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SOLUTION MANUAL FOR
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Daladier
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LIBROS UNIVERISTARIOS 
Y SOLUCIONARIOS DE 
MUCHOS DE ESTOS LIBROS

LOS SOLUCIONARIOS 
CONTIENEN TODOS LOS 
EJERCICIOS DEL LIBRO
RESUELTOS Y EXPLICADOS
DE FORMA CLARA

VISITANOS PARA
DESARGALOS GRATIS.
TABLE OF CONTENTS 
 
1. Introduction. 
Engineering and Mechanics. Learning Mechanics. Fundamental Concepts. Units. Newtonian 
Gravitation. 
 
2. Vectors. 
Vector Operations and Definitions. Scalars and Vectors. Rules for Manipulating Vectors. 
Cartesian Components. Components in Two Dimensions. Components in Three Dimensions. 
Products of Vectors. Dot Products. Cross Products. Mixed Triple Products. 
 
3. Forces. 
Types of Forces. Equilibrium and Free-Body Diagrams. Two-Dimensional Force Systems. 
Three-Dimensional Force Systems. 
 
4. Systems of Forces and Moments. 
Two-Dimensional Description of the Moment. The Moment Vector. Moment of a Force About a 
Line. Couples. Equivalent Systems. Representing Systems by Equivalent Systems. 
 
5. Objects in Equilibrium. 
The Equilibrium Equations. Two-Dimensional Applications. Statically Indeterminate Objects. 
Three-Dimensional Applications. Two-Force and Three-Force. 
 
6. Structures in Equilibrium. 
Trusses. The Method of Joints. The Method of Sections. Space Trusses. Frames and Machines. 
 
7. Centroids and Centers of Mass 316. 
Centroids. Centroids of Areas. Centroids of Composite Areas. Distributed Loads. Centroids of 
Volumes and Lines. The Pappus-Guldinus Theorems. Centers of Mass. Definition of the Center 
of Mass. Centers of Mass of Objects. Centers of Mass of Composite Objects. 
 
8. Moments of Inertia. 
Areas. Definitions. Parallel-Axis Theorems. Rotated and Principal Axes. Masses. Simple 
Objects. Parallel-Axis Theorem. 
 
9. Friction. 
Theory of Dry Friction. Applications. 
 
10. Internal Forces and Moments. 
Beams. Axial Force, Shear Force, and Bending Moment. Shear Force and Bending Moment 
Diagrams. Relations Between Distributed Load, Shear Force, and Bending Moment. Cables. 
Loads Distributed Uniformly Along Straight Lines. Loads Distributed Uniformly Along Cables. 
Discrete Loads. Liquids and Gasses. Pressure and the Center of Pressure. Pressure in a Stationary 
Liquid. 
 
11. Virtual Work and Potential Energy. 
Virtual Work. Potential Energy. 
 
Problem 1.1 The value of � is 3.14159265. . . If C is
the circumference of a circle and r is its radius, deter-
mine the value of r/C to four significant digits.
Solution: C D 2�r ) r
C
D 1
2�
D 0.159154943.
To four significant digits we have
r
C
D 0.1592
Problem 1.2 The base of natural logarithms is e D
2.718281828 . . .
(a) Express e to five significant digits.
(b) Determine the value of e2 to five significant digits.
(c) Use the value of e you obtained in part (a) to deter-
mine the value of e2 to five significant digits.
[Part (c) demonstrates the hazard of using rounded-off
values in calculations.]
Solution: The value of e is: e D 2.718281828
(a) To five significant figures e D 2.7183
(b) e2 to five significant figures is e2 D 7.3891
(c) Using the value from part (a) we find e2 D 7.3892 which is
not correct in the fifth digit.
Problem 1.3 A machinist drills a circular hole in a
panel with a nominal radius r D 5 mm. The actual radius
of the hole is in the range r D 5š 0.01 mm. (a) To what
number of significant digits can you express the radius?
(b) To what number of significant digits can you express
the area of the hole?
Solution:
a) The radius is in the range r1 D 4.99 mm to r2 D 5.01 mm. These
numbers are not equal at the level of three significant digits, but
they are equal if they are rounded off to two significant digits.
Two: r D 5.0 mm
b) The area of the hole is in the range from A1 D �r12 D 78.226 m2
to A2 D �r22 D 78.854 m2. These numbers are equal only if rounded
to one significant digit:
One: A D 80 mm2
Problem 1.4 The opening in the soccer goal is 24 ft
wide and 8 ft high, so its area is 24 ftð 8 ft D 192 ft2.
What is its area in m2 to three significant digits?
Solution:
A D 192 ft2
(
1 m
3.281 ft
)2
D 17.8 m2
A D 17.8 m2
Problem 1.5 The Burj Dubai, scheduled for comple-
tion in 2008, will be the world’s tallest building with a
height of 705 m. The area of its ground footprint will be
8000 m2. Convert its height and footprint area to U.S.
customary units to three significant digits.
Solution:
h D 705 m
(
3.281 ft
1 m
)
D 2.31ð 103 ft
A D 8000 m2
(
3.218 ft
1 m
)2
D 8.61ð 104 ft2
h D 2.31ð 103 ft, A D 8.61ð 104 ft2
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1
Problem 1.6 Suppose that you have just purchased
a Ferrari F355 coupe and you want to know whether
you can use your set of SAE (U.S. Customary Units)
wrenches to work on it. You have wrenches with widths
w D 1/4 in, 1/2 in, 3/4 in, and 1 in, and the car has nuts
with dimensions n D 5 mm, 10 mm, 15 mm, 20 mm,
and 25 mm. Defining a wrench to fit if w is no more
than 2% larger than n, which of your wrenches can you
use?
n
Solution: Convert the metric size n to inches, and compute the
percentage difference between the metric sized nut and the SAE
wrench. The results are:
5 mm
(
1 inch
25.4 mm
)
D 0.19685.. in,
(
0.19685� 0.25
0.19685
)
100
D �27.0%
10 mm
(
1 inch
25.4 mm
)
D 0.3937.. in,
(
0.3937� 0.5
0.3937
)
100 D �27.0%
15 mm
(
1 inch
25.4 mm
)
D 0.5905.. in,
(
0.5905� 0.5
0.5905
)
100 D C15.3%
20 mm
(
1 inch
25.4 mm
)
D 0.7874.. in,
(
0.7874� 0.75
0.7874
)
100 D C4.7%
25 mm
(
1 inch
25.4 mm
)
D 0.9843.. in,
(
0.9843� 1.0
0.9843
)
100 D �1.6%
A negative percentage implies that the metric nut is smaller than the
SAE wrench; a positive percentage means that the nut is larger then
the wrench. Thus within the definition of the 2% fit, the 1 in wrench
will fit the 25 mm nut. The other wrenches cannot be used.
Problem 1.7 Suppose that the height of Mt. Everest is
known to be between 29,032 ft and 29,034 ft. Based on
this information, to how many significant digits can you
express the height (a) in feet? (b) in meters?.
Solution:
a) h1 D 29032 ft
h2 D 29034 ft
The two heights are equal if rounded off to four significant digits.
The fifth digit is not meaningful.
Four: h D 29,030 ft
b) In meters we have
h1 D 29032 ft
(
1 m
3.281 ft
)
D 8848.52 m
h2 D 29034 ft
(
1 m
3.281 ft
)
D 8849.13 m
These two heights are equal if rounded off to three significant
digits. The fourth digit is not meaningful.
Three: h D 8850 m
Problem 1.8 The maglev (magnetic levitation) train
from Shanghai to the airport at Pudong reaches a speed
of 430 km/h. Determine its speed (a) in mi/h; (b) ft/s.
Solution:
a) v D 430 km
h
(
0.6214 mi
1 km
)
D 267 mi/h v D 267 mi/h
b) v D 430km
h
(
1000 m
1 km
) (
1 ft
0.3048 m
) (
1 h
3600 s
)
D 392 ft/s
v D 392 ft/s
Problem 1.9 In the 2006 Winter Olympics, the men’s
15-km cross-country skiing race was won by Andrus
Veerpalu of Estonia in a time of 38 minutes, 1.3 seconds.
Determine his average speed (the distance traveled
divided by the time required) to three significant digits
(a) in km/h; (b) in mi/h.
Solution:
a) v D 15 km(
38C 1.3
60
)
min
(
60 min
1 h
)
D 23.7 km/h v D 23.7 km/h
b) v D �23.7 km/h�
(
1 mi
1.609 km
)
D 14.7 mi/h v D 14.7 mi/h
2
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 1.10 The Porsche’s engine exerts 229 ft-lb
(foot-pounds) of torque at 4600 rpm. Determine the
value of the torque in N-m (Newton-meters).
Solution:
T D 229 ft-lb
(
1 N
0.2248 lb
) (
1 m
3.281 ft
)
D 310 N-m T D 310N-m
Problem 1.11 The kinetic energy of the man in Active
Example 1.1 is defined by 12mv
2, where m is his
mass and v is his velocity. The man’s mass is 68 kg
and he is moving at 6 m/s, so his kinetic energy is
1
2 (68 kg)(6 m/s)
2 D 1224 kg-m2/s2. What is his kinetic
energy in U.S. Customary units?
Solution:
T D 1224 kg-m2/s2
(
1 slug
14.59 kg
) (
1 ft
0.3048 m
)2
D 903 slug-ft2/s
T D 903 slug-ft2/s
Problem 1.12 The acceleration due to gravity at sea
level in SI units is g D 9.81 m/s2. By converting units,
use this value to determine the acceleration due to
gravity at sea level in U.S. Customary units.
Solution: Use Table 1.2. The result is:
g D 9.81
( m
s2
)( 1 ft
0.3048 m
)
D 32.185 . . .
(
ft
s2
)
D 32.2
(
ft
s2
)
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 nights). If you walk to class
at 2 m/s, what is your speed in furlongs per fortnight to
three significant digits?
Solution:
v D 2 m/s
(
1 ft
0.3048 m
)(
1 furlong
660 ft
) (
3600 s
hr
) (
24 hr
1 day
) (
14 day
1 fortnight
)
v D 12,000 furlongs
fortnight
Problem 1.14 Determine the cross-sectional area of
the beam (a) in m2; (b) in in2.
120 mm x
y
40 mm
40 mm
40
mm
200 mm
Solution:
A D �200 mm�2 � 2�80 mm��120 mm� D 20800 mm2
a) A D 20800 mm2
(
1 m
1000 mm
)2
D 0.0208 m2 A D 0.0208 m2
b) A D 20800 mm2
(
1 in
25.4 mm
)2
D 32.2 in2 A D 32.2 in2
Problem 1.15 The cross-sectional area of the C12ð30
American Standard Channel steel beam is A D 8.81 in2.
What is its cross-sectional area in mm2?
x
y
A
Solution:
A D 8.81 in2
(
25.4 mm
1 in
)2
D 5680 mm2
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
3
Problem 1.16 A pressure transducer measures a value
of 300 lb/in2. Determine the value of the pressure in
pascals. A pascal (Pa) is one newton per meter squared.
Solution: Convert the units using Table 1.2 and the definition of
the Pascal unit. The result:
300
(
lb
in2
) (
4.448 N
1 lb
) (
12 in
1 ft
)2 ( 1 ft
0.3048 m
)2
D 2.0683 . . . �106�
(
N
m2
)
D 2.07�106� Pa
Problem 1.17 A horsepower is 550 ft-lb/s. A watt is
1 N-m/s. Determine how many watts are generated by
the engines of the passenger jet if they are producing
7000 horsepower.
Solution:
P D 7000 hp
(
550 ft-lb/s
1 hp
)(
1 m
3.281 ft
) (
1 N
0.2248 lb
)
D 5.22ð 106 W
P D 5.22ð 106 W
Problem 1.18 Chapter 7 discusses distributed loads
that are expressed in units of force per unit length. If
the value of a distributed load is 400 N/m, what is its
value in lb/ft?.
Solution:
w D 400 N/m
(
0.2248 lb
1 N
) (
1 m
3.281 ft
)
D 27.4 lb/ft w D 27.4 lb/ft
Problem 1.19 The moment of inertia of the rectan-
gular area about the x axis is given by the equation
I D 13bh3.
The dimensions of the area are b D 200 mm and h D
100 mm. Determine the value of I to four significant
digits in terms of (a) mm4; (b) m4; (c) in4.
h
b
x
y
Solution:
(a) I D 1
3
�200 mm��100 mm�3 D 66.7ð 106 mm4
(b) I D 66.7ð 106 mm4
(
1 m
1000 mm
)4
D 66.7ð 10�6 m4
(c) I D 66.7ð 106 mm4
(
1 in
25.4 mm
)4
D 160 in4
Problem 1.20 In Example 1.3, instead of Einstein’s
equation consider the equation L D mc, where the mass
m is in kilograms and the velocity of light c is in meters
per second. (a) What are the SI units of L? (b) If the
value of L in SI units is 12, what is its value in U.S.
Customary base units?
Solution:
a) L D mc) Units �L� D kg-m/s
b) L D 12 kg-m/s
(
0.0685 slug
1 kg
) (
3.281 ft
1 m
)
D 2.70 slug-ft/s
L D 2.70 slug-ft/s
4
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 1.21 The equation
� D My
I
is used in the mechanics of materials to determine
normal stresses in beams.
(a) When this equation is expressed in terms of SI base
units, M is in newton-meters (N-m), y is in meters
(m), and I is in meters to the fourth power (m4).
What are the SI units of �?
(b) If M D 2000 N-m, y D 0.1 m, and I D 7ð
10�5 m4, what is the value of � in U.S. Customary
base units?
Solution:
(a) � D My
I
D (N-m)m
m4
D N
m2
(b)
� D My
I
D �2000 N-m��0.1 m�
7ð 10�5 m4
(
1 lb
4.448 N
) (
0.3048 m
ft
)2
D 59,700 lb
ft2
Problem 1.22 The acceleration due to gravity on the
surface of the moon is 1.62 m/s2. (a) What would the
mass of the C-clamp in Active Example 1.4 be on the surface
of the moon? (b) What would the weight of the C-clamp
in newtons be on the surface of the moon?
Solution:
a) The mass does not depend on location. The mass in kg is
0.0272 slug
(
14.59 kg
1 slug
)
D 0.397 kg mass D 0.397 kg
b) The weight on the surface of the moon is
W D mg D �0.397 kg��1.62 m/s2� D 0.643 N W D 0.643N
Problem 1.23 The 1 ft ð 1 ft ð 1 ft cube of iron
weighs 490 lb at sea level. Determine the weight in
newtons of a 1 m ð 1 m ð 1 m cube of the same
material at sea level. 1 ft
1 ft 1 ft
Solution: The weight density is � D 490 lb
1 ft3
The weight of the 1 m3 cube is:
W D �V D
(
490 lb
1 ft3
)
�1 m�3
(
1 ft
0.3048 m
)3 ( 1 N
0.2248 lb
)
D 77.0 kN
Problem 1.24 The area of the Pacific Ocean is
64,186,000 square miles and its average depth is 12,925 ft.
Assume that the weight per unit volume of ocean water
is 64 lb/ft3. Determine the mass of the Pacific Ocean
(a) in slugs; (b) in kilograms
Solution: The volume of the ocean is
V D �64,186,000 mi2��12,925 ft�
(
5,280 ft
1 mi
)2
D 2.312ð 1019 ft3
(a) m D �V D
(
64 lb/ft3
32.2 ft/s2
)
�2.312ð 1019 ft3� D 4.60ð 1019 slugs
(b) m D �4.60ð 1019 slugs�
(
14.59 kg
1 slug
)
D 6.71ð 1020 kg
Problem 1.25 The acceleration due to gravity at
sea level is g D 9.81 m/s2. The radius of the earth
is 6370 km. The universal gravitational constant is
G D 6.67ð 10�11 N-m2/kg2. Use this information to
determine the mass of the earth.
Solution: Use Eq (1.3) a D GmE
R2
. Solve for the mass,
mE D gR
2
G
D
�9.81 m/s2��6370 km�2
(
103
m
km
)2
6.67�10�11�
(
N-m2
kg2
)
D 5.9679 . . . �1024� kg D 5.97�1024� kg
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
5
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).
WDmg
(
RE
r
)2
D
(
WE
g
)
g
(
RE
RE CH
)2
DWE
(
3960
3960C 200
)2
D �180��0.90616� D 163 lb
Problem 1.27 The acceleration due to gravity on the
surface of the moon is 1.62 m/s2. The moon’s radius is
RM D 1738 km.
(a) What is the weight in newtons on the surface of
the moon of an object that has a mass of 10 kg?
(b) Using the approach described in Example 1.5, deter-
mine the force exerted on the object by the gravity
of the moon if the object is located 1738 km above
the moon’s surface.
Solution:
a) W D mgM D �10 kg��1.26 m/s2� D 12.6 N W D 12.6 N
b) Adapting equation 1.4 we have aM D gM
(
RM
r
)2
. The force is
then
F D maM D �10 kg��1.62 m/s2�
(
1738 km
1738 kmC 1738 km
)2
D 4.05 N
F D 4.05 N
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 acceler-
ation due to gravity at sea level is g D 9.81 m/s2. The
radius of the earthis 6370 km. The weight of an object
at sea level is mg, where m is its mass. At what height
above the earth does the weight of the object decrease
to 0.99 mg?
Solution: Use a variation of Eq (1.5).
W D mg
(
RE
RE C h
)2
D 0.99 mg
Solve for the radial height,
h D RE
(
1p
0.99
� 1
)
D �6370��1.0050378 � 1.0�
D 32.09 . . . km D 32,100 m D 32.1 km
Problem 1.29 The planet Neptune has an equatorial
diameter of 49,532 km and its mass is 1.0247ð 1026 kg.
If the planet is modeled as a homogeneous sphere, what
is the acceleration due to gravity at its surface? (The
universal gravitational constant is G D 6.67ð 10�11
N-m2/kg2.)
Solution:
We have: W D GmNm
rN2
D
(
G
mN
r2
)
m) gN D GmN
rN2
Note that the radius of Neptune is rN D 12 �49,532 km� D 24,766 km
Thus gN D
(
6.67ð 10�11 N-m
2
kg2
) (
1.0247ð 1026 kg
�24766 km�2
) (
1 km
1000 m
)2
D 11.1 m/s2 gN D 11.1 m/s2
Problem 1.30 At a point between the earth and the
moon, the magnitude of the force exerted on an object
by the earth’s gravity equals the magnitude of the force
exerted on the object by the moon’s gravity. What is
the distance from the center of the earth to that point
to three significant digits? The distance from the center
of the earth to the center of the moon is 383,000 km,
and the radius of the earth is 6370 km. The radius of the
moon is 1738 km, and the acceleration due to gravity at
its surface is 1.62 m/s2.
Solution: Let rEp be the distance from the Earth to the point where
the gravitational accelerations are the same and let rMp be the distance
from the Moon to that point. Then, rEp C rMp D rEM D 383,000 km.
The fact that the gravitational attractions by the Earth and the Moon
at this point are equal leads to the equation
gE
(
RE
rEp
)2
D gM
(
RM
rMp
)2
,
where rEM D 383,000 km. Substituting the correct numerical values
leads to the equation
9.81
( m
s2
) (6370 km
rEp
)2
D 1.62
( m
s2
) ( 1738 km
rEM � rEp
)2
,
where rEp is the only unknown. Solving, we get rEp D 344,770 km D
345,000 km.
6
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 2.1 In Active Example 2.1, suppose that the
vectors U and V are reoriented as shown. The vector
V is vertical. The magnitudes are jUj D 8 and jVj D 3.
Graphically determine the magnitude of the vector
UC 2V.
VU
45�
Solution: Draw the vectors accurately and measure the resultant.
R D jUC 2Vj D 5.7 R D 5.7
Problem 2.2 Suppose that the pylon in Example 2.2 is
moved closer to the stadium so that the angle between
the forces FAB and FAC is 50°. Draw a sketch of the
new situation. The magnitudes of the forces are jFABj D
100 kN and jFACj D 60 kN. Graphically determine the
magnitude and direction of the sum of the forces exerted
on the pylon by the cables.
Solution: Accurately draw the vectors and measure the magnitude
and direction of the resultant
jFAB C FACj D 146 kN
˛ D 32°
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
7
Problem 2.3 The magnitude jFAj D 80 lb and the
angle ˛ D 65°. The magnitude jFA C FBj D 120 lb.
Graphically determine the magnitude of FB.
FB
FA
a
FC
�
Solution: Accurately draw the vectors and measure the magnitude
of FB.
jFBj D 62 lb
Problem 2.4 The magnitudes jFAj D 40 N, jFBj D
50 N, and jFCj D 40 N. The angle ˛ D 50° and ˇ D 80°.
Graphically determine the magnitude of FA C FB C FC.
FB
FA
a
FC
�
Solution: Accurately draw the vectors and measure the magnitude
of FA C FB C FC.
R D jFA C FB C FCj D 83 N
8
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 2.5 The magnitudes jFAj D jFBj D jFCj D
100 lb, and the angles ˛ D 30°. Graphically determine
the value of the angle ˇ for which the magnitude
jFA C FB C FCj is a minimum and the minimum value
of jFA C FB C FCj.
FB
FA
a
FC
�
Solution: For a minimum, the vector FC must point back to the
origin.
R D jFA C FB C FCj D 93.2 lb
ˇ D 165°
Problem 2.6 The angle � D 50°. Graphically determine
the magnitude of the vector rAC. 60 mm 150 mm
A C
B
rAB rBC
rAC
�
Solution: Draw the vectors accurately and then measure jrACj.
jrACj D 181 mm
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9
Problem 2.7 The vectors FA and FB represent
the forces exerted on the pulley by the belt.
Their magnitudes are jFAj D 80 N and jFBj D 60 N.
Graphically determine the magnitude of the total force
the belt exerts on the pulley.
45�
FA
FB
10�
Solution: Draw the vectors accurately and then measure jFA C
FBj.
jFA C FBj D 134 N
Problem 2.8 The sum of the forces FA C FB C
FC D 0. The magnitude jFAj D 100 N and the angle ˛ D
60°. Graphically determine the magnitudes jFBj and jFCj.
30 
FB
FA
FC
a
Solution: Draw the vectors so that they add to zero.
jFBj D 86.6 N, jFCj D 50.0 N
Problem 2.9 The sum of the forces FA C FB C
FC D 0. The magnitudes jFAj D 100 N and jFBj D
80 N. Graphically determine the magnitude jFCj and the
angle ˛.
30 
FB
FA
FC
a
Solution: Draw the vectors so that they add to zero.
jFCj D 50.4 N, ˛ D 52.5°
10
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Problem 2.10 The forces acting on the sailplane are
represented by three vectors. The lift L and drag D
are perpendicular. The magnitude of the weight W is
500 lb. The sum of the forces WC LC D D 0. Graph-
ically determine the magnitudes of the lift and drag.
W
D
L
25�
Solution: Draw the vectors so that they add to zero. Then measure
the unknown magnitudes.
jLj D 453 lb
jDj D 211 lb
Problem 2.11 A spherical storage tank is suspended
from cables. The tank is subjected to three forces, the
forces FA and FB exerted by the cables and its weight W.
The weight of the tank is jWj D 600 lb. The vector sum
of the forces acting on the tank equals zero. Graphically
determine the magnitudes of FA and FB.
40˚
FA
W
FB
20˚ 20˚
Solution: Draw the vectors so that they add to zero. Then measure
the unknown magnitudes.
jFAj D jFBj D 319 lb
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11
Problem 2.12 The rope ABC exerts forces FBA and
FBC of equal magnitude on the block at B. The
magnitude of the total force exerted on the block by
the two forces is 200 lb. Graphically determine jFBAj.
20�
FBC
FBA
B
C
A
B
Solution: Draw the vectors accurately and then measure the
unknown magnitudes.
jFBAj D 174 lb
Problem 2.13 Two snowcats tow an emergency shelter
to a new location near McMurdo Station, Antarctica.
(The top view is shown. The cables are horizontal.)
The total force FA C FB exerted on the shelter is in
the direction parallel to the line L and its magnitude
is 400 lb. Graphically determine the magnitudes of FA
and FB.
L
Top View
FA
FB50� 30�
Solution: Draw the vectors accurately and then measurethe
unknown magnitudes.
jFAj D 203 lb
jFBj D 311 lb
Problem 2.14 A surveyor determines that the horizon-
tal distance from A to B is 400 m and the horizontal
distance from A to C is 600 m. Graphically determine
the magnitude of the vector rBC and the angle ˛.
East
North
60�
20�
C
B
A
rBC
a
Solution: Draw the vectors accurately and then measure the
unknown magnitude and angle.
jrBCj D 390 m
˛ D 21.2°
12
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Problem 2.15 The vector r extends from point A to
the midpoint between points B and C. Prove that
r D 12 �rAB C rAC�.
A
C
B
rAC
r
rABrAB
Solution: The proof is straightforward:
r D rAB C rBM, and r D rAC C rCM.
Add the two equations and note that rBM C rCM D 0, since the two
vectors are equal and opposite in direction.
Thus 2r D rAC C rAB, or r D
( 1
2
)
�rAC C rAB�
rAC
rAB
r
A B
M
C
Problem 2.16 By drawing sketches of the vectors,
explain why
UC �VCW� D �UC V�CW.
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 VCW, and then the addition
of U. The result is the vector UC �VCW�.
(b) shows the addition of UC V, and then the addition of W, leading
to the result �UC V�CW.
The final vector in the two sketches is the same vector, illustrating that
associativity of vector addition is intuitively reasonable.
(a)
U
W
V
U
W
V
V+W
U+V
U+[V+W]
[U+V]+W
(b)
Problem 2.17 A force F D 40 i� 20 j �N�. What is
its magnitude jFj?
Strategy: The magnitude of a vector in terms of its
components is given by Eq. (2.8).
Solution: jFj D p402 C 202 D 44.7 N
Problem 2.18 An engineer estimating the components
of a force F D Fx iC Fy j acting on a bridge abutment
has determined that Fx D 130 MN, jFj D 165 MN, and
Fy is negative. What is Fy?
Solution:
jFj D
√
jFx j2 C jFy j2
jFy j D
√
jFj2 � jFx j2 D
√
�165 MN�2 � �130 MN�2 D 101.6 MN
Fy D �102 MN
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13
Problem 2.19 A support is subjected to a force F D
Fx iC 80j (N). If the support will safely support a force
of 100 N, what is the allowable range of values of the
component Fx?
Solution: Use the definition of magnitude in Eq. (2.8) and reduce
algebraically.
100 ½
√
�Fx�2 C �80�2, from which �100�2 � �80�2 ½ �Fx�2.
Thus jFx j �
p
3600, or �60 � �Fx� � C60 (N)
Problem 2.20 If FA D 600i� 800j (kip) and FB D
200i� 200j (kip), what is the magnitude of the force
F D FA � 2FB?
Solution: Take the scalar multiple of FB, add the components of
the two forces as in Eq. (2.9), and use the definition of the magnitude.
F D �600� 2�200��iC ��800� 2��200��j D 200i� 400j
jFj D
√
�200�2 C ��400�2 D 447.2 kip
Problem 2.21 The forces acting on the sailplane are its
weight W D �500j�lb�, the drag D D �200iC 100j(lb)
and the lift L. The sum of the forces WC LC D D 0.
Determine the components and the magnitude of L.
y
x
W
D
L
Solution:
L D �W� D D ���500j�� ��200iC 100j� D 200iC 400j�lb�
jLj D
√
�200 lb�2 C �400 lb�2 D 447 lb
L D 200iC 400j�lb�, jLj D 447 lb
14
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Problem 2.22 Two perpendicular vectors U and V lie
in the x-y plane. The vector U D 6i� 8j and jVj D 20.
What are the components of V? (Notice that this problem
has two answers.)
Solution: The two possible values of V are shown in the sketch.
The strategy is to (a) determine the unit vector associated with U,
(b) express this vector in terms of an angle, (c) add š90° to this
angle, (d) determine the two unit vectors perpendicular to U, and
(e) calculate the components of the two possible values of V. The
unit vector parallel to U is
eU D 6i√
62 C ��8�2
� 8j√
62 C ��8�2
D 0.6i� 0.8j
Expressed in terms of an angle,
eU D i cos˛� j sin˛ D i cos�53.1°�� j sin�53.1°�
Add š90° to find the two unit vectors that are perpendicular to this
unit vector:
ep1 D i cos�143.1°�� j sin�143.1°� D �0.8i� 0.6j
ep2 D i cos��36.9°�� j sin��36.9°� D 0.8iC 0.6j
Take the scalar multiple of these unit vectors to find the two vectors
perpendicular to U.
V1 D jVj��0.8i� 0.6j� D �16i� 12j.
The components are Vx D �16, Vy D �12
V2 D jVj�0.8iC 0.6j� D 16iC 12j.
The components are Vx D 16, Vy D 12
y
x6
8
U
V2
V1
Problem 2.23 A fish exerts a 10-lb force on the line
that is represented by the vector F. Express F in terms
of components using the coordinate system shown.
y
x
F
7
11
Solution: We can use similar triangles to determine the
components of F.
F D �10 lb�
(
7p
72 C 112 i�
11p
72 C 112 j
)
D �5.37i� 8.44j� lb
F D �5.37i� 8.44j� lb
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15
Problem 2.24 A man exerts a 60-lb force F to push a
crate onto a truck. (a) Express F in terms of components
using the coordinate system shown. (b) The weight of
the crate is 100 lb. Determine the magnitude of the sum
of the forces exerted by the man and the crate’s weight.
F 20�
x
y
Solution:
(a) F D �60 lb��cos 20°iC sin 20°j� D �56.4iC 20.5j� lb
F D �56.4iC 20.5j�lb
(b) W D ��100 lb�j
FCW D �56.4iC [20.5� 100]j� lb D �56.4i� 79.5j� lb
jFCWj D
√
�56.4 lb�2 C ��79.5 lb�2 D 97.4 lb
jFCWj D 97.4 lb
Problem 2.25 The missile’s engine exerts a 260-kN
force F. (a) Express F in terms of components using the
coordinate system shown. (b) The mass of the missile
is 8800 kg. Determine the magnitude of the sum of the
forces exerted by the engine and the missile’s weight.
y
x
F
3
4
Solution:
(a) We can use similar triangles to determine the components of F.
F D �260 kN�
(
4p
42 C 32 iC
3p
42 C 32 j
)
D �208iC 156j� kN
F D �208iC 156j� kN
(b) The missile’s weight W can be expressed in component and then
added to the force F.
W D ��8800 kg��9.81 m/s2�j D ��86.3 kN�j
FCW D �208iC [156� 86.3]j� kN D �208i� 69.7j� kN
jFCWj D
√
�208 kN�2 C ��69.7 kN�2 D 219 kN
jFCWj D 219 kN
Problem 2.26 For the truss shown, express the
position vector rAD from point A to point D in terms of
components. Use your result to determine the distance
from point A to point D.
A
C
y
x
B
0.4 m
0.6 m
0.7 m
D
0.6 m 1.2 m
Solution: Coordinates A(1.8, 0.7) m, D(0, 0.4) m
rAD D �0� 1.8 m�iC �0.4 m� 0.7 m�j D ��1.8i� 0.3j� m
rAD D
√
��1.8 m�2 C ��0.3 m�2 D 1.825 m
16
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Problem 2.27 The points A, B, . . . are the joints of
the hexagonal structural element. Let rAB be the position
vector from joint A to joint B, rAC the position vector
from joint A to joint C, and so forth. Determine the
components of the vectors rAC and rAF.
2 m
x
y
A B
E
F C
D
Solution: Use the xy coordinate system shown and find the
locations of C and F in those coordinates. The coordinates of the
points inthis system are the scalar components of the vectors rAC and
rAF. For rAC, we have
rAC D rAB C rBC D �xB � xA�iC �yB � yA�j
C �xC � xB�iC �yC � yB�j
or rAC D �2m� 0�iC �0� 0�jC �2m cos 60° � 0�i
C �2m cos 60° � 0�j,
giving
rAC D �2mC 2m cos 60°�iC �2m sin 60°�j. For rAF, we have
rAF D �xF � xA�iC �yF � yA�j
D ��2m cos 60°xF � 0�iC �2m sin 60° � 0�j.
Problem 2.28 For the hexagonal structural element in
Problem 2.27, determine the components of the vector
rAB � rBC.
Solution: rAB � rBC.
The angle between BC and the x-axis is 60°.
rBC D 2 cos�60°�iC 2�sin�60°�j �m�
rBC D 1iC 1.73j �m�
rAB � rBC D 2i� 1i� 1.73j �m�
rAB � rBC D 1i� 1.73j �m�
Problem 2.29 The coordinates of point A are (1.8,
3.0) ft. The y coordinate of point B is 0.6 ft. The vector
rAB has the same direction as the unit vector eAB D
0.616i� 0.788j. What are the components of rAB?
x
y
A
B
rABSolution: The vector rAB can be written two ways.
rAB D jrABj�0.616i� 0.788j� D �Bx � Ax�iC �By � Ay�j
Comparing the two expressions we have
�By � Ay� D �0.6� 3.0�ft D ��0.788�jrABj
jrABj D �2.4 ft�0.788 D 3.05 ft
Thus
rAB D jrABj�0.616i� 0.788j� D �3.05 ft��0.616i� 0.788j� D �1.88i� 2.40j� ft
rAB D �1.88i� 2.40j� ft
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17
Problem 2.30 (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.
45 in
98 in
50 in55 in
35 in
A
50 in
y
x
C
B
Solution: The coordinates are A(50, 35); B(98, 50); C(45, 55).
(a) The vector from point A to B:
rAB D �98� 50�iC �50� 35�j D 48iC 15j (in)
(b) The vector from point B to C is
rBC D �45� 98�iC �55� 50�j D �53iC 5j (in).
(c) The distance from A to C is the magnitude of the sum of the
vectors,
rAC D rAB C rBC D �48� 53�iC �15C 5�j D �5iC 20j.
The distance from A to C is
jrACj D
√
��5�2 C �20�2 D 20.62 in
18
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Problem 2.31 In Active Example 2.3, the cable AB
exerts a 900-N force on the top of the tower. Suppose
that the attachment point B is moved in the horizontal
direction farther from the tower, and assume that the
magnitude of the force F the cable exerts on the top
of the tower is proportional to the length of the cable.
(a) What is the distance from the tower to point B
if the magnitude of the force is 1000 N? (b) Express
the 1000-N force F in terms of components using the
coordinate system shown.
A
B
80 m
40 m
A
B
80 m
F
40 m
x
y
Force
 exerted on
 the tower
 by cable
 AB
Solution: In the new problem assume that point B is located a
distance d away from the base. The lengths in the original problem
and in the new problem are given by
Loriginal D
√
�40 m�2 C �80 m�2 D
√
8000 m2
Lnew D
√
d2 C �80 m�2
(a) The force is proportional to the length. Therefore
1000 N D �900 N�
√
d2 C �80 m�2√
8000 m2
d D
√
�8000 m2�
(
1000 N
900 N
)2
� �80 m�2 D 59.0 m
d D 59.0 m
(b) The force F is then
F D �1000 N�
(
d√
d2 C �80 m�2
i� 80 m√
d2 C �80 m�2
j
)
D �593i� 805j� N
F D �593i� 805j� N
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19
Problem 2.32 Determine the position vector rAB in
terms of its components if (a) � D 30°, (b) � D 225°.
60 mm 150 mm
x
y
A
C
B
rAB
rBC
θ
Solution:
(a) rAB D �60� cos�30°�iC �60� sin�30°�j, or
rAB D 51.96iC 30j mm. And
(b) rAB D �60� cos�225°�iC �60� sin�225°�j or
rAB D �42.4i� 42.4j mm.
FBCFAB
60
mm
150
mm
y
x
C F
A
B
θ
Problem 2.33 In Example 2.4, the coordinates of the
fixed point A are (17, 1) ft. The driver lowers the bed of
the truck into a new position in which the coordinates
of point B are (9, 3) ft. The magnitude of the force F
exerted on the bed by the hydraulic cylinder when the
bed is in the new position is 4800 lb. Draw a sketch of
the new situation. Express F in terms of components.
A
F
x
30�
y
B
30� A
B
Solution:
� D tan�1
(
2 ft
8 ft
)
D 14.04°
F D 4800 lb�� cos �iC sin �j�.
F D ��4660iC 1160j� lb
20
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 2.34 A surveyor measures the location of
point A and determines that rOA D 400iC 800j (m). He
wants to determine the location of a point B so that
jrABj D 400 m and jrOA C rABj D 1200 m. What are the
cartesian coordinates of point B?
x
y
A
B
O
rAB
rOA
Proposed
roadway
N
Solution: Two possibilities are: The point B lies west of point A,
or point B lies east of point A, as shown. The strategy is to determine
the unknown angles ˛, ˇ, and �. The magnitude of OA is
jrOAj D
√
�400�2 C �800�2 D 894.4.
The angle ˇ is determined by
tanˇ D 800
400
D 2, ˇ D 63.4°.
The angle ˛ is determined from the cosine law:
cos˛ D �894.4�
2 C �1200�2 � �400�2
2�894.4��1200�
D 0.9689.
˛ D 14.3°. The angle � is � D ˇ š ˛ D 49.12°, 77.74°.
The two possible sets of coordinates of point B are
{
rOB D 1200�i cos 77.7C j sin 77.7� D 254.67iC 1172.66j (m)
rOB D 1200�i cos 49.1C j sin 49.1� D 785.33iC 907.34j (m)
The two possibilities lead to B(254.7 m, 1172.7 m) or B(785.3 m,
907.3 m)
B
y
x
0
θ
α
α
β
BA
Problem 2.35 The magnitude of the position vector
rBA from point B to point A is 6 m and the magnitude of
the position vector rCA from point C to point A is 4 m.
What are the components of rBA?
A
B C
x
y
3 m
Solution: The coordinates are: A�xA, yA�, B�0, 0�, C�3 m, 0�
Thus
rBA D �xA � 0�iC �yA � 0�j) �6 m�2 D xA2 C yA2
rCA D �xA � 3 m�iC �yA � 0�j) �4 m�2 D �xA � 3 m�2 C yA2
Solving these two equations, we find xA D 4.833 m, yA D š3.555 m.
We choose the “-” sign and find
rBA D �4.83i� 3.56j� m
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21
Problem 2.36 In Problem 2.35, determine the compo-
nents of a unit vector eCA that points from point C toward
point A.
Strategy: Determine the components of rCA and then
divide the vector rCA by its magnitude.
Solution: From the previous problem we have
rCA D �1.83i� 3.56j� m, rCA D
√
1.832 C 3.562 m D 3.56 m
Thus
eCA D rCA
rCA
D �0.458i� 0.889j�
Problem 2.37 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
x
B (4, 13) m
C
(9, 1) m
A
(0, 1.2) m
Solution:
rAB D �xB � xA�iC �yB � yA�j
rCB D �xB � xC�iC �yC � yB�j
Points are: A (0, 1.2), B (4, 13) and C (9, 1)
Substituting, we get
rAB D 4iC 11.8j �m�, jrABj D 12.46 �m�
rCB D �5iC 12j �m�, jrCBj D 13 �m�
The unit vectors are given by
eAB D rABjrABj and eCB D
rCB
jrCBj
Substituting, we get
eAB D 0.321iC 0.947j
eCB D �0.385iC 0.923j
22
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 2.38 The length of the bar AB is 0.6 m.
Determine the components of a unit vector eAB that
points from point A toward point B. B
A
x
y
0.3 m
0.4 m
Solution: We need to find the coordinates of point B�x, y�
We have the two equations
�0.3 mC x�2 C y2 D �0.6 m�2
x2 C y2 D �0.4 m�2
Solving we find
x D 0.183 m, y D 0.356 m
Thus
eAB D rAB
rAB
D �0.183 m� [�0.3 m]�iC �0.356 m�j√
�0.183 mC 0.3 m�2 C �0.356 m�2
D �0.806iC 0.593j�
A O
B
y
x0.3 m
0.6
 m
0.
4 
m
Problem 2.39 Determine the components of a unit
vector that is parallel to the hydraulic actuator BC and
points from B toward C.
1 m
0.6 m Scoop
A B
D
C
0.15 m
0.6 m
1 m
x
y
Solution: Point B is at (0.75, 0) and point C is at (0, 0.6). The
vector
rBC D �xC � xB�iC �yC � yB�j
rBC D �0� 0.75�iC �0.6� 0�j �m�
rBC D �0.75iC 0.6j �m�
jrBCj D
√
�0.75�2 C �0.6�2 D 0.960 �m�
eBC D rBCjrBCj D
�0.75
0.96
iC 0.6
0.96
j
eBC D �0.781iC 0.625j
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23
Problem 2.40 The hydraulic actuator BC in Problem
2.39 exerts a 1.2-kN force F on the joint at C that is
parallel to the actuator and points from B toward C.
Determine the components of F.
Solution: From the solution to Problem 2.39,
eBC D �0.781iC 0.625j
The vector F is given by F D jFjeBC
F D �1.2���0.781iC 0.625j� �k РN�
F D �937iC 750j �N�
Problem 2.41 A surveyor finds that the length of the
line OA is 1500 m and the length of line OB is 2000 m.
(a) 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.
x
y
60�
B
A
O
Proposed bridge
River30�
N
Solution: We need to find the coordinates of points A and B
rOA D 1500 cos 60°iC 1500 sin 60°j
rOA D 750iC 1299j �m�
Point A is at (750, 1299) (m)
rOB D 2000 cos 30°iC 2000 sin 30°j �m�
rOB D 1732iC 1000j �m�
Point B is at (1732, 1000) (m)
(a) The vector from A to B is
rAB D �xB � xA�iC �yB � yA�j
rAB D 982i� 299j �m�
(b) The unit vector eAB is
eAB D rABjrABj D
982i� 299j
1026.6
eAB D 0.957i� 0.291j
24
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 2.42 The magnitudes of the forces exerted
by the cables are jT1j D 2800 lb, jT2 D 3200 lb, jT3j D
4000 lb, and jT4j D 5000 lb. What is the magnitude of
the total force exerted by the four cables?
x
y
29�
9�
40�51�
T4 T3
T2
T1
Solution: The x-component of the total force is
Tx D jT1j cos 9° C jT2j cos 29°jT3j cos 40° C jT4j cos 51°
Tx D �2800 lb� cos 9° C �3200 lb� cos 29° C �4000 lb� cos 40° C �5000 lb� cos 51°
Tx D 11,800 lb
The y-component of the total force is
Ty D jT1j sin 9° C jT2j sin 29° C jT3j sin 40° C jT4j sin 51°
Ty D �2800 lb� sin 9° C �3200 lb� sin 29° C �4000 lb� sin 40° C �5000 lb� sin 51°
Ty D 8450 lb
The magnitude of the total force is
jTj D
√
Tx2 C Ty2 D
√
�11,800 lb�2 C �8450 lb�2 D 14,500 lb jTj D 14,500 lb
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25
Problem 2.43 The tensions in the four cables are equal:
jT1j D jT2j D jT3j D jT4j D T. Determine the value of
T so that the four cables exert a total force of 12,500-lb
magnitude on the support.
x
y
29�
9�
40�51�
T4 T3
T2
T1
Solution: The x-component of the total force is
Tx D T cos 9° C T cos 29° C T cos 40° C T cos 51°
Tx D 3.26T
The y-component of the total force is
Ty D T sin 9° C T sin 29° C T sin 40° C T sin 51°
Ty D 2.06T
The magnitude of the total force is
jTj D
√
Tx2 C Ty2 D
√
�3.26T�2 C �2.06T�2 D 3.86T D 12,500 lb
Solving for T we find T D 3240 lb
26
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 2.44 The rope ABC exerts forces FBA and
FBC on the block at B. Their magnitudes are equal:
jFBAj D jFBC j. The magnitude of the total force exerted
on the block at B by the rope is jFBA C FBC j D 920 N.
Determine jFBAj by expressing the forces FBA and FBC
in terms of components.
20°
FBC
FBA
B
C
A
B
Solution:
FBC D F�cos 20°iC sin 20°j�
FBA D F��j�
FBC C FBA D F�cos 20°iC [sin 20° � 1]j�
Therefore
�920 N�2 D F2�cos2 20° C [sin 20° � 1]2�) F D 802 N
20°
FBC
FBA
Problem 2.45 The magnitude of the horizontal force
F1 is 5 kN and F1 C F2 C F3 D 0. What are the magni-
tudes of F2 and F3?
F1
F2
F3
30˚
45˚
x
y
Solution: Using components we have
∑
Fx : 5 kNC F2 cos 45° � F3 cos 30° D 0
∑
Fy : �F2 sin 45° C F3 sin 30° D 0
Solving simultaneously yields:
) F2 D 9.66 kN, F3 D 13.66 kN
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27
Problem 2.46 Four groups engage in a tug-of-war. The
magnitudes of the forces exerted by groups B, C, and D
are jFBj D 800 lb, jFCj D 1000 lb, jFDj D 900 lb. If the
vector sum of the four forces equals zero, what are the
magnitude of FA and the angle ˛?
FB
x
y
70°
30°
20°α
FC
FA
FD
Solution: The strategy is to use the angles and magnitudes to
determine the force vector components, to solve for the unknown force
FA and then take its magnitude. The force vectors are
FB D 800�i cos 110° C j sin 110°� D �273.6iC 751.75j
FC D 1000�i cos 30° C j sin 30°� D 866iC 500j
FD D 900�i cos��20°�C j sin��20°�� D 845.72i� 307.8j
FA D jFAj�i cos�180C ˛�C j sin�180C ˛��
D jFAj��i cos˛� j sin˛�
The sum vanishes:
FA C FB C FC C FD D i�1438.1� jFAj cos˛�
C j�944� jFAj sin˛� D 0
From which FA D 1438.1i C 944j. The magnitude is
jFAj D
√
�1438�2 C �944�2 D 1720 lb
The angle is: tan˛ D 944
1438
D 0.6565, or ˛ D 33.3°
28
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Problem 2.47 In Example 2.5, suppose that the attach-
ment point of cable A is moved so that the angle between
the cable and the wall increases from 40° to 55°. Draw
a sketch showing the forces exerted on the hook by the
two cables. If you want the total force FA C FB to have
a magnitude of 200 lb and be in the direction perpen-
dicular to the wall, what are the necessary magnitudes
of FA and FB?
A
20�
40�
B
40� FA
FB
20�
Solution: Let FA and FB be the magnitudes of FA and FB.
The component of the total force parallel to the wall must be zero. And
the sum of the components perpendicular to the wall must be 200 lb.
FA cos 55° � FB cos 20° D 0
FA sin 55° C FB sin 20° D 200 lb
Solving we find
FA D 195 lb
FB D 119 lb
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29
Problem 2.48 The bracket must support the two forces
shown, where jF1j D jF2j D 2 kN. An engineer deter-
mines that the bracket will safely support a total force
of magnitude 3.5 kN in any direction. Assume that 0 �
˛ � 90°. What is the safe range of the angle ˛?
F2
F1
α
Solution:∑
Fx : �2 kN�C �2 kN� cos˛ D �2 kN��1C cos˛�
∑
Fy : �2 kN� sin˛
Thus the total force has a magnitude given by
F D 2 kN
√
�1C cos ˛�2 C �sin˛�2 D 2 kNp2C 2 cos ˛ D 3.5 kN
Thus when we are at the limits we have
2C 2 cos ˛ D
(
3.5 kN
2 kN
)2
D 49
16
) cos˛ D 17
32
) ˛ D 57.9°
In order to be safe we must have
57.9° � ˛ � 90°
α
β αF1
F1 + F2
F2
30
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Problem 2.49 The figure shows three forces acting on
a joint of a structure. The magnitude of Fc is 60 kN, and
FA C FB C FC D 0. What are the magnitudes of FA and
FB?
y
x
FB
FC
FA
15°
40°
Solution: We need to write each force in terms of its components.
FA D jFAj cos 40iC jFAj sin 40j �kN�
FB D jFBj cos 195°iC jFBj sin 195j �kN�
FC D jFCj cos 270°iC jFCj sin 270°j �kN�
Thus FC D �60j kN
Since FA C FB C FC D 0, their components in each direction must also
sum to zero.
{
FAx C FBx C FCx D 0
FAy C FBy C FCy D 0
Thus,
{ jFAj cos 40° C jFBj cos 195° C 0 D 0
jFAj sin 40° C jFBj sin 195° � 60 �kN� D 0
Solving for jFAj and jFBj, we get
jFAj D 137 kN, jFBj D 109 kN
FB
FC
FA195°
270°
40°
x
Problem 2.50 Four forces act on a beam. The vector
sum of the forces is zero. The magnitudes jFBj D
10 kN and jFCj D 5 kN. Determine the magnitudes of
FA and FD.
FD
30°
FB FC
FA
Solution: Use the angles and magnitudes to determine the vectors,
and then solve for the unknowns. The vectors are:
FA D jFAj�i cos 30° C j sin 30°� D 0.866jFAjiC 0.5jFAjj
FB D 0i� 10j, FC D 0iC 5j, FD D �jFDjiC 0j.
Take the sum of each component in the x- and y-directions:
∑
Fx D �0.866jFAj � jFDj�i D 0
and
∑
Fy D �0.5jFAj � �10� 5��j D 0.
From the second equation we get jFAj D 10 kN . Using this value in
the first equation, we get jFDj D 8.7 kN
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31
Problem 2.51 Six forces act on a beam that forms part
of a building’s frame. The vector sum of the forces
is zero. The magnitudes jFBj D jFEj D 20 kN, jFCj D
16 kN, and jFDj D 9 kN. Determine the magnitudes of
FA and FG.
50�70� 40� 40�
FEFB
FGFC FDFA
Solution: Write each force in terms of its magnitude and direction
as
F D jFj cos �iC jFj sin �j
where � is measured counterclockwise from the Cx-axis.
Thus, (all forces in kN)
FA D jFAj cos 110°iC jFAj sin 110°j �kN�
FB D 20 cos 270°iC 20 sin 270°j �kN�
FC D 16 cos 140°iC 16 sin 140°j �kN�
FD D 9 cos 40°iC 9 sin 40°j �kN�
FE D 20 cos 270°iC 20 sin 270°j �kN�
FG D jFGj cos 50°iC jFGj sin 50°j �kN�
We know that the x components and y components of the forces must
add separately to zero.
Thus
{
FAx C FBx C FCx C FDx C FEx C FGx D 0
FAy C FBy C FCy C FDy C FEy C FGy D 0
{ jFAj cos 110° C 0� 12.26C 6.89C 0C jFGj cos 50° D 0
jFAj sin 110° � 20C 10.28C 5.79� 20C jFGj sin 50° D 0
Solving, we get
jFAj D 13.0 kN jFGj D 15.3 kN
x
y
θ
32
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Problem 2.52 The total weight of the man and parasail
is jWj D 230 lb. The drag force D is perpendicular to
the lift force L. If the vector sum of the three forces is
zero, what are the magnitudes of L and D?
2
5
x
y
L
D
W
Solution: Let L and D be the magnitudes of the lift and drag
forces. We can use similar triangles to express the vectors L and D
in terms of components. Then the sum of the forces is zero. Breaking
into components we have
2p
22 C 52 L �
5p
22 C 52D D 0
5p
22 C 52 L �
2p
22 C 52D� 230 lb D 0
Solving we find
jDj D 85.4 lb, jLj D 214 lb
Problem 2.53 The three forces acting on the car are
shown. The force T is parallel to the x axis and the
magnitude of the force W is 14 kN. If TCWC N D 0,
what are the magnitudes of the forces T and N?
20�
20�W
T
N
y
x
Solution:
∑
Fx : T�N sin 20° D 0
∑
Fy : N cos 20° � 14 kN D 0
Solving we find
N D 14.90 N, T D 5.10 N
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33
Problem 2.54 The cables A, B, and C help support a
pillar that forms part of the supports of a structure. The
magnitudes of the forces exerted by the cables are equal:
jFAj D jFBj D jFCj. The magnitude of the vector sum of
the three forces is 200 kN. What is jFAj?
FA
FB
FC
4 m
A B C
6 m
4 m
4 m
Solution: Use the angles and magnitudes to determine the vector
components, take the sum, and solve for the unknown. The angles
between each cable and the pillar are:
�A D tan�1
(
4 m
6 m
)
D 33.7°,
�B D tan�1
(
8
6
)
D 53.1°
�C D tan�1
(
12
6
)
D 63.4°.
Measure the angles counterclockwise form the x-axis. The force vec-
tors acting along the cables are:
FA D jFAj�i cos 303.7° C j sin 303.7°� D 0.5548jFAji� 0.8319jFAjj
FB D jFBj�i cos 323.1° C j sin 323.1°� D 0.7997jFBji� 0.6004jFBjj
FC D jFCj�i cos 333.4° C j sin 333.4°� D 0.8944jFCji�0.4472jFCjj
The sum of the forces are, noting that each is equal in magnitude, is
∑
F D �2.2489jFAji� 1.8795jFAjj�.
The magnitude of the sum is given by the problem:
200 D jFAj
√
�2.2489�2 C �1.8795�2 D 2.931jFAj,
from which jFAj D 68.24 kN
Problem 2.55 The total force exerted on the top of the
mast B by the sailboat’s forestay AB and backstay BC is
180i� 820j (N). What are the magnitudes of the forces
exerted at B by the cables AB and BC ?
y
x
B (4, 13) m
C
(9, 1) m
A
(0, 1.2) m
Solution: We first identify the forces:
FAB D TAB ��4.0 mi� 11.8 mj�√
��4.0 m�2 C �11.8 m�2
FBC D TBC �5.0 mi� 12.0 mj�√
�5.0 m�2 C ��12.0 m�2
Then if we add the force we find
∑
Fx : � 4p
155.24
TAB C 5p
169
TBC D 180 N
∑
Fy : � 11.8p
155.24
TAB � 12p
169
TBC D �820 N
Solving simultaneously yields:
) TAB D 226 N, TAC D 657 N
34
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Problem 2.56 The structure shown forms part of a
truss designed by an architectural engineer to support
the roof of an orchestra shell. The members AB, AC,
and AD exert forces FAB, FAC, and FAD on the joint A.
The magnitude jFABj D 4 kN. If the vector sum of the
three forces equals zero, what are the magnitudes of FAC
and FAD?
FAB
A
(– 4, 1) m
B
C
D
x
y
FAC
FAD
(–2, –3) m
(4, 2) m
Solution: Determine the unit vectors parallel to each force:
eAD D �2p
22 C 32 iC
�3p
22 C 32 j D �0.5547i� 0.8320j
eAC D �4p
42 C 12 iC
1p
42 C 12 j D �0.9701iC 0.2425j
eAB D 4p
42 C 22 iC
2p
42 C 22 j D 0.89443i C 0.4472j
The forces are FAD D jFADjeAD, FAC D jFACjeAC,
FAB D jFABjeAB D 3.578iC 1.789j. Since the vector sum of the forces
vanishes, the x- and y-components vanish separately:
∑
Fx D ��0.5547jFADj � 0.9701jFACj C 3.578�i D 0, and
∑
Fy D ��0.8320jFADj C 0.2425jFACj C 1.789�j D 0
These simultaneous equationsin two unknowns can be solved by any
standard procedure. An HP-28S hand held calculator was used here:
The results: jFACj D 2.108 kN , jFADj D 2.764 kN
A
B
C
D
Problem 2.57 The distance s D 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
x
s
A
B
C
(14, 45) in
(75, 12) in
Solution:
(a) The unit vector is the position vector from B to A divided by its
magnitude
rBA D �[14� 75]iC [45� 12]j�in D ��61iC 33j� in
jrBAj D
√
��61 in�2 C �33 in�2 D 69.35 in
eBA D 169.35 in ��61iC 33j� in D ��0.880iC 0.476j�
eBA D ��0.880iC 0.476j�
(b) To find the coordinates of point C we will write a vector from
the origin to point C.
rC D rA C rAC D rA C seBA D �75iC 12j� inC �45 in���0.880i
C 0.476j�
rC D �35.4i�C 33.4j� in
Thus the coordinates of C are C �35.4, 33.4� in
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35
Problem 2.58 In Problem 2.57, determine the x and y
coordinates of the collar C as functions of the distance s.
Solution: The coordinates of the point C are given by
xC D xB C s��0.880� and yC D yB C s�0.476�.
Thus, the coordinates of point C are xC D 75� 0.880s in and yC D
12C 0.476s in. Note from the solution of Problem 2.57 above, 0 �
s � 69.4 in.
Problem 2.59 The position vector r goes from point
A to a point on the straight line between B and C. Its
magnitude is jrj D 6 ft. Express r in terms of scalar
components.
x
y
A
(7, 9) ft
(12, 3) ft
(3, 5) ft
B
C
r
Solution: Determine the perpendicular vector to the line BC from
point A, and then use this perpendicular to determine the angular orien-
tation of the vector r. The vectors are
rAB D �7� 3�iC �9� 5�j D 4iC 4j, jrABj D 5.6568
rAC D �12� 3�iC �3� 5�j D 9i� 2j, jrACj D 9.2195
rBC D �12� 7�iC �3� 9�j D 5i� 6j, jrBCj D 7.8102
The unit vector parallel to BC is
eBC D rBCjrBCj D 0.6402i� 0.7682j D i cos 50.19
° � j sin 50.19°.
Add š90° to the angle to find the two possible perpendicular vectors:
eAP1 D i cos 140.19° � j sin 140.19°, or
eAP2 D i cos 39.8° C j sin 39.8°.
Choose the latter, since it points from A to the line.
Given the triangle defined by vertices A, B, C, then the magnitude of
the perpendicular corresponds to the altitude when the base is the line
BC. The altitude is given by h D 2�area�
base
. From geometry, the area of
a triangle with known sides is given by
area D ps�s� jrBCj��s� jrACj��s� jrABj�,
where s is the semiperimeter, s D 12 �jrACj C jrABj C jrBCj�. Substi-
tuting values, s D 11.343, and area D 22.0 and the magnitude of the
perpendicular is jrAPj D 2�22�7.8102 D 5.6333. The angle between the
vector r and the perpendicular rAP is ˇ D cos�1 5.63336 D 20.1°. Thus
the angle between the vector r and the x-axis is ˛ D 39.8š 20.1 D
59.1° or 19.7°. The first angle is ruled out because it causes the vector
r to lie above the vector rAB, which is at a 45° angle relative to the
x-axis. Thus:
r D 6�i cos 19.7° C j sin 19.7°� D 5.65iC 2.02j
y
x
B[7,9]
A[3,5]
C[12,3]
P
r
36
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 2.60 Let r be the position vector from point
C to the point that is a distance s meters along the
straight line between A and B. Express r in terms of
components. (Your answer will be in terms of s).
x
y
A
(9, 3) m
(10, 9) m
(3, 4) m
B
C
r
s
Solution: First define the unit vector that points from A to B.
rB/A D �[10� 3]iC [9� 4]j� m D �7iC 5j� m
jrB/Aj D
√
�7 m�2 C �5 m�2 D
p
74 m
eB/A D 1p
74
�7iC 5j�
Let P be the point that is a distance s along the line from A to B. The
coordinates of point P are
xp D 3 mC s
(
7p
74
)
D �3C 0.814s� m
yp D 4 mC s
(
5p
74
)
D �4C 0.581s� m.
The vector r that points from C to P is then
r D �[3C 0.814s � 9]iC [4C 0.581s � 3]j� m
r D �[0.814s � 6]iC [0.581s C 1]j� m
Problem 2.61 A vector U D 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
jUj D
√
32 C ��4�2 C ��12�2 D 13
Problem 2.62 The vector e D 13 iC 23 jC ezk is a unit
vector. Determine the component ez. (Notice that there
are two answers.)
Solution:
e D 1
3
iC 2
3
jC ezk)
(
1
3
)2
C
(
2
3
)2
C ez2 D 1) e2 D 49
Thus
ez D 23 or ez D �
2
3
Problem 2.63 An engineer determines that an attach-
ment point will be subjected to a force F D 20iC Fyj�
45k �kN�. If the attachment point will safely support a
force of 80-kN magnitude in any direction, what is the
acceptable range of values for Fy?
y
z
x
F
Solution:
802 ½ F2x C F2y C F2z
802 ½ 202 C F2y C �45�2
To find limits, use equality.
F2yLIMIT D 802 � 202 � �45�2
F2yLIMIT D 3975
FyLIMIT D C63.0,�63.0 �kN�
jFyLIMIT j � 63.0 kN� 63.0 kN � Fy � 63.0 kN
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
37
Problem 2.64 A vector U D UxiCUyjCUzk. Its
magnitude is jUj D 30. Its components are related by
the equationsUy D �2Ux and Uz D 4Uy . Determine the
components. (Notice that there are two answers.)
Solution: Substitute the relations between the components, deter-
mine the magnitude, and solve for the unknowns. Thus
U D UxiC ��2Ux�jC �4��2Ux��k D Ux�1i� 2j� 8k�
where Ux can be factored out since it is a scalar. Take the magnitude,
noting that the absolute value of jUxj must be taken:
30 D jUxj
p
12 C 22 C 82 D jUxj�8.31�.
Solving, we get jUx j D 3.612, or Ux D š3.61. The two possible
vectors are
U D C3.61iC ��2�3.61��jC �4��2��3.61��k
D 3.61i� 7.22j� 28.9k
U D �3.61iC ��2��3.61��j
C 4��2���3.61�k D �3.61iC 7.22jC 28.9k
Problem 2.65 An object is acted upon by two
forces F1 D 20iC 30j� 24k (kN) and F2 D �60iC
20jC 40k (kN). What is the magnitude of the total force
acting on the object?
Solution:
F1 D �20iC 30j� 24k� kN
F2 D ��60iC 20jC 40k� kN
F D F1 C F2 D ��40iC 50jC 16k� kN
Thus
F D
√
��40 kN�2 C �50 kN�2 C �16 kN�2 D 66 kN
Problem 2.66 Two vectors U D 3i� 2jC 6k and
V D 4iC 12j� 3k.
(a) Determine the magnitudes of U and V.
(b) Determine the magnitude of the vector 3UC 2V.
Solution: The magnitudes:
(a) jUj D p32 C 22 C 62 D 7 and jVj D p42 C 122 C 32 D 13
The resultant vector
3UC 2V D �9C 8�iC ��6C 24�jC �18� 6�k
D 17iC 18jC 12k
(b) The magnitude j3UC 2Vj D p172 C 182 C 122 D 27.51
38
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Problem 2.67 In Active Example 2.6, suppose that
you want to redesign the truss, changing the position
of point D so that the magnitude of the vector rCD from
point C to point D is 3 m. To accomplish this, let the
coordinates of point D be �2, yD, 1� m, and determine
the value of yD so that jrCDj D 3 m. Draw a sketch of
the truss with point D in its new position. What are the
new directions cosines of rCD?
rCD
(2, 3, 1) m
(4, 0, 0) m
(a)
C
x
y
D
z
Solution: The vector rCD and the magnitude jrCDj are
rCD D �[2 m� 4 m]iC [yD � 0]jC [1 m� 0]k� D ��2 m�iC �yD�j
C �1 m�k
jrCDj D
√
��2 m�2 C �yCD�2 C �1 m�2 D 3 m
Solving we find yCD D
√
�3 m�2 � ��2 m�2 � �1 m�2 D 2 m
yCD D 2 m
The new direction cosines of rCD.
cos �x D �2/3 D�0.667
cos �y D 2/3 D 0.667
cos �z D 1/3 D 0.333
Problem 2.68 A force vector is given in terms of its
components by F D 10i� 20j� 20k (N).
(a) What are the direction cosines of F?
(b) Determine the components of a unit vector e that
has the same direction as F.
Solution:
F D �10i� 20j� 20k� N
F D
√
�10 N�2 C ��20 N�2 C ��20 N�2 D 30 N
(a)
cos �x D 10 N30 N D 0.333, cos �y D
�20 N
30 N
D �0.667,
cos �z D �20 N
30 N
D �0.667
(b) e D �0.333i� 0.667j� 0.667k�
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39
Problem 2.69 The cable exerts a force F on the hook
at O whose magnitude is 200 N. The angle between the
vector F and the x axis is 40°, and the angle between
the vector F and the y axis is 70°.
(a) What is the angle between the vector F and the
z axis?
(b) Express F in terms of components.
Strategy: (a) Because you know the angles between
the vector F and the x and y axes, you can use Eq. (2.16)
to determine the angle between F and the z axis.
(Observe from the figure that the angle between F and
the z axis is clearly within the range 0 < �z < 180°.) (b)
The components of F can be obtained with Eqs. (2.15).
y
x
z
O
70°
40°
F
Solution:
(a) �cos 40°�2 C �cos 70°�2 C �cos �z�2 D 1) �z D 57.0°
(b)
F D 200 N�cos 40°iC cos 70°jC cos 57.0°k�
F D �153.2iC 68.4jC 108.8k� N
Problem 2.70 A unit vector has direction cosines
cos �x D �0.5 and cos �y D 0.2. Its z component is posi-
tive. Express it in terms of components.
Solution: Use Eq. (2.15) and (2.16). The third direction cosine is
cos �z D š
√
1� �0.5�2 � �0.2�2 D C0.8426.
The unit vector is
u D �0.5iC 0.2jC 0.8426k
Problem 2.71 The airplane’s engines exert a total thrust
force T of 200-kN magnitude. The angle between T and
the x axis is 120°, and the angle between T and the y axis
is 130°. The z component of T is positive.
(a) What is the angle between T and the z axis?
(b) Express T in terms of components.
130�
120�
y
x
z
T
y
x
z
Solution: The x- and y-direction cosines are
l D cos 120° D �0.5, m D cos 130° D �0.6428
from which the z-direction cosine is
n D cos�z D š
√
1� �0.5�2 � �0.6428�2 D C0.5804.
Thus the angle between T and the z-axis is
(a) �z D cos�1�0.5804� D 54.5° , and the thrust is
T D 200��0.5i� 0.6428jC 0.5804k�, or:
(b) T D �100i� 128.6jC 116.1k (kN)
40
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Problem 2.72 Determine the components of the posi-
tion vector rBD from point B to point D. Use your result
to determine the distance from B to D.
B (5, 0, 3) m
C (6, 0, 0) m
D (4, 3, 1) m
z
y
x
A
Solution: We have the following coordinates: A�0, 0, 0�,
B�5, 0, 3� m, C�6, 0, 0� m, D�4, 3, 1� m
rBD D �4 m� 5 m�iC �3 m� 0�jC �1 m� 3 m�k
D ��iC 3j� 2k� m
rBD D
√
��1 m�2 C �3 m�2 C ��2 m�2 D 3.74 m
Problem 2.73 What are the direction cosines of the
position vector rBD from point B to point D?
Solution:
cos �x D �1 m3.74 m D �0.267, cos �y D
3 m
3.74 m
D 0.802,
cos �z D �2 m3.74 m D 0.535
Problem 2.74 Determine the components of the unit
vector eCD that points from point C toward point D.
Solution: We have the following coordinates: A�0, 0, 0�,
B�5, 0, 3� m, C�6, 0, 0� m, D�4, 3, 1� m
rCD D �4 m� 6 m�iC �3 m� 0�jC �1 m� 0�k D ��2iC 3jC 1k�
rCD D
√
��2 m�2 C �3 m�2 C �1 m�2 D 3.74 m
Thus
eCD D 1
3.74 m
��2iC 3jC k� m D ��0.535iC 0.802jC 0.267k�
Problem 2.75 What are the direction cosines of the
unit vector eCD that points from point C toward point D?
Solution: Using Problem 2.74
cos �x D �0.535, cos �y D 0.802, cos �z D 0.267
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41
Problem 2.76 In Example 2.7, suppose that the
caisson shifts on the ground to a new position. The
magnitude of the force F remains 600 lb. In the new
position, the angle between the force F and the x axis
is 60° and the angle between F and the z axis is 70°.
Express F in terms of components.
40� F
y
x54�
zSolution: We need to find the angle �y between the force F and
the y axis. We know that
cos2 �x C cos2 �y C cos2 �z D 1
cos �y D š
√
1� cos2 �x � cos2 �z D š
√
1� cos2 60° � cos2 70° D š0.7956
�y D š cos�1�0.7956� D 37.3° or 142.7°
We will choose �y D 37.3° because the picture shows the force pointing
up. Now
Fx D �600 lb� cos 60° D 300 lb
Fy D �600 lb� cos 37.3° D 477 lb
Fz D �600 lb� cos 70° D 205 lb
Thus F D �300iC 477jC 205k� lb
Problem 2.77 Astronauts on the space shuttle use radar
to determine the magnitudes and direction cosines of the
position vectors of two satellites A and B. The vector rA
from the shuttle to satellite A has magnitude 2 km, and
direction cosines cos �x D 0.768, cos �y D 0.384, cos �z D
0.512. The vector rB from the shuttle to satellite B has
magnitude 4 km and direction cosines cos �x D 0.743,
cos �y D 0.557, cos �z D �0.371. What is the distance
between the satellites?
x
rB
z
B
A rA
y
Solution: The two position vectors are:
rA D 2�0.768iC 0.384jC 0.512k�D1.536iC 0.768jC 1.024k (km)
rB D 4�0.743iC 0.557j� 0.371k�D2.972iC 2.228j� 1.484k (km)
The distance is the magnitude of the difference:
jrA � rBj
D
√
�1.536�2.927�2 C �0.768�2.228�2 C �1.024���1.484��2
D 3.24 (km)
42
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Problem 2.78 Archaeologists measure a pre-Colum-
bian 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.
4 m
y
10 m
z
b
x
A
C
10 m
B
4 m
8 m
8 m
Solution:
(a) The coordinates are A (0, 16, 14) m and B (10, 8, 4) m.
rAB D �[10� 0]iC [8� 16]jC [4� 14]k�m D �10i� 8j� 10k�m
jrABj D
√
102 C 82 C 102 m D
p
264 m D 16.2 m
jrABj D 16.2 m
(b)
cos �x D 10p
264
D 0.615
cos �y D �8p
264
D �0.492
cos �z D �10p
264
D �0.615
Problem 2.79 Consider the structure described in
Problem 2.78. After returning to the United States,
an archaeologist discovers that a graduate student has
erased the only data file containing the dimension b.
But from recorded GPS data he is able to calculate that
the distance from point B to point C is 16.61 m.
(a) What is the distance b?
(b) Determine the direction cosines of the position
vector from B to C.
4 m
y
10 m
z
b
x
A
C
10 m
B
4 m
8 m
8 m
Solution: We have the coordinates B (10 m, 8 m, 4 m), C (10 mC
b, 0 18 m).
rBC D �10 mC b� 10 m�iC �0� 8 m�jC �18 m� 4 m�k
rBC D �b�iC ��8 m�jC �14 m�k
(a) We have �16.61 m�2 D b2 C ��8 m�2 C �14 m�2 ) b D 3.99 m
(b) The direction cosines of rBC are
cos �x D 3.99 m16.61 m D 0.240
cos �y D �8 m16.61 m D �0.482
cos �z D 14 m
16.61 m
D 0.843
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43
Problem 2.80 Observers at A and B use theodolites to
measure the direction from their positions to a rocket
in flight. If the coordinates of the rocket’s position at a
given instant are (4, 4, 2) km, determine the direction
cosines of the vectors rAR and rBR that the observerswould measure at that instant.
B (5, 0, 2) km
A
rAR
rBR
x
y
z
Solution: The vector rAR is given by
rAR D 4iC 4jC 2k km
and the magnitude of rAR is given by
jrARj D
√
�4�2 C �4�2 C �2�2 km D 6 km.
The unit vector along AR is given by
uAR D rAR/jrARj.
Thus, uAR D 0.667iC 0.667jC 0.333k
and the direction cosines are
cos �x D 0.667, cos �y D 0.667, and cos �z D 0.333.
The vector rBR is given by
rBR D �xR � xB�iC �yR � yB�jC �zR � zB�k km
D �4� 5�iC �4� 0�jC �2� 2�k km
and the magnitude of rBR is given by
jrBRj D
√
�1�2 C �4�2 C �0�2 km D 4.12 km.
The unit vector along BR is given by
eBR D rBR/jrBRj.
Thus, uBR D �0.242iC 0.970jC 0k
and the direction cosines are
cos �x D �0.242, cos �y D 0.970, and cos �z D 0.0.
44
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Problem 2.81 In Problem 2.80, suppose that the coor-
dinates of the rocket’s position are unknown. At a given
instant, the person at A determines that the direction
cosines of rAR are cos �x D 0.535, cos �y D 0.802, and
cos �z D 0.267, and the person at B determines that the
direction cosines of rBR are cos �x D �0.576, cos �y D
0.798, and cos �z D �0.177. What are the coordinates of
the rocket’s position at that instant.
Solution: The vector from A to B is given by
rAB D �xB � xA�iC �yB � yA�jC �zB � zA�k or
rAB D �5� 0�iC �0� 0�jC �2� 0�k D 5iC 2k km.
The magnitude of rAB is given by jrABj D
√
�5�2 C �2�2 D 5.39 km.
The unit vector along AB, uAB, is given by
uAB D rAB/jrABj D 0.928iC 0jC 0.371k km.
The unit vector along the line AR,
uAR D cos �x iC cos �y jC cos �zk D 0.535iC 0.802jC 0.267k.
Similarly, the vector along BR, uBR D �0.576iC 0.798� 0.177k.
From the diagram in the problem statement, we see that rAR D
rAB C rBR. Using the unit vectors, the vectors rAR and rBR can be
written as
rAR D 0.535rARiC 0.802rARjC 0.267rARk, and
rBR D �0.576rBRiC 0.798rBRj� 0.177rBRk.
Substituting into the vector addition rAR D rAB C rBR and equating
components, we get, in the x direction, 0.535rAR D �0.576rBR , and
in the y direction, 0.802rAR D 0.798rBR . Solving, we get that rAR D
4.489 km. Calculating the components, we get
rAR D rAReAR D 0.535�4.489�iC 0.802�4.489�jC 0.267�4.489�k.
Hence, the coordinates of the rocket, R, are (2.40, 3.60, 1.20) km.
Problem 2.82* The height of Mount Everest was orig-
inally measured by a surveyor in the following way.
He first measured the altitudes of two points and the
horizontal distance between them. For example, suppose
that the points A and B are 3000 m above sea level
and are 10,000 m apart. He then used a theodolite to
measure the direction cosines of the vector rAP from
point A to the top of the mountain P and the vector rBP
from point B to P. Suppose that the direction cosines of
rAP are cos �x D 0.5179, cos �y D 0.6906, and cos �z D
0.5048, and the direction cosines of rBP are cos �x D
�0.3743, cos �y D 0.7486, and cos �z D 0.5472. Using
this data, determine the height of Mount Everest above
sea level.
P
y
A
z
B x
Solution: We have the following coordinates A�0, 0, 3000� m,
B�10, 000, 0, 3000� m, P�x, y, z�
Then
rAP D xiC yjC �z � 3000 m�k D rAP �0.5179iC 0.6906jC 0.5048k�
rBP D �x � 10,000 m�iC yjC �z � 3000 m�k
D rBP ��0.3743iC 0.7486jC 0.5472k�
Equating components gives us five equations (one redundant) which
we can solve for the five unknowns.
x D rAP 0.5179
y D rAP 0.6906
z � 3000 m D rAP 0.5048 ) z D 8848 m
x � 10000 m D �rBP � 0.7486
y D rBP 0.5472
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45
Problem 2.83 The distance from point O to point A is
20 ft. The straight line AB is parallel to the y axis, and
point B is in the x-z plane. Express the vector rOA in
terms of scalar components.
Strategy: You can resolve rOA into a vector from O to
B and a vector from B to A. You can then resolve the
vector form O to B into vector components parallel to
the x and z axes. See Example 2.8.
y
x
z
30°
B
O
A
rOA
60°
Solution: See Example 2.8. The length BA is, from the right triangle
OAB,
jrABj D jrOAj sin 30° D 20�0.5� D 10 ft.
Similarly, the length OB is
jrOBj D jrOAj cos 30° D 20�0.866� D 17.32 ft
The vector rOB can be resolved into components along the axes by the
right triangles OBP and OBQ and the condition that it lies in the x-z
plane.
Hence,
rOB D jrOBj�i cos 30° C j cos 90° C k cos 60°� or
rOB D 15iC 0jC 8.66k.
The vector rBA can be resolved into components from the condition
that it is parallel to the y-axis. This vector is
rBA D jrBAj�i cos 90° C j cos 0° C k cos 90°� D 0iC 10jC 0k.
The vector rOA is given by rOA D rOB C rBA, from which
rOA D 15iC 10jC 8.66k (ft)
P
rOA
B
Q
O
A
z
y
x
60°
30°
Problem 2.84 The magnitudes of the two force vectors
are jFAj D 140 lb and jFBj D 100 lb. Determine the mag-
nitude of the sum of the forces FA C FB.
x
y
z
40�
50�30�
60�
FB
FA
Solution: We have the vectors
FA D 140 lb�[cos 40° sin 50°]iC [sin 40°]jC [cos 40° cos 50°]k�
FA D �82.2iC 90.0jC 68.9k� lb
FB D 100 lb�[� cos 60° sin 30°]iC [sin 60°]jC [cos 60° cos 30°]k�
FB D ��25.0iC 86.6jC 43.3k� lb
Adding and taking the magnitude we have
FA C FB D �57.2iC 176.6jC 112.2k� lb
jFA C FBj D
√
�57.2 lb�2 C �176.6 lb�2 C �112.2 lb�2 D 217 lb
jFA C FBj D 217 lb
46
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Problem 2.85 Determine the direction cosines of the
vectors FA and FB.
x
y
z
40�
50�30�
60�
FB
FA
Solution: We have the vectors
FA D 140 lb�[cos 40° sin 50°]iC [sin 40°]jC [cos 40° cos 50°]k�
FA D �82.2iC 90.0jC 68.9k� lb
FB D 100 lb�[� cos 60° sin 30°]iC [sin 60°]jC [cos 60° cos 30°]k�
FB D ��25.0iC 86.6jC 43.3k� lb
The direction cosines for FA are
cos �x D 82.2 lb
140 lb
D 0.587, cos �y D 90.0 lb
140 lb
D 0.643,
cos �z D 68.9 lb
140 lb
D 0.492
The direction cosines for FB are
cos �x D �25.0 lb100 lb D �0.250, cos �y D
86.6 lb
100 lb
D 0.866,
cos �z D 43.3 lb100 lb D 0.433
FA : cos �x D 0.587, cos �y D 0.643, cos �z D 0.492
FB : cos �x D �0.250, cos �y D 0.866, cos �z D 0.433
Problem 2.86 In Example 2.8, suppose that a change
in the wind causes a change in the position of the balloon
and increases the magnitude of the force F exerted on
the hook at O to 900 N. In the new position, the angle
between the vector component Fh and F is 35°, and
the angle between the vector components Fh and Fz is
40°. Draw a sketch showing the relationship of these
angles to the components of F. Express F in terms of its
components. O
y
x
B
F
z
O
A
Solution: We have
jFy j D �900 N� sin 35° D 516 N
jFhj D �900 N� cos 35° D 737 N
jFx j D jFhj sin 40° D 474 N
jFzj D jFhj cos 40° D 565 N
Thus
F D �474iC 516jC 565k� N
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
47
Problem 2.87 An engineer calculates that the magni-
tude of the axial force in one of the beams of a geodesic
dome is jPj D 7.65 kN. The cartesian coordinates of
the endpoints A and B of the straight beam are (�12.4,
22.0, �18.4) m and (�9.2, 24.4, �15.6) m, respectively.
Express the force P in terms of scalar components.
B
P
A
Solution: The components of the position vector from B to A are
rBA D �xA� xB�iC �yA � yB�jC �zA � zB�k
D ��12.4C 9.2�iC �22.0� 24.4�j
C ��18.4C 15.6�k
D �3.2i� 2.4j� 2.8k �m�.
Dividing this vector by its magnitude, we obtain a unit vector that
points from B toward A:
eBA D �0.655i� 0.492j� 0.573k.
Therefore
P D jPjeBA
D 7.65 eBA
D �5.01i� 3.76j� 4.39k �kN�.
Problem 2.88 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
A
C (3, 0, 4) m
z
x
F
Solution:
(a) eBC D rBCjrBCj D
�xC � xB�iC �yC � yB�jC �zC � zB�k√
�xC � xB�2 C �yC � yB�2 C �zC � zB�2
eBC D �2i� 6jC 3kp
22 C 62 C 32 D �
2
7
i� 6
7
jC 3
7
k
eBC D �0.286i� 0.857jC 0.429k
(b) F D jFjeBC D 8eBC D �2.29i� 6.86jC 3.43k �kN�
48
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Problem 2.89 A cable extends from point C to
point E. It exerts a 50-lb force T on plate C that is
directed along the line from C to E. Express T in terms
of components.
D x
CB
A
20�
4 ft
4 ft
6 ft
2 ft
E
z
y
T
Solution: Find the unit vector eCE and multiply it times the magni-
tude of the force to get the vector in component form,
eCE D rCEjrCEj D
�xE � xC�iC �yE � yC�jC �zE � zC�k√
�xE � xC�2 C �yE � yC�2 C �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 D �0� 4�iC �2� ��1.37��jC �6� 3.76�kp
42 C 3.372 C 2.242
eCE D �0.703iC 0.592jC 0.394k
T D 50eCE �lb�
T D �35.2iC 29.6jC 19.7k �lb�
D x
C
B
A
20°
4 ft
4 ft
6 ft
2 ft
E
z
y
T
T
Problem 2.90 In Example 2.9, suppose that the metal
loop at A is moved upward so that the vertical distance to
A increases from 7 ft to 8 ft. As a result, the magnitudes
of the forces FAB and FAC increase to jFABj D jFACj D
240 lb. What is the magnitude of the total force F D
FAB C FAC exerted on the loop by the rope?
7 ft
B
C
A
x
z
y
FAB FAC
6 ft
10 ft2 ft
4 ft 6 ft
6 ft
7 ft
10 ft2 ft
4 ft 6 ft
A
C
B
Solution: The new coordinates of point A are (6, 8, 0) ft. The
position vectors are
rAB D ��4i� 8jC 4k� ft
rAC D �4i� 8jC 6k� ft
The forces are
FAB D �240 lb� rABjrABj D ��98.0i� 196jC 98.0k� lb
FAC D �240 lb� rACjrACj D �89.1i� 178jC 134.0k� lb
The sum of the forces is
F D FAB C FAC D ��8.85i� 374jC 232k� lb
The magnitude is jFj D 440 lb
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49
Problem 2.91 The cable AB exerts a 200-lb force FAB
at point A that is directed along the line from A to B.
Express FAB in terms of components.
A (6, 0, 10) ft
B
C
8 ft
x
z
y
6 ft
FAB
FAC
8 ft
Solution: The coordinates of B are B(0,6,8). The position vector
from A to B is
rAB D �0� 6�iC �6� 0�jC �8� 10�k D �6iC 6j� 2k
The magnitude is jrABj D
p
62 C 62 C 22 D 8.718 ft.
The unit vector is
uAB D �68.718 iC
6
8.718
j� 2
8.718
k
or
uAB D �0.6882iC 0.6882j� 0.2294k.
FAB D jFABjuAB D 200��0.6882iC 0.6882j� 0.2294k�
The components of the force are
FAB D jFABjuAB D 200��0.6882iC 0.6882j� 0.2294k� or
FAB D �137.6iC 137.6j� 45.9k
Problem 2.92 Consider the cables and wall described
in Problem 2.91. Cable AB exerts a 200-lb force FAB
at point A that is directed along the line from A to B.
The cable AC exerts a 100-lb force FAC at point A that
is directed along the line from A to C. Determine the
magnitude of the total force exerted at point A by the
two cables.
Solution: Refer to the figure in Problem 2.91. From Problem 2.91
the force FAB is
FAB D �137.6iC 137.6j� 45.9k
The coordinates of C are C(8,6,0). The position vector from A to C is
rAC D �8� 6�iC �6� 0�jC �0� 10�k D 2iC 6j� 10k.
The magnitude is jrACj D
p
22 C 62 C 102 D 11.83 ft.
The unit vector is
uAC D 211.83 iC
6
11.83
j� 10
11.83
k D 0.1691iC 0.5072j� 0.8453k.
The force is
FAC D jFACjuAC D 100uAC D 16.9iC 50.7j� 84.5k.
The resultant of the two forces is
FR D FAB C FAC D ��137.6C 16.9�iC �137.6C 50.7�j
C ��84.5� 45.9�k.
FR D �120.7iC 188.3j� 130.4k.
The magnitude is
jFRj D
p
120.72 C 188.32 C 130.42 D 259.0 lb
50
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Problem 2.93 The 70-m-tall tower is supported by
three cables that exert forces FAB, FAC, and FAD on it.
The magnitude of each force is 2 kN. Express the total
force exerted on the tower by the three cables in terms
of components.
A
x
y
40 m
60 m
40 m
40 m
60 m
B
C
D
z
FABFAC
FAD
A
Solution: The coordinates of the points are A (0, 70, 0),
B (40, 0, 0), C (�40, 0, 40) D (�60, 0, �60).
The position vectors corresponding to the cables are:
rAD D ��60� 0�iC �0� 70�jC ��60� 0�k
rAD D �60i� 70k� 60k
rAC D ��40� 0�iC �0� 70�jC �40� 0�k
rAC D �40i� 70jC 40k
rAB D �40� 0�iC �0� 70�jC �0� 0�k
rAB D 40i� 70jC 0k
The unit vectors corresponding to these position vectors are:
uAD D rADjrADj D
�60
110
i� 70
110
j� 60
110
k
D �0.5455i� 0.6364j� 0.5455k
uAC D rACjrACj D �
40
90
i� 70
90
jC 40
90
k
D �0.4444i� 0.7778jC 0.4444k
uAB D rABjrABj D
40
80.6
i� 70
80.6
jC 0k D 0.4963i� 0.8685j C 0k
The forces are:
FAB D jFABjuAB D 0.9926i� 1.737jC 0k
FAC D jFACjuAC D �0.8888i� 1.5556jC 0.8888
FAD D jFADjuAD D �1.0910i� 1.2728j � 1.0910k
The resultant force exerted on the tower by the cables is:
FR D FAB C FAC C FAD D �0.9875i� 4.5648j� 0.2020k kN
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51
Problem 2.94 Consider the tower described in Pro-
blem 2.93. The magnitude of the force FAB is 2 kN. The
x and z components of the vector sum of the forces
exerted on the tower by the three cables are zero. What
are the magnitudes of FAC and FAD?
Solution: From the solution of Problem 2.93, the unit vectors are:
uAC D rACjrACj D �
40
90
i� 70
90
jC 40
90
k
D �0.4444i� 0.7778jC 0.4444k
uAD D rADjrADj D
�60
110
i� 70
110
j� 60
110
D �0.5455i� 0.6364j� 0.5455k
From the solution of Problem 2.93 the force FAB is
FAB D jFABjuAB D 0.9926i� 1.737jC 0k
The forces FAC and FAD are:
FAC D jFACjuAC D jFACj��0.4444i� 0.7778j C 0.4444k�
FAD D jFADjuAD D jFADj��0.5455i� 0.6364j� 0.5455k�
Taking the sum of the forces:
FR D FAB C FAC C FAD D �0.9926� 0.4444jFACj � 0.5455jFADj�i
C ��1.737� 0.7778jFACj � 0.6364jFADj�j
C �0.4444jFACj � 0.5455jFADj�k
The sum of the x- and z-components vanishes, hence the set of simul-
taneous equations:
0.4444jFACj C 0.5455jFADj D 0.9926 and
0.4444jFACj � 0.5455jFADj D 0
These can be solved by means of standard algorithms, or by the use of
commercial packages such as TK Solver Plus  or Mathcad. Here
a hand held calculator was used to obtain the solution:
jFACj D 1.1163 kN jFADj D 0.9096 kN
Problem 2.95 In Example 2.10, suppose that the
distance from point C to the collar A is increased from
0.2 m to 0.3 m, and the magnitude of the force T
increases to 60 N. Express T in terms of its components.
0.5 m
0.2 m
z
D
O
0.25 m
0.3 m0.2 m
x
0.4 m
0.15 m
y
A
T
CB
Solution: The position vector from C to A is now
rCA D �0.3 m�eCD D ��0.137i� 0.205jC 0.171k�m
The position vector form the origin to A is
rOA D rOC C rCA D �0.4iC 0.3j� mC ��0.137i� 0.205jC 0.171k� m
rOA D �0.263iC 0.0949jC 0.171k�m
The coordinates of A are (0.263, 0.0949, 0.171) m.
The position vector from A to B is
rAB D �[0� 0.263]iC [0.5� 0.0949]j C [0.15� 0.171]k� m
rAB D ��0.263iC 0.405j� 0.209k� m
The force T is
T D �60 N� rABjrABj D ��32.7iC 50.3j� 2.60k� N
T D ��32.7iC 50.3j� 2.60k� N
52
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Problem 2.96 The cable AB exerts a 32-lb force T on
the collar at A. Express T in terms of components.
x
y
A
6 ft
B
4 ft
4 ft
7 ft
4 ft
z
T
Solution: The coordinates of point B are B (0, 7, 4). The vector
position of B is rOB D 0iC 7jC 4k.
The vector from point A to point B is given by
rAB D rOB � rOA.
From Problem 2.95, rOA D 2.67iC 2.33jC 2.67k. Thus
rAB D �0� 2.67�iC �7� 2.33�jC �4� 2.67�j
rAB D �2.67iC 4.67jC 1.33k.
The magnitude is
jrABj D
p
2.672 C 4.672 C 1.332 D 5.54 ft.
The unit vector pointing from A to B is
uAB D rABjrABj D �0.4819iC 0.8429jC 0.2401k
The force T is given by
TAB D jTABjuAB D 32uAB D �15.4iC 27.0jC 7.7k (lb)
Problem 2.97 The circular bar has a 4-m radius and
lies in the x-y plane. Express the position vector from
point B to the collar at A in terms of components.
z
y
x
A
B
4 m
4 m
3 m
20°
Solution: From the figure, the point B is at (0, 4, 3) m. The coor-
dinates of point A are determined by the radius of the circular bar
and the angle shown in the figure. The vector from the origin to A
is rOA D 4 cos�20°�iC 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 D �xA �
xB�iC �yA � yB�jC �zA � zB�k D 3.76i� 2.63j� 3k m. Finally, the
scalar components of the vector from B to A are (3.76, �2.63, �3) m.
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53
Problem 2.98 The cable AB in Problem 2.97 exerts a
60-N force T on the collar at A that is directed along the
line from A toward B. Express T in terms of components.
Solution: We know rBA D 3.76i� 2.63j� 3k m from Problem
2.97. The unit vector uAB D �rBA/jrBAj. The unit vector is uAB D
�0.686iC 0.480jC 0.547k. Hence, the force vector T is given by
TDjTj��0.686iC 0.480jC 0.547k� ND�41.1iC 28.8jC 32.8k N
Problem 2.99 In Active Example 2.11, suppose that
the vector V is changed to V D 4i� 6j� 10k.
(a) What is the value of UžV?
(b) What is the angle between U and V when they are
placed tail to tail?
Solution: From Active Example 2.4 we have the expression for U.
Thus
U D 6i� 5j� 3k,V D 4i� 6k� 10k
U РV D �6��4�C ��5���6�C ��3���10� D 84
cos � D U Ð VjVjjVj D
84√
62 C ��5�2 C ��3�2
√
42 C ��6�2 C ��10�2
D 0.814
� D cos�1�0.814� D 35.5°
�a� U Ð V D 84, �b� � D 35.5°
Problem 2.100 In Example 2.12, suppose that the coor-
dinates of point B are changed to (6, 4, 4) m. What is
the angle � between the lines AB and AC?
x
y
z
(8, 8, 4) m
(6, 1, �2) m(4, 3, 2) m
A B
C
u
Solution: Using the new coordinates we have
rAB D �2iC jC 2k� m, jrABj D 3 m
rAC D �4iC 5jC 2k� m, jrACj D 6.71 m
cos � D rAB Ð rACjrABjjrACj D
��2��4�C �1��5�C �2��2�� m2
�3 m��6.71 m�
D 0.845
� D cos�1�0.845� D 32.4°
� D 32.4°
Problem 2.101 What is the dot product of the position
vector r D �10iC 25j (m) and the force vector
F D 300iC 250jC 300k �N�?
Solution: Use Eq. (2.23).
F Рr D �300���10�C �250��25�C �300��0� D 3250 N-m
Problem 2.102 Suppose that the dot product of two
vectors U and V is U Ð V D 0. If jUj 6D 0, what do you
know about the vector V?
Solution:
Either jVj D 0 or V ? U
54
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Problem 2.103 Two perpendicular vectors are given
in terms of their components by
U D Uxi� 4jC 6k
and V D 3iC 2j� 3k.
Use the dot product to determine the component Ux.
Solution: When the vectors are perpendicular, U Ð V � 0.
Thus
U Ð V D UxVx CUyVy CUzVz D 0
D 3Ux C ��4��2�C �6���3� D 0
3Ux D 26
Ux D 8.67
Problem 2.104 Three vectors
U D UxiC 3jC 2k
V D �3iC VyjC 3k
W D �2iC 4jCWzk
are mutually perpendicular. Use the dot product to deter-
mine the components Ux,Vy , and Wz.
Solution: For mutually perpendicular vectors, we have three
equations, i.e.,
U Ð V D 0
U ÐW D 0
V ÐW D 0
Thus
�3Ux C 3Vy C 6 D 0
�2Ux C 12C 2Wz D 0
C6C 4Vy C 3Wz D 0


3 Eqns
3 Unknowns
Solving, we get
Ux D 2.857
Vy D 0.857
Wz D �3.143
Problem 2.105 The magnitudes jUj D 10 and jVj D
20.
(a) Use the definition of the dot product to determine
U Ð V.
(b) Use Eq. (2.23) to obtain U Ð V.
x
y
V
45�
U
30�Solution:
(a) The definition of the dot product (Eq. (2.18)) is
U Ð V D jUjjVj cos �. Thus
U Ð V D �10��20� cos�45° � 30°� D 193.2
(b) The components of U and V are
U D 10�i cos 45° C j sin 45°� D 7.07iC 7.07j
V D 20�i cos 30° C j sin 30°� D 17.32iC 10j
From Eq. (2.23) U РVD �7.07��17.32�C �7.07��10� D 193.2
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55
Problem 2.106 By evaluating the dot product U Ð V,
prove the identity cos��1 � �2� D cos �1 cos �2 C sin �1
sin �2.
Strategy: Evaluate the dot product both by using
Eq. (2.18) and by using Eq. (2.23).
x
y
VU
u1
u2Solution: The strategy is to use the definition Eq. (2.18) and the
Eq. (2.23). From Eq. (2.18) and the figure,
U РV D jUjjVj cos��1 � �2�. From Eq. (2.23) and the figure,
U D jUj�i cos �1 C j sin �2�, V D jVj�i cos �2 C j sin �2�,
and the dot product is U Ð V D jUjjVj�cos �1 cos �2 C sin �1 sin �2�.
Equating the two results:
U РV D jUjjVj cos��1 � �2� D jUjjVj�cos �1 cos �2 C sin �1 sin �2�,
from which if jUj 6D 0 and jVj 6D 0, it follows that
cos��1 � �2� D cos �1 cos �2 C sin �1 sin �2 , Q.E.D.
Problem 2.107 Use the dot product to determine the
angle between the forestay (cable AB) and the backstay
(cable BC).
y
x
B (4, 13) m
C
(9, 1) m
A
(0, 1.2) m
Solution: The unit vector from B to A is
eBA D rBAjrBAj D �0.321i� 0.947j
The unit vector from B to C is
eBC D rBCjrBCj D 0.385i� 0.923j
From the definition of the dot product, eBA Ð eBC D 1 Ð 1 Ð cos �, where
� is the angle between BA and BC. Thus
cos � D ��0.321��0.385�C ��0.947���0.923�
cos � D 0.750
� D 41.3°
56
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Problem 2.108 Determine the angle � between the
lines AB and AC (a) by using the law of cosines (see
Appendix A); (b) by using the dot product.
(4, 3, �1) m
B
y
xA
z
C
(5, �1, 3) m
u
Solution:
(a) We have the distances:
AB D
√
42 C 32 C 12 m D
p
26 m
AC D
√
52 C 12 C 32 m D
p
35 m
BC D
√
�5� 4�2 C ��1� 3�2 C �3C 1�2 m D
p
33 m
The law of cosines gives
BC2 D AB2 C AC2 � 2�AB��AC� cos �
cos � D AB
2 C AC2 � BC2
2�AB��AC�
D 0.464 ) � D 62.3°
(b) Using the dot product
rAB D �4iC 3j� k� m, rAC D �5i� jC 3k� m
rAB РrAC D �4 m��5 m�C �3 m���1 m�C ��1 m��3 m� D 14 m2
rAB РrAC D �AB��AC� cos �
Therefore
cos � D 14 m
2
p
26 m
p
35 m
D 0.464) � D 62.3°
Problem 2.109 The ship O measures the positions of
the ship A and the airplane B and obtains the coordinates
shown. What is the angle � between the lines of sightOA and OB?
x
y
z
A
B
O
(6, 0, 3) km
(4, 4, �4) km
u
Solution: From the coordinates, the position vectors are:
rOA D 6iC 0jC 3k and rOB D 4iC 4j� 4k
The dot product: rOA РrOB D �6��4�C �0��4�C �3���4� D 12
The magnitudes: jrOAj D
p
62 C 02 C 32 D 6.71 km and
jrOAj D
p
42 C 42 C 42 D 6.93 km.
From Eq. (2.24) cos � D rOA Ð rOBjrOAjjrOBj D 0.2581, from which � D š75
°.
From the problem and the construction, only the positive angle makes
sense, hence � D 75°
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57
Problem 2.110 Astronauts on the space shuttle use
radar to determine the magnitudes and direction cosines
of the position vectors of two satellites A and B. The
vector rA from the shuttle to satellite A has magnitude
2 km and direction cosines cos �x D 0.768, cos �y D
0.384, cos �z D 0.512. The vector rB from the shuttle
to satellite B has magnitude 4 km and direction cosines
cos �x D 0.743, cos �y D 0.557, cos �z D �0.371. What
is the angle � between the vectors rA and rB?
x
rB
z
B
A rA
y
θ
Solution: The direction cosines of the vectors along rA and rB
are the components of the unit vectors in these directions (i.e.,
uA D cos �x iC cos �y jC cos �zk, where the direction cosines are those
for rA). Thus, through the definition of the dot product, we can find
an expression for the cosine of the angle between rA and rB.
cos � D cos �xA cos �xB C cos �yA cos �yB C cos �zA cos �zB .
Evaluation of the relation yields
cos � D 0.594) � D 53.5° .
Problem 2.111 In Example 2.13, if you shift your
position and the coordinates of point A where you apply
the 50-N force become (8, 3, �3) m, what is the vector
component of F parallel to the cable OB?
x
y
(6, 6, –3) m
(10, �2, 3) m
F
A
B
O
z
Solution: We use the following vectors to define the force F.
rOA D �8iC 3j� 3k� m
eOA D rOAjrOAj D �0.833iC 0.331j� 0.331k�
F D �50 N�eOA D �44.2iC 16.6j� 16.6k� N
Now we need the unit vector eOB.
rOB D �10i� 2jC 3k� m
eOB D rOBjrOBj D �0.941i� 0.188jC 0.282k�
To find the vector component parallel to OB we use the dot product
in the following manner
F РeOB D �44.2 N��0.941�C �16.6 N���0.188�C ��16.6 N��0.282� D 33.8 N
Fp D �F РeOB�eOB D �33.8 N��0.941i� 0.188jC 0.282k�
Fp D �31.8i� 6.35jC 9.53k� N
58
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 2.112 The person exerts a force F D 60i�
40j (N) on the handle of the exercise machine. Use
Eq. (2.26) to determine the vector component of F that
is parallel to the line from the origin O to where the
person grips the handle.
O
150 mm
250 mm
200 mm
F
z
y
x
Solution: The vector r from the O to where the person grips the
handle is
r D �250iC 200j� 150k� mm,
jrj D 354 mm
To produce the unit vector that is parallel to this line we divide by the
magnitude
e D rjrj D
�250iC 200j� 150k� mm
354 mm
D �0.707iC 0.566j� 0.424k�
Using Eq. (2.26), we find that the vector component parallel to the
line is
Fp D �e РF�e D [�0.707��60 N�C �0.566���40 N�]�0.707i
C 0.566j� 0.424k�
Fp D �14.0iC 11.2jC 8.4k� N
Problem 2.113 At the instant shown, the Harrier’s
thrust vector is T D 17,000iC 68,000j� 8,000k (N)
and its velocity vector is v D 7.3iC 1.8j� 0.6k (m/s).
The quantity P D jTpjjvj, where Tp is the vector
component of T parallel to v, is the power currently
being transferred to the airplane by its engine. Determine
the value of P.
x
y
T
v
Solution:
T D �17,000iC 68,000j� 8,000k� N
v D �7.3iC 1.8j� 0.6k� m/s
Power D T Рv D �17,000 N��7.3 m/s�C �68,000 N��1.8 m/s�
C ��8,000 N���0.6 m/s�
Power D 251,000 Nm/s D 251 kW
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
59
Problem 2.114 Cables extend from A to B and from
A to C. The cable AC exerts a 1000-lb force F at A.
(a) What is the angle between the cables AB and AC?
(b) Determine the vector component of F parallel to
the cable AB. F
A (0, 7, 0) ft
B
C
x
y
z (14, 0, 14) ft
(0, 0, 10) ft
Solution: Use Eq. (2.24) to solve.
(a) From the coordinates of the points, the position vectors are:
rAB D �0� 0�iC �0� 7�jC �10� 0�k
rAB D 0i� 7jC 10k
rAC D �14� 0�iC �0� 7�jC �14� 0�k
rAC D 14i� 7jC 14k
The magnitudes are:
jrABj D
p
72 C 102 D 12.2 (ft) and
jrABj D
p
142 C 72 C 142 D 21.
The dot product is given by
rAB РrAC D �14��0�C ��7���7�C �10��14� D 189.
The angle is given by
cos � D 189
�12.2��21�
D 0.7377,
from which � D š42.5°. From the construction: � D C42.5°
(b) The unit vector associated with AB is
eAB D rABjrABj D 0i� 0.5738jC 0.8197k.
The unit vector associated with AC is
eAC D rACjrAC j D 0.6667i� 0.3333j C 0.6667k.
Thus the force vector along AC is
FAC D jFjeAC D 666.7i� 333.3jC 666.7k.
The component of this force parallel to AB is
�FAC Ð eAB�eAB D �737.5�eAB D 0i� 422.8jC 604.5k (lb)
60
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Problem 2.115 Consider the cables AB and AC shown
in Problem 2.114. Let rAB be the position vector from
point A to point B. Determine the vector component of
rAB parallel to the cable AC.
Solution: From Problem 2.114, rAB D 0i� 7jC 10k, and eAC D
0.6667i� 0.3333jC 0.6667k. Thus rAB Ð eAC D 9, and �rAB Ð eAC�eAC
D �6i� 3jC 6k� ft.
Problem 2.116 The force F D 10iC 12j� 6k �N�.
Determine the vector components of F parallel and nor-
mal to line OA.
y
x
z
(0, 6, 4) m
O
A
F
Solution: Find eOA D rOAjrOAj
Then
FP D �F Ð eOA�eOA
and FN D F� FP
eOA D 0iC 6jC 4kp
62 C 42 D
6jC 4kp
52
eOA D 6
7.21
jC 4
7.21
k D 0.832jC 0.555k
FP D [�10iC 12j� 6k� Ð �0.832jC 0.555k�]eOA
FP D [6.656]eOA D 0iC 5.54jC 3.69k �N�
FN D F� FP
FN D 10iC �12� 5.54�jC ��6� 3.69k�
FN D 10iC 6.46j� 9.69k N
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61
Problem 2.117 The rope AB exerts a 50-N force T on
collar A. Determine the vector component of T parallel
to the bar CD.
0.4 m
0.5 m
0.15 m
0.3 m0.2 m
0.25 m
0.2 m
z
x
y
A
B C
D
O
T
Solution: We have the following vectors
rCD D ��0.2i� 0.3jC 0.25k� m
eCD D rCDjrCDj D ��0.456i� 0.684jC 0.570k�
rOB D �0.5jC 0.15k� m
rOC D �0.4iC 0.3j� m
rOA D rOC C �0.2 m�eCD D �0.309iC 0.163jC 0.114k� m
rAB D rOB � rOA D ��0.309iC 0.337jC 0.036k� m
eAB D rABjrABj D �0.674iC 0.735jC 0.079k�
We can now write the force T and determine the vector component
parallel to CD.
T D �50 N�eAB D ��33.7iC 36.7jC 3.93k� N
Tp D �eCD Ð T�eCD D �3.43iC 5.14j� 4.29k� N
Tp D �3.43iC 5.14j� 4.29k� N
Problem 2.118 In Problem 2.117, determine the vector
component of T normal to the bar CD.
0.4 m
0.5 m
0.15 m
0.3 m0.2 m
0.25 m
0.2 m
z
x
y
A
B C
D
O
T
Solution: From Problem 2.117 we have
T D ��33.7iC 36.7jC 3.93k� N
Tp D �3.43iC 5.14j� 4.29k� N
The normal component is then
Tn D T� Tp
Tn D ��37.1iC 31.6jC 8.22k� N
62
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Problem 2.119 The disk A is at the midpoint of the
sloped surface. The string from A to B exerts a 0.2-lb
force F on the disk. If you express F in terms of vector
components parallel and normal to the sloped surface,
what is the component normal to the surface?
y
z
A
B
F
(0, 6, 0) ft
2 ft
8 ft
10 ft
x
Solution: Consider a line on the sloped surface from A perpendic-
ular to the surface. (see the diagram above) By SIMILAR triangles we
see that one such vector is rN D 8jC 2k. Let us find the component
of F parallel to this line.
The unit vector in the direction normal to the surface is
eN D rNjrNj D
8jC 2kp
82 C 22 D 0.970jC 0.243k
The unit vector eAB can be found by
eAB D �xB � xA�iC �yB � yA�jC �zB � zA�h√
�xB � xA�2 C �yB � yA�2 C �zB � zA�2
Point B is at (0, 6, 0) (ft) and A is at (5, 1, 4) (ft).
Substituting, we get
eAB D �0.615iC 0.615j� 0.492k
Now F D jFjeAB D �0.2�eAB
F D �0.123iC 0.123j� 0.0984k �lb�
The component of F normal to the surface is the component parallel
to the unit vector eN.
FNORMAL D �F Ð eN�eN D �0.955�eN
FNORMAL D 0iC 0.0927j C 0.0232k lb
y
z
8
8
2
2
Problem 2.120 In Problem 2.119, what is the vector
component of F parallel to the surface?
Solution: From the solution to Problem 2.119,
F D �0.123iC 0.123j� 0.0984k �lb� and
FNORMAL D 0iC 0.0927j C 0.0232k �lb�
The component parallel to the surface and the component normal to
the surface add to give F�F D FNORMAL C Fparallel�.
Thus
Fparallel D F� FNORMAL
Substituting, we get
Fparallel D �0.1231iC 0.0304j� 0.1216k lb
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63
Problem 2.121 An astronaut in a maneuvering unit
approaches a space station. At the present instant, the
station informs him that his position relative to the origin
of the station’s coordinate system is rG D 50iC 80jC
180k (m) and his velocity is v D �2.2j� 3.6k (m/s).
The position of the airlock is rA D �12iC 20k (m).
Determine the angle between his velocity vector and the
line from his position to the airlock’s position.
Solution: Points G and A are located at G: (50, 80, 180) m
and A: (�12, 0, 20) m. The vector rGA is rGA D �xA � xG�iC �yA �
yG�jC �zA � zG�k D ��12� 50�iC �0� 80�jC �20� 180�k m. The
dot product between v and rGA is v ž rGA D jvjjrGAj cos � D vxxGA C
vyyGA C vzzGA, where � is the angle between v and rGA. Substituting
in the numerical values, we get � D 19.7°.
y
z
x
A
G
Problem 2.122 In Problem 2.121, determine the vec-
tor component of the astronaut’s velocity parallel to the
line from his position to the airlock’s position.
Solution: The coordinates are A (�12, 0, 20) m, G (50, 80, 180) m.
Therefore
rGA D ��62i� 80j� 160k� m
eGA D rGAjrGAj D ��0.327i� 0.423j� 0.845k�
The velocity is given as
v D ��2.2j� 3.6k� m/s
The vector component parallel to the line is now
vp D �eGA Рv�eGA D [��0.423���2.2�C ��0.845���3.6�]eGA
vp D ��1.30i� 1.68j� 3.36k� m/s
64
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Problem 2.123 Point P is at longitude 30°W and lati-
tude 45°N on the Atlantic Ocean between Nova
Scotia and France. 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.
Equator
y
z
x
P
N
O
45�
30� 60�
G
20�
Q
Solution: The distance is the product of the angle and the radius of
the sphere, d D RE�, where � is in radian measure. From Eqs. (2.18)
and (2.24), the angular separation of P and Q is given by
cos � D
(
P ÐQ
jPjjQj
)
.
The strategy is to determine the angle � in terms of the latitude and
longitude of the two points. Drop a vertical line from each point P and
Q to b and c on the equatorial plane. The vector position of P is the sum
of the two vectors: P D rOB C rBP. The vector rOB D jrOBj�i cos �P C
0jC k sin �P�. From geometry, the magnitude is jrOBj D RE cos �P.
The vector rBP D jrBPj�0iC 1jC 0k�. From geometry, the magnitude
is jrBPj D RE sin �P. Substitute and reduce to obtain:
P D rOB C rBP D RE�i cos �P cos �P C j sin �P C k sin �P cos �P�.
A similar argument for the point Q yields
Q D rOC C rCQ D RE�i cos �Q cos �Q C j sin �Q C k sin �Q cos �Q�
Using the identity cos2 ˇC sin2 ˇ D 1, the magnitudes are
jPj D jQj D RE
The dot product is
P ÐQ D R2E�cos��P � �Q� cos �P cos �Q C sin �P sin �Q�
Substitute:
cos � D P ÐQjPjjQj D cos��P � �Q� cos �P cos �Q C sin �P sin �Q
Substitute �P D C30°, �Q D �60°, �p D C45°, �Q D C20°, to obtain
cos � D 0.2418, or � D 1.326 radians. Thus the distance is d D
1.326RE
y
x
30°
45°
60°
20°
RE
θ
Q
P
G
b c
N
Problem 2.124 In Active Example 2.14, suppose that
the vector V is changed to V D 4i� 6j� 10k.
(a) Determine the cross product Uð V. (b) Use the dot
product to prove that Uð V is perpendicular to V.
Solution: We have U D 6i� 5j� k,V D 4k� 6j� 10k
(a) Uð V D
∣∣∣∣∣∣
i j k
6 �5 �1
4 �6 �10
∣∣∣∣∣∣ D 44iC 56j� 16k
Uð V D 44iC 56j� 16k
(b) �Uð V� Ð V D �44��4�C �56���6�C ��16���10� D 0)
�Uð V� ? V
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65
Problem 2.125 Two vectors U D 3iC 2j and V D 2i
C 4j.
(a) What is the cross product Uð V?
(b) What is the cross product Vð U?
Solution: Use Eq. (2.34) and expand into 2 by 2 determinants.
Uð V D
∣∣∣∣∣∣
i j k
3 2 0
2 4 0
∣∣∣∣∣∣ D i��2��0�� �4��0��� j��3��0�� �2��0��
C k��3��4�� �2��2�� D 8k
Vð U D
∣∣∣∣∣∣
i j k
2 4 0
3 2 0
∣∣∣∣∣∣ D i��4��0�� �2��0��� j��2��0�� �3��0��
C k��2��2�� �3��4�� D �8k
Problem 2.126 The two segments of the L-shaped bar
are parallel to the x and z axes. The rope AB exerts
a force of magnitude jFj D 500 lb on the bar at A.
Determine the cross product rCA ð F, where rCA is the
position vector form point C to point A.
F
y
x
z
4 ft
4 ft
5 ft
(6, 0, 4) ft
C
A
B
Solution: We need to determine the force F in terms of its
components. The vector from A to B is used to define F.
rAB D �2i� 4j� k� ft
F D �500 lb� rABjrABj D �500 lb�
�2i� 4j� k�√
�2�2 C ��4�2 C ��1�2
F D �218i� 436j� 109k� lb
Also we have rCA D �4iC 5k� ft
Therefore
rCA ð F D
∣∣∣∣∣∣
i j k
4 0 5
218 �436 �109
∣∣∣∣∣∣ D �2180iC 1530j� 1750k� ft-lb
rCA ð F D �2180iC 1530j � 1750k� ft-lb
66
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Problem 2.127 The two segments of the L-shaped bar
are parallel to the x and z axes. The rope AB exerts
a force of magnitude jFj D 500 lb on the bar at A.
Determine the cross product rCB ð F, where rCB is the
position vector form point C to point B. Compare your
answers to the answer to Problem 2.126.
F
y
x
z
4 ft
4 ft
5 ft
(6, 0, 4) ft
C
A
B
Solution: We need to determine the force F in terms of its compo-
nents. The vector from A to B is used to define F.
rAB D �2i� 4j� k� ft
F D �500 lb� rABjrABj D �500 lb�
�2i� 4j� k�√
�2�2 C ��4�2 C ��1�2
F D �218i�436j� 109k� lb
Also we have rCB D �6i� 4jC 4k� ft
Therefore
rCB ð F D
∣∣∣∣∣∣
i j k
6 �4 4
218 �436 �109
∣∣∣∣∣∣ D �2180iC 1530j� 1750k� ft-lb
rCB ð F D �2180iC 1530j � 1750k� ft-lb
The answer is the same for 2.126 and 2.127 because the position
vectors just point to different points along the line of action of the
force.
Problem 2.128 Suppose that the cross product of two
vectors U and V is Uð V D 0. If jUj 6D 0, what do you
know about the vector V?
Solution:
Either V D 0 or VjjU
Problem 2.129 The cross product of two vectors U
and V is Uð V D �30iC 40k. The vector V D 4i�
2jC 3k. The vector U D �4iCUyjCUzk�. Determine
Uy and Uz.
Solution: From the given information we have
Uð V D
∣∣∣∣∣∣
i j k
4 Uy Uz
4 �2 3
∣∣∣∣∣∣
D �3Uy C 2Uz�iC �4Uz � 12�jC ��8� 4Uy�k
Uð V D ��30iC 40k�
Equating the components we have
3Uy C 2Uz D �30, 4Uz � 12 D 0, �8� 4Uy D 40.
Solving any two of these three redundant equations gives
Uy D �12,Uz D 3.
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67
Problem 2.130 The magnitudes jUj D 10 and jVj D
20.
(a) Use the definition of the cross product to determine
Uð V.
(b) Use the definition of the cross product to determine
Vð U.
(c) Use Eq. (2.34) to determine Uð V.
(d) Use Eq. (2.34) to determine Vð U.
U
V
x
y
45°
30°
Solution: From Eq. (228) Uð V D jUjjVj sin �e. From the sketch,
the positive z-axis is out of the paper. For Uð V, e D �1k (points into
the paper); for Vð U, e D C1k (points out of the paper). The angle
� D 15°, hence (a) Uð V D �10��20��0.2588��e� D 51.8e D �51.8k.
Similarly, (b) Vð U D 51.8e D 51.8k (c) The two vectors are:
U D 10�i cos 45° C j sin 45� D 7.07iC 0.707j,
V D 20�i cos 30° C j sin 30°� D 17.32iC 10j
Uð V D
∣∣∣∣∣∣
i j k
7.07 7.07 0
17.32 10 0
∣∣∣∣∣∣ D i�0�� j�0�C k�70.7� 122.45�
D �51.8k
(d) Vð U D
∣∣∣∣∣∣
i j k
17.32 10 0
7.07 7.07 0
∣∣∣∣∣∣ D i�0�� j�0�C k�122.45� 70.7�
D 51.8k
Problem 2.131 The force F D 10i� 4j (N). Deter-
mine the cross product rAB ð F.
y
x
B
A
rAB
(6, 3, 0) m
(6, 0, 4) m
F
z
Solution: The position vector is
rAB D �6� 6�iC �0� 3�jC �4� 0�k D 0i� 3jC 4k
The cross product:
rAB ð F D
∣∣∣∣∣∣
i j k
0 �3 4
10 �4 0
∣∣∣∣∣∣ D i�16�� j��40�C k�30�
D 16iC 40jC 30k (N-m)
y
x
B
A
rAB
(6, 3, 0)
(6, 0, 4) F
z
68
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Problem 2.132 By evaluating the cross product
Uð V, prove the identity sin��1 � �2� D sin �1 cos �2 �
cos �1 sin �2.
x
y
V
θ1
θ2
U
Solution: Assume that both U and V lie in the x-y plane. The
strategy is to use the definition of the cross product (Eq. 2.28) and the
Eq. (2.34), and equate the two. From Eq. (2.28) Uð V D jUjjVj sin��1
� �2�e. Since the positive z-axis is out of the paper, and e points into
the paper, then e D �k. Take the dot product of both sides with e, and
note that k Ð k D 1. Thus
sin��1 � �2� D �
(
�Uð V� Ð k
jUjjVj
)
The vectors are:
U D jUj�i cos �1 C j sin �2�, and V D jVj�i cos �2 C j sin �2�.
The cross product is
Uð V D
∣∣∣∣∣∣
i j k
jUj cos �1 jUj sin �1 0
jVj cos �2 jVj sin �2 0
∣∣∣∣∣∣
D i�0�� j�0�C k�jUjjVj��cos �1 sin �2 � cos �2 sin �1�
Substitute into the definition to obtain: sin��1 � �2� D sin �1 cos �2 �
cos �1 sin �2. Q.E.D.
y
x
U
V
θ
θ
1
2
Problem 2.133 In Example 2.15, what is the minimum
distance from point B to the line OA?
B
(6, 6, �3) m
x
y
z
O
A (10, �2, 3) m
Solution: Let � be the angle between rOA and rOB. Then the
minimum distance is
d D jrOBj sin �
Using the cross product, we have
jrOA ð rOBj D jrOAjjrOBj sin � D jrOAjd) d D jrOA ð rOBjjrOAj
We have
rOA D �10i� 2jC 3k� m
rOB D �6iC 6j� 3k� m
rOA ð rOB D
∣∣∣∣∣∣
i j k
10 �2 3
6 6 �3
∣∣∣∣∣∣ D ��12iC 48jC 72k� m2
Thus
d D
√
��12 m2�C �48 m2�2 C �72 m2�2√
�10 m�2 C ��2 m�2 C �3 m�2
D 8.22 m
d D 8.22 m
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69
Problem 2.134 (a) What is the cross product rOA ð
rOB? (b) Determine a unit vector e that is perpendicular
to rOA and rOB.
A (6, –2, 3) m
(4, 4, –4) mB
x
y
z
O
rOB
rOA
Solution: The two radius vectors are
rOB D 4iC 4j� 4k, rOA D 6i� 2jC 3k
(a) The cross product is
rOA ð rOB D
∣∣∣∣∣∣
i j k
6 �2 3
4 4 �4
∣∣∣∣∣∣ D i�8� 12�� j��24� 12�
C k�24C 8�
D �4iC 36jC 32k �m2�
The magnitude is
jrOA ð rOBj D
p
42 C 362 C 322 D 48.33 m2
(b) The unit vector is
e D š
(
rOA ð rOB
jrOA ð rOBj
)
D š��0.0828iC 0.7448jC 0.6621k�
(Two vectors.)
Problem 2.135 For the points O, A, and B in Pro-
blem 2.134, use the cross product to determine the length
of the shortest straight line from point B to the straight
line that passes through points O and A.
Solution:
rOA D 6i� 2jC 3k (m)
rOB D 4iC 4j� 4k �m�
rOA ð rOB D C
(C is ? to both rOA and rOB)
C D
∣∣∣∣∣∣
i j k
6 �2 3
4 4 �4
∣∣∣∣∣∣ D
�C8� 12�i
C�12C 24�j
C�24C 8�k
C D �4iC 36jC 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. Cð rOA is such a
line. We then need to find the component of rOB in this direction and
compute its magnitude.
Cð rOA D
∣∣∣∣∣∣
i j k
�4 C36 32
6 � 2 3
∣∣∣∣∣∣
C D 172iC 204j� 208k
The unit vector in the direction of C is
eC D CjCj D 0.508iC 0.603j� 0.614k
(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 D rOB Ð eC
D �4iC 4j� 4k� Ð eC
P D 6.90 m
A(6, –2, 3) m
(4, 4, –4) mB
x
y
z
O
rOB
rOA
P
70
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Problem 2.136 The cable BC exerts a 1000-lb force F
on the hook at B. Determine rAB ð F.
rAB
A
B
C
z
x
y
rAC
F
6 ft
8 ft
4 ft
4 ft 12 ft
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
rOA D 16iC 0jC 12k, rOB D 4iC 6jC 0k, rOC D 4iC 0jC 8k.
The force F acts along the unit vector
eBC D rBCjrBCj D
rOC � rOB
jrOC � rOBj D
rAB
jrABj
Noting rOC � rOB D �4� 4�iC �0� 6�jC �8� 0�k D 0i� 6jC 8k
jrOC � rOBj D
p
62 C 82 D 10. Thus
eBC D 0i� 0.6jC 0.8k, and F D jFjeBC D 0i� 600jC 800k (lb).
The vector
rAB D �4� 16�iC �6� 0�jC �0� 12�k D �12iC 6j� 12k
Thus the cross product is
rAB ð F D
∣∣∣∣∣∣
i j k
�12 6 �12
0 �600 800
∣∣∣∣∣∣ D �2400iC 9600jC 7200k (ft-lb)
r
A
B
C
x
y
6 ft
8 ft
4 ft
4 ft 12 ft
Problem 2.137 The force vector F points along the
straight line from point A to point B. Its magnitude
is jFj D 20 N. The coordinates of points A and B
are xA D 6 m, yA D 8 m, zA D 4 m and xB D 8 m, yB D
1 m, zB D �2 m.
(a) Express the vector F in terms of its components.
(b) Use Eq. (2.34) to determine the cross products
rA ð F and rB ð F. x
y
z
rB
rA
A
B 
F
Solution: We have rA D �6iC 8jC 4k� m, rB D �8iC j� 2k� m,
(a)
F D �20 N� �8� 6� miC �1� 8� mjC ��2� 4� mk√
�2 m�2 C ��7 m�2 C ��6 m�2
D 20 Np
89
�2i� 7j� 6k�
(b)
rA ð F D 20 Np
89
∣∣∣∣∣∣∣
i j k
6 m 8 m 4 m
2 �7 �6
∣∣∣∣∣∣∣
D ��42.4iC 93.3j� 123.0k� Nm
rB ð F D 20 Np
89
∣∣∣∣∣∣∣
i j k
8 m 1 m �2 m
2 �7 �6
∣∣∣∣∣∣∣
D ��42.4iC 93.3j� 123.0k� Nm
Note that both cross products give the same result (as they must).
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71
Problem 2.138 The rope AB exerts a 50-N force T
on the collar at A. Let rCA be the position vector from
point C to point A. Determine the cross product rCA ð T.
0.4 m
0.5 m
0.15 m
0.3 m0.2 m
0.25 m
0.2 m
z
x
y
A
B C
D
O
T
Solution: We define the appropriate vectors.
rCD D ��0.2i� 0.3jC 0.25k� m
rCA D �0.2 m� rCDjrCDj D ��0.091i� 0.137jC 0.114k� m
rOB D �0.5jC 0.15k� m
rOC D �0.4iC 0.3j� m
rAB D rOB � �rOC C rCA� D �0.61i� 1.22j� 0.305k� m
T D �50 N� rABjrABj D ��33.7iC 36.7jC 3.93k� N
Now take the cross product
rCA ð T D
∣∣∣∣∣∣
i j l
�0.091 �0.137 0.114
�33.7 36.7 3.93
∣∣∣∣∣∣ D ��4.72i� 3.48jC�7.96k� N-m
rCA ð T D ��47.2i� 3.48jC�7.96k� N-m
Problem 2.139 In Example 2.16, suppose that the
attachment point E is moved to the location (0.3, 0.3,
0) m and the magnitude of T increases to 600 N. What
is the magnitude of the component of T perpendicular
to the door?
x
y
z
A (0.5, 0, 0) m
E
B
C
DT
(0, 0.2, 0) m
(0.35, 0, 0.2) m
(0.2, 0.4, �0.1) m
Solution: We first develop the force T.
rCE D �0.3iC 0.1j� m
T D �600 N� rCEjrCEj D �569iC 190j� N
From Example 2.16 we know that the unit vector perpendicular to the
door is
e D �0.358iC 0.894jC 0.268k�
The magnitude of the force perpendicular to the door (parallel to e) is
then
jTnj D T Рe D �569 N��0.358�C �190 N��0.894� D 373 N
jTnj D 373 N
72
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Problem 2.140 The bar AB is 6 m long and is perpen-
dicular to the bars AC and AD. Use the cross product to
determine the coordinates xB, yB, zB of point B.
C
A
B
(0, 0, 3) m (4, 0, 0) m x
y
(0, 3, 0) m
(xB, yB, zB)
D
z
Solution: The strategy is to determine the unit vector perpendic-
ular to both AC and AD, and then determine the coordinates that will
agree with the magnitude of AB. The position vectors are:
rOA D 0iC 3jC 0k, rOD D 0iC 0jC 3k, and
rOC D 4iC 0jC 0k. The vectors collinear with the bars are:
rAD D �0� 0�iC �0� 3�jC �3� 0�k D 0i� 3jC 3k,
rAC D �4� 0�iC �0� 3�jC �0� 0�k D 4i� 3jC 0k.
The vector collinear with rAB is
R D rAD ð rAC D
∣∣∣∣∣∣
i j k
0 �3 3
4 �3 0
∣∣∣∣∣∣ D 9iC 12jC 12k
The magnitude jRj D 19.21 (m). The unit vector is
eAB D RjRj D 0.4685iC 0.6247j C 0.6247k.
Thus the vector collinear with AB is
rAB D 6eAB D C2.811iC 3.75jC 3.75k.
Using the coordinates of point A:
xB D 2.81C 0 D 2.81 (m)
yB D 3.75C 3 D 6.75 (m)
zB D 3.75C 0 D 3.75 (m)
Problem 2.141* Determine the minimum distance
from point P to the plane defined by the three points
A, B, and C.
A
(3, 0, 0) m
(0, 5, 0) mB
x
y
z
C
(0, 0, 4) m
P
(9, 6, 5) m
Solution: The strategy is to find the unit vector perpendicular to
the plane. The projection of this unit vector on the vector OP: rOP Ð e is
the distance from the origin to P along the perpendicular to the plane.
The projection on e of any vector into the plane (rOA Ð e, rOB Ð e, or
rOC Ð e) is the distance from the origin to the plane along this same
perpendicular. Thus the distance of P from the plane is
d D rOP Ð e� rOA Ð e.
The position vectors are: rOA D 3i, rOB D 5j, rOC D 4k and rOP D
9iC 6jC 5k. The unit vector perpendicular to the plane is found
from the cross product of any two vectors lying in the plane. Noting:
rBC D rOC � rOB D �5jC 4k, and rBA D rOA � rOB D 3i� 5j. The
cross product:
rBC ð rBA D
∣∣∣∣∣∣
i j k
0 �5 4
3 �5 0
∣∣∣∣∣∣ D 20iC 12jC 15k.
The magnitude is jrBC ð rBAj D 27.73, thus the unit vector is e D
0.7212iC 0.4327jC 0.5409k. The distance of point P from the plane
is d D rOP Ð e� rOA Ð e D 11.792� 2.164 D 9.63 m. The second term
is the distance of the plane from the origin; the vectors rOB, or rOC
could have been used instead of rOA.
y
z
x
B[0,5,0]
P[9,6,5]
A[3,0,0]
C[0,0,4]
O
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73
Problem 2.142* The force vector F points along
the straight line from point A to point B. Use
Eqs. (2.28)–(2.31) to prove that
rB ð F D rA ð F.
Strategy: Let rAB be the position vector from point A
to point B. Express rB in terms of of rA and rAB . Notice
that the vectors rAB and F are parallel. x
y
z
rB
rA
A
B 
F
Solution: We have
rB D rA C rAB.
Therefore
rB ð F D �rA C rAB�ð F D rA ð FC rAB ð F
The last term is zero since rABjjF.
Therefore
rB ð F D rA ð F
Problem 2.143 For the vectors U D 6iC 2j� 4k,
V D 2iC 7j, and W D 3iC 2k, evaluate the following
mixed triple products: (a) U Ð �VðW�; (b) W Ð �Vð
U�; (c) V Ð �Wð U�.
Solution: Use Eq. (2.36).
(a) U Ð �VðW� D
∣∣∣∣∣∣
6 2 �4
2 7 0
3 0 2
∣∣∣∣∣∣
D 6�14�� 2�4�C ��4���21� D 160
(b) W Ð �Vð U� D
∣∣∣∣∣∣
3 0 2
2 7 0
6 2 �4
∣∣∣∣∣∣
D 3��28�� �0�C 2�4� 42� D �160
(c) V Ð �Wð U� D
∣∣∣∣∣∣
2 7 0
3 0 2
6 2 �4
∣∣∣∣∣∣
D 2��4�� 7��12� 12�C �0� D 160
74
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Problem 2.144 Use the mixed triple product to calcu-
late the volume of the parallelepiped.
x
y
z
(140, 90, 30) mm
(200, 0, 0) mm
(160, 0, 100) mm
Solution: We are given the coordinates of point D. From the geom-
etry, we need to locate points A and C. The key to doing this is to note
that the length of side OD is 200 mm and that side OD is the x axis.
Sides OD, AE, and CG are parallel to the x axis and the coordinates
of the point pairs (O and D), (A and E), and (C and D) differ only by
200 mm in the x coordinate. Thus, the coordinates of point A are (�60,
90, 30) mm and the coordinates of point C are (�40, 0, 100) mm.
Thus, the vectors rOA, rOD, and rOC are rOD D 200i mm, rOA D
�60iC 90jC 30k mm, and rOC D �40iC 0jC 100k mm. The mixed
triple product of the three vectors is the volume of the parallelepiped.
The volume is
rOA Ð �rOC ð rOD� D
∣∣∣∣∣∣
�60 90 30
�40 0 100
200 0 0
∣∣∣∣∣∣
D �60�0�C 90�200��100� C �30��0� mm3
D 1,800,000 mm3
y
z
A
B
C
O
F
G
D
E
x
(140, 90, 30)
mm
(200, 0, 0)
mm
(160, 0, 100)
mm
Problem 2.145 By using Eqs. (2.23) and (2.34), show
that
U Ð �VðW� D
∣∣∣∣∣
Ux Uy Uz
Vx Vy Vz
Wx Wy Wz
∣∣∣∣∣
.
Solution: One strategy is to expand the determinant in terms of
its components, take the dot product, and then collapse the expansion.
Eq. (2.23) is an expansion of the dot product: Eq. (2.23): U Ð V D
UXVX CUYVY CUZVZ. Eq. (2.34) is the determinant representation
of the cross product:
Eq. (2.34) Uð V D
∣∣∣∣∣∣
i j k
UX UY UZ
VX VY VZ
∣∣∣∣∣∣
For notational convenience, write P D �Uð V�. Expand the determi-
nant about its first row:
P D i
∣∣∣∣UY UZVY VZ
∣∣∣∣� j
∣∣∣∣UX UZVX VZ
∣∣∣∣C k
∣∣∣∣UX UZVX VZ
∣∣∣∣
Since the two-by-two determinants are scalars, this can be written in
the form: P D iPX C jPY C kPZ where the scalars PX, PY, and PZ are
the two-by-two determinants. Apply Eq. (2.23) to the dot product of
a vector Q with P. Thus Q Ð P D QXPX C QYPY C QZPZ. Substitute
PX, PY, and PZ into this dot product
Q Ð P D QX
∣∣∣∣UY UZVY VZ
∣∣∣∣� QY
∣∣∣∣UX UZVX VZ
∣∣∣∣C Qz
∣∣∣∣UX UZVX VZ
∣∣∣∣
But this expression can be collapsed into a three-by-three determinant
directly, thus:
Q Ð �Uð V� D
∣∣∣∣∣∣
QX QY QZ
UX UY UZ
VX VY VZ
∣∣∣∣∣∣. This completes the demonstration.
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75
Problem 2.146 The vectors U D iCUYjC 4k, V D
2iC j� 2k, and W D �3iC 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 pro-
duct of any two will produce a vector perpendicular to the plane, and
the dot product with the third will vanish, by definition of the dot
product. Thus U Ð �VðW� D 0, for example.
U Ð �VðW� D
∣∣∣∣∣∣
1 UY 4
2 1 �2
�3 1 �2
∣∣∣∣∣∣
D 1��2C 2�� �UY���4� 6�C �4��2C 3�
D C10UY C 20 D 0
Thus UY D �2
Problem 2.147 The magnitude of F is 8 kN. Express
F in terms of scalar components.
F
x
y
(7, 2) m
(3, 7) m
Solution: The unit vector collinear with the force F is developed
as follows: The collinear vector is r D �7� 3�iC �2� 7�j D 4i� 5j
The magnitude: jrj D p42 C 52 D 6.403 m. The unit vector is
e D rjrj D 0.6247i� 0.7809j. The force vector is
F D jFje D 4.998i� 6.247j D 5i� 6.25j (kN)
Problem 2.148 The magnitude of the vertical force W
is 600 lb, and the magnitude of the force B is 1500 lb.
Given that AC BCW D 0, determine the magnitude of
the force A and the angle ˛.
50°
B W
A
α
Solution: The strategy is to use the condition of force balance to
determine the unknowns. The weight vector is W D �600j. The vector
B is
B D 1500�i cos 50° C j sin 50°� D 964.2iC 1149.1j
The vector A is A D jAj�i cos�180C ˛�C j sin�180C ˛��
A D jAj��i cos˛� j sin˛�. The forces balance, hence AC BC
W D 0, or �964.2� jAj cos˛�i D 0, and �1149.1� 600� jAj sin˛�j D
0. Thus jAj cos˛ D 964.2, and jAj sin˛ D 549.1. Take the ratio of the
two equations to obtain tan˛ D 0.5695, or ˛ D 29.7°. Substitute this
angle to solve: jAj D 1110 lb
76
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 2.149 The magnitude of the vertical force
vector A is 200 lb. If AC BC C D 0, what are the mag-
nitudes of the force vectors B and C?
C
70 in. 100 in.
50 in.
E
F
DB
A
Solution: The strategy is to express the forces in terms of scalar
components, and then solve the force balance equations for the un-
knowns. C D jCj��i cos˛� j sin˛�, where
tan˛ D 50
70
D 0.7143, or ˛ D 35.5°.
Thus C D jCj��0.8137i� 0.5812j�. Similarly, B D CjBji, and A D
C200j. The force balance equation is AC BC C D 0. Substituting,
��0.8137jCj C jBj�i D 0, and ��0.5812jCj C 200�j D 0. Solving,
jCj D 344.1 lb, jBj D 280 lb
Problem 2.150 The magnitude of the horizontal force
vector D in Problem 2.149 is 280 lb. If DC EC F D 0,
what are the magnitudes of the force vectors E and F?
Solution: The strategy is to express the force vectors in terms of
scalar components, and then solve the force balance equation for the
unknowns. The force vectors are:
E D jEj�i cosˇ � j sinˇ�, where tan ˇD 50
100
D0.5, or ˇD26.6°.
Thus
E D jEj�0.8944i� 0.4472j�
D D �280i, and F D jFjj.
The force balance equation is DC EC F D 0. Substitute and resolve
into two equations:
�0.8944jEj � 280�i D 0, and ��0.4472jEj C jFj�j D 0.
Solve: jEj D 313.1 lb, jFj D 140 lb
Problem 2.151 What are the direction cosines of F?
Refer to this diagram when solving Problems 2.151–
2.157.
x
y
z
A
(4, 4, 2) ft
B (8, 1, �2) ft
F � 20i � 10j � 10k (lb)
u
Solution: Use the definition of the direction cosines and the
ensuing discussion.
The magnitude of F: jFj D p202 C 102 C 102 D 24.5.
The direction cosines are cos �x D FxjFj D
20
24.5
D 0.8165,
cos �y D FyjFj D
10
24.5
D 0.4082
cos �z D FzjFj D
�10
24.5
D �0.4082
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77
Problem 2.152 Determine the scalar components of
a unit vector parallel to line AB that points from A
toward B.
Solution: Use the definition of the unit vector, we get
The position vectors are: rA D 4iC 4jC 2k, rB D 8iC 1j� 2k. The
vector from A to B is rAB D �8� 4�iC �1� 4�jC ��2� 2�k D
4i� 3j� 4k. The magnitude: jrABj D
p
42 C 32 C 42 D 6.4. The unit
vector is
eAB D rABjrABj D
4
6.4
i� 3
6.4
j� 4
6.4
k D 0.6247i� 0.4685j� 0.6247k
Problem 2.153 What is the angle � between the line
AB and the force F?
Solution: Use the definition of the dot product Eq. (2.18), and
Eq. (2.24):
cos � D rAB Ð FjrABjjFj .
From the solution to Problem 2.130, the vector parallel to AB is rAB D
4i� 3j� 4k, with a magnitude jrABj D 6.4. From Problem 2.151, the
force is F D 20iC 10j� 10k, with a magnitude of jFj D 24.5. The dot
product is rAB РF D �4��20�C ��3��10�C ��4���10� D 90. Substi-
tuting, cos � D 90
�6.4��24.5�
D 0.574, � D 55°
Problem 2.154 Determine the vector component of F
that is parallel to the line AB.
Solution: Use the definition in Eq. (2.26): UP D �e Ð U�e, where e
is parallel to a line L. From Problem 2.152 the unit vector parallel to
line AB is eAB D 0.6247i� 0.4688j� 0.6247k. The dot product is
e РF D �0.6247��20�C ��0.4688��10�C ��0.6247���10� D 14.053.
The parallel vector is
�e Ð F�e D �14.053�e D 8.78i� 6.59j� 8.78k (lb)
Problem 2.155 Determine the vector component of F
that is normal to the line AB.
Solution: Use the Eq. (2.27) and the solution to Problem 2.154.
FN D F� FP D �20� 8.78�iC �10C 6.59�jC ��10C 8.78�k
D 11.22iC 16.59j� 1.22k (lb)
Problem 2.156 Determine the vector rBA ð F, where
rBA is the position vector from B to A.
Solution: Use the definition in Eq. (2.34). Noting rBA D �rAB,
from Problem 2.155 rBA D �4iC 3jC 4k. The cross product is
rBA ð F D
∣∣∣∣∣∣
i j k
�4 3 4
20 10 �10
∣∣∣∣∣∣ D ��30� 40�i� �40� 80�j
C ��40� 60�
D �70iC 40j� 100k (ft-lb)
78
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 2.157 (a) Write the position vector rAB from
point A to point B in terms of components.
(b) A vector R has magnitude jRj D 200 lb and is
parallel to the line from A to B. Write R in terms of
components.
x
y
z
A
(4, 4, 2) ft
B (8, 1, �2) ft
F � 20i � 10j � 10k (lb)
u
Solution:
(a) rAB D �[8� 4]iC [1� 4]jC [�2� 2]k� ft
rAB D �4i� 3j� 4k� ft
(b) R D �200 N� rABjrABj D �125i� 93.7j� 125k� N
R D �125i� 96.3j� 125k� N
Problem 2.158 The rope exerts a force of magnitude
jFj D 200 lb on the top of the pole at B.
(a) Determine the vector rAB ð F, where rAB is the
position vector from A to B.
(b) Determine the vector rAC ð F, where rAC is the
position vector from A to C.
x
y
B (5, 6, 1) ft 
C (3, 0, 4) ft
z
A
F
Solution: The strategy is to define the unit vector pointing from B
to A, express the force in terms of this unit vector, and take the cross
product of the position vectors with this force. The position vectors
rAB D 5iC 6jC 1k, rAC D 3iC 0jC 4k,
rBC D �3� 5�iC �0� 6�jC �4� 1�k D �2i� 6jC 3k.
The magnitude jrBCj D
p
22 C 62 C 32 D 7. The unit vector is
eBC D rBCjrBCj D �0.2857i� 0.8571jC 0.4286k.
The force vector is
F D jFjeBC D 200eBC D �57.14i� 171.42jC 85.72k.
The cross products:
rAB ð F D
∣∣∣∣∣∣
i j k
5 6 1
�57.14 �171.42 85.72
∣∣∣∣∣∣
D 685.74i� 485.74j� 514.26k
D 685.7i� 485.7j� 514.3k (ft-lb)
rAC ð F D
∣∣∣∣∣∣
i j k
3 0 4
�57.14 �171.42 85.72
∣∣∣∣∣∣
D 685.68i� 485.72j� 514.26k
D 685.7i� 485.7j� 514.3k (ft-lb)
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79
Problem 2.159The pole supporting the sign is parallel
to the x axis and is 6 ft long. Point A is contained in the
y–z plane. (a) Express the vector r in terms of compo-
nents. (b) What are the direction cosines of r?
Bedford
Falls
Bedford
Falls
45�
60�
O
x
y
r
z
A
Solution: The vector r is
r D jrj�sin 45°iC cos 45° sin 60°jC cos 45° cos 60°k�
The length of the pole is the x component of r. Therefore
jrj sin 45° D 6 ft) jrj D 6 ft
sin 45°
D 8.49 ft
(a) r D �6.00iC 5.20jC 3.00k� ft
(b) The direction cosines are
cos �x D rxjrj D 0.707, cos �y D
ry
jrj D 0.612, cos �z D
rz
jrj D 0.354
cos �x D 0.707, cos �y D 0.612, cos �z D 0.354
Problem 2.160 The z component of the force F is
80 lb. (a) Express F in terms of components. (b) what
are the angles �x, �y , and �z between F and the positive
coordinate axes?
y
z
x
A
F
20�
60�
O
Solution: We can write the force as
F D jFj�cos 20° sin 60°iC sin 20°jC cos 20° cos 60°k�
We know that the z component is 80 lb. Therefore
jFj cos 20° cos 60° D 80 lb) jFj D 170 lb
(a) F D �139iC 58.2jC 80k� lb
(b) The direction cosines can be found:
�x D cos�1
(
139
170
)
D 35.5°
�y D cos�1
(
58.2
170
)
D 70.0°
�z D cos�1
(
80
170
)
D 62.0°
�x D 35.5°, �y D 70.0°, �z D 62.0°
80
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Problem 2.161 The magnitude of the force vector FB
is 2 kN. Express it in terms of scalar components.
(4, 3, 1) m
y
x
(6, 0, 0) m
(5, 0, 3) mB
C
D
A
FA
FC
FB
F
z
Solution: The strategy is to determine the unit vector collinear
with FB and then express the force in terms of this unit vector.
The radius vector collinear with FB is
rBD D �4� 5�iC �3� 0�jC �1� 3�k or rBD D �1iC 3j� 2k.
The magnitude is
jrBDj D
p
12 C 32 C 22 D 3.74.
The unit vector is
eBD D rBDjrBDj D �0.2673iC 0.8018j� 0.5345k
The force is
FB D jFBjeBD D 2eBD (kN) FB D �0.5345iC 1.6036j � 1.0693k
D �0.53iC 1.60j� 1.07k (kN)
F
D (4,3,1)
C(6,0,0)
B (5,0,3)
x
z
y
A
FA
FB
FC
Problem 2.162 The magnitude of the vertical force
vector F in Problem 2.161 is 6 kN. Determine the vector
components of F parallel and normal to the line from B
to D.
Solution: The projection of the force F onto the line from B
to D is FP D �F Ð eBD�eBD. The vertical force has the component
F D �6j (kN). From Problem 2.139, the unit vector pointing from
B to D is eBD D �0.2673iC 0.8018j � 0.5345k. The dot product is
F Ð eBD D �4.813. Thus the component parallel to the line BD is FP D
�4.813eBD D C1.29i� 3.86jC 2.57k (kN). The component perpen-
dicular to the line is: FN D F� FP. Thus FN D �1.29i� 2.14j�
2.57k (kN)
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81
Problem 2.163 The magnitude of the vertical force
vector F in Problem 2.161 is 6 kN. Given that FC FA C
FB C FC D 0, what are the magnitudes of FA, FB, and
FC?
Solution: The strategy is to expand the forces into scalar compo-
nents, and then use the force balance equation to solve for the un-
knowns. The unit vectors are used to expand the forces into scalar
components. The position vectors, magnitudes, and unit vectors are:
rAD D 4iC 3jC 1k, jrADj D
p
26 D 5.1,
eAD D 0.7845iC 0.5883jC 0.1961k.
rBD D �1iC 3j� 2k, jrBDj D
p
14 D 3.74,
eBD D �0.2673iC 0.8018j� 0.5345k.
rCD D �2iC 3jC 1k, jrCDj D
p
14 D 3.74,
eCD D �0.5345iC 0.8018jC 0.2673k
The forces are:
FA D jFAjeAD,FB D jFBjeBD,FC D jFCjeCD,F D �6j (kN).
Substituting into the force balance equation
FC FA C FB C FC D 0,
�0.7843jFAj � 0.2674jFBj � 0.5348jFCj�i D 0
�0.5882jFAj C 0.8021jFBj C 0.8021jFCj � 6�j
D 0�0.1961jFAj � 0.5348jFBj C 0.2674jFCj�k D 0
These simple simultaneous equations can be solved a standard method
(e.g., Gauss elimination) or, conveniently, by using a commercial
package, such as TK Solver, Mathcad, or other. An HP-28S hand
held calculator was used here: jFAj D 2.83 (kN), jFBj D 2.49 (kN),
jFCj D 2.91 (kN)
Problem 2.164 The magnitude of the vertical force W
is 160 N. The direction cosines of the position vector from
A to B are cos �x D 0.500, cos �y D 0.866, and cos �z D
0, and the direction cosines of the position vector from
B to C are cos �x D 0.707, cos �y D 0.619, and cos �z D
�0.342. Point G is the midpoint of the line from B to C.
Determine the vector rAG ðW, where rAG is the position
vector from A to G.
x
y
W
600
 m
m
60
0 
m
m
G
C
B
A
z
Solution: Express the position vectors in terms of scalar compo-
nents, calculate rAG, and take the cross product. The position vectors
are: rAB D 0.6�.5iC 0.866jC 0k� rAB D 0.3iC 0.5196jC 0k,
rBG D 0.3�0.707iC 0.619j� 0.342k�,
rBG D 0.2121iC 0.1857j � 0.1026k.
rAG D rAB C rBG D 0.5121iC 0.7053j� 0.1026k.
W D �160j
rAG ðW D
∣∣∣∣∣∣
i j k
0.5121 0.7053 �0.1026
0 �160 0
∣∣∣∣∣∣
D �16.44iC 0j� 81.95k D �16.4iC 0j� 82k (N m)
A
B
G
W
C
600 mm
600 mm
x
82
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Problem 2.165 The rope CE exerts a 500-N force T
on the hinged door.
(a) Express T in terms of components.
(b) Determine the vector component of T parallel to
the line from point A to point B.
x
y
z
A (0.5, 0, 0) m
E
B
C
DT
(0, 0.2, 0) m
(0.35, 0, 0.2) m
(0.2, 0.4, �0.1) m
Solution: We have
rCE D �0.2iC 0.2j� 0.1k� m
T D �500 N� rCEjrCEj D �333iC 333j� 167k� N
(a) T D �333iC 333j� 167k� N
(b) We define the unit vector in the direction of AB and then use this
vector to find the component parallel to AB.
rAB D ��0.15iC 0.2k� m
eAB D rABjrABj D ��0.6iC 0.8k�
Tp D �eAB РT�eAB D �[�0.6][333 N]C [0.8][�167 N]���0.6iC 0.8k�
Tp D �200i� 267k� N
Problem 2.166 In Problem 2.165, let rBC be the posi-
tion vector from point B to point C. Determine the cross
product rBC ð T.
x
y
z
A (0.5, 0, 0) m
E
B
C
DT
(0, 0.2, 0) m
(0.35, 0, 0.2) m
(0.2, 0.4, �0.1) m
Solution: From Problem 2.165 we know that
T D �333iC 333j� 167k� N
The vector rBC is
rBC D ��035iC 0.2j� 0.2k� m
The cross product is
rBC ð T D
∣∣∣∣∣∣
i j k
�0.35 0.2 �0.2
333 333 �137
∣∣∣∣∣∣ D �33.3i� 125j� 183k� Nm
rBC ð T D �33.3i� 125j� 183k� Nm
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83
Problem 3.1 In Active Example 3.1, suppose that the
angle between the ramp supporting the car is increased
from 20° to 30°. Draw the free-body diagram of the car
showing the new geometry. Suppose that the cable from
A to B must exert a 1900-lb horizontal force on the car
to hold it in place. Determine the car’s weight in pounds.
A B
20�
Solution: The free-body diagram is shown to the right.
Applying the equilibrium equations
∑
Fx : T�N sin 30° D 0,
∑
Fy : N cos 30° � mg D 0
Setting T D 1900 lb and solving yields
N D 3800 lb, mg D 3290 lb
Problem 3.2 The ring weighs 5 lb and is in equilib-
rium. The force F1 D 4.5 lb. Determine the force F2 and
the angle ˛.
x
y
30�
F2
F1a
Solution: The free-body diagram is shown below the drawing. The
equilibrium equations are
∑
Fx : F1 cos 30° � F2 cos˛ D 0
∑
Fy : F1 sin 30° C F2 sin˛� 5 lb D 0
We can write these equations as
F2 sin˛ D 5 lb� F1 sin 30°
F2 cos˛ D F1 cos 30°
Dividing these equations and using the known value for F1 we have.
tan˛ D 5 lb� �4.5 lb� sin 30°
�4.5 lb� cos 30°D 0.706) ˛ D 35.2°
F2 D �4.5 lb� cos 30
°
cos˛
D 4.77 lb
F2 D 4.77 lb, ˛ D 35.2°
84
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Problem 3.3 In Example 3.2, suppose that the attach-
ment point C is moved to the right and cable AC is
extended so that the angle between cable AC and the
ceiling decreases from 45° to 35°. The angle between
cable AB and the ceiling remains 60°. What are the
tensions in cables AB and AC?
B
C
A
60�
45�
Solution: The free-body diagram is shown below the picture.
The equilibrium equations are:
∑
Fx : TAC cos 35° � TAB cos 60° D 0
∑
Fy : TAC sin 35° C TAB sin 60° � 1962 N D 0
Solving we find
TAB D 1610 N, TAC D 985 N
Problem 3.4 The 200-kg engine block is suspended
by the cables AB and AC. The angle ˛ D 40°. The free-
body diagram obtained by isolating the part of the system
within the dashed line is shown. Determine the forces
TAB and TAC.
B
C
A A
x
y
a a
TAB TAC
(200 kg) (9.81 m/s2)
Solution:
˛ D 40°
∑
Fx : TAC cos ˛� TAB cos˛ D 0
∑
Fy : TAC sin˛C TAB sin˛� 1962 N D 0
Solving: TAB D TAC D 1.526 kN
TAB TAC
α α
 1962 Ν
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85
Problem 3.5 A heavy rope used as a mooring line for
a cruise ship sags as shown. If the mass of the rope is
90 kg, what are the tensions in the rope at A and B?
B 40�
55� A
Solution: The free-body diagram is shown.
The equilibrium equations are
∑
Fx : TB cos 40° � TA cos 55° D 0
∑
Fy : TB sin 40° C TA sin 55° � 90�9.81� N D 0
Solving: TA D 679 N, TB D 508 N
Problem 3.6 A physiologist estimates that the
masseter muscle of a predator, Martes, is capable of
exerting a force M as large as 900 N. Assume that
the jaw is in equilibrium and determine the necessary
force T that the temporalis muscle exerts and the force
P exerted on the object being bitten.
T
22�
P
M
36�
Solution: The equilibrium equations are
∑
Fx : T cos 22° �M cos 36° D 0
∑
Fy : T sin 22° CM sin 36° � P D 0
Setting M D 900 N, and solving, we find
T D 785 N, P D 823 N
86
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Problem 3.7 The two springs are identical, with un-
stretched lengths 250 mm and spring constants k D
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?
B
A
300 mm
280 mm
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.
TUA D 0iC
(
1200
N
m
)
�0.3 m� 0.25 m�j D 0iC 60j N
Similarly, the tension in the lower spring acts on block A in the nega-
tive Y direction
TLA D 0i�
(
1200
N
m
)
�0.28 m� 0.25 m�j D 0i� 36j N
The weight is WA D 0i� jWAjj
The equilibrium conditions are
∑
F D
∑
Fx C
∑
Fy D 0,
∑
F DWA C TUA C TLA D 0
Collect and combine like terms in i, j
∑
Fy D ��jWAj C 60� 36�j D 0
Solve jWAj D �60� 36� D 24 N
The mass of A is
mA D jWL jjgj D
24 N
9.81 m/s2
D 2.45 kg
The free body diagram for block B is shown.
The tension in the lower spring TLB D 0iC 36j
The weight: WB D 0i� jWBjj
Apply the equilibrium conditions to block B.
∑
F DWB C TLB D 0
Collect and combine like terms in i, j:
∑
Fy D ��jWBj C 36�j D 0
Solve: jWBj D 36 N
The mass of B is given by mB D jWBjjgj D
36 N
9.81 m/s2
D 3.67 kg
B
A
300 mm
280 mm
A
Tension,
upper spring
Tension,
lower
spring
Weight,
mass A
B
Tension,
lower spring
Weight,
mass B
x
y
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87
Problem 3.8 The two springs in Problem 3.7 are iden-
tical, with unstretched lengths of 250 mm. Suppose that
their spring constant k is unknown and the sum of the
masses of blocks A and B is 10 kg. Determine the value
of k and the masses of the two blocks.
Solution: All of the forces are in the vertical direction so we will
use scalar equations. First, consider the upper spring supporting both
masses (10 kg total mass). The equation of equilibrium for block the
entire assembly supported by the upper spring is A is TUA � �mA C
mB�g D 0, where TUA D k��U � 0.25� N. The equation of equilibrium
for block B is TUB � mBg D 0, where TUB D k��L � 0.25� N. The
equation of equilibrium for block A alone is TUA C TLA � mAg D 0
where TLA D �TUB. Using g D 9.81 m/s2, and solving simultane-
ously, we get k D 1962 N/m, mA D 4 kg, and mB D 6 kg .
Problem 3.9 The inclined surface is smooth (Remem-
ber that “smooth” means that friction is negligble). The
two springs are identical, with unstretched lengths of
250 mm and spring constants k D 1200 N/m. What are
the masses of blocks A and B?
B
A
300 mm
280 mm
30�
Solution:
F1 D �1200 N/m��0.3� 0.25�m D 60 N
F2 D �1200 N/m��0.28� 0.25�m D 36 N
∑
FB &: �F2 C mBg sin 30° D 0
∑
FA &: �F1 C F2 C mAg sin 30° D 0
Solving: mA D 4.89 kg, mB D 7.34 kg
mAg
mBg
F1
F2
F2
NA
NB
88
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Problem 3.10 The mass of the crane is 20,000 kg. The
crane’s cable is attached to a caisson whose mass is
400 kg. The tension in the cable is 1 kN.
(a) Determine the magnitudes of the normal and
friction forces exerted on the crane by the
level ground.
(b) Determine the magnitudes of the normal and
friction forces exerted on the caisson by the
level ground.
Strategy: To do part (a), draw the free-body diagram
of the crane and the part of its cable within the
dashed line.
45°
Solution:
(a)
∑
Fy : Ncrane � 196.2 kN� 1 kN sin 45° D 0
∑
Fx : �Fcrane C 1 kN cos 45° D 0
Ncrane D 196.9 kN, Fcrane D 0.707 kN
(b)
∑
Fy : Ncaisson � 3.924 kNC 1 kN sin 45° D 0
∑
Fx : �1 kN cos 45° C Fcaisson D 0
Ncaisson D 3.22 kN, Fcaisson D 0.707 kN
Ncrane
Fcrane
196.2 kN
1 kN
x
y
45°
45°
1 kN
3.924 kN
Ncaisson
Fcaisson
Problem 3.11 The inclined surface is smooth. The
100-kg crate is held stationary by a force T applied to
the cable.
(a) Draw the free-body diagram of the crate.
(b) Determine the force T.
60�
T
Solution:
(a) The FBD
60°
 Ν
T
981 Ν
(b)
∑
F-: T� 981 N sin 60° D 0
T D 850 N
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89
Problem 3.12 The 1200-kg car is stationary on the
sloping road.
(a) If ˛ D 20°, what are the magnitudes of the total
normal and friction forces exerted on the car’s tires
by the road?
(b) The car can remain stationary only if the total
friction force necessary for equilibrium is not
greater than 0.6 times the total normal force.
What is the largest angle ˛ for which the car can
remain stationary?
a
Solution:
(a) ˛ D 20°
∑
F% : N� 11.772 kN cos ˛ D 0
∑
F- : F� 11.772 kN sin˛ D 0
N D 11.06 kN, F D 4.03 kN(b) F D 0.6 N
∑
F% : N� 11.772 kN cos ˛ D 0 ) ˛ D 31.0°
∑
F- : F� 11.772 kN sin˛ D 0
N
α
F
11.772 kN
Problem 3.13 The 100-lb crate is in equilibrium on the
smooth surface. The spring constant is k D 400 lb/ft. Let
S be the stretch of the spring. Obtain an equation for S
(in feet) as a function of the angle ˛.
a
Solution: The free-body diagram is shown.
The equilibrium equation in the direction parallel to the inclined
surface is
kS� �100 lb� sin˛ D 0
Solving for S and using the given value for k we find
S D �0.25 ft� sin˛
90
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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 ˛ D 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
A
B
Solution: Isolate the box. Resolve the forces into scalar compo-
nents, 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� ˛� (or 270C ˛). The weight vector is
W D jWj�i sin˛� j cos˛� D �600��i sin˛� j cos˛�
The projections of the rope tension and the normal force are
T D �jTxjiC 0j N D 0iC jNy jj
The equilibrium conditions are
∑
F DWC NC T D 0
Substitute, and collect like terms
∑
Fx D �600 sin ˛� jTx j�i D 0
∑
Fy D ��600 cos˛C jNy j�j D 0
Solve for the unknown tension when
For ˛ D 25°
jTxj D 600 sin˛ D 253.6 lb.
For a tension of 400 lb, (600 sin˛� 400� D 0. Solve for the unknown
angle
sin˛ D 400
600
D 0.667 or ˛ D 41.84°
α
A
B
T
N
W
y
x
α
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91
Problem 3.15 The 80-lb box is held in place on
the smooth inclined surface by the rope AB. Determine
the tension in the rope and the normal force exerted
on the box by the inclined surface.
A
B
30�
50�
Solution: The equilibrium equations (in terms of a coordinate
system with the x axis parallel to the inclined surface) are
∑
Fx : �80 lb� sin 50° � T cos 50 D 0
∑
Fx : N� �80 lb� cos 50° � T sin 50 D 0
Solving: T D 95.34 lb, N D 124 lb
Problem 3.16 The 1360-kg car and the 2100-kg tow
truck are stationary. The muddy surface on which the
car’s tires rest exerts negligible friction forces on them.
What is the tension in the tow cable?
18�10�
26�
Solution: FBD of the car being towed
∑
F- : T cos 8° � 13.34 kN sin 26° D 0
T D 5.91 kN
13.34 kN
T
N
18°
26°
92
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 3.17 Each box weighs 40 lb. The angles are
measured relative to the horizontal. The surfaces are
smooth. Determine the tension in the rope A and the
normal force exerted on box B by the inclined surface.
D
C
B
A
70�
20�
45�
Solution: The free-body diagrams are shown.
The equilibrium equations for box D are
∑
Fx : �40 lb� sin 20° � TC cos 25° D 0
∑
Fy : ND � �40 lb� cos 20° C TC sin 25° D 0
The equilibrium equations for box B are
∑
Fx : �40 lb� sin 70° C TC cos 25° � TA D 0
∑
Fy : NB � �40 lb� cos 70° C TC sin 25° D 0
Solving these four equations yields:
TA D 51.2 lb, TC D 15.1 lb, NB D 7.30 lb, ND D 31.2 lb
Thus TA D 51.2 lb, NB D 7.30 lb
Problem 3.18 A 10-kg painting is hung with a wire
supported by a nail. The length of the wire is 1.3 m.
(a) What is the tension in the wire?
(b) What is the magnitude of the force exerted on the
nail by the wire? 1.2 m
Solution:
(a)
∑
Fy : 98.1 N� 2 513T D 0
T D 128 N
(b) Force D 98.1 N
98.1 N
T T5
12 12
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93
Problem 3.19 A 10-kg painting is hung with a wire
supported by two nails. The length of the wire is 1.3 m.
(a) What is the tension in the wire?
(b) What is the magnitude of the force exerted on each
nail by the wire? (Assume that the tension is the
same in each part of the wire.)
Compare your answers to the answers to Problem 3.18.
0.4 m0.4 m 0.4 m
Solution:
(a) Examine the point on the left where the wire is attached to the
picture. This point supports half of the weight
∑
Fy : T sin 27.3° � 49.05 N D 0
T D 107 N
(b) Examine one of the nails∑
Fx : �Rx � T cos 27.3° C T D 0
∑
Fy : Ry � T sin 27.3° D 0
R D
√
Rx2 C Ry2
R D 50.5 N
27.3°R
T
49.05 N
Ry
Rx
T
T
27.3°
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� 15�
Solution:


∑
Fx D TR cos�15°�� TL cos�14°� D 0∑
Fy D TR sin�15°�C TL sin�14°�� 150 D 0
Solving, we get TL D 299 lb, TR D 300 lb
y
x
TR
TL
14° 15°
150 lb
94
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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 D TR cos�15°�� TL cos�14°� D 0∑
Fy D TR sin�15°�C TL sin�14°�� mg D 0
Solving, we get
TL D 1.56 kN, TR D 1.57 kN
14° 15°
TL
TR
y
x
mg = (80) (9.81) N
Problem 3.22 The construction worker exerts a 20-lb
force on the rope to hold the crate in equilibrium in the
position shown. What is the weight of the crate?
5�
30�
Solution: The free-body diagram is shown.
The equilibrium equations for the part of the rope system where the
three ropes are joined are
∑
Fx : �20 lb� cos 30° � T sin 5° D 0
∑
Fy : ��20 lb� sin 30° C T cos 5° �W D 0
Solving yields W D 188 lb
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95
Problem 3.23 A construction worker on the moon,
where the acceleration due to gravity is 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 to hold
the crate ub equilibrium (a) in newtons; (b) in pounds?
5�
30�
Solution: The free-body diagram is shown.
From Problem 3.22 we know that the weight is W D 188 lb. Therefore
its mass is
m D 188 lb
32.2 ft/s2
D 5.84 slug
m D 5.84 slug
(
14.59 kg
slug
)
D 85.2 kg
The equilibrium equations for the part of the rope system where the
three ropes are joined are
∑
Fx : F cos 30° � T sin 5° D 0
∑
Fy : �F sin 30° C T cos 5° � mgm D 0
where gm D 1.62 m/s2.
Solving yields F D 3.30 lb D 14.7 N
�a� F D 14.7 N, �b� F D 3.30 lb
96
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Problem 3.24 The person wants to cause the 200-lb
crate to start sliding toward the right. To achieve this,
the horizontal component of the force exerted on the
crate by the ropemust equal 0.35 times the normal force
exerted on the crate by the floor. In Fig.a, the person
pulls on the rope in the direction shown. In Fig.b, the
person attaches the rope to a support as shown and pulls
upward on the rope. What is the magnitude of the force
he must exert on the rope in each case?
10�
20�
(a)
(b)
Solution: The friction force Ffr is given by
Ffr D 0.35N
(a) For equilibrium we have
∑
Fx : T cos 20° � 0.35N D 0
∑
Fy : T sin 20° � 200 lbCN D 0
Solving: T D 66.1 lb
(b) The person exerts the force F. Using the free-body diagram of
the crate and of the point on the rope where the person grabs the
rope, we find
∑
Fx : TL � 0.35N D 0
∑
Fy : N� 200 lb D 0
∑
Fx : �TL C TR cos 10° D 0
∑
Fy : F� TR sin 10° D 0
Solving we find F D 12.3 lb
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97
Problem 3.25 A traffic engineer wants to suspend a
200-lb traffic light above the center of the two right
lanes of a four-lane thoroughfare as shown. Determine
the tensions in the cables AB and BC.
30 ft
CA
B
10 ft
20 ft
80 ft
Solution:
∑
Fx : � 6p
37
TAB C 2p
5
TBC D 0
∑
Fy :
1p
37
TAB C 1p
5
TBC � 200 lb D 0
Solving: TAB D 304 lb, TBC D 335 lb
6
1
2
1
TBC
TAB
200 lb
Problem 3.26 Cable AB is 3 m long and cable BC is
4 m long. The mass of the suspended object is 350 kg.
Determine the tensions in cables AB and BC.
C
B
A
5m
Solution:
∑
Fx : � 35TAB C
4
5
TBC D 0
∑
Fy :
4
5
TAB C 35TBC � 3.43 kN D 0
TAB D 2.75 kN, TBC D 2.06 kN
4
4
3
3
TAB
TAC
3.43 kN
98
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Problem 3.27 In Problem 3.26, the length of cable AB
is adjustable. If you don’t want the tension in either cable
AB or cable BC to exceed 3 kN, what is the minimum
acceptable length of cable AB?
Solution: Consider the geometry:
We have the constraints
LAB
2 D x2 C y2, �4 m�2 D �5 m� x�2 C y2
These constraint imply
y D
√
�10 m�x � x2 � 9 m2
L D
√
�10 m�x � 9 m2
Now draw the FBD and write the equations in terms of x
∑
Fx : � xp
10x � 9TAB C
5� x
4
TBC D 0
∑
Fy :
p
10x � x2 � 9p
10x � 9 TAB C
p
10x � x2 � 9
4
TBC � 3.43 kN D 0
If we set TAB D 3 kN and solve for x we find x D 1.535, TBC D
2.11 kN < 3 kN
Using this value for x we find that LAB D 2.52 m
x
y
5−x
4 m
LAB
TAB
TBC
x
5−x
y
y
4
3.43 kN
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99
Problem 3.28 What are the tensions in the upper and
lower cables? (Your answers will be in terms of W.
Neglect the weight of the pulley.)
45° 30°
W
Solution: Isolate the weight. The frictionless pulley changes the
direction but not the magnitude of the tension. The angle between the
right hand upper cable and the x axis is ˛, hence
TUR D jTUj�i cos˛C j sin˛�.
The angle between the positive x and the left hand upper pulley is
�180° � ˇ�, hence
TUL D jTUj�i cos�180� ˇ�C j sin�180� ˇ��
D jTUj��i cosˇC j sinˇ�.
The lower cable exerts a force: TL D �jTLjiC 0j
The weight: W D 0i� jWjj
The equilibrium conditions are
∑
F DWC TUL C TUR C TL D 0
Substitute and collect like terms,
∑
Fx D ��jTUj cosˇ C jTUj cos ˛� jTL j�i D 0
∑
Fy D �jTUj sin ˛C jTUj sin ˇ� jWj�j D 0.
Solve: jTUj D
( jWj
�sin˛C sinˇ�
)
,
jTL j D jTUj�cos˛� cosˇ�.
From which jTL j D jWj
(
cos˛� cosˇ
sin˛C sinˇ
)
.
For ˛ D 30° and ˇ D 45°
jTUj D 0.828jWj,
jTL j D 0.132jWj
TU
TL
TU
y
x
W
β α
100
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Problem 3.29 Two tow trucks lift a 660-lb motorcycle
out of a ravine following an accident. If the motorcycle
is in equilibrium in the position shown, what are the
tensions in cables AB and AC? B
A
C
(12, 32) ft
(36, 36) ft
(26, 16) ft
x
y
Solution: The angles are
˛ D tan�1
(
32� 16
26� 12
)
D 48.8°
ˇ D tan�1
(
36� 16
36� 26
)
D 63.4°
Now from equilibrium we have
∑
Fx : �TAB cos˛C TAC cos ˇ D 0
∑
Fy : TAB sin˛C TAC sinˇ � 660 lb D 0
Solving yields TAB D 319 lb, TAC D 470 lb
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101
Problem 3.30 An astronaut candidate conducts exper-
iments on an airbearing platform. While she carries out
calibrations, the platform is held in place by the hori-
zontal 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?
3.0 m 1.5 m
B C
TOP VIEW
D
4.0 m
3.5 m
A
Solution: Isolate the platform. The angles ˛ and ˇ are
tan ˛ D
(
1.5
3.5
)
D 0.429, ˛ D 23.2°.
Also, tanˇ D
(
3.0
3.5
)
D 0.857, ˇ D 40.6°.
The angle between the tether AB and the positive x axis is �180° � ˇ�,
hence
TAB D jTABj�i cos�180° � ˇ�C j sin�180° � ˇ��
TAB D jTABj��i cosˇ C j sinˇ�.
The angle between the tether AC and the positive x axis is �180° C ˛�.
The tension is
TAC D jTACj�i cos�180° C ˛�C j sin�180° C ˛��
D jTACj��i cos˛� j sin˛�.
The tether AD is aligned with the positive x axis, TAD D jTADjiC 0j.
The equilibrium condition:
∑
F D TAD C TAB C TAC D 0.
Substitute and collect like terms,
∑
Fx D ��jTABj cos ˇ� jTACj cos˛C jTADj�i D 0,
∑
Fy D �jTABj sinˇ � jTACj sin˛�j D 0.
3.5
m
1.5 m
3.0 m
4.0
m
A
B
C
D
B
β
α
A
C
D
x
y
Solve: jTABj D
(
sin˛
sinˇ
)
jTACj,
jTADj D
( jTACj sin�˛C ˇ�
sinˇ
)
.
For jTACj D 2 N, ˛ D 23.2° and ˇ D 40.6°,
jTABj D 1.21 N, jTADj D 2.76 N
102
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Problem 3.31 The bucket contains concrete and
weighs 5800 lb. What are the tensions in the cables AB
and AC?
B C
A
(20, 34) ft(5, 34) ft
(12, 16) ft
y
x
Solution: The angles are
˛ D tan�1
(
34� 16
12� 5
)
D 68.7°
ˇ D tan�1
(
34� 16
20� 12
)
D 66.0°
Now from equilibrium we have
∑
Fx : �TAB cos˛C TAC cos ˇ D 0
∑
Fy : TAB sin˛C TAC sinˇ � 660 lb D 0
Solving yields TAB D 319 lb, TAC D 470 lb
Problem 3.32 The slider A is in equilibrium and the
bar is smooth. What is the mass of the slider? 20�
45�
200 NA
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 D T sin 20° CN cos 45° D 0 �T D 200 N�∑
Fy D N sin 45° C T cos 20° � mg D 0 g D 9.81 m/s2
Substituting for T and g, we have two eqns in two unknowns
(N and m).
Solving, we get N D �96.7 N, m D 12.2 kg.
y
45°
20°
N
x
T = 200 N
mg = (9.81) g
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103
Problem 3.33 The 20-kg mass is suspended from threecables. Cable AC is equipped with a turnbuckle so that
its tension can be adjusted and a strain gauge that allows
its tension to be measured. If the tension in cable AC is
40 N, what are the tensions in cables AB and AD?
B C
A
D
0.4 m0.4 m 0.48 m
0.64 m
Solution:
TAC D 40 N
∑
Fx : � 5p
89
TAB C 5p
89
TAC C 11p
185
TAD D 0
∑
Fy :
8p
89
TAB C 8p
89
TAC C 8p
185
TAD � 196.2 N D 0
Solving: TAB D 144.1 N, TAD D 68.2 N
TAB
TAC
TAD
11
8
8 8
5
5
196.2 N
Problem 3.34 The structural joint is in equilibrium. If
FA D 1000 lb and FD D 5000 lb, what are FB and FC?
FB
FC
FA FD
80�
65�
35�
Solution: The equilibrium equations are
∑
Fx : FD � FC cos 65° � FB cos 35° � FA D 0
∑
Fy : �FC sin 65° C FB sin 35° D 0
Solving yields FB D 3680 lb, FC D 2330 lb
104
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Problem 3.35 The collar A slides on the smooth
vertical bar. The masses mA D 20 kg and mB D 10 kg.
When h D 0.1 m, the spring is unstretched. When the
system is in equilibrium, h D 0.3 m. Determine the
spring constant k.
B
A
h
k
0.25 m
Solution: The triangles formed by the rope segments and the hori-
zontal line level with A can be used to determine the lengths Lu and
Ls. The equations are
Lu D
√
�0.25�2 C �0.1�2 and Ls D
√
�0.25�2 C �0.3�2.
The stretch in the spring when in equilibrium is given by υ D Ls � Lu.
Carrying out the calculations, we get Lu D 0.269 m, Ls D 0.391 m,
and υ D 0.121 m. The angle, �, between the rope at A and the hori-
zontal when the system is in equilibrium is given by tan � D 0.3/0.25,
or � D 50.2°. From the free body diagram for mass A, we get two
equilibrium equations. They are
∑
Fx D �NA C T cos � D 0
and
∑
Fy D T sin � � mAg D 0.
We have two equations in two unknowns and can solve. We get NA D
163.5 N and T D 255.4 N. Now we go to the free body diagram for B,
where the equation of equilibrium is T� mBg� kυ D 0. This equation
has only one unknown. Solving, we get k D 1297 N/m
Lu
Lu
Ls
0.1 m
0.3 m
0.25 m
0.25 m
NA A
T
mAg
mBg
T
B
Kδ
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105
Problem 3.36* Suppose that you want to design a
cable system to suspend an object of weight W from
the ceiling. The two wires must be identical, and the
dimension b is fixed. The ratio of the tension T in each
wire to its cross-sectional area A must equal a specified
value T/A D �. The “cost” of your design is the total
volume of material in the two wires, V D 2Apb2 C h2.
Determine the value of h that minimizes the cost.
W
b
h
b
Solution: From the equation
∑
Fy D 2T sin � �W D 0,
we obtain T D W
2 sin �
D W
p
b2 C h2
2h
.
Since T/A D �, A D T
�
D W
p
b2 C h2
2�h
and the “cost” is V D 2Apb2 C h2 D W�b
2 C h2�
�h
.
To determine the value of h that minimizes V, we set
dV
dh
D W
�
[
� �b
2 C h2�
h2
C 2
]
D 0
and solve for h, obtaining h D b.
T T
W
θ θ
106
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Problem 3.37 The system of cables suspends a
1000-lb bank of lights above a movie set. Determine
the tensions in cables AB, CD, and CE.
D
C
A
E
45° 30°
20 ft
B
18 ft
Solution: Isolate juncture A, and solve the equilibrium equations.
Repeat for the cable juncture C.
The angle between the cable AC and the positive x axis is ˛. The
tension in AC is TAC D jTACj�i cos˛C j sin˛�
The angle between the x axis and AB is �180° � ˇ�. The tension is
TAB D jTABj�i cos�180� ˇ�C j sin�180� ˇ��
TAB D ��i cosˇ C j sin ˇ�.
The weight is W D 0i� jWjj.
The equilibrium conditions are
∑
F D 0 DWC TAB C TAC D 0.
Substitute and collect like terms,
∑
Fx D �jTACj cos˛� jTABj cosˇ�i D 0
∑
Fy D �jTABj sinˇ C jTACj sin ˛� jWj�j D 0.
Solving, we get
jTABj D
(
cos˛
cos ˇ
)
jTACj and jTACj D
( jWj cosˇ
sin�˛C ˇ�
)
,
jWj D 1000 lb, and ˛ D 30°, ˇ D 45°
jTACj D �1000�
(
0.7071
0.9659
)
D 732.05 lb
jTABj D �732�
(
0.866
0.7071
)
D 896.5 lb
Isolate juncture C. The angle between the positive x axis and the cable
CA is �180° � ˛�. The tension is
TCA D jTCAj�i cos�180° C ˛�C j sin�180° C ˛��,
or TCA D jTCAj��i cos ˛� j sin ˛�.
The tension in the cable CE is
TCE D ijTCEj C 0j.
The tension in the cable CD is TCD D 0iC jjTCDj.
The equilibrium conditions are
∑
F D 0 D TCA C TCE C TCD D 0
Substitute t and collect like terms,
∑
Fx D �jTCEj � jTCAj cos˛�i D 0,
∑
Fy D �jTCDj � jTCAj sin˛�j D 0.
Solve: jTCEj D jTCAj cos ˛,
jTCDj D jTCAj sin ˛;
for jTCAj D 732 lb and ˛ D 30°,
jTABj D 896.6 lb,
jTCEj D 634 lb,
jTCDj D 366 lb
B
C
A
W
y
x
β α
D
C
A
α
E
y
x
90°
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107
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 D 20 ft
cos 45°
D 28.284 ft.
The length AC is given by
AC D 18 ft
cos 30°
D 20.785 ft.
The altitude of the triangle for which AC is the hypotenuse is
18 tan 30° D 10.392 ft. The distance CD is given by 20� 10.392 D
9.608 ft.
The distance AD is given by
AD D ACCCD D 20.784C 9.608 D 30.392
The new angles are given by the cosine law
AB2 D 382 C AD2 � 2�38��AD� cos˛.
Reduce and solve:
cos˛ D
(
382 C �30.392�2 � �28.284�2
2�38��30.392�
)
D 0.6787, ˛ D 47.23°.
cosˇ D
(
�28.284�2 C �38�2 � �30.392�2
2�28.284��38�
)
D 0.6142, ˇ D 52.1°.
Isolate the juncture A. The angle between the cable AD and the positive
x axis is ˛. The tension is:
TAD D jTADj�i cos˛C j sin˛�.
The angle between x and the cable AB is �180° � ˇ�. The tension is
TAB D jTABj��i cosˇ C j sinˇ�.
The weight is W D 0i� jWjj
The equilibrium conditions are
∑
F D 0 DWC TAB C TAD D 0.
Substitute and collect like terms,
∑
Fx D �jTADj cos˛� jTABj cosˇ�i D 0,
∑
Fy D �jTABj sinˇ C jTADj sin˛� jWj�j D 0.
20 ft 18 ft
B
C
A
D
αβ
B
B
W
y
x
A
A
D
D
38
β α
α
β
β
α
28.284 20.784+9.608
= 30.392
Solve: jTABj D
(
cos˛
cosˇ
)
jTADj,
and jTADj D
( jWj cosˇ
sin�˛C ˇ�
)
.
For jWj D 1000 lb, and ˛ D 51.2°, ˇ D 47.2°
jTADj D �1000�
(
0.6142
0.989
)
D 621.03 lb,
jTABj D �622.3�
(
0.6787
0.6142
)
D 687.9 lb
108
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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.
A
B
C
D
70°
50°
30°
Solution: Isolate each cable juncture,beginning with A and solve
the equilibrium equations at each juncture. The angle between the
cable AB and the positive x axis is ˛ D 70°; the tension in cable AB
is TAB D jTABj�i cos ˛C j sin ˛�. The weight is W D 0i� jWjj. The
tension in cable AT is T D �jTjiC 0j. The equilibrium conditions are
∑
F DWC TC TAB D 0.
Substitute and collect like terms
∑
Fx�jTABj cos˛� jTj�i D 0,
∑
Fy D �jTABj sin˛� jWj�j D 0.
Solve: the tension in cable AB is jTABj D
( jWj
sin˛
)
.
For jWj D �1250 kg�
(
9.81
m
s2
)
D 12262.5 N and ˛ D 70°
jTABj D
(
12262.5
0.94
)
D 13049.5 N
Isolate juncture B. The angles are ˛ D 50°, ˇ D 70°, and the tension
cable BC is TBC D jTBCj�i cos˛C j sin˛�. The angle between the
cable BA and the positive x axis is �180C ˇ�; the tension is
TBA D jTBAj�i cos�180C ˇ�C j sin�180C ˇ��
D jTBAj��i cosˇ � j sinˇ�
The tension in the left horizontal cable is T D �jTjiC 0j. The equi-
librium conditions are
∑
F D TBA C TBC C T D 0.
Substitute and collect like terms
∑
Fx D �jTBCj cos˛� jTBAj cos ˇ� jTj�i D 0
∑
Fy D �jTBCj sin˛� jTBAj sin ˇ�j D 0.
Solve: jTBCj D
(
sinˇ
sin˛
)
jTBAj.
For jTBAj D 13049.5 N, and ˛ D 50°, ˇ D 70°,
jTBCj D �13049.5�
(
0.9397
0.7660
)
D 16007.6 N
Isolate the cable juncture C. The angles are ˛ D 30°, ˇ D 50°. By
symmetry with the cable juncture B above, the tension in cable CD is
jTCDj D
(
sinˇ
sin˛
)
jTCBj.
Substitute: jTCDj D �16007.6�
(
0.7660
0.5
)
D 24525.0 N.
This completes the problem solution.
y
x
T
B
A
C
α
β
y
x
T
W
B
A α
y
x
C
B
T
D
α
β
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
109
Problem 3.40 A truck dealer wants to suspend a 4000-
kg truck as shown for advertising. The distance b D
15 m, and the sum of the lengths of the cables AB and
BC is 42 m. Points A and C are at the same height. What
are the tensions in the cables?
CA
b
40 m
B
Solution: Determine the dimensions and angles of the cables. Iso-
late the cable juncture B, and solve the equilibrium conditions. The
dimensions of the triangles formed by the cables:
b D 15 m, L D 25 m, ABC BC D S D 42 m.
Subdivide into two right triangles with a common side of unknown
length. Let the unknown length of this common side be d, then by the
Pythagorean Theorem b2 C d2 D AB2, L2 C d2 D BC2.
Subtract the first equation from the second to eliminate the unknown d,
L2 � b2 D BC2 � AB2.
Note that BC2 � AB2 D �BC� AB��BCC AB�.
Substitute and reduce to the pair of simultaneous equations in the
unknowns
BC� AB D
(
L2 � b2
S
)
, BCC AB D S
Solve: BC D
(
1
2
) (
L2 � b2
S
C S
)
D
(
1
2
) (
252 � 152
42
C 42
)
D 25.762 m
and AB D S� BC D 42� 25.762 D 16.238 m.
The interior angles are found from the cosine law:
cos˛ D
(
�L C b�2 C BC2 � AB2
2�L C b��BC�
)
D 0.9704 ˛ D 13.97°
cosˇ D
(
�L C b�2 C AB2 � BC2
2�L C b��AB�
)
D 0.9238 ˇ D 22.52°
Isolate cable juncture B. The angle between BC and the positive x axis
is ˛; the tension is
TBC D jTBCj�i cos˛C j sin˛�
The angle between BA and the positive x axis is �180° � ˇ�; the
tension is
TBA D jTBAj�i cos�180� ˇ�C j sin�180� ˇ��
D jTBAj��i cosˇ C j sinˇ�.
The weight is W D 0i� jWjj.
The equilibrium conditions are
∑
F DWC TBA C TBC D 0.
15 m 25 m
b
A
A
C
B
B
W
C
y
x
L
α
α
β
β
Substitute and collect like terms
∑
Fx D �jTBCj cos ˛� jTBAj cosˇ�i D 0,
∑
Fy D �jTBCj sin ˛C jTBAj sinˇ � jWj�j D 0
Solve: jTBCj D
(
cosˇ
cos˛
)
jTBAj,
and jTBAj D
( jWj cos˛
sin�˛C ˇ�
)
.
For jWj D �4000��9.81� D 39240 N,
and ˛ D 13.97°, ˇ D 22.52°,
jTBAj D 64033 D 64 kN,
jTBCj D 60953 D 61 kN
110
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Problem 3.41 The distance h D 12 in, and the tension
in cable AD is 200 lb. What are the tensions in cables
AB and AC?
12 in.
12 in.
12 in.
8 in.
8 in.
h
D
B
A
C
Solution: Isolated the cable juncture. From the sketch, the angles
are found from
tan ˛ D
(
8
12
)
D 0.667 ˛ D 33.7°
tanˇ D
(
4
12
)
D 0.333 ˇ D 18.4°
The angle between the cable AB and the positive x axis is �180° � ˛�,
the tension in AB is:
TAB D jTABj�i cos�180� ˛�C j sin�180� ˛��
TAB D jTABj��i cos ˛C j sin ˛�.
The angle between AC and the positive x axis is �180C ˇ�. The
tension is
TAC D jTACj�i cos�180C ˇ�C j sin�180C ˇ��
TAC D jTACj��i cos ˇ� j sinˇ�.
The tension in the cable AD is
TAD D jTADjiC 0j.
The equilibrium conditions are
∑
F D TAC C TAB C TAD D 0.
Substitute and collect like terms,
∑
Fx D ��jTABj cos˛� jTACj cosˇ C jTADj�i D 0
∑
Fy D �jTABj sin˛� jTACj sinˇ�j D 0.
Solve: jTABj D
(
sinˇ
sin˛
)
jTACj,
and jTACj D
(
sin˛
sin�˛C ˇ�
)
jTADj.
For jTADj D 200 lb, ˛ D 33.7°, ˇ D 18.4°
jTACj D 140.6 lb, jTABj D 80.1 lb
y
x
C
D
B
A
α
β
12 in
8 in
4 in
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111
Problem 3.42 You are designing a cable system to
support a suspended object of weight W. Because your
design requires points A and B to be placed as shown,
you have no control over the angle ˛, but you can choose
the angle ˇ by placing point C wherever you wish. Show
that to minimize the tensions in cables AB and BC, you
must choose ˇ D ˛ if the angle ˛ ½ 45°.
Strategy: Draw a diagram of the sum of the forces
exerted by the three cables at A.
W
CB
A
βα
Solution: Draw the free body diagram of the knot at point A. Then
draw the force triangle involving the three forces. Remember that ˛ is
fixed and the force W has both fixed magnitude and direction. From
the force triangle, we see that the force TAC can be smaller than
TAB for a large range of values for ˇ. By inspection, we see that the
minimum simultaneous values for TAC and TAB occur when the two
forces are equal. This occurs when ˛ D ˇ. Note: this does not happen
when ˛ < 45°.
In this case, we solved the problem without writing the equations of
equilibrium. For reference, these equations are:
∑
Fx D �TAB cos˛C TAC cos ˇ D 0
and
∑
Fy D TAB sin ˛C TAC sinˇ �W D 0.
W
A
B
x
y
α
TAB
TAC
Possible locations
for C lie on line
B
W
α
Candidate β
C? C?
TAB
Candidate values
for TAC
Fixed direction for
line AB
Problem 3.43* The length of the cable ABC is 1.4 m.
The 2-kN force is applied to a small pulley. The system
is stationary. What is the tension in the cable?
C
B
A
1 m
0.75 m
15�
2 kN
Solution: Examine the geometry
√
h2 C �0.75 m�2 C
√
h2 C �0.25 m�2 D 1.4 m
tan˛ D h
0.75 m
, tan ˇ D h
0.25 m
) h D 0.458 m, ˛ D 31.39°, ˇ D 61.35°
Now draw a FBD and solve for the tension. We can use either of the
equilibrium equations
∑
Fx : �T cos˛C T cosˇC �2 kN� sin 15° D 0
T D 1.38 kN
βα
0.75 m 0.25 m
h
βα
2 kN
15°
T
T
112
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 3.44 The masses m1 D 12 kg and m2 D 6 kg
are suspended by the cable system shown. The cable BC
is horizontal. Determine the angle ˛ and the tensions in
the cables AB, BC, and CD.
C
A
α
B
D
70�
m2
m1
117.7 N
TAB
TBCα
 B
Solution: We have 4 unknowns and 4 equations
∑
FBx : �TAB cos˛C TBC D 0
∑
FBy : TAB sin˛� 117.7 N D 0
∑
FCx : �TBC C TCD cos 70° D 0
∑
FCy : TCD sin 70° � 58.86 N D 0
Solving we find
˛ D 79.7°, TAB D 119.7 N, TBC D 21.4N, TCD D 62.6 N
70°
TCD
TBC
58.86 N
C
Problem 3.45 The weights W1 D 50 lb and W2 are
suspended by the cable system shown. Determine the
weight W2 and the tensions in the cables AB, BC,
and CD.
A D
B
C
W1
W2
30 in 30 in 30 in
20 in 16 in
Solution: We have 4 unknowns and 4 equilibrium equations to use
∑
FBx : � 3p
13
TAB C 15p
229
TBC D 0
∑
FBy :
2p
13
TAB C 2p
229
TBC � 50 lb D 0
∑
FCx : � 15p
229
TBC C 15
17
TCD D 0
∑
FCy : � 2p
229
TBC C 8
17
TCD �W2 D 0
)
W2 D 25 lb, TAB D 75.1 lb
TBC D 63.1 lb, TCD D 70.8 lb
TAB
TBC
50 lb
B
2
3
2
15
TCD
TBC
W2
2 
15
8
15 
C
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113
Problem 3.46 In the system shown in Problem 3.45,
assume that W2 DW1/2. If you don’t want the tension
anywhere in the supporting cable to exceed 200 lb, what
is the largest acceptable value of W1?
Solution:
∑
FBx : � 3p
13
TAB C 15p
229
TBC D 0
∑
FBy :
2p
13
TAB C 2p
229
TBC �W1 D 0
∑
FCx : � 15p
229
TBC C 15
17
TCD D 0
∑
FCy : � 2p
229
TBC C 817TCD �
W1
2
D 0
TAB D 1.502W1, TBC D 1.262W1, TCD D 1.417W1
AB is the critical cable
200 lb D 1.502W1 )W1 D 133.2 lb
W1
TAB
TBC
 B
2
2
15
3
C
TCD
W2 = W1/2
15 
2 
15 
8 
TBC
114
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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.
1 m
0.6 m Scoop
A B
D
C
0.15 m
0.6 m
1 m
Hydraulic�
cylinder
Solution: From the figure, if C is at the origin, then points A, B,
and D are located at
A�0.15,�0.6�
B�0.75,�0.6�
D�1.00, 0.4�
and forces FCA, FBC, and FCD are parallel to CA, BC, and CD, respec-
tively.
We need to write unit vectors in the three force directions and express
the forces in terms of magnitudes and unit vectors. The unit vectors
are given by
eCA D rCAjrCAj D 0.243i� 0.970j
eCB D rCBjrCBj D 0.781i� 0.625j
eCD D rCDjrCDj D 0.928iC 0.371j
Now we write the forces in terms of magnitudes and unit vectors. We
can write FBC as FCB D �8eCB kN or as FCB D 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 D �6.25iC 5.00j kN
Similarly,
FCA D 0.243FCAi� 0.970FCAj
FCD D 0.928FCDiC 0.371FCDj
For equilibrium, FCA C FCB C FCD D 0
In component form, this gives


∑
Fx D 0.243FCA C 0.928FCD � 6.25 (kN) D 0∑
Fy D �0.970FCA C 0.371FCD C 5.00 (kN) D 0
Solving, we get
FCA D 7.02 kN, FCD D 4.89 kN
y
C
x
D
B
A
FBC
FCD
FCA
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115
Problem 3.48 The 50-lb cylinder rests on two smooth
surfaces.
(a) Draw the free-body diagram of the cylinder.
(b) If ˛ D 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 �90C ˇ�. The normal force is
NR D jNRj�i cos�90C ˇ�C j sin�90C ˇ��
NR D jNRj��i sinˇ C j cos ˇ�.
The angle between the positive x axis and the left hand force is normal
�90� ˛�; the normal force is NL D jNL j�i sin˛C j cos˛�. The weight
is W D 0i� jWjj. The equilibrium conditions are
∑
F DWC NR C NL D 0.
Substitute and collect like terms,
∑
Fx D ��jNRj sinˇ C jNL j sin˛�i D 0,
∑
Fy D �jNRj cosˇ C jNL j cos˛� jWj�j D 0.
y
x
W
βα
NL
NR
Solve: jNRj D
(
sin˛
sinˇ
)
jNL j,
and jNLj D
( jWj sinˇ
sin�˛C ˇ�
)
.
For jWj D 50 lb, and ˛ D 30°, ˇ D 45°, the normal forces are
jNL j D 36.6 lb, jNRj D 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 ˛ 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).
jNRj D
(
sin˛
sinˇ
)
jNL j jNL j D
( jWj sin ˇ
sin�˛C ˇ�
)
.
Part (b): The isolated cylinder with the coordinate system is shown.
The angle between the right hand normal force and the positive x axis
is 180°. The normal force: NR D �jNRjiC 0j. The angle between the
left hand normal force and the positive x is 180� �˛C ˇ�. The normal
force is NL D jNL j��i cos�˛C ˇ�C j sin�˛C ˇ��.
The angle between the weight vector and the positive x axis is �ˇ.
The weight vector is W D jWj�i cosˇ � j sinˇ�.
The equilibrium conditions are
∑
F DWC NR C NL D 0.
y
x
α β
NL
NR
W
Substitute and collect like terms,
∑
Fx D ��jNRj � jNL j cos�˛C ˇ�C jWj cosˇ�i D 0,
∑
Fy D �jNL j sin�˛C ˇ�� jWj sinˇ�j D 0.
Solve: jNL j D
( jWj sinˇ
sin�˛C ˇ�
)
116
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Problem 3.50 The two springs are identical, with
unstretched length 0.4 m. When the 50-kg mass is
suspended at B, the length of each spring increases to
0.6 m. What is the spring constant k?
C
B
A
k
k
0.6 m
Solution:
F D k�0.6 m� 0.4 m�
∑
Fy : 2F sin 60° � 490.5 N D 0
k D 1416 N/m
60° 60°
FF
490.5 N
Problem 3.51 The cable AB is 0.5 m in length. The
unstretched length of the spring is 0.4 m. When the
50-kg mass is suspended at B, the length of the spring
increases to 0.45 m. What is the spring constant k?
C
B
A
k
0.7 m
Solution: The Geometry
Law of Cosines and Law of Sines
0.72 D 0.52 C 0.452 � 2�0.5��0.45� cosˇ
sin �
0.45 m
D sin�
0.5 m
D sinˇ
0.7 m
ˇ D 94.8°, � D 39.8° � D 45.4°
Now do the statics
F D k�0.45 m� 0.4 m�
∑
Fx : �TAB cos � C F cos� D 0
∑
Fy : TAB sin � C F sin� � 490.5 N D 0
Solving: k D 7560 N/m
θ φ
β
0.45 m0.5 m
0.7 m
θ φ
490.5 N
FTAB
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117
Problem 3.52 The small sphere of mass m is attached
to a string of length L and rests on the smooth surface
of a fixed sphere of radius R. The center of the sphere
is directly below the point where the string is attached.
Obtain an equation for the tension in the string in terms
of m, L, h, and R.
R
h L
m
Solution: From the geometry we have
cos � D RC h� y
L
, sin � D x
L
cos� D y
R
, sin� D x
R
Thus the equilibrium equations can be written
∑
Fx : � x
L
TC x
R
N D 0
∑
Fy :
RC h� y
L
TC y
R
N� mg D 0
Solving, we find T D mgL
RC h
Problem 3.53 The inclined surface is smooth. Deter-
mine the force T that must be exerted on the cable to
hold the 100-kg crate in equilibrium and compare your
answer tothe answer of Problem 3.11.
T
60�
Solution: ∑
F- : 3 T� 981 N sin 60° D 0
T D 283 N
60°
N
981 N
3T
118
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Problem 3.54 In Example 3.3, suppose that the mass
of the suspended object is mA and the masses of the
pulleys are mB D 0.3mA, mC D 0.2mA, and mD D 0.2mA.
Show that the force T necessary for the system to be in
equilibrium is 0.275mAg.
T
A
BBB
DD
CC
Solution: From the free-body diagram of pulley C
TD � 2T� mCg D 0) TD D 2TC mCg
Then from the free-body diagram of pulley B
TC TC 2TC mCg� mBg� mAg D 0
Thus
T D 1
4
�mA C mB � mC�g
T D 1
4
�mA C 0.3mA � 0.2mA�g D 0.275mAg
T D 0.275mAg
T
T T
TD
T T
(a)
TD � 2T � mg
(b)
mAg
A
D
CC
C
B
B
mg
mg
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119
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.
T
A
Solution: Draw free body diagrams of each pulley and the object
A. Each pulley and the object A must be in equilibrium. The weights
of the pulleys and object A are W D mg and WA D mAg. The equilib-
rium equations for the weight A, the lower pulley, second pulley, third
pulley, and the top pulley are, respectively, B�WA D 0, 2C� B�
W D 0, 2D �C�W D 0, 2T� D�W D 0, and FS � 2T�W D 0.
Begin with the first equation and solve for B, substitute for B in the
second equation and solve for C, substitute for C in the third equation
and solve for D, and substitute for D in the fourth equation and solve
for T, to get T in terms of W and WA. The result is
B DWA, C D WA
2
C W
2
,
D D WA
4
C 3W
4
, and T D WA
8
C 7W
8
,
or in terms of the masses,
T D g
8
�mA C 7m�.
Fs
WA
T
T T
T
WW
W
C
CC
B
B
W
D
DD
120
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Problem 3.56 The suspended mass m1 D 50 kg. Neg-
lecting the masses of the pulleys, determine the value
of the mass m2 necessary for the system to be in
equilibrium.
A
B
C
m1
m2
Solution:∑
FC : T1 C 2m2g� m1g D 0
∑
FB : T1 � 2m2g D 0
m2 D m14 D 12.5 kg
B
T1
T = m2 g
T
m1 g
T
T1
T
C
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
121
Problem 3.57 The boy is lifting himself using the
block and tackle shown. If the weight of the block and
tackle is negligible, and the combined weight of the boy
and the beam he is sitting on is 120 lb, what force
does he have to exert on the rope to raise himself at
a constant rate? (Neglect the deviation of the ropes from
the vertical.)
Solution: A free-body diagram can be obtained by cutting the four
ropes between the two pulleys of the block and tackle and the rope
the boy is holding. The tension has the same value T in all five of
these ropes. So the upward force on the free-body diagram is 5T and
the downward force is the 120-lb weight. Therefore the force the boy
must exert is
T D �120 lb�/5 D 24 lb
T D 24 lb
122
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Problem 3.58 Pulley systems containing one, two, and
three pulleys are shown. Neglecting the weights of the
pulleys, determine the force T required to support the
weight W in each case.
T T T
W
W
W
(a) One pulley
(b) Two pulleys
(c) Three pulleys
Solution:
(a)
∑
Fy : 2T�W D 0) T D W
2
(b)
∑
Fupper : 2T� T1 D 0
∑
Flower : 2T1 �W D 0
T D W
4
(c)
∑
Fupper : 2T� T1 D 0
∑
Fmiddle : 2T1 � T2 D 0
∑
Flower : 2T2 �W D 0
T D W
8
(a) For one pulleys
W
T T
(b) For two pulleys
W 
T
T
T1 T1
(c) For three pulleys
T
T
T1T1
T2
T2
T2
W
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123
Problem 3.59 Problem 3.58 shows pulley systems
containing one, two, and three pulleys. The number of
pulleys in the type of system shown could obviously be
extended to an arbitrary number N.
(a) Neglecting the weights of the pulleys, determine
the force T required to support the weight W as a
function of the number of pulleys N in the system.
(b) Using the result of part (a), determine the force T
required to support the weight W for a system with
10 pulleys.
Solution: By extrapolation of the previous problem
(a) T D W
2N
(b) T D W
1024
Problem 3.60 A 14,000-kg airplane is in steady flight
in the vertical plane. The flight path angle is � D 10°,
the angle of attack is ˛ D 4°, and the thrust force exerted
by the engine is T D 60 kN. What are the magnitudes
of the lift and drag forces acting on the airplane? (See
Example 3.4).
Solution: Let us draw a more detailed free body diagram to see
the angles involved more clearly. Then we will write the equations of
equilibrium and solve them.
W D mg D �14,000��9.81� N
The equilibrium equations are


∑
Fx D T cos ˛� D�W sin � D 0∑
Fy D T sin ˛C L �W cos � D 0
T D 60 kN D 60000 N
Solving, we get
D D 36.0 kN, L D 131.1 kN
y
T
L
D
W
Horizon
γ
α
Path
x
y
α = 4°
γ = 10°
L
T
D
x
W
γ
α
γ
124
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Problem 3.61 An airplane is in steady flight, the angle
of attack ˛ D 0, the thrust-to-drag ratio T/D D 2, and
the lift-to-drag ratio L/D D 4. What is the flight path
angle �? (See Example 3.4).
Solution: Use the same strategy as in Problem 3.52. The angle
between the thrust vector and the positive x axis is ˛,
T D jTj�i cos˛C j sin˛�
The lift vector: L D 0iC jLjj
The drag: D D �jDjiC 0j. The angle between the weight vector and
the positive x axis is �270� ��;
W D jWj��i sin � � j cos ��.
The equilibrium conditions are
∑
F D TC LC DCW D 0.
Substitute and collect like terms
∑
Fx D �jTj cos ˛� jDj � jWj sin ��i D 0,
and
∑
Fy D �jTj sin ˛C jLj � jWj cos ��j D 0
Solve the equations for the terms in � :
jWj sin � D jTj cos˛� jDj,
and jWj cos � D jTj sin˛C jLj.
Take the ratio of the two equations
tan � D
( jTj cos˛� jDj
jTj sin˛C jLj
)
.
Divide top and bottom on the right by jDj.
For ˛ D 0, jTjjDj D 2,
jLj
jDj D 4, tan � D
(
2� 1
4
)
D 1
4
or � D 14°
y
x
L
T
D
W
α
γ
Path
Horizontal
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125
Problem 3.62 An airplane glides in steady flight (T D
0), and its lift-to-drag ratio is L/D D 4.
(a) What is the flight path angle �?
(b) If the airplane glides from an altitude of1000 m
to zero altitude, what horizontal distance does it
travel?
(See Example 3.4.)
Solution: See Example 3.4. The angle between the thrust vector
and the positive x axis is ˛:
T D jTj�i cos˛C j sin˛�.
The lift vector: L D 0iC jLjj.
The drag: D D �jDjiC 0j. The angle between the weight vector and
the positive x axis is �270� ��:
W D jWj��i sin � � j cos ��.
The equilibrium conditions are
∑
F D TC LC DCW D 0.
Substitute and collect like terms:
∑
Fx D �jTj cos˛� jDj � jWj sin ��i D 0
∑
Fy D �jTj sin˛C jLj � jWj cos ��j D 0
Solve the equations for the terms in � ,
jWj sin � D jTj cos˛� jDj,
and jWj cos � D jTj sin˛C jLj
Part (a): Take the ratio of the two equilibrium equations:
tan � D
( jTj cos˛� jDj
jTj sin ˛C jLj
)
.
Divide top and bottom on the right by jDj.
For ˛ D 0, jTj D 0, jLjjDj D 4, tan � D
(�1
4
)
� D �14°
Part (b): The flight path angle is a negative angle measured from the
horizontal, hence from the equality of opposite interior angles the angle
� is also the positive elevation angle of the airplane measured at the
point of landing.
tan � D 1
h
, h D 1
tan �
D 1(
1
4
) D 4 km
y
x
L
T
D
W
α
γ
Path
Horizontal
1 km
h
γ
γ
126
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Problem 3.63 In Active Example 3.5, suppose that the
attachment point B is moved to the point (5,0,0) m. What
are the tensions in cables AB, AC, and AD?
(�3, 0, 3) m
(�2, 0, �2) m
z
D
(0, �4, 0) m
(4, 0, 2) m
B
A
100 kg
yC
x
Solution: The position vector from point A to point B can be used
to write the force TAB.
rAB D �5iC 4j� m
TAB D TAB rABjrABj D TAB�0.781iC 0.625j�
Using the other forces from Active Example 3.5, we have
∑
Fx : 0.781TAB � 0.408TAC � 0.514TAD D 0
∑
Fy : 0.625TAB C 0.816TAC C 0.686TAD � 981 N D 0
∑
Fz : �0.408TAC C 0.514TAD D 0
Solving yields TAB D 509 N, TAC D 487 N, TAD D 386 N
Problem 3.64 The force F D 800iC 200j (lb) acts at
point A where the cables AB, AC, and AD are joined.
What are the tensions in the three cables?
y
z
(12, 4, 2) ft
(0, 4, 6) ft
(6, 0, 0) ftB
C
A
F
(0, 6, 0) ftD
x
Solution: We first write the position vectors
rAB D ��6i� 4j� 2k� ft
rAC D ��12iC 6k� ft
rAD D ��12iC 2j� 2k� ft
Now we can use these vectors to define the force vectors
TAB D TAB rABjrABj D TAB��0.802i� 0.535j� 0.267k�
TAC D TAC rACjrACj D TAC��0.949iC 0.316k�
TAD D TAD rADjrADj D TAD��0.973iC 0.162j� 0.162k�
The equilibrium equations are then
∑
Fx : �0.802TAB � 0.949TAC � 0.973TAD C 800 lb D 0
∑
Fy : �0.535TAB C 0.162TAD C 200 lb D 0
∑
Fz : �0.267TAB C 0.316TAC � 0.162TAD D 0
Solving, we find TAB D 405 lb, TAC D 395 lb, TAD D 103 lb
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127
Problem 3.65 Suppose that you want to apply a
1000-lb force F at point A in a direction such that the
resulting tensions in cables AB, AC, and AD are equal.
Determine the components of F.
y
z
(12, 4, 2) ft
(0, 4, 6) ft
(6, 0, 0) ftB
C
A
F
(0, 6, 0) ftD
x
Solution: We first write the position vectors
rAB D ��6i� 4j� 2k� ft
rAC D ��12iC 6k� ft
rAD D ��12iC 2j� 2k� ft
Now we can use these vectors to define the force vectors
TAB D T rABjrABj D T��0.802i� 0.535j� 0.267k�
TAC D T rACjrACj D T��0.949iC 0.316k�
TAD D T rADjrADj D T��0.973iC 0.162j� 0.162k�
The force F can be written F D �FxiC FyjC Fzk�
The equilibrium equations are then
∑
Fx : �0.802T � 0.949T � 0.973T C Fx D 0) Fx D 2.72T
∑
Fy : �0.535TC 0.162T C Fy D 0) Fy D 0.732T
∑
Fz : �0.267TC 0.316T � 0.162T C Fz D 0) Fz D 0.113T
We also have the constraint equation
√
Fx2 C Fy2 C Fz2 D 1000 lb
) T D 363 lb
Solving, we find Fx D 990 lb, Fy D 135 lb, Fz D 41.2 lb
128
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Problem 3.66 The 10-lb metal disk A is supported by
the smooth inclined surface and the strings AB and AC.
The disk is located at coordinates (5,1,4) ft. What are
the tensions in the strings?
(0, 6, 0) ft
(8, 4, 0) ftC
y
B
z
2 ft
8 ft
10 ft
A
x
Solution: The position vectors are
rAB D ��5iC 5j� 4k� ft
rAC D �3iC 3j� 4k� ft
The angle ˛ between the inclined surface the horizontal is
˛ D tan�1�2/8� D 14.0°
We identify the following force:
TAB D TAB rABjrABj D TAB��0.615iC 0.615j� 0.492k�
TAC D TAC rACjrACj D TAC�0.514iC 0.514j� 0.686k�
N D N�cos˛jC sin˛k� D N�0.970jC 0.243k�
W D ��10 lb�j
The equilibrium equations are then
∑
Fx : �0.615TAB C 0.514TAC D 0
∑
Fy : 0.615TAB C 0.514TAC C 0.970N� 10 lb D 0
∑
Fz : �0.492TAB � 0.686TAC C 0.243N D 0
Solving, we find N D 8.35 lb TAB D 1.54 lb, TAC D 1.85 lb
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129
Problem 3.67 The bulldozer exerts a force FD2i (kip)
at A. What are the tensions in cables AB, AC, and AD?
y
C
A
Dz
4 ft
3 ft
2 ft
x
8 ft
8 ft
6 ft
B
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:
A�8, 0, 0�, B�0, 3, 8�, C�0, 2,�6�, D�0,�4, 0�.
The radius vectors for these points are
rA D 8iC 0jC 0k, rB D 0iC 3jC 8k,
rC D 0iC 2j� 6k, rD D 0iC 4jC 0k.
By definition, the unit vector parallel to the tension in cable AB is
eAB D rB � rAjrB � rAj .
Carrying out the operations for each of the cables, the results are:
eAB D �0.6835iC 0.2563jC 0.6835k,
eAC D �0.7845iC 0.1961j� 0.5883k,
eAD D �0.8944i� 0.4472jC 0k.
The tensions in the cables are expressed in terms of the unit vectors,
TAB D jTABjeAB, TAC D jTACjeAC, TAD D jTADjeAD.
The external force acting on the juncture is F D 2000iC 0jC 0k. The
equilibrium conditions are
∑
F D 0 D TAB C TAC C TAD C F D 0.
Substitute the vectors into the equilibrium conditions:
∑
FxD ��0.6835jTABj� 0.7845jTACj� 0.8944jTADjC2000�iD0
∑
Fy D �0.2563jTABj C 0.1961jTACj � 0.4472jTADj�j D 0
∑
Fz D �0.6835jTABj � 0.5883jTACj C 0jTADj�k D 0
The commercial program TK Solver Plus was used to solve these
equations. The results are
jTABj D 780.31 lb , jTACj D 906.49 lb , jTADj D 844.74 lb .
130
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Problem 3.68 Prior to its launch, a balloon carrying a
set of experiments to high altitude is held in place by
groups of student volunteers holding the tethers at B, C,
and D. The mass of the balloon, experiments package,
and the gas it contains is 90 kg, and the buoyancy force
on the balloon is 1000 N. The supervising professor
conservatively estimates that each student can exert at
least a 40-N tension on the tether for the necessary length
of time. Based on this estimate, what minimum numbers
of students are needed at B, C, and D?
y
z B
x
C (10,0, –12) m
(16, 0, 16) m
D
(–16, 0, 4) m
 (0, 8, 0) mA
Solution:
∑
Fy D 1000� �90��9.81�� T D 0
T D 117.1 N
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 D 0.667i� 0.333jC 0.667k
eAC D 0.570i� 0.456j� 0.684k
eAD D �0.873i� 0.436jC0.218k
We now write the forces in terms of magnitudes and unit vectors


FAB D 0.667FABi� 0.333FABjC 0.667FABk
FAC D 0.570FACi� 0.456FACj� 0.684FACk
FAD D �0.873FADi� 0.436FACjC 0.218FACk
T D 117.1j (N)
The equations of equilibrium are
∑
Fx D 0.667FAB C 0.570FAC � 0.873FAD D 0
∑
Fy D �0.333FAB � 0.456FAC � 0.436FAC C 117.1 D 0
∑
Fz D 0.667FAB � 0.684FAC C 0.218FAC D 0
Solving, we get
FAB D 64.8 N ¾ 2 students
FAC D 99.8 N ¾ 3 students
FAD D 114.6 N ¾ 3 students
y
T
(0, 8, 0)
(−16, 0, 4)
(10, 0, −12) m
A FAC
FAD
D
z
x
C
(16, 0, 16) mB
1000 N
(90) g
T
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131
Problem 3.69 The 20-kg mass is suspended by cables
attached to three vertical 2-m posts. Point A is at
(0, 1.2, 0) m. Determine the tensions in cables AB, AC,
and AD.
B
A
C
D
y
z
x
2 m
0.3 m
1 m
1 m
Solution: Points A, B, C, and D are located at
A�0, 1.2, 0�, B��0.3, 2, 1�,
C�0, 2,�1�, D�2, 2, 0�
Write the unit vectors eAB, eAC, eAD
eAB D �0.228iC 0.608jC 0.760k
eAC D 0iC 0.625j� 0.781k
eAD D 0.928iC 0.371jC 0k
The forces are
FAB D �0.228FABiC 0.608FABjC 0.760FABk
FAC D 0FACiC 0.625FACj� 0.781FACk
FAD D 0.928FADiC 0.371FADjC 0k
W D ��20��9.81�j
The equations of equilibrium are


∑
Fx D �0.228FAB C 0C 0.928FAD D 0∑
Fy D 0.608FAB C 0.625FAC C 0.371FAD � 20�9.81� D 0∑
Fz D 0.760FAB � 0.781FAC C 0 D 0
We have 3 eqns in 3 unknowns solving, we get
FAB D 150.0 N
FAC D 146.1 N
FAD D 36.9 N
y
C
A
FAD
FAC
FAB
D
W
B
z
x
(20) (9.81) N
132
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Problem 3.70 The weight of the horizontal wall sec-
tion is W D 20,000 lb. Determine the tensions in the
cables AB, AC, and AD.
CB6 ft
10 ft
4 ft 8 ft
7 ft
14 ft
W
D
A
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 DW
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
eAI D �xI � xA�iC �yI � yA�jC �zI � zA�k√
�xI � xA�2 C �yI � yA�2 C �zI � zA�2
,
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 D �0.324i� 0.487j� 0.811k,
eAC D 0.566i� 0.424j� 0.707k,
and eAD D 0iC 0.625j� 0.781k.
The forces are
TAB D TABeAB, TAC D TACeAC, and TAD D TADeAD.
The equilibrium equation for the knot at point A is
TC TAB C TAC C TAD D 0.
From the vector equilibrium equation, write the scalar equilibrium
equations in the x, y, and z directions. We get three linear equations
in three unknowns. Solving these equations simultaneously, we get
TAB D 9393 lb, TAC D 5387 lb, and TAD D 10,977 lb
CB6 ft
10 ft
4 ft 8 ft
7 ft
14 ft
W
D
A
T
W
X
D
B
A
C
z
TB
TD
TC
y
7 ft
14 ft
4 ft 8 ft
6 ft
10 ft
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133
Problem 3.71 The car in Fig. a and the pallet
supporting it weigh 3000 lb. They are supported by
four cables AB, AC, AD, and AE. The locations of the
attachment points on the pallet are shown in Fig. b.
The tensions in cables AB and AE are equal. Determine
the tensions in the cables.
C
A
y
B
E
(a)
x
8 ft 6 ft
6 ft 5 ft
5 ft
5 ft
4 ft
D C
B
z
z
x
E
(b)
(0, 10, 0) ft
Solution: Isolate the knot at A. Let TAB,TAC,TAD and TAE be
the forces exerted by the tensions in the cables. The force exerted by
the vertical cable is (3000 lb)j. We first find the position vectors and
then express all of the forces as vectors.
rAB D �5i� 10jC 5k� ft
rAC D �6i� 10j� 5k� ft
rAD D ��8i� 10j� 4k� ft
rAE D ��6i� 10jC 5k� ft
TAB D TAB rABjrABj D TAB�0.408i� 0.816jC 0.408k�
TAC D TAC rACjrACj D TAC�0.473i� 0.788j� 0.394k�
TAD D TAD rADjrADj D TAD��0.596i� 0.745j� 0.298k�
TAE D TAE rAEjrAEj D TAE��0.473i� 0.788jC 0.394k�
The equilibrium equations are
∑
Fx : 0.408TAB C 0.473TAC � 0.596TAD � 0.473TAE D 0
∑
Fy : �0.816TAB � 0.788TAC � 0.745TAD � 0.788TAE C 3000 lb D 0
∑
Fz : 0.408TAB � 0.394TAC � 0.298TAD C 0.394TAE D 0
Solving, we find
TAB D 896 lb, TAC D 1186 lb, TAD D 843 lb, TAE D 896 lb
134
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Problem 3.72 The 680-kg load suspended from the
helicopter is in equilibrium. The aerodynamic drag force
on the load is horizontal. The y axis is vertical, and cable
OA lies in the x-y plane. Determine the magnitude of the
drag force and the tension in cable OA.
y A
10°
xO
B
C D
Solution:
∑
Fx D TOA sin 10° � D D 0,
∑
Fy D TOA cos 10° � �680��9.81� D 0.
Solving, we obtain D D 1176 N, TOA D 6774 N.
y
xD
10°
TOA
(680) (9.81) N
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135
Problem 3.73 In Problem 3.72, the coordinates of the
three cable attachment points B, C, and D are (�3.3,
�4.5, 0) m, (1.1, �5.3, 1) m, and (1.6, �5.4, �1) m,
respectively. What are the tensions in cables OB, OC,
and OD?
Solution: The position vectors from O to pts B, C, and D are
rOB D �3.3i� 4.5j (m),
rOC D 1.1i� 5.3jC k (m),
rOD D 1.6i� 5.4j� k (m).
Dividing by the magnitudes, we obtain the unit vectors
eOB D �0.591i� 0.806j,
eOC D 0.200i� 0.963jC 0.182k,
eOD D 0.280i� 0.944j� 0.175k.
Using these unit vectors, we obtain the equilibrium equations
∑
Fx D TOA sin 10° � 0.591TOB C 0.200TOC C 0.280TOD D 0,
∑
Fy D TOA cos 10° � 0.806TOB � 0.963TOC � 0.944TOD D 0,
∑
Fz D 0.182TOC � 0.175TOD D 0.
From the solution of Problem 3.72, TOA D 6774 N. Solving these
equations, we obtain
TOB D 3.60 kN, TOC D 1.94 kN, TOD D 2.02 kN.
y
x
TOA
TOD
TOC
TOB
10°
136
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Problem 3.74 If the mass of the bar AB is negligible
compared to the mass of the suspended object E, the
bar exerts a force on the “ball” at B that points from A
toward B. The mass of the object E is 200 kg. The y-axis
points upward. Determine the tensions in the cables BC
and CD.
Strategy: Draw a free-body diagram of the ball at B.
(The weight of the ball is negligible.)
x
y
z
C
BD
A
(0, 5, 5) m
(0, 4, �3) m
(4, 3, 1) m
E
Solution:
FAB D FAB
(�4i� 3j� kp
26
)
,TBC D TBC
(�4iC j� 4kp
33
)
,
The forces
TBD D TBD
(�4iC 2jC 4k
6
)
,W D ��200 kg��9.81 m/s2�j
The equilibrium equations
∑
Fx : � 4p
26
FAB � 4p
33
TBC � 4
6
TBD D 0
∑
Fy : � 3p
26
FAB C 1p
33
TBC C 26TBD � 1962 N D 0
∑
Fz : � 1p
26
FAB � 4p
33
TBC C 46TBD D 0
)
TBC D 1610 N
TBD D 1009 N
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137
Problem 3.75* The 3400-lb car is at rest on the plane
surface. The unit vector en D 0.456iC 0.570jC 0.684k
is perpendicular to the surface. Determine the magni-
tudes of the total normal force N and the total friction
force f exerted on the surface by the car’s wheels.
y
z
x
en
Solution: The forces on the car are its weight, the normal force,
and the friction force.
The normal force is in the direction of the unit vector, so it can be
written
N D Nen D N�0.456iC 0.570jC 0.684k�
The equilibrium equation is
Nen C f� �3400 lb�j D 0
The friction force f is perpendicular to N, so we can eliminate the
friction force from the equilibrium equation by taking the dot product
of the equation with en.
�Nen C f� �3400 lb�j� Рen D N� �3400 lb��j Рen� D 0
N D �3400 lb��0.57� D 1940 lb
Now we can solve for the friction force f.
f D �3400 lb�j�Nen D �3400 lb�j� �1940 lb��0.456iC 0.570jC 0.684k�
f D ��884iC 2300j � 1330k� lb
jfj D
√
��884 lb�2 C �2300 lb�2 C ��1330 lb�2 D 2790 lb
jNj D 1940 lb, jfj D 2790 lb
138
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Problem 3.76 The system shown anchors a stanchion
of a cable-suspended roof. If the tension in cable AB is
900 kN, what are the tensions in cables EF and EG?
y
z
x
F
(0, 1.4, 1.2) m
(0, 1.4, �1.2) m
(1, 1.2, 0) m
(2, 1, 0) m
(2.2, 0, 1) m
(2.2, 0, �1) m
(3.4, 1, 0) m
G
E
B A
C
D
Solution: Using the coordinates for the points we find
rBA D [�3.4� 2�iC �1� 1�jC �0� 0�k] m
rBA D �1.4 m�i eBA D rBAjrBAj D i
Using the same procedure we find the other unit vectors that we need.
eBC D 0.140i� 0.700jC 0.700k
eBD D 0.140i� 0.700jC 0.700k
eBE D �0.981iC 0.196j
eEG D �0.635iC 0.127j� 0.762k
eEF D �0.635iC 0.127jC 0.762k
We can now write the equilibrium equations for the connections at B
and E.
�900 kN�eBA C TBCeBC C TBDeBD C TBEeBE D 0, TBE��eBE�C TEFeEF C TEGeEG0
Breaking these equations into components, we have the following six
equations to solve for five unknows (one of the equations is redundant).
�900 kN�C TBC�0.140�C TBD�0.140�C TBE��0.981� D 0
TBC��0.700�C TBD��0.700�C TBE�0.196� D 0
TBC�0.700�C TBD��0.700� D 0
TBE�0.981�C TEG��0.635�C TEF��0.635� D 0
TBE��0.196�C TEG�0.127�C TEF�0.127� D 0
TEG��0.726�C TEF�0.726� D 0
Solving, we find TBC D TBD D 134 kN, TBE D 956 kN
TEF D TEG D 738 kN
Problem 3.77* The cables of the system will each
safely support a tension of 1500 kN. Based on this
criterion, what is the largest safe value of the tension
in cable AB?
y
z
x
F
(0, 1.4, 1.2) m
(0, 1.4, �1.2) m
(1, 1.2, 0) m
(2, 1, 0) m
(2.2, 0, 1) m
(2.2, 0, �1) m
(3.4, 1, 0) m
G
E
B A
C
D
Solution: From Problem 3.76 we know that if the tension in AB
is 900 kN, then the largest force in the system occurs in cable BE and
that tension is 956 kN. To solve this problem, we can just scale the
results from Problem 3.76
TAB D
(
1500 kN
956 kN
)
�900 kN� TAB D 1410 kN
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139
Problem 3.78 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.
z
y
x
A
B
2 m
5 m
2 m
2 m
Solution: The coordinates of the points A, B are A�2, 2, 0�,
B�0, 5, 2�. The vector positions
rA D 2iC 2jC 0k, rB D 0iC 5jC 2k
The equilibrium conditions are:
∑
F D TC NCW D 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 D 0iC 1jC
0k. The dot product of the unit vector and the normal force vanishes:
eB Ð N D 0. Take the dot product of eB with the equilibrium conditions:
eB Ð N D 0.
∑
eB Ð F D eB Ð TC eB ÐW D 0.
The weight is
eB ÐW D 1j Ð ��jjWj� D �jWj D ��200��9.81� D �1962 N.
The unit vector parallel to the cable is by definition,
eAB D rB � rAjrB � rAj .
Substitute the vectors and carry out the operation:
eAB D �0.4851iC 0.7278jC 0.4851k.
(a) The tension in the cable is T D jTjeAB. Substitute into the modi-
fied equilibrium condition
∑
eBF D �0.7276jTj � 1962� D 0.
Solve: jTj D 2696.5 N from which the tension vector is
T D jTjeAB D �1308iC 1962jC 1308k.
(b) The equilibrium conditions are
∑
F D 0 D TC NCW D �1308iC 1308k C N D 0.
Solve for the normal force: N D 1308i� 1308k. The magnitude
is jNj D 1850 N.
T
N
W
Note: For this specific configuration, the problem can be solved with-
out eliminating the slider bar normal force, since it does not appear in
the y-component of the equilibrium equation (the slider bar is parallel
to the y-axis). However, in the general case, the slider bar will not be
parallel to an axis, and the unknown normal force will be projected
onto all components of the equilibrium equations (see Problem 3.79
below). In this general situation, it will be necessary to eliminate the
slider bar normal force by some procedure equivalent to that used
above. End Note.
140
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Problem 3.79 In Example 3.6, suppose that the cable
AC is replaced by a longer one so that the distance from
point B to the slider C increases from 6 ft to 8 ft. Deter-
mine the tension in the cable.
6 ft
O
y
C
A
B
D
z
4 ft
7 ft
4 ft
4 ft
x
Solution: The vector from B to C is now
rBC D �8 ft� eBD
rBC D �8 ft�
(
4
9
i� 7
9
jC 4
9
k
)
rBC D �3.56i� 6.22jC 3.56k� ft
We can now find the unit vector form C to A.
rCA D rOA � �rOB C rBC� D [�7jC 4k�
� f�7j�C �3.56i� 6.22jC 3.56k�g] ft
rCA D ��3.56iC 6.22jC 0.444k� ft
eCA D rCAjrCAj D ��0.495iC 0.867jC 0.0619k�
Using N to stand for the normal force between the bar and the slider,
we can write the equilibrium equation:
TeCA C N� �100 lb�j D 0
We can use the dot product to eliminate N from the equation
[TeCA C N� �100 lb�j] РeBD D T�eCA РeBD�� �100 lb��j РeBD� D 0
T
([
4
9
]
[�0.495]C
[
� 7
9
]
[0.867]C
[
4
9
]
[0.0619]
)
� �100 lb���0.778� D 0
T��0.867�C �77.8 lb� D 0) T D 89.8 lb
T D 89.8 lb
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141
Problem 3.80 The cable AB keeps the 8-kg collar A in
place on the smooth bar CD. The y axis points upward.
What is the tension in the cable?
0.4 m
0.5 m
0.15 m
0.3 m0.2 m
0.25 m
0.2 m
z
y
B
C
D
O x
A
Solution: We develop the following position vectors and unit
vectors
rCD D ��0.2i� 0.3jC 0.25k� m
eCD D rCDjrCDj D ��0.456i� 0.684jC 0.570k�
rCA D �0.2 m�eCD D ��0.091i� 0.137jC 0.114k� m
rAB D rOB � �rOC C rCA�
rAB D [�0.5jC 0.15k�� �f0.4iC 0.3jg C f�0.091i� 0.137jC 0.114kg�] m
rAB D ��0.309iC 0.337jC 0.036k� m
eAB D rABjrABj D ��0.674iC 0.735jC 0.079k�
We can now write the equilibrium equation for the slider using N to
stand for the normal force between the slider and the bar CD.
TeAB C N� �8 kg��9.81 m/s2�j D 0
To eliminate the normal force N we take a dot product with eCD.
[TeAB C N� �8 kg��9.81 m/s2�j] РeCD D 0
T�eAB РeCD�� �78.5 N��j РeCD�D 0
T�[�0.674][�0.456]C [0.735][�0.684] C [0.079][0.570]� � �78.5 N���0.684� D 0
T��0.150�C 53.6 N D 0
T D 357 N
Problem 3.81 Determine the magnitude of the normal
force exerted on the collar A by the smooth bar.
0.4 m
0.5 m
0.15 m
0.3 m0.2 m
0.25 m
0.2 m
z
y
B
C
D
O x
A
Solution: From Problem 3.81 we have
eAB D ��0.674iC 0.735jC 0.079k�
T D 357 N
The equilibrium equation is
TeAB C N� �78.5 N�j D 0
We can now solve for the normal force N.
N D �78.5 N�j� �357 N���0.674iC 0.735jC 0.079k�
N D �240i� 184j� 28.1k� N
The magnitude of N is
jNj D
√
�240 N�2 C ��184 N�2 C ��28.1 N�2
jNj D 304 N
142
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Problem 3.82* The 10-kg collar A and 20-kg collar B
are held in place on the smooth bars by the 3-m cable
from A to B and the force F acting on A. The force F
is parallel to the bar. Determine F.
z
x
y
B
A
F
3 m
(4, 0, 0) m
(0, 0, 4) m
(0, 3, 0) m
(0, 5, 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 D erCDx iC eCDyjC eCDzk,
and eEG D erEGx iC eEGyjC eEGzk.
Using the coordinates of points C, D, E, and G from the picture, the
unit vectors are
eCD D �0.625iC 0.781jC 0k,
and eEG D 0iC 0.6jC 0.8k.
The location of point A is given by
xA D xC CCAeCDx, yA D yC C CAeCDy,
and zA D zC CCAeCDz,
where CA D 3 m. From these equations, we find that the location of
point A is given by A (2.13, 2.34, 0) m. Once we know the location
of point A, we can proceed to find the location of point B. We have
two ways to determine the location of B. First, B is 3 m from point A
along the line AB (which we do not know). Also, B lies on the line
EG. The equations for the location of point B based on line AB are:
xB D xA C ABeABx, yB D yA C ABeABy,
and zB D zA C ABeABz.
The equations based on line EG are:
xB D xE C EBeEGx, yB D yE C EBeEGy,
and zB D zE C EBeEGz.
We have six new equations in the three coordinates of B and the
distance EB. Some of the information in the equations is redundant.
However, we can solve for EB (and the coordinates of B). We get that
the length EB is 2.56 m and that point B is located at (0, 1.53, 1.96) m.
We next write equilibrium equations for bodies A and B. From the free
body diagram for A, we get
NAx C TABeABx C FeCDx D 0,
NAy C TABeABy C FeCDy � mAg D 0,
and NAz C TABeABz C FeCDz D 0.
From the free body diagram for B, we get
NBx � TABeABx D 0,
Nby � TABeABy � mBg D 0,
and NBz � TABeABz D 0.
y
F
B x
z
3 m
NA
TAB
TAB
A
NB
D (0, 5, 0)
C (4, 0, 0) m
E (0, 0, 4) m
m
G (0, 3, 0)
m
mAg
mBg
We now have two fewer equation than unknowns. Fortunately, there
are two conditions we have not yet invoked. The bars at A and B
are smooth. This means that the normal force on each bar can have
no component along that bar. This can be expressed by using the dot
product of the normal force and the unit vector along the bar. The two
conditions are
NA Ð eCD D NAxeCDx CNAyeCDy CNAzeCDz D 0
for slider A and
NB Ð eEG D NBxeEGx CNByeEGy CNBzeEGz D 0.
Solving the eight equations in the eight unknowns, we obtain
F D 36.6 N .
Other values obtained in the solution are EB D 2.56 m,
NAx D 145 N, NAy D 116 N, NAz D �112 N,
NBx D �122 N, NBy D 150 N, and NBz D 112 N.
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143
Problem 3.83 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.
45�
A
B
30�
Solution: The free-body diagram is sketched. The equilibrium
equations are
∑
Fx- : T cos�45° � 30°�� �100 lb� sin 30° D 0
∑
F% : T sin�45° � 30°�� �100 lb� cos 30° CN D 0
Solving, we find
T D 51.8 lb, N D 73.2 lb
Problem 3.84 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
W
A
B
20�
25�
60�
x
Solution: The force in the cable is W everywhere. The free-body
diagram of the leg is shown. The downward force is given, but the
horizontal force FH is unknown.
The equilibrium equation in the vertical direction is
∑
Fy : W sin 25° CW sin 60° � 32.2 lb D 0
Thus
W D 32.2 lb
sin 25° C sin 60°
W D 25.0 lb
144
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Problem 3.85 The 400-lb engine block is suspended
by the cables AB and AC. If you don’t want either TAB
or TAC to exceed 400 lb, what is the smallest acceptable
value of the angle ˛?
B
C
A A
x
y
a a
TAB TAC
400 lb
Solution: The equilibrium equations are
∑
Fx : �TAB cos˛C TAC cos ˛ D 0
∑
Fy : TAB sin˛C TAC sin˛� �400 lb� D 0
Solving, we find
TAB D TAC D 400 lb2 sin˛
If we limit the tensions to 400 lb, we have
400 lb D 400 lb
2 sin˛
) sin˛ D 1
2
) ˛ D 30°
Problem 3.86 The cable AB is horizontal, and the box
on the right weighs 100 lb. The surface 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: We have the following equilibrium equations
∑
FyB : NB cos 40° � 100 lb D 0
∑
FxB : NB sin 40° � T D 0
∑
FxA : T�NA sin 20° D 0
∑
FyA : NA cos 20° �WA D 0
Solving these equations sequentially, we find
NB D 131 lb, T D 83.9 lb
NA D 245 lb,WA D 230.5 lb
Thus we have T D 83.9 lb,WA D 230.5 lb
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145
Problem 3.87 Assume that the forces exerted on the
170-lb climber by the slanted walls of the “chimney”
are perpendicular to the walls. If he is in equilibrium
and is exerting a 160-lb force on the rope, what are the
magnitudes of the forces exerted on him by the left and
right walls?
4�
10�
3�
Solution: The forces in the free-body diagram are in the directions
shown on the figure. The equilibrium equations are:
∑
Fx : �T sin 10° CNL cos 4° �NR cos 3° D 0
∑
Fy : T cos 10° � �170 lb�CNL sin 40° CNR sin 3° D 0
where T D 160 lb. Solving we find
NL D 114 lb, NR D 85.8 lb
Left Wall: 114 lb
Right Wall: 85.8 lb
146
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Problem 3.88 The mass of the suspended object A is
mA and the masses of the pulleys are negligible. Deter-
mine the force T necessary for the system to be in
equilibrium.
T A
Solution: Break the system into four free body diagrams as shown.
Carefully label the forces to ensure that the tension in any single cord
is uniform. The equations of equilibrium for the four objects, starting
with the leftmost pulley and moving clockwise, are:
S� 3T D 0, R� 3S D 0, F� 3R D 0,
and 2TC 2SC 2R� mAg D 0.
We want to eliminate S, R, and F from our result and find T in
terms of mAand g. From the first two equations, we get S D 3T,
and R D 3S D 9T. Substituting these into the last equilibrium equation
results in 2TC 2�3T�C 2�9T� D mAg.
Solving, we get T D mAg/26 .
R
R
R
R R
R
F
S
S
S
S
S
S
T
T
T
T
T
A
mAg
Note: We did not have to solve for F to find the appropriate value of T.
The final equation would give us the value of F in terms of mA and g.
We would get F D 27mAg/26. If we then drew a free body diagram
of the entire assembly, the equation of equilibrium would be F� T�
mAg D 0. Substituting in the known values for T and F, we see that
this equation is also satisfied. Checking the equilibrium solution by
using the “extra” free body diagram is often a good procedure.
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147
Problem 3.89 The assembly A, including the pulley,
weighs 60 lb. What force F is necessary for the system
to be in equilibrium?
F
A
Solution: From the free body diagram of the assembly A, we have
3F� 60 D 0, or F D 20 lb
F
F
F
F
F F
60 lb.
F
148
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Problem 3.90 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
45�
A
20�
F
Solution: Isolate the top block. Solve the equilibrium equations.
The weight is. The angle between the normal force N1 and the posi-
tive x axis is. The normal force is. The force N2 is. The equilibrium
conditions are
∑
F D N1 C N2 CW D 0
from which
∑
Fx D �0.7071jN1j � jN2j�i D 0
∑
Fy D �0.7071jN1j � 490.5�j D 0.
Solve: N1 D 693.7 N, jN2j D 490.5 N
Isolate the bottom block. The weight is
W D 0i� jWjj D 0i� �42��9.81�j D 0i� 412.02j (N).
The angle between the normal force N1 and the positive x axis is
�270° � 45°� D 225°.
The normal force:
N1 D jN1j�i cos 225° C j sin 225°� D jN1j��0.7071i� 0.7071j�.
The angle between the normal force N3 and the positive x-axis is
�90° � 20°� D 70°.
The normal force is
N1 D jN3j�i cos 70° C j sin 70°� D jN3j�0.3420iC 0.9397j�.
The force is . . .F D jFjiC 0j. The equilibrium conditions are
∑
F DWC N1 C N3 C F D 0,
from which:
∑
Fx D ��0.7071jN1j C 0.3420jN3j C jFj�i D 0
∑
Fy D ��0.7071jN1j C 0.9397jN3j � 412�j D 0
For jN1j D 693.7 N from above: jFj D 162 N
y
x
B
W
N2
N1 α
x
y
W
F A
N3
N1
α
β
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149
Problem 3.91 The climber A is being helped up an icy
slope by two friends. His mass is 80 kg, and the direction
cosines of the force exerted on him by the slope are
cos �x D �0.286, cos �y D 0.429, cos �z D 0.857. The y
axis is vertical. If the climber is in equilibrium in the
position shown, what are the tensions in the ropes AB
and AC and the magnitude of the force exerted on him
by the slope?
A
(3, 0, 4) m
B
(2, 2, 0) m
C
(5, 2, �1) m
y
x
z
Solution: Get the unit vectors parallel to the ropes using the coor-
dinates of the end points. Express the tensions in terms of these unit
vectors, and solve the equilibrium conditions. The rope tensions, the
normal force, and the weight act on the climber. The coordinates
of points A, B, C are given by the problem, A�3, 0, 4�, B�2, 2, 0�,
C�5, 2,�1�.
The vector locations of the points A, B, C are:
rA D 3iC 0jC 4k, rB D 2iC 2jC 0k, rC D 5iC 2j� 1k.
The unit vector parallel to the tension acting between the points A, B
in the direction of B is
eAB D rB � rAjrB � rAj
The unit vectors are
eAB D �0.2182iC 0.4364j� 0.8729k,
eAC D 0.3482iC 0.3482j � 0.8704k,
and eN D �0.286iC 0.429jC 0.857k.
where the last was given by the problem statement. The forces are
expressed in terms of the unit vectors,
TAB D jTABjeAB, TAC D jTACjeAC, N D jNjeN.
The weight is
W D 0i� jWjjC 0k D 0i� �80��9.81�jC 0k� 0i� 784.8jC 0k.
The equilibrium conditions are
∑
F D 0 D TAB C TAC C NCW D 0.
A
C
W
N
B
Substitute and collect like terms,
∑
Fx D ��0.2182jTABj C 0.3482jTACj � 0.286jNj�i D 0
∑
Fy D �0.4364jTABj C 0.3482jTACj C 0.429jNj � 784.8�j D 0
∑
Fz D �0.8729jTABj C 0.8704jTACj � 0.857jNj�k D 0
We have three linear equations in three unknowns. The solution is:
jTABj D 100.7 N , jTACj D 889.0 N , jNj D 1005.5 N .
150
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Problem 3.92 Consider the climber A being helped by
his friends in Problem 3.91. To try to make the tensions
in the ropes more equal, the friend at B moves to the
position (4, 2, 0) m. What are the new tensions in the
ropes AB and AC and the magnitude of the force exerted
on the climber by the slope?
Solution: Get the unit vectors parallel to the ropes using the coor-
dinates of the end points. Express the tensions in terms of these
unit vectors, and solve the equilibrium conditions. The coordinates
of points A, B,C are A�3, 0, 4�, B�4, 2, 0�, C�5, 2,�1�. The vector
locations of the points A, B,C are:
rA D 3iC 0jC 4k, rB D 4iC 2jC 0k, rC D 5iC 2j� 1k.
The unit vectors are
eAB D C0.2182iC 0.4364j� 0.8729k,
eAC D C0.3482iC 0.3482j� 0.8704k,
eN D �0.286iC 0.429jC 0.857k.
where the last was given by the problem statement. The forces are
expressed in terms of the unit vectors,
TAB D jTABjeAB, TAC D jTACjeAC, N D jNjeN.
The weight is
W D 0i� jWjjC 0k D 0i� �80��9.81�jC 0k� 0i� 784.8jC 0k.
The equilibrium conditions are
∑
F D 0 D TAB C TAC C NCW D 0.
Substitute and collect like terms,
∑
Fx D �C0.281jTABj C 0.3482jTACj � 0.286jNj�i D 0
∑
Fy D �0.4364jTABj C 0.3482jTACj C 0.429jNj � 784.8�j D 0
∑
Fz D �0.8729jTABj C 0.8704jTACj � 0.857jNj�k D 0
The HP-28S hand held calculator was used to solve these simultaneous
equations. The solution is:
jTABj D 420.5 N , jTACj D 532.7 N , jNj D 969.3 N .
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151
Problem 3.93 A climber helps his friend up an icy
slope. His friend is hauling a box of supplies. If the
mass of the friend is 90 kg and the mass of the supplies
is 22 kg, what are the tensions in the ropes AB and CD?
Assume that the slope is smooth. That is, only normal
forces are exerted on the man and the box by the slope.
D
60�
75 
40�
C
B
20�
A
Solution: Isolate the box. The weight vector is
W2 D �22��9.81�j D �215.8j (N).
The angle between the normal force and the positive x axis is �90° �
60°� D 30°.
The normal force is NB D jNBj�0.866i� 0.5j�.
The angle between the rope CD and the positive x axis is �180° �
75°� D 105°; the tension is:
T2 D jT2j�i cos 105° C j sin 105°� D jT2j��0.2588iC 0.9659j�
The equilibrium conditions are
∑
Fx D �0.866jNBj C 0.2588jT2j�i D 0,
∑
Fy D �0.5jNBj C 0.9659jT2j � 215.8�j D 0.
Solve: NB D 57.8 N, jT2j D 193.5 N.
Isolate the friend. The weight is
W D ��90��9.81�j D �882.9j (N).
The angle between the normal force and the positive x axis is �90° �
40°� D 50°. The normal force is:
NF D jNFj�0.6428iC 0.7660j�.
The angle between the lower rope and the x axis is �75°; the tension is
T2 D jT2j�0.2588iC 0.9659j�.The angle between the tension in the upper rope and the positive x
axis is �180° � 20°� D 160° , the tension is
T1 D jT1j�0.9397iC 0.3420j�.
The equilibrium conditions are
∑
F DWC T1 C T2 C NF D 0.
From which:
∑
Fx D �0.6428jNFj C 0.2588jT2j � 0.9397jT1j�i D 0
∑
Fy D ��0.7660jNFj � 0.9659jT2j C 0.3420jT1j � 882.9�j D 0
Solve, for jT2j D 193.5 N. The result:
jNFj D 1051.6 N , jT1j D 772.6 N .
y
x
N
W
α
β
T
y
xW
N
T1
T2
20°
40°
75°
152
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Problem 3.94 The 2800-lb car is moving at constant
speed on a road with the slope shown. The aerodynamic
forces on the car the drag D D 270 lb, which is parallel
to the road, and the lift L D 120 lb, which is perpendic-
ular to the road. Determine the magnitudes of the total
normal and friction forces exerted on the car by the road.
D
L
15�
Solution: The free-body diagram is shown. If we write the equi-
librium equations parallel and perpendicular to the road, we have:
∑
F- : N� �2800 lb� cos 15° C �120 lb� D 0
∑
F% : f� �270 lb�� �2800 lb� sin 15° D 0
Solving, we find
N D 2580 lb, f D 995 lb
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153
Problem 3.95 An engineer doing preliminary design
studies for a new radio telescope envisions a triangular
receiving platform suspended by cables from three equ-
ally spaced 40-m towers. The receiving platform has
a mass of 20 Mg (megagrams) and is 10 m below the
tops of the towers. What tension would the cables be
subjected to?
65 m
20 m
TOP VIEW
Solution: Isolate the platform. Choose a coordinate system with
the origin at the center of the platform, with the z axis vertical, and
the x,y axes as shown. Express the tensions in terms of unit vectors,
and solve the equilibrium conditions. The cable connections at the
platform are labeled a, b, c, and the cable connections at the towers
are labeled A, B,C. The horizontal distance from the origin (center of
the platform) to any tower is given by
L D 65
2 sin�60�
D 37.5 m.
The coordinates of points A, B, C are
A�37.5, 0, 10�, B�37.5 cos�120°�, 37.5 sin�120°�.10�,
C�37.5 cos�240°�, 37.5 sin�240°�, 10�,
The vector locations are:
rA D 37.5iC 0jC 10k, rB D 18.764iC 32.5jC 10k,
rC D 18.764iC�32.5jC 10k.
The distance from the origin to any cable connection on the platform is
d D 20
2 sin�60°�
D 11.547 m.
The coordinates of the cable connections are
a�11.547, 0, 0�, b�11.547 cos�120°�, 11547 sin�120°�, 0�,
c�11.547 cos�240°�, 11.547 sin�240°�, 0�.
The vector locations of these points are,
ra D 11.547iC 0jC 0k, rb D 5.774iC 10jC 0k,
rc D 5.774iC 10jC 0k.
The unit vector parallel to the tension acting between the points A, a
in the direction of A is by definition
eaA D rA � rajrA � ra .
Perform this operation for each of the unit vectors to obtain
eaA D C0.9333iC 0j� 0.3592k
ebB D �0.4667iC 0.8082j� 0.3592k
ecC D �0.4667iC 0.8082jC 0.3592k
A
C
B
x y
z
a b
c
The tensions in the cables are expressed in terms of the unit vectors,
TaA D jTaAjeaA, TbB D jTbBjebB, TcC D jTcCjecC.
The weight is W D 0i� 0j� �20000��9.81�k D 0iC 0j� 196200k.
The equilibrium conditions are
∑
F D 0 D TaA C TbB C TcC CW D 0,
from which:
∑
Fx D �0.9333jTaAj � 0.4666jTbBj � 0.4666jTcCj�i D 0
∑
Fy D �0jTaAj C 0.8082jTbBj � 0.8082jTcCj�j D 0
∑
Fz D �0.3592jTaAj � 0.3592jTbBj
C 0.3592jTcC � 196200j�k D 0
The commercial package TK Solver Plus was used to solve these
equations. The results:
jTaAj D 182.1 kN , jTbBj D 182.1 kN , jTcCj D 182.1 kN .
Check: For this geometry, where from symmetry all cable tensions may
be assumed to be the same, only the z-component of the equilibrium
equations is required:
∑
Fz D 3jTj sin � � 196200 D 0,
where � D tan�1
(
10
37.5� 11.547
)
D 21.07°,
from which each tension is jTj D 182.1 kN. check.
154
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Problem 3.96 To support the tent, the tension in the
rope AB must be 35 lb. What are the tensions in the
ropes AC, AD, and AE?
x
A B
C
D
(0, 5, 0) ft
(6, 4, 3) ft
(8, 4, 3) ft
 E 
(3, 0, 3) ft
(0, 6, 6) ft
z
y
Solution: We develop the following position vectors
rAB D �2i� ft
rAC D ��6iC j� 3k� ft
rAD D ��6iC 2jC 3k� ft
rAE D ��3i� 4j� ft
If we divide by the respective magnitudes we can develop the unit
vectors that are parallel to these position vectors.
eAB D 1.00i
eAC D �0.885iC 0.147j� 0.442k
eAD D �0.857iC 0.286jC 0.429k
eAE D �6.00i� 0.800j
The equilibrium equation is
TABeAB C TACeAC C TADeAD C TAEeAE D 0.
If we break this up into components, we have
∑
Fx : TAB � 0.885TAC � 0.857TAD � 0.600TAE D 0∑
Fy : 0.147TAC C 0.286TAD � 0.800TAE D 0∑
Fz : �0.442TAC C 0.429TAD D 0
If we set TAB D 35 lb, we cans solve for the other tensions. The result
is
TAC D 16.7 lb, TAD D 17.2 lb, TAE D 9.21 lb
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155
Problem 3.97 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?
x
3 m
y
A
B
D
O
C
(6, 2, 0) m
z
12 m
4 m
5 m
5 m
8 m
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 coordi-
nates of the points are A�6, 2, 0�, B�12, 3, 0�, C�0, 8, 5�, D�0, 4,�5�,
O�0, 0, 0�.
The vector locations of these points are:
rA D 6iC 2jC 0k, rB D 12iC 3jC 0k, rC D 0iC 8jC 5k,
rD D 0iC 4j� 5k, rO D 0iC 0jC 0k.
The unit vector parallel to the tension acting between the points A, B
in the direction of B is by definition
eAB D rB � rAjrB � rAj .
Perform this for each of the unit vectors
eAB D C0.9864iC 0.1644jC 0k
eAC D �0.6092iC 0.6092jC 0.5077k
eAD D �0.7442iC 0.2481j� 0.6202k
eAO D �0.9487i� 0.3162jC 0k
The tensions in the cables are expressed in terms of the unit vectors,
TAB D jTABjeAB D 50eAB, TAC D jTACjeAC,
TAD D jTADjeAD, TAO D jTAOjeAO.
The equilibrium conditions are
∑
F D 0 D TAB C TAC C TAD C TAO D 0.
Substitute and collect like terms,
∑
Fx D �0.9864�50�� 0.6092jTACj � 0.7422jTADj
� 0.9487jTAOj�i D 0
∑
Fy D �0.1644�50�C 0.6092jTACj C 0.2481jTADj
� 0.3162jTAOj�j D 0
∑
Fz D �C0.5077jTACj � 0.6202jTADj�k D 0.
This set of simultaneous equations in the unknown forces may be
solved using any of several standard algorithms. The results are:
jTAOj D 43.3 kN, jTACj D 6.8 kN, jTADj D 5.5 kN.
y
A
D
O
C
(6, 2, 0) m
4 m
5 m
5 m
8 m
156
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Problem 3.98* The 1350-kg car is at rest on a
plane surface with its brakes locked. The unit vector
en D 0.231iC 0.923jC 0.308k is perpendicular to the
surface. The y axis points upward. The direction cosines
of the cable from A to B are cos �x D �0.816, cos �y D
0.408, cos �z D �0.408, and the tension in the cableis
1.2 kN. Determine the magnitudes of the normal and
friction forces the car’s wheels exert on the surface.
en
ep
y
z
x
B
Solution: Assume that all forces act at the center of mass of the
car. The vector equation of equilibrium for the car is
FS C TAB CW D 0.
Writing these forces in terms of components, we have
W D �mgj D ��1350��9.81� D �13240j N,
FS D FSx iC FSy jC FSzk,
and TAB D TABeAB,
where
eAB D cos �x iC cos �y jC cos �zk D �0.816iC 0.408j� 0.408k.
Substituting these values into the equations of equilibrium and solving
for the unknown components of FS, we get three scalar equations of
equilibrium. These are:
FSx � TABx D 0, FSy � TABy �W D 0,
and FSz � TABz D 0.
Substituting in the numbers and solving, we get
FSx D 979.2 N, FSy D 12, 754 N,
and FSz D 489.6 N.
The next step is to find the component of FS normal to the surface.
This component is given by
FN D FN Ð en D FSxeny C FSxeny C FSzenz.
Substitution yields
FN D 12149 N .
From its components, the magnitude of FS is FS D 12800 N. Using
the Pythagorean theorem, the friction force is
f D
√
F2S � F2N D 4033 N.
z
x
y
W
en
FN
FS
F
TAB
"car"
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157
Problem 3.99* The brakes of the car in Problem 3.98
are released, and the car is held in place on the
plane surface by the cable AB. The car’s front wheels
are aligned so that the tires exert no friction forces
parallel to the car’s longitudinal axis. The unit vector
ep D �0.941iC 0.131jC 0.314k is parallel to the plane
surface and aligned with the car’s longitudinal axis.
What is the tension in the cable?
Solution: Only the cable and the car’s weight exert forces in the
direction parallel to ep. Therefore
ep �T� mgj� D 0: ��0.941iC 0.131jC 0.314k�
Р[T��0.816iC 0.408j� 0.408k�� mgj] D 0,
�0.941��0.816�T
C �0.131��0.408T � mg�C �0.314���0.408T� D 0.
Solving, we obtain T D 2.50 kN.
158
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Problem 4.1 In Active Example 4.1, the 40-kN force
points 30° above the horizontal. Suppose that the force
points 30° below the horizontal instead. Draw a sketch
of the beam with the new orientation of the force. What
is the moment of the force about point A?
6 m
40 kN
30°A
Solution: The perpendicular distance from A to the line of action
of the force is unchanged
D D �6 m� sin 30° D 3 m
The magnitude of the moment is therefore unchanged
M D �3 m��40 kN� D 120 kN-m
However, with its new orientation, the force would tend to cause the
beam to rotate about A in the clockwise direction. The moment is
clockwise
M D 120 kN-m clockwise
Problem 4.2 The mass m1 D 20 kg. The magnitude of
the total moment about B due to the forces exerted on
bar AB by the weights of the two suspended masses is
170 N-m. What is the magnitude of the total moment
due to the forces about point A?
0.35 m
A
m1 m2
B
0.35 m 0.35 m
Solution: The total moment about B is
MB D m2�9.81 m/s2��0.35 m�C �20 kg��9.81 m/s2��0.7 m�
D 170 N-m
Solving, we find m2 D 9.51 kg
The moment about A is then
jMAj D �20 kg��9.81 m/s2��0.35 m�C �9.51 kg��9.81 m/s2��0.7 m�
jMAj D 134 N-m
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159
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.
A B C D
25 ft10 ft 15 ft
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 D
10 ft, and DAC D 35 ft. The moments are clockwise (negative).
Hence,
∑
MA D �10�40�� 35�44� D �1940 ft-kip .
(b) The normal distances from D to the lines of action are DDB D
40 ft, and DDC D 15 ft. The actions are positive; hence
∑
MD D C�40��40�C �15��44� D 2260 ft-kip
A
B C
D
10 ft 25 ft 15 ft
Problem 4.4 What force F applied to the pliers is
required to exert a 4 N-m moment about the center of
the bolt at P?
42�
165
mm
P
F
Solution:
MP D 4 N-m D F�0.165 m sin 42°�) F D 4 N-m0.165 m sin 42°
D 36.2 N
160
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Problem 4.5 Two forces of equal magnitude F are
applied to the wrench as shown. If a 50 N-m moment is
required to loosen the nut, what is the necessary value
of F?
F
F
300 mm
380 mm
F
F
30�
20�
Solution:
∑
Mnut center D �F cos 30°��0.3 m�C �F cos 20°��0.38 m�
D 50 N-m
F D 50 N-m
�0.3 m� cos 30° C �0.38 m� cos 20° D 81.1 N
Problem 4.6 The force F D 8 kN. What is the
moment of the force about point P?
F
x
y
(7, 2) m
(3, 7) m
(3, 2) m
(8, 5) m
P
QSolution: The angle between the force F and the x axis is
˛ D tan�1�5/4� D 51.3°
The force can then be written
F D �8 kN��cos˛i� sin˛j� D �5.00i� 6.25j� kN
The line of action of the j component passes through P, so it exerts no
moment about P. The moment of the i component about P is clockwise,
and its magnitude is
MP D �5 m��5.00 kN� D 25.0 kN-m MP D 25.0 kN-m clockwise
Problem 4.7 If the magnitude of the moment due to
the force F about Q is 30 kN-m, what is F?
F
x
y
(7, 2) m
(3, 7) m
(3, 2) m
(8, 5) m
P
Q
Solution: The angle between the force F and the x axis is
˛ D tan�1�5/4� D 51.3°
The force can then be written
F D F�cos˛i� sin ˛j� D F�0.625i� 0.781j�
Treating counterclockwise moment as positive, the total moment about
point Q is
MQ D �0.781F��5 m�� �0.625F��2 m� D 30 kN-m
Solving, we find F D 11.3 kN
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161
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?
25°
A
B
F 30°
Solution:
MA D L Ð F sin 30° D L
(
15
2
)
MA D 7.5 L kN Ðm
set MA D MAmax D 18 kN Ðm D 7.5 Lmax
Lmax D 2.4 m 25°
30°
30°
L 
F = 15 kN
Problem 4.9 The length of the bar AP is 650 mm. The
radius of the pulley is 120 mm. Equal forces T D 50 N
are applied to the ends of the cable. What is the sum of
the moments of the forces (a) about A; (b) about P.
T
T
A
P 45�
45�
30�
Solution:
(a)
∑
MA D �50 N��0.12 m�� �50 N��0.12 m� D 0
MA D 0
(b)
∑
MP D �50 N��0.12 m�
� �50 N cos 30°��0.65 m sin 45° C 0.12 m cos 30°�
� �50 N sin 30°��0.65 m cos 45° C 0.12 m sin 30°�
MP D �31.4 N-m or MP D 31.4 N-m CW
162
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Problem 4.10 The force F D 12 kN. A structural
engineer determines that the magnitude of the moment
due to F about P should not exceed 5 kN-m.What
is the acceptable range of the angle ˛? Assume that
0 � ˛ � 90°.
2 m
1 m
F
P
α
Solution: We have the moment about P
MP D �12 kN sin˛��2 m�� �12 kN cos˛��1 m�
MP D 12�2 sin˛� cos ˛� kN-m
The moment must not exceed 5 kN-m
Thus 5 kN-m ½ j12�2 sin˛� cos˛�jkN-m
The limits occur when
12�2 sin˛� cos˛� D 5 ) ˛ D 37.3
12�2 sin˛� cos˛� D �5 ) ˛ D 15.83°
So we must have 15.83° � ˛ � 37.3°
2 m
P
1 m
12 kN
α
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163
Problem 4.11 The length of bar AB is 350 mm.
The moments exerted about points B and C by the
vertical force F are MB D �1.75 kN-m and MC D
�4.20 kN-m. Determine the force F and the length of
bar AC.
B
C
20°
30°
A
F
Solution: We have
1.75 kN-m D F�0.35 m� sin 30° ) F D 10 kN
4.20 kN-m D F�LAC� cos 20° ) LAC D 0.447 m
In summary F D 10 kN, LAC D 447 mm
C
F
B
30°
20°
d2
d1
20°
30°
0.450 m
0.350 m
600 N
164
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Problem 4.12 In Example 4.2, suppose that the 2-kN
force points upward instead of downward. Draw a sketch
of the machine part showing the orientations of the
forces. What is the sum of the moments of the forces
about the origin O?
300 mm
4 kN
2 kN
3 kNO
30�
5 kN
400 mm300 mm
Solution: If the 2-kN force points upward, the magnitude of its
moment about O does not change, but the direction of the moment
changes from clockwise to counterclockwise. Treating counterclock-
wise moments as positive, the moment due to the 2-kN force is
�0.3 m��2 kN� D 0.6 kN-m
The moments due to the other forces do not change, so the sum of the
moments of the four forces is∑
MO D �0.6� 1.039C 1.400� kN-m
∑
MO D 0.961 kN-m
Problem 4.13 Two equal and opposite forces act on
the beam. Determine the sum of the moments of the two
forces (a) about point P; (b) about point Q; (c) about the
point with coordinates x D 7 m, y D 5 m.
2 m 2 m
40 N
30�
P
y
Q
40 N
30�
x
Solution:
(a)
MP D ��40 N cos 30°��2 m�C �40 N cos 30°��4 m�
D 69.3 N-m �CCW�
(b) MQ D �40 N cos 30°��2 m� D 69.3 N-m �CCW�
(c)
M D �40 N sin 30°��5 m�C �40 N cos 30°��5 m�
� �40 N sin 30°��5 m�� �40 N cos 30°��3 m�
D 69.3 N-m �CCW�
40 N
40 N
30° 30°
y
x
P 2 m2 m Q
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165
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
30°
E 40°
12 in.
13
 in
.
Solution: The key is the geometry
From trigonometry,
cos 40° D d2
13 in
, cos 30° D d1
12 in
Thus d1 D �12 in� cos 30°
d1 D 10.3900
and d2 D �13 in� cos 40°
d2 D 9.9600
We are given that
299 in-lb D d2W D 9.96 W
W D 30.0 lb
Now,
Ms D �d1 C d2�W
Ms D �20.35��30.0�
Ms D 611 in-lb clockwise
12 in
30°
40°
S
d1
d2E
13 in
W
Problem 4.15 The magnitudes of the forces exerted on
the pillar at D by the cables A, B, and C are equal: FA D
FB D FC. The magnitude of the total moment about E
due to the forces exerted by the three cables at D is
1350kN-m. What is FA? FA
FB
FC
4 m
A
E
D
D
B C
6 m
4 m
4 m
Solution: The angles between the three cables and the pillar are
˛A D tan�1�4/6� D 33.7°
˛B D tan�1�8/6� D 53.1°
˛C D tan�1�12/6� D 63.4°
The vertical components of each force at point D exert no moment
about E. Noting that FA D FB D FC, the magnitude of the moment
about E due to the horizontal components is
∑
ME D FA�sin˛A C sin˛B C sin˛C��6 m� D 1350 kN-m
Solving for FA yields FA D 100 kN
166
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Problem 4.16 Three forces act on the piping. Deter-
mine the sum of the moments of the three forces about
point P.
0.2 m
P
2 kN
0.2 m
0.2 m
2 kN
4 kN
0.2 m
20�
Solution:
∑
MP D ��4 kN��0.2 m�C �2 kN��0.6 m�� �2 kN cos 20°��0.2 m�
C �2 kN sin 20°��0.4 m� D 10.18 kN-m
MP D 0.298 kN-m CCW
2 kN
2 kN4 kN
0.
2 
m
0.2 m0.2 m0.2 m
P
20°
Problem 4.17 The forces F1 D 30 N, F2 D 80 N, and
F3 D 40 N. What is the sum of the moments of the
forces about point A?
F1
F2
F3
30�
45�
A
B
C
x
y
8 m
2 m
Solution: The moment about point A due to F1 is zero. Treating
counterclockwise moments as positive the sum of the moments is
∑
MA D F3 sin 30°�8 m�C F2 cos 45°�2 m�
∑
MA D 273 N-m counterclockwise
Problem 4.18 The force F1 D 30 N. The vector sum
of the forces is zero. What is the sum of the moments
of the forces about point A?
F1
F2
F3
30�
45�
A
B
C
x
y
8 m
2 m
Solution: The sums of the forces in the x and y directions equal
zero:∑
Fx : F1 C F2 cos 45° � F3 cos 30° D 0
∑
Fy : �F2 sin 45° C F3 sin 30° D 0
Setting F1 D 30 N and solving yields
F2 D 58.0 N, F3 D 82.0 N.
The sum of the moments about point A is
∑
MA D F2 sin 30°�8 m�C F2 cos 45°�2 m� D 410 N-m
∑
MA D 410 N-m counterclockwise
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167
Problem 4.19 The forces FA D 30 lb, FB D 40 lb, FC D
20 lb, and FD D 30 lb. What is the sum of the
moments of the forces about the origin of the coordinate
system? FD30� 
FB FC
FA
y
x
6 ft 4 ft
Solution: The moment about the origin due to FA and FD is
zero. Treating counterclockwise moments as positive, the sum of the
moments is∑
M D �FB�6 ft�C FC�10 ft�
D ��40 lb��6 ft�C �20 lb��10 ft� D �40 ft-lb
∑
M D 40 ft-lb clockwise
Problem 4.20 The force FA D 30 lb. The vector sum
of the forces on the beam is zero, and the sum of the
moments of the forces about the origin of the coordinate
system is zero.
(a) Determine the forces FB, FC, and FD.
(b) Determine the sum of the moments of the forces
about the right end of the beam.
FD30� 
FB FC
FA
y
x
6 ft 4 ft
Solution:
(a) The sum of the forces and the sum of the moments equals zero
∑
Fx : FA cos 30° � FD D 0
∑
Fy : FA sin 30° � FB C FC D 0
∑
Morigin : �FB�6 ft�C FC�10 ft� D 0
Setting FA D 30 lb and solving yields FB D 37.5 lb, FC D 22.5 lb, FD D 26.0 lb
(b) The sum of the moments about the right end is
∑
MRight End : FB�4 ft�� FA sin 30°�10 ft�
D �37.5 lb��4 ft�� �30 lb��10 ft�
D 0
∑
MRight End D 0
168
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Problem 4.21 Three forces act on the car. The sum of
the forces is zero and the sum of the moments of the
forces about point P is zero.
(a) Determine the forces A and B.
(b) Determine the sum of the moments of the forces
about point Q.
B
P Q
A
y
x
2800 lb
6 ft 3 ft
Solution:
∑
Fy : AC B� 2800 lb D 0
∑
MP : ��2800 lb��6 ft�C A�9 ft� D 0
(a) Solving we find
A D 1867 lb, B D 933 lb
(b)
∑
MQ D �2800 lb��3 ft�� B�9 ft� D 0
MQ D 0
6 ft 3 ft
P
B A
Q
2800 lb
Problem 4.22 Five forces act on the piping. The vector
sum of the forces is zero and the sum of the moments
of the forces about point P is zero.(a) Determine the forces A, B, and C.
(b) Determine the sum of the moments of the forces
about point Q.
2 ft
P
80 lb
2 ft
2 ft
20 lb
2 ft
45�
A
B
y
x
Q
C
Solution:
(a) The conditions given in the problem are:
∑
Fx : �AC 80 lb cos 45° D 0
∑
Fy : �B�C� 20 lbC 80 lb sin 45° D 0
∑
MP : ��20 lb��2 ft��C�6 ft�� �80 lb cos 45°��2 ft�
C �80 lb sin 45°��4 ft� D 0
Solving we have
A D 56.6 lb, B D 24.4 lb, C D 12.19 lb
(b)
∑
MQ : ��80 lb cos 45°��2 ft�� �80 lb sin 45°��2 ft�
C�20 lb��4 ft�C B�6 ft� D 0
20 lb
y
x
P
A
B C
Q
80 lb
2 ft 2 ft
2 
ft
2 ft
45°
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169
Problem 4.23 In Example 4.3, suppose that the attach-
ment point B is moved upward and the cable is length-
ened so that the vertical distance from C to B is 9 ft.
(the positions of points C and A are unchanged.) Draw a
sketch of the system with the cable in its new position.
What is the tension in the cable?
2 ft 2 ft
4 ft
C
W
7 ft
B
A
Solution: The angle ˛ between the cable AB and the horizontal is.
˛ D tan�1�5/4� D 51.3°
The sum of the moments about C is∑
MC : �W�2 ft�C T cos˛�4 ft�C T sin˛�4 ft� D 0
Solving yields
T D 106.7 lb
Problem 4.24 The tension in the cable is the same on
both sides of the pulley. The sum of the moments about
point A due to the 800-lb force and the forces exerted
on the bar by the cable at B and C is zero. What is the
tension in the cable?
A
30 in 800 lb
B
C
30 in 30 in
30 
Solution: Let T be the tension in the cable. The sum of the
moments about A is∑
MA : T�30 in�C T sin 30°�90 in�� �800 lb��60in� D 0
Solving yields T D 640 lb
170
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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. 600 
mm
60
0 m
m
40°
20°
160 N
40 N
160 N
C
A
150 
mm
B
Solution: The strategy is to find the perpendicular distance from
the points to the line of action of the forces, and determine the sum
of the moments, using the appropriate sign of the action.
The distance from A to the action line of the weight of the arm AB is:
dAB D �0.300� cos 40° D 0.2298 m
The distance from A to the action line of the weight of the arm BC is
dBC D �0.600��cos 40°�C �0.300��cos 20°� D 0.7415 m.
The distance from A to the line of action of the force is
dF D �0.600��cos 40°�C �0.600��cos 20°�C �0.150��cos 20°�
D 1.1644 m.
The sum of the moments about A is
∑
MA D �dAB�160�� dBC�160�� dF�40� D �202 N-m
Problem 4.26 The space shuttle’s attitude thrusters
exert two forces of magnitude F D 7.70 kN. What
moment do the thrusters exert about the center of
mass G?
18 m 12 m
5° 6°
2.2 m
2.2 m
G
F F
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)
CMFRONTý D �18�F sin 5° � �2.2�F cos 5°
CMREARý D �2.2�F cos 6° � �12�F sin 6°
CMTOTAL DMFRONT CMREAR
CMTOTAL D �4.80C 7.19 N-m
CMTOTAL D 2.39 N-m
5˚
F sin 5°
F cos 5°
2.2 m 2.2 m
18 m 12 m
F cos 6°
F sin 6°
FRONT REAR
6°c
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171
Problem 4.27 The force F exerts a 200 ft-lb counter-
clockwise moment about A and a 100 ft-lb
clockwise moment about B. What are F and �?
 A
(–5, 5) ft
 B
(3, –4) ft
(4, 3) ft
F
θ
y
x
Solution: The strategy is to resolve F into x- and y-components,
and compute the perpendicular distance to each component from A
and B. The components of F are: F D iFX C jFY. The vector from A
to the point of application is:
rAF D �4� ��5��iC �3� 5�j D 9i� 2j.
The perpendicular distances are dAX D 9 ft, and dAY D 2 ft, and the
actions are positive. The moment about A is MA D �9�FY C �2�FX D
200 ft-lb. The vector from B to the point of application is rBF D
�4� 3�iC �3� ��4��j D 1iC 7j; the distances dBX D 1 ft and dBY D
7 ft, the action of FY is positive and the action of FX is nega-
tive. The moment about B is MB D �1�FY � �7�FX D �100 ft-lb. The
two simultaneous equations have solution: FY D 18.46 lb and FX D
16.92 lb. Take the ratio to find the angle:
� D tan�1
(
FY
FX
)
D tan�1
(
18.46
16.92
)
D tan�1�1.091� D 47.5°.
From the Pythagorean theorem
jFj D
√
F2Y C F2X D
p
18.462 C 16.922 D 25.04 lb
A
(–5, 5) ft
B
(3, –4) ft
(4, 3) ft
F
θ
y
x
172
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Problem 4.28 Five forces act on a link in the gear-
shifting 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.
650 mm
650 mm 350 mm
450 mm30 kN
45°
20°
B
Ax
Ay
25 kN
Solution: The strategy is to resolve the forces into x- and
y-components, determine the perpendicular distances from B to the line
of action, determine the sign of the action, and compute the moments.
The angles are measured counterclockwise from the x axis. The
forces are
F2 D 30�i cos 135° C j sin 135°� D �21.21iC 21.21j
F1 D 25�i cos 20° C j sin 20°� D 23.50iC 8.55j.
(a) The sum of the forces is
∑
F D AC BC F1 C F2 D 0.
Substituting:
∑
FX D �AX C BX C 23.5� 21.2�i D 0,
and
∑
FY D �AY C 21.2C 8.55�j D 0.
Solve the second equation: AY D �29.76 kN. The distances of
the forces from A are: the triangle has equal base and altitude,
hence the angle is 45°, so that the line of action of F1 passes
through A. The distance to the line of action of B is 0.65 m,
with a positive action. The distance to the line of action of the
y-component of F2 is �0.650C 0.350� D 1 m, and the action is
positive. The distance to the line of action of the x-component
of F2 is �0.650� 0.450� D 0.200 m, and the action is positive.
The moment about A is
∑
MA D �8.55��1�C �23.5��0.2�C �BX��0.65� D 0.
Solve: BX D �20.38 kN. Substitute into the force equation to
obtain AX D 18.09 kN
(b) The distance from B to the line of action of the y-component of
F1 is 0.350 m, and the action is negative. The distance from B
to the line of action of AX is 0.650 m and the action is negative.
The distance from B to the line of action of AY is 1 m and the
action is positive. The distance from B to the line of action of
the x-component of F2 is 0.450 m and the action is negative. The
sum of the moments about B:
∑
MB D ��0.350��21.21�� �0.650��18.09�
C �1��29.76�� �0.450��23.5� D 0
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173
Problem 4.29 Five forces act on a model truss built by
a civil engineering student as part of a design project.
The dimensions are b D 300 mm and h D 400 mm; FD
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?
b b b b b b
h
60°
F
60°
F
Ax
Ay B
Solution: The x- and y-components of the force F are
F D �jFj�i cos 60° C j sin 60°� D �jFj�0.5iC 0.866j�.
The distance from A to the x-component is h and the action is positive.
The distances to the y-component are 3b and 5b. The distance to B is
6b. The sum of the moments about A is
∑
MA D 2jFj�0.5��h�� 3bjFj�0.866� � 5bjFj�0.866� C 6bB D 0.
Substitute and solve: B D 1.6784jFj
1.8
D 93.2 N
Problem 4.30 Consider the truss shown in Problem
4.29. The dimensions are b D 3 ft and h D 4 ft; F D
300 lb. The vector sum of the forces acting on the truss
is zero, and the sum of the moments of the forces about
the point where Ax and Ay act is zero.
(a) 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:
F D �300�i cos 60° C j sin 60°� D �150i� 259.8j.
(a) The sum of the forces:
∑
F D 2FC AC B D 0.
The x- and y-components:
∑
Fx D �Ax � 300�i D 0,
∑
Fy D ��519.6C Ay C B�j D 0.
Solve the first: Ax D 300 lb. The distance from point A to the
x-components of the forces is h, and the action is positive. The
distances between the point A and the lines of action of the 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
∑
MA D 2�150� h� 3b�259.8� � 5b�259.8�C 6b B D 0.
Substitute and solve: B D 279.7 lb. Substitute this value into the
force equation and solve: Ax D 519.6� 279.7 D 239.9 lb
(b) The distances from B and the line of action of AY is 6b and the
action is negative. The distance between B and the x-component
of the forces is h and the action is positive. The distance between
B and the y-components of the forces is b and 3b, and the action
is positive. The sum of the moments about B:
∑
MB D �6b�239.9� C 2�150� hC b�259.8�C 3b�259.8� D 0
174
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Problem 4.31 The mass m D 70 kg. What is the mom-
ent about A due to the force exerted on the beam at B
by the cable?
3 m
30°
m
B
A
45°
Solution: The strategy is to resolve the force at B into components
parallel to and normal to the beam, and solve for the moment using
the normal component of the force. The force at B is to be determined
from the equilibrium conditions on the cable juncture O. Angles are
measured from the positive x axis. The forces at the cable juncture are:
FOB D jFOBj�i cos 150° C j sin 150°� D jFOBj��0.866iC 0.5j�
FOC D jFOCj�i cos 45° C j sin 45°� D jFOCj�0.707iC 0.707j�.
W D �70��9.81��0i� 1j� D �686.7j (N).
The equilibrium conditions are:
∑
Fx D ��0.866jFOBj C 0.7070jFOCj�i D 0
∑
FY D �0.500jFOBj C �.707jFOCj�� 686.7�j D 0.
Solve: jFOBj D 502.70 N. This is used to resolve the cable tension at B:
FB D 502.7�i cos 330° C j sin 330°� D 435.4i� 251.4j. The distance
from A to the action line of the y-component at B is 3 m, and the
action is negative. The x-component at passes through A, so that the
action line distance is zero. The moment at A is MA D �3�251.4� D
�754.0 N-m
FOB
FOC
O
W
Problem 4.32 The weights W1 and W2 are suspended
by the cable system shown. The weightW1 D 12 lb. The
cable BC is horizontal. Determine the moment about
point P due to the force exerted on the vertical post at
D by the cable CD.
A D
B C
P
W2W1
6 ft
50�
Solution: Isolate part of the cable system near point B. The equi-
librium equations are
∑
Fx : TBC � TAC cos 50° D 0∑
Fy : TAB � 12 lb D 0
Solving yields TAB D 15.7 lb, TBC D 10.1 lb
Let ˛ be the angle between the cable CD and the horizontal. The
magnitude of the moment about P due to the force exerted at D by
cable CD is
M D TCD cos˛�6 ft�
Isolate part of the cable system near point C. From the equilibrium
equation
∑
Fx : TCD cos˛� TBC D 0) TCD cos˛ D TBC D 10.1 lb
Thus M D �10.1 lb��6 ft� M D 60.4 lb
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175
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
A
O
4 m
1 m
1 m 3 m
Solution: The strategy is to resolve the force at B into components
parallel to and normal to the wall, determine the perpendicular distance
from O to the line of action, and compute the moment about O in terms
of the magnitude of the force exerted by the bar.
By inspection, the bar forms a 3, 4, 5 triangle. The angle the bar makes
with the horizontal is cos � D 35 D 0.600, and sin � D 45 D 0.800. The
force at B is FB D jFBj��0.600iC 0.800j�. The perpendicular distance
from O to the line of action of the x-component is �4C 1� D 5 m, and
the action is positive. The distance from O to the line of action of
the y-component is 1 m, and the action is positive. The moment about
O is
∑
MO D 5�0.600�jFBj C 1�0.800�jFBj D 3.8jFBj D 38 kN, from
which jFBj D 10 kN
B
FB
A
O
4 m
1 m
1 m 3 m
θ
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?
4 ft
6 ft
H
7 ft 15 ft
Solution: The strategy is to resolve the line tension into a compo-
nent normal to the rod; use the length from H to tip as the perpen-
dicular distance; determine the sign of the action, and compute the
moment.
The line and rod form two right triangles, as shown in the sketch. The
angles are:
˛ D tan�1
(
2
7
)
D 15.95°
ˇ D tan�1
(
6
15
)
D 21.8°.
The angle between the perpendicular distance line and the fishing line
is � D ˛C ˇ D 37.7°. The force normal to the distance line is F D
5�sin 37.7°� D 3.061 lb. The distance is d D p22 C 72 D 7.28 ft, and
the action is negative. The moment about H is MH D �7.28�3.061� D
�22.3 ft-lb Check: The tension can be resolved into x and y compo-
nents,
Fx D F cosˇ D 4.642 lb, Fy D �F sinˇ D �1.857 lb.
The moment is
M D �2Fx C 7Fy D �22.28 D �22.3 ft-lb. check.
α
α
β
β
2 ft
6 ft
15 ft
7 ft
176
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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 aboutO 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?
20 m
45° 60°
A
BOC
Solution: The strategy is to resolve the cable tensions into compo-
nents 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 D 5�cos 60°� D
2.5 kN. The action is negative. The moment about O is MOA D
�2.5�20� D �50 kN-m
(b) By a similar process, the normal component of the tension in
the cable AC is FCN D jFCj cos 45° D 0.707jFCj. The action is
positive. If the sum of the momentsis zero,
∑
MO D �0.707�20�jFCj � 50� D 0,
from which
jFCj D 50 kN m
�0.707��20 m�
D 3.54 kN
A AFN FN
60° 45°
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177
Problem 4.36 The cable from B to A (the sailboat’s
forestay) exerts a 230-N force at B. The cable from B to
C (the backstay) exerts a 660-N force at B. The bottom
of the sailboat’s mast is located at x D 4 m, y D 0. What
is the sum of the moments about the bottom of the mast
due to the forces exerted at B by the forestay and back-
stay?
y
x
B (4,13) m
C
(9,1) m
A
(0,1.2) m
Solution: Triangle ABP
tan ˛ D 4
11.8
, ˛ D 18.73°
Triangle BCQ
tan ˇ D 5
12
, ˇ D 22.62°
CMO D �13��230� sin˛� �13��660� sinˇ
CMO D �2340 N-m
230 N 660 N
A (0,1.2)
O (4,0)
C (9,1)
B (4,13)
α
β
P
Q
α
α
β
13 m
O
230 sin β660 sin 
178
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 4.37 The cable AB exerts a 290-kN force on
the building crane’s boom at B. The cable AC exerts a
148-kN force on the boom at C. Determine the sum of
the moments about P due to the forces the cables exert
on the boom.
8 m
16 m
38 m
56 m
PG
A
B C
Boom
Solution:
∑
MP D � 8p
3200
�290 kN��56 m�� 8p
320
�148 kN��16 m�
D �3.36 MNm
∑
MP D 3.36 MN-m CW
290 
 kN
14
8 
kN 16
88
56
B C
A
P40 m 16 m
Problem 4.38 The mass of the building crane’s boom
in Problem 4.37 is 9000 kg. Its weight acts at G. The
sum of the moments about P due to the boom’s weight,
the force exerted at B by the cable AB, and the force
exerted at C by the cable AC is zero. Assume that the
tensions in cables AB and AC are equal. Determine the
tension in the cables.
Solution:
∑
MP D � 8p
3200
TAB�56 m�� 8p
320
TAC�16 m�
C �9000 kg��9.81 m/s2��38 m� D 0
using TAB D TAC we solve and find
TAB D TAC D 223 kN
TAB
A
B PC
56
8
18 m 22 m 16 m
16
8
T A
C
9000 kg (9.81 m/s2)
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179
Problem 4.39 The mass of the luggage carrier and the
suitcase combined is 12 kg. Their weight acts at A. The
sum of the moments about the origin of the coordinate
system due to the weight acting at A and the vertical
force F applied to the handle of the luggage carrier
is zero. Determine the force F (a) if ˛ D 30°; (b) if
˛ D 50°.
F
1.2 m
0.28 m 0.14 m
A
C
y
x
a
Solution: O is the origin of the coordinate system
∑
MO D F�1.2 m cos ˛�
� �12 kg��9.81 m/s2��0.28 cos˛� 0.14 sin˛� D 0
Solving we find
F D �12 kg��9.81 m/s
2��0.28 cos˛� 0.14 sin˛�
1.2 m cos˛
(a) For ˛ D 30° We find F D 19.54 N
(b) For ˛ D 50° We find F D 11.10 N
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?
1.8 m 1.2 m
7 m
2.4 m
1 m
A
B
C
Solution: The strategy is to resolve the force exerted by the hydra-
ulic cylinder into the normal component about the crane; determine the
distance; determine the sign of the action, and compute the moment.
Two right triangles are constructed: The angle formed by the hydraulic
cylinder with the horizontal is
ˇ D tan�1
(
2.4
1.2
)
D 63.43°.
The angle formed by the crane with the horizontal is
˛ D tan�1
(
1.4
3
)
D 25.02°.
The angle between the hydraulic cylinder and the crane is � D ˇ � ˛ D
38.42° . The normal component of the force is: FN D �300��sin 38.42°�
D 186.42 kN. The distance from point A is d Dp1.42 C 32 D 3.31 m.
The action is positive. The moment about A isMO D C3.31�186.42� D
617.15 kN-m Check: The force exerted by the actuator can be resolved
into x- and y-components, Fx D F cosˇ D 134.16 kN, Fy D F sinˇ
D 268.33 kN. The moment about the point A is M D �1.4Fx C 3.0
Fy D 617.15 kN m. check.
α
α
β
β
3 m
1.4 m
2.4 m
1.2 m
180
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Problem 4.41 The hydraulic piston AB exerts a 400-lb
force on the ladder at B in the direction parallel to the
piston. The sum of the moments about C due to the force
exerted on the ladder by the piston and the weight W of
the ladder is zero. What is the weight of the ladder?
A B
C
3 ft
6 ft 3 ft
6 ft
WSolution: The angle between the piston AB and the horizontal is
˛ D tan�1�3/6� D 26.6°
The sum of the counterclockwise moment about C is∑
MC : W�6 ft�� �400 lb� cos˛�3 ft�� �400 lb� sin˛�3 ft� D 0
Solving yields W D 268 lb
Problem 4.42 The hydraulic cylinder exerts an 8-kN
force at B that is parallel to the cylinder and points from
C toward B. Determine the moments of the force about
points A and D.
1 m
0.6 m Scoop
A B
D
C
0.15 m
0.6 m
1 m
Hydraulic
cylinder
Solution: Use x, y coords with origin A. We need the unit vector
from C to B, eCB. From the geometry,
eCB D 0.780i� 0.625j
The force FCB is given by
FCB D �0.780�8i� �0.625�8j kN
FCB D 6.25i� 5.00j kN
For the moments about A and D, treat the components of FCB as two
separate forces.
CMA D �5, 00��0.15�� �0.6��6.25� kN Ðm
CMA D �3.00 kN Ðm
For the moment about D
C
∑
MD D �5 kN��1 m�C �6.25 kN��0.4 m�
CMD D 7.5 kN Ðm
5.00 kN
6.25 kN
0.6 m
0.15 m
C (−0.15, + 0.6)
A (0 , 0)
5.0 kN
0,4 m
6.25 kN
C
D
m
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181
Problem 4.43 The structure shown in the diagram is
one of the two identical structures that support the scoop
of the excavator. The bar BC exerts a 700-N force at C
that points from C toward B. What is the moment of this
force about K?
H
L K
100
mm
160
mm
Shaft
260
mm
320
mm
380
mm
C
D
1040
mm 1120
mm
260
mm
Scoop
B
180
mm
J
Solution:
MK D � 320p
108800
�700 N��0.52 m� D �353 Nm
MK D 353 Nm CW
52
0 
m
m
320 
80 
700 N 
K 
Problem 4.44 In the structure shown in Problem 4.43,
the bar BC exerts a force at C that points from C
toward B. The hydraulic cylinder DH exerts a 1550-N
force at D that points from D toward H. The sum of the
moments of these two forces about K is zero. What is
the magnitude of the force that bar BC exerts at C?
Solution:
∑
MK D 1120p
1264400
�1550 N��0.26 m�� 320p
108800
F�0.52� D 0
Solving we find
F D 796 N
26
0 
m
m
26
0 
m
m
320 
1120 
100 
1550 N
80 
BC 
K 
182
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Problem 4.45 In Active Example 4.4, what is the
moment of F about the origin of the coordinate system?
(0, 6, 5) ft
A
C (7, 7, 0) ft
y
x
z
B
F
(11, 0, 4) ft
Solution: The vector from the origin to point B is
r D �11iC 4k� ft
From Active Example 4.4 we know that the force F is
F D ��40iC 70j� 40k� lb
The moment of F about the origin is
M D rð F D
∣∣∣∣∣∣
i j k
11 0 4
�40 70 �40
∣∣∣∣∣∣ D ��280iC 280jC 770k� ft-lb
M D ��280iC 280jC 770k� ft-lb
Problem 4.46 Use Eq. (4.2) to determine the momentof the 80-N force about the origin O letting r be the
vector (a) from O to A; (b) from O to B.
80j (N)
(6, 4, 0) mB
O
y
x
(6, 0, 0) mA
Solution:
(a) MO D rOA ð F
D 6ið 80j D 480k �N-m�.
(b) MO D rOB ð F
D �6iC 4j�ð 80j
D 480k �N-m�.
Problem 4.47 A bioengineer studying an injury sus-
tained in throwing the javelin estimates that the magni-
tude of the maximum force exerted was jFj D 360 N and
the perpendicular distance from O to the line of action of
F was 550 mm. The vector F and point O are contained
in the x�y plane. Express the moment of F about the
shoulder joint at O as a vector.
x
y
F
O
Solution: The magnitude of the moment is jFj�0.55 m� D �360 N�
�0.55 m� D 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 direc-
tion. Therefore MO D 198k �N-m�.
550 mm
F
y
x
O
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183
Problem 4.48 Use Eq. (4.2) to determine the moment
of the 100-kN force (a) about A, (b) about B.
6 m
A
y
8 m
12 m
B
x
100j (kN)
Solution: (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 D �8� 0�iC
�0� 6�j D 8i� 6j. The force is F D 100j (kN). The cross product is
rAF ð F D
∣∣∣∣∣∣
i j k
8 �6 0
0 100 0
∣∣∣∣∣∣ D 800k (kN-m)
(b) The coordinates of B are (12,0,0). The position vector from B to
the point of application of the force is rBF D �8� 12� i D �4i. The
cross product is:
rBF ð F D
∣∣∣∣∣∣
i j k
�4 0 0
0 100 0
∣∣∣∣∣∣ D �400k (kN-m)
Problem 4.49 The cable AB exerts a 200-N force on
the support at A that points from A toward B. Use
Eq. (4.2) to determine the moment of this force about
point P in two ways: (a) letting r be the vector from P
to A; (b) letting r be the vector from P to B.
y
x
(1, 0.2) m
(0.3, 0.5) m
(0.9, 0.8) m
B
P
ASolution: First we express the force as a vector. The force points
in the same direction as the position vector AB.
AB D �1� 0.3� miC �0.2� 0.5� mj D �0.7i� 0.3j� m
jABj D
√
�0.7 m�2 C �0.3 m�2 D
p
0.58 m
F D 200 Np
0.58
�0.7i� 0.3j�
(a) MP D PAð F D ��0.6 mi� 0.3 mj�ð 200 Np
0.58
�0.7i� 0.3j�
Carrying out the cross product we find
MP D 102.4 N-mk
(b) MP D PBð F D �0.1 mi� 0.6 mj�ð 200 Np
0.58
�0.7i� 0.3j�
Carrying out the cross product we find
MP D 102.4 N-mk
184
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Problem 4.50 The line of action of F is contained in
the x�y plane. The moment of F about O is 140k (N-
m), and the moment of F about A is 280k (N-m). What
are the components of F?
A
O
y
x
(5, 3, 0) m
(0, 7, 0) m
F
Solution: The strategy is to find the moments in terms of the
components of F and solve the resulting simultaneous equations. The
position vector from O to the point of application is rOF D 5iC 3j.
The position vector from A to the point of application is rAF D �5�
0�iC �3� 7�j D 5i� 4j. The cross products:
rOF ð F D
∣∣∣∣∣∣
i j k
5 3 0
FX FY 0
∣∣∣∣∣∣ D �5FY � 3FX�k D 140k, and
rAF ð F D
∣∣∣∣∣∣
i j k
5 �4 0
FX FY 0
∣∣∣∣∣∣ D �5FY C 4FX�k D 280k.
Take the dot product of both sides with k to eliminate k. The simul-
taneous equations are:
5FY � 3FX D 140, 5FY C 4FX D 280.
Solving: FY D 40, FX D 20, from which F D 20iC 40j (N)
A
(0,7,0) F
(5,3,0)
O
y
x
Problem 4.51 Use Eq. (4.2) to determine the sum of
the moments of the three forces (a) about A, (b) about B.
3 kN 3 kN
0.2 m 0.2 m 0.2 m 0.2 m
A B x
6 kN
y
Solution:
(a) MA D 0.2ið ��3j�C 0.4ið 6jC 0.6ið ��3j�
D O.
(b) MB D ��0.2i�ð ��3j�C ��0.4i�ð 6jC ��0.6i�ð ��3j�
D O.
Problem 4.52 Three forces are applied to the plate.
Use Eq. (4.2) to determine the sum of the moments of
the three forces about the origin O.
200 lb
6 ft 4 ft
3 ft
3 ft
200 lb
500 lb
O x
y
Solution: The position vectors from O to the points of applica-
tion of the forces are: rO1 D 3j, F1 D �200i; rO2 D 10i, F2 D �500j;
rO3 D 6iC 6j, F3 D 200i.
The sum of the moments about O is
∑
MO D
∣∣∣∣∣∣
i j k
0 3 0
�200 0 0
∣∣∣∣∣∣C
∣∣∣∣∣∣
i j k
10 0 0
0 �500 0
∣∣∣∣∣∣C
∣∣∣∣∣∣
i j k
6 6 0
200 0 0
∣∣∣∣∣∣ lb
D 600k� 5000k� 1200k D �5600k ft-lb
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185
Problem 4.53 Three forces act on the plate. Use
Eq. (4.2) to determine the sum of the moments of the
three forces about point P.
0.18 m P
0.10 m
0.12 m
0.28 m 12 kN
3 kN
4 kN
20�
30�
45�
x
y
Solution:
r1 D ��0.12iC 0.08j� m, F1 D �4 cos 45°iC 4 sin 45°j� kN
r2 D �0.16i� m, F2 D �3 cos 30°iC 3 sin 30°j� kN
r3 D �0.16i� 0.1j� m, F3 D �12 cos 20°i� 12 sin 20°j� kN
MP D r1 ð F1 C r2 ð F2 C r3 ð F3
MP D �0.145 kN-m�k D �145 N-m�k
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.
z
A
y
x
150k (N)
(0, 6, 0) m
(6, 0, 0) mSolution:
(a) The perpendicular from A to the line of action of the force lies
in the x�y plane
d D p62 C 62 D 8.485 m
jMj D dF D �8.485��150� D 1270 N-m
(b) M D ��6iC 6j�ð �150k� D �900jC 900i N-m
jMj D p9002 C 9002 D 1270 N-m
186
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Problem 4.55 (a) Determine the magnitude of the
moment of the 600-N force about A by calculating the
perpendicular distance from A to the line of action of
the force.
(b) Use Eq. (4.2) to determine the magnitude of the
moment of the 600-N force about A.
z
y
x
(0.6, 0.5, 0.4) m
0.8 m
A
600i (N)
Solution:
(a) Choose some point P�x, 0, 0.8 m�. on the line of action of the
force. The distance from A to P is then
d D
√
�x � 0.6 m�2 C �0� 0.5 m�2 C �0.8 m� 0.4 m�2
The perpendicular distance is the shortest distance d which occurs
when x D 0.6 m. We have d D 0.6403 m. Thus the magnitude of
the moment is
M D �600 N��0.6403 m� D 384 N-m
(b) Define the point on the end of the rod to be B. Then AB D
��0.6i� 0.5jC 0.4k� m we have
M D ABð F D ��0.6i� 0.5jC 0.4k� mð �600 N�i
M D �240jC 300k� N-m
Thus the magnitude is
M D
√
�240 Nm�2 C �300 Nm�2 D 384 N-m
Problem 4.56 what is the magnitude of the moment of
F about point B?
x
y
z
A
(4, 4, 2) ft
B (8, 1, �2) ft
F � 20i � 10j � 10k (lb)
Solution: The position vector from B to A is
rBA D [�4� 8�iC �4� 1�jC �2� ��2��k] ft
rBA D ��4iC 3jC 4k� ft
The moment of F about B is
MB D rBA ð F D
∣∣∣∣∣∣
i j k
�4 3 4
20 10 �10
∣∣∣∣∣∣ D ��70iC 40j� 100k� ft-lb
Its magnitude is
jMBj D
√
��70 ft-lb�2 C �40 ft-lb�2 C ��100 ft-lb�2 D 128 ft-lb
jMBj D 128 ft-lb
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187
Problem 4.57 In Example 4.5, suppose that the attach-
ment point C is moved to the location (8,2,0) m and the
tension in cable AC changes to 25 kN. What is the sum
of the moments about O due to the forces excertedon
the attachment point A by the two cables?
y
x
z
A
(6, 3, 0) m
(0, 4, 8) m O
(4, 0, 6) m
B
C
Solution: The position vector from A to C is
rAC D [�8� 4�iC �2� 0�jC �0� 6�k] m
rAC D �4iC 2j� 6k�
The force exerted at A by cable AC can be written
FAC D �25 kN� rACjrACj D �13.4iC 6.68j� 20.0k� kN
The total force exerted at A by the two cables is
F D FAB C FAC D �6.70iC 13.3j� 16.7k� kN
The moment about O is
MO D rAB ð F D
∣∣∣∣∣∣
i j k
4 0 6
6.70 13.3 �16.7
∣∣∣∣∣∣ D ��80.1iC 107jC 53.4k� kN-m
MO D ��80.1iC 107jC 53.4k� kN-m
Problem 4.58 The rope exerts a force of magnitude
jFj D 200 lb on the top of the pole at B. Determine the
magnitude of the moment of F about A.
x
y B (5, 6, 1) ft
C (3, 0, 4) ft
z
A
FSolution: The position vector from B to C is
rBC D [�3� 5�iC �0� 6�jC �4� 1�k] ft
rBC D ��2i� 6jC 3k� ft
The force F can be written
F D �200 lb� rBCjrBCj D ��57.1i� 171jC 85.7k� lb
The moment of F about A is
MA D rAB ð F D
∣∣∣∣∣∣
i j k
5 6 1
�57.1 �171 85.7
∣∣∣∣∣∣
D �686i� 486j� 514k� ft-lb
Its magnitude is
jMAj D
√
�686 ft-lb�2 C ��486 ft-lb�2 C ��514 ft-lb�2 D 985 ft-lb
jMAj D 985 ft-lb
188
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Problem 4.59 The force F D 30iC 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
x
z
(4, 3, 3) m
A (8, 2, –4) m
F
Solution: The vector from A to the point of application of F is
r D 4i� 1j� 7k m
and
jrj D p42 C 12 C 72 D 8.12 m
(a) The moment of F about A is
MA D rð F D
∣∣∣∣∣∣
i j k
4 �1 �7
30 20 �10
∣∣∣∣∣∣D150i� 170jC 110k N-m
jMAj D
p
1502 C 1702 C 1102 D 252 N-m
(b) The maximum moment occurs when r ? F. In this case
jMAmax j D jrjjFj
Hence, we need jFj.
jFj D p302 C 202 C 102 D 37.4 �N�
Thus,
jMAmax j D �8.12��37.4� D 304 N-m
Problem 4.60 The direction cosines of the force F are
cos �x D 0.818, cos �y D 0.182, and cos �z D �0.545.
The support of the beam at O will fail if the magnitude of
the moment of F about O exceeds 100 kN-m. Determine
the magnitude of the largest force F that can safely be
applied to the beam. z
y
O
x
F
3 m
Solution: The strategy is to determine the perpendicular distance
from O to the action line of F, and to calculate the largest magnitude of
F from MO D DjFj. The position vector from O to the point of appli-
cation of F is rOF D 3i (m). Resolve the position vector into compo-
nents parallel and normal to F. The component parallel to F is rP D
�rOF Ð eF�eF, where the unit vector eF parallel to F is eF D i cos �X C
j cos �Y C k cos �Z D 0.818iC 0.182j� 0.545k. The dot product is
rOF Ð eF D 2.454. The parallel component is rP D 2.007iC 0.4466j�
1.3374k. The component normal to F is rN D rOF � rP D �3� 2�i�
0.4466jC 1.3374k. The magnitude of the normal component is the
perpendicular distance: D D p12 C 0.44662 C 1.3372 D 1.7283 m.
The maximum moment allowed is MO D 1.7283jFj D 100 kN-m,
from which
jFj D 100 kN-m
1.7283 m
D 57.86 ¾D 58 kN
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189
Problem 4.61 The force F exerted on the grip of the
exercise machine points in the direction of the unit vector
e D 23 i� 23 jC 13 k and its magnitude is 120 N. Determine
the magnitude of the moment of F about the origin O.
x
y
z
O
F
150 mm
200 mm
250 mm
Solution: The vector from O to the point of application of the
force is
r D 0.25iC 0.2j� 0.15k m
and the force is F D jFje
or
F D 80i� 80jC 40k N.
The moment of F about O is
MO D rð F D
∣∣∣∣∣∣
i j k
0.25 0.2 �0.15
80 �80 40
∣∣∣∣∣∣ N-m
or
MO D �4i� 22j� 36k N-m
and
jMOj D
p
42 C 222 C 362 N-m
jMOj D 42.4 N-m
Problem 4.62 The force F in Problem 4.61 points in
the direction of the unit vector e D 23 i� 23 jC 13 k. The
support at O will safely support a moment of 560 N-m
magnitude.
(a) Based on this criterion, what is the largest safe
magnitude of F?
(b) If the force F may be exerted in any direction, what
is its largest safe magnitude?
Solution: See the figure of Problem 4.61.
The moment in Problem 4.61 can be written as
MO D
∣∣∣∣∣∣
i j k
0.25 0.2 �0.15
2
3F � 23F C 13F
∣∣∣∣∣∣ where F D jFj
MO D ��0.0333i� 0.1833j� 0.3k�F
And the magnitude of MO is
jMOj D �
p
0.03332 C 0.18332 C 0.32�F
jMOj D 0.353 F
If we set jMOj D 560 N-m, we can solve for jFmaxj
560 D 0.353jFmaxj
jFmaxj D 1586 N
(b) If F can be in any direction, then the worst case is when r ? F.
The moment in this case is jMOj D jrjjFworstj
jrj D p0.252 C 0.22 C 0.152 D 0.3536 m
560 D �0.3536�jFWORSTj
jFworstj D 1584 N
190
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Problem 4.63 A civil engineer in Boulder, Colorado
estimates that under the severest expected Chinook
winds, the total force on the highway sign will be
F D 2.8i� 1.8j (kN). Let MO be the moment due to F
about the base O of the cylindrical column supporting
the sign. The y component of MO is called the torsion
exerted on the cylindrical column at the base, and the
component of MO parallel to the x�z plane is called
the bending moment. Determine the magnitudes of the
torsion and bending moment.
y
O
F
z
x
8 m
8 m
Solution: The total moment is
M D �8jC 8k� mð �2.8i� 1.8j� kN
D �14.4iC 22.4j� 22.4k� kN-m
We now identify
Torsion D My D 22.4 kN-m
Bending moment D
√
Mx2 CMz2
D
√
�14.4 kNm�2 C �22.4 kNm�2 D 26.6 kN-m
Problem 4.64 The weights of the arms OA and AB of
the robotic manipulator act at their midpoints. The direc-
tion cosines of the centerline of arm OA are cos �x D
0.500, cos �y D 0.866, and cos �z D 0, and the direction
cosines of the centerline of arm AB are cos �x D 0.707,
cos �y D 0.619, and cos �z D �0.342. What is the sum
of the moments about O due to the two forces?
x
z
y
160 N
600
 m
m
B
A
O
200 N
60
0 
m
m
Solution: By definition, the direction cosines are the scalar compo-
nents of the unit vectors. Thus the unit vectors are e1 D 0.5iC 0.866j,
and e2 D 0.707iC 0.619j� 0.342k. The position vectors of the mid-
points of the arms are
r1 D 0.3e1 D 0.3�0.5iC 0.866j� D 0.15iC 0.2598j
r2 D 0.6e1 C 0.3e2 D 0.512iC 0.7053j� 0.1026k.
The sum of moments is
M D r1 ðW1 C r2 ðW2
D
∣∣∣∣∣∣
i j k
0.15 0.2598 0
0 �200 0
∣∣∣∣∣∣C
∣∣∣∣∣∣
i j k
0.512 0.7053 �0.1026
0 �160 0
∣∣∣∣∣∣
D �16.42i� 111.92k (N-m)
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191
Problem 4.65 The tension in cable AB is 100 lb. If
you want the magnitude of the moment about the base
O of the tree due to the forces exerted on the tree by the
two ropes to be 1500 ft-lb, what is the necessary tension
in rope AC ?
B
O
C
x
z
(0, 0, 10) ft
(14, 0, 14) ft
y
A (0, 8, 0) ft
Solution: We have the forces
F1 D 100 lbp
164
��8jC 10k�, F2 D TACp
456
�14i� 8jC 14k�
Thus the total moment is
M D �8 ft�jð �F1 C F2� D �625 ft lbC 5.24 ft TAC�i
� �5.24 ft�TACK
The magnitude squared is then
�625 ft lbC 5.24 ft TAC�2 C �5.24 ft TAC�2 D �1500 ft lb�2
Solving we find
TAC D 134 lb
Problem 4.66* A force F actsat the top end A of the
pole. Its magnitude is jFj D 6 kN and its x component
is Fx D 4 kN. The coordinates of point A are shown.
Determine the components of F so that the magnitude
of the moment due to F about the base P of the pole is
as large as possible. (There are two answers.)
(4, 3, �2) m
x
y
P
A
F
z
Solution: The force is given by F D �4 kNiC Fy jC Fzk�.
Since the magnitude is constrained we must have
�4 kN�2 C Fy2 C Fz2 D �6 kN�2 ) Fz D
√
20 kN2 � Fy2
Thus we will use (suppressing the units)
F D
(
4iC FyjC
√
20� Fy2k
)
The moment is now given by
M D �4iC 3j� 2k�ð F
M D
[
2Fy C 3
√
20� Fy2
]
i�
[
8C 4
√
20� Fy2
]
jC [�12C 4Fy] k
The magnitude is
M2 D 708� 5Fy2 C 64
√
20� Fy2 C 12Fy
(
�8C
√
20� Fy2
)
To maximize this quantity we solve
dM2
dFy
D 0 for the critical values
of Fy .
There are three solutions Fy D �4.00,�3.72,�3.38.
The first and third solutions produce the same maximum moment.
The second answer corresponds to a local minimum and is there-
fore discarded.
So the force that produces the largest moment is
F D �4i� 4jC 2k� or F D �4i� 3.38jC 2.92k�
192
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Problem 4.67 The force F D 5i (kN) acts on the ring
A where the cables AB, AC, and AD are joined. What is
the sum of the moments about point D due to the force
F and the three forces exerted on the ring by the cables?
Strategy: The ring is in equilibrium. Use what you
know about the four forces acting on it.
(12, 4, 2) m
(6, 0, 0) m
(0, 4, 6) m
(0, 6, 0) m
y
B
x
z
C
D
A F
Solution: The vector from D to A is
rDA D 12i� 2jC 2k m.
The sum of the moments about point D is given by
∑
MD D rDA ð FAD C rDA ð FAC C rDA ð FAB C rDA ð F
∑
MD D rDA ð �FAD C FAC C FAB C F�
However, we are given that ring A is in equilibrium and this
implies that
�FAD C FAC C FAB C F� D O D 0
Thus,
∑
MD D rDA ð �O� D 0
FAD
FAC
FAB
A F
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193
Problem 4.68 In Problem 4.67, determine the mo-
ment 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 com-
ponent form, and then solve the resulting three equations for the three
unknown magnitudes. The unit vectors are of the form
eAP D �xP � xA�iC �yP � yA�jC �zP � zA�kjrAPj
Where P takes on values B, C, and D
Calculating the unit vectors, we get


eAB D �0.802i� 0.535j� 0.267k
eAC D �0.949iC 0jC 0.316k
eAD D �0.973iC 0.162j� 0.162k
From equilibrium, we have
FABeAB C FACeAC C FADeAD C 5i �kN� D 0
In component form, we get


i: �0.802FAB � 0.949FAC � 0.973FAD C 5 D 0
j: �0.535FAB C �0�FAC C 0.162FAD D 0
k: �0.267FAB C 0.316FAC � 0.162FAD D 0
Solving, we get
FAB D 779.5 N, FAC D 1976 N
FAD D 2569 N
The vector from D to A is
rDA D 12i� 2jC 2k m
The force FAB is given by
FAB D FABeAB
FAB D �0.625i� 0.417j� 0.208k �kN�
The moment about D is given by
MD D rDA ð FAB D
∣∣∣∣∣∣
i j k
12 �2 2
�0.625 �0.417 �0.208
∣∣∣∣∣∣
MD D 1.25iC 1.25j� 6.25k �kN-m�
FAC
FAD
D(0, 6, 0)
C(0, 4, 6) m
B(6, 0, 0) m
A(12, 4, 2) m F = 5i (kN)
194
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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.
35 m
35 m
40 m
40 m
40 m
A
y
x
z
B
O
C
D
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:
rAD D ��35� 0�iC �0� 70�jC ��35� 0�k
rAD D �35i� 70k� 35k
rAC D ��40� 0�iC �0� 70�jC �40� 0�k
rAC D �40i� 70jC 40k
rAB D �40� 0�iC �0� 70�jC �0� 0�k
rAB D 40i� 70jC 0k
The unit vectors corresponding to these position vectors are:
eAD D rADjrADj D
�35
85.73
i� 70
85.73
j� 35
85.73
D �0.4082i� 0.8165j� 0.4082k
eAC D rACjrACj D �
40
90
i� 70
90
jC 40
90
k
D �0.4444i� 0.7778jC 0.4444k
eAB D rABjrABj D
40
80.6
i� 70
80.6
jC 0k D 0.4962i� 0.8682jC 0k
The forces at point A are
TAB D 4eAB D 1.9846i� 3.4729j C 0k
TAC D 2eAB D �0.8889i� 1.5556jC 0.8889k
TAD D 2eAD D �0.8165i� 1.6330j � 0.8165k.
The sum of the forces acting at A are
TA D 0.2792i� 6.6615jC 0.07239k (kN-m)
The position vector of A is rOA D 70j. The moment about O is M D
rOA ð TA
M D
∣∣∣∣∣∣
i j k
0 70 0
0.2792 �6.6615 0.07239
∣∣∣∣∣∣
D �70��0.07239�i� j0� k�70��0.2792� D 5.067i� 19.54k
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195
Problem 4.70 Consider the 70-m tower in Prob-
lem 4.69. Suppose that the tension in cable AB is 4 kN,
and you want to adjust the tensions in cables AC and
AD so that the sum of the moments about the origin O
due to the forces exerted by the cables at point A is zero.
Determine the tensions.
Solution: From Varignon’s theorem, the moment is zero only if
the resultant of the forces normal to the vector rOA is zero. From
Problem 4.69 the unit vectors are:
eAD D rADjrADj D
�35
85.73
i� 70
85.73
j� 35
85.73
D �0.4082i� 0.8165j� 0.4082k
eAC D rACjrACj D �
40
90
i� 70
90
jC 40
90
k
D �0.4444i� 0.7778jC 0.4444k
eAB D rABjrABj D
40
80.6
i� 70
80.6
jC 0k D 0.4963i� 0.8685j C 0k
The tensions are TAB D 4eAB, TAC D jTACjeAC, and TAD D jTADjeAD.
The components normal to rOA are
∑
FX D ��0.4082jTADj � 0.4444jTACj C 1.9846�i D 0
∑
FZ D ��0.4082jTADj C 0.4444jTACj�k D 0.
The HP-28S calculator was used to solve these equations:
jTACj D 2.23 kN, jTADj D 2.43 kN
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.
C
y
x
D
z
A
B
5 m
5 m
8 m
8 m
4 m
Solution: The coordinates of the points A, B, C are A (8, 0, 0),
B (0, 4, �5), C (0, 8, 5), D(0, 0, 5). The point A is the intersection of
the lines of action of the forces. The position vector DA is
rDA D 8iC 0j� 5k.
The position vectors AB and AC are
rAB D �8iC 4j� 5k, rAB D
p
82 C 42 C 52 D 10.247 m.
rAC D �8iC 8jC 5k, rAC D
p
82 C 82 C 52 D 12.369 m.
The unit vectors parallel to the cables are:
eAB D �0.7807iC 0.3904j� 0.4879k,
eAC D �0.6468iC 0.6468jC 0.4042k.
The tensions are
TAB D 150eAB D �117.11iC 58.56j� 73.19k,
TAC D 100eAC D �64.68iC 64.68jC 40.42k.
The sum of the forces exerted by the wall on A is
TA D �181.79iC 123.24j� 32.77k.
The force exerted on the wall by the cables is �TA. The moment about
D is MD D �rDA ð TA,
MD D
∣∣∣∣∣∣
i j k
8 0 �5
181.79 �123.24 C32.77
∣∣∣∣∣∣ D ��123.24��5�i
� ��8��C32.77�� ��5��181.79��jC �8���123.24�k
MD D �616.2i� 117.11j� 985.9k (N-m)
(Note: An alternate method of solution is to express the moment in
terms of the sum: MD D �rDC ð TC C �rDB ð TB�.�
C
y
x
D Fz
A
B
5 m
5 m
8 m
8 m
4 m
196
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Problem 4.72 Consider the wall shown in Prob-
lem 4.71. The total force exerted by the two cables in
the direction perpendicular to the wall is 2 kN. The
magnitude of the sum of the moments about D due to
the forces exerted on the wall by the cables is 18 kN-m.
What are the tensions in the cables?
Solution: From the solution of Problem 4.71, we have rDA D 8iC
0j� 5k. Forces in both cables pass through point A and we can use
this vector to determine moments of both forces about D. The position
vectors AB and AC are
rAB D �8iC 4j� 5k, jrABj D
p
82 C 42 C 52 D 10.247 m.
rAC D �8iC 8jC 5k, jrACj D
p
82 C 82 C 52 D 12.369 m.
The unit vectors parallel to the cables are:
eAB D �0.7807iC 0.3904j� 0.4879k,
eAC D �0.6468iC 0.6468jC 0.4042k.
The tensions are
TBA D �TBAeAB D �TBA��0.7807iC 0.3904j � 0.4879k�, and
TCA D �TCAeAC D �TCA��0.6468iC 0.6468jC 0.4042k�.
The sum of the forces exerted by the cables perpendicular to the wall
is given by
TPerpendicular D TAB�0.7807� C TAC�0.6468� D 2 kN.
The moments of these two forces about D are given by
MD D �rDA ð TCA�C �rDA ð TBA� D rDA ð �TCA C TBA�.
The sum of the two forces is given by
MD D
∣∣∣∣∣∣
i j k
8 0 �5
�TCA C TCB�X �TCA C TCB�Y �TCA C TCB�Z
∣∣∣∣∣∣ .
This expression can be expanded to yield
MD D 5�TCA C TCB�YiC [�8�TCA C TCB�Z � 5�TCA C TCB�X]j
C 8�TCA C TCB�Yk.
The magnitude of this vector is given as 18 kN-m. Thus, we obtain
the relation
jMDj D
√
25�TCA C TCB�2Y C [�8�TCA C TCB�Z
�5�TCA C TCB�X]2 C 64�TCA C TCB�2Y D 18 kN-m.
We now have two equations in the two tensions in the cables. Either
algebraic substitution or a numerical solver can be used to give
TBA D 1.596 kN, and TCA D 1.166 kN.
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197
Problem 4.73 The tension in the cable BD is 1 kN. As
a result, cable BD exerts a 1-kN force on the “ball” at
B that points from B toward D. Determine the moment
of this force about point A.
x
y
z
C
BD
A
(0, 5, 5) m
(0, 4, �3) m
(4, 3, 1) m
E
Solution: We have the force and position vectors
F D 1 kN
6
��4iC 2jC 4k�, r D AB D �4iC 3jC k� m
The moment is then
M D rð F D �1.667i� 3.33jC 3.33k� kN-m
Problem 4.74* Suppose that the mass of the sus-
pended object E in Problem 4.73 is 100 kg and the mass
of the bar AB is 20 kg. Assume that the weight of the
bar acts at its midpoint. By using the fact that the sum
of the moments about point A due to the weight of the
bar and the forces exerted on the “ball” at B by the three
cables BC, BD, and BE is zero, determine the tensions
in the cables BC and BD.
Solution: We have the following forces applied at point B.
F1 D ��100 kg��9.81 m/s2�j, F2 D TBCp
33
��4iC j� 4k�,
F3 D TBD6 ��4iC 2jC 4k�
In addition we have the weight of the bar F4 D ��20 kg��9.81 m/s2�j
The moment around point A is
MA D �4iC 3jC k� mð �F1 C F2 C F3�
C �2iC 1.5jC 0.5k� mð F4 D 0
Carrying out the cross products and breaking into components we find
Mx D 1079� 2.26TBC C 1.667TBD D 0
My D 2.089TBC � 3.333TBD D 0
Mz D �4316C 2.785TBC C 3.333TBD D 0
Only two of these three equations are independent. Solving we find
TBC D 886 N, TBD D 555 N
198
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Problem 4.75 The 200-kg slider at A is held in place
on the smooth vertical bar by the cable AB. Determine
the moment about the bottom of the bar (point C with
coordinates x D 2 m, y D z D 0) due to the force exerted
on the slider by the cable.
z
y
x
A
C
B
2 m
5 m
2 m
2 m
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
Thus, rAB D �2iC 3jC 2k m
eAB D �0.485iC 0.728jC 0.485k
and FAB D FABeAB
The horizontal force exerted by the bar on the slider is
H D HxiCHzk
Equilibrium requires HC FAB � mgj D 0
i: Hx � 0.485FAB D 0 m D 200 kg
j: 0.728FAB � mg D 0 g D 9.81 m/s2
k: Hz C 0.485FAB D 0
Solving, we get
FAB D 2697N D 2, 70 kN
Hx D 1308N D 1.31 kN
Hz D �1308N D �1.31 kN
rCA D 2j m
FAB D FABeAB
FAB D �1308iC 1962jC 1308k N
Mc D
∣∣∣∣∣∣
i j k
0 2 0
�1308 1962 1308
∣∣∣∣∣∣
Mc D 2616iC 0jC 2616k N-m
Mc D 2.62iC 2.62i kN-m
FAB
H
−mg j
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199
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
x
A
B
D
O
C
(6, 2, 0) m
z
12 m
3 m
4 m
5 m
5 m
8 m
Solution: To find the moment, we must find the force in cable AB.
In order to do this, we must find the forces in cables AO and AD also.
This requires that we solve the equilibrium problem at A.
Our first task is to write unit vectors eAB, eAO, eAC, and eAD. Each
will be of the form
eAi D �xi � xA�iC �yi � yA�jC �zi � zA�k√
�xi � xA�2 C �yi � yA�2 C �zi � zA�2
where i takes on the values B, C, D, and O. We get
eAB D 0.986iC 0.164jC 0k
eAC D �0.609iC 0.609jC 0.508k
eAD D �0.744iC 0.248j� 0.620k
eAO D �0.949i� 0.316jC 0k
We now write the forces as
TAB D TABeAB
TAC D TACeAC
TAD D TADeAD
TAO D TAOeAO
We then sum the forces and write the equilibrium equations in compo-
nent form.
For equilibrium at A,
∑
FA D 0
∑
FA D TAB C TAC C TAD C TAO D 0.
TAC
TAO
TAD
TAB
A (6, 2, 0) m
C (0, 8, 5) m
D (0, 4, −5) m
O (0, 0, 0) m
B(12, 3, 0) m
In component form,


TABeABx C TACeACx C TADeADx C TAOeAOx D 0
TABeABy C TACeACy C TADeADy C TAOeAOy D 0
TABeABz C TACeACz C TADeADz C TAOeAOz D 0
We know TAC D 22 kN. Substituting this in, we have 3 eqns in 3
unknowns. Solving, we get
TAB D 163.05 kN, TAD D 18.01 kN TAO D 141.28 kN
We now know that TAB is given as
TAB D TABeAB D 160.8iC 26.8j �kN�
and that the force acting at B is ��TAB�.
The moment about the bottom of the post is given by
MBOTTOM D rð ��TAB� D 3jð ��TAB�
Solving, we get
MBOTTOM D 482k �kN-m�
200
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Problem 4.77 The force F D 20iC 40j� 10k (N). Use
both of the procedures described in Example 4.7 to deter-
mine the moment due to F about the z axis.
y
x
F
(8, 0, 0) m
z
Solution: First Method: We can use Eqs. (4.5) and (4.6)
r D �8i� m
F D �20iC 40j� 10k� N
Mz�axis D [k Ð �rð F�]k
jMz�axisj D k Ð �rð F� D
∣∣∣∣∣∣
0 01
8 m 0 0
20 N 40 N �10 N
∣∣∣∣∣∣ D 320 N-m
Mz�axis D �320 N-m�k
Second Method: The y-component of the force is perpendicular to the
plane containing the z axis and the position vector r. The perpendicular
distance from the z axis to the y-component of the force is 8 m.
Therefore
jMz�axisj D �40 N��8 m� D 320 N-m
Using the right-hand rule we see that the moment is about the Cz axis.
Thus
Mz�axis D �320 N-m�k
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 ifyou can write
down the results without using the equations.)
(7, 4, 0) m
20k (N)
z
y
x
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
p
42 C 72 D p65 m.
By inspection, the moment about the x axis is
Mx D �4��20�i (N-m)
Mx D 80i �N-m�
By inspection, the moment about the y axis is My D �7��20���j� N-m
My D �140j (N-m)
By inspection, the moment about the z axis is zero since F is parallel
to the z axis.
Mz D 0 �N-m�
Now for the calculations using (4.5) and (4.6)
ML D [e Ð �rð F�]e
Mx D
∣∣∣∣∣∣
1 0 0
7 4 0
0 0 20
∣∣∣∣∣∣ i D 80i �N-m�
My D
∣∣∣∣∣∣
0 1 0
7 4 0
0 0 20
∣∣∣∣∣∣ j D �140j �N-m�
Mz D
∣∣∣∣∣∣
0 0 1
7 4 0
0 0 20
∣∣∣∣∣∣ k D 0k �N-m�
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201
Problem 4.79 Three forces parallel to the y axis act
on the rectangular plate. Use Eqs. (4.5) and (4.6) to
determine the sum of the moments of the forces about
the x axis. (First see if you can write down the result
without using the equations.)
3 kN
x
y
z
6 kN2 kN
900 mm
600 mm
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.
∑
Mx D [�2��0.6�� �6��0.6�]i kN
∑
Mx D �2.4i kN
Calculating the result:
M3 kN D
∣∣∣∣∣∣
1 0 0
0 0 0
0 �3 0
∣∣∣∣∣∣ i D 0i kN
M2 kN D
∣∣∣∣∣∣
1 0 0
0 0 .6
0 �2 0
∣∣∣∣∣∣ i D 1.2i kN
M6 kN D
∣∣∣∣∣∣
1 0 0
0 0 .6
0 6 0
∣∣∣∣∣∣ i D �3.6i kN
∑
Mx DM3 kN CM2 kN CM6 kN
∑
Mx D 0C 1.2i� 3.6i �kN�
∑
Mx D �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 D eY Ð �rð F�. The position vectors of the three forces are given
in the solution to Problem 4.79. The magnitude for each force is:
eY Ð �rð F� D
∣∣∣∣∣∣
0 1 0
0.9 0 0
0 �3 0
∣∣∣∣∣∣ D 0,
eY Ð �rð F� D
∣∣∣∣∣∣
0 1 0
0.9 0 0.6
0 6 0
∣∣∣∣∣∣ D 0,
eY Ð �rð F� D
∣∣∣∣∣∣
0 1 0
0 0 0.6
0 �2 0
∣∣∣∣∣∣ D 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 Ð �rð F� D
∣∣∣∣∣∣
0 1 0
0.9 0 0
0 �3 0
∣∣∣∣∣∣ D �2.7,
eZ Ð �rð F� D
∣∣∣∣∣∣
0 0 1
0.9 0 0.6
0 C 6 0
∣∣∣∣∣∣ D 5.4,
eZ Ð �rð F� D
∣∣∣∣∣∣
0 0 1
0 0 0.6
0 �2 0
∣∣∣∣∣∣ D 0.
Thus the moment about the z axis is
∑
MZ D �2.7eZ C 5.4eZ D 2.7k (kN-m)
202
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Problem 4.81 The person exerts a force F D 0.2i�
0.4jC 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?
2 ft
y
x
3.5 ft
A
B
C
Solution:
M D [e Ð �rð F�]e
e D j, r D 2i ft, F is given
MY D
∣∣∣∣∣∣
0 1 0
2 0 0
.2 �.4 1.2
∣∣∣∣∣∣ j D �2.4j �ft-lb�
Problem 4.82 Four forces act on the plate. Their
components are
FA D �2iC 4jC 2k (kN),
FB D 3j� 3k (kN),
FC D 2jC 3k (kN),
FD D 2iC 6jC 4k (kN).
Determine the sum of the moments of the forces
(a) about the x axis; (b) about the z axis.
FA
FD FC
FB
x
y
z
3 m
2 m
Solution: Note that FA acts at the origin so no moment is generated
about the origin. For the other forces we have
MO D
∣∣∣∣∣∣∣
i j k
3 m 0 0
0 3 kN �3 kN
∣∣∣∣∣∣∣C
∣∣∣∣∣∣∣
i j k
3 m 0 2 m
0 2 kN 3 kN
∣∣∣∣∣∣∣
C
∣∣∣∣∣∣∣
i j k
0 0 2 m
2 kN 6 kN 4 kN
∣∣∣∣∣∣∣
MO D ��16iC 4jC 15k� kN-m
Now we find
Mx DMO Ð i D �16 kN-m, Mz DMO Ð k D 15 kN-m
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203
Problem 4.83 The force F D 30iC 20j� 10k (lb).
(a) What is the moment of F about the y axis?
(b) Suppose that you keep the magnitude of F fixed,
but you change its direction so as to make the
moment of F about the y axis as large as possible.
What is the magnitude of the resulting moment?
z
x
(4, 2, 2) ft
F
y
Solution:
(a) My D j Ð [�4iC 2jC 2k� ftð �30iC 20j� 10k� lb]
My D
∣∣∣∣∣∣∣
0 1 0
4 ft 2 ft 2 ft
30 lb 20 lb �10 lb
∣∣∣∣∣∣∣ D 100 ft lb
)My D �100 ft-lb�j
(b)
Mymax D Fd D �
p
302 C 202 C 102 lb��p42 C 22 ft�
D 167.3 ft-lb
Note that d is the distance from the y axis, not the distance from
the origin.
Problem 4.84 The moment of the force F shown in
Problem 4.83 about the x axis is�80i (ft-lb), the moment
about the y axis is zero, and the moment about the z axis
is 160k (ft-lb). If Fy D 80 lb, what are Fx and Fz?
Solution: The magnitudes of the moments:
e ž �rð F� D
∣∣∣∣∣∣
eX eY eZ
rX rY rZ
FX FY FZ
∣∣∣∣∣∣ ,
eZ Ð �rð F� D
∣∣∣∣∣∣
0 0 1
4 2 �2
FX 80 FZ
∣∣∣∣∣∣ D 320� 2FX D 160
Solve: FX D 80 lb, FZ D 40 lb, from which the force vector is F D
80iC 80jC 40k
204
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Problem 4.85 The robotic manipulator is stationary.
The weights of the arms AB and BC act at their
midpoints. The direction cosines of the centerline of arm
AB are cos �x D 0.500, cos �y D 0.866, cos �z D 0, and
the direction cosines of the centerline of arm BC are
cos �x D 0.707, cos �y D 0.619, cos �z D �0.342. What
total moment is exerted about the z axis by the weights
of the arms?
x
z
y
600
 m
m
C
B
A
200 N
60
0 
m
m
160 N
Solution: The unit vectors along AB and AC are of the form
e D cos �x iC cos �yjC cos �zk.
The unit vectors are
eAB D 0.500iC 0.866jC 0k and eBC D 0.707iC 0.619j� 0.342k.
The vector to point G at the center of arm AB is
rAG D 300�0.500iC 0.866jC 0k� D 150iC 259.8jC 0k mm,
and the vector from A to the point H at the center of arm BC is
given by
rAH D rAB C rBH D 600eAB C 300eBC
D 512.1iC 705.3j� 102.6k mm.
The weight vectors acting at G and H are WG D �200j N, and WH D
�160j N. The moment vectors of these forces about the z axis are of
the form
e ž �rð F� D
∣∣∣∣∣∣
eX ey ez
rX rY rZ
FX FY FZ
∣∣∣∣∣∣ .
Here, WG and WH take on the role of F, and e D k.
Substituting into the form for the moment of the force at G, we get
e ž �r ð F� D
∣∣∣∣∣∣
0 0 1
0.150 0.260 0
0 �200 0
∣∣∣∣∣∣ D �30 N-m.
Similarly, for the moment of the force at H, we get
e ž �rð F� D
∣∣∣∣∣∣
0 0 1
0.512 0.705 �0.103
0 �160 0
∣∣∣∣∣∣ D �81.9 N-m.
The total moment about the z axis is the sum of the two moments.
Hence, Mz axis D �111.9 N-m
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205
Problem 4.86 In Problem 4.85, what total moment is
exerted about the x axis by the weights of the arms?
Solution: The solution is identical to that of Problem 4.85 except
that e D i. Substituting into the form for the moment of the force at
G, we get
e Ð �rð F� D
∣∣∣∣∣∣
1 0 0
0.150 0.260 0
0 �200 0
∣∣∣∣∣∣ D 0 N-m.
Similarly, for the moment ofthe force at H, we get
e Ð �rð F� D
∣∣∣∣∣∣
1 0 0
0.512 0.705 �0.103
0 �160 0
∣∣∣∣∣∣ D �16.4 N-m.
The total moment about the x axis is the sum of the two moments.
Hence, Mx axis D �16.4 N-m
Problem 4.87 In Active Example 4.6, suppose that the
force changes to F D �2iC 3jC 6k (kN). Determine
the magnitude of the moment of the force about the axis
of the bar BC.
y
F � �2i � 6j � 3k (kN)
x
B
C
A (4, 2, 2) m
(0, 0, 3) m
(0, 4, 0) m
z
Solution: We have the following vectors
rBA D �4iC 2j� 1k� m
F D ��2iC 3jC 6k� kN
rBC D �4j� 3k� m
eBC D rBCjrBCj D �0.8j� 0.6k�
The moment of F about the axis of the bar is
jMBCj D eBC Ð �rð F� D
∣∣∣∣∣∣
0 0.8 �0.6
4 2 �1
�2 3 6
∣∣∣∣∣∣ D �27.2 kN-m
Thus MBC D ��27.2 kN-m�eBC, jMBCj D 27.2 kN-m
206
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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.
(–4, 0, 0) m
B
A (0, 5, 0) m
(7, 4, 0) m
20k (N)
y
x
z
Solution: First, we need the unit vector
eAB D �xB � xA�iC �yB � yA�jC �zB � zA�k√
�xB � xA�2 C �yB � yA�2 C �zB � zA�2
eAB D �0.625i� 0.781j D �eBA
Now, the moment of the 20k (N) force about AB is given as
ML D
∣∣∣∣∣∣
ex ey ez
rx ry rz
Fx Fy Fz
∣∣∣∣∣∣ e 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.
r D �7� 0�iC �4� 5�jC �0� 0�k m
r D 7i� 1jC 0k m
Using eAB
ML D
∣∣∣∣∣∣
�0.625 �0.781 0
7 �1 0
0 0 20
∣∣∣∣∣∣ ��0.625i� 0.781j�
ML D �76.1i� 95.1j �N-m�
Using eBA
ML D
∣∣∣∣∣∣
0.625 0.781 0
7 �1 0
0 0 20
∣∣∣∣∣∣ �0.625iC 0.781j�
ML D �76.1i� 95.1j �N-m�
ŁResults are the same
Problem 4.89 The force F D �10iC 5j� 5k (kip).
Determine the moment of F about the line AB. Draw
a sketch to indicate the sense of the moment.
(6, 0, 0) ft
A
x
(6, 6, 0) ft
B
y
F
z
Solution: The moment of F about pt. A is
MA D �6ið F
D
∣∣∣∣∣∣
i j k
�6 0 0
�10 5 �5
∣∣∣∣∣∣
D �30j� 30k �ft-kip�.
The unit vector j is parallel to line AB, so the moment about AB is
MAB D �j ÐMA�j
D �30j �ft-kip�.
z
x
y
A
B
Direction of moment
−30j (ft-kip)
F
z
y
x
A
B
(6, 0, 0) ft
(6, 6, 0) ft
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207
Problem 4.90 The force F D 10iC 12j� 6k (N).
What is the moment of F about the line OA? Draw a
sketch to indicate the sense of the moment.
y
x
z
(0, 6, 4) m
(8, 0, 6) m
O
A
F
Solution: The strategy is to determine a unit vector parallel to OA
and to use this to determine the moment about OA. The vector parallel
to OA is rOA D 6jC 4k. The magnitude: F. The unit vector parallel
to OA is eOA D 0.8321jC 0.5547k. The vector from O to the point
of application of F is rOF D 8iC 6k. The magnitude of the moment
about OA is
jMOj D eOA Ð �rOF ð F� D
∣∣∣∣∣∣
0 0.8321 0.5547
8 0 6
10 12 �6
∣∣∣∣∣∣
D 89.8614 C 53.251 D 143.1 N-m.
The moment about OA is MOA D jMOAjeOA D 119.1jC 79.4k (N-m).
The sense of the moment is in the direction of the curled fingers of
the right hand when the thumb is parallel to OA, pointing to A.
Problem 4.91 The tension in the cable AB is 1 kN.
Determine the moment about the x axis due to the force
exerted on the hatch by the cable at point B. Draw a
sketch to indicate the sense of the moment.
y
z
x
B
A (400, 300, 0) mm
600 mm
1000 mm
Solution: The vector parallel to BA is
rBA D �0.4� 1�iC 0.3j� 0.6k D �0.6iC 0.3j� 0.6k.
The unit vector parallel to BA is
eBA D �0.6667iC 0.3333j� 0.6667k.
The moment about O is
MO D rOB ð T D
∣∣∣∣∣∣
i j k
1 0 0.6
�0.6667 0.3333 �0.66667
∣∣∣∣∣∣
MO D �0.2iC 0.2667jC 0.3333k.
The magnitude is
jMXj D eX ÐMO D �0.2 kN-m.
The moment is MX D �0.2i kN-m. The sense is clockwise when
viewed along the x axis toward the origin.
208
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Problem 4.92 Determine the moment of the force app-
lied at D about the straight line through the hinges A
and B. (The line through A and B lies in the y�z plane.)
20i – 60j (lb)
D
x
CB
A
20°
E
z
y
6 ft
2 ft
4 ft
4 ft
Solution: From the figure, we see that the unit vector along the
line from A toward B is given by eAB D � sin 20°jC cos 20°k. The
position vector is rAD D 4i ft, and the force vector is as shown in the
figure. The moment vector of a force about an axis is of the form
e ž �rð F� D
∣∣∣∣∣∣
eX ey ez
rX rY rZ
FX FY FZ
∣∣∣∣∣∣ .
For this case,
e ž �rð F� D
∣∣∣∣∣∣
0 � sin 20° cos 20°
4 0 0
20 �60 0
∣∣∣∣∣∣ D �240 cos 20° ft-lb
D �225.5 ft-lb.
The negative sign is because the moment is opposite in direction to
the unit vector from A to B.
Problem 4.93 In Problem 4.92, the tension in the
cable CE is 160 lb. Determine the moment of the force
exerted by the cable on the hatch at C about the straight
line through the hinges A and B.
Solution: From the figure, we see that the unit vector along the line
from A toward B is given by eAB D � sin 20°jC cos 20°k. The position
vector is rBC D 4i ft. The coordinates of point C are (4,�4 sin 20°,
4 cos 20°). The unit vector along CE is �0.703iC 0.592jC 0.394k
and the force vector is as shown acting at point D.
The moment vector is a force about an axis is of the form
e ž �rð F� D
∣∣∣∣∣∣
eX ey ez
rX rY rZ
FX FY FZ
∣∣∣∣∣∣ .
For this case,
rCE D �4iC 3.368jC 2.242k
TCE D 160eCE D �112.488iC 94.715jC 63.049k
e ž �rð F� D
∣∣∣∣∣∣
0 � sin 20° cos 20°
4 0 0
�112.488 94.715 63.049
∣∣∣∣∣∣ D �240 cos 20° ft-lb
D 701 ft-lbs.
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209
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
z
A
BSolution: The position vectors:
rOA D �2.4i� 0.6k (m), rOB D �2.2iC 0.7j� 1.2k (m),
rBA D ��2.4C 2.2�iC �0� 0.7�jC ��0.6C 1.2�k (m)
D �0.2i� 0.7jC 0.6k (m).
The magnitude is jrBAj D 0.9434 m.
The unit vector parallel to BA is
eBA D �0.2120i� 0.7420jC 0.6360k.
The tension is TBA D 130eBA .
The moment of TBA about the origin is
MO D rOB ð TBA D
∣∣∣∣∣∣
i j k
�2.2 0.7 �1.2
�27.56 �96.46 82.68
∣∣∣∣∣∣ ,
or MO D �57.88iC 214.97j C 231.5k.
The magnitude of the moment about the y axis is
jMYj D eY ÐMO D 214.97 N-m.
The moment is MY D eY�214.97� D 214.97j N-m.
210
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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.
(2, 5, –2) ftA
(10, –2, 3) ftB
x
y
z
Solution: The position vector from B to A is
rBA D �2� 10�iC [5� ��2�]jC ��2� 3�k
D �8iC 7j� 5k �ft�,
So the force exerted on B is
F D 200 rBAjrBAj D �136.2iC 119.2j� 85.1k�lb�.
The moment of F about the origin O is
rOB ð F D
∣∣∣∣∣∣
i j k
10 �2 3
�136.2 119.2 �85.1
∣∣∣∣∣∣
D �187iC 443jC 919k �ft-lb�.
The moments about the x, y, and z axes are
[�rOB ð F� Ð i]i D �187i �ft-lb�,
[�rOB ð F� Ð j]j D 443j �ft-lb�,
[�rOB ð F� Ð k]k D 919k �ft-lb�.
z
y
x
443 ft-lb
187 ft-lb
919 ft-b
Problem 4.96 The total force exerted on the blades
of the turbine by the steam nozzle is F D 20i� 120jC
100k (N), and it effectively acts at the point (100, 80,
300) mm. What moment is exerted about the axis of the
turbine (the x axis)?
y
x
z
Fixed
Rotating
Solution: The moment about the origin is
MO D
∣∣∣∣∣∣
i j k
0.1 0.08 0.3
20 �120 100
∣∣∣∣∣∣
D 44.0i� 4.0j� 13.6k �N-m�.
The moment about the x axis is
�MO Ð i�i D 44.0i �N-m�.
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211
Problem 4.97 The pneumatic support AB holds a trunk
lid in place. It exerts a 35-N force on the fixture at B that
points in the direction from A toward B. Determine the
magnitude of the moment of the force about the hinge
axis of the lid, which is the z axis.
z
y
x
(480, �40, 40) mm
A
(60, 100, �30) mmB
O
Solution: The vector from A to B is
rAB D [�60� 480�iC �100� ��40��jC ��30� 40�k] mm
rAB D ��420iC 140j� 70k� mm
The 35-N force can be written
F D �35 N� rABjrABj D ��32.8iC 10.9j� 5.47k� N
The moment about point O is
MO D rOB ð F D
∣∣∣∣∣∣
i j k
60 100 �30
�32.8 10.9 �5.47
∣∣∣∣∣∣
D ��219iC 1310jC 3940k� N-mm
The magnitude of the moment about the z axis is
Mz DMO Ð k D 3940 N-mm D 3.94 N-m
Mz D 3.94 N-m
212
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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
x
3 ft
8 ft
6 ft
B
C
D
A (6, 0, 10) ftz
Solution: The strategy is to find the moment about the point C
exerted by the force at B, and then to find the component of that
moment acting along the line CD. The coordinates of the points B,
C, D are B (8, 6, 0), C (3, 6, 0), D(3, 0, 0). The position vectors
are: rOB D 8iC 6j, rOC D 3iC 6j, rOD D 3i. The vector parallel to
CD is rCD D rOD � rOC D �6j. The unit vector parallel to CD is
eCD D �1j. The vector from point C to B is rCB D rOB � rOC D 5i.
The position vector of A is rOA D 6iC 10k. The vector parallel to
BA is rBA D rOA � rOB D �2i� 6jC 10k. The magnitude is jrBAj D
11.832 ft. The unit vector parallel to BA is
eBA D �0.1690i� 0.5071jC 0.8452k.
The tension acting at B is
TBA D 80eBA D �13.52i� 40.57jC 67.62k.
The magnitude of the moment about CD due to the tension acting at
B is
jMCDj D eCD Ð �rCB ð TBA� D
∣∣∣∣∣∣
0 �1 0
5 0 0
�13.52 �40.57 67.62
∣∣∣∣∣∣
D 338.1 (ft lb).
The moment about CD is MCD D 338.1eCD D �338.1j (ft lb). The
sense of the moment is along the curled fingers of the right hand when
the thumb is parallel to CD, pointing toward D.
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213
Problem 4.99 The magnitude of the force F is 0.2 N
and its direction cosines are cos �x D 0.727, cos �y D
�0.364, and cos �z D 0.582. Determine the magnitude
of the moment of F about the axis AB of the spool.
y
x
z
(160, 475, 290) mm
(200, 400, 0) mm
(�100, 500, 400) mm
B
PA
F
Solution: We have
rAB D �0.3i� 0.1j� 0.4k� m,
rAB D
√
�0.3�2 C �0.12�C �0.4�2 m D
p
0.26 m
eAB D 1p
0.26
�0.3i� 0.1j� 0.4k�
F D 0.2 N�0.727i� 0.364jC 0.582k�
rAP D �0.26i� 0.025j� 0.11k� m
Now the magnitude of the moment about the spool axis AB is
MAB D 0.2 Np
0.26
∣∣∣∣∣∣∣
0.3 �0.1 �0.4
0.26 m �0.025 m �0.11 m
0.727 �0.364 0.582
∣∣∣∣∣∣∣ D 0.0146 N-m
Problem 4.100 A motorist applies the two forces
shown to loosen a lug nut. The direction cosines of
F are cos �x D 413 , cos �y D 1213 , and cos �z D 313 . 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?
–F F
16 in16 in
x
z
y
Solution: The unit vectors for the forces are the direction cosines.
The position vector of the force F is rOF D �1.333k ft. The magnitude
of the moment due to F is
jMOFj D eX Ð �rOF ð F� D
∣∣∣∣∣∣
1 0 0
0 0 �1.333
0.3077F 0.9231F 0.2308F
∣∣∣∣∣∣
jMOFj D 1.230F ft lb.
The magnitude of the moment due to �F is
jM�OFj D eX Ð �r�OF ð�F�
D
∣∣∣∣∣∣
1 0 0
0 0 1.333
�.3077F �0.9231F �0.2308F
∣∣∣∣∣∣ D 1.230F ft lb.
The total moment about the x axis is
∑
MX D 1.230FiC 1.230Fi D 2.46Fi,
from which, for a total magnitude of 32 ft lb, the force to be applied is
F D 32
2.46
D 13 lb
214
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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.
B
3 m
1 m
D
2 m
C
A
2 m
Solution: The strategy is to determine the moment about C due
to A, and determine the component parallel to CD. The moment is
determined from the distance CA and the components of the tension,
which is to be found from the magnitude of the tension and the unit
vector parallel to AB. The coordinates of the points A, B, C, and
D are: A (2, 2, 0), B (3, 0, 1), C (0, 2, 0), and D (0,0,0). The unit
vector parallel to CD is by inspection eCD D �1j. The position vectors
parallel to DC, DA, and DB:
rDC D 2j, rDA D 2iC 2j, rDB D 3iC 1k.
The vector parallel to CA is rCA D 2i. The vector parallel to AB is
rAB D rDB � rDA D 1i� 2jC 1k.
The magnitude: jrABj D 2.4495 m. The unit vector parallel to AB is
eAB D 0.4082i� 0.8165jC 0.4082k.
The tension is
TAB D 2eAB D 0.8165i� 1.633jC 0.8165k.
The magnitude of the moment about CD is
jMCDj D eCD Ð �rCA ð TAB� D
∣∣∣∣∣∣
0 �1 0
2 0 0
0.8164 �1.633 0.8165
∣∣∣∣∣∣
D 1.633 kN-m.
The moment about CD is
MCD D eCDjMCDj D �1.633j (kN-m).
The sense is in the direction of the curled fingers of the right hand
when the thumb is parallel to DC, pointed toward D.
Problem 4.102 The axis of the car’s wheel passes
through the origin of the coordinate system and
its direction cosines are cos �x D 0.940, cos �y D 0,
cos �z D 0.342. The force exerted on the tire by the road
effectively acts at the point x D 0, y D �0.36 m, z D 0
and has components F D �720iC 3660jC 1240k (N).
What is the moment of F about the wheel’s axis?
x
y
z
Solution: We have to determine the moment about the axle where
a unit vector along the axle is
e D cos �x iC cos �y jC cos �zk
e D 0.940iC 0jC 0.342k
The vector from the origin to the point of contact with the road is
r D 0i� 0.36jC 0k m
The force exerted at the point of contact is
F D �720iC 3660jC 1240k N
The moment of the force F about the axle is
MAXLE D [e Ð �rð F�]e
MAXLE D
∣∣∣∣∣∣
0.940 0 0.342
0 �0.36 0
�720 C3660 C1240
∣∣∣∣∣∣ �0.940iC 0.342k� �N-m�
MAXLE D ��508.26��0.940iC 0.342k� �N-m�
MAXLE D �478i� 174k �N-m�
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Problem 4.103 The direction cosines of the centerline
OA are cos �x D 0.500, cos �y D 0.866, and cos �z D 0,
and the direction cosines of the line AG are cos �x D
0.707, cos �y D 0.619, and cos �z D �0.342. What is the
moment about OA due to the 250-N weight? Draw a
sketch to indicate the sense of the moment about the
shaft.
x
z
y
750
 m
m
A
O
60
0 
m
m
250 N
G
Solution: By definition, the direction cosines are the scalar compo-
nents of the unit vectors. Thus the unit vectors are
e1 D 0.5iC 0.866j, and e2 D 0.707iC 0.619j� 0.341k.
The force is W D 250j (N). The position vector of the 250 N weight is
rW D 0.600e1 C 0.750e2 D 0.8303iC 0.9839j � 0.2565k
The moment about OA is
MOA D eOA�eOA Ð �rW ðW��
D
∣∣∣∣∣∣
0.5 0.866 0
0.8303 0.9839 �0.2565
0 �250 0
∣∣∣∣∣∣ e1 D �32.06e1
D �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.
Problem 4.104 The radius of the steering wheel is
200 mm. The distance from O to C is 1 m. The center C
of the steering wheel lies in the x � y plane. The driver
exerts a force F D 10iC 10j� 5k (N) on the wheel at A.
If the angle ˛ D 0, what is the magnitude of the moment
about the shaft OC? Draw a sketch to indicate the sense
of the moment about the shaft.
O
z
x
y
20°
F
A
C
α
Solution: The strategy is to determine the moment about C, and
then determine its component about OC. The radius vectors parallel
to OC and CA are:
rOC D 1�i cos 20° C j sin 20° C 0k� D 0.9397iC 0.3420j.
The line from C to the x axis is perpendicular to OC since it lies in
the plane of the steering wheel. The unit vector from C to the x axis is
eCX D i cos�20� 90�C j sin�20� 90� D 0.3420i� 0.9397j,
where the angle is measured positive counterclockwise from the x axis.
The vector parallel to CA is
rCA D 0.2eCX D C0.0684i� 0.1879j (m).
The magnitude of the moment about OC
jMOCj D eOC Ð �rCA ð F� D
∣∣∣∣∣∣
0.9397 0.3420 0
0.0684 �0.1879 0
10 10 �5
∣∣∣∣∣∣
D 0.9998 D 1 N-m.
The sense of the moment is in the direction of the curled fingers of
the right hand if the thumb is parallel to OC, pointing from O to C.
216
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Problem 4.105* The magnitude of the force F is 10 N.
Suppose that you want to choose the direction of the
force F so that the magnitude of its moment about the
line L is a maximum. Determine the components of F
and the magnitude of its moment about L. (There are
two solutions for F.)
z
y
A (3, 8, 0) m
L
B
(0, 2, 6) m (12, 4, 4) mP
F
x
Solution: The moment of the general force F D Fx iC FyjC Fzk
about the line is developed by
eBA D 3iC 6j� 6k9 D
1
3
�iC 2j� 2k�,
rBP D �12iC 2j� 2k� m,
MBA D eBA Ð �rBP ð F�
This expression simplifies to MBA D � 22 m
3
�Fy C Fz�
We also have the constraint that �10 N�2 D Fx2 C Fy2 C Fz2
Since Fx does not contribute to the moment we set it equal to zero.
Solving the constraint equation for Fz and substituting this into the
expression for the moment we find
MBA D � 223 �Fy š
√
100� Fy2�.) dMBA
dFy
D 0
) Fy D š5
p
2N) Fz D š5
p
2
We thus have two answers:
F D �7.07jC 7.07k� N or F D ��7.07jC 7.07k�
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217
Problem 4.106 The weight W causes a tension of
100 lb in cable CD. If d D 2 ft, what is the moment
about the z axis due to the force exerted by the cable
CD at point C?
(3, 0, 10) ft
(12, 10, 0) ft
(0, 3, 0) ft
W
D
C
d
y
x
z
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:
rOB D 3iC 0jC 10k, rOT D 12iC 10jC 0k.
The vector from the bottom to the top of the bar is
rBT D rOT � rOB D 9iC 10j� 10k.
The magnitude:
jrBTj D
p
92 C 102 C 102 D 16.763 ft.
The unit vector parallel to the bar, pointing toward the top, is
eBT D 0.5369iC 0.5965j � 0.5965k.
The position vector of the point C relative to the bottom of the bar is
rBC D 2eBT D 1.074iC 1.193j� 1.193k.
The position vector of point C relative to the origin is
rOC D rOB C rBC D 4.074iC 1.193jC 8.807k.
The position vector of point D is
rOD D 0iC 3jC 0k.
The vector parallel to CD is
rCD D rOD � rOC D �4.074iC 1.807j� 8.807k.
The magnitude is
jrCDj D
p
4.0742 C 1.8072 C 8.8072 D 9.87 ft.
The unit vector parallel to CD is
eCD D �0.4127iC 0.1831j� 0.8923k.
The tension is
TCD D 100eCD D �41.27iC 18.31j� 89.23k lb.
The magnitude of the moment about the z axis is
jMOj D eZ Ð �rOC ð TCD� D
∣∣∣∣∣∣
0 0 1
4.074 1.193 8.807
�41.27 18.31 �89.23
∣∣∣∣∣∣
D 123.83 ft lb
218
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Problem 4.107* The y axis points upward. The weight
of the 4-kg rectangular plate acts at the midpoint G of
the plate. The sum of the moments about the straight
line through the supports A and B due to the weight of
the plate and the force exerted on the plate by the cable
CD is zero. What is the tension in the cable?
A
(100, 250, 0) mm
B
D
G
C (200, 55, 390) mm
(0, 180, 360) mm
(100, 500, 700) mm
x
z
y
Solution: Note that the coordinates of point G are
(150, 152.5, 195).
We calculate the moment about the line BA due to the two forces
as follows.
eBA D 0.1iC 0.07j� 0.36kp
0.1445
r1 D �0.2i� 0.125jC 0.03k� m,
F1 D TCD ��0.1iC 0.445jC 0.31k�p
0.304125
r2 D �0.15i� 0.0275j� 0.165k� m,
F2 D ��4 kg��9.81 m/s2�j
MBA D eBA Ð �r1 ð F1 C r2 ð F2�
The moment reduces to
MBA D 3.871 N-m� �0.17793 m�TCD D 0) TCD D 21.8 N
Problem 4.108 In Active Example 4.9, suppose that
the point of application of the force F is moved from
(8, 3, 0) m to (8, 8, 0) m. Draw a sketch showing the new
position of the force. From your sketch, will the moment
due to the couple be clockwise or counterclockwise?
Calculate the moment due to the couple. Represent the
moment by its magnitude and a circular arrow indicating
its direction.
y
x
(6, 6, 0) m
(8, 3, 0) m
�F
F
Solution: From Active Example 4.9 we know that
F D �10i� 4j� N
From the sketch, it is evident that the moment will be clockwise.
The moment due to the couple is the sum of the moments of the two
forces about any point. If we determine the sum of the moments about
the point of application of one of the forces, the moment due to that
force is zero and we only need to determine the moment due to the
other force.
Let us determine the moment about the point of application of the
force F. The vector from the point of application of F to the point of
application of the force -F is
r D [�6� 8�iC �6� 8�j] m D ��2i� 2j� m
The sum of the moments of the two forces is
M D rð ��F� D
∣∣∣∣∣∣
i j k
�2 �2 0
�10 4 0
∣∣∣∣∣∣ D �28k N-m
The magnitude of the moment is 28 N-m. Pointing the thumb of the
right hand into the page, the right-hand rule indicates that the moment
is clockwise.
M D 28 N-m clockwise
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219
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?
1000 lb 1000 lb
60°60°
20 ft 20 ft
x
y
Solution: The right hand force is
F D [1000 �lb�]�cos 60°i� sin 60°j�
F D C500i� 867j lb.
The vector from the x intercept of the left force to that of the right
force is r D 40i ft.
The moment is MC D rð F
MC D 40ið �500i� 867j� �ft-lb�
MC D �34700 �ft-lb� k
or MC D �34700 �ft-lb) clockwise
Problem 4.110 The moment of the couple is 600 k
(N-m). What is the angle ˛?
100 N
100 N
(0, 4) m
(5, 0) m
a
a
x
y
Solution:
M D �100 N cos˛��4 m�C �100 N sin˛��5 m� D 600 N-m
Solving yields two answers:
˛ D 30.9° or ˛ D 71.8°
Problem 4.111 Point P is contained in the x�y
plane, jFj D 100 N, and the moment of the couple is
�500k (N-m). What are the coordinates of P?
x
F
30°
P
70°
y
–F
Solution: The force is
F D 100�i cos��30°�C j sin��30°�� D 86.6i� 50j.
Let r be the distance OP. The vector parallel to OP is
r D r�i cos 70° C j sin 70°� D r�0.3420iC 0.9397j�.
The moment is
M D rð F D
∣∣∣∣∣∣
i j k
0.3420r 0.9397r 0
86.6 �50.0 0
∣∣∣∣∣∣ D �98.48rk.
From which, r D 500
98.48
D 5.077 m. From above,
r D 5.077�0.3420i C 0.9397j�.
The coordinates of P are
x D 5.077�0.3420� D 1.74 m, y D 5.077�0.9397� D 4.77 m
220
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Problem 4.112 Three forces of equal magnitude are
applied parallel to the sides of an equilateral triangle.
(a) Show that the sum of the moments of the forces is
the same about any point. (b) Determine the magnitude
of the sum of the moments.
LF
F
F
Solution:
(a) Resolving one of the forces into vector components parallel to
the other two forces results in two equal and opposite forces with
the same line of action and one couple. Therefore the moment
due to the forces is the same about any point.
(b) Determine the moment about one of the vertices of the triangle.
A vertex lies on the line of action of two of the forces, so the
moment due to them is zero. The perpendicular distance to the
line of action of the third force is L cos 30°, so the magnitude of
the moment due to the three force is
M D FL cos 30°
Problem 4.113 In Example 4.10, suppose that the 200
ft-lb couple is counterclockwise instead of clockwise.
Draw a sketch of the beam showing the forces and
couple acting on it. What are the forces A and B?
4 ft 4 ft
x
y
A B
200 ft-lb
Solution: In Example 4.10 we are given that the sum of the forces
is zero and the sum of the moments is zero. Thus
Fy D AC B D 0
MA D B �4 ft�C 200 ft-lb D 0
Solving we find A D 50 lb, B D �50 lb
Problem 4.114 The moments of two couples are
shown. What is the sum of the moments about point P?
50 ft-lb
y
10 ft-lb
x
(–4, 0, 0) ft
P
Solution: The moment of a couple is the same anywhere in the
plane. Hence the sum about the point P is
∑
M D �50kC 10k D �40k ft lb
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221
Problem 4.115 Determine the sum of the moments
exerted on the plate by the two couples.
x
20 lb
30 lb
30 lb
3 ft
2 ft
20 lb
5 ft 4 ft
y
Solution: The moment due to the 30 lb couple, which acts in a
clockwise direction is
M30 D �3�30�k D �90k ft lb.
The moment due to the 20 lb couple, which acts in a counterclockwise
direction, is
M20 D 9�20�k D 180k ft lb.
The sum of the moments is
∑
M D �90kC 180k D C90k 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
900 ft-lb
400 lb
BA
3 ft 4 ft 3 ft 4 ft
Solution: Let the x axis point to the right and the y axis point
upward in the plane of the page. The moments of the forces are
M100 D ��3i�ð �100j� D �300k (ft-lb),
and M400 D �7i�ð ��400j� D �2800k (ft-lb).
The moment of the couple is MC D 900k (ft-lb). Summing the
moments, we get
MTotal D �2200k (ft-lb)
Problem 4.117 Determine the sum of the moments
exerted about A by the couple and the two forces.
0.2 m
30°
100 N
0.2 m 0.2 m0.2 m
300 N-m
200 N
A
Solution:
∑
MA D �0.2i�ð ��200j�C �0.4iC 0.2j�
ð �86.7iC 50j�C 300k �N-m�
∑
MA D �40kC 2.66kC 300k �N-m�
∑
MA D 262.7k �N-m� ' 263k �N-m�
Problem 4.118 The sum of the moments about point
A due to the forces and couples acting on the bar is zero.
(a) What is the magnitude of the couple C?
(b) Determine the sum of the moments about point B
due to the forces and couples acting on the bar.
3 m
3 m5 m
A
B
C
4 kN
4 kN
3 kN2 kN5 kN
20 kN-m
Solution:
(a)
∑
MA D 20 kN-m� �2 kN��5 m�� �4 kN��3 m�
� �3 kN��8�CC D 0
C D 26 kN-m
(b)
∑
MB D ��3 kN��3 m�� �4 kN��3 m�� �5 kN��5 m�
C 20 kN-mC 26 kN-m D 0
222
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Problem 4.119 In Example 4.11, suppose that instead
of acting in the positive z direction, the upper 20-N force
acts in the positive x axis direction. Instead of acting in
the negative z axis direction, let the lower 20-N force
act in the negative x axis direction. Draw a sketch of the
pipe showing the forces acting on it. Determine the sum
of the moments exerted on the pipe by the two couples. x
y
60�
30 N
20 N
20 N
30 N2 m
4 m 4 m
60�
z
Solution: The magnitude of the moment of the 20-N couple is
unchanged,
�2 m��20 N� D 40 N-m.
The direction of the moment vector is perpendicular to eh x-y plane,
and the right-hand rule indicates that it points in the negative z axis
direction. The moment of the 20-N couple is (�40 N-m) k.
The sum of the moments exerted on the pipe by the two couples is
M D ��40 N-m� kC �30 N� cos 60°�4 m� j� �30 N� sin 60°�4 m� k
M D �60j� 144k� N-m
Problem 4.120 (a) What is the moment of the couple?
(b) Determine the perpendicular distance between the
lines of action of the two forces.
z
y
(0, 4, 0) m
(0, 0, 5) m
�2i � 2j � k (kN)
2i � 2j � k (kN)
x
Solution:
(a)
M D �4j� 5k� mð �2i� 2j� k� kN
D ��14i� 10j� 8k� kN-m
(b) M D
√
��14�2 C ��10�2 C ��8�2 kN-m D 18.97 kN-m
F D
√
�2�2 C ��2�2 C ��1�2 kN D 3 kN
M D Fd) d D M
F
D 18.97 kN-m
3 kN
D 6.32 m
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223
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.)
z 60° 80 lb 60° 80 lb
40 lb
40 lb
8 ft
x
y
3 ft
20 lb 20 lb
3 ft
Solution: The moments of two of the couples can be determined
from inspection:
M1 D ��3��20�k D �60k ft lb.
M2 D �8��40�j D 320j ft lb
The forces in the 3rd couple are resolved:
F D �80��i sin 60° C k cos 60°� D 69.282iC 40k
The two forces in the third couple are separated by the vector
r3 D �6iC 8k�� �8k� D 6i
The moment is
M3 D r3 ð F3 D
∣∣∣∣∣∣
i j k
6 0 0
69.282 0 40
∣∣∣∣∣∣ D �240j.
The sum of the moments due to the couples:
∑
M D �60kC 320j� 240j D 80j� 60k ft lb
Problem 4.122 What is themagnitude of the sum of
the moments exerted on the T-shaped structure by the
two couples?
z
y
x
50i + 20j – 10k (lb)
–50i – 20j + 10k (lb)
50j (lb)
–50j (lb)
3 ft
3 ft
3 ft
3 ft
Solution: The moment of the 50 lb couple can be determined by
inspection:
M1 D ��50��3�k D �150k ft lb.
The vector separating the other two force is r D 6k. The moment is
M2 D rð F D
∣∣∣∣∣∣
i j k
0 0 6
50 20 �10
∣∣∣∣∣∣ D �120iC 300j.
The sum of the moments is
∑
M D �120iC 300j� 150k.
The magnitude is
jMj D p1202 C 3002 C 1502 D 356.23 ft lb
z
y
x
50 j (lb)
–50 j (lb)
3 ft
3 ft
3 ft
3 ft
F
–F
224
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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?
3 m
3 m
A (0, 2, 0) m
y
z
C
D
B
x
(6, –2, 3) m
Solution: One condition for a couple is that the sum of a pair
of forces vanish; another is for a non-zero moment to be the same
anywhere. The first condition is demonstrated by determining the unit
vectors parallel to the action lines of the forces. The vector position of
point B is rB D 3i m. The vector position of point A is rA D 2j. The
vector parallel to cable AB is
rBA D rA � rB D �3iC 2j.
The magnitude is:
jrABj D
p
32 C 22 D 3.606 m.
The unit vector:
eAB D rABjrABj D �0.8321iC 0.5547j.
The tension is
TAB D jTABjeAB D �416.05iC 277.35j.
The vector position of points C and D are:
rC D 3iC 3k, rD D 6i� 2jC 3k.
The vector parallel to the cable CD is rCD D rD � rC D 3i� 2j. The
magnitude is jrCDj D 3.606 m, and the unit vector parallel to the cable
CD is eCD D C0.8321i� 0.5547j. The magnitude of the tension in
the two cables is the same, and eBA D �eCD, hence the sum of the
tensions vanish on the plate. The second condition is demonstrated by
determining the moment at any point on the plate. By inspection, the
distance between the action lines of the forces is
rCB D rB � rC D 3i� 3i� 3k D �3k.
The moment is
M D rCB ð TAB D
∣∣∣∣∣∣
i j k
0 0 �3
�416.05 277.35 0
∣∣∣∣∣∣
D 832.05i� 1248.15j (N-m).
The moment about the origin is
MO D �rB � rC�ð TAB D rCB ð TAB,
which is identical with the above expression for the moment. Let rPC
and rPB be the distances to points C and B from an arbitrary point
P on the plate. Then MP D �rPB � rPC�ð TAB D rCB ð TAB which
is identical to the above expression. Thus the moment is the same
everywhere on the plate, and the forces form a couple.
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225
Problem 4.124 The cables AB and CD exert a couple
on the vertical pipe. The tension in each cable is 8 kN.
Determine the magnitude of the moment the cables exert
on the pipe.
x
z
y
D
C
(�1.6, 2.2, �1.2) m
(0.2, 0.6, 0.2) m
(0.2, 1.6, �0.2) m
(1.6, 0, 1.2) m
A
B
Solution:
FAB D 8 kN �1.4i� 0.6jC 1.0k�p
3.32
, rDB D �3.2i� 2.2jC 2.4k� m
M D rBD ð FAB D ��3.34iC 0.702jC 5.09k� kN-m
) M D 6.13 kN-m
Problem 4.125 The bar is loaded by the forces
FB D 2iC 6jC 3k (kN),
FC D i� 2jC 2k (kN),
and the couple
MC D 2iC j� 2k (kN-m).
Determine the sum of the moments of the two forces
and the couple about A.
1 m
1 m
z
y
x
B
FB
FC
A
C
MC
Solution: The moments of the two forces about A are given by
MFB D �1i�ð �2iC 6jC 3k� (kN-m) D 0i� 3jC 6k (kN-m) and
MFC D �2i�ð �1i� 2jC 2k� (kN-m) D 0i� 4j� 4k (kN-m).
Adding these two moments and
MC D 2iC 1j� 2k (kN-m),
we get MTOTAL D 2i� 6jC 0k (kN-m)
Problem 4.126 In Problem 4.125, the forces
FB D 2iC 6jC 3k (kN),
FC D i� 2jC 2k (kN),
and the couple
MC DMCyjCMCzk (kN-m).
Determine the values for MCy and MCz, so that the sum
of the moments of the two forces and the couple about
A is zero.
Solution: From the solution to Problem 4.125, the sum of the
moments of the two forces about A is
MForces D 0i� 7jC 2k (kN-m).
The required moment, MC, must be the negative of this sum.
Thus MCy D 7 (kN-m), and MCz D �2 (kN-m).
226
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Problem 4.127 Two wrenches are used to tighten an
elbow fitting. The force F D 10k (lb) on the right
wrench is applied at (6, �5, �3) in, and the force �F
on the left wrench is applied at (4, �5, 3) in.
(a) 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.
x
y
F
–F
z
Solution: The position vector of the force on the right wrench is
rR D 6i� 5j� 3k. The magnitude of the moment about the x axis is
jMRj D eX Ð �rR ð F� D
∣∣∣∣∣∣
1 0 0
6 �5 �3
0 0 10
∣∣∣∣∣∣ D �50 in lb
(a) The moment about the x axis is
MR D jMRjeX D �50i (in lb).
(b) The moment of the couple is
MC D �rR � rL�ð FR D
∣∣∣∣∣∣
i j k
2 0 �6
0 0 10
∣∣∣∣∣∣ D �20j in lb
(c) 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:
jMXj D eX Ð �rL ð FL� D
∣∣∣∣∣∣
1 0 0
4 �5 3
0 0 �10
∣∣∣∣∣∣ D 50 in lb
from which MXL D 50i in lb, which is opposite in direction and
equal in magnitude to the moment exerted on the x axis by the
right wrench. The left wrench force is applied 2 in nearer the
origin than the right wrench force, hence the moment must be
absorbed by the space between, where it is wanted.
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227
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.
1 m
y
x
50 N
System 1
100 N
1 m
2 m
y
x
System 2
50 N
Solution: The strategy is to check the two conditions for equiv-
alence: (a) the sums of the forces must be equal and (b) the sums of
the moments about an arbitrary point must be equal. The sums of the
forces of the two systems:
∑
FX D 0, (both systems) and
∑
FY1 D �100jC 50j D �50j (N)
∑
FY2 D �50j (N).
The sums of the forces are equal. The sums of the moments about the
left end are:
∑
M1 D ��1��100�k D �100k (N-m)
∑
M2 D ��2��50�k D �100k (N-m).
The sums of the moments about the left end are equal. Choose any
point P at the same distance r D xi from the left end on each beam.
The sums of the moments about the point P are
∑
M1 D ��50x C 100�x � 1��k D �50x � 100�k (N-m)
∑
M2 D ��50�2� x��k D �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
228
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Problem 4.129 Two systems of forces and moments
act on the beam. Are they equivalent?
20 lb
x
y
10 lb
50 ft-lb
30 ft-lb
System 1
System 2
10 lb
2 ft
x
y
2 ft
20 lb
2 ft2 ft
Solution: The sums of the forces are:
∑
FX D 0 (both systems)
∑
FY1 D 10j� 20j D �10j (lb)
∑
FY2 D �20jC 10j D �10j (lb)
Thus the sums of the forces are equal. The sums of the moments about
the left end are:
∑
M1 D ��20��4�kC 50k D �30k (ft lb)
∑
M2 D �C10�2��k� 30k D �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?
10 kN
8 m
System 1
80 kN-m
System 2
System 3
20 kN
10 kN
System 4
4 m4 m
10 kN
10 kN
20 kN
80 kN-m
8 m
8 m
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.
x
y
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229
Problem 4.131 The four systems shown in Problem
4.130 can be made equivalent by adding a
couple to one of the systems. Which system is it, and
what couple must be added?
Solution: From the solution to 4.130, all systems have
∑
F D 10j kN
and systems 1, 2, and 4 have
∑
ML D 80k �kN-m�
system 3 has
∑
ML D 160k �kN-m�.
Thus, we need to add a couple M D �80k �kN-m� to system 3 (clock-
wise 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 O0
along the same line of action. Explain why these systems
are equivalent. (This simple result is called the principle
of transmissibility.)
O
F
System 1 System 2
F
O'
O
Solution: The sum of forces is obviously equal for both systems.
Let P be any point on the dashed line. The moment about P is the
cross product of the distance from P to the line of action of a force
times the force, that is, M D rPL ð F, where rPL is the distance from
P to the line of action of F. Since both systems have the same line of
action, and the forces are equal, the systems are equivalent.
230
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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. 12 m
12 m
16 m
B
A
20 m
D
C
E
6 m
Solution: The angle formed by the single cable with the positive
x axis is
� D 180° � tan�1
(
12
16
)
D 143.13°.
The single cable tension is
T1 D jTj�i cos 143.13° C j sin 143.13°�
D jTj��0.8iC 0.6j�.
The position vector to the center of the log from the left end is rc D 10i.
The moment about the end of the log is
M D rð T1 D jT1j
∣∣∣∣∣∣
i j k
10 0 0
�0.8 0.6 0
∣∣∣∣∣∣ D jTj�6�k (N-m).
For the two cables, the angles relative to the positive x axis are
�1 D 180° � tan�1
(
12
6
)
D 116.56°, and
�2 D 180� tan�1
(
12
26
)
D 155.22°.
The two cable vectors are
TL D jTLj�i cos 116.56° C j sin 116.56°�
D jTLj��0.4472iC 0.8945j�,
TR D jTRj�i cos 155.22° C j sin 155.22°�
D jTRj��0.9079iC 0.4191j�.
Since the vector sum of the forces in the two systems is equal, two
simultaneous equations are obtained:
0.4472jTL j C 0.9079jTRj D 0.8jT1j, and
0.8945jTL j C 0.4191jTRj D 0.6jT1j
Solve:
jTL j D 0.3353jT1j, and
jTRj D 0.7160jT1j.
The tension in the right hand cable is TR D jT1j�0.7160���0.9079iC
0.4191j� D jT1j��0.6500iC 0.3000�. The position vector of the right
end of the log is rR D 20i m relative to the left end. The moments
about the left end of the log for the second system are
M2 D rR ð TR D jT1j
∣∣∣∣∣∣
i j k
20 0 0
�0.6500 0.3000 0
∣∣∣∣∣∣ D jT1j�6�k (N-m).
This is equal to the moment about the left end of the log for System
1, hence the systems are equivalent.
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231
Problem 4.134 Systems 1 and 2 each consist of a
couple. If they are equivalent, what is F?
System 1 System 2
30°
30°
5 m 200 N
x
y
4 m
20°
20°
2 m
(5, 4, 0) m
x
y
F
200 N
F
Solution: For couples, the sum of the forces vanish for both sys-
tems. For System 1, the two forces are located at r11 D 4i, and r12 D
C5j. The forces are F1 D 200�i cos 30° C j sin 30°� D 173.21iC 100j.
The moment due to the couple in System 1 is
M1 D �r11 � r12�ð F1 D
∣∣∣∣∣∣
i j k
4 �5 0
173.21 100 0
∣∣∣∣∣∣ D 1266.05k (N-m).
For System 2, the positions of the forces are r21 D 2i, and r22 D
5iC 4j. The forces are
F2 D F�i cos��20°�C j sin��20°�� D F�0.9397i� 0.3420j�.
The moment of the couple in System 2 is
M2 D �r21 � r22�ð F2 D F
∣∣∣∣∣∣
i j k
�3 �4 0
0.9397 �0.3420 0
∣∣∣∣∣∣ D 4.7848Fk,
from which, if the systems are to be equivalent,
F D 1266
4.7848
D 264.6 N
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.
3 m 3 m
50 N
60 N
System 1
120 N-m
6 m
40 N
System 2
3 m MFA
FB
3 m
Solution: The sums of the forces for System 1 are
∑
FX D 50, and
∑
FY D �FA C 60.
The sums of the forces for System 2 are
∑
FX D FB, and
∑
FY D 40.
For equivalent systems: FB D 50 N, and FA D 60� 40 D 20 N.
The sum of the moments about the left end for
System 1 is
∑
M1 D ��3�FA � 120 D �180 N-m.
The sum of the moments about the left end for
System 2 is
∑
M2 D ��3�FB CM D �150CM N-m.
Equating the sums of the moments, M D 150� 180 D �30 N-m
232
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Problem 4.136 Two equivalent systems of forces and
moments act on the plate. Determine the force F and
the couple M.
50 lb
5 in
8 in
30 lb 10 lb
M
System 1
5 in
8 in
30 lb
System 2
30 lb
F
100 in-lb
Solution: The sums of the forces for System 1 are
∑
FX D 30 lb,
∑
FY D 50� 10 D 40 lb.
The sums of the forces for System 2 are
∑
FX D 30 lb,
∑
FY D F� 30 lb.
For equivalent forces, F D 30C 40 D 70 lb. The sum of the moments
about the lower left corner for System 1 is
∑
M1 D ��5��30�� �8��10�CM D �230CM in lb.
The sum of the moments about the lower left corner for System 2 is
∑
M2 D �100 in lb.
Equating the sum of moments, M D 230� 100 D 130 in lb
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233
Problem 4.137 In Example 4.13, suppose that the 30-
kN vertical force in system 1 is replaced by a 230-kN
vertical force. Draw a sketch of the new system 1. If
you represent system 1 by a single force F as in system
3, at what position D on the x axis must the force be
placed?
y
O
30j (kN) 20i � 20j (kN)
System 1
210 kN-m
x
3 m 2 m
Solution: The first step is to represent system 1 by a single force
F acting at the origin and a couple M (system 2). The force F must
equal the sum of the forcesin system 1:
�F�2 D �F�1
F D �230 kN� jC �20iC 20j� kN
F D �20iC 250j� kN
The moment about the origin in system 2 is M. Therefore M must
equal the sum of the moments about the origin due to the forces and
moments in system 1:
�M�2 D �M�1
M D �230 kN��3 m�C �20 kN��5 m�
C �210 kN-m� D 1000 kN-m
The next step is to represent system 2 by system 3.
The sums of the forces in the two systems are equal. The sums of
the moments about the origin must be equal. The j component of F is
250 kN, so
�M�3 D �M�2
�1000 kN-m� D �250 kN�D
D D 1000 kN-m
250 kN
D 4m
D D 4 m
234
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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?
40 lb
20 lb
x
y
System 2
System 1
x
y
2 ft
30 ft-lb
30 lb
M
A
F
System 3
x
y
F
A
D
2 ft
A
Solution: The sum of the forces in System 1 is
∑
FX D 0i,
∑
FY D ��20C 40� 30�j D �10j lb.
The sum of the moments about the left end for System 1 is
∑
M1 D �2�40�� 4�30�C 30�k D �10k ft lb.
(a) For System 2, the force at A is F D �10j lb
The moment at A is M2 D �10k ft lb
(b) For System 3 the force at D is F D �10j lb. The distance D is
the ratio of the magnitude of the moment to the magnitude of the
force, where the magnitudes are those in System 1:
D D 10
10
D 1 ft
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235
Problem 4.139 Represent the two forces and couple
acting on the beam by a force F. Determine F and deter-
mine where its line of action intersects the x axis.
60 i + 60 j (N)
y
–40 j (N)
280 N-m
3 m 3 m
x
Solution: We first represent the system by an equivalent system
consisting of a force F at the origin and a couple M:
This system is equivalent if
F D �40jC 60iC 60j
D 60iC 20j �N�,
M D �280C �6��60�
D 80 N-m.
We then represent this system by an equivalent system consisting of
F alone:
For equivalence, M D d�Fy�, so
d D M
Fy
D 80
20
D 4 m.
x
y
F
M
d
x
y
F
Problem 4.140 The bracket is subjected to three forces
and a couple. If you represent this system by a force F,
what is F, and where does its line of action intersect the
x axis?
0.65 m
0.4 m
0.2 m
140 N-m
400 N
180 N
200 N
y
x
Solution: We locate a single equivalent force along the x axis a
distance d to the right of the origin. We must satisfy the following
three equations:
∑
Fx D 400 N� 200 N D Rx
∑
Fy D 180 N D Ry
∑
MO D ��400 N��0.6 m�C �200 N��0.2 m�C �180 N��0.65 m�
C 140 Nm D Ryd
Solving we find
Rx D 200 N, Ry D 180 N, d D 0.317 m
236
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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?
Ax
Ay 200 N
x
30 N-m
y
600 N
MA
380 mm 180 mm
Solution: (a) The sum of the forces is
∑
FX D AXi D 0 and
∑
FY D �AY � 600C 200�j D 0,
from which AY D 400 N. The sum of the moments is
∑
ML D �MA � 0.38�600� � 30C 0.560�200��k D 0,
from which MA D 146 N-m. (b) The sum of the moments about the
right end of the beam is
∑
MR D 0.18�600� � 30C 146� 0.56�400� D 0.
(c) The sum of the forces for the new system is
∑
FY D �AY C F�j D 0,
from F D �AY D �400 N, or F D �400j N. The sum of the moments
for the new system is
∑
M D �MA CM� D 0,
from which M D �MA D �146 N-m
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
4 kip
6 kip
x
6 ft
3 ft
3 ft
3 ft
Ax O
Ay B
y
Solution: (a) The sum of the forces is
∑
FX D �AX � 2� 4� 6�i D 0,
from which AX D 12 kip
∑
FY D �AY C B�j D 0.
The sum of the moments about the origin is
∑
MO D �3��6�C �6��4�C �9��2�C 6�B� D 0,
from which B D �10j kip. (b) Substitute into the force balance eq-
uation to obtain AY D �B D 10 kip. (b) The force in the new system
will replace the 2, 4, and 6 kip forces, F D ��2� 4� 6�i D �12i kip.
The force must match the moment due to these forces: FD D 3�6�C
�6��4�C �9��2� D 60 kip ft, from which D D 60
12
D 5 ft, or the 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.
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237
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?
40 kN
85 kN80 kN
30 kN35 kN
y
x
3 m 3 m 3 m 3 m
Solution: The equivalent force must equal the sum of the forces
exerted by the soil:
F D �80C 35C 30C 40C 85�j D 270j kN
The sum of the moments about any point must be equal for the two
systems. The sum of the moments are
∑
M D 3�35�C 6�30�C 9�40�C 12�85� D 1665 kN-m.
Equating the moments for the two systems FD D 1665 kN-m from
which
D D 1665 kN-m
270 kN
D 6.167 m.
Thus the action line intersects the x axis at a distance D D 6.167 m to
the right of the origin.
Problem 4.144 At a particular instant, aerodynamic
forces distributed over the airplane’s surface exert the
88-kN and 16-kN vertical forces and the 22 kN-m
counterclockwise couple shown. If you represent these
forces and couple by a system consisting of a force F
acting at the center of mass G and a couple M, what are
F and M?
9 m
5.7 m
5 m
88 kN
22 kN-m
y
x
16 kN
G
Solution:
∑
Fy D 88 kNC 16 kN D Ry
∑
MG D ��88 kN��0.7 m�C �16 kN��3.3 m�C 22 kN-m D M
Solving we find
Ry D 104 kN, M D 13.2 kN-m
Problem 4.145 If you represent the two forces and
couple acting on the airplane in Problem 4.144 by a
force F, what is F, and where does its line of action
intersect the x axis?
Solution:
∑
Fy D 88 kNC 16 kN D Ry
∑
MOrigin D �88 kN��5 m�C �16 kN��9 m�C 22 kN-m D Ryx
Solving we find
F D Ryj D 104 kNj, x D 5.83 m
238
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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?
100 lb
A
100 lbA
B C60° 40°
y
FAB FAC
x
Solution: The sum of the forces acting at A is in opposition to the
weight, or F D jWjj D 100j lb.
The moment about point A is zero.
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
x
5 m
6 m 4 m30 N
30 N
50 N
O
Solution: (a) The sum of the forces is
∑
FX D 30i N, and
∑
FY D �30C 50�j D 80j N.
The equivalent at O is F D 30iC 80j (N). The sum of the moments
about O:
∑
M D ��5�30�C 10�50�� D 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 D 350
80
D 4.375 m
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239
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?
400 mm
300 mm 300 mm
800 mm
O
A
C
D
B
y
x
Solution: From the right triangle, the angle between the positive
x axis and the cable AB is
� D � tan�1
(
400
800
)
D �26.6°.
The tension in AB is
TAB D 400�i cos��26.6°�Cj sin��26.6°�� D 357.77i� 178.89j (N).
The angle between the positive x axis and the cable CD is
˛ D � tan�1
(
300
800
)
D �20.6°.
The tension in CD is
TCD D 600�i cos��20.6°�C j sin��20.6°�� D 561.8i� 210.67j.
The equivalent force acting at the origin O is the sum of the forces
acting on the left post:
F D �357.77C 561.8�iC ��178.89� 210.67�j
D 919.6i� 389.6j (N).
The sum of the moments acting on the left post is the product of the
moment arm and the x-component of the tensions:
∑
M D �0.7�357.77�k� 0.3�561.8�k D �419k N-m
Check: The position vectors at the point of application are rAB D 0.7j,
and rCD D 0.3j. The sum of the moments is
∑
M D �rAB ð TAB�C �rCD ð TCD�
D
∣∣∣∣∣∣
i j k
0 0.7 0
357.77 �178.89 0
∣∣∣∣∣∣C
∣∣∣∣∣∣
i j k
0 0.3 0
561.8 �210.67 0
∣∣∣∣∣∣
D �0.7�357.77�k� 0.3�561.8�k D �419k
Check. (b) The equivalent single force retains the same scalar compo-
nents, 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
D D 419
919.6
D 0.456 m
Check: The moment is
M D rF ð F D
∣∣∣∣∣∣
i j k
0 D 0
919.6 �389.6 0
∣∣∣∣∣∣ D �919.6Dk D �419k,
from which D D 419
919.6
D 0.456 m, Check.
240
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Problem 4.149 Consider the system shown in Problem
4.148. The tension in each of the cables AB and CD is
400 N. If you represent the forces exerted on the right
post by the cables by a force F, what is F, and where
does its line of action intersect the y axis?
Solution: From the solution of Problem 4.148, the tensions are
TAB D �400�i cos��26.6°�Cj sin��26.6°�� D �357.77iC 178.89j,
and
TCD D �400�i cos��20.6°�Cj sin��20.6°�� D �374.42iC 140.74j.
The equivalent force is equal to the sum of these forces:
F D ��357.77� 374.42�iC �178.77C 140.74�j
D �732.19iC 319.5j (N).
The sum of the moments about O is
∑
M D 0.3�357.77� C 0.8�140.74C 178.89�k D 363k (N-m).
The intersection is D D 363
732.19
D 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
x
z
500 lb
500 lb
800 lb
6 in
6 in
Solution: The sum of the forces is
∑
FX D �500� 500�i D 0.
∑
FY D 800j.
Thus a force and a couple with moment M D 500k ft lb act on the
cross section. The equivalent force is F D 800j which acts at a positive
x axis location of D D 500
800
D 0.625 ft D 7.5 in to the right of the
origin.
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241
Problem 4.151 In Active Example 4.12, suppose that
the force FB is changed to FB D 20i� 15jC 30k (kN),
and you want to represent system 1 by an equivalent
system consisting of a force F acting at the point P with
coordinates (4, 3, �2) m and a couple M (system 2).
Determine F and M.
y
x
z
(6, 0, 0) m
FA
P
(4, 3, �2) m
System 1
MC
FB
Solution: From Active Example 4.12 we know that
FA D ��10iC 10j� 15k� kN
MC D ��90iC 150jC 60k� kN-m
The force F must equal the sum of the forces in system 1:
�F�2 D �F�1 :
F D FA C FB D �10i� 5jC 15k� kN
In system 2, the sum of the moments about P is M. Therefore equiva-
lence requires that M be equal to the sum of the moments about point
P due to the forces and moments in system 1:
�MP�2 D �MP�1 :
M D


∣∣∣∣∣∣
i j k
�4 �3 2
�10 10 �15
∣∣∣∣∣∣C
∣∣∣∣∣∣
i j k
2 �3 2
20 �15 30
∣∣∣∣∣∣C ��90iC 150jC 60k�

 kN-m
M D ��125iC 50jC 20k� kN-m
Thus F D �10i� 5jC 15k� kN, M D ��125iC 50jC 20k� kN-m
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?
12 in
z
y
x
O 10i – 30j + 3k (lb)
Solution:
(a) The moment about the z axis is negative,
MZ D �1�30� D �30 ft lb,
(b) The moment about the y axis is negative,
MY D �1�3� D �3 ft lb
(c) The equivalent force at O must be equal to the force at x D 12 in,
thus FEQ D 10i� 30jC 3k (lb)
The couple moment must equal the moment exerted by the force at x D
12 in. This moment is the product of the moment arm and the y- and z-
components of the force: M D �1�30�k� 1�3�j D �3j� 30k (ft lb).
242
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Problem 4.153 A basketball player executes a “slam
dunk” shot, then hangs momentarily on the rim, exerting
the two 100-lb forces shown. The dimensions are h D
14 12 in, and r D 9 12 in, and the angle ˛ D 120°.
(a) 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 jMj > 4000 in-
lb. Does it break?
y
O
z
x
h
–100j (lb)
r
–100j (lb)
α
Solution: The equivalent force at the origin must equal the sum of
the forces applied: FEQ D �200j. The position vectors of the points of
application of the forces are r1 D �hC r�i, and r2 D i�hC r cos˛��
kr sin˛. The moments about the origin are
M D �r1 ð F1�C �r2 ð F2� D �r1 C r2�ð F
D
∣∣∣∣∣∣
i j k
2hC r�1C cos˛� 0 �r sin ˛
0 �100 0
∣∣∣∣∣∣
D �100�r sin˛�i� 100�2h C r�1C cos˛��k.
For the values of h, r, and ˛ given, the moment is M D �822.72i�
3375k in lb. This is the couple moment required. (b) The magnitude
of the moment is jMj D p822.722 C 33752 D 3473.8 in lb. The back-
board does not break.
Problem 4.154 In Example 4.14, suppose that the 30-
lb upward force in system 1 is changed to a 25-lb upward
force. If you want to represent system 1 by a single force
F (system 2), where does the line of action of F intersect
the x�zplane?
y
x
z
System 1
(2, 0, 4) ft
(6, 0, 2) ft
(�3, 0, –2) ft
y
x
z
System 2
O
O
20j (lb)
30j (lb)
�10j (lb)
F
P
Solution: The sum of the forces in system 2 must equal the sum
of the forces in system 1:
�F�2 D �F�1
F D �20C 25� 10�j lb
F D 35j lb
The sum of the moments about a point in system 2 must equal the sum
of the moments about the same point is system 1. We sum moments
about the origin.
�M�2 D �M�1
∣∣∣∣∣∣
i j k
x y z
0 35 0
∣∣∣∣∣∣ D
∣∣∣∣∣∣
i j k
6 0 2
0 25 0
∣∣∣∣∣∣C
∣∣∣∣∣∣
i j k
2 0 4
0 �10 0
∣∣∣∣∣∣C
∣∣∣∣∣∣
i j k
�3 0 �2
0 20 0
∣∣∣∣∣∣
Expanding the determinants results in the equations
�35z D �50C 40C 40
35x D 150� 20� 60
Solving yields x D 2.00 ft, z D �0.857 ft
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243
Problem 4.155 The normal forces exerted on the car’s
tires by the road are
NA D 5104j (N),
NB D 5027j (N),
NC D 3613j (N),
ND D 3559j (N).
If you represent these forces by a single equivalent force
N, what is N, and where does its line of action intersect
the x�z plane?
1.4 m
x
z
y
0.8 m
1.4 m
D B
C A
0.8 m
x
Solution: We must satisfy the following three equations
∑
Fy :5104 NC 5027 NC 3613 NC 3559 N D Ry
∑
Mx :�5104 NC 3613 N��0.8 m�
� �5027 NC 3559 N��0.8 m� D �Ryz
∑
Mz :�5104 NC 5027 N��1.4 m�
� �3613 NC 3559 N��1.4 m� D Ryx
Solving we find
Ry D 17303 N, x D 0.239 m, z D �0.00606 m
Problem 4.156 Two forces act on the beam. If you
represent them by a force F acting at C and a couple M,
what are F and M?
z
y
x
100 N
3 m
80 N
C
Solution: The equivalent force must equal the sum of forces: F D
100jC 80k. The equivalent couple is equal to the moment about C:
∑
M D �3��80�j� �3��100�k D 240j� 300k
244
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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?
z
x
y
O
P i
b
h
Solution: The equivalent force at the origin is equal to the applied
force F D Pi. The position vector of the applied force is r D �hjC bk.
The moment is
M D �rð P� D
∣∣∣∣∣∣
i j k
0 �h Cb
P 0 0
∣∣∣∣∣∣ D bPjC hPk.
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 F0 acting at a point P with coordinates
�xP, yP, zP� and a couple M0, what are F0 and M0? x
z
B
B
y
O
A
h
h
r
A
1
2
A
1
2
Solution: (a) Equivalent force at the origin O has the same value
as the sum of forces,
∑
FX D �B� B�i D 0,
∑
FY D
(�AC 12AC 12A) j D 0,
thus F D 0. The equivalent couple moment has the same value as the
moment exerted on the brace by the forces,
∑
MO D �rA�i.
Thus the couple at O has the moment M D rAi. (b) The equivalent
force at �xP, yP, zP� has the same value as the sum of forces on the
brace, and the equivalent couple at �xP, yP, zP� has the same moment
as the moment exerted on the brace by the forces: F D 0, M D rAi.
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245
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
x
z
1 m
4i – 4j + 4k (kN-m)
MC =
2i – j (kN)
FB =
P
– i + j + k (kN)
FA =
Solution: The equivalent force at P has the value of the sum of
forces,
∑
F = (2 − 1)i + (1 − 1)j + k, FP = i + k (kN).
The equivalentcouple at P has the moment exerted by the forces and
moment about P. The position vectors of the forces relative to P are:
rA D �i� jC k, and rB D Ck. The moment of the couple:
∑
M D �rA ð FA�C �rB ð FB�CMC
D
∣∣∣∣∣∣
i j k
�1 �1 1
�1 1 1
∣∣∣∣∣∣C
∣∣∣∣∣∣
i j k
0 0 1
2 �1 0
∣∣∣∣∣∣CMC
D 3i� 2jC 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?
x
y
z
4 kN-m
6 kN-m
40°
30°
Solution: The equivalent force at the origin is zero, F D 0 since
there is no resultant force on the system. Represent the couples of
4 kN-m and 6 kN-m magnitudes by the vectors M1 and M2. The
couple at the origin must equal the sum:
∑
M DM1 CM2.
The sense of M1 is (see sketch) negative with respect to both y and
z, and the sense of M2 is positive with respect to both x and y.
M1 D 4��j sin 30° � k cos 30°� D �2j� 3.464k,
M2 D 6�i cos 40° C j sin 40°� D 4.5963iC 3.8567j.
Thus the couple at the origin is MO D 4.6iC 1.86j� 3.46k (kN-m)
(b) The magnitude of the total moment exerted by the two couples is
jMOj D
p
4.62 C 1.862 C 3.462 D 6.05 (kN-m)
246
c� 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 4.161 The two systems of forces and
moments acting on the bar are equivalent. If
FA D 30iC 30j� 20k (kN),
FB D 40i� 20jC 25k (kN),
MB D 10iC 40j� 10k (kN-m),
what are F and M?
z
y
x
z
y
x
MBFA
FB
System 1
M
F
System 2
A
B
2 m
2 m
Solution:
F D FA C FB D �70iC 10jC 5k� kN
M D �2 mi�ð FA C �4 mi�ð FB CMB
D �10i� 20j� 30k� kNm
Problem 4.162 Point G is at the center of the block.
The forces are
FA D �20iC 10jC 20k (lb),
FB D 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?
30 in
10 in
y
z
x
FA
FB
G
20 in
Solution: The equivalent force is the sum of the forces:∑
F D ��20�iC �10C 10�jC �20� 10�k
D �20iC 20jC 10k (lb).
The equivalent couple is the sum of the moments about G. The position
vectors are:
rA D �15iC 5jC 10k (in),
rB D 15iC 5j� 10k.
The sum of the moments:∑
MG D �rA ð FA�C �rB ð FB�
D
∣∣∣∣∣∣
i j k
�15 5 10
�20 10 20
∣∣∣∣∣∣C
∣∣∣∣∣∣
i j k
15 5 �10
0 10 �10
∣∣∣∣∣∣
D 50iC 250jC 100k (in lb)
c� 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
247
Problem 4.163 The engine above the airplane’s fuse-
lage exerts a thrust T0 D 16 kip, and each of the engines
under the wings exerts a thrust TU D 12 kip. The dimen-
sions are h D 8 ft, c D 12 ft, and b D 16 ft. If you repre-
sent the three thrust forces by a force F acting at the
origin O and a couple M, what are F and M?
y
x
b b
O
z
y
2 TU
T0
c
h
O
Solution: The equivalent thrust at the point G is equal to the sum
of the thrusts:
∑
T D 16C 12C 12 D 40 kip
The sum of the moments about the point G is
∑
M D �r1U ð TU�C �r2U ð TU�C �rO ð TO�
D �r1U C r2U�ð TU C �rO ð TO�.
The position vectors are r1U D Cbi� hj, r2U D �bi� hj, and rO D
Ccj. For h D 8 ft, c D 12 ft, and b D 16 ft, the sum of the moments
is
∑
M D
∣∣∣∣∣∣
i j k
0 �16 0
0 0 12
∣∣∣∣∣∣C
∣∣∣∣∣∣
i j k
0 12 0
0 0 16
∣∣∣∣∣∣ D ��192C 192�i D 0.