.8 kN-m.
j
2
−19.60
k 0 = 0.
9.80 Problem 5.80 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 θx = 0.174, cos θy =
0.985, cos θz = 0, and the direction cosines of the centerline of arm BC are cos θx = 0.743, cos θy = 0.557,
cos θz = −0.371. The support at A behaves like a builtin support.
(a) What is the sum of the moments about A due to the
weights of the two arms?
(b) What are the reactions at A?
y
600 mm
C
160 N
B
600 mm
200 N
A
z
Denote the center of mass of arm AB as D1 and that
of BC as D2 . We need
Solution:
rAD ,
rAB ,
and rBD2 .
To get these, use the direction cosines to get the unit vectors eAB and
eBC . Use the relation
e = cos θX i + cos θY j + cos θZ k
eAB = 0.174i + 0.985j + 0k
eBC = 0.743i + 0.557j − 0.371k
rAD1 = 0.3eAB m
rAB = 0.6eAB m
rBC = 0.6eBC m
rBD2 = 0.3eBC m
WAB = −200j N
WBC = −160j N
Thus rAD1 = 0.0522i + 0.2955j m
rAB = 0.1044i + 0.5910j m
rBD2 = 0.2229i + 0.1671j − 0.1113k m
rBC = 0.4458i + 0.3342j − 0.2226k m
and rAD2 = rAB + rBD2
rAD2 = 0.3273i + 0.7581j − 0.1113k m
x
(a)
We now have the geometry determined and are ready to determine the
moments of the weights about A.
MW = rAD1 × W1 + rAD2 × W2
where
i
rAD1 × W1 = 0.0522
0
j
0.2955
−200
k
0
0
rAD1 × W1 = −10.44k N-m
and
i
rAD2 × W2 = 0.3273
0
j
0.7581
−160
k
−0.1113 0
rAD2 × W2 = −17.81i − 52.37k
Thus,
MW = −17.81i − 62.81k (N-m)
(b) Equilibrium Eqns
FX : AX = 0
FY : AY − W1 − W2 = 0
FZ : AZ = 0
Sum Moments about
A : MA +
MW = 0
MX :
MAx − 17.81 = 0 (N-m)
MY :
MAy + 0 = 0
MZ :
MZ − 62.81 = 0 (N-m)
5.80 Contd.
Thus: AX = 0, AY = 360 (N), AZ = 0,
MAx = 17.81 (N-m), MAy = 0, MAz = 62.81 (N-m)
y
600 mm
B
C
160 N
600 mm
200 N
A
z
x
C
W2
W1
D2
B
D1
MA
MA = MAXi + MAYj + MAZk
W1 = 200 N
W2 = 160 N
AX
AZ
AY
Problem 5.81 The force exerted on the grip of the
exercise machine is F = 260i − 130j (N). What are the
reactions at the built-in support at O?
150
mm
y
F
O
200
mm
z
250
mm
x
Solution:
MO = MOx i + MOy j + MOz k
Equilibrium (Forces)
FX :
OX + FX = OX + 260 = 0 (N)
FY :
OY + FY = OY − 130 = 0 (N)
FZ :
OZ + FZ = OZ = 0 (N)
F
O
MF = rOP
i
× F = 0.25
260
j
0.2
−130
k −0.15 0 MF = −19.5i − 39j − 84.5k (N-m)
and from the moment equilibrium eqns,
MOX = 19.5 (N-m)
MOY = 39.0 (N-m)
MOZ = 84.5 (N-m)
200
mm
z
250
mm
Thus, OX = −260 N, OY = 130 N, OZ = 0
Summing Moments about O
MX :
M O X + M FX = 0
MY :
M O Y + M FY = 0
MZ :
MOZ + MFZ = 0
where
150
mm
y
rOP = 0.25i + 0.2j − 0.15k
0.15 P
m
y
MO
OX
z
OZ
x
OY
0.2
F = 260 i – 130 j (N)
5m
0.2 m
x
Problem 5.82 The designer of the exercises machine
in Problem 5.85 assumes that the force F exerted on
the grip will be parallel to the x − y 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.
150
mm
y
MO = M O X i + M OY j + M O Z k
F = F cos θi + F sin θj
F
O ≤ θ ≤ 360◦
O
200
mm
|F| = 900 N
z
rOP = 0.25i + 0.2j − 0.15k m
FX :
OX + FX = OX + F cos θ = 0
FY :
OY + FY = OY + F sin θ = 0
FZ :
OZ + F Z = OZ + 0 = 0
250
mm
y
MO
5m
0.1
OY = −F sin θ
θ
OX
θ
O
OZ = 0
O X + OY
2
0.2 m
= |F| = 900 N
The moment equilibrium equations are
MX :
M O X + M FX = 0
MY :
M O Y + M FY = 0
MZ :
MOZ + MFZ = 0
where MF = rOP
i
× F = 0.25
F cos θ
j
0.2
F sin θ
k −0.15 O +(0.25 F sin θ − 0.2 F cos θ)k
and MO = −MF
320
300
280
260
240
220
200
180
160
140
120
0
50
100
150 200 250
Theta (deg)
300
350
400
Moment Componente (N-m) vs. Theta (deg)
300
Moment Mag (N-m)
The first plot on the next page shows |MO | vs θ for O ≤ θ ≤ 360◦ .
The second plot shows the three components of MO as functions of
θ.
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 x − y plane and it must be able
to exert a moment |MO | ≥ 320 (N-m) From the component plots, we
see that the support must provide
−288 (N-m) ≤ MOZ ≤ 288 (N-m)
5m
Moment (N-m) vs. Theta (deg)
−0.15 F cos θj
−135 (N-m) ≤ MOY ≤ 135 (N-m)
0.2
x
MF = +0.15 F sin θi
−135 (N-m) ≤ MOX ≤ 135 (N-m)
OZ
z
Moment Magnitude (N-m)
2
F
P
OY
Thus OX = −F cos θ
x
MOZ
200
MOY
100
0
−100
−200
−300
MOX
0
50
100
150 200 250
Theta (deg)
300
350
400
Problem 5.83 The boom ABC is subjected to a force
F = −8j (kN) at C and is supported by a ball and socket
at A and the cables BD and BE.
(a) Draw the free-body diagram of the boom.
(b) Determine the tension in the cables and the reactions
at A.
y
2m
1.5 m
D
E
1m
2m
A
B
z
2m
C
2m
x
F
Solution:
e=
r
|r|
First, we need the unit vectors eBD and eBE , use
Eqns (1), (2), (3), (5) and (6) are 5 eqns in the 5 unknowns AX ,
AY , AZ , TBD , and TBE (considering the definitions of TBD
and TBE in terms of their unit vectors. Solving, we get
eBD = −0.625i + 0.625j + 0.469k
AX = 20.4 kN, AY = −8 kN, AZ = 0
eBE = −0.667i + 0.333j − 0.667k
TBD = 18.6 kN, TBE = 13.1 kN
TBD = TBD eBD , TBE = TBE eBE
TBD = −0.625TBD i + 0.625TBD j + 0.469TBD k
y
TBE = −0.667TBE i + 0.333TBE j − 0.667TBE k
A = AX i + AY j + AZ k
2m
1.5 m
F = 0i − 8j + 0k (kN)
E
D
Force Equilibrium requires:
1m
A + TBD + TBE + F = 0
FX :
AX + TBDX + TBEX = 0
FY :
AY + TBDY + TBEY − 8 = 0 (2)
FZ :
AZ + TBDZ + TBEZ = 0
2m
(1)
A
B
z
2m
C
(3)
2m
Moment equilibrium (about A) requires
M = rAB × TBD + rAB × TBE + rAC × F = 0
y
or rAB × (TBD + TBE ) + rAC × F = 0
E
D
rAB × (TBD + TBE )
i
=
2
(T
+T
)
BDX
BEX
(TBDY
j
k
0
0
+ TBEY ) (TBDZ + TBEZ
)
rAB × (TBD + TBE ) = −2(TBDZ + TBEZ )j + 2(TBDY + TBEY )k
rAC × F = 4i × (−8j) = −32k
MX :
0=0
(4)
MY :
−2TBDZ − 2TBEZ = 0
(5)
MZ :
2TBDY + 2TBEY − 32 = 0 (6)
x
F
AX
AY
A
AZ
2m
z
TBD
TBE
B
2m
E(0, 1, –2)
D(0, 2, 1.5)
B(2, 0, 0)
C(4, 0, 0)
A(0, 0, 0)
C
F
x
y
Problem 5.84 The cables supporting the boom ABC
in Problem 5.87 will each safely support a tension of
25 kN. Based on this criterion, what is the largest safe
magnitude of the downward force F?
2m
1.5 m
E
D
Solution:
Note that the solution of Problem 5.87 is linear in all
of the force and moment components. This means that we can find
the largest cable tension, scale it up to 25 kN, and then scale the F of
Problem 5.87 up by the same factor. In 5.87, TBD = 18.6 kN was
25
the largest tension. Let F = 18.6
Fmax = (−8)f = −10.7 kN
1m
2m
A
B
z
2m
|Fmax | = 10.7 kN
C
2m
x
F
Problem 5.85 The suspended load exerts a force F =
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
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
(0, 10, 4) ft
–Fj
eAP = (xP − xA )i + (yP − yA )j + (zP − zA )k,
where P can be either A or B. The forces in cables AB and AC are
x
O
TAB = TAB eAB = TABX i + TABY j + TABZ k,
and TAC = TAC eAB = TACX i + TACY j + TACZ k.
The weight force is
z
F = 0i − 600j + 0k,
and the support force at the ball joint is
S = SX i + SY j + SZ k.
The vector form of the force equilibrium equation (which gives three
scalar equations) for the bar is
|TAB | = 474.1 lb,
TAC = −154.8i + 0j − 193.5k lb,
TAB + TAC + F + S = 0.
Let us take moments about the origin. The moment equation, in vector
form, is given by
MO = rOA × TAB + rOA × TAC
+rOA × F = 0,
|TAC | = 247.9 lb,
MAB = rOA × TAB = 1161i − 1548j + 3871k ft-lb,
MAC = rOA × TAC = −1161i + 1548j + 929k ft-lb,
and S = 541.9i + 406.5j + 0k lb
where rOA = 8i + 6j + 0k.
The cross products are evaluated using the form
i
j
k M=r×H= 8
6
0 ,
H
H
H X
Y
Z
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. If we carry through these operations
in the sequence described, we get the following vectors:
eAB = −0.816i + 0.408j + 0.408k,
eAC = −0.625i + 0j − 0.781k,
TAB = −387.1i + 193.5j + 193.5k lb,
y
B
(0, 10, 4)
Sy
ft
C
TAB
(0, 6, –10) ft
TAC
A
(8, 6, 0) ft
−Fj
x
O
Sz
z
Sx
Problem 5.86 In Problem 5.89, suppose that the suspended load exerts a force F = 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
y
C
(0, 6, –10) ft
TAC
A
(8, 6, 0) ft
rOG = 4i + 3j + 0k ft.
The weight of the bar is
B
TAB
(0, 10, 4) ft
WB = 0i − 200j + 0k lb,
Sy
G
and its moment around the origin is
WB
MW B = 0i + 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 + TAC + F + S + WB = 0.
The new moment equilibrium equation is
MO = rOA × TAB + rOA × TAC
+rOA × F + rOG × WB = 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 = −0.816i + 0.408j + 0.408k,
eAC = −0.625i + 0j − 0.781k,
TAB = −451.6i + 225.8j + 225.8k lb,
|TAB | = 553.1 lb,
TAC = −180.6i + 0j − 225.8k lb,
|TAC | = 289.2 lb,
MAB = rOA × TAB = 1355i − 1806j + 4516k ft-lb,
MAC = rOA × TAC = −1354i + 1806j + 1084k ft-lb,
and S = 632.3i + 574.2j + 0k lb
O
Sz
z
Sx
–Fj
x
Problem 5.87 The 158,000-kg airplane is at rest on
the ground (z = 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:
FY = (NL + NR ) + NF − W = 0
MR = −3 mg + 21NF = 0
21 m
Solving,
NF = 221.4 kN (1)
6m
B
(NL + NR ) = 1328.6 kN (2)
A
G
FY = NR + NL + NF − W = 0
x
C
6m
(same equation as before)
+
MO = 0.5 W − 6(NR ) + 6(NL ) = 0 (3)
Solving (1), (2), and (3), we get
y
NF = 221.4 kN
NR = 728.9 kN
NL = 599.7 kN
mg
3m
21 m
Side
View
x
R
F
(NL + NR)
NF
Z
0.5 m
W
Front
View
y
6
6
NF
NR
z
NL
Problem 5.88 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
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 + B + C − W = 0. The
moments are calculated from
MB = rOB × Bk,
MC = rOC × Ck,
and MG = rOG × (−W )k.
The moment equilibrium equation is
MO = MB + MC + MG = 0.
Carrying out these operations, we get
A = 3717 N,
B = 2596 N,
C = 1534 N,
and W = 7848 N.
z
y
B
7m
C
A
7m
6m
O
7m
4m
x
8m
mg
Problem 5.89 The cables in Problem 5.92 will each
safely support a tension of 10 kN. Based on this criterion, what is the largest safe mass of the horizontal wall
section?
Solution: The solution for this problem involves exactly the same
equations as Problem 5.92. The unknowns and knowns just shift a bit.
We no longer know the mass of the wall section, but we do know
the tension in one of the cables. From the solution to Problem 5.92,
we see that the cable carrying the largest load is the cable at A. We
set A = 10 kN, make m (or W ) an unknown, and solve the exact
same equations as above. The result is A = 10 kN, B = 6.984 kN,
C = 4.127 kN, W = 21.114 kN, and m = 2152 kg.
7m
Problem 5.90 An engineer designs a system of pulleys
to pull his model trains up and out of the way when they
aren’t in use. What are the tensions in the three ropes
when the system is in equilibrium?
B
2 ft
1.6 ft
C
A
3.5 ft
2 ft
8 ft
250 lb
Solution: Take the origin as the point C on the platform. The
position vectors are
rA = −2i + 8k,
rB = −4i,
rG = −1.6i + 3.5k.
The sum of the moments about C is
i
j k i
j k
MC = −2 0 8 + −4 0 0 0 A 0 0 B 0
i
j
k + −1.6
0
3.5 = 0
0
−250 0 B
1.6 ft
3.5 ft
C
8 ft
A
2 ft
2 ft
250 lb
MC = −(8Ai + 2Ak) − 4Bk + (875i + 400k) = 0.
B
Collecting like terms:
C
(−8A + 875)i = 0,
(−2A − 2B + 400)k = 0,
from which A =
B =
875
= 109.375 lb,
8
200 − A
= 45.3125 lb.
2
The value of the reaction at C is found from the sum of the vertical
forces:
FY = (−250 + A + B + C)j = 0,
from which C = 95.3125 lb. Two pulley ropes support each point, so
the tension in each rope is one-half the reaction at the point supported:
TA =
A
= 54.67 lb,
2
TB =
B
= 22.65 lb,
2
TC =
C
= 47.66 lb
2
A
2 ft
2 ft
1.6 ft
t
2.508f ft
3.5 ft
Problem 5.91 The L-shaped bar is supported by a
bearing at A and rests on a smooth horizontal surface
at B. The vertical force F = 4 kN and the distance
b = 0.15 m. Determine the reactions at A and B.
y
F
b
A
x
B
0.2 m
0.3 m
z
Solution:
Equilibrium Eqns:
FX :
O=0
FY :
AY + B − F = 0
FZ :
AZ = 0
y
F
b
A
Sum moments around A
x:
F b − 0.3B = (4)(0.15) − 0.3B = 0
y:
M AY = 0
z:
MAZ + 0.2F − 0.2B = 0
x
B
0.2 m
0.3 m
z
Solving,
AX = 0,
y
AY = 2 (kN),
AZ = 0
F
b
MAX = 0,
MAY = 0,
B
MAZ = −0.4 (kN-m)
3
0.
0.2 m
m
MY
(MAX = 0)
AX = 0
x
AY
b = 0.15 m
F = 4 kN
AZ
B
z
MZ
A
Problem 5.92 In Problem 5.95, the vertical force F =
4 kN and the distance b = 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 x − z plane?
We want to represent the forces at A & B by a single
force. From Prob. 5.96
Solution:
A = +2j (kN),
B = +2j (kN)
zR (4) = (+0.3)(2)
zR = +0.15 m
xR (4) = 0.2(2)
xR = 0.1 m
MA = −0.4k (kN-m)
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 .
y
R
R = A + B = 4j (kN)
F
b
Let R pierce the x, z plane at (xR , zR )
MX :
−zR R = −0.3B
MZ :
−xR R = 0.2AY
A
x
B
z
0.2 m
0.3 m
Problem 5.93 In Problem 5.95, the vertical force F =
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?
We make AY unknown, b unknown, and B unknown (F = 4 kN, MAY =
+0.5 (kN-m), and solve we get AY = −2.5 at b = 0.4875 m However, 0.3 is
the physical limit of the device.
Thus, 0.1125 m ≤ b ≤ 0.3 m
y
Solution: The solution to Prob. 5.95 produced the relations
AY + B − F = 0
(F = 4 kN )
F
F b − 0.3B = 0
b
A
MAZ + 0.2F − 0.2B = 0
x
AX = AZ = MAX = MAY = 0
Set the force at A to its limit of 2.5 kN and solve for b. In this case,
MAZ = −0.5 (kN-m) which is at the moment limit. The value for b
is b = 0.1125 m
B
0.2 m
0.3 m
z
Problem 5.94 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) Draw the free-body diagram of the bar.
(b) Determine the reactions at A and B.
y
B
400 mm
D
Solution:
(a)
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.
(b) 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:
√
0.62 + 0.72
α = cos−1
= 33.1◦ .
1.1
The coordinates of the points are: A (0.7, 0, 0.6), B (0, 1.1
(sin 33.1◦ ), 0) = (0, 0.6, 0), from which the vector parallel to
the bar is
rAB = rB − rA = −0.7i + 0.6j − 0.6k (m).
Expand and collect like terms:
MA = (0.6BZ − 0.3819)i − (0.6BX − 0.7BZ )j
From which,
BZ =
0.3819
= 0.6365 kN,
0.6
BX =
0.4455
= 0.7425 kN.
0.6
The reactions at A are determined from the sums of the forces:
FX = (BX + AX )i = 0, from which AX = −0.7425 kN.
FY = (AY − 1)j = 0, from which AY = 1 kN.
FZ = (BZ + AZ )k = 0, from which AZ = −0.6365 kN
= (1.1 − 0.4)eAB = 0.7eAB
y
= −0.4455i + 0.3819j − 0.3819k.
FB
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 = rAB × B + rAD × D = 0 = −0.7 0.6 −0.6 B
0
B i
+ −0.4455
0
X
j
0.3819
−1
600 mm
+(−0.6BX + 0.4455)k = 0.
The vector location of the point D relative to A is
rAD
rAB
=
= −0.6364i + 0.5455j − 0.5455k.
1.1
A
700 mm
z
The unit vector parallel to the bar is
eAB
x
C
k
−0.3819 = 0
0
B
400 m
T
D
Z
600 m
C
z
700 m
FY
x
A
FZ
FX
Problem 5.95 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 = −50k (lb). Determine the tension in the cable BD
and the reactions at A and C.
y
A
3 ft
B
Solution: The strategy is to determine the sum of the moments
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.
The angle formed by the rod is
3
α = sin−1
= 22◦ .
8
z
4 ft
F
D
2 ft
C
x
The vector positions are:
rA = 3j,
rD = 4i + 2k
from which |T| =
◦
and rC = (8 cos 22 )i = 7.4162i.
The vector parallel to the rod is
rAC = rC − rA = 7.4162i − 3j.
The unit vector parallel to the rod is
CY =
75
1.192
= 62.92 lb
2.036|T|
= 17.27 lb.
7.4162
The reaction at A is determined from the sums of forces:
FX = (AX + 0.1160|T|)i = 0,
from which AX = −7.29 lb,
FY = (AY − 0.5960|T| + CY )j = 0,
eAC = 0.9270i − 0.375j.
The location of B is
The vector parallel to the cable is
from which AY = 20.23 lb
FZ = (AZ + 0.7946|T| − 50)k = 0,
rBD = rD − (rA + rAB ) = 0.2919i − 1.5j + 2k.
from which AZ = 0 lb
rAB = 4eAC = 3.7081i − 1.5j.
The unit vector parallel to the cable is
y
eBD = 0.1160i − 0.5960j + 0.7946k.
The tension in the cable is T = |T|eBD . The reaction at the roller
support C is normal to the x − z plane. The sum of the moments about
A
MA = rAB × F + rAB × T + rAC × C = 0
i
j
k = 3.7081 −1.5
0 0
0
−50 A
F
C
D
z
4 ft
2 ft
i
+|T| 3.7081
0.1160
i
+ 7.4162
0
B
3 ft
j
k −1.5
0 −0.5960 0.7946 j
k
−3 0 = 0
C
0
x
y
AY
AZ
Y
AX
= 75i + 185.4j + |T|(−1.192i − 2.9466j − 2.036k)
+7.4162CY k = 0,
T
F
z
D
CY
x
Problem 5.96 Consider the 8-ft bar in Problem 5.99.
The force F = 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
j
k MA = 3.7081 −1.5
0 0
F
−50 Y
i
+|T| 3.7081
0.1160
i
+ 7.4162
0
j
k −1.5
0 −0.5960 0.7946 j
k
−3 0 = 0
C
0
Y
= 75i + 185.4j + 3.7081FY k
+|T|(−1.192i − 2.9466j − 2.036k)
+7.4162CY k = 0,
75
from which, |T| = 1.192
= 62.92 lb.
Collecting terms in k, 3.7081FY + 2.384|T| − 7.4162CY = 0.
For CY = 0, FY = 128.11
= 34.54 lb
3.708
Problem 5.97 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
C
40 m
50 m
E
A
40 m
D
50 m
z
The strategy is to determine moments about A due to
the cables. This requires the unit vectors parallel to the cables.
The coordinates of the points are:
Solution:
A(0, 0, 0), B(0, 70, 0), C(−50, 0, 0),
D(20, 0, 50), E(40, 0, −40).
The unit vectors parallel to the cables, directed from B to the points
E, D, and C
20 m
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.)
FX = (AX + 0.17932)i = 0,
from which AX = −0.179 kN,
FY = (AY − 4.7682)j = 0,
rBE = 40i − 70j − 40k,
from which AY = 4.768 kN,
FZ = (AZ + 0.2434)k = 0,
rBD = 20i − 70j + 50k,
from which AZ = −0.2434 kN
rBC = −50i − 70j.
B
The unit vectors parallel to the cables, pointing from B, are:
eBE = 0.4444i − 0.7778j − 0.4444k,
eBD = 0.2265i − 0.7926j + 0.5661k,
C
eBC = −0.5812i − 0.8137j + 0k.
40 m
The tensions in the cables are:
50 m
TBD = 2eBD = 0.4529i − 1.5852j + 1.1323k (kN),
TBE = 2eBE = 0.8889i − 1.5556j − 0.8889k (kN),
TBC = 2eBC = −1.1625i − 1.6275j − 0k.
20 m
50 m
MA
i
=M + 0
0.1793
j
70
−4.7682
B
TBC
from which
A
MX
= −17.038 kN-m,
MYA = 0,
A
MZ
= 12.551 kN-m.
TBE
TBD
k 0 =0
0.2434 A
= (MX
+ 17.038)i + (MYA + 0)j
A
+(MZ
− 12.551)k = 0
x
y
rAB × TBD + rAB × TBC = 0
A
40 m
D
z
= MA + rAB × (TBE + TBC + TBD )
E
A
The sum of the moments about A is
MA = MA + rAB × TBE
+
x
AY ,
C
A
MY
AZ ,MA
EA
A X , MA
Z
z
D
X
x
Problem 5.98 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 = MA + rAB × (TBE + TBC + TBD ) = 0.
If the couple MA = 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 · (TBC + |TBE |eBE + |TBD |eBD ) = 0,
and ez · (TBC + |TBE |eBE + |TBD |eBD ) = 0.
Two simultaneous equations in two unknowns result;
0.4444|TBE | + 0.2265|TBD | = 1.1625
−0.4444|TBE | + 0.5661|TBD | = 0.
Solve:
|TBE | = 1.868 kN,
|TBD | = 1.467 kN.
The reactions at A oppose the sum of the cable tensions in the x-, y-,
and z-directions.
AX = 0,
AY = 4.243 kN,
AZ = 0.
(These results are to be expected if there is no moment about A.)
Problem 5.99 The space truss has roller supports at B,
C, and D and is subjected to a vertical force F = 20 kN
at A. What are the reactions at the roller supports?
y
F
A (4, 3, 4) m
B
D (6, 0, 0) m
x
z
C (5, 0, 6) m
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
y
F
A(4, 3, 4) m, B(0, 0, 0) m,
C(5, 0, 6) m, D(6, 0, 0) m
A (4, 3, 4) m
we need eAB , eAC , eAD , eBC , eBD , and eCD . Use the form
eP Q =
B
(xQ − xP )i + (yQ − yP )j + (zQ − zP )k
[(xQ − xP )2 + (yQ − yP )2 + (zQ − zP )2 ]1/2
D (6, 0, 0) m
x
eAB = −0.625i − 0.469j − 0.625k
eAC = 0.267i − 0.802j + 0.535k
z
C (5, 0, 6) m
eAD = 0.371i − 0.557j − 0.743k
eBC = 0.640i + 0j + 0.768k
eBD = 1i + 0j + 0k
F
Joint A :
eCD = 0.164i + 0j − 0.986k
TAB = TAB eAB , TAC = TAC eAC
TAD = TAD eAD , TBC = TBC eBC
TAD
TAB
We will write each force as a magnitude times the appropriate unit
vector.
TAC
D
B
TBD = TBD eBD , TCD = TCD eCD
Each force will be written in component form, i.e.

TABX = TAB eABX 
TABY = TAB eABY
etc.

TABZ = TAB eABZ
Joint A:
C
Joint B :
–TAB
TBD
TAB + TAC + TAD + F = 0
NBJ
TABX + TACX + TADX = 0
Joint C :
TBC
–TAC
TABY + TACY + TADY − 20 = 0
TABZ + TACZ + TADZ = 0
Joint B:
−TAB + TBC + TBD + NB j = 0
Joint C:
−TAC − TBC + TCD + NC j = 0
Joint D:
−TAD − TBD − TCD + ND j = 0
Solving for all the unknowns, we get
NB = 4.44 kN
NC = 2.22 kN
ND = 13.33 kN
Also, TAB = −9.49 kN, TAC = −16.63 kN
TAD = −3.99 kN, TBC = 7.71 kN
TBD = 0.99 kN, TCD = 3.00 kN
TCD
–TBC
NCJ
Joint D :
–TAD
–TBD
–TCD
NDJ
Problem 5.100 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 = 13.33 kN. The maximum value is 15 kN. Scaling the force
by the factor 15/13.33 gives
15
Fmax =
(20 kN) = 22.5 kN
13.33
y
F
A (4, 3, 4) m
B
D (6, 0, 0) m
x
z
C (5, 0, 6) m
Fmax = 22.5 kN
Problem 5.101 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?
Solution:
y
4 ft
1 ft
B
The position vector of the midpoint of the door:
5 ft
rCM = (2 cos 50◦ )i + 3.5j + (2 cos 40◦ )k
= 1.2856i + 3.5j + 1.532k.
The position vectors of the hinges:
rA = j,
A
1 ft
rB = 6j.
x
The forces are: W = −40j,
40°
A = AX i + AY j + AZ k,
z
B = BX i + BZ k.
The position vectors relative to A are
y
rACM = rCM − rA = 1.2856i + 2.5j + 1.532k,
4 ft
rAB = rB − rA = 5j.
1 ft
The sum of the moments about A
MA = rACM × W + rAB × B
i
j
k i
= 1.2856 2.5 1.532 + 0
0
−40
0 B
X
5 ft
j
5
0
k 0 =0
BZ A
1 ft
x
z
40°
MA = (5BZ + 40(1.532))i + (−5BX − 40(1.285))k = 0,
from which BZ =
−40(1.532)
= −12.256 lb
5
BX =
−40(1.285)
= −10.28 lb.
5
and
B
y
BX
BZ
The reactions at A are determined from the sums of forces:
FX = (AX + BX )i = 0,
AY
from which AX = 10.28 lb,
FY = (AY − 40)j = 0,
from which AY = 40 lb,
FZ = (AZ + BZ )k = 0,
from which AZ = 12.256 lb
AZ
z
W
AX
x
Problem 5.102 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 = 10i − 30j − 10k (N).
y
200 mm
100 mm
Solution:
The position vector of the point of application of the
100 mm
force is
rF = 0.2i − 0.2k.
B
200 mm
The position vector of the bearing is
rB = 0.1i.
F
z
The position vector of the cable attachment to the wheel is
x
A
rC = 0.1k.
he position vectors relative to B are:
rBC = rC − rB = −0.1i + 0.1k,
y
rBF = rF − rB = 0.1i − 0.2k.
The sum of the moments about the bearing B is
MB = MB + rBF × F + rBC × C = 0,
i
j
k i
or
MB = MB + 0.1
0
−0.2 + −0.1
10 −30 −10 0
200 mm
100 mm
100 mm
j
0
−T
k 0.1 0 B 200 mm
F
z
BZ
= (−6 + 0.1T )i + (MBY − 1)j
By
x
A
+(MBZ − 3 + 0.1T )k = 0,
from which T =
6
= 60 N,
0.1
MBY = +1 N-m,
MBZ = −0.1T + 3 = −3 N-m.
The force reactions at the bearing are determined from the sums of
forces:
FX = (BX + 10)i = 0,
from which BX = −10 N.
FY = (BY − 30 − 60)j = 0,
from which BY = 90 N.
FZ = (BZ − 10)j = 0,
from which BZ = 10 N.
Problem 5.103 In Problem 5.106, suppose that the where FZ = 0 is given in the problem statement. The moment equations can
z component of the force F is zero, but otherwise F be developed by inspection of the figure also. They are
is unknown. If the couple exerted on the shaft by the
Mx = MBX + MAX + MF X = 0,
bearing at B is MB = 6j − 6k N-m, what are the force
MY = MBY + MAY + MF Y = 0,
F and the tension in the cable?
MZ = MBZ + MAZ + MF Z = 0,
Solution: From the diagram of Problem 5.106, the force equilib-
and
rium equation components are
Fx = BX + FX = 0,
Fy = BY + FY = 0,
and
Fz = BZ + FZ = 0,
where MB = 6j − 6k N-m. Note that MBX = 0 can be inferred. The
moments which need to be substituted into the moment equations are
MA = (0.1)Ai + 0j + (0.1)Ak N-m,
and MF = (0.2)FY i − (0.2)FX j + (0.1)FY k N-m.
Substituting these values into the equilibrium equations, we get F = 30i −
60j + 0k N, and A = 120 N.
Problem 5.104 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 = 10i − 30j − 10k (N), what are the tension in the
vertical cable and the reactions at the bearings B and C?
y
200 mm
50 mm
100 mm
50 mm
200 mm
B
C
F
z
x
A
The position vectors relative to the bearing B are: the
position vector of the cable attachment to the wheel is
Solution:
50 mm
100 mm
50 mm
200 mm
rBT = −0.05i + 0.1k.
The position vector of the bearing C is:
B
rBC = 0.1i.
C
200 mm
The position vector of the point of application of the force is:
rBF = 0.15i − 0.2k.
A
The sum of the moments about B is
MB = rBT × T + rBC × C + rBF × F = 0
i
j
k i
j
k MB = −0.05
0
0.1 + 0.1
0
0 0
−T
0
0 C
C i
+ 0.15
10
j
0
−30
k −0.2 = 0
−10 Y
+(0.05T + 0.1CY − 4.5)k = 0.
From which: T = 60 N,
CZ =
−0.5
= −5 N,
0.1
CY =
4.5 − 0.05T
= 15 N.
0.1
The reactions at B are found from the sums of forces:
FX = (BX + 10)i = 0,
from which BX = −10 N.
FY = (BY + CY − T − 30)j = 0,
from which BY = 75 N.
FZ = (BZ + CZ − 10)k = 0,
from which BZ = 15 N
y
Z
MB = (0.1T − 6)i + (−0.1CZ + 1.5 − 2)j
F
x
BY
BX CY
BZ
F
CZ
z
T
x
Problem 5.105 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 = 30 kip acts at (4.5,
5, 2) ft. What is the magnitude of the reaction on the
launcher at E?
The position vectors of the points D, E and W are
Solution:
rD = 7i + 7k,
rE = 9i + 6j + 4k (ft),
rW = 4.5i + 5j + 2k (ft).
The vector parallel to DE is
rDE = rE − rD = 2i + 6j − 3k.
The unit vector parallel to DE is
eDE = 0.2857i + 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
0
0
MX = 4.5
5
2 +|FDE | 9
6
4
=0
0 −30 0 0.2857 0.8571 −0.4286 = 60 − 6|FDE | = 0.
From which
|FDE | =
60
= 10 kip
6
y
B
A
3 ft 3 ft
E
W
D
x
y
30 kip
BY
AY
AX
AZ
BZ
x
FDE
z
y
W
E
A
B
x
3 ft 3 ft
D
Problem 5.106
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
The components of the moment eq. are
can be written
−5.9997FDE + 60 = 0,
FDE = FDE (0.2857i + 0.8571j − 0.4286k).
−3AZ − 6BZ + 5.0001FDE = 0,
The equilibrium equations are
FX = AX + 0.2857FDE = 0,
FY = AY + BY + 0.8571FDE − 30 = 0,
FZ = AZ + BZ − 0.4286FDE = 0,
i
j
k i
j
k M(origin) = 3
0
0 + 6
0
0 A
A
A 0 B
B X
Y
i
+FDE 7
0.2857
i
+ 4.5
0
Z
j
0
0.8571
j
k
5
2=0
−30 0 Y
3AY + 6BY + 5.9997FDE − 135 = 0.
Solving, we obtain
FDE = 10.00 kip, AX = −2.86 kip,
AY = 17.86 kip, AZ = −8.09 kip,
BY = 3.57 kip, BZ = 12.38 kip.
Z
k
7
−0.4286 Problem 5.107 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.)
Solution: The position vector from C to D is
y
rCD = 3i − 6j − 3k (m),
C
so we can write the force exerted at C by cable CD as
TCD = TCD
rCD
= TCD (0.408i − 0.816j − 0.408k).
|rCD |
A
The coordinates of pt. B are x = 46 (3) = 2 m, y = 4 m.
The moment about the origin due to the forces exerted by the two
cables is
i
j k i
j
k
MO = 2 4 0 + 3
6
0
−8 0 0 0.408T
−0.816T
−0.408T
CD
CD
= 32k − 2.448TCD i + 1.224TCD j − 4.896TCD k.
The moment about the z axis is
k · MO = 32 − 4.896TCD = 0,
so
TCD = 6.54 kN.
CD
B
F
E
z
2m
2m
D
3m
x
Problem 5.108
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 .
Solution: See the solution of Problem 5.111. The force exerted
at C can be written
y
TCD = TCD (0.408i − 0.816j − 0.408k)
and the coordinates of pt. B are (2, 4, 0) m.
The equilibrium equations are
FX = EX + FX − 8 + 0.408TCD = 0,
FY = EY + FY − 0.816TCD = 0,
FZ = EZ − 0.408TCD = 0,
i
j
k i
j
k i
MO = 0
0
2 + 0
0 −2 + 2
E
E
E F
F
0 −8
X
Y
i
+
3
0.408T
CD
Z
X
j
6
−0.816TCD
Y
k
0
−0.408TCD
C
8 kN
EY
j k
4 0
0 0
= 0.
The components of the moment equation are
MX = −2EY + 2FY − 2.448TCD = 0,
MY = 2EX − 2FX + 1.224TCD = 0,
MZ = 32 − 4.896TCD = 0.
From the last equation, TCD = 6.54 kN. Solving the other eqs, we
obtain
EX = 667 N,
EY = −1,333 N,
EZ = 2,667 N,
FX = 4,667 N,
FY = 6,667 N.
TCD
B
FY
z
EZ
O
FX
EX
x
Problem 5.109 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 y − z 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
B
z
C
20°
2m
Solution:
y
F = A + B + FD + TCE = 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.
Point C is at (2, −2 sin 20◦ , 2 cos 20◦ ) Point E is at (0, 1, 3)
eCE =
3m
2i – 6j (kN)
E
rCE
|rCE |
eCE = −0.703i + 0.592j + 0.394k
We also need the unit
B(0, −2 sin 20◦ , 2 cos 20◦ )
vector
x
A
D
2m
1m
eAB .
A(0,
0,
x
0),
B
z
C
20°
eAB = 0i − 0.342j + 0.940k
2m
The moment of FD about A (a point on AB) is
y
MFD = rAD × FD1 = (2i) × (2i − 6j)
MFD = −12k
3m
The moment of TCE about B (another point on line CE) is
E
MTCE = rBC × TCE eCE = 2i × TCE eCE ,
1m
z
MF DAB = MF D · eAB
MF DAB = −11.27 N-m
The moment of TCE about line AB is
MCEAB = TCE (2i × eCE ) · eAB
MCEAB = TCE (−0.788j + 1.184k) · eAB
MCEAB = 1.382TCE
The sum of the moments about line AB is zero. Hence
MF DAB + MCEAB = 0
−11.27 + 1.382TCE = 0
TCE = 8.15 kN
AX
A
where eCE is given above.
The moment of FD about line AB is
BZ
BY
AZ
B
20°
FD = 2i – 6j
AY
D
x
TCE
2m
BX
2m
C
Problem 5.110 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 = 8.15 kN
Also, from that solution,
y
eAB = 0i − 0.342j + 0.940k
3m
We are given that the force at force at hinge B does not exert a force
parallel to AB at B. This implies
2i – 6j (kN)
B · eAB = 0.
A
E
D
B · eAB = −0.342BY + 0.940BZ = 0 (1)
2m
1m
We also had, in the solution to Problem 5.113
eCE = −0.703i + 0.592j + 0.394k
z
and TCE = TCE eCE (kN)
B
C
20°
2m
For Equilibrium,
F = A + B + TCE + F = 0
y
FX :
AX + BX + TCE eCEX + 2 = 0 (kN) (2)
FY :
AY + BY + TCE eCEY − 6 = 0 (kN) (3)
FZ :
AZ + BZ + TCE eCEZ = 0 (kN)
AX
BY
BZ
rAC × TCE = (−2 sin θTZ − 2 cos θTY )i
+(2 cos θTX − 2TZ )j
+(2TY + 2TX sin θ)k
rCE × B = (−2BZ sin θ − 2BY cos θ)i
+(2BX cos θ)j + (+2BX sin θ)k
MA = 0,
Hence
−2 sin θTZ − 2 cos θTY − 2BZ sin θ
−2BY cos θ = 0
(5)
y:
2 cos θTX − 2TZ + 2BX cos θ = 0
(6)
z:
−12 + 2TY + 2TX sin θ + 2BX sin θ = 0 (7)
Solving Eqns (1)–(7), we get
|A| = 8.53 (kN), |B| = 10.75 (kN)
x
TCE
2
BX
C (2, –2sinθ , + 2cosθ)
rAD × F = 2i × (2i − 6j) = −12k (kN)
x:
θ
z
rAD × F + rAC × TCE + rAB × B = 0
D
AZ
(4)
Summing Moments about A, we have
F = + 2i – 6j (kN)
AY
x
Problem 5.111 The bar ABC is supported by ball
and socket supports at A and C and the cable BD, and
is loaded by the 200-lb suspended weight. What is the
tension in cable BD?
y
(–2, 2, –1) ft
D
2 ft
4 ft
B
A
C
x
4 ft
z
Solution: The strategy is to take the moments about A. Note that
a ball and socket cannot support a couple reaction. The vector parallel
to the cable is
From which:
•rBD = −2i + 2j − k.
|T| =
CY = 0,
800
= 200 lb
4
The unit vector parallel to the cable is
y
eBD = −0.6667i + 0.6667j − 0.3333k.
(–2, 2, –1) ft
The vector positions of the weight, point B and point C relative to
point A are:
2 ft
D
4 ft
A x
B
rAW = −4i,
C
rAB = −6i,
z
and rAC = −6i + 4k.
4 ft
The sum of the moments about A is
MA = rAW × W + rAB × T
MA
+rAC × C = 0
i
j
k
i
= −4
0
0 + |T| −6
0 −200 0 −0.6667
i
j
k + −6
0
4 =0
C
C
C X
Y
y
T
j
0
0.6667
k
0
−0.3333 AY
CY
z
CZ CX
W
AZ
AX
x
Z
= −4CY i + (−2|T| + 4CX + 6CZ )j
+(800 − 4|T| − 6CY )k = 0.
Problem 5.112 In Problem 5.115, determine the y
components of the reactions exerted on the bar ABC by
the ball and socket supports at A and C.
Solution: From the figure in Problem 5.115, we see that there
are seven unknowns (3 reaction components each at A and C plus
the magnitude of the tension force in the cable). Equilibrium in three
dimensions gives us only six equations. Thus, statics alone will not
give us information sufficient to find all of the unknowns. However,
we are often able to find values for some of the variables. In Problem
5.115, we found that the cable tension was 200 lb and the value for the
y component of the force at C is zero. We are now asked to determine
the values of the y components of the forces at A and C. We have
already determined CY = 0, and
TY = T eBDY = 200(0.667) = 133.3 lb.
Since we also know the weight, we know all of the vertical forces except the
vertical support force at A. The vertical equilibrium equation is
CY + AY + TY − W = 0.
Substituting and solving results in
AY = W − TY = 200 − 133.33 = 66.67 lb.
Problem 5.113 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.
y
200 i (N)
300 mm
x
C
180 mm
B
z
A
100 k (N)
150 mm
Solution: The strategy is to take the moments about A and solve
the resulting simultaneous equations. The position vectors of the bearings relative to A are:
y
300 mm
200 i (N)
rAB = −0.15i + 0.15j,
rAC = −0.15i + 0.33j + 0.3k.
180 mm
C
B
Denote the lower force by subscript 1, and the upper by subscript 2:
rA1 = −0.15i,
z
rA2 = −0.15i + 0.33j.
A
x
The sum of the moments about A is:
MA = rA1 × F1 + rAB × B + rA2 × F2 + rAC × C = 0
i
j
k i
j
k MA = −0.15 0
0 + −0.15 0.15
0 0
0 100
B
0
B i
+ −0.15
200
X
j
k i
0.33 0 + −0.15
0
0 CX
100 k (N)
j
0.33
CY
MA = (0.15BZ − 0.3CY )i + (15 + 0.15BZ + 0.3CX )j
+(−0.15BX − 66 − 0.15CY − 0.33CX )k = 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 = (BX + CX + 200)i = 0.
The four equations in four unknowns:
0BX + 0.15BZ + 0CX − 0.3CZ = 0
0BX + 0.15BZ + 0.3CX + 0CY = −15
−0.15BX + 0BZ − 0.33CX − 0.15CY = 66
BX + 0BZ + CX + 0CZ = −200.
(The HP-28S hand held calculator was used to solve these equations.)
The solution:
BX = 750 N,
BZ = 1800 N,
CX = −950 N,
CY = 900 N.
The reactions at A are determined by the sums of forces:
FY = (AY + CY )j = 0, from which AY = −CY = −900 N
FZ = (AZ + BZ + 100)k = 0, from which AZ = −1900 N
150
mm
y
CY
Z
k 0.3 = 0
0 150 mm
200 N
x
z
CX
BX
BZ
100 N
AY
AZ
150 mm
Problem 5.114 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 AF G 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.
y
A
A
Spinnaker
50 ft
C
C
F
P
P
6 ft
x
E
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 = 50j,
The sum of the moments about the base of the mast is
FP = 50(−0.707i + 0.707k) = −35.35i + 35.35k.
MQ = rA × FA + rA × TAB + rA × TAC + rC × TCE
rB = −21i,
rP = 6j,
rC = 25j − 7.5k.
+rP × FP = 0
MQ = rA × (FA + TAB + TAC ) + rC × TCE + rP × FP = 0.
The vector parallel to the backstay AB is
From above,
rAB = rB − rA = −21i − 50j.
FA + TAB + TAC = FT X i + FT Y j + FT Z k
The unit vector parallel to backstay AB is
= (193.2 − (0.3872)|TAB |)i + (−0.922|TAB |
eAB = −0.3872i − 0.9220j.
−0.9578|TAC |)j + (51.76 − 0.2873|TAC |)k
i
j
k i j
k
= 0
50
50 + 0 25
−7.5
F
FT Z 0 0 0.2873|TAC | T X FT Y
i
j
k +
0
6
6 =0
−35.35 0 35.35 The vector parallel to AC is
rAC = rC − rA = −25j − 7.5k.
MQ
The forces acting on the mast are: (1) The force due to the spinnaker
at the top of the mast:
FA = 200(i cos 15◦ + k cos 75◦ ) = 193.19i + 51.76k.
= (50FT Z + (25)(0.2873)|TAC | + 212.1)i
(2) The reaction due to the backstay:
TAB = |TAB |eAB
(3) The reaction due to the shroud:
TAC = |TAC |eAC
(4) The force acting on the cross spar CE:
TCE = −(k · TAC )k = 0.2873|TAC |k.
+(−50FT X + 212.1)k = 0.
Substituting and collecting terms:
(2800 − 7.1829|TAC |)i + (−9447.9 + 19.36|TAB |)k = 0,
from which
|TAC | =
2800
= 389.81 lb,
7.1829
|TAB | = 488.0 lb.
5.114 Contd.
The tension in cable CD is the vertical component of the tension
in AC,
|TCD | = |TAC |(j · eAC ) = |TAC |(0.9578) = 373.37 lb.
The reaction at the base is found from the sums of the forces:
FX = (QX + 193.19 − 35.35 − |TAB |(0.3872)) = 0,
from which QX = 31.11 lb
FY = (QY − 0.922|TAB | − 0.9578|TAC |)j = 0,
from which QY = 823.24 lb
FZ = (QZ + 51.76 + (0.2873|TAC |
−0.2873|TAC | + 35.35))k = 0,
from which QZ = −87.11 lb
Collecting the terms, the reaction is
Q = 31.14i + 823.26j − 87.12k (lb)
A
A
SPINNAKER
50 ft
C
C
6 ft
E
P
P
B
Side View
21 ft
15 ft
Aft View
z
A
P
C
50 lb DTop View
15°
45°
G
D
D
200 lb
E
B
y
y
FA
TAC
FA
TCD TCE
TAB
z
FP
QX
QY
TCD
FP
QY
QB
SIDE VIEW z AFT VIEW
FA
QB
x
QX
FP TAB
TOP VIEW
z
Problem 5.115 The door is supported by the cable DE
and hinges at A and B, and is subjected to a 2-kN force
at C. The door’s weight is negligible. The hinges do
not exert couples on the door, and their axes are aligned
with the line from A to B. Determine the tension in the
cable.
y
D (0.6, 0.4, – 0.3) m
(0, 0.6, 0.6) m
E
A
x
C (1.2, 0.2, 0.6) m
z
–2j (kN)
B (0.6, – 0.2, 0.9) m
Solution: We will sum moments around the origin. There are two
unit vectors that we need to find. We need the unit vector along the
hinge line AB and the unit vector along the cable DE. These unit
vectors are
(0, 0.6, 0.6) m
We also need to know the vectors from the origin, A, to each of the
points involved. Each of these vectors is of the form
A
TDE = TDE eDE = TDEX i + TDEY j + TDEZ k N.
The force at C is of the form C = −2j kN, and the forces at A and
B are
A = AX i + AY j + AZ k
and B = BX i + BY j + BZ k.
With these definitions, the force and moment vector equations of equilibrium are
A+B+C+T =0
and MT + MB + MC = 0.
Writing these in scalar form yields six equations in seven unknowns.
To find the magnitude T , we need another condition. If we look at
the picture, we see that the moments around the hinge line AB must
be balanced or the door will move. Furthermore, this equation does
not involve forces at either A or B. We get one equation in the one
unknown we are looking for. To get the equation for the moment about
line AB, we find the sum of the moments of the forces around any point
on AB (choose A) and then dot this into the unit vector along AB.
The condition is
(MT + MB + MC ) • eAB = 0.
If we write the equations in scalar form and add this to our equations
and solve, we get T = 2.44 N.
x
C (1.2, 0.2, 0.6) m
C
Ax
z
Az
–2j (kN)
Bx
B (0.6, – 0.2, 0.9) m
rAP = xP i + yP j + zp k,
where the point is at (xP , yP , zP ). The force in the cable is of the
form
D (0.6, 0.4, – 0.3) m
E
eAB = 0.545i − 0.182j + 0.818k
and eDE = −0.545i + 0.182j + 0.818k.
Ay
T
y
By
Bz
Problem 5.116 Determine the reactions at the hinges
supporting the door in Problem 5.l19. Assume that the
hinge at B exerts no force parallel to the hinge axis.
Strategy: Express the reactions at the hinges as
A = Ax i + Ay j + Az k
and B = Bx i + By j + Bz k.
Solution:
We have done much of this problem in Problem 5.119. The extra
equation that we found, summing moments around the hinge axis, did not give
us an additional equation to count in the equations versus unknowns balancing
process. It was merely a handy linear combination of the equations we already
had. — However, now we have been given additional information. The force
B does not act parallel to the hinge axis, and it a hinge axis coordinate system,
would have only two components. This, in effect, removes one unknown (or
adds a new equation), balancing equations and unknowns. The equation that we
add is
Let eAB be a unit vector parallel to the hinge axes. Since eAB • B = eABX BX + eABY BY + eABZ BZ = 0.
the hinge at B exerts no force parallel to the hinge axis,
Adding this to the force and moment equations from above and solving, we get
you know that eAB • B = 0.
A = 0.474i − 0.825j − 1.956k N and
B = 0.860i + 2.380j − 0.044k N.
Note that in these coordinates, the force at B has components parallel to all three
axes. If we rotated into coordinates parallel and perpendicular to the hinge axis
AB, one of the components of the force at B would disappear.
Problem 5.117 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 α, 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
one-force supports. The reaction at A is
A = A(i cos 60◦ + j sin 60◦ ) = A(0.5i + 0.866j).
The reaction at B is
B = B(i cos α + j sin α).
The sum of the moments about B is
MB = +W (1) − A(0.866)(2) = 0,
from which
W
(9.81)(10)
A =
=
= 56.64 N
(0.866)(2)
1.732
The sums of the forces:
FX = A(0.5) + B(cos α) = 0,
Combining:
tan α =
49.05
= −1.732,
−28.32
from which
α = 120◦
or α = 300◦ .
Since
cos α ≤ 0
and sin α ≥ 0,
the angle is in the second quadrant, hence α = 120◦ , and B =
56.64 N
from which
B cos α = −56.64(0.5) = −28.32.
FY = A(0.866) − W + B sin α = 0,
B sin α = 98.1 − 45.09 = 49.05.
α
B
30°
from which
C
2m
A
α
60°
W
49.05
sin α
=
Problem 5.118 The horizontal bar is of negligible
weight. Use the fact that the bar is a three-force member
to determine the angle α 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
4
h =
= 8.
cos 60◦
The vertical distance to the point P is D =
The angle α is:
6.9282
90◦ − α = tan−1
,
9
√
F
4
A
9
P
30°
α
F
82 − 42 = 6.9282.
90 − α
60°
from which α = 90◦ − 37.589◦ = 52.4◦
Problem 5.119 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 = 5B − 10(1000) = 0, from which the magnitude at B is
B = 2000 lb. The sums of the forces:
FX = AX + B = AX + 2000 = 0, from which AX = −2000 lb.
FY = AY − 1000 = 0, from which AY = 1000 lb.
√
The magnitude at A is A = 20002 + 10002 = 2236 lb
A
5 ft
B
10 ft
1000 lb
AX
AY
5 ft
B
1000 lb
10 ft
Problem 5.120 The weight W = 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°
W
Solution: Denote the magnitude of the reaction at the pinned joint
by B. The sums of the forces are:
FX = BX − T sin 60◦ = 0,
and
FY = BY + T cos 60◦ − W = 0.
60°
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 = −RBY +RT = 0, from which BY = T .
Substitute into the sum of the forces to obtain: T + T (0.5) − W = 0,
from which
T =
2
W = 33.33 lb.
3
W
T
60°
R
Substitute into the sum of forces to obtain
BX
W
BY
BX = T sin 60◦ = 28.86 lb.
The magnitude of the reaction at the pinned joint is
B = 33.332 + 28.862 = 44.1 lb
Problem 5.121 The weight W = 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 = R cos α, where α is given by (see sketch)
R−h
150 − 50
α = sin−1
= sin−1
= 41.81◦ .
R
150
F
150 mm
50 mm
W
α
The perpendicular distance to the action line of the force is
D = 2R − h = 300 − 50 = 250 mm.
The sum of the moments about the contact point is
F
MB = −(R cos α)W + (2R − h)F = 0,
from which F =
(150 cos 41.81◦ )W
250
= 0.4472W = 17.88 N
W
h
b
Problem 5.122 Use the fact that the horizontal bar is
a three-force member to determine the angle α and the
magnitudes of the reactions at A and B.
3 kN
α
A
B
30°
1m
The sum of the moments about B is
Solution:
MB = −(3)(3 sin α) + (2)A = 0,
α
from which A = 9 sin
= 4.5 sin α.
2
The sum of the forces:
FX = 3 cos α − B sin 30◦ = 0,
and
FY = 3 sin α − A + B cos 30◦ = 0.
Eliminate B as follows: solve each equation for B, and substitute the
value for A:
B =
and B =
1.5 sin α
,
cos 30◦
3 cos α
,
sin 30◦
from which
1.5 sin α
3 cos α
=
,
cos 30◦
sin 30◦
or
tan α =
3 cos 30◦
1.5 sin 30◦
= 3.4641,
α = 73.9◦ .
Substitute into the force equations to obtain
B = 1.66 kN,
A = 4.32 kN
3 kN
α
A
B
30°
1m
2m
30°
3 kN
α
1m
A
B
2m
2m
Problem 5.123 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.
D
3 ft
A
B
C
3 ft
3 ft
Solution: Isolate the member ABC. The angle ABD is 45◦
since the base and altitude of the triangle are equal. The sum of the
moments about A is
MA = +3B sin 45◦ − 6(600) = 0
6(600)
from which B = 3 sin 45◦ = 1697.1 lb.
The sum of the forces
FX = AX − B cos 45◦ = 0,
from which AX = 1199.8 lb.
FY = AY + B sin 45◦ − 600 = 0,
from which AY = 600 − 1199.82 = −599.82.
The magnitude at A is
A = 1199.82 + 599.82 = 1341.4 lb
D
3 ft
B
A
C
3 ft
3 ft
B
AY
AX
600 lb
3 ft
3 ft
Problem 5.124 (a) Is the L-shaped bar a three-force
member?
(b) Determine the magnitudes of the reactions at A
and B.
(c) Are the three forces acting on the L-shaped bar concurrent?
2 kN
3 kN-m
B
300 mm
150 mm
700 mm
A
250
mm
(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
150
α = tan−1
= 30.96◦ .
250
Solution:
The sum of the moments about A is
MA = −3 − 0.5(2) + 0.7B cos α = 0,
from which B = 6.6637 kN. The sum of forces:
FX = AX + B cos α = 0,
from which AX = −5.7143 kN.
FY = AY − B sin α − 2 = 0,
from which AY = 5.4281 kN. The magnitude at A:
A = 5.712 + 5.432 = 7.88 kN (c) No, by inspection.
2 kN
3 kN-m
300 mm
B
150 mm
700 mm
A
500
mm
250
mm
0.5
m
α
3 kN-m
2 kN
0.3 m
B
0.7 m
Ay
Ax
500
mm
Problem 5.125 The bucket of the excavator is supported by the two-force member AB and the pin support
at C. Its weight is W = 1500 lb. What are the reactions
at C?
14 in.
16 in.
B
A
4 in.
C
W
8 in.
The angle of the member AB relative to the positive x
Solution:
axis is
α = tan−1
12
14
= 40.6◦ .
The moment about the point C is
MC = 4A cos 40.6◦ + 16A sin 40.6◦ + 8W = 0,
from which A = −0.5948W = −892.23 lb. The sum of forces:
FX = Cx − A cos 40.6◦ = 0,
from which: CX = −677.4 lb
FY = CY − A sin 40.6◦ − W = 0,
from which CY = 919.4 lb
14 in.
B
16 in. A
4 in.
C
W
8 in.
8 in.
A
α
4 in
CY
CX
W
8 in 8 in
8 in.
Problem 5.126 The member ACG of the front-end
loader is subjected to a load W = 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) Draw a free-body diagrams of the hydraulic cylinder
and the member ACG.
(b) Determine the reactions on the member ACG.
A
0.75 m
B
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 + CX = 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
Fx = AX + CX = 0,
Fy = AY − W = 0,
and
MA = (0.75)CX − (3)W = 0.
1.5 m
A
0.75 m
B
C
1m
G
0.5 m
1.5
m
W
1.5
m
Using the given value for W and solving these equations, we get
BX
AX = −8 kN,
CX
B
AY = 2 kN,
CX = 8 kN,
and BX = −8 kN.
AY
AX
0.75 m
CX
1m
0.5 m
1.5 m
1.5 m
W
Problem 5.127 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.
tan θ = 0.75/3,
◦
or θ = 14.04 .
Using this, we can write force equilibrium equations in the form
Fx = −A cos θ + CX = 0, and
Fy = A sin θ − W = 0.
Solving these equations, we get A = 8.246 kN, and CX = 8 kN.
The components of A are as calculated in Problem 5.130.
y
A
A
0.75 m
1m
θ
CX C
1.5 m
x
1.5 m
G
W = 2 kN
Problem 5.128 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
FX = AX + BX = 0,
from which AX = −BX
FY = AY + BY = 0,
from which AY = −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
MA = Bx (h) − bBy = 0,
from which the angle of direction of B is
BY
h
tan−1
= tan−1
= αB .
BX
b
or (180 + αB ). Similarly, by substituting A:
AY
h
tan−1
= tan−1
= αA ,
AX
b
or (180 + αA ). But
h
α = tan−1
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.
B
h
A
By
y
Ay
Fig a
Ax
Bx
Fig b
x
Problem 5.129 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
F1
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 :
M = r1 × F1 + r2 × F2 + r3 × F3 = 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 = r2 − r1 ,
and d13 = r3 − r1 .
Then the sum of the moments can be rewritten,
M = r1 × (F1 + F2 + F3 )
+d12 × F2 + d13 × F3 = 0.
Since the system is in equilibrium,
F1 + F2 + F3 = 0,
and the sum of moments reduces to
M = d12 × F2 + d13 × F3 = 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 · (d12 × F2 )e + e · (d13 × F3 )e = 0,
or F2 · (d12 × e)e + F3 · (d13 × e)e = 0.
But by definition, F2 lies in the same plane as the line L, hence it is
normal to the cross product d12 × e = 0, and the term
F2 · (d12 × e) = 0.
But this means that
F3 · (d13 × e)e = 0,
which implies that F3 also lies in the same plane as F2 , since
d13 × e = 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.
F2
F3
F1
P
L
Problem 5.130 (a) Draw the free-body diagram of the
50-lb plate, and explain why it is statically indeterminate.
(b) Determine as many of the reactions at A and B as
possible.
y
A
12 in
8 in
B
x
20 in
50 lb
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.
Sums the forces:
A
12 in
8 in
B
FX = AX + BX = 0,
20 in
or AX = −BX , and
AY
FY = AY + BY − 50 = 0.
The sum of the moments about B
MB = −20AX − 50(20) = 0,
from which AX = −50 lb,
AX
12 in.
BY
8 in.
BX
50 lb
x
20 in.
and from the sum of forces BX = 50 lb.
Problem 5.131 The mass of the truck is 4 Mg. Its
wheels are locked, and the tension in its cable is T =
10 kN.
(a) Draw the free-body diagram of the truck.
(b) Determine the normal forces exerted on the truck’s
wheels by the road.
x
50 lb
(003) 676-5942
30°
AL's
Tow i n g
2m
T
2.5 m
2.2 m
mg
Solution: The weight is 4000(9.81) = 39.24 kN. The sum of
the moments about B
MB = −3T sin 30◦ − 2.2T cos 30◦ + 2.5W − 4.5AN = 0
30°
from which
AN =
2.5W − T (3 sin 30◦ + 2.2 cos 30◦ )
4.5
64.047
=
= 14.23 N
4.5
The sum of the forces:
FY = AN − W + BN − T cos 30◦ = 0,
from which BN = T cos 30◦ − AN + W = 33.67 N
AX
A
AN
W
BX
B
BN
T
3m
Problem 5.132 Assume that the force exerted on the
head of the nail by the hammer is vertical, and neglect
the hammer’s weight.
(a) Draw the free-body diagram of the hammer.
(b) If F = 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.
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 = 11F − 2FN = 0,
from which the force exerted by the nail head is FN = 11F
= 5.5F .
2
The sum of the forces:
FX = −F cos 25 + Hx = 0,
Solution:
from which the friction force exerted on the hammer is HX =
0.9063F .
FY = NH − FN + F sin 25◦ = 0,
from which the normal force exerted by the floor on the hammer is
NH = 5.077F
If the force on the handle is
F = 10 lb,
then FN = 55 lb,
HX = 9.063 lb,
and NH = 50.77 lb
F
11 in.
65°
2 in.
F
11 in.
65°
HX
B NH
FN
Problem 5.133 (a) Draw the free-body diagram of
the beam.
(b) Determine the reactions at the supports.
300 N
A
200 N-m
B
1m
1m
Solution:
(a) The free body diagram is shown.
(b) The sum of the moments about B
MB = +200 − (1)AY + (1)300 = 0,
from which AY = 200 + 300 = 500 N-m . The sum of the
forces:
FX = B X = 0
and
FY = AY + BY + 300 = 0,
from which BY = −AY − 300 = −800 N
Problem 5.134
Consider the beam shown in
Problem 5.145. First represent the loads (the 300-N
force and the 200-N-m couple) by a single equivalent
force; and then determine the reactions at the supports.
Solution: The equivalent force is equal to the applied forces: F =
300 N. Measure the distance x from the point B. The moment about
B due to the loads:
MB = 300x + 200 + 1(300) = 0,
from which
x =−
500
= −1.6667 m,
300
or 1.6667 m to the left of B. The reactions: The sum of the moments
about B
MB = 300(1.6667) − (1)AY = 0
from which AY = 500 N. The sum of the forces
FX = BX = 0,
FY = AY + BY + 300 = 0,
from which BY = −AY − 300 = −800 N
300 N
200 N-m
A
1m
200 N-m
1m
B
1m
A
AY
BY
BX
1m
1m
300 N
B
1m
1m
Problem 5.135 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 = B(400) − W (1100) = 0,
from which B =
1100W
= 2.75W
400
B = 2.75(90)(9.81) = 2427.975 = 2.43 kN.
The sum of the forces:
FX = AX = 0
FY = AY + B − W = 0,
from which AY = W − B = 882.9 − 2427.975 = −1.545 kN
400
mm
700
mm
300 mm
A
B
AX
W
B
AY
400 mm
700 mm
Problem 5.136 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
1.4 ft
H
870 lb
1.6 ft
A
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 = 1.4i + 2.8j ft and
rH = 8i + 1.6j.
The angle of the weight vector realtive to the positive x-axis is
α = 270◦ − 15◦ = 255◦ .
The weight has the components
W = W (i cos 255◦ + j sin 255◦ ) = 870(−0.2588i − 0.9659j)
= −225.173i − 840.355j (lb).
The sum of the moments about H is
MH = (rW − rH ) × W + (rA − rH ) × A,
i
j
k i
j
k
MH = −6.6
1.2
0 + −8 −1.6 0 = 0
−225.355 −840.355 0 0
AY
0
= 5816.55 − 8AY = 0,
from which AY =
5816.55
= 727.1 lb.
8
The sum of the forces is
FX = (HX − 225.173)i = 0, from which HX = 225.2 lb,
FY = (AY + HY − 840.355)j = 0, from which HY = 113.3 lb
y
1.4 ft
H
870 lb
1.6 ft
2.8 ft
8 ft
A
15°
1.4 ft
1.2 ft
W
15°
AY
8 ft
HX
HY
8 ft
x
Problem 5.137 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 α, obtaining the
following results:
α
Ay (kN) B (kN)
10◦
10.134
4.357
20◦
10.150
3.677
What are the distances b and h?
Solution:
The position vectors of the cm and the point B are
rCM = (2.7 − b)i + hj,
y
h
B
W
α
Ax
b
Ay
2.7 m
These two simultaneous equations in two unknowns were solved using the HP28S hand held calculator.
b = 1.80 m,
rB = 2.7i.
The angle between the weight and the positive x-axis is β = 270 − α.
The weight vector at each of the two angles is
h = 0.50 m
W10 = W (i cos 260◦ + j sin 260◦ )
h
y
W10 = W (−0.1736i − 0.9848j)
x
W20 = W (i cos 250◦ + j sin 250◦ ) or
B
W20 = W (−0.3420i − 0.9397j)
W
The weight W is found from the sum of forces:
FY = AY + BY + W sin β = 0,
from which Wβ
α A
y
AY + BY
=
.
sin β
Taking the values from the table of measurements:
W10 = −
10.134 + 4.357
= 14.714 kN,
sin 260◦
[check :W20 = −
10.150 + 3.677
= 14.714 kN check ]
sin 250◦
The moments about A are
MA = rCM × W + rB × B = 0.
Taking the values at the two angles:
i
j
k i
j
10
MA = 2.7 − b
h
0 + 2.7
0
−2.5551 −14.4910 0 0 4.357
M20
A
= 14.4903b + 2.5551h − 27.3618 = 0
i
j
k i
j
= 2.7 − b
h
0 + 2.7
0
−5.0327 −13.8272 0 0 3.677
= 013.8272b + 5.0327h − 27.4054 = 0
k
0=0
0
k
0
0
x
Ax
b
2.7 m
Problem 5.138 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 α, the tension in the cable, and the magnitude
of the reaction at A.
C
A
B
α
W
L/2
L/2
The sum of the moments about B is
L
= −LAY +
W = 0,
2
Solution:
MB
from which AY = W
. The sum of the forces:
2
FX = T cos α = 0,
from which T = 0 or cos α = 0. The choice is made from the sum
of forces in the y-direction:
FY = AY − W + T sin α = 0,
from which T sin α = W − AY = W
. This equation cannot be
2
satisfied if T = 0, hence cos α = 0, or α = 90◦ , and T = W
2
C
L
2
L
2
α
A
B
W
T
α
AY
W
L/2
L/ 2
Problem 5.139 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
T
T − 2TB sin 35◦ = 0,
from which TB =
T
= 0.8717T.
2 sin 35◦
Take the origin of the system to be at A. The position vector of the
point of attachment of B is rB = 45j (mm). The position vector of
the point of attachment of the cable is rC = 40i + 85j (mm).
The force exerted by the brake pad is B = −Bi. The force vector due
the cable tension is
35°
40 mm
B
TB = TB (i cos 145◦ + j sin 145◦ ) = TB (−0.8192i + 0.5736j).
45 mm
A
The moment about A is
MA = r B × B + r C × T B = 0
i
j
k i
MA = 0
45 45 + 40
−B 0
0 −0.8192
j
85
0.5736
30
mm
k
85 TB = 0
0 TB
MA = (45B + 92.576TB )k = 0,
from which B =
92.576TB
= 2.057TB .
45
35°
40
mm
B
Substitute the expression for the cable tension:
B = (2.057)(0.8717)T = 1.793T
AY
AX
40
mm
45
mm
Problem 6.1 Determine the axial forces in the members of the truss and indicate whether they are in tension
(T) or compression (C).
Strategy: Draw free-body diagram of joint A. By
writing the equilibrium equations for the joint, you can
determine the axial forces in the two members.
800 N
A
30°
B
60°
Solution:
C
800 N
A
y
60°
30°
B
800 N
C
x
A
30°
60°
Solving: We get
FAB
FAC
FAB = 693 N (tension)
FAC = −400 N (compression)
Assume the forces are in the directions shown (both in tension). If a
force turns up negative, that force will be in compression.
Equilibrium Eqns.
Fx :
−FAB cos 30◦ + FAC cos 60◦ + 800 = 0
Fy :
−FAB sin 30◦ − FAC sin 60◦ = 0
Problem 6.2 The truss supports a 10-kN load at C.
(a) Draw the free-body diagram of the entire truss, and
determine the reactions at its supports.
(b) Determine the axial forces in the members. Indicate
whether they are in tension (T) or compression (C).
B
3m
A
C
10 kN
4m
Solution: (a) The free-body diagram of the system is shown. The
sum of the moments about B is: MB = 3Ax − 4(10) = 0, from
which Ax = 13.33 kN. The sums of the forces:
Fx = Ax + Bx = 0,
B
3m
A
from which Bx = −Ax = −13.33 kN .
Fy = By − 10 = 0,
4m
from which By = 10 kN . (b) The interior angle ACB is α =
tan−1 (0.75) = 36.87◦ . (b) Assume that the unknown forces act
away from the joint. Denote the axial force in the member I, K by
IK. The axial forces are FCB = BC(−i cos α + j sin α), and
FCA = −ACi. Summing the forces:
Fy = BC sin α − 10 = 0,
from which BC = 16.67 kN (T ) .
Fx = −BC cos α − AC = 0,
from which AC = −13.33 kN (C) . For the joint A,
BX
C
10 kN
BY
3m
AX
4m
Fy = AB = 0,
from which AB = 0
W
Problem 6.3 In Example 6.1, suppose that the 2-kN
load is applied at D in the horizontal direction, pointing from D toward B. What are the axial forces in the
members?
Solution: First, solve for the support forces and then use the
method of joints at each joint to solve for the forces.
AY
AX
6m
10 m
BX
θ
C
3m
D
D 2 kN
x
B
3m
tan θ = 6
10
θ = 30.96
θ
y
A
Fx :
Bx + Ax − 2 kN = 0
Fy :
Ay = 0
MB :
5m
B
Fy :
5m
0
AC /sin θ − BC sin θ − CD sin θ = 0
−BC − CD = 0
−6Ax = 0
Solving, we get BC = CD = 0
Joint B:
Solving, we get
Ax = Ay = 0,
AB
Bx = 2 kN
BC
θ
Joint A:
AY = 0
BX
θ = 30.96
θ
AB
AC cos θ = 0
Fy :
−AB − AC sin θ = 0
θ
Fy : (all forces zero) = 0
we have
AB = AC = BC = CD = 0
BD = −2 kN (c)
Note that we did not have to use joint D as we had already solved for the forces
there. The F BD at D is BD, with the (−) sign, is opposite the direction shown.
AC = 0 from
above
θ = 30.96°
AC
0
BC /cos θ + BD + Bx = 0
BD = −2 kN (compression)
Fx :
CD = 0
θ
C
CD
BC
Fx :
BD = −2 kN (c)
θ
θ
x
BD + 2 kN = 0
Solving, we get AB = AC = 0
Joint C:
Fx :
AC
θ = 30.96◦
BD
We already know AB = BC = 0 and Bx = 2 kN
AX = 0
2 kN
0
AC /cos θ − BC cos θ + CD cos θ = 0
−BC + CD = 0
D
2 kN
Problem 6.4 Determine the axial forces in the members of the truss.
A
0.3 m
2 kN
B
0.4 m
C
0.6 m
1.2 m
Solution: First, solve for the support reactions at B and C, and
then use the method of joints to solve for the forces in the members.
A
0.3 m
BY
B
A
2 kN
0.3 m
0.4 m
BX
C
0.4 m
0.6 m
0.6 m
tan γ =
Fx :
Bx + 2 kN = 0
Fy :
B y + Cy = 0
MB :
7
12
(BC = −0.961 kN)
γ = 30.26◦
0, 6Cy − (0, 3)(2 kN) = 0
Solving, Bx = −2 kN Cy = 1 kN By = −1 kN
Joint B:
BY (−1 kN)
1.2 m
1.2 m
CY
+
2 kN
Fx = −BC cos θ + AC cos γ = 0
Fy = BC sin θ + AC sin γ + 1 = 0
Solving, we get AC = −0.926 kN
y
We have
AB = 2.839 kN (T )
BC = −0.961 kN (C)
AC = −0.926 kN (C)
3
AB
BX (−2 kN)
φ
x
18
θ 6
Check: Look at Joint A
y
4
2 kN
BC
Fx :
AB cos φ + BC cos θ − 2 = 0
Fy :
AB sin φ − BC sin θ − 1 = 0
AB
Solving, we get AB = 2.839 kN, BC = −0.961 kN
Joint C:
γ
θ
12
1 kN
7
AC
γ
AC
BC
x
φ
Fx :
−AB cos φ − AC cos γ + 2 = 0
Fy :
−AB sin φ − AC sin γ = 0
Substituting in the known values, the equations are satisfied: ∴ Check!
Problem 6.5 (a) Let the dimension h = 0.1 m. Determine the axial forces in the members, and show that in
this case this truss is equivalent to the one in Problem 6.4.
(b) Let the dimension h = 0.5 m. Determine the axial
forces in the members. Compare the results to (a), and
observe the dramatic effect of this simple change in design on the maximum tensile and compressive forces to
which the members are subjected.
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 TP Q = TP Q eP Q = TP QX i + TP QY j where P and Q take
on the designations A, B, C, and D as needed.
Equilibrium yields
At joint A:
Fx = TABX + TACX = 0,
and
Fy = TABY + TACY − 1 kN = 0.
y
At joint D:
Fx = −TCDX − TBDX + DX = 0,
and
Fy = −TCDY − TBDY + DY = 0.
B
A
h
DX D
0.4 m
1 kN
0.7 m
DY
x
C
At joint B:
Fx = −TABX + TBCX + TBDX = 0,
and
Fy = −TABY + TBCY + TBDY = 0.
At joint C:
Fx = −TBCX − TACX + TCDX = 0,
and
Fy = −TBCY − TACY + TCDY + CY = 0.
1.2 m
0.6 m
CY
y
h
0.4
m
B
1.2 m
−TAB
TAB
−TBD
TBD T
BC
TAC
DX D
−TBC
T
−TCD CD
−TAC
DY
C
CY
0.6 m
A
1 kN
0.7 m
x
1.2 m
Solve simultaneously to get
TAB = TBD = 2.43 kN,
eAB = −0.986i + 0.164j,
TAC = −2.78 kN,
eAC = −0.864i − 0.504j,
TBC = 0, TCD = −2.88 kN.
eBC = 0i − 1j,
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.
(b) For this part of the problem, we set h = 0.5 m. The unit vectors
change because h is involved in the coordinates of point B. The new
unit vectors are
eBD = −0.768i − 0.640j,
and eCD = −0.832i + 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
TAB = 1.35 kN,
TAC = −1.54 kN,
TBC = −1.33,
TBD = 1.74 kN,
and TCD = −1.60 kN.
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.
Problem 6.6 The load F = 10 kN. Determine the
axial forces in the members.
B
3m
C
A
D
F
4m
Solution:
The free-body diagram of joint D is
TBD
α
4m
From the equations
Fx = −TAB cos α + TBD cos α = 0,
Fy = −TBC − TAB sin α − TBD sin α = 0,
we obtain
TAB = 1.67F = 16.7 kN,
TCD
F
TBC = −2F = −20 kN.
Joint C
where α = arctan(3/4) = 36.9◦ . From the equations
Fx = −TCD − TBD cos α = 0,
Fy = TBD sin α − F = 0,
TBC
TAC
TCD
we obtain
TBD = 1.67F = 16.7 kN,
C
TCD = −1.33F = −13.3 kN.
Joint B
we see that
TAB
TAC = TCD = −1.33F = −13.3 kN.
α
α
TBD
TBC
Problem 6.7 Consider the truss in Problem 6.6. Each
member will safely support a tensile force of 150 kN
and a compressive force of 30 kN. What is the largest
downward load F that the truss will safely support at D?
Solution:
See the solution of Problem 6.6. The largest tensile
load is 1.67F in members BD and AB. Setting
1.67F = 150 kN
gives F = 90 kN. The largest compressive load is 2F in member
BC. Setting
2F = 30 kN
gives F = 15 kN. The largest load is F = 15 kN.
Problem 6.8 The Howe and Pratt bridge trusses are
subjected to identical loads.
(a) In which truss does the largest tensile force occur?
In what member(s) does it occur, and what is its
value?
(b) In which truss does the largest compressive force
occur? In what member(s) does it occur, and what
is its value?
L
L
L
B
L
C
D
L
A
E
G
H
I
F
F
F
Howe
Solution: (a) Howe Bridge: The moment about A is MA =
−6F + 4E = 0, from which E = 32 F . Denote the axial force in the
member I, K by IK.
√
(1) Joint E: DE = − sinE45◦ = − 3 2 2 F (C)
EI = DE cos 45◦ =
(2)
3
F (T )
2
Joint I: CI =
3
F (T )
2
F −|DI|
sin 45◦
=−
L
G
F
2
F (C),
2
L
√
L
B
√
(3)
Joint E: DE = − 3 2 2 F (C), EI =
Joint D: DH = − 2DI −
DH
√
2
−DE
√
2
=
EI = 32 F (T )
√
DE = 22 F (T ),
= −2F (C)
G
F
F
L
members DE and AB = −
2
E
F
L
C
B
L
L
D
G
H
F
I
F
E
F
L
L
L
L
DC =
DE
√
2
AX
−
, and the highest compressive force occurs in
2
I
L
Joint H: GH = HI = 32 F (T ). The axial forces in the remaining members are determined from symmetry. In the Pratt Bridge,
the highest tensile force occurs in members EI, HI, GH, and
√
L
D
3
F (T )
2
(5)
3
L
C
H
A
Joint C: BC = DC = −2F (C), CH = 0
AG =
F
L
(4)
3
F (T )
2
F
L
A
(b) Pratt Bridge: The moment about A is MA = −6F +4E =
0, from which E = 32 F .
√
I
√
Joint H: CH = F (T ), GH = HI = 2F (T ).
By symmetry (the reaction at A has no x-component) the axial
forces in the other members are HG = HI, CG = CI, BG =
DI, CD = BC, AG = EI, and AB = DE. In the Howe
truss, the members HI and GH have the highest tensile force
Joint I: DI = F (T ), HI =
H
Pratt
highest compressive force AB = DE = − 3 2 2 F (C)
(2)
D
E
GH = HI = 2F (T ) and the members DE and AB have the
(1)
C
A
HI = EI − CI sin 45◦ = 2F (T )
(4)
L
L
B
Joint D: CD = DE cos 45◦ = − 32 F (C),
DI = −DE sin 45◦ =
(3)
L
L
F (C) . Thus (a) the Howe
bridge has the highest tensile force in a member, and (b) the
value of the
compressive force is the same in members DE and
√
AB = 3 2 2 F (C) for both bridges.
F
AY
F
F
E
Howe Bridge
CI
CD
DE
DE
45 °
45°
EI
E
(1) Joint E
DI
(2) Joint D
DI
CH
45°
HI
EI
F
(3) Joint I
GH
F
HI
(4) Joint H
Pratt Bridge
DE
DC
DC
DH 45° 45° DE BC
EI
EI
CH
DI
E HI F
(1) Joint E (2) Joint I (3) Joint D (4) Joint C
BH 45° 45°
HI
GH
F
(5) Joint H
45°
DI
Problem 6.9 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
300 mm
Solution: First, solve for the support reactions and then use the
method of joints to solve for the reactions in the members.
8 kN
B
400 mm
14 kN
0.4 m
0.4 m
0.4 m
+
C
E
H
G
FY
300 mm
B
Fx :
Bx = 0
Fy :
By + Fy − 8 − 14 = 0 (kN)
MB :
400 mm
14 kN
A
BY
400 mm
8 kN
0.8 m
400 mm
0.4 m
0.3 m
BX
F
D
(0.4)(8) + 0.8Fy − 1.2(14) = 0
Solving, we get Bx = 0, By = 5.00 kN Fy = 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.
Joint A:
400 mm
F
D
400 mm
400 mm
Fx :
−BC = 0
Fy :
−AC + CE = 0
400 mm
BC = 0,
Solving, we get
CE = 10.67 kN (T )
Joint B:
y
y
8 kN
4
θ
θ
3
5
BE
x
θ
AB
BC
AB
AC
BD
sin θ = 0.6 cos θ = 0.8 θ = 36.87◦
x
BY
Fx = AC + AB cos θ = 0
We know AB = −13.33 kN BC = 0 By = 5.00 kN.
We know 3 of the 5 forces at B Hence, we can solve for the other two.
BD + BE cos θ − AB cos θ = 0
Fx :
Fy :
BC + By + BE sin θ + AB sin θ = 0
Fy = −8 − AB sin θ = 0
Solving, we get AC = 10.67 kN (T )
AB = −13.33 kN (C)
Joint C: (Again, assume all forces are in tension)
Solving, we get
y
BD = −14.67 kN (C)
BE = 5.00 kN (T )
AC
From Joint C, we had BC = 0
CE
x
BC
[AC = 10.67 kN (T )]
Thus
BC = 0, BD = −14.67 kN (C)
BE
=
5.00 kN (T )
Problem 6.10 For the truss in Problem 6.9, determine
the axial forces in members DF , EF , and F G.
Solution: First, solve for the support reactions and then use the
method of joints to solve for the reactions in the members.
8 kN
14 kN
0.4 m
0.4 m
0.4 m
8 kN
14 kN
C
A
E
H
G
0.4 m
300 mm
0.3 m
BX
0.8 m
400 mm
400 mm
400 mm
Joint G:
y
Fx :
Bx = 0
Fy :
By + Fy − 8 − 14 = 0 (kN)
MB :
F
D
400 mm
FY
BY
+
B
EG
GH
x
G
(0.4)(8) + 0.8Fy − 1.2(14) = 0
FG
Solving, we get Bx = 0,
By = 5.00 kN
[GH = 18.67 kN (T )]
Fy = 17.00 kN
The forces we are seeking are involved with joints D, E, F , and G
The method of joints allows us to solve for two unknown forces at a
joint. We need to start with a joint with only two unknowns. Joints A
and H qualify. Joint H is nearest to the members we want to know
about, so let us start there. Assume all unknown forces are tensions. If
we get a negative force in a solution, this will then imply compression
in that member.
Fx :
GH − EG = 0
Fy :
FG = 0
Solving
EG = 18.67 kN (T )
FG = 0
Joint F:
y
y
14 kN
FG
EF
GH
H
θ
FH
x
FH
θ
θ
x
DF
Joint H
FY
3
sin θ = = 0.6
5
cos θ =
4
= 0.8
5
Fx :
−GH − F H cos θ = 0
Fy :
−14 − F H sin θ = 0
Solving GH = 18.67 kN (T ),
F H = −23.33 kN (C)
We know
F H = −23.33 kN (C)
FG = 0
Fy = 17.00 kN
Fx :
F H cos θ − EF cos θ − DF = 0
Fy :
F G + Fy + F H sin θ + EF sin θ = 0
Solving, we get
EF = −5.00 kN (C)
DF = −14.67 kN (C)
and from above F G = 0
Problem 6.11 The loads F1 = F2 = 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).
External loads:
3m
B
C
4m
F1 = 8 kN
D
y
G
A
E
4m
F2 = 8 kN
F1
D
G
A
AX
3m
x
C
AY
+
8m
3m
F2
B
E
GY
3m
Fx :
Ax + F1 + F2 = 0 (kN)
Fy :
Ay + Gy = 0
MA :
G
A
C
8Gy − 3F2 − 6F1 = 0
4m
Solving for the external loads, we get
Ax = −16 kN (to the left)
Solving,
Ay = −9 kN (downward)
4m
BD = 10 kN (T )
DE = −6 kN (C)
Gy = 9 kN (upward)
Joint E:
Now use the method of joints to determine BD, BE, and BG.
Start with joint D.
Joint D:
y
DE
y
F2 = 8 kN
BE
D
F1 = 8 kN
DE
θ
x
x
EG
BD
cos θ = 0.8
DE = −6 kN
sin θ = 0.6
θ = 36.87◦
Fx :
F1 − BD cos θ = 0
Fy :
−BD sin θ − DE = 0
Fx = DE − EG = 0
Fy = −BE + F2 = 0
Solving: EG = −6 kN (C)
BE = 8 kN (T )
6.11 Contd.
Joint G:
y
Fx :
−CG − BG cos θ = 0
Fy :
BG sin θ + EG + Gy = 0
Solving, we get
BG = −5 kN (C)
EG
BG
CG = 4 kN (T )
Thus, we have
θ
x
BD = 10 kN (T )
BE = 8 kN (T )
BG = −5 kN (C)
CG
GY
(EG = −6 kN (C))
Gy = 9 kN
Problem 6.12 If the loads on the truss shown in Problem 6.11 are F1 = 6 kN and F2 = 10 kN, what are the
axial forces in members AB, BC, and BD?
Solution:
Find the external support loads and then use the method
of joints to determine loads in members. (Assume all loads in members
to be tensions).
External Loads:
F1
D
3m
D
F1 = 6 kN
F2
B
y
E
B
3m
F2 = 10 kN
G
A
3m
θ
A
3m
C
x
AX
8m
4m
GY
AY
Joint A:
y
sin θ = 0.6 cos θ = 0.8 θ = 36.87◦
+
Fx :
Ax + F1 + F2 = 0
Fy :
Ay + Gy = 0
MA :
8Gy − 3F2 − 6F1 = 0
4m
AB
θ
AX
A
Solving, the external loads are
Ax = −16 kN,
AC
AY
Ay = −8.25 kN,
Gy = 8.25 kN.
Ay = −8.25 kN
Now use the method of joints to determine AB, BC, and BD.
Start with Joint A:
Ax = −16 kN
Fx :
AC + Ax + AB cos θ = 0
Fy :
Ay + AB sin θ = 0
x
6.12 Contd.
Solving,
AC = 5 kN (T )
AB = 13.75 kN (T )
Joint C:
y
BC
C
AC
x
CG
(AC = 5 kN)
Fx :
CG − AC = 0
Fy :
BC = 0
Solving,
BC = 0,
CG = 5 kN (T )
Joint D:
y
F1
x
θ
DE
BD
F1 = 6 kN
Fx :
F1 − BD cos θ = 0
Fy :
−BD sin θ − DE = 0
Solving, we get
DE = −4.5 kN (C)
and BD = 7.5 kN (T )
Thus, we have
AB = 13.75 kN (T )
BC = 0
BD = 7.5 kN (T )
Problem 6.13 The truss supports loads at C and E.
If F = 3 kN, what are the axial forces in members BC
and BE?
1m
1m
A
Solution:
The moment about A is
1m
B
D
1m
MA = −1F − 4F + 3G = 0,
G
C
5
F
3
from which G =
= 5 kN. The sums of forces:
FY = AY − 3F + G = 0,
E
F
from which AY = 43 F = 4 kN.
FX = AX = 0,
2F
from which AX = 0. The interior angles GDE, EBC are 45◦ ,
1m
1
from which sin α = cos α = √ .
2
A
Denote the axial force in a member joining I, K by IK.
(1) Joint G:
DG
Fy = √ + G = 0,
2
from which
1m
B
1m
G
C
AY
AX
1m
5
DG
EG = − √ = F = 5kN (T ).
3
2
F
1m
(2) Joint D:
DG
Fy = −DE − √ = 0,
2
1m
BD
EG
G
Joint G
5
F = 5 kN (T ).
3
5
BD = − F = −5 kN (C).
3
from which BE = 2 2F −
2DE =
BE
Fx = −CE − √ + EG = 0,
2
from which
4
BE
CE = EG − √ = F = 4 kN (T ).
3
2
DE
EG
Joint E
(4) Joint A:
AC
Fy = Ay − √ = 0,
2
(3) Joint E:
BE
Fy = √ − 2F + DE = 0,
2
√
BE
45°
CE
CE
F
Joint C
45
Joint A
from which
G
45° BC
AC
AB
AC
DG
DE 45°
Joint D
AY
DG
Fx = −BD + √ = 0,
2
√
2F
1m
DG
45°
from which
2F
√
from which
E
F
√
√
5 2
DG = − 2G = −
F = −5 2 kN (C).
3
DG
Fx = − √ − EG = 0,
2
DE =
1m
D
√
2
F
3
=
√
2 kN (T ).
from which AC =
4
√
3
2
√
F = 4 2 kN (T ).
AC
Fx = AB + √ = 0,
2
from which AB = − 43 F = −4 kN (C).
(5) Joint C:
AC
Fy = BC + √ − F = 0,
2
from which BC = F −
AC
√
2
= − 13 F = −1 kN (C).
Problem 6.14 Consider the truss in Problem 6.13.
Each member will safely support a tensile force of 28 kN
and a compressive force of 12 kN. Taking this criterion
into account, what is the largest safe (positive) value
of F ?
Solution: From the solution to Problem 6.14, the member with
the largest tensile force is EG = 53 F , from which F = 35 EG =
16.8 kN. The member with the largest compressive force is DG,
DG =
√
−5 2
F,
3
from which F =
5
3
√
2
DG =
36
√
5 2
= 5.09 kN
is the largest safe value.
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 360 lb at A and at B.
What are the resulting axial forces in members CF , DF ,
and F G?
Solution: Assume loads of 360 lbs at A and at B. Use the method
1 ft
1 ft
G
1 ft
F
2 ft
E
of joints, starting with A and B, to work through the structure.
Joint A:
D
C
8 in
B
y
A
AC
AX
x
A
360 lb
1 ft
1 ft
G
Fy :
1 ft
F
AC − 360 lb = 0
2 ft
AC = 360 lb
Fx :
Ax = 0
E
D
C
If Ax = 0, then Bx = 0 because the stretcher must be in equilibrium
Joint B:
8 in
B
BC
BD
φ
BX
B
360 lb
Fx :
Bx + BC cos φ = 0
Fy :
BC sin φ + BD − 360 = 0
Solving, BD = 360 lb, BC = 0
tan φ =
8
24
φ = 18.43◦
Bx = 0
A
6.15 Contd.
Joint C:
y
CF
θ
CD
C
x
φ
AC
BC
2
1
tan θ =
θ = 63.43◦
BC = 0,
AC = 360 lb
Fx :
−CD − BC cos φ − CF cos θ = 0
Fy :
CF sin θ − BC sin φ − AC = 0
Solving, CD = −180 lb (C)
CF = 402 lb (T )
Joint F:
y
F
FG
x
θ
θ
DF
CF
θ = 63.43◦
(CF = 402 lbs (T ))
Fx :
CF cos θ − DF cos θ − F G = 0
Fy :
−CF sin θ − DF sin θ = 0
Solving; we get
DF
FG
and
CF
= −402 lb (C)
= 360 lb (T )
from earlier
= 402 lb (T )
Problem 6.16 Upon learning of an upgrade in the helicopter’s engine, the engineer designing the truss shown
in Problem 6.15 does new simulations and concludes that
the downward forces the stretcher will exert at A and at
B may be as large as 400 lb. What are the resulting axial
forces in members DE, DF , and DG?
Solution: Assume loads of 400 lb at A and B. Use the method
of Joints, starting with A and B, and work through the structure.
Joint A:
1 ft
1 ft
G
y
1 ft
F
AY
2 ft
AX
E
x
A
Fx :
Ax = 0
Fy :
Ay − F1 = 0
Fx :
Bx + BC cos φ = 0
Fy :
BC sin φ + BD − F2 = 0
Solving, BC = 0, BD = 400 lb(T )
Joint C:
Ay = 400 lb.
Ax = 0
CF
If Ax = 0, then Bx = 0 for the stretcher not to move horizontally.
(Ax + Bx = 0)
Joint B:
y
θ
y
CD
BD
C
x
φ
AC
BC
BC
φ
BX
B
A
B
F1 = 400 lb
Solving,
C
8 in
F1 = 400 lb
D
x
tan θ =
F2
2
1
θ = 63.43◦
BC = 0, AC = 400 lb.
Bx = 0
F2 = 400 lb
8
tan φ =
24
φ = 18.43◦
Fx :
−CD − BC cos φ − CF cos θ = 0
Fy :
CF sin θ − BC sin φ − AC = 0
Solving, CD = −200 lb (C)
6.16 Contd.
Joint F:
y
FG
F
x
θ
θ
CF
DF
θ = 63.43◦
CF = 44.7 lb (T )
Fx :
CF cos θ − DF cos θ − F G = 0
Fy :
−CF sin θ − DF sin θ = 0
Solving, we get
CF = 447 lb (T )
DF = −447 lb (C)
F G = 400
Joint D:
y
DF
DG
θ
θ
DE
CD
x
BD
CD = −200 lb (C)
BD = 400 lb (T )
DF = −447 lb (C)
Fx :
CD − DE + DF cos θ − DG cos θ = 0
Fy :
DF sin θ + DG sin θ − BD = 0
Solving DE = −968 lb (C)
DG = 894 lb (T )
Thus,
DE = −800 lb (C)
DF = −447 lb (C)
DG = 894 lb (T )
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
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 α = tan−1 0.8 = 38.66◦ . The angle formed by member DB
with the positive x axis is 90◦ + α. The angle formed by member AB
with the positive x-axis is α.
Joint E:
Fy = −DE cos α − W = 0,
from which BE = 0.8W (T )
Joint D:
Fx = DE cos α + BD cos α − CD cos α = 0,
from which BD − CD = −DE.
Fy = −BD sin α + DE sin α − CD sin α = 0,
from which BD + CD = DE.
Problem 6.18 Consider the truss in Problem 6.17.
Each member will safely support a tensile force of 6 kN
and a compressive force of 2 kN. Use this criterion to
determine the largest weight W the truss will safely support.
Solution:
From the solution to Problem 6.17, the largest tensile force is in member AB, AB = 1.28W (T ), from which
6
W = 1.28
= 4.69 kN is the maximum safe load for tension.
The largest compressive forces occur in members DE and CD,
is the largest safe load for compression.
CD = DE = −1.28W (C) , BD = 0
Joint B:
Fx = BE − AB sin α − BD sin α = 0,
from which AB =
BE
sin α
= 1.28W (T )
Fy = −AB cos α − BC = 0,
from which BC = −AB cos α = −W (C)
Fy = −BE − DE sin α = 0,
DE = CD = 1.28W (C), from which W =
0.8 m
Solving these two equations in two unknowns:
from which DE = −1.28W (C) .
0.8 m
2
1.28
= 1.56 kN
Problem 6.19 The loads F1 = 600 lb and F2 =
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
G
MG = +6E − 4F1 − 8F2 = 0,
D
F2
from which
4F1 + 8F2
E =
= 800 lb.
6
α
6 ft
E
A
4 ft
4 ft
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.
Joint E:
Fy = E + AE = 0,
F1
GX
F2
GY
6 ft
from which AE = −E = −800 lb (C) .
E
4 ft
Fy = EG = 0,
EG
from which EG = 0.
Joint A:
Fy = −AE − AC cos α = 0,
from which AC = −
3 ft
α
The interior angle EAG is
6
α = tan−1
= 36.87◦ .
8
B
C
AE
= 1000 lb(T ).
0.8
Fy = AC sin α + AB = 0,
from which AB = −AC(0.6) = −600 lb(C).
Joint B:
Fy = BD sin α − AB − F1 = 0,
E
AC
AE
Joint E
from which BD =
α
AE
AB
4 ft
BD
α
BC
Joint A
F2 +AB
0.6
F2
AB
−300
0.6
α
CD
Joint B
=
F1
DG
Joint D
= −500 lb(C) .
Fx = −BC − BD cos α = 0,
from which BC = −BD(0.8) = 400 lb(T ).
Joint D:
Fy = −BD sin α − CD − F1 = 0,
from which CD = −F1 − BD(0.6) = −300 lb(C)
BD
Problem 6.20 Consider the truss in Problem 6.19. The
loads F1 = 450 lb and F2 = 150 lb. Determine the axial
forces in members AB, AC, and BC.
Solution: From the solution to Problem 6.19 the angle α =
2
36.87◦ and the reaction at E is E = 4F1 +8F
= 500 lb. Denote the
6
axial force in a member joining two points I, K by IK.
Joint E:
E
Fx = AE + E = 0,
Joint A:
Fx = −AE − AC cos α = 0,
from which AC = − AE
= 625 lb(T ) .
0.8
Fy = AC sin α + AB = 0,
from which AB = −AC(0.6) = −375 lb(C)
Joint B:
Fy = BD sin α − F2 − AB = 0,
from which BD =
AE
Joint E
Fy = EG = 0.
from which AE = −E = −500 lb(C).
EG
F2 +AB
0.6
= −375 lb(C)
Fx = −BC − BD cos α = 0,
from which BC = −BD(0.8) = 300 lb(T )
AC
AB
α
AE
Joint A
BD
α
BC
F2
AB
Joint B
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
AB
Solution: The common interior angle BAC = DCE =
EF D = CDB is α = tan−1 (1) = 45◦ .
Note cos α = sin α = √1 . Denote the axial force in a member
2
joining two points I, K by IK.
Joint F:
DF
Fy = − √ − W = 0,
2
√
from which DF = − 2W (C).
Fx
DF
= −EF − √ = 0,
2
1m
1m
1m
E
F
1m
C
D
m
1m
from which EF = W (T ).
A
B
Joint E:
CE
Fx = − √ + EF = 0
2
√
from which CE = 2W (T ).
EF
W
Joint F
CE
α
CD
CE
Fy = −ED − √ = 0,
2
from which ED = −W (C).
AC
DF
BD
FY = ED + √ − √ = 0,
2
2
√
from which BD = −2 2W (C).
DF
BD
FX = √ − √ − CD = 0,
2
2
from which CD = W (T )
Joint C:
AC
CE
Fx = − √ + √ + CD = 0,
2
2
√
from which AC = 2 2W (T )
CE α
DF
Joint D:
ED
α
AC
CE
Fy = − √ + √ − BC = 0,
2
2
from which BC = −W (C)
BC
Joint C
EF
ED
Joint E
BC α BD
AB
CD
α
DF
BD
Joint D
B
Joint B
Joint B:
BD
Fx = −AB + √ = 0,
2
from which AB = −2W (C)
This completes the determination of the axial forces in √
all nine members. The
maximum tensile force occurs in√member AC, AC = 2 2W (T ), from which
4
the safe load is W = √
= 2 = 1.414 kN. The maximum compression
2 2
√
occurs in member BD, BD = −2 2W (C), from which the maximum safe
1
√
load is W =
= 0.3536 kN. The largest mass m that can be safely
2
supported is m =
2
353.6
9.81
= 36.0 kg
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
E
G
I
6m
6m
6m
6m
B
D
F
H
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.
AY
50 kN
50 kN
6m
6m
6m
3m
50 kN
3m
Fy :
2m
x
A
IY
Ay + Iy − 4(50) = 0 (kN)
MA :
Solving
50 kN
−3(50) − 9(50) − 15(50) − 21(50) + 24 Iy = 0
C
6m
E
G
6m
6m
I
6m
AB = −180.3 kN
θ = 33.69◦
Ay = 100 kN
Iy = 100 kN
Joint A:
Fx :
BC cos θ + BD − AB cos θ = 0
Fy :
−50 − AB sin θ − BC sin θ = 0
Solving, BC = 90.1 kN (T )
y
BD = −225 kN (C)
AB
Joint C:
θ
y
A
x
AC
AY
tan θ =
θ
2
3
C
CE
AC = 150 kN (T )
AB cos θ + AC = 0
Fy :
AB sin θ + Ay = 0
AB = −180.3 kN (C)
BC = 90.1 kN (T )
AC = 150 kN (T )
Fx :
CE − AC + CD cos θ − BC cos θ = 0
Fy :
CD sin θ + BC sin θ = 0
Solving,
Joint B:
CE = 300 kN (T )
y
CD = −90.1 kN (C)
50 kN
Hence
B
θ
BD
x
θ
BC
AB
x
θ = 33.69◦
Fx :
Solving,
θ
AC
θ = 33.69◦
CD
BC
BC = 90.1 kN (T )
CD = −90.1 kN (C)
CE = 300 kN (T )
Problem 6.23 For the Warren truss in Problem 6.22,
determine the axial forces in members DF , EF ,
and F G.
Solution: In the solution to Problem 6.22, we solved for the forces
in AB, AC, BC, BD, CD, and CE. Let us continue the process.
We ended with Joint C. Let us continue with Joint D.
Joint D:
y
B
H
F
D
50 kN
2m
A
D
BD
DF
EG = 300 kN (T )
Note: The results are symmetric to this point!
Joint F:
BD = −225 kN (C)
y
CD = −90.1 kN (C)
50 kN
Fx :
DF − BD + DE cos θ − CD cos θ = 0
Fy :
−50 − CD sin θ − DE sin θ = 0
Solving,
I
6m
EF = 0
θ = 33.69◦
G
6m
Solving, we get
DE
CD
E
6m
x
θ
θ
C
6m
DF
x
θ
θ
DF = −300 kN (C)
FH
F
DE = 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:
y
EF
θ = 33.69◦
DF = −300 kN (C)
EF = 0
EF
DE
θ
θ
CE
E
FG
EG
x
Fx :
F H − DF + F G cos θ − EF cos θ = 0
Fy :
−50 − EF sin θ − F G sin θ = 0
Solving: F H = −225 kN (C)
F G = −90.1 kN (C)
θ = 33.69◦
CE = 300 kN (T )
DE = 0
Fx :
EG − CE + EF cos θ − DE cos θ = 0
Fy :
DE sin θ + EF sin θ = 0
Thus, we have
DF = −300 kN (C)
EF
=
0
F G = −90.1 kN (C)
Note-symmetry holds!
Problem 6.24 The Pratt bridge truss supports five
forces (F = 300 kN). The dimension L = 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
L
A
H
F
Solution:
diagram,
and
F
F
F
F
Find support reactions at A and H. From the free body
L
Fx = AX = 0,
Fy = AY + HY − 5(300) = 0,
L
L
G
I
J
K
L
M
G
θ I
A
L
8
AY
J
L
8
F
F
L=8m
and TAI = 750 kN.
L
K
L
8
L
8
F
y
TBI
θ
I
x
TBC = −1200 kN
and TBJ = 636 kN.
TIJ
TAI
x
F
AY
and TIJ = 750 kN.
From these equations,
HY
Joint I
TAB
TAI
L
8
L
H
F
F
F = 300 kN
y
A
M
L
8
Joint A
From these equations,
Joint B: From the free body diagram,
Fx = TBC + TBJ cos θ − TAB cos θ = 0,
Fy = −TBI − TBJ sin θ − TAB sin θ = 0.
L
E
B
TAB = −1061 kN
TBI = 300 kN
L
D
H
From these equations,
Joint I: From the free body diagram,
Fx = TIJ − TAI = 0,
Fy = TBI − 300 = 0.
L
C
A
MA = 6(8)HY − 300(8 + 16 + 24 + 32 + 40) = 0.
From these equations, AY = HY = 750 kN.
From the geometry, the angle θ = 45◦
Joint A: From the free body diagram,
Fx = AX + TAB cos θ + TAI = 0,
Fy = TAB sin θ + AY = 0.
L
B
Joint B
y
TBC
θ
θ
x
TBJ
TAB
TBI
Problem 6.25 For the Pratt bridge truss in Problem 6.24, determine the axial forces in members CD,
CJ, and CK.
Solution: Use all of the known values from Problem 6.24, and
start with Joint J.
Joint J: From the free body diagram,
Fx = TJK − TBJ cos θ − TIJ = 0,
Fy = TCJ + TBJ sin θ − 300 = 0.
From these equations,
TJK = 1200 kN
and TCJ = −150 kN.
Joint C: From the free body diagram,
Fx = TCD + TCK cos θ − TBC = 0,
Fy = −TCJ − TCK sin θ = 0.
From these equations,
TCD = −1350 kN
and TCK = 212 kN.
Joint J
y
TBJ
θ
TIJ
Joint C
y
TBC
TCJ
TJK
F
TCD
θ
x
TCJ
TCK
x
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
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.
The interior angles HIB and HJC differ. The pitch angle is
8
αPitch = tan−1
= 33.7◦ .
12
The length of the vertical members:
8
BH = 4
= 2.6667 ft,
12
from which the angle
2.6667
αHIB = tan−1
= 33.7◦ .
4
CI = 8
4 ft
J
K
4 ft
L
4 ft
4 ft
800 lb
600 lb
400 lb
D
600 lb
C
E
8 ft
400 lb
F
B
G
A
H
I
4 ft
4 ft
J
4 ft
8
= 5.3333 ft,
12
K
4 ft
L
4 ft
4 ft
800 lb
600 lb
400 lb
600 lb
400 lb
from which the angle
5.333
αIJC = tan−1
= 53.1◦ .
4
A
The moment about G:
G
4 ft 4 ft 4 ft 4 ft 4 ft 4 ft
MG = (4 + 20)(400) + (8 + 16)(600) + (12)(800) − 24A = 0,
from which A = 33600
= 1400 lb. Check: The total load is 2800 lb.
24
From left-right symmetry each support A, G supports half the total
load. check.
The method of joints: Denote the axial force in a member joining two
points I, K by IK.
Joint A:
Fy = AB sin αP + 1400 = 0,
1400
from which AB = − sin
= −2523.9 lb (C)
α
AB
1400 lb
AH
Joint A
BI
α Pitch
BH
α Pitch
CI
HI
IJ
Joint I
AH
HI
400 lb
α Pitch
AB
Joint H
600 lb
CD
α Pitch
α IJC
BC CI
CJ
Joint C
p
4 ft
Fx = AB cos αPitch + AH = 0,
from which AH = (2523.9)(0.8321) = 2100 lb (T )
Joint H:
Fy = BH = 0, or, BH = 0.
Fx = −AH + HI = 0,
from which HI = 2100 lb (T )
Joint B:
Fx = −AB cos αPitch + BC cos αPitch
+BI cos αPitch = 0,
from which BC + BI = AB
BC
α Pitch
BH BI
Joint B
6.26 Contd.
Fy = −400 − AB sin αPitch + BC sin αPitch
−BI sin αPitch = 0,
from which BC − BI = AB +
400
.
sin αPitch
Solve the two simultaneous equations in unknowns BC, BI:
BI = −
400
= −360.56 lb (C),
2 sin αPitch
BC = AB − BI = −2163.3 lb (C)
and
Joint I:
Fx = −BI cos αPitch − HI + IJ = 0,
from which IJ = 1800 lb (T )
Fy = +BI sin αPitch + CI = 0,
from which CI = 200 lb (T )
Joint C:
Fx = −BC cos αPitch + CD cos αPitch + CJ cos αIJC = 0,
from which CD(0.8321) + CJ(0.6) = −1800
Fy = −600 − CI − BC sin αPitch + CD sin αPitch
−CJ sin αIJC = 0,
from which CD(0.5547) − CJ(0.8) = −400
Solve the two simultaneous equations to obtain CJ
−666.67 lb (C),
and
CD = −1682.57 lb (C)
=
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 α = 45◦ . Determine the axial
forces in the members of the truss.
C
B
F
2.2 m
The included angles
4
γ = tan−1
= 49.64◦ ,
3.4
2.2
β = tan−1
= 32.91◦ ,
3.4
1.8
θ = tan−1
= 27.9◦ .
3.4
G
H
A
α
E
3.4 m
Solution:
D
1.8 m
3.4 m
3.4 m
3.4 m
The complete structure as a free body: The sum of the moments about
A is
D
B
C
1.8 m
2.2 m
F
E
G H
α
A
A 3.4 m 3.4 m 3.4 m 3.4 m
MA = −(75)(3.4)(1 + 2 + 3) + (4)(3.4)Ey = 0.
75 kN 75 kN 75 kN
with this relation and the fact that Ex cos 45◦ + Ey cos 45◦ = 0, we
obtain Ex = −112.5 kN and Ey = 112.5 kN. From
FxA = Ax + Ex = 0, AX = −EX = 112.5 kN.
FyA = Ay − 3(75) + Ey = 0,
AY 3.4
m
from which Ay = 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.
The method of joints: Denote the axial force in a member joining two
points I, K by IK.
Joint A:
Fx = AB cos γ + Ax + AF cos β = 0,
Fy = AB sin γ + Ay + AF sin β = 0,
AB
AX
and
AF = −44.67 kN (C) ,
AB = −115.8 kN (C)
Joint E:
Fy = −DE cos γ + Ex − EH cos β = 0.
Fy = DE sin γ + Ey + EH sin β = 0,
β γ
EH
Solve:
and
γ β
CD
θ
DG
EY
Joint E
75 kN
γ
DH
Joint D
DE
EH = −44.67 kN(C) ,
DE = −115.8 kN(C)
Joint F:
Fx = −AF cos β + F G = 0,
from which two simultaneous equations are obtained.
from which F G = −37.5 kN (C)
3.4
m
3.4
m
BF
DE
EX
AF
AY
Joint A
75 kN
BC
θ
γ
BF BG
AB
Joint B
from which two simultaneous equations are obtained.
Solve:
EX
AX
Fy = −AF sin β + BF = 0,
from which BF = −24.26 kN (C)
Joint H:
Fx = EH cos β − GH = 0,
β
AF
EY
3.4
m
DH
FG
Joint F
75 kN
BC
CD
CG
Joint C
GH
β
EH
Joint H
6.27 Contd.
from which GH = −37.5 kN (C)
from which DG = 80.1 kN (T )
Fy = −EH sin β + DH = 0,
Fx = DE cos γ − CD − DG cos θ = 0,
from which DH = −24.26 kN (C)
Joint B:
Fy = −AB sin γ − BF + BG sin θ − 75 = 0,
from which CD = −145.8 kN (C)
Joint C:
Fx = CD − BC = 0,
from which BG = 80.1 kN (T )
from which CD = BC Check.
Fy = −CG − 75 = 0,
Fx = −AB cos γ + BC + BG cos θ = 0,
from which BC = −145.8 kN (C)
Joint D:
Fy = −DE sin γ − DH − DG sin θ − 75 = 0,
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)
from which CG = −75 kN (C)
1000 lb
C
2000 lb
D
4 ft
A
5 ft
Solution:
2 ft
B
5 ft
5 ft
Problem 6.28 don’t have unique solution
Problem 6.29 (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).
3m
2m
A
2m
B
1m
C
D
2 kN
Solution:
Problem 6.29 don’t have unique solution
2 kN
Problem 6.30 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.
(b) Use the method of sections to determine the axial
force in member DG.
A
B
C
D
E
F
G
H
1m
J
100 kN
Solution:
(a) Start with Joint J
A
B
C
D
E
F
G
H
DJ
1m
45°
J
HJ
1m
1m
1m
J
1m
100 kN
100 kN
◦
Fx :
−HJ − DJ cos 45 = 0
Fy :
DJ sin 45◦ − 100 = 0
Solving
Fx :
−CD − DG cos 45◦ + DJ cos 45◦ = 0
Fy :
−DG sin 45◦ − DJ sin 45◦ − DH = 0
CD = 200 kN
Solving,
DG = −141.4 kN (C)
DJ = 141.4 kN (T )
(b) Method of Sections
HJ = −100 kN (C)
CD
D
Joint H:
45°
1m
DG
DH
J
HJ
GH
GH
1m
H
Fx :
HJ − GH = 0
Fy :
DH = 0
DH = 0,
GH = −100 kN (C)
Joint D
CD
H
D
x
45°
45°
Fx :
−CD − DG cos 45◦ − GH = 0
Fy :
−DG sin 45◦ − 100 = 0
MD :
Solving,
−(1)GH − (1)(100) = 0
GH = −100 kN (C)
CD = 200 kN (T )
DG = −141.4 kN (C)
DJ
DH
DG
100 kN
Problem 6.31 For the truss in Problem 6.30, use the
method of sections to determine the axial forces in members BC, CF , and F G.
A
B
C
D
E
F
G
H
1m
Solution:
BC
C
J
D
100 kN
45°
1m
CF
Solving
CF = −141.4 kN (C)
J
F
FG
G
H
1m
BC = 300 kN (T )
F G = −200 kN (C)
1m
100 kN
Fx :
−BC − CF cos 45 − F G = 0
Fy :
−CF sin 45◦ − 100 = 0
MC :
−(1)F G − 2(100) = 0
Problem 6.32 Use the method of sections to determine
the axial forces in members AB, BC, and CE.
1m
1m
A
Solution:
B
1m
D
First, determine the forces at the supports
B
θ = 45°
D
1m
AX
G
AY
C
1m
F
E
1m
C
1m
θ
E
F
1m
GY
2F
2F
+
1m
Fx :
Ax = 0
Fy :
Ay + Gy − 3F = 0
MA :
1m
A
B
1m
D
−1(F ) − 2(2F ) + 3Gy = 0
1m
Solving
Ax = 0
Gy = 1.67F
G
Ay = 1.33F
C
Method of Sections:
E
F
y
AY = 1. 33 F
2F
AX = 0
AX = 0
AB
B
BC
+
1m
AY
Fx :
CE + AB = 0
Fy :
BC + Ay − F = 0
MB :
(−1)Ay + (1)CE = 0
Solving, we get
1m
C
F
CE
x
AB = −1.33F (C)
CE = 1.33F (T )
BC = −0.33F (C)
Problem 6.33 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
Solution:
F
D
400 mm
400 mm
400 mm
First find the external support loads on the truss
24 kN
18 kN
18 kN
24 kN
C
E
C
A
E
G
H
G
H
300 mm
0.8 m
B
F
D
BX
D
FY
BY
400 mm
0.4 m
1.2 m
Fx :
Bx = 0
Fy :
By + Fy − 18 − 24 = 0 (kN)
MB :
+
0.8Fy − (1.2)(24) + (0.4)18 = 0
Solving,
Solving:
Bx = 0
By = 15 kN
Fy = 27 kN
Method of sections:
18 kN
C
CE
E
BE
0.4 m
BD
BY
tan θ =
0.3 m
θ
θ
3
4
θ = 36.87◦
By = 15 kN
400 mm
400 mm
400 mm
Fx :
CE + BE cos θ + BD = 0
Fy :
By − 18 + BE sin θ = 0
MB :
+(0.4)(18) − (0.3)(CE) = 0
CE = 24 kN (T )
BE = 5 kN (T )
BD = −28 kN (C)
Problem 6.34 For the truss in Problem 6.33, use the
method of sections to determine the axial forces in members EG, EF , and DF .
Solution:
From the solution to Problem 6.33, the external forces
at B and F are
Bx = 0,
By = 15 kN,
Fy = 27 kN.
24 kN
18 kN
C
A
E
G
H
300 mm
B
F
D
400 mm
400 mm
400 mm
400 mm
24 kN
G
EG
E
H
EF
0.3 m
θ = 36.87°
θ
D
DF
F
FY
0.4 m
Fx :
−EG − EF cos θ − DF = 0
Fy :
−24 + Fy + EF sin θ = 0
MF :
Solving:
−(0.4)(24) + (0.3)EG
EG = 32 kN (T )
EF = −5 kN (C)
DF = −28 kN (C)
Problem 6.35 For the Howe and Pratt trusses, use
the method of sections to determine the axial force in
member BC.
L
L
L
B
L
C
D
L
A
E
G
H
I
2F
2F
F
Howe
L
L
L
L
B
C
D
L
A
E
G
H
I
2F
2F
F
Pratt
Solution: From the free body diagram of the whole truss, the
equations of equilibrium are
Fx = AX = 0,
Fy = AY + EY − 5F = 0,
and
MA = 4LEY − LF − (2L)2F − (3L)2F = 0.
L
L
D
E
G
F
L
L
L
I
2F
H
2F
Howe
L
B
L
C
D
L
A
TBCHowe = −2.25F (compression).
E
G
F
Pratt Section: From the Pratt truss section we see that summing moments about H is advantageous. Hence,
MH = −2LAY + LF − LTBCPratt = 0,
or
L
C
A
From these equations, we get AX = 0, AY = 2.25F, and EY =
2.75F . Note that the support forces are the same for the Howe and
Pratt trusses.
Howe Section: From the Howe truss section, we see that if we sum
moments about G, we get one equation in one unknown, i.e.,
MG = −LAY − LTBCHowe = 0,
or
L
B
H
2F
I
2F
Pratt
TBCPratt = −3.5F (compression).
y
B
L
C
D
L
L
L
AX
A
AY
y
AX
AY
B
G
G
F
I
2F
2F
TBC
TGC
TGH x
F
HOWE
H
y
B
L
E
EY x
TBC
TBH
AX
AY
G T H x
GH
PRATT
Problem 6.36 For the Howe and Pratt trusses in Problem 6.35, determine the axial force in member HI.
Solution: Howe Section: From the Howe truss section, we see
that if we sum moments about C, we get one equation in one unknown, i.e.,
MC = 2LEY − 2LF − LTHIHowe = 0,
or
THIHowe = 3.5F (tension).
C TCD D
TCI
I
E
2F
EY
x
H THI
Pratt Section: From the Pratt truss section we see that summing moments about D is advantageous. Hence,
MD = LEY − LTHIPratt = 0
or
y
HOWE
THIPratt = 2.75F (tension).
TCD
y
D
THD I
E
H THI
EY
2F
x
PRATT
Problem 6.37 The Pratt bridge truss supports five
forces F = 340 kN. The dimension L = 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
L
A
H
F
Solution:
F
F
F
First determine the external support forces.
L
AX
F
L
L
L
L
L
L
L
L
L
L
B
C
D
E
G
I
J
K
L
M
L
L
F
AY
F
F
F
F
HY
A
H
F = 340 kN, L = 8 M
+
Solving:
F
Fx :
Ax = 0
Fy :
Ay − 5F + Hy = 0
MA :
F = 340 kN
Ay = 850 kN
Ay = 850 kN
Hy = 850 kN
Note the symmetry:
Method of sections to find axial force in member JK.
C
θ
A
L
CD
D
CK
K
AY
F
+
Fx :
CD + JK + CK cos θ = 0
Fy :
Ay − 2F − CK sin θ = 0
MC :
L(JK) + L(F ) − 2L(Ay ) = 0
Solving,
JK = 1360 kN (T )
Also,
CK = 240.4 kN (T )
CD = −1530 kN (C)
J
L
F
F
L = 8M
Ax = 0,
I
F
θ = 45◦
6LHy − LF − 2LF − 3LF − 4LF − 5LF = 0
B
F
JK
F
Problem 6.38 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 6.37, the support forces
are Ax = 0, Ay = Hy = 850 kN.
Method of Sections to find axial force in EK.
E
DE
L
L
E
G
I
J
K
L
M
H
F
L
F
F
HY
Fx :
−DE − EK cos θ − KL = 0
Fy :
Hy − 2F − EK sin θ = 0
ME :
L
D
A
KL
L
C
L
EK
L
B
G
θ
L
F
F
Solution:
EK = 240.4 kN (T )
Also,
KL = 1360 kN (T )
F
F
DE = −1530 kN (C)
−(L)(KL) − (L)(F ) + (2L)Hy = 0
Problem 6.39 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
E
G
I
6m
6m
6m
6m
B
D
F
H
Solution: First, find the external support forces. By symmetry,
Ay = Iy = 100 kN (we solved this problem earlier by the method of
joints).
50 kN
B
y
D
BD
2m
A
6m
C
CD
θ
CE
2m
A
x
C
6m
E
6m
AY
Solving: CE = 300 kN (T )
2
tan θ =
3
Also,
θ = 33.69◦
Fx :
BD + CD cos θ + CE = 0
Fy :
Ay − 50 + CD sin θ = 0
MC :
−6Ay + 3(50) − 2BD = 0
BD = −225 kN (C)
CD = −90.1 kN (C)
G
6m
I
6m
Problem 6.40 The walkway in Problem 6.39 exerts
equal vertical loads on the Warren truss at B, D, F ,
and H. Use the method of sections to determine the
maximum allowable value of each vertical load if the
magnitude of the axial force in member F G is not to
exceed 100 kN.
B
D
F
H
2m
C
A
6m
E
6m
G
6m
I
6m
Solution: Let the loads at B, D, F , and H be denoted by W . By
summetry Ay = Iy = 2W .
Method of Sections
W
H
FH
F
FG
θ
2m
EG
G
tan θ =
3m
I
3m
IY
2
3
θ = 33.69◦
Fx :
−EG − F H − F G cos θ = 0
Fy :
Iy − W + F G sin θ = 0
MG :
2F H + 6Iy − 3W = 0
We set F G = ±100 kN and solve:
For F G = +100 kN, W = −55.5 kN (this implies an upward load
on the bridge)
For F G = −100 kN (in compression)
W = 55.5 kN.
This is the load limit on the bridge based on the load in member F G.
Problem 6.41 The mass m = 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
AB
Solution: First, find the support reactions using the first free body
diagram. Then use the section shown in the second free body diagram
to determine the forces in the three members.
Support Reactions: Equilibrium equations are
Fx = AX = 0,
Fy = AY + BY − mg = 0,
1m
F
1m
120 g
C
D
1m
AX
B
BY
AY
and, summing moments around C,
MC = (1)TBD cos(45◦ ) − (1)AY + (1)AX = 0.
1m
Solving, we get TBD = −3.30 kN,
1m
1m
1m
1m
E
F
TCE
C
TCD = 1.18 kN,
1m
E
and summing moments around A,
MA = −3mg + (1)BY = 0.
Thus, AX = 0, AY = −2.35kN, and BY = 3.53 kN
Section: From the second free body diagram, the equilibrium equations
for the section are
Fx = AX + TCD + TCE cos(45◦ ) + TBD cos(45◦ ) = 0,
Fy = AY + BY + TCE sin(45◦ ) + TBD sin(45◦ ) = 0,
1m
D
TCD
TBD
B
AX
TCE = 1.66 kN.
BY
AY
Problem 6.42 For the truss in Problem 6.41, use the
method of sections to determine the axial forces in members AC, BC, and BD.
Solution: Use the support reactions found in Problem 6.41. The
free body diagram for the section necessary to find the three unknowns
is shown at right. The equations of equilibrium are
Fx = AX + TAC cos(45◦ ) + TBD cos(45◦ ) = 0,
Fy = AY + BY + TBC + TAC sin(45◦ ) + TBD sin(45◦ ) = 0,
and, summing moments around B,
MB = (−1)AY − (1)TAC sin(45◦ ) = 0.
The results are
TAC = 3.30 kN,
TBC = −1.18 kN,
and TBD = −3.30 kN.
1m
1m
E
1m
1m
1m
TAC
C
TBC
AX
AY
BY
D
TBD
F
Problem 6.43 The Howe truss helps support a roof.
Model the supports at A and G as roller supports.
(a) Use the method of joints to determine the axial force
in member BI.
(b) Use the method of sections to determine the axial
force in member BI.
2 kN
2 kN
2 kN
D
2 kN
2 kN
C
E
4m
B
F
A
G
H
2
m
I
J
K
L
2m
2m
2m
2m
2m
Solution:
The pitch of the roof is
4
α = tan−1
= 33.69◦ .
6
2 kN
2 kN
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
2 kN
MA = −2(2)(1 + 2 + 3 + 4 + 5) + 6(2)G = 0,
B
30
6
from which G =
= 5 kN. The sum of the forces:
FY = A − 5(2) + G = 0,
from which AB =
=
−5
0.5547
4m
G
H
I
2m
2m
J
2m
K
2m
L
2m
2m
F
F
F
F = 2 kN
F
= −9.01 kN (C).
Fx = AB cos α + AH = 0,
G
A
from which AH = −AB cos α = 7.5 kN (T ).
Joint H:
2m 2m 2m 2m 2m 2m
Fy = BH = 0.
Joint B:
BH
AB
Fx = −AB cos α + BI cos α + BC cos α = 0,
Fy = −2 − AB sin α − BI sin α + BC sin α = 0.
(a)
A
α
AH
AH
HI
2 kN
from which HI =
3
A
2
= 7.5 kN(T ). The sum of the forces:
BH
Joint B
Joint H
Fx = BC cos α + BI cos α + HI = 0,
Fy = A − F + BC sin α − BI sin α = 0.
Solve: BI = −1.803 kN (C) .
F
B
α
(b)
BC
α
BI
α
AB
Solve: BI = −1.803 kN (C) , BC = −7.195 kN (C)
Make the cut through BC, BI and HI. The section as a free
body: The sum of the moments about B:
MB = −A(2) + HI(2 tan α) = 0,
2 kN
F
Joint A
(b)
E
C
A
from which A = 10 − 5 = 5 kN.
The method of joints: Denote the axial force in a member joining I,
K by IK.
(a) Joint A:
Fy = A + AB sin α = 0,
−A
sin α
2 kN
D
A
BC
α
α
BI
HI
2m
Problem 6.44 Consider the truss in Problem 6.43. Use
the method of sections to determine the axial force in
member EJ.
Solution:
From the solution to Problem 6.43, the pitch angle is
α = 36.69◦ , and the reaction G = 5 kN. The length of member EK
is
LEK = 4 tan α =
16
= 2.6667 m.
6
DE
β
F
E
F
EJ
α
JK
The interior angle KJE is
β = tan−1
LEK
2
2m
= 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 = +4G − 2(F ) − JK(2.6667) = 0,
from which JK =
20−4
2.6667
= 6 kN (T ).
The sum of the forces:
Fx = −DE cos α − EJ cos β − JK = 0.
Fy = DE sin α − EJ sin β − 2F + G = 0,
from which the two simultaneous equations:
0.8321DE + 0.6EJ = −6,
0.5547DE − 0.8EJ = −1.
Solve: EJ = −2.5 kN (C) .
2m
G
Problem 6.45 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:
α = tan−1
The included angle at the apex BAC is
12
= 36.87◦ .
16
The interior angles BCA, DEC, F GE, HIG are γ = 90◦ − α =
53.13◦ . The length of the member ED is LED = 8 tan α = 6 ft.
The interior angle DEF is
4
β = tan−1
= 33.69◦ .
LED
The complete structure as a free body: The moment about H is MH =
−10(12) − 10(16) + I(12) = 0, from which I = 280
= 23.33 kip.
12
The sum of forces:
Fy = Hy + I = 0,
from which Hy = −I = −23.33 kip.
Fx = Hx + 20 = 0,
from which Hx = −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 = −10(4) − 10(8) + DF (LED ) = 0,
from which DF = 120
= 20 kip.
6
The sum of forces:
Fy = −EF sin β − EG sin γ − DF = 0,
10 kip
A
4 ft
10 kip
C
B
4 ft
E
D
4 ft
G
F
4 ft
I
H
12 ft
F = 10 kip
F = 10 kip
Fx = −EF cos β + EG cos γ + 20 = 0,
from which the two simultaneous equations: 0.5547EF + 0.8EG =
−20, and 0.8320EF − 0.6EG = 20. Solve: EF = 4.0 kip (T )
DF
β
EF
E
γ
EG
Problem 6.46 Consider the truss in Problem 6.45. Use
the method of sections to determine the axial force in
member F G.
From the solution of Problem 6.45, the apex A included
angle is α = 36.87◦ . The length of the member F G is LF G =
12 tan α = 9 ft. Make the cut through EG, GF , and F H, and
consider the upper section. Denote the axial force in a member joining
I, K by IK. The section as a free body: The cut in EG is made
very near the point G; the moment about this cut by MG = −(8 +
12)F + LF G F H = 0 (where F = 10 kip from Problem 6.45), from
which F H = 22.22 kip (T ). The sum of the forces, from which
EG = −27.78 kip(C).
Fx = F G + 2F + EG cos β = 0,
Solution:
F = 10
F = 10
γ
EG
9 ft
FH
from which F G = −3.33 kip (C) .
Problem 6.47 The load F = 20 kN and the dimension
L = 2 m. Use the method of sections to determine the
axial force in member HK.
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.
G
FG
L
L
A
B
C
F
L
E
D
F
G
L
I
Solution: The complete structure as a free body: The sum of the
moments about K is MK = −F L(2 + 3) + M L(2) = 0, from
which M = 5F
= 50 kN. The sum of forces:
2
FY = KY + M = 0,
H
J
K
M
L
from which KY = −M = −50 kN.
FX = KX + 2F = 0,
from which KX = −2F = −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:
Fx = Kx − HI + IJ = 0,
from which HI − IJ = Kx . Sum moments about K to get MK =
M (L)(2) + JM (L)(2) − IJ(L) + HI(L) = 0.
Substitute HI − IJ = Kx , to obtain JM = −M − K2x =
−30 kN (C).
Fy = Ky + M + JM + HK = 0,
L
L
B
A
C
L
E
D
G
L
I
J
H
L
M
K
from which HK = −JM = 30 kN(T )
2L
F
L
F
2L
KX
M
KY
HI
HK
KX
HJ
JM
M
KY
2L
L
Problem 6.48 The weight of the bucket is W =
1000 lb. The cable passes over pulleys at A and D.
(a) Determine the axial forces in member F G and HI.
(b) By drawing free-body diagrams of sections, explain
why the axial forces in members F G and HI are
equal.
D
A
C
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 α = 35◦ relative to the horizontal.
The angles of the members F G and HI relative to the horizontal are
β = 45◦ + 35◦ = 80◦ . (a) Make the cut through F H, F G, and EG,
and consider the upper section. Denote the axial force in a member
joining, α, β by αβ.
The section as√a free body: The perpendicular distance from point F
is LF W = 3 2 sin β + 3.5 = 7.678 ft.
The sum of the moments about F is MF = −W LF W + W (3.25) −
|EG|(3) = 0, from which EG = −1476.1 lb (C).
The sum of the forces:
FY = −F G sin β − F H sin α − EG sin α − W sin α − W = 0,
FX = −F G cos β − F H cos α − EG cos α − W cos α = 0,
from which the two simultaneous equations:
−0.9848F G − 0.5736F H = 726.9, and −0.1736F G −
0.8192F H = −389.97.
Solve: F G = −1158.5 lb (C) , and F H = 721.64 lb (T ). Make
the cut through JH, HI, and GI, and consider the upper section.
The section as a free body: The perpendicular distance from point
H√ to the line of action of the weight is LHW = 3 cos α +
3 2 sin β + 3.5 = 10.135 ft. The sum of the moments about
H is MH = −W (L) − |GI|(3) + W (3.25) = 0, from which
|GI| = −2295 lb (C).
FY = −HI sin β − JH sin α − GI sin α − W sin α − W = 0,
FX = −HI cos β − JH cos α − GI cos α − W cos α = 0,
D
F
J
3 ft
E
G
L
3 ft 3 in
I
35°
3 ft
K
W
FH
β
3.25 ft
W
α
3 ft
FG
EG
W
HI = −1158.5 lb(C) ,
3.5 ft
W
−0.9848HI − 0.5736JH = 257.22,
Solve:
B
3 ft 6 in
H
from which the two simultaneous equations:
and −0.1736HI − 0.8192JH = −1060.8.
A
C
JH
HI
GI
and
JH = 1540.6 lb(T ) .
(b) Choose a coordinate system with the y-axis parallel to JH. Isolate a section
by making cuts through F H, F G, 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 F G and HI, the two axial forces must be equal:
Fx = HI cos 45◦ − F G cos 45◦ = 0,
from which HI = F G
Problem 6.49 Consider the truss in Problem 6.48. The
weight of the bucket is W = 1000 lb. The cable passes
over pulleys at A and D. Determine the axial forces in
members IK and JL.
W
W
β
JL
α
3.5 ft
3.25 ft
Make a cut through JL, JK, and IK, and consider the
upper section. Denote the axial force in a member joining, α, β by αβ.
The section as a free body: The perpendicular distance
√ from point J to
the line of action of the weight is L = 6 cos α + 3 2 sin β + 3.5 =
12.593 ft. The sum of the moments about J is MJ = −W (L) +
W (3.25) − IK(3) = 0, from which IK = −3114.4 lb(C).
The sum of the forces:
Fx = JL cos α − IK cos α
Solution:
and
Fy = −JL sin α − IK sin α
IK
from which two simultaneous equations:
0.8192JL + 0.1736JK = −1732
and
−W cos α − JK cos β = 0,
JK
3 ft
0.5736JL + 0.9848JK = 212.75.
JL = 2360 lb(T ) ,
Solve:
−W sin α − W − JK sin β = 0,
JK = −1158.5 lb(C) .
and
Problem 6.50 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
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, α, β by αβ.
The section as a free body: The moment about M is
Solution:
MM = −IL − 2(1) − 4(2) − 6(1) = 0,
H
F
D
The angle of member IM is α =
The sums of the forces:
Fy = −IM sin α − 4 = 0,
=
6m
KM
α
from which KM = 24 kN(T )
1m
IM
IL
1 kN
2m
2m 2m 2m 2m 2m
K
J
I
2m
M
L
O
Q
N
P
R
G
H
2m
1 kN 2 kN 1 kN
F
2m
D
2m
A
E
C
B
6m
B
26.57◦ .
from which IM = − sin4 α = −8.944 kN (C).
Fx = −KM − IM cos α − IL = 0,
1m
C
2m
A
tan−1 (0.5)
E
2m
IL = −16 kN (C) .
from which
G
2m
2 kN
2m
1 kN
2m
1 kN
Problem 6.51 Consider the truss in Problem 6.50. 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 structure as afree
body: The angle formed by AJ with
the vertical is α = tan−1 48 = 26.57◦ . The moment about B is
MB = 6AJ cos α − 24 = 0, from which AJ = 4.47 kN (T ).
The section as a free body: The angles of members HJ and HI
relative to the vertical are β = tan−1 28 = 14.0◦ , and γ =
tan−1 1.5
= 36.87◦ respectively. Make a cut through the four
2
members AJ, HJ, HI, and GI, and consider the upper section. The
moment about the point I is MI = −24+2AJ cos α+2HJ cos β =
1m
AJ
HJ
αβ
γ
HI
2m
I
1 kN
GI
2m
2m
2 kN
2m
1 kN
2m
0. From which HJ = 8.25 kN (T ) . The sums of the forces:
Fx = −AJ sin α + HJ sin β − HI sin γ = 0,
sin β
2−2
from which HI = AJ sin α−HJ
= sin
= 0.
sin γ
γ
FY = −AJ cos α − HJ cos β − HI cos γ − GI − 4 = 0,
from which GI = −16 kN (C)
Problem 6.52 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
Solution: Since BC is a two force member, the force in BC must
be a long the line between B and C.
0.3 m
AX
0.3 m
B
FBC
400 mm
200 N
B
A
y
200 N
300 mm
x
400 mm
AY
0.4 m
C
45°
0.4 m
Fx :
Ax + FBC cos 45◦ = 0
Fy :
Ay − FBC sin 45◦ − 200 = 0
MA :
Solving:
−(0.3)(200) − (0.6)CFBC sin 45◦ = 0
Ax = 100 N,
Ay = 100 N
FBC = 141.2 N (compression)
300 mm
300 mm
400 mm
Problem 6.53 (a) Determine the forces and couples
on member AB for cases (1) and (2).
(b) You know that the moment of a couple is the same
about any point. Explain why the answers are not the
same in cases (1) and (2).
200 N-m
A
B
C
1m
1m
(1)
200 N-m
A
B
C
1m
1m
(2)
Solution: Case (a) Element BC: The moment about B is
MB = (1)Cy = 0, hence Cy = 0 . The sum of the forces:
200 N-m
A
B
C
Fy = By + Cy = 0,
from which By = 0 .
1m
Fx = Bx = 0.
Element AB: The sum of the moments about A: M = MA −200 = 0,
from which MA = 200 N-m . The sum of forces:
Fy = By + Ay = 0,
(a)
A
from which Cy = 200 N. The sum of the forces:
Fy = ByBC + Cy = 0,
from which ByBC = −200 N.
Fx = BxBC = 0.
Element AB: The moments about A:
M = MA + (1)ByAB = 0,
from which, since the reactions across the joint are equal and opposite:
ByAB = −ByBC = 200 N , MA = −200 N-m . The sum of the
forces:
Fy = Ay + ByAB = 0,
from which Ay = −200 N .
1m
Fx = Ax + BxAB = 0,
from which Ax = 0.
Explanation of difference: The forces are equal and opposite across
the joint B, so it matters on which side of B the couple is applied.
C
1m
(b)
Fx = Ax = 0.
Case (b) Element BC: The sum of the moments about B:
MB = (1)Cy − 200 = 0,
200 N-m
B
from which Ay = 0 ,
1m
AY
(a)
AX
(b) AX
AB
BY
AB
BC
MA 200 N-M BX BX
AB
AY
BY
AB BC
BX BX
MA
BC
BY
BC
CY
BY
200 N-M
CY
Problem 6.54 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
Solution:
and
3 ft
3 ft
Element AB: The equilibrium equations are:
FX = AX + BX = 0,
A
200 lb
B
FY = AY + BY = 0,
MA = NA + (6)BY = 0.
6 ft
C
Element BC: The equilibrium conditions are
FX = −BX − C sin(20◦ ) = 0,
FY = −BY − 200 + C cos(20◦ ) = 0,
6 ft
and, summing moments around B,
MB = −(3)200 − (6)C sin(20◦ ) + (6)C cos(20◦ ) = 0.
We have six equations in six unknowns. Solving simultaneously yields
AX = 57.2 lb, AY = 42.8 lb, BX = −57.2 lb, BY = −42.8 lb,
C = 167.3 lb, and NA = 256.6 ft-lb.
NA
AX
20°
3 ft 3 ft
AY
BY
A
B
BX
6 ft
6 ft
BY
BX
B
200 lb
6 ft
C
6 ft
3 ft 3 ft
C
20°
20°
Problem 6.55 The force F = 10 kN. Determine the
forces on member ABC, presenting your answers as
shown in Fig. 6.35.
F
E
D
A
B
C
1m
Solution: The complete structure as a free body: The sum of the
moments about G:
MG = +3F − 5A = 0,
from which A = 3F
= 6 kN which is the reaction of the floor. The
5
sum of the forces:
Fy = Gy − F + A = 0,
1m
C
B
1m
2m
1m
F
GY
GX
A
2m
F −4G
3m
F
GY
D
1m
from which D = F − E − Gy = 10 + 2 − 4 = 8 kN.
Element ABC: Noting that the reactions are equal and opposite:
2m
E
1m
F
A
1m
and
G
E
1m
y
from which E =
= 10−16
= −2 kN.
3
3
The sum of the forces:
Fy = Gy − F + E + D = 0,
B = −D = −8 kN ,
1m
F
from which Gy = F − A = 10 − 6 = 4 kN.
Fx = Gx = 0.
Element DEG: The sum of the moments about D
M = −F + 3E + 4Gy = 0,
2m
D
A
C = −E
B= D
C = −E = 2 kN .
The sum of the forces:
Fy = A + B + C = 0,
A
1m
from which A = 8 − 2 = 6 kN. Check
6 kN
G
8 kN
2 kN
B
C
3m
Problem 6.56 Consider the frame in Problem 6.55.
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?
F −4G
y
From the solution to Problem 6.55: E =
,
3
3
F
Gy = F − A, and A = 5 F . Back substituting, E = − 5 or
F = −5E, from which, for E = 10 kN, F = −50 kN
Solution:
Problem 6.57 The hydraulic actuator BD exerts a 6kN force on member ABC. The force is parallel to BD,
and the actuator is in compression. Determine the forces
on member ABC, presenting your answers as shown in
Fig. 6.35.
A
B
C
0.5 m
0.5 m
D
0.5 m
Solution: The surface at C is smooth.
Element ABC: The sum of the moments about A is
M = (0.5)B sin 45◦ + (1)C = 0,
A
0.5 m
D
0.5 m
C
AY
from which Ay = −B (0.707) − C = −2.121 kN .
Fx = Ax − B cos 45◦ = 0,
C
0.5 m
from which C = −3(0.707) = −2.121 kN .
The sum of the forces:
Fy = Ay + B sin 45◦ + C = 0,
B
AX
B
0.5
m
from which Ax = 4.24 kN
45°
0.5
m
2.12 kN
4.24
kN
A
B
C
45°
2.12 kN
6 kN
Problem 6.58 The simple hydraulic jack shown in
Problem 6.57 is designed to exert a vertical force at point
C. The hydraulic actuator BD exerts a force on the beam
ABC that is parallel to BD. The largest lifting force the
jack can exert is limited by the pin support A, which will
safely support a force of magnitude 20 kN. What is the
largest lifting force the jack can exert at C, and what is
the resulting axial force in the hydraulic actuator?
Solution:
Ay
From the solution to Problem 6.??
B
= − √ − C,
2
B
Ax = √ ,
2
B
and C = − √ .
2 2
Substituting, Ay = −
and
|A| =
A2x
2
+
B
√
2
,
A2y
B
= √
2
12 +
√
2
1
5B
= √ .
2
2 2
For |A| = 20 kN,
B=
and
2
√
2(20)
√
5
C=−
2
B
√
2
= 25.3 kN is the largest axial force,
= −8.944 kN is the largest lifting force.
Problem 6.59 Determine the forces on member BC
and the axial force in member AC.
0.3 m
0.5 m
800 N
B
0.4 m
C
A
Solution:
Element BC: The sum of the moments about B:
M = −(0.3)800 + (0.8)C = 0,
300 m
from which C = 300 N . The sum of the forces
Fy = B − 800 + C = 0,
from which B = 500 N .
Fx = Cx = 0.
B
500 m
800 N
400 m
(The roller support prevents an x-direction reaction in C.) Element
AC: The sum of the forces
Fx = Ax = 0
C
A
800 N
B
C
300
mm
500
mm
C
Ax
Problem 6.60 An athlete works out with a squat thrust
machine. To rotate the bar ABD, he 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
0.42 m
A
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.
0.6 m
0.6 m
C
y
Ay
FBC
0.6 m
θ
0.6 m
C
Dy 0.42 m
D
Dx
x
(FBC = 1800 N)
0.42 m
A
B
D
(FBC = 1800 N)
tan θ =
0.42
θ = 34.99◦ .
0.6
+
Fx :
Dx − FBC cos θ = 0
Fy :
Ay − FBC sin θ + Dy = 0
MD :
−1.2Ay + 0.6FBC sin θ = 0
Solving, we get Dx = 1475 N
Dy = 516 N
Ay = 516 N
1.65 m
E
Problem 6.61 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
0.8 m
Note that members BE and CF are two force members. Consider the 6 kN load as being applied to member ABC.
0.4 m
Solution:
Ay
Ax
0.4 m
A
6 kN
1.0 m
B
0.4 m
6 kN
1.0 m
B
C
C
FCF
FBE
θ
θ
0.5 m
E
D
F
0.8 m
tan θ =
0.5
0.4
θ = 51.34◦
tan φ =
0.5
0.2
φ = 68.20◦
Member DEF
FBE
θ
Dx
FCF
E
0.8 m
F
φ
0.4 m
Dy
Equations of equilibrium:
Member ABC:
Fx :
Ax + FBE cos θ − FCF cos φ = 0
Fy :
Ay − FBE sin θ − FCF sin φ − 6 = 0
+
MA :
−(0.4)FBE sin θ − (1.4)FCF sin φ − 1.4(6) = 0
Member DEF :
Fx :
Fy :
+
MD :
Dx − FBE cos θ + FCF cos φ = 0
Dy + FBE sin θ + FCF sin φ = 0
(0.8)(FBE sin θ) + 1.2FCF sin φ = 0
Unknowns Ax , Ay , Dx , Dy , FBE , FCF we have 6 eqns in 6
unknowns.
Solving, we get
Ax = −16.8 kN
Ay = 11.25 kN
Dx = 16.3 kN
Dy = −5.25 kN
Also, FBE = 20.2 kN (T )
FCF = −11.3 kN (C)
0.4 m
Problem 6.62 The mass m = 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:
m
200 mm
The equations of equilibrium for the entire frame are
Ay
A
FX = AX + EX = 0,
Ax
FY = AY − 2mg = 0,
and summing moments at A,
MA = (0.3)EX − (0.2)mg − (0.4)mg = 0.
Solving yields AX = −2354 N, AY = 2354 N, and EX = 2354 N.
Member ABC: The equilibrium equations are
FX = AX + CX = 0,
FY = AY − BY + CY = 0,
and
MA = −(0.2)BY + (0.4)CY = 0.
We have three equations in the three unknowns BY , CX , and CY .
Solving, we get BY = 4708 N, CX = 2354 N, and CY = 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.
B
C
300 m
D
Ex
E
AY
mg
m
200 m
AX
m
200 m
mg
BY
CY
CX
BY
CX
DY
CY
DY
EX
2354 N
B
2354 N A
4708 N
4708 N 2354 N
C
2354 N
B
2354 N
2354 N
4708 N
4708 N
E
2354 N
C
D
1177 N
1177 N
Problem 6.63 The tension in cable BD is 500 lb.
Determine the reactions at A for cases (1) and (2).
G
E
6 in
D
6 in
A
B
C
300 lb
8 in
8 in
(1)
Solution: Case (a) The complete structure as a free body: The
sum of the moments about G:
MG = −16(300) + 12Ax = 0,
E
G
6 in
D
from which Ax = 400 lb . The sum of the forces:
Fx = Ax + Gx = 0,
6 in
from which Gx = −400 lb.
Fy = Ay − 300 + Gy = 0,
A
300 lb
ments about E:
ME = −16Gy = 0,
from which Ax = 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 = −8B sin α − 16Ay = 0,
noting that B = 500 lb and α = tan−1 68 = 36.87◦ ,
Ay = −150 lb.
Ay
E
B
α
8 in
Cy
Cx
300 lb
6 in
D
6 in
6 in
A
C
300 lb
8 in
E
G
6 in
D
A
B
8 in
8 in
(a)
8 in
(b)
Gy
(a) 12 in
Gx
Ax
Ax
8 in
(2)
G
from which Gy = 0, and from above Ay = 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
MC = −16(300) + 12Ax = 0,
(b)
C
8 in
8 in
from which Ay = 300 − Gy . Element GE: The sum of the mo-
then
B
Gy
Ay
Gx
16 in
300 lb
Ey
Ex
C
300 lb
Problem 6.64
Determine the forces on member
ABCD, presenting your answers as shown in Fig. 6.35.
E
3 ft
400 lb
3 ft
B
A
D
C
4 ft
4 ft
Solution: The complete structure as a free body: The sum of the
moments about A:
MA = −400(3) + 12Dy = 0,
4 ft
E
3 ft
400 lb
from which Dy = 100 lb . The sum of the forces:
Fx = Ax + 400 = 0,
A
B
from which Ax = −400 lb .
Fy = Ay + Dy = 0,
4 ft
from which the cable tension is E = 360.6 lb . The sum of the
forces:
Fx = −Cx + 400 − E cos α = 0,
from which Cx = 200 lb.
Fy = −Cy − E sin α = 0,
from which Cy = −300 lb.
Element ABCD: The tension in the cable acts on element ABCD
with equal and opposite tension to the reaction on element EB, up
and to the right at an angle of 56.31◦ , Cx = 200 lb to the right, and
Cy = −300 lb downward.
4 ft
E
F
Cy
Ay
The angle of cable element EB is
6
α = tan−1
= 56.31◦ ,
4
D
4 ft
from which Ay = −100 lb .
Element EC: The sum of the moments about C:
MC = 6E cos α − 3(400) = 0.
C
B
Cx
Cy
Ax
Cx
Dy
360
lb
400 lb
A
100 lb
56.3°
300 lb
200 lb
100 lb
D
3 ft
Problem 6.65 The mass m = 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
F
Solution: The weight of the mass hanging is W = mg =
50(9.81) = 490.5 N The complete structure as a free body: The
sum of the moments about A:
MA = −2W + Fy = 0,
from which Fy = 981 N. The sum of the forces:
Fy = Ay + Fy − W = 0,
1m
D
C
from which Ax = −Fx . Element BF: The sum of the moments
about F :
MF = −Bx − By = 0,
1m
from which By = −Bx . The sum of the forces:
Fy = By + Fy = 0,
1m
W
B
Dy
from which Fx = −981 N, and from above, Ax = 981 N ,
Element DE: The sum of the moments about D:
MD = −Ey − 2W = 0,
Ex
Dx
Dx
Dy
Cy
Cx
Cy
Bx
By
Ey
W
Ex
By
Ax
Fx = −Dx − Ex = 0,
from which Ex = −981 N, and from above Dx = 981 N .
Fy = Ey + Cy = 0,
from which Cy = 981 N.
Fx = Ex + Cx = 0,
from which Cx = 981 N, and
Element ABCD: All reactions on ABCD have been determined
above.
The components at B and C have the magnitudes
√
B = C = 9812 + 9812 = 1387 N , at angles of 45◦ .
Ey
Cx
Bx
Fy
Ay
from which Dy = 490.5 N .
from which Dx = −Ex . Element CE: The sum of the moments
about C:
MC = Ey − Ex = 0,
F
A
from which By = −981 N, and Bx = 981 N.
Fx = Bx + Fx = 0,
E
1m
from which Ay = −490.5 N,
Fx = Ax + Fx = 0,
from which Ey = −981 N. The sum of the forces:
Fy = −Dy − Ey − W = 0,
1m
490.5 N
D
C
981 N
1387 N
45°
B
A
45°
1387 N
981 N
490.5 N
Fx
Problem 6.66
ber BCD.
Determine the forces on mem-
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 = −(6)400 + 8Ey = 0,
B
from which Ey = 300 lb. The sum of the forces:
Fx = Dx = 0.
Fy = Ey + Dy − 400 = 0,
from which Dy = 100 lb. Element AB: The sum of the moments
about A:
MA = −8By − (6)400 = 0,
4 ft
C
4 ft
E
D
from which By = −300 lb. The sum of forces:
Fy = −By − Ay − 400 = 0,
from which Ay = −100 lb.
Fx = −Ax − Bx = 0,
from which (1) Ax + Bx = 0 Element ACE: The sum of the moments
about E:
ME = −8Ax + 4Cx − 8Ay + 4Cy = 0,
from which (2) −2Ax +Cx −2Ay +Cy = 0. The sum of the forces:
Fy = Ay + Ey − Cy = 0,
from which Cy = 200 lb .
Fx = Ax − Cx = 0,
from which (3) Ax = Cx . The three numbered equations are solved:
Ax = −400 lb, Cx = 400 lb , and Bx = −400 lb .
Element BCD:
The reactions are now known:
By = −300 lb , Bx = −400 lb , Cy = 200 lb ,
Dx = 0 , Dy = 100 lb ,
where negative sign means that the force is reversed from the direction
shown on the free body diagram.
400 lb
6 ft
A
8 ft
400 lb
Ay
Ax
Ax
By
Bx
Cy
Cx E
Ay
Bx
Cy
Dy
Cx
Dx
By
Problem 6.67
Determine the forces on member ABC.
E
6 kN
1m
C
D
1m
A
B
2m
Solution:
2m
The frame as a whole: The equations of equilibrium are
E
FX = AX + EX = 0,
EX
FY = AY + EY − 6000 N = 0,
1m
and, with moments about E,
ME = 2AX − (5)6000 = 0.
Solving for the support reactions, we get AX = 15,000 N and EX =
−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 = AX − BX = 0,
FY = AY − BY − CY = 0,
and summing moments about A,
MA = −2BY − 4CY = 0.
Member BDE: The equations of equilibrium are
FX = EX + DX + BX = 0,
FY = EY + DY + BY = 0,
and, summing moments about E,
ME = (1)DY + (1)DX + (2)BY + (2)BX = 0.
Member CD: The equations of equilibrium are
FX = −DX = 0,
FY = −DY + CY − 6000 = 0,
and summing moments about D,
MD = −(4)6000 + 3CY = 0.
Solving these equations simultaneously gives values for all of the forces
in the frame. The values are AX = 15,000 N, AY = −8,000 N,
BX = 15,000 N, BY = −16,000 N, CY = 8,000 N, DX = 0,
and DY = 2,000 N.
6 kN
D
EY
C
1m
AX
B
2m
1m
2m
AY
EX
EY
E DY
DX D
DX D
DY
BX B
BY BX
B
AX A
BY
AY
CY
C
CY
C
6 kN
1m
Problem 6.68
ber ABD.
Determine the forces on mem8 in
8 in
8 in
A
60 lb
8 in
60 lb
B
E
8 in
C
Solution:
and
D
The equations of equilibrium for the truss as a whole are
AY
FX = AX + CX = 0,
FY = AY − 60 − 60 = 0,
8 in.
AX
8 in.
8 in.
A
MA = 16CX − 16(60) − 24(60) = 0.
60 lb
8 in.
60 lb
Solving these three equations yields
B
AX = −150 lb,
AY = 120 lb,
8 in.
C
and CX = 150 lb.
Member ABD: The equilibrium equations for this member are:
FX = AX − BX − DX = 0,
FY = AY − BY − DY = 0,
and
MA = −8BY − 8DY − 8BX − 16DX = 0.
Member BE: The equilibrium equations for this member are:
FX = BX + EX = 0,
FY = BY + EY − 60 − 60 = 0,
and
MB = −8(60) − 16(60) + 16EY = 0.
Member CDE: The equilibrium equations for this member are:
FX = CX + DX − EX = 0,
FY = DY − EY = 0,
and
MD = 8EX − 16EY = 0.
Solving these equations, we get BX = −180 lb, BY = 30 lb,
DX = 30 lb, DY = 90 lb, EX = 180 lb, and EY = 90 lb. Note
that we have 12 equations in 9 unknowns. The extra equations provide
a check.
E
D
CX
AY
AX
BX
DX
BY
60 lb
BY
BX B
EX
EY
DY
DY
CX
60 lb
DX
EY
EX
Problem 6.69 The mass m = 12 kg. Determine the
forces on member CDE.
A
200 mm
100 mm
E
B
200 mm
C
D
200 mm
Solution:
and
m
400 mm
The equations of equilibrium for the entire truss are:
AY
FX = AX + CX = 0,
FY = AY − mg = 0,
MA = 0.4CX − 0.7mg = 0.
AX
A
200 mm
From these equations we get
100 mm
E
B
AX = −206.0 N, AY = 117.7 N,
200 mm
and Cx = 206.0 N.
Member ABD: The equations are
FX = AX + BX + DX + T = 0,
FY = AY + BY + DY = 0,
and
MA = 0.2BY + 0.2BX + 0.2DY + 0.4DX + 0.1T = 0.
C
CX
The Pulley: The equations are
FX = −T − PX = 0,
and
FY = −T − PY = 0.
The Weight: The equation is
FY = T − mg = 0.
Solving the equations simultaneously, we get
BX = 117.7 N, BY = 0, DX = −29.4 N, DY = −117.7 N,
EX = −235.4 N, EY = −117.7 N, T = 117.7 N,
PX = −117.7 N, PY = −117.7 N
200 mm
m
400 mm
AY
AX
T
PY
PX
T
Member CDE: The equations are
FX = CX − DX + EX = 0,
FY = −DY + EY = 0,
and
MD = 0.4EY − 0.2EX = 0.
Member BE: The equations are
FX = −BX + PX − EX = 0,
FY = −BY + PY − EY = 0,
and
ME = 0.4BY = 0.
F
D
BX
DX
T
BY
DY
CX
DX
BX
BY
EX
EY
PY PX
EY
T
EX
DY
mg
Problem 6.70 The weight W = 80 lb. Determine the
forces on member ABCD.
11 in
5 in
12 in
3 in
A
B
D
C
8 in
W
F
E
Solution: The complete structure as a free body: The sum of the
moments about A:
MA = −31W + 8Ex = 0,
from which Ex = 310 lb. The sum of the forces:
Fx = Ex + Ax = 0,
3 in.
from which Cx = 310 lb . The sum of the moments about E:
ME = 8F − 16Cy + 8Cx = 0.
For frictionless pulleys, F = W , and thus Cy = 195 lb . The sum
of forces parallel to y:
Fy = Ey − Cy + F = 0,
from which Ey = 115 lb .
Equation (1) above is now solvable: Ay = −35 lb .
Element ABCD: The forces exerted by the pulleys on element
ABCD are, by inspection: Bx = W = 80 lb , By = 80 lb ,
Dx = 80 lb , and Dy = −80 lb , where the negative sign means
that the force is reversed from the direction of the arrows shown on the
free body diagram.
B
A
D
C
8 in.
from which Ax = −310 lb .
Fy = Ey + Ay − W = 0,
from which (1) Ey + Ay = W .
Element CFE: The sum of the forces parallel to x:
Fx = Ex − Cx = 0,
12 in.
5 in.
11 in.
W
F
E
Ay
Ax
Cx
By
Cx
Ey
F
Ex
Dy
Dx
Cy
Bx
Cy
W
Problem 6.71 The man 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
1 ft 6 in
2 ft
D
B
C
A
9 in
60°
6 ft
80 lb
E
Solution: The complete structure as a free body: The sum of the
moments about E:
M = −26W − 68W sin 60◦ + 50Fy − 81W cos 60◦ + ME = 0
from which (1) 50Fy + ME = 10031. The sum of the forces:
Fx = Fx + W cos 60◦ + Ex = 0,
F
2 ft 2 in
2 ft
1 ft 6 in D
B
9 in
A
C
60°
from which (2) Fx + Ex = −40.
Fy = −W − W sin 60◦ + Ey + Fy = 0,
from which (3) Ey + Fy = 149.28
Element CF: The sum of the moments about F :
M = −72Cx = 0,
from which Cx = 0. The sum of the forces:
Fx = Cx + Fx = 0,
6 ft
80 lb
F
E
from which Fx = 0 . From (2) above, Ex = −40 lb
Element AE: The sum of the moments about E:
M = ME − 72Ax = 0, .
26
in
42
in
from which (4) ME = 72Ax . The sum of the forces:
Fy = Ey + Ay = 0,
from which (5) Ey + Ay = 0.
Fx = Ax + Ex = 0;
from which Ax
=
ME = 2880 in lb = 240 ft lb .
60°
W
W
81 in
ME
40
lb,
and from (4)
From (1) Fy = 143.0 lb ,
Ey
Ex
and from (2) Ey = 6.258 lb . This completes the determination of
Fy
Fx
50 in
the 5 reactions on E and F .
Ay
Cy
Ax
Cx
ME
Fx
Fy
Ex
Ey
Problem 6.72 The frame supports a horizontal load F
at C. The resulting compressive axial force in the twoforce member CD is 2400 N. Determine the magnitude
of the reaction exerted on member ABC at B.
C
F
100 mm
B
100 mm
Solution: First, write eqns to determine the support reactions at
A and E (CD is a two force member)
C
D
F
100 mm
A
B
E
0.3 m
100 mm
200 mm
D
100 mm
E
AX
EY
0.4 m
C
F
AY
+
100 mm
Fx :
Ax + F = 0
(1)
Fy :
Ay + Ey = 0
(2)
MA :
B
100 mm
0.4Ey − 0.3F = 0 (3)
D
We have these eqns in four Unknowns (Ax , Ay , Ey , and F ) Now
write eqns for member ABC
100 mm
A
E
CY
F
BY
200 mm
100 mm
100 mm
0.1 m
BY
BX
0.2 m
AX
BX
DY
AY
0.1 m
CY = 2400 N
0.2 m
0.1
m
Cy = 2400 N
+
Fx :
Ax + Bx + F = 0
(4)
Fy :
Ay + By + Cy = 0
(5)
MA :
0.2By − 0.2Bx + 0.3Cy − 0.3F = 0 (6)
We now have 6 eqns in unknowns: (Ax , Ay , Bx , By Ey , F ) Next,
write the equations for member BDE.
Solving, we get Bx = 0 By = −1200 N
|B| = 1200 N
0.1
m
EY
Cy = −Dy
(two force member)
Fx :
−Bx = 0
(7)
Fy :
−By + Dy + Ey = 0 (8)
MB :
0.1Dy + 0.2Ey = 0 (9)
We now have 9 equations in 8 unknowns. Obviously, if they are compatible,
one is a linear combination of the others. We could also have more than one
redundant equation and still need another equation.
Combining Eqs. (4) and (7) gives Eq. (1). Thus, one of these three equations is
not need.
Problem 6.73 The two-force member CD of the
frame shown in Problem 6.72 will safely support a compressive axial load of 3 kN. Based on this criterion, what
is the largest safe magnitude of the horizontal load F ?
Solution: In the solution to Problem 6.72, we derived the equation’s listed below for the loads shown on the frame.
Ax + F = 0
Ay + Ey = 0
0.4Ey − 0.3F = 0
Ax + Bx + F = 0
Ay + By + Cy = 0
0.2By − 0.2Bx + 0.3Cy − 0.3F = 0
C










F
100 mm
B
Entire Frame
100 mm
Member ABC
D
100 mm
A
Cy = −Dy − 2 force member
E
Set Cy = 3000 N and solve.
200 mm
We get F = 2000 N = 2 kN
Ax = −2 kN
Ay = −1.5 kN
Ey = 1.5 kN
Bx = 0
By = −1.5 kN
100 mm
100 mm
Problem 6.74 The unstretched length of the string is
LO . Show that when the system is in equilibrium the
angle α satisfies the relation sin α = 2(LO − 2F/k)L.
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 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 α for any non-zero force F .
The complete structure as a free body:
The sum of the moments about A
MA = −F L sin α + EL sin α = 0,
F
1− L
4
1− L
4
C
from which E = F . The sum of forces:
Fx = Ax = 0,
1− L
2
from which Ax = 0.
from which Ay = 0, which completes a demonstration that F does
not exert a moment on the system. The spring C: The elongation of the
spring is ∆s = 2 L
sin α − LO , from which the force in the spring is
4
L
T =k
sin α − LO
2
Element BE: The strategy is to determine Cy , which is the spring force
on BE. The moment about E is
L
L
L
ME = − Cy cos α − By cos α − Bx cos α = 0,
4
2
2
α
F
L
α
Ax
Ay
E
C
from which 2y + By = −Bx . The sum of forces:
Fx = Bx = 0,
from which Bx = 0.
Fy = Cy + By + E = 0,
from which k
Solve: sin α =
L
sin α − LO
2
2(LO − 2F
k )
L
By
Cy
Bx
L
4
from which Cy +By = −E = −F . The two simultaneous equations
are solved: Cy = −2F , and By = F .
The solution for angle α: The spring force is
L
Cy = T = k
sin α − LO ,
2
= −2F .
D
α
A
Fy = Ay + E − F = 0,
k
B
α
L
4
E
E
Problem 6.75 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 = mg = 9.81 g
The complete structure as a free body:
Sum the forces in the x-direction:
Fx = Ax = 0,
from which Ax = 0
Element ABC: The sum of the moments about A:
MA = +0.3Bx + 0.9Cx − 0.4W = 0,
4
W.
7
from which By =
The magnitude of the reaction at B is
2
5 2
4
|B| = W
+
= 1.0104W.
6
7
24
1.0104
For a safe value of |B| = 24 kN, W =
= 23.752 kN is the
maximum load that can be carried. Thus, the largest mass that can be
supported is m = W/g = 23752 N/9.81 m/s2 = 2421 kg.
400 mm
400 mm
C
500 mm
100 mm
300 mm
E
D
B
F
W
A
from which (1) 0.3Bx + 0.9Cx = 0.4W . The sum of the forces:
Fx = −Bx − Cx + W + Ax = 0,
from which (2) Bx + Cx = W . Solve the simultaneous equations (1)
and (2) to obtain Bx = 56 W
Element BE: The sum of the moments about E:
ME = 0.4W − 0.7By = 0,
F
300 mm 400 mm 400 mm
Cy
Cy
Cx
Cx
W
By
W
Bx
By Bx
W
Ay
Ax
Ey
Ex
Ey
Ex
F
Problem 6.76
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:
MA = −4(18) + 3(36) + 12Cy − 72 = 0,
8 ft
A
C
from which Cy = 3 lb. The sum of the forces:
Fy = Ay + Cy − 18 = 0,
3 ft
72 ft-lb
36 lb
from which Ay = 15 lb.
Fx = Ax + Cx + 36 = 0,
18 lb
from which (1) Cx = −Ax − 36
Element AB: The sum of the forces:
Fy = Ay − By − 18 = 0,
4 ft
from which By = −3 lb. The sum of the moments:
MA = 6Bx − 4(18) − 4By − 72 = 0,
from which Bx = 22 lb. The sum of the forces:
Fx = Ax + Bx = 0,
3 ft
B
8 ft
Cy
Ay
Ax
Cx 3 ft
72 ft-lb
36 lb
from which Ax = −22 lb From equation (1) Cx = −14 lb
18 lb
8 ft
4 ft
Ay
Ax
72 ft-lb
By 6 ft
Bx
18 lb
Problem 6.77
Determine the forces on member AD.
200 N
130 mm
D
400 mm
C
A
400 N
B
400 mm
Denote the reactions of the support by Rx and Ry . The
complete structure as a free body:
Fx = Rx − 400 = 0,
400 mm
Solution:
from which Rx = 400 N. The sum of moments:
MA = 800C − 400(800) − 400(400) − 400(200) = 0,
200 N
D
400 mm
from which C = 300 N.
Fy = C + Ry − 400 − 200 = 0,
A
400 mm
Ax = −200 N, and Dx = 200 N.
Element BD: The sum of forces:
Fx = Bx − Dx − 400 = 0
from which Bx = 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.
400 mm
Dy
200 N
from which By = 600 N. Element BD: The sum of the forces:
Fy = By − Dy − 400 = 0,
from which Ay = 0: Element AD: The sum of the forces:
Fx = Ax + Dx = 0
and
MA = −400(200) + 800Dy − 400Dx = 0
400 N
C
B
from which Ry = 300 N. Element ABC: The sum of the moments:
MA = −4By + 8C = 0,
from which Dy = 200 N.
Element AD: The sum of the forces:
Fy = Ay + Dy − 200 = 0,
130 mm
Ay
Ay
Ax
Ry
Dy
400 N
By
400 N
Ax
Rx
Bx
By
Dx
Dx
Bx
C
Problem 6.78 The frame shown is used to support
high-tension wires. If b = 3 ft, α = 30◦ , and W =
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 = b tan α = 1.732 ft
Member ABC. The sum of the forces:
Fx = Ax + B sin α = 0,
Fy = Ay − W − B cos α = 0.
The sum of the moments about B:
MB = bAy − hAx + bW = 0.
These three equations have the solution: Ax = 173.21 lb, Ay =
−100 lb, and B = −346.4 lb.
Member BDEF: The sum of the forces:
Fx = Dx − B sin α − E sin α = 0,
Fy = Dy − W + B cos α − E cos α = 0.
The sum of the moments about H:
MH = bGy − hGx + bW + 2bE cos α − 2hE sin α = 0.
These three equations have the solution: Gx = 346.4 lb, Gy =
200 lb, and H = 300 lb. This is the axial force in HJ.
b
b
A
B
C
D
α
α
E
G
W
F
H
I
W
J
α
α
W
The sum of the moments about D:
MD = −2bW − bE cos α − hE sin α − bB cos α + hB sin α = 0.
These three equations have the solution: Dx = −259.8 lb, Dy =
350 lb, E = −173.2 lb.
Member EGHI: The sum of the forces:
Fx = Gx + E sin α − H cos α = 0,
Fy = Gy − W + E cos α + H sin α = 0.
b
b
B
b
Ay
h
Ax
B
Dy
Dx E
b G
y
W
Gx
H
W
b
W
b
Problem 6.79 What are the magnitudes of the forces
exerted by the pliers on the bolt at A when 30-lb forces
are applied as shown? (B is a pinned connection.)
6 in
45°
2
in
30 lb
B
A
30 lb
Solution:
Element AB: The sum of the moments about B:
MB = 2F − (6)30 = 0,
from which F = 90 lb.
6 in
45°
30 lb
2 in
B
A
30 lb
6 in
By
30 lb
2 in
F
Bx
Problem 6.80 The weight W = 60 kip. What is the
magnitude of the force the members exert on each other
at D?
A
3 ft
B
C
2 ft
D
3 ft
W
3 ft
3 ft
Solution: Assume that a tong half will carry half the weight, and
denote the vertical reaction to the weight at A by R. The complete
structure as a free body: The sum of the forces:
Fy = R − W = 0,
A
from which R = W
3 ft
C
B
Tong-Half ACD:
2 ft
Element AC: The sum of the moments about A:
(1)
MA = 3Cy + 3Cx = 0.
D
(3)
The sum of the forces:
R
Fy =
+ Cy + Ay = 0, and
2
Fx = Cx + Ax = 0.
W
(4)
Element CD: The sum of the forces:
Fx = Dx − P − Cx = 0, and
(2)
(5)
(6)
(7)
(8)
3 ft
3 ft
3 ft
W
= 0.
Fy = Dy − Cy −
2
The sum of the moments:
3
MD = 2Cx − 3Cy − 3P + W = 0
2
Element AB: The sum of the forces:
R
Fy = −Ay +
− By = 0, and
2
Fx = −Ax − Bx = 0.
Element BD: The sum of the forces:
W
(9)
Fy = By − Dy −
= 0, and
2
(10)
Fx = Bx − Dx + P = 0.
These are ten equations in ten unknowns. These have the solution
R = 60 kip. Check, Ax = −30 kip, Ay = 0, Bx = 30 kip,
By = 30 kip, Cx = 30 kip, Cy = −30 kip, Dx = 110 kip,
Dy = 0, and P = 80 kip. The magnitude of the force the
members exert on each other at D is D = 110 kip.
Ay
R
2
R
2
Ay
Ax
Ax
By
Cy
Bx
Dy
Cx
Dy
Bx
Cx
Dx
By
Dx
Cy
P
P
W
2
W
2
Problem 6.81 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 = 2 kg.
(a) Determine the tension in the biceps muscle AB.
(b) 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
9N
m
200 mm
150 mm
(b)
Solution: Make a cut through AB and BC just above the elbow
joint C. The angle formed
muscle with respect to the
by the biceps
forearm is α = tan−1 290
= 80.2◦ . The weight of the mass is
50
W = 2(9.81) = 19.62 N.
The section as a free body: The sum of the moments about C is
MC = −50T sin α + 150(9) + 350W = 0,
from which T = 166.76 N is the tension exerted by the biceps muscle
AB. The sum of the forces on the section is
FX = Cx + T cos α = 0,
(a)
from which Cx = −28.33 N.
FY = Cy + T sin α − 9 − W = 0,
B
from which Cy = −135.72. The magnitude of the force exerted by
the forearm on the upper arm at joint C is
F = Cx2 + Cy2 = 138.65 N
290
mm
(b)
A
W 200 mm
C
150 mm
50
9N
mm
T
W
9N
200
mm
α Cy
50
150 mm
mm
Cx
C
Problem 6.82 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
125 mm
B
50 mm
E
A
50 mm
C
50 mm
D
Solution: Consider the upper jaw only.
The section ABC as a free body:
The sum of the moments about C is
MC = 100B − 250A = 0,
from which, for A = 200 N, B = 500 N. The sum of the forces:
Fx = Cx − B = 0,
from which Cx = 500 N,
Fy = Cy + A = 0,
from which Cy = −200 N. The magnitude of the reaction at C:
C = Cx2 + Cy2 = 538.52 N
125
mm
125
mm
125
mm
B
50
mm
E
A
50
mm
C
50
mm
D
B
Cy
100 mm
A
Cx
250 mm
Problem 6.83 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)
where d = 0.3 cos(45◦ ). Thedistance b can be determined from the
Pythagorean Theorem as b = (0.35)2 − d2 . From the diagram, the
angle θ = 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.
The unit vector from C toward B is uCB = −0.795i + 0.606j.
The equations of force equilibrium at C are
FX = FBC cos θ − 2000 = 0,
and
FY = N − FBC sin θ = 0.
B
d
45°
A
d
θ
b
C
y
FBC
FBCY
c
2000 N
x
FBCX
Solving these equations, we get N = 1524 Newtons(N), and FBC =
2514 N.
N
The force acting at B due to member BC is FBC uBC = −2000i +
1524j N.
The position vector from A to B is rAB = 0.212i + 0.212j m, and
the moment of the force acting at B about A, calculated from the cross
product, is given by MF BC = 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 = −747.6k N-m (clockwise).
0.35 m
0.3 m
B
y
FBC uCB
M
rAB
A
x
Problem 6.84 In Problem 6.83, determine the forces
on member AB at A and B.
Solution: In the solution of Problem 6.83, we found that the force
acting at point B of member AB was FBC uBC = −2000i +
1524j N, and that the moment acting on member BC about point
A was given by M = −747.6k N-m (clockwise). Member AB must
be in equilibrium, and we ensured moment equilibrium in solving
Problem 6.83.
From the free body diagram, the equations for force equilibrium are
FX = AX + FBC uBCX = AX − 2000 N = 0,
and
FY = AY + FBC uBCY = AY + 1524 N = 0.
Thus, AX = 2000 N, and AY = −1524 N.
FBC uCB
y
B
M
AX
A
AY
x
Problem 6.85 The mechanism is used to weigh mail.
A package placed at A causes the weighted point to rotate
through an angle α. Neglect the weights of the members
except for the counterweight at B, which has a mass of
4 kg. If α = 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 cos(30 − α). The restoring moment of the counter weight is
100 mg sin α. Thus the sum of the moments is
M = 100 mB g sin α − 100 mA g cos(30 − α) = 0.
Define the ratio of the masses of the counter weight to the mass of the
B
package to be RM = m
. The sum of moments equation reduces to
mA
M = RM sin α − cos(30 − α) = 0,
A
100
mm
100
mm
cos(30−α)
from which RM =
= 2.8794, and the mass of the packsin α
age is mA = R4 = 1.3892 = 1.39 kg
M
B
30°
α
Problem 6.86 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 =
1.6 kN. Determine the axial force in the actuator BD
and the magnitude of the reaction at A.
A
C
0.8 m
D
B
F
0.2 m
1m
1m
Solution: The section ABC as a free body: The sum of the moments about A:
MA = 0.8BD − 2F = 0,
from which BD = 4 kN.
The sum of the forces:
Fx = Ax + BD = 0,
B
from which Ax = −4 kN.
Fy = Ay − F = 0,
from which Ay = 1.6 kN. The magnitude of the reaction at A is
A = A2x + A2y = 4.308 kN
C
80 cm
D
F
20 cm
1m
1m
Ay
Ax
0.8 m
BD
F
2m
Problem 6.87 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
0.6 m
0.15 m
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 = 0.928i + 0.371j, and the force acting on the scoop at D is
given by D = DX i + DY j = 0.928Di + 0.371Dj. From the free
body diagram of the scoop, the equilibrium equations are
FX = BX + DX = 0,
FY = BY + DY − mg = 0,
and
MB = −0.3 mg + xBD DY − yBD DX = 0.
From the geometry, xBD = 0.25 m, and yBD = 1 m. Solving the
equations of equilibrium, we obtain BX = 719.4 N, BY = 2246 N,
and D = −774.8 N (member CD is in tension).
D
C
1m
0.6 m
G
B
A
0.15 m
0.6 m
DY
y
D
DX
C
mg
B
x
BX
BY
D uCD
DY
DX
Problem 6.88 In Problem 6.87, determine the axial
force in the hydraulic actuator BC.
Solution:
Solving these equations, we get TBC = −1267 N(compression), and
TAC = 1112 N(tension).
Scoop
1m
A
The unit vectors in the directions of the forces acting
at C are uCD = 0.928i + 0.371j, uCA = 0.243i − 0.970j, and
uCB = 0.781i − 0.625j. The force equilibrium equations at C are
FX = TBC uCBX + TAC uCAX + TCD uCDX = 0,
and
FY = TBC uCBY + TAC uCAY + TCD uCDY = 0.
0.3 m
D
TAC
TCD
C
C
TCD
TBC
A
TBC
TAC
TCD
C
TBC
TAC
0.3 m Scoop
100 N
Problem 6.89 Determine the force exerted on the bolt
by the bolt cutters.
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.
Member AB: The equations of equilibrium are:
FX = AX + BX = 0,
FY = AY + BY = 0,
and
MB = 90F − 75AX − 425(100) = 0
100 N
F
90 mm 60 mm 65 mm
300 mm
λ
40 mm
B
55 mm
75 mm
A
BX
BY
90 mm
60 mm 65 mm
AY
AX
300 mm
CY
40 mm
75 mm
90 mm
C
55 mm X
C
D
F
60 mm 65 mm
300 mm
BY
DX
B
BX
100 N
100 N
AY
AX
F
Member CD: The equations are:
FX = −CX − DX = 0,
FY = −CY − DY = 0.
Solving the equations simultaneously (we have extra (but compatible)
equations, we get F = 1051 N, AX = 695 N, AY = 1586 N,
BX = −695 N, BY = −435 N, CX = 695 N, CY = 535 N,
DX = −695 N, and Dy = −535 N
40 mm
C 55 mm
B
D
Member BD: The equations are
FX = −BX + DX = 0,
FY = −BY + DY + 100 = 0,
and
MB = 15DX + 60DY + 425(100) = 0.
Member AC: The equations are
FX = −AX + CX = 0,
FY = −AY + CY + F = 0,
and
MA = −90F + 125CY + 40CX = 0.
A
75
mm
D
DY
60 mm 65 mm
300 mm
100 N
CY
C
DX
D
DY
CX
Problem 6.90 For the bolt cutters in Problem 6.89,
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 6.107, we know BX = −695 N,
and BY = −435 N. We also know that CX = 695 N, and CY =
535 N, from which the axial load in member
CD can be calculated.
2 + C 2 = 877 N
CX
Y
Problem 6.91 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 α = 80◦ , what horizontal force is exerted on
the horizontal bar at A?
90°
D
0
25
α
m
B
mm
25
0m
The load in CD is given by TCD =
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
90°
Ry = 250 sin α = 246.2 mm,
The angle formed by member DE with the positive x axis is θ =
Ry
180 − tan−1 400−R
= 145.38◦ . The components of the
x
force produced by DE are Fx = F cos θ = −658.3 N, and
Fy = F sin θ = 454.5 N. The angle of the element AB with the
positive x axis is β = 180 − 90 − α = 10◦ , and the components
of the force for this member are Px = P cos β and Py = P sin β,
where P is to be determined. The angle of the arm BC with the positive x axis is γ = 90 + α = 170◦ . The projection of point B is
Lx = 250 cos γ = −246.2 mm, and Ly = 250 sin γ = 43.4 mm.
Sum the moments about C:
MC = Rx Fy − Ry Fx + Lx Py − Ly Px = 0.
Substitute and solve: P = 2126.36 N, and Px = P cos β = 2094 N
is the horizontal force exerted at A.
α
D
and Rx = 250 cos α = 43.4 mm.
B
250 mm
C
A
250 mm
250 mm
400 mm
Fy
B
Py
D
Px
Fx
Cx
Cy
E
Problem 6.92 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 = 2.5 ft and h = 1.5 ft.
If W = 300 lb, what force must the actuator exert to
hold the load in equilibrium?
b
W
A
B
h
D
C
The angle ADC is α = sin−1
distance CD is d = 4 cos α.
Solution:
h
4
E
= 22.02◦ . The
b
The complete structure as a free body: The sum of the forces:
Fy = −W + Cy + Dy = 0.
Fx = Cx + Dx = 0.
W
F
B
A
The sum of the moments about C:
MC = −bW + dDy = 0.
h
These have the solution:
Cy = 97.7 lb,
D
C
E
Dy = 202.3 lb,
and Cx = −Dx .
Divide the system into three elements: the platform carrying the
weight, the member AB, and the member BC.
W
The Platform: (See Free body diagram) The moments about the
point A:
MA = −bW − dB = 0.
A
B
Ex
The sum of the forces:
Fy = A + B + W = 0.
B
A
Cy
Ey
Cx E
These have the solution:
B = −202.3 lb,
Ex
Dx
and A = −97.7 lb.
Element BC: The sum of the moments about E is
h
d
d
MC = −
Cy +
Cx +
B = 0, from which
2
2
2
(3)
Dy
Cy − Ey + B = 0
Element AD: The sum of the moments about E:
d
h
d
ME =
Dy +
Dx −
A = 0,
2
2
2
(1)
dCx − hCy − dB = 0. The sum of the forces:
Fx = Cx − Ex = 0, from which
(2)
Ex − Cx = 0,
Fy = Cy − Ey + B = 0,
These are four equations in the four unknowns: EX , EY , Dx , CX and
DX
from which
Solving, we obtain Dx = −742 lb.
from which
(4)
dDy + hDx − dA = 0.
Problem 6.93 The linkage is in equilibrium under the
action of the couples MA and MB . If αA = 60◦ and
αB = 70◦ , what is the ratio MA /MB ?
250 mm
MB
αB
MA
αA
150 mm
350 mm
Solution: Make a cut through the linkage connecting the two
cranks, and treat each system as a free body. The equilibrium condition occurs when the reaction forces in the linkage are equal and
opposite.
200 mm
The position vector of the end of the system B crank is
αB
MB
rB = RB (i cos αB + i sin αB ) = 85.51i + 234.92j (mm).
MA
The position vector at the end of the system A crank is
150 mm
rA = RA (i cos αA + j sin αA ) = 75i + 129.9j (mm).
The angle of the linkage from the end of the system B crank with
respect to the horizontal is
yA − yB
β = tan−1
= −17.19◦ .
xA − xB + 350
350 mm
The unit vector parallel to the linkage, originating at the B crank, is
|F|
eBA = i cos β + j sin β = 0.9553i − 0.2955j.
The unit vector originating at A crank is eAB = −eBA . The components of the forces in the linkage are |F|eAB , and |F|eBA .
System B: When the system is in equilibrium,
0
0
1
MB + |F| 85.5
234.9
0 = 0,
0.9553 −0.2955 0 from which MB = 249.7|F|.
System A: When the system is in equilibrium:
0
0
1
MA + |F| 75
129.9 0 = 0,
−0.9553 0.2955 0 from which MA = −146.27|F|.
Complete system: Both systems are in equilibrium for the value |F|.
Take the ratio of the two moments to eliminate |F|.
MA
146.27
=−
= −0.5858
MB
249.7
αA
MB
|F|
MA
Problem 6.94 A load W = 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
Solution:
1.5 m
The sum of the moments about A is
MA = 0.75BC − 3(2) = 0,
from which BC = 8 kN is the axial force. The sum of the forces
FX = AX + BC = 0,
from which AX = −8 kN.
FY = AY − 2 = 0,
A
0.75 m
B
C
1.0 m
from which AY = 2 kN.
G
0.5 m
W
1.5 m
1.5 m
AY
0.75 m
AX
BC
W
3m
Problem 6.95 The dimensions are a = 260 mm, b =
300 mm, c = 200 mm, d = 150 mm, e = 300 mm,
and f = 520 mm. The ground exerts a vertical force
F = 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.
Solution: The free-body diagram of the shovel is from which we
obtain the equations
Fx = Cx − T cos β = 0,
(1)
Fy = Cy + T sin β + F − mg = 0, (2)
M(ptC) = f F − emg + (b − c)T sin β
+dT cos β = 0.
b
P
Shovel
A
D
a
d
B
C
(3)
G
c
The angle β = arctan[(a − d)/b].
e
From the free-body diagram of joint A,
B
P
f
F
T
T
we obtain the equation
F = P + T cos β = 0. (4)
β
d
b−c
Substituting the given information into Eqs. (1)–(4) and solving, we
obtain
T = −19, 260 N,
CX
CY
mg
P = 18, 080 N,
Cx = −18, 080 N,
and Cy = 513 N.
e
f
F
Problem 6.96 The truss supports a load F = 10 kN.
Determine the axial forces in the members AB, AC,
and BC.
B
3m
C
A
D
4m
3m
F
Solution:
+
Solving,
Find the support reactions at A and D.
Fx :
Ax = 0
Fy :
Ay + Dy − 10 = 0
MA :
B
(−4)(10) + 7Dy = 0
Ax = 0,
3m
C
A
D
Ay = 4.29 kN
Dy = 5.71 kN
4m
Joint A:
tan θ =
3m
3
4
F
θ = 36.87◦
(Ay = 4.29 kN)
Fx :
FAB cos θ + FAC = 0
Fy :
Ay + FAB sin θ = 0
Solving,
3m
AX
4m
FAB = −7.14 kN (C)
3m
AY
10 kN
FAC = 5.71 kN (T )
Joint C:
Fx :
Fy :
FAB
FCD − FAC = 0
y
FBC − 10 kN = 0
Solving FBC = 10 kN (T )
θ
FAC
x
FCD = +5.71 kN (T )
AY
FBC
FAC
FCD
10 kN
DY
Problem 6.97 Each member of the truss shown in
Problem 6.96 will safely support a tensile force of 40 kN
and a compressive force of 32 kN. Based on this criterion,
what is the largest downward load F that can safely be
applied at C?
B
2
5
1
3m
C
A
D
3
4
4m
3m
F
Solution: Assume a unit load F and find the magnitudes of the
tensile and compressive loads in the truss. Then scale the load F
up (along with the other loads) until either the tensile limit or the
compressive limit is reached.
External Support Loads:
Fx :
Ax = 0
(1)
Fy :
Ay + Dy − F = 0 (2)
MA :
−4F + 7Dy = 0 (3)
AX
F
AY
Joint A:
tan θ =
3
2
y
FAB
θ = 36.87◦
Fx :
FAC + FAB cos θ = 0 (4)
Fy :
FAB sin θ + Ay = 0
θ
(5)
x
FAC
Joint C
Fx :
Fy :
FCD − FAC = 0 (6)
FBC − F = 0
AY
(7)
Joint D
tan φ =
3
3
y
FBD
φ = 45◦
Fx :
−FCD − FBD cos φ = 0 (8)
Fy :
FBD sin φ + Dy = 0
φ
(9)
Setting F = 1 and solving, we get the largest tensile load of 0.571 in
AC and CD. The largest compressive load is 0.808 in member BD.
x
FCD
Largest Tensile is in member BC. BC = F = 1
DY
The compressive load will be the limit
Fmax
32
=
1
0.808
y
FBC
Fmax = 40 kN
FCD
FAC
F
x
DY
Problem 6.98 The Pratt bridge truss supports loads at
F , G, and H. Determine the axial forces in members
BC, BG, and F G.
Solution:
The angles of the cross-members are α = 45◦ .
The complete structure as a free body:
B
C
D
4m
E
A
F
G
The sum of the moments about A:
MA = −60(4) − 80(8) − 20(12) + 16E = 0,
60 kN
80 kN
20 kN
4m
4m
4m
4m
from which E = 70 kN. The sum of the forces:
Fx = Ax = 0.
Fy = Ay − 60 − 80 − 20 + E = 0,
from which Ay = 90 kN
H
B
C
D
The method of joints: Joint A:
FY = Ay + AB sin α = 0,
from which AB = −127.3 kN (C),
Fx = AB cos α + AF = 0,
4m
A
E
F
from which AF = 90 kN (T ). Joint F:
Fx = −AF + F G = 0,
G
H
60 kN
80 kN
20 kN
4m
4m
4m
4m
from which F G = 90 kN (T ) .
Fy = BF − 60 = 0,
from which BF = 60 kN (C). Joint B:
Fx = −AB cos α + BC + BG cos α = 0,
and
Fy = −AB sin α − BF − BG sin α = 0,
4m
Ay
−AB sin α − BF − BG sin α = 0.
and − AB cos α + BC + BG cos α = 0,
from which BC = −120 kN (C)
4m
4m
4m
Ax
from which:
Solve: BG = 42.43 kN (T ) ,
4m
Ay
AB
α
AF
Joint A
60 kN 80 kN 20 kN
BF
AF FG
60 kN
Joint F
E
α
α BC
AB BF BG
Joint B
Problem 6.99 Consider the truss in Problem 6.98. Determine the axial forces in members CD, GD, and GH.
BC
CD
BG
α
CG
Joint C
Solution:
CG
GD
α
GH
80 kN
Joint G
Use the results of the solution of Problem 6.98:
BC = −120 kN (C),
BG = 42.43 kN (T ),
and F G = 90 kN (T ).
The angle of the cross-members with the horizontal is α = 45◦ .
Joint C:
Fx = −BC + CD = 0,
from which CD = −120 kN (C)
FY = −CG = 0,
from which CG = 0.
Joint G:
Fy = BG sin α + GD sin α + CG − 80 = 0,
from which GD = 70.71 kN (T ) .
Fy = −BG cos α + GD cos α − F G + GH = 0,
from which GH = 70 kN (T )
Problem 6.100 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
moments about A:
MA = −900(400) + 600B = 0,
from which B = 600 N. The sum of forces:
Fx = Ax + B = 0,
from which BD = −632.5 N (C)
Fy = AB + BD sin θ = 0,
from which AB = 200 N (T )
Joint A:
Fy = Ay − AD sin αAD − AB = 0,
from which AD = 233.2 N (T )
Fx = Ax + AC + AD cos αAD = 0,
from which AC = 480 N (T )
E
G
300 mm
H
F
D
B
300
mm
300
mm
300
mm
900 mm
Ay
400 N
Ax
600 mm
The angle from the horizontal of element AD is
300
αAD = 90 − tan−1
= 59.04◦ .
600 − 300 tan θ
Joint B:
Fx = B + BD cos θ = 0,
C
600 mm
from which Ay = 400 N.
The angle from the horizontal of element CF is
300
αCF = 90 − tan−1
= 53.13◦ .
600(1 − tan θ)
300 mm
400 N
A
from which Ax = −600 N.
Fy = Ay − 400 = 0,
The method of joints: The angle from the horizontal of element BD
is
300
θ = tan−1
= 18.43◦ .
900
300 mm
B
AB
B
θ
BD
AC
αAD
AD
AB
Joint A
Joint B
CD
AD
αAD
DF
θ
BD
AY
AX
AC
Joint D
CE
αCF
CF
CD
Joint C
Joint D:
Fx = −AD cos αAD − BD cos θ + DF cos θ = 0,
from which DF = −505.96 N (C)
Fy = AD sin αAD + CD − BD sin θ + DF sin θ = 0,
from which CD = −240 N (C)
Joint C:
Fy = −CD − CF sin αCF = 0,
from which CF = 300 N (T )
Problem 6.101 Consider the truss in Problem 6.100.
Determine the axial forces in members CE, EF ,
and EH.
Solution:
Use the results of the solution of Problem 6.100:
AC = 480 N (T ),
AC
CF = 300 N (T ),
DF = −505.96 N (C),
CD
αCF = 53.1◦ .
The method of joints: The angle from the horizontal of element EH
is
300
αEH = 90 − tan−1
= 45◦
600 − 900 tan θ
CF
Joint C
θ = 18.4◦ ,
EF
CF
αCF
θ
DF
CE
αCF
FH
EF
Joint F
Joint E
from which F H = −316.2 N (C)
Fy = EF + CF sin αCF − DF sin θ + F H sin θ = 0,
from which EF = −300 N (C)
from which CE = 300 N (T )
Joint E:
Fy = −EH sin αEH − EF = 0,
from which EH = 424.3 N (T )
Problem 6.102 The mass m = 120 kg. Determine
the forces on member ABC.
The weight of the hanging mass is given by
m
W = mg = 120 kg 9.81 2 = 1177 N.
s
A
B
C
300 mm
D
The complete structure as a free body: The equilibrium equations are:
FX = AX + EX = 0,
FY = AY − W = 0,
and
MA = 0.3EX − 0.4W = 0.
200 mm
A
AX = −1570 N,
200 mm
B
AY = 1177 N,
Element ABC: The equilibrium equations are
FX = Ax + CX = 0,
FY = AY + CY − BY − W = 0,
and:
MA = −0.2BY + 0.4cY − 0.4W = 0.
Solution gives BY = 2354 N (member BD is in tension),
CX = 1570 N,
m
E
Solving, we get
and EX = 1570 N.
EH
Joint F:
Fy = −CF cos αCF − DF cos θ + F H cos θ = 0,
Joint C:
Fx = −AC + CE + CF cos αCF = 0,
Solution:
EG
αEH
CE
300 mm
D
E
200
mm
W
200
mm
Cy
Ay
Ax
Cx
B
W
B
and CY = 2354 N.
Cx
B
Cy
B
Ex
Problem 6.103
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
Solution: The complete structure as a free body: The sum of the
moments:
MA = 100(1) − 400(6) − 200 + 4E = 0,
D
from which E = 625 lb. The sum of the forces:
Fy = Ay + E − 400 = 0,
2 ft
from which Ay = −225 lb.
Fx = Ax + 100 = 0,
1 ft
400 lb
A
B
2 ft
2 ft
2 ft
Dy
Dx
By
Bx
Ay
Ax
Dy
Bx
By
Cx
Dx
400 lb
Cy200 ft-lb
from which (6) By − Dy = 0
Element ABC: The sum of the moments about A:
MA = −2By − 4Cy − 200 − 6(400) = 0,
from which (7) By + 2Cy = −1300.
Cx
E
Element BD: The sum of the moments about B:
MB = 2Dx − 2Dy = 0,
from which (5) Bx − Dx = 0.
Fy = By − Dy = 0,
Cy
100 lb
thus (3) Dy + Cy + E = 0.
from which (4) Dx − Dy = 0. The sum of the forces:
Fx = Bx − Dx = 0,
200
ft-lb
E
Element ECD: (See the free body diagram.) The sum of the moments
about E:
ME = −4Dx − 2Cx − 100 = 0,
from which (2) Dx + Cx = −100.
Fy = E + Cy + Dy = 0,
C
100 lb
1 ft
from which Ax = −100 lb. These results are used as a check on the
solution below.
from which (1) 4Dx + 2Cx = −100. The sum of the forces:
Fx = Dx + Cx + 100 = 0,
2 ft
100 lb
675 lb
400 lb
150
lb
200
ftÐlb
50 lb
225
lb
50 lb
The sum of the forces:
Fx = Ax − Bx − Cx = 0,
from which (8) Ax − Bx − Cx = 0.
Fy = Ay − By − Cy − 400 = 0,
from which (9) Ay − By − Cy = 400. These nine equations are solved for
the nine reactions The reactions are DX = 50 lb, DY = 50 lb,:
CX = −150 lb, CY = −675 lb, BX = 50 lb,
BY = 50 lb, AX = −100 lb, AY = −225 lb,
and E = 625 lb.
Problem 6.104 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
Solution:
Element AB: The moment about A is
MA = −10B − 54F = 0,
80
mm
160
mm
540 mm
100
mm
90 N
where F = 90 N. From which B = −486 N. The sum of the forces:
Fy = A + B − F = 0,
from which A = 576 N
Element BC: The moment about C:
MC = −16B − 8FC = 0,
from which the cutting force is FC = 972 N
Problem 6.105 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.
C
FC
8
cm
B
90 N
A
B
16
cm
54 cm
10
cm
B
C
2 ft
A
B
C
2 ft
A
5 ft
1 ft 6 in
E
2 ft 6 in
D
uBC = 0.981i − 0.196j,
1 ft
and uBD = −0.447i − 0.894j.
The scoop: The equilibrium equations for the scoop are
FX = −TCB uBCX + EX = 0,
FY = −TCB uBCY + EY − 600 = 0,
and
MC = 1.5EX − 1.5(600 lb) = 0.
Solving, we get
1 ft 6"
TCB
C
1.5 ft
1.5 ft
G
EX
EX = 600 lb,
E
EY
EY = 480 lb,
600 lb
and TCB = 611.9 lb.
Joint B: The equilibrium equations for the scoop are
FX = TBA uBAX + TBD uBDX + TCB uBCX = 0,
and
FY = TBA uBAY + TBD uBDY + TCB uBCY = 0.
Solving, we get
TBA = 835 lb,
and TBD = −429 lb.
2 ft 6 in
1 ft
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 = −0.949i − 0.316j,
E
D
5 ft
y
TBA
x
TCB
TBD
1 ft 6 in
600 lb
Scoop
Problem 7.1 If a = 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
x
a
Solution:
y
x dA
x = x =
(1, 1)
dA
a
0
x(y dx)
a
y = x2
y dx
0
Substituting y = x2 , we get
a
4 a
x
x3 dx
3a
4 x = 0a
= 3 =
x
4
2
x dx
3
y
y = x2
0
0
For
x
a
a =2
3
2
x =
dA = y dx
x
a
Problem 7.2 Determine the y coordinate of the centroid of the area shown in Problem 7.1 if a = 3.
Solution:
y
y dA
y = y =
(1, 1)
dA
a
0
1
y(y dx)
2
a
y = x2
(y dx)
0
Substituting y = x2 , we get
a
a
1 4
1 x5
x dx
2
5
2
0
y = a
= 3 a0 =
x
2
x dx
3
0
0
a
a5
2.5
a3
3
x
y
dA = y dx
3 2
y =
a
10
For a = 3,
y=
27
10
Midpoint
y/2
a
dx
x
Problem 7.3 If the x coordinate of the centroid of the
area is x = 2, what is the value of a?
y
y = x3
0
Solution:
x dA
x = A
dA
A
a
x = 0
x(y dx)
a
.
(y dx)
0
Substituting y = x3 ,
a

x4 dx
0
x = a
=
x3 dx
0
If x = 2,
2=
4a
5
x5
5
x4
4
and
a
 = 4a
5
0
a=
10
4
= 2.5
y
y = x3
0
a
x
y
y = x3
dA = y dx
a
x
a
x
Problem 7.4 The x coordinate of the centroid of the
area shown in Problem 7.3 is x = 2. What is the y
coordinate of the centroid?
From Problem 7.4, a = 2.5
2.5
2.5
1
1 6
y(y dx)
x dx
y dA
2
2
0
0
y = =
= 2.5
2.5
dA
y dx
x3 dx
Solution:
0
2.5
2.5
4x3 y = 4 =
= 4.46
x
14 0
4
y
0
y = x3
x7
14
0
y = 4.46
y = x3
a
0
x
y
y/2
a
x
Problem 7.5 Consider the area in Problem 7.3. The
“center of the area” is defined to be the point for which
there is as much area to the right of the point as to the
left of it and as much area above the point as below it.
If a = 4, what are the x coordinate of the center of area
and the x coordinate of the centroid?
Center of Area: Let X be the coordinate of the center
of the area. The area to the left of X is
X
X4
AL =
x3 dx =
.
4
0
Solution:
The area to the right is
a
a4
X4
AR =
x3 dx =
−
.
4
4
X
Equating the two areas,
a4
X4
X4
−
=
,
4
4
4
from which a4 = 2X 4
or
1
X = a(2)− 4 .
For a = 4, X = 0.8408(4) = 3.3636
The centroid: The centroid is
a
x4 dx
4a
0
x= a
=
.
5
x3 dx
0
For a = 4, x =
16
5
= 3.200
Problem 7.6 Determine the x coordinate of the centroid of the area and compare your answer to the value
given in Appendix B.
y
y = cx n
0
Solution:
a
x(y dx)
=
a
y dx
0
x =
a
xn+2
(n+2) 0
n+1 a
x
(n+1) 0
x =a
a
x
y = cx n
dA
0
x =
x
y
x dA
x = a
a
0
a
0
=
xC/xn dx
0
C/xn dx
y = cxn
y
(n + 1)
a
(n + 2)
dA = y dx
(n + 1)
(n + 2)
Checks with result in Appendix
x
a
dx
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:
y
y dA
y = a
0
y =
1
y
y dx
2
a
y dx
a
0
2
c x
dx
cxn dx
c2
=
c
0
y =
0
a
x
0
2 2n
a
a
y 2 dx
1 0
a
=
2
y dx
0
y =
y = cx n
dA
x2n+1
(2n+1)
xn+1
(n+1)
a
0
c (n + 1) n
a
2 (2n + 1)
y/2
Checks with Appendix
a
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
Solution:
A =
1
The area:
x2
x4
+
2
4
(x + x3 ) dx =
0
The x-coordinate:
1
x(x + x3 ) dx =
x3
x5
+
3
5
0
Divide by the area: x =
32
45
1
1
=
0
=
0
3
.
4
8
.
15
= 0.711
The y-coordinate: The element of area is dA = (1 − x) dy. Note that
dy = (1 + 3x2 ) dx, hence dA = (1 − x)(1 + 3x2 ) dx. Thus
4
yA =
y dA =
(x + x3 )(1 − x)(1 + 3x2 ) dx,
A
0
from which
1
(x − x2 + 4x3 − 4x4 + 3x5 − 3x6 ) dx
0
=
1
1
4
4
3
3
− + − + − = 0.4381.
2
3
4
5
6
7
Divide by A y = 0.5841
y
y = x + x3
0
1 ft
x
1 ft
x
Problem 7.9 The y coordinate of the centroid of the
area is y = 1.063. Determine the value of the constant
c and the x coordinate of the centroid.
y
y = cx 2
0
2
x
4
Solution:
y
y dA
y = dA
dA = y dx
4
y = 2
y = cx 2
y
y dx
2
4
y dx
2
4
24
C 2 x4 dx
Cx2 dx
0
2
c2/
x5
5
2C/
x3
3
y =
=
1
2
4
C 1024
− 32
2
5
5
64
4 =
2 3 − 83
2
4
x
y = cx2
y
2
y = 5.314C
But y = 1.063
∴ C = 0.200
Now we have C known and y = Cx2
4
x4
C/x3 dx
x dA
4
2
x = = 4
=
x3
dA
C/x2 dx
3
2
x =
(256−16)
4
(64−8)
3
x = 3.214
= 3.214
4
2
4
2
y/2
1
2
3
4
x
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
Let dA bea vertical strip:
The area dA = y dx = 4 − 14 x2 dx. The curve intersects the x
1 2
axis where 4 − 4 x = 0, or x = ±4.
Therefore
4
1
4 4
4x − x3 dx
x dA
2x2 − x16
4
−4
−4
x = A
= 4
= 4 = 0.
1 2
x3
4x − 12
dA
4− x
dx
−4
4
A
−4
Solution:
To determine y, let y in equation (7.7) be the height of the midpoint of
the vertical strip:
4
1
1
1
4 − x2
4 − x2 dx
y dA
4
4
−4 2
=
y = A
4
1 2
dA
4− x
dx
4
A
−4
4
4
1 4
3
x5
8 − x2 +
x
dx
8x − x3 + 5(32)
32
−4
−4
=
=
4
4
x2
x3
4x − 12
4−
dx
−4
4
−4
=
34.1
= 1.6 ft.
21.3
y
1
y = 4 – – x 2 ft
4
x
y
dA
y
x
x
dx
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
10
y dx =
0
4
(2 + 0.02x2 ) dx
3
0
Area = 2x + 0.02
10
x3
3
y, m
Area =
= 26.67 m2
y = 2 + 0.02x2
2
0
1
dA = y dx = (2 + 0.02x2 ) dx
0
0
2
4
6
x, m
Y
X
10
0
x = 10
x dA
=
10
26.67
dA
0
2
x =
(2x + 0.02x3 ) dx
0
10
4
2/ x2/ + 0.02 x4
0
26.67
m
4
x =
100 + (0.02) 104
26.67
x = 5.62 m
10
y =
0
y
y dx
2
10
dA
=
150
26.67
1
=
(2)
10
(2 + 0.02x2 )2 dx
0
(26.67)
0
y =
1
2(26.67)
10
4x + 0.08
y =
y =
(4 + 0.08x2 + 0.0004x4 ) dx
0
x3
3
+ 0.0004
2(26.67)
74.67
53.34
y = 1.40 m
x5
5
10
0
8
10
Problem 7.12 Determine the x coordinate of the centroid of the area.
y
y = – x 2 + 8x – 12
x
Solution: First, we must determine where the curve intersects the
x-axis. These will be the limits of our integration.
y
Set y = 0
y = – x 2 + 8x – 12
0 = −x2 + 8x − 12
or
x2 − 8x + 12 = 0
(x − 6)(x − 2) = 0
Thus, y = 0 at x = 2 and x = 6.
6
6
x dA
x(y dx)
x = 2 6
= 2 6
dA
(y dx)
2
6
2
x =
x
2
(−x3 + 8x2 − 12x) dx
6
2
(−x2 + 8x − 12) dx
4
3
2 6
− x4 + 8 x3 − 12 x2
42.67
2
x = =4
6 =
10.67
x3
x2
− 3 + 8 2 − 12x
2
Note: Once we had the limits of integration, the result was apparent
due to symmetry.
Problem 7.13 Determine the y coordinate of the centroid of the area shown in Problem 7.12.
From Problem 7.12, the limits of integration are x = 2
and x = 6. The area is 10.67 units.
6
6
y
1
dA
y 2 dx
2
2
2
2
y =
=
Area
Area
6
(−x2 + 8x − 12)2 dx
1
2
y =
2
Area
6
(x4 − 16x3 + 88x2 − 192x + 144) dx
1 2
y =
2
Area
5
6
4
x
x
x3
x2
−
16
+
88
−
192
+
144x
4
3
2
1 5
2
y =
2
(10.67)
Solution:
y =
34.13
= 1.6
21.33
y = 1.6
y
y = – x 2 + 8x – 12
x
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
x dA
x(x − x3 ) dx
0
0
x = 1
= 1
dA
(x − x3 ) dx
x = 1
0
3
x
3
x2
2
5
−
x
5
−
x4
4
y
y = x3
0
1
y=x
1
0
3
1 = 1
2
0
−
−
1
5
1
4
=
2
15
1
4
= 0.533
dA
x = 0.533
x
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
1
(x + x3 ) (x − x3 ) dx
y dA
2
y = A
= 0
1
dA
(x − x3 ) dx
A
y =
1
2
6
(x − x ) dx
1 0
=
1
2
(x − x3 ) dx
1
2
y = x3
0
y =
y
0
1
7
1
−
2
4
31
−
y = 0.381
=
4
21
2
=
y=x
7 1
− x7
0
2
4 1
2 x2 − x4
0
x3
3
8
= 0.381
21
½(x+x3)
x
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 =
x1/2 − x2 dx, so
1
5/2
4 1
x
x dA
x3/2 − x3 dx
− x4
5/2
0
x = A
= 01
= 1 = 0.45.
x3/2
x3
−
dA
x1/2 − x2 dx
3/2
3
A
y
y = x 1/2
(1, 1)
0
0
If we use a horizontal strip to obtain y, we obtain
1
y dA
y 3/2 − y 3 dy
y = A
= 01
= 0.45
dA
y 1/2 − y 2 dy
A
x
0
y = x2
x
y
y = x1/2
(1, 1)
y = x2
dA
x
dx
x
Problem 7.17 Determine the x coordinate of the centroid of the area.
y
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 = x, and
y = x2 − 20. This is equivalent to the solution of the quadratic
x2 − x − 20 = 0, x1 = −4, and x2 = 5. These establish the limits
on the integration. The area: Choose a vertical strip dx wide. The
length of the strip is (x − x2 + 20), which is the distance between the
straight line y = x and the parabola y = x2 − 20. Thus the element
of area is dA = (x − x2 + 20) dx and
+5
+5
x2
x3
A =
(x − x2 + 20) dx =
−
+ 20x
= 121.5.
2
3
−4
−4
The x-coordinate:
xA =
x dA =
A
=
x=
+5
−4
= 0.5
y = x 2 – 20
y=x
2
3
(x − x + 20x) dx
x3
x4
−
+ 10x2
3
4
60.75
121.5
y
+5
= 60.75.
−4
x
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:
1
2
y =x−
yA =
1
2
(x − x2 + 20) =
y dA =
A
=
=
1
2
1
2
5
−4
(x + x2 − 20)(x − x2 + 20) dx
+5
−4
1
2
(x + x2 − 20).
(−x4 + 41x2 − 400) dx
x5
41x3
+
− 400x
5
3
−
5
−4
= −923.4.
y = − 923.4
= −7.6
121.5
Problem 7.19 Determine the y coordinate of the centroid of the area.
y
(3, 7)
Solution: The area: The area element is the horizontal strip xlong and dy wide. The length is determined by the straight line, which
has the equation y = mx + b, where
5
y 1 − y2
(7 − 0)
=
= −2.3333,
x1 − x 2
(3 − 6)
m =
and b = y1 − mx1 = 7 − (2.3333)3 = 14.
2
The length of the strip is
1
m
x =
(y − b).
0
The element of area is dA =
A =
1
m
5
2
A
=
1
m
(y − b) dy, from which
1
m
(y − b) dy =
The y-coordinate:
yA =
y dA =
y2
− by
2
y
5
2
1
m
5
2
y(y − b) dy
5
= 46.2857.
2
y = 3.4286
Problem 7.20 Determine the x coordinate of the centroid of the area in Problem 7.19.
Solution: From the solution to Problem 7.19, the area A = 13.5
is bounded by the straight line y = mx + b, where m = −2.3333,
and b = 14. The horizontal strip is x-wide and dy high, where
x=
1
m
(y − b).
The centroid of x is one half this strip, hence
1
1
xA =
(y − b)2 dy =
[(y − b)3 ]52 = 30.583,
2m2
6m2
from which x =
(3, 7)
= 13.5.
5
y3
y2
−b
3
2
1
m
x
6
30.583
13.5
= 2.265
2
0
6
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?
y
0.5 mi
0.3 mi
0.3 mi
x
0.5 mi
Solution: The area: The element of area is the vertical strip (yt −
yb ) long and dx wide, where yt = mt x + bt and yb = mb x + bb
are the two straight lines bounding the area, where
0.5
miles
and bt = 0.8 − 1.3 mt = 0.3.
(0.3 − 0)
= 0.2308,
(1.3 − 0)
0.5 miles
and bb = 0.
The element of area is
dA = (yt − yb ) dx = ((mt − mb )x + bt − bb ) dx
= (0.1538x + 0.3) dx,
from which
1.1
A =
(0.1538x + 0.3) dx
0.5
= 0.1538
x2
+ 0.3x
2
1.1
= 0.2538 sq mile.
0.5
The x-coordinate:
1.1
x dA =
(0.1538x + 0.3)x dx
0.5
= 0.1538
x3
x2
+ 0.3
3
2
1.1
= 0.2058.
0.5
x = 0.8109 mi
The y-coordinate: The y-coordinate of the centroid of the elemental
area is
1
1
y = yb + ( )(yt − yb ) = ( )(yt + yb ) = 0.3077x + 0.15.
2
2
Thus, yA =
y dA
=
=
A
1.1
(0.3077x + 0.15)(0.1538x + 0.3) dx
0.5
1.1
(0.0473x2 + 0.1154x + 0.045) dx
0.5
= 0.0471
x3
x2
+ 0.1153
+ 0.045x
3
2
Divide by the area: y =
0.3
miles
x
0.3
miles
Similarly:
mb =
0.1014
0.2538
0.2 mi
y
(0.8 − 0.3)
= 0.3846,
(1.3 − 0)
mt =
A
0.6 mi
= 0.3995 mi
1.1
= 0.1014.
0.5
0.6 miles
0.2
miles
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 = ax − bx3 . Note that y = 0 at x = 100,
thus b = a × 10−4 . Thus
100
A =
dA = a
(x − (10−4 )x3 ) dx
A
y
y = ax – bx3
0
2
= (0.5a)[x − (0.5 × 10−4 )x4 ]100
0
x
= 0.5a × 104 − 0.25b × 108 ,
100 m
and the area is A = 0.25a×104 . The y-coordinate: The y-coordinate
of the centroid of the elemental area is
y = (0.5)(ax − bx3 ) = (0.5a)(x − (10−4 )x3 ),
from which
yA =
y dA
A
= (0.5)a2
= (0.5)a2
100
0
100
0
= (0.5a2 )
(x − (10−4 )x3 )2 dx
(x2 − 2(10−4 )x4 + (10−8 )x6 ) dx
x3
2x5
x7
− (10−4 )
+ (10−8 )
3
5
7
100
0
= 3.81a2 × 104 .
Divide by the area:
y =
3.810a2 × 104
= 15.2381a.
0.25a × 104
For y = 10, a = 0.6562 , and b = 6.562 × 10−5 m−2
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
y2 = 1
x2 + —
—
a2
b2
x
2b
a
Solution:
y
x2 + —
y2 =
—
y
a2
1
b2
x2
y2
a2
b2
— + — =
1
b
x
b
x
2b
a
By symmetry, y = 0.
a
From the equation of the ellipse,
b 2
y=
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 ±
values for y.
y = ab a2 – x2
y
b
dA = y dx
0
a
dx
x dA
x = =
dA
b
a
b
a
a
0
x
a
0
a2 − x2 dx
a2 − x2 dx
Using integral tables
(a2 − x2 )3/2
x a2 − x2 dx = −
3
√
2 − x2
x
a
a2
a2 − x2 dx =
+
sin−1
2
2
Substituting, we get
3/2 a
− a 2 − x2
/3
0 x =
√
a
x a2 −x2
a2
−1 x
+
sin
2
2
a
0
x = x =
−0 + a3 /3
a3 /3
=
2
a2 π/4
0 + a2 π2 − 0 − 0
4a
3π
x
a
x
Problem 7.24 Determine the coordinates of the centroid of the area.
Strategy: Write the equation for the circular boundary
in the form y = (R2 − x2 )1/2 and use a vertical “strip”
of width dx as the element of area dA.
y
Solution: The area: The equation of the circle is x2 +√y2 = R2 .
2
2
Take the elemental area to be a vertical strip of height
√ y = R −x
and width dx, hence the element of area is dA = R2 − x2 dx. The
2
area is A = Acircle
= πR
. The x-coordinate:
4
4
xA =
x=
R
x dA =
A
x
R2
x2
−
0
(R2 − x2 )3/2
dx = −
3
R
x
R
=
0
R3
:
3
y
4R
3π
The y-coordinate: The y-coordinate of the centroid of the element of
area is at the midpoint:
1 y = ( ) R 2 − x2 ,
2
hence yA =
=
y=
A
1
2
1
2
y dA =
0
x3
3
R2 x −
R
x
(R2 − x2 ) dx
R
=
0
R
R3
3
4R
3π
Problem 7.25 Determine the x coordinate of the centroid of the area. By setting h = 0, confirm the answer
to Problem 7.24.
y
Solution:
Use a vertical strip:
R
x dA
x(R2 − x2 )1/2 dx
A
h
= R
.
x = dA
(R2 − x2 )1/2 dx
A
R
x
h
h
The upper integral is
R
x(R2 − x2 )1/2 dx = R
h
R
h
1/2
x 2
x 1− 2
R
1
=R −
(R2 − x2 )3/2
3R
The area is
A =
dA =
A
R
h
R
=R
h
1−
R
=
2
2 1/2
(R − x )
x2
R2
R
x2
=
x 1− 2
2
R
2
dx = R
h
dx
R
h
dA
1
= (R2 − h2 )3/2 .
3
2 1/2
(1 − x )
dx
1/2
x
+ R arcsin
R
− R arcsin
The centroid is x = (R2 − h2 )3/2 /(3A).
R3
πR
3( R
2 )( 2 )
=
R
1/2
4R
3π
h
y = (R2 – x2)½
x
2
1/2
πR
h2
−h 1− 2
2
R
If h = 0, x =
R
y
h
h
R
.
dx
x
dx
Problem 7.26 Determine the y coordinate of the centroid of the area in Problem 7.25.
Let y in Equation (7.7) be the height of the midpoint
of a vertical strip:
R
1 2
y dA
(R − x2 )1/2 (R2 − x2 )1/2 dx
2
y = A
= h
.
dA
dA
Solution:
A
y
h
y = ½(R2 – x2)½
x
A
The upper integral is
R
R
1 2
1
x3
(R − x2 ) dx =
R2 x −
2
3 h
h 2
1
=
2
x
2R3
h3
− R2 h +
3
3
dx
From the solution of Problem 7.25,
R πR
h2
A =
dA =
−h 1− 2
2
2
R
A
.
The centroid is y =
Problem 7.27
troids.
Solution:
1
2A
2R3
3
− R2 h +
1/2
− R arcsin
h3
3
Determine the coordinates of the ceny
Let us solve this by parts.
40 mm
A1
h
A2
b
b = 60 mm
l = 40 mm
h = 40 mm
x
60 mm
40 mm
l
y
b = 60 mm
l = 40 mm
h = 40 mm
A1 =
1
1
bh = (60)(40) = 1200 mm2
2
2
A2 = lh = (40)(40) = 1600 mm
40 mm
2
A1 + A2 = 2800 mm2
From the tables and inspection
x1 =
2
b x2 = b + l/z
3
y1 =
1
h
3
y2 =
x1 = 40 mm
y1 = 13.33 mm
1
h
2
x2 = 80 mm
y2 = 20 mm
For the composite, substituting,
x =
x1 A1 + x2 A2
= 62.9 mm
A1 + A2
y =
y1 A1 + y2 A2
= 17.1 mm
A1 + A2
x
60 mm
40 mm
h
R
.
Problem 7.28
troids.
Determine the coordinates of the ceny
20 mm
60 mm
x
30 mm
70 mm
Solution: Let us solve this problem by using symmetry and by
breaking the composite shape into parts.
l1
20 mm
y
A1
20 mm
h1
l1 = 70 mm
h1 = 70 mm
l2 = 70 mm
h2 = 70 mm
A2
h2
60 mm
60 mm
x
l2
h1 = 20 mm
l2 = 30 mm
h2 = 60 mm
A1 = l1 h1 = 1400 mm2
A2 = l2 h2 = 1800 mm2
By symmetry,
y1 = 70 mm
x2 = 0
y2 = 30 mm
For the composite,
x =
0+0
x1 A1 + x2 A2
=
=0
A1 + A2
320 mm2
y =
y1 A1 + y2 A2
A1 + A2
y =
(70)(1400) + (30)(1800)
152000
=
3200
3200
y = 47.5 mm
x =0
30 mm
70 mm
l1 = 70 mm
x1 = 0
x
Problem 7.29
troids.
Determine the coordinates of the cen-
y
Solution: Divide the shape up into a rectangle, a semicircle, and
a circular cutout as shown. Note that the y coordinates of the centroids
of all three component areas lie on the x axis. Thus, y = 0 for the
combined area.
20 mm
x
40 mm
Rectangle:
2
Area1 = a(2R) = 9600 mm ,
120 mm
and x1 = a/2 = 60 mm.
Semicircle: See example 7.3 or 7.4 for the value of the x coordinate
of the centroid of a semicircle. Also note the x displacement of the
centroid relative to the y axis.
y
Area2 = πR2 /2 = 2513 mm2 ,
20 mm
x2 = a + (4R)/(3π) = 137.0 mm.
x
Cutout:
40 mm
Area3 = πr 2 = 1257 mm2 ,
120 mm
x3 = 120 mm.
Combined Area:
R
x = (x1 Area1 + x2 Area2 − x3 Area3 )/(Area1 + Area2 − Area3 )
1
= 70.9 mm
Determine the coordinates of the cen-
r
3
y
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 of the circle, y = ± R2 − x2 . √
The height of the strip will
be twice the positive value, so that dA = 2 R2 − x2 dx, from which
R
A=
dA = 2
(R2 − x2 )1/2 dx
0
√
x R 2 − x2
R2
=2
+
sin−1
2
2
x
R
R
=
0
10 in
x
πR2
2
20 in
The x-coordinate:
R x dA = 2
x R2 − x2 dx
A
2
a
Problem 7.30
troids.
A
+
2R
0
=2 −
Divide by A: x =
(R2 − x2 )3/2
3
R
=
0
2R3
.
3
y
10 in
4R
3π
The y-coordinate: From symmetry, the y-coordinate is zero.
x
4(20)
The composite: For a complete half circle x1 = 3π = 8.488 in..
For the inner half circle x2 = 4.244 in. The areas are
A1 = 628.32 in.2
2
and A2 = 157.08 in .
20 in
Problem 7.31
troids.
Solution:
Determine the coordinates of the cen-
y
Use Appendix B:
800
mm
y
B
a
600 m
m
x
h
m
m
0
80
600
mm
400
mm
θ
C
X = (a + b) / 3
Y = h/ 3
400 mm
x
D
y
b
We need to know h and a. This is equivalent to knowing the coordinates
of point B.
We can use the law of cosines to find the angle θ and then use θ to find
(xB , yB ).
800
mm
600
mm
B
x
d
c
C
b
D
b = 400 mm
c = 600 mm
d = 800 mm
From the law of cosines
c2 = b2 + d2 − 2 bd cos θ
Substituting, θ = 46.57◦
xB = d cos θ = 550.0 mm
yB = d sin θ = 580.9 mm
∴ h = 580.9 mm
a = 550.0 mm b = 400 mm
x = (a + b)/3 = 316.67 mm
y = h/3 = 193.6 mm
400
mm
Problem 7.32 Determine the coordinates of the centroids.
Solution: The results for a half-circle of radius R:
y
πR2
4R
, x1 =
, y1 = 0.
2
3π
A1 =
Consider three figures: The complete circle, (1) the half circle cut out,
and (2) the composite figure. The centroid of the complete circle is at
the origin; x = 0 and y = 0. The product of its centroid coordinates
and its area is zero. From the composite algorithm, it follows that
12 in
x
20 in
0 = A1 x1 + A2 x2 ,
from which
x2 = −
A1
A2
x1 ,
and y2 = −
A1
A2
y1 .
y
12 in
The areas:
A1 =
πR12
,
2
A2 =
πR22
x
πR12
−
.
2
For R1 = 12 in. and R2 = 20 in., A1 = 226.2 in.2 , and A2 =
1030.4 in.2 , and x1 = 5.093 in.. Thus
x2 = −
226.2
1030.4
20 in
(5.093) = −1.18 in.,
and y2 = 0 , since y1 = 0 .
Problem 7.33
troids.
Determine the coordinates of the cen-
y
Solution: Divide the object into three areas: (1) A rectangle on
the left, 100 mm by 60 mm. (2) A rectangle at the lower right, 80 mm
by 40 mm. (3) A semi circle to the far lower right, radius 20 mm.
60 mm
The areas and centroid coordinates are
(1)
A1 = 6 × 103 mm2 ,
100 mm
x1 = 30 mm,
40 mm
y1 = 50 mm
(2)
x
A2 = 3.2 × 103 mm2 ,
140 mm
x2 = 100 mm,
y2 = 20 mm, and
(3)
y
A3 = 628.32 mm2 ,
60
mm
x3 = 148.49 mm,
y3 = 20 mm.
100 mm
(1) The composite area is
A =
3
1
Ai = 9.828 × 103 .
The centroid coordinates for the composite are
3
x=
1
3
and
y=
1
Ai xi
A
= 60.37 mm ,
Ai yi
A
= 38.31 mm
140
mm
40
mm
x
Problem 7.34
troids.
Determine the coordinates of the cen-
y
18 in
x
6 in
6 in
6 in
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.
The areas and their centroids are determined by inspection:
(1)
(2)
(3)
A1 = 182 = 324 in.2 , x1 = 9 in., y1 = 9 in.
A2 = ( 12 )(18)(6) = 54 in.2 , x2 = 9 in., y2 = 6 in.
A3 = π9
2
21.8 in.
2
= 127.2 in.2 , x3 = 9 in., y3 = 18 +
4(9)
3π
The composite area: A = A1 − A2 + A3 = 397.2 in.2 .
The composite centroid:
x=
A1 x1 −A2 x2 +A3 x3
A
= 9 in.
y=
A1 y1 −A2 y2 +A3 y3
A
= 13.51 in.
y
18 in.
x
6 in. 6 in. 6 in.
=
Problem 7.35
troids.
Determine the coordinates of the cen-
y
20 mm
30 mm
20 mm
10
mm
30 mm
x
90 mm
Solution:
Determine this result by breaking the compound object
y
into parts
y
m
20 m
m
10 m
30 mm
=
A1
A2
+
40 mm
30 mm
30 mm
10
mm
20 mm
A1 = (30)(90) = 2700 mm
x
90 mm
2
For the composite:
y1 = 15 mm
x
=
x1 A1 + x2 A2 + x3 A3 − x4 A4
(A1 + A2 + A3 − A4 )
x
=
155782
= 35.3 mm
4414.2
y
=
y1 A1 + y2 A2 + y3 A3 − y4 A4
A1 + A2 + A3 − A4
y
=
146675
= 33.2 mm
4414.2
(sits on top of A1 )
A2
= (40)(50) = 2000 mm2
x2
= 20 mm
y2
= 30 + 25 = 55 mm
A3 =
1 2
π
πr = (20)2 = 628.3 mm2
2 0
2
x3 = 20 mm
y3 = 80 mm +
A4 :
A4
x
x1 = 45 mm
A3 :
–
30 mm
90 mm
A2 :
A3
+
20 mm
A1 :
20 mm
m
50 m
4r0
= 88.49 mm
3π
A4 = (30)(20) + πri2
A4 = 600 + π(10)2 = 914.2 mm2
x4 = 20 mm
y4 = 50 + 15 = 65 mm
Area (composite)
= A1 + A2 + A3 − A4
= 4414.2 mm2
The value for y is not the same as in the new problem statement. This value
seems correct. (The x value checks).
Problem 7.36
troids.
Determine the coordinates of the cen-
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 = 25 mm, R = 15 mm, and r = 5 mm. If we do this,
we find that for this shape, measuring from the y axis, x = 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
y
1
15
15 mm
Area 1: Area1 = 1025 mm2 , x1 = 18.04 mm, and y1 = 50 mm.
50 mm
y 4
Area 2: Area2 = 1025 mm2 , x2 = 18.04 mm, and y2 = 0 mm.
Area 3: Area3 = 1025 mm2 , x3 = −18.04 mm, and y3 = 0 mm.
5 mm
5
5 mm
3
2
15
5 mm
2
Area 4: Area4 = 600 mm , x4 = 0 mm, and y4 = 25 mm.
Area 5: Area5 = 450 mm2 , x5 = −7.5 mm, and y5 = 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 = Area1 + Area2 + Area3 + Area4 + Area5
= 4125 mm2 .
Then, x = (x1 Area1 + x2 Area2 + x3 Area3 + x4 Area4
+x5 Area5 )/AreaTOTAL = 3.67 mm,
and y = (y1 Area1 + y2 Area2 + y3 Area3 + y4 Area4
+y5 Area5 )/AreaTOTAL = 21.52 mm.
15 mm
15 mm
10 15 15 10
mm mm mm mm
x
Problem 7.37 The dimensions b = 42 mm and h =
22 mm. Determine the y coordinate of the centroid of
the beam’s cross section.
y
200 mm
h
120 mm
x
b
Solution:
Work as a composite shape
y
y
100 mm
100 mm
A2
h
b = 42 mm
h = 42 mm
A
1
200 mm
h
120 mm
120 mm
x
b
x
b = 42 mm
b
h = 22 mm
A1 = 120 b mm2 = 5040 mm2
x1 = 0
by symmetry
y1 = 60 mm
A2 = 200 h = 4400 mm2
x2 = 0
y2 = 120 +
x =
h
= 131 mm
2
A1 x1 + A2 x2
0+0
=
A1 + A2
9440
x =0
y =
A1 y1 + A2 y2
= 93.1 mm
A1 + A2
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 = 90 mm, what are the
dimensions b and h?
Solution: From the solution to Problem 7.37
A1 = 120 b, A2 = 200 h
and y =
y =
y1 A1 + y2 A2
A1 + A2
(60)(120 b) + 120 +
h
2
(200 h)
120 b + 200 h
where y1 = 60 mm
y = 90 mm
A1 + A2 = 8400 mm2
Also, y2 = 120 + h/2
Solving these equations simultaneously we get
h = 18.2 mm
b = 39.7 mm
y
200 mm
h
120 mm
x
b
200 mm
h
A2
A1
b
120 mm
Problem 7.39 Determine the x coordinate of the centroid of the Boeing 747’s vertical stabilizer.
y
11 m
48°
x
70°
12.5 m
Solution: We can treat the stabilizer as a rectangular area (1) with
two triangular cutouts (2 and 3): The dimensions a and b are
y
•
a = 11 tan 48 = 12.22 m
and b =
11
= 4.00 m.
tan 70•
The areas are
11 m
48°
70°
x
12.5 m
A1 = (11)(12.5 + b) = 181.5,
A2 =
1
(11)a = 67.2 m,
2
A3 =
1
(11)b = 22.0 m.
2
3
The x coordinate of the centroid is
A1 x1 − A2 x2 − A3 x3
x =
A1 − A2 − A3
− A2 a3 − A3 (12.5 + 2b/3)
A1 12.5+b
2
=
= 9.64 m
A1 − A2 − A3
Problem 7.40 Determine the y coordinate of the centroid of the vertical stabilizer in Problem 7.39.
Solution:
Treating the stabilizer as a rectangular area (1) with
triangular cutouts (2 and 3) as shown in the sol of Problem 7.39, the y
coordinate of the centroid is
y =
=
A1 y1 − A2 y2 − A3 y3
A1 − A2 − A3
A1 (11/2) − A2 [2(11/3)] − A3 (11/3)
.
A1 − A2 − A3
From the solution of Problem 7.39,
A1 = 181.5 m,
A2 = 67.2 m,
and A3 = 22.0 m,
giving the result y = 4.60 m.
y
a
2
11 m
48°
70°
x
b
12.5 m
Problem 7.41 The area has elliptical boundaries. If
a = 30 mm, b = 15 mm, and ε = 6 mm, what is the x
coordinate of the centroid of the area?
Solution:
x2
(a + ε)2
y
The equation of the outer ellipse is
y2
=1
(b + ε)2
+
and for the inner ellipse
x2
a2
+
y2
b2
b
=1
x
We will handle the problem by considering two solid ellipses
For any ellipse
x dA
x = =
dA
β
/
α
β
/
α
α
0
x
α2 − x2 dx
α2
−
x2
y
dx
From integral tables
(α2 − x2 )3/2
x α2 − x2 dx = −
3
√
x α 2 − x2
α2
α2 − x2 dx =
+
sin−1
2
2
3/2 α
− α 2 − x2
0 Substituting x =
√
x α
x α2 −x2
α2
+ 2 sin α
3
x
α
a
−0 + α3 /3
α3 /3
=
x = 2 π/4
α2 π
α
0+ 2 2 −0−0
4α
3π
Also Area =
dA =
β
=
α
Area =
β
α
β
α
α
α2
2
π
2
y
ε
b
α2 + x2 dx
0
√
x α 2 − x2
α2
+
sin−1
2
2
b
x
0
x =
a
x
α
α
ε
a
= παβ/4
A1
=
–
(The area of a full ellipse is παβ so this checks.
Now for the composite area.
y
For the outer ellipse, α = a + ε β = b + ε and for the inner ellipse
α=a β=b
β
y = α α 2 – x2
dA = y dx
β
Outer ellipse
x1
4(a + ε)
=
3π
A1 =
π(a + ε)(b + ε)
4
Inner Ellipse
x2 =
A2
4a
3π
πab
=
4
x
0
x
α
For the composite
x =
x1 A1 − x2 A2
A1 − A2
Substituting, we get
x1 = 15.28 mm
A1 = 2375 mm
and x = 19.0 mm
2
x2 = 12.73 mm
A2 = 1414 mm2
A2
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
4a(a + 2b)
.
3π(a + b)
x=
Solution:
From the solution to 7.41, we have
x1 =
4(a + ε)
3π
x2 =
4a
3π
so x1 A1
x2 A2
A1 − A2
A1 − A2
A1 =
(x1 A1 − x2 A2 ) =
π(a + ε)(b + ε)
4
A2 =
1 2
(a b + 2abε + bε2
3
+a2 ε + 2aε2 + ε3 − a2 b)
πab
4
(x1 A1 − x2 A2 ) =
1
((2ab + a2 )ε
3
+(2a + b)ε2 + ε3 )
(a + ε)2 (b + ε)
=
3
Finally x =
a2 b
=
3
π
= (ab + aε + bε + ε2 − ab)
4
π
= (aε + bε + ε2 )
4
x =
x =
x1 A1 − x2 A2
A1 − A2
1
(2ab + a2 ) + (2a + b)ε + ε2 ε/
3
π
4
[(a + b) + ε] ε/
4a(a + 2b)
4(2a + b)ε
4 2
+
+
ε
3π(a + b)
3π
3π
Taking the limit as ε → 0
x=
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) = 4ft,
and (3) the composite sail. The areas and coordinates are:
(1)
2
A1 = 210 ft2 ,
x1 =
2
3
20 = 13.33 ft,
y1 =
1
3
21 = 7 ft.
(2)
1
4a(a + 2b)
3π(a + b)
A2 = 42 ft2 ,
x2 = 16 +
3
2
3
4 = 18.67 ft,
y2 = 7 ft.
(a)
y
(3)
y
y
(14, 29)
(12.5, 23)
(20, 21)
(3, 20)
1
2
3
x
(16, 0)
(3.5, 21)
x
(10, 0)
(b)
x
(23, 0)
The composite area: A = A1 − A2 = 168 ft2 .
The composite centroid:
x=
A1 x1 −A2 x2
A
= 12 ft ,
y=
A1 y1 −A2 y2
A
= 7 ft
Problem 7.44
lem 7.43.
Determine the centroid of sail 2 in Prob-
Solution:
Divide the object into five areas: (1) a triangle on the
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 =
3(20)
2
= 30 ft2 ,
x1 =
2
3
3 = 2 ft,
y1 =
1
3
20 = 6.67 ft.
(2)
y2 =
9.5
2
= 7.75 ft,
23
= 11.5 ft
2
Problem 7.45
lem 7.43.
2
(3)(9.5) = 14.25 ft2 ,
1
3
x3 = 3 +
9.5 = 6.167 ft,
2
3 = 22 ft
3
A4 = 12 (2.5)(23) = 28.75 ft2 ,
y3 = 20 +
(4)
y4 =
2
3
1
3
(2.5) = 11.67 ft,
23 = 7.66 ft
The composite area: A = A1 + A2 − A3 − A4 = 205.5 ft2 .
The composite centroid:
(5)
x=
A1 x1 +A2 x2 −A3 x3 −A4 x4
A
= 6.472 ft ,
y=
A1 y1 +A2 y2 −A3 y3 −A4 y4
A
= 10.603 ft
Determine the centroid of sail 3 in Prob-
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)
1
x4 = 10 +
A2 = (23)(9.5) = 218.5 ft2 ,
x2 = 3 +
A3 =
(3)
a
f
g
b
e
2
A1 = 36.75 ft ,
x1 = 1.167 ft,
y1 = 14 ft.
(2)
A2 = 406 ft2 ,
o
x2 = 7 ft,
(5)
y2 = 14.5 ft.
(3)
(6)
The composite area:
A = −A1 + A2 − A3 + A4 − A5 = 429.75 ft2 .
2
A4 = 130.5 ft ,
y4 = 9.67 ft.
A5 = 28 ft2 ,
y5 = 25 ft.
y3 = 26.33 ft
x4 = 17 ft,
d
x5 = 1.75 ft,
A3 = 42 ft2 ,
x3 = 7 ft,
(4)
c
The composite centroid:
x=
−A1 x1 +A2 x2 −A3 x3 +A4 x4 −A5 x5
A
= 10.877 ft
y=
−A1 y1 +A2 y2 −A3 y3 +A4 y4 −A5 y5
A
= 11.23 ft
Problem 7.46 The value of the distributed load w at
x = 6 m is 240 N/m.
(a) The equation for the loading curve is w = 40x N/m.
Use Eq. (7.10) to determine the magnitude of the
total force exerted on the beam by the distributed
load.
(b) 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?
(c) Determine the reactions at A and B.
Solution:
(a) ω(x) = 40x N/m
6
F =
ω(x) dx =
0
6
0
6
x2 40x dx = 40 2 0
F = 720 N
(b)
Using the area analogy
6
6
ω(x)x dx
40x2 dx
0
0
x= 6
=
720
ω(x) dx
x=
0
3
40 x3
6
720
0
= 4.00 m
x = 4.00 m (720 N force)
(c)
Fx :
Ax = 0
Fy :
Ay + By − 720 = 0
MA :
(−4)(720) + (6)By = 0
Solving,
By = 480 N
Ay = 240 N
Ax = 0
y
240 N/m
A
B
x
6m
y
720 N
4m
AX
2m
x
AY
BY
y
240 N/m
A
B
6m
x
Problem 7.47 In a preliminary design study for a
pedestrian bridge, an engineer models the combined
weight of the bridge and maximum expected load due
to traffic by the distributed load shown.
y
= 50 kN/m
(a) Use Eq. (7.10) to determine the magnitude of the
total force exerted on the bridge by the distributed
load.
(b) 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?
(c) Determine the reactions at A and B.
Solution:
(a) F =
10
ω(x) dx =
0
10
0
A
10 m
10
50 dx = 50x kN
0
F = 500 kN
(b)
Area analogy
10
10
xω(x) dx
50x dx
x = 0 10
= 0
500 kN
ω(x) dx
0
10
25x2 0
x=
500 kN
=
2500 kN · m
500 kN
x = 5 m (500 kN force)
(c)
Fx :
Ax = 0
Fy :
Ay + By − 500 = 0
MA :
(−5)(500) + 10By = 0
By = 250 kN
Ax = 0
Ay = 250 kN
y
= 50 kN/m
x
B
A
10 m
y
500 kN
5m
5m
Ax
x
Ay
By
B
x
Problem 7.48
support A.
Determine the reactions at the built-in
y
200 N/m
x
A
3m
Total distributed force acting on the beam ω(x) =
− 3) for 3 ≤ x ≤ 6. The total force acting on the beam
due to the distributed load is
6
6
200
200 x2
F =
(x − 3) dx =
− 3x
3
3
2
3
3
200 36
9
200 9
F =
− 18 − + 9 =
·
3
2
2
3
2
Solution:
200
(x
3
F = 300 N.
Using the area analogy and the fact that the load is triangular, x = 5 m.
300 N
mA
5m
Ax
Ay
Fx :
Ax = 0
Fy :
Ay − 300 N = 0
MA :
MA − (5)(300) = 0
Ax = 0
Ay = 300 N
MA = 1500 N-m
y
200 N/m
x
A
3m
3m
3m
Problem 7.49
Determine the reactions at A and B.
y
x
A
B
L/2
Solution:
L/2
Let us break the load into two parts and use the area
analogy.
y
x
A
L1
A
B
L2
L
2
L
2
B
L/2
L/2
x
Now we can find the support reactions
3L
4
For Load L1
2ω0
x for (0 ≤ x ≤ L/2)
L
ω(x) =
L
3
L
≤x≤L
2
L1 =
L1 =
L/2
0
L
L/2
L
ω0 dx = ω0 x
L/2
=
ω0 L
2
And from the area analogy, L2 acts half way between L/2 and L.
3L
x2 =
.
4
x1 = L/3
By
Ay
using the area analogy, load L1 acts 2/3 of the distance from the origin
to L/2. Thus
L2 =
L
2
L/2
2ω0
2/ω0 x2 x dx =
L
L 2/ 0
ω0 L2
Lω0
=
L 4
4
Load 2
4
Ax
Load 1
2
L
For Load L2
ω(x) = ω0 for
L
Fx :
Ax = 0
Fy :
Ay + By −
MA :
By
L
2
Lω0
Lω0
−
=0
4
2
−
Lω0
4
Solving the third eqn.
By =
Lω0
3Lω0
11
+
=
Lω0
6
4
12
From the second eqn,
Ay + By =
Hence Ay =
3
Lω0
4
3
lω0
4
0
− By = − Lω
6
Ax = 0 Ay = −Lω0 /6
By = 11 Lω0 /12
L
3
−
Lω0
2
3L
4
=0
Problem 7.50
support A.
Determine the reactions at the built-in
y
Solution: The free-body diagram of the beam is: The downward
force exerted by the distributed load is
5
x2
w dx =
3 1−
dx
25
L
0
5
x3
=3 x−
= 10 kN.
75 0
= 3(1 – x 2/25) kN/m
The clockwise moment about the left end of the beam due to the distributed load is
5
x3
xw dx =
3 x−
dx
25
L
0
5
x2
x4
= 18.75 kN-m.
=3
−
2
100 0
x
A
5m
= 3(1 – x 2/25) kN/m
From the equilibrium equations
Fx = Ax = 0,
Fy = Ay − 10 = 0,
m(leftend) = Ma + 5Ay − 18.75 = 0,
x
A
5m
= 3(1 – x 2/25) kN/m
we obtain
Ma
Ax = 0,
Ay = 10 kN,
Ax
5m
x
Ay
and Ma = −31.25 kN-m.
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 = −10x − x2 + 0.2x3 kN[/]m.
(a) Determine the magnitude of the total force exerted
on the foundation by the distributed load.
(b) 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
F
=−
(10x + x2 − 0.2x3 ) dx
0
= −5x2 −
|F |
x3
3
+
0.2 4
x
4
y
10
A
0
x
= 333.3 kN
(b)
The moment about the origin is
10
M
=−
(10x + x2 − 0.2x3 )x dx
0
= −
|M |
10 3
1
0.2 5
x − x4 +
x
3
4
5
10
,
0
= 1833.33 kN.
The distance from the origin to the equivalent force is
d=
10 m
2m
|M |
= 5.5 m,
F
from which
|MA | = (d + 2)F = 2500 kN m.
w
Problem 7.52 The distributed load is w = 6x +
0.4x2 N/m. Determine the reactions at A and B.
A
B
2m
4m
Solution:
6
F =
The total distributed load is
6
w(x) dx =
(6x + 0.4x3 ) dx
0
F = 6
0
x2
2
+ 0.4
6
x4
4
= 3.36 +
0
0.4(36)2
4
F = 237.6 N
A
Using the area analogy, the point of application of the equivalent concentrated force F is
6
6
xw(x) dx
(6x2 + 0.4x4 ) dx
x = 0 6
= 0
237.6
w(x) dx
x =
0
3
5
6 x3 + 0.4 x5
6
237.6
0
= 4.436 m
Now to determine the support reactions
y
x
F
Ax
x
4m
Ay
By
Fx :
Ax = 0
Fy :
Ay + By − F = 0
MA :
4By − xF = 0
Solving, we get
Ax = 0
Ay = −25.9 N
By = 263.5 N
B
4m
2m
Problem 7.53 The aerodynamic lift of the wing is
described by the distributed load
w = −300 1 − 0.04x2 N/m.
y
The mass of the wing is 27 kg, and its center of mass is
located 2 m from the wing root R.
x
R
(a) Determine the magnitudes of the force and the moment about R exerted by the lift of the wing.
(b) Determine the reactions on the wing at R.
2m
5m
Solution:
y
(a)
The force due to the lift is
5
F
= −w =
300(1 − 0.04x2 )1/2 dx,
w
0
300 5
=
(25 − x2 )1/2 dx
5 0
√
x 25 − x2
25
= 60
+
sin−1
2
2
F
F
|F |
R
x
5
2m
5
5m
= 375π N,
0
= 1178.1 N.
w
The moment about the root due to the lift is
5
M
= 300
(1 − 0.04x2 )1/2 x dx,
0
(25 − x2 )3/2
= −60
3
M
|M |
5
0
MR
mg
FR
60(25)3/2
=
= 2500
3
2m
3m
= 2500 Nm.
(b) The sum of the moments about the root:
M = M R + 2500 − 27g(2) = 0,
from which M R = −1970 N-m. The sum of forces
Fy = FR + 1178.1 − 27g = 0,
from which FR = −1178.1 + 27g = −913.2 N
Problem 7.54 The force F = 2000 lb. Determine the
reactions at A and B.
Solution:
The free-body diagram of the beam is: The downward
force exerted by the distributed load is
3
3
x3
w dx =
400x2 dx = 400
= 3600 lb.
3 0
L
0
The clockwise moment about the left end of the beam due to the distributed load is
3
3
x4
xw dx =
400x3 dx = 400
= 8100 ft-lb.
4 0
L
0
From the equilibrium equations
Fx = Ax = 0,
Fy = Ay + B − 3600 − 2000 = 0,
m(leftend) = 3Ay − 6(2000) + 8B − 8100 = 0,
we obtain Ax = 0, Ay = 4940 lb, B = 660 lb.
y
w = 400x2 lb/ft
F
A
B
x
3 ft
3 ft
w = 400x2 lb/ft
2 ft
2000 lb
AX
x
3 ft
AY
3 ft
2 ft
B
Determine the reactions at A and B.
Problem 7.55
y
4 kN/m
20 kN-m
A
6 kN
Solution: Break the load into two parts and find the equivalent
concentrated load for each part. Then find the reactions at A and B
x
6
12
6 kN
18
w1 (x) = 4 kN/m (6 m ≤ × ≤ 12 m)
F1 =
12
6
kN/m (12 m ≤ × ≤ 18 m)
w1 (x) dx =
12
F1 = 4x = 24 kN
12
4 dx
6
6
By symmetry, F1 is applied at x = 9 m
18
18
2
F2 =
w2 (x) dx =
− x + 12
3
12
12
18
x2
F2 = −
+ 12x
= 108 − 96 kN
3
12
dx
F2 = 12 kN
By the area analogy, this load is applied at x = 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 = 0
Fy :
Ay + By − 24 − 12 − 6 = 0
MA :
Solving
x
B
w2 (x)
4 kN/m
2
− x + 12
3
6m
4 kN/m
A
w1 (x)
w2 (x) =
6m
y
20 k N-m
y
6m
x
B
(6)(6) + 20 − 3(24) + 6By − 8(12) = 0
Ax = 0, Ay = 23.3 kN, By = 18.7 kN
6m
6m
6m
Problem 7.56 Determine the reactions on member
AB at A and B.
600 N/m
C
A
B
1m
1m
Solution: Divide the beams at the pin, B. Find the equivalent
concentrated load on AB and find two equivalent concentrated loads
on BC. Then find the support reactions
600 N/m
Load 1 w1 (x) = 300x N/m
1
1
F1 =
300x dx = 150x2 = 150 N
0
C
A
B
1m
0
1m
Similarly,
w2 (x)
F2 = 150 N
2
2
F3 =
300 dx = 300x = 300 N
1
300 N/m
w1 (x)
1m
A
1
300 N/m
x
B
w3 (x) B
using the area analogy,
F1
F3
MA
5
m
3
AX
3
= 300 N applied at x = m
2
MA :
For BC
Fx :
Fy :
MB :
Ay + By − 150 = 0
2
3
(150) + MA = 0
−Bx = 0
−By − 300 − 150 + Cy = 0
(1)Cy − (0.5)(300) −
Solving, we get, acting on AB
Ax = 0
Ay = 350 N
MA = 300 N-m
By = −200 N
Bx = 0
also Cy = 250 N
C
1m
BY
300 N
BX
0.5 m
BX
Ax + Bx = 0
(1)By −
2m
3
AY
we now need to write the equilibrium equations.
For AB
Fx :
Fy :
1m
150 N
2
= 150 N applied at x = m
3
F2 = 150 N applied at x =
300 N/m
x
2
3
(150) = 0
BY
1m
150 N
1 m
3
CY
Problem 7.57 Determine the reactions on member
ABCD at A and D.
2 kN/m
2 kN/m
D
E
1m
1m
C
1m
B
1m
A
F
1m
Solution: First, replace the distributed forces with equivalent concentrated forces, then solve for the loads. Note that BF and CE are
two force members.
Distributed Load on ABCD, F1
By area analogy, concentrated load is applied at y = ±m. The load
is 12 (2)(3) kN
E
D
1m
1m
C
F1 = 3 kN
By the area analogy,
1m
F2 = 4 kN applied at x = 1 m
Assume FCE and FBF are tensions
For ABCD:
Fx :
Ax + FBF cos 45◦ + FCE cos 45◦ + Dx + 3 kN = 0
Fy :
Ay + Dy − FBF sin 45◦ + FCE sin 45◦ = 0
MA :
−1(FBF cos 45◦ ) − 2(FCE cos 45◦ ) − 2(3) − 3Dx = 0
For DE:
Fx :
Fy :
ME :
2 kN/m
2 kN/m
B
1m
A
F
1m
2 kN / m
y
−Dx − FCE cos 45◦ = 0
−Dy − FCE sin 45◦ − 4 = 0
(1)Dy = 0
3m
x
y
2 kN/ m
2m
6.57 Contd.
Solving, we get
Ax = 7 kN
Ay = −6 kN
Dx = 4 kN
Dy = 0
also FBF = −14.14 kN(c)
FCE = −5.66 kN(c)
DY
DX
FCE
45
3 kN
°
C
B
2m
FBF
45°
AX
AY
1
4 kN
1m
DX
E
DY
45°
FCE
2
Problem 7.58
of the frame.
Determine the forces on member ABC
A
1m
3 kN/m
B
1m
C
2m
2m
1m
Solution: The free body diagram of the member on which the
distributed load acts is
From the equilibrium equations
Fx = Bx = 0,
Fy = By + E − 12 = 0,
m(leftend) = 3E − (2)(12) = 0,
we find that Bx = 0, By = 4 kN, and E = 8 kN. From the lower
fbd, writing the equilibrium equation
m(leftend) = −2Cy − (4)(8) = 0,
we obtain Cy = −16 kN. Then from the middle free body diagram,
we write the equilibrium equations
Fx = Ax + Cx = 0,
Fy = Ay − 4 − 16 = 0,
m(rightend) = −2Ax − 2Ay + (1)(4) = 0
A
1m
3 kN/m
B
1m
C
2m
1m 1m
1m
(4 m)(3 kN/m) = 12 kN
BX
2m
BY
1m
E
AX
4 kN
AY
obtaining Ax = −18 kN, Ay = 20 kN, Cx = 18 kN.
CX
8 kN
CY
CX
DX
DY
2m
CY
2m
Problem 7.59 Determine the coordinates of the centroid of the truncated conical volume.
Strategy: Use the method described in Example 7.8.
y
z
R
x
h–
2
Solution:
Refer to Example 7.8.
h–
2
y
y
dV
z
z
R
x
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.
xdV
x = V
dV
= V
x =
h
x4 /4 h/2
[x3 /3]h
h/2
x =
45
h
56
h
h/2
h
h/2
=
π
R2 3
x dx
h2
π
R2 2
x dx
h2
h4 /4 − h4 /64
[h3 /3 − h3 /24]
h–
2
h–
2
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:
y
O
y
y = ax1/2
1/2
y = ax
7m
y
dx
dV = π y2dx
10 m
x
x
dV = πy 2 dx
7
7
χπy 2 dx
χπa2 x dx
x = 0 7
= 0 7
πy 2 dx
πa2 x dx
x =
0
7
x3 /3 0
[x2 /2]70
0
= 4.67 m
The height of the centroid above the ground is 7 m − x
h = 2.33 m
The volume is
7
V =
πa2 x dx = πa2
0
49
2
To determine a,
y = 5, m when x = 7 m.
√
y = ax1/2 , 5 = a 7
√
a = 5/ 7a2 = 25/7
V =π
25
7
V = 275 m3
49
2
= 275 m3
m3
Problem 7.61 The object shown, designed to serve
as a pedestal for a speaker, has a profile obtained by
revolving the curve y = 0.167x2 about the x axis. What
is the x coordinate of the centroid of the object?
y
z
x
0.75 m
0.75 m
Solution:
y
y = 0.167 x2
dV = π y2dx
x
dv
xdV
x = V
V
2 2
xπ(0.167x ) dx
= 0.75
1.50
dV
π(0.167x2 )2 dx
x =
z
1.50
0.75
1.5
π(0.167)2
· 0.75
1.5
π(0.167)2
0.75
x = 1.27 m
x
0.75 m
0.75 m
x5 dx
=
4
x dx
x6 /6
1.5
0.75
[x5 /5]1.5
0.75
Problem 7.62
the pyramid.
Determine the volume and centroid of
y
400 mm
z
600 mm
x
400 mm
Solution:
The volume: The element of volume is a square of
thickness dx. The length of a side is a linear function of the height of
the pyramid. Thus L = Ax + B. For x = 0, L = 0, and therefore
B = 0. For x = 600, L = 400, therefore A = 23 . The area is
L2 = 49 x2 . The volume element is dV = 49 x2 dx, from which
600
4
V =
dV =
x2 dx
9
V
0
=
4
27
[x3 ]600
= 3.2 × 107 mm3 = 0.032 m3
0
The x-coordinate:
600
4
x dV =
x3 dx
9
V
0
=
1
9
[x4 ]600
= 1.44 × 1010 .
0
Divide by V : x = 0.45 m.
By symmetry, the y- and z-coordinates of the centroid are zero.
y
y
x
600
mm
z
400
mm
400
mm
Problem 7.63 Determine the centroid of the hemispherical volume.
y
R
z
x
Solution:
z 2 = R2 .
The equation of the surface of a sphere is x2 + y 2 +
The volume: The element of volume is a disk of radius ρ and thickness
dx. The radius of the disk at any point within the hemisphere is ρ2 =
y 2 + z 2 . From the equation of the surface of the sphere, ρ2 = (R2 −
x2 ). The area is πρ2 , and the element of volume is dV = π(R2 −
x2 ) dx, from which
V =
Vsphere
2π 3
=
R .
2
3
The x-coordinate is:
R
x dV = π
(R2 − x2 )x dx
V
0
=π
R 2 x2
x4
−
2
4
R
0
π
= R4 .
4
Divide by the volume:
x =
πR4
4
3
2πR3
=
3
R.
8
By symmetry, the y- and z-coordinates of the centroid are zero.
y
x
R
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
ρ and thickness dx. The area of the disk is πρ2 , and the element of
volume is πρ2 dx. From the equation of the surface of a sphere (see
solution to Problem 7.63) ρ2 = R2 − x2 , from which the element of
volume is dV = π(R2 − x2 ) dx. Thus
R
dV = π
(R2 − x2 ) dx
V =
V
R/2
= π R2 x −
x3
3
R
5π
24
=
R/2
R3 .
The x-coordinate:
R
x dV = π
(R2 − x2 )x dx
V
R/2
=π
R 2 x2
x4
−
2
4
R
=
R/2
9π 4
R .
64
Divide by the volume:
x =
9πR4
64
24
5πR3
=
27
R = 0.675R.
40
By symmetry the y- and z-coordinates are zero.
y
R
–
2
x
R
Problem 7.65 A volume of revolution is obtained by
2
2
revolving the curve xa2 + yb2 = 1 about the x axis. Determine its centroid.
y
y2
x 2 + ––
––
=1
2
a
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 πy 2 . From the equation
for the surface of the ellipse,
x2
1− 2
a
πy 2 = πb2
and dV = πy 2 dx = πb2
from which
V =
dV = πb2
V
a
0
= πb2 x −
x3
3a2
The x-coordinate:
x dV = πb2
V
= πb2
a
a
=
0
1−
1−
x2
a2
x2
a2
πb2 a2
4
x
dx
2πb2 a
.
3
x2
x dx
a2
0
a
x2
x4
πb2 a2
− 2
=
.
2
4a 0
4
1−
3
2πb2 a
=
3
8
x2 + y2 = 1
a2 b2
dx,
Divide by volume:
x =
y
a.
By symmetry, the y- and z-coordinates of the centroid are zero.
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 π(y 2 − R2 ). The radius
a
is y = h
x + R. The volume element is
dV = π
a
x+R
h
2
y
dx − πR2 dx.
R
Denote
m=
a
, dV = π(m2 x2 + 2mRx) dx,
h
from which
V =
dV = πm
V
h
(mx2 + 2Rx) dx
0
= πm m
x3
+ Rx2
3
The x-coordinate:
x dV = πm
V
h
h
= πmh2
0
mh
+R .
3
(mx3 + 2Rx2 ) dx
0
= πm m
x4
2Rx3
+
4
3
= πmh3
m
h
2R
+
4
3
h
0
.
Divide by the volume:
x =h
mh
+
4
2R
3
mh
+R
3
a
4
= h a
3
+
2R
3
+R
.
By symmetry, the y- and z-coordinates of the centroid are zero.
h
R+a
Problem 7.67 Determine the y coordinate of the centroid of the line (see Example 7.9).
y
(1, 1)
y = x2
L
x
Solution:
line is
The length of the line: The elementary length of the
y
dL =
1+
dy
dx
2
(1, 1)
dx.
dy
Noting that dx
= 2x, the element of length is dL = (1+4x2 )1/2 dx,
from which
1
L=
dL =
(1 + 4x2 )1/2 dx
L
0
1
1
(1 + 4x2 )1/2 x + loge (2x + (1 + 4x2 )1/2 )
2
2
√
√
5
1
=
+ loge (2 + 5) = 1.4789.
2
4
=
y = x2
L
x
1
0
The y-coordinate: The coordinate of the centroid of the length element
is y = y = x2 , from which
1
x(1 + 4x2 )3/2
x(1 + 4x2 )1/2
y =
(1 + 4x2 )1/2 x2 dx =
−
16
32
0
−
1
loge (2x + (1 + 4x2 )1/2 )
64
1
0
√
1
1
1
=
(5)3/2 −
(5)1/2 −
loge (2 + 5) = 0.6063.
16
32
64
Divide by the length: y =
0.6063
1.4789
= 0.410
Problem 7.68 Determine the x coordinate of the centroid of the line.
Solution:
y
dL =
is
The length: Noting that
dy
dx
1+
2
dx =
√
L
L
0
Divide by the length: x =
0
x
5
2 5/2
x
5
21.961
6.7869
5
5
= 6.7869.
1
= 21.961.
1
= 3.2357
2
y = – (x – 1)3/2
3
y
0
Problem 7.69 Determine the x coordinate of the centroid of the line.
2
(x)3/2
3
1
The x-coordinate:
5
x dL =
x3/2 dx =
= (x−1)1/2 , the element of length
x dx
from which
5
L =
dL =
(x)1/2 dx =
2
y = – (x – 1)3/2
3
dy
dx
x
5
y
2
y = – x 3/2
3
0
Solution:
length is
dy
dx
2
dx =
√
= x1/2 the element of
2
y = – x 3/2
3
1 + x dx
from which
2
L =
dL =
(1 + x)1/2 dx =
L
dy
dx
y
1+
dL =
The length: Noting that
2
0
2
(1 + x)3/2
3
2
= 2.7974
0
The x-coordinate:
2
2
(1 + x)5/2
(1 + x)3/2
x dL =
x(1 + x)1/2 dx = 2
−
5
3
L
0
0
35/2
33/2
1
1
=2
−
−
+
= 3.0379.
5
3
5
3
Divide by the length: x = 1.086
0
2
x
x
Problem 7.70
arc.
Determine the centroid of the circular
y
α
x
R
Solution:
y=
The length: From the equation for the circle,
R2 − x2 and
dy
= −(R2 − x2 )−1/2 x.
dx
The element of length
dy 2
dL = 1 +
dx = R(R2 − x2 )−1/2 dx,
dx
from which
L =
dL =
L
= R sin−1
R
R cos α
R(R2 − x2 )−1/2 dx
x R
R R cos α
π
− sin−1 (cos α)
2
π
π
=R
− sin−1 sin
−α
2
2
=R
= Rα
Check: L = Rα from the definition of α. check.
The x-coordinate:
R
x dL = R
L
R cos α
x(R2 − x2 )−1/2 dx
R
= R −(R2 − x2 )1/2
R cos α
Divide by the length: x =
R
α
= R2 sin α
sin α.
The y-coordinate:
The y-coordinate of the centroid of each element is
√
y = y = R2 − x2 . Hence
R
y dL = R
(R2 − x2 )1/2 (R2 − x2 )−1/2 dx
L
=R
R cos α
Rc
dx
R cos α
= R2 (1 − cos α).
Divide by the length:
y =
R
α
(1 − cos α)
y
α
x
R
Problem 7.71
Determine the centroids of the volumes.
y
60 mm
40 mm
x
40 mm
40 mm
60 mm
z
Solution: Divide the object into two volumes: (1) The left-most
volume with dimensions 40 by 80 by 60 mm. (2) The right-most
volume with dimensions 60 by 60 by 40 mm. The volumes and centroids are:
(1)
y
V1 = 1.92 × 105 mm3 ,
60 mm
x1 = 20 mm,
y1 = 40 mm,
40 mm
z1 = 30 mm.
(2)
V2 = 1.44 × 105 mm3 ,
x2 = 70 mm,
x
40 mm
y2 = 20 mm,
z2 = 30 mm.
(3)
The composite volume: V = V1 + V2 = 3.36 × 105 mm3 .
The composite centroid:
x=
V1 x1 + V2 x2
= 41.43 mm.
V
y=
V1 y1 + V2 y2
= 31.43 mm.
V
z=
V1 z1 + V2 z2
= 30 mm
V
z
40 mm
60 mm
Problem 7.72
Determine the centroids of the volumes.
y
15 in
20 in
15 in
x
15 in
35 in
z
Solution:
The Rectangle:
AreaR = 30 × 35 = 1050 in2 ,
y
15 in
xR = 35/2 = 17.5 in,
yR = 30/2 = 15 in
The Triangle:
20 in
15 in
AreaT = (20)(15)/2 = 150 in2 ,
xT = 15 +
2
(20) = 28.33 in,
3
yT = 30 − 15/3 = 25 in
x
15 in
35 in
z
The Solid:
x = (xR AreaR − xT AreaT )/(AreaR − AreaT ) = 15.7 in,
y
y = y = (yR AreaR − yT AreaT )/(AreaR − AreaT ) = 13.3 in,
15 in
20 in
and from symmetry, z = 0.
15 in
30 in
15 in
35 in
x
Problem 7.73
Determine the centroids of the volumes.
y
160
mm
80 mm
60 mm
40 mm
40 mm
50 mm
80 mm
z
Solution: For the block L = 240 mm, H = 160 mm, and
D = 50 mm. For each hole, r = 20 mm. Centroids of parts are at
the geometric centers.
Component
Block
Hole 1
Hole 2
Hole 3
Hole 4
V
LHD
πr 2
πr 2
πr 2
πr 2
Component Properties
V (mm3 )
x (mm)
2 × 106
120
62832
80
62832
80
62832
180
62832
180
x =
x1 V1 − x2 V2 − x3 V3 − x4 V4 − x5 V5
V1 − V2 − V3 − V4 − V5
x =
2.073 × 108 mm4
= 118.56 mm
1.748 × 106 mm3
y (mm)
80
40
120
100
40
z (mm)
25
25
25
25
25
y = SIMILAR Eqn = 80.72 mm
z = 25 mm
y
2
3
160
mm
80 mm
60 mm
1
40 mm
4
40 mm
50 mm
80 mm
100 mm
60 mm
z
Holes are 40 mm in diameter.
x
100 mm
60 mm
Holes are 40 mm in diameter.
x
Problem 7.74
Determine the centroids of the volumes.
y
120 mm
100
mm
40 mm
z
30 mm
20 mm
Solution: Divide the object into four volumes: (1) The left-most
cylinder, with diameter 80 mm and length 120 mm, (2) the cylinder to
the right, with diameter 60 mm and length 100 mm, (3) the cylinder
bored through the center, with diameter 40 mm and length 220 mm,
and (4) the composite cylinder. The volumes and centroids are:
(1)
V1 = 6.032 × 105 mm3 ,
x1 = 60 mm,
y1 = z1 = 0,
V2 = 2.83 × 105 mm3 ,
(2)
x2 = 170 mm,
y2 = z2 = 0,
(3)
V3 = 2.76 × 105 mm3 ,
x3 = 110 mm,
y3 = z3 = 0.
(4)
The composite volume: V = V1 + V2 − V3 = 6.095 ×
105 mm3 . The composite centroid:
x=
V1 x1 + V2 x2 − V3 x3
= 88.4 mm,
V
y=z=0
y
z
x
y
60
40 mm
mm
x
80
mm
x
120 mm
100 mm
Problem 7.75
Determine the centroids of the volumes.
y
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
πR2 L
1
πr 2 h
3
Volume (mm3 )
1.1706 × 107
1.3572 × 106
R = 90 mm
L = 460 mm
x
L/2
L-h/4
y
x (mm)
230
370
z
r = 60 mm
60 mm
90 mm
h = 360 mm
x =
XCyL VCyL − XCONE VCONE
VCyL − VCONE
x = 211.6 mm
y = z = 0 mm
360 mm
460 mm
x
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
1
2
3
4
Volume (V)
LW H
hW D
x
0
0
y
H/2
(H + h/2)
z
L/2
D/2
πR2 D/2
πr 2 D
0
0
H + h + 4R
3π
(H + h)
D/2
D/2
y
20 mm
25 mm
75 mm
where R = W/2. For the composite,
x=
x1 V1 + x2 V2 + x3 V3 − x4 V4
V1 + V2 + V3 − V4
x
with similar eqns for y and z
120 mm
25 mm
The dimensions, from the figure, are
100 mm
L = 120 mm
z
W = 100 mm
y
H = 25 mm
x
r = 20 mm
h = 75 mm
(H)
25 mm
D = 25 mm
+
R = 50 mm
12
V mm3
300000
187500
98175
31416
x (mm)
0
0
0
0
y (mm)
12.5
62.5
121.2
100
z (mm)
60
12.5
12.5
12.5
0m
(L)
m
100
mm (W)
y
z
1
+
50
mm
Object
+1
+2
+3
−4
–
3
x
7.76 Contd.
Substituting into the formulas for the composite, we get
y
x =0
y = 43.7 mm
z = 38.2 mm
mm
25
mm
100
(D)
mm
75
x
(h)
2
H
y
z
r = 20 mm
x
4
z
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 2(1.75) = 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:
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
The composite volume is V = V1 − V2 + V3 − V4 + V5 = 46.4 in3 .
The composite centroid:
x =
V1 x1 − V2 x2 + V3 x3 − V4 x4 + V5 x5
= 1.02 in.,
V
y =
V1 y1 − V2 y2 + V3 y3 − V4 y4 + V5 y5
= 1.9 in.,
V
z =0
1 in
x
1.75 in
z
z
x 5 in
4 in
y
1 in
x
Problem 7.78
Determine the centroids of the volumes.
y
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
Object
Cylinder
Cone 1
Cone 2
Cylinder
Cone 1
Cone 2
V
πr 2 L/2
1
πR2 L
3
1
πr 2 L
3
2
(mm3 )
5.089 × 105
1.357 × 106
1.696 × 105
180
mm
y
x
L/4
3L/4
30 mm
3(L/2)/4
(mm)
90
270
135
60 mm
z
x
180
mm
L = 360 mm
180
mm
r = 30 mm
R = 60 mm
y
For the composite shape
x = 229.5 mm
2R
xCyl VCyL + x1 V1 − x2 V2
VCyL + V1 − V2
2r
L/2
L/2
y
60 mm
Cylinder
1
y
x
360 mm
cone
+
120 mm
x =
2
y
cone
–
60 mm
3
180 mm
x
x
Problem 7.79 The dimensions of the Gemini spacecraft (in meters) are a = 0.70, b = 0.88, c = 0.74,
d = 0.98, e = 1.82, f = 2.20, g = 2.24, and h = 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
(R2 −R1 )
(1)
H=
(2)
2
πR2
H
.
3
η=
(4)
2
πR1
η
.
3
(5)
(6)
(7)
(8)
The length of the truncated portion is
R1 L
(R2 −R1 )
(3)
with volume
with volume
The volume of the truncated cone is the difference of the
two volumes,
V =
πL
3
3
3
R2
−R1
R2 −R1
. The centroid of the removed part of the
cone is
xη = 34 η, and the centroid of the complete cone is
xh = 34 H, measured from the pointed end. From the composite theorem, the centroid of the truncated cone is
V x −V x
x = h hV η η −η+x, where x is the x-coordinate of the left
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.
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.
Problem 7.80 Two views of a machine element are
shown. Determine the centroid of its volume.
y
y
24 mm
8 mm
18 mm
60 mm
8 mm
z
x
20
mm
16
mm
50 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
y
V1 = (60)(48)(50) = 144,000 mm3 ,
V2 =
2
1
Π(24)2 (50) = 45, 239 mm3 ,
2
2
3
3
V3 = Π(8) (50) = 10, 053 mm ,
V4 = (16)(36)(20) = 11, 520 mm3 ,
V5 =
1
Π(18)2 (20) = 10, 179 mm3 ,
2
1
6
5
4
z
V6 = Π(8)2 (20) = 4021 mm3 .
4(18)
= 47.6 mm,
3Π
The coordinates of the centroids are
z5 = 24 + 16 +
x1 = 25 mm,
x6 = 10 mm,
y1 = 30 mm,
y6 = 18 mm,
z1 = 0,
z6 = 24 + 16 = 40 mm.
x2 = 25 mm,
The x coordinate of the centroid is
y2
4(24)
= 60 +
= 70.2 mm,
3Π
z2 = 0,
x3 = 25 mm,
y3 = 60 mm,
z3 = 0,
x4 = 10 mm,
y4 = 18 mm,
z4 = 24 + 8 = 32 mm,
x5 = 10 mm,
y5 = 18 mm,
x =
x1 V1 + x2 V2 − x3 V3 + x4 V4 + x5 V5 − x6 V6
= 23.65 mm.
V1 + V2 − V3 + V4 + V5 − V6
Calculating the y and z coordinates in the same way, we obtain y = 36.63 mm
and z = 3.52 mm
Problem 7.81
Determine the centroids of the lines.
y
4m
Solution:
Part 1
Part 2
Break the line into two parts
xi
4
11
yi
2
2
x
L
√i
√80 = 8.94 m
52 = 7.21 m
6m
8m
y
x1 L1 + x2 L2
x =
= 7.12 m
L1 + L2
By inspection, since y1 = y2 = 2 m we get
y = 2 m.
4m
x
8m
Problem 7.82
Determine the centroids of the lines.
Solution:
(1)
(2)
(3)
y
6m
The object is divided into two lines and a composite.
L1 = 6 m, x1 = 3 m, y1 = 0.
L2 = 3π m, x2 = 6 + π6 m (Note: See Example 7.13) y2 = 3.
The composite length: L = 6 + 3π m. The composite centroid:
x=
L1 x1 + L2 x2
= 6 m,
L
y=
3π
= 1.83 m
2+π
3m
y
3m
x
6m
x
6m
Problem 7.83
Determine the centroids of the lines.
Solution: Break the composite line into two parts (the quarter circle and
the straight line segment). (see Appendix B)
y
xi
Part 1
Part 2
yi
2R/π
3m
Li
2R/π
0
πR/2
2m
(R = 2 m)
2R
= 1.273 πR/2 = 3.14 m
π
2m
x =
x1 L1 + x2 L2
= 1.94 m
L1 + L2
y =
y1 L1 + y2 L2
= 0.778 m
L1 + L2
2m
x
2m
2m
2m
2m
x
2m
2m
Problem 7.84 The semicircular part of the line lies in
the x − z plane. Determine the centroid of the line.
y
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:
L =
3
i=1
100 mm
160 mm
Li .
x
120 mm
The composite coordinates are:
3
x =
i=1
and y =
,
L
3
i=1
Segment
L1
L2
L3
Composite
z
Li xi
y
Li yi
L
100 mm
Length, mm
120π
100
188.7
665.7
x, mm
240
π
0
80
65.9
y, mm
0
50
50
21.7
z, mm
120
0
0
68.0
2
3
160 mm
z
1
120 mm
x
Problem 7.85 The mass of the homogeneous flat plate
is 450 kg. What are the reactions at A and B?
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.
2m
1m
A
B
5m
Solution: To find the location of the center of mass, find the centroid by breaking the plate into a triangle and a rectangle.
X1
2m
1m
1m
A
5m
B
X2
5m
1m
x
5m
Area1 = 5/2 m2
x1 = 5/3 m
Area2 = 5 m2
x2 =
x =
5
m
2
x1 A1 + x2 A2
= 2.22 m
A1 + A2
m = 450 kg
x = 2.22 m
mg
X
AX
5m
AY
Fx :
Ax = 0
Fy :
Ay + By − mg = 0
MA :
−x mg + 5By = 0
Solving:
Ax = 0,
Ay = 2.45 kN
By = 1.96 kN
BY
Problem 7.86 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
x-coordinate. The areas and the x-coordinate of the centroid are tabulated. The last row is the composite area and x-coordinate of the
centroid.
Area
Rectangle
Circle
Triangle
Composite
A, sq mm
3.2 × 105
3.14 × 104
1.2 × 105
4.09 × 105
400
mm
200
mm
100 mm
A
x
400
600
1000
561
600 mm
B
600 mm
800 mm
600 mm
The composite area is A = Arect − Acirc + Atriang . The composite
x-coordinate of the centroid is
x =
Arect xrect − Acirc xcirc + Atriang xtriang
.
A
The sum of the moments about A:
Fy = Ay + B − 500 = 0,
from which Ay = 300 N.
X
B
AY
1400 mm
MA = −500(561) + 1400B = 0,
from which B = 200 N. The sum of the forces:
500 N
AX
Fx = Ax = 0
Problem 7.87 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
D
y = 1 + 0.0625x2
y
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 = y dx = (1 + ax2 ) dx, where
a = 0.0625. Thus
4
4
ax3
A =
dA =
(1 + ax2 ) dx = x +
= 5.3333 sq ft.
3 0
A
0
The x-coordinate:
4
x dA =
x(1 + ax2 ) dx =
A
0
Divide A: x =
12
5.3333
C
4m
1m
E
B
D
y
x2
2
+
4
ax4
4
y = 1 + 0.0625x2
= 12.
0
= 2.25 ft.
The equilibrium conditions: The angle of the member CE is
1
α = tan−1 ( ) = 14.04◦ .
4
The weight of the sign is W = 130(9.81) = 1275.3 N. The sum of
the moments about D is
MD = −2.25W + 4CE sin α = 0,
from which CE = 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
1
β = tan−1 ( ) = 26.57◦ .
2
The section as a free body: The sum of the vertical forces:
FY = AD sin β − W = 0
from which AD = 2851.7 N (T )
2m
A
x
Problem 7.88 The bar has a mass of 80 kg. What are
the reactions at A and B?
A
2m
2m
B
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
A
L = L1 + L2 = 2 m + 2πR/4(R = 2 m)
z
L =2+πm
2m
Part
Lengthi (m)
1
2
2
π
Massi (kg)
2
2+π
π
2+π
m1 = 31.12 kg
x1 = 1 m
m2 = 48.88 kg
x2 = 3.27 m
xi (m)
80
1
80
2+
B
2R
π
y
X1
m1g
m2g
AX
Fx :
Ax = 0
Fy :
Ay + By − m1 g − m2 g = 0
MA :
2m
X2
AY
−x1 m1 g − x2 m2 g + 4By = 0
x
4m
Solving
BY
Ax = 0, Ay = 316 N, B = 469 N
Problem 7.89 The semicircular part of the homogeneous slender bar lies in the x − z plane. Determine the
center of mass of the bar.
y
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:
L =
3
i=1
3
i=1
12 in
z
i=1
Segment
L1
L2
L3
Composite
x
Li xi
,
L
3
and y =
16 in
Li .
The composite coordinates are:
x =
10 in
y
Li yi
10 in
L
16 in
Length, in.
12π
10
18.868
66.567
x, in.
24
π
0
8
6.594
y, in.
0
5
5
2.168
z, in.
12
0
0
6.796
z
x
12 in
Problem 7.90 When the truck is unloaded, the total
reactions at the front and rear wheels are A = 54 kN
and B = 36 kN. The density of the load of gravel is
ρ = 1600 kg/m3 . The dimension of the load in the z
direction is 3 m, and its surface profile, given by the
function shown, does not depend on z. What are the
total reactions at the front and rear wheels of the loaded
truck?
y
y = 1.5 – 0.45x + 0.062x2
x
A
B
2.8 m
3.6 m
5.2 m
Solution: First, find the location of the center of mass of the unloaded truck (and its mass). Then find the center of mass and mass of
the load. Combine to find the wheel loads on the loaded truck.
Unloaded Truck
Fx :
no forces
Fy :
54000 + 36000 − mT g = 0(N )
MA :
−xT mT g + 5.2(36) = 0
y
y = 1.5 – 0.45x + 0.062 x 2
x
Solving xT = 2.08 m, mT = 9174 kg
A
B
Next, find xL and mL (for the load)
x dm
Num
m
xL = L
=
mL
dm
mL
where mL =
mL =
3.6
mL
2.8 m
dm, Num =
3ρy dx = 3ρ
0
3.6
0
mL = 3ρ 1.5x − 0.45
x2
2
3.6 m
5.2 m
XT
x dm
mTg
mL
5.2 m
(1.5 − 0.45x + 0.062x2 ) dx
+ 0.062
x3
3
B
36 kN
54 kN
3.6
mL g
y
0
mL = 16551 kg
Num = 3ρ
3.6
0
Num = 3ρ 1.5
y = 1.5 − 0.45x − 0.062x2
dm = p(3)y dx
(1.5x − 0.45x2 + 0.062x3 ) dx
x2
2
− 0.45
x3
3
+ 0.062
XL
x4
4
3.6
0
Num = 25560 kg · m
xL
0
Num
=
= 1.544 m
mL
dL
measured from the front of the load
XT
The horizontal distance from A to the center of mass of the load is
dL = xL + 2.8 m = 4.344 m
Now we can find the wheel loads on the loaded truck
Fx :
no forces
Fy :
Ay + By − mT g − mL g = 0
MA :
5.2By − xT mT g − dL mL g = 0
Solving Ay = 80.7 kN, By = 171.6 kN
3.6 m
AY
mL g
mT g
5.2 m
BY
x
Problem 7.91 The 10-ft horizontal cylinder with 1-ft
radius is supported at A and B. Its weight density is
γ = 100(1 − 0.002x2 ) lb/ft3 . What are the reactions at
A and B?
y
10 ft
A
1 ft
z
x
B
Solution: The weight: Denote a = 0.002. The element of volume is a disk of radius R = 1 ft, thickness dx, and weight
y
1 ft
2
dW = γπR dx,
from which
W =
dW = 100πR2
x
L
0
= 100πR2
ax3
x−
3
(1 − ax2 ) dx
L
= 2932.15 lb
0
The x-coordinate of the mass center:
L
x dW = 100πR2
(1 − ax2 )x dx
W
0
L
25πR2 (1 − ax2 )2 0 = 14137.17.
=−
a
Divide by W : x = 4.8214 ft.
The equilibrium conditions: The sum of the moments about A is
MA = −W x + 10B = 0,
from which
B=
Wx
(2932.15)(4.8214)
=
L
10
= 1413.7 lb .
The sum of the vertical forces:
FY = A + B − W = 0,
from which A = 1518.4 lb . The horizontal components of the reactions are zero:
FX = 0
A
B
10 ft
Problem 7.92 A horizontal cone with 800-mm length
and 200-mm radius has a built-in support at A. Its mass
density is ρ = 6000(1 + 0.4x2 ) 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 = 0.25x. Denote a = 0.4. The mass
of the disk is
y
200
mm
x
A
dm = ρπy 2 dx = 6000π(1 + ax2 )(0.25x)2 dx
= 375π(1 + ax2 )x2 dx,
800 mm
from which
m = 375π
0.8
(1 + ax2 )x2 dx = 375π
0
x3
x5
+a
3
5
0.8
0
= 231.95 kg
The x-coordinate of the mass center:
0.8
0.8
x4
x6
x dm = 375π
(1 + ax2 )x3 dx = 375π
+a
4
6 0
m
0
= 141.23.
Divide by the mass: x = 0.6089 m
The equilibrium conditions: The sum of the moments about A:
M = MA − mgx = 0,
from which
MA = mgx = 231.94(9.81)(0.6089)
= 1385.4 N-m .
The sum of the vertical forces:
FY = AY − mg = 0
from which AY = 2275.4 N . The horizontal component of the
reaction is zero,
FX = 0.
Problem 7.93 The circular cylinder is made of aluminum (Al) with mass density 2700 kg/m3 and iron (Fe)
with mass density 7860 kg/m3 .
y
(a) Determine the centroid of the volume of the cylinder.
(b) Determine the center of mass of the cylinder.
Al
z
Fe
200 mm
600 mm
x
600 mm
Solution:
(a)
y
The volume of the cylinder is
AI
Fe
V = π(0.1)2 (1.2) = 0.0377 m3 .
The volume of the parts:
VAl = VF e
V
=
= 0.0188 m3 .
2
The centroid of the first part is xAl = 0.3 m, yAl = zAl = 0.
The centroid of the iron part is
xF e = 0.6 + 0.3 = 0.9 m,
yF e = zF e = 0.
The composite centroid
x=
VAl (0.3) + VF e (0.9)
1.2
=
= 0.6 m,
V
2
y = z = 0.
(b)
The mass center: The mass of the aluminum part is mAl =
VAl (2700) = 50.89 kg. The mass of the iron part is mF e =
VF e (7860) = 148.16 kg. The composite mass is m = mAl +
mF e = 199.05 kg. The composite center of mass is
xm =
(50.89)(0.3) + (148.16)(0.9)
= 0.7466 m
199.05
ym = zm = 0
200 mm
x
600 mm
600 mm
z
y
Problem 7.94 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
y
VAl = π(0.0352 − 0.022 )(0.2) = 5.18 × 10−4 m3 .
The mass of the aluminum tube is mAl = (2700)VAl = 1.4 kg. The
centroid of the aluminum tube is xAL = 0.1 m, yAl = zAl = 0.
The volume of the steel plug is VF e = π(0.02)2 (0.1) = 1.26 ×
10−4 m3 . The mass of the steel plug is mF e = (7800)VF e =
0.9802 kg. The centroid of the steel plug is xF e = 0.15 m, yF e =
zF e = 0.
The composite mass is m = 2.38 kg. The composite centroid is
mAl (0.1) + mF e (0.15)
x =
= 0.121 m
m
y =z=0
Problem 7.95 A machine consists of three parts. The
masses and the locations of the centers of mass of the
parts are:
Part
1
2
3
Mass (kg)
2.0
4.5
2.5
x (mm)
100
150
180
y (mm)
50
70
30
z (mm)
−20
0
0
Determine the coordinates of the center of mass of the
machine.
Solution:
The composite mass is 9 kg.
x =
2(100) + 4.5(150) + 2.5(180)
= 147 mm
9
y =
2(50) + 4.5(70) + 2.5(30)
= 54.4 mm
9
z =
2(−20)
= −4.4 mm
9
z
x
y
y
A
x
100 100
mm mm
Section A-A
20 mm
z
A
35 mm
Problem 7.96 A machine consists of three parts. The Solution: The composite mass is m = 2.0 + 4.5 + 2.5 = 9 kg. The
masses and the locations of the centers of mass of two location of the third part is
of the parts are:
120(9) − 2(100) − 4.5(150)
x3 =
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 = 120 mm, y = 80 mm, and z = 0.
Determine the necessary position of the center of mass
of part 3.
Problem 7.97 Two views of a machine element are
shown. Part 1 is aluminum alloy with mass density
2800 kg/m3 , and part 2 is steel with mass density
7800 kg/m3 . Determine the x coordinate of its center
of mass.
= 82 mm
2.5
y3 =
80(9) − 2(50) − 4.5(70)
= 122 mm
2.5
z3 =
2(20)
= 16 mm
2.5
y
y
1
24 mm
2
8 mm
18 mm
60 mm
8 mm
z
x
20
mm
16
mm
50 mm
Solution:
V1
V2
The volumes of the parts are
1
= (60)(48) + π(24)2 − π(8)2 (50)
2
y
y
1
24 mm
= 179, 186 mm3 = 17.92 × 10−5 m3 ,
1
= (16)(36) + π(18)2 − π(8)2 (20) = 17, 678 mm3
2
2
= 1.77 × 10−5 m3 ,
so their masses are
x
m1 = S1 V1 = (2800)(17.92 × 10−5 ) = 0.502 kg,
m2 = S2 V2 = (7800)(1.77 × 10−5 ) = 0.138 kg.
The x coordinates of the centers of mass of the parts are x1 = 25 mm,
x2 = 10 mm, so
x =
x1 m1 + x2 m2
= 21.8 mm
m1 + m2
8 mm
18 mm
20
mm
50 mm
60 mm
8 mm 8 mm
z
16
mm
Problem 7.98 Determine the y and z coordinates of
the center of mass of the machine element in Problem 7.97.
The z coordinate of the center of mass of part 1 is
z1 = 0. To determine the y coordinate, we divide it into three parts
a, b, and c. (Part c is the “hole” with radius 8 mm). The volumes are
Solution:
y
Va = (60)(48)(50) = 144,000 mm3 ,
Vb =
y
24 mm
b
c
1
π(24)2 (50) = 45, 239 mm3 ,
2
a
60 mm
Vc = π(8)2 (50) = 10, 053 mm3
50
mm
z
so
y1
ya Va + yb Vb − yc Vc
=
Va + Vb − Vc
4(24)
(30)Va + 60 + 3π Vb − (60)Vc
=
Va + Vb − Vc
y
Vd = (16)(36)(20) = 11,520 mm3 ,
Ve =
1
π(18)2 (20) = 10,179 mm3 ,
2
Vf = π(8)2 (20) = 4021 mm3
so
z2 =
=
zd Vd + ze Ve − zf Vf
Vd + Ve − Vf
(24 + 8)Vd + 24 + 16 +
4(18)
3π
Ve − (24 + 16)Vf
Vd + Ve + Vf
= 39.2 mm.
From the solution of Problem 7.97, the masses are m1 = 0.502 kg,
m2 = 0.138 kg, Therefore,
y =
y1 m1 + y2 m2
= 34.05 mm,
m1 + m2
z =
z1 m1 + z2 m2
= 8.45 mm.
m1 + m2
y
16 mm
e
f
= 38.5 mm.
The y coordinate of the center of mass of part 2 is y2 = 18 mm. To
determine the z coordinate, we divide it into three parts d, e, and f .
The volumes are
x
d
z
x
18 mm 24 mm
20 mm
Problem 7.99 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.
(a) 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?
(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?
E
C
0.6 m
0.45 m
A
B
1.14 m
b
2.60 m
Solution:
(a)
The composite mass is m = mC + mE = 1320 kg. The
x-coordinate of the composite center of mass is given:
2.6
x=
= 1.3 m,
2
from which the x-coordinate of the center of mass of the engine is
xE = b =
(1.3 m − 1.14 mC )
= 2.1 m.
mE
The y-coordinate of the composite center of mass is
y=
(b)
0.45 mC + 0.6 mE
= 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 = 2.6A + ymg sin(15◦ )
−(2.6 − x)mg cos(15◦ ) = 0,
from which A = 5641.7 N. This is the normal force exerted by
the road on A. The normal force exerted on B is obtained from;
FN = A − mg cos(15◦ ) + B = 0,
from which B = 6866 N
C
E
0.6 m
0.45 m
A
1.14 m
b
2.60 m
B
Problem 7.100 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 = 30+140+146 =
316 kN. The center of mass of the airplane:
My−axis = 30(10) − xA WA = 0,
x
A
from which xA = 0.949 m.
Mx−axis = (140 − 146)(6) + yA WA = 0,
from which yA = 0.114 m. The weight of the loaded plane:
10 m
B
W = 31 + 142 + 147 = 320 kN.
The center of mass of the loaded plane:
My−axis = (31)10 − xW = 0,
from which x = 0.969 m.
Mx−axis = (142 − 147)(6) + yW = 0,
from which y = 0.0938 m. The weight of the crate is Wc = W −
WA = 4 kN. The center of mass of the crate:
xc =
W x − WA xA
= 2.5 m,
Wc
yc =
W y − WA yA
= −1.5 m.
Wc
The mass of the crate:
mc =
Wc × 103
= 407.75 kg
9.81
6m
C
6m
y
Determine the Iy and ky .
Problem 7.101
y
h
x
b
Let dA = x dy = b dy. A =
Solution:
Iy =
x2 dA = h
A
b
0
x3
x2 dx = h
3
h
b dy = hb.
0
b
0
b3 h
=
ky =
3
y
Iy
b
= √
A
3
h
x
b
Problem 7.102 Determine Ix and kx by letting dA be
(a) a horizontal strip of height dy; (b) a vertical strip of
width dx.
Solution:
(See figure in Problem 7.101.)
(a) A
h
=
Ix
b dy = hb.
0
y 2 dA = b
=
A
kx
=
(b) A
0
=h
kx
=
bh3
,
3
h dx = hb.
0
Ix
y 2 dy =
Ix
h
= √
A
3
b
=
h
b
y 2 dx = h
0
h
0
y2
b
h
dy =
bh3
,
3
Ix
h
= √
A
3
Problem 7.103
Determine Ixy .
Solution:
(See figure in Problem 7.101.)
b h
dA =
dx dy = hb.
A =
A
0
Ixy =
0
xy dA =
A
b
0
h
0
xy dx dy =
y2
2
h
0
x2
2
b
=
0
h2 b2
4
Problem 7.104 Determine Ix , kx , Iy , and ky for the
beam’s rectangular cross section.
y
60 mm
x
40 mm
Solution:
Ix =
y
y 2 dA
A
30
Let dA = 40 dy = 60 mm
30
y3
=
40y dy = 40
3
−30
27000
27000
= 40
+
3
3
Ix
2
dA = 40 dy
30
−30
60 mm
= 40[18000] mm4
–30
Ix = 7.2 × 105 mm4
A = 2400 mm2
kx =
Ix
=
A
40 mm
7.2 × 105
mm
2.4 × 103
kx = 17.3 mm
y
–20
(mm)
120
30
x
–30
dA = 60 dx
Iy =
Iy =
x2 dA
A
20
x2 (60) dx =
−20
Iy = 60
x3
3
20
ky =
ky =
√
60x2 dx
−20
20
= 60
−20
8000
8000
+
3
3
Iy = 3.2 × 105 mm4
Iy
=
A
x
3.2 × 105
2.4 × 103
133.3 = 11.5 mm
mm4
Problem 7.105 Determine Ixy and JO for the beam’s
rectangular cross section.
Solution:
Ixy =
y
xy dA
Ixy =
A
20
xy dx dy
−30
−20
30
x2
y
2
Ixy =
30
−30
+20
30
dy =
dA
0 dy = 0
60 mm
−30
−20
x
Ixy = 0
JO =
r2 dA =
A
−30
30
=
−30
30
=
JO =
30
−30
30
−30
x3
+ y2 x
3
20
(x2 + y 2 ) dx dy
−20
20
40 mm
dy
−20
8000
8000
+ 20y 2 +
+ 20y 2 dy
3
3
16000
+ 40y 2 dy
3
16000
y3
y + 40
3
3
JO =
(16000)
(60) + 80
3
b
dx
0
−h
x+h
b
h
− x+h
b
0
Iy =
x2 dA =
A
ky =
b
dx = −
x2 dx
mm4
y
hx2
+ hx
2b
−h
x+h
b
b
=
0
h
hb
2
x
dy
b
0
h
− x3 + hx2
b
0
dy
0
b
=
303
3
0
b
=
−30
JO = 10.4 × 105 mm4
Problem 7.106 Determine Iy and ky .
Solution: y = − hb x + h, dA = dy dx, therefore
A =
30
JO =
dx = −
hx4
hx3
+
4b
3
b
=
0
hb3
12
y
Iy
b
= √
A
6
h
x
b
Determine JO and kO .
Problem 7.107
kx =
Solution: (See figure in Problem 7.106.)
h
b
y =−
Ix =
=
=
x + h, dA = dx dy
y 2 dA =
A
1
3
b
dx
b
0
bh3
h
− x+h
b
3
−h
x+h
b
dx = −
bh
(b2 + h2 )
12
(b2 + h2 )
2
2
= kx + ky =
6
JO = Ix + Iy =
y 2 dy
0
12
Ix
h
= √
A
6
b
12h
h
− x+h
b
4 b
0
kO
Determine Ixy .
Problem 7.108
Solution:
(See figure for Problem 7.106.)
Ixy =
b
xy dA =
A
1
2
=
0
b
0
−h
x+h
b
x dx
2
h
x − x+h
b
dx
h 2 x4
h 2 x3
x2
−2
+ h2
2
b 4
b 3
2
2
2
2
2
h b
1
2
1
h b
=
− +
=
2
4
3
2
24
1
2
=
y dy
0
b
0
Determine Iy .
Problem 7.109
y
y = cxn
x
O
b
Solution:
Iy =
Iy =
x2 dA
b
Iy
x2 (cxn ) dx
y = cx n
0
Iy =
y
A
b
cxn+3
(n + 3)
cx(n+2) dx =
0
b
0
cb(n+3)
=
(n + 3)
x
b
dA = ydx
= cx ndx
Determine Ix .
Problem 7.110
Solution:
Ix =
y 2 dA =
A
Ix =
b
0
Ix =
Ix =
b
0
3 0
cx
y3 b
0
cxn
y
y 2 dy dx
0
n
3n c3 x
3
c3
x3n+1
3
(3n + 1)
dx
dx =
b
=
0
Ix = c3 b3n+1 /(9n + 3)
y = cx n
c3
3
b
dA = dydx
x3n dx
0
c3 b3n+1
3(3n + 1)
x
b
Determine JO .
Problem 7.111
Solution:
JO =
2
r dA =
A
JO =
JO =
b
0
b
x2 y +
0
b
y3
3
cxn
0
2
(x + y ) dy dx
0
cxn
dx
0
c3 x3n
3
cxn+2 +
y
2
y = cx n
dx
dA
b
cxn+3
c3 x3n+1 +
(n + 3)
3 (3n + 1) 0
JO =
cb(n+3)
JO =
(n + 3)
+
x
c3 b(3n+1)
b
(9n + 3)
dA = dy dx
Determine Ixy .
Problem 7.112
Solution:
Ixy =
A
Ixy =
b
0
cx
xy 2
b
2
0
dx
xc2 x2n
dx =
2
0
b
c2 x2n+2 =
2 (2n + 2) 0
Ixy = c b
xy dy dx
0
0
2 (2n+2)
y
cxn
n
b
Ixy =
Ixy
xy dA =
b
0
y = cx n
c2 (2n+1)
x
dx
2
dA
x
/(4n + 4)
b
Determine Iy and ky .
Problem 7.113
y
y = x3
Solution:
Iy =
Iy =
x2 dA =
A
0
1
0
1
[x2 y]x
dx =
x3
x
x3
1
0
y=x
x2 dy dx
(x3 − x5 ) dx
1
x4
x6 1
1
−
= −
4
6 0
4
6
Iy =
x
Iy = 0.0833
Area =
1
0
x
x3
dy dx =
1
0
x
y
dx
x3
1
1
x2
x4 Area =
(x − x3 ) dx =
−
= 0.25
2
4 0
0
ky =
Iy
=
Area
ky = 0.577
0.0833
0.25
y=x
y
(1, 1)
y = x3
dA = dy dx
x
Determine Ix and kx .
Problem 7.114
Solution:
Ix =
Ix =
A
1
y
x
x
1
y 3 x3
x9
dx
=
−
dx
3 x3
3
3
0
1
x4
x10 1 1
1
−
=
−
4
10 0
3 4
10
1
1
3
y=x
y 2 dy dx
x3
0
0
Ix =
y 2 dA =
(1, 1)
y = x3
dA = dx dy
Ix = 0.0500
1
Area =
x
x3
0
dy dx =
0
−
= 0.25
2
4 0
Ix
0.050
=
=
= 0.447
A
0.25
Determine JO and kO .
Problem 7.115
Solution:
JO =
JO =
JO =
JO =
JO =
JO
0
x
1
x4
x2
Area =
ky
x
1
1 x
y dx =
(x − x3 ) dx
3
r2 dA =
A
1
0
1
x2 y +
0
1
x3 +
0
y3
3
x3
3
x
x
x3
y
(x2 + y 2 ) dy dx
dx
(1, 1)
x3
x9
− x5 −
3
9
1
0
4 3
x
x − x5 −
3
3
4/
3
x4
4/
−
x6
x10
−
6
30
dx
y = x3
dx
1
x
0
Area = 0.25
JO
0.133
kO =
=
= 0.730
Area
0.25
1
1
1
= − −
3
6
30
JO = 0.133
Area =
0
=
1
1
0
x
x3
y=x
dy dx =
(x − x3 ) dx =
1
0
kO = 0.730
x
[y] dx
3
x
x2
2
−
x4
4
1
0
Determine Ixy .
Problem 7.116
Solution:
Ixy =
Ixy =
1
x
x3
0
1
x
0
y
xy dy dx
y2
2
x
x3
dx =
0
1
x4
x8 −
=
8
16 0
Ixy =
1
x3
x7
−
2
2
y=x
1
(1, 1)
dx
1
1
−
8
16
Ixy = 0.0625
x
1
Problem 7.117 Determine the moment of inertia Iy
of the metal plate’s cross-sectional area.
y
1
y = 4 – – x 2 ft
4
x
Solution:
Iy =
Iy =
Iy =
x2 dA =
A
4x2 −
−4
x3
3
43
3
43
−
−
−
Iy = 68.3 ft4
20
y
x2 dy dx
dx
1
y = 4 – – x2
4
1 4
x dx
4
5 4
1x
4 5
45
(4− 1
x2 )
4
0
[x2 y]0
4
3
(4− 1
x2 )
4
−4
Iy = 4
4
−4
4
Iy = 4
Iy = 8
−
−4
4(−43 )
(−4)5
+
3
20
2(45 )
= 170.67 − 102.4
20
4
dA = dy dx
x
–4
4
Problem 7.118 Determine the moment of inertia Ix
and the radius of gyration kx of the cross-sectional area
of the metal plate.
Solution:
Ix =
y 2 dA =
A
Ix =
(4− 1
x2 )
4
y
y 2 dy dx
1
y = 4 – – x2
4
0
(4− 14 x2 )
y 3 dx
3 0
−4
1
=
3
−4
4
1
=
3
4
4
3
1
−4 − x2
4
−4
dA = dy dx
dx
x
4
–4
3
1 6
64 − 12x2 + x4 −
x
dx
4
64
−4
4
1
12x3
3 x5
1 x7 =
64x −
+
−
3
3
4 5
64 7 −4
4
1 2
1 x3
x dx = 4x −
4
4 3 −4
−4
16
16
Area = 16 −
+ 16 −
= 21.33 ft2
3
3
Area =
Ix = 78.0 ft4
Area =
dA =
A
Area =
4
−4
4
−4
(4− 1
x2 )
4
4
dy dx
kx =
0
4− 1 x2
4
[y]
dx
4−
Ix
=
Area
78.02
ft
21.33
kx = 1.91 ft
0
Problem 7.119 (a) Determine Iy and ky by letting dA
be a vertical strip of width dx.
(b) The polar moment of inertia of a circular area with
its center at the origin is JO = 12 πR4 . Explain how you
can use this information to confirm your answer to (a).
y
The equation of the circle is x2 +y 2 = R2 , from which
y = ± R2 √
− x2 . The strip dx wide and y long has the elemental
area dA = 2 R2 − x2 dx. The area of the semicircle is
R πR2
A=
Iy =
x2 dA = 2
x2 R2 −x2 dx
2
A
0
R
x(R2 −x2 )3/2 R2 x(R2 −x2 )1/2 R4
x
=2 −
+
+
sin−1
4
8
8
R
x
Solution:
√
0
=
4
πR4
8
Iy R
ky =
=
A
2
R
(b) If the integration were done for a circular area with the center at the origin,
the limits of integration for the variable x would be from −R to R, doubling
the result. Hence, doubling the answer above,
Iy =
πR4
.
4
By symmetry, Ix = Iy , and the polar moment would be
y
JO = 2Iy =
x
R
πR4
,
2
which is indeed the case. Also, since kx = ky by symmetry for the full circular
area,
Ix
JO
Iy
Iy
kO =
+
= 2
=
A
A
A
A
as required by the definition. Thus the result checks.
Problem 7.120 (a) Determine Ix and kx for the area
in Problem 7.119 by letting dA be a horizontal strip of
height dy.
(b) The polar moment of inertia of a circular area with
its center at the origin is JO = 12 πR4 . Explain how you
can use this information to confirm your answer to (a).
Use the results of the solution to Problem 7.119, A =
The equation for the circle is x2 + y 2 = R2 , from which
x = ± R2 − y 2 . The horizontal strip is from 0 to R, hence the
element of area is
dA = R2 −y 2 dy.
+R Ix =
y 2 dA =
y 2 R2 −y 2 dy
Solution:
πR2
.
2
A
y(R2 −y 2 )3/2 R2 y(R2 −y 2 )1/2 R4
= −
+
+
sin−1
4
8
8
R4
R4
y
R
R
−R
πR4
π
π
+
=
8 2
8 2
8
kx =
Ix =
πR4
.
4
By symmetry Iy = Ix ,
and JO = 2Ix =
−R
=
(b) If the area were circular, the strip would be twice as long, and the moment
of inertia would be doubled:
πR4
,
2
which is indeed the result. Since kx = ky by symmetry for the full circular
area, the
Ix
JO
Iy
Ix
kO =
+
= 2
=
A
A
A
A
as required by the definition. This checks the answer.
Ix R
= .
A
2
Problem 7.121 Determine the moments of inertia Ix
and Iy .
Strategy: Use the procedure described in Example 8.2
to determine JO , then use the symmetry of the area to
determine Ix and Iy .
y
Ro
x
Ri
Solution:
JO =
Ro
Ri
JO = 2π
Let dA = 2πr dr
Ro
r2 dA = 2π
r 2 r dr
y
Ri
RO
R
Ri4
r4 o
Ro4
=
2π
−
4 R
4
4
i
From symmetry Ix = Iy
Also
JO = Ix + Iy
∴ Ix = Iy =
π 4
(R − Ri4 )
4 0
x
Ri
Problem 7.122 If a = 5 m and b = 1 m, what are the
values of Iy and ky for the elliptical area of the airplane’s
wing?
y
x2
y2
— + — =1
a2
b2
x
2b
a
Solution:
Iy =
A
Iy = 2
0
Iy = 2
[x
a
a
x
2b
a
2
1−
x2
a2
dx
1/2
dx
a
y = b 1– x2
a2
x2
dx
a2
a
x2
0
0
a2 − x2 dx
√
a4
sin−1
8
x
a
a
0
(from the integral tables)

0
√ 0

a3 a2 − a2
2b  a(a2 − a2 )3/2
Iy =
/+
/
−
a 
4
8


a4
+
8
sin−1

a  
−
a
0√
+
0
0(a2 )3/2
0
a2 · 0 a2 a4
/+
sin−1
8
8
Iy =
2/b a4 π
a 8 2/
Iy =
2a3 bπ
8
4
The area of the ellipse (half ellipse) is
A =2
0
2 1/2
b 1− xa
0
/
2b
a
b 1−
a
0
dy dx
x2
a
dx
(a2 − x2 )1/2 dx
√
a
x a 2 − x2
a2
x
+
sin−1
2
2
a
0
$ √
%
2
2b
a 0
a
a
=
+
sin−1
a
2
2
a
√
0 a
a2
0
−
+
sin−1
2
2
a
=
2b
a
2/b a2/ π
πab
=
a/ 2/ 2
2
Evaluating, we get
A = 7.85 m2
Finally
ky =
Iy
=
A
ky = 2.5 m
Iy = 49.09 m4
a
=
A =


0 
/


a

Evaluating, we get
a
A =2
2b
x(a2 − x2 )3/2
a 2 x a 2 − x2
=
−
+
a
4
8
+
x
2
b(1− x2 )1/2
a
y]0
0
Rewriting
x2 + y2 = 1
a2 b2
−y
x2 dy dx
x2 b 1 −
0
Iy = 2b
y
x2 dy dx
2b
2
0
Iy = 2
Iy
y
y
0
a
a
−0
a
Iy =
x2 dA =
49.09
7.85
Problem 7.123 What are the values of Ix and kx
for the elliptical area of the airplane’s wing in Problem 7.122?
Solution:
Ix =
a
y dA = 2
A
Ix = 2
a
y3 a
b
3
0
√
0
b
y= a
√
a2 −x2
y
y dy dx
0
a2 −x2
dx
2b
x
0
y = b a2 – x2
a
a
0
√
a(0)
3a3 0
3 π
+
+ a4
4
8
8 2
√
0(a2 )
3a2 · 0 a2
−
−
+0
4
8
Ix =
x2
y2
— + — =1
a2
b2
2
b3
(a2 − x2 )3/2 dx
3
0 3a
√
2b3 x(a2 − x2 )3/2
3a2 x a2 − x2
=
+
3a3
4
8
a
3
x + a4 sin−1
8
a Ix = 2
Ix
2
2b3
3a3
a
Ix =
2/b3
·
3/a3/
Ix =
3/ab3 π
ab3 π
=
3/.8
8
3/
8
π
2/
Evaluating (a = 5, b = 1)
Ix =
5π
= 1.96 m4
8
From Problem 7.122, the area of the wing is A = 7.85 m2
Ix
1.96
kx =
=
kx = 0.500 m
A
7.85
Determine Iy and ky .
Problem 7.124
a4/
y
y = x 2 – 20
Solution: The straight line and curve intersect where x = x2 −
20. Solving this equation for x, we obtain
√
1 ± 1 + 80
x =
= −4, 5.
2
y=x
x
If we use a vertical strip: the area
dA = [x − (x2 − 20)] dx.
Therefore
Iy =
x2 dA =
A
5
−4
x2 (x − x2 + 20) dx
x4
x5
20x3
−
+
4
5
3
=
The area is
dA =
A =
A
5
= 522.
−4
y
5
−4
(x − x2 + 20) dx
5
So
ky
x2
x3
=
−
+ 20x
= 122.
2
3
−4
Iy
522
=
=
= 2.07.
A
122
y = x2 – 20
y=x
x
dA
dx
Problem 7.125
Problem 7.124.
Determine Ix and kx for the area in
Solution: Let us determine the moment of inertia about the x axis
of a vertical strip holding x and dx fixed:
x
x
y3
(Ix )(strip) =
y 2 dAs =
y 2 (dx dy) = dx
3 x2 −20
As
x2 −20
y
y=x
y = x2 – 20
dx
=
(−x6 + 60x4 + x3 − 1200x2 + 8000).
3
dAs
x
Integrating this value from x = −4 to x = 5 (see the solution to
Problem 7.124), we obtain Ix for the entire area:
5
1
Ix =
(−x6 + 60x4 + x3 − 1200x2 + 8000) dx
−4 3
5
x7
x4
400x3
8000x
= −
+ 4x5 +
−
+
= 10,900.
21
12
3
3
−4
x
dx
From the solution to Problem 7.124, A = 122 so
Ix
10,900
kx =
=
= 9.48.
A
122
Problem 7.126 A vertical plate of area A is beneath
the surface of a stationary body of water. The pressure of
the water subjects each element dA of the surface of the
plate to a force (p0 + γy) dA, where p0 is the pressure
at the surface of the water and γ is the weight density
of the water. Show that the magnitude of the moment
about the x axis due to the pressure on the front face of
the plate is
x
A
y
M(x axis) = p0 yA + γIx ,
where y is the y coordinate of the centroid of A and Ix
is the moment of inertia of A about the x axis.
The moment about the x-axis is dM = y(p0 +γy) dA
integrating over the surface of the plate:
M =
(p0 + γy)y dA.
Solution:
x
A
Noting that p0 and γ are constants over the area,
M = p0
y dA + γ y 2 dA.
A
By definition,
y dA
A
y =
and Ix =
A
y 2 dA,
A
then M = p0 yA + γIX , which demonstrates the result.
A
y
Problem 7.127 Determine Ix and kx for the composite
area by dividing it into rectangles 1 and 2 as shown, and
compare your results to those of Example 8.4.
y
1m
1
4m
2
1m
x
3m
Solution:
Ix =
Ix =
A1
1
4
A2
3
y3
3
0
1
1
0
3
dx +
1m
0
4
y3
3
0
y
y 2 dA2
y 2 dy dx +
1
1
y 2 dy dx
1
dx
0
A1
3m
4m
3
1
64
1
=
−
dx +
dx
3
3
0
0 3
63 1
1 3
63
3
=
x + x =
+
3
3 0
3
3
0
Ix
0
Ix =
Ix
y 2 dA1 +
1
A2
1m
x
3m
Ix = 21 + 1 = 22 ft4
Area = 3 + 3 = 6 ft2
Ix
22
kx =
=
= 1.91 ft
Area
6
Determine Iy and ky for the composite
Problem 7.128
area.
Solution:
Iy =
Iy =
Iy =
Iy =
=
A1
x2 dA1 +
1
0
[x
0
1
2
A2
x2 dy dx +
1
1
0
4
y]41
3
dx +
[x
0
(4x2 − x2 ) dx +
1
3x3
3
0
+
3
x3
3
3
0
0
Iy = 1 + 9 = 10.00 m4
10
ky =
= 1.29 m
6
y
x2 dA2
x2 dy dx
0
2
1
1m
y]10
3
0
A1
dx
Area = 6m2
3m
4m
x2 dx
A2
1m
x
3m
Determine Ix and kx .
Problem 7.129
y
6m
x
2m
3m
12 m
Solution:
Ix =
y 2 dA =
A
Ix =
3
9
Ix =
9
−3
y
4
y 2 dy dx
Area = (6)(12) m2
Area = 72 m2
−2
4m
dx
−2
(4)3
(−2)3
−
3
3
−3
−3
4
y3
9
−3
Ix =
9
72
dx = 24
3
9
dx =
−3
9
−3
64
8
+
3
3
x
2m
dx
3m
9
dx = 24x
9m
−3
Ix = 24[9 − (−3)] = (24)(12)
Ix = 288 m4
√
Ix
288
ky =
=
= 4=2m
Area
72
Determine Iy and ky .
Problem 7.130
Solution:
Iy =
x2 dA =
A
Iy =
−3
Iy =
9
9
9
−3
4
[x2 y]
x2 dy dx
−2
dx =
9
−3
−2
6x2 dx = 6
−3
3
y
4
x3
3
9
−3
x2 (4 − (−2)) dx
= [2x3 ]9−3
3
Iy = 2[9 − (−3) ]
Iy = 1512 m4
ky =
Iy
=
Area
ky = 4.58 m
√
1512
= 21
72
Area = (6)(12) m2
Area = 72 m2
4m
x
2m
3m
9m
Determine Ix and kx .
Problem 7.131
Solution:
Ix =
Ix =
A1
−15
−45
70
y3
3
−45
45
Ix =
−15
Ix
15
703 =
x
3
703
Ix =
3
703
100
y 2 dy dx
70
70
mm
100
y3
3
dx
70
dx
x
0
45
1003
−45
3
dx +
3
−
703
30 mm
dx
3
90 mm
dx
y
45
x
1003
703
−
3
3
−45
45
x
A2
−15
1003
703
−
3
3
(30) +
3
70
+
−45
+
−45
70
3
15
45
dx +
0
3
45
45
y 2 dA3
A3
100
mm
703
−45
+
y 2 dy dx
70
y3
3
15
y 2 dy dx +
70
0
−15
+
A2
−45
15
Ix =
y 2 dA2 +
0
45
+
y 2 dA1 +
y
Area = (100)(90)
–(30)(70)
= 9000 – 2100
Area = 6900 mm2
703
90 +
(30)
3
100 mm
70 mm
A1
30 mm
Ix = 2.66 × 107 mm4
kx =
Ix
=
Area
A3
30 mm
2.66 × 107
= 62.1 mm
6900
30 mm
x
90 mm
kx = 62.1 mm
Determine Iy and ky .
Problem 7.132
Solution:
Iy =
Iy =
x2 dA1 +
A1
−15
−45
15
Iy =
−15
−45
15
Iy =
Iy
−15
−45
x2 dy dx +
45
−45
70
A3
[x2 y]70
0 dx +
100
45
−45
0
70x dx +
100 mm
A1
45
−45
(15)3
(45)3
−
3
3
A3
70
30
mm
2
(100 − 70)x dx +
(−15)3
(−453 )
−
3
3
+ 70
Area = 9000 mm2
– 2100 mm2
Area = 6900 mm2
100
(x2 y)
dx
−15
45
45
x3 x3 x3 = 70 + 30 + 70 3 −45
3 −45
3 15
Iy = 70
A2
x2 dy dx
70
70
2 (x y) dx
2
y
x2 dA3
x2 dy dx
0
45
+
A2
x2 dA2 +
0
45
+
70
+ 30
45
2
70x dx
15
453
(−45)3
−
3
3
Iy = 140
453
153
−
3
3
+ 60
30
mm
453
3
Iy = 5.92 × 106 mm4
Iy
5.92 × 106
ky =
=
= 29.3 mm
Area
6900
ky = 29.3 mm
30
mm
70 mm
x
Determine JO and kO .
Problem 7.133
Solution:
JO =
JO =
r2 dA =
A
A
2
x dA +
A
y
(x2 + y 2 ) dA
A
y 2 dA = Ix + Iy
From the solutions to 8.31 and 8.32
7
Ix = 2.66 × 10 mm
7
Area = 9000 mm2
– 2100 mm2
Area = 6900 mm2
4
and Iy = 0.592 × 10 mm
4
100
mm
70 mm
Hence,
JO = 3.25 × 107 mm4
kO =
JO
=
A
30
mm
3.25 × 107
6900
kO = 68.6 mm
Problem 7.134 If you design the beam cross section
so that Ix = 6.4×105 mm4 , what are the resulting values
of Iy and JO ?
y
h
x
Solution:
The area moment of inertia for a triangle about the
base is
h
1
12
Ix =
bh3 ,
1
12
from which Ix = 2
30
mm
(60)h3 = 10h3 mm4 ,
30
mm
Ix = 10h3 = 6.4 × 105 mm4 ,
y
from which h = 40 mm.
Iy = 2
1
12
(2h)(303 ) =
1
3
h(303 )
h
x
from which Iy =
1
3
3
5
(40)(30 ) = 3.6 × 10 mm
4
and JO = Ix + Iy = 3.6 × 105 + 6.4 × 105 = 1 × 106 mm4
h
30 mm 30 mm
30
mm
30
mm
x
Problem 7.135
Determine Iy and ky .
y
160
mm
40 mm
Solution:
Divide the area into three parts:
200
mm
40
mm
Part (1): The top rectangle.
40 mm
A1 = 160(40) = 6.4 × 103 mm2 ,
dx1 =
160
= 80 mm,
2
1
12
Iyy1 =
x
120
mm
(40)(1603 ) = 1.3653 × 107 mm4 .
y
From which
160
mm
Iy1 = d2x1 A1 + Iyy1 = 5.4613 × 107 mm4 .
Part (2): The middle rectangle:
40 mm
A2 = (200 − 80)(40) = 4.8 × 103 mm2 ,
200
mm
dx2 = 20 mm,
1
12
Iyy2 =
40
mm
(120)(403 ) = 6.4 × 105 mm4 .
40 mm
x
From which,
120
mm
Iy2 = d2x2 A2 + Iyy2 = 2.56 × 106 mm4 .
Part (3) The bottom rectangle:
A3 = 120(40) = 4.8 × 103 mm2 ,
dx3 =
120
= 60 mm,
2
1
12
Iyy3 =
40(1203 ) = 5.76 × 106 mm4
Problem 7.136
Solution:
Use the solution to Problem 7.135. Divide the area into
Part (1): The top rectangle.
A1 = 6.4 × 103 mm2 ,
dy1 = 200 − 20 = 180 mm,
1
12
(160)(403 ) = 8.533 × 105 mm4 .
From which
Ix1 = d2y1 A1 + Ixx1 = 2.082 × 108 mm4
Part (2): The middle rectangle:
A2 = 4.8 × 103 mm2 ,
dy2 =
Ixx2 =
120
+ 40 = 100 mm,
2
1
12
Iy3 = d2X3 A3 + Iyy3 = 2.304 × 107 mm4
The composite:
Iy = Iy1 + Iy2 + Iy3 = 8.0213 × 107 mm4
Iy
ky =
= 70.8 mm.
(A1 + A2 + A3 )
Determine Ix and kx .
three parts:
Ixx1 =
From which
(40)(1203 ) = 5.76 × 106 mm4
from which
Ix2 = d2y2 A2 + Ixx2 = 5.376 × 107 mm4
Part (3) The bottom rectangle:
A3 = 4.8 × 103 mm2 ,
dy3 = 20 mm,
Ixx3 =
1
12
120(403 ) = 6.4 × 105 mm4
and Ix3 = d2y3 A3 + Ixx3 = 2.56 × 106 mm4 .
The composite:
Ix = Ix1 + Ix2 + Ix3 = 2.645 × 108 mm4
Ix
kx =
= 128.6 mm
(A1 + A2 + A3 )
Problem 7.137
Determine Ixy .
Solution: (See figure in Problem 7.135). Use the solutions in
Problems 7.135 and 8.36. Divide the area into three parts:
Part (1): A1 = 160(40) = 6.4 ×
dx1
103
mm2 ,
Ixy2 = dx2 dy2 A2 = 9.6 × 106 mm4 .
Part (3): A3 = 120(40) = 4.8 × 103 mm2 ,
160
=
= 80 mm,
2
dx3 =
dy1 = 200 − 20 = 180 mm,
120
= 60 mm,
2
dy3 = 20 mm,
Ixxyy1 = 0,
from which
from which
7
4
Ixy1 = dx1 dy1 A1 + Ixxyy1 = 9.216 × 10 mm .
Part (2) A2 = (200 − 80)(40) = 4.8 × 103 mm2 ,
dx2 = 20 mm,
dy2 =
from which
Ixy3 = dx3 dy3 A3 = 5.76 × 106 .
The composite:
Ixy = Ixy1 + Ixy2 + Ixy3 = 1.0752 × 108 mm4
120
+ 40 = 100 mm,
2
Problem 7.138
Determine Ix and kx .
y
160
mm
Solution: The strategy is to use the relationship Ix = d2 A+Ixc ,
where Ixc is the area moment of inertia about the centroid. From this
Ixc = −d2 A + Ix . Use the solutions to Problems 7.135, 7.136, and
7.137. Divide the area into three parts and locate the centroid relative
to the coordinate system in the Problems 7.135, 7.136, and 7.137.
40 mm
x
200
mm
40
mm
40 mm
Part (1) A1 = 6.4 × 103 mm2 ,
120
mm
dy1 = 200 − 20 = 180 mm.
Part (2) A2 = (200 − 80)(40) = 4.8 × 103 mm2 ,
y
dx1 =
160
= 80 mm,
2
dy2 =
120
+ 40 = 100 mm,
2
Part (3) A3 = 120(40) = 4.8 × 103 mm2 ,
dx3 =
120
= 60 mm,
2
160
mm
dx2 = 20 mm,
40 mm
200
mm
x
40 mm
40 mm
dy3 = 20 mm.
120 mm
The total area is
A = A1 + A2 + A3 = 1.6 × 104 mm2 .
from which
The centroid coordinates are
Ixc = −y2 A + Ix = −1.866 × 108 + 2.645 × 108
x =
A1 dx1 + A2 dx2 + A3 dx3
= 56 mm,
A
y =
A1 dy1 + A2 dy2 + A3 dy3
= 108 mm
A
Problem 7.139
Determine Iy and ky .
Solution: The strategy is to use the relationship Iy = d2 A+Iyc ,
where Iyc is the area moment of inertia about the centroid. From this
Iyc = −d2 A + Iy . Use the solution to Problem 7.138. The centroid
coordinates are x = 56 mm, y = 108 mm, from which
Iyc = −x2 A + Iy = −5.0176 × 107 + 8.0213 × 107
kyc
= 3.0 × 107 mm4 ,
Iyc
=
= 43.33 mm
A
kxc
= 7.788 × 107 mm4
Ixc
=
= 69.77 mm
A
Determine Ixy .
Problem 7.140
Solution:
Use the solution to Problem 7.137. The centroid coor-
dinates are
x = 56 mm,
y = 108 mm,
from which Ixyc = −xyA + Ixy = −9.6768 × 107 + 1.0752 × 108
= 1.0752 × 107 mm4
Problem 7.141 Determine Ix and kx .
Solution: Divide the area into two parts:
y
Part (1): a triangle and Part (2): a rectangle. The area moment of
inertia for a triangle about the base is
Ix =
1
12
3 ft
4 ft
bh3 .
3 ft
The area moment of inertia about the base for a rectangle is
Ix =
1
3
x
bh3 .
Part (1) Ix1 =
Part (2) Ix2 =
1
12
1
3
4(33 ) = 9 ft2 .
3(33 )
= 27.
y
4 ft
3 ft
The composite: Ix = Ix1 + Ix2 = 36 ft4 . The area:
3 ft
1
4(3) + 3(3) = 15 ft2 .
2
Ix
=
= 1.549 ft.
A
A =
kx
Problem 7.142
x
Determine JO and kO .
Solution: (See Figure in Problem 7.141.) Use the solution to
Problem 7.141.
from which
Part (1): The area moment of inertia about the centroidal axis parallel
to the base for a triangle is
where A2 = 9 ft2 .
Iyc =
1
36
bh3 =
1
36
3(43 ) = 5.3333 ft4 ,
from which
Iy1 =
8
3
2
A1 + Iyc = 48 ft4 .
where A1 = 6 ft2 .
Part (2): The area moment of inertia about a centroid parallel to the
base for a rectangle is
Iyc =
1
12
bh3 =
1
12
3(33 ) = 6.75 ft4 ,
Iy2 = (5.5)2 A2 + Iyc = 279 ft4 ,
The composite: Iy = Iy1 + Iy2 = 327 ft4 , from which, using a result from
Problem 7.141,
JO = Ix + Iy = 327 + 36 = 363 ft4
JO
and kO =
= 4.92 ft
A
Problem 7.143
Determine Ixy .
Solution: (See Figure in Problem 7.141.) Use the results of the
solutions to Problems 7.141 and 7.142. The area cross product of
the moment of inertia about centroidal axes parallel to the bases for a
1 2 2
triangle is Ix y = 72
b h , and for a rectangle it is zero. Therefore:
Ixy1 =
1
72
(42 )(32 ) +
8
3
3
3
A1 = 18 ft4
and Ixy2 = (1.5)(5.5)A2 = 74.25 ft4 ,
Ixy = Ix y 1 + Ixy2 = 92.25 ft4
Problem 7.144
Determine Ix and kx .
y
4 ft
3 ft
x
3 ft
Solution: Use the results of Problems 7.141, 7.142, and 7.143.
The strategy is to use the parallel axis theorem and solve for the area
moment of inertia about the centroidal axis. The centroidal coordinate
y =
A1 (1) + A2 (1.5)
= 1.3 ft.
A
From which
Ixc = −y2 A + Ix = 10.65 ft4
Ixc
and kxc =
= 0.843 ft
A
Problem 7.145
Determine JO and kO .
Solution: Use the results of Problems 7.141, 7.142, and 7.143.
The strategy is to use the parallel axis theorem and solve for the area
moment of inertia about the centroidal axis. The centroidal coordinate:
A1 83 + A2 (5.5)
x =
= 4.3667 ft,
A
from which
IY C = −x2 A + IY = 40.98 ft4 .
Using a result from Problem 7.144,
JO = IXC + IY C = 10.65 + 40.98 = 51.63 ft4
JO
and kO =
= 1.855 ft
A
y
4 ft
3 ft
3 ft
x
Determine IXY .
Problem 7.146
Solution: Use the results of Problems 7.141–7.145. The strategy
is to use the parallel axis theorem and solve for the area moment of
inertia about the centroidal axis. Using the centroidal coordinates
determined in Problems 7.144 and 7.145,
Ixyc = −xyA + Ixy = −85.15 + 92.25 = 7.1 ft4
Determine Ix and kx .
Problem 7.147
y
120
mm
20
mm
80
mm
x
40 mm
80 mm
Solution: Let Part 1 be the entire rectangular solid without the
hole and let part 2 be the hole.
Ix1 =
1 3
bh
3
y
m
20 m
where b = 80 mm
h = 120 mm
Ix1
y′
1
= (80)(120)3 = 4.608 × 107 mm4
3
For Part 2,
Ix 2
40 mm
1
1
= πR4 = π(20)4 mm4
4
4
Ix 2 = 1.257 × 105 mm4
Area = (80)(120)
– π (20)2
= 8343 mm2
120 mm
40 mm
Part 2
x′
Ix2 = Ix 2 + d2y A
dy = 80 mm
where A = πR2 = 1257 mm2
d = 80 mm
Ix2 = 1.257 × 105 + π(20)2 (80)2
6
Part 1
6
x
4
Ix2 = 0.126 × 10 + 8.042 × 10 mm
= 8.168 × 106 mm4 = 0.817 × 107 mm4
7
4
Ix = Ix1 − Ix2 = 3.79 × 10 mm
Area = hb − πR2 = (80)(120) − πR2
Area = 8343 mm2
Ix
kx =
= 67.4 mm
Area
80 mm
Determine JO and kO .
Problem 7.148
Solution: For the rectangle,
JO1 = Ix1 + Iy1 =
1 3
1
bh + hb3
3
3
7
7
JO1 = 4.608 × 10 + 2.048 × 10 mm
y
40 mm
4
y′
JO1 = 6.656 × 107 mm4
R = 20 mm
A1 = bh = 9600 mm2
For the circular cutout about x y x′
1
1
= πR4 + πR4
4
4
JO2
= Ix 2 + Iy
2
120 mm
A2
JO2
= 1.257 × 105 + 1.257 × 105 mm4
JO2
= 2.513 × 105 mm2
(h) 80 mm
A1
Using the parallel axis theorem to determine JO2 (about x, y)
JO2 = J0 2 + (d2x + d2y )A2
x
A2 = πR2 = 1257 mm2
JO2 = 1.030 × 107 mm4
80 mm
JO = JO1 − JO2
(b)
7
7
JO = 6.656 × 10 − 1.030 × 10 mm
4
JO = 5.63 × 107 mm4
JO
JO
kO =
=
Area
A1 − A2
kO = 82.1 mm
Determine Ixy .
Problem 7.149
Solution:
y
A1 = (80)(120) = 9600 mm2
80 mm
A2 = πR2 = π(20)2 = 1257 mm2
For the rectangle (A1 )
Ixy1 =
R = 20 mm
y′
1 2 2
1
b h = (80)2 (120)2
4
4
x′
Ixy1 = 2.304 × 107 mm2
A2
A1
For the cutout
Ix y 2 = 0
120 mm
A2
dy = 80 mm
A1
and by the parallel axis theorem
Ixy2 = Ix y 2 + A2 (dx )(dy )
Ixy2 = 0 + (1257)(40)(80)
6
4
Ixy2 = 4.021 × 10 mm
Ixy = Ixy1 − Ixy2
Ixy = 2.304 × 107 − 0.402 × 107 mm4
Ixy = 1.90 × 107 mm4
x
dx = 40 mm
Determine Ix and kx .
Problem 7.150
y
20
mm
120
mm
x
80
mm
40 mm
80 mm
Solution: We must first find the location of the centroid of the
total area. Let us use the coordinates XY to do this. Let A1 be
the rectangle and A2 be the circular cutout. Note that by symmetry
Xc = 40 mm
Rectangle1
Circle2
Area
9600 mm2
1257 mm2
Xc
40 mm
40 mm
y
Y
80 mm
Yc
60 mm
80 mm
R = 20 mm
A1 = 9600 mm2
120 mm
A2 = 1257 mm2
For the composite,
A1 Xc1 − A2 Xc2
Xc =
= 40 mm
A1 − A2
Yc =
x
80 mm
A1 Yc1 − A2 Yc2
= 57.0 mm
A1 − A2
Now let us determine Ix and kx about the centroid of the composite body.
X
40 mm
40 mm
Rectangle about its centroid (40, 60) mm
Ix1 =
1 3
1
bh =
(80)(120)3
12
12
7
3
Now to C → dy2 = 80 − 57 = 23 mm
Ixc2 = Ix2 + (dy2 )2 A2
Ix1 = 1.152 × 10 mm ,
Ixc2 = 7.91 × 105 mm4
Now to C
For the composite about the centroid
Ixc1 = Ix1 + (60 − Yc )2 A1
7
4
Ixc1 = 1.161 × 10 mm
Circular cut out about its centroid
A2 = πR2 = (20)2 π = 1257 mm2
Ix2 =
1
πR4 = π(20)4 /4
4
Ix2 = 1.26 × 105 mm4
Ix = Ixc1 − Ixc2
Ix = 1.08 × 107 mm4
The composite Area = 9600 − 1257 mm2
kx
= 8343 mm2
Ix
=
= 36.0 mm
A
Problem 7.151
Determine Iy and ky .
Solution: From the solution to Problem 7.150, the centroid of the
composite area is located at (40, 57.0) mm.
y
The area of the rectangle, A1 , is 9600 mm2 .
The area of the cutout, A2 , is 1257 mm2 .
The area of the composite is 8343 mm2 .
(1)
Rectangle about its centroid (40, 60) mm.
Iy1 =
1 3
1
hb =
(120)(80)3
12
12
+
Iy1 = 5.12 × 106 mm4
dx1 = 0
(2)
x
Circular cutout about its centroid (40, 80)
80 mm
Iy2 = πR4 /4 = 1.26 × 105 mm4
dx2 = 0
40 mm
Since dx1 and dx2 are zero. (no translation of axes in the xdirection), we get
80 mm
Iy = Iy1 − Iy2
6
4
Iy = 4.99 × 10 mm
Finally,
Iy
=
A1 − A2
ky
=
ky
= 24.5 mm
Problem 7.152
4.99 × 106
8343
Determine JO and kO .
Solution: From the solutions to Problems 7.151 and 7.152,
y
Ix = 1.07 × 107 mm4
Iy = 4.99 × 106 mm4
and A = 8343 mm2
JO = Ix + Iy = 1.57 × 107 mm4
JO
kO =
= 43.4 mm
A
20
mm
x
120 mm
80 mm
Problem 7.153
Determine Iy and ky .
y
12 in
x
20 in
Solution: Treat the area as a circular area with a half-circular
cutout: From Appendix B,
(Iy )1 =
and (Iy )2
1
π(20)4 in4
4
1
1
1
π(20)4 − π(12)4 = 1.18 × 105 in4 .
4
8
The area is A = π(20)2 − 12 π(12)2 = 1030 in2
Iy
1.18 × 105
so, ky =
=
A
1.03 × 103
= 10.7 in
Problem 7.154
Determine JO and kO .
Solution: Treating the area as a circular area with a half-circular
cutout as shown in the solution of Problem 7.153, from Appendix B,
(JO )1 = (Ix )1 + (Iy )1 =
and (JO )2 = (Ix )2 + (Iy )2 =
Therefore JO =
y
y
12
in.
1
= π(12)4 in4 ,
8
so Iy =
y
1
π(20)4 in4
2
1
π(12)4 in4 .
4
1
1
π(20)4 − π(12)4
2
4
= 2.35 × 105 in4 .
From the solution of Problem 7.153,
JO
2
A = 1030 in Ro =
A
2.35 × 105
=
= 15.1 in.
1.03 × 103
20 in.
2
x
x
20 in.
2 in.
x
Determine Iy and ky if h = 3 m.
Problem 7.155
Solution:
y
Break the composite into two parts, a rectangle and a
semi-circle.
1.2 m
For the semi-circle
Ix c =
Iy c =
9
π
−
8
8π
1
πR4
8
R4
d=
4R
3π
h
y′
x
y
d
x′
1.2 m
AC
d = 4R
3π
To get moments about the x and y axes, the (dxc , dyc ) for the semicircle are
dxc = 0,
dyc = 3 m +
4R
3π
AR
3m = h
and Ac = πR2 /2 = 2.26 m2
1
πR4
8
Iy c =
x
and Iyc = Iy c + d2xc A (dx = 0)
To get moments of area about the x, y axes, dxR = 0, dyR = 1.5 m
Iyc = Iy c = π(1.2)4 /8
0
IyR = Iy R + (dxR )2 (/ bh)
Iyc = 0.814 m4
For the Rectangle
Ix R =
1 3
bh
12
Iy R =
1 3
hb
12
IyR = Iy R =
1
(3)(2, 4)3 m4
12
IyR = 3.456 m2
AR = bh = 7.2 m2
AR = bh
Iy = Iyc + IyR
Iy = 4.27 m2
y′ y
To find ky , we need the total area, A = AR + Ac
A = 7.20 + 2.26 m2
2.4 m
h
3m
b
A = 9.46 m2
Iy
ky =
= 0.672 m
A
x′
x
Determine Ix and kx if h = 3 m.
Problem 7.156
Solution: Break the composite into two parts, the semi-circle and
the rectangle. From the solution to Problem 7.155,
Ix c =
π
9
−
8
8π
dyc =
3+
4R
3π
y
R4
Ac
m
Ac = 2.26 m2
Ixc = Ix c + Ac d2yc
Substituting in numbers, we get
R = 1.2 m
h=3m
b = 2.4 m
AR
3m
h
Ix c = 0.0717 m4
dyc = 3.509 m
and Ixc = Ix c + Ac d2y
Ixc = 27.928 m2
For the Rectangle h = 3 m, b = 2.4 m
x
2.4 m
yc′
Area: AR = bh = 7.20 m2
Ix R =
1 3
bh , dyR = 1.5 m
12
IxR = Ix R +
d2yR AR
xc′
R
Substituting, we get
Ix R = 5.40 m4
IxR = 21.6 m4
For the composite,
Ix = IxR + Ixc
Ix = 49.5 m4
Also kx =
Ix
= 2.29 m
AR + Ac
kx = 2.29 m
4R
3π
Problem 7.157
Determine the centroid of the area.
y
60 cm
x
60 cm
80 cm
Solution: The strategy is to develop useful general results for the
triangle and the rectangle.
y
The rectangle: The area of the rectangle of height h and width w is
w
A =
h dx = hw = 4800 cm2 .
0
60 cm
The x-coordinate:
w
w
x2
hx dx = h
=
2 0
0
Divide by the area: x =
w
2
1
2
80 cm
= 40 cm
Divide by the area: y =
The y-coordinate:
w
1
h2 dx =
2
0
1
h2 w.
2
Divide by the area: y = 12 h = 30 cm
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
b
b
a
ax2
A =
y(x) dx =
− x + a dx = −
− ax
b
2b
0
0
0
ab
ab
= −
+ ab =
= 1800 cm2
2
2
Check: This is the familiar result. check.
The x-coordinate:
b
b
a
ax3
ax2
ab2
− x + a x dx = −
+
=
.
b
3b
2 0
6
0
Divide by the area: x =
x
hw2 .
b
3
= 20 cm
The y-coordinate:
b
2
1
a
y dA =
− x+a
dx
2
b
A
0
3 b
b
a
ba2
=−
− x+a
=
.
6a
b
6
0
x =
a
20
3
60 cm
cm. The composite:
xR AR + xT AT
40(4800) + 100(1800)
=
AR + AT
4800 + 1800
= 56.36 cm
y =
(30)(4800) + (20)(1800)
4800 + 1800
= 27.27 cm
Problem 7.158
Determine the centroid of the area.
y
40 mm
20 mm
40 mm
Solution:
Divide the object into five areas:
80 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:
(1)
A1 = 6400 mm2 ,
x1 = 40 mm, y1 = 40 mm,
(2)
A2 = 9600 mm2 ,
x2 = 120 mm, y2 = 60 mm,
(3)
A3 = 2513.3 mm2 ,
x3 = 120 mm, y3 = 136.98 mm,
(4)
A4 = 1256.6 mm2 ,
x4 = 120 mm, y4 = 120 mm.
40 mm
(5)
The composite area: A = A1 + A2 + A3 − A4 =
17256.6 mm2 . The composite centroid:
80 mm
x
120 mm
160 mm
y
x=
A1 x1 +A2 x2 +A3 x3 −A4 x4
A
= 90.3 mm .
y=
A1 y1 +A2 y2 +A3 y3 −A4 y4
A
= 59.4 mm
20 mm
40 mm
x
120 mm
160 mm
Problem 7.159 The cantilever beam is subjected to a
triangular distributed load. What are the reactions at A?
y
200 N/m
x
A
10 m
line with intercept
Solution: The load distribution is a straight
w = 200 N/m at x = 0, and slope − 200
= −20 N/m2 . The sum
10
of the moments is
10
M = MA −
(−20x + 200)x dx = 0,
y
200
N/m
0
from which
MA = −
20 3
x + 100x2
3
x
10
10 m
= 3333.3 Nm.
0
The sum of the forces:
10
Fy = Ay −
(−20x + 200) dx = 0,
0
from which
and
10
Ay = −10x2 + 200x 0 = 1000 N,
Fx = Ax = 0
200 N/m
AX
MA
AY
10 m
Problem 7.160
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 = 0 at y = 0 and the slope is
+ 100
= 20.
5
Over the interval 5 ≤ y ≤ 10, the load is a constant w = 100 N/m.
The moment about the origin E due to the load is
5
10
ME =
(20y)y dy +
100y dy,
0
from which
ME =
20 3
y
3
100 N/m
C
5m
B
D
5m
5
E
A
10 m
5
0
+
100 2
y
2
10
= 4583.33 N-m.
Cy
Cx
5
Cx
Check: The area of the triangle is
1
F1 = ( )(5)(100) = 250 N.
2
The area of the rectangle: F2 = 500 N. The centroid distance for the
triangle is
d1
2
= ( )5 = 3.333 m.
3
The centroid distance of the rectangle is d2 = 7.5 m. The moment
about E is
ME = d1 F1 + d2 F2 = 4583.33 Nm check.
The Complete Structure: The sum of the moments about E is
M = −10AR + ME = 0,
where AR is the reaction at A, from which AR = 458.33 N.
The element ABC: Element BD is a two force member, hence By = 0.
The sum of the moments about C:
MC = −5Bx − 10Ay = 0,
where Ay is equal and opposite to the reaction of the support, from
which
Bx = −2Ay = 2AR = 916.67 N.
Since the reaction in element BD is equal and opposite, Bx =
−916.67 N, which is a tension in BD.
By
Bx
Ay
Cy
Bx
By
Dy
Dx
Dx Dy
Ey
Ex
Problem 7.161 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 = 400 N/m, and slope −5. The moment about the base of the tower
due to the wind load is
40
MW =
(−5y + 400)y dy,
0
5
MW = − y 3 + 200y 2
3
40
θ = 90◦ − tan−1
15
40
200 N/m
B
40 m
A
15 m
= 213.33 kN-m,
0
clockwise about the base, looking North. The angle formed by the
cable with the horizontal at the top of the tower is
= 69.44◦ .
The sum of the moments about the base of the tower is
M = −MW + 40TA cos θ = 0,
(a)
400 N/m
(b)
200 N/m
θ
40 m
TA
from which
TA =
1
40 cos θ
MW = 15.19 kN
Fx
Fy
Problem 7.162 If the wind in Problem 7.161 blows
from the east and cable A is slack, what are the tensions
in cables B and C?
Solution: From Problem 7.161, the moment about the base of the
tower is MW = 213.33 kN-m, counterclockwise if the wind is from
the east and the observer is looking North. The angle in the horizontal
plane between the cables and the east is 60◦ . The sum of the moments
about the base is
M = MW − 40TB cos θ cos 60◦ − 40TC cos θ cos 60◦ = 0.
From symmetry, the tensions in the two cables are equal, from which
TB = TC =
1
80 cos θ cos 60◦
MW = 15.19 kN
(c)
(b)
400 N/m
N
C
(c)
Problem 7.163 Determine the y coordinate of the center of mass of the homogeneous steel plate.
y
20 mm
Solution: Divide the object into five areas: (1) The lower rectangle 20 by 80 mm, (2) an upper rectangle, 20 by 40 mm, (3) the
semicircle of radius 20 mm, (4) the circle of radius 10 mm, and (5)
the composite part. The areas and the centroids are tabulated. The last
row is the composite and the centroid of the composite. The composite
area is
A =
3
1
10 mm
20 mm
20 mm
Ai − A4 .
x
80 mm
The centroid:
3
x =
1
,
A
3
and y =
Ai xi − A4 x4
1
10 mm
y
20 mm
Ai yi − A4 y4
.
A
20 mm
The following relationships were used for the centroids: For a rectangle: the centroid is at half the side and half the base. For a semicircle,
the centroid is on the centerline and at 4R
from the base. For a circle,
3π
the centroid is at the center.
A, sq mm
1600
800
628.3
314.2
2714
Area
A1
A2
A3
A4
Composite
Problem 7.164
x, mm
40
60
60
60
48.2
20 mm
x
y, mm
10
30
48.5
40
21.3
80 mm
Determine Iy and ky .
Solution: Divide the section into two parts: Part (1) is the upper
rectangle 40 mm by 200 mm, Part (2) is the lower rectangle, 160 mm
by 40 mm.
Part (1) A1 = 0.040(0.200) = 0.008 m2 ,
y
40
mm
y1 = 0.180 m
x1 = 0,
Iy1 =
1
12
160
mm
0.04(0.2)3 = 2.6667 × 10−5 m4 .
Part (2): A2 = (0.04)(0.16) = 0.0064 m2 ,
x
80
mm
y2 = 0.08 m,
40
mm
x2 = 0,
Iy2 =
1
12
(0.16)(0.04)3 = 8.5 × 10−7 m4 .
The composite:
y
40 mm
2
A = A1 + A2 = 0.0144 m ,
160 mm
Iy = Iy1 + Iy2 ,
Iy = 2.752 × 10−5 m4 = 2.752 × 107 mm4 ,
Iy
and ky =
= 0.0437 m = 43.7 mm
A
80 40 80
mm mm mm
80
mm
Problem 7.165
Problem 7.164.
Solution:
Determine Ix and kx for the area in
Use the results in the solution to Problem 7.164. Part (1)
A1 = 0.040(0.200) = 0.008 m2 ,
y1 = 0.180 m,
1
12
Ix1 =
0.2(0.043 ) + (0.18)2 A1 = 2.603 × 10−4 m4 .
Part (2):
A2 = (0.04)(0.16) = 0.0064 m2 ,
y2 = 0.08 m,
Ix2 =
1
12
(0.04)(0.16)3 + (0.08)2 A2 = 5.461 × 10−5 m4 .
The composite: A = A1 + A2 = 0.0144 m2 , The area moment of
inertia about the x axis is
Ix = Ix1 + Ix2 = 3.15 × 10−4 m4 = 3.15 × 108 mm4 ,
Ix
and kx =
= 0.1479 m = 147.9 mm
A
Problem 7.166
y
Determine Ix and kx .
40
mm
x
160
mm
80
mm
Solution:
Use the results of the solutions to Problems 7.164–
7.165. The centroid is located relative to the base at
xc =
x1 A1 + x2 A2
= 0,
A
yc =
y1 A1 + y2 A2
= 0.1356 m.
A
The moment of inertia about the x-axis is
2
Ixc = −yC
A + IX = 5.028 × 107 mm4
Ixc
and kxc =
= 59.1 mm
A
Problem 7.167
Problem 7.166.
Determine JO and kO for the area in
Solution: Use the results of the solutions to Problems 7.164–
7.165. The area moments of inertia about the centroid are
Ixc = 5.028 × 10−5 m4
and Iyc = Iy = 2.752 × 10−5 m4 ,
from which
JO = Ixc + Iyc = 7.78 × 10−5 m4 = 7.78 × 107 mm4
JO
and kO =
= 0.0735 m
A
= 73.5 mm
40
mm
80
mm
y
40 mm
x
160 mm
80 40 80
mm mm mm
Problem 8.1 The coefficients of static and kinetic friction between the 0.4-kg book and the table are µs = 0.30
and µk = 0.28. A person exerts a horizontal force on
the book as shown.
(a) If the magnitude of the force is 1 N and the book
remains stationary, what is the magnitude of the friction force exerted on the book by the table?
(b) What is the largest force the person can exert without
causing the book to slip?
(c) If the person pushes the book across the table at a
constant speed, what is the magnitude of the friction
force?
Solution:
(a)
(b)
Fx :
F −f =0
Fy :
N −W =0
F =1N
Solving f = 1 N
fmax = µs N, µs = 0.3
N = 3.92 N, fmax = 1.18 N
Fmax = 1.18 N
(c)
For constant speed,
f = µk N = (0.28)(3.92)
f = 1.10 N
Thus, since F − f = 0
F = 1.10 N
W = mg = (0.4) 9.81 = 3.92 N
F
N
f
Problem 8.2 The 10.5-kg Sojourner rover, placed on
the surface of Mars by the Pathfinder Lander on July
4, 1997, was designed to negotiate a 45◦ slope without
tipping over.
(a) What minimum static coefficient of friction between
the wheels of the rover and the surface is necessary
for it to rest on a 45◦ slope? The acceleration due
to gravity at the surface of Mars is 3.69 m/s2 .
(b) Engineers testing the Sojourner on Earth want to
confirm that it will negotiate a 45◦ slope without
tipping over. What minimum static coefficient of
friction between the wheels of the rover and the surface is necessary for it to rest on a 45◦ slope on
Earth?
(a) Assume that slip is impending when the rover is on a 45◦
slope. The friction force f = µs N , and the free-body diagram is: The equilibrium equations are:
Fx = −µs N + mg sin 45◦ = 0,
Fy = N − mg cos 45◦ = 0.
Solution:
Summing the two equations, we obtain N − µs N = 0 so µs = 1 is the
minimum static coefficient of friction. (b) the solution in (a) is independent of
the value of g so µs = 1 is the minimum on earth also.
y
45°
µsN
mg
N
x
Problem 8.3 The coefficient of static friction between
the tires of the 8000-kg truck and the road is µs = 0.6.
(a) If the truck is stationary on the incline and α =
15◦ , what is the magnitude of the total friction force
exerted on the tires by the road?
(b) What is the largest value of α for which the truck
will not slip?
Solution:
y
α
W = mg
x
α
α
f
(a)
Fy :
Fx :
N
f − mg sin α = 0
N − mg cos α = 0
g = 9.81 m/s2 , α = 15◦ , m = 8000 kg
Solving, f = 20.3 kN also N = 75.8 kN
(b) Set f = µs N and solve for α in the basic eqns.
Fx : µs N − mg sin α = 0
Fy : N − mg cos α = 0
m = 8000 kg, g = 9.81 m/s2
Solving, α = 30.96◦
Problem 8.4 The coefficient of static friction between
the 5-kg box and the inclined surface is µs = 0.3. The
force F is horizontal and the box is stationary.
(a) If F = 40 N, what friction force is exerted on the
box by the inclined surface?
(b) What is the largest value of F for which the box will
not slip?
F
30°
Solution:
30°
F
y
mg
30°
x
30°
F
30°
f
N
(a) F = 40 N
Fx : f + F cos 30◦ − mg sin 30◦ = 0
Fy : N − F sin 30◦ − mg cos 30◦ = 0
(b)
f = −10.1 N (down the plane)
In the equilibrium eqns, set f = −µs N and treat F as unknown.
F is negative to resist F .
Solving, F = 52.0 N
Problem 8.5 In Problem 9.4, what is the smallest value
of the force F for which the box will not slip?
Solution: For this problem, facts up the plane
30°
mg
F
y
x
F
30°
30°
f
N
Fx :
f + F cos 30◦ − mg sin 30◦ = 0
Fy :
= N − mg cos 30◦ − F sin 30◦ = 0
Solving, F = 11.6 N
Problem 8.6 The device shown is designed to position
pieces of luggage on a ramp. It exerts a force parallel
to the ramp. The mass of the suitcase S is 9 kg. The
coefficients of friction between the suitcase and ramp are
µs = 0.20 and µk = 0.18.
(a) Will the suitcase remain stationary on the ramp
when the device exerts no force on it?
(b) What force must the device exert to start the suitcase
moving up the ramp?
(c) What force must the device exert to move the suitcase up the ramp at a constant speed?
S
20°
Solution:
y
(a)
20°
mg
x
f
F
(b)
(c)
Fx :
F − f − mg sin(20◦ ) = 0
Fy :
N − mg cos(20◦ ) = 0
Set f = −µs N (up the plane). If F ≤ 0, it will remain stationary.
Solving, we get F = 13.6 N (required to hold it stationary. No, it will not
remain stationary if F = 0.
Set f = µs N (down the plane) Solving, we get F = 46.8 N
Set f = µk N (down the plane) Solving, F = 45.1 N
20°
20°
N
µs = 0.20
µk = 0.18
m = 9 kg
Problem 8.7 The mass of the stationary crate is 40 kg.
The length of the spring is 180 mm, its unstretched length
is 200 mm, and the spring constant is k = 2500 N/m.
The coefficient of static friction between the crate and the
inclined surface is µs = 0.6. Determine the magnitude
of the friction force exerted on the crate.
20°
Solution: The magnitude of the force exerted on the crate by the
compressed spring is. (2500 N/m)(0.2 m − 0.18 m) = 50 N. The free
body diagram of the crate is shown. From the equilibrium equations
Fx = 50 − f + (40)(9.81) sin 20◦ = 0,
Fy = N − (40)(9.81) cos 20◦ = 0,
20°
we obtain N = 369 N, f = 184 N.
y
50 N
20°
mg
f
N
x
Problem 8.8 The coefficient of kinetic friction between the 40-kg crate and the floor is µk = 0.3. If the
angle α = 20◦ , what tension must the person exert on
the rope to move the crate at constant speed?
α
Solution:
mg
Fx :
T cos 20◦ − f = 0
Fy :
N + T sin 20◦ − mg = 0
Solving, T = 112.94 N
T
also f = 106.13 N
N = 353.77 N
α
α
f
N
m = 40 kg
α = 20◦
f = µk N
Problem 8.9 In Problem 9.8, for what angle α is the
tension necessary to move the crate at constant speed a
minimum? What is the necessary tension?
Solution: From the solution to Problem 9.8, we have
Differentiating with respect to α, we get
dT
T (sin α − µk cos α)
=
dα
(cos α + µk sin α)
mg
T
α
= 0, we get tan α = µk
Setting dT
dα
Solving, α = 16.7◦ . Substituting back into the equilibrium equations, we can
now solve for N and T .
T = 112.76 N,
N = 360 N
f = µkN
f = 108 N
N
Fx :
T cos α − µk N = 0
Fy :
+ T sin α + N − mg = 0
Solving the second eqn. for N and substituting into the first, we get
T cos α − µk mg + T µk sin α = 0.
α
Problem 8.10 Box A weighs 100 lb, and box B weighs
30 lb. The coefficients of friction between box A and
the ramp are µs = 0.30 and µk = 0.28. What is the
magnitude of the friction force exerted on box A by the
ramp?
A
B
30°
Solution:
The sum of the forces parallel to the inclined surface is
F = −A sin α + B + f = 0,
from which f = A sin α − B = 100 sin 30◦ − 30 = 20 lb
A
B
30°
B
α
A
f
N
Problem 8.11 In Problem 9.10, box A weighs 100 lb,
and the coefficients of friction between box A and the
ramp are µs = 0.30 and µk = 0.28. For what range of
the weights of the box B will the system remain stationary?
Solution: The upper and lower limits on the range are determined
by the weight required to move the box up the ramp, and the weight that
will allow the box to slip down the ramp. Assume impending slip. The
friction force opposes the impending motion. For impending motion
up the ramp the sum of forces parallel to the ramp are
F = A sin α − BMAX + µS A cos α = 0,
from which
BMAX
A
α
N
µ sN
BMAX = A(sin α + µs cos α)
= 100(sin 30◦ + 0.3 cos 30◦ ) = 75.98 lb
BMIN
from which
B = A(sin α − µs cos α)
= 100(sin 30◦ − 0.3 cos 30◦ ) = 24.02 lb
α
A
For impending motion down the ramp:
F = A sin α − BMIN − µs A cos α = 0,
µ sN
N
Problem 8.12 The mass of the box on the left is 30 kg,
and the mass of the box on the right is 40 kg. The coefficient of static friction between each box and the inclined
surface is µs = 0.2. Determine the minimum angle α
for with the boxes will remain stationary.
α
Solution: If the boxes slip when α is decreased, they will slip
toward the right. Assume that slip toward the right impends, the free
body diagrams are as shown.
The equilibrium equations are
Fx = T − 0.2 NA − (30)(9.81) sin α = 0,
(1)
Fy = NA − (30)(9.81) cos α = 0,
(2)
◦
Fx = −T − 0.2 NB + (40)(9.81) sin 30 = 0, (3)
Fy = NB − (40)(9.81) cos 30◦ = 0,
(4)
Summing Equations (1) and (3), we obtain −0.2 NA − 0.2 NB −
(30)(9.81) sin α + (40)(9.81) sin 30◦ = 0. Solving Equation (2)
for NA and Equation (4) for NB and substituting the results into
Equation (5) gives 15sin α + 3 cos α = 10 − 4 cos 30◦ . (6) Using
the identity cos α = 1 − sin2 α and solving Equation (6) for sin α,
we obtain sin α = 0.242, so α = 14.0◦
30°
α
30°
y
y
T
T
x
α
30°
(40)(9.81)
(30)(9.81)
0.2 NB
0.2 NA
NA
NB
x
Problem 8.13 In Problem 9.12, determine the maximum angle α for with the boxes will remain stationary.
If the boxes slip when α is increased, they will slip
toward the left. When slip toward the left impends, The free body
diagrams are as shown.
The equilibrium equations are
Fx = T + 0.2 NA − (30)(9.81) sin α = 0,
(1)
Fy = NA − (30)(9.81) cos α = 0,
(2)
Fx = −T + 0.2 NB + (40)(9.81) sin 30◦ = 0, (3)
Fy = NB − (40)(9.81) cos 30◦ = 0,
(4)
Solution:
Summing Equations (1) and (3), we obtain
0.2NA + 0.2 NB − (30)(9.81) sin α + (40)(9.81) sin 30◦ = 0. (5)
Solving Equation (2) for NA and Equation (4) for NB and substituting the results into Equation (5) gives 15 sin α − 3 cos α =
10 + 4 cos 30◦ . (6) Using the identity cos α =
1 − sin2 α
and solving Equation (6) for sin α, we obtain sin α = 0.956, so
α = 73.0◦
y
y
α
(30)(9.81)
T
T
30°
x
(40)(9.81)
0.2 NA
NA
NB
0.2 NB
x
Problem 8.14 The box is stationary on the inclined
surface. The coefficient of static friction between the
box and the surface is µs .
(a) If the mass of the box is 10 kg, α = 20◦ , β = 30◦ ,
and µs = 0.24, what force T is necessary to start
the box sliding up the surface?
(b) Show that the force T necessary to start the box
sliding up the surface is a minimum when tan β =
µs .
T
β
α
Solution:
T
T
β
mg
y
α
β
α
α
f
x
N
α = 20◦
µs = 0.24
m = 10 kg
g = 9.81 m/s2
(a)
Fx :
− T cos β + f + mg sin α = 0
Fy :
N + T sin β − mg cos α = 0
β = 30◦ , f = µs N
Substituting the known values and solving, we get
T = 56.5 N,
N = 64.0 N,
f = 15.3 N
Solving the 2nd equilibrium eqn for N and substituting for
f (f = µs N ) in the first eqn, we get
−T cos β + µs mg cos α − µs T sin β + mg sin α = 0
Differentiating with respect to β, we get
dT
T (sin β − µs cos β)
=
dβ
(cos β + µs sin β)
Setting
tan β = µs
dT
dβ
= 0, we get
Problem 8.15 To explain observations of ship launchings at the port of Rochefort in 1779, Coulomb analyzed
the system shown in Problem 9.14 to determine the minimum force T necessary to hold the box stationary on
the inclined surface. Show that the result is
(sin α − µs cos α)mg
T =
.
cos β − µs sin β
T
β
Solution:
mg
y
T
β
α
α
α is fixed, β is variable. Solve the second eqn for N and substitute into the first.
We get
0 = T (µs sin β − cos β) = mg(sin α − µs cos α)
α
or T =
x
mg(sin α − µs cos α)
(cos β − µs sin β)
To get the conditions for the minimum, set
N
f = µsN
dT
dβ
dT
T (sin β + µs cos β)
=
=0
dβ
(cos β − µs sin β)
Fx :
− T cos β + mg sin α − µs N = 0
For the min.
Fy :
N + T sin β − mg cos α = 0
tan β = −µs .
Note β is negative!
Problem 8.16 Two sheets of plywood A and B lie on
the bed of a truck. They have the same weight W , and
the coefficient of static friction between the two sheets
of wood and between sheet B and the truck bed is µs .
(a) If you apply a horizontal force to sheet A and apply
no force to sheet B, can you slide sheet A off the
truck without causing sheet B to move? What force
is necessary to cause sheet A to start moving?
(b) If you prevent sheet A from moving by applying a
horizontal force on it, what horizontal force on sheet
B is necessary to start it moving?
A
B
Solution:
(a)
The friction force exerted by sheet A on B at impending motion
is fAB = µs W . The friction force exerted by sheet B on the bed
of the truck is fBT = µs (2W ), since the normal force is due to
the weight of both sheets. Since fBT > fAB , the top sheet will
begin moving before the bottom sheet. Yes
The force required to start sheet A to move is
fAB
(a)
The force on B is the friction between A and B and the friction
between B and the truck bed. Thus the force required to start B
in motion is
fBT
FB = fAB + fBT = 3µs W.
(b)
fAB
W
F = fAB = µs W.
(b)
F
W
fAB
2W
W
FB
fBT
2W
=0
Problem 8.17 Suppose that the truck in Problem 9.16
is loaded with N sheets of plywood of the same weight
W , labeled (from the top) sheets 1, 2, . . . , N . The coefficient of static friction between the sheets of wood and
between the bottom sheet and the truck bed is µs . If you
apply a horizontal force to the sheets above it to prevent them from moving, can you pull out the ith sheet,
1 ≤ i ≤ N , without causing any of the sheets below it
to move? What force must you apply to cause it to start
moving?
The force holding the sheets below the ith sheet from
moving is the friction force between the bed of the truck and the bottom
sheet. The weight is N W , hence the force opposing motion of the
sheets below the ith sheet is FN = µs N W . The force causing motion
to start is the friction between the ith sheet and those below it, which
is FB = µs (iW ). The resultant force is FR = µs W (N − i). The
bottom sheets will begin to move when the resultant is zero, which
can only occur for the last sheet, i = N . Thus the ith sheet can be
extracted without the sheets below it moving. Yes The force required
to move the ith sheet is the force required to overcome the friction
force due to sheet above it and the sheet below it. The weight of the
sheets above is (i − 1)W and the weight on the sheet below it is iW .
The maximum total friction force opposing motion is
Solution:
(i−1)W
µs(i−1)W
Fi
µsiW
iW
Fi = µs (iW + (i − 1)W ) = µs W (2i − 1)
Problem 8.18 The masses of the two boxes are m1 =
45 kg and m2 = 20 kg. The coefficients of friction
between the left box and the inclined surface are µs =
0.12 and µk = 0.10. Determine the tension the man
must exert on the rope to pull the boxes upward at a
constant rate.
m1
30°
30°
Solution:
m1g
m2
y
30° y
Tman
30°
m1
N1
f = µ kN1
m1
T2
x
30°
T2
m2
m2
m2g
Equilibrium Eqns:
Mass 2:
F:
T2 − m2 g = 0
Mass 1:
Fx :
Fy :
T2 − TMAN + µk N1 + m1 g sin 30◦ = 0
N1 − m1 g cos 30◦ = 0
m1 = 45 kg,
m2 = 20 kg,
g = 9.81 m/s2 ,
and µk = 0.1.
Substituting these values into the eqns. and solving,
TMAN = 455 N
Problem 8.19 In Problem 9.18, for what range of tensions exerted on the rope by the man will the boxes remain stationary?
Solution:
(1)
(2)
We must look at two cases.
Impending slip up the plane
Impending slip down the plane.
In either case, |f | = µs N1 , only the sign changes
30°
m1
30°
m2
y
m1g
Tman
30°
30°
f2
T2
N1
f1
x
T2 = m2 g
f1 is used in case 1
f2 is used in case 2
Equilibrium Eqns:
Fx :
T2 − Tman ± f + m1 g sin(30◦ ) = 0
Fy :
N1 − m1 g cos(30◦ ) = 0
f = µs N1 = 0.12 N1 .
The (+) is used for case 1 and the (−) is used for case 2
Solving case 1, we get
Tman = 462.8 N
Solving case 2, we get
Tman = 371.0 N
Thus, the boxes are stationary for
371 N ≤ Tman ≤ 463 N
Problem 8.20 The coefficient of static friction between the two boxes is µs = 0.2, and between the lower
box and the inclined surface it is µs = 0.32. What is the
largest angle α for which the lower box will not slip?
W
W
α
Solution:
We need free body diagrams of both boxes
W
W
α
y
y
α
m 2g
NU
T
x
x
fU
fU
1
α
fL
α NL
m1g
fU = 0.2 NU
fL = 0.32 NL
m1 = m2 = m
Lower Box
Fx :
fL + fU − mg sin α = 0
Fy :
NL − NU − mg cos α = 0
Upper Box
m1 = m2 = m
Fx :
T − fU − mg sin α = 0
Fy :
NU − mg cos α = 0
Substituting in known values and solving, we get
α = 40.0◦
NU
Problem 8.21 The coefficient of static friction between the two boxes and between the lower box and
the inclined surface is µs . What is the largest force F
that will not cause the boxes to slip?
W
F
2W
α
Solution: At impending motion, the sum of the forces parallel to
the inclined surface for the upper box is
FP = F − µs W cos α + W sin α − T = 0,
where T is the tension in the string, and the friction force opposes
impending motion in the direction of F . For the lower box,
F = 2W sin α + µs (3W cos α) + µs W cos α − T = 0,
W
F
2W
α
where the friction force opposes impending motion in the direction of
T . Combining:
F = µs (5W cos α) + W sin α = W (5µs cos α + sin α)
W
F
Wcos α
µsWcosα
T
µsWcosα
Wcos α
T
2W
3µsWcosα
3Wcosα
Problem 8.22 Consider the system shown in Problem 9.21. The coefficient of static friction between the
two boxes and between the lower box and the inclined
surface is µs . If F = 0, the lower box will slip down the
inclined surface. What is the smallest force F for which
the boxes will not slip?
Solution: The solution is obtained by the same procedure as in
Problem 10.21, with the exception that the friction forces now oppose
impending motion in the direction of T for the upper box, and impending motion down the surface, for the lower box. The sums of forces
parallel to the inclined surface for the two boxes are:
FP = F + µs W cos α + W sin α − T = 0
and
F = 2W sin α − µs (3W cos α) − µs W cos α − T = 0.
Combining:
F = −µs W cos α + W sin α − 4µs W cos α = W (sin α − 5µs cos α)
F
µsWcosα
T
W
Wcos α
Wcos α
µsWcosα
T
2W
3Wcosα
3µsWcosα
Problem 8.23 A sander consists of a rotating disk with
sandpaper bonded to the outer surface. The normal force
exerted on the workpiece A by the sander is 30 lb. The
workpiece A weighs 50 lb. The coefficients of friction
between the sander and the workpiece A are µs = 0.65
and µk = 0.60. The coefficients of friction between the
workpiece A and the table are µs = 0.35 and µk = 0.30.
Will the workpiece remain stationary while it is being
sanded?
Solution:
(1)
(2)
Two possible situations can cause impending motion:
before the contact surface has begun slipping with respect to the
workpiece (that is, as the sanding wheel is brought into contact
with the work piece), so that the static coefficient of friction
applies between the sanding wheel and the workpiece.
After the sanding wheel has begun slipping relative to the workpiece, when the kinetic coefficient of friction applies between
the sander and the workpiece. In the first situation, the force
inducing motion of the workpiece is
F = µs 30 = 19.5 lb.
The force resisting motion is FA = µs (30 + 50) = 28 lb. Thus
the workpiece will not slip. The second situation is less severe,
since the force inducing motion is F = 30µk = 18 lb, and the
force opposing motion is the same. The workpiece will remain
stationary.
A
30 lb
µ (30)
50 lb
f
(30 + 50) lb
A
Problem 8.24 Suppose that you want the bar of length
L to act as a simple brake that will allow the workpiece
A to slide to the left but will not allow it to slide to the
right no matter how large a horizontal force is applied
to it. The weight of the bar is W , and the coefficient
of static friction between it and the workpiece A is µs .
You can neglect friction between the workpiece and the
surface it rests on.
(a) What is the largest angle α for which the bar will
prevent the workpiece from moving to the right?
(b) If α has the value determined in (a), what horizontal
force is required to start the workpiece A toward the
left at a constant rate?
L
α
A
Solution:
(a)
To resist motion to the right no matter how large the horizontal
force requires a very large friction force. For impending motion
to the right, the sum of the moments about the bar hinge is
M =+
L
W
L sin α − FN L sin α + f L cos α = 0,
2
α
where FN is the normal force and f is the friction force resisting
the impending motion.
Noting that f = µs FN , the sum of moments yields
FN =
W sin α
.
2(sin α − µs cos α)
If the denominator vanishes, the normal force and hence the friction force become as large as required. Thus sin α − µs cos α =
0, from which µs = tan α, or α = tan−1 (µs )
(b) For impending motion to the left, the sum of moments about the
bar hinge is
L
M = +W
sin α − FN L sin α − f L cos α = 0,
2
where the friction force opposes the impending motion. Noting
FN = µf , then
s
f =
2
A
W sin α
sin α
µs
+ cos α
=
W sin α
µs W
=
4 cos α
4
is the force required to start motion to the left.
α
α
W
f
FN
W
f
FN
(a)
(b)
Problem 8.25 The coefficient of static friction between the 20-lb bar and the floor is µs = 0.3. Neglect
friction between the bar and the wall.
(a) If α = 20◦ , what is the magnitude of the friction
force exerted on the bar by the floor?
(b) What is the maximum value of α for which the bar
will not slip?
L
α
Solution:
The sum of the moments about the upper end of the
bar is
WL
sin α + FN L sin α − f L cos α = 0,
2
−W
+ FN sin α
2
from which f =
.
cos α
M =−
L
α
The sum of the forces in the vertical direction
FY = −W + FN = 0,
from which FN = W.
Substitute:
f =
W sin α
= 10 tan 20◦ = 3.64 lb
2 cos α
(b) At impending slip, f = µs FN = µs W , from which, substituting
above,
1
µs =
tan α, or α = tan−1 (2µs ) = tan−1 (0.6) = 31◦
2
R
α
W
f
FN
Problem 8.26 The masses of the ladder and person
are 18 kg and 90 kg, respectively. The center of mass
of the 4-m ladder is at its midpoint. If α = 30◦ , what
is the minimum coefficient of static friction between the
ladder and the floor necessary for the person to climb
to the top of the ladder? Neglect friction between the
ladder and the wall.
α
x
Solution: The weight of the ladder is W = 18 g = 176.58 N.
The weight of the person is P = 90 g = 882.9 N. Let h be the
distance along the ladder of the person’s center of mass, and L be
the length of the ladder. The horizontal distance is. The sum of the
moments about the top of the ladder:
M = P (L sin α − x) − FN L sin α + W
L
sin α + f L cos α = 0.
2
From the sum of the forces,
FY = FN − W − P = 0, from
which the normal force at the foot of the ladder is FN = W + P .
Substitute, solve for the friction force, and reduce algebraically:
h
1
f =
P + W tan α.
L
2
α
h
At the top of the ladder, L
= 1, hence
W
f = P+
tan α = (883 + 88.3)(0.5774) = 560.72 N
2
x
At impending slip, f = µs FN = µs (P + W ), from which
µs =
f
560.72
=
= 0.5292
P +W
1059.48
R
P
α
W
f
FN
x
Problem 8.27 In Problem 9.26, the coefficient of static
friction between the ladder and the floor is µs = 0.6. The
masses of the ladder and the person are 18 kg and 100 kg,
respectively. The center of mass of the 4-m ladder is at
its midpoint. What is the maximum value of α for which
the person can climb to the top of the ladder? Neglect
friction between the ladder and the wall.
Solution: The solution in Problem 9.26 for the friction force is
f =
h
1
P+ W
L
2
tan α and f = µs (W + P ).
At impending slip
µs (W + P )
.
tan α = h
P + 12 W
L
At the top of the ladder
h
= 1, and tan α = 0.6495, or α = tan−1 (0.6495) = 33◦
L
Problem 8.28 In Problem 9.26, the coefficient of static
friction between the ladder and the floor is µs = 0.6, and
α = 35◦ . The center of mass of the 4-m ladder is at its
midpoint, and its mass is 18 kg.
(a) If a football player with a mass of 140 kg attempts
to climb the ladder, what maximum value of x will
he reach? Neglect friction between the ladder and
the wall.
(b) What minimum friction coefficient would be required for him to reach the top of the ladder?
The weight of the football player is P = 140 g =
1373.4 N, and the weight of the ladder is W = 176.58 N. From the
solution to Problem 9.26,
h
1
f =
P + W tan α
L
2
Solution:
and f = µs (W + P ).
Substitute and solve for
µs (W + P ) − W
tan α
h
2
=
= 0.90277.
L
P tan α
From 9.26, we see that x = h sin α, from which
h
L sin α = 2.07 m
x =
L
(b) Solve the above for the static friction coefficient:
h
P + 12 W tan α
L
µs =
.
(W + P )
h
At the top of the ladder, L
= 1, and
1
P + 2 W tan α
µs =
,
(P + W )
µs = 0.66
Problem 8.29 The disk weighs 50 lb. Neglect the
weight of the bar. The coefficients of friction between
the disk and the floor are µs = 0.6 and µk = 0.4.
(a) What is the largest couple M that can be applied to
the stationary disk without causing it to start rotating?
(b) What couple M is necessary to rotate the disk at a
constant rate?
Solution: The normal force at the point of contact is found from
the sum of moments about the pin support.
M = −8(100) − 20(50) + 20FN = 0,
8 in
12 in
100 lb
5 in
M
100 lb
8 in
12 in
5 in
from which FN = 90 lb. The friction force is f = µs FN . The
moment exerted by the friction force is
MF = µs RFN = 0.6(5)(90) = 270 in lb
M
This is the moment to be overcome at impending slip.
(b) The moment required to rotate the disk at a constant rate is
MK = µk RFN = 0.4(5)(90) = 180 in lb
8 in
12 in
5 in
100 lb
M
f
Problem 8.30 The cylinder has weight W . The coefficient of static friction between the cylinder and the floor
and between the cylinder and the wall is µs . What is the
largest couple M that can be applied to the stationary
cylinder without causing it to rotate?
R
M
Solution: Assume impending slip. The force opposing rotation
is the sum of the friction force at the wall and at the floor. Denote the
normal force at the wall by FN W and the normal force on the floor by
FN F . From the sum of forces:
Fy = µs FN W + FN F − W = 0,
and
Fx = FN W − µs FN F = 0.
R
M
Solve these two simultaneous equations to obtain:
FN F =
W
,
1 + µ2s
and FN W =
µs W
.
1 + µ2s
µsFNW
The sum of moments about the center of the cylinder is
MC = Mapp − µs RFN W − µs RFN F = 0.
Substitute and solve:
1 + µs
Mapp = µs RW
.
1 + µ2s
At impending slip, this is the maximum moment that can be applied
to the cylinder.
FN
M
FNW
W
µsFNF
FNF
Problem 8.31 The cylinder has weight W . The coefficient of static friction between the cylinder and the floor
and between the cylinder and the wall is µs . What is the
largest couple M that can be applied to the stationary
cylinder without causing it to rotate?
R
α
M
Solution: Assume impending slip. Denote the normal force at
the wall by FN W and the normal force at the floor by FN F . The
projection of the friction force at the wall on an x-y coordinate system
is
f W = µs |FN W |(i cos(90 − α) + j sin(90 − α))
= µs |FN W |(i sin α + j cos α).
R
M
α
The projection of the normal force at the wall on an x-y coordinate
system is
FN W = |FN W |(i cos α − j sin α).
The sum of the forces:
Fy = |FN W |(µs cos α − sin α) + |FN F | − |W| = 0,
and
Fx = |FN W |(cos α + µs sin α) − µs |FN F | = 0.
Solve these simultaneous equations to obtain:
|FN F | =
and |FN W | =
|W|(cos α + µs sin α)
,
(1 + µ2s ) cos α
FNW
α
W
µs |W|
.
(1 + µ2s ) cos α
The sum of the moments about the center of the cylinder is
MC = Mapp − µs R|FN F | − µs R|FN W | = 0.
Substitute and reduce algebraically:
Mapp =
fW
µs RW (cos α + µs sin α + µs )
.
(1 + µ2s ) cos α
This is the maximum applied moment at impending slip. Check: This
reduces to the solution of Problem 9.30 when α = 0, as it should.
check.
µs |FNF |
FNF
Problem 8.32 Suppose that α = 30◦ in Problem 9.31
and that a couple M = 0.5RW is required to turn the
cylinder at a constant rate. What is the coefficient of
kinetic friction?
Solution:
Substitute the angle and the moment into the solution
of Problem 9.31 to obtain:
√
√
µk 23 + µ2k + µk
µk (1 + 3µk )
√
0.5 =
=
.
(1 + µ2k )
(1 + µ2 ) 3
k
2
This reduces to the quadratic equation µ2k + 2bµk − c = 0, where
b =c=
1
√
.
(2 3 − 1)
√
The solution is µk = −b ± b2 + c. Substitute numerical values:
µk = 0.3495, or µk = −1.1612. The negative value has no meaning here.
Problem 8.33 The disk of weight W and radius R is
held in equilibrium on the circular surface by a couple
M . The coefficient of static friction between the disk
and the surface is µs . Show that the largest value M can
have without causing the disk to slip is
M
µs RW
M =
.
1 + µ2s
This is an inclined plane problem. Let α be the angle
at the point of contact. From the sum of forces: the normal force is
FN = W cos α, and the friction force is f = µs FN = W sin α,
from which µs = tan α. The sum of moments about the center of the
disk yields M = f R = µs RW cos α. Noting that
Solution:
cos α = √
1
1 + tan2 α
M
,
µs RW
,
then M = 1 + µ2s
which is the moment at impending slip.
M
f
W
FN
α
Problem 8.34 The coefficient of static friction between the jaws of the pliers and the gripped object is
µs . What is the largest value of the angle α for which
the gripped object will not slip? (Neglect the object’s
weight.)
α
Solution: Choose an x-y coordinate system such that the x-axis
bisects the angle α. Define
β =
α
.
2
The projection of the normal forces on the x-y system is, for the top:
FN T = |FN T |(−i sin β − j cos β).
For the bottom:
α
FN B = |FN B |(−i sin β + j cos β).
The projection of the friction forces on the x-y system is, for the top:
fT = µs |FN T |(i cos β − j sin β).
For the bottom:
fB = µs |FN B |(i cos β + j sin β).
The forces tending to expel the gripped object are the components
of the normal forces in the negative x direction, and the components
tending to retain the gripped object are the friction forces in the positive
x-direction. These must balance:
Fx = −|FN B | sin β − |FN T | sin β + µs |FN T | cos β
+ µs |FN B | cos β = 0,
from which
|FN B |(− sin β + µs cos β) + |FN T |(− sin β + µs cos β) = 0.
From symmetry, |FN B | = |FN t |, since the weight of the object
is neglected. For non-trivial values of the normal forces, − sin β +
µs cos β = 0, from which µs = tan β, or β = tan−1 (µs ). Noting
β =
α
,
2
α = 2 tan−1 (µs )
β
FNT
fT
fB
β
FNB
Problem 8.35 The stationary disk, of 300-mm radius,
is attached to a pin support at D. The disk is held in place
by the brake ABC in contact with the disk at C. The
hydraulic actuator BE exerts a horizontal 400-N force
on the brake at B. The coefficients of friction between
the disk and the brake are µs = 0.6 and µk = 0.5. What
couple must be applied to the stationary disk to cause it
to slip in the counterclockwise direction?
C 300
mm
200 mm
E
D
B
200 mm
A
200
mm
Solution: Assume impending slip. For counterclockwise motion
the friction force f = µs FN opposes the impending slip, so that it acts
on the brake in a downward direction, producing a negative moment
(clockwise) about A. The sum of the moments about A:
MA = −0.2(400) + (0.4 − 0.2µs )FN ,
from which FN = 285.7 N. The sum of the moments about the center
of the disk:
MD = M − 0.3(µs )FN = 0,
C
D
200 mm
E
B
200 mm
300
mm
A
200
mm
from which M = 51.43 N m.
f
M
FN
FN
200 mm
400 N
f
200 mm
200
mm
Problem 8.36 What couple must be applied to the
stationary disk in Problem 9.35 to cause it to slip in a
clockwise direction?
Solution: Assume impending slip. For clockwise motion the friction force f = µs FN opposes the impending slip, so that it acts on
the brake in an upward direction, producing a positive moment (counterclockwise) about A. The sum of the moments about A:
MA = −0.2(400) + (0.4 + 0.2µs )FN = 0,
from which FN = 153.85 N. The sum of the moments about the
center of the disk:
MD = M − 0.3µs FN = 0,
from which M = 27.69 N m
300
mm
Problem 8.37 The mass of block B is 8 kg. The
coefficient of static friction between the surfaces of the
clamp and the block is µs = 0.2. When the clamp is
aligned as shown, what minimum force must the spring
exert to prevent the block from slipping out?
45°
160 mm
200 mm
B
100
mm
Solution: The free-body diagram of the block when slip is impending is shown. From the equilibrium equation
45°
µs FT + µs (FT + W cos α) − W cos α = 0,
160 mm
we obtain
FT =
=
w(1 − µs ) cos α
2µs
200 mm
(8)(9.81)(1 − 0.2) cos 45◦
2(0.2)
= 111 N.
B
100
mm
The free-body diagram of the upper arm of the clamp is shown. Summing moments about the upper end,
0.16Fs + 0.1µs FT − 0.36FT = 0,
the force exerted by the spring is
Fs
W
0.36FT − 0.1µs FT
=
0.16
=
µsFT
FT
FT +Wcosα
100 mm
[0.36 − 0.1(0.2)]111
0.16
= 236 N.
µsFT
Problem 8.38 By altering its dimensions, redesign
the clamp in Problem 9.37 so that the minimum force
the spring must exert to prevent the block from slipping
out is 180 N. Draw a sketch of your new design.
Solution:
This problem does not have a unique solution.
µs (FT + Wcos α )
FS
FT
160
200 mm
mm
Problem 8.39 The horizontal bar is attached to a collar
that slides on the smooth vertical bar. The collar at P
slides on the smooth horizontal bar. The total mass of
the horizontal bar and the two collars is 12 kg. The
system is held in place by the pin in the circular slot. The
pin contacts only the lower surface of the slot, and the
coefficient of static friction between the pin and the slot
is 0.8. If the system is in equilibrium and y = 260 mm,
what is the magnitude of the friction force exerted on the
pin by the slot?
P
y
300 mm
Solution: The free body diagram of the horizontal bar and right
collar is as shown, where m1 is the mass of the horizontal bar and right
collar, N1 is the normal force exerted by the vertical bar, and N2 is
the force exerted by the left collar. From the equilibrium equations
Fx = −N1 = 0,
Fx = N2 − m1 g = 0,
P
we see that N2 = m1 g. The free body diagram of the left collar is
as shown, where m2 is the mass of the left collar and N , f are the
normal and friction forces exerted by the curved slot.
y
300 mm
y = 260 mm = (300 mm) sin θ,
so the angle θ = 60.1◦ .
From the equilibrium equations,
Fx = −f + m2 g cos θ + N2 cos θ = 0,
Fy = N − m2 g sin θ − N2 sin θ = 0,
y
N2
N1
m1g
we obtain
f = (m2 g + N2 ) cos θ = (m2 + m1 ) cos θ = (12)(9.81) cos 60.1◦
y
N2
f
= 58.7 N.
θ
N
x
m2g
Problem 8.40 In Problem 9.39, what is the minimum
height y at which the system can be in equilibrium?
Solution: From the solution of Problem 9.39, the friction and
normal forces exerted on the pin by the circular slot are
f = (m2 g + N2 ) cos θ,
N = (m2 g + N2 ) sin θ,
so
f
N
= cot θ. When slip impends,
f = µs N = 0.8 N,
so 0.8 = cot θ and θ = 51.3◦ . The height y = 300 sin θ = 234 mm.
x
Problem 8.41 The rectangular 100-lb plate is supported by the pins A and B. If friction can be neglected at
A and the coefficient of static friction between the pin at
B and the slot is µs = 0.4, what is the largest angle α
for which the plate will not slip?
α
2 ft 3 in
B
A
2 ft
2 ft
Choose a coordinate system with the x- axis parallel to
the rail. The sum of the moments about A is
MA = −2W cos α − 2.25W sin α + 4B = 0,
Solution:
from which
α
B
A
W
(2 cos α + 2.25 sin α).
B =
4
The component of weight causing the plate to slide is F = W sin α.
This must be balanced by the friction force: 0 = −W sin α + µs B,
from which
W sin α
W
=
(2 cos α + 2.25 sin α).
µs
4
Reduce algebraically to obtain
µs
α = tan−1
= 14.47◦
2 − 1.125µs
2 ft 3 in
2 ft
2 ft
2 ft
2 ft
µ sB
2.25 ft
B
A
W
Problem 8.42 If you can neglect friction at B in Problem 9.41 and the coefficient of static friction between the
pin at A and the slot is µs = 0.4, what is the largest angle
α for which the plate will not slide?
Solution: The normal force acts normally to the slots, and the
friction force acts parallel to the slot. Choose a coordinate system
with the x-axis parallel to the slots. The normal component of the
reaction at A is found from the sum of the moments about B:
MB = −2.25W sin α + 2W cos α − 4A = 0,
from which
AN =
W
(−2.25 sin α + 2 cos α).
4
The force tending to make the plate slide is F = −W sin α. This is
balanced by the friction force at A,
0 = −W sin α + µs AN ,
from which
W sin α
W
=
(−2.25 sin α + 2 cos α).
µs
4
Reduce algebraically to obtain
µs
= 9.27◦
α = tan−1
2 − 1.125µs
Check: The normal reactions at A and B are unequal: as the slots are inclined
from the horizontal, the parallel component of the gravity force reduces the
normal force at A, and increases the normal force at B. check.
Check: The sum of the
reactions at A and B are AN + BN = W cos α.
check. The magnitude (AN + BN )2 + (µs AN )2 = W , hence the system
is in equilibrium at impending slip. check.
2 ft
2 ft
µs AN
2.25 ft
BN
AN
W
Problem 8.43 The airplane’s weight is W = 2400 lb.
Its brakes keep the rear wheels locked, and the coefficient
of static friction between the wheels and the runway is
µs = 0.6. The front (nose) wheel can turn freely and
so exerts only a normal force on the runway. Determine
the largest horizontal thrust force T the plane’s propeller
can generate without causing the rear wheels to slip.
Solution: The free body diagram when slip of the rear wheels
impends is shown. From the equilibrium equations
fx = −T + µs B = 0,
fy = A + B − W = 0,
MptA = 4T − 5W + 7B = 0,
we obtain
A = 1120 lb,
B = 1280 lb,
and T = 766 lb.
T
4 ft
W
A
5 ft
B
2 ft
y
T
W
4 ft
µsB
A
B
5 ft
2 ft
x
T
4 ft
W
A
5 ft
B
2 ft
Problem 8.44 The refrigerator weighs 350 lb. The
distances h = 60 in. and b = 14 in. The coefficient of
static friction at A and B is µs = 0.24.
(a) What force F is necessary for impending slip?
(b) Will the refrigerator tip over before it slips?
F
h
A
B
b
Solution: The normal forces on the right and left supports are
found: The sum of moments about support A:
MA = −bW − hF + 2bB = 0,
b
F
from which
B =
bW + hF
.
2b
h
The sum of forces:
FY = A + B − W = 0,
b
b
from which
A =W −B =
B
A
bW − hF
.
2b
(a)
The friction forces must balance the applied force at impending
slip:
bW − hF
0 = F − µs A − µ s B = F − µ s
2b
bW + hF
− µs
,
2b
from which F = µs W = 0.24(350) = 84 lb.
If the normal force at A approaches zero before motion occurs,
the refrigerator will start to tip:
b
14
Ftip =
W =
350 = 81.67 lb.
h
60
(b)
F
h
W
µs A
A
b
µs B
B
b
Since Ftip < Fmove , the refrigerator will tip.
Yes.
Problem 8.45 If you want the refrigerator in Problem 9.44 to slip before it tips over, what is the maximum
height h at which you can push it?
Solution:
From the solution to Problem 9.44, the tipping force must be
equal to or greater than the moving force, Fmove = 84 lb. Thus when the
normal force at A approaches zero, the tipping force must equal or exceed 84 lb,
from which
W
350
h≤b
= 14
≤ 58.33 in
Fmove
84
Problem 8.46 To obtain a preliminary evaluation of
the stability of a turning car, imagine subjecting the stationary car to an increasing lateral force F at the height
of its center of mass, and determine whether the car will
slip (skid) laterally before it tips over. Show that this will
be the case if b/h > 2µs . (Notice the importance of the
height of the center of mass relative to the width of the
car. This reflects on recent discussions of the stability of
sport utility vehicles and vans that have relatively high
centers of mass.)
F
h
b_
2
b_
2
b_
2
b_
2
Solution:
y
F
mg
h
F
h
A
fL
NL
B
b
2
b
2
fR
NR
EQUILIBRIUM Eqns:
Fx :
F − fL − fR = 0
Fy :
NL + NR − mg = 0
MA :
−hF + bNR −
b
mg = 0
2
Assume skid and tip simultaneously.
f L = µs N L ,
fR = µs NR (skid)
and NL = 0 (tip), ∴ fL = 0.
fR = µs mg.
The equilibrium eqns become
F = fR = µs NR = µs mg
and the moment eqn. uses
−h(µs mg) + b(mg) −
or
b
h
For
b
h
b
h
b
h
b
mg = 0
2
= 2µs
b
h
> 2µs , slip before tip
< 2µs , tip before slip
big N low cm, relative to track width
small N high cm, relative to track width
x
Problem 8.47 The man exerts a force P on the car at an
angle α = 20◦ . The 1760-kg car has front wheel drive.
The driver spins the front wheels, and the coefficient of
kinetic friction is µk = 0.02. Snow behind the rear tires
exerts a horizontal resisting force S. Getting the car to
move requires overcoming a resisting force S = 420 N.
What force P must the man exert?
Solution:
Fx :
S − µk NF − P cos α = 0
Fy :
NR + NF − mg − P sin α = 0
MA :
−(1.62)mg + 2.55 NF
+(0.90)P cos α − (3.40)P sin α = 0
α = 20◦ ,
m = 1760 kg,
g = 9.81 m/s2
S = 420 N,
µk = 0.02
3 eqns in 3 unknowns (NR , NF , and P )
Solving the equations, we get P = 213 N
P
α
NR = 6.34 kN
NF = 11.00 kN
0.90 m
S
1.62 m
2.55 m
3.40 m
P
α
0.90 m
S
1.62 m
2.55 m
3.40 m
y
mg
F
α
1.62 m
S
0.90 m
B
A
2.55 m
µ k NF
NR
NF
x
Problem 8.48 In Problem 9.47, what value of the angle α minimizes the magnitude of the force P the man
must exert to overcome the resisting force S = 420 N
exerted on the rear tires by the snow? What force must
he exert?
Solution: From the solution to Problem 9.47, we have
Use this eqn to find
S − µk NF − P cos α = 0
(1)
NR + NF − mg − P sin α = 0
(2)
−3.40
dP
sin α − 3.40 P cos α = 0
dα
or
dP
2.55
cos α
0.90 cos α − 3.40 sin α −
dα
µk
µk = 0.02,
S = 420 N,
m = 1760 kg,
From Eqn (1),
1
(S − P cos α) (a)
µk
NR = −NF + mg + P sin α
1
(S − P cos α) + mg + P sin α (b)
µk
Substitute (a) and (b) into (3)
We get
−1.62 mg + 2.55
k
tan α = From Eqn (2),
or NR = −
2.55
sin α − 0.90 sin α − 3.40 cos α = 0
µk
2.55
−P
− 0.90 sin α − 3.40 cos α
dP
µk
=0=
dα
cos α
0.90 cos α − 3.40 sin α − 2.55
µ
+P
and g = 9.81 m/s2 .
NF =
and set it to zero.
2.55
dP
dP
−
cos α + P sin α + 0.90
cos α − 0.90 P sin α
µk
dα
dα
−1.62 mg + 2.55 NF + 0.90P cos α − 3.40P sin α = 0 (3)
where
dP
dα
1
µk
(S − P cos α)
+0.90 P cos α − 3.40 P sin α = 0
3.40
2.55
µk
− 0.90
Solving, α = 1.54◦
Substituting this back into eqns (1), (2), and (3), and solving, we get
P = 202 N
Problem 8.49 The coefficient of static friction between the 3000-lb car’s tires and the road is µs = 0.5.
Determine the steepest grade (the largest value of the
angle α) the car can drive up at constant speed if the car
has (a) rear-wheel drive; (b) front-wheel drive; (c) fourwheel drive.
n
19 i
n
35 i
n
72 i
α
Solution: The friction force acts parallel to the incline, and the
normal force is normal to the incline. Choose a coordinate system
with the x-axis parallel to the incline. The component of the weight
that acts parallel to the incline is W sin α, and the component acting
normally to the incline is W cos α.
(a) For rear wheel drive: The moment about the point of contact of
the front wheels:
MF W = 35W cos α + 19W sin α − 107R = 0,
from which the normal reaction of the two rear wheels is
R=
W
(35 cos α + 19 sin α).
107
(c)
For four wheel drive: Use the reactions of the front and rear wheels obtained in Parts (a) and (b). The sum of the forces parallel to the incline
is
FX = −W sin α + µs R + µs F = 0,
from which
W sin α
W
=
(35 cos α + 19 sin α + 72 cos α − 19 sin α).
µs
107
Reduce and solve: α = tan−1 (µs ) = 26.57◦
Check: This result is the same as if the Mercedes with four wheel drive
were a box on an incline, as it should be.
The force causing impending slip is W sin α, which is balanced
by the friction force: 0 = W sin α − µs R, from which
W sin α
W
=
(35 cos α + 19 sin α).
µs
107
Reduce and solve:
α = tan−1
(b)
35
− 19
19 in
35 in
= 10.18◦
107
µs
72 in
is the maximum angle at impending slip.
For front wheel drive: The moments about the point of contact
of the rear wheels is
α
MRW = −72W cos α + 19W sin α + 107F = 0,
α
from which the normal reaction of the two front wheels is
F =
µsF
W
(72 cos α − 19 sin α).
107
The friction force balances the component of gravity parallel to
the incline: 0 = −W sin α + µs F , from which
W sin α
W
=
(72 cos α − 19 sin α).
µs
107
Reduce and solve:
α = tan−1
72
107
+ 19
µ
s
= 17.17◦
W
µsR
R
F
35 in
72 in
Problem 8.50 The stationary cabinet has weight W .
Determine the force F that must be exerted to cause it to
move if (a) the coefficient of static friction at A and B
is µs ; (b) if the coefficient of static friction at A is µsA
and the coefficient of static friction at B is µsB .
F
G
h
H
A
B
b
—
2
(a) The sum of the moments about B is
b
= −hF +
W − bA = 0,
2
b
—
2
Solution:
MB
from which
W
A =
−
2
F
h
h
F.
b
H
A
The sum of forces:
Fy = −W + A + B = 0,
W
h
B =W −A=
+
F.
2
b
Fx = F − µs A − µs B = 0,
from which
W
h
W
h
F = µs
+
F+
−
F = µs W
2
b
2
b
(b) Use the normal reactions found in Part (a). From the sum of forces
parallel to the floor,
W
h
W
h
F = µsA A + µsB B = µsA
−
F + µsB
+
.
2
b
2
b
Reduce and solve:
W
F = 2
1+
b
—
2
B
F
from which
b
—
2
(µsA + µsB )
h
b
(µsA − µsB )
h
W
µSAA A
B
b
2
b
2
µSBB
Problem 8.51 A force F = 200 N is necessary to raise
the block A at a constant rate. The mass of the wedge
B is negligible. Between all of the contacting surfaces,
µs = 0.28 and µk = 0.26. What is the mass of block
A?
A
F
B
10°
Solution: The friction at all surfaces is kinetic. Draw a free body
diagram of each block and write the equilibrium equations.
f1 = µk N1 (1)
f2 = µk N2 (2)
A
f3 = µk N3 (3)
F
F = 200 N
10°
µk = 0.26
Block A:
Fx :
Fy :
Wedge B:
Fx :
Fy :
B
N1 − f2 cos 10◦ − N2 sin 10◦ = 0
y
(4)
mAg
f1
−f1 − mA g − f2 sin 10◦ + N2 cos 10◦ = 0 (5)
f1 = µ KN1 (1)
A
N1
f2 = µ KN2 (2)
f3 − F + f2 cos 10◦ + N2 sin 10◦ = 0 (6)
N3 + f2 sin 10◦ − N2 cos 10◦ = 0
(7)
10°
f3 = µ KN3 (3)
x
f2
N2
N2
f2
Unknowns:
f1 , N1 , f2 , N2 , f3 , N3 , mA
We have 7 eqns. in 7 unknowns.
Solving, we get
mA = 25.0 kg
Also,
f1 = 33.2 N,
N1 = 127.5 N
f2 = 77.2 N,
N2 = 296.7 N
f3 = 72.5 N,
N3 = 278.8 N
B
F = 200 N
µK = 0.26
f3
N3
10°
F
Problem 8.52 In Problem 8.51, suppose that the mass
of block A is 30 kg and the mass of the wedge B is 5 kg.
What force F is necessary to start the wedge B moving
to the left?
Solution:
The solution to this problem is very similar to that of
Problem 8.51.
mA = 30 kg
f1 = µs N1 (1)
A
f2 = µs N2 (2)
f3 = µs N3 (3)
F
B
10°
(impending slip)
mB = 5 kg
y
Block A:
Fx :
Fy :
N1 − f2 cos 10◦ − N2 sin 10◦ = 0
Wedge B:
Fx :
Fy :
f3 − F + f2 cos 10◦ + N2 sin 10◦ = 0
mAg
(4)
−f1 − mA g − f2 sin 10◦ + N2 cos 10◦ = 0 (5)
(6)
N1
A
N3 − mB g + f2 sin 10◦ − N2 cos 10◦ = 0 (7)
f1
Unknowns:
10°
f1 , N1 , f 2 , N 2 , f 3 , N 3 , F
N2
F = 272 N
N2
Also
f1 = 45.7 N
x
f2
We have 7 eqns. in 7 unknowns.
Solving, we get
f2
N1 = 163.2 N
10°
f2 = 101.7 N N2 = 363.2 N
f3 = 108.9 N N3 = 389.0 N
mBg
B
F
f3
N3
Problem 8.53 The wedge shown is being used to split
the log. The wedge weighs 20 lb and the angle α equals
30◦ . The coefficient of kinetic friction between the faces
of the wedge and the log is 0.28. If the normal force
exerted by each face of the wedge must equal 150 lb to
split the log, what vertical force F is necessary to drive
the wedge into the log at a constant rate?
Solution:
µk = 0.28
f = 0.28 N (1)
N = 150 lb (2)
Fx : (N − N ) cos 15◦ + (f − f ) sin 15◦ = 0
(no information here)
Fy : 2f cos 15◦ + 2 N sin 15◦ − F = 0 (3)
Unknowns: N, f, F
We have 3 eqns. in 3 unknowns
Solving
F = 159 lb, f = 42 lb
F
α
F
y
F
15°
15°
f
f
30°
15°
15°
N
N
x
F
F
α
Problem 8.54 The coefficient of static friction between the faces of the wedge and the log in Problem 8.53
is 0.30. Will the wedge remain in place in the log when
the vertical force F is removed?
Solution: For this problem, remove F and solve for the minimum
µs necessary for equilibrium.
The required µs
F
α
F
f = µs N
f
µs =
(1)
N
Fy :
−2f cos 15◦ + 2 N sin 15◦ = 0 (2)
f
= µs = tan 15◦ = 0.268 < 0.30.
N
15°
Yes—the wedge will stay in place
15°
f
f
15°
15°
N
N
Problem 8.55 The masses of A and B are 42 kg and
50 kg, respectively. Between all contacting surfaces,
µs = 0.05. What force F is required to start A moving
to the right?
B
45°
F
A
20°
Solution: If F is decreased until slip of A to the left impends, the
free body diagrams are as shown. The equilibrium equations are left
block:
Fx = F + N sin 20◦ + 0.05 N cos 20◦ − P cos 45◦
+ 0.05P cos 45◦ = 0,
Fy = N cos 20◦ − 0.05 N sin 20◦ − P cos 45◦
Solving, we obtain
N = 955 N,
P = 632 N,
Q = 425 N,
and F = 53.0 N.
− 0.05P cos 45◦ − (42)(9.81) = 0.
Right block:
Fx = P cos 45◦ − 0.05P cos 45◦ − Q = 0,
Fy = P cos 45◦ + 0.05P cos 45◦ + 0.05Q − (50)(9.81) = 0.
P
0.05 Q
0.05 P
(42)(9.81) 0.05 P
F
(50)(9.81)
P
20°
N
0.05 N
45°
Q
Problem 8.56 The stationary blocks A, B, and C each
have a mass of 200 kg. Between all contacting surfaces,
µs = 0.6. What force F is necessary to start B moving
downward?
F
B
A
C
80°
80°
Solution: The wedge angle is 10◦ for each side. The block A
cannot move, hence the friction contact surfaces are the wedge surfaces
plus the bottom surface of block C. Assuming that downward slip of
B impends, the free body diagrams of blocks B and C are as shown.
The equilibrium equations are Block B:
Fx = N sin 80◦ − µs N cos 80◦ − P sin 80◦ + µs P cos 80◦
= 0,
◦
◦
Fy = N cos 80 + µs N sin 80 + P cos 80 + µs P sin 80
Block C:
Fx = P sin 80◦ − µs P cos 80◦ − µs Q = 0,
(3)
◦
◦
Fy = Q − P cos 80 − µs P sin 80 − (200)(9.81) = 0. (4)
Solving them with µs = 0.6, we obtain
Q = 4100 N,
and F = 2300 N
80°
C
80°
◦
− F − (200)(9.81) = 0.
P = 2790 N,
B
A
(1)
◦
N = 2790 N,
F
F
(2)
µsN
(20.0)(9.81)
µsP
P
N
80°
(20.0)(9.81)
B
80°
P
µsP
80°
µsQ
C
Q
Problem 8.57 Small wedges called shims can be used
to hold an object in place. The coefficient of kinetic
friction between the contacting surfaces is 0.4. What
force F is needed to push the shim downward until the
horizontal force exerted on the object A is 200 N?
F
Shims
5°
A
5°
Solution:
F
fL = µk NL (1)
fR = µk NR (2)
NL = 200 N (3)
Fx :
Fy :
Shims
NL − NR cos 5◦ + fR sin 5◦ = 0
◦
(4)
5°
◦
−F + fL + fR cos 5 + NR sin 5 = 0 (5)
Unknowns: fL , NL , FR , NR , F
(5 eqns. in 5 unknowns)
Solving,
A
5°
F = 181 N
y
F
5°
fL
NL
fR
NR
x
Problem 8.58 The coefficient of static friction between the contacting surfaces in Problem 8.57 is 0.44. If
the shims are in place and exert a 200-N horizontal force
on the object A, what upward force must be exerted on
the left shim to loosen it?
Solution:
FL = µs NL (1)
FR = µs NR (2)
µs = 0.44
NL = 200 N
Fx :
NL − NR cos 5◦ − fR sin 5◦ = 0
Fy :
F − fL − fR cos 5◦ + NR sin 5◦ = 0 (4)
(3)
Unknowns F, fL , fR , NR
Solving, F = 156 N
F
Shims
5°
A
5°
5°
y
F
NL
NR
fL
fR
x
Problem 8.59 The crate A weighs 600 lb. Between all
contacting surfaces, µs = 0.32 and µk = 0.30. Neglect
the weights of the wedges. What force F is required to
move A to the right at a constant rate?
F
5°
A
5°
Solution: The active sliding contact surfaces are between the wall
and the left wedge, between the wedges, between the floor and the
bottom of the right wedge, and between the crate and the floor. Leftmost
wedge: Denote the normal force exerted by the wall by Q, and the
normal force between the wedges by N . The equilibrium conditions
for the left wedge moving at a constant rate are:
Fy = −F + µk N cos α + N sin α + µk Q = 0.
Fx = Q − N cos α + µk N sin α = 0.
For the right wedge: Denote the normal force exerted by the crate by
A, and the normal force exerted by the floor by P .
Fy = −N sin α − µk N cos α + P = 0.
Fx = N cos α − µk N sin α − µk P − A = 0.
For the crate: Denote the weight of the crate by W .
Fx = A − µk W = 0.
These five equations are solved for the five unknowns by iteration:
Q = 204.4 lb,
N = 210.7 lb
P = 81.34 lb,
A = 180 lb,
and F = 142.66 lb
F
5°
A
5°
F
µkQ
µkN
Q
N
N
A
W
A
µkN
P
µkP
µkW
W
Problem 8.60 Suppose that between all contacting
surfaces in Problem 8.59, µs = 0.32 and µk = 0.30.
Neglect the weights of the 5◦ wedges. If a force F =
800 N is required to move A to the right at a constant
rate, what is the mass of A?
Solution: The free body diagrams of the left wedge and the combined right wedge and crate are as shown. The equilibrium equations are
Wedge:
Fx = N − P cos 5◦ + 0.3P sin 5◦ = 0,
Fy = 0.3 N + P sin 5◦ + 0.3P cos 5◦ − F = 0,
P = 1180 N,
N = 1150 N,
Q = 3820 N,
and m = 343 kg.
F
Wedge and box:
Fx = P cos 5◦ − 0.3P sin 5◦ − 0.3Q = 0,
Fy = Q − P sin 5◦ − 0.3P cos 5◦ − 9.81 m = 0.
0.3 N
N
0.3 P
P
P
Solving them, we obtain
m (9.81)
0.3 P
5°
0.3 Q
Q
Problem 8.61 The box A has a mass of 80 kg, and the
wedge B has a mass of 40 kg. Between all contacting
surfaces, µs = 0.15 and µk = 0.12. What force F is
required to raise A at a constant rate?
A
10°
F
B
10°
Solution:
From the free-body diagrams shown, the equilibrium
equations are
Box A:
A
Q − N sin 10◦ − µk N cos 10◦ = 0,
N cos 10◦ − µk N sin 10◦ − µk Q − W = 0.
Wedge B:
◦
◦
◦
10°
B
F
10°
◦
P sin 10 + µk P cos 10 + N sin 10 + µk N cos 10 − F = 0
P cos 10◦ − µk P sin 10◦ − N cos 10◦ + µk N sin 10◦ − Ww = 0.
Solving with
W = (80)(9.81) N,
Ww = (40)(9.81) N,
and µk = 0.12,
we obtain
N = 845 N,
Q = 247 N,
P = 1252 N,
and F = 612 N.
Q
N
A
W
µkQ
B WW
P
µkN
N
µkN
µkP
F
Problem 8.62 Suppose that in Problem 8.61, A weighs
800 lb and B weighs 400 lb. The coefficients of friction
between all of the contacting surfaces are µs = 0.15 and
µk = 0.12. Will B remain in place if the force F is
removed?
Solution:
The equilibrium conditions are: For the box A: Denote
the normal force exerted by the wall by Q, and the normal force exerted
by the wedge by N . The friction forces oppose motion.
Fy = −W + N cos α + µs N sin α + µs Q = 0,
Fx = +µs N cos α − N sin α + Q = 0.
(A comparison with the equilibrium conditions for Problem 8.61 will show
that the friction forces are reversed, since for slippage the box A will move
downward, and the wedge B to the right.) The strategy is to solve these equations
for the required µs to keep the wedge B in place when F = 0. The solution
Q = 0, N = 787.8 lb, P = 1181.8 lb and µs = 0.1763. Since the value of
µs required to hold the wedge in place is greater than the value given, the wedge
will slip out.
For the wedge B. Denote the normal force on the lower surface by P .
Fx = −µs N cos α − µs P cos α + P sin α + N sin α = 0.
Fy = −N cos α + P cos α − µs N sin α + µs P sin α − Ww = 0.
µSN
µSQ
WW
W
Q
µSP
N
P
µSN
Problem 8.63 Between A and B, µs = 0.20, and
between B and C, µs = 0.18. Between C and the wall,
µs = 0.30. The weights WB = 20 lb and WC = 80 lb.
What force F is required to start C moving upward?
C
F
B
15°
A
Solution: The active contact surfaces are between the wall and C,
between the wedge B and C, and between the wedge B and A. For the
weight C: Denote the normal force exerted by the wall by Q, and the
normal force between B and C by N . Denote the several coefficients
of static friction by subscripts. The equilibrium conditions are:
Fy = −WC + N − µCW Q = 0,
Fx = −Q + µBC N = 0.
C
F
B
µ sQ
These four equations in four unknowns are solved:
and F = 66.9 lb
µsN
Q
WC
F
N = 84.6 lb,
P = 114.4 lb,
15°
A
For the wedge B: Denote the normal force between A and B by P .
Fy = −N + P cos α − µAB P sin α − WB = 0.
Fx = F − µBC N − µAB P cos α − P sin α = 0.
Q = 15.2 lb,
N
N
µsN
N
WB
µsP
P
Problem 8.64 The masses of A, B, and C are 8 kg,
12 kg, and 80 kg, respectively. Between all contacting
surfaces, µs = 0.4. What force F is required to start C
moving upward?
F
Solution: The active contact surfaces are between A and B, between A and the wall, between B and the floor, and between B and
C. Assume that the roller supports between C and the wall exert no
friction forces. For the wedge A: Denote the normal force exerted by
the wall as Q and the normal force between A and B as N . The weight
is WA = 8 g = 78.48 N. The equilibrium conditions:
Fy = −F + µs Q + µs N cos α + N sin α − WA = 0
Fy = Q − N cos α + µs N sin α = 0.
C
A
10°
B
12°
For wedge B: Denote the normal force exerted on B by the floor by
P , and the normal exerted by the weight C as S. The weight of B is
WB = 12 g = 117.72 N. The equilibrium conditions:
Fy = −N sin α − S cos β + P − µs N cos α + µs S sin β − WB
= 0.
C
F
Fx = N cos α − µs N sin α − µs P − µs S cos β − S sin β = 0.
For the weight C: The weight is WC = 80 g = 784.8 N. The
equilibrium conditions:
Fy = −WC + S cos β − µs S sin β = 0.
A
B
10°
12°
These five equations in five unknowns are solved:
Q = 1157.6 N,
N = 1293.5 N,
F
S = 857.4 N,
µsQ
P = 1677.5,
Q
and F = 1160 N
Problem 8.65 The vertical threaded shaft fits into a
mating groove in the tube C. The pitch of the threaded
shaft is p = 0.1 in., and the mean radius of the thread
is r = 0.5 in. The coefficients of friction between the
thread and the mating groove are µs = 0.15 and µk =
0.10. The weight W = 200 lb. Neglect the weight of
the threaded shaft.
(a) Will the stationary threaded shaft support the weight
if no couple is applied to the shaft?
(b) What couple must be applied to the threaded shaft
to raise the weight at a constant rate?
α
WA
R
µsN
N
N µ
sN µ P
s
W
C
(a) The angle of static friction is θs = tan−1 (0.15) =
8.53◦ . The pitch angle is
p 0.1
α = tan−1
= tan−1
= 1.82◦ .
2πr
2π(0.5)
Solution:
From Eq. (10.14) the moment necessary for the shaft to be on the
verge of rotating is M = rF tan(θs − α). For a zero moment,
θs = α, which is not satisfied. Therefore the shaft will support the
weight when no moment is applied. (b) The angle of kinetic friction
is θk = tan−1 (0.10) = 5.71◦ . From Eq. (9.9) the moment required
to raise the weight at a constant rate is
M = rW tan(θk + α) = 0.5(200) tan(7.533) = 13.2 in lb.
µsS
W
C
S
WB
P
WC
β
S
µsS
Problem 8.66 Suppose that in Problem 8.65, the pitch
of the threaded shaft is p = 2 mm and the mean radius
of the thread is r = 20 mm. The coefficients of friction
between the thread and the mating groove are µs = 0.22,
and µk = 0.20. The weight W = 500 N. Neglect
the weight of the threaded shaft. What couple must be
applied to the threaded shaft to lower the weight at a
constant rate?
Solution:
θk = tan
−1
The angle of kinetic friction is
(0.2) = 11.31◦ .
The angle of pitch is
p 2
α = tan−1
= tan−1
= 0.9118◦ .
2πr
2π(20)
The moment required to lower the weight at a constant rate is
M = 0.02(500) tan(11.31 − 0.9118) = 1.835 N-m.
Problem 8.67 The position of the horizontal beam can
be adjusted by turning the machine screw A. Neglect the
weight of the beam. The pitch of the screw is p = 1 mm,
and the mean radius of the thread is r = 4 mm. The
coefficients of friction between the thread and the mating
groove are µs = 0.20 and µk = 0.18. If the system is
initially stationary, determine the couple that must be
applied to the screw to cause the beam to start moving
(a) upward; (b) downward.
Solution:
400 N
A
100 mm
300 mm
The sum of the moments about the pin support is
M = −0.4F + (0.3)400 = 0,
400 N
from which the force exerted by the screw is F = 300 N. The pitch
angle is
1
α = tan−1
= 2.28◦ .
2π(4)
The static friction angle is θs = tan−1 (0.2) = 11.31◦ . (a) The
moment required to start motion upward is
300
mm
100
mm
M = 0.004(300) tan(11.31◦ + 2.28◦ ) = 0.29 N-m
(b) The moment required to start motion downward is
400 N
M = 0.004(300) tan(11.31◦ − 2.28◦ ) = 0.19 N-m
F
100
mm
300
mm
Problem 8.68 Suppose that in Problem 8.67, the pitch
of the machine screw is p = 1 mm and the mean radius
of the thread is r = 4 mm. What minimum value of the
coefficient of static friction between the thread and the
mating groove is necessary for the beam to remain in the
position shown with no couple applied to the screw?
Solution:
From the solution to Problem 8.67 the force applied to
the screw is F = 300 N. The pitch angle is
1
α = tan−1
= 2.28◦ .
2π(4)
The moment required to start motion downward is M = 0.004(300)
tan(θs − α). For M = 0, tan(θs − α) = 0, from which
θs = α = 2.279◦ ,
and µs = tan(2.279◦ ) = 0.0398
Problem 8.69 The mass of block A is 60 kg. Neglect
the weight of the 5◦ wedge. The coefficient of kinetic
friction between the contacting surfaces of the block A,
the wedge, the table, and the wall is µk = 0.4. The pitch
of the threaded shaft is 5 mm, the mean radius of the
thread is 15 mm, and the coefficient of kinetic friction
between the thread and the mating groove is 0.2. What
couple must be exerted on the threaded shaft to raise the
block A at a constant rate?
A
5°
Denote the wedge angle by β = 5◦ and the normal
force on the top by N and on the lower surface by P . The free body
diagrams of the wedge and block are as shown. The equilibrium equations for wedge:
Fx = F − µk P − N sin 5◦ − µk N cos 5◦ = 0,
Fy = P − N cos 5◦ + µk N sin 5◦ = 0.
Solution:
A
5°
For the Block:
Fx = N sin 5◦ + µk N cos 5◦ − Q = 0,
Fy = N cos 5◦ − µk N sin 5◦ − µk Q − W = 0.
Solving them, we obtain F = 668 N. From Equation (9.9), the couple
necessary to rotate the threaded shaft when it is subjected to the axial
force F is M = rF tan(θk + α) r is the radius 15 mm = 0.015 m.
θk is the angle of kinetic friction θk = arctan(0.2) = 11.31◦ .
From Equation (9.7), the slope is given in terms of the pitch by
P
5
α = arctan
= arctan
= 3.04◦ .
2πr
2π(15)
The couple is
M = (0.015 m)(668 N) tan(11.31◦ + 3.04◦ ) = 2.56 N-m.
µkQ
W
µkN
N
µkP
P
Q
F
N
µkN
Problem 8.70 The vise exerts 80-lb forces on A. The
threaded shafts are subjected only to axial loads by the
jaws of the vise. The pitch of their threads is p = 1/8 in.,
the mean radius of the threads is r = 1 in., and the
coefficient of static friction between the threads and the
mating grooves is 0.2. Suppose that you want to loosen
the vise by turning one of the shafts. Determine the
couple you must apply (a) to shaft B; (b) to shaft C.
A
4 in
B
4 in
C
Solution:
Isolate the left jaw. The sum of the moments about C:
MC = −4B + 8(80) = 0,
A
from which B = 160 lb (T ). The sum of the forces:
Fx = −80 + B − C = 0,
4 in
B
from which C = 80 lb (C). The pitch angle is
1
α = tan−1
= 1.14◦ .
16π
C
The static friction angle is θs = tan−1 (0.2) = 11.31◦ . The moments required to loosen the vise are
1
MB =
(160) tan(11.31◦ − 1.14◦ ) = 2.39 ft lb,
12
80 lb
and MC = rC tan(θs − α) = 1.2 ft-lb.
C
Problem 8.71 Suppose that you want to tighten the
vise in Problem 8.70 by turning one of the shafts. Determine the couple you must apply (a) to shaft B; (c) to
shaft C.
Solution: Use the solution to Problem 8.70. (a) The moment on
shaft B required to tighten the vise is MB = rB tan(θs + α). Note
1
that r = 12
, B = 160 lb,
1
α = tan−1
= 1.14◦
16π
and θs = tan−1 (0.2) = 11.31◦ ,
then MB = 2.94 ft lb (b) For shaft C, MC = rC tan(θs + α),
where C = 80 lb, MC = 1.47 ft-lb.
B
4 in
4 in
4 in
Problem 8.72 The threaded shaft has a ball and socket
support at B. The 400-lb load A can be raised or lowered
by rotating the threaded shaft, causing the threaded collar at C to move relative to the shaft. Neglect the weights
of the members. The pitch of the shaft is p = 14 in., the
mean radius of the thread is r = 1 in., and the coefficient of static friction between the thread and the mating
groove is 0.24. If the system is stationary in the position shown, what couple is necessary to start the shaft
rotating to raise the load?
9 in
C
A
12 in
B
9 in
Solution: Denote the lower right pin support
by D. The length
√
of the connecting member CD is LCD = 92 + 122 = 15 in. The
angle between the threaded shaft and member CD is
9
β = 2 tan−1
= 73.74◦ .
12
The sum of the moments about D is
MD = LCD F cos(90 − β) − 18W = 0,
9 in
12 in
9 in
from which F = 500 lb. The pitch angle is
p α = tan−1
= 2.28◦ .
2πr
The angle of static friction is θs = tan−1 (0.24) = 13.5◦ . The
moment needed to start the threaded collar in motion is
1
M = rF tan(θs + α) =
(500) tan(13.5◦ + 2.28◦ )
12
= 11.77 ft-lb
9 in
9 in
W
C
F
Dy
β
LCD
Dx
18 in
Problem 8.73 In Problem 8.72, if the system is stationary in the position shown, what couple is necessary
to start the shaft rotating to lower the load?
Solution: Use the results of the solution to Problem 8.72. The
moment is M = rF tan(θs − α), where
1
r =
ft,
12
F = 500 lb,
θs = 13.5◦ ,
and α = 2.28◦ ,
from which M = 8.26 ft lb
18 in
18 in
Problem 8.74 The car jack is operated by turning the
threaded shaft at A. The threaded shaft fits into a mating
groove in the collar at B, causing the collar to move
relative to the shaft as the shaft turns. As a result, points
B and D move closer together or farther apart, causing
point C (where the jack is in contact with the car) to
move up or down. The pitch of the threaded shaft is
p = 5 mm, the mean radius of the thread is r = 10 mm,
and the coefficient of kinetic friction between the thread
and the mating groove is 0.15. What couple is necessary
to turn the shaft at a constant rate and raise the jack when
it is in the position shown if F = 6.5 kN?
F
C
150 mm
D
150 mm
300 mm
Solution: Isolate members BC and BD. Assume that half the
car load is carried by these members. The equilibrium conditions for
member BC are:
Fx = Cx − Bx = 0,
F
MB = 0.3
− 0.15Cx = 0.
2
These equations are solved for
F
6.5
=
= 3.25 kN
2
2
C
150 mm B
B
150 mm
D
300 mm
300 mm
F
α
C
C
θs = tan−1 (0.15) = 8.53◦ .
2C sin α = F
C
The moment required to rotate the shaft at a constant rate is
D = 2Ccos α
F = 2F
tan α
D
M = (0.01)(6.5) tan(8.53◦ + 4.55◦ ) = 0.0151 kN
m = 15.1 N m
300 mm
F
to obtain Bx = 6.5 kN, which is the force on the collar to be balanced
by the rotating threaded shaft. The pitch angle is
5
α = tan−1
= 4.55◦ .
2π10
The angle of kinetic friction is
=
C
F
2
CX
BY
BX
BX
DY
BY
DX
Problem 8.75 In Problem 8.74, what couple is necessary to turn the threaded shaft at a constant rate and
lower the jack when it is in the position shown if the
force F = 6.5 kN?
Solution: Use the results of the solution of Problem 8.74. The moment required to lower the jack at a constant rate is M = rB tan(θk −
α), where r = 0.01 m, B = 6500 N, θk = 8.53◦ , α = 4.55◦ , from
which M = 4.52 N-m
A
B
A
Problem 8.76 A turnbuckle, used to adjust the length
or tension of a bar or cable, is threaded at both ends.
Rotating it draws threaded segments of a bar or cable
together or moves them apart. Suppose that the pitch
of the threads is p = 3 mm their mean radius is r =
25 mm, and the coefficient of static friction between the
threads and the mating grooves is 0.24. If T = 800 N,
what couple must be exerted on the turnbuckle to start
tightening it?
T
T
T
T
Solution:
θs = a tan(µs ) = 13.49◦
p α = a tan
= 1.09◦
2πr
M = rT tan(θs + α)
since M tends to create motion opposite to the direction of T .
M = (0.025)(800 N) tan(14.59◦ )
M = 5.21 N-m
for each screw.
There are two screws in the turnbuckle.
∴ M = 10.42 N-m
Problem 8.77 The horizontal shaft is supported by
two journal bearings. The coefficient of kinetic friction
between the shaft and the bearings is µk = 0.2. The
radius of the shaft is 20 mm, and its mass is 5 kg. Determine the couple M necessary to rotate the shaft at a
constant rate.
Strategy: You can obtain the moment necessary to rotate the shaft at a constant rate by replacing θs by θk in
Eq. (9.12).
Solution: The weight of the shaft is W = mg = 5(9.81) =
49 N, divided between two bearings. The angle of kinetic friction is
θk = tan−1 (0.2) = 11.31◦ . The moment per bearing is
W
M =
(0.02) sin θk = 0.096 N m.
2
The total moment is Mt = 0.192 N m
M
M
Problem 8.78 The horizontal shaft is supported by
two journal bearings. The coefficient of static friction
between the shaft and the bearings is µs = 0.3. The
radius of the shaft is 20 mm, and its mass is 5 kg. Determine the largest mass m that can be suspended as
shown without causing the stationary shaft to slip in the
bearings.
m
Solution: The weight of the shaft is W = mg = 5(9.81) =
49 N. This weight is divided between two bearings. The angle of
static friction is θs = tan−1 (0.3) = 16.7◦ . The load per bearing is
F =
W + Wm
,
2
and the moment required to start rotation is
Mm = (W + Wm )r sin θs
where Wm is the suspended weight,
Wm =
Mm
.
r
From which
Wm r = (W + Wm )r sin θs ,
from which
W
sinθs
m =
= 2.02 kg.
g
1 − sin θs
Problem 8.79 Suppose that in Problem 8.78 the mass
m = 8 kg and the coefficient of kinetic friction between
the shaft and bearings is µk = 0.26. What couple must
be applied to the shaft to raise the mass at a constant
rate?
Solution:
From the solution to Problem 8.78 the moment re-
quired is
Mapplied = (W + Wm )r sin θk ,
where Wm = mg is the weight of the suspended mass. The moment
required to raise the suspended mass is Mm = Wm r. The total
moment is the sum of the moment required to turn the shaft and the
moment required to raise the mass:
Mtotal = (W + mg)r sin θk + Wm r
= (49 + 78.5)(0.02) sin(14.6◦ ) + 1.57 = 2.21 N m
m
Problem 8.80 The pulley is mounted on a horizontal shaft supported by journal bearings. The coefficient
of kinetic friction between the shaft and the bearings is
µk = 0.3. The radius of the shaft is 20 mm, and the
radius of the pulley is 150 mm. The mass m = 10 kg.
Neglect the masses of the pulley and shaft. What force
T must be applied to the cable to move the mass upward
at a constant rate?
m
T
The angle of kinetic friction is θk = tan−1 (µk ) =
16.7◦ . The moment required to turn the shaft is M = (mg +
T )r sin θk . The applied moment is M = (T − mg)R where R
is the radius of the pulley. Equating and reducing:
r
1+ R
sin θk
1.0383
T = mg
=
(98.1)
= 105.92 N
r
1− R
sin θk
0.9617
Solution:
m
T
Problem 8.81 In Problem 8.80, what force T must be
applied to the cable to lower the mass at a constant rate?
Solution: Form the solution to Problem 8.80, θk =
tan−1 (µk ) = 16.7◦ , and M = (mg + T )r sin θk . The applied
moment is M = (mg − T )R. Substitute and reduce:
r
1− R
sin θk
0.9617
T = mg
=
(98.1)
= 90.86 N
r
1+ R
sin θk
1.0383
Problem 8.82 The pulley of 8-in. radius is mounted
on a shaft of 1-in. radius. The shaft is supported by
two journal bearings. The coefficient of static friction
between the bearings and the shaft is µs = 0.15. Neglect
the weights of the pulley and shaft. The 50-Ib block A
rests on the floor. If sand is slowly added to the bucket
B, what do the bucket and sand weigh when the shaft
slips in the bearings?
(See Problem 8.80). The angle of static friction is θs =
tan−1 (µs ) = 8.53◦ . The moment required to start rotation for both bearings is M = r(B + W ) sin θs . The applied moment is M = (B − W )R,
where R is the radius of the pulley. Substitute and reduce:
r
1+ R
sin θs
1.0185
B =W
=
(50)
= 51.9 lb
r
1− R
sin θs
0.9815
Solution:
8 in
1 in
8 in
B
A
B
A
Problem 8.83 The pulley of 50-mm radius is mounted
on a shaft of 10-mm radius. The shaft is supported by
two journal bearings. The mass of the block A is 8 kg.
Neglect the weights of the pulley and shaft. If a force
T = 84 N is necessary to raise the block A at a constant
rate, what is the coefficient of kinetic friction between
the shaft and the bearings?
50 mm
20°
10 mm
T
A
The weight is W = mg = 78.5 N. The force on the
pulley is
F = (W + T sin α)2 + (T cos α)2 ,
Solution:
where α = 20◦ .
F = 107.22 + 78.92 = 133.13 N.
The moment required to raise the mass at constant rate for both bearings
is M = rF sin θk = 1.33 sin θk . The applied moment is M =
(T − W )R = 0.276 N m. Substitute and reduce:
sin θk =
(T − W )R
0.276
=
= 0.2073,
rF
1.33
from which
θk = 11.96◦
and µk = tan(11.96◦ ) = 0.2119
50 mm
10 mm
20°
T
A
Problem 8.84 The mass of the suspended object is
4 kg. The pulley has a 100-mm radius and is rigidly
attached to a horizontal shaft supported by journal bearings. The radius of the horizontal shaft is 10 mm and the
coefficient of kinetic friction between the shaft and the
bearings is 0.26. What tension must the person exert on
the rope to raise the load at a constant rate?
25°
100 mm
Solution:
R = 0.1 m
25°
µk = 0.26
Shaft radius 0.01 m
10 mm
µk (shaft) = 0.26
tan θk = µk
θk = 14.57◦
Ms = rF sin θk
R = 0.1 m
m = 4 kg
To Find F , we must find the forces acting on the shaft.
Fx :
Ox − T cos 25◦ = 0
(1)
Fy :
Oy − T sin 25◦ − mg = 0 (2)
F = Ox2 + Oy2
(3)
Mo :
RT − Rmg − Ms = 0
(4)
Ms = rF sin θk
Unknowns: Ox , Oy , T, Ms , F
Solving, we get
T = 40.9 N
Also,
F = 67.6 N,
Ms = 0.170 N-m
Ox = 37.1 N,
Oy = 56.5 N
(5)
25°
T
µ K = 0.26
Shaft radius 0.01 m
µK (shaft) = 0.26
MS
OX
OY
mg
Problem 8.85 The circular flat-ended shaft is pressed
into the thrust bearing by an axial load of 100 N. Neglect
the weight of the shaft. The coefficients of friction between the end of the shaft and the bearing are µs = 0.20
and µk = 0.15. What is the largest couple M that can
be applied to the stationary shaft without causing it to
rotate in the bearing?
100 N
30 mm
M
Solution: The bearing meets the conditions for Eq. (9.14). For
impending rotation, the moment is
2
2
M = µs F r =
(0.2)(100)(0.03) = 0.4 N m
3
3
100 N
30 mm
M
Problem 8.86 In Problem 8.85, what couple M is
required to rotate the shaft at a constant rate?
Solution: The bearing meets the conditions for Eq. (10.17). The
moment required to sustain a constant rate of rotation is
2
2
M =
µk F r =
(0.15)(100)(0.03) = 0.3 N m
3
3
Problem 8.87 Suppose that the end of the shaft in
Problem 8.85 is supported by a thrust bearing of the
type shown in Fig. 9.22, where ro = 30 mm, ri =
10 mm, α = 30◦ , and µk = 0.15. What couple M is
required to rotate the shaft at a constant rate?
Solution: The bearing meets the conditions for Eq (9.13). The
moment required to sustain a constant rate of rotation is
3
ro − r13
2µk F
M =
= 0.3753 N m
3 cos α
ro2 − r12
Problem 8.88 The disk D is rigidly attached to the
vertical shaft. The shaft has flat ends supported by thrust
bearings. The disk and the shaft together have a mass
of 220 kg and the diameter of the shaft is 50 mm. The
vertical force exerted on the end of the shaft by the upper
thrust bearing is 440 N. The coefficient of kinetic friction
between the ends of the shaft and the bearings is 0.25.
What couple M is required to rotate the shaft at a constant
rate?
Solution:
M
M
D
D
There are two thrust bearings, one at the top and one at
the bottom
FU = 440 N
m = 220 kg
Fy :
FL − FU − mg = 0
M
M
D
D
FL = 2598.2 N.
The couple necessary to turn D at a constant rate is the sum of the
couples for the two bearings.
MU =
2
µk F U r
3
ML =
2
µk F L r
3
FU
r = 0.025 m
µk = 0.25
Solving,
MU = 1.833 N-m
D
ML = 10.826
MTOTAL = 12.7 N-m
mg
FL
Problem 8.89 Suppose that the ends of the shaft in
Problem 8.88 are supported by thrust bearings of the
type shown in Fig. 9.22, where ro = 25 mm, ri =
6 mm, α = 45◦ , and µk = 0.25. What couple M is
required to rotate the shaft at a constant rate?
Solution:
There are two thrust bearings, one at the top and one at
the bottom.
FU = 440 N
m = 220 kg
M
Fy :
FL − FU − mg = 0
M
D
D
FL = 2598.2 N.
The couple necessary to turn D at a constant rate is the sum of the
couples for the two bearings.
For the bearings used
m =
2µk F (ro3 − ri3 )
3 cos α (ro2 − ri2 )
FU
α = 45◦ , ro = 0.025 m
µk = 0.25, ri = 0.006 m
Thus,
MU =
2µk FU (ro3 − ri3 )
= 2.7 N-m
3 cos α (ro2 − ri2 )
ML =
2µk FL (ro3 − ri3 )
= 16.0 N-m
3 cos α (ro2 − ri2 )
D
mg
MTOTAL = MU + ML = 18.7 N-m
FL
γi
α
γo
Problem 8.90 The shaft is supported by thrust bearings that subject it to an axial load of 800 N. The coefficients of kinetic friction between the shaft and the left
and right bearings are 0.20 and 0.26, respectively. What
couple is required to rotate the shaft at a constant rate?
15 mm
38 mm
38
mm
Solution:
The left bearing: The parameters are
ro = 38 mm,
38 mm
ri = 0,
15 mm
α = 45◦ ,
µk = 0.2,
and F = 800 N.
The moment required to sustain a constant rate of rotation is
ro3 − ri3
2µk F
Mleft =
= 5.73 N m.
3 cos α ro2 − ri2
38
mm
The right bearing: This is a flat-end bearing. The parameters are
µk = 0.26, r = 15 mm, and F = 800 N. The moment required to
sustain a constant rate of rotation is
Mright =
2µk F r
= 2.08 N m.
3
The sum of the moments: M = 5.73 + 2.08 = 7.81 N m
Problem 8.91 A motor is used to rotate a paddle for
mixing chemicals. The shaft of the motor is coupled to
the paddle using a friction clutch of the type shown in
Fig. 9.25. The radius of the disks of the clutch is 120 mm,
and the coefficient of static friction between the disks is
0.6. If the motor transmits a maximum torque of 15 N-m
to the paddle, what minimum normal force between the
plates of the clutch is necessary to prevent slipping?
Solution:
M =
Clutch
Paddle
The moment necessary to prevent slipping is
2µs F r
2(0.6)(0.12)F
=
= 15 N m.
3
3
Solve: F = 312.5 N
Clutch
Paddle
Problem 8.92 The thrust bearing is supported by contact of the collar C with a fixed plate. The area of contact
is an annulus with an inside diameter D1 = 40 mm and
an outside diameter D2 = 120 mm. The coefficient
of kinetic friction between the collar and the plate is
µk = 0.3. The force F = 400 N. What couple M is
required to rotate the shaft at a constant rate?
F
F
M
M
C
C
D1
D2
Solution:
This is a thrust bearing with parameters
µk = 0.3,
F
F
α = 0,
ro = 60 mm,
ri = 20 mm,
and F = 400 N.
The moment required to sustain rotation at a constant rate is
2µk F ro3 − ri3
M =
= 5.2 N m
3
ro2 − ri2
Problem 8.93 Suppose that you want to lift a 50-lb
crate off the ground by using a rope looped over a tree
limb as shown. The coefficient of static friction between
the rope and the limb is 0.4, and the rope is wound 120◦
around the limb. What force must you exert to lift the
crate?
Strategy: The tension necessary to cause impending
slip of the rope on the limb is given by Eq. (9.17), with
T1 = 50 lb, µs = 0.4, and β = (π/180)(120) rad.
Solution:
radians is
β = (120◦ )
This meets the conditions for Eq. (9.17). The angle in
π = 2.094 radians.
180
The force required is T = 50eµs β = 115.6 lb
C
M
M
C
D1
D2
Problem 8.94 In Problem 8.93, once you have lifted
the crate off the ground, what is the minimum force you
must exert on the rope to keep it suspended?
If T is decreased until slip of the rope toward the left
impends, Equation (9.17) is 50 = T eµs β where µs = 0.4 and β =
(π/180)(120) rad.
Solving for T , we obtain T = 21.6 lb.
Solution:
Problem 8.95 Winches are used on sailboats to help
support the forces exerted by the sails on the ropes
(sheets) holding them in position. The winch shown is a
post that will rotate in the clockwise direction (seen from
above), but will not rotate in the counterclockwise direction. The sail exerts a tension TS = 800 N on the sheet,
which is wrapped two complete turns around the winch.
The coefficient of static friction between the sheet and
the winch is µs = 0.2. What tension TC must the crew
member exert on the sheet to prevent it from slipping on
the winch?
TC
TS
Solution:
Ts = Tc eµs β
Ts = 800 N
µs = 0.2
β = 4π
Solving,
Tc = 64.8 N
TC
TS
TC
TS
Problem 8.96 The coefficient of kinetic friction between the sheet and the winch in Problem 8.95 is µk =
0.16. If the crew member wants to let the sheet slip at a
constant rate, releasing the sail, what initial tension TC
must he exert on the sheet as it begins slipping?
Solution:
Ts = Tc eµk β
Ts = 800 N, µk = 0.16
β = 4π
Solving
Tc = 107.1 N
C
TS
TC
TS
Problem 8.97 The mass of the block A is 18 kg. The
rope is wrapped one and one-fourth turns around the
fixed wooden post. The coefficients of friction between
the rope and post are µs = 0.15 and µk = 0.12. What
force would the person have to exert to raise the block at
a constant rate?
Solution:
β = 2.5π
µk = 0.12
m = 18 kg.
T = (mg)eµk β
A
Solving,
T = 453 N
T
mg
A
Problem 8.98 The weight of the block A is W . The
disk is supported by a smooth bearing. The coefficient of
kinetic friction between the disk and the belt is µk . What
couple M is necessary to turn the disk at a constant rate?
Solution: The angle is β = π radians. The tension in the left belt when
the belt is slipping on the disk is Tleft = W eµk β . The tension in the right belt
is Tright = W . The moment applied to the disk is
M = R(Tleft − Tright ) = R(W eµk β − W ) = RW (eµk π − 1).
This is the moment that is required to rotate the disk at a constant rate.
r
M
r
M
A
A
Problem 8.99 The couple required to turn the wheel of
the exercise bicycle is adjusted by changing the weight
W . The coefficient of kinetic friction between the wheel
and the belt is µk . Assume the wheel turns clockwise.
(a) Show that the couple M required to turn the wheel
is M = W R(1 − e−3.4µk ).
(b) If W = 40 lb and µk = 0.2, what force will the
scale S indicate when the bicycle is in use?
S
15°
R
30°
Solution: Let β be the angle in radians of the belt contact with
wheel. The tension in the top belt when the belt slips is Tupper =
W e−µk β . The tension in the lower belt is Tlower = W . The moment
applied to the wheel is
S
M = R(Tlower − Tupper ) = RW (1 − e−µk β ).
This is the moment required to turn the wheel at a constant rate. The
angle β in radians is
π β = π + (30 − 15)
= 3.40 radians,
180
15°
R
30°
from which M = RW (1 − e−3.4µk ). (b) The upper belt tension is
W
Tupper = 40e−3.4(0.2) = 20.26 lb.
This is also the reading of the scale S.
Problem 8.100 The box B weighs 50 lb. The coefficient of friction between the cable and the fixed round
supports are µs = 0.4 and µk = 0.3.
(a) What is the minimum force F required to support
the box?
(b) What force F is required to move the box upward
at a constant rate?
B
F
Solution: The angle of contact between the cable and each round
support is β = π2 radians.
(a) Denote the tension in the horizontal part of the cable by H. The
tension in H is H = W e−µs β . The force F is
F = He−µs β = W e−2µs β ,
(b)
from which F = 14.23 lb is the force necessary to hold the box
stationary.
As the box is being raised,
H
and F
= W eµk β ,
= Heµk β = W e2µk β ,
from which F = 128.32 lb
B
F
W
Problem 8.101 The 20-kg box A is held in equilibrium
on the inclined surface by the force T acting on the rope
wrapped over the fixed cylinder. The coefficient of static
friction between the box and the inclined surface is 0.1.
The coefficient of static friction between the rope and the
cylinder is 0.05. Determine the largest value of T that
will not cause the box to slip up the inclined surface.
45°
A
20°
T
Solution: Assuming that slip of the box up the surface impends.
The free body diagrams of the box and rope around the cylinder are as
shown.
From the equilibrium equations
Fx = TA cos 45◦ − N sin 20◦ − 0.1 N cos 20◦ = 0,
Fy = TA sin 45◦ + N cos 20◦ − 0.1 N sin 20◦ − (20)(9.81) = 0
45°
A
we obtain TA = 90.2 N. Equation (9.17) is T = TA eµs B where
µs = 0.05 and β = (π/180)(135) rad. Solving for T we obtain
T = 101 N.
T
20°
y
TA
45°
20°
x
(20)(9.81)
0.1 N
N
135°
TA
T
Problem 8.102
In Problem 8.101, determine the
smallest value of T necessary to hold the box in equilibrium on the inclined surface.
Solution: In this case, we assume that slip of the box down the
surface impends. This requires reversing the direction of the friction
force in the free body diagram of Problem 8.101. The friction now
acts up the surface and the friction on the drum is reversed. See the
free body diagrams. From the equilibrium equations,
FX = TA cos(45◦ ) − N sin(20◦ ) + 0.1 N cos(20◦ ) = 0,
and
FY = TA sin(45◦ ) + N cos(20◦ ) + 0.1 N sin(20◦ )
y
TA
20°
45°
(20)(9.81)
− (20)(9.81) = 0.
x
0.1 N
N
Solving, we obtain TA = 56.3 N. We can now use this to find the
force T that must be applied to the rope to keep the box from slipping
down the plane. Eq. (9.17) is TA = T eµs β , where µs = 0.05 and
β = (π/180)(135) rad. Solving for T , we obtain T = 50.1 N.
135°
TA
T
Problem 8.103 The mass of the block A is 14 kg. The
coefficient of kinetic friction between the rope and the
cylinder is 0.2. If the cylinder is rotated at a constant
rate, first in the counterclockwise direction and then in
the clockwise direction, the difference in the height of
block A is 0.3 m. What is the spring constant k?
k
A
Solution:
k
T1 = T2 eµk β
Case 1: Clockwise Rotation
µk = 0.2
m = 14 kg
β = π/2
mg = Ts1 e(0.2)(π/2)
Ts1 = 100.31 N
A
Case 2: Counterclockwise Rotation
Case 1: Clockwise Rotation
Ts2 = mge(0.2)(π/2)
Ts2 = 188.03 N
k = 0.2
m = 14 kg
= /2
TS1
we know
Ts1 = kδ1
mg = TS1 e(0.2)(π/ 2)
Ts2 = kδ2
TS1 = 100.31 N
Ts2 − Ts1 = k(δ2 − δ1 )
mg
and δ2 − δ1 = 0.3 m
k = (Ts2 − Ts1 )/(δ2 − δ1 )
k = 292 N/m
Case 2: Counterclockwise Rotation
TS2
TS2 = mg e(0.2)(π/ 2)
TS2 = 188.03 N
mg
Problem 8.104 The weight of the box is W = 30 lb,
and the force F is perpendicular to the inclined surface.
The coefficient of static friction between the box and the
inclined surface is µs = 0.2.
(a) If F = 30 lb, what is the magnitude of the friction
force exerted on the stationary box?
(b) If F = 10 lb, show that the box cannot remain at
rest on the inclined surface.
F
W
20°
Solution: The maximum friction force is defined to be f = µs N ,
where N is the normal force.
(a) The box is stationary, hence the friction force is equal to the force
acting to move the box down the plane:
FP = f − WP = 0,
(b)
from which f = WP = W sin α = 10.26 lb
The component of force parallel to the surface is WP =
W sin α = 10.26 lb acting to move the box down the plane.
The friction force is f = µs (10 + 30 cos α) = 7.638 lb, acting
to hold the box in place. Since WP > f , the box will move.
F
W
α
F
20°
W
f
N
Problem 8.105 In Problem 8.104, what is the smallest force F necessary to hold the box stationary on the
inclined surface?
Solution: At impending slip, the sum of the forces parallel to the
surface is
FP = f − WP = 0,
from which f = WP . The friction force is f = µs (F + W cos α),
and WP = W sin α. Equate and solve:
sin α
sin 20◦
F =W
− cos α = 30
− cos 20◦ = 23.1 lb
µs
0.2
Problem 8.106 The mass of the van is 2250 kg, and
the coefficient of static friction between its tires and the
road is 0.6. If its front wheels are locked and its rear
wheels can turn freely, what is the largest value of α for
which it can remain in equilibrium?
1m
1.2 m
α
3m
Choose a coordinate system with the x-axis parallel to
the incline. The weight of the van is W = mg = 22072.5 N. The
moment about the point of contact of the rear wheels is
MR = (3 − 1.2)W cos α + 1W sin α − 3 N = 0,
Solution:
from which the normal force at the front wheels is
W (1.8 cos α + sin α)
N =
.
3
1m
1.2
α
m
3m
The sum of the forces parallel to the inclined surface is
Fx = +µs N − W sin α = 0.
Combine and reduce:
µ
1.8
s
µs
cos α +
− 1 sin α = 0,
3
3
from which
α = tan−1
1.8µs
3 − µs
1m
W
R
= tan−1 (0.45) = 24.2◦
µSN
N
1.8 m
1.2 m
Problem 8.107 In Problem 8.106, what is the largest
value of α for which the van can remain in equilibrium
if it points up the slope?
Solution: The sum of the moments about the point of contact of
the rear wheels is
MR = −1.8W cos α + 1W sin α + 3 N = 0.
The normal force is
The sum of forces parallel to the incline is
Fx = +µs N − W sin α = 0.
Combine and reduce:
µ
1.8µs
s
cos α −
+ 1 sin α = 0,
3
3
from which
α = tan−1
1.8µs
µs + 3
µsN
W
W (1.8 cos α − sin α)
N =
.
3
= 16.7◦
1m
N
1.2 m
R
1.8 m
A
Problem 8.108 Each of the uniform 1-m bars has a
mass of 4 kg. The coefficient of static friction between
the bar and the surface at B is 0.2. If the system is in
equilibrium, what is the magnitude of the friction force
exerted on the bar at B?
45°
O
B
30°
Solution:
The free body diagrams of the bars are as shown. The
equilibrium equations are
Left bar:
Fx = Cx + Ax = 0,
Fy = Cy + Ay − mg = 0,
M(leftend) = (1) cos 45◦ Ay − (1) cos 45◦ Ax − (0.5) cos 45◦ mg
A
Right bar:
B
45°
O
= 0,
30°
Ay
Fx = −Ax + f cos 30◦ − N sin 30◦ = 0,
45°
Fy = −Ay − mg + f sin 30◦ + N cos 30◦ = 0,
Cy
M(rightend) = (1) cos 45◦ Ax + (1) cos 45◦ Ay + (0.5) cos 45◦ mg
Ay
Ax
45°
Ax
mg
mg
f
= 0,
Cx
30°
N
Solving, we obtain N = 43.8 N and f = 2.63 N.
Problem 8.109 In Problem 8.108, what is the mini- Solution: From the solution of Problem 8.108, the normal and friction
are N = 43.8 N and f = 2.63 N. Slip impends when f = µs N so,
mum coefficient of static friction between the bar and forces 2.63
µs = 43.8 = 0.06.
the surface at B necessary for the system to be in equilibrium?
Problem 8.110 The clamp presses two pieces of wood
together. The pitch of the threads is p = 2 mm, the mean
radius of the thread is r = 8 mm, and the coefficient
of kinetic friction between the thread and the mating
groove is 0.24. What couple must be exerted on the
thread shaft to press the pieces of wood together with a
force of 200 N?
125 mm
125 mm
125 mm
B
50 mm
E
A
50 mm
C
50 mm
D
Solution: The free-body diagram of the upper arm of the clamp
is shown.
From the equilibrium equation
M(ptc) = −(0.25)(200) − 0.1BE = 0,
we find that BE = −500 N. The compressive load in BE is 500 N.
The slope of the thread is
P
α = arctan
2πr
= arctan
From Eq. (9.9) with θs = θk , the required couple is
M = rF tan(θk + α)
= (0.008)(500) tan(13.496◦ + 2.279◦ )
= 1.13 N-m.
BE
0.002
2π(0.008)
= 2.279◦ .
Cy
200 N
Cx
The angle of friction is
θk = arctan(0.24) = 13.496◦ .
0.1 m
0.25 m
Problem 8.111 In Problem 8.110, the coefficient of
static friction between the thread and the mating groove
is 0.28. After the threaded shaft is rotated sufficiently to
press the pieces of wood together with a force of 200 N,
what couple must be exerted on the shaft to loosen it?
Solution:
First, find the forces in the parts of the clamp. Then
analyze the threaded shaft. BE is a two force member
Fx :
BE + Cx = 0
(1)
Fy :
200 + Cy = 0
(2)
MA :
−0.05BE + 0.05Cx + 0.25Cy = 0 (3)
tan α = P/2πr =
2
2π(8)
α = 2.28◦
F = |BE| = 500 N
Solving
M = 0.950 N-m
Solving, we get
BE = −500 N
(compression)
Cy = −200 N
We don’t have to solve for additional forces because we used the fact
that member BE was a two force member.
From Problem 8.110, P = 2 mm, r = 8 mm. We have µs = 0.28.
We want to loosen the clamp (Turn the clamp such that the motion is
in the direction of the axial force.
To do this,
125 mm
B
50 mm
E
A
50 mm
C
50 mm
D
M = rF tan(θs − α)
y
where
tan θs = µs = 0.28
125 mm
125 mm
Cx = 500 N
0.125 m
B
0.05 m
θs = 15.64◦
A
0.125 m
200 N
BE
x
0.05 m
C
CX
CY
Problem 8.112 The axles of the tram are supported
by journal bearing. The radius of the wheels is 75 mm,
the radius of the axles is 15 mm, and the coefficient of
kinetic friction between the axles and the bearings is
µk = 0.14. The mass of the tram and its load is 160 kg.
If the weight of the tram and its load is evenly divided
between the axles, what force P is necessary to push the
tram at a constant speed?
Solution: Assume that there are two bearings per axle. The weight of the
tram is W = mg = 1569.6 N. This load is divided between four bearings:
F =
W
= 392.4 N.
4
The angle of kinetic friction is θk = tan−1 (µk ) = 7.97◦ . The moment required to turn each bearing at a constant rate is M = F r sin θk = 0.8161 N m,
and the force per wheel is
Pw =
M
0.8161
=
= 10.88 N.
R
0.075
The total force required to push the tram is P = 4Pw = 43.5 N
P
P
Problem 8.113 The two pulleys have a radius of 6 in.
and are mounted on shafts of 1-in. radius supported by
journal bearings. Neglect the weights of the pulleys and
shafts. The coefficient of kinetic friction between the
shafts and the bearings is µk = 0.2. If a force T = 200 lb
is required to raise the man at a constant rate, what is his
weight?
T
Solution: Denote the tension in the horizontal portion of the cable
by H. The angle of kinetic friction is θk = tan−1 (0.2) = 11.31◦
Consider the right Pulley: The force on the right pulley is
F = T 2 + H2.
The magnitude
√ of the moment required to turn the shaft in the bearing is
Mright = r T 2 + H 2 sin θk . The applied moment is Mapplied =
√
(T − H)R, from which (T − H)R = r T 2 + H 2 sin θk . Square
both sides and reduce to obtain the quadratic:
1
H 2 − 2T H
+ T 2 = 0,
r 2
1− R
sin2 θk
or H 2 − 2(200.214)H + 40000 = 0.
√
This has the solutions: H = 200.214 ± 200.2142 − 40000 =
209.46, 190.95. The lesser root corresponds to the horizontal tension,
H = 190.97 = 191 lb.
Consider
the left Pulley: The force on the pulley is Fleft =
√
W 2 + H 2 . The applied moment is Mapplied = −(H − W )R,
√
from which (H − W )R = r W 2 + H 2 sin θk . Square both sides
and reduce to the quadratic:
H
2
W −2
W + H 2 = 0,
r 2
1− R
sin2 θk
or W 2 − 2(191.1)W + 36431.9 = 0.
√
This has the solutions: W1,2 = 191.166± 191.1662 − 36431.9 =
199.9 lb, 182.25 lb. By an analogous argument to that used in
Problem ??.??, the lesser root corresponds to the weight of the man,
Wraised = 182.3 lb
Problem 9.1 The prismatic bar has a circular cross
section with 50-mm radius and is subjected to 4-kN axial
loads. Determine the average normal stress at the plane
P.
Free Body Diagrams:
Solution:
The cross-sectional area of the bar is:
A = πr 2 = π(0.05 m)2 = 0.007854 m2
The average normal stress is the load distributed across the face of the
cross section.
σAV = P/A = (4000 N)/(0.007854 m2 )
ANS:
σAV = 509 kPa
Problem 9.2 In Problem 9.1, what is the average shear
stress at the plane P ?
Solution:
Free Body Diagram:
We see from the FBD that the shearing force is the only force with any
vertical component. Summing vertical forces:
ΣFy = 0 = −τ A = −τ [π(0.025 m)2 ]
ANS:
τ =0
Problem 9.3 The prismatic bar has a cross-sectional
area A = 30 in2 and is subjected to axial loads. Determine the average normal stress (a) at plane P1 ; (b) at
plane P2 .
Free Body Diagrams:
Solution:
At Plane P1 , the average normal stress is:
(σAV )1 = P/A = (2000 lb)/(30 in2 )
ANS: (σAV )1 = 66.7 psi
At Plane P2 , the average normal stress is:
(σAV )2 = P/A = (2000 lb)/(30 in2 )
ANS:
(σAV )2 = 66.7 psi
Problem 9.4 The prismatic bar has a solid circular
cross section with 2-in. radius. Determine the average
normal stress (a) at plane P1 ; (b) at plane P2 .
Free Body Diagram:
Solution:
The cross-sectional area of the prismatic bar is:
A = πr 2 = π(2 in)2 = 12.566 in2
Note that the loads are NOT the same at P1 and P2 . The bar is in
tension at P1 and in compression at P2 .
(a) The average normal stress at P1 is:
(σAV )1 = P/A = (4000 lb)/(12.566 in2 )
ANS: (σAV )1 = 318.3 psi
(b) The average normal stress at P2 is:
(σAV )2 = P/A = (−8000 lb)/(12.566 in2 )
ANS:
(σAV )2 = −636.6 psi
Problem 9.5 A prismatic bar with cross-sectional area
A is subjected to axial loads. Determine the average
normal and shear stresses at the plane P if A = 0.02 m2 ,
P = 4 kN, and θ = 25◦ .
Free Body Diagram:
Solution:
The cross-sectional area of this section is larger than 0.02 m2 because
the bar is cut at an angle. The actual area of the angled cut is:
A = (0.02 m2 )/(cos 25◦ ) = 0.02206 m2
Summing horizontal forces on the bar:
ΣFx = 0 = −4000 N+[σAV (0.02206 m2 )](cos 25◦ )−τAV (0.02206 m2 )(sin 25◦ )
[1]
τAV = −429, 048 N + 2.145σAV
Summing vertical forces on the bar:
Σy = 0 = σAV (0.02206 m2 )(sin 25◦ )+τAV (0.02206 m2 )(cos 25◦ )
[2]
τAV = −0.4663σAV
Solving Equations [1] and [2] together:
ANS: σAV = 164.3 kPa
ANS: τAV = −76.6 kPa Note: The (−) sign indicates the
wrong direction assumed for the FBD.
Problem 9.6 Suppose that the prismatic bar shown in
Problem 9.5 has cross-sectional area A = 0.024 m2 . If
the angle θ = 35◦ and the average normal stress on the
plane P is σAV = 200 kPa, what are τAV and the axial
force P ?
Free Body Diagram:
Solution:
The cross-section area at plane P is:
A = (0.024 m2 )/(cos 35◦ ) = 0.0293 m2
Summing vertical forces on the FBD:
ΣFy = 0 = τAVG A(sin 55◦ ) + σAVG A(sin 35◦ )
τAVG = −0.7σAVG = −(0.7)(200 kPa)
τAVG = −140 kPa (−) indicates opposite direction.
Summing horizontal forces on the FBD:
ANS:
ΣFx = 0 = −P + σAVG A(cos 35◦ ) − τAVG A(cos 55◦ )
0 = −P +(200 kPa)(0.0293 m2 )(cos 35◦ )−(−0.7)(200 kPa)(0.0293 m2 )(sin 35◦ )
ANS:
P = 7.15 kN
Problem 9.7 For the prismatic bar in Problem 9.5, derive equations for the average normal and shear stresses
at the plane P as functions of θ.
Free Body Diagram:
σAV A
Solution:
Summing forces in the horizontal direction:
ΣFx = 0 = [A/(cos θ)]σAV (cos θ)−[A/(cos θ)]τAV (sin θ)−P
[1]
τAV = σAV (cos θ)/(sin θ) − [P (cos θ)]/A(sin θ)
Summing forces in the vertical direction:
ΣFy = 0 = σAV A(sin θ) + τAV A(cos θ)
[2]
τAV = −σAV (sin θ)/(cos θ)
Solving Equations [1] and [2] together:
−σAV (sin θ)/(cos θ) = σAV (cos θ)/(sin θ)−[P (cos θ)]/A(sin θ)
−σAV [(sin θ)/(cos θ)]−[(cos θ)/(sin θ)] = −[P (cos θ)]/A(sin θ)
−[σAV (sin2 θ) + (cos2 θ)] = −[P (cos2 θ)]/A
ANS: σAV = P (cos2 θ)/A
From Equation [2]:
τAV = −σAV (sin θ)/(cos θ) = P [(cos2 θ)/A][(sin θ)/(cos θ)]
ANS:
τAV = (P/A)(sin θ)(cos θ)
θ
Problem 9.8 The prismatic bar has a solid circular Free Body Diagram:
cross section with 30-mm radius. It is suspended from
one end and is loaded only by its own weight. The mass
density of the homogeneous material is 2800 kg/m3 . Determine the average normal stress at the plane P , where
x is the distance from the bottom of the bar in meters.
Strategy: Draw a free-body diagram of the part of the
bar below the plane P .
Solution:
The weight of the cylinder for any distance x from its bottom is:
W = ρgπr2 x = (2800 kg/m3 )(9.81 m/sec2 )π(0.03 m)2 x
W = 77.66x N
The stress produced in supporting this weight is:
σ = W/A = (77.66x N)/π(0.03 m)2
ANS:
σ = 27, 467x Pa = 27.5x kPa
Problem 9.9 The beam has cross-sectional area A =
0.0625 m2 . What are average normal stress and the magnitude of the average shear stress at the plane P ?
Strategy: Draw the free-body diagram of the entire
beam and determine the reactions at the pin and roller
supports. Then determine the average normal and shear
stresses by drawing the free-body diagram of the part of
the beam to the left of plane P .
Free Body Diagram:
Solution:
To find Ax , sum horizontal forces.
ΣFx = 0 = 4 kN − Ax
Ax = 4 kN ←
To find Ay , sum moments about point B:
ΣMB = 0 = (2 kN)(2 m) − (6 kN − m) + Ay (6 m)
Ay = 333.33 N ↓
Σfy = 0 = −333.3 − 2000 + By = 0
By = 2333.33 N ↑
Now cut the beam through the plane P . Using the left-hand portion of
the beam:
ΣFx = 0 = σA − Ax = σAV G (0.0625 m2 ) − 4, 000 N
ANS: σAV G = 64 kPa
Summing vertical forces on the left-hand portion of the beam:
ΣFy = 0 = −Ay + τAV G A = −333.3 lb + τAV G (0.0625 m2 )
ANS:
τAV G = 5.33 kPa
Problem 9.10 Determine the average normal stress
and the magnitude of the average shear stress at the plane
P of the beam in Problem 9.9 by drawing the free-body
diagram of the part of the beam to the right of plane P
and compare your answers to those of Problem 9.9.
Free Body Diagram:
Solution:
Sum moments about the left-hand end of the beam to find By :
ΣMA = 0 = −(2, 000 N)(4 m) − 6, 000 N − m + By (6 m)
By = 2, 333 N ↑
Summing vertical forces on the FBD of the right-hand portion of the
beam:
ΣFy = 0 = 2, 333 N − 2, 000 N − τAV G (0.0625 m2 )
ANS: τAV G = 5.33 kPa
Summing horizontal forces on the FBD of the right-hand portion of
the beam:
ΣFx = 0 = 4, 000 N − σAV G (0.0625 m2 )
ANS:
σAV G = 64 kPa
Problem 9.11 The beams have cross-sectional area
A = 60 in2 . What are the average normal stress and
the magnitude of the average shear stress at the plane P
in cases (a) and (b)?
Solution:
18 kip
Ax
20 kip
Ay
(a)
By
Summing moments about point B on the beam:
ΣMB = 0 = −(18, 000 lb)(4 ft) + Ay (12 ft)
Ay = 6, 000 lb ↑
F.2 = 1/2(12)(3000)
8'
Ax
4'
Ay
20 kip
By
Summing horizontal forces on the beam:
ΣFx = 0 = −20, 000 lb + Ax
Ax = 20, 000 lb →
Summing vertical forces on the left-hand portion of the beam:
ΣFy = 0 = Ay − τavg (area) = 6, 000 lb − τavg (60 in2 )
ANS: τavg = 100 psi
Summing horizontal forces on the left-hand portion of the beam:
ΣFx = 0 = Ax − σavg (area) = 20, 000 lb − σavg (60 in2 )
ANS:
(b)
σavg = +333.3 psi(C)
Summing moments about point B on the beam:
ΣMB = 0 = −1/2(3, 000 lb/ft)(12 ft)(4 ft) + Ay (12 ft)
Ay = 6, 000 lb ↑
Summing horizontal forces on the beam:
ΣFx = 0 = −20, 000 lb + Ax
Ax = 20, 000 lb →
Summing vertical forces on the left-hand portion of the beam:
ΣFy = 0 = Ay −1/2(1, 500 lb/ft)(6 ft)−τavg (area) = 6, 000 lb−4, 500 lb+τavg (60 in2 )
τavg = 650 psi
Summing horizontal forces on the left-hand portion of the beam:
ANS:
ΣFx = 0 = Ax − σavg (area) = 20, 000 lb − σavg (60 in2 )
ANS:
σavg = −333.33 psi(C)
Problem 9.12 Figure (a) is a diagram of the bones and
biceps muscle of a person’s arm supporting a mass. Figure (b) is a biomedical model of the arm in which the
biceps muscle AB is represented by a bar with pin supports. The suspended mass is m = 2 kg and the weight
of the forearm is 9 N. If the cross-sectional area of the tendon connecting the biceps to the forearm at A is 28 mm2 ,
what is the average normal stress in the tendon.
Free Body Diagram:
Solution:
Summing moments about point C:
ΣMc = 0 = −(19.62 N)(0.35 m)−(9 N)(0.15 m)+σavg (28×10−6 m2 )(0.05 m(sin 80.2))
ANS:
σavg = 5.96 MPa
Problem 9.13 The force F exerted on the bar is 20i −
20j − 10k (lb). The plane P is parallel to the y-z plane
and is 5 in. from the origin O. The bar’s cross sectional
area at P is 0.65 in2 . What is the average normal stress
in the bar at P ?
Free Body Diagram:
Solution:
The average normal stress at point P is exerted only by the xcomponent of the applied force. Summing forces in the x-direction:
ΣFx = 0 = 20 lb − σavg (area) = 20 lb − σavg (0.65 in2 )
ANS:
σavg = 30.8 psi
Problem 9.14 In Problem 9.13, what is the magnitude
of the average shear stress in the bar at P ?
Free Body Diagram:
Solution:
The average shear stress at point P is exerted only by the y- and zcomponents of the applied force. Summing forcers in the y-direction:
ΣFy = 0 = −20 lb + Py
Py = 20 lb
Summing forces in the z-direction:
ΣFz = 0 = −10 lb + Pz
Pz = 10 lb
The y- and z-components of the reaction at P are added vectorially.
RP = Py2 + Pz2 = (20 lb)2 + (10 lb)2
RP = 22.36 lb
The average shear stress at point P is:
τavg = RP /A = (22.36 lb)/(0.65 in2 )
ANS:
τavg = 34.4 psi
Problem 9.15 The plane P is parallel to the y-z plane
of the coordinate system. The cross-sectional area of
the tennis racquet at P is 400 mm2 . Including the force
exerted on the racquet by the ball and inertial effects,
the total force on the racquet above the plane P is 35i −
16j − 85k N. What is the average normal stress on the
racquet at P ?
Free Body Diagram:
Solution:
The average normal stress at point P is exerted only by the xcomponent of the applied force. Summing forces in the x-direction:
ΣFy = 0 = 35 N − σavg (area) = 35 N − σavg (400 × 10−6 m2 )
ANS:
σavg = 87, 500 Pa = 87.5 kPa
Problem 9.16 The fixture shown connects a 50-mm Free Body Diagram:
diameter bridge cable to a flange that is attached to the
bridge. A 60-mm diameter circular pin connects the
fixture to the flange. If the average normal stress in the
cable is σAV = 120 MPa, what average shear stress
τAV must the pin support?
Solution:
The average normal stress of 120 MPa in the cable means that the axial
load in the cable is:
P = σavg A = (120 × 106 N/m2 )(π)(0.025 m)2
P = 236 kN
We see that the shear load in the pin is divided between two areas, one
on either side of the flange. The average shear stress which must be
supported by the pin is:
τavg = P/(2A) = (236, 000 N)/(2)[π(0.03 m2 )]
ANS:
τavg = 41.7 MPa
Problem 9.17 Consider the fixture shown in Problem 9.16. The cable will safely support an average normal stress of 700 MPa and the circular pin will safely
support an average shear stress of 220 MPa. Based on
these criteria, what is the largest tensile load the cable
will safely support?
Solution:
Calculating the largest load which can be supported by the cable:
(FMAX )CABLE = (σAVG )ALLOW A = (700×106 N/m2 )[π(0.025 m)2 ]
(FMAX )CABLE = 1.37 MN
Calculating the largest load which can be supported by the pin:
(FMAX )PIN = 2[(τAVG )MAX A] = 2[(220×106 N/m2 )(π)(0.03 m)2 ]
(FMAX )PIN = 1.24 MN
We see that the cable can support a larger load than the pin. Any load
larger than 1.24 MN will cause the pin to fail in shear. The largest load
which can be supported by the cable and pin is:
ANS: FMAX = 1.24 MN
Problem 9.18 The truss is made of prismatic bars with Free Body Diagram:
cross-sectional area A = 0.25 ft2 . Determine the average normal stress in member BE acting on a plane
perpendicular to the axis of the member.
Solution:
We can determine the axial load in member BE directly by summing
vertical forces on the FBD.
ΣFy = 0 = −8, 000 lb − 10, 000 lb + PBE (sin 45◦ )
PBE = 25.5 kip
The average normal stress in member BE of the truss is:
σavg = (PBE )/A = (25, 500 lb)/(0.25 ft2 )
ANS:
σavg = 102 ksf = 708 psi
Problem 9.19 For the truss in Problem 9.18, determine Free Body Diagram:
the average normal stress in member BD acting on a
plane perpendicular to the axis of the member.
Solution:
To solve directly for the axial load in member BD of the truss, sum
moments on the FBD about point E.
ΣME = 0 = −(8, 000 lb)(20 ft)−(10, 000 lb)(10 ft)+PBD (10 ft)
PBD = 26, 000 lb (C) = 26 kip (C)
The average normal stress in member BD of the truss is:
σavg = PBD /A = (26, 000 in)/(0.25 ft2 )
σavg = −104, 000 psf(C) = −722.2 psi NOTE: The
negative sign indicates compressive stress.
ANS:
Problem 9.20 Three views of joint A of the truss in
Problem 9.18 are shown. The joint is supported by a
cylindrical pin 2 in. in diameter. What is the magnitude
of the average shear stress in the pin?
Free Body Diagram:
Solution:
Sum vertical forces at joint A to determine the axial load in member
AC.
ΣFy = 0 = −8, 000 lb + PAC (sin 45◦ )
PAC = 11, 313.7 lb (T) = 11.3 kip (T)
Sum horizontal forces at joint A to find the axial load in member AB.
ΣFx = 0 = PAB −PAC (cos 45◦ ) = PAB −(11, 300 lb)(cos 45◦ )
PAB = 8, 000 lb (C) = 8 kip (C)
When the 8,000 lb load and PAB are added vectorially, we see the load
which must be supported by the pin (right-hand free body diagram).
The average shear stress in the pin is:
τavg = (11, 313.7 lb)/[(π)(1 in)2 ]
ANS:
τavg = 3597 psi
Problem 9.21 The top view of pin A of the pliers is
shown. The cross-sectional area of the pin is 4.5 mm2 .
What is the average shear stress in the pin when 150-N
forces are applied to the pliers as shown?
Free Body Diagram:
Solution:
The easier way in which to find the load in member AB is to draw the
FBD of the lower handle and sum moments about point D.
ΣMD = 0 = (150 N)(0.13 m) − PAB (sin 23.2◦ )(0.03 m)
PAB = 1650 N (C)
The shear load which must be supported by the pin at joint A is the
load in member AB. The average shear stress is:
τavg = (PAB )/(A) = (1650 N)/(4.5 × 10−6 m2 )
ANS:
τavg = 367 × 106 Pa = 367 MPa
Problem 9.22 In Problem 9.21 the vertical plane P is Free Body Diagrams:
30 mm to the left of C. The cross-sectional area of member AC of the pliers at plane P is 50 mm2 . Determine
the average normal stress and the magnitude of the average shear stress at P when 150-N forces are applied to
the pliers as shown.
Solution:
The easier way in which to find the load in member AB is to draw the
FBD of the lower handle and sum moments about point D.
ΣMD = 0 = (150 N)(0.13 m) − PAB (sin 23.2◦ )(0.03 m)
PAB = 1650 N (C)
To find the average normal stress at plane P , sum horizontal forces on
the FBD of the upper handle.
ΣFx = 0 = −PAB (cos 23.2◦ )+σavg A = −(1650 N)(cos 23.2◦ )+σavg (50×10−6 m2 )
ANS: σavg = 30.3 × 106 Pa = 30.3 MPa
To find the average shear stress at plane P , sum vertical forces on the
FBD of the upper handle.
ΣFy = 0 = −150 N+(1650 N)(sin 23.2◦ )−τavg (50×10−6 m2 )
ANS:
τavg = 10 × 106 Pa = 10 MPa
Problem 9.23 The suspended crate weighs 2000 lb and
the angle α = 30◦ . The top view of the pin support A
of the crane’s boom is shown. The cross-sectional area
of the pin is 23 in2 . What is the average shear stress in
the pin?
Free Body Diagram:
Solution:
The Pythagorean Theorem will help to determine the distance, x.
(9 ft)2 = (4.5 ft)2 + (6 ft + x)2
x = 1.79 ft
From the geometric construction
tan−1 (1.79 ft/4.5 ft), or:
we
see
that
β
=
β = 21.7◦
Summing moments about point A:
ΣMA = 0 = −(2000 lb)(15 ft)(cos 30◦ )+PBC (cos 21.7◦ )[(9 ft)(cos 30◦ )]−PBC (sin 21.7◦ )[(9 ft)(sin 30◦ )]
PBC = 4658 lb (C)
To find the reaction at point A, sum horizontal and vertical forces on
the boom.
ΣFx = 0 = PBC (sin 21.7◦ ) − Ax = (4657 lb)(sin 21.7◦ ) − Ax
Ax = 1722 lb ←
ΣFy = 0 = −2, 000 lb+PBC (cos 21.7◦ )−Ay = −2, 000 lb+(4657 lb)(cos 21.7◦ )−Ay
Ay = 2327 lb ↓
Total shearing force at joint A is:
FA = A2x + A2y = (1722 lb)2 + (2327 lb)2
FA = 2895 lb
We see that the shear load is divided between the cross-sections on
either side of the pin where it connects member BC to the boom. The
average shearing load is:
τavg = FA /2A = (2895 lb)/(2)(23 in2 )
ANS:
τavg = 62.9 psi
Problem 9.24 In Problem 9.23, the plane P is 3 ft
from end D of the crane’s boom and is perpendicular to
the boom. The cross-sectional area of the boom at P
is 15 in2 . Determine the average normal stress and the
magnitude of the average shear stress in the boom at P .
Free Body Diagram:
Solution:
Note the directions for the x- and y-axes on the free body diagram.
The average normal stress can be found by summing forces in the
x-direction.
ΣFx = 0 = −(2, 000 lb)(sin 30◦ )+σavg A = −(2, 000 lb)(sin 30◦ )+σavg (15 in2 )
σavg = −66.7 psi NOTE: The negative sign indicates that
the normal stress is compressive.
The average shear stress can be determined by summing forces in the
y-direction.
ANS:
ΣFy = 0 = −(2, 000 lb)(cos 30◦ )+τavg A = −(2, 000 lb)(cos 30◦ )+τavg (15 in2 )
ANS:
τavg = 115.5 psi
Problem 9.25 Three rectangular boards are glued together and subjected to axial loads as shown. What is
the average shear stress on each glued surface?
Free Body Diagram of the Upper Board
Solution:
The glued area of the upper board is:
A = lw = (0.09 m)(0.135 m)
A = 0.01215 m2
The average shear stress across the glued joint is:
τavg = P/A = (1, 000 N)/(0.01215 m2 )
ANS:
τavg = 82, 300 N = 82.3 kPa
Problem 9.26 Two boards with 4 in. × 4 in. square
cross sections are mitered and glued together as shown.
If the axial forces P = 600 lb, what average shear stress
must the glue support?
Free Body Diagram:
Solution:
Total glued area which is subjected to shear stress is:
A = 2[(4 in)(4 in)]
A = 32 in2
The average shear stress across the glued surfaces is:
τavg = P/A = (600 lb)/(32 in2 )
ANS:
τavg = 18.75 psi
Problem 9.27 A 1/8-in. diameter punch is used to cut
blanks out of a 1/16-in. thick plate of aluminum. If an
average shear stress of 20,000 psi must be induced in the
plate to create a blank, what force F must be applied?
Free Body Diagram:
Solution:
The average shear stress in the material to be punched will be exerted
around the entire cylindrical area which is left after the hole has been
punched. The cylindrical area for this process is:
A = 2πrh = 2π(1/16 in)(1/16 in)
A = 0.0245 in2
The force required to produce the needed average shear stress is:
P = τavg A = (20, 000 lb/in2 )(0.0245 in2 )
ANS: P = 490.9 lb
or as per Example 9.3
F = τAV πtD = 20, 000(π)(
1 1
)( ) = 490.9 lbs
16 8
Problem 9.28 Two pipes are connected by bolted
flanges. The bolts are 20 mm in diameter. One pipe has
a built-in support and the other is subjected to a torque
T = 6 kN − m about its axis.
Estimate the resulting average shear stress in each bolt.
Why is the result an estimate?
Free Body Diagram:
Solution:
The free body diagram illustrates the manner in which the bolts resist
an applied counterclockwise torque. The moment exerted by each of
the bolts is:
M = τavg Ar → τavg = M/6 Ar
τavg = (6, 000 N − m)/(6)[(π)(0.01 m)2 (0.15 m)]
ANS: τavg = 21.2 × 106 Pa = 21.2 MPa
The result assumes that the load is evenly distributed among the six
bolts. The load supported by each bolt can be affected by (among
other factors) the degree to which each bolt is (or is NOT) tightened,
the accuracy of their radii from the axis of the flange, the uniformity of
material quality among the bolts, and the precision of their diameters.
Problem 9.29 The bolts in Problem 9.28 will each
safely support an average shear stress of 130 MPa. Based
on this criteria, estimate the largest safe torque that can
be applied.
Free Body Diagram:
Solution:
The free body diagram illustrates the manner in which the bolts resist
an applied counterclockwise torque. The moment exerted by each of
the bolts is:
M = 6(τavg Ar) = (6)[(130×106 N/m2 )[π(0.01 m)2 ](0.15 m)]
ANS:
M = 36, 800 N − m = 36.8 kN − m
Problem 9.30 “Shears”, such as the familiar scissors,
have two blades which subject a material to shear stress.
For the shearing process shown, draw a suitable freebody diagram and determine the average shear stress the
blades exert on the sheet of material of thickness t and
width b.
Strategy: Obtain a free-body diagram by passing a
vertical plane through the material between the two
blades.
Free Body Diagram:
Solution:
The average shear stress is assumed to be evenly distributed across the
entire cross-section of the material.
The area over which the shear load is distributed is:
A = bt
The magnitude of the average shear stress is:
τavg = F/A
ANS:
τavg = F/(bt)
Problem 9.31 A 2-in. diameter cylindrical steel bar
is attached to a 3-in. thick fixed plate by a cylindrical
rubber grommet. If the axial load P = 60 lb, what is the
average shear stress on the cylindrical surface of contact
between the bar and the grommet?
Solution:
The area of contact between the bar and the grommet is:
A = 2πrh = 2π(1 in)(3 in) = 18.8 in2
The average shear stress between the bar and the grommet is:
τavg = P/A = (60 lb)/(18.8 in2 )
ANS:
τavg = 3.19 psi
Problem 9.32 The outer diameter of the cylindrical
rubber grommet in Problem 9.31 is 3.5 in. What is the
average shear stress on the cylindrical surface of contact
between the grommet and the fixed plate?
Solution:
The cylindrical area of contact between the grommet and the fixed
plate is:
A = πdh = π(3.5 in)(3 in) = 32.99 in2
Average shear stress between the grommet and the fixed plate is:
τAVG =
ANS:
P
60 lb
=
A
32.99 in2
τAVG = 1.82 psi
Problem 9.33 The steel bar described in Problem 9.31
is subjected to a torque T = 100 in − lb about its axis.
What is the average shear stress on the cylindrical surface
of contact between the bar and the grommet?
Solution:
The area of contact between the bar and the grommet is:
A = 2πrh = 2π(1 in)(3 in) = 18.8 in2
The force resulting from the shear stress between the bar and the grommet is:
F = τavg A = τavg (2π)(1 in)(3 in)
F = 6πτavg lb
The force exerted by the average shear stress is located at a radius from
the axis of the bar of:
r = 1 in.
The average shear stress which is produced by the applied moment is:
M = F r = (6π)(τavg )(1 in) = 100 in − lb
ANS:
τavg = 5.31 psi
Problem 9.34 A traction distribution t acts on a plane
surface A. The value of t at a given point on A is t =
45i + 40j − 30k kPa.
The unit vector i is perpendicular to A and points away
from the material. What is the normal stress σ at the
given point?
Solution:
The i-component of the applied stress is the only portion which contributes to NORMAL stress. The j- and k-components act in the plane
of the material and so contribute to shear stress.
The normal stress at any point on the plane surface is:
ANS: σavg = 45 kPa
Problem 9.35 In Problem 9.34, what is the magnitude
of the shear stress τ at the given point?
Solution:
The components of the applied stress which produce shear are the jand k-components. The resultant of these two orthogonal components
is:
R = t2j + t2k = (40 kPa)2 + (−30 kPa)2
ANS:
R = 50 kPa
Problem 9.36 A traction distribution t acts on a plane
surface A. The value of t at a given point on A is t =
3000i − 2000j + 6000k psi.
The unit vector e = (6/7)i + (3/7)j + (2/7)k is perpendicular to A and points away from the material. What is
the normal stress σ at the given point?
Solution:
To find that portion of the applied stress which is normal to the plane
surface it is necessary to find the scalar product between the applied
stress and the unit vector which is normal to the plane.
Finding the scalar product between the vectors:
σavg = t·e = (3000i−2000j+6000k)·(6/7i+3/7j+2/7k) = [(3000 psi)(6/7)]+[(−2000 psi)(3/7)]+[(6000 psi)(2/7)]
ANS:
σavg = 3430 psi
Problem 9.37 A line of length dL at a particular point
of a material in a reference state has length dL = 1.2 dL
in a deformed state. What is the extensional strain corresponding to that particular point and the direction of
the line dL?
Solution:
Extensional strain is the ratio of the change in length to the length In
the reference state.
ε = (dL − dL)/dL = (1.2dL − dL)/dL
ANS:
ε = 0.2
Problem 9.38 The extensional strain corresponding to
a point of a material and the direction of a line of length
dL in the reference state is ε = 0.15. What is the length
dL of the line in the deformed state?
Solution:
The definition of extensional strain can be used in this solution.
Using Equation 9.3 from the text:
dL = (1 + ε)dL = (1 + 0.15)dL
ANS:
dL = 1.15dL
Problem 9.39 A straight line within a reference state
of an object is 50 mm long.
In a deformed state, the line is 54 mm long. If the extensional strain ε in the direction tangent to the line is
uniform throughout the line’s length, what is ε?
Solution:
Using Equation 9.3 from the text:
dL = (1 + ε)dL
ε = (dL − dL)/dL = (54 mm − 50 mm)/(50 mm)
ANS:
ε = 0.08
Problem 9.40 The length of the curved line within the
material in the reference state is L = 0.2 m. The material then undergoes a deformation such that the value
of the extensional strain ε in the direction tangent to the
curved line is ε = 0.03 at each point of the line.
What is the length L of the line in the deformed state?
Solution:
Using Equation 9.3 from the text:
L = (1 + ε)L
L = (1 + 0.03)(0.2 m)
ANS:
L = 0.206 m
Problem 9.41 The length of the curved line L in the
reference state shown in Problem 9.40 is 0.5 m. Its length
in the deformed state is L = 0.488 m. If the value of
the extensional strain ε in the direction tangent to the line
is the same at each point of the line, what is ε?
Solution:
Using the definition of strain:
ε=
ANS:
L − L
0.488 m − 0.5 m
=
L
0.5 m
ε = −0.024
Problem 9.42 The coordinate s measures distance
along the curved line in the reference state. The length of
the line is L = 0.2 m. The material then undergoes a deformation such that the value of the extensional strain ε in
the direction tangent to the curved line is ε = 0.03+2s2 .
What is the length L of the line in the deformed state?
Solution:
Since the extensional strain rate is not constant, we need to integrate.
Using Equation 9.4 from the text:
0.2
L = 00.2 (1 + ε)ds = 00.2 1 + 0.03 + 2s2 ds = 00.2 1.03 + 2s2 ds = 1.03s + 23 s3 0
L = 0.211m
Problem 9.43 In Problem 9.42, suppose that the material undergoes a deformation that induces an extensional strain tangent to the line given by the equation
ε = 0.01[1 + (s/L)3 ]. What is the length L of the line
in the deformed state?
Solution:
Since the extensional strain rate is not constant, we need to integrate.
Using Equation 9.4 from the text:
3 s 3 s
L = 0L (1 + ε) ds = 0L 1 + 0.01 1 + L
ds = 0L 1.01 + 0.01 L
ds
3
4 L
4
(L)
s
= 1.01L + 0.01 4L3
L = 1.01s + 0.01 4L
3 L = 1.0125L
0
Problem 9.44 In a shock-wave experiment, the left
side of a 100-mm thick plate of steel is subjected to a
constant velocity of 1.5 km/s to the right at time t = 0.
As a result, a shock wave travels across the plate with
a constant velocity U > 1.5 km/s. To the right of the
shock wave, the material of the plate is stationary and
undeformed. To the left of the shock wave, the material is moving with a uniform velocity of 1.5 km/s and is
subject to a homogeneous (uniform) extensional strain ε.
If optical instrumentation indicates that the shockwave
arrives at the right side of the plate at t = 16 × 10−6 s,
what is ε?
Solution:
The distance traveled by the shock wave during the elapsed time is:
D = RT = (1500 m/sec)(16 × 10−6 sec) = 0.024 m
The strain generated by the shock wave is the distance traveled by the
wave divided by the thickness of the material:
ε = D/t = (0.024 m)/(0.100 m)
ANS:
ε = 0.24
Problem 9.45 In Problem 9.44, suppose that the time
at which the shock wave arrives at the right side of the
plate is unknown, but an embedded strain gage indicates
that the homogeneous extensional strain to the left of the
shock wave is ε = −0.3.
What is the velocity U of the shock wave?
Solution:
The change in dimension of the left-hand portion of the plate is:
∆L = (0.1 m)(−0.3) = −0.03
Calculating the time required for the shock wave to travel through the
plate:
−0.03 = (−1500 m/s) t → t = 2 × 10−5 sec
The speed of the shock wave is:
v=
ANS:
D
0.1 m
=
= 5000 m/s
t
2 × 10−5 sec
v = 5 km/sec
Problem 9.46 When it is unloaded, the nonprismatic
bar is 12 in. long. The loads cause axial strain given by
the equation ε = 0.04/(x + 12), where x is the distance
from the left end of the bar in inches. What is the change
in length of the bar?
Solution:
The change in length is the deformed length minus the reference length.
δ=
12
0
(1 + ε)dx − L =
12
0
(1 +
0.04
x+12
dx − L = [x + 0.04 ln (x + 12)]|12
0 − 12in
δ = [12 + 0.04 ln (12 + 12) − (0 + 0.04 ln (0 + 12))] − 12in
δ = 0.028in
Problem 9.47 The force F causes point B to move Drawing:
downward 0.002 m. If you assume the resulting extensional strain ε parallel to the axis of the bar AB is uniform
along the bar’s length, what is ε?
Solution:
The original length of the bar is:
L = (0.6 m)2 + (0.8 m)2 = 1.0 m
The deformed length of the bar is:
L = (0.6 m)2 + (0.798 m)2 = 0.9984 m
The strain is:
−L
m−1.0 m
ε = LL
= 0.99841.0
m
ε = −0.00160
Problem 9.48 When the truss is subjected to the vertical force F , joint A moves a distance v = 0.3 m vertically and a distance u = 0.1 m horizontally. If the
extensional strain εAB in the direction parallel to member AB is uniform throughout the length of the member,
what is εAB ?
Diagram:
Solution:
The reference length of member AB is:
2m
= 2.309 m
sin 60◦
L=
The deformed length of member AB is:
L = (2.2996 m)2 + (1.2545 m)2 = 2.6195
The strain for member AB is:
−L
εAB = L L
=
εAB = 0.1344
2.6195 m−2.309 m
2.309 m
Problem 9.49 In Problem 9.48, if the extensional
strain εAC in the directional parallel to member AC is
uniform throughout the length of the member, what is
εAC ?
Diagram:
Solution:
The reference length of member AC is:
L = (2 m)2 + (2 m)2 = 2.828 m
The deformed length of member AC is:
L = (2.299 m)2 + (1.899 m)2 = 2.982 m
The strain for member AC is:
−L
ε = LL
=
ε = 0.0548
2.982 m−2.828 m
2.828 m
Problem 9.50 Suppose that a downward force is applied at point A of the truss, causing point A to move
0.360 in. downward and 0.220 in. to the left. If the
resulting extensional strain εAB in the direction parallel
to the axis of bar AB is uniform, what is εAB ?
Drawing:
Solution:
The reference length of member AB is:
L = (16 in)2 + (16 in)2 = 22.627 in
The deformed length of member AB is:
L = (16.36 in)2 + (15.78 in)2 = 22.73 in
The strain in member AB is:
−L
ε = LL
=
ε = 0.00455
22.73 in−22.627 in
22.627 in
Problem 9.51 In Problem 9.50, if the resulting extensional strain εAC in the direction parallel to the axis of
bar AC is uniform, what is εAC ?
Drawing:
Solution:
The reference length for member AC is:
L = (16 in)2 + (24 in)2 = 28.844 in
The deformed length for member AC is:
L = (24.22 in)2 + (16.36 in)2 = 29.228 in
The strain in member AC is:
−L
ε = LL
=
ε = 0.0133
29.228 in−28.844 in
28.844 in
Problem 9.52 A steel tube (a) has an outer radius
r = 20 mm. The tube is then pressurized, increasing
its outer radius to r = 20.04 mm. What is the resulting
extensional strain of the bar’s outer circumference in the
direction tangent to the circumference?
Solution:
The reference circumference of the tube is:
C = 2πr = 2π(20 mm) = 125.664 mm
The deformed circumference of the tube is:
C = 2πr = 2π(20.04 mm) = 125.915 mm
The strain in the bar’s outer circumference is:
mm−125.664 mm
ε = C C−C = 125.915125.664
mm
ε = 0.00199 ≈ 0.002
Problem 9.53 The angle between two infinitesimal
lines dL1 and dL2 which are perpendicular in a reference state is 120◦ in a deformed state. What is the shear
strain at this point corresponding to the directions dL1
and dL2 ?
Solution:
The angle between the two reference lines has increased by 30◦ . The
shear strain between the two lines is this angle measured in radians.
π
τ = −γ
2
τ =
ANS:
π
2
π
− π = − = −0.524
2
3
6
γ = −0.524
Problem 9.54 When the airplane’s wing is unloaded
(the reference state), the perpendicular lines L1 and L2
on the upper surface of the right wing are each 600 mm
long. In the loaded state shown, L1 is 600.2 mm long
and L2 is 595 mm long. If you assume that they are
uniform, what are the longitudinal strains in the L1 and
L2 directions?
Solution:
The longitudinal strain in the direction of L1 is:
L −L
ε1 = 1L 1 =
1
ε1 = 0.000333
600.2 mm−600 mm
600 mm
The longitudinal strain in the direction of L2 is:
L −L
mm
ε2 = 2L 2 = 595 mm−600
600 mm
2
ε2 = −0.008333
Problem 9.55 In Problem 9.54, the angle between the
lines L1 and L2 at the point where they intersect is 90.2◦ .
What is the shear strain referred to the directions L1 and
L2 at that point?
Solution:
The change in the angle between L1 and L2 increases by 0.2◦ . This
angle, measured in radians, is:
π
τ = − γ = 90 − 90.2 = −0.2◦ = −0.00349 rad
2
ANS:
γ = −0.00349
Problem 9.56 Two infinitesimal lines dL1 and dL2
are shown in a reference state and in a deformed state.
(The lines dL1 , dL2 , dL1 and dL2 are contained in
the x-y plane.) What is the shear strain at this point
corresponding to the directions dL1 and dL2 ?
Solution:
The angle between L1 and L2 , originally 90◦ , has been reduced by
40◦ due to deformation. The shear strain in the direction of L1 and
L2 is:
ANS: γ = 40◦ = 0.698
Problem 9.57 Two infinitesimal lines dL1 and dL2
within a material are parallel to the x and y axes in a
reference state (Figure a). After a motion and deformation of the material, dL1 points in the direction of the
unit vector e1 = and dL2 points in the direction of the
unit vector e2 = −0.408i + 0.816j − 0.408k (Figure b).
What is the shear strain referred to the directions dL1
and dL2 ?
Solution:
We need to find the angle between the two unit vectors. The scalar
product is of use:
e1 · e2 = |e1 ||e2 |(cos θ)
The magnitude of each of the unit vectors is obviously 1. The cosine
of the angle between he two unit vectors is:
cos θ = e1 · e2 = [(0.667) (−0.408)] + [(0.667) (0.816)] + [(0.333) (−0.408)]
cos θ = 0.1363
θ = 82.17◦
The original angle of 90◦ between the two unit vectors has been reduced by an amount of:
θ = 90◦ − γ
82.17◦ = 90◦ − γ
γ = 7.83◦ or
ANS:
γ = 0.137 in radians
Problem 9.58 In Problem 9.57, suppose that after the
motion and deformation of the material dL1 points in
the direction of the unit vector e1 = 0.667i + 0.667j +
0.333k and dL2 = −0.514i + 0.686j + 0.514k. What is
the shear strain referred to the directions dL1 and dL2 ?
Solution:
We need to find the angle between the two unit vectors.
The scalar product is of use: e1 · e2 = |e1 ||e2 |(cos θ)
The magnitude of each of the unit vectors is obviously 1.
The cosine of the angle between he two unit vectors is:
cos θ = e1 · e2 = [(0.667) (−0.514) + (0.667) (0.686) + (0.333) (0.514)]
cos θ = 0.2859
θ = 73.4◦
The original angle of 90◦ between the two unit vectors has been reduced by an amount of:
θ = 90◦ − γ
73.4◦ = 90◦ − γ
γ = 16.6◦
ANS:
γ = 0.290 in radians
Problem 9.59 An infinitesimal rectangle at a point in a
reference state of a material is shown. In a deformed state
the extensional strains in the dL1 and dL2 directions are
ε1 = 0.04 and ε2 = −0.02 and the shear strain referred
to the dL1 and dL2 is γ = 0.02. What is the extensional
strain in the dL direction? (See Example 9.6)
Solution:
Using the derivation from Example 2.6, the equation we will use is:
ε=
π
(1 + ε1 )2 cos2 θ + (1 + ε2 )2 sin2 θ − 2 (1 + ε1 ) (1 + ε2 ) (cos θ) (sin θ) cos
+ γ −1
2
Substituting the given values for ε1 , ε2 , and γ:
ε = (1 + 0.04)2 cos2 (40◦ ) + (1 − 0.02)2 sin2 (40◦ ) − 2 (1 + 0.04) (1 − 0.02) cos (40◦ ) sin (40◦ ) cos π2 + 0.02
ε = 0.0255 − 1
Problem 9.60 For the infinitesimal rectangle at a point
in a reference state of a material shown in Problem X.XX,
suppose that in a deformed state the extensional strains
in the dL1 , dL2 , and dL directions are ε1 = 0.030,
ε2 = 0.020, and ε = 0.038. What is the shear strain
referred to the dL1 and dL2 directions?
Solution:
We use the equation derived in Example 9.6:
π
(1 + ε1 )2 cos2 θ + (1 + ε2 )2 sin2 θ − 2 (1 + ε1 ) (1 + ε2 ) (cos θ) (sin θ) cos
+ γ −1
2
ε=
Solving this equation for γ:
cos
cos
π
2
π
2
+γ =
+γ =
−(1+ε)2 +(1+ε1 )2 (cos2 θ )+(1+ε2 )2 (sin2 θ )
2(1+ε1 )(1+ε2 )(cos θ)(sin θ)
−(1+2ε+ε2 )+(1+ε1 )2 (cos2 θ )+(1+ε2 )2 (sin2 θ )
2(1+ε1 )(1+ε2 )(cos θ)(sin θ)
Substituting the given values for ε, ε1 , and ε2 :
π
− 1 + 2 (0.038) + 0.0382 + (1 + 0.03)2 cos2 40◦ + (1 + 0.02)2 sin2 40◦
cos
+γ =
= −0.0242
2
2 (1 + 0.03) (1 + 0.02) (cos 40◦ ) (sin 40◦ )
π
+ γ = 91.39◦
2
γ = 1.39◦ = 0.0242
Problem 9.61 The bar is made of material 1-in. thick.
Its width varies linearly from 2 in. at its left end to 4 in.
at its right end. If the axial load P = 200 lb, what is the
average normal stress (a) at plane P1 ; (b) at plane P2 .
Free Body Diagram:
Solution:
The width of the material at P1 and P2 is:
W1 = 2 in + (1/3)(2 in) = 2.667 in
W2 = 2 in + (2/3)(2 in) = 3.333 in
The cross-sectional areas at P1 and P2 are:
A1 = (2.667 in)(1 in) = 2.667 in2
A2 = (3.333 in)(1 in) = 3.333 in2
The average normal stress at P1 and P2 are:
σ1 = (200 lb)/(2.667 in2 )
σ2 = (200 lb)/(3.333 in2 )
ANS:
ANS:
σ1 = 75 psi
σ2 = 60 psi
Problem 9.62 If the average normal stress at plane P1
of the bar described in Problem 9.61 is σAV = 300 psi,
what is the axial load P and the average normal stress at
plane P2 ?
Solution:
Free Body Diagrams:
The width of the material at P1 and P2 is:
W1 = 2 in+(2 in)(x/12 in) = 2 in+(2 in)(4 in/12 in) = 2.667 in
W2 = 2 in+(2 in)(x/12 in) = 2 in+(2 in)(8 in/12 in) = 3.333 in
The cross-sectional areas at P1 and P2 are:
A1 = W1 t = (2.667 in)(1 in) = 2.667 in2
A2 = W2 t = (3.333 in)(1 in) = 3.333 in2
Summing horizontal forces on the left-hand FBD:
ΣFx = 0 = −P + σAV G A1 = −P + (300 lb/in2 )(2.667 in2 )
ANS:
P = 800 lb
Summing horizontal forces on the right-hand FBD:
ΣFx = 0 = −P + σAV G A2 = −800 lb + (σAV G )2 (3.333 in2 )
ANS:
(σAV G )2 = 240 psi
Problem 9.63 The beam has cross-sectional area A =
0.1 m2 . What are the average normal stress and the magnitude of the average shear stress at the plane P ?
Free Body Diagram:
160 = 40(4)
2m
Ax
2m
Ay
20
By
Solution:
Summing horizontal forces:
ΣFx = 0 = 20 kN − Ax
Ax = 20 kN
Summing moments about the right-hand end of the beam:
ΣM = 0 = (40 kN/m)(4 m)(2 m) − Ay (4 m)
Ay = 80 kN ↑
Cut the beam through plane P and sum forces on the free body diagram:
ΣFy = 0 = −(80 kN)+Ay +τ (0.1 m2 ) = −(80 kN)+(80 kN)+τ (0.1 m2 )
ANS: τ = 0
Summing horizontal forces on the free body diagram:
ΣFx = 0 = −Ax + σA = −20, 000 N + σ(0.1 m2 )
ANS:
σ = 200, 000 N/m2 = 200 kN/m2 = 200 kPa
Problem 9.64 In Problem 9.63, what are the average
normal stress and the magnitude of the average shear
stress at the plane P if the plane is 1 m from the left end
of the beam?
Free Body Diagram:
160 = 40(4)
2m
Ax
2m
Ay
20
By
Solution:
Summing moments about the right-hand end of the beam in order to
find Ay :
ΣMB = 0 = (40, 000 N/m)(4 m)(2 m) − Ay (4 m)
Ay = 80, 000 N
Summing horizontal forces on the FBD:
ΣFx = 0 = −20, 000 N + σAV G (0.1 m2 )
ANS: σAV G = 200 kPa
Summing vertical forces on the FBD:
ΣFy = 0 = 40, 000 N − τAV G (0.1 m2 ) = 0
ANS:
τAV G = 400, 000 N/m2 = 400 kPa
Problem 9.65 The prismatic bar AB has cross- Free Body Diagram:
sectional area A = 0.01 m2 .
If the force F = 6 kN, what is the average normal stress
at the plane P ?
Solution:
Determine the axial load in the prismatic bar by summing vertical
forces at point B.
ΣFy = 0 = −6, 000 N + Paxial (cos 36.9◦ )
Paxial = 7502.95 N
The average normal stress at the plane, P , is:
σavg = (Paxial )/A = (7500 N)/(0.01 m2 )
σavg = −750, 000 Pa = −750 kPa NOTE: The negative
sign indicates a compressive stress.
ANS:
Problem 9.66 The prismatic bar AB in Problem 9.65 Free Body Diagram:
will safely support an average compressive normal stress
of 1.2 MPa on the plane P .
Based on this criterion, what is the largest downward
force F that can safely be applied?
Solution:
For an average compressive normal stress of 1.2 MPa at plane P , the
axial load in the prismatic bar is:
F = σavg A = (1.2 × 106 n/m2 )(0.01 m2 )
F = 12, 000 N = 12 kN
Now the load, F , can be found by summing vertical forces at point B.
ΣFy = 0 = (12 kN)(sin 53.1◦ ) − F
ANS:
F = 9.6 kN ↓
Problem 9.67 The jaws of the bolt cutter are connected
by two links AB.
The cross-sectional area of each link is 750 mm2 . What
average normal stress is induced in each link by the 90-N
forces exerted on the handles?
Free Body Diagram:
FBD Handle
FBD Jaw
Solution:
Inspection of the FBD of jaw A reveals that Cx must be zero. This
information is necessary in the analysis of the handle.
Summing moments about point D on the handle:
ΣMD = 0 = −(90 N)(0.54 m) + Cy (0.1 m)
Cy = 486 N
Summing moments about point P on the jaw:
ΣMP = 0 = Cy (0.24 m)−Ey (0.08 m) = (486 N)(0.24 m)−Ey(0.08 m)
Ey = 1458 N
Here we must realize that there are TWO links connecting the jaws of
the bolt cutter, so HALF of the force Ey is exerted by each of the links.
The average normal stress in link AB is:
σavg = (0.5)Ey /A = (0.5)(1458 N)/(750 × 10−6 m2 )
ANS:
σavg = 972, 000 Pa = 972 kPa
Problem 9.68 The pins connecting the links AB to
the jaws of the bolt cutter in Problem 9.67 are 20-mm
in diameter. What average shear stress is induced in the
bolts by the 90-N forces exerted on the handles?
Free Body Diagrams:
Solution:
From the solution to Problem 9.67 we have seen that the total downward
force on the upper jaw is 1458 N. We see that the shear load produced
in the pin by this downward force is divided between the two portions
of the pin through jaw A which extend from the jaw into the connecting
links.
The average shear stress in each of the pins is:
τavg = (1/2)(1458 N)/[π(0.01 m)2 ]
ANS:
τavg = 2, 320, 000 Pa = 2.32 MPa
Problem 9.69 Suppose that you subject a 2-m prismatic bar to compressive axial forces that cause a uniform extensional strain ε = −0.003 in the axial direction. What is the bar’s length in the deformed state?
Solution:
Using Equation 9.3 from the text:
L = (1 + ε)L
L = (1 − 0.003)(2m)
L = 1.994m
Problem 9.70 A prismatic bar is subjected to loads that
cause uniform axial strains ε = 0.002 in its left half and
ε = −0.004 in its right half.
What is the resulting change in length of the 28-in. bar?
Free Body Diagram:
Solution:
The change in length over the left half of the prismatic bar is:
δL = (0.002)(14 in) = 0.028 in
The change in length of the right half of the prismatic bar is:
δR = (−0.004)(14 in) = −0.056 in
Total change in length for the prismatic bar is:
δ = δL + δR = 0.028 in − 0.056 in
ANS:
δ = −0.028 in
Problem 9.71 The prismatic bar in Problem 9.70 is
subjected to loads that cause a uniform axial strain εL =
0.006 in its left half and a uniform axial strain εR = in
its right half. As a result, the length of the 28-in. bar
increases by 0.032 in. What is εR ?
Solution:
Total change in length of the deformed bar is:
∆L = 0.032 = εL (14) + εR (14)
0.032 = 0.006(14) + εR (14)
Solving for εR :
ANS: εR = −0.00371
Problem 10.1 A prismatic bar with cross-sectional
area A = 0.1 m2 is loaded at the ends in two ways:
(a) by 100-Pa uniform normal tractions; (b) by 10-N axial
forces acting at the centroid of the bar’s cross section.
What are the normal and shear stress distributions at the
plane P in the two cases?
Diagram:
Solution:
(a) Because the applied loads are in the axial direction and the plane
is normal to the axis of the bar, there is no shear stress. In the first
case, the applied load in a uniform normal traction, so the normal stress
MUST be:
ANS: σ = 100 Pa
As stated above, since the applied load is in the axial direction and the
plane is normal to the axis of the bar:
ANS: τ = 0
(b) In the second case, the normal stress distribution is:
σ = P/A = (10 N)/(0.1 m2 )
ANS: σ = 100 Pa
As stated above, since the applied load is in the axial direction:
ANS: τ = 0
Problem 10.2 A prismatic bar with cross-sectional
area A = 4 in2 is subjected to tensile axial loads P . It
consists of a material that will safely support a tensile
normal stress of 60 ksi. Based on this criterion, what is
the largest safe value of P ?
Diagram:
Solution:
Using the definition of normal stress:
PALLOW = σALLOW A = (60, 000 lb/in2 )(4 in2 )
ANS:
PALLOW = 240, 000 lb = 240 kip
Problem 10.3 A prismatic bar has a solid circular
cross section with 20-mm diameter. It consists of a material that will safely support a tensile normal stress of
300 MPa. Based on this criterion, what is the largest
tensile load P to which the bar can be subjected?
Free Body Diagram:
Solution:
The cross-sectional area of the bar is:
A = πr 2 = π(0.01 m)2 = 3.142 × 10−4 m2
For a maximum normal stress of 300 × 106 Pa:
300 × 106 N/m2 =
ANS:
PM AX
3.142 × 10−4 m2
PMAX = 94.2 kN
Problem 10.4 The cross-sectional area of bar AB is
0.5 in2 . If the force F = 3 kips, what is the normal
stress on a plane perpendicular to the axis of bar AB?
Free Body Diagrams:
Solution:
Summing moments about point C:
ΣMC = 0 = (3000 lb)(6 ft) − FAB (4 ft)
FAB = 4500 lb (T)
The normal stress in member AB is:
σAB = FAB /AAB = (4500 lb)/(0.5 in2 )
ANS:
σAB = 9000 psi = 9 ksi
Problem 10.5 Bar AB of the frame in Problem 10.4 Free Body Diagram:
consists of material that will safely support a tensile normal stress of 20 ksi. If you want to design the frame to
support forces F as large as 8 kip, what is the minimum
required cross-sectional area of bar AB?
Solution:
The maximum safe load which can be supported by member AB is:
[1]
(FAB )MAX = σALLOW AAB = (20, 000 lb/in2 )(AAB )
Summing moments about point C:
ΣMC = 0 = F (6 ft)−(FAB )MAX (4 ft) = (8000 lb)(6 ft)−(FAB )MAX (4 ft)
(FAB )MAX = 12, 000 lb
Using Equation [1] to find the cross-sectional area of member AB:
12, 000 lb = (20, 000 lb/in2 )(AAB )
ANS:
AAB = 0.6 in2
Problem 10.6
Free Body Diagram:
The mass of the suspended box is 800 kg. The mass
of the crane’s arm (not including the hydraulic actuator
BC) is 200 kg, and its center of mass is 2 m to the right
of A. The cross-sectional area of the upper part of the
hydraulic actuator is 0.004 m2 . What is the normal stress
on a plane perpendicular to the axis of the upper part of
the actuator?
Solution:
The weight of the box is:
WB = mg = (800 kg)(9.81 m/sec2 ) = 7848 N
The weight of the crane’s arm is:
WARM = mg = (200 kg)(9.81 m/sec2 ) = 1962 N
To get the axial load in the actuator, sum moments about point A.
ΣMA = 0 = −WB (7 m)−WARM (2 m)+FACT (sin 63.4◦ )(3 m)
= −(7848 N)(7 m)−(1962 N)(2 m)+FACT (sin 63.4◦ )(3 m)−FACT cos 63.4(1.4 m)
FACT = 28, 634 N (C)
The normal stress on the upper portion of the actuator is:
σACT = FACT /AACT = (28, 634 N)/(0.004 m2 )
ANS:
σACT = 7.15 MPa
Problem 10.7 The cross-sectional area of the lower
part of the hydraulic actuator in Problem 10.6 is
0.010 m2 . What is the normal stress on a plane perpendicular to the axis of the lower part of the actuator?
Free Body Diagram:
Solution:
Summing moments about point A:
ΣMA = 0 = −(800kg)(9.81 m/sec2 )(7 m)−(200kg)(9.81 m/sec2 )(2 m)+FBC (cos 63.4◦ )(3 m)−FBC (cos 63.4◦ )(1.4 m)
FBC = 28, 634 N
The normal stress on the lower part of the actuator is:
σ=
FBC
28, 634 N
=
Note: The negative sign indicates compression.
A
0.010 m2
ANS:
σ = −2.86 MPa
Problem 10.8 The cross-sectional area of each bar is
A. What is the normal stress on a plane perpendicular
to the axis of one of the bars?
Free Body Diagram:
Solution:
Because the support is symmetrical, the axial load is the same in each
member. Draw the FBD where the load is applied and sum vertical
forces.
ΣFy = 0 = −F + 2[FAXIAL (sin β)]
FAXIAL = F/(2)(sin β)
The normal stress in the support members is:
σAXIAL = FAXIAL /A = [F/(2)(sin β)]/A
ANS:
σAXIAL = F/[2A sin β]
Problem 10.9 The angle β of the system in Prob- Free Body Diagram:
lem 10.8 is 60◦ . The bars are made of a material that will
safely support a tensile normal stress of 8 ksi. Based on
this criterion, if you want to design the system so that it
will support a force F = 3 kip, what is the minimum
necessary value of the cross-sectional area A?
Solution:
The maximum load which can be safely supported by EACH of the
support members is:
FMAX = σMAX (A) = (8000 lb/in2 )(A)
Summing vertical forces on the FBD:
ΣFy = 0 = −3000 lb + 2(8000 lb/in2 )(A)(sin 60◦ )
ANS:
A = 0.217 in2
Problem 10.10 Suppose that the horizontal distance
between the supports of the system in Problem 10.8 and
the load F are specified, and the prismatic bars are made
of a material that will safely a tensile normal stress σ0 .
You want to choose the angle β and the cross-sectional
area A of the bars so that the total volume of material
used is a minimum. What are β and A?
Free Body Diagram:
L = d/(cos β)
[1]
Solution:
Summing vertical forces on the FBD to find the force supported by the
two prismatic bars:
ΣFy = 0 = −F +2 [σ0 A(sin β)] where σ0 is the maximum allowable average normal stress.
From the above equation, the cross-sectional area of one of the prismatic bars may be expressed as:
A=
F
2σ0 (sin β)
[2]
Recall: sin 2β = 2 sin β cos β
Using Equations [1] and [2], the volume of a single bar is:
V = AL =
Fd
Fd
=
2σ0 sin β cos β
σ0 sin 2β
We see that the volume will be minimum when sin2β is maximum,
which is when sin 2β = 1, so:
ANS: β = 45◦ [3]
Using this value for β in Equation [2] to find A:
A=
ANS:
A = 0.707 σF
0
F
2σ0 (0.707)
Problem 10.11 The cross-sectional area of each bar
is 60 mm2 . If F = 4 kN, what are the normal stresses
on planes perpendicular to the axes of the bars?
Free Body Diagram:
Solution:
Summing horizontal forces on the FBD:
ΣFx = 0 = −FAB (cos 60◦ ) + FAC (cos 45◦ )
FAB = 1.414FAC
[1]
Summing vertical forces on the FBD:
ΣFy = 0 = −40, 000 N + FAB (sin 60◦ ) + FAC (sin 45◦ )
Substituting Equation [1] into Equation [2]:
40, 000 N = [1.414FAC ](sin 60◦ ) + FAC (sin 45◦ )
FAC = 20, 708 lb
Therefore:
FAB = 29, 281 lb
The normal stresses in the two supporting members are:
σAB = FAB /AAB = (29, 281 lb)/(60 × 10−6 m2 )
ANS:
σAB = 488 MPa
σAC = FAC /AAC = (20, 708 lb)/(60 × 10−6 m2 )
ANS:
σAC = 345 MPa
[2]
Problem 10.12 The bars of the truss in Problem 10.11
are made of material that will safely support a tensile
normal stress of 600 MPa. Based on this criterion, what
is the largest safe value of the force F ?
Free Body Diagram:
Solution:
Summing horizontal forces on the FBD:
ΣFx = 0 = −FAB (cos 60◦ ) + FAC (cos 45◦ )
FAB = 1.414FAC
[1]
Summing vertical forces on the FBD:
ΣFy = 0 = −F + FAB (sin 60◦ ) + FAC (sin 45◦ )
[2]
Substituting Equation [1] into Equation [2]:
F = [1.414FAC ](sin 60◦ ) + FAC (sin 45◦ )
FAC = 0.5177F
Therefore:
FAB = 0.732F
The normal stresses in the two supporting members are:
σAB = FAB /AAB = (0.732F )/(60×10−6 m2 ) = 600×106 N/ m2
(FMAX )AB = 49, 180 N
σAC = FAC /AAC = 0.5177F )/(60×10−6 m2 ) = 600×106 N/ m2
(FMAX )AC = 69, 538 N
We see that member AB will fail if subjected to (FMAX )AC , so the
maximum allowable load is:
ANS: FMAX = 49, 180 N = 49.2 kN
Problem 10.13 The cross-sectional area of each bar of
the truss is 400 mm2 . If F = 30 kN, what is the normal
stress on a plane perpendicular to the axis of member
BE?
Free Body Diagram:
Solution:
To find the axial load in member BE directly, sum vertical forces on
the FBD.
ΣFy = 0 = −30, 0000 N + FBE (sin 45◦ )
FBE = 42, 400 N (T)
The normal stress in member BE is:
σBE = FBE /ABE = (42, 400 N)/(400 × 10−6 )
ANS:
σBE = 106 MPa
Problem 10.14
In Problem 10.13, what is the normal stress on a plane
perpendicular to the axis of member BC?
Free Body Diagram:
Fig (A)
Fig (B)
Solution:
Summing vertical forces at point A:
ΣFY = 0 = −30, 000 N + PAC (sin 45◦ )
PAC = 42, 426 N (T)
Summing vertical forces at point C:
ΣFY = 0 = −PAC (sin 45◦ )+PBC = −(42, 426 N)(sin 45◦ )+PBC
PBC = 30, 000 N(C)
The normal stress in the lower part of the actuator is:
σ=
ANS:
30, 000 N
PBC
=
ABC
400 × 10−6 m2
σ = −75 MPa NOTE: (−) indicates compressive stress.
Problem 10.15 The truss in Problem 10.13 is made of
a material that will safely support a normal stress (tension
or compression) of 340 MPa. Based on this criterion,
what is the largest safe value of the force F ?
Solution:
We must first determine the axial load in EACH member of the truss.
Summing vertical forces at joint A:
ΣFy = 0 = −F + FAC (sin 45◦ )
FAC = 1.414F (T )
Summing horizontal forces at joint A:
ΣFx = 0 = FAB − FAC (cos 45◦ ) = FAB − (1.414F )(cos 45◦ )
Fig (A)
FAB = F (C)
Summing vertical forces at joint C:
ΣFy = 0 = FBC − FAC (sin 45◦ ) = FBC − (1.414F )(sin 45◦ )
FBC = F (C)
Summing horizontal forces at joint C:
ΣFx = −FCE + FAC (cos 45◦ ) = −FCE + (1.414F )(cos 45◦ )
FCE = F (T)
Fig (B)
Summing horizontal forces at joint E:
ΣFx = 0 = FCE + FBE (cos 45◦ ) = F + FBE (cos 45◦ )
FBE = 1.414F (C)
Summing vertical forces at joint E:
+ ↓ ΣFy = 0 = −FDE +FBE (sin 45◦ ) = −FDE −1.414F sin 45◦
FDE = F (T)
Summing moments about joint E:
Fig (C)
ΣME = 0 = −F (0.5 m) + Dx (0.25 m)
Ey
Dx = 2F
Therefore:
FBD = 2F (C)
Ex
We see that member BD is carrying the greatest load (2F ), and so we
recognize that this member controls the safe load which can be carried
by the truss. The largest load which can be safely carried by member
BD is:
FBD = σALLOW A = (340×106 N/ m2 )(400×10−6 m2 ) = 2F
ANS:
FMAX = 68, 000 N = 68 kN
Dx
F
Problem 10.16 The cross-sectional area of the pris- Free Body Diagram:
matic bar is A = 2 in2 and the axial force P = 20 kip.
Determine the normal and shear stresses on the plane P .
Draw a diagram isolating the part of the bar to the right
of plane P and show the stresses.
Solution:
The area of plane P is:
AP = (2 in2 )/(cos 70◦ )
AP = 5.848 in2
Summing vertical forces on the FBD:
ΣFy = 0 = −σAP (sin 70◦ ) + τ AP (cos 70◦ )
[1]
τ = 2.75σ
Summing horizontal forces on the FBD:
[2]
ΣFx = 0 = 20, 000 lb − σAP (cos 70◦ ) − τ AP (sin 70◦ )
Substituting Equation [1] into Equation [2]:
20, 000 lb = σAP (cos 70◦ )+(2.75σ)AP (sin 70◦ ) = σ(5.848 in2 )(cos 70◦ )+(2.75σ)(5.848 in2 )(sin 70◦ )
ANS: σ = 1169 psi
Using the determined value of σ in Equation [1]:
τ = 2.75σ = −3215 psi
τ = 2.75σ = −3215 psi Note: The negative sign conforms to the sign convention given in Figure 3.15
ANS:
Problem 10.17
Free Body Diagram:
If the normal stress on the plane P in Problem 10.16 is
6000 psi, what is the axial force P ?
Solution:
The area of plane P is:
AP = (2 in2 )/(cos 70◦ )
AP = 5.848 in2
Summing vertical forces on the FBD:
ΣFy = 0 = −σAP (sin 70◦ ) + τ AP (cos 70◦ )
[1]
τ = 2.75σ
Summing horizontal forces on the FBD:
[2]
ΣFx = 0 = P − σAP (cos 70◦ ) − τ AP (sin 70◦ )
Subtsituting Equation [1] into Equation [2]:
P = σAP (cos 70◦ )+(2.75σ)(AP )(sin 70◦ ) = (6000 lb)(5.848 in2 )(cos 70◦ )+(2.75)(6000 lb)(5.848 in2 )(sin 70◦ )
ANS:
P = 102, 600 lb = 102.6 kip
Problem 10.18 The cross-sectional area of the prismatic bar is 0.02 m2 . If the normal and shear stresses
on the plane P are σθ
=
1.25 MPa and τθ =
−1.5 MPa, what are the angle θ and the axial force P ?
Free Body Diagram:
θ
Solution:
The area of the slanted plane is:
AP = A/ cos θ
Summing vertical forces on the FBD:
ΣFy = 0 = τθ AP (cos θ) − σθ AP (sin θ)
τ0
=σ
0 (tan θ)
We can determine the magnitude of θ:
τθ
tan θ =
= (1.5 × 106 P a)/(1.25 × 106 Pa) = 1.2
σθ
ANS:
θ = 50.2◦
So the area of the plane is:
AP = A/ cos θ = (0.02 m2 )/(cos 50.2◦ )
AP = 0.0312 m2
Summing horizontal forces on the FBD:
ΣFx = 0 = P − τθ AP (sin θ) − σθ AP (cos θ)
P = (1.5×106 N/ m2 )(0.0312 m2 )(sin 50.2◦ )+(1.25×106 N/ m2 )(0.0312 m2 )(cos 50.2◦ )
ANS:
P = 61 kN
Problem 10.19 The cross-sectional area of the bar is
A = 0.5 in2 and the force F = 3000 lb. Determine the
normal stresses and the magnitudes of the shear stresses
on the planes (a) and (b).
Solution:
(a) The area of the plane is:
AP = A/ cos 30◦ = 0.5 in2 / cos 30◦
AP = 0.577 in2
To find the axial load in each member (note the symmetry), sum vertical
forces at the point where the load is applied.
ΣFy = 0 = −3000 lb + 2[P (sin 60◦ )
Fig (A)
P = 1732 lb
Summing forces in the y-direction (note the direction of the x- and
y-axes) on the FBD:
ΣFx = 0 = σAP (sin θ)−τ AP (cos θ) = σ(0.577 in2 )(sin 30◦ )−τ (0.577 in2 )(cos 30◦ )
[1]
τ = 0.577σ
Summing forces in the x-direction on the FBD:
ΣFx = 0 = 1732 lb − σAP (cos 30◦ ) − τ AP (sin 30◦ )
2
[2]
Fig (B)
2
1732 lb = σ(0.577 in )(0.866) + τ (0.577 in )(0.5)
Substituting Equation [1] into Equation [2]:
1732 lb = σ(0.577 in2 )(0.866) + (0.577σ)(0.577 in2 )(0.5)
ANS: σ = 2600 psi
Using this result in Equation [1]:
τ = 0.577σ
ANS: τ = 1500 psi
(b) The area of the plane is:
AP = A/ cos 60◦ = 0.5 in2 / cos 60◦
AP = 1 in2
In part (a) above, the axial load in the member was found to be P =
1732 lb. Summing forces in the y-direction (note the directions of the
x- and y-axes) on the FBD:
ΣFy = 0 = σAP (sin 60◦ ) − τ AP (cos 60◦ )
[1]
τ = 1.732σ
Summing forces in the x-direction:
ΣFx = 0 = −1732 lb + σAP (cos 60◦ ) + τ AP (sin 60◦ )
[2]
1732 lb = σ(1 in2 )(0.5) + τ (1 in2 )(0.866)
Substituting Equation [1] into Equation [2]:
1732 lb = σ(1 in2 )(0.5) + (1.732σ)(1 in2 )(0.866)
ANS: σ = 866 psi
Using this result in Equation [1]:
τ = 1.732σ
ANS:
τ = 1500 psi
Fig (C)
Problem 10.20 The truss in Problem 10.19 is constructed of a material that will safely support a normal
stress of 8 ksi and a shear stress of 3 ksi. Based on these
criteria, what is the largest force F that can safely be
applied
Solution:
We have seen from the text that maximum normal stress occurs when
θ = 0 and that maximum shear stress occurs when θ = 45◦ . Determining the axial load in each member of the truss while supporting the
load F :
ΣFy = 0 = −F + 2[P (sin 60◦ )]
P = F/(2 sin 60◦ )
[1]
P = 0.577F
Using the maximum normal stress as the design criterion:
Fig (A)
8000 lb/in2 = PMAX /0.5 in2
PMAX = 4000 lb
Using this value of PMAX in Equation [1]:
FMAX = PMAX /0.577 = 4000 lb/0.577
FMAX = 6932 lbThis is the load which produces the limiting normal stress.
Using the maximum shear stress as the design criteria, we pass a plane
through the cross-section at an angle of 45◦ .
The area of the plane is:
AP = A/ sin 45◦ = (0.5 in2 )/(0.707)
AP = 0.707 in2
Summing forces in the x-direction on the FBD:
ΣFx = 0 = σAP (cos 45◦ ) − τ AP (sin 45◦ )
[2]
σ=τ
Summing forces in the y-direction on the FBD:
[3]
Fig (B)
ΣFy = 0 = −P + σAP (sin 45◦ ) + τ AP (cos 45◦ )
Substituting Equation [1] into Equation [2]:
P = τ AP (0.707)+τ AP (0.707) = τ (0.707 in2 )(0.707)+τ (0.707 in2 )(0.707)
P = τ lb
Using the limiting shear stress as the design criterion:
[4]
P = 3000 lb
Now using Equation [1] in Equation [4]:
FM AX = P/0.577 = 3000 lb/0.577
ANS:
FMAX = 5196 lb
Fig (C)
Problem 10.21 Two marks are made 2 inches apart
on an unloaded bar. When the bar is subjected to axial
forces P , the marks are 2.004 inches apart. What is the
axial strain of the loaded bar?
Solution:
From the definition of strain:
ε=
L −L
L
=
2.004 in−2.000 in
2.000 in
ε = 0.002
Problem 10.22 The total length of the unloaded bar in
Problem 10.21 is 10 in. Use the result of Problem 10.21
to determine the total length of the loaded bar. What
assumption are you making when you do so?
Solution:
From the definition of strain:
ε=
L − L
2.004 in − 2.000 in
=
L
2.000 in
ε = 0.002
Total length of the deformed bar will be:
L = L(1 + ε) = (10 in)(1 + 0.002)
ANS:
L = 10.02 in
Problem 10.23 If the forces exerted on the bar in Problem 10.21 are P = 20 kip and the bar’s cross-sectional
area is A = 1.5 in2 , what is the modulus of elasticity of
the material?
Solution:
From Problem 10.21 we determined that the observed strain is 0.002.
The normal stress in the bar is:
σ=
P
20, 000 lb
=
= 13, 333 psi
A
1.5 in2
From the definition of the modulus of elasticity:
E=
σ
ε
=
13,333 lb/in2
0.002
E = 6.67 × 106 psi
Problem 10.24 A prismatic bar with length L = 6 m
and a circular cross section with diameter D = 0.02 m
is subjected to 20-kN compressive forces at its ends.
The length and diameter of the deformed bar are measured and determined to be L = 5.940 m and D =
0.02006 m. What are the modulus of elasticity and Poisson’s ration of the material?
Solution:
The strain in the bar is:
ε=
L − L
5.94 m − 6.0 m
=
= −0.01
L
6.0 m
The compressive stress in the bar is:
P
−20, 000 N
=
= −63.7 MPa
A
π(0.01 m)2
σ=
The modulus of elasticity for the material is:
E=
σ
63.7 MPa
=
ε
0.01
ANS: E = 6.37 GPa
Poison’s ration for the material is:
υ=
−εLAT
ε
=
(D −D)/D
ε
=
−(0.02006 m−0.02 m)/0.02 m
−0.01
υ = 0.3
Problem 10.25
The bar has modulus of elasticity E = 30 × 106 psi and
Poisson’s ration ν = 0.32. It has a circular cross section
with diameter D = 0.75 in. What compressive force
would have to be exerted on the right end of the bar to
increase its diameter to 0.752 in?
Solution:
From the definition of Poisson’s ratio:
0.32 =
−(0.752 in − 0.75 in)/0.75 in
ε
The strain which will be produced by the applied load is:
σ
P
1
ε=
=
2
2
6
E
π(0.375 in)
30 × 10 lb/in
Substituting the above expression for the strain into the expression for
Poisson’s ratio:
0.32 =
ANS:
−(0.752 in − 0.75 in)/0.75 in
(P/(π(0.375 in)2 )) 30×1061 lb/in2
P = 110.4 kip
Problem 10.26 What tensile force would have to be
exerted on the right end of the bar in Problem 10.25 to
increase its length to 9.02 in.? What is the bar’s diameter
after this load is applied?
Solution:
The strain in the bar will be:
ε=
L − L
9.02 in − 9.00 in
=
= 0.00222
L
9.00 in
The stress required to produce this strain is:
σ = Eε = (30 × 106 lb/in2 )(0.00222) = 66, 667 lb/in2
The load required to produce the stress is:
π(0.75 in)2
P = σA = (66, 667 lb/in2 )
4
ANS: P = 29.5 kip
The radial strain in the deformed bar is:
εLAT = εν = (0.00222)(−0.32) = 7.104 × 10−4
The diameter of the deformed bar is:
D = D(1 − εLAT ) = (0.75 in)(1 − 7.104 × 10−4 )
ANS:
D = 0.7495 in
Problem 10.27 A prismatic bar is 300 mm long and
has a circular cross section with 20-mm diameter. Its
modulus of elasticity is 120 Gpa and its Poisson’s ratio
is 0.33. Axial forces P are applied to the ends of the
bar which cause its diameter to decrease to 19.948 mm.
(a) What is the length of the loaded bar? (b) What is the
value of P ?
Solution:
We can use Poisson’s ratio to determine the extensional strain in the
material.
0.33 =
−(19.948 mm − 20 mm)/(20 mm)
→
ε
ε = 0.0079
(a) The length of the loaded bar is:
L = L(1 + ε) = (300 mm)(1 + 0.0079)
L = 302.37 mm
(b) The value of P is determined using the definition of the modulus
of elasticity.
ANS:
P = εEA = (0.0079)(120 × 109 N/m2 )(π)(0.01 m)2
ANS:
P = 297.8 kN
Problem 10.28 When unloaded, bars AB and AC are
each 36 in. in length and have a cross sectional area of
2 in2 . Their modulus of elasticity is E = 1.6 × 106 psi.
When the weight W is suspended at A, bar AB increases
its length by 0.01 in. What is the change in length of bar
AC?
FAB
30˚
60˚
20˚
FAC
W
Given the strain in member AB, the load W can be determined.
Solution:
We start by establishing the relationship between the loads in member
AB and AC.
Summing vertical forces at point A:
ΣFy = 0 = −W + FAB (cos 30◦ ) + FAC (sin 20◦ )
W = 0.866FAB + 0.342FAC
1.6 × 106 psi =
W = 9317 lb
FAC = 4733 lb (C)
The strain in member AC will be:
◦
ΣFx = 0 = FAC (cos 20 ) − FAB (sin 30 )
δAC =
[2]
(4733 lb)(36 in)
PL
=
= −0.532
AE
(2 in2 )(1.6 × 106 psi)
εAC = −0.532 Note: The negative strain results from the compressive load.
Solving Equations [1] and [2] together:
FAB = 0.954W (T)
FAC = 0.508W (C)
Problem 10.29 If a weight W = 12, 000 lb is suspended from the truss in Problem 10.28, what are the
changes in length of the two bars?
FAB
30˚
Solution:
Establish the relationship between the loads in member AB and AC.
Summing vertical forces at point A:
ΣFy = 0 = −12, 000 lb + FAB (cos 30◦ ) + FAC (sin 20◦ )
12, 000 lb = 0.866FAB + 0.342FAC
20˚
W = 12,000
[1]
Summing horizontal forces at point A:
ΣFx = 0 = FAC (cos 20◦ ) − FAB (sin 30◦ )
[2]
Solving Equations [1] and [2] together:
FAB = 0.954(12, 000 lb) = 11, 448 lb (T)
FAC = 0.508(12, 000 lb) = 6, 096 lb (C)
The stresses in the two bars are:
σAB =
FaB
11, 448 lb
=
= 5, 724 lb/in2
A
2 in2
σAC =
FAC
6, 096 lb
=
= 3, 048 lb/in2
A
2 in2
Strains in the two bars are:
εAB =
σAB
5, 724 lb/in2
=
= 0.00358
E
1.6 × 106 lb/in2
εAC =
σAC
−3, 048 lb/in2
=
= −0.00191
E
1.6 × 106 lb/in2
Changes in length for the two bars are:
δAB = εAB LAB = (0.00358)(36 in)
ANS:
δAB = 0.1288 in
0.954W/(2 in2 )
(0.1 in/36 in)
FAC = 0.508W = (0.508)(9317 lb)
[1]
◦
FAC = 0.532FAB
=
The load in member AC is:
Summing horizontal forces at point A:
FAC = 0.532FAB
FAB /A
ε
δAC = εAC LAC = (−0.00191)(36 in)
δAC = −0.0688 in
Problem 10.30 Bars AB and AC are each 300 mm
in length, have a cross-sectional area of 500 mm2 , and
have modulus of elasticity E = 72 Gpa. If a 24 kN
downward force is applied at A, what is the resulting
Solution:
displacement of point A?
FAB = P
FAC = P
30˚
Because the truss is symmetrical, the displacement will be ALL VERTICAL. Summing vertical forces at point A:
30˚
24,000 N
ΣFy = 0 = −24, 000 N + 2 [(P ) (sin 30◦ )]
P = 24, 000 N
Knowing the load in each member, we can calculate the strain in each
member.
24,000 N/(500×10−6 m2 )
72×109 N/m2
P/A
ε= E =
ε = 66.7 × 10−5
The new length for each of the two truss members is:
L = (300 mm)(1 + 66.7 × 10−5 ) = 300.2 mm
The new vertical distance from the overhead to point A is:
d = (300.2 mm)2 − [(300 mm) (cos 30◦ )]2 = 150.4 mm
The original vertical distance from the overhead to point A was:
d0 = (300 mm)2 − [(300 mm) (cos 30◦ )]2
d0 = 150 mm
The vertical displacement is:
D = d − d0 = 150.4 mm − 150 mm
D = 0.4 mm ↓
ANS:
Problem 10.31 Bars AB and AC of the truss shown in
Problem 10.30 are each 300 mm in length, have a crosssectional area of 500 mm2 , and are made of the same
material. When a 30-kN downward force is applied at
Solution:
point A, it deflects downward 0.4 mm. What is the
Summing vertical forces at point A:
modulus of elasticity of the material?
FAB = P
ΣFy = 0 = −30, 000 N + 2 [(P ) (sin 30◦ )]
P = 30, 000 N
The stress in each of the members is:
σ=
P
30, 000 N
=
= 60 × 106 N/m2 = 60 MPa
A
500 × 10−6 m2
The deformed length of each of the bars is:
L = [(0.3 m) (cos 30◦ )]2 + [(0.3 m) (sin 30◦ ) + 0.0004 m]2 = 0.3002 m
The strain in each of the bars is:
ε=
L − L
.3002 m − 0.3 m
=
= 6.667 × 10−4
L
0.3 m
The modulus of elasticity is:
E=
ANS:
E = 90 Gpa
σ
60 × 106 N/m2
=
ε
6.667 × 10−4
FAC = P
30,000 N
Problem 10.32
Bar AB has cross-sectional area A = 100 mm2 and
modulus of elasticity E = 102 Gpa. The distance H =
400 mm. If a 200-kN downward force is applied to
bar CD at D, through what angle in degrees does bar
CD rotate? (You can neglect the deformation of bar
CD.) Strategy: Because the bar’s change in length is
small, you can assume that the downward displacement
v of point B is vertical, and that the angle (in radians)
through which bar CD rotates is v/H.
Free Body Diagram:
Solution:
We can determine the angle of rotation by finding the vertical displacement at point B. Find the axial load in member AB by summing
moments about point C:
ΣMc = 0 = −(200 kN)(0.6 m) + FAB (sin 60◦ )(0.4 m)
FAB = 346.4 kN (C)
The strain in member AB will be:
P/A
ε= E =
ε = 0.034
(346,400 N)/(100×10−6 m)
102×109 N/m2
The original length of member AB is:
L = (300 mm)/(sin 60◦ ) = 346.3 mm
The new length of member AB is:
L = L(1 + ε) = (346.3 mm)(1 − 0.034) = 334.5 mm
The original height of point B is h = 300 mm.
The deformed height of point B is:
h = (L )2 − [300 mm/tan 60◦ ]2 = (334.5 mm)2 − (173.2 mm)2 = 286.2 mm
The change in vertical height at point B is:
v = h − h = 300 mm − 286.2 mm = 13.8 mm ↓
The angle, in radians, through which the bar CD rotates is:
ANS: θ = 13.8 mm/400 mm = 0.0345 radians = 1.98◦ clockwise
Problem 10.33 Bar AB in Problem 10.32 is made of
a material that will safely support a normal stress (in
tension or compression) of 5 GPa. Based on this criterion, through what angle in degrees can bar CD safely
be rotated relative to the position shown.
Free Body Diagram:
Solution:
Maximum allowable strain in the material is:
ε=
σ
5 × 109 N/m2
=
= 0.049
E
102 × 109 N/m2
Maximum allowable change in length for H is:
∆H = Lε = (0.4 m) (.049) = 0.0196 m
In the diagram, maximum allowable distance d is:
d = (∆H)(cos 30◦ ) = (0.0196 m)(cos 30◦ )
d = 0.01697 m
The angle through which bar AB may rotate is:
θ=
ANS:
d
±0.01697 m
=
= ±0.049 rad
LAB
(0.3/ sin 60◦ )
θ = ±2.8◦
Problem 10.34 If an upward force is applied at H
that causes bar GH to rotate 0.02 degrees in the counterclockwise direction, what are the axial strains in bars
AB, DC, and EF ? (You can neglect the deformation
of bar GH.)
Solution:
The vertical displacement at point H is:
θ = (0.02◦ /180◦ )(3.14159 radians) = 349 × 10−6 radians
The vertical displacement of points B, D and F are:
VB = (349 × 10−6 radians)(400 mm) = 0.14 mm
VD = (349 × 10−6 radians)(800 mm) = 0.28 mm
VF = (349 × 10−6 radians)(1200 mm) = 0.42 mm
The strains in each of the vertical bars is:
mm
= 0.00035 εCD =
ANS: εAB = 0.14
400 mm
0.28 mm
400 mm
= 0.00070
εEF =
0.42 mm
400 mm
= 0.00105
Problem 10.35 The bar has cross-sectional area A and
modulus of elasticity E. The left end of the bar is fixed.
There is initially a gap b between the right end of the bar
and the rigid wall (Figure 1). The bar is stretched until
it comes into contact with the rigid wall and is welded
to it (Figure 2). Notice that this problem is statically
indeterminate because the axial force in the bar after it
is welded to the wall cannot be determined from statics
alone. (a) What is the compatibility condition in this
problem? (b) What is the axial force in the bar after it is
welded to the wall?
Solution:
The compatibility condition requires that the bar’s change in length
must be limited to the amount of the gap, b. Using the relationship
δ = P L/AE to find the axial load:
= bAE
ANS: P = δAE
L
L
Problem 10.36 The bar has cross-sectional area A and
modulus of elasticity E. If an axial force F directed toward the right is applied at C, what is the normal stress in
the part of the bar to the left of C? (Strategy: Draw the
free-body diagram of the entire bar and write the equilibrium equation. Then apply the compatibility condition
that the increase in length of the part of the bar to the left
of C must equal the decrease in length of the part to the
right of C.)
Free Body Diagram:
Solution:
Summing horizontal forces on the FBD:
[1]
ΣFx = 0 = F − RL − RR
The compatibility condition requires that the change in length of the
left portion of the bar must equal the change in length for the right
portion of the bar.
RL LL
RR LR
=
AL EL
AR ER
Since the denominators of the above equation are identical, we need
only consider the numerators.
[2]
RR = (LL /LR )(RL ) = [(L/3)/(2L/3)](RL ) = RL /2
Substituting Equation [2] into Equation [1]:
F = RL + RR = RL + (RL /2) = 3RL /2
or RL = (2/3)F
The stress in the left-hand portion of the bar is:
σ=
ANS:
σ = 2F/3A
RL
P
=
A
A
Problem 10.37 In Problem 10.36, what is the resulting Free Body Diagram:
displacement of point C?
Solution:
Substituting Equation [2] into Equation [1]:
We can use either the left-hand or right-hand portion of the bar to
determine the displacement of point C. The left-hand portion of the
bar is chosen.
Summing horizontal forces on the FBD:
[1]
ΣFx = 0 = F − RL − RR
F = RL + RR = RL + (RL /2) = 3RL /2
or RL = (2/3)F
The displacement of point C is:
The compatibility condition requires that the change in length of the
left portion of the bar must equal the change in length for the right
portion of the bar.
RL LL
RR LR
=
AL EL
AR ER
Since the denominators of the above equation are identical, we need
only consider the numerators.
[2]
RR = (LL /LR )(RL ) = [(L/3)/(2L/3)](RL ) = RL /2
Problem 10.38 The bar in Problem 10.36 has crosssectional area A = 0.005 m2 , modulus of elasticity
E = 72 GPa, and L = 1 m. It is made of a material that will safely support a normal stress (in tension
and compression) of 120 MPa. Based on this criterion,
what is the largest axial force that can be applied at C?
Free Body Diagram:
Solution:
We can use either the left-hand or right-hand portion of the bar to
determine the displacement of point C. The left-hand portion of the
bar is chosen.
Summing horizontal forces on the FBD:
[1]
ΣFx = 0 = F − RL − RR
The compatibility condition requires that the change in length of the
left portion of the bar must equal the change in length for the right
portion of the bar.
RL LL
RR LR
=
AL EL
AR ER
Since the denominators of the above equation are identical, we need
only consider the numerators.
[2]
RR = (LL /LR )(RL ) = [(L/3)/(2L/3)](RL ) = RL /2
Substituting Equation [2] into Equation [1]:
F = RL + RR = RL + (RL /2) = 3RL /2
or RL = (2/3)F
We see that the greater strain is in the left-hand portion of the bar
(RL > RR ). Using the given maximum allowable stress:
120 × 106 N/m2 =
ANS:
F = 900 kN
(2/3) F
(2/3) F
=
A
0.005 m2
PL
δ=
=
AE
ANS:
δ = 2F L/9AE
2 L
F
3
3
AE
Problem 10.39 In Example 3-7, determine the normal
stresses in parts A and B of the bar if the force applied at
the joint between parts A and B is (a) 40 kip; (b) 200 kip.
Free Body Diagram:
Solution:
(a) Assuming no reaction from the right-hand wall, total displacement
of the right-hand end of the bar is:
δ=
FL
(40, 000 lb) (10 in)
= 0.0106 in
= (2 in)2
AE
π 4
12 × 106 lb/in2
We see that the displacement is not sufficient to close the 0.02-in gap,
so there can be no reaction from the wall at the right.
The stress in the left-hand portion of the bar is:
σ=
P
40, 000 lb
=
(2 in)2
A
π 4
ANS: σ = 12, 732 lb/in2
(b) Again assuming no reaction from the right wall, we calculate the
displacement of the right-hand end of the bar.
δ=
PL
(200, 000 lb) (10 in)
= 0.053 in
= (2 in)2
AE
π 4
12 × 106 lb/in2
We see that this displacement is MORE than enough to close the 0.02in gap. The reaction at the right-hand wall must be sufficient to limit
the displacement of the right end of the bar to 0.02 in. This means that
the reaction at the right wall must be sufficient to displace the right end
of the bar a distance of:
δ = 0.053 in − 0.02 in = 0.033 in
The reaction at the right-hand wall is:
δ=
π
(R
R )(10 in)
(12×106 lb/in2 )
(2 in)2
4
+
(R
R )(8 in)
(4 in)2
π
(12×106 lb/in2 )
4
= 0.033 in
RR = 103, 672 lb
The stress in the right-hand portion of the bar is:
σR =
P
A
=
103,672 lb
π
σR = 8250psi
(4 in)2
4
Summing horizontal forces on the bar:
ΣFx = 0 = 200, 000 lb−RR −RL = 200, 000 lb−103, 672 lb−RL
RL = 96, 328 lb
The stress in the left hand portion of the bar is:
σL =
ANS:
σL = 30, 600 psi
P
96, 238 lb
=
(2 in)2
A
π 4
Problem 10.40 The bar has a circular cross section
and modulus of elasticity E = 70 GPa. Parts A and C
are 40 mm in diameter and part B is 80 mm in diameter.
If F1 = 60 kN and F2 = 30 kN, what is the normal
stress in part B?
Free Body Diagram:
Solution:
We must first determine the reactions at the left and right walls. We
allow the right-hand side of the bar to “float.”The displacement of the
“free” right-hand side of the bar is:
δR =
δR =
F1 L A
F2 L A
− AF2 LEB − A
AA EA
B B
A EA
(60,000N)(0.2 m)
π
(0.04 m)2
4
(70×106
N/m2 )
−
(30,000N)(0.4 m)
(0.08 m)2
π
(70×106 N/m2 )
4
−
(30,000N)(0.2 m)
(0.04 m)2
π
(70×106 N/m2 )
4
δR = 0.0341 m = 34.1 mm
The reaction at the right wall must be sufficient to prevent ANY displacement. Its magnitude is:
 
RR
0.2
m
0.4 m
 +
0.0341 m = 70×106 lb/in2  +
2
(0.08 m)2
π
(0.04 m)
4
π
4
0.2 m
π
(0.04 m)2
4


RR = 6000 N
The reaction at the left wall is:
ΣFx = 0 = 60, 000 N − 6, 000 N − 30, 000 N − RL
RL = 24, 000 N ←
The stress in section A is:
σB =
P
A
=
−24,000 N+60,000 N
π
(0.08 m)2
4
Note: The negative sign indicates a compressive stress.
ANS: σA = −7.16 MPa
Problem 10.41 In Problem 10.40, if F1 = 60 kN,
what force F2 will cause the normal stress in part C to
be zero?
Free Body Diagram:
Solution:
For the normal stress in section C to be zero, we see that the displacement of the intersection of sections B and C must be zero.
The equation for the displacement of the intersection of sections B and
C is:
δBC = 0 =
ANS:
(60, 000 N)(0.2 m)
F2 (0.2 m)
F2 (0.4 m)
−
−
2
2
2
9
2
9
2
π ((0.04 m) /4) 70 × 10 N/m
π ((0.04 m) /4) 70 × 10 N/m
π ((0.08 m) /4) 70 × 109 N/m2
F2 = 40, 000 N = 40 kN
Problem 10.42 The bar in Problem 10.40 consists of Free Body Diagram:
a material that will safely support a normal stress of
40 MPa. If F2 = 20 kN, what is the largest safe value
of F1 ?
Solution:
We see that section B has a cross-sectional area which is four times
that of sections A and C.
The cross-sectional areas are:
AA = AC = π (0.02 m)2 = 0.00126 m2
AB = π (0.04 m)2 = 0.00503 m2
With both ends of the bar restrained, total displacement must be zero.
R (0.2 m)
(F −R )(0.4 m)
(F −R −20,000 N)(0.2 m)
1
1
L
L
L
−
−
(0.00126 m2 )(70×109 n/m2 )
(0.00503 m2 )(70×109 N/m2 )
(0.00126 m2 )(70×109 N/m2 )
2.268 × 10−9 RL + 1.136 × 10−9 RL − 1.137 × 10−9 F1 + 2.268 × 10−9 RL − 2.268 × 10−9 F1 + 4.535 × 10−5 RL = 0
δR = 0 =
RL = 0.6F1 − 8, 000 N
Summing horizontal forces to find RR :
ΣFx = 0 = −RL +F1 −20, 000 N−RR = −0.6F1 +8004 N+F1 −20, 000 N−RR
RR = 0.4F1 − 12, 000 N
The axial loads in sections A, B and C are:
σa = 40 × 106 =
F1 = 97.3 kN
σb = 40 × 106 =
F1 = 523 kN
σc = 40 × 106 =
F1 = 156 kN
0.6f1 −8,000
0.00126
0.6f1 −8,000−F1
0.00503
0.4f1 −12,000
0.00126
The smallest of these three values for F1 is the highest allowable value.
ANS: F1 = 97.3 kN
Problem 10.43 Two aluminum bars (EAL = 10.0 × Free Body Diagram:
106 psi) are attached to a rigid support at the left and
a cross-bar at the right. An iron bar (EF E = 28.5 ×
106 psi) is attached to the rigid support at the left and
there is a gap b between the right end of the iron bar and
the cross-bar. The cross-sectional area of each bar is A =
0.5 in2 and L = 10 in. The iron bar is stretched until it
contacts the cross-bar and welded to it. Afterward, the
axial strain of the iron bar is measured and determined
to be εF E = 0.002. What was the size of the gap b?
Solution:
The foreshortening of the two aluminum bars plus the lengthening of
the steel bar must equal the gap, b. The lengthening of the steel bar is
(approximately):
δF E = (0.002)(10 in) = 0.02 in
Calculating the force in the steel bar:
PF E = εEA = (0.002)(28.5×106 lb/in2 )(0.5 in2 ) = 28, 500 lb
This same force is compressing the TWO aluminum bars. The aluminum bars are shortened by an amount of:
δAL =
PL
(28, 500 lb)(10 in))
= 0.0285 in
= AE
2 0.5 in2 10 × 106 lb/in2
The total original gap is:
b = δAL + δF E = 0.0285 in + 0.02 in
ANS:
b = 0.0485 in
Problem 10.44 In Problem 10.43, the iron will safely Free Body Diagram:
support a tensile stress of 100 ksi and the aluminum will
safely support a compressive stress of 40 ksi. What is
the largest safe value of the gap b?
Solution:
We see from the FBD that FF e = 2FAl .
Maximum allowable load in the aluminum bars is:
(σM AX )Al = 40, 000 lb/in2 =
FAl
→ (FAl )MAX = 20, 000 lb
0.5 in2
Maximum allowable load in the iron bar is:
(σM AX )F e = 100, 000 lb/in2 =
(FF e )M AX
→ (FF e )MAX = 50, 000 lb
0.5 in2
The maximum allowable stresses show that the allowable load in the
aluminum is the controlling criterion.
Using a load of 20, 000 lb in the aluminum and 40,000 in the iron:
b=
FAl L FF e L
(20, 000 lb)(10 in)
(40, 000 lb)(10 in)
+
=
+
AEAl AEF e
(0.5 in2 )(10 × 106 lb/in2 ) (0.5 in2 )(28.5 × 106 lb/in2 )
ANS:
b = 0.068 in
Problem 10.45 Bars AB and AC each have cross- Free Body Diagram:
sectional area A and modulus of elasticity E. If a downward force F is applied at A, show that the resulting
Fh
1
downward displacement of point A is EA
1+cos3 θ .
Solution:
Summing vertical forces at point A;
[1]
ΣFy = 0 = −F + PAB (cos θ) + PAC
We see that the VERTICAL movement of the A-end of member AB
will be:
[2]
vA = δAB (1/ cos θ)
From Equation [2], we see that the vertical movement of point A must
be the same for BOTH members of the truss, so:
h
1
PAB cos
PAC h
cos θ
=
AE
AE
[3]
PAB = PAC (cos2 θ)
Substituting Equation [3] into Equation [1]:
F = PAB (cos θ) + PAC
= PAC cos2 θ (cos θ) + PAC
F = PAC 1 + cos3 θ
F
PAC = 1+cos
3θ
Substituting this value for PAC into the expression for change in length
for member AC:
F
h
PAC h
1+cos3 θ
δAB =
=
AE
AE
Fh
1
ANS: δAB = AE 1+cos3 θ
Problem 10.46 If a downward force F is applied at
point A of the system shown in Problem 10.45, what are
the resulting normal stresses in bars AB and AC?
Free Body Diagram:
Solution:
Summing vertical forces at point A;
[1]
ΣFy = 0 = −F + PAB (cos θ) + PAC
We see that the VERTICAL movement of the A-end of member AB
will be:
[2]
vA = δAB (1/ cos θ)
From Equation [2], we see that the vertical movement of point A must
be the same for BOTH members of the truss, so:
h
1
PAB cos
PAC h
cos θ
=
AE
AE
[3]
PAB = PAC (cos2 θ)
Substituting Equation [3] into Equation [1]:
2
F = PAB (cos θ) + PAC
= PAC cos θ (cos θ) + PAC
3
F = PAC 1 + cos θ
[4]
PAC =
F
1 + cos3 θ
Using Equation [4] to determine the stress in member AC:
ANS:
[5]
σAC =
PAC
A
=
F
A(1+cos3 θ )
Combining Equations [3] and [4] to calculate the normal stress in
member AB:
F
cos2 θ
PAB
1+cos3 θ
σAB =
=
A
A
ANS:
σAB =
F cos2 θ
A[1+cos3 θ]
Problem 10.47 Each bar has a 500 − m m2 crosssectional area and modulus of elasticity E = 72 GPa.
If a 160-kN downward force is applied at A, what is the
resulting displacement of point A?
Free Body Diagram:
Solution:
The symmetry of the truss dictates that ALL of the displacement will
be vertical. We see that the axial loads in the two angled members will
be equal. In other words, δB = δC .
Summing vertical forces where the load is applied:
[1]
ΣFy = 0 = −160, 000 N + PA + 2PB (sin 60◦ )
The vertical displacement for each of the two angled members is:
vangle = δangle (1/ sin 60◦ )
Using the relationship between displacements for the straight member
and the two angled members:
δA = δB sin160◦ = δC sin160◦
0.3 m
1 PB ( sin
PA (0.3 m)
60◦ )
=
AE
AE
sin 60◦
[2]
PA = 1.333PB = 1.333PC
Substituting Equation [2] into Equation [1]:
0 = −160, 000 N + 1.333PB + 2PB (sin 60◦ )
PB = PC = 52, 201 N
From Equation [2] we get the axial load in member A:
PA = 1.333(52, 201 N) = 69, 584 N
The displacement of point A is:
(69,584 N)(0.3 m)
LA
=
= 0.580 mm ↓
ANS: δA = PAAE
(500×10−6 m2 )(72×109 N/m2 )
Problem 10.48 The bars in Problem 10.47 are made
of material that will safely support a tensile stress of
270 MPa. Based on this criterion, what is the largest
downward force that can safely be applied at A?
Free Body Diagram:
Solution:
The symmetry of the truss dictates that ALL of the displacement will
be vertical. We see that the axial loads in the two angled members will
be equal. In other words, δB = δC . Summing vertical forces where
the load is applied:
ΣFy = 0 = −F + PA + 2PB (sin 60◦ )
[1]
The vertical displacement for each of the two angled members is:
vangle = δangle (1/ sin 60◦ )
Using the relationship between displacements for the straight member
and the two angled members:
δA = δB sin160◦ = δC sin160◦
0.3 m
1 PB ( sin
PA (0.3 m)
60◦ )
=
AE
AE
sin 60◦
PA = 1.333PB = 1.333PC → PB = PC = 0.75PA
[2]
We see that the largest load is supported by the central member. Calculating the maximum allowable load in the members of the truss:
PA = PMAX = σMAX A = (270×106 N/m2 )(500×10−6 m2 ) = 135, 000 N
Using Equations [2] and [3] in Equation [1]:
F = 135, 000 N + 2[(0.75)(135, 000 N)](sin 60◦ )
ANS:
F = 310 kN
[3]
Problem 10.49 Each bar has a 500 − m m2 crosssectional area and modulus of elasticity E = 72 GPa.
If there is a gap h = 2 mm between the hole in the
vertical bar and the pin A connecting bars AB and AD,
what are the normal stresses in the three bars after the
vertical bar is connected to the pin at A?
Free Body Diagram:
Solution:
The downward displacement of bar C plus the upward displacement
of bars B and D equal 2 mm.
[1]
δA + δB / sin 60◦ = 2 × 10−3 m
The symmetry of the structure indicates that PB = PD . We also see
that the sum of vertical forces on point A must be zero after the bars
are connected.
ΣFy = 0 = −PA + 2(PB )(sin 60◦ )
[2]
PA = 1.732PB
Combining Equations [1] and [2]:
(1.732PB )(0.3 m)
(500×10−6 m2 )(72×109
PBC = 78, 295 N
N/m2 )
+
0.3 m
PB ( sin
60◦ )
(500×10−6 m2 )(72×109
Therefore: PA = (1.732)(78, 295 N)
PA = 135, 600 N
The normal stresses in the truss members are:
σA =
PA
A
=
135,600 N
500×10−6 m2
σA = 271 MPa
σB =
ANS:
PB
A
σB = 157 MPa
=
78295 N
500×10−6 m2
N/m2 )
1
sin 60◦
= 2 × 10−3 m
Problem 10.50 The bars in Problem 10.49 are made of
material that will safely support a normal stress (tension
or compression) of 400 MPa. Based on this criterion,
what is the largest safe value of the gap h?
Free Body Diagram:
Solution:
Summing horizontal forces at point A:
ΣFx = 0 = FB (cos 60◦ ) − FD (cos 60◦ ) → FB = FD
Summing vertical forces at point A:
ΣFy = 0 = FC +FB (sin 60◦ )+FD (sin 60◦ ) = FC +2FB (sin 60◦ )
Fc = −1.732FB = −1.732FD
[1]
From Equation [1] we see that the greatest load exists in member AC.
The magnitude of the maximum allowable force in member AC is:
(FC )MAX = σM AX A = (400 × 106 N/m2 )(500 × 10−6 m2 )
(FC )MAX = 200, 000 N (T)
From Equation [1] we see that the load in members AB and AD is:
FB = FD =
(FC )MAX
200, 000 N
=
1.732
1.732
FB = FD = 115, 473 N (C)
From the diagram we see that:
vAD = δAD / cos 30◦
[2]
The change in length for member AC is:
δAC =
δAC =
FC (0.3 m)
(500 × 10−6 m2 )(72 × 109 N/m2 )
(200, 000 N)(0.3 m)
(500 × 10−6 m2 )(72 × 109 N/m2 )
δAC = 0.00167 m = 1.67 mm ↑
The change in length for members AB and AD is:
δAB =
(115, 473 N) (0.3 m/sin 60◦ )
(500 × 10−6 m2 )(72 × 109 N/m2 )
δAB = δAD = 0.00111 m = 1.11 mm
The maximum allowable gap h is:
h = δAC +
ANS:
δAB
1.11 mm
= 1.67 mm +
cos 30◦
cos 30◦
h = 2.95 mm
Problem 10.51 The bar’s cross-sectional area is A −
(1 + 0.1x) in2 and the modulus of elasticity of the material is E = 12 × 106 psi. If the bar is subjected to tensile
axial forces P = 20 kip at its ends, what is the normal
stress at x = 6 in?
Free Body Diagram:
Solution:
At x = 6 in, the cross-sectional area of the bar is:
A = (1 + 0.1(6)) in2 = 1.6 in2
The normal stress in the bar at x = 6 in is:
σ = P/A = (20, 000 lb/(1.6 in2 )
ANS:
σ = 12, 500 psi = 12.5 ksi
Problem 10.52 What is the change in length of the bar
in Problem 10.51?
Free Body Diagram:
Solution:
The normal stress at any point in the bar is:
σ = P/A = (20, 000 lb)/(1 + 0.1x) in2
The strain at any point in the bar is:
ε=
σ
(20, 000 lb)/(1 + 0.1x) in2
0.001667 in2
=
=
2
6
E
(1
+ 0.1x) in2
12 × 10 lb/in
The change in length of the bar is:
10
10
10
0.001667 in2
1
0.001667 in 10
0.1
δ=
ε=
dx = 0.001667 in3
dx =
dx
2
2
0.1
(1 + 0.1x)
(1 + 0.1x) in
(1 + 0.1x)in
0
0
0
0
δ = 0.01667 [ln (1 + 0.1(10)) − ln (1 + 0.1(0))]
ANS:
δ = 0.01155 in
Problem 10.53 The cross-sectional area fo the bar in
Problem 10.51 is A = (1+az) in2 , where a is a constant,
and the modulus of elasticity of the material is E =
8 × 106 psi. When the bar is subjected to tensile axial
forces P = 14 kip at its ends, its change in length is δ =
0.01 in. What is the value of the constant a? (Strategy:
Estimate the value of a by drawing a graph of δ as a
function of a.
Free Body Diagram:
Solution:
The normal stress at any point in the bar is:
σ=
14, 000 lb
P
=
A
(1 + ax) in2
The strain at any point in the bar is:
ε=
σ
(14, 000/1 + ax) lb/in2
0.00175
=
=
2
6
E
1 + ax
8 × 10 lb/in
Total change in length of the bar may be expressed as:
10
10
0.00175
0.00175 10 a dx
0.00175
δ=
dx =
=
ln(1 + ax) +C
1
+
ax
a
1
+
ax
a
0
0
0
δ=
0.00175
[ln(1 + 10a)] + C
a
At x = 0, we see that δ = 0, so C = 0.
For the given change in length:
0.01 in =
0.00175
[ln(1 + 10a)]
a
Rearranging the terms in the equation and using a graphing calculator
to find the value of a:
ln(1 + 10a) − 5.714a = 0
ANS:
a = 0.1805 in−1
Problem 10.54 From x = 0 to x = 100 mm, the bar’s
height is 20 mm. From x = 100 mm to x = 200 mm,
its height varies linearly from 20 mm to 40 mm. From
x = 200 mm to x = 300 mm, its height is 40 mm. The
flat var’s thickness is 20 mm. The modulus of elasticity
of the material is E = 70 GPa. If the bar is subjected
to tensile axial forces P = 50 kN at its ends, what is its
change in length?
Free Body Diagrams:
Solution:
The problem is solved by considering each of the three sections of the
bar separately.
Elongation of the left-hand section of the bar is:
δL =
PL
(50, 000 N)(0.1 m)
=
= 0.1785 mm
AE
(0.02 m)(0.02 m)(70 × 109 N/m2
Elongation of the right-hand section of the bar is:
δR =
PL
(50, 000 N)(0.1 m)
=
= 0.0893 mm
AE
(0.02 m)(0.04 m)(70 × 109 N/m2
Determining the elongation of the center section of the bar will require
integration. The cross-sectional area at any point in the center section
is:
AC = (0.02 m)[0.02 m+((0.02 m)/(0.1 m)(x)] = 0.0004+0.004x
The stress at any point in the center section of the bar is:
σ=
50, 000
P
=
n/m2
A
0.0004 + 0.004x
Total elongation of the center section of the bar is:
σ
50, 000 N/(0.0004 + 0.004x) m2
dδC =
dx =
dx
2
E
70 × 109 N/m
δC =
0.1
0
7.143 × 10−7
7.143 × 10−7
dx =
0.0004 + 0.004x
0.004
0.1
0
0.004
dx
0.0004 + 0.004x
δC = 1.786×10−4 [ln (0.0004 + 0.004(0.1)) − ln (0.0004 + 0.004(0))]
δC = 0.1238 mm
Total change in length of the bar is:
δ = δL + δC + δR = 0.1785 mm + 0.1238 mm + 0.0893 mm
ANS:
δ = 0.392 mm
Problem 10.55 Fro x = 0 to x = 10in, the bar’s
cross-sectional area is A = 1 in2 . Fro x = 10 in to
x = 20 in, A = (0.1x) in2 . The modulus of elasticity
of the material is 12×106 psi. There is a gap b = 0.02 in
between the right end of the bar and the rigid wall. If the
bar is stretched so that is contacts the rigid wall and is
welded to it, what is the axial force in the bar afterward?
Free Body Diagrams:
Solution:
The deformation of the right-hand portion of the bar is:
dδR =
δR =
σ
P/A
P/0.1x in2
10P lb/in2
dx =
dx =
dx =
dx in
E
E
12 × 106 lb/in2
(12 × 106 lb/in2 )x
20
10
10P
10P
dx =
(12 × 106 )x
12 × 106
20
10
dx
P
=
[ln 20−ln 10] = (5.776×10−7 )P in
x
12 × 105
The expression for the total deformation of the bar is:
P
0.02 in = δL +δR =
+(5.77×10−7 )P
2
2
(1 in )(12 × 106 lb/in
ANS:
P = 3.0288 × 104 lbs
Problem 10.56 From x = 0 to x = 10 in, the crosssectional area of the bar in Problem 10.55 is A = 1 in2 .
The modulus of elasticity of the material is E = 12 ×
106 psi. There is a gap b = 0.02 in between the right
end of the bar and the rigid wall. If a 40 kip axial force
toward the right is applied to the bar at x = 10 in, what
is the resulting normal stress in the left half of the bar?
Free Body Diagram:
Solution:
We first determine whether the 40, 000−lb load is sufficient to close
the 0.02−in gap.
δ=
(40, 000 lb)(10 in)
(1 in2 )(12 × 106 lb/in2 )
= 0.0333 in,
which is larger than the 0.02−in gap, so there will be reaction at the
right-hand end of the bar.
Deformation of the left-hand portion of the bar is:
δL =
(40, 000 lb − RR )(10 in)
(1 in2 )(12 × 106 lb/in2 )
= 0.0333 in − 8.333 × 10−7 RR
Deformation of the right-hand portion of the bar is:
20
20
−RR dx
−RR
dx
−RR
δR =
=
=
[ln(20) − ln(10)] = −5.776×10−7 RR in
2
5
2
6
12 × 10 10 x
12 × 105
10 (0.1x in )(12 × 10 lb/in )
The expression for total deformation of the bar is:
δL + δR = 0.02 in
(0.0333 in − 8.333 × 10−7 RR ) − 5.776 × 10−7 RR = 0.02 in
RR = +9, 430 lb
RL = +30, 570 lb
Normal stress in the left-hand portion of the bar is:
σ=
ANS:
RL
30, 570 lb
=
1 in2
1 in2
σ = 30, 570lb/in2
Problem 10.57 The diameter of the bar’s circular
cross-section varies linearly from 10 mm at its left end
to 20 mm at its right end. The modulus of elasticity of
the material is E = 45 GPa. If the bar is subjected to
tensile axial forces P = 6 kN at its ends, what is the
normal stress at x = 80 mm?
Free Body Diagram:
Solution:
The function which describes the diameter of the bar is:
0.02 m − 0.01 m
D = 0.01 m+
(x) m = 0.01 m+(0.0667x) m
0.150 m
The diameter of the bar at x = 80 mm is:
D80 = 0.01 m + (0.0667)(0.08 m) = 0.01534 m
The cross-sectional area of the bar at x = 80 mm is:
A80 =
2
πD80
π(0.01534 m)2
=
= 0.0001847 m2
4
4
The normal stress in the bar at x = 80 mm is:
σ=
ANS:
P
6, 000 N
=
A
0.0001847 m2
σ = 32.5 MPa
Problem 10.58 What is the change in length of the bar
in Problem 10.57?
Free Body Diagram:
Solution:
The radius of the cross-sectional area at any point along the length of
the bar is:
0.02 − 0.01
d = 0.01 +
x m = (0.01 + 0.0667x) m
0.15
The cross-sectional area of the bar at any point along its length is:
A=π
d2
(0.01 + 0.0667x)2
=π
4
4
The change in length of the bar is:
0.15
0.15
0.15
σ
P
6, 000 N
δ=
dx =
dx =
dx
(0.01+0.0667x)2
E
AE
0
0
0
π
m2 (45 × 109 N/m2 )
4
Using Mathcad to evaluate the itegral:
ANS: δ = 0.127 mm
Problem 10.59 The bar is fixed at the left and is subjected to a uniformly distributed axial force. It has crosssectional area A and modulus of elasticity E. (a) Determine the internal axial force P in the bar as a function
of x. (b) What is the bar’s change in length?
Free Body Diagram:
Solution:
(a) The internal axial force at any point in the bar is:
P (x) = qL − qx = q(L − x)
ANS:
(b) The change in length of the bar is:
L
L
L
L
σ
P (x)
q(L − x)
qLx
qx2 δ=
dx =
dx =
dx =
−
AE
AE
AE
2AE 0
0 E
0
0
δ=
ANS:
qL2
2AE
Problem 10.60 The bar shown in Problem 10.59 has
length L = 2 m, cross-sectional area A = 0.03 m2 , and
modulus of elasticity E = 200 GPa. It is subjected to a
distributed axial force q = 12(1 + 0.4x) MN/m. What
is the bar’s change in length?
Free Body Diagram:
Solution:
The reaction at the wall is:
2
2
R=
12(1 + 0.4x) dx = (12x + 2.4x2 )0 = 33.6 MN ←
0
Total change in length for the bar is:
2
(33.6 − 12 − 4.8x) MN dx
δ=
2
2
9
0 (0.03 m )(200 × 10 N/m )
ANS:
δ = 0.0056 m = 5.6 mm
Problem 10.61 A cylindrical bar with 1−in diameter
fits tightly into a circular hole in a 5−in thick plate. The
modulus of elasticity of the marerial is E = 14×106 psi.
A 1000−lb tensile force is applied at the left end of the
bar, causing it to begin slipping out of the hole. At the
instant slipping begins, determine (a) the magnitude of
the uniformly distributed axial force exerted on the bar
by the plate; (b) the total change in the bar’s length.
Free Body Diagram:
Solution:
The distributed load along the 5−inch section of the bar is:
lb
ANS: q = 1,000
= 200 lb/in
5 in
The elongation of the 10−inch section of the bar is:
δ10 =
PL
(1, 000 lb)(10 in)
=
= 0.0009095 in
AE
π(0.5) in)2 (14 × 106 lb/in2 )
The elongation of the 5−inch section of the bar is:
5
(1000 − 200x) lb/π(0.5 in2 )2
σ
δ5 = εL = L =
dx = 0.0002273 in
E
14 × 106 lb/in2
0
Total change in length of the bar is:
δ = δ10 + δ5 = 0.0009095 in + 0.0002273 in
ANS:
δ = 0.00114 in
Problem 10.62 The bar has a circular cross section
with 0.002−m diameter and its modulus of elasticity is
E = 86.6 GPa. The bar is fixed at both ends and is
subjected to a distributed axial force q = 75 kN/m and
an axial force F = 15 kN. What is its change in length?
Free Body Diagram:
Solution:
Cross-sectional area of the bar is:
0.002 m 2
A=π
= 3.142 × 10−6 m2
2
Summing horizontal forces on the FBD to find the magnitude of R:
ΣFx = 0 = −R − 15, 000 N + (75, 000 N/m)(0.8 m)
R = 45, 000 N ←
Total change in length of the bar is:
0.8
(45, 000 N − 75, 000(x) N) dx
δ=
(3.142 × 10−6 m2 )(86.6 × 109 m2 )
0
ANS:
δ = 0.0441 m = 44.1 mm
Problem 10.63 In Problem 10.62, what axial force F
would cause the bar’s change in length to be zero?
Free Body Diagram:
Solution:
Summing horizontal forces on the FBD to find the reaction at the wall:
ΣFx = 0 = −R − F + (75, 000 N/m)(0.8 m)
R = (F − 60, 000 N
Setting the expression for total change in length equal to zero:
0.8
(F − 75, 000(x)) dx
δ=0=
(3.142 × 10−6 m2 )(86.6 × 109 N/m2
0
0.8
0 = F x − 37, 500x2 0 0.8F − 24, 000 N
ANS:
F = 30, 000 N
Problem 10.64 If the bar in Problem 10.62 is subjected
to a distributed force q = 75(1 + 0.2x) kN/m and an
axial force F = 15 kN, what is its change in length?
Free Body Diagram:
Solution:
The reaction at the wall can be found by summing horizontal forces:
0.8
ΣFx = 0 = −15, 000 N +
(75, 000)(1 + 0.2x) dx − RW
0
RW = 49, 800 N ←
Starting at the left-hand end of the bar, the function which describes
the axial force in the bar is:
x
P = 49, 800 N−
(75, 000)(1+0.2x) dx = 49, 800 N−75, 000x−7, 500x2
0
The cross-sectional area of the bar is:
A = π(0.001 m)2 = 3.1416 × 10−6 m2
Stress at any point in the bar is:
σ=
P
(49, 800 − 75, 000x − 7, 500x2 )
N/m2
=
A
3.1416 × 10−6
The change in length for the bar is:
0.8
σ
49, 800 − 75, 000x − 7, 500x2 ) N
δ= L=
dx
E
(3.1416 × 10−6 m2 )(86.6 × 109 N/m2 )
0
ANS:
δ = 0.0535m
Problem 10.65 The bar is fixed at A and B and is subjected to a uniformly distributed axial force. It has crosssectional area A and modulus of elasticity E. What are
the reactions at A and B?
Free Body Diagram:
Solution:
We recognize that the sum of horizontal forces on the bar must be zero.
ΣFx = 0 = −RA − RB + qL [1]
Removing the fixed structure at the right-hand end of the bar, the distributed load will produce a change in length of:
L
L
qx
qx2 qL2
δ=
dx =
=
2AE 0
2AE
0 AE
RB =
qL
2
Substituting this value for RB in Equation [1]:
RA = −RB + qL = −(qL/2) + qL
ANS:
RA = qL/2
Problem 10.66 What point of the bar in Problem 10.65
undergoes the largest displacement, and what is the displacement?
Free Body Diagram:
Solution:
We regognize that the sum of horizontal forces on the bar must be zero.
L
L
qx
qx2 qL2
Σ=
dx =
=
[1]
2AE 0
2AE
0 AE
The reaction at the right-hand support must be sufficient to produce a
reduction in length of −(qL2 )/(2AE).
−qL2
RB L
=
2AE
AE
RB =
qL
←
2
Substituting this value for RB in Equation [1]:
RA = −RB + qL = −(ql/2) + qL
RA = qL/2 ←
The expression for displacement of any point on the bar is:
L
(qL/2) − qx
δ=
dx [2]
AE
0
We know that maximum deflection occurs where the expression
dδ/dx = 0. Setting the derivative of Equation [2] equal to zero:
(qL/2) − qx
=0
AE
ANS: x = L/2 [3]
Using the value of x from Equation [3] in Equation [2] and evaluating:
L/2
L/2
(qL/2) − qx
(qLx/2) − qx2 /2
dx =
δ=
AE
AE
0
0
ANS:
δMAX =
qL2
8AE
Problem 10.67 A line L within an unconstrained sample of material is 200 mm long. The coefficient of thermal expansion of the material is α = 22 × 10−6◦ C−1 .
If the temperature of the material is increased by 30◦ C,
what is the length of the line?
Solution:
The new length of the line will be:
δ = Lα(∆T ) = (200 mm)(22 × 10−6◦ C−1 )(30◦ C) = 0.132
ANS:
L = 200.132 mm
Problem 10.68 The length of the line L within the
unconstrained sample of material shown in Problem 67 is
2 in. The coefficient of thermal expansion of the material
is α = 8 × 10−6◦ F−1 . After the temperature of the
material is increased, the length of the line is 2.002 in.
How much was the temperature increased?
Solution:
The thermal strain produced by the change in temperature is:
εT =
L − L
2.002 in − 2.000 in
=
= 0.001
L
2.000 in
The temperature increase required to produce the above thermal strain
is:
εT = α(∆T )
0.001 = (8 × 10−6◦ F−1 )(∆T )
ANS:
∆T = 125◦ F
Problem 10.69 Consider a 1 in. × 1 in. × 1 in. cube
within an unconstrained sample of material. The coefficient of thermal expansion of the material is α =
14 × 10−6◦ F−1 . If the temperature of the material is
decreased by 40◦ F, what is the volume of the cube?
Solution:
As the material warms, EVERY dimension (length, width and breadth)
is increased. The new length of one side of the cube will be:
L = L+∆L = L+(L)(α)(∆T ) = 1 in+(1 in)(14×10−6◦ C−1 )(−40◦ C)
L = 0.99944 in
The new volume of the cube will be:
V = (L )3 = (0.99944 in)3
ANS:
V = 0.998 in3
Problem 10.70 A prismatic bar is 200 mm long and
has a circular cross section with 30-mm diameter. After
the temperature of the unconstrained bar is increased, its
length is measured and determined to be 200.160 mm.
What is the bar’s diameter after the increase in temperature?
Solution:
Strain in the material will be the same in every direction.
The thermal strain in the direction of the length of the bar is:
εT =
L − L
200.160 mm − 200 mm
=
= 0.0008
L
200 mm
The same thermal strain in the direction of the radius will produce a
diameter of:
D = εD = (0.0008)(30 mm)
ANS:
D = 30.024 mm
Problem 10.71 If the increase in temperature in Problem 10.70 is 20◦ C, what is the coefficient of thermal
expansion of the bar?
Solution:
The original length of the bar is 200 mm, and the length of the bar after
thermal strain is 200.160 mm. The coefficient of thermal expansion
is:
∆L = L − L = L(α)(∆T )
200.160 mm − 200 mm = (200 mm)(α)(20◦ C)
ANS:
α = 40 × 10−6◦ C−1
Problem 10.72 If the increase in temperature in Problem 10.70 is 20◦ C and the modulus of elasticity of the
material is E = 72 GPa, what is the normal stress on a
plane perpendicular to the bar’s axis after the increase in
temperature? Strategy: Obtain a free-body diagram
by passing a plane through the bar.
Solution:
Since the bar is unconstrained, there will be no normal stress in the bar
regardless of the change in temperature.
ANS:
σ=0
Problem 10.73 The prismatic bar is made of material with modulus of elasticity E = 28 × 106 psi and
coefficient of thermal expansion α = 8 × 10−6◦ F−1 .
The temperature of the unconstrained bar is increased
by 50◦ F above its initial temperature T . (a) What is the
change in the bar’s length? (b) What is the change in
the bar’s diameter? (c) What is the normal stress on a
plane perpendicular to the bar’s axis after the increase in
temperature?
Strategy: To answer part (c), obtain a free-body diagram by passing a plane through the bar.
Solution:
(a) The new (heated) length of the bar will be:
∆L(α)(∆T ) = (15 in.)(8 × 10−6◦ F−1 )(50◦ F)
ANS:
L = 0.006 in.
(b) The change in the diameter of the bar is:
∆D = D(α)(∆T ) = (2 in.)(8 × 10−6◦ F−1 )(50◦ F)
ANS:
∆D = 0.0008 in.
(c) Because the bar is unconstrained, no forces are exerted on the ends
of the bar. The normal stress on EVERY plane passed through the bar
is
ANS:
σ=0
Problem 10.74 Suppose that the temperature of the
unconstrained bar in Problem 10.73 is increased by 50◦ F
above its initial temperature T and the bar is also subjected to 30,000-lb tensile axial forces at the ends. What
is the resulting change in the bar’s length?
Determine the change in length assuming that (a) the
temperature is first increased and then the axial forces
are applied; (b) the axial forces are applied first and then
the temperature is increased.
Solution:
Considering first the case in which the temperature changes first, then
the load is applied:
(∆L)T = Lα(∆T ) = (15 in.)(8×10−6◦ F−1 )(50◦ F) = 0.006 in.
(∆L)F =
PL
(30, 000 lb)(15.006 in)
=
= 0.0051 in.
AE
π(1 in)2 (28 × 106 lb/in2 )
Total change in length in this scenario is:
∆L = 0.006 in. + 0.0051 in.
ANS: ∆L = 0.0111 in.
Considering now the case in which the load is applied first, then the
temperature is increased:
(∆L)F =
PL
(30, 000 lb)(15 in)
=
= 0.0051 in.
AE
π(1 in)2 (28 × 106 lb/in2 )
(∆L)T = Lα(∆T ) = (15.005 in.)(8×10−6◦ F−1 )(50◦ F) = 0.006 in.
Total change in length in this scenario is:
∆L = 0.0051 in. + 0.006 in.
ANS:
∆L = 0.0111 in.
Problem 10.75 The prismatic bar is made of material
with modulus of elasticity E = 28 × 106 psi and coefficient of thermal expansion α = 8 × 10−6◦ F−1 . It is
constrained between rigid walls. If the temperature is
increased by 50◦ F above the bar’s initial temperature T ,
what is the normal stress on a plane perpendicular to the
bar’s axis?
Solution:
Because the bar is constrained, its length cannot change (∆L = 0).
∆L = Lα(∆T )−
ANS:
PL
P (15 in)
= (15 in)(8×10−6◦ F−1 )(50◦ F)−
AE
A(28 × 106 lb/in2 )
P/A = σ = −11, 200 lb/in2
NOTE: The (−) sign indicates a compressive stress.
Problem 10.76 The walls between which the prismatic bar in Problem 10.75 is constrained will safely support a compressive normal stress of 30, 000 psi. Based
on this criterion, what is the largest safe temperature increase to which the bar can be subjected?
Solution:
The maximum stress to which the wall can be subjected is also the
maximum stress to which the bar may be subjected.
∆L = 0 = Lα(∆T )−
ANS:
PL
15 in
= (15 in)(8×10−6◦ F−1 )(∆T )−(30, 000 lb/in2 )
2
AE
28 × 106 lb/in
∆T = 134◦ F
Problem 10.77 The prismatic bar in Problem 10.75
has a cross-sectional area A = 3 in2 and is made of
material with modulus of elasticity E = 28 × 106 psi
and coefficient of thermal expansion α = 8×10−6◦ F−1 .
It is constrained between rigid walls. The temperature is
increased by 50◦ F above the bar’s initial temperature T
and a 20,000-lb axial force to the right is applied midway
between the two walls. What is the normal stress on a
plane perpendicular to the bar’s axis to the right of the
point where the force is applied?
Solution:
Because the bar is constrained, the total change in length must be zero.
The change in length due to the applied load would be:
(∆L)F =
(20, 000 lb)(7.5 in)
(3 in2 )(28 × 106 lb/in2 )
= 0.0018 in.
The change in length due to the temperature change would be:
(∆L)T = Lα(∆T ) = (15in.)(8×10−6◦ F−1 )(50◦ F) = 0.006 in.
The reaction at the right-hand end of the bar will be sufficient to hold
the change in length to zero.
∆L = 0 = 0.0018 in. + 0.006 in. −
RR = 43, 680 lb ←
The stress in the right-hand portion of the bar is:
σR =
ANS:
RR
−43, 680 lb
=
A
3 in2
σR = 14, 560 lb/in2
RR (15 in)
(3 in2 )(28 × 106 lb/in2 )
Problem 10.78 The prismatic bar is made of material
with modulus of elasticity E = 28 × 106 psi and coefficient of thermal expansion α = 8×10−6◦ F−1 . It is fixed
to a rigid wall at the left. There is a gap B = 0.0002 in.
between the bar’s right end and the rigid wall.
If the temperature is increased by 50◦ F above the bar’s
initial temperature T , what is the normal stress on a plane
perpendicular to the bar’s axis?
Solution:
The rigid wall will exert an axial force on the bar sufficient to restrict
the bar’s elongation to 0.002 inches.
∆L = 0.002 in. = Lα(∆T )−
ANS:
RR L
15 in
= (15 in)(8×10−6◦ F−1 )(50◦ F)−σ
2
AE
28 × 106 lb/in
σ = −7, 467 lb/in2
NOTE: The (−) sign indicates a compressive stress.
Problem 10.79 Bar A has a cross-sectional area of
0.04 m2 , modulus of elasticity E = 70 GPa, and coefficient of thermal expansion α = 14 × 10−6◦ C−1 . Bar B
has a cross-sectional area of 0.01 m2 , modulus of elasticity E = 120 GPa, and coefficient of thermal expansion
α = 16 × 10−6◦ C−1 . There is a gap B = 0.4 mm
between the ends of the bars. What minimum increase
in the temperature of the bars above their initial temperature T is necessary to cause them to come into contact?
Solution:
The sum of the expansions of the two bars will be 0.0004 m.
εT = LA αA (∆T )+LB αB (∆T ) = (1 m)(14×10−6 /◦ C)(∆T )+(1 m)(16×10−6 /◦ C)(∆T )
ANS:
∆T = 13.3◦ C
Problem 10.80 If the temperature of the bars in Problem 10.79 is increased by 40◦ C above their initial temperature T , what are the normal stresses in the bars?
Solution:
The reaction at the walls will be sufficient to restrict the total elongation
of the bars to 0.0004 m.
∆L = LA αA (∆T ) + LB αB (∆T ) − σA
LA
LB
− σB
EA
EB
[1]
The relationship between the stress in the two bars is:
PA = PB → σB = 4σA
[2]
Substituting Equation [2] into Equation [1]:
0.0004 m = (1 m)(14×10−6 /◦ C)(40◦ C)+(1 m)(16×10−6 /◦ C)(40◦ C)−σA
ANS:
σA = −16.8 MPa σB = −67.2 MPa
NOTE: The (−) indicates a compressive stress.
1m
70 ×
109
N/m
2
−4σA
1m
120 ×
109
N/m
2
Problem 10.81 Each bar has a 2 − in2 cross-sectional
area, modulus of elasticity E = 14 × 106 psi, and coefficient of thermal expansion α = 11 × 10−6◦ F−1 . The
normal stresses in the bars are initially zero. If their
temperature is increased by 40◦ F from their initial temperature T , what is the resulting displacement of point
A.
Solution:
The original length of each bar is:
L = (36 in)2 + (18 in)2 = 40.25 in
The change in length for each bar due to the temperature increase is:
(∆L)T = Lα(∆T ) = (41.25 in)(11×10−6 /◦ F)(40◦ F) = 0.0177in.
The new vertical distance from the fixed surface to point A is:
y = (40.25 in + 0.0177 in)2 − (18 in)2 = 36.021 in
The vertical displacement of point A is:
v = y − y = 36.021 in − 36 in
ANS:
v = 0.021 in
Problem 10.82 If the temperature of the bars in Problem 10.81 is decreased by 30◦ F from their initial temperature T , what force would need to be applied at A
so that the total displacement of point A caused by the
temperature change and the force is zero?
Solution:
The original length of each bar is:
L = (36 in)2 + (18 in)2 = 40.25 in
The amount by which the bars’ length decreases due to the decrease
in temperature is:
(∆L)T = Lα(∆T ) = (40.25 in)(11×10−6 /◦ F)(−30◦ ) = −0.0133 in
The force required to return the bars to their original length is:
0.0133 in =
P (40.25 in)
(2 in2 )(14 × 106 lb/in2 )
→ P = 9, 252 lb
The load, F , required to produce this axial load in BOTH bars is:
ΣFy = 0 = −F + 2P (sin 60◦ ) = −F + 2(9.252 lb)(sin 60◦ )
ANS:
F = 16, 000 lb
Problem 10.83 You are designing a bar with a solid
circular cross section that is to support a 4-kN tensile
axial load. You have decided to use 6061-T6 aluminum
alloy (See Appendix D), and you want the factor of safety
to be S = 2. Based on this criterion, what should the
bar’s diameter be?
Solution:
From Appendix B, the yield stress of the material is σy = 270 ×
106 N/ m2 .
Calculating the required cross-sectional area for the circular bar:
σy
P
=
→
S
A
ANS:
D = 6.14 mm
270 × 106 N/m2
4, 000 N
= 2
2
π D
2
Problem 10.84 You are designing a bar with a solid
circular cross section with 5-mm diameter that is to support a 4-kN tensile axial load, and you want the factor of
safety to be at least S = 2. Choose an aluminum alloy
from Appendix D that satisfies this requirement.
Solution:
Calculating the required cross-sectional area for the circular bar:
σy
P
=
→
S
A
σy
4, 000 N
=
2
π(0.0025 m)2
From Appendix D, the yield stress for the material is σy = 407 ×
106 N/ m2 .
ANS:
σy = 407 × 106 N/m2
Either of aluminum alloys 7075-T6 or 2014-T6 will support the load.
Problem 10.85 You are designing a bar with a solid
circular cross section that is to support a 4000-lb tensile
axial load. You have decided to use ASTM-A572 structural steel (See Appendix D), and you want the factor
of safety to be S = 1.5. Based on this criterion, what
should the bar’s diameter be?
Solution:
From Appendix, the yield stress for the material is σy
50, 000 lb/in2 .
Calculating the required cross-sectional area for the circular bar:
σy
P
=
→
S
A
ANS:
=
50, 000 N/m2
4, 000 lb
= 2
1.5
π D
2
D = 0.391 in
Problem 10.86 You are designing a bar with a solid
circular cross section with 1/2-in. diameter that is to
support a 4000-lb tensile axial load, and you want the
factor of safety to be at least S = 3. Choose a structural
steel from Appendix D that satisfies this requirement.
Solution:
Calculating the required cross-sectional area for the circular bar:
σy
P
=
→
S
A
σy
4, 000 lb
=
3
π(1/4 in)2
σy = 61, 115 lb/in2
ANS:
ASTM - A 514 will support the load.
Problem 10.87 The horizontal beam of length L =
2 m supports a load F = 30-kN. The beam is supported
by a pin support and the brace BC. The dimension
h = 0.54 m. Suppose that you want to make the brace
out of existing stock that has cross-sectional area A =
0.0016 m2 and yield stress σY = 400 MPa. If you want
the brace to have a factor of safety S = 1.5, what should
the angle θ be?
Free Body Diagram:
Solution:
The maximum allowable force in the brace is:
σy
400 × 106 N/m2
(FBC )M AX =
A=
(0.0016 m2 ) = 426.7×103 N
S
1.5
Summing moments about the pin connection at the wall (at D):
ΣMD = 0 = (30, 000 N)(2 m)−FBC (sin θ)(0.54 m) = (30, 000 N)(2 m)−(426.7×103 N)(cos θ)(0.54 m)
cos θ = 0.2604
ANS:
θ=
74.9◦
Problem 10.88 Consider the system shown in Problem 10.87. The horizontal beam of length L = 4 ft
supports a load F = 20 kip. The beam is supported
by a pin support and the brace BC. The dimension
h = 1 ft and the angle θ = 60◦ . Suppose that you want
to make the brace out of existing stock that has yield
stress σy = 50 ksi. If you want to design the brace
BC to have a factor of safety S = 2, what should its
cross-sectional area be?
Free Body Diagram:
Solution:
Summing moments about the pin connection at the wall (at D):
ΣMD = 0 = (20, 000 lb)(4 ft) − FBC (cos 60◦ )(1.732 ft)
FBC = 92, 380 lb
Calculating the cross-sectional area of the brace required to support
the load:
FBC
σy
=
→ :
A
S
ANS:
A = 3.69 in2
92, 380 lb
50, 000 lb/in2
=
A
2
Problem 10.89 The horizontal beam shown in Fig- Free Body Diagram:
ure P10-87 is of length L and supports a load F . The
beam is supported by a pin support and the brace BC.
Suppose that the brace is to consist of a specified material
for which you have chosen an allowable stress σALLOW ,
and you want to design the brace so that its weight is a
minimum. You can do this by assuming that the brace is
subjected to the allowable stress and choosing the angle
θ so that the volume of the brace is a minimum. What is
the necessary angle θ?
Solution:
Let P be the compressive axial load in the bar. Summing moments
about the pin connection at the wall (at D):
ΣMD = 0 = LF − (P )(cos θ)(b) → P = LF/(b cos θ)
[1]
The maximum allowable value of P is:
P = σALLOW A [2]
We see that the volume of the bar is:
V = LBAR A = (LBAR P )/(σALLOW )
[3]
Using Equation [1] in Equation [3]:
V σALLOW =
LF
lBAR
LF
1
·
=
·
[4]
cos θ
b
cos θ sin θ
V σALLOW
1
=
LF
cos θ sin θ
We see from Equation [4] that V is minimum where the product
(cos θ)(sin θ) is maximum. Using a graphing calculator to find the
angle at which the product is a maximum:
ANS:
θ = 45◦
Problem 10.90 In Problem 10.89, draw a graph showing the dependence of the volume of the brace on the
angle θ for 5◦ < θ < 85◦ . Notice that the graph is
relatively flat near the optimum angle, meaning that the
designer can choose θ within a range of angles near the
optimum value and still obtain a near-optimum design.
Solution:
Using Equation [4] from the solution to Problem 10.89:
V σALLOW =
LF
lBAR
LF
1
·
=
·
[4]
cos θ
b
cos θ sin θ
V σALLOW
1
=
LF
cos θ sin θ
We note that the quantities L and F are constant. We draw the graph
of the function
f (θ) =
1
sin(θ) cos(θ)
We see that the graph is practically flat from approximately θ = 39◦
to approximately θ = 51◦ , giving a designer considerable latitude in
choosing the angle for member BC.
Problem 10.91 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 360 lb. At A and
at B. Assume that the members of the truss have the
same cross-sectional area. (Choose a material from Appendix D) and determine the cross-sectional area so that
the structure has a factor of safety S = 2.5.
Free Body Diagrams:
Solution:
The method of joints is used to determine the axial loads in each member of the truss.
At joint B:
ΣFy = 0 = −360 lb + FBD → FBD = 360 lb (T)
ΣFx = 0 = FBC (cos 18.4◦ ) → FBC = 0
At joint A:
ΣFy = 0 = −360 lb + FAC → FAC = 360 lb (T)
At joint C:
ΣFy = 0 = −360 lb + FCF (sin 63.4◦ ) → FCF = 402.6 lb (T)
ΣFx = 0 = −(402.6 lb)(cos 63.4◦ )+FCD → FCD = 180.3 lb (C)
At joint F : From symmetry we see that
FDF = FCF → FDF = 402.6 lb (C)
ΣFx = 0 = (402.6 lb)(cos 63.4◦ )+FDF (cos 63.4◦ )−FF G → FF G = 360.5 lb (T)
The largest axial load is 402.6 lb.
From Appendix D we see that σy for 2014-T6 Al is 60, 000 lb/in2 .
Calculating the required cross-sectional area for the truss members:
σy
PM AX
=
→
S
A
ANS:
A = 0.017 in2
60, 000 lb/in2
402.6 lb
=
2.5
A
Problem 10.92 Upon learning of an upgrade in the helicopter’s engine, the engineer designing the truss shown
is Problem 10.91 does new simulations and concludes
that the downward forces on the stretcher will exert at A
and B may be as large as 400 lb. He also decides the
truss will be made of existing stock with cross-sectional
area A = 0.1 in2 . Choose an aluminum alloy from Appendix B so that the structure will have a factor of safety
of at least S = 5.
Free Body Diagrams:
Solution:
The method of joints is used to determine the axial loads in each member of the truss.
At joint A:
ΣFy = 0 = −400 lb + FAC → FAC = 400 lb (T)
At joint B:
ΣFy = 0 = −400 lb + FBD → FBD = 400 lb (T)
ΣFx = 0 → FBC = 0
At joint C:
ΣFy = 0 = −400 lb + FCF (sin 63.4◦ ) → FCF = 447.4 lb (T)
ΣFx = 0 = −(447.4 lb)(cos 63.4◦ )+FCD → FCD = 200.3 lb (C)
At joint F : From symmetry we see that
FDF = FCF → FDF = 447.4 lb (C)
ΣFx = 0 = (447.4 lb)(cos 63.4◦ )+(447.4(cos 63.4◦ )−FF G → FF G = 400.7 lb (T)
The largest axial load is 447.4 lb.
Calculating the required cross-sectional area for the truss members:
σy
PM AX
=
→
S
A
σy
447.4 lb
=
5
0.1 in2
σy = 22, 370 lb/in2
The aluminum alloys in Appendix D which exceed this minimum yield
stress are:
ANS: 2014-T6, 6061-T6 and 7075-T6.
Problem 10.93 Two candidate truss designs to sup- Free Body Diagrams:
port the load F are shown. Members of a given crosssectional area A and yield stress σY = are to be used.
Ay
Compare the factors of safety and weights of the two designs and discuss reasons that might lead you to choose
Ax
one design over the other. (The weights can be compared
by calculating the total lengths of their members.)
Bx
Solution:
C
Considering candidate truss (a);
At joint C:
(a)
ΣFy = 0 = −F + FAC (sin 26.6◦ ) → FAC = 2.23F (T)
ΣFx = 0 = −(2.23F )(cos 26.6◦ )+FBC → FBC = 1.994F (C)
Summing vertical forces at point B:
Ay
D
Ax
→ FAB = 0
So the largest load in truss (a) is:
FMAX = 2.23F
ANS:
The factor of safety for truss (a) is:
σy
σy
S=
=
σMAX
2.23F/A
ANS:
The length of truss (a) is:
La = 2h + h + (2h)2 + h2 = 5.24h
Considering candidate truss (b):
At joint C:
ΣFy = 0 = −F + FCD (sin 45◦ ) → FCD = 1.414F
ΣFx = 0 = −(1.414F )(sin 45◦ ) + FBC → FBC = F
At joint D:
ΣFy = 0 = FBD (sin 45◦ ) − FCD (sin 45◦ ) → FBD = 1.414F
ΣFx = 0 = −2(1.414F )(cos 45◦ ) + FAD → FAD = 2F
At joint B:
ΣFy = 0 = FBD (sin 45◦ ) − FAB → FAB = F
The largest load in truss (b) is:
FMAX = 2F
ANS:
S=
The factor of safety for truss (b) is:
σy
σMAX
=
σy
2F/A
ANS:
The length of truss (b) is:
√
Lb = h + h + 2h + 2( 2h) = 6.83h
If the design constraint is cost, truss (a) contains less material and fewer
joints. If the design constraint is strength, truss (b) is the better choice.
Bx
C
(b)
Problem 10.94 The cross-sectional area of bar AB is
0.015 m2 . If the force F = 20 kN, what is the normal
stress on a plane perpendicular to the axis of bar AB?
Diagram:
Solution:
Sum moments about point C to find the load in member AB.
ΣMC = 0 = (20, 000 N)(3 m) − FAB (sin 60◦ )(2 m)
FAB = 34, 641 N
The normal stress in member AB is:
σAB = FAB /AAB = (34, 641 N)/(0.015 m2 )
ANS:
σAB = 2.31 MPa
Problem 10.95 Bar AB of the frame in Problem 10.94
consists of a material that will safely support a tensile
normal stress of 20 MPa. Based on this criterion, what
is the largest safe value of the force F ?
Solution:
The maximum safe load which can be supported by member AB is:
(FAB )MAX = (20 × 106 N/ m2 )(0.015 m2 )
(FAB )MAX = 300, 000 N = 300 kN
Summing moments about point C to find FMAX :
ΣMC = 0 = −(FAB )MAX (sin 60◦ )(2 m)+FMAX (3 m) = −(300, 000 N)(sin 60◦ )(2 m)+FMAX (3 m)
ANS:
FMAX = 173, 200 N = 173.2 kN
Problem 10.96 The system shown supports half of the
weight of the 680-kg excavator. The cross-sectional area
of member AB is 0.0012 m2 . If the system is stationary,
what normal stress acts on a plane perpendicular to the
axis of member AC?
Free Body Diagrams:
Solution:
The weight of the excavator is:
W = mg = (680 kg)(9.81 m/sec2 )
W = 6671 N
The weight which must be supported by EACH SIDE of the system is
W/2 = 3335 N.
The axial load in member AB can be determined directly by summing
moments about point D.
ΣMD = 0 = (3335 N)(0.2 m)−FAB (sin 41.2◦ )(0.45 m)−FAB (cos 41.2◦ )(0.1 m)
FAB = 1795 N
The normal stress in member AB is:
σAB = FAB /AAB = (1795 N)/(0.0012 m2 )
ANS:
σAB = 1.496 MPa = 1.5 MPa
Problem 10.97
Free Body Diagram:
Member AC in Problem 10.96 has a cross-sectional area
of 0.0014 m2 . If the system is stationary, what normal
stress acts on a plane perpendicular to the axis of member
AC?
Solution:
The load in member AB was (1795 N) was calculated in the solution
for Problem 10.96. The axial load in member AC can be determined
directly by summing vertical forces at joint A.
ΣFy = 0 = −FAB (sin 41.2◦ )+FAC (sin 48.4◦ ) = −(1795 N)(sin 41.2◦ )+FAC (sin 48.4◦ )
FAC = 1581 N (C)
The normal stress in member AC is:
σAC = FAC /AAC = (1581 N)/(0.0014 m2 )
ANS: σAC = −1.13 MPa Note: The negative sign indicates a
compressive stress.
Problem 10.98 The bar has modulus of elasticity E =
30 × 106 psi, Poisson’s ration ν = 0.32, and a circular
cross section with diameter D = 0.75 in. There is a gap
b = 0.02 in. between the right end of the bar and the
rigid wall. If the bar is stretched so that it contacts the
rigid wall and is welded to it, what is the bar’s diameter
afterward?
Solution:
The strain which must be produced in order to close the gap is: ε =
L −L
in−9.00 in
= 9.02 9.00
= 0.0022
L
in
We can use the definition of Poisson’s ratio to establish the change in
diameter.
−(D −D)/D
0.32 =
ε
D = 0.7495 in
=
−(D−0.75 in)/(0.75 in)
0.0022
Problem 10.99 After the bar in Problem 10.98 is
welded to the rigid wall, what is the normal stress on
a plane perpendicular to the bar’s axis?
Solution:
A strain of ε = 0.0022 was determined in the solution to Problem 10.98.
The stress required to produce this strain is found using the modulus
of elasticity.
E=
σ
ε
σ = εE = (0.00222) 30 × 106 lb/in2
σ = 66.7 ksi
Problem 10.100 The link AB of the pliers has a
cross-sectional area of 40 mm2 and elastic modulus
E = 210 GPa. If forces F = 150 N are applied to
the pliers, what is the change in length of link AB?
Free Body Diagram:
Solution:
Summing moments about point D on the lower handle allows us to
solve directly for PAB .
ΣMD = 0 = −(150 N)(0.13 m) + (PAB )(sin 23.2◦ )(0.03 m)
PAB = 1650 N (C)
The length of link AB is:
LAB = (70 mm)2 + (30 mm)2 = 76.16 mm
The change in length for link AB is:
δ=
ANS:
PL
AE
=
(−1650 N)(76.16 mm)
N/m2 )
(40×10−6 m2 )(210×109
δ = −0.015 mm
Problem 10.101
Suppose that you want to design the pliers in Problem 10.100 so that forces F as large as 450 N can be
applied. The link AB is to be made of a material that
will support a compressive normal stress of 200 MPa.
Based on this criterion, what minimum cross-sectional
area must link AB have?
Free Body Diagram:
Solution:
With an applied load of 450 N to the lower handle, we can solve
directly for the load in link AB by summing moments about point D.
ΣMD = 0 = PAB (sin 23.2◦ ) (0.03 m) − (450 N) (0.13 m)
PAB = 4950 N
Determining the minimal safe cross-sectional area for link AB:
A=
ANS:
P
4950 N
=
σALLOW
200 × 106 N/m2
A = 24.8 × 10−6 m2 = 24.8 mm2
Problem 10.102 Each bar has a cross-sectional area
2
of 3 in2 and modulus of elasticity E = 12 × 106 lb/in .
If a 40-kip horizontal force directed toward the right is
applied at A, what are the normal stresses in the bars?
Free Body Diagram:
Solution:
Step 1—Equilibrium:
Members AB and AD are assumed to be in tension. Member AC is
assumed to be in compression. Summing vertical forces at point A:
ΣFy = 0 = PAB (sin 40◦ ) − PAC (sin 50◦ ) + PAD (sin 70◦ )
PAC = +0.839PAB + 1.227PAD
[1]
Summing horizontal forces at point A:
ΣFx = 0 = 40, 000 lb−PAB (cos 40◦ )−PAD (cos 70◦ )−PAC (cos 50◦ )
PAC = +62, 229 lb − 1.192PAB − 0.532PAD
[2]
Step 2—Lengths:
LAB =
60
= 93.343
sin 40◦
LAC =
60
= 78.324
sin 50◦
LAD =
60
= 63.857
sin 70◦
Step 3—Force–Deformation:
δAB
=
=
PAB LAB
AE
PAB (93.343)
(3)(12 × 106 )
δAB = 2.59286 × 106 PAB
δAC
=
=
PAC LAC
AE
PAC (78.324)
(3)(12 × 106 )
δAC = 2.17567 × 106 PAC
δAD
=
=
[3]
[4]
PAD LAD
AE
PAD (63.857)
(3)(12 × 106 )
δAD = 1.77364 × 106 PAD
[5]
Step 4—Compability: Changes in length related to the horizontal (u)
and vertical (v) displacements
δAB = u cos 40◦ + v sin 40◦
δAB = 0.76604u + 0.64288v
[6]
δAC = u cos 50◦ − v sin 50◦
δAC = 0.64278u − 0.76604v
[7]
δAD = u cos 70◦ + v sin 70◦
δAD = 0.34202u + 0.93969v
[8]
Step 5—Unknowns:
PAB , PAC , PAC , δAB , δAC , δAC , u, v
There are eight unknowns and eight equations. Solve the system simultaneously to find
Loads:
PAB = 22, 336.227
PAC = 30, 495.968
PAD = 9581.85
Horizontal displacement:
u = 0.087
Vertical displacement:
v = 0.01358
Normal stresses:
σAB =
ANS:
σAB = 7.445 ksi
σAC =
ANS:
22, 336.227
= 7445.409
3
30, 495.968
= 10, 165.32
3
σAC = 10.17 ksi
σAD =
ANS:
σAD = 3.19 ksi
9581.85
= 3193.95
3
Problem 10.103 The bars of the system in Problem 10.102 consist of a material that will safely support
a tensile normal stress of 20 ksi. Based on this criterion,
what is the largest downward force that can safely be
applied at A?
Free Body Diagram:
Solution:
Summing horizontal forces at point A:
ΣFx = 0 = FAC (cos 50◦ )−FAD (cos 70◦ )−FAB (cos 40◦ )
[1]
Summing vertical forces at point A:
ΣFy = 0 = −F +FAB (sin 40◦ )+FAD (sin 70◦ )+FAC (sin 50◦ )
The lengths of the bars are:
LAB = (60 in)/(sin 40◦ ) = 93.3 in.
[3]
LAC = (60 in)/(sin 50◦ ) = 78.3 in
[4]
LAD = (60 in)/(sin 70◦ ) = 63.9 in
[5]
The stresses in the three bars are:
σAB = FAB /3 in2
[6]
σAC = FAC /3 in2
[7]
σAD = FAD /3 in2
[8]
The horizontal displacement may be expressed in three different ways.
u = (LAB + δAB )(cos θAB ) − LAB (cos 40◦ )
[9]
u = (LAC + δAC )(cos θAC ) − LAC (cos 50◦ )
[10]
u = (LAD + δAD )(cos θAD ) − LAD (cos 70◦ )
[11]
The vertical displacement may be expressed in three different ways.
v = (LAB + δAB )(cos θAB ) − LAB (sin 40◦ )
[12]
v = (LAC + δAC )(cos θAC ) − LAC (sin 50◦ )
[13]
v = (LAD + δAD )(cos θAD ) − LAD (sin 70◦ )
[14]
Assuming that bar AD carries the largest load among the three bars
and will thus be the limiting stress consideration, we solve this system
of 14 equations together.
ANS:
F = 112.3 kip
[2]
Problem 11.1 A cube of material is subjected to a pure
shear stress τ = 9 MPa. The angle β is measured ad
determined to be 89.98◦ . What is the shear modulus G
of the material?
Diagram:
Solution:
Converting the shear strain angle into radians:
γ=
(90◦ − 89.98◦ )
π = 3.49 × 10−4 radians
180◦
Using the definition of the shear modulus:
G=
ANS:
τ
9 × 106 N/m2
=
γ
3.49 × 10−4
G = 25.8 GPa
Problem 11.2 If the cube in Problem 11.1 consists of
material with shear modulus G = 4.6 × 106 psi and the
shear stress τ = 8000 psi, what is the angle β in degrees?
Free Body Diagram:
Solution:
The shear strain will be:
γ=
τ
8, 000 lb/in2
=
= 1.739×10−3 radians = 0.0996◦
G
4.6 × 106 lb/in2
The angle β is:
β = 90◦ − γ = 90◦ − 0.0996◦
ANS:
β = 89.9◦
Problem 11.3 If the cube in Problem 11.1 consists of
aluminum alloy that will safely support a pure stress of
270 MPa and G = 26.3 GPa, what is the largest shear
strain to which the cube can safely be subjected?
Solution:
The shear strain will be:
γ=
ANS:
τ
270 × 106 N/m2
= 0.010266
=
G
26.3 × 109 N/m2
γ = 0.0103
Problem 11.4 The cube of material is subjected to a
pure shear stress τ = 12 MPa. What are the normal
stress and the magnitude of the shear stress on the plane
P?
Free Body Diagram:
Solution:
Summing vertical forces on the free body diagram:
ΣFY = 0 = −(12×106 N/m2 )A(cos 30◦ )+τP A(cos 30◦ )+σA(sin 30◦ )
[1]
0.866τP + 0.5σP = 10.39 × 106 N/m2
Summing horizontal forces on the free body diagram:
ΣFX = 0 = −(12×106 N/m2 )A(sin 30◦ )−τP A(sin 30◦ )+σP A(cos 30◦ )
[2]
− 0.5τP + 0.866σP = 6 × 106 N/m2
Solving Equations [1] and [2] together:
ANS:
ANS:
τP = 6 MPa
σP = 10.4 MPa
Problem 11.5 In Problem 11.4, what are the magnitudes of the maximum tensile, compressive, and shear
stresses to which the material is subjected?
Free Body Diagram:
Solution:
Summing forces in the x-direction on the element:
ΣFx = 0 = σA A(cos θ)−(12×106 N/m2 )A(sin θ)−τ A(sin θ)[1]
Solving Equation [1] for σA :
τ =
σA (cos θ) − (12 × 106 N/m2 )(sin θ)
= σA (cot θ)−12×106 N/m2
sin θ
We see that σA is maximum when cot θ is minimum (θ = 0◦ ), or:
ANS:
σMAX = τ = 12 MPa
Summing forces in the y-direction on the element:
ΣFy = 0 = −(12×106 N/m2 )(A)(cos θ)+τ A(cos θ)+σA(sin θ)
Solving Equation [2] for τ
τ = 12 MPa − σ(tan θ)
We see that τ is maximum when tan θ is minimum (θ = 0), or:
ANS:
τMAX = τ = 12 MPa
[2]
Problem 11.6 The cube of material shown in Problem 11.4 is subjected to a pure shear stress τ . If the
normal stress on the plane P is 14 MPa, what is τ ?
Free Body Diagram:
Solution:
Summing forces in the x-direction on the element:
ΣFx = 0 = (14×106 N/m2 )A(cos 30◦ )−τ A(sin 30◦ )−τ A(sin 30◦ ) → τ = 24.25×106 N/m2 −τ
[1]
Summing forces in the y-direction on the element:
ΣFy = 0 = (14×106 N/m2 )A(sin 30◦ )+τ A(cos 30◦ )−τ A(cos 30◦ ) → τ = −8.083×106 N/m2 +τ
Solving Equations [1] and [2] together:
24.25 × 106 N/m2 − τ = −8.083 × 106 N/m2 + τ
ANS:
τ = 16.17 MPa
Problem 11.7 The cube of material shown in Problem 11.4 is subjected to a pure shear stress τ . The shear
modulus of the material is G = 28 GPa. If the normal
stress on the plane P is 80 MPa, what is the shear strain
of the cube.
Free Body Diagram:
Solution:
Summing forces in the x-direction on the element:
ΣFx = 0 = (80×106 N/m2 )A(cos 30◦ )−τ A(sin 30◦ )−τ A(sin 30◦ ) → τ = 138.6 MPa−τ
[1]
Summing forces in the y-direction on the element:
ΣFy = 0 = (80×106 N/m2 )A(sin 30◦ )+τ A(cos 30◦ )−τ A(cos 30◦ ) → τ = −46.19 MPa+τ
Solving Equations [1] and [2] together:
τ = 92.4 MPa
The shear strain is:
γ=
ANS:
τ
92.4 × 106 N/m2
=
= 0.003299
G
28 × 109 N/m2
γ = 0.0033
[2]
[2]
Problem 11.8 The cube of material is subjected to a
pure shear stress τ = 20 ksi.
(a) What are the normal stress and the magnitude of the
shear stress on the plane P ? (b) What are the magnitudes
of the maximum tensile, compressive, and shear stresses
to which the material is subjected?
Free Body Diagram:
Solution:
Summing vertical forces on the free body diagram:
ΣFY = 0 = (20, 000 lb/in2 )A(cos 30◦ )−τθ A(cos 30◦ )−σθ A(sin 30◦ )
[1] τθ (0.866) + σθ (0.5) = 17, 320 lb/in2
ΣFX = 0 = −(20, 000 lb/in2 )A(sin 30◦ )−τθ A(sin 30◦ )+σθ A(cos 30◦ )
[2] τθ (0.5) − σθ (0.866) = −10, 000 lb/in2
Solving Equations [1] and [2] together:
σθ = −17, 316 lb/in2
(a) ANS:
ANS: τθ = 10, 000 lb/in2
(b) ANS: 20 ksi
Problem 11.9 The cube of material shown in Problem 11.4 is subjected to a pure shear stress τ . If the
normal stress on the plane P is -20 ksi, what is τ ?
Free Body Diagram:
Solution:
Summing forces in the x-direction on the element:
ΣFx = 0 = (20, 000 lb/in2 )A(cos 30◦ )+τ A(sin 30◦ )−τ A(sin 30◦ ) → τ = 34, 641 lb/in2 +τ
[1]
Summing forces in the y-direction on the element:
ΣFy = 0 = −(20, 000 lb/in2 )A(sin 30◦ )+τ A(cos 30◦ )−τ A(cos 30◦ ) → τ = −11, 547 lb/in2 −τ
Solving Equations [1] and [2] together:
−34, 641 lb/in2 − τ = 11, 547 lb/in2 + τ
ANS:
τ = −23, 094 lb/in2
[2]
Problem 11.10 The cube of material shown in Problem 11.4 is subjected to a pure shear stress τ . The shear
modulus of the material is G = 4 × 106 psi. If the normal stress on the plane P is -12 ksi, what is the shear
strain of the cube?
Free Body Diagram:
Solution:
Summing forces in the x-direction on the element:
ΣFx = 0 = −(12, 000 lb/in2 )A(cos 30◦ )+τ A(sin 30◦ )−τ A(sin 30◦ ) → τ = 20, 784 lb/in2 +τ
[1]
Summing forces in the y-direction on the element:
ΣFy = 0 = −(12, 000 lb/in2 )A(sin 30◦ )+τ A(cos 30◦ )−τ A(cos 30◦ ) → τ = −6, 928 lb/in2 −τ
Solving Equations [1] and [2] together:
20, 784 lb/in2 − τ = −6, 928 lb/in2 + τ
τ = 13, 856 lb/in2
The shear strain is:
γ=
ANS:
τ
13, 856 lb/in2
=
G
4 × 106 lb/in2
γ = 0.00346
Problem 11.11 If a bar has a solid circular cross section with 15-mm diameter, what is the polar moment of
inertia of its cross section in m4 ?
Solution:
The polar moment of inertia for the cross section is:
π
π 0.015 m 4
J = r4 =
2
2
2
ANS:
J= 4.97 × 10−9 m4
Problem 11.12 If a bar has a hollow circular cross
section with 2-in. outer radius and 1-in. inner radius,
what is the polar moment of inertia of its cross section?
Solution:
The polar moment of inertia for the cross section is:
π
π
J=
(ro )4 − (ri )4 =
(2in)4 − (1in)4
2
2
ANS:
J= 23.56 in4
[2]
Problem 11.13 The bar has a circular cross section
with 15-mm diameter and the shear modulus of the material is G = 26 GPa. If the torque T = 10 N − m, determine (a) the magnitude of the maximum shear stress
in the bar; (b) the angle of twist of the end of the bar in
degrees.
Free Body Diagram:
Solution:
The polar moment of inertia for the shaft is:
J = π2 c4 = π2 (0.0075m)4
J = 4.97 × 10−9 m4
Maximum shear stress in the bar is:
τMAX =
ANS:
Tρ
(10 N − m) (0.0075 m)
=
J
4.97 × 10−9
(a) τ MAX = 15.1 × 106 N/m2
The angle of twist for the bar is:
φ=
ANS:
LT
(0.8m)(10 N − m)
=
JG
(4.97 × 10−9 m4 )(26 × 109 N/m2 )
φ = 0.0619 rad = 3.547◦
Problem 11.14 If the bar in Problem 11.13 is subjected Free Body Diagram:
to a torque T that causes the end of the bar to rotate 4◦ ,
what is the magnitude of the maximum shear stress in
the bar?
Solution:
The polar moment of inertia for the cross section is:
π
π
J = c4 = (0.0075)4 = 4.97 × 10−9
2
2
Using the angle of rotation at the end of the bar to determine the applied
torque:
◦ 4
TL
(0.8 m)(T )
(π) rad =
=
→ T = 11.27 N − m
180◦
JG
(4.97 × 10−9 m4 )(26 × 109 N/m2 )
Maximum shear stress in the cross section is:
τMAX =
ANS:
Tc
(11.27 N − m)(0.0075 m)
=
J
4.97 × 10−9 m4
τ MAX = 17.01 MPa
Problem 11.15 The bar in Problem 11.13 is to be used
in an application that requires that it be subjected to an
angle of twist no greater than 1◦ . What is the maximum
allowable value of the torque T ?
Free Body Diagram:
Solution:
The polar moment of inertia for the shaft is:
J = π2 c4 = π2 (0.0075 m)4
J = 4.97 × 10−9 m4
Converting the angle of twist into radians:
φ=
1◦
π = 0.0175 radians
180◦
The torque which will produce this angle of twist is:
(0.0175) 4.97 × 10−9 m4 26 × 109 N/m2
φJG
T =
=
L
0.8 m
ANS:
T = 2.82 N − m
Problem 11.16 The solid circular shaft that connects
the turbine blades of the hydroelectric power unit to the
generator has a 0.4-m radius and supports a torque T =
2 MN − m. What is the maximum shear stress in the
shaft?
Free Body Diagram:
Solution:
The polar moment of inertia for the shaft is:
π
π
J = c4 = (0.4 m)4 = 0.0402 m4
2
2
Maximum shear stress in the shaft is:
τMAX =
ANS:
Tρ
(2 × 106 N − m)(0.4 m)
=
J
0.0402 m4
τ MAX = 19.9 MPa
Problem 11.17 Consider the solid circular shaft in
Problem 11.16. The shear modulus of the material is
G = 80 GPa. What angle of twist per unit meter of
length is caused by the 2-MN-m torque?
Free Body Diagram:
Solution:
The polar moment of inertia for the shaft is:
π
π
J = c4 = (0.4 m)4 = 0.0402 m4
2
2
The angle of twist per meter of length is:
φ
T
2 × 106 N − m
=
=
L
JG
(0.0402 m4 )(80 × 109 N/m2 )
ANS:
φ
L
= 0.00062 rad/m = 0.0356 degrees/m
Problem 11.18 If the shaft in Problem 11.16 has a
hollow circular cross section with 0.5-m outer radius and
0.3-m inner radius, what is the maximum shear stress?
Free Body Diagram:
Solution:
The polar moment of inertia for the hollow shaft is:
π
π
J = (ro − ri ) =
(0.5 m)4 − (0.3 m)4
2
2
J = 0.0855 m4
Maximum shear stress in the shaft is:
2 × 106 N − m (0.5 m)
Tρ
τMAX =
=
J
0.0855 m4
ANS:
τ MAX = 11.7 MPa
Problem 11.19 The propeller of the wind generator is
supported by a hollow circular shaft with 0.4-m outer
radius and 0.3-m inner radius. The shear modulus of the
material is G = 80 GPa. If the propeller exerts an 840kN-m torque on the shaft, what is the resulting maximum
shear stress?
Free Body Diagram:
Solution:
The polar moment of inertia for the shaft is:
J = π2 (0.4 m)4 − (0.3 m)4
J = 0.0275 m4
Maximum shear stress in the shaft is:
τMAX =
ANS:
Tc
(840, 000 N − m) (0.4 m)
=
J
0.0275 m4
τ MAX = 12.2 MPa
Problem 11.20 In Problem 11.19, what is the angle of
twist of the propeller shaft per meter of length?
Free Body Diagram:
Solution:
The polar moment of inertia for the shaft is:
J = π2 (0.4 m)4 − (0.3 m)4
J = 0.0275 m4
Angle of twist for the shaft is:
φ=
ANS:
LT
(1 m) (840, 000 N − m)
=
JG
(0.0275 m4 ) (80 × 109 N/m2 )
φ = 0.000382 rad = 0.0219◦
Problem 11.21 In designing a new shaft for the wind
generator in Problem 11.19, the engineer wants to limit
the maximum shear stress in the shaft to 10 MPa, but design constraints require retaining the 0.4-m outer radius.
What new inner radius should she use?
Solution:
The polar moment of inertia for the shaft is:
J = π2 (0.4 m)4 − (ri )4
J = 0.0402 − 1.571ri4
Maximum shear stress in the shaft is:
τMAX =
ANS:
Tc
(840, 000 N − m) (0.4 m)
=
= 10 × 106 N/m2
J
(0.0402 − 1.571ri4 )
ri = 0.2546 m
Problem 11.22 The bar has a circular cross section
with 1-in. diameter and the shear modulus of the material
is G = 5.8 × 106 psi. If the torque T = 1000 in − lb,
determine (a) the magnitude of the maximum shear stress
in the bar; (b) the magnitude of the angle of twist of the
right end of the bar relative to the wall in degrees.
Free Body Diagram:
Solution:
Maximum torque in the shaft is 1,000 in-lb.
The polar moment of inertia for the shaft is:
J = π2 (0.5 in)4
J = 0.0982 in4
(a) Maximum shear stress in the shaft is:
TC
(1, 000 in − lb) (0.5 in)
=
J
0.0982 in4
τMAX =
ANS:
τMAX = 5092.958 lb/in2
(b) The angle of twist in the 8-inch section of the bar is:
φ8 in =
φ8 in
(8 in)(500 in−lb))
(0.0982 in4 )(5.8×106 in4 )
= 0.00702 rad = 0.402◦
The angle of twist in the 6-inch section of the bar is:
φ6 in =
φ6 in
(6 in)(1,000 in−lb)
(0.0982 in4 )(5.8×106 lb/in2 )
= 0.0105 rad = 0.604◦
Total angle of twist for the bar is:
φ = φ8 in + φ6 in = 0.402◦ + 0.604◦
ANS:
φ = 1.006◦
Problem 11.23 For the bar in Problem 11.22, what
value of the torque T would cause the angle of twist of
the end of the bar to be zero?
Free Body Diagram:
Solution:
The torque in the 8-inch section of the bar is (T − 500 in − lb).
The torque in the 6-inch section of the bar is T .
The equation for total angle of twist for the bar is:
0=
L8 in T8 in L6 in T6 in
(8 in) (T − 500 in − lb) (6 in) T
+
=
+
JG
JG
JG
JG
Solving the equation for T :
ANS:
T = 286 in − lb
Problem 11.24 Part A of the bar has a solid circular cross section and Part B has a hollow circular
cross section. The shear modulus of the material is
G = 3.8 × 106 psi. Determine the magnitudes of the
maximum shear stresses in parts A and B of the bar.
Free Body Diagram:
Solution:
The torque in the solid section of the bar is 250,000 in-lb. The torque
in the hollow section of the bar is 100,000 in-lb. Polar moment of
inertia for the solid section of the bar is:
π
JS = (2 in)4 = 25.13 in4
2
Polar moment of inertia for the hollow section of the shaft is:
π
JH =
(2 in)4 − (1 in)4 = 23.56 in4
2
Maximum shear stress in the solid section of the bar is:
(τMAX )S =
ANS:
TS cS
(250, 000 in − lb) (2 in)
=
JS
25.13 in − lb
(τMAX )S = 19, 896.54 lb/in2 = 19.89 ksi
Maximum shear stress in the hollow section of the bar is:
(τMAX )H =
ANS:
TH cH
(100, 000 in − lb) (2 in)
=
JH
23.56 in4
(τMAX )H = 8, 488.96 lb/in2 = 8.49 ksi
Problem 11.25 For the bar in Problem 11.24, deter- Free Body Diagram:
mine the magnitude of the angle of twist of the end of
the bar in degrees.
Solution:
The torque in the solid section of the bar is 250,000 in-lb. The torque
in the hollow section of the bar is 100,000 in-lb. Polar moment of
inertia for the solid section of the bar is:
π
JS = (2 in)4 = 25.13 in4
2
Polar moment of inertia for the hollow section of the shaft is:
π
JH =
(2 in)4 − (1 in)4 = 23.56 in4
2
The angle of twist for the solid section of the shaft is:
φ=
LS TS
(7 in) (250, 000 in − lb)
= JS GS
25.13 in4 3.8 × 106 lb/in2
φS = 0.0183 rad = 1.05◦
The angle of twist for the hollow section of the shaft is:
φ=
LH TH
(14 in) (100, 000 in − lb)
= JH GH
23.56 in4 3.8 × 106 lb/in2
φH = 0.0156 rad = 0.896◦
Total angle of twist for the shaft is:
φt = φS + φH = 1.05◦ + 0.896◦
φ = 1.95◦
ANS:
Problem 11.26 For the bar in Problem 11.24, deter- Free Body Diagram:
mine the magnitude of the maximum shear stresses in
parts A and B of the bar and the magnitude of the angle
of twist of the end of the bar in degrees if the 150 in-kip
couple acts in the opposite direction.
Solution:
Polar moments of inertia for the two sections of the bar are:
π
π
JA = (2in)4 = 25.13 in4 JB =
(2in)4 − (1in)4 = 23.56 in4
2
2
From the FBD we see that the torque in the sections of the bar is:
TA = −50, 000 in − lb
TB = 100, 000 in − lb
Maximum shear stresses in the sections of the bar are:
(τA )MAX =
(50, 000 in − lb)(2 in)
25.13 in4
(τB )MAX =
(100, 000 in − lb)(2 in)
23.56 in4
(τA )MAX = 3, 980 lb/in2 (τB )MAX = 8, 488.96 lb/in2 ≈ 8.49 ksi
ANS:
The angles of twist in each of the sections of the bar are:
φA = −
(50, 000 in − lb)(7 in)
(25.13 in4 )(3.8 × 106 lb/in2 )
φA = −0.00367 radians
φB =
(100, 000 in − lb)(14 in)
(23.56 in4 )(3.8 × 106 lb/in2 )
φB = 0.01564 radians
Total angle of twist is:
φ = φA + φB = −0.00367 rad + 0.01564 rad
ANS:
φ = 0.01197 rad = 0.686◦
Problem 11.27 The lengths LA = LB = 200 mm and
LC = 240 mm. The diameter of parts A and C of the bar
is 25 mm and the diameter of part B is 50 mm. The shear
modulus of the material is G = 80 GPa.If thetorqueT
= 2.2 kN-m, determine the magnitude of the angle of
twist of the right end of the bar relative to the wall.
Free Body Diagram:
Solution:
The torques in the three sections of the blade are:
TA = 2, 200 N − m−8, 000 N − m+4, 000 N − m = −1, 800 N − m
TB = 2, 200 N − m − 8, 000 N − m = −5, 800 N − m
TC = 2, 200 N − m
The polar moment of inertia for sections A ad C is:
π
JA = JC = (0.0125 m)4 = 38.35 × 10−9 m4
2
The polar moment of inertia for section B of the bar is:
π
JB = (0.025 m)4 = 614 × 10−9 m4
2
The angle of twist in each section of the bar is:
φA =
L A TA
JA GA
=
φB =
L B TB
JB GB
=
φC =
L C TC
JC GC
=
(0.2 m)(−1,800 N−m)
m4 )(80×109 N/m2 )
(38.35×10−9
(0.2 m)(−5,800 N−m)
m4 )(80×109 N/m2 )
(614×10−9
= −0.0236 rad = −1.353◦
(0.24 m)(2,200 N−m)
m4 )(80×109 N/m2 )
(38.35×10−9
= −0.1173 rad = −6.723◦
= 0.172 rad = 9.861◦
Total angle of twist I the bar is:
φ = φA + φB + φC = −6.723◦ − 1.353◦ + 9.861◦
ANS:
φ = 1.78◦
Problem 11.28 For the bar in Problem 11.27, what
value of the torque T would cause the angle of twist of
the right end of the bar relative to the wall to be zero?
Free Body Diagram:
Solution:
The torques in the three sections of the blade are:
TA = T − 8, 000 N − m + 4, 000 N − m = T − 4, 000 N − m
TB = T − 8, 000 N − m
TC = T
The equation which expresses the angle of twist in the bar is:
0=
LA TA LB TB LC TC
(0.2 m) (T − 4, 000 N − m) (0.2 m) (T − 8, 000 N − m) (0.24 m) (T )
+
+
=
+
+
JA GA JB GB JC GC
JG
JG
JG
Solving the above equation for T :
ANS:
T = 1, 988.84 N − m = 1.99 kN − m
Problem 11.29 The bar in Problem 11.27 is made of
a material that can safely support a pure shear stress of
1.1 GPa. Based on this criterion, what is the range of
positive values of the torque T that can safely be applied?
Free Body Diagram:
Solution:
We see from the FBD that the torques in the three sections of the bar
are:
TA = −4, 000 N − m−T
TB = −8, 000 N − m+T
TC = T
Note: As the torque T increases, the torque in section A initially
decreases. We can see that the center section, B, will NOT be
factor in determining the limiting torque (large J and relatively small
T ).
The torque, T , which will result in maximum shear stress in section A
is:
τA = 1.1×109 N/m2 =
(−4, 000 N − m + T )(0.0125 m)
→ T = 7, 375 N − m
π
(0.0125 m)4
2
The torque, T , which will result in maximum shear stress in section C
is:
τC = 1.1 × 109 N/m2 =
T (0.0125 m)
→ T = 3, 375 N − m
m)4
π
(0.0125
2
The range of allowable torques, T , is:
ANS:
3, 375 N − m ≤ T ≤ 7, 375 N − m
Problem 11.30 The bars AB and CD each have a solid
circular cross section with 30-mm diameter and consist
of a material with a shear modulus G = 28 GPa. The
ratio of the gears are rB = 120 mm and rC = 90 mm.
If the torque TA = 200 N − m, what are the maximum
shear stresses in the bars?
Diagram:
Solution:
The polar moment of inertia for the two shafts is:
π
J = (0.015 m)4 = 79.52 × 10−9 m4
2
Maximum shear stress in bar AB is:
(τMAX )AB =
ANS:
TAB cAB
(200 N − m) (0.015 m)
=
JAB
79.52 × 10−9 m4
(τMAX )AB = 37.7 MPa
The torque in shaft CD is:
rc
0.045 m
TCD =
TAB =
200 N − m = 150 N − m
rb
0.060 m
Maximum shear stress in bar CD is:
(τMAX )CD =
ANS:
TCD cCD
(150 N − m) (0.015 m)
=
JCD
79.52 × 10−9 m4
(τMAX )CD = 28.3 MPa
Problem 11.31 In Problem 11.30, what is the angle of
twist at A? (Assume that the deformations of the gears
are negligible.)
Diagram:
Solution:
The polar moment of inertia for the two shafts is:
π
J = (0.015m)4 = 79.52 × 10−9 m4
2
The torques on the two shafts are:
90 mm
TA = 200 N − m TC =
TA = (0.75)(200 N − m) = 150 N − m
120 mm
Total angle of twist at A is:
φ = φCD +φAB =
ANS:
(1m)(150 N − m)
(1m)(200 N − m)
+
(79.52 × 10−9 m4 )(28 × 109 N/m2 ) (79.52 × 10−9 m4 )(28 × 109 N/m2 )
φ = 0.157 rad = 9◦
Problem 11.32 Consider the system shown in Prob- Diagram:
lem 11.30. The bars AB and CD each have a solid
circular cross section with 30-mm diameter. The radii
of the gears must satisfy the relation rB +rC = 210 mm.
If the torque TA = 200 kN − m and the bars are made
of a material that will safely support a pure shear stress
of 40 MPa, what is the largest safe value of the radius
rC ?
Solution:
The torque in bar AB is TAB = 200 N − m. The shear stress in bar
AB is:
τAB =
(200 N − m)(0.015 m)
= 37.73 MPa
π
(0.015 m)4
2
The shear stress in bar CD, limited to 40 MPa, can be expressed as:
rC
rC
τAB =
(37.73 MPa) [1]
τCD = 40×106 N/m2 =
rB
rB
A second equation we can use is:
rB + rC = 0.21 m
[2]
Solving equations [1] and [2] together:
ANS:
rC = 0.108 m = 108 mm
Problem 11.33 The bar has a circular cross section Free Body Diagram:
with 1-in. diameter. If the torque TO = 1000 in − lb,
determine the magnitudes of the maximum shear stresses
in parts A and B of the bar.
Solution:
The polar moment of inertia for the bar is:
π
π
J = c4 = (0.5 in)4 = 0.0982 in4
2
2
Since the sum of moments on the bar must be zero:
MA + MB = 1, 000 in − lb
[1]
We see that φA = φB , so we have:
(8 in) (MA )
(6 in) MB
=
JG
JG
MA = 0.75MB
[2]
Solving equations [1] and [2] together, we get:
MA = 428.6 in − lb
MB = 571.4 in − lb
Maximum shear stress in each section of the bar is:
(τMAX )A =
ANS:
(τMAX )A = 2180 lb/in2
(τMAX )B =
ANS:
M A cA
(428.6 in − lb) (0.5 in)
=
JA
0.0982 in4
M B cB
(571.4 in − lb) (0.5 in)
=
JB
0.0982 in4
(τMAX )B = 2910 lb/in2
Problem 11.34 Suppose that the bar in Problem 11.33 Free Body Diagram:
consists of a material that will safely support a maximum
shear stress of 40 ksi. Based on this criterion, what is
the maximum safe magnitude of the torque TO ?
Solution:
The polar moment of inertia for the bar is:
π
π
J = c4 = (0.5 in)4 = 0.0982 in4
2
2
We see that:
MA + MB = T
[1]
Since φA = φB, we also see that:
MA (8 in)
MB (6 in)
=
JG
JG
MA = 0.75MB
[2]
Solving equations [1] and [2] together, we get:
MA = 0.429T
MB = 0.571T
We see that the maximum torque is in section B. With a maximum
allowable shear stress of 40,000 psi:
40, 000 lb/in2 =
(0.571T ) (0.5 in)
0.0982 in4
Maximum allowable torque is:
ANS:
T = 13.8 in − kip
Problem 11.35 Suppose that the bar in Problem 11.33 Free Body Diagram:
is subjected to a torque T0 = 10, 000 in − lb and consists
of a material that will safely support a maximum shear
stress of 40 ksi. Based on this criterion, what is the
largest distance from the left end of the bar at which the
torque can safely be applied?
Solution:
The polar moment of inertia for the bar is:
π
π
J = c4 = (0.5 in)4 = 0.0982 in4
2
2
To find the maximum allowable moment:
40, 000 lb/in2 =
MMAX (0.5 in)
0.0982 in4
MMAX = 7856 in − lb
We see that:
MA + MB = 10, 000 in = lb
As the applied moment moves from left-to-right, the moment at the
right-hand end increases. Knowing that φA = φB :
((10, 000 in − lb) − MMAX )(14in − LR )
MMAX LR
=
JG
JG
((10, 000 in − lb) − 7856 in − lb)(14in − LR )
(7856 in − lb)(LR )
=
JG
JG
LR = 3 in
ANS:
LL = 11 in
Problem 11.36 The bar is fixed at both ends. It consists Free Body Diagram:
of material with shear modulus G = 28 GPa and has a
solid circular cross section. Part A is 40 mm in diameter
and part B is 20 mm in diameter. Determine the torques
exerted on the bar by the walls.
Solution:
Polar moments of inertia for the two sections are:
π
π
π
π
JA = c4 = (0.02 m)4 = 251×10−9 m4 JB = c4 = (0.01 m)4 = 15.7×10−9 m4
2
2
2
2
We see that:
MA + MB = 1, 200 N − m
[1]
Since φA = φB , we also see that:
(0.16 m) (MA )
(0.12 m) (MB )
=
(251 × 10−9 m4 ) G
(15.7 × 10−9 m4 ) G
MA = 12MB
[2]
Solving equations [1] and [2] together:
MA = 1107.7 N − m
ANS:
ANS:
MB = 92.3 N − m
Problem 11.37 Determine the magnitudes of the max- Free Body Diagram:
imum shear stresses in parts A and B of the bar in Problem 11.36.
Solution:
JA =
π 4
c
2
=
π
2
(0.02 m)4 = 251 × 10−9 m4
JB =
We see that:
MA + MB = 1, 200 N − m
[1]
Since φA = φB , we also see that:
(0.16 m) (MA )
(0.12 m) (MB )
=
(251 × 10−9 m4 ) G
(15.7 × 10−9 m4 ) G
MA = 12MB
[2]
Solving equations [1] and [2] together:
MA = 1107.7 N − m
MB = 92.3 N − m
Maximum shear stress in the two sections is:
(τMAX )A =
ANS:
M A cA
(1107.7 N − m) (0.02 m)
=
JA
251 × 10−9 m4
(τMAX )A = 88.3 MPa
(τMAX )B =
ANS:
M B cB
(92.3 N − m) (0.01 m)
=
JB
15.7 × 10−9 m4
(τMAX )B = 58.8 MPa
π 4
π
c = (0.01 m)4 = 15.7×10−9 m4
2
2
Problem 11.38 Each bar is 10 in. long and has a solid Free Body Diagram:
circular cross section. Bar A has a diameter of 1 in. and
its shear modulus is 6 × 106 psi. Bar B has a diameter
of 2 in. and its shear modulus is 3.8 × 106 psi. The ends
of the bars are separated by a small gap. The free end of
bar A is rotated 2◦ about the bar’s axis and the bars are
welded together. What are the magnitudes of the angles
of twist (in degrees) of the two bars afterward?
Solution:
Polar moments of inertia for the two bars are:
π
π
π
π
JA = r 4 = (0.5 in)4 = 0.0982 in4 JB = r 4 = (1 in)4 = 1.571 in4
2
2
2
2
The moment required to produce an angle of twist of two degrees in
bar A is:
(0.0349 rad) 0.0982 in4 6 × 106 lb/in2
φJG
M =
=
= 2056 in − lb
L
10 in
After the two bars are welded together, each of the welded ends will
rotate until equilibrium is achieved. The total of the two deflection
angles will be 2◦ . Because the bars, after welding, are in contact with
each other, the moments exerted by each of the bars are equal. We
have φA = φB and MA = MB , so:
MA (10 in)
MA (10 in)
+
= (2◦ )
0.0982 in4 6 × 106 lb/in2
1.571 in4 3.8 × 106 lb/in2
3.14159 rad
180◦
MA = MB = 1872 in − lb
The angle of twist in each bar is:
ANS:
φA =
MA LA
JA GA
=
ANS:
φB =
MB LB
JB GB
=
(1872 in−lb)(10 in)
lb/in2 )
= 0.03177 rad = 1.82◦
(0.0982 in4 )(6×106
(1872 in−lb)(10 in)
lb/in2 )
(1.571 in4 )(3.8×106
= 0.00314 rad = 0.18◦
Problem 11.39 In Problem 11.38, the ends of the bars
are separated by a small gap. Suppose that the free end
of bar A is rotated 2◦ about its axis. The fee end of bar B
is rotated 2◦ about its axis in the opposite direction, and
the bars are welded together. What are the magnitudes of
the maximum shear stresses in the two bars afterward?
Solution:
Polar moments of inertia for the two bars:
π
π
JA = (0.5 in)4 = 0.0982 in4 JB = (1 in)4 = 1.571 in4
2
2
We see that the resulting moments in the two bars will have the same
magnitude. We also see that the total angle of twist for the two bars
will be 4◦ (0.0698 rad) when the bars achieve equilibrium.
MA = MB
[1]
φA + φB = 0.0698 rad
[2]
MA (10 in)
MA (10 in)
+
= 0.0698
(0.0982 in4 )(6 × 106 lb/in2 ) (1.571 in4 )(3.8 × 106 lb/in2 )
MA = MB = 3743 in − lb
Calculating maximum shear stress in each bar:
(3743 in−lb)(0.5 in)
0.0982 in4
ANS:
(τMAX )A =
MA rA
JA
ANS:
(τMAX )B =
(3742 in−lb)(1 in)
1.571 in4
=
= 2.38 ksi
= 19.06 ksi
Problem 11.40 The lengths LA = LB = 200 mm Free Body Diagram:
and LC = 240 mm. The diameter of parts A and C is
25 mm and the diameter of part B is 50 mm. The shear
modulus of the material is G = 80 GPa. What is the
magnitude of the maximum shear stress in the bar?
Solution:
Polar moments of inertia for the sections of the bar are:
π
π
JA = JC = (0.0125 m)4 = 3.835×10−8 m4 JB = (0.025 m)4 = 6.136×10−7 m4
2
2
We see that the sum of the three angles of twist from A to C must be
zero, so we have:
φA + φB + φC = 0
(0.2 m)(−MC + (8000 N − m) − (4000 N − m)) (0.2 m)(−MC + (8000 N − m)
(−MC )(0.24 m)
+
+
=0
(3.835 × 10−8 m4 )(G)
(6.136 × 10−7 m4 )(G)
(3.835 × 10−8 m4 )(G)
MC = 1989 N − m
Because the sum of moments on the bar is zero:
−4000 N − m + 8000 N − m − 1989 N − m − MA = 0
MA = 2011 N − m
Maximum shear stresses in each of the three sections are:
N−m)(0.0125 m)
= 655 MPa
ANS: (τMAX )A = (2011
3.835×10−8 m4
(τMAX )B =
((8000 N − m) − (1989 N − m))(0.025 m)
= 245 MPa
6.136 × 10−7 m4
(τMAX )C =
(1989 N − m)(0.0125 m)
= 648 MPa
3.835 × 10−8 m4
Problem 11.41 In Problem 11.40, through what angle
does the bar rotate at the position where the 8 kN-m
couple is applied?
Solution:
Polar moments of inertia for the sections of the bar are:
π
π
JA = JC = (0.0125 m)4 = 3.835×10−8 m4 JB = (0.025 m)4 = 6.136×10−7 m4
2
2
We see that the sum of the three angles of twist from A to C must be
zero, so we have:
φA + φB + φC = 0
(0.2 m)(−MC + (8000 N − m) − (4000 N − m)) (0.2 m)(−MC + (8000 N − m)
(−MC )(0.24 m)
+
+
=0
(3.835 × 10−8 m4 )(G)
(6.136 × 10−7 m4 )(G)
(3.835 × 10−8 m4 )(G)
MC = 1989 N − m
Because the sum of moments on the bar is zero
−4000 N − m + 8000 N − m − 1989 N − m − MA = 0
MA = 2011 N − m
The simplest means of determining the angle of twist is to start from
the right-hand end.
φ=
ANS:
LC TC
(0.24 m)(1989 N − m)
=
JC GC
(3.835 × 10−8 m4 )(80 × 109 N/m2 )
φ = 0.1556 rad = 8.91◦
Problem 11.42 The collar is rigidly attached to bar A.
The cylindrical bar A is 80 mm in diameter and its shear
modulus is G = 66 GPa. There are gaps b = 2 mm
between the arms of the collar and the ends of the identical bars B and C. Bars B and C are 30 mm in diameter
and their modulus of elasticity is E = 170 GPa. If the
bars B and C are extended so that they come into contact
with the arms of the collar and are welded to them, what
is the magnitude of the maximum shear stress in bar A
afterward?
Free Body Diagram:
Solution:
The polar moment of inertia for bar A is:
π
π
J = r4 = (0.04 m)4 = 4.021 × 10−6 m4
2
2
The torque exerted upon bar A by the bars B and C is:
T = 2[P (0.3 m)]
[1]
The compatibility condition for the gap between the collar and bars B
and C is:
PL
0.002 m = rφ + AE
0.002 m = (0.3 m) φ +
P (0.4 m)
π(0.015 m)2 (66×109 N/m2 )
0.002 m = 0.3φ + 8.57 × 10−9 P
[2]
Combining Equations [1] and [2]:
0.002 m = (0.3 m)
(0.6P N − m) (0.8 m)
P (0.4 m)
+
(4.021 × 10−6 m4 ) (66 × 109 N/m2 )
π (0.015 m)2 (170 × 109 N/m2 )
P = 3663 N
Using Equation [1] to determine the torque:
T = 2198 N − m
Maximum shear stress in bar A is:
(τMAX )A =
ANS:
Tr
(2198 N − m) (0.04 m)
=
J
4.021 × 10−6 m4
(τMAX )A = 21.9 MPa
Problem 11.43 In Example 11.2, what is the magnitude of the maximum shear stress in the bar?
Free Body Diagram:
Solution:
Maximum shear stress occurs at the wall (smallest cross-section for
the bar). From the given function for J, the polar moment of inertia at
the wall (x = 0) is:
J = 0.00016 m4
The radius of the bar at the wall is needed.
J = (π/2)(r)4 → r =
2J
π
1/4
=
1/4
2(0.00016 m4 )
π
= 0.1005 m
Maximum shear stress in the bar at the wall is:
τMAX =
(200, 000 N − m)(0.1005m)
0.00016 m4
τMAX = 125.6 MPa
ANS:
Problem 11.44 In Example 4.2, suppose that the
torque T is applied to the bar at x = 1 m. What is
the magnitude of the angle of twist of the entire bar?
Free Body Diagram:
Solution:
From the given function for J, the polar moment of inertia at x = 1 m
is:
J = 0.00016 + 0.0006(1) m4 = 0.00076 m4
The radius of the bar at x = 1 m is:
r=
2J
π
1/4
=
2(0.00076 m4 )
π
1/4
= 0.148 m
The angle of twist for the entire bar is:
1
φ=
0
ANS:
(200, 000 N − m)dx
=
(0.00016 + 0.0006x2 )(47 × 109 N/m2 )
φ = 0.015207 radians =
0.8609◦
≈ 0.861
200, 000 N − m
47 × 109 N/m2
1
0
dx
(0.00016 + 0.0006x2 )
Problem 11.45 The bar has a solid circular crosssection. Its polar moment of inertia is given by J =
(0.1 + 0.15x) in4 , where x is the axial position in
inches, and the shear modulus of the material is G =
4.6 × 106 psi. If the bar is subjected to an axial torque
T = 20 in − kip, what is the magnitude of the maximum
shear stress at x = 6 in?
Free Body Diagram:
Solution:
At x = 6 in., the polar moment of inertia is:
J = (0.1 + 0.15(6)) in4 = 1 in4
The radius of the bar at x = 6 in. is:
2 1 in4
2J 1/4
r=
=
π
π
1/4
= 0.893 in.
Maximum shear stress in the bar at x = 6 in. is:
τMAX =
ANS:
Tr
(20, 000 in − lb) (0.893 in)
=
J
1 in4
τMAX = 17, 860 lb/in2 = 17.86 ksi
Problem 11.46 What is the angle of twist (in degrees)
of the entire bar in Problem 11.45?
Free Body Diagram:
Solution:
The angle of twist is determine by integrating over the length of the
bar.
10
φ=
0
(20, 000 in − lb)dx
=
(0.1 + 0.15x)(4.6 × 106 lb/in2 )
20, 000 in − lb
4.6 × 106 lb/in2
φ = 0.029 [ln(1.6) − ln(0.1)]
ANS:
φ = 0.080365 rad = 4.6045◦
10
0
dx
20, 000 in − lb
=
(0.1 + 0.15x)
4.6 × 106 lb/in2
1
0.15
1
0
0.15dx
(0.1 + 0.15x)
Problem 11.47 Suppose that an axial hole is drilled
through the bar in Problem 11.45 so that it has a hollow
circular cross section with inner radius ri = 0.3 in. What
is the angle of twist (in degrees) of the entire bar due to
the 20-in-kip torque?
Free Body Diagram:
Solution:
The new expression for the polar moment of inertia is:
π
J = (0.1 + 0.15x) − (0.3 in)4 = (0.08728 + 0.15x) in4
2
The angle of twist is determined by integrating from x = 0 to x = 10.
10
φ=
0
φ=
T dx
=
JG
10
0
(20, 000 in − lb)dx
=
(0.08728 + 0.15x)(4.6 × 106 lb/in2 )
20, 000 in − lb
4.6 × 106 lb/in2
ANS:
1
0.15
10
0
20, 000 in − lb
4.6 × 106 lb/in2
10
0
dx
(0.0873 + 0.15x)
0.15dx
= 0.02899 [ln(1.587) − ln(0.0873)]
(0.0873 + 0.15x)
φ = 0.084077 rad = 4.817◦
Problem 11.48 The radius of the bar’s circular cross Free Body Diagram:
section varies linearly from 10 mm at x = 0 to 5 mm
at x = 150 mm. The shear modulus of the material is
G = 17 GPa. What torque T would cause a maximum
shear stress of 10 MPa at x = 80 mm?
Solution:
An expression for the radius of the bar for any value of x is:
5 mm
r = 0.010 m −
x = (0.010 − 0.0333x) m
150 mm
At x = 80 mm, the radius of the bar is:
r = 0.010 m − (0.0333)(0.08 m) = 0.0073333 m
The polar moment of inertia at x = 80 mm is:
π
J = (0.007336 m)4 = 4.5428 × 10−9 m4
2
The torque required to produce a maximum shear stress of 10 MPa at
x = 80 mm is:
10 × 106 N/m2 4.549 × 10−9 m4
τJ
T =
=
r
0.007336 m
ANS:
T = 6.19 N − m
Problem 11.49 In Problem 11.48, what torque T
would cause the end of the bar to rotate one degree?
Free Body Diagram:
Solution:
An expression for the radius of the bar for any value of x is:
5 mm
r = 0.010 m −
x = (0.010 − 0.0333x) m
150 mm
The expression for the polar moment of inertia at any point on the bar
is:
π
J = (0.010 − 0.0333x)4 m4
2
The integral expression for the angle of twist in the bar is:
0.15
φ=
0
T dx
=
JG
0.15
0
T dx
π
2
(0.010 − 0.0333x)4 m4 (17 × 109 N/m2 )
= 0.01745 rad
Recognizing that T , φ, π, and G are constant, the expression for the
angle of twist reduces to:
π(0.01745 rad)(17×109 N/m2 )
2T
4.66×108
T
ANS:
(−0.0333) =
0.15
0
=
0.15
0
dx
(0.010−0.0333x)4
(−0.0333)dx
(0.010−0.0333x)4
=
(0.010−0.0333x)−3
−3
T = 6.67 N − m
Problem 11.50 In Problem 11.48, suppose that the Free Body Diagram:
torque T at the end of the bar is 20 N-m and you want to
apply a torque in the opposite direction at x = 75 mm
so that the angle through which the end of the bar rotates
is zero. What is the magnitude of the torque you must
apply?
Solution:
An expression for the radius of the bar for any value of x is:
5 mm
r = 0.010 m −
x = (0.010 − 0.0333x) m
150 mm
The expression for the polar moment of inertia at any point on the bar
is:
π
J = (0.010 − 0.0333x)4 m4
2
The angle of twist is calculated by integrating over each of the two
sections if the bar.
0.075
0=
0
0=
π
2
((20 N − m) − T )dx
+
[(0.01 − 0.0333x)4 ] m4 (17 × 109 N/m2 )
((20 N − m) − T )
π
(17 × 109 N/m2 )
2
ANS:
1
−0.033
T = 101.902 N − m
0.075
0
0.15
π
0.075 2
(20 N − m)dx
[(0.01 − 0.0333x)4 ] m4 (17 × 109 N/m2 )
(−0.033)dx
+
(0.01 − 0.0333x)4
20 N − m
π
(17
× 109 N/m2 )
2
1
−0.033
0.15
0.075
dx
(0.01 − 0.0333x)4
Problem 11.51 Bars A and B have solid circular cross
sections and consist of material with shear modulus G =
17 GPa. Bar A is 150 mm long and its radius varies
linearly from 10 mm at its left to 5 mm at its right end.
The prismatic bar B is 100 mm long and its radius is
5 mm. There is a small gap between the bars. The end
of bar A is given an axial rotation of one degree and the
bars are welded together. What is the torque in the bars
afterward?
Free Body Diagram:
Solution:
An expression for the radius of bar A is:
.005
rA = 0.010 −
x m = ((0.010 − 0.0333x) m)
0.150
An expression for the polar moment of inertia for bar A is:
π
π
J = r 4 = (0.01 m − (0.0333x) m)4
2
2
After joining, the resulting bar will be subjected to a moment at end A
and a moment at end B, or:
MA − MB = 0
MA = MB
[1]
When the two bars are joined, the angle of twist for bar A will be
reduced and an angle of twist will be introduced into bar B. The sum
of these two angles of twist will be one degree (0.0175 radians), or:
φA + φB = 0.01745 radians
[2]
The angle of twist in bar A can be described by:
0.15
φA =
0
π
2
MA dx
(0.01m − (0.033x) m)4 (17 × 109 N/m2 )
=
2MA
π (17 × 109 N/m2 )
φA = 2.615 × 10−3 MA
Using this expression for φA and Equation [1] in Equation [2]:
2.615 × 10−3 MA +
ANS:
π
2
MA (0.1 m)
(0.005 m)4 (17 × 109 N/m2 )
MA = 2.027 N − m
= 0.01745 rad
0.15
0
dx
(0.01 m − (0.033x) m)4
Problem 11.52 The aluminum alloy bar has a circular Free Body Diagram:
cross section with 20-mm diameter, length L = 120 mm,
and a shear modulus of 28 GPa. If the distributed torque
is uniform and causes the end of the bar to rotate 0.5◦ ,
what is the magnitude of the maximum shear stress in
the bar?
Solution:
The expression for the angle of twist at the end of the bar is:
0.12
0.0087266 rad =
(T dx)x
π
2
0
(0.01 m)4 (28 × 109 N/m2 )
=
T
π
2
(0.01 m)4 (28 × 109 N/m2 )
0.12
xdx
0
T = 533.078 N − m/m
Maximum torque in the bar is at the wall and has a magnitude of:
TMAX = (533 N − m/m)(0.12 m) = 63.969 N − m
Maximum shear stress in the bar is:
τMAX =
ANS:
(64 N − m)(0.01 m)
π
(0.01 m)4
2
τMAX = 40.7 MPa
Problem 11.53 If the distributed torque in Prob- Free Body Diagram:
lem 11.48 is given by the equation c = co (x/L)3 and
causes the end of the bar to rotate 0.5◦ , what is the magnitude of the maximum shear stress in the bar?
Solution:
The polar moment of inertia for the bar is:
π
J = (0.01 m)4 = 1.571 × 10−8 m4
2
The moment applied at any point on the bar is:
x 3
M = C0
dx
0.12
The angle through which the end of the bar rotates is:
π φ = (0.5◦ )
= 0.00873 rad
180◦
The expression for the angle of twist is:
0.12
φ=
C0
3
x
0.12
xdx
JG
0
0.00873 rad =
→ 0.00873 rad =
(1.571 × 10−8
C0
JG(0.12 m)3
The total torque applied to the bar is:
0.12
0
x 3
dx = 39.99M − n
0.12
Maximum shear stress in the bar is:
τMAX =
ANS:
Tρ
(39.99 N − m)(0.01 m)
=
J
1.571 × 10−8 m4
τMAX = 25.5 MPa
x4 dx
0
C0
m4 )(28 × 109 N/m2 )(0.12)3
C0 = 1333 N − m/m
T = (1333 N − m)
0.12
0.12
x4 dx
0
Problem 11.54 A cylindrical bar with 1-in. diameter
fits tightly in a circular hole in a 5-in. thick plate. The
shear modulus of the material is G = 5.6 × 106 psi. A
12,000 in-lb axial torque is applied at the left end of the
bar. The distributed torque exerted on the bar by the plate
is given by the equation c = c0 [1−(x/5)1/2 ] in − lb/in,
where c0 is a constant and x is the axial position in inches
measured from the left side of the plate. Determine the
constant c0 and the magnitude of the maximum shear
stress in the bar at x = 2 in.
Free Body Diagram:
Solution:
The polar moment of inertia for the bar is:
π
J = (0.5 in)4 = 0.0982 in4
2
We know that the sum of moments on the bar must equal zero. That
is, the distributed torque must be equal and opposite to the 12,00 in-lb
applied torque.
5
12, 000 in − lb =
c0 1 −
0
ANS:
x 1/2
5
dx = c0 x −
x3/2
√
3
5
2
5
= c0 (1.667)
0
c0 = 7200 in − lb/in
Starting from the left-hand end of the bar, we calculate the torque at
x = 2 in:
2
T = (12, 000 in − lb) − (7200)
0
x1/2
1− √
5
T = 3669.6 in − lb
Maximum shear stress at x = 2 in is:
τ =
ANS:
Tr
(3669.6 in − lb)(0.5 in)
=
J
0.0982 in4
τ = 18.684 ksi
dx
Problem 11.55 In Problem 11.54, what is the magnitude of the angle of twist of the left end of the bar relative
to its right end?
Free Body Diagram:
Solution:
For the 10-inch section, the angle of twist is:
φ10 =
LT
(10in)(12, 000 in − lb)
=
= 0.2182 rad = 12.5◦
JG
(0.0982 in4 )(5.6 × 106 lb/in2 )
For the 5-inch section, the angle of twist is:
LT
φ5 =
=
JG
5
0
(7200 in − lb/in) 1 −
(0.0982 in4 )(5.6 ×
1/2
x√
dx(x)
5
6
10 lb/in2 )
φ5 = 0.02182 rad = 1.25◦
Total angle of twist for the bar is:
φ = φ10 + φ5 = 12.5◦ + 1.25◦
ANS:
φ = 13.75◦
Problem 11.56 The aluminum alloy bar has a circular Free Body Diagram:
cross section with 20-mm diameter and a shear modulus
of 28 GPa. What is the magnitude of the maximum shear
stress in the bar due to the uniformly distributed torque?
Solution:
The polar moment of inertia for the bar is:
π
J = (0.01 m)4 = 1.571 × 10−8 m4
2
From the symmetry of the loading, we see that the angle of twist will
be maximum at x = 60 mm. Total moment applied to the bar is:
T = (8, 500 N − m/m)(0.12 m) = 1, 020 N − m
As a result of the symmetric loading, the reactions at the ends of the
bar are:
ML = MR = T /2 = (1, 020 N − m)/2 = 510 N − m
Maximum shear stress is found at the extreme ends of the bar, where
the torque is maximum (510 N-m).
τMAX =
ANS:
Tr
(510 N − m)(0.01 m)
=
J
1.571 × 10−8 m4
τMAX = 324.6 MPa
Problem 11.57 One type of high-strength steel drill
pipe used in drilling oil wells has a 5-in. outside diameter
and 4.28-in. inside diameter. If the steel will safely
support a shear stress of 95 ksi, what is the largest torque
to which the pipe can safely be subjected?
Free Body Diagram:
Solution:
The polar moment of inertia for the pipe cross-section is:
π 4
π
J=
r − ri4 =
(2.5 in)4 − (2.14 in)4
2 o
2
J = 28.42 in4
Maximum permissible torque for the pipe is:
95, 000 lb/in2 28.4 in4
τMAX J
TMAX =
=
ρ
2.5 in
ANS:
T MAX = 1.08 × 106 in − lb
Problem 11.58 The drill pipe described in Problem 11.57 has a shear modulus G = 12 × 106 psi. If it
is used to drill an oil well 20,000 ft deep and the drilling
operation subjects the bottom of the pipe to a torque
T = 7500 in − lb, what is the resulting angle of twist
(in degrees) of the 20,000-ft pipe?
Free Body Diagram:
Solution:
The polar moment of inertia for the cross-section is:
J = π2 (2.5 in)4 − (2.14 in)4
J = 28.4 in4
Being careful of units, we apply the equation for the angle of twist:
(20, 000 ft) 12 in
(7500 in − lb)
LT
ft
φ=
= JG
28.4 in4 12 × 106 lb/in2
ANS:
φ= 5.28 rad = 302.5◦
Problem 11.59 The radius R = 200 mm. The in- Free Body Diagram:
finitesimal element is at the surface of the bar. What are
the normal stress and the magnitude of the shear stress
on the plane P ?
Solution:
The polar moment of inertia for the bar is:
π
J = (0.2 m)4 = 0.002513 m4
2
The shear stress at the surface of the bar (which is maximum shear
stress) is:
τ =
TR
(400 N − m) (0.2 m)
=
= 31, 834 N/m2
J
0.0025 m4
Summing vertical forces on the sectioned element:
ΣFy = 0 = (32, 000 N/m) (A) (sin 35◦ )−τ A (sin 35◦ )+σA (cos 35◦ )
[1]
τ − 0.7002σ = −32, 000 N/m2
Summing horizontal forces on the sectioned element:
ΣFX = 0 = − 32, 000 N/m2 A (sin 35◦ )−τ A (sin 35◦ )+σA (cos 35◦ )
[2]
τ − 1.428σ = −32, 000 N/m2
Solving equations [1] and [2] together:
ANS: σ = −30 kPa Note: The negative sign indicates a compressive stress.
ANS:
τ = 10.94 kPa
Problem 11.60 For the element in Problem 11.59, determine the normal stress and the magnitude of the shear
stress on the plane P shown.
Free Body Diagram:
Solution:
Summing horizontal forces on the sectioned element:
ΣFX = 0 = 32, 000 N/m2 A (cos 20◦ )−τ A (cos 20◦ )−σA (sin 20◦ )
[1]
τ + 0.364σ = 32, 000 N/m2
Summing vertical forces on the sectioned element:
ΣFY = 0 = 32, 000 N/m2 A (sin 20◦ )+τ A (sin 20◦ )−σA (cos 20◦ )
[2] − τ + 2.747σ = 32, 000 N/m2
Solving equations [1] and [2] together:
ANS:
ANS:
σ = 20.5 kPa
τ = 24.5 kPa
Problem 11.61 Part A of the bar has a solid circular Free Body Diagram:
cross section and part B has a hollow circular cross section. The bar is fixed at both ends and the shear modulus
of the material is G = 3.8 × 106 psi. Determine the
torques exerted on the bar by the walls.
Solution:
The polar moments of inertia are:
JA =
π 4
π
c = (2 in)4 = 25.13 in4
2
2
JB =
π 4
π
ro − ri4 =
(2 in)4 − (1 in)4 = 23.56 in4
2
2
We see that:
MA + MB = 150, 000 in − lb
[1]
Knowing that φA = φB , we also see that:
(7 in) (MA )
(14 in)(MB )
=
(23.56 in4 )G
25.13 in4 G
MA = 2.133MB
[2]
Solving Equations [1] and [2] together:
ANS:
ANS:
MA = 102.1 in − kip
MB = 47.9 in − kip
Problem 11.62 Determine the magnitudes of the max- Free Body Diagram:
imum shear stresses in parts A and B of the bar in Problem 11.61.
Solution:
The polar moments of inertia are:
π
π
JA = c4 = (2 in)4 = 25.13 in4
2
2
JB =
π 4
π
ro − ri4 =
(2 in)4 − (1 in)4 = 23.56 in4
2
2
We see that:
MA + MB = 150, 000 in − lb
We also see that:
(7 in)(MA )
(25.13 in4 )G
=
[1]
(14 in)(MB )
(23.56 in4 )G
MA = 2.133MB
[2]
Solving equations [1] and [2] together:
MA = 102.1 in − kipMB = 47.9 in − kip
Maximum torques in each section of the shaft:
(τMAX )A =
ANS:
(τMAX )A = 8.13 ksi
(τMAX )B =
ANS:
M A cA
(102.1 in − kip) (2 in)
=
JA
25.13 in4
M B cB
(47, 900 in − lb) (2 in)
=
JB
23.56 in4
(τMAX )B = 4.07 ksi
Problem 11.63 Suppose that you want to decrease the
weight of the bar in Problem 11.61 by increasing the
inside diameter of Part B. The bar is made of material
that will safely support a pure shear stress of 10 ksi.
Based on this criterion, what is the largest safe value of
the inside diameter?
Solution:
The polar moments of inertia for the two sections of the bar are:
π
π
π JA = (2 in)4 = 25.13 in4 JB =
(2 in)4 − ri4 = 25.13 in4 − ri4
2
2
2
We see that:
MA + MB = 150, 000 in − lb
[1]
Using the given limit of 10,000 ksi in shear stress:
10, 000 lb/in2 =
MB (2 in)
(25.13 in4 − π2 ri4 )
MB = 125, 650 in − lb − 7854ri4
[2]
Since the angle of twist must be the same for each section of the bar
(using equation [1]:
MA LA
MB LB
=
JA GA
JB GB
((150, 000 in − lb) − MB ) (7 in)
MB (14in)
= (25.13 in4 )(G)
25.13 in4 − π2 ri4 (G)
MB = 75, 000 in − lb−0.5MB −4688 in − lb+0.0313MB ri4
Solving Equations [2] and [3] together:
0 = −118, 163 in4 + 15, 714ri4 − 245.8ri8
Using the quadratic equation to solve for ri4 :
−15, 714 ± (15, 714)2 − 4(−245.8)(−118, 163)
ri4 =
2(−245.8)
ri4 = 8.704, 55.23
(clearly the 55,23 result is impossible, so it is discarded)
ri = 1.718
ANS:
di = 3.44 in
[2]
Problem 11.64 The bar has a circular cross section
with polar moment of inertia J and shear modulus G.
The distributed torque c = c0 (x/L)2 . What are the
magnitudes of the torques exerted on the bar by the left
and right walls?
Free Body Diagram:
Solution:
The total torque exerted by the distributed torque is:
L
T =
c0
x 2
L
0
c0
L2
dx =
L
x2 dx =
0
c0
L2
L3
3
=
c0 L
3
The equation of equilibrium for the bar is:
TA + TB − T = 0 → TA + TB = c0 L/3
[1]
We see that the angle of twist at each end of the bar must be zero. An
expression for the angle of twist in the bar is:
3
L
L
x 2
x
x c0 L
dx
c0 L
dx
2
TA L
TA L
0=
−
=
−
JG
JG
JG
JG
0
0
The equation can be simplified by multiplying both sides of the equation by the product JG.
0 = TA L −
ANS:
TA =
c0
L2
L
x3 dx = TA L −
0
c0 L
4
Using the above value of TA in Equation [1]:
ANS:
TB =
c0 L
12
c0
L2
L4
4
Problem 11.65 In Problem 11.64, at what axial position x is the bar’s angle of twist the greatest and what is
its magnitude?
Free Body Diagram:
Solution:
The total torque exerted by the distributed torque is:
L
T =
c0
0
x 2
L
c0
dx = 2
L
L
x2 dx =
0
c0
L2
L3
3
=
c0 L
3
The equation of equilibrium for the bar is:
TA + TB − T = 0 → TA + TB = c0 L/3
[1]
We see that the angle of twist at each end of the bar must be zero. An
expression for the angle of twist in the bar is:
3
L
L
x 2
x
x c0 L
dx
c0 L
dx
2
TA L
TA L
0=
−
=
−
JG
JG
JG
JG
0
0
The equation can be simplified by multiplying both sides of the equation by the product JG.
0 = TA L −
c0
L2
L
x3 dx = TA L −
0
TA =
c0
L2
L4
4
c0 L
4
Using the above value of TA in Equation [1]:
TB =
c0 L
12
Considering the loading from end A to end B, we see that the load at
any point on the bar is:
x2
L
C0 2 dx x − C0
x = M [2]
L
4
The angle of maximum twist occurs where the applied moment is
maximum. We find the location of maximum moment by setting the
differential of Equation [2] equal to zero.
x3
L
=0
C 0 2 − C0
L
4
The value of x where the maximum angle of twist occurs is:
ANS:
x=
L
41/3
= 0.63L
Calculating the magnitude of the maximum angle of twist:
0.63L
0.63L
x2
C0 L
2 dx x
C0
C0
φ=
=
x3 dx =
JG
JGL2
JGL2
0
ANS:
0
φ=
2
0L
0.0394 CJG
0.63L
x4 4 0
Problem 11.66 If the bar in Problem 11.64 is acted
upon by the distributed load C = C0 (x/L)2 and is free
of external torque from x = L/2 to x = L, what are the
magnitudes of the torque exerted on the bar by the left
and right walls?
Free Body Diagram:
Solution:
The total moment applied to the bar is:
L/2
M =
C0
0
L/2
x2
x3 L
dx
=
C
= C0
0
L2
3L2 0
24
The reaction at the right-hand end of the bar must be sufficient to off-set
the angle of twist which results from the applied moment.
L/2
x2
C0 L
2 dx x
MR L
−
=0
JG
JG
0
Multiplying through by JG:
L/2
C0
0
C0
ANS:
MR =
x3
dx − MR L = 0
L2
L2
− MR L = 0
64
C0 L
64
Since the sum of moments acting on the bar must equal zero:
ML + MR =
ML +
ANS:
ML =
5
C L
192 0
C0 L
24
C0 L
C0 L
=
64
24
Problem 12.1 The components of plain stress at a point
p of a material are σx = 20 MPa, σy = 0 and τxy = 0.
If θ = 45◦ , what are the stresses σx , σy and τxy
at point
p?
Solution:
Using Equation (12-7) to find σx :
σx +σy
σ −σ
+ x 2 y (cos 2θ) + τxy (sin 2θ)
2
σx == 20 MPa+0
+ 20 MPa−0
(cos 90◦ ) + 0
2
2
σx =
ANS:
σx = 10 MPa
Using Equation (12-9) to find σy :
σy =
σy =
ANS:
σx +σy
σ −σ
− x 2 y (cos 2θ) − τxy (sin 2θ)
2
20 MPa+0
− 20 MPa−0
(cos 90◦ ) − 0
2
2
σy = 10 MPa
:
Using Equation (12-8) to find τxy
σx −σy
(sin 2θ) + τxy (cos 2θ)
2
− 20 MPa−0
(sin 90◦ ) + 0
2
=−
τxy
=
τxy
ANS:
= −10 MPa
τxy
Problem 12.2 The components of plane stress at a
point p of a material are σx = 0, σy = 0 and τxy =
25 ksi. If θ = 45◦ , what are the stresses σx , σy and τxy
at point p?
Solution:
Using Equation (12-7) to find σx :
σx =
σx
ANS:
=
σx +σy
σ −σ
+ x 2 y (cos 2θ) + τxy (sin 2θ)
2
0+0
0−0
+ 2 (cos 90◦ ) + (25 ksi) (sin 90◦ )
2
σx = 25 ksi
Using Equation (12-9):
σy =
σy
ANS:
=
σx +σy
σ −σ
− x 2 y cos 2θ − τxy sin 2θ
2
0+0
0−0
− 2 cos(90◦ ) − 25 sin(90◦ )
2
σy = −25 ksi
:
Using Equation (12-8) to find τxy
σx −σy
(sin 2θ) + τxy (cos 2θ)
2
− 0−0
(sin
90◦ ) + (25 ksi) (cos 90◦ )
2
=−
τxy
τxy
ANS:
=
=0
τxy
Problem 12.3 The components of plane stress at a
point p of a material are σx = −8 ksi, σy = 6 ksi
and τxy = −6 ksi. If θ = 30◦ , what are the stresses σx ,
σy and τxy
at point p?
Solution:
Using Equation (12-7) to find σx :
σx =
σx
=
ANS:
σx +σy
σ −σ
+ x 2 y (cos 2θ) + τxy (sin 2θ)
2
−8 ksi+6 ksi
ksi
+ −8 ksi−6
(cos 60◦ ) + (−6
2
2
ksi) (sin 60◦ )
σx = −9.7 ksi
Using Equation (12-9):
σy =
σy
=
ANS:
σx +σy
σ −σ
− x 2 y (cos 2θ) − τxy (sin 2θ)
2
−8 ksi+6 ksi
ksi
− −8 ksi−6
(cos 60◦ ) − (−6
2
2
ksi) (sin 60◦ )
σy = 7.7 ksi
:
Using Equation (12-8) to find τxy
σx −σy
(sin 2θ) + τxy (cos 2θ)
2
(−8−6)
− 2
sin 60◦ + (−6) cos 60◦
=−
τxy
=
τxy
ANS:
= 3.062 ksi
τxy
Problem 12.4 During liftoff, strain gauges attached to
one of the Space Shuttle main engine nozzles determine
that the components of plane stress σx = 66.46 MPa,
σy = 82.54 MPa, and τxy
= 6.75 MPa at θ = 20◦ .
What are the stresses σx , σy and τxy at that point?
Solution:
Adding Equations (12-7) and (12-9):
σx + σy = σx + σy → σx + σy = 149 MPa
[1]
Using Equation [1] in Equation (12-9):
82.54 MPa =
149 MPa σx − σy
−
[cos (40◦ )] − τxy [sin (40◦ )]
2
2
8.04 MPa = −(σx − σy )(0.383) − (0.643)τxy
τxy = −12.5 MPa − (σx − σy )(0.596)
[2]
Substituting the given information into Equation (12-8):
6.75 MPa = −
σx − σy
[sin (40◦ )] + τxy [cos (40◦ )]
2
6.75 MPa = −(0.321)(σx − σy ) + τxy (0.766)
τxy = 8.81 MPa + (0.419)(σx − σy )
[3]
Subtracting Equations [2] and [3]:
0 = −21.31 − 1.015(σx − σy ) → σx − σy = −20.99 MPa
Adding Equations [1] and [4]:
2σx = 128.01 MPa
ANS:
σx = 64 MPa
σy = 85 MPa
τxy = 0 MPa
[4]
Problem 12.5 The components of plane stress at a
point p of a material are σx = 240 MPa, σy =
−120 MPa, and τxy = 240 MPa and the components
referred to x y z coordinate system are σx = 347 MPa,
σy = −227 MPa and τxy
= −87 MPa. What is the angle θ? [Strategy: Solve Equations (12-7) and (12-8) for
sin 2θ and cos 2θ. Knowing these two quantities, you
can determine θ. (Why are the values of both sin 2θ and
cos 2θ needed to uniquely determine θ?)]
Solution:
Using Equation (12-7) to find σx :
σx =
σx =
σx +σy
σ −σ
+ x 2 y (cos 2θ) + τxy (sin 2θ)
2
240 MPa−(−120 MPa)
240 MPa−120 MPa
+
2
2
(cos 2θ) + (240 MPa) (sin 2θ)
347 = 60 MPa + 180 MPa cos 2θ + 240 MPa sin 2θ
σy =
σy =
σx +σy
σ −σ
− x 2 y (cos 2θ) − τxy (sin 2θ)
2
240 MPa+(−120 MPa)
240 MPa−(−120 MPa)
−
2
2
[1]
(cos 2θ) − (240 MPa)(sin 2θ)
−227 MPa = 60 MPa−(180 MPa)(cos 2θ)−(240 MPa)(sin 2θ)
[2]
Using Equation (12-8) to solve for sin(2θ):
sin 2θ =
240 cos 2θ + 87
= 1.333(cos 2θ) + 0.483
180
[3]
Substituting Equation [3] into Equation [1]:
347 MPa = 60 MPa+(180 MPa)(cos 2θ)+(240 MPa)[1.333(cos 2θ)+0.483]
ANS:
θ = 35◦
Problem 12.6 The components of plane stress at a
point p of a bit during a drilling operation are σx =
40 ksi, σy = −30 ksi, and τxy = 30 ksi, and the
components referred to the x y z coordinate system are
σx = 12.5 ksi, σy = −2.5 ksi, and τxy
= 45.5 ksi.
What is the angle θ?
Solution:
Using the given information in Equation (12-7):
12.5 ksi =
40 ksi − 30 ksi 40 ksi − (−30 ksi)
+
cos(2θ)+(30 ksi)(sin(2θ))
2
2
cos(2θ) =
12.5 ksi
5 ksi
30 ksi
−
−
sin(2θ)
35 ksi
35 ksi
35 ksi
[1]
Using the given information in Equation (12-8):
45.5 ksi = −
40 ksi − (−30 ksi)
sin(2θ) + (30 ksi)(cos(2θ))
2
cos(2θ) =
45.5 ksi
35 ksi
+
sin(2θ)
30 ksi
30 ksi
Solving Equations [1] and [2] together:
ANS:
θ = −20◦
[2]
Problem 12.7 The components of plane stress at a
point p of a material referred to the x y z coordinate system are σx = −8 MPa, σy = 6 MPa and
τxy
= −16 MPa. If θ = 20◦ , what are the stresses
σx , σy , and τxy at point p?
Solution:
Adding Equations (12-7) and (12-9):
σx + σy = σx + σy
−8 MPa + 6 MPa = σx + σy
σx = −2 MPa − σy
[1]
Using Equation [1] in Equation (12-8):
=−
τxy
σx −σy
2
(sin 2θ) + τxy (cos 2θ)
(−2 MPa−σy )−σy
−16 MPa = −
(sin 40◦ ) + τxy (cos 40◦ )
2
−16 MPa = (1 MPa + σy )(sin 40◦ ) + τxy (cos 40◦ )
τxy = −21.73 MPa − 0.839σy
[2]
Using Equations [1] and [2] in Equation (12-9):
σx +σy
σ −σ
− x 2 y (cos 2θ) − τxy (sin 2θ)
2
(−2 MPa−σy )+σy
(−2 MPa−σy )−σy
MPa =
−
2
2
σy =
6
(cos 40◦ ) − [−21.73 MPa − 0.839σy ] (sin 40◦ )
−7.736 MPa = 1.305σy
ANS:
σy = −5.93 MPa
σx = 3.93 MPa
τxy = −16.75 MPa
Problem 12.8 A point p of the car’s frame is subjected
to the components of plane stress σx = 32 MPa, σy =
−16 MPa and τxy
= −24 MPa. If θ = 20◦ , what are
the stresses σx , σy , and τxy at point p?
Solution:
Adding Equations (12-7) and (12-9):
σx + σy = σx + σy
32 MPa + (−16 MPa) = σx + σy
[1]σy = 16 MPa − σx
Using Equation [1] in Equation (12-8):
=−
τxy
σx −σy
2
(sin 2θ) + τxy (cos 2θ)
σ −(16 MPa−σx )
−24 MPa = − x
(sin 40◦ ) + τxy (cos 40◦ )
2
−24 MPa = (σx + 8 MPa)(sin 40◦ ) + τxy (cos 40◦ )
[2]τxy = −38.04 MPa − 0.839σx
Using Equations [1] and [2] in Equation (12-8):
σx +σy
σ −σ
− x 2 y (cos 2θ) − τxy (sin 2θ)
2
σ +(16 MPa−σx )
σ −(16 MPa−σx )
MPa = x
− x
2
2
σy =
32
(cos 40◦ ) + [−38.04 MPa + 0.839σx ] (sin 40◦ )
32 MPa = 8 MPa+0.766σx −6.128 MPa−24.45 MPa+0.539σx
ANS:
σx = 41.8 MPa
σy = −25.8 MPa τxy = −2.97 MPa
Problem 12.9 In Problem 12.8, what are the stresses
σx , σy , and τxy at point p if θ = −40◦ ?
Solution:
Adding Equations (12-7) and (12-9):
σx + σy = σx + σy
32 MPa + (−16 MPa) = σx + σy
σy = 16 MPa − σx
[1]
Using Equation [1] in Equation (12-8):
=−
τxy
σx −σy
2
−24 MPa = −
(sin 2θ) + τxy (cos 2θ)
σx −(16 MPa−σx )
(sin(−80◦ )) + τxy (cos −80◦ )
2
−24 MPa = 0.985σx − 7.878 MPa + 0.174τxy
τxy = −92.66 MPa + 5.66σx
[2]
Using Equations [1] and [2] in Equation (12-7):
σx +σy
σ −σ
− x 2 y (cos 2θ) − τxy (sin 2θ)
2
σ +(16 MPa−σx )
σ −(16 MPa−σx )
MPa = x
− x
2
2
σy =
32
(cos(−80◦ )) + [−92.66 MPa + 5.66σx ] (sin(−80◦ ))
32 MPa = 8 MPa+0.174σx −1.388 MPa+91.25 MPa+5.574σx
ANS:
ANS:
ANS:
σx = −11.45 MPa
σy = 27.45 MPa
τxy = −157.46 MPa
Problem 12.10 The components of plane stress at a
point p of the material shown are σx = 4 ksi, σy =
−2 ksi, and τxy = 2 ksi. What are the normal stress and
the magnitude of the shear stress on the plane p at point
p?
Solution:
σy =
σy =
ANS:
σx +σy
σ −σ
− x 2 y (cos 2θ) − τxy (sin 2θ)
2
4 ksi+(−2 ksi)
4 ksi−(−2 ksi)
−
(cos 60◦ ) −
2
2
(2 ksi)(sin 60◦ )
σy = −2.23 ksi
:
Using Equation (12-8) to find τxy
σx −σy
(sin 2θ) + τxy (cos 2θ)
2
4 ksi−(−2 ksi)
−
(sin 60◦ ) + (2 ksi) (cos 60◦ )
2
=−
τxy
=
τxy
ANS:
|τxy | = 1.6 ksi
Problem 12.11 The components of plane stress at a
point p of the material shown in Problem 12.10 are σx =
−10.5 MPa, σy = 6.0 MPa, and τxy = −4.5 MPa.
What are the normal stress and magnitude of the shear
stress on the plane p at point p?
Solution:
σx +σy
σ −σ
− x 2 y (cos 2θ) − τxy (sin 2θ)
2
(−10.5 MPa−6.0 MPa)
−10.5 MPa+6 MPa
−
2
2
σy =
σy =
(cos 60◦ ) − (−4.5 MPa)(sin 60◦ )
σy = 5.77 MPa
ANS:
:
Using Equation (12-8) to find τxy
σx −σy
(sin 2θ) + τxy
2
(−10.5 MPa−6.0 MPa)
−
2
=−
τxy
(cos 2θ)
=
τxy
(sin 60◦ ) + (−4.5 MPa) (cos 60◦ )
= 4.89 MPa
τxy
ANS:
Problem 12.12 Determine the stresses σ and τ : (a) by Free Body Diagram:
writing equilibrium equations for the element shown;
(b) by using Equations (12-7) and (12-8).
Solution:
(a) Summing horizontal and vertical forces:
ΣFx = 0 = −σA cos(55◦ )+τ A sin(55◦ )−(12 MPa)(A sin(55◦ ))−(10 MPa)(A cos(55◦ ))
τ =
σ(cos 55◦ ) + 12 MPa)(sin 55◦ ) + (10 MPa)(cos 55◦ )
sin 55◦
τ = 0.7σ + 19 MPa
[1]
ΣFy = 0 = −σA sin(55◦ )−τ A cos(55◦ )−(12 MPa)(A cos(55◦ ))+(10 MPa)(A sin(55◦ ))
τ =
−σ(sin 55◦ ) − (12 MPa)(cos 55◦ ) + (10 MPa)(sin 55◦ )
cos 55◦
τ = −1.428σ + 2.28 MPa
[2]
Solving Equations [1] and [2] together:
ANS:
= 13.5 MPa
σx = −7.86 MPaτxy
(b) Using Equation (12-7) to find σx :
σx =
σx =
ANS:
σx +σy
σ −σ
+ x 2 y (cos 2θ) + τxy (sin 2θ)
2
(−10 MPa−10 MPa)
−10 MPa+10 MPa
+
(cos 110◦ )
2
2
+ (−12 MPa) (sin 110◦ )
σx = −7.86 MPa
:
Using Equation (12-8) to find τxy
σx −σy
(sin 2θ) + τxy (cos 2θ)
2
−10 MPa−10 MPa
−
(sin 110◦ ) +
2
=−
τxy
=
τxy
ANS:
= 13.5 MPa
τxy
(−12 MPa) (cos 110◦ )
Problem 12.13 The stress τxy = 14 MPa and the angle θ = 25◦ . Determine the components of stress on the
right element.
Solution:
Using Equation (12-7) to find σx :
σx =
σx
=
σx +σy
σ −σ
+ x2 y
2
8 MPa+(−6 MPa)
2
σy
=
+
8 MPa−(−6 MPa)
2
(cos 50◦ ) + (14 MPa) (sin 50◦ )
σx = 16.22 MPa
ANS:
σy =
(cos 2θ) + τxy (sin 2θ)
σx +σy
σ −σ
− x2 y
2
8 MPa+(−6 MPa)
2
(cos 2θ) − τxy (sin 2θ)
−
8 MPa−(−6 MPa)
2
(cos 50◦ ) − (14 MPa)(sin 50◦ )
σy = −14.22 MPa
ANS:
:
Using Equation (12-8) to find τxy
σx −σy
(sin 2θ) + τxy (cos 2θ)
2
8 MPa−(−6 MPa)
−
(sin 50◦ ) + (14
2
=−
τxy
=
τxy
MPa) (cos 50◦ )
= 3.64 MPa
τxy
ANS:
Problem 12.14 On the elements shown in Problem 12.13, the stresses τxy = 12 MPa, σx = 14 MPa,
σy = −12 MPa. Determine the stress τxy
and the angle
θ.
Solution:
Using Equation (12-7) to find θ:
σx +σy
σ −σ
+ x 2 y (cos 2θ) + τxy (sin 2θ)
2
8 MPa+(−6 MPa)
8 MPa−(−6 MPa)
MPa =
+
2
2
σx =
14
(cos 2θ) + (14 MPa) (sin 2θ)
A graphing calculator reveals two angles at which these conditions
exist. The angles are:
θ1 = 19.5◦ θ2 = 40.2◦
ANS:
:
Using Equation (12-8) to find the two corresponding values of τxy
σx −σy
(sin 2θ) + τxy (cos 2θ)
2
8 MPa−(−6 MPa)
−
(sin 2 ∗ 19.5◦ )
2
=−
τxy
=
τxy
+ (12 MPa) (cos(2 ∗ 19.5◦ )
ANS:
) = 4.9 MPa
(τxy
1
τxy
=−
8 MPa − (−6 MPa)
(sin 2 ∗ 40.2◦ )+(12 MPa) (cos(2 ∗ 40.2◦ )
2
ANS:
) = −4.9 MPa
(τxy
2
Problem 12.15 On the elements shown in Prob
lem 12.13, the stresses τxy
= 12 MPa and the angle
◦
θ = 35 . Determine σx , σy , and τxy .
Solution:
Using Equation (12-8) to find τxy :
=−
τxy
12 MPa
ANS:
σx −σy
(sin 2θ) + τxy
2
8 MPa−(−6 MPa)
=−
2
(cos 2θ)
(sin 70◦ ) + τxy (cos 70◦ )
τxy = 54.3 MPa
Using Equation (12-7) to find σx :
σx =
σx =
ANS:
σy =
σy
=
ANS:
σx +σy
σ −σ
+ x2 y
2
8 MPa+(−6 MPa)
2
(cos 2θ) + τxy (sin 2θ)
+
8 MPa−(−6 MPa)
2
(cos 70◦ ) + (54.3 MPa) (sin 70◦ )
σx = 54.4 MPa
σx +σy
σ −σ
− x2 y
2
8 MPa+(−6 MPa)
2
(cos 2θ) − τxy (sin 2θ)
−
8 MPa−(−6 MPa)
2
(cos 70◦ ) − (54.3 MPa)(sin 70◦ )
σy = −52.4 MPa
Problem 12.16 A point p of the airplane’s wing is subjected to plane stress. When θ = 55◦ , σx = 100 psi,
σy = −200 psi, and σx = −175 psi. Determine the
stresses τxy and τxy
at p.
Solution:
Using Equation (12-7) to find σx :
σx =
−175
ANS:
σx +σy
σ −σ
+ x 2 y (cos 2θ) + τxy (sin 2θ)
2
100 psi−(−200 psi)
100 psi−200 psi
psi =
+
2
2
(cos 110◦ ) + τxy (sin 110◦ )
τxy = −78.4 psi
:
Using Equation (12-8) to find τxy
σ −σ
y
=− x
τxy
(sin 2θ) + τxy (cos 2θ)
2
100 psi−(−200 psi)
τxy = −
(sin 110◦ ) + (−78.4 psi) (cos 110◦ )
2
ANS:
= −114.1 psi
τxy
Problem 12.17 Under a different flight condition of
the airplane in Problem 12.16, the stress components
σx = 80 psi, σy = −120 psi, and τxy = −100 psi,
σx = −80 psi, and σy = 40 psi. Determine the stresses
τxy
and the angle θ.
Solution:
Using Equation (12-7) to find σx :
σx +σy
σ −σ
+ x 2 y (cos 2θ) + τxy (sin 2θ)
2
80 psi+(−120 psi)
80 psi−(−120 psi)
+
psi =
2
2
σx =
−80
(cos 2θ) + (−100 psi)(sin 2θ)
A graphing calculator reveals that there two angles at which these
stresses occur.
ANS:
θ1 = 35.1◦ θ2 = −80.1◦
:
Using Equation (12-8) to find the two corresponding values of τxy
σx −σy
(sin 2θ) + τxy (cos 2θ)
2
psi
− 80 psi+120
(sin 70◦ ) + (−100
2
=−
τxy
=
τxy
ANS:
) = −128.2 psi
(τxy
1
(τxy
)2 = −
ANS:
psi) (cos 70◦ )
80 psi + 120 psi
(sin(−160.2◦ ))+(−100 psi)(cos(−160.2◦ ))
2
) = 128.1 psi
(τxy
2
Problem 12.18 Equations (12-7)–(12-9) apply to
plane stress, but they also apply to states of stress of
the form


σx τxy 0


0 .
 τyx σy
0
0 σz
Show that for a fluid at rest, σx = −p and τxy
= 0. See
Equation (12-2). That is, the normal stress at a point is
the negative of the pressure and the shear stress is zero
for any plane through the point.
Solution:
By definition, shear stress for a fluid is low (negligible).
From Equation (12-7):
σx =
σx =
ANS:
σx +σy
σ −σ
+ x 2 y (cos 2θ) + τxy (sin 2θ)
2
−p+(−p)
−p−(−p)
+
(cos 2θ) + 0
2
2
σx = −p for ANY value of θ
From Equation (12-8):
ANS:
σx −σy
(sin 2θ) + τxy
2
−p−(−p)
−
(sin 2θ) + 0
2
=−
τxy
(cos 2θ)
τxy
= 0(sin 2θ)
=
= 0 for ANY value of θ
τxy
Problem 12.19 By substituting the trigonometric identities (12-6) into Equations (12-4) and (12-5), derive
Equations (12-7) and (12-8).
Solution:
Equations (12-6) are: (a) 2 cos2 θ = 1 + cos 2θ
(a) 2 sin2 θ = 1 − cos 2θ
(b) 2 sin θ cos θ = sin 2θ
(c) cos2 θ − sin2 θ = cos 2θ
Equation (12-4) is:
σx = σx cos2 θ + σy sin2 θ + 2τxy sin θ cos θ
Equation (12-5) is:
τxy
= −(σx − σy )(sin θ)(cos θ) + τxy (cos2 θ − sin2 θ)
Using Equations (a) and (b) in Equation (12-4), we get:
2θ
2θ
+ σy 1−cos
+ τxy (sin 2θ)
σx = σx 1+cos
2
2
σx =
σx
2
σx =
Equation (12-7)
ANS:
+
σx cos 2θ
2
σx +σy
2
+
+
σy
2
σy cos 2θ
2
−
σx −σy
2
+ τxy (sin 2θ)
cos 2θ + τxy (sin 2θ) ← This is
Using Equation (c) in Equation (12-5):
sin 2θ
= − (σx − σy )
+ τxy cos2 θ − sin2 θ
τxy
2
Now using Equations (a) and (b) in the above result:
σ x − σy
1 + cos 2θ
1 − cos 2θ
τxy
=−
(sin 2θ) + τxy
−
2
2
2
=−
ANSτxy
tion (12-8)
σx −σy
2
(sin 2θ) + τxy (cos 2θ)
← This is Equa-
Problem 12.20 The components of plane stress acting
on an element of a bar subjected to axial loads are shown
in Figure 12-7. Assuming the stress σx to be known,
determine the principal stresses and the maximum inplane stress and show them acting on properly oriented
elements.
Solution:
Equation (12-15) is used to find the principal stresses.
ANS:
σ1,2 =
σx +σy
2
σ1,2 =
σx +0
2
±
±
σx −σy
2
σx −0
2
2
2
2
+ τxy
+0
σ1 = σx σ2 = 0
Using Equation (12-19) to find τMAX :
τMAX =
τMAX =
ANS:
τMAX =
σx
2
σx −σy
2
σx −0
2
2
2
2
+ τxy
+0
Problem 12.21 The components of plane stress acting
on an element of a bar subjected to torsion are shown in
Figure 12-8. Assuming the stress τxy to be known, determine the principal stresses and the maximum in-plane
shear stress and show them acting on properly oriented
elements.
Solution:
Equation (12-15) is used to find the principal stresses.
ANS:
σ1,2 =
σx +σy
2
σ1,2 =
0+0
2
±
σx −σy
2
0−0 2
2
±
2
2
+ τxy
2
+ τxy
σ1 = τxy σ2 = −τxy
Using Equation (12-18) to find τMAX :
τMAX =
0−0 2
2
τMAX =
ANS:
σx −σy
2
2
2
+ τxy
2
+ τxy
τMAX = |τxy |
Problem 12.22 Determine the principal stresses and
the maximum in-plane shear stress and show them acting
on properly oriented elements. σx = 20 MPa, σy =
10 MPa, and τxy = 0.
Solution:
Equation (12-15) is used to find the principal stresses.
σ1,2 =
σx +σy
2
σ1,2 =
20 MPa+10 MPa
2
ANS:
±
σ1 = 20 MPa
σx −σy
2
±
2
2
+ τxy
20 MPa−10 MPa
2
2
+0
σ2 = 10 MPa
Using Equation (12-19) to find τMAX :
τMAX =
τMAX =
ANS:
σx −σy
2
2
2
+ τxy
20 MPa−10 MPa
2
2
+0
τMAX = 5 MPa
The angles for the principal stresses and the maximum shear stress are:
tan 2θp =
2τxy
0
=
=0
σx − σy
20 MPa − 10 MPa
tan 2θs =
σ x − σy
20 MPa − 10 MPa
=
= ∞ (Equation 12-16)
2τxy
0
ANS:
θp = 0
θs = 45◦
(Equation 12-12)
Problem 12.23 Determine the principal stresses and
the maximum in-plane shear stress and show them acting
on properly oriented elements. σx = 25 ksi, σy = 0 ksi,
and τxy = −25 ksi.
Solution:
Equation (12-15) is used to find the principal stresses.
ANS:
σ1,2 =
σx +σy
2
σ1,2 =
25 ksi+0
2
±
±
σ1 = 40.45 ksi
2
σx −σy
2
25 ksi−0
2
2
+ τxy
2
+ (−25 ksi)2
σ2 = −15.45 ksi
Using Equation (12-19) to find τMAX :
τMAX =
τMAX =
ANS:
σx −σy
2
2
25 ksi−0
2
2
+ τxy
2
+ (−25 ksi)2
τMAX = 27.95 MPa
The angles for the principal stresses and the maximum shear stress are:
tan 2θp =
tan 2θs =
ANS:
2τxy
2(−25 ksi)
=
= −2
σ x − σy
25 ksi − o
σ x − σy
25 ksi − 0
=
= −0.5
2τxy
2(−25 ksi)
θp = −31.7◦
(Equation 12-12)
(Equation 12-16)
θs = −13.3◦
Problem 12.24 Determine the principal stresses and
the maximum in-plane shear stress and show them acting
on properly oriented elements. σx = −8 ksi, σy = 6 ksi,
and τxy = −6 ksi.
Solution:
Equation (12-15) is used to find the principal stresses.
ANS:
σ1,2 =
σx +σy
2
σ1,2 =
−8+6
2
σ1 = 8.22 ksi
±
±
σx −σy
2
−8−6 2
2
2
2
+ τxy
+ (−6)2
σ2 = −10.22 ksi
Using Equation (12-18) to find τMAX :
τMAX =
τMAX =
ANS:
σx −σy
2
2
2
+ τxy
−8 ksi−6 ksi
2
2
+ (−6 ksi)2
τMAX = |9.22 MPa|
The angles for the principal stresses and the maximum shear stress are:
tan 2θp =
tan 2θs =
ANS:
2τxy
2(−6 ksi)
=
= 0.857
σx − σ y
−8 ksi − 6 ksi
−σx − σy
(−8 − 6)
=−
= −1.1667
2τxy
2(−6)
θp = 20.3◦
θs = −24.7◦
Problem 12.25 Determine the principal stresses and
the maximum in-plane shear stress and show them acting
on properly oriented elements. σx = 240 MPa, σy =
−120 MPa, and τxy = 240 MPa.
Solution:
Equation (12-15) is used to find the principal stresses.
ANS:
σ1,2 =
σx +σy
2
σ1,2 =
240+(−120)
2
±
±
σx −σy
2
2
2
+ τxy
240−(−120) 2
2
+ (240)2
σ1 = 360 MPaσ2 = −240 MPa
Using Equation (12-19) to find τMAX :
τMAX =
τMAX =
ANS:
σx −σy
2
2
2
+ τxy
240−(−120)
2
2
+ (240)2
τMAX = 300 MPa
The angles for the principal stresses and the maximum shear stress are:
tan 2θp =
tan 2θs =
ANS:
2τxy
2(240)
=
= 1.333
σx − σy
240 − (−120)
σx − σy
240 MPa − (−120 MPa)
=
= 0.75
2τxy
2(240 MPa)
θp = 26.6◦ θs = 18.4◦
Problem 12.26 For the state of plane stress σx =
20 MPa, σy = 10 MPa, and τxy = 0, what is the
absolute maximum shear stress?
Solution:
Since τxy = 0, we can use σx and σy as σ1 and σ2 .
Using Equations (12-24) to find the absolute maximum shear stress:
σ1 − σ2
20 MPa − 10 MPa
=
= 5 MPa
2
2
σ1
20 MPa
=
= 10 MPa
2
2
σ2
10 MPa
=
= 5 MPa
2
2
We see that the largest of these values is:
ANS: τMAX = 10 MPa
Problem 12.27 For the state of plane stress σx =
25 ksi, σy = 0, and τxy = −25 ksi, what is the absolute maximum shear stress?
Solution:
Using Equation (12-15):
σ1,2 =
σx +σy
2
σ1,2 =
25+0
2
±
±
σ1 = 40.451
2
σx −σy
2
25−0 2
2
2
+ τxy
+ (−25)2
σ2 = −15.451
The absolute maximum shear stress is given by the largest of the three
values:
ANS:
40.451−(−15.451)
2
σ1 −σ2
2
=
σ1
2
σ2
2
40.451
= 20.23
2
−15.451
= 7.73
2
=
=
= 27.95
τMAX = 27.95 ksi
Problem 12.28 For the state of plane stress σx = 8 ksi,
σy = 6 ksi, and τxy = −6 ksi, what is the absolute
maximum shear stress?
Solution:
Using Equation (12-15):
σ1,2 =
σx +σy
2
σ1,2 =
8+6
2
σ1 = 13.08
±
±
σx −σy
2
8−6 2
2
2
2
+ τxy
+ (−6)2
σ2 = 0.917
The absolute maximum shear stress is given by the largest of the three
values:
σ1 −σ2
= 13.08−0.917
2
2
σ1
= 13.08
= 6.54
2
2
σ2
= 0.917
= 0.46
2
2
ANS:
τMAX = 6.54 ksi
= 6.08
Problem 12.29 The components of plane stress at a
point p of a material are σx = 20 MPa, σy = 0, and
τxy = 0, and the angle θ = 45◦ . Use Mohr’s circle to
determine the stresses σx , σy , and τxy
at point p.
Solution:
The center of the circle is at:
σc =
ANS:
σx + σy
20 MPa + 0 MPa
=
= 10 MPa
2
2
τc = 0 MPa
The radius of the circle is:
r = σx − σc = 20 MPa − 10 MPa = 10 MPa
For an element oriented at θ = 45◦ , we move counter-clockwise
around Mohr’s circle from point P through an angle of 2(45◦ ) = 90◦ .
to find σx . We move 90◦ counter-clockwise from point Q to find σy .
The shear stress at θ = 45◦ is the second (or τ ) coordinate of points
P and Q .
The normal on both the x- and y-faces of an element oriented 45◦
from the element shown is:
ANS:
σx = 10 MPa
σy = 10 MPa
The shear stress on an element oriented 45◦ from the element shown
is:
ANS:
τxy
= 10 MPa
Problem 12.30 The components of plane stress at a
point p of a material are σx = 0, σy = 0, and τxy =
25 ksi, and the angle θ = 45◦ . Use Mohr’s circle to
determine the stresses σx , σy , and τxy
at point p.
Solution:
The center of the circle is at:
σc =
σx + σy
0 ksi + 0 ksi
=
= 0 ksi
2
2
τc = 0 ksi
The radius of the circle is:
r = τxy − τc = 25 ksi − 0 ksi = 25 ksi
For an element oriented at θ = 45◦ , we move counter-clockwise
around Mohr’s circle from point P through an angle of 2(45◦ ) = 90◦ .
To find σx . We move 90◦ counter-clockwise from point Q to find σy .
The shear stress at θ = 45◦ is the second (or τ ) coordinate of points
P and Q .
σx = 25 ksi
ANS:
σy = −25 ksi
= 0 ksi
τxy
Problem 12.31 The components of plane stress at
a point p of a material are σx = 240 MPa, σy =
−120 MPa, and τxy = 240 MPa, and the components referred to the x y z coordinate system are σx =
347 MPa, σy = −227 MPa, and τxy = −87 MPa Use
Mohr’s circle to determine the angle θ.
Solution:
The center of the circle is at :
σc =
σx + σy
240 MPa + (−120 MPa)
=
= 60 MPa
2
2
τc = 0 MPa
The radius of the circle is:
r=
2 =
(σx − σc )2 + τxy
(240 MPa − 60 MPa)2 + (240 MPa)2 = 300 MPa
Angle θ1 has a magnitude of θ1 = tan−1
Angle θ2 has a magnitude of θ2 = tan−1
τxy
σx − σ c
τxy
σx
− σc
= tan−1
= tan−1
We add angles θ1 and θ2 to get the angle 2θ. 53.1◦ + 18.9◦ = 72◦
The angle θ is 72◦ /2, or
ANS:
θ = 36◦
240 MPa
240 MPa − 60 MPa
97 MPa
344 MPa − 60 MPa
= 53.1◦
= 18.9◦
Problem 12.32 Use Mohr’s circle to determine the
stress at a point of the Space shuttle’s main engine in
Problem 12.4.
Solution:
The data from Problem 12.4 is:
σx = 66.46 MPa
σy = 82.54 MPa
τxy
= 6.75 MPa
The center of the circle is located at:
σC =
66.46 MPa + 82.54 MPa
= 74.5 MPa
2
σC (74.5, 0)
The radius of the circle is:
r=
(74.5M P P a − 66.46 MPa)2 − (6.75 MPa)2 = 10.5 MPa
Calculating σx :
σx = 74.5 MPa + 10.5 MPa
ANS:
σx = 85 MPa
Calculating σy :
σy = 74.5 MPa − 10.5 MPa
ANS:
σy = 64 MPa
Problem 12.33 The components of plane stress at a
point p of a material referred to the x y z coordinate
system are σx = −8 MPa, σy = 6 MPa, and τxy
=
−16 MPa, and the angle θ = 20◦ . Use Mohr’s circle to
determine the stresses σx , σy , and τxy at point p.
Solution:
The center of the circle is at :
σc =
σx + σy
2
=
−8 MPa + 6 MPa
= −1 MPa
2
τc = 0 MPa
The radius of the circle is:
r=
)2 =
(σx − σc )2 + (τxy
(−8 MPa − (−1 MPa))2 + (−16 MPa)2 = 17.5 MPa
The angle between the horizontal axis and the line P Q is:
θ = 180◦ − 66.3◦ = 73.7◦
The coordinates of point P are:
σx = −1 MPa+r(cos 73.7◦ ) = −1 MPa+(17.5 MPa)(cos 73.7◦ )
ANS:
σx = 3.91 MPa
τx = −r(sin 73.7◦ ) = −(17.5 MPa)(sin 73.7◦ )
ANS:
τx = −16.8 MPa
The normal stress on the y-face of the element (σ-coordinate of point
Q) is:
σy = −1 MPa−r(cos 73.7◦ ) = −1 MPa−(17.5 MPa)(cos 73.7◦ )
ANS:
σy = −5.91 MPa
Problem 12.34 The components of plane stress at a
point p of a bit during a drilling operation are σx =
40 ksi, σy = −30 ksi, and τxy = 30 ksi, and the
components referred to the x y z coordinate system are
σx = 12.5 ksi, σy = −2.5 ksi, and τxy
= 45.5 ksi. Use
Mohr’s circle to estimate the angle θ.
Solution:
The center of the circle is located at:
σC =
40 ksi+(−30 ksi)
2
= 5 ksi
τc = 0 ksi
The radius of the circle is:
r=
(40 ksi − 5 ksi)2 + (30 ksi)2 = 46.1 ksi
θ for σx is:
θσ = tan−1 [30 ksi/(40 ksi − 5 ksi)] = 40.6◦
θ for σx is:
θσ = tan−1 [45.5 ksi/(12.5 ksi − 5 ksi)] = 80.6◦
The angle between the two orientations is:
2θ = θσ − θσ = 40.6◦ − 80.6◦
ANS:
θ = −20◦
Problem 12.35 The components of plane stress at a
point p of a material are σx = 4 ksi, σy = −2 ksi, and
τxy = 2 ksi. Use Mohr’s circle to determine the normal
stress and the magnitude of the shear stress on the plane
p at point p.
Solution:
The center of the circle is at :
σc =
σx + σy
2
=
4 ksi + (−2 ksi)
= 1 ksi
2
τc = 0 ksi
The radius of the circle is:
r=
(σx − σc )2 + (τxy )2 =
(4 ksi − 1 ksi)2 + (2 ksi)2 = 3.61 ksi
Angle θ1 has a magnitude of
τxy
2 ksi
θ1 = tan−1
= tan−1
= 33.7◦
σ x − σc
(4 ksi − 1 ksi)
Angle θ2 has a magnitude of 2(30◦ ) − 33.7◦ = 26.3◦ .
The normal stress at an angle of θ = 30◦ is:
σy = 1 MPa − r(cos 26.3◦ ) = 1 MPa − (3.61 MPa)(cos 26.3◦ )
ANS:
σy = −2.23 ksi
τx = r(sin 26.3) = (3.61 MPa)(sin 26.3◦ )
ANS:
τx = 4.23 ksi
Problem 12.36 The components of plane stress at a
point p of the materialshown in Problem 12.35 are σx =
−10.5 MPa, σy = 6.0 MPa, and τxy = −4.5 MPa.
Use Mohr’s circle to determine the normal stress and the
magnitude of the shear stress on the plane p at the point
p.
Solution:
The center of the circle is at :
σc =
σx + σy
−10.5 MPa + 6 MPa
=
= −2.25 MPa
2
2
τc = 0 ksi
The radius of the circle is:
(−10.5 MPa − (−2.25 MPa)2 + (−4.5 MPa)2 = 9.4 MPa
τxy
Angle θ1 has a magnitude of: θ1 = tan−1 σ −σ
=
x
c
−4.5 MPa
−1
◦
tan
= 28.6
−10.5 MPa−(−2.25 MPa)
r=
(σx − σc )2 + (τxy )2 =
Angle θ2 has a magnitude of 2(30◦ ) − 28.6◦ = 31.4◦ .
The normal stress at an angle of θ = 30◦ is:
σy = −2.25 MPa−r(cos 31.4◦ ) = −2.25 MPa+(9.4 MPa)(cos 31.4◦ )
ANS:
σy = 5.77 ksi
τx = r(sin 31.4) = (9.4 MPa)(sin 31.4◦ )
ANS:
τx = 4.89 ksi
Problem 12.37 Determine the stresses σ and τ : (a) by
using Mohr’s circle; (b) by using Equations (12-7) and
(12-8).
Solution:
The center of the circle is at :
σc =
σ x + σy
−300 psi + (−200 psi)
=
= −250 psi
2
2
τc = 0 psi
The radius of the circle is:
r=
(σx − σc )2 + (τxy )2 =
(−300 − (−250 psi)2 + (−100 psi)2 = 111.8 psi
Angle θ1 has a magnitude of:
τxy
−100 psi
θ1 = tan−1
= tan−1
= 63.4◦
σx − σ c
−300 psi − (−250 psi)
Angle θ2 has a magnitude of 63.4◦ − 2(20◦ ) = 23.4◦ .
The normal stress at an angle of θ = 20◦ is:
σx = −250 psi−r(cos 23.4◦ ) = −250 psi−(111.8 psi)(cos 23.4◦ )
σx = −352.6 psi
ANS:
τx = r(sin 23.4) = (111.8 psi(sin 23.4◦ )
τx = −44.4 psi
Using Equations (12-6) and (12-7) to find σ and τ :
ANS:
σx =
σx =
σx +σy
σ −σ
+ x 2 y cos 2θ + τxy sin 2θ
2
−300 psi+(−200 psi)
−300 psi−(−200 psi)
+
2
2
cos 40◦ + (−100 psi) sin 40◦
σx = −352.6 psi
ANS:
σx −σy
sin 2θ + τxy cos 2θ
2
−300 psi−(−200 psi)
−
sin 40◦
2
=−
τxy
=
τxy
ANS:
σy =
σy =
ANS:
+ (−100 psi) cos 40◦
= −44.4 psi
τxy
σx +σy
σ −σ
− x 2 y cos 2θ − τxy sin 2θ
2
−300+(−200)
(−300)−(−200)
−
cos 40◦
2
2
σy = −147.42 psi
− (−100) sin 40◦
Problem 12.38 Solve Problem 12.37 if the 300-psi
stress on the element is in tension instead of compression.
Solution:
The center of the circle is located at:
σC =
300 psi + (−200 psi)
= 50 psi
2
τC = 0
The radius of the circle is:
(300 psi − 50 psi)2 + (100 psi)2 = 269.3 psi
r=
The coordinates at point P are:
σx = 50 psi + (269.3 psi)(cos 61.8◦ )
σx = 177.3 psi
ANS:
τxy
= (269.3 psi)(sin(61.8◦ ))
= −237.3 psi
τxy
ANS:
Using Equations (12-7) and (12-8):
σx =
σx =
σx +σy
σ −σ
+ x 2 y cos 2θ + τxy sin 2θ
2
300 psi+(−200 psi)
300 psi−(−200 psi)
+
2
2
ANS:
cos(40◦ ) + (−100 psi) sin(40◦ )
σx = 177.2 psi
σx −σy
sin 2θ + τxy cos 2θ
2
300 psi−(−200 psi)
−
sin 40◦ +
2
=−
τxy
=
τxy
ANS:
σy =
σy =
ANS:
(−100 psi) cos 40◦
= −237.3 psi
τxy
σx +σy
σ −σ
− x 2 y cos 2θ + τxy sin 2θ
2
300+(−200)
300−(−200)
−
cos 40◦ −
2
2
σy = −77.2 psi
(−100) sin 40◦
Problem 12.39 Determine the stresses σ and τ : (a) by
using Mohr’s circle; (b) by using Equations (12-7) and
(12-8).
Solution:
The center of the circle is at :
σc =
σ x + σy
−10 MPa + 10 MPa
=
= 0 MPa
2
2
τc = 0 MPa
The radius of the circle is:
r=
(σx − σc )2 + (τxy )2 =
(−10 MPa − 0 MPa)2 + (−12 MPa)2 = 15.6 MPa
Angle θ1 has a magnitude of:
τxy
−12 MPa
θ1 = tan−1
= tan−1
= 50.2◦
σx − σ c
−10 MPa − 0 MPa)
Angle θ2 has a magnitude 180◦ − 50.2◦ − 2(35◦ ) = 59.8◦ .
The normal stress at an angle of θ = −35◦ is:
σy = 0 MPa − r(cos 59.8◦ ) = 0 MPa − (15.6 MPa)(cos 59.8◦ )
ANS:
σx = −7.85 MPa
τx = −r(sin 59.8◦ ) = −(15.6 MPa)(sin 59.8◦ )
ANS:
τx = −13.5 MPa
Using Equations (12-6) and (12-7) to find σ and τ :
σx =
σx =
ANS:
σx +σy
σ −σ
+ x 2 y cos 2θ + τxy sin 2θ
2
−10 MPa−(10 MPa)
−10 MPa+10 MPa
+
2
2
cos(−70◦ ) + (−12 MPa) sin(−70◦ )
σx = 7.86 MPa
σx −σy
sin 2θ + τxy cos 2θ
2
−10 MPa−10 MPa
−
sin(−70◦ )
2
=−
τxy
=
τxy
ANS:
= 13.5 MPa
τxy
+ (−12 MPa) cos(−70◦ )
Problem 12.40 Use Mohr’s circle to determine the
principal stresses and the maximum in-plane shear stress
and show them acting on properly oriented elements.
σx = 20 MPa, σy = 10 MPa, and τxy = 0
Solution:
The center of the circle is at :
σc =
σ x + σy
20 MPa + 10 MPa
=
= 15 MPa
2
2
τc = 0
The radius of the circle is:
r=
(σx − σc )2 + (τxy )2 =
(20 MPa − 15 MPa)2 + 02 = 5 MPa
We see from Mohr’s circle that:
ANS:
σ1 = 20 MPa
σ2 = 10 MPa
τMAX = 5 MPa
Problem 12.41 Use Mohr’s circle to determine the
principal stresses and the maximum in-plane shear stress
and show them acting on properly oriented elements.
σx = 25 ksi, σy = 0, and τxy = −25 ksi.
Solution:
The center of the circle is at :
σc =
σ x + σy
25 ksi + 0
=
= 12.5 ksi
2
2
τc = 0
The radius of the circle is:
r=
(σx − σc )2 + (τxy )2 =
(25 ksi − 12.5 ksi)2 + (−25 ksi)2 = 27.95 ksi
We see from Mohr’s circle that:
ANS:
ANS:
ANS:
σ1 = σc + r = 12.5 ksi + 27.95 ksi = 40.45 ksi
σ2 = σc − r = 12.5 ksi − 27.95 ksi = −15.45 ksi
|τMAX | = r = 27.95 ksi
Problem 12.42 Use Mohr’s circle to determine the
principal stresses and the maximum in-plane shear stress
and show them acting on properly oriented elements.
σx = −8 ksi, σy = 6 ksi, and τxy = −6 ksi.
Solution:
The center of the circle is at :
σc =
σ x + σy
−8 ksi + 6 ksi
=
= −1 ksi
2
2
τc = τxy
The radius of the circle is:
(σx − σc )2 + (τxy )2 =
r=
(−8 ksi − 6 ksi)2 + (−6 ksi)2 = 9.22 ksi
We see from Mohr’s circle that:
σ1 = σC + r = −1 ksi + 9.22 ksi
ANS:
σ1 = 8.22 ksi
σ2 = σC − r = −1 ksi − 9.22 ksi
ANS:
ANS:
σ2 = −10.22 ksi
|τMAX | = r = 9.22 ksi
Problem 12.43 At touchdown, a point p of the space
shuttle’s landing gear is subjected to the state of plane
stress . σx = −120 MPa, σy = 80 MPa, and τxy =
−50 MPa. Use Mohr’s circle to determine the principal
stresses and the maximum in-plane shear stress.
Solution:
The center of the circle is at :
σc =
σx + σy
−120 MPa + 80 MPa
=
= −20 MPa
2
2
τc = 0
The radius of the circle is:
r=
(σx − σc )2 + (τxy )2 =
(−120 MPa − (−20 MPa))2 + (−50 MPa)2 = 111.8 MPa
We see from Mohr’s circle that:
ANS:
ANS:
ANS:
σ1 = σc − r = −20 MPa + 111.8 MPa = 91.8 MPa
σ2 = σc − r = −20 MPa − 111.8 MPa = −131.8 MPa
|τMAX | = r = 111.8 MPa
Problem 12.44 For the state of plane stress σx = 8 ksi,
σy = 6 ksi, and τxy = −6 ksi, use Mohr’s circle to determine the absolute maximum shear stress. [Strategy:
Use Mohr’s circle to determine the principal stresses and
then determine the absolute maximum shear stress from
the expressions (12-24).]
Solution:
The center of the circle is located at:
σC =
σx + σy
8 ksi + 6 ksi
=
= 7 ksi
2
2
τxy = 0
The radius of the circle is:
r=
(8 ksi − 7 ksi)2 + (−6 ksi)2 = 6.08 ksi
The principal stresses are:
σ1 = σC + r = 7 ksi + 6.08 ksi = 13.08 ksi
σ2 = σC − r = 7 ksi − 6.08 ksi = 0.92 ksi
Using Equations (12-24) to determine the absolute maximum shear
stress:
σ1
13.08 ksi
=
= 6.54 ksi
2
2
σ2
0.92 ksi
=
= 0.46 ksi
2
2
σ1 − σ2
13.08 ksi − (0.92 ksi)
=
= 6.08 ksi
2
2
ANS:
τABS MAX = 6.54 ksi
Problem 12.45 At a point p a material is subjected to
the state of plane stress σx = 20 MPa, σy = 10 MPa,
τxy = 0. Use Equation (12-25) to determine the principal stresses and use Equation (12-27) to determine
the absolute maximum shear stress. Confirm the absolute maximum shear stress by drawing the superimposed
Mohr’s circle as shown in Figure 12-36.
Solution:
From Equations (12-26):
I1 = σx + σy + σz
2 − τ2 − τ2
I2 = σx σy + σy σz + σz σx − τxy
yz
zx
2 − σ τ 2 − σ τ 2 + 2τ τ τ
I3 = σx σy σz − σx τyz
y xz
z xy
xy yz zx
I1 = 20 + 10 + 0 = 30
I2 = (20)(10) + 10)(0) + (0)(20) − (0)2 − (0)2 − (0)2 = 200
I3 = ((20)(10)(0) − (20)(0) − (10)(0) − (0)(0) + 2(0)(0)(0) = 0
From Equation (12-25):
σ 3 − I1 σ 2 + I2 σ − I3 = 0
The roots of the equation are:
σ1 = 20 MPa
ANS:
σ2 = 10 MPa
σ3 = 0 MPa
Using Equations (12-27) to fid the absolute maximum shear stress:
σ1 − σ2
20 MPa − 10 MPa
=
= 5 MPa
2
2
σ1 − σ3
20 MPa − 0 MPa
=
= 10 MPa
2
2
σ2 − σ3
10 MPa − 0
=
= 5 MPa
2
2
The largest value from Equation (12-27) is:
τMAX = 10 MPa
ANS:
Using Mohr’s circle to solve the problem:
The center of the circle is at :
σc =
σ x + σy
20 MPa + 10 MPa
=
= 15 MPa
2
2
τc = 0
The radius of the circle is:
r=
(σx − σc )2 + (τxy )2 =
(20 MPa − 15 MPa)2 + (0)2 = 5 MPa
The radius of the superimposed Mohr’s circle is:
r =
(σx − σz )2 =
(20 MPa − 10 MPa)2 = 10 MPa
We see from Mohr’s circle that:
ANS:
|τMAX | = r = 10 MPa
Problem 12.46 At a point p a material is subjected
to the state of plane stress σx = 25 ksi, σy = 0,
τxy = −25 ksi. Use Equation (12-25) to determine the
principal stresses and use Equation (12-27) to determine
the absolute maximum shear stress.
Solution:
From Equations (12-26):
I1 = σx + σy + σz
2 − τ2 − τ2
I2 = σx σy + σy σz + σz σx − τxy
yz
zx
2 − σ τ 2 − σ τ 2 + 2τ τ τ
I3 = σx σy σz − σx τyz
y xz
z xy
xy yz zx
I1 = 25 + 0 + 0 = 25
I2 = (25)(0) + (0)(0) + (0)(25) − (−25)2 − (0)2 − (0)2 = −625
I3 = ((25)(0)(0) − (0)(0) − (0)(0) − (0)(−25) + 2(−25)(0)(0) = 0
From Equation (12-25):
σ 3 − I1 σ 2 + I2 σ − I3 = 0
σ 3 − 25σ 2 − 625σ − 0 = 0
The roots of the equation are:
ANS:
σ1 = 40.45 ksi
σ2 = −15.4 ksi
σ3 = 0 ksi
Using Equations (12-27) to fid the absolute maximum shear stress:
σ1 − σ2
40.45 ksi − (−15.4 ksi)
=
= 27.9 ksi
2
2
σ1 − σ3
40.45 ksi − 0
=
= 20.225 ksi
2
2
σ2 − σ3
−15.4 − 0
=
= 7.7 ksi
2
2
The largest value from Equation (12-26) is:
ANS:
τMAX = 27.9 ksi
Using Mohr’s circle to solve the problem:
The center of the circle is at :
σc =
σ x + σy
25 ksi + 0 ksi
=
= 12.5 ksi
2
2
τc = 0
The radius of the circle is:
r=
(σx − σc )2 + (τxy )2 =
(25 ksi − 12.5 ksi)2 + (−25 ksi)2 = 27.95 ksi
The radius of the superimposed Mohr’s circle is the same as the twodimensional Mohr’s circle.
r = 27.95 ksi
We see from Mohr’s circle that:
ANS:
ANS:
σ1 = σc + r = 12.5 ksi + 27.95 ksi = 40.45 ksi
|τMAX | = r = 10 MPa
σ2 = σc − r = 12.5 ksi − 27.95 ksi = −15.4 ksi
Problem 12.47 At a point p a material is subjected
to the state of plane stress σx = 240 MPa, σy =
−120 MPa, τxy = 240 MPa. Use Equation (12-24) to
determine the principal stresses and use Equation (12-27)
to determine the absolute maximum shear stress. Confirm the absolute maximum shear stress by drawing the
superimposed Mohr’s circle as shown in Figure 12-36.
Solution:
From Equations (12-26):
I1 = σx + σy + σz
2 − τ2 − τ2
I2 = σx σy + σy σz + σz σx − τxy
yz
zx
2 − σ τ 2 − σ τ 2 + 2τ τ τ
I3 = σx σy σz − σx τyz
y xz
z xy
xy yz zx
I1 = (240) + (−120) + 0 = 120
I2 = (240)(−120) + (−120)(0) + (0)(240) − (240)2 − (0)2 − (0)2 = −86, 400
I3 = (240)(0)(0) − (240)(0)2 − (−120)(0)2 − (0)(240) + 2(−240)(0)(0) = 0 − 0 − 0 − 0 − 0 + 0 = 0
From Equation (12-25):
σ 3 − I1 σ 2 + I2 σ − I3 = 0
σ 3 − 120σ 2 + −86, 400σ − 0 = 0
The roots of the equation are:
σ1 = 360 MPa
ANS:
σ2 = 0 MPa
σ3 = −240 MPa
Using Equations (12-27) to fid the absolute maximum shear stress:
σ1 − σ 2
360 MPa − (−240 MPa)
=
= 300 MPa
2
2
σ1 − σ 3
360 MPa − (0 MPa)
=
= 180 MPa
2
2
σ2 − σ3
= 120 MPa
2
The largest value from Equation (12-27) is:
τMAX = 300 MPa
ANS:
Using Mohr’s circle to solve the problem:
The center of the circle is at :
σc =
σx + σy
240 MPa + (−120 MPa)
=
= 60 MPa
2
2
τc = 0
The radius of the circle is:
r=
(σx − σc )2 + (τxy )2 =
(240 MPa − (−120 MPa))2 + (240 MPa)2 = 300 MPa
The radius of the superimposed Mohr’s circle is the same as the twodimensional Mohr’s circle.
r = 300 MPa
We see from Mohr’s circle that:
ANS:
ANS:
σ1 = σc + r = 60 MPa + 300 MPa = 360 MPa
|τMAX | = r = 300 MPa
σ2 = 0 MPa
σ3 = σc − r = 60 MPa − 300 MPa = −240 MPa
Problem 12.48 Strain gauges attached to one of the
Space Shuttle main engine nozzles determine that the
components of plane stress are σx = 67.34 MPa, σy =
82.66 MPa, and τxy = 6.43 MPa. Use Equation (12-25)
to determine the principal stresses and use Equation (1227) to determine the absolute maximum shear stress.
Solution:
From Equations (12-26):
I1 = σx + σy + σz
2 − τ2 − τ2
I2 = σx σy + σy σz + σz σx − τxy
yz
zx
2 − σ τ 2 − σ τ 2 + 2τ τ τ
I3 = σx σy σz − σx τyz
y xz
z xy
xy yz zx
I1 = (67.34) + (82.66) + 0 = 150
I2 = (67.34)(82.66) + (82.66)(0) + (0)(67.34) − (6.43)2 − (0)2 − (0)2 = 5525
I3 = (67.34)(82.66)(0) − (67.34)(0)2 − (82.66)(0)2 − (0)(6.43)2 − 2(6.43)(0)(0) = 0
From Equation (12-25):
σ 3 − I1 σ 2 + I2 σ − I3 = 0
σ 3 − 150σ 2 + 5525σ = 0
The roots of the equation are:
ANS:
σ1 = 85 MPa
σ2 = 65 MPa
σ3 = 0 MPa
Using Equations (12-27) to fid the absolute maximum shear stress:
σ1 − σ2
85 MPa − (65 MPa)
=
= 10 MPa
2
2
σ1 − σ3
85 MPa − 0 MPa
=
= 42.5 MPa
2
2
σ2 − σ3
65 MPa − 0 MPa
=
= 32.5 MPa
2
2
The absolute maximum shear stress is:
ANS:
τMAX = 42.5 MPa
Problem 12.49 Use Equation (12-25) to determine the
principal stresses for an arbitrary state of plane stress σx ,
σy , τxy and confirm Equation (12-15).
Solution:
From Equations (12-26):
I1 = σx + σy
2 − τ2 − τ2 = σ σ + 0 + 0 − τ2 = σ σ − τ2
I2 = σx σy + σy σz + σz σx − τxy
x y
x y
yz
zx
xy
xy
I3 = σx σy (0) = 0
From Equation (12-25):
σ 3 − I1 σ 2 + I2 σ − I3 = 0
2
σ 2 − (σx + σy )σ + σx σy − τxy
=0
Using the quadratic Equation to solve for the values of σ:
σ1,2 =
ANS:
(σx +σy )± (σx +σy )2 −4(1)
2
σx σy −τxy
2
σ1,2 =
σx +σy
2
±
σx −σy
2
2
2
− τxy
=
σx +σy
2
±
2 +2σ σ +σ 2
σx
x y
y
4
−
2
4σx σy −4τxy
4
Problem 12.50 At a point p a material is subjected
to the state of triaxial stress σx = 240 MPa, σy =
−120 MPa, σz = 240 MPa. Determine the principal
stresses and the absolute maximum shear stress.
Solution:
From Equations (12-26):
I1 = σx + σy + σz = 240 MPa − 120 MPa + 240 MPa = 360 MPa
2 − τ2 − τ2
I2 = σx σy + σy σz + σz σx − τxy
yz
zx
= (240 MPa)(−120 MPa) + (−120 MPa)(240 MPa) + (240 MPa)(240 MPa) − (0)2 − (0)2 − (0)2 = 0
I3 = σx σy σz = (240)(−120)(240) = −6, 912, 000
From Equation (12-26):
σ 3 − I1 σ 2 + I2 σ − I3 = 0
σ 3 − 360σ 2 + (0)σ + 6, 912, 000 = 0
Using a graphing calculator to solve for the values of σ:
ANS:
σ1 = σ2 = 240 MPa
σ3 = −120 MPa
Using Equations (12-27) to find the maximum shear stress:
σ1 −σ2
2
=
σ1 −σ3
2
=
σ2 −σ3
2
=
240 MPa−240 MPa
=
2
240 MPa−(−120 MPa)
2
240 MPa−(−120 MPa)
2
0
= 180 MPa
= 180 MPa
Maximum shear stress is:
ANS:
τMAX = 180 MPa
Problem 12.51 At a point p a material is subjected to
the state of triaxial stress σx = 40 ksi, σy = 80 ksi,
σz = −20 ksi. Determine the principal stresses and the
absolute maximum shear stress.
Solution:
From Equations (12-26):
I1 = σx + σy + σz = 40 + 80 − 20 = 100
2 − τ 2 − τ 2 = (40)(80) + (80)(−20) + (−20)(40) − 0 − 0 − 0 = 800
I2 = σx σy + σy σz + σz σx − τxy
yz
zx
I3 = σx σy σz = (40)(80)(−20) = −64, 000
From Equation (12-25):
σ 3 − I1 σ 2 + I2 σ − I3 = 0
σ 3 − 100σ 2 + 800σ + 64, 000 = 0
Using a graphing calculator to solve for the values of σ:
ANS:
σ1 = 80 ksi
σ2 = 40 ksi
σ3 = −20 ksi
Using Equations (12-27) to determine the maximum shear stress:
σ1 −σ2
2
=
σ1 −σ3
2
=
σ2 −σ3
2
=
80 ksi−40 ksi
=
2
80 ksi−(−20 ksi)
2
40 ksi−(−20 ksi)
2
The maximum shear stress is:
ANS:
τMAX = 50 ksi
20 ksi
= 50 ksi
= 30 ksi
Problem 12.52 At a point p a material is subjected to
a state of stress (in ksi)
Determine the principal stresses and the absolute maximum shear stress. Confirm the absolute maximum shear
stress by drawing the superimposed Mohr’s circle as
shown in Figure 12.36.

σx

 τyx
τzx
τxy
σy
τzy
 
τxz
300
 
τyz  =  150
σz
−100
150
200
100

−100

100 
−200
Solution:
From Equations (12-26):
I1 = σx + σy + σz = 300 + 200 − 200 = 300
2 − τ 2 − τ 2 = (300)(200) + (200)(−200) + (−200)(300) − (150)2 − (100)2 − (−100)2 = −82, 500
I2 = σx σy + σy σz + σz σx − τxy
yz
zx
2 − σ τ 2 − σ τ 2 + 2τ τ τ
I3 = σx σy σz − σx τyz
y xz
z xy
xy yz zx
I3 = (300)(200)(−200) − (300)(−100)2 − (200)(−100)2 − (−200)(150)2 + 2(150)(−100)(100) = −15, 500, 000
From Equation (12-25):
σ 3 − I1 σ 2 + I2 σ − I3 = 0
σ 3 − 300σ 2 − 82, 500σ + 15, 500, 000 = 0
Using a graphing calculator to solve for the values of σ:
ANS:
σ1 = 409 ksi
σ2 = 148 ksi
σ3 = −257 ksi
Using Equations (12-27) to determine the maximum shear stress:
σ1 −σ2
2
=
σ1 −σ3
2
=
σ2 −σ3
2
=
409 ksi−148 ksi
=
2
409 ksi−(−257 ksi)
2
130.5 ksi
148 ksi−(−257 ksi)
2
= 333 ksi
= 202.5 ksi
The maximum shear stress is:
ANS:
τMAX = 333 ksi
Problem 12.53 A finite element analysis of a bearing
housing indicates that at a point p the material is subjected to the state of stress (in MPa)

 

σx τxy τxz
20 20
0

 

 τyx σy τyz  =  20 −30 −10 
τzx τzy σz
0 −10 40
Determine the principal stresses and the absolute maximum shear stress.
Solution:
From Equations (12-26):
I1 = σx + σy + σz = 20 + (−30) + 40 = 30
2 − τ2 − τ2
I2 = σx σy + σy σz + σz σx − τxy
yz
zx
I2 = (20)(−30) + (−30)(40) + (40)(20) − (20)2 − (−10)2 − (0)2 = −1500
2 − σ τ 2 − σ τ 2 + 2τ τ τ
I3 = σx σy σz − σx τyz
y xz
z xy
xy yz zx
I3 = (20)(−30)(40) − (20)(−10)2 − (−30)(0)2 − (40)(20)2 + 2(20)(−10)(0) = −42, 000
From Equation (12-25):
σ 3 − 30σ 2 + (−1500)σ − (−42, 000) = 0
ANS:
σ1 = 41.86 ksi
σ2 = 26.29 ksi
σ3 = −38.16 ksi
Using Equations (12-26):
ANS:
σ1 −σ2
2
=
σ1 −σ3
2
=
σ2 −σ3
2
=
41.86−26.29
=
2
41.86−(−38.16)
2
26.29−(−38.16)
2
τMAX = 40 ksi
7.785 ksi
= 40.01 ksi
= 32.225 ksi
Problem 12.54 The components of plane stress at a
point p of a material are σx = −8 ksi, σy = 6 ksi,
and τxy = −6 ksi. Use Equations (12-7)–(12-9) to
determine the components of stress σx , σy , and τxy
corresponding to a coordinate system x y z oriented
at θ = 30◦ . (a) Determine the principal stresses from
Equation (12-25) using the components of stress σx , σy ,
and τxy . (b) Determine the principal stress from Equation (12-25) using the components of stress σx , σy , and
τxy
.
Solution:
Using Equation (12-7) to determine σx :
σx =
σx =
σx + σy
σ x − σy
+
cos(2θ) + τxy sin(2θ)
2
2
−8 ksi + 6 ksi −8 ksi − 6 ksi
+
cos(60◦ )+(−6 ksi) sin(60◦ ) = −9.7 ksi
2
2
Using Equation (12-9) to determine σy :
σy =
σx =
σx + σy
σx − σy
−
cos(2θ) − τxy sin(2θ)
2
2
−8 ksi + 6 ksi2 −8 ksi − 6 ksi
−
cos(60◦ )−(−6 ksi) sin(60◦ ) = 7.7 ksi
2
2
:
Using Equation (12-8) to determine τxy
τxy
=−
τxy
=−
σx − σy
sin(2θ) + τxy cos(2θ)
2
−8 ksi − 6 ksi
sin(60◦ ) + (−6 ksi) cos(60◦ ) = 3.06 ksi
2
Using Equations (12-26) and σx , σy and τxy to determine the coefficients I1 , I2 and I3 :
I1 = σx + σy + σz = −8 + 6 + 0 = −2
2 − τ 2 − τ 2 = (−8)(6) + (6)(0) + (0)(−8) − (−6)2 + 02 + 02 = −84
I2 = σx σy + σy σz + σz σx − τxy
yz
zx
2 − σ τ 2 − σ τ 2 + 2τ τ τ
2
2
2
I3 = σx σy σz − σx τyz
y xz
z xy
xy yz zx = (−8)(6)(0) − (−8)(0) − (6)(0) − (0)(−6) + 2(−6)(0)(0) = 0
From Equation (12-26):
σ 3 − I1 σ 2 + I2 σ − I3 = 0
σ 3 + 2σ 2 − 84σ + 0 = 0
ANS:
σ1 = 8.22 ksi,
σ2 = −10.22 ksi,
σ3 = 0 ksi
to determine the coeffiUsing Equations (12-26) and σx , σy and τxy
cients I1 , I2 and I3 :
I1 = σx + σy + σz = −9.7 + 7.7 + 0 = −2
2 − τ 2 − τ 2 = (−9.7)(7.7) + (7.7)(0) + (0)(−9.7) − (3.06)2 − (0)2 − (0)2 = −84
I2 = σx σy + σy σz + σz σx − τxy
yz
zx
2 − σ τ 2 − σ τ 2 + 2τ τ τ = 0
I3 = σx σy σz − σx τyz
y xy
z xy
xy yz zx
We see that the coefficients I1 , I2 , and I3 are the same as for the
previous calculation using σx , σy and τxy . The resulting values of
σ1 , σ2 and σ3 MUST be the same as in the previous calculation.
Problem 12.55 A spherical pressure vessel has a 2.5-m
radius and a 5-mm wall thickness. It contains a gas with
pressure pi = 6 × 105 Pa and the outer wall is subjected
to atmospheric pressure p0 = 1 × 105 Pa. Determine
the maximum normal stress in the vessel wall.
Solution:
The projected area of half of the sphere is:
AP = πr 2 = π(2.5 m)2 = 19.6 m2
The net force on the hemisphere is:
F = (Pi − P0 )(AP ) = (6 × 105 N/m2 − 1 × 105 N/m2 )(19.6 m2 )
F = 9.8 × 106 N
The area over which the circumferential load is exerted is:
A = 2πrt = 2π(2.5 m)(0.005 m) = 0.0785 m2
The normal (circumferential) stress in the material is:
σ=
ANS:
F
9.8 × 106 N
=
A
0.0785 m2
σ = 125 MPa
Problem 12.56 A spherical pressure vessel has a 24in. radius and a 1/64-in. wall thickness. It contains a
gas with pressure pi = 200 psi and the outer wall is subjected to atmospheric pressure p0 = 14.7 psi. Determine
the maximum normal stress and the absolute maximum
shear stress at the vessel’s inner surface.
Solution:
The projected area of the cross-section is:
AP = πr 2 = π(24 in)2 = 1809.6 in2
The area of material which resists the pressure is:
A = 2πrt = 2π(24 in)(1/64 in) = 2.356 in2
Net force on the projected area of the hemisphere is:
F = (200 lb/in2 − 14.7 lb/in2 )(1809.6 in2 ) = 335, 319 lb
The normal stress in the material is:
σx = σy =
335, 319 lb
= 142.3 ksi
2.356 in2
The shear stress in the sphere is:
τxy = (Pi −PATM ) = (200 lb/in2 −14.7 lb/in2 ) = 185.3 lb/in2
Using Equations (12-26) to determine I1 , I2 , and I3 :
I1 = σx + σy + σz = 142.3 + 142.3 + 0 = 284.6
2 − τ 2 + τ 2 = (142.3)(142.3) + (142.3)(0) + (0)(142.3) − (0.185)2 − (0)2 − (0)2 = 20, 250
I2 = σx σy + σy σz + σz σx − τxy
yz
zx
2 − σ τ 2 − σ τ 2 + 2τ τ τ
2
2
I3 = σx σy σz − σx τyz
y xz
z xy
xy yz yz = (142.3)(142.3)(0) − (142.3)(0) − (142.3)(0) − (0)(0.185) + 2(0.185)(0)(0) = 0
Using Equation (12-29) to determine the absolute maximum shear
stress:
σ 3 −I1 σ 2 +I2 σ−I3 = 0 → σ 3 −284.6σ 2 +20, 250σ+0 = 0
The roots of Equation [1] are:
σ1 = 142.3
σ2 = 142.3
σ3 = 0
Absolute maximum shear stress on the sphere is:
τMAX =
ANS:
σ1 − σ3
142.3 MPa − 0 MPa
=
2
2
τMAX = 71.15 ksi
[1]
Problem 12.57 Suppose that the spherical pressure
vessel described in Problem 12.56 is made of material
with a yield shear stress τY = 100 ksi. If the vessel
is designed to contain gas with a maximum pressure
pi = 150 psi and the outer wall is subjected to atmospheric pressure p0 = 14.7 psi, what is the factor of
safety?
Solution:
The projected area of the cross-section is:
AP = πr 2 = π(24 in)2 = 1809.6 in2
The area of material which resists the pressure is:
A = 2πrt = 2π(24 in)(1/64 in) = 2.356 in2
Net force on the projected area of the hemisphere is:
F = (150 lb/in2 − 14.7 lb/in2 )(1809.6 in2 ) = 244, 839 lb
The normal stress in the material is:
σx = σy =
244, 839 lb
= 103.9 ksi
2.356 in2
The shear stress in the sphere is:
τxy = (Pi −PATM ) = (150 lb/in2 −14.7 lb/in2 ) = 135.3 lb/in2
Using Equations (12-26) to determine I1 , I2 , and I3 :
I1 = σx + σy + σz = 103.9 + 103.9 + 0 = 207.8
2 − τ 2 + τ 2 = (103.9)(103.9) + (103.9)(0) + (0)(103.9) − (0.135)2 − (0)2 − (0)2 = 10, 795
I2 = σx σy + σy σz + σz σx − τxy
yz
zx
2 − σ τ 2 − σ τ 2 + 2τ τ τ
2
2
I3 = σx σy σz − σx τyz
y xz
z xy
xy yz yz = (142.3)(142.3)(0) − (142.3)(0) − (142.3)(0) − (0)(0.135) + 2(0.135)(0)(0) = 0
Using Equation (12-29) to determine the absolute maximum shear
stress:
σ 3 −I1 σ 2 +I2 σ−I3 = 0 → σ 3 −207.8σ 2 +10, 795σ+0 = 0
The roots of Equation [1] are:
σ1 = 104.4
σ2 = 104.4
σ3 = 0
Absolute maximum shear stress is the sphere is:
τMAX =
ANS:
σ1 − σ3
104.4 MPa − 0 MPa
=
2
2
τMAX = 52.2 MPa
τALLOW =
τYIELD
100 × 103
=
S
S
As per the book, absolute maximum shear stress is:
σ + Pi
2
Therefore:
σ+Pi
2
= τALLOW =
103.9×103 +150
2
ANS:
S = 1.92
=
100×103
S
100×103
S
[1]
Problem 12.58 A cylindrical pressure vessel with
hemispherical ends has a 2.5-m radius and a 5-mm wall
thickness. It contains a gas with pressure pi = 6×105 Pa
and the outer wall is subjected to atmospheric pressure
p0 = 1 × 105 Pa. Determine the maximum normal
stress in the vessel wall. Compare your answer with the
answer to Problem 12.55.
Solution:
The area over which the axial load is distributed is:
AA = 2πrt = 2π(2.5 m)(0.005 m) = 0.0785 m2
The area over which the circumferential load is distributed is:
AC = 2rL = 2(2.5 m)(L) = (5L) m2
The projected area in the axial direction is:
(APROJ )A = πr 2 = π(2.5 m)2 = 19.63 m2
The projected area for a given section of the cylinder is:
(APROJ )C = 2tL = 2(0.005 m)(L) = (0.01L) m2
The force exerted in the axial direction is:
FA = (Pi −Po )(APROJ )A = (6×105 N/m2 −1×105 N/m2 )(19.63 m2 )
FA = 9.81 MN
The force exerted in the circumferent