# Bedford - Engineering Mechanics - Statics - Solution Manual (5th Ed)

```SOLUTION MANUAL FOR
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.
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, determine the value of r/C to four significant digits.
Problem 1.2 The base of natural logarithms is e D
2.718281828 . . .
(a)
(b)
(c)
Express e to five significant digits.
Determine the value of e2 to five significant digits.
Use the value of e you obtained in part (a) to determine the value of e2 to five significant digits.
Solution: C D 2r )
r
1
D
D 0.159154943.
C
2
To four significant digits we have
r
D 0.1592
C
Solution: The value of e is: e D 2.718281828
(a)
To five significant figures e D 2.7183
(b)
e2 to five significant figures is e2 D 7.3891
(c)
Using the value from part (a) we find e2 D 7.3892 which is
not correct in the fifth digit.
[Part (c) demonstrates the hazard of using rounded-off
values in calculations.]
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 5 š 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 r1 2 D 78.226 m2
to A2 D r2 2 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 ft &eth; 8 ft D 192 ft2 .
What is its area in m2 to three significant digits?
Solution:
A D 192 ft2
1m
3.281 ft
2
D 17.8 m2
A D 17.8 m2
Problem 1.5 The Burj Dubai, scheduled for completion 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:
3.281 ft
D 2.31 &eth; 103 ft
1m
3.218 ft 2
A D 8000 m2
D 8.61 &eth; 104 ft2
1m
h D 705 m
h D 2.31 &eth; 103 ft,
A D 8.61 &eth; 104 ft2
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1
Problem 1.6 Suppose that you have just purchased
a Ferrari F355 coupe and you want to know whether
you can use your set of SAE (U.S. Customary Units)
wrenches to work on it. You have wrenches with widths
w D 1/4 in, 1/2 in, 3/4 in, and 1 in, and the car has nuts
with dimensions n D 5 mm, 10 mm, 15 mm, 20 mm,
and 25 mm. Defining a wrench to fit if w is no more
than 2% larger than n, which of your wrenches can you
use?
Solution: Convert the metric size n to inches, and compute the
percentage difference between the metric sized nut and the SAE
wrench. The results are:
5 mm
1 inch
25.4 mm
D 0.19685.. in,
0.19685 0.25
0.19685
100
D 27.0%
10 mm
15 mm
n
20 mm
25 mm
1 inch
25.4 mm
1 inch
25.4 mm
1 inch
25.4 mm
1 inch
25.4 mm
D 0.3937.. in,
D 0.5905.. in,
D 0.7874.. in,
D 0.9843.. in,
0.3937 0.5
0.3937
0.5905 0.5
0.5905
100 D 27.0%
100 D C15.3%
0.7874 0.75
0.7874
0.9843 1.0
0.9843
100 D C4.7%
100 D 1.6%
A negative percentage implies that the metric nut is smaller than the
SAE wrench; a positive percentage means that the nut is larger then
the wrench. Thus within the definition of the 2% fit, the 1 in wrench
will fit the 25 mm nut. The other wrenches cannot be used.
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
1m
D 8848.52 m
h1 D 29032 ft
3.281 ft
1m
D 8849.13 m
h2 D 29034 ft
3.281 ft
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.
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.
2
Solution:
a)
v D 430
b)
v D 430
km
h
0.6214 mi
1 km
D 267 mi/h
v D 267 mi/h
1 ft
1h
km 1000 m
D 392 ft/s
h
1 km
0.3048 m
3600 s
v D 392 ft/s
Solution:
15 km
1.3
38 C
min
60
a)
vD b)
v D 23.7 km/h
60 min
1h
1 mi
1.609 km
D 23.7 km/h
v D 23.7 km/h
D 14.7 mi/h
v D 14.7 mi/h
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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:
Problem 1.11 The kinetic energy of the man in Active
Example 1.1 is defined by 12 mv2 , 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
2 2
2 (68 kg)(6 m/s) D 1224 kg-m /s . What is his kinetic
energy in U.S. Customary units?
Solution:
T D 229 ft-lb
1N
0.2248 lb
T D 1224 kg-m2 /s2
1m
3.281 ft
1 slug
14.59 kg
D 310 N-m
1 ft
0.3048 m
Solution: Use Table 1.2. The result is:
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:
Solution:
A D 200 mm2 280 mm120 mm D 20800 mm2
2
1m
a) A D 20800 mm2
D 0.0208 m2 A D 0.0208 m2
1000 mm
b)
A D 20800 mm2
1 in
25.4 mm
g D 9.81
m
v D 12,000
D 903 slug-ft2 /s
1 ft
0.3048 m
1 ft
0.3048 m
D 32.185 . . .
1 furlong
660 ft
ft
s2
3600 s
hr
D 32.2
24 hr
1 day
ft
s2
14 day
1 fortnight
furlongs
fortnight
y
40 mm
x
120 mm
40 mm
40
mm
200 mm
2
A D 32.2 in
Problem 1.15 The cross-sectional area of the C12&eth;30
American Standard Channel steel beam is A D 8.81 in2 .
What is its cross-sectional area in mm2 ?
s2
v D 2 m/s
2
D 32.2 in2
2
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.
Problem 1.14 Determine the cross-sectional area of
the beam (a) in m2 ; (b) in in2 .
T D 310 N-m
y
A
Solution:
A D 8.81 in2
25.4 mm
1 in
2
D 5680 mm2
x
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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
D 2.0683 . . . 106 12 in
1 ft
N
m2
2 1 ft
0.3048 m
2
D 2.07106 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
1m
3.281 ft
1N
0.2248 lb
D 5.22 &eth; 106 W
P D 5.22 &eth; 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?.
Problem 1.19 The moment of inertia of the rectangular area about the x axis is given by the equation
I D 13 bh3 .
The dimensions of the area are b D 200 mm and h D
100 mm. Determine the value of I to four significant
digits in terms of (a) mm4 ; (b) m4 ; (c) in4 .
Solution:
w D 400 N/m
0.2248 lb
1N
1m
3.281 ft
D 27.4 lb/ft
w D 27.4 lb/ft
Solution:
1
200 mm100 mm3 D 66.7 &eth; 106 mm4
3
(a)
ID
(b)
I D 66.7 &eth; 106 mm4
(c)
I D 66.7 &eth; 106 mm4
y
1m
1000 mm
1 in
25.4 mm
4
D 66.7 &eth; 106 m4
4
D 160 in4
h
x
b
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?
4
Solution:
a)
L D mc )
b)
L D 12 kg-m/s
Units L D kg-m/s
0.0685 slug
1 kg
3.281 ft
1m
D 2.70 slug-ft/s
L D 2.70 slug-ft/s
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Problem 1.21 The equation
D
Solution:
My
I
(a)
is used in the mechanics of materials to determine
normal stresses in beams.
(a)
(b)
D
(N-m)m
My
N
D
D 2
I
m4
m
D
2000 N-m0.1 m
My
D
I
7 &eth; 105 m4
(b)
When this equation is expressed in terms of SI base
units, M is in newton-meters (N-m), y is in meters
(m), and I is in meters to the fourth power (m4 ).
What are the SI units of ?
If M D 2000 N-m, y D 0.1 m, and I D 7 &eth;
105 m4 , what is the value of in U.S. Customary
base units?
D 59,700
1 lb
4.448 N
0.3048 m
ft
2
lb
ft2
Problem 1.22 The acceleration due to gravity on the
Solution:
a) The mass does
surface of the moon is 1.62 m/s2 . (a) What would the
on location. The mass in kg is
not depend
14.59 kg
mass of the C-clamp in Active Example 1.4 be on the surface
mass D 0.397 kg
D 0.397 kg
0.0272 slug
1 slug
of the moon? (b) What would the weight of the C-clamp
in newtons be on the surface of the moon?
b)
The weight on the surface of the moon is
W D mg D 0.397 kg1.62 m/s2 D 0.643 N
Problem 1.23 The 1 ft &eth; 1 ft &eth; 1 ft cube of iron
weighs 490 lb at sea level. Determine the weight in
newtons of a 1 m &eth; 1 m &eth; 1 m cube of the same
material at sea level.
Solution: The weight density is D
The weight of the 1 m3 cube is:
W D V D
490 lb
1 ft3
1 m3
W D 0.643N
1 ft
490 lb
1 ft3
1 ft
0.3048 m
3 1N
0.2248 lb
1 ft
1 ft
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
64 lb/ft3
32.2 ft/s2
2
D 2.312 &eth; 1019 ft3
(a)
m D V D
(b)
m D 4.60 &eth; 1019 slugs
2.312 &eth; 1019 ft3 D 4.60 &eth; 1019 slugs
Problem 1.25 The acceleration due to gravity at
sea level is g D 9.81 m/s2 . The radius of the earth
is 6370 km. The universal gravitational constant is
G D 6.67 &eth; 1011 N-m2 /kg2 . Use this information to
determine the mass of the earth.
5,280 ft
1 mi
Solution: Use Eq (1.3) a D
14.59 kg
1 slug
D 6.71 &eth; 1020 kg
GmE
. Solve for the mass,
R2
m 2
9.81 m/s2 6370 km2 103
gR2
mE D
D
km
G
N-m2
11
6.6710 kg2
D 5.9679 . . . 1024 kg D 5.971024 kg
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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).
W D mg
RE
r
2
D
WE
g
g
RE
RE C H
2
D WE
3960
3960 C 200
2
D 1800.90616 D 163 lb
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, determine 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:
Problem 1.28 If an object is near the surface of the
earth, the variation of its weight with distance from the
center of the earth can often be neglected. The acceleration due to gravity at sea level is g D 9.81 m/s2 . The
radius of the earth is 6370 km. The weight of an object
at sea level is mg, where m is its mass. At what height
above the earth does the weight of the object decrease
to 0.99 mg?
Solution: Use a variation of Eq (1.5).
W D mgM D 10 kg1.26 m/s2 D 12.6 N
a)
b)
Adapting equation 1.4 we have aM D gM
then
F D maM D 10 kg1.62 m/s2 W D 12.6 N
RM
r
2
. The force is
1738 km
1738 km C 1738 km
2
D 4.05 N
F D 4.05 N
W D mg
RE
RE C h
2
D 0.99 mg
h D RE
p
1
0.99
1 D 63701.0050378 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 1.0247 &eth; 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 6.67 &eth; 1011
N-m2 /kg2 .)
Solution:
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
mN m mN mN
D G 2 m ) gN D G 2
rN 2
r
rN
Note that the radius of Neptune is rN D 12 49,532 km D 24,766 km
N-m2
1.0247 &eth; 1026 kg
1 km 2
Thus gN D 6.67 &eth; 1011
24766 km2
1000 m
kg2
We have: W D G
D 11.1 m/s2
gE
RE
rEp
gN D 11.1 m/s2
2
D gM
RM
rMp
2
,
where rEM D 383,000 km. Substituting the correct numerical values
9.81
m 1738 km 2
m 6370 km 2
D
1.62
,
s2
rEp
s2
rEM rEp
where rEp is the only unknown. Solving, we get rEp D 344,770 km D
345,000 km.
6
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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
U C 2V.
45⬚
U
V
Solution: Draw the vectors accurately and measure the resultant.
R D jU C 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&deg; . Draw a sketch of the
new situation. The magnitudes of the forces are jFAB j D
100 kN and jFAC j 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 FAC j D 146 kN
˛ D 32&deg;
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7
Problem 2.3 The magnitude jFA j D 80 lb and the
angle ˛ D 65&deg; . The magnitude jFA C FB j D 120 lb.
Graphically determine the magnitude of FB .
FB
FC
␤
a
FA
Solution: Accurately draw the vectors and measure the magnitude
of FB .
jFB j D 62 lb
Problem 2.4 The magnitudes jFA j D 40 N, jFB j D
50 N, and jFC j D 40 N. The angle ˛ D 50&deg; and ˇ D 80&deg; .
Graphically determine the magnitude of FA C FB C FC .
FB
FC
␤
a
FA
Solution: Accurately draw the vectors and measure the magnitude
of FA C FB C FC .
R D jFA C FB C FC j D 83 N
8
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Problem 2.5 The magnitudes jFA j D jFB j D jFC j D
100 lb, and the angles ˛ D 30&deg; . Graphically determine
the value of the angle ˇ for which the magnitude
jFA C FB C FC j is a minimum and the minimum value
of jFA C FB C FC j.
FB
FC
␤
a
FA
Solution: For a minimum, the vector FC must point back to the
origin.
R D jFA C FB C FC j D 93.2 lb
ˇ D 165&deg;
Problem 2.6 The angle D 50&deg; . Graphically determine
the magnitude of the vector rAC .
150 mm
60 mm
B
rAB
A
␪
rBC
rAC
C
Solution: Draw the vectors accurately and then measure jrAC j.
jrAC j D 181 mm
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9
Problem 2.7 The vectors FA and FB represent
the forces exerted on the pulley by the belt.
Their magnitudes are jFA j D 80 N and jFB j D 60 N.
Graphically determine the magnitude of the total force
the belt exerts on the pulley.
Solution: Draw the vectors accurately and then measure jFA C
FB j.
jFA C FB j D 134 N
FB
45⬚
FA
10⬚
Problem 2.8 The sum of the forces FA C FB C
FC D 0. The magnitude jFA j D 100 N and the angle ˛ D
60&deg; . Graphically determine the magnitudes jFB j and jFC j.
Solution: Draw the vectors so that they add to zero.
jFB j D 86.6 N, jFC j D 50.0 N
FB
30
a
FA
FC
Problem 2.9 The sum of the forces FA C FB C
FC D 0. The magnitudes jFA j D 100 N and jFB j D
80 N. Graphically determine the magnitude jFC j and the
angle ˛.
FB
30
a
FA
FC
Solution: Draw the vectors so that they add to zero.
jFC j D 50.4 N, ˛ D 52.5&deg;
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 W C L C D D 0. Graphically determine the magnitudes of the lift and drag.
L
25⬚
D
W
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 .
FA
40˚
Solution: Draw the vectors so that they add to zero. Then measure
the unknown magnitudes.
jFA j D jFB j D 319 lb
FB
20˚
20˚
W
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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 jFBA j.
Solution: Draw the vectors accurately and then measure the
unknown magnitudes.
jFBA j D 174 lb
FBC
C
20⬚
B
B
FBA
A
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 .
Solution: Draw the vectors accurately and then measure the
unknown magnitudes.
jFA j D 203 lb
jFB j D 311 lb
L
FA
30⬚
50⬚
FB
Top View
Problem 2.14 A surveyor determines that the horizontal 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 ˛.
Solution: Draw the vectors accurately and then measure the
unknown magnitude and angle.
jrBC j D 390 m
˛ D 21.2&deg;
North
B
a
rBC
C
60⬚
20⬚
A
12
East
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 2.15 The vector r extends from point A to
the midpoint between points B and C. Prove that
C
r D 12 rAB C rAC .
rAC
r
rAB
B
A
Solution: The proof is straightforward:
C
r D rAB C rBM , and r D rAC C rCM .
rAC
r
Add the two equations and note that rBM C rCM D 0, since the two
vectors are equal and opposite in direction.
Thus 2r D rAC C rAB , or r D
1
2
rAC C rAB A
M
B
rAB
Problem 2.16 By drawing sketches of the vectors,
explain why
U C V C W D U C V C W.
Solution: Additive associativity for vectors is usually given as an
axiom in the theory of vector algebra, and of course axioms are not
subject to proof. However we can by sketches show that associativity
for vector addition is intuitively reasonable: Given the three vectors to
of U. The result is the vector U C V C W.
V
U
V+W
(a)
W
U+[V+W]
V
to the result U C V C W.
U
W
The final vector in the two sketches is the same vector, illustrating that
associativity of vector addition is intuitively reasonable.
Problem 2.17 A force F D 40 i 20 j N. What is
its magnitude jFj?
U+V
(b)
[U+V]+W
p
Solution: jFj D 402 C 202 D 44.7 N
Strategy: The magnitude of a vector in terms of its
components is given by Eq. (2.8).
Problem 2.18 An engineer estimating the components
of a force F D Fx i C Fy j acting on a bridge abutment
has determined that Fx D 130 MN, jFj D 165 MN, and
Fy is negative. What is Fy ?
Solution:
jFj D
jFy j D
jFx j2 C jFy j2
jFj2 jFx j2 D
165 MN2 130 MN2 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 i C 80j (N). If the support will safely support a force
of 100 N, what is the allowable range of values of the
component Fx ?
Solution: Use the definition of magnitude in Eq. (2.8) and reduce
algebraically.
100 &frac12;
Fx 2 C 802 , from which 1002 802 &frac12; Fx 2 .
Thus jFx j Problem 2.20 If FA D 600i 800j (kip) and FB D
200i 200j (kip), what is the magnitude of the force
F D FA 2FB ?
Solution: Take the scalar multiple of FB , add the components of
the two forces as in Eq. (2.9), and use the definition of the magnitude.
F D 600 2200i C 800 2200j D 200i 400j
jFj D
Problem 2.21 The forces acting on the sailplane are its
weight W D 500jlb, the drag D D 200i C 100j(lb)
and the lift L. The sum of the forces W C L C D D 0.
Determine the components and the magnitude of L.
p
3600, or 60 Fx C60 (N)
2002 C 4002 D 447.2 kip
y
L
Solution:
D
L D W D D 500j 200i C 100j D 200i C 400jlb
jLj D
200 lb2 C 400 lb2 D 447 lb
W
L D 200i C 400jlb, jLj D 447 lb
x
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
Solution: The two possible values of V are shown in the sketch.
y
The strategy is to (a) determine the unit vector associated with U,
(b) express this vector in terms of an angle, (c) add š90&deg; to this
angle, (d) determine the two unit vectors perpendicular to U, and
(e) calculate the components of the two possible values of V. The
unit vector parallel to U is
eU D 6i
62
C 82
8j
D 0.6i 0.8j
2
6 C 82
V2
6
V1
U
x
8
Expressed in terms of an angle,
eU D i cos ˛ j sin ˛ D i cos53.1&deg; j sin53.1&deg; Add š90&deg; to find the two unit vectors that are perpendicular to this
unit vector:
ep1 D i cos143.1&deg; j sin143.1&deg; D 0.8i 0.6j
ep2 D i cos36.9&deg; j sin36.9&deg; D 0.8i C 0.6j
Take the scalar multiple of these unit vectors to find the two vectors
perpendicular to U.
V1 D jVj0.8i 0.6j D 16i 12j.
The components are Vx D 16, Vy D 12
V2 D jVj0.8i C 0.6j D 16i C 12j.
The components are Vx D 16, Vy D 12
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.
Solution: We can use similar triangles to determine the
components of F.
F D 10 lb
y
11
7
p
i p
j D 5.37i 8.44j lb
72 C 112
72 C 112
F D 5.37i 8.44j lb
7
11
F
x
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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.
Solution:
(a) F D 60 lbcos 20&deg; i C sin 20&deg; j D 56.4i C 20.5j lb
F D 56.4i C 20.5jlb
(b) W D 100 lbj
y
F C W D 56.4i C [20.5 100]j lb D 56.4i 79.5j lb
F
20⬚
jF C Wj D
56.4 lb2 C 79.5 lb2 D 97.4 lb
jF C Wj D 97.4 lb
x
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
F
3
4
Solution:
(a)
We can use similar triangles to determine the components of F.
F D 260 kN
p
3
iC p
j D 208i C 156j kN
42 C 32
42 C 32
4
x
F D 208i C 156j kN
(b)
The missile’s weight W can be expressed in component and then
W D 8800 kg9.81 m/s2 j D 86.3 kNj
F C W D 208i C [156 86.3]j kN D 208i 69.7j kN
jF C Wj D
208 kN2 C 69.7 kN2 D 219 kN
jF C Wj 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.
y
rAD D 0 1.8 mi C 0.4 m 0.7 mj D 1.8i 0.3j m
rAD D 1.8 m2 C 0.3 m2 D 1.825 m
B
A
0.6 m
D
0.7 m
0.4 m
C
0.6 m
16
Solution: Coordinates A(1.8, 0.7) m, D(0, 0.4) m
x
1.2 m
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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 .
Solution: Use the xy coordinate system shown and find the
locations of C and F in those coordinates. The coordinates of the
points in this system are the scalar components of the vectors rAC and
rAF . For rAC , we have
rAC D rAB C rBC D xB xA i C yB yA j
y
C xC xB i C yC yB j
E
D
2m
rAC D 2m 0i C 0 0j C 2m cos 60&deg; 0i
or
C 2m cos 60&deg; 0j,
F
giving
C
rAC D 2m C 2m cos 60&deg; i C 2m sin 60&deg; j. For rAF , we have
A
B
x
rAF D xF xA i C yF yA j
D 2m cos 60&deg; xF 0i C 2m sin 60&deg; 0j.
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&deg; .
rBC D 2 cos60&deg; i C 2sin60&deg; j m
rBC D 1i C 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 ?
y
Solution: The vector rAB can be written two ways.
A
rAB
rAB D jrAB j0.616i 0.788j D Bx Ax i C By Ay j
Comparing the two expressions we have
By Ay D 0.6 3.0ft D 0.788jrAB j
jrAB j D
B
x
2.4 ft
D 3.05 ft
0.788
Thus
rAB D jrAB j0.616i 0.788j D 3.05 ft0.616i 0.788j D 1.88i 2.40j ft
rAB D 1.88i 2.40j ft
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17
y
Problem 2.30 (a) Express the position vector from
point A of the front-end loader to point B in terms of
components.
98 in
45 in
C
(b) Express the position vector from point B to point C
in terms of components.
A
55 in
(c) Use the results of (a) and (b) to determine the
distance from point A to point C.
B
50 in
35 in
x
50 in
Solution: The coordinates are A(50, 35); B(98, 50); C(45, 55).
(a)
The vector from point A to B:
rAB D 98 50i C 50 35j D 48i C 15j (in)
(b)
The vector from point B to C is
rBC D 45 98i C 55 50j D 53i C 5j (in).
(c)
The distance from A to C is the magnitude of the sum of the
vectors,
rAC D rAB C rBC D 48 53i C 15 C 5j D 5i C 20j.
The distance from A to C is
jrAC j D
18
52 C 202 D 20.62 in
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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
80 m
40 m
y
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
Lnew D
(a)
40 m2 C 80 m2 D
A
Force
exerted on
the tower
by cable
AB
8000 m2
d2 C 80 m2
80 m
F
The force is proportional to the length. Therefore
d2 C 80 m2
1000 N D 900 N 8000 m2
dD
B
8000 m2 1000 N
900 N
B
x
40 m
2
80 m2 D 59.0 m
d D 59.0 m
(b)
The force F is then
F D 1000 N
d
d2 C 80 m2
i 80 m
d2 C 80 m2
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&deg; , (b) D 225&deg; .
y
150 mm
60 mm
B
rAB
rBC
C
θ
x
A
Solution:
(a)
y
rAB D 60 cos30&deg; i C 60 sin30&deg; j, or
150
mm
60
mm
rAB D 51.96i C 30j mm. And
B
(b)
FAB
rAB D 60 cos225&deg; i C 60 sin225&deg; j or
A
rAB D 42.4i 42.4j mm.
θ
FBC
C
F
x
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.
y
B
B
30⬚
F
A
30⬚
A
x
Solution:
D tan1
2 ft
8 ft
D 14.04&deg;
F D 4800 lb cos i C sin j.
F D 4660i C 1160j lb
20
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
y
Problem 2.34 A surveyor measures the location of
point A and determines that rOA D 400i C 800j (m). He
wants to determine the location of a point B so that
jrAB j D 400 m and jrOA C rAB j D 1200 m. What are the
cartesian coordinates of point B?
B
A
N
rAB
rOA
Proposed
x
O
Solution: Two possibilities are: The point B lies west of point A,
or point B lies east of point A, as shown. The strategy is to determine
the unknown angles ˛, ˇ, and . The magnitude of OA is
jrOA j D
B
4002 C 8002 D 894.4.
α
The angle ˇ is determined by
tan ˇ D
A
B
y
α
θ
β
800
D 2, ˇ D 63.4&deg; .
400
0
x
The angle ˛ is determined from the cosine law:
cos ˛ D
894.42 C 12002 4002
D 0.9689.
2894.41200
˛ D 14.3&deg; . The angle is D ˇ š ˛ D 49.12&deg; , 77.74&deg; .
The two possible sets of coordinates of point B are
rOB D 1200i cos 77.7 C j sin 77.7 D 254.67i C 1172.66j (m)
rOB D 1200i cos 49.1 C j sin 49.1 D 785.33i C 907.34j (m)
The two possibilities lead to B(254.7 m, 1172.7 m) or B(785.3 m,
907.3 m)
Problem 2.35 The magnitude of the position vector
rBA from point B to point A is 6 m and the magnitude of
the position vector rCA from point C to point A is 4 m.
What are the components of rBA ?
y
Thus
rBA D xA 0i C yA 0j ) 6 m2 D xA 2 C yA 2
rCA D xA 3 mi C yA 0j ) 4 m2 D xA 3 m2 C yA 2
3m
B
Solution: The coordinates are: AxA , yA , B0, 0, C3 m, 0
x
C
Solving these two equations, we find xA D 4.833 m, yA D š3.555 m.
We choose the “-” sign and find
rBA D 4.83i 3.56j m
A
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21
Problem 2.36 In Problem 2.35, determine the components of a unit vector eCA that points from point C toward
point A.
Strategy: Determine the components of rCA and then
divide the vector rCA by its magnitude.
Solution: From the previous problem we have
rCA D 1.83i 3.56j m,
rCA D
1.832 C 3.562 m D 3.56 m
Thus
eCA D
rCA
D 0.458i 0.889j
rCA
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.
Solution:
rAB D xB xA i C yB yA j
rCB D xB xC i C yC yB j
Points are: A (0, 1.2), B (4, 13) and C (9, 1)
Substituting, we get
rAB D 4i C 11.8j m, jrAB j D 12.46 m
y
B (4, 13) m
rCB D 5i C 12j m, jrCB j D 13 m
The unit vectors are given by
eAB D
rAB
rCB
and eCB D
jrAB j
jrCB j
Substituting, we get
eAB D 0.321i C 0.947j
eCB D 0.385i C 0.923j
A
(0, 1.2) m
22
C
(9, 1) m
x
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Problem 2.38 The length of the bar AB is 0.6 m.
Determine the components of a unit vector eAB that
points from point A toward point B.
y
B
0.4 m
A
0.3 m
x
Solution: We need to find the coordinates of point Bx, y
B
We have the two equations
m
y
m
0.6
0.4
0.3 m C x2 C y 2 D 0.6 m2
x 2 C y 2 D 0.4 m2
Solving we find
x D 0.183 m,
y D 0.356 m
A
0.3 m
O
x
Thus
eAB D
rAB
0.183 m [0.3 m]i C 0.356 mj
D rAB
0.183 m C 0.3 m2 C 0.356 m2
D 0.806i C 0.593j
y
Problem 2.39 Determine the components of a unit
vector that is parallel to the hydraulic actuator BC and
points from B toward C.
1m
D
C
Solution: Point B is at (0.75, 0) and point C is at (0, 0.6). The
vector
1m
rBC D xC xB i C yC yB j
0.6 m
B
A
x
rBC D 0 0.75i C 0.6 0j m
rBC D 0.75i C 0.6j m
jrBC j D
eBC D
0.15 m
0.6 m
Scoop
0.752 C 0.62 D 0.960 m
0.75
0.6
rBC
D
iC
j
jrBC j
0.96
0.96
eBC D 0.781i C 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 0.781i C 0.625j
The vector F is given by F D jFjeBC
F D 1.20.781i C 0.625j k &ETH; N
F D 937i C 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.
y
N
(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.
A Proposed bridge
B
Solution: We need to find the coordinates of points A and B
rOA D 1500 cos 60&deg; i C 1500 sin 60&deg; j
60⬚
30⬚
rOA D 750i C 1299j m
Point A is at (750, 1299) (m)
River
x
O
rOB D 2000 cos 30&deg; i C 2000 sin 30&deg; j m
rOB D 1732i C 1000j m
Point B is at (1732, 1000) (m)
(a)
The vector from A to B is
rAB D xB xA i C yB yA j
rAB D 982i 299j m
(b)
The unit vector eAB is
eAB D
982i 299j
rAB
D
jrAB j
1026.6
eAB D 0.957i 0.291j
24
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 jT1 j D 2800 lb, jT2 D 3200 lb, jT3 j D
4000 lb, and jT4 j D 5000 lb. What is the magnitude of
the total force exerted by the four cables?
y
T4
51⬚
T3
40⬚
T2
29⬚
T1
9⬚
x
Solution: The x-component of the total force is
Tx D jT1 j cos 9&deg; C jT2 j cos 29&deg; jT3 j cos 40&deg; C jT4 j cos 51&deg;
Tx D 2800 lb cos 9&deg; C 3200 lb cos 29&deg; C 4000 lb cos 40&deg; C 5000 lb cos 51&deg;
Tx D 11,800 lb
The y-component of the total force is
Ty D jT1 j sin 9&deg; C jT2 j sin 29&deg; C jT3 j sin 40&deg; C jT4 j sin 51&deg;
Ty D 2800 lb sin 9&deg; C 3200 lb sin 29&deg; C 4000 lb sin 40&deg; C 5000 lb sin 51&deg;
Ty D 8450 lb
The magnitude of the total force is
jTj D
Tx 2 C Ty 2 D
11,800 lb2 C 8450 lb2 D 14,500 lb
jTj D 14,500 lb
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
25
Problem 2.43 The tensions in the four cables are equal:
jT1 j D jT2 j D jT3 j D jT4 j D T. Determine the value of
T so that the four cables exert a total force of 12,500-lb
magnitude on the support.
y
T4
51⬚
T3
40⬚
T2
29⬚
T1
9⬚
x
Solution: The x-component of the total force is
Tx D T cos 9&deg; C T cos 29&deg; C T cos 40&deg; C T cos 51&deg;
Tx D 3.26T
The y-component of the total force is
Ty D T sin 9&deg; C T sin 29&deg; C T sin 40&deg; C T sin 51&deg;
Ty D 2.06T
The magnitude of the total force is
jTj D
Tx 2 C Ty 2 D
Solving for T we find
26
3.26T2 C 2.06T2 D 3.86T D 12,500 lb
T D 3240 lb
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:
jFBA j D jFBC j. The magnitude of the total force exerted
on the block at B by the rope is jFBA C FBC j D 920 N.
Determine jFBA j by expressing the forces FBA and FBC
in terms of components.
FBC
C
20&deg;
B
B
FBA
A
Solution:
FBC
FBC D Fcos 20&deg; i C sin 20&deg; j
20&deg;
FBA D Fj
FBC C FBA D Fcos 20&deg; i C [sin 20&deg; 1]j
Therefore
920 N2 D F2 cos2 20&deg; C [sin 20&deg; 1]2 ) F D 802 N
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 magnitudes of F2 and F3 ?
y
F3
30˚
F1
Solution: Using components we have
Fx : 5 kN C F2 cos 45&deg; F3 cos 30&deg; D 0
45˚
Fy :
F2 sin 45&deg;
C F3 sin 30&deg;
D0
F2
Solving simultaneously yields:
x
) F2 D 9.66 kN,
F3 D 13.66 kN
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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 jFB j D 800 lb, jFC j D 1000 lb, jFD j D 900 lb. If the
vector sum of the four forces equals zero, what are the
magnitude of FA and the angle ˛?
y
FB
FC
70&deg;
30&deg;
Solution: The strategy is to use the angles and magnitudes to
determine the force vector components, to solve for the unknown force
FA and then take its magnitude. The force vectors are
FB D 800i cos 110&deg; C j sin 110&deg; D 273.6i C 751.75j
FC D 1000i cos 30&deg; C j sin 30&deg; D 866i C 500j
α
20&deg;
FD
FA
x
FD D 900i cos20&deg; C j sin20&deg; D 845.72i 307.8j
FA D jFA ji cos180 C ˛ C j sin180 C ˛
D jFA ji cos ˛ j sin ˛
The sum vanishes:
FA C FB C FC C FD D i1438.1 jFA j cos ˛
C j944 jFA j sin ˛ D 0
From which FA D 1438.1i C 944j. The magnitude is
jFA j D
14382 C 9442 D 1720 lb
The angle is: tan ˛ D
28
944
D 0.6565, or ˛ D 33.3&deg;
1438
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 2.47 In Example 2.5, suppose that the attachment point of cable A is moved so that the angle between
the cable and the wall increases from 40&deg; to 55&deg; . 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 perpendicular to the wall, what are the necessary magnitudes
of FA and FB ?
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.
A
40⬚
20⬚
B
FA cos 55&deg; FB cos 20&deg; D 0
FA sin 55&deg; C FB sin 20&deg; D 200 lb
Solving we find
FA
40⬚
FA D 195 lb
FB D 119 lb
20⬚
FB
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
29
Problem 2.48 The bracket must support the two forces
shown, where jF1 j D jF2 j D 2 kN. An engineer determines that the bracket will safely support a total force
of magnitude 3.5 kN in any direction. Assume that 0 ˛ 90&deg; . What is the safe range of the angle ˛?
F2
α
F1
F2
Solution:
Fx : 2 kN C 2 kN cos ˛ D 2 kN1 C cos ˛
F1
β
α
α
F1 + F2
Fy : 2 kN sin ˛
Thus the total force has a magnitude given by
F D 2 kN
p
1 C cos ˛2 C sin ˛2 D 2 kN 2 C 2 cos ˛ D 3.5 kN
Thus when we are at the limits we have
2 C 2 cos ˛ D
3.5 kN
2 kN
2
D
49
17
) cos ˛ D
) ˛ D 57.9&deg;
16
32
In order to be safe we must have
57.9&deg; ˛ 90&deg;
30
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 2.49 The figure shows three forces acting on
a joint of a structure. The magnitude of Fc is 60 kN, and
FA C FB C FC D 0. What are the magnitudes of FA and
FB ?
y
FC
FB
15&deg;
x
40&deg;
FA
Solution: We need to write each force in terms of its components.
FA
195&deg;
FA D jFA j cos 40i C jFA j sin 40j kN
40&deg;
x
FB D jFB j cos 195&deg; i C jFB j sin 195j kN
FC D jFC j cos 270&deg; i C jFC j sin 270&deg; j kN
FB
Thus FC D 60j kN
270&deg;
Since FA C FB C FC D 0, their components in each direction must also
sum to zero.
FC
FAx C FBx C FCx D 0
FAy C FBy C FCy D 0
Thus,
jFA j cos 40&deg; C jFB j cos 195&deg; C 0 D 0
jFA j sin 40&deg; C jFB j sin 195&deg; 60 kN D 0
Solving for jFA j and jFB j, we get
jFA j D 137 kN, jFB j D 109 kN
Problem 2.50 Four forces act on a beam. The vector
sum of the forces is zero. The magnitudes jFB j D
10 kN and jFC j D 5 kN. Determine the magnitudes of
FA and FD .
FD
30&deg;
FA
FB
FC
Solution: Use the angles and magnitudes to determine the vectors,
and then solve for the unknowns. The vectors are:
FA D jFA ji cos 30&deg; C j sin 30&deg; D 0.866jFA ji C 0.5jFA jj
FB D 0i 10j, FC D 0i C 5j, FD D jFD ji C 0j.
Take the sum of each component in the x- and y-directions:
Fx D 0.866jFA j jFD ji D 0
and
Fy D 0.5jFA j 10 5j D 0.
From the second equation we get jFA j D 10 kN . Using this value in
the first equation, we get jFD j D 8.7 kN
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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 jFB j D jFE j D 20 kN, jFC j D
16 kN, and jFD j D 9 kN. Determine the magnitudes of
FA and FG .
FA
70⬚
FC
FD
40⬚
50⬚
40⬚
FB
Solution: Write each force in terms of its magnitude and direction
FG
FE
y
as
F D jFj cos i C jFj sin j
where is measured counterclockwise from the Cx-axis.
Thus, (all forces in kN)
θ
FA D jFA j cos 110&deg; i C jFA j sin 110&deg; j kN
FB D 20 cos 270&deg; i C 20 sin 270&deg; j kN
x
FC D 16 cos 140&deg; i C 16 sin 140&deg; j kN
FD D 9 cos 40&deg; i C 9 sin 40&deg; j kN
FE D 20 cos 270&deg; i C 20 sin 270&deg; j kN
FG D jFG j cos 50&deg; i C jFG j sin 50&deg; j kN
We know that the x components and y components of the forces must
Thus
FAx C FBx C FCx C FDx C FEx C FGx D 0
FAy C FBy C FCy C FDy C FEy C FGy D 0
jFA j cos 110&deg; C 0 12.26 C 6.89 C 0 C jFG j cos 50&deg; D 0
jFA j sin 110&deg; 20 C 10.28 C 5.79 20 C jFG j sin 50&deg; D 0
Solving, we get
jFA j D 13.0 kN
32
jFG j D 15.3 kN
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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?
y
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
p
L
5
p
2
2
22 C 52
5
22 C 52
L p
L p
5
22 C 52
2
22 C 52
DD0
D 230 lb D 0
Solving we find
D
jDj D 85.4 lb, jLj D 214 lb
x
W
Problem 2.53 The three forces acting on the car are
shown. The force T is parallel to the x axis and the
magnitude of the force W is 14 kN. If T C W C N D 0,
what are the magnitudes of the forces T and N?
Solution:
Fx : T N sin 20&deg; D 0
Fy : N cos 20&deg; 14 kN D 0
Solving we find
N D 14.90 N,
T D 5.10 N
20⬚
y
T
W
x
20⬚
N
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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:
jFA j D jFB j D jFC j. The magnitude of the vector sum of
the three forces is 200 kN. What is jFA j?
Solution: Use the angles and magnitudes to determine the vector
components, take the sum, and solve for the unknown. The angles
between each cable and the pillar are:
A D tan1
B D tan1
FC
FA
FB
D 33.7&deg; ,
8
D 53.1&deg;
6
C D tan1
4m
6m
12
6
D 63.4&deg; .
6m
A
B
Measure the angles counterclockwise form the x-axis. The force vectors acting along the cables are:
C
FA D jFA ji cos 303.7&deg; C j sin 303.7&deg; D 0.5548jFA ji 0.8319jFA jj
4m
4m
4m
FB D jFB ji cos 323.1&deg; C j sin 323.1&deg; D 0.7997jFB ji 0.6004jFB jj
FC D jFC ji cos 333.4&deg; C j sin 333.4&deg; D 0.8944jFC ji0.4472jFC jj
The sum of the forces are, noting that each is equal in magnitude, is
F D 2.2489jFA ji 1.8795jFA jj.
The magnitude of the sum is given by the problem:
200 D jFA j
2.24892 C 1.87952 D 2.931jFA j,
from which jFA j D 68.24 kN
Problem 2.55 The total force exerted on the top of the
mast B by the sailboat’s forestay AB and backstay BC is
180i 820j (N). What are the magnitudes of the forces
exerted at B by the cables AB and BC ?
Solution: We first identify the forces:
FAB D TAB 4.0 mi 11.8 mj
4.0 m2 C 11.8 m2
FBC D TBC y
B (4, 13) m
5.0 mi 12.0 mj
5.0 m2 C 12.0 m2
Then if we add the force we find
4
5
TAB C p
TBC D 180 N
Fx : p
155.24
169
11.8
12
TAB p
TBC D 820 N
Fy : p
155.24
169
Solving simultaneously yields:
) TAB D 226 N,
A
(0, 1.2) m
34
C
(9, 1) m
TAC D 657 N
x
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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 jFAB j D 4 kN. If the vector sum of the
three forces equals zero, what are the magnitudes of FAC
y
B
(– 4, 1) m
FAB
FAC
(4, 2) m
C
x
A
D
(–2, – 3) m
Solution: Determine the unit vectors parallel to each force:
B
2
3
iC p
j D 0.5547i 0.8320j
22 C 32
22 C 32
C
A
4
1
iC p
j D 0.9701i C 0.2425j
eAC D p
42 C 12
42 C 12
D
4
2
iC p
j D 0.89443i C 0.4472j
eAB D p
42 C 22
42 C 22
FAB D jFAB jeAB D 3.578i C 1.789j. Since the vector sum of the forces
vanishes, the x- and y-components vanish separately:
Fx D 0.5547jFAD j 0.9701jFAC j C 3.578i D 0, and
Fy D 0.8320jFAD j C 0.2425jFAC j C 1.789j D 0
These simultaneous equations in two unknowns can be solved by any
standard procedure. An HP-28S hand held calculator was used here:
The results: jFAC j D 2.108 kN , jFAD j D 2.764 kN
Problem 2.57 The distance s D 45 in.
Solution:
(a)
(a)
(b)
Determine the unit vector eBA that points from B
toward A.
Use the unit vector you obtained in (a) to determine
the coordinates of the collar C.
The unit vector is the position vector from B to A divided by its
magnitude
rBA D [14 75]i C [45 12]jin D 61i C 33j in
jrBA j D
y
eBA D
A
(14, 45) in
61 in2 C 33 in2 D 69.35 in
1
61i C 33j in D 0.880i C 0.476j
69.35 in
eBA D 0.880i C 0.476j
C
s
(b)
B
(75, 12) in
x
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 75i C 12j in C 45 in0.880i
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 s0.880 and yC D yB C s0.476.
Thus, the coordinates of point C are xC D 75 0.880s in and yC D
12 C 0.476s in. Note from the solution of Problem 2.57 above, 0 s 69.4 in.
Problem 2.59 The position vector r goes from point
A to a point on the straight line between B and C. Its
magnitude is jrj D 6 ft. Express r in terms of scalar
components.
y
B
(7, 9) ft
r
A (3, 5) ft
C
(12, 3) ft
x
Solution: Determine the perpendicular vector to the line BC from
y
point A, and then use this perpendicular to determine the angular orientation of the vector r. The vectors are
B[7,9]
P
rAB D 7 3i C 9 5j D 4i C 4j,
jrAB j D 5.6568
r
rAC D 12 3i C 3 5j D 9i 2j,
jrAC j D 9.2195
rBC D 12 7i C 3 9j D 5i 6j,
jrBC j D 7.8102
A[3,5]
C[12,3]
x
The unit vector parallel to BC is
eBC D
rBC
D 0.6402i 0.7682j D i cos 50.19&deg; j sin 50.19&deg; .
jrBC j
Add š90&deg; to the angle to find the two possible perpendicular vectors:
eAP1 D i cos 140.19&deg; j sin 140.19&deg; , or
eAP2 D i cos 39.8&deg; C j sin 39.8&deg; .
Choose the latter, since it points from A to the line.
Given the triangle defined by vertices A, B, C, then the magnitude of
the perpendicular corresponds to the altitude when the base is the line
2area
. From geometry, the area of
BC. The altitude is given by h D
base
a triangle with known sides is given by
area D
p
ss jrBC js jrAC js jrAB j,
where s is the semiperimeter, s D 12 jrAC j C jrAB j C jrBC j. Substituting values, s D 11.343, and area D 22.0 and the magnitude of the
222
D 5.6333. The angle between the
perpendicular is jrAP j D
7.8102
5.6333
vector r and the perpendicular rAP is ˇ D cos1
D 20.1&deg; . Thus
6
the angle between the vector r and the x-axis is ˛ D 39.8 š 20.1 D
59.1&deg; or 19.7&deg; . The first angle is ruled out because it causes the vector
r to lie above the vector rAB , which is at a 45&deg; angle relative to the
x-axis. Thus:
r D 6i cos 19.7&deg; C j sin 19.7&deg; D 5.65i C 2.02j
36
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
y
B
(10, 9) m
s
r
Solution: First define the unit vector that points from A to B.
A (3, 4) m
rB/A D [10 3]i C [9 4]j m D 7i C 5j m
jrB/A j D
C (9, 3) m
p
7 m2 C 5 m2 D 74 m
x
1
eB/A D p 7i C 5j
74
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 m C s
yp D 4 m C s
7
p
74
5
p
74
D 3 C 0.814s m
D 4 C 0.581s m.
The vector r that points from C to P is then
r D [3 C 0.814s 9]i C [4 C 0.581s 3]j m
r D [0.814s 6]i C [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).
Problem 2.62 The vector e D 13 i C 23 j C ez k is a unit
vector. Determine the component ez . (Notice that there
Solution: Use definition given in Eq. (14). The vector magnitude is
jUj D
32 C 42 C 122 D 13
Solution:
eD
2
1
i C j C ez k )
3
3
2 2
1
2
4
C
C ez 2 D 1 ) e2 D
3
3
9
Thus
ez D
Problem 2.63 An engineer determines that an attachment point will be subjected to a force F D 20i C Fy j 45k kN. If the attachment point will safely support a
force of 80-kN magnitude in any direction, what is the
acceptable range of values for Fy ?
y
2
3
or
ez D 2
3
Solution:
802 &frac12; Fx2 C Fy2 C F2z
802 &frac12; 202 C Fy2 C 452
To find limits, use equality.
Fy2LIMIT D 802 202 452
F
Fy2LIMIT D 3975
Fy LIMIT D C63.0, 63.0 kN
jFy LIMIT j 63.0 kN 63.0 kN Fy 63.0 kN
z
x
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 Ux i C Uy j C Uz k. Its
magnitude is jUj D 30. Its components are related by
the equations Uy D 2Ux and Uz D 4Uy . Determine the
components. (Notice that there are two answers.)
Solution: Substitute the relations between the components, deter-
U D C3.61i C 23.61j C 423.61k
mine the magnitude, and solve for the unknowns. Thus
D 3.61i 7.22j 28.9k
U D Ux i C 2Ux j C 42Ux k D Ux 1i 2j 8k
where Ux can be factored out since it is a scalar. Take the magnitude,
noting that the absolute value of jUx j must be taken:
U D 3.61i C 23.61j
C 423.61k D 3.61i C 7.22j C 28.9k
p
30 D jUx j 12 C 22 C 82 D jUx j8.31.
Solving, we get jUx j D 3.612, or Ux D š3.61. The two possible
vectors are
Problem 2.65 An object is acted upon by two
forces F1 D 20i C 30j 24k (kN) and F2 D 60i C
20j C 40k (kN). What is the magnitude of the total force
acting on the object?
Solution:
F1 D 20i C 30j 24k kN
F2 D 60i C 20j C 40k kN
F D F1 C F2 D 40i C 50j C 16k kN
Thus
FD
Problem 2.66 Two vectors U D 3i 2j C 6k and
V D 4i C 12j 3k.
(a) Determine the magnitudes of U and V.
(b) Determine the magnitude of the vector 3U C 2V.
40 kN2 C 50 kN2 C 16 kN2 D 66 kN
Solution: The magnitudes:
(a)
jUj D
p
p
32 C 22 C 62 D 7 and jVj D 42 C 122 C 32 D 13
The resultant vector
3U C 2V D 9 C 8i C 6 C 24j C 18 6k
D 17i C 18j C 12k
(b)
38
The magnitude j3U C 2Vj D
p
172 C 182 C 122 D 27.51
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
y
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 jrCD j D 3 m. Draw a sketch of
the truss with point D in its new position. What are the
new directions cosines of rCD ?
D (2, 3, 1) m
rCD
(4, 0, 0) m
Solution: The vector rCD and the magnitude jrCD j are
C
rCD D [2 m 4 m]i C [yD 0]j C [1 m 0]k D 2 mi C yD j
z
C 1 mk
(a)
jrCD j D
x
2 m2 C yCD 2 C 1 m2 D 3 m
Solving we
yCD D 2 m
find
yCD D
3 m2 2 m2 1 m2 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).
Solution:
F D 10i 20j 20k N
(a)
(b)
What are the direction cosines of F?
Determine the components of a unit vector e that
has the same direction as F.
FD
10 N2 C 20 N2 C 20 N2 D 30 N
cos x D
(a)
10 N
D 0.333,
30 N
cos z D
(b)
cos y D
20 N
D 0.667,
30 N
20 N
D 0.667
30 N
e D 0.333i 0.667j 0.667k
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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&deg; , and the angle between
the vector F and the y axis is 70&deg; .
y
70&deg;
(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 &lt; z &lt; 180&deg; .) (b)
The components of F can be obtained with Eqs. (2.15).
F
40&deg;
x
O
z
Solution:
(a)
cos 40&deg; 2 C cos 70&deg; 2 C cos z 2 D 1 ) z D 57.0&deg;
F D 200 Ncos 40&deg; i C cos 70&deg; j C cos 57.0&deg; k
(b)
F D 153.2i C 68.4j C 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 positive. Express it in terms of components.
Solution: Use Eq. (2.15) and (2.16). The third direction cosine is
cos z D š 1 0.52 0.22 D C0.8426.
The unit vector is
u D 0.5i C 0.2j C 0.8426k
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&deg; , and the angle between T and the y axis
is 130&deg; . The z component of T is positive.
(a) What is the angle between T and the z axis?
(b) Express T in terms of components.
l D cos 120&deg; D 0.5, m D cos 130&deg; D 0.6428
from which the z-direction cosine is
n D cosz D š
1 0.52 0.64282 D C0.5804.
Thus the angle between T and the z-axis is
y
y
Solution: The x- and y-direction cosines are
(a)
z D cos1 0.5804 D 54.5&deg; , and the thrust is
T D 2000.5i 0.6428j C 0.5804k, or:
130⬚
x
x
(b)
T D 100i 128.6j C 116.1k (kN)
120⬚
T
z
40
z
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 2.72 Determine the components of the position vector rBD from point B to point D. Use your result
to determine the distance from B to D.
Solution: We have the following
B5, 0, 3 m, C6, 0, 0 m, D4, 3, 1 m
coordinates:
A0, 0, 0,
rBD D 4 m 5 mi C 3 m 0j C 1 m 3 mk
y
D (4, 3, 1) m
rBD
D i C 3j 2k m
D 1 m2 C 3 m2 C 2 m2 D 3.74 m
A
C (6, 0, 0) m
x
z
B (5, 0, 3) 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 m
D 0.267,
3.74 m
cos z D
Problem 2.74 Determine the components of the unit
vector eCD that points from point C toward point D.
cos y D
3m
D 0.802,
3.74 m
2 m
D 0.535
3.74 m
Solution: We have the following
B5, 0, 3 m, C6, 0, 0 m, D4, 3, 1 m
coordinates:
A0, 0, 0,
rCD D 4 m 6 mi C 3 m 0j C 1 m 0k D 2i C 3j C 1k
rCD D
2 m2 C 3 m2 C 1 m2 D 3.74 m
Thus
eCD D
1
2i C 3j C k m D 0.535i C 0.802j C 0.267k
3.74 m
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
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
41
Problem 2.76 In Example 2.7, suppose that
caisson shifts on the ground to a new position.
magnitude of the force F remains 600 lb. In the
position, the angle between the force F and the x
is 60&deg; and the angle between F and the z axis is
Express F in terms of components.
the
The
new
axis
70&deg; .
y
40⬚
F
x
54⬚
Solution: We need to find the angle y between the force F and
the y axis. We know that
z
cos2 x C cos2 y C cos2 z D 1
cos y D š
1 cos2 x cos2 z D š
1 cos2 60&deg; cos2 70&deg; D š0.7956
y D š cos1 0.7956 D 37.3&deg; or 142.7&deg;
We will choose y D 37.3&deg; because the picture shows the force pointing
up. Now
Fx D 600 lb cos 60&deg; D 300 lb
Fy D 600 lb cos 37.3&deg; D 477 lb
Fz D 600 lb cos 70&deg; D 205 lb
Thus F D 300i C 477j C 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?
B
rB
x
Solution: The two position vectors are:
A
y
rA
rA D 20.768iC 0.384jC 0.512k D 1.536i C 0.768j C 1.024k (km)
rB D 40.743iC 0.557j 0.371k D 2.972i C 2.228j 1.484k (km)
z
The distance is the magnitude of the difference:
jrA rB j
D
1.5362.9272 C 0.7682.2282 C 1.0241.4842
D 3.24 (km)
42
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 2.78 Archaeologists measure a pre-Columbian ceremonial structure and obtain the dimensions
shown. Determine (a) the magnitude and (b) the
direction cosines of the position vector from point A to
point B.
y
4m
10 m
4m
10 m
A
8m
Solution:
(a)
B
The coordinates are A (0, 16, 14) m and B (10, 8, 4) m.
b
rAB D [10 0]i C [8 16]j C [4 14]k m D 10i 8j 10k m
jrAB j D
p
102 C 82 C 102 m D 264 m D 16.2 m
8m
z
C
x
jrAB j D 16.2 m
(b)
10
D 0.615
cos x D p
264
8
D 0.492
cos y D p
264
10
cos z D p
D 0.615
264
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.
y
10 m
4m
4m
10 m
A
8m
B
(a)
(b)
What is the distance b?
Determine the direction cosines of the position
vector from B to C.
b
8m
z
C
x
Solution: We have the coordinates B (10 m, 8 m, 4 m), C (10 m C
b, 0 18 m).
rBC D 10 m C b 10 mi C 0 8 mj C 18 m 4 mk
rBC D bi C 8 mj C 14 mk
(a)
(b)
We have 16.61 m2 D b2 C 8 m2 C 14 m2 ) b D 3.99 m
The direction cosines of rBC are
3.99 m
D 0.240
16.61 m
8 m
cos y D
D 0.482
16.61 m
14 m
cos z D
D 0.843
16.61 m
cos x D
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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 observers
would measure at that instant.
y
rAR
Solution: The vector rAR is given by
rBR
A
rAR D 4i C 4j C 2k km
x
and the magnitude of rAR is given by
jrAR j D
z
42
C 42
C 22
B (5, 0, 2) km
km D 6 km.
The unit vector along AR is given by
uAR D rAR /jrAR j.
Thus, uAR D 0.667i C 0.667j C 0.333k
and the direction cosines are
cos x D 0.667, cos y D 0.667, and cos z D 0.333.
The vector rBR is given by
rBR D xR xB i C yR yB j C zR zB k km
D 4 5i C 4 0j C 2 2k km
and the magnitude of rBR is given by
jrBR j D
12 C 42 C 02 km D 4.12 km.
The unit vector along BR is given by
eBR D rBR /jrBR j.
Thus, uBR D 0.242i C 0.970j C 0k
and the direction cosines are
cos x D 0.242, cos y D 0.970, and cos z D 0.0.
44
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 2.81 In Problem 2.80, suppose that the coordinates of the rocket’s position are unknown. At a given
instant, the person at A determines that the direction
cosines of rAR are cos x D 0.535, cos y D 0.802, and
cos z D 0.267, and the person at B determines that the
direction cosines of rBR are cos x D 0.576, cos y D
0.798, and cos z D 0.177. What are the coordinates of
the rocket’s position at that instant.
Solution: The vector from A to B is given by
rAB D xB xA i C yB yA j C zB zA k or
Similarly, the vector along BR, uBR D 0.576i C 0.798 0.177k.
From the diagram in the problem statement, we see that rAR D
rAB C rBR . Using the unit vectors, the vectors rAR and rBR can be
written as
rAB D 5 0i C 0 0j C 2 0k D 5i C 2k km.
The magnitude of rAB is given by jrAB j D
The unit vector along AB, uAB , is given by
rAR D 0.535rAR i C 0.802rAR j C 0.267rAR k, and
52 C 22 D 5.39 km.
uAB D rAB /jrAB j D 0.928i C 0j C 0.371k km.
The unit vector along the line AR,
uAR D cos x i C cos y j C cos z k D 0.535i C 0.802j C 0.267k.
rBR D 0.576rBR i C 0.798rBR j 0.177rBR k.
Substituting into the vector addition rAR D rAB C rBR and equating
components, we get, in the x direction, 0.535rAR D 0.576rBR , and
in the y direction, 0.802rAR D 0.798rBR . Solving, we get that rAR D
4.489 km. Calculating the components, we get
rAR D rAR eAR D 0.5354.489i C 0.8024.489j C 0.2674.489k.
Hence, the coordinates of the rocket, R, are (2.40, 3.60, 1.20) km.
Problem 2.82* The height of Mount Everest was originally measured by a surveyor in the following way.
He first measured the altitudes of two points and the
horizontal distance between them. For example, suppose
that the points A and B are 3000 m above sea level
and are 10,000 m apart. He then used a theodolite to
measure the direction cosines of the vector rAP from
point A to the top of the mountain P and the vector rBP
from point B to P. Suppose that the direction cosines of
rAP are cos x D 0.5179, cos y D 0.6906, and cos z D
0.5048, and the direction cosines of rBP are cos x D
0.3743, cos y D 0.7486, and cos z D 0.5472. Using
this data, determine the height of Mount Everest above
sea level.
z
P
y
B
x
A
Solution: We have the following coordinates A0, 0, 3000 m,
B10, 000, 0, 3000 m, Px, y, z
Then
rAP D xi C yj C z 3000 mk D rAP 0.5179i C 0.6906j C 0.5048k
rBP D x 10,000 mi C yj C z 3000 mk
D rBP 0.3743i C 0.7486j C 0.5472k
Equating components gives us five equations (one redundant) which
we can solve for the five unknowns.
x D rAP 0.5179
y D rAP 0.6906
z 3000 m D rAP 0.5048
) z D 8848 m
x 10000 m D rBP 0.7486
y D rBP 0.5472
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
45
y
Problem 2.83 The distance from point O to point A is
20 ft. The straight line AB is parallel to the y axis, and
point B is in the x-z plane. Express the vector rOA in
terms of scalar components.
A
rOA
Strategy: You can resolve rOA into a vector from O to
B and a vector from B to A. You can then resolve the
vector form O to B into vector components parallel to
the x and z axes. See Example 2.8.
O
x
30&deg;
60&deg;
B
z
Solution: See Example 2.8. The length BA is, from the right triangle
The vector rOA is given by rOA D rOB C rBA , from which
OAB,
rOA D 15i C 10j C 8.66k (ft)
jrAB j D jrOA j sin 30&deg; D 200.5 D 10 ft.
Similarly, the length OB is
A
jrOB j D jrOA j cos 30&deg; D 200.866 D 17.32 ft
The vector rOB can be resolved into components along the axes by the
right triangles OBP and OBQ and the condition that it lies in the x-z
plane.
Hence,
rOB D
jrOB ji cos 30&deg;
C j cos 90&deg;
C k cos 60&deg; rOA
y
30&deg;
O
x
Q
z
P
60&deg;
B
or
rOB D 15i C 0j C 8.66k.
The vector rBA can be resolved into components from the condition
that it is parallel to the y-axis. This vector is
rBA D jrBA ji cos 90&deg; C j cos 0&deg; C k cos 90&deg; D 0i C 10j C 0k.
y
Problem 2.84 The magnitudes of the two force vectors
are jFA j D 140 lb and jFB j D 100 lb. Determine the magFB
nitude of the sum of the forces FA C FB .
FA
Solution: We have the vectors
60⬚
FA D 140 lb[cos 40&deg; sin 50&deg; ]i C [sin 40&deg; ]j C [cos 40&deg; cos 50&deg; ]k
30⬚
40⬚
x
50⬚
FA D 82.2i C 90.0j C 68.9k lb
z
FB D 100 lb[ cos 60&deg; sin 30&deg; ]i C [sin 60&deg; ]j C [cos 60&deg; cos 30&deg; ]k
FB D 25.0i C 86.6j C 43.3k lb
Adding and taking the magnitude we have
FA C FB D 57.2i C 176.6j C 112.2k lb
jFA C FB j D
57.2 lb2 C 176.6 lb2 C 112.2 lb2 D 217 lb
jFA C FB j D 217 lb
46
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
y
Problem 2.85 Determine the direction cosines of the
vectors FA and FB .
FB
Solution: We have the vectors
FA
FA D 140 lb[cos 40&deg; sin 50&deg; ]i C [sin 40&deg; ]j C [cos 40&deg; cos 50&deg; ]k
FA D 82.2i C 90.0j C 68.9k lb
60⬚
FB D 100 lb[ cos 60&deg; sin 30&deg; ]i C [sin 60&deg; ]j C [cos 60&deg; cos 30&deg; ]k
30⬚
FB D 25.0i C 86.6j C 43.3k lb
40⬚
x
50⬚
z
The direction cosines for FA are
cos x D
82.2 lb
90.0 lb
D 0.587, cos y D
D 0.643,
140 lb
140 lb
cos z D
68.9 lb
D 0.492
140 lb
The direction cosines for FB are
cos x D
25.0 lb
86.6 lb
D 0.250, cos y D
D 0.866,
100 lb
100 lb
cos z D
43.3 lb
D 0.433
100 lb
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&deg; , and
the angle between the vector components Fh and Fz is
40&deg; . Draw a sketch showing the relationship of these
angles to the components of F. Express F in terms of its
components.
y
B
F
O
O
z
x
A
Solution: We have
jFy j D 900 N sin 35&deg; D 516 N
jFh j D 900 N cos 35&deg; D 737 N
jFx j D jFh j sin 40&deg; D 474 N
jFz j D jFh j cos 40&deg; D 565 N
Thus
F D 474i C 516j C 565k N
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47
Problem 2.87 An engineer calculates that the magnitude 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.
Solution: The components of the position vector from B to A are
rBA D xA xB i C yA yB j C zA zB k
D 12.4 C 9.2i C 22.0 24.4j
C 18.4 C 15.6k
D 3.2i 2.4j 2.8k m.
B
P
A
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.
y
Problem 2.88 The cable BC exerts an 8-kN force F
on the bar AB at B.
B (5, 6, 1) m
(a)
Determine the components of a unit vector that
points from B toward point C.
(b) Express F in terms of components.
F
A
Solution:
(a)
eBC D
x
xC xB i C yC yB j C zC zB k
rBC
D jrBC j
xC xB 2 C yC yB 2 C zC zB 2
C (3, 0, 4) m
z
2i 6j C 3k
2
6
3
eBC D p
D i jC k
7
7
7
22 C 62 C 32
eBC D 0.286i 0.857j C 0.429k
(b)
48
F
D jFjeBC D 8eBC D 2.29i 6.86j C 3.43k kN
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
y
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.
6 ft
E
A
D
z
4 ft
T
2 ft
20⬚
B
x
C
4 ft
y
Solution: Find the unit vector eCE and multiply it times the magnitude of the force to get the vector in component form,
eCE D
6 ft
xE xC i C yE yC j C zE zC k
rCE
D jrCE j
xE xC 2 C yE yC 2 C zE zC 2
A
E
The coordinates of point C are 4, 4 sin 20&deg; , 4 cos 20&deg; or
4, 1.37, 3.76 (ft) The coordinates of point E are (0, 2, 6) (ft)
D
T
2 ft
eCE
0 4i C 2 1.37j C 6 3.76k
p
D
42 C 3.372 C 2.242
x
4 ft
T
z
B
20&deg;
C
4 ft
eCE D 0.703i C 0.592j C 0.394k
T D 50eCE lb
T D 35.2i C 29.6j C 19.7k lb
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 jFAB j D jFAC j D
240 lb. What is the magnitude of the total force F D
FAB C FAC exerted on the loop by the rope?
Solution: The new coordinates of point A are (6, 8, 0) ft. The
position vectors are
rAB D 4i 8j C 4k ft
rAC D 4i 8j C 6k ft
The forces are
6 ft
A
FAB D 240 lb
rAB
D 98.0i 196j C 98.0k lb
jrAB j
FAC D 240 lb
rAC
D 89.1i 178j C 134.0k lb
jrAC j
7 ft
4 ft
6 ft
B
The sum of the forces is
C
2 ft
F D FAB C FAC D 8.85i 374j C 232k lb
10 ft
The magnitude is jFj D 440 lb
y
6 ft
A
FAB
FAC
7 ft
4 ft
2 ft
x
B
6 ft
C
10 ft
z
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49
y
Problem 2.91 The cable AB exerts a 200-lb force FAB
at point A that is directed along the line from A to B.
Express FAB in terms of components.
8 ft
C
8 ft
Solution: The coordinates of B are B(0,6,8). The position vector
from A to B is
6 ft
B
x
rAB D 0 6i C 6 0j C 8 10k D 6i C 6j 2k
The magnitude is jrAB j D
p
62 C 62 C 22 D 8.718 ft.
FAB
The unit vector is
uAB D
6
2
6
iC
j
k
8.718
8.718
8.718
z
FAC
A (6, 0, 10) ft
or
uAB D 0.6882i C 0.6882j 0.2294k.
FAB D jFAB juAB D 2000.6882i C 0.6882j 0.2294k
The components of the force are
FAB D jFAB juAB D 2000.6882i C 0.6882j 0.2294k or
FAB D 137.6i C 137.6j 45.9k
Problem 2.92 Consider the cables and wall described
in Problem 2.91. Cable AB exerts a 200-lb force FAB
at point A that is directed along the line from A to B.
The cable AC exerts a 100-lb force FAC at point A that
is directed along the line from A to C. Determine the
magnitude of the total force exerted at point A by the
two cables.
Solution: Refer to the figure in Problem 2.91. From Problem 2.91
the force FAB is
The force is
FAC D jFAC juAC D 100uAC D 16.9i C 50.7j 84.5k.
FAB D 137.6i C 137.6j 45.9k
The resultant of the two forces is
The coordinates of C are C(8,6,0). The position vector from A to C is
FR D FAB C FAC D 137.6 C 16.9i C 137.6 C 50.7j
rAC D 8 6i C 6 0j C 0 10k D 2i C 6j 10k.
The magnitude is jrAC j D
The unit vector is
uAC D
C 84.5 45.9k.
p
22 C 62 C 102 D 11.83 ft.
6
10
2
iC
j
k D 0.1691i C 0.5072j 0.8453k.
11.83
11.83
11.83
FR D 120.7i C 188.3j 130.4k.
The magnitude is
jFR j D
50
p
120.72 C 188.32 C 130.42 D 259.0 lb
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 2.93 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
y
A
FAB
FAC
D
Solution: The coordinates of the points are A (0, 70, 0),
B (40, 0, 0), C (40, 0, 40) D (60, 0, 60).
60 m
60 m
B
The position vectors corresponding to the cables are:
x
40 m
rAD D 60 0i C 0 70j C 60 0k
C
40 m
40 m
z
rAC D 40 0i C 0 70j C 40 0k
rAC D 40i 70j C 40k
rAB D 40 0i C 0 70j C 0 0k
rAB D 40i 70j C 0k
The unit vectors corresponding to these position vectors are:
60
70
60
D
i
j
k
110
110
110
D 0.5455i 0.6364j 0.5455k
uAC D
40
70
40
rAC
D i
jC
k
jrAC j
90
90
90
D 0.4444i 0.7778j C 0.4444k
uAB D
40
70
rAB
D
i
j C 0k D 0.4963i 0.8685j C 0k
jrAB j
80.6
80.6
The forces are:
FAB D jFAB juAB D 0.9926i 1.737j C 0k
FAC D jFAC juAC D 0.8888i 1.5556j C 0.8888
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 Problem 2.93. The magnitude of the force FAB is 2 kN. The
x and z components of the vector sum of the forces
exerted on the tower by the three cables are zero. What
are the magnitudes of FAC and FAD ?
Solution: From the solution of Problem 2.93, the unit vectors are:
uAC D
40
70
40
rAC
D i
jC
k
jrAC j
90
90
90
Taking the sum of the forces:
FR D FAB C FAC C FAD D 0.9926 0.4444jFAC j 0.5455jFAD ji
C 1.737 0.7778jFAC j 0.6364jFAD jj
D 0.4444i 0.7778j C 0.4444k
60
70
60
D
i
j
110
110
110
The sum of the x- and z-components vanishes, hence the set of simultaneous equations:
D 0.5455i 0.6364j 0.5455k
From the solution of Problem 2.93 the force FAB is
FAB D jFAB juAB D 0.9926i 1.737j C 0k
The forces FAC and FAD are:
FAC D jFAC juAC D jFAC j0.4444i 0.7778j C 0.4444k
0.4444jFAC j C 0.5455jFAD j D 0.9926 and
0.4444jFAC j 0.5455jFAD j D 0
These can be solved by means of standard algorithms, or by the use of
commercial packages such as TK Solver Plus  or Mathcad. Here
a hand held calculator was used to obtain the solution:
jFAC j D 1.1163 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.
y
0.15 m
0.4 m
B
C
Solution: The position vector from C to A is now
rCA D 0.3 meCD D 0.137i 0.205j C 0.171km
The position vector form the origin to A is
T
A
0.2 m
0.5 m
rOA D rOC C rCA D 0.4i C 0.3j m C 0.137i 0.205j C 0.171k m
O
rOA D 0.263i C 0.0949j C 0.171k m
The coordinates of A are (0.263, 0.0949, 0.171) m.
The position vector from A to B is
0.3 m
x
D
0.25 m
0.2 m
z
rAB D [0 0.263]i C [0.5 0.0949]j C [0.15 0.171]k m
rAB D 0.263i C 0.405j 0.209k m
The force T is
T D 60 N
rAB
D 32.7i C 50.3j 2.60k N
jrAB j
T D 32.7i C 50.3j 2.60k N
52
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
y
Problem 2.96 The cable AB exerts a 32-lb force T on
the collar at A. Express T in terms of components.
4 ft
B
Solution: The coordinates of point B are B (0, 7, 4). The vector
T
6 ft
position of B is rOB D 0i C 7j C 4k.
The vector from point A to point B is given by
7 ft
rAB D rOB rOA .
A
x
From Problem 2.95, rOA D 2.67i C 2.33j C 2.67k. Thus
4 ft
rAB D 0 2.67i C 7 2.33j C 4 2.67j
4 ft
z
rAB D 2.67i C 4.67j C 1.33k.
The magnitude is
jrAB j 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
rAB
D 0.4819i C 0.8429j C 0.2401k
jrAB j
The force T is given by
TAB D jTAB juAB D 32uAB D 15.4i C 27.0j C 7.7k (lb)
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.
y
Solution: From the figure, the point B is at (0, 4, 3) m. The coordinates of point A are determined by the radius of the circular bar
and the angle shown in the figure. The vector from the origin to A
is rOA D 4 cos20&deg; i C 4 sin20&deg; j m. Thus, the coordinates of point A
are (3.76, 1.37, 0) m. The vector from B to A is given by rBA D xA xB i C yA yB j C zA zB k D 3.76i 2.63j 3k m. Finally, the
scalar components of the vector from B to A are (3.76, 2.63, 3) m.
3m
B
A
4m
20&deg;
4m
x
z
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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 /jrBA j. The unit vector is uAB D
0.686i C 0.480j C 0.547k. Hence, the force vector T is given by
T D jTj0.686iC 0.480jC 0.547k N D 41.1i C 28.8j C 32.8k N
Problem 2.99 In Active Example 2.11, suppose that
the vector V is changed to V D 4i 6j 10k.
(a)
(b)
What is the value of UžV?
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 &ETH; V D 64 C 56 C 310 D 84
cos D
84
U&ETH;V
D D 0.814
jVjjVj
62 C 52 C 32 42 C 62 C 102
D cos1 0.814 D 35.5&deg;
a U &ETH; V D 84, b D 35.5&deg;
Problem 2.100 In Example 2.12, suppose that the coordinates of point B are changed to (6, 4, 4) m. What is
the angle between the lines AB and AC?
Solution: Using the new coordinates we have
rAB D 2i C j C 2k m, jrAB j D 3 m
rAC D 4i C 5j C 2k m, jrAC j D 6.71 m
y
C
(8, 8, 4) m
u
A
(4, 3, 2) m
B
(6, 1, ⫺2) m
x
cos D
24 C 15 C 22 m2
rAB &ETH; rAC
D
D 0.845
jrAB jjrAC j
3 m6.71 m
D cos1 0.845 D 32.4&deg;
D 32.4&deg;
z
Problem 2.101 What is the dot product of the position
vector r D 10i C 25j (m) and the force vector
Solution: Use Eq. (2.23).
F &ETH; r D 30010 C 25025 C 3000 D 3250 N-m
F D 300i C 250j C 300k N?
Problem 2.102 Suppose that the dot product of two
vectors U and V is U &ETH; V D 0. If jUj 6D 0, what do you
54
Solution:
Either jVj D 0 or V ? U
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Problem 2.103 Two perpendicular vectors are given
in terms of their components by
Solution: When the vectors are perpendicular, U &ETH; V 0.
Thus
U D Ux i 4j C 6k
U &ETH; V D Ux Vx C Uy Vy C Uz Vz D 0
and V D 3i C 2j 3k.
Use the dot product to determine the component Ux .
D 3Ux C 42 C 63 D 0
3Ux D 26
Ux D 8.67
Problem 2.104 Three vectors
Solution: For mutually perpendicular vectors, we have three
equations, i.e.,
U D Ux i C 3j C 2k
U&ETH;VD0
V D 3i C Vy j C 3k
W D 2i C 4j C Wz k
are mutually perpendicular. Use the dot product to determine the components Ux , Vy , and Wz .
U&ETH;WD0
V&ETH;WD0
Thus

3Ux C 3Vy C 6 D 0

3 Eqns
2Ux C 12 C 2Wz D 0

 3 Unknowns
C6 C 4Vy C 3Wz D 0
Solving, we get
Ux
Vy
Wz
Problem 2.105 The magnitudes jUj D 10 and jVj D
20.
(a)
(b)
Use the definition of the dot product to determine
U &ETH; V.
Use Eq. (2.23) to obtain U &ETH; V.
Solution:
(a)
D 2.857
D 0.857
D 3.143
y
V
U
45⬚
The definition of the dot product (Eq. (2.18)) is
30⬚
x
U &ETH; V D jUjjVj cos . Thus
U &ETH; V D 1020 cos45&deg; 30&deg; D 193.2
(b)
The components of U and V are
U D 10i cos 45&deg; C j sin 45&deg; D 7.07i C 7.07j
V D 20i cos 30&deg; C j sin 30&deg; D 17.32i C 10j
From Eq. (2.23)
U &ETH; V D 7.0717.32 C 7.0710 D 193.2
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55
Problem 2.106 By evaluating the dot product U &ETH; V,
prove the identity cos1 2 D cos 1 cos 2 C sin 1
sin 2 .
y
Strategy: Evaluate the dot product both by using
Eq. (2.18) and by using Eq. (2.23).
U
V
u1
Solution: The strategy is to use the definition Eq. (2.18) and the
u2
x
Eq. (2.23). From Eq. (2.18) and the figure,
U &ETH; V D jUjjVj cos1 2 . From Eq. (2.23) and the figure,
U D jUji cos 1 C j sin 2 , V D jVji cos 2 C j sin 2 ,
and the dot product is U &ETH; V D jUjjVjcos 1 cos 2 C sin 1 sin 2 .
Equating the two results:
U &ETH; V D jUjjVj cos1 2 D jUjjVjcos 1 cos 2 C sin 1 sin 2 ,
from which if jUj 6D 0 and jVj 6D 0, it follows that
cos1 2 D cos 1 cos 2 C sin 1 sin 2 ,
Q.E.D.
Problem 2.107 Use the dot product to determine the
angle between the forestay (cable AB) and the backstay
(cable BC).
y
B (4, 13) m
Solution: The unit vector from B to A is
eBA D
rBA
D 0.321i 0.947j
jrBA j
The unit vector from B to C is
eBC D
rBC
D 0.385i 0.923j
jrBC j
From the definition of the dot product, eBA &ETH; eBC D 1 &ETH; 1 &ETH; cos , where
is the angle between BA and BC. Thus
A
(0, 1.2) m
C
(9, 1) m
x
cos D 0.3210.385 C 0.9470.923
cos D 0.750
D 41.3&deg;
56
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y
Problem 2.108 Determine the angle between the
lines AB and AC (a) by using the law of cosines (see
Appendix A); (b) by using the dot product.
B
(4, 3, ⫺1) m
Solution:
(a)
A
We have the distances:
AB D
AC D
BC D
42 C 32 C 12 m D
x
u
p
26 m
(5, ⫺1, 3) m
C
z
p
52 C 12 C 32 m D 35 m
5 42 C 1 32 C 3 C 12 m D
p
33 m
The law of cosines gives
BC2 D AB2 C AC2 2ABAC cos cos D
(b)
AB2 C AC2 BC2
D 0.464
2ABAC
) D 62.3&deg;
Using the dot product
rAB D 4i C 3j k m,
rAC D 5i j C 3k m
rAB &ETH; rAC D 4 m5 m C 3 m1 m C 1 m3 m D 14 m2
rAB &ETH; rAC D ABAC cos Therefore
14 m2
p
D 0.464 ) D 62.3&deg;
cos D p
26 m 35 m
Problem 2.109 The ship O measures the positions of
the ship A and the airplane B and obtains the coordinates
shown. What is the angle between the lines of sight
OA and OB?
y
B
(4, 4, ⫺4) km
Solution: From the coordinates, the position vectors are:
u
x
rOA D 6i C 0j C 3k and rOB D 4i C 4j 4k
O
The dot product: rOA &ETH; rOB D 64 C 04 C 34 D 12
The magnitudes: jrOA j D
p
62 C 02 C 32 D 6.71 km and
A
z
(6, 0, 3) km
p
jrOA j D 42 C 42 C 42 D 6.93 km.
rOA &ETH; rOB
D 0.2581, from which D š75&deg; .
jrOA jjrOB j
From the problem and the construction, only the positive angle makes
sense, hence D 75&deg;
From Eq. (2.24) cos D
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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 ?
B
rB
x
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 i C cos y j C cos z k, where the direction cosines are those
for rA ). Thus, through the definition of the dot product, we can find
an expression for the cosine of the angle between rA and rB .
y
rA
A
z
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&deg; .
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?
y
A
(6, 6, –3) m
F
Solution: We use the following vectors to define the force F.
O
rOA D 8i C 3j 3k m
eOA
z
x
(10, ⫺2, 3) m
B
rOA
D 0.833i C 0.331j 0.331k
D
jrOA j
F D 50 NeOA D 44.2i C 16.6j 16.6k N
Now we need the unit vector eOB .
rOB D 10i 2j C 3k m
eOB D
rOB
D 0.941i 0.188j C 0.282k
jrOB j
To find the vector component parallel to OB we use the dot product
in the following manner
F &ETH; eOB D 44.2 N0.941 C 16.6 N0.188 C 16.6 N0.282 D 33.8 N
Fp D F &ETH; eOB eOB D 33.8 N0.941i 0.188j C 0.282k
Fp D 31.8i 6.35j C 9.53k N
58
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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.
r D 250i C 200j 150k mm,
jrj D 354 mm
To produce the unit vector that is parallel to this line we divide by the
magnitude
150 mm
y
Solution: The vector r from the O to where the person grips the
handle is
eD
r
250i C 200j 150k mm
D
D 0.707i C 0.566j 0.424k
jrj
354 mm
Using Eq. (2.26), we find that the vector component parallel to the
line is
F
O
Fp D e &ETH; Fe D [0.70760 N C 0.56640 N]0.707i
200 mm
z
C 0.566j 0.424k
250 mm
Fp D 14.0i C 11.2j C 8.4k N
x
Problem 2.113 At the instant shown, the Harrier’s
thrust vector is T D 17,000i C 68,000j 8,000k (N)
and its velocity vector is v D 7.3i C 1.8j 0.6k (m/s).
The quantity P D jTp jjvj, where Tp is the vector
component of T parallel to v, is the power currently
being transferred to the airplane by its engine. Determine
the value of P.
Solution:
y
v
T
T D 17,000i C 68,000j 8,000k N
v D 7.3i C 1.8j 0.6k m/s
x
Power D T &ETH; v D 17,000 N7.3 m/s C 68,000 N1.8 m/s
C 8,000 N0.6 m/s
Power D 251,000 Nm/s D 251 kW
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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.
y
A
(a) What is the angle between the cables AB and AC?
(b) Determine the vector component of F parallel to
the cable AB.
(0, 7, 0) ft
F
x
Solution: Use Eq. (2.24) to solve.
(a)
From the coordinates of the points, the position vectors are:
rAB D 0 0i C 0 7j C 10 0k
B
(0, 0, 10) ft
z
C
(14, 0, 14) ft
rAB D 0i 7j C 10k
rAC D 14 0i C 0 7j C 14 0k
rAC D 14i 7j C 14k
The magnitudes are:
jrAB j D
p
72 C 102 D 12.2 (ft) and
jrAB j D
p
142 C 72 C 142 D 21.
The dot product is given by
rAB &ETH; rAC D 140 C 77 C 1014 D 189.
The angle is given by
cos D
189
D 0.7377,
12.221
from which D š42.5&deg; . From the construction: D C42.5&deg;
(b)
The unit vector associated with AB is
eAB D
rAB
D 0i 0.5738j C 0.8197k.
jrAB j
The unit vector associated with AC is
eAC D
rAC
jrAC j
D 0.6667i 0.3333j C 0.6667k.
Thus the force vector along AC is
FAC D jFjeAC D 666.7i 333.3j C 666.7k.
The component of this force parallel to AB is
FAC &ETH; eAB eAB D 737.5eAB D 0i 422.8j C 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 7j C 10k, and eAC D
0.6667i 0.3333j C 0.6667k. Thus rAB &ETH; eAC D 9, and rAB &ETH; eAC eAC
D 6i 3j C 6k ft.
Problem 2.116 The force F D 10i C 12j 6k N.
Determine the vector components of F parallel and normal to line OA.
y
A
(0, 6, 4) m
Solution: Find eOA
F
O
rOA
D
jrOA j
x
Then
z
FP D F &ETH; eOA eOA
and FN D F FP
eOA D
0i C 6j C 4k
6j C 4k
p
D p
52
62 C 42
eOA D
6
4
jC
k D 0.832j C 0.555k
7.21
7.21
FP D [10i C 12j 6k &ETH; 0.832j C 0.555k]eOA
FP D [6.656]eOA D 0i C 5.54j C 3.69k N
FN D F FP
FN D 10i C 12 5.54j C 6 3.69k
FN D 10i C 6.46j 9.69k N
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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.
y
0.15 m
Solution: We have the following vectors
0.4 m
B
C
rCD D 0.2i 0.3j C 0.25k m
eCD D
T
rCD
D 0.456i 0.684j C 0.570k
jrCD j
0.2 m 0.3 m
A
0.5 m
rOB D 0.5j C 0.15k m
O
x
0.25 m
D
rOC D 0.4i C 0.3j m
0.2 m
rOA D rOC C 0.2 meCD D 0.309i C 0.163j C 0.114k m
z
rAB D rOB rOA D 0.309i C 0.337j C 0.036k m
eAB D
rAB
D 0.674i C 0.735j C 0.079k
jrAB j
We can now write the force T and determine the vector component
parallel to CD.
T D 50 NeAB D 33.7i C 36.7j C 3.93k N
Tp D eCD &ETH; TeCD D 3.43i C 5.14j 4.29k N
Tp D 3.43i C 5.14j 4.29k N
Problem 2.118 In Problem 2.117, determine the vector
component of T normal to the bar CD.
y
0.15 m
Solution: From Problem 2.117 we have
0.4 m
B
C
T D 33.7i C 36.7j C 3.93k N
T
Tp D 3.43i C 5.14j 4.29k N
A
The normal component is then
0.5 m
O
Tn D T T p
x
D
Tn D 37.1i C 31.6j C 8.22k N
62
0.2 m 0.3 m
0.25 m
0.2 m
z
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y
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?
B
(0, 6, 0) ft
F
2 ft
A
x
8 ft
10 ft
z
2
Solution: Consider a line on the sloped surface from A perpendicular to the surface. (see the diagram above) By SIMILAR triangles we
see that one such vector is rN D 8j C 2k. Let us find the component
of F parallel to this line.
The unit vector in the direction normal to the surface is
eN D
y
8
rN
8j C 2k
D 0.970j C 0.243k
D p
jrN j
82 C 22
2
The unit vector eAB can be found by
z
xB xA i C yB yA j C zB zA h
eAB D xB xA 2 C yB yA 2 C zB zA 2
8
Point B is at (0, 6, 0) (ft) and A is at (5, 1, 4) (ft).
Substituting, we get
eAB D 0.615i C 0.615j 0.492k
Now F D jFjeAB D 0.2eAB
F D 0.123i C 0.123j 0.0984k lb
The component of F normal to the surface is the component parallel
to the unit vector eN .
FNORMAL D F &ETH; eN eN D 0.955eN
FNORMAL D 0i C 0.0927j C 0.0232k lb
Problem 2.120 In Problem 2.119, what is the vector
component of F parallel to the surface?
Solution: From the solution to Problem 2.119,
Thus
F D 0.123i C 0.123j 0.0984k lb and
Fparallel D F FNORMAL
FNORMAL D 0i C 0.0927j C 0.0232k lb
Substituting, we get
The component parallel to the surface and the component normal to
the surface add to give FF D FNORMAL C Fparallel .
Fparallel D 0.1231i C 0.0304j 0.1216k lb
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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 50i C 80j C
180k (m) and his velocity is v D 2.2j 3.6k (m/s).
The position of the airlock is rA D 12i C 20k (m).
Determine the angle between his velocity vector and the
line from his position to the airlock’s position.
Solution: Points G and A are located at G: (50, 80, 180) m
and A: (12, 0, 20) m. The vector rGA is rGA D xA xG i C yA yG j C zA zG k D 12 50i C 0 80j C 20 180k m. The
dot product between v and rGA is v ž rGA D jvjjrGA j cos D vx xGA C
vy yGA C vz zGA , where is the angle between v and rGA . Substituting
in the numerical values, we get D 19.7&deg; .
y
G
A
z
x
Problem 2.122 In Problem 2.121, determine the vector component of the astronaut’s velocity parallel to the
line from his position to the airlock’s position.
Solution: The coordinates are A (12, 0, 20) m, G (50, 80, 180) m.
Therefore
rGA D 62i 80j 160k m
eGA D
rGA
D 0.327i 0.423j 0.845k
jrGA j
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 &ETH; veGA D [0.4232.2 C 0.8453.6]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&deg; W and latitude 45&deg; N on the Atlantic Ocean between Nova
Scotia and France. Point Q is at longitude 60&deg; E and
latitude 20&deg; 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 .
y
N
P
Q
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.
45⬚
z
20⬚
O
30⬚
60⬚
G
Equator
x
Solution: The distance is the product of the angle and the radius of
the sphere, d D RE , where is in radian measure. From Eqs. (2.18)
and (2.24), the angular separation of P and Q is given by
cos D
P&ETH;Q
jPjjQj
.
The strategy is to determine the angle in terms of the latitude and
longitude of the two points. Drop a vertical line from each point P and
Q to b and c on the equatorial plane. The vector position of P is the sum
of the two vectors: P D rOB C rBP . The vector rOB D jrOB ji cos P C
0j C k sin P . From geometry, the magnitude is jrOB j D RE cos P .
The vector rBP D jrBP j0i C 1j C 0k. From geometry, the magnitude
is jrBP j D RE sin P . Substitute and reduce to obtain:
The dot product is
P &ETH; Q D RE2 cosP Q cos P cos Q C sin P sin Q Substitute:
cos D
P&ETH;Q
D cosP Q cos P cos Q C sin P sin Q
jPjjQj
Substitute P D C30&deg; , Q D 60&deg; , p D C45&deg; , Q D C20&deg; , to obtain
cos D 0.2418, or D 1.326 radians. Thus the distance is d D
1.326RE
y
P D rOB C rBP D RE i cos P cos P C j sin P C k sin P cos P .
N
P
θ
A similar argument for the point Q yields
45&deg;
Q D rOC C rCQ D RE i cos Q cos Q C j sin Q C k sin Q cos Q Q
RE
b
30&deg;
60&deg;
Using the identity cos2 ˇ C sin2 ˇ D 1, the magnitudes are
x
20&deg;
c
G
jPj D jQj D RE
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 U &eth; V. (b) Use the dot
product to prove that U &eth; V is perpendicular to V.
Solution: We have U D 6i 5j k, V D 4k 6j 10k
(a)
i
U &eth; V D 6
4
j
5
6
k 1 D 44i C 56j 16k
10 U &eth; V D 44i C 56j 16k
(b)
U &eth; V &ETH; V D 444 C 566 C 1610 D 0 )
U &eth; V ? V
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65
Problem 2.125 Two vectors U D 3i C 2j and V D 2i
C 4j.
(a) What is the cross product U &eth; V?
(b) What is the cross product V &eth; U?
Solution: Use Eq. (2.34) and expand into 2 by 2 determinants.
i
U &eth; V D 3
2
j
2
4
k 0 D i20 40 j30 20
0
j
4
2
k 0 D i40 20 j20 30
0
C k34 22 D 8k
i
V &eth; U D 2
3
C k22 34 D 8k
y
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 &eth; F, where rCA is the
position vector form point C to point A.
4 ft
C
5 ft
Solution: We need to determine the force F in terms of its
components. The vector from A to B is used to define F.
4 ft
rAB D 2i 4j k ft
A
x
F
2i 4j k
rAB
D 500 lb F D 500 lb
jrAB j
22 C 42 C 12
F D 218i 436j 109k lb
B
z
(6, 0, 4) ft
Also we have rCA D 4i C 5k ft
Therefore
i
j
k rCA &eth; F D 4
0
5 D 2180i C 1530j 1750k ft-lb
218 436 109 rCA &eth; F D 2180i C 1530j 1750k ft-lb
66
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y
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 &eth; F, where rCB is the
position vector form point C to point B. Compare your
4 ft
C
5 ft
Solution: We need to determine the force F in terms of its compo4 ft
nents. The vector from A to B is used to define F.
A
rAB D 2i 4j k ft
F D 500 lb
x
F
2i 4j k
rAB
D 500 lb jrAB j
22 C 42 C 12
B
z
F D 218i 436j 109k lb
(6, 0, 4) ft
Also we have rCB D 6i 4j C 4k ft
Therefore
i
j
k rCB &eth; F D 6
4
4 D 2180i C 1530j 1750k ft-lb
218 436 109 rCB &eth; F D 2180i C 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 U &eth; V D 0. If jUj 6D 0, what do you
Solution:
Either V D 0
or
VjjU
Problem 2.129 The cross product of two vectors U
and V is U &eth; V D 30i C 40k. The vector V D 4i 2j C 3k. The vector U D 4i C Uy j C Uz k. Determine
Uy and Uz .
Solution: From the given information we have
i
U &eth; V D 4
4
j
Uy
2
k Uz 3
D 3Uy C 2Uz i C 4Uz 12j C 8 4Uy k
U &eth; V D 30i C 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.
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
67
Problem 2.130 The magnitudes jUj D 10 and jVj D
20.
y
V
(a)
Use the definition of the cross product to determine
U &eth; V.
(b) Use the definition of the cross product to determine
V &eth; U.
(c) Use Eq. (2.34) to determine U &eth; V.
(d)
U
30&deg;
45&deg;
x
Use Eq. (2.34) to determine V &eth; U.
Solution: From Eq. (228) U &eth; V D jUjjVj sin e. From the sketch,
the positive z-axis is out of the paper. For U &eth; V, e D 1k (points into
the paper); for V &eth; U, e D C1k (points out of the paper). The angle
D 15&deg; , hence (a) U &eth; V D 10200.2588e D 51.8e D 51.8k.
Similarly, (b) V &eth; U D 51.8e D 51.8k (c) The two vectors are:
U D 10i cos 45&deg; C j sin 45 D 7.07i C 0.707j,
V D 20i cos 30&deg; C j sin 30&deg; D 17.32i C 10j
i
U &eth; V D 7.07
17.32
j
k 7.07 0 D i0 j0 C k70.7 122.45
10 0 D 51.8k
i
(d) V &eth; U D 17.32
7.07
j
k 10 0 D i0 j0 C k122.45 70.7
7.07 0 D 51.8k
Problem 2.131 The force F D 10i 4j (N). Determine the cross product rAB &eth; F.
y
(6, 3, 0) m
A
rA B
x
z
(6, 0, 4) m
B
F
Solution: The position vector is
y
A (6, 3, 0)
rAB D 6 6i C 0 3j C 4 0k D 0i 3j C 4k
The cross product:
i
j k rAB &eth; F D 0 3 4 D i16 j40 C k30
10 4 0 rA B
x
z
B (6, 0, 4)
F
D 16i C 40j C 30k (N-m)
68
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 2.132 By evaluating the cross product
U &eth; V, prove the identity sin1 2 D sin 1 cos 2 cos 1 sin 2 .
y
U
V
θ1
θ2
Solution: Assume that both U and V lie in the x-y plane. The
strategy is to use the definition of the cross product (Eq. 2.28) and the
Eq. (2.34), and equate the two. From Eq. (2.28) U &eth; V D jUjjVj sin1
2 e. Since the positive z-axis is out of the paper, and e points into
the paper, then e D k. Take the dot product of both sides with e, and
note that k &ETH; k D 1. Thus
sin1 2 D U &eth; V &ETH; k
jUjjVj
x
y
U
V
θ1
θ2
x
The vectors are:
U D jUji cos 1 C j sin 2 , and V D jVji cos 2 C j sin 2 .
The cross product is
i
U &eth; V D jUj cos 1
jVj cos 2
j
jUj sin 1
jVj sin 2
k 0 0
D i0 j0 C kjUjjVjcos 1 sin 2 cos 2 sin 1 Substitute into the definition to obtain: sin1 2 D sin 1 cos 2 cos 1 sin 2 . Q.E.D.
Problem 2.133 In Example 2.15, what is the minimum
distance from point B to the line OA?
y
Solution: Let be the angle between rOA and rOB . Then the
minimum distance is
d D jrOB j sin Using the cross product, we have
B
(6, 6, ⫺3) m
jrOA &eth; rOB j D jrOA jjrOB j sin D jrOA jd ) d D
jrOA &eth; rOB j
jrOA j
We have
O
z
x
A (10, ⫺2, 3) m
rOA D 10i 2j C 3k m
rOB D 6i C 6j 3k m
rOA &eth; rOB
i
D 10
6
j
k 2 3 D 12i C 48j C 72k m2
6 3 Thus
dD
12 m2 C 48 m2 2 C 72 m2 2
D 8.22 m
10 m2 C 2 m2 C 3 m2
d D 8.22 m
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69
Problem 2.134 (a) What is the cross product rOA &eth;
rOB ? (b) Determine a unit vector e that is perpendicular
to rOA and rOB .
y
B ( 4, 4, –4) m
Solution: The two radius vectors are
rOB
rOB D 4i C 4j 4k, rOA D 6i 2j C 3k
(a)
The cross product is
i
D 6
4
rOA &eth; rOB
O
j
2
4
k 3 D i8 12 j24 12
4 x
rOA
A (6, –2, 3) m
z
C k24 C 8
D 4i C 36j C 32k m2 The magnitude is
jrOA &eth; rOB j D
(b)
p
42 C 362 C 322 D 48.33 m2
The unit vector is
eDš
rOA &eth; rOB
jrOA &eth; rOB j
D š0.0828i C 0.7448j C 0.6621k
(Two vectors.)
Problem 2.135 For the points O, A, and B in Problem 2.134, use the cross product to determine the length
of the shortest straight line from point B to the straight
line that passes through points O and A.
Solution:
(The magnitude of C is 338.3)
rOA D 6i 2j C 3k (m)
rOB D 4i C 4j 4k m
We now want to find the length of the projection, P, of line OB in
direction ec .
P D rOB &ETH; eC
rOA &eth; rOB D C
D 4i C 4j 4k &ETH; eC
(C is ? to both rOA and rOB )
i
C D 6
4
j
2
4
C8 12i
k 3 D C12 C 24j
C24 C 8k
4 P D 6.90 m
y
B ( 4, 4, –4) m
C D 4i C 36j C 32k
C is ? to both rOA and rOB . Any line ? to the plane formed by C and
rOA will be parallel to the line BP on the diagram. C &eth; rOA is such a
line. We then need to find the component of rOB in this direction and
compute its magnitude.
C &eth; rOA
i
D 4
6
j
C36
2
rOB
k 32 3
C D 172i C 204j 208k
O
x
rOA
P
z
A(6, –2, 3) m
The unit vector in the direction of C is
eC D
70
C
D 0.508i C 0.603j 0.614k
jCj
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Problem 2.136 The cable BC exerts a 1000-lb force F
on the hook at B. Determine rAB &eth; F.
y
Solution: The coordinates of points A, B, and C are A (16, 0, 12),
B (4, 6, 0), C (4, 0, 8). The position vectors are
B
F
6 ft
rAB
rOA D 16i C 0j C 12k, rOB D 4i C 6j C 0k, rOC D 4i C 0j C 8k.
x
8 ft
The force F acts along the unit vector
eBC D
rAC
4 ft
rBC
rOC rOB
rAB
D
D
jrBC j
jrOC rOB j
jrAB j
4 ft
p
62 C 82 D 10. Thus
y
eBC D 0i 0.6j C 0.8k, and F D jFjeBC D 0i 600j C 800k (lb).
B
The vector
6 ft
8 ft
C
Thus the cross product is
k 12 D 2400i C 9600j C 7200k (ft-lb)
800 4 ft
4 ft
A
12 ft
y
Problem 2.137 The force vector F points along the
straight line from point A to point B. Its magnitude
is jFj D 20 N. The coordinates of points A and B
are xA D 6 m, yA D 8 m, zA D 4 m and xB D 8 m, yB D
1 m, zB D 2 m.
(a)
(b)
r
x
rAB D 4 16i C 6 0j C 0 12k D 12i C 6j 12k
i
j
6
rAB &eth; F D 12
0
600
A
12 ft
z
Noting rOC rOB D 4 4i C 0 6j C 8 0k D 0i 6j C 8k
jrOC rOB j D
C
A
F
B
rA
Express the vector F in terms of its components.
Use Eq. (2.34) to determine the cross products
rA &eth; F and rB &eth; F.
rB
x
z
Solution: We have rA D 6i C 8j C 4k m, rB D 8i C j 2k m,
F D 20 N
(a)
8 6 mi C 1 8 mj C 2 4 mk
2 m2 C 7 m2 C 6 m2
20 N
D p 2i 7j 6k
89
20 N
rA &eth; F D p
89
(b)
i
6 m
2
j
8m
7
k 4 m
6 D 42.4i C 93.3j 123.0k Nm
i
j
k 20 N rB &eth; F D p 8 m 1 m 2 m 89 2
7
6 D 42.4i C 93.3j 123.0k Nm
Note that both cross products give the same result (as they must).
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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 &eth; T.
y
0.15 m
0.4 m
B
C
Solution: We define the appropriate vectors.
T
rCD D 0.2i 0.3j C 0.25k m
A
0.2 m
0.3 m
0.5 m
rCA
rCD
D 0.091i 0.137j C 0.114k m
D 0.2 m
jrCD j
O
x
D
0.25 m
0.2 m
rOB D 0.5j C 0.15k m
z
rOC D 0.4i C 0.3j m
rAB D rOB rOC C rCA D 0.61i 1.22j 0.305k m
T D 50 N
rAB
D 33.7i C 36.7j C 3.93k N
jrAB j
Now take the cross product
i
j
rCA &eth; T D 0.091 0.137
33.7
36.7
l 0.114 D 4.72i 3.48j C 7.96k N-m
3.93 rCA &eth; T D 47.2i 3.48j C 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?
Solution: We first develop the force T.
rCE D 0.3i C 0.1j m
T D 600 N
E
(0.2, 0.4, ⫺0.1) m
From Example 2.16 we know that the unit vector perpendicular to the
door is
e D 0.358i C 0.894j C 0.268k
y
T
D
The magnitude of the force perpendicular to the door (parallel to e) is
then
C
(0, 0.2, 0) m
A (0.5, 0, 0) m
B
(0.35, 0, 0.2) m
rCE
D 569i C 190j N
jrCE j
x
jTn j D T &ETH; e D 569 N0.358 C 190 N0.894 D 373 N
jTn j D 373 N
z
72
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
y
Problem 2.140 The bar AB is 6 m long and is perpendicular to the bars AC and AD. Use the cross product to
determine the coordinates xB , yB , zB of point B.
B
Solution: The strategy is to determine the unit vector perpendicular to both AC and AD, and then determine the coordinates that will
agree with the magnitude of AB. The position vectors are:
(0, 3, 0) m
(xB, yB, zB)
A
rOA D 0i C 3j C 0k, rOD D 0i C 0j C 3k, and
rOC D 4i C 0j C 0k. The vectors collinear with the bars are:
rAD D 0 0i C 0 3j C 3 0k D 0i 3j C 3k,
rAC D 4 0i C 0 3j C 0 0k D 4i 3j C 0k.
The vector collinear with rAB is
i
j k R D rAD &eth; rAC D 0 3 3 D 9i C 12j C 12k
4 3 0 C
D
(0, 0, 3) m
x
(4, 0, 0) m
z
The magnitude jRj D 19.21 (m). The unit vector is
eAB D
R
D 0.4685i C 0.6247j C 0.6247k.
jRj
Thus the vector collinear with AB is
rAB D 6eAB D C2.811i C 3.75j C 3.75k.
Using the coordinates of point A:
xB D 2.81 C 0 D 2.81 (m)
yB D 3.75 C 3 D 6.75 (m)
zB D 3.75 C 0 D 3.75 (m)
Problem 2.141* Determine the minimum distance
from point P to the plane defined by the three points
A, B, and C.
y
B
(0, 5, 0) m
P
(9, 6, 5) m
Solution: The strategy is to find the unit vector perpendicular to
the plane. The projection of this unit vector on the vector OP: rOP &ETH; 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 &ETH; e, rOB &ETH; e, or
rOC &ETH; e) is the distance from the origin to the plane along this same
perpendicular. Thus the distance of P from the plane is
A
(3, 0, 0) m
C
d D rOP &ETH; e rOA &ETH; e.
The position vectors are: rOA D 3i, rOB D 5j, rOC D 4k and rOP D
9i C 6j C 5k. The unit vector perpendicular to the plane is found
from the cross product of any two vectors lying in the plane. Noting:
rBC D rOC rOB D 5j C 4k, and rBA D rOA rOB D 3i 5j. The
cross product:
rBC &eth; rBA
i
D 0
3
(0, 0, 4) m
z
y
P[9,6,5]
j k 5 4 D 20i C 12j C 15k.
5 0 The magnitude is jrBC &eth; rBA j D 27.73, thus the unit vector is e D
0.7212i C 0.4327j C 0.5409k. The distance of point P from the plane
is d D rOP &ETH; e rOA &ETH; e D 11.792 2.164 D 9.63 m. The second term
is the distance of the plane from the origin; the vectors rOB , or rOC
could have been used instead of rOA .
x
B[0,5,0]
x
O
A[3,0,0]
z
C[0,0,4]
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
73
y
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
A
F
rB &eth; F D rA &eth; F.
rA
Strategy: Let rAB be the position vector from point A
to point B. Express rB in terms of of rA and rAB . Notice
that the vectors rAB and F are parallel.
B
rB
x
z
Solution: We have
rB D rA C rAB .
Therefore
rB &eth; F D rA C rAB &eth; F D rA &eth; F C rAB &eth; F
The last term is zero since rAB jjF.
Therefore
rB &eth; F D rA &eth; F
Problem 2.143 For the vectors U D 6i C 2j 4k,
V D 2i C 7j, and W D 3i C 2k, evaluate the following
mixed triple products: (a) U &ETH; V &eth; W; (b) W &ETH; V &eth;
U; (c) V &ETH; W &eth; U.
Solution: Use Eq. (2.36).
6
(a) U &ETH; V &eth; W D 2
3
2
7
0
4 0 2
D 614 24 C 421 D 160
3
(b) W &ETH; V &eth; U D 2
6
0
7
2
2 0 4 D 328 0 C 24 42 D 160
2
(c) V &ETH; W &eth; U D 3
6
7
0
2
0 2 4 D 24 712 12 C 0 D 160
74
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 2.144 Use the mixed triple product to calculate the volume of the parallelepiped.
y
(140, 90, 30) mm
(200, 0, 0) mm
x
(160, 0, 100) mm
z
Solution: We are given the coordinates of point D. From the geometry, we need to locate points A and C. The key to doing this is to note
that the length of side OD is 200 mm and that side OD is the x axis.
Sides OD, AE, and CG are parallel to the x axis and the coordinates
of the point pairs (O and D), (A and E), and (C and D) differ only by
200 mm in the x coordinate. Thus, the coordinates of point A are (60,
90, 30) mm and the coordinates of point C are (40, 0, 100) mm.
Thus, the vectors rOA , rOD , and rOC are rOD D 200i mm, rOA D
60i C 90j C 30k mm, and rOC D 40i C 0j C 100k mm. The mixed
triple product of the three vectors is the volume of the parallelepiped.
The volume is
60
rOA &ETH; rOC &eth; rOD D 40
200
90
0
0
30 100 0 y
(140, 90, 30)
mm
E
A
B
F
O
D
G
C
x
(200, 0, 0)
mm
(160, 0, 100)
mm
z
D 600 C 90200100 C 300 mm3
D 1,800,000 mm3
Problem 2.145 By using Eqs. (2.23) and (2.34), show
that
Ux Uy Uz U &ETH; V &eth; W D Vx Vy Vz W W W x
y
z
.
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 &ETH; V D
UX VX C UY VY C UZ VZ . Eq. (2.34) is the determinant representation
of the cross product:
i
Eq. (2.34) U &eth; V D UX
VX
j
UY
VY
k UZ VZ U
Q &ETH; P D QX Y
VY
For notational convenience, write P D U &eth; V. Expand the determinant about its first row:
U
P D i Y
VY
UX
UZ j
VX
VZ UX
UZ C
k
VX
VZ Since the two-by-two determinants are scalars, this can be written in
the form: P D iPX C jPY C kPZ where the scalars PX , PY , and PZ are
the two-by-two determinants. Apply Eq. (2.23) to the dot product of
a vector Q with P. Thus Q &ETH; P D QX PX C QY PY C QZ PZ . Substitute
PX , PY , and PZ into this dot product
UZ VZ UX
UZ Q
Y
VZ VX
UX
UZ C
Q
z
VZ VX
UZ VZ But this expression can be collapsed into a three-by-three determinant
directly, thus:
QX
Q &ETH; U &eth; V D UX
VX
QY
UY
VY
QZ UZ . This completes the demonstration.
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
75
Problem 2.146 The vectors U D i C UY j C 4k, V D
2i C j 2k, and W D 3i C j 2k are coplanar (they
lie in the same plane). What is the component Uy ?
Solution: Since the non-zero vectors are coplanar, the cross product of any two will produce a vector perpendicular to the plane, and
the dot product with the third will vanish, by definition of the dot
product. Thus U &ETH; V &eth; W D 0, for example.
1 UY
1
U &ETH; V &eth; W D 2
3 1
4 2 2 D 12 C 2 UY 4 6 C 42 C 3
D C10UY C 20 D 0
Thus UY D 2
Problem 2.147 The magnitude of F is 8 kN. Express
F in terms of scalar components.
Solution: The unit vector collinear with the force F is developed
as follows: The collinear vector is r D 7 3i C 2 7j D 4i 5j
y
The magnitude: jrj D
(3, 7) m
eD
F
p
42 C 52 D 6.403 m. The unit vector is
r
D 0.6247i 0.7809j. The force vector is
jrj
F D jFje D 4.998i 6.247j D 5i 6.25j (kN)
(7, 2) m
x
Problem 2.148 The magnitude of the vertical force W
is 600 lb, and the magnitude of the force B is 1500 lb.
Given that A C B C W D 0, determine the magnitude of
the force A and the angle ˛.
B
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
W
50&deg;
α
A
B D 1500i cos 50&deg; C j sin 50&deg; D 964.2i C 1149.1j
The vector A is A D jAji cos180 C ˛ C j sin180 C ˛
A D jAji cos ˛ j sin ˛. The forces balance, hence A C B C
W D 0, or 964.2 jAj cos ˛i D 0, and 1149.1 600 jAj sin ˛j D
0. Thus jAj cos ˛ D 964.2, and jAj sin ˛ D 549.1. Take the ratio of the
two equations to obtain tan ˛ D 0.5695, or ˛ D 29.7&deg; . Substitute this
angle to solve: jAj D 1110 lb
76
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 A C B C C D 0, what are the magnitudes of the force vectors B and C?
Solution: The strategy is to express the forces in terms of scalar
components, and then solve the force balance equations for the unknowns. C D jCji cos ˛ j sin ˛, where
tan ˛ D
100 in.
70 in.
50 in.
50
D 0.7143, or ˛ D 35.5&deg; .
70
C
E
B
D
A
F
Thus C D jCj0.8137i 0.5812j. Similarly, B D CjBji, and A D
C200j. The force balance equation is A C B C C D 0. Substituting,
0.8137jCj C jBji D 0, and 0.5812jCj C 200j D 0. Solving,
jCj D 344.1 lb, jBj D 280 lb
Problem 2.150 The magnitude of the horizontal force
vector D in Problem 2.149 is 280 lb. If D C E C F D 0,
what are the magnitudes of the force vectors E and F?
Solution: The strategy is to express the force vectors in terms of
scalar components, and then solve the force balance equation for the
unknowns. The force vectors are:
E D jEji cos ˇ j sin ˇ, where tan ˇ D
50
D 0.5, or ˇ D 26.6&deg; .
100
Thus
E D jEj0.8944i 0.4472j
D D 280i, and F D jFjj.
The force balance equation is D C E C F D 0. Substitute and resolve
into two equations:
0.8944jEj 280i D 0, and 0.4472jEj C jFjj D 0.
Solve: jEj D 313.1 lb, jFj D 140 lb
Problem 2.151 What are the direction cosines of F?
y
Refer to this diagram when solving Problems 2.151–
2.157.
A
(4, 4, 2) ft
Solution: Use the definition of the direction cosines and the
ensuing discussion.
p
The magnitude of F: jFj D 202 C 102 C 102 D 24.5.
F ⫽ 20i ⫹ 10j ⫺ 10k (lb)
u
B (8, 1, ⫺2) ft
x
z
Fx
20
D
D 0.8165,
The direction cosines are cos x D
jFj
24.5
cos y D
Fy
10
D
D 0.4082
jFj
24.5
cos z D
10
Fz
D
D 0.4082
jFj
24.5
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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 4i C 4j C 2k, rB D 8i C 1j 2k. The
vector from A to B is rAB D 8 p4i C 1 4j C 2 2k D
4i 3j 4k. The magnitude: jrAB j D 42 C 32 C 42 D 6.4. The unit
vector is
eAB D
Problem 2.153 What is the angle between the line
AB and the force F?
rAB
4
3
4
D
i
j
k D 0.6247i 0.4685j 0.6247k
jrAB j
6.4
6.4
6.4
Solution: Use the definition of the dot product Eq. (2.18), and
Eq. (2.24):
cos D
rAB &ETH; F
.
jrAB jjFj
From the solution to Problem 2.130, the vector parallel to AB is rAB D
4i 3j 4k, with a magnitude jrAB j D 6.4. From Problem 2.151, the
force is F D 20i C 10j 10k, with a magnitude of jFj D 24.5. The dot
product is rAB &ETH; F D 420 C 310 C 410 D 90. Substi90
tuting, cos D
D 0.574, D 55&deg;
6.424.5
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 &ETH; Ue, where e
is parallel to a line L. From Problem 2.152 the unit vector parallel to
line AB is eAB D 0.6247i 0.4688j 0.6247k. The dot product is
e &ETH; F D 0.624720 C 0.468810 C 0.624710 D 14.053.
The parallel vector is
e &ETH; Fe D 14.053e 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 8.78i C 10 C 6.59j C 10 C 8.78k
D 11.22i C 16.59j 1.22k (lb)
Problem 2.156 Determine the vector rBA &eth; F, where
rBA is the position vector from B to A.
Solution: Use the definition in Eq. (2.34). Noting rBA D rAB ,
from Problem 2.155 rBA D 4i C 3j C 4k. The cross product is
i
rBA &eth; F D 4
20
j
3
10
k 4 D 30 40i 40 80j
10 C 40 60
D 70i C 40j 100k (ft-lb)
78
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Problem 2.157 (a) Write the position vector rAB from
point A to point B in terms of components.
y
A
(4, 4, 2) ft
(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.
F ⫽ 20i ⫹ 10j ⫺ 10k (lb)
u
B (8, 1, ⫺2) ft
x
Solution:
(a)
rAB D [8 4]i C [1 4]j C [2 2]k ft
z
rAB D 4i 3j 4k ft
(b)
R D 200 N
rAB
D 125i 93.7j 125k N
jrAB j
R D 125i 96.3j 125k N
y
Problem 2.158 The rope exerts a force of magnitude
jFj D 200 lb on the top of the pole at B.
(a)
(b)
Determine the
position vector
Determine the
position vector
vector rAB &eth; F, where rAB is the
from A to B.
vector rAC &eth; F, where rAC is the
from A to C.
F
A
Solution: The strategy is to define the unit vector pointing from B
to A, express the force in terms of this unit vector, and take the cross
product of the position vectors with this force. The position vectors
rAB D 5i C 6j C 1k, rAC D 3i C 0j C 4k,
B (5, 6, 1) ft
x
C (3, 0, 4) ft
z
rBC D 3 5i C 0 6j C 4 1k D 2i 6j C 3k.
The magnitude jrBC j D
eBC D
p
22 C 62 C 32 D 7. The unit vector is
rBC
D 0.2857i 0.8571j C 0.4286k.
jrBC j
The force vector is
F D jFjeBC D 200eBC D 57.14i 171.42j C 85.72k.
The cross products:
i
rAB &eth; F D 5
57.14
j
6
171.42
k 1 85.72 D 685.74i 485.74j 514.26k
D 685.7i 485.7j 514.3k (ft-lb)
i
rAC &eth; F D 3
57.14
j
0
171.42
k 4 85.72 D 685.68i 485.72j 514.26k
D 685.7i 485.7j 514.3k (ft-lb)
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
79
Problem 2.159 The 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 components. (b) What are the direction cosines of r?
Solution: The vector r is
r D jrjsin 45&deg; i C cos 45&deg; sin 60&deg; j C cos 45&deg; cos 60&deg; k
The length of the pole is the x component of r. Therefore
y
jrj sin 45&deg; D 6 ft ) jrj D
A
(a)
Bedford
Falls
(b)
r D 6.00i C 5.20j C 3.00k ft
The direction cosines are
cos x D
r
45⬚
6 ft
D 8.49 ft
sin 45&deg;
rx
ry
rz
D 0.707, cos y D
D 0.612, cos z D
D 0.354
jrj
jrj
jrj
cos x D 0.707, cos y D 0.612, cos z D 0.354
60⬚
x
O
z
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
F
Solution: We can write the force as
x
60⬚
We know that the z component is 80 lb. Therefore
jFj cos 20&deg; cos 60&deg;
(a)
(b)
20⬚
O
F D jFjcos 20&deg; sin 60&deg; i C sin 20&deg; j C cos 20&deg; cos 60&deg; k
A
D 80 lb ) jFj D 170 lb
F D 139i C 58.2j C 80k lb
z
The direction cosines can be found:
139
D 35.5&deg;
x D cos1
170
y D cos1
z D cos1
58.2
170
80
170
D 70.0&deg;
D 62.0&deg;
x D 35.5&deg; , y D 70.0&deg; , z D 62.0&deg;
80
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y
Problem 2.161 The magnitude of the force vector FB
is 2 kN. Express it in terms of scalar components.
F
D
(4, 3, 1) m
FC
FA
FB
A
z
C
x
(6, 0, 0) m
B
Solution: The strategy is to determine the unit vector collinear
with FB and then express the force in terms of this unit vector.
F
y
The radius vector collinear with FB is
D (4,3,1)
rBD D 4 5i C 3 0j C 1 3k or rBD D 1i C 3j 2k.
FA
The magnitude is
p
jrBD j D
12 C 32 C 22 D 3.74.
The unit vector is
eBD
(5, 0, 3) m
FC
A
z
x
C(6,0,0)
FB
B (5,0,3)
rBD
D
D 0.2673i C 0.8018j 0.5345k
jrBD j
The force is
FB D jFB jeBD D 2eBD (kN) FB D 0.5345i C 1.6036j 1.0693k
D 0.53i C 1.60j 1.07k (kN)
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 &ETH; eBD eBD . The vertical force has the component
F D 6j (kN). From Problem 2.139, the unit vector pointing from
B to D is eBD D 0.2673i C 0.8018j 0.5345k. The dot product is
F &ETH; eBD D 4.813. Thus the component parallel to the line BD is FP D
4.813eBD D C1.29i 3.86j C 2.57k (kN). The component perpendicular to the line is: FN D F FP . Thus FN D 1.29i 2.14j 2.57k (kN)
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81
Problem 2.163 The magnitude of the vertical force
vector F in Problem 2.161 is 6 kN. Given that F C FA C
FB C FC D 0, what are the magnitudes of FA , FB , and
FC ?
Solution: The strategy is to expand the forces into scalar components, and then use the force balance equation to solve for the unknowns. The unit vectors are used to expand the forces into scalar
components. The position vectors, magnitudes, and unit vectors are:
The forces are:
FA D jFA jeAD , FB D jFB jeBD , FC D jFC jeCD , F D 6j (kN).
Substituting into the force balance equation
p
26 D 5.1,
F C FA C FB C FC D 0,
eAD D 0.7845i C 0.5883j C 0.1961k.
rBD D 1i C 3j 2k, jrBD j D
0.7843jFA j 0.2674jFB j 0.5348jFC ji D 0
p
14 D 3.74,
0.5882jFA j C 0.8021jFB j C 0.8021jFC j 6j
eBD D 0.2673i C 0.8018j 0.5345k.
rCD D 2i C 3j C 1k, jrCD j D
p
D 00.1961jFA j 0.5348jFB j C 0.2674jFC jk D 0
14 D 3.74,
These simple simultaneous equations can be solved a standard method
(e.g., Gauss elimination) or, conveniently, by using a commercial
package, such as TK Solver, Mathcad, or other. An HP-28S hand
held calculator was used here: jFA j D 2.83 (kN), jFB j D 2.49 (kN),
jFC j D 2.91 (kN)
eCD D 0.5345i C 0.8018j C 0.2673k
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 &eth; W, where rAG is the position
vector from A to G.
Solution: Express the position vectors in terms of scalar components, calculate rAG , and take the cross product. The position vectors
are: rAB D 0.6.5i C 0.866j C 0k rAB D 0.3i C 0.5196j C 0k,
rBG D 0.30.707i C 0.619j 0.342k,
rBG D 0.2121i C 0.1857j 0.1026k.
rAG D rAB C rBG D 0.5121i C 0.7053j 0.1026k.
W D 160j
y
0
60
mm
C
i
rAG &eth; W D 0.5121
0
j
0.7053
160
k
0.1026 0
G
D 16.44i C 0j 81.95k D 16.4i C 0j 82k (N m)
B
W
600 mm
600 mm
C
600 mm
G
B
W
A
A
z
82
x
x
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Problem 2.165 The rope CE exerts a 500-N force T
on the hinged door.
E
(0.2, 0.4, ⫺0.1) m
y
(a)
(b)
Express T in terms of components.
Determine the vector component of T parallel to
the line from point A to point B.
T
D
C
(0, 0.2, 0) m
Solution: We have
A (0.5, 0, 0) m
rCE D 0.2i C 0.2j 0.1k m
T D 500 N
(a)
(b)
x
B
(0.35, 0, 0.2) m
rCE
D 333i C 333j 167k N
jrCE j
z
T D 333i C 333j 167k N
We define the unit vector in the direction of AB and then use this
vector to find the component parallel to AB.
rAB D 0.15i C 0.2k m
eAB D
rAB
D 0.6i C 0.8k
jrAB j
Tp D eAB &ETH; TeAB D [0.6][333 N] C [0.8][167 N]0.6i C 0.8k
Tp D 200i 267k N
Problem 2.166 In Problem 2.165, let rBC be the position vector from point B to point C. Determine the cross
product rBC &eth; T.
E
(0.2, 0.4, ⫺0.1) m
y
T
Solution: From Problem 2.165 we know that
D
C
(0, 0.2, 0) m
A (0.5, 0, 0) m
T D 333i C 333j 167k N
x
B
(0.35, 0, 0.2) m
The vector rBC is
rBC D 035i C 0.2j 0.2k m
The cross product is
i
j
rBC &eth; T D 0.35 0.2
333
333
z
k 0.2 D 33.3i 125j 183k Nm
137 rBC &eth; 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&deg; to 30&deg; . 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&deg; D 0,
Fy : N cos 30&deg; 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 equilibrium. The force F1 D 4.5 lb. Determine the force F2 and
the angle ˛.
y
F2
a
F1
30⬚
x
Solution: The free-body diagram is shown below the drawing. The
equilibrium equations are
Fx : F1 cos 30&deg; F2 cos ˛ D 0
Fy : F1 sin 30&deg; C F2 sin ˛ 5 lb D 0
We can write these equations as
F2 sin ˛ D 5 lb F1 sin 30&deg;
F2 cos ˛ D F1 cos 30&deg;
Dividing these equations and using the known value for F1 we have.
tan ˛ D
F2 D
5 lb 4.5 lb sin 30&deg;
D 0.706 ) ˛ D 35.2&deg;
4.5 lb cos 30&deg;
4.5 lb cos 30&deg;
D 4.77 lb
cos ˛
F2 D 4.77 lb, ˛ D 35.2&deg;
84
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Problem 3.3 In Example 3.2, suppose that the attachment 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&deg; to 35&deg; . The angle between
cable AB and the ceiling remains 60&deg; . What are the
tensions in cables AB and AC?
B
60⬚
45⬚
C
A
Solution: The free-body diagram is shown below the picture.
The equilibrium equations are:
Fx : TAC cos 35&deg; TAB cos 60&deg; D 0
Fy : TAC sin 35&deg; C TAB sin 60&deg; 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&deg; . The freebody diagram obtained by isolating the part of the system
within the dashed line is shown. Determine the forces
TAB and TAC .
y
TAB
B
TAC
C
a
A
a
A
x
(200 kg) (9.81 m/s2)
Solution:
TAB
TAC
˛ D 40&deg;
Fx : TAC cos ˛ TAB cos ˛ D 0
α
α
Fy : TAC sin ˛ C TAB sin ˛ 1962 N D 0
Solving:
TAB D TAC D 1.526 kN
1962 Ν
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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?
55⬚
A
B
40⬚
Solution: The free-body diagram is shown.
The equilibrium equations are
Fx : TB cos 40&deg; TA cos 55&deg; D 0
Fy : TB sin 40&deg; C TA sin 55&deg; 909.81 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.
22⬚
T
P
M
Solution: The equilibrium equations are
Fx : T cos 22&deg; M cos 36&deg; D 0
36⬚
Fy : T sin 22&deg; C M sin 36&deg; 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 unstretched 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?
300 mm
A
280 mm
B
Solution: The tension in the upper spring acts on block A in the
positive Y direction, Solve the spring force-deflection equation for
the tension in the upper spring. Apply the equilibrium conditions to
block A. Repeat the steps for block B.
300 mm
N
0.3 m 0.25 mj D 0i C 60j N
TUA D 0i C 1200
m
Similarly, the tension in the lower spring acts on block A in the negative Y direction
TLA D 0i 1200
A
280 mm
N
m
0.28 m 0.25 mj D 0i 36j N
B
The weight is WA D 0i jWA jj
The equilibrium conditions are
FD
Fx C
Fy D 0,
Tension,
upper spring
F D WA C TUA C TLA D 0
A
Collect and combine like terms in i, j
Solve
Fy D jWA j C 60 36j D 0
Tension,
lower
spring
Weight,
mass A
jWA j D 60 36 D 24 N
The mass of A is
mA D
24 N
jWL j
D
D 2.45 kg
jgj
9.81 m/s2
The free body diagram for block B is shown.
The tension in the lower spring TLB D 0i C 36j
The weight: WB D 0i jWB jj
Apply the equilibrium conditions to block B.
Tension,
lower spring
y
B
x
Weight,
mass B
F D WB C TLB D 0
Collect and combine like terms in i, j:
Solve:
Fy D jWB j C 36j D 0
jWB j D 36 N
The mass of B is given by mB D
36 N
jWB j
D
D 3.67 kg
jgj
9.81 m/s2
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87
Problem 3.8 The two springs in Problem 3.7 are identical, with unstretched lengths of 250 mm. Suppose that
their spring constant k is unknown and the sum of the
masses of blocks A and B is 10 kg. Determine the value
of k and the masses of the two blocks.
Solution: All of the forces are in the vertical direction so we will
use scalar equations. First, consider the upper spring supporting both
masses (10 kg total mass). The equation of equilibrium for block the
entire assembly supported by the upper spring is A is TUA mA C
mB g D 0, where TUA D kU 0.25 N. The equation of equilibrium
for block B is TUB mB g D 0, where TUB D kL 0.25 N. The
equation of equilibrium for block A alone is TUA C TLA mA g D 0
where TLA D TUB . Using g D 9.81 m/s2 , and solving simultaneously, we get k D 1962 N/m, mA D 4 kg, and mB D 6 kg .
Problem 3.9 The inclined surface is smooth (Remember 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?
300 mm
A
280 mm
B
30⬚
Solution:
F1 D 1200 N/m0.3 0.25m D 60 N
mAg
F1
F2 D 1200 N/m0.28 0.25m D 36 N
F2
FB &amp;: F2 C mB g sin 30&deg; D 0
m Bg
F2
FA &amp;: F1 C F2 C mA g sin 30&deg; D 0
NA
Solving: mA D 4.89 kg, mB D 7.34 kg
NB
88
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 3.10 The mass of the crane is 20,000 kg. The
crane’s cable is attached to a caisson whose mass is
400 kg. The tension in the cable is 1 kN.
(a)
(b)
Determine the
friction forces
level ground.
Determine the
friction forces
level ground.
magnitudes of the normal and
exerted on the crane by the
45&deg;
magnitudes of the normal and
exerted on the caisson by the
Strategy: To do part (a), draw the free-body diagram
of the crane and the part of its cable within the
dashed line.
Solution:
(a)
45&deg;
Fy : Ncrane 196.2 kN 1
kN sin 45&deg;
D0
196.2 kN
1 kN
Fx : Fcrane C 1 kN cos 45&deg; D 0
y
Ncrane D 196.9 kN, Fcrane D 0.707 kN
(b)
Fy : Ncaisson 3.924 kN C 1 kN sin 45&deg; D 0
x
Fcrane
Fx : 1 kN cos 45&deg; C Fcaisson D 0
Ncrane
Ncaisson D 3.22 kN, Fcaisson D 0.707 kN
1 kN
3.924 kN
45&deg;
Fcaisson
Ncaisson
Problem 3.11 The inclined surface is smooth. The
100-kg crate is held stationary by a force T applied to
the cable.
Solution:
(a)
The FBD
T
(a)
(b)
Draw the free-body diagram of the crate.
Determine the force T.
981 Ν
T
Ν
60&deg;
60⬚
(b)
F -: T 981 N sin 60&deg; D 0
T D 850 N
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89
Problem 3.12 The 1200-kg car is stationary on the
If ˛ D 20&deg; , what are the magnitudes of the total
normal and friction forces exerted on the car’s tires
(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)
a
Solution:
11.772 kN
(a)
˛D
20&deg;
F% : N 11.772 kN cos ˛ D 0
F- : F 11.772 kN sin ˛ D 0
N D 11.06 kN, F D 4.03 kN
(b)
α
F D 0.6 N
F
F% : N 11.772 kN cos ˛ D 0
) ˛ D 31.0&deg;
N
F- : F 11.772 kN sin ˛ D 0
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&deg; , 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 ˛?
B
A
α
Solution: Isolate the box. Resolve the forces into scalar components, and solve the equilibrium equations.
A
B
The external forces are the weight, the tension in the rope, and the
normal force exerted by the surface. The angle between the x axis and
the weight vector is 90 ˛ (or 270 C ˛). The weight vector is
α
W D jWji sin ˛ j cos ˛ D 600i sin ˛ j cos ˛
The projections of the rope tension and the normal force are
y
T D jTx ji C 0j
N D 0i C jNy jj
T
The equilibrium conditions are
x
FDWCNCTD0
Substitute, and collect like terms
N
W
α
Fx D 600 sin ˛ jTx ji D 0
Fy D 600 cos ˛ C jNy jj D 0
Solve for the unknown tension when
For ˛ D 25&deg;
jTx j D 600 sin ˛ D 253.6 lb.
For a tension of 400 lb, (600 sin ˛ 400 D 0. Solve for the unknown
angle
sin ˛ D
400
D 0.667 or ˛ D 41.84&deg;
600
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91
A
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.
30⬚
B
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&deg; T cos 50 D 0
Fx : N 80 lb cos 50&deg; 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&deg; 13.34 kN sin 26&deg; D 0
13.34 kN
T
18&deg;
T D 5.91 kN
26&deg;
N
92
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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.
A
B
C
70⬚
45⬚
D
20⬚
Solution: The free-body diagrams are shown.
The equilibrium equations for box D are
Fx : 40 lb sin 20&deg; TC cos 25&deg; D 0
Fy : ND 40 lb cos 20&deg; C TC sin 25&deg; D 0
The equilibrium equations for box B are
Fx : 40 lb sin 70&deg; C TC cos 25&deg; TA D 0
Fy : NB 40 lb cos 70&deg; C TC sin 25&deg; 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)
(b)
What is the tension in the wire?
What is the magnitude of the force exerted on the
nail by the wire?
1.2 m
Solution:
(a)
Fy : 98.1 N 2
98.1 N
5
TD0
13
T D 128 N
T
(b)
T
5
Force D 98.1 N
12
12
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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.)
0.4 m
0.4 m
0.4 m
T
Solution:
(a)
Examine the point on the left where the wire is attached to the
picture. This point supports half of the weight
R
27.3&deg;
Fy : T sin 27.3&deg; 49.05 N D 0
T D 107 N
(b)
49.05 N
Examine one of the nails
Fx : Rx T cos 27.3&deg; C T D 0
Ry
Fy : Ry T sin 27.3&deg; D 0
Rx
27.3&deg;
RD
Rx 2 C Ry 2
T
T
R D 50.5 N
Problem 3.20 Assume that the 150-lb climber is in
equilibrium. What are the tensions in the rope on the
left and right sides?
15⬚
14⬚
y
Solution:


Fx D TR cos15&deg; TL cos14&deg; D 0

Fy D TR sin15&deg; C TL sin14&deg; 150 D 0
14&deg;
TR
TL
15&deg;
Solving, we get TL D 299 lb, TR D 300 lb
x
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?
y
Solution:


Fx D TR cos15&deg; TL cos14&deg; D 0

Fy D TR sin15&deg; C TL sin14&deg; mg D 0
TR
TL
14&deg;
15&deg;
Solving, we get
x
TL D 1.56 kN,
TR D 1.57 kN
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&deg; T sin 5&deg; D 0
Fy : 20 lb sin 30&deg; C T cos 5&deg; 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
mD
188 lb
D 5.84 slug
32.2 ft/s2
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&deg; T sin 5&deg; D 0
Fy : F sin 30&deg; C T cos 5&deg; 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 rope must equal 0.35 times the normal force
exerted on the crate by the floor. In Fig.a, the person
pulls on the rope in the direction shown. In Fig.b, the
person attaches the rope to a support as shown and pulls
upward on the rope. What is the magnitude of the force
he must exert on the rope in each case?
20⬚
(a)
10⬚
(b)
Solution: The friction force Ffr is given by
Ffr D 0.35N
(a)
For equilibrium we have
Fx : T cos 20&deg; 0.35N D 0
Fy : T sin 20&deg; 200 lb C N 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&deg; D 0
Fy : F TR sin 10&deg; 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.
Solution:
80 ft
20 ft
A
C
6
2
Fx : p TAB C p TBC D 0
37
5
1
1
Fy : p TAB C p TBC 200 lb D 0
37
5
Solving:
TAB D 304 lb, TBC D 335 lb
10 ft
B
TBC
30 ft
6
1
1
TAB
2
200 lb
Problem 3.26 Cable AB is 3 m long and cable BC is
4 m long. The mass of the suspended object is 350 kg.
Determine the tensions in cables AB and BC.
5m
A
C
B
Solution:
TAB
TAC
3
4
Fx : TAB C TBC D 0
5
5
4
4
3
Fy : TAB C TBC 3.43 kN D 0
5
5
4
3
3
TAB D 2.75 kN, TBC D 2.06 kN
3.43 kN
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:
x
5−x
We have the constraints
LAB 2 D x 2 C y 2 , 4 m2 D 5 m x2 C y 2
y
LAB
4m
These constraint imply
yD
10 mx x2 9 m2
TAB
L D 10 mx 9 m2
Now draw the FBD and write the equations in terms of x
x
5x
TBC D 0
Fx : p
TAB C
4
10x 9
Fy :
TBC
y
4
y
x
5−x
p
p
10x x2 9
10x x2 9
p
TBC 3.43 kN D 0
TAB C
4
10x 9
If we set TAB D 3 kN and solve for x we find x D 1.535, TBC D
2.11 kN &lt; 3 kN
3.43 kN
Using this value for x we find that LAB D 2.52 m
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99
Problem 3.28 What are the tensions in the upper and
Neglect the weight of the pulley.)
45&deg;
30&deg;
W
Solution: Isolate the weight. The frictionless pulley changes the
TU
direction but not the magnitude of the tension. The angle between the
right hand upper cable and the x axis is ˛, hence
TU
β
TUR D jTU ji cos ˛ C j sin ˛.
α
y
The angle between the positive x and the left hand upper pulley is
180&deg; ˇ, hence
TUL D jTU ji cos180 ˇ C j sin180 ˇ
TL
W
x
D jTU ji cos ˇ C j sin ˇ.
The lower cable exerts a force: TL D jTL ji C 0j
The weight: W D 0i jWjj
The equilibrium conditions are
F D W C TUL C TUR C TL D 0
Substitute and collect like terms,
Fx D jTU j cos ˇ C jTU j cos ˛ jTL ji D 0
Fy D jTU j sin ˛ C jTU j sin ˇ jWjj D 0.
Solve:
jTU j D
jWj
sin ˛ C sin ˇ
,
jTL j D jTU jcos ˛ cos ˇ.
From which
For
jTL j D jWj
cos ˛ cos ˇ
sin ˛ C sin ˇ
.
˛ D 30&deg; and ˇ D 45&deg;
jTU j D 0.828jWj,
jTL j D 0.132jWj
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?
(36, 36) ft
y
(12, 32) ft
C
B
(26, 16) ft
A
x
Solution: The angles are
˛ D tan1
ˇ D tan1
32 16
26 12
36 16
36 26
D 48.8&deg;
D 63.4&deg;
Now from equilibrium we have
Fx : TAB cos ˛ C TAC cos ˇ D 0
Fy : TAB sin ˛ C TAC sin ˇ 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 experiments on an airbearing platform. While she carries out
calibrations, the platform is held in place by the horizontal tethers AB, AC, and AD. The forces exerted by
the tethers are the only horizontal forces acting on the
platform. If the tension in tether AC is 2 N, what are the
tensions in the other two tethers?
TOP VIEW
D
4.0 m
A
3.5 m
B
C
3.0 m
Solution: Isolate the platform. The angles ˛ and ˇ are
tan ˛ D
Also,
tan ˇ D
1.5
3.5
3.0
3.5
D 0.429,
˛ D 23.2&deg; .
B
3.0 m
A
D 0.857,
ˇ D 40.6&deg; .
C 3.5
m
4.0
m
y
B
x
TAB D jTAB ji cos180&deg; ˇ C j sin180&deg; ˇ
β
α
TAB D jTAB ji cos ˇ C j sin ˇ.
The angle between the tether AC and the positive x axis is 180&deg; C ˛.
The tension is
D jTAC ji cos ˛ j sin ˛.
C
Solve:
jTAB j D
The equilibrium condition:
F D TAD C TAB C TAC D 0.
Substitute and collect like terms,
102
D
A
TAC D jTAC ji cos180&deg; C ˛ C j sin180&deg; C ˛
D
1.5 m
The angle between the tether AB and the positive x axis is 180&deg; ˇ,
hence
1.5 m
sin ˛
sin ˇ
jTAC j,
jTAC j sin˛ C ˇ
sin ˇ
.
For jTAC j D 2 N, ˛ D 23.2&deg; and ˇ D 40.6&deg; ,
jTAB j D 1.21 N, jTAD j D 2.76 N
Fx D jTAB j cos ˇ jTAC j cos ˛ C jTAD ji D 0,
Fy D jTAB j sin ˇ jTAC j sin ˛j D 0.
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Problem 3.31 The bucket contains concrete and
weighs 5800 lb. What are the tensions in the cables AB
and AC?
(5, 34) ft
y
B
C
(12, 16) ft
(20, 34) ft
A
x
Solution: The angles are
˛ D tan1
ˇ D tan1
34 16
12 5
34 16
20 12
D 68.7&deg;
D 66.0&deg;
Now from equilibrium we have
Fx : TAB cos ˛ C TAC cos ˇ D 0
Fy : TAB sin ˛ C TAC sin ˇ 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⬚
200 N
A
45⬚
Solution: The pulley does not change the tension in the rope that
passes over it. There is no friction between the slider and the bar.
y
T = 200 N
20&deg;
Eqns. of Equilibrium:


Fx D T sin 20&deg; C N cos 45&deg; D 0 T D 200 N

Fy D N sin 45&deg; C T cos 20&deg; mg D 0 g D 9.81 m/s2
x
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.
N
45&deg;
mg = (9.81) g
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103
Problem 3.33 The 20-kg mass is suspended from three
cables. Cable AC is equipped with a turnbuckle so that
its tension can be adjusted and a strain gauge that allows
its tension to be measured. If the tension in cable AC is
40 N, what are the tensions in cables AB and AD?
0.4 m
0.4 m
B
0.48 m
C
D
0.64 m
A
Solution:
TAC
TAB
5
TAC D 40 N
5
5
11
Fx : p TAB C p TAC C p
89
89
185
8
11
5
8
8
8
8
8
Fy : p TAB C p TAC C p
89
89
185
Solving: TAB D 144.1 N, TAD D 68.2 N
196.2 N
Problem 3.34 The structural joint is in equilibrium. If
FA D 1000 lb and FD D 5000 lb, what are FB and FC ?
FC
80⬚
FB
65⬚
35⬚
FA
FD
Solution: The equilibrium equations are
Fx : FD FC cos 65&deg; FB cos 35&deg; FA D 0
Fy : FC sin 65&deg; C FB sin 35&deg; 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.
0.25 m
h
A
B
k
Solution: The triangles formed by the rope segments and the horizontal line level with A can be used to determine the lengths Lu and
Ls . The equations are
Lu D
0.252 C 0.12 and Ls D
0.252 C 0.32 .
The stretch in the spring when in equilibrium is given by υ D Ls Lu .
Carrying out the calculations, we get Lu D 0.269 m, Ls D 0.391 m,
and υ D 0.121 m. The angle, , between the rope at A and the horizontal when the system is in equilibrium is given by tan D 0.3/0.25,
or D 50.2&deg; . From the free body diagram for mass A, we get two
equilibrium equations. They are
and
T
NA
A
mA g
Fx D NA C T cos D 0
T
Fy D T sin mA g D 0.
We have two equations in two unknowns and can solve. We get NA D
163.5 N and T D 255.4 N. Now we go to the free body diagram for B,
where the equation of equilibrium is T mB g kυ D 0. This equation
has only one unknown. Solving, we get k D 1297 N/m
Lu
0.1 m
B
mBg
Kδ
0.25 m
Ls
Lu
0.3 m
0.25 m
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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 ispthe total
volume of material in the two wires, V D 2A b2 C h2 .
Determine the value of h that minimizes the cost.
b
b
h
W
Solution: From the equation
T
T
θ
θ
Fy D 2T sin W D 0,
we obtain T D
p
W
W b2 C h2
D
.
2 sin 2h
Since T/A D , A D
p
W b2 C h2
T
D
2h
W
p
Wb2 C h2 .
and the “cost” is V D 2A b2 C h2 D
h
To determine the value of h that minimizes V, we set
dV
W
b2 C h2 D
C2 D0
dh
h2
and solve for h, obtaining h D b.
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.
20 ft
18 ft
B
D
Solution: Isolate juncture A, and solve the equilibrium equations.
C
Repeat for the cable juncture C.
E
The angle between the cable AC and the positive x axis is ˛. The
tension in AC is TAC D jTAC ji cos ˛ C j sin ˛
45&deg;
30&deg;
A
The angle between the x axis and AB is 180&deg; ˇ. The tension is
TAB D jTAB ji cos180 ˇ C j sin180 ˇ
TAB D i cos ˇ C j sin ˇ.
The weight is W D 0i jWjj.
The equilibrium conditions are
Solve: jTCE j D jTCA j cos ˛,
F D 0 D W C TAB C TAC D 0.
jTCD j D jTCA j sin ˛;
Substitute and collect like terms,
for jTCA j D 732 lb and ˛ D 30&deg; ,
Fx D jTAC j cos ˛ jTAB j cos ˇi D 0
jTAB j D 896.6 lb,
Fy D jTAB j sin ˇ C jTAC j sin ˛ jWjj D 0.
jTCE j D 634 lb,
Solving, we get
jTAB j D
cos ˛
cos ˇ
and
jTAC j
jTAC j D
jWj cos ˇ
sin˛ C ˇ
jTCD j D 366 lb
,
jWj D 1000 lb, and ˛ D 30&deg; , ˇ D 45&deg;
jTAC j D 1000
jTAB j D 732
0.7071
0.9659
0.866
0.7071
B
C
A
β
α
D 732.05 lb
y
x
W
D 896.5 lb
Isolate juncture C. The angle between the positive x axis and the cable
CA is 180&deg; ˛. The tension is
D
TCA D jTCA ji cos180&deg; C ˛ C j sin180&deg; C ˛,
C
90&deg;
E
or TCA D jTCA ji cos ˛ j sin ˛.
The tension in the cable CE is
α
y
A
TCE D ijTCE j C 0j.
x
The tension in the cable CD is TCD D 0i C jjTCD j.
The equilibrium conditions are
F D 0 D TCA C TCE C TCD D 0
Substitute t and collect like terms,
Fx D jTCE j jTCA j cos ˛i D 0,
Fy D jTCD j jTCA j sin ˛j D 0.
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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.
18 ft
20 ft
D
B
C
The lengths are calculated as follows: The vertical interior distance
in the triangle is 20 ft, since the angle is 45 deg. and the base and
altitude of a 45 deg triangle are equal. The length AB is given by
α
β
A
AB D
20 ft
D 28.284 ft.
cos 45&deg;
The length AC is given by
AC D
18 ft
D 20.785 ft.
cos 30&deg;
38
B
The altitude of the triangle for which AC is the hypotenuse is
18 tan 30&deg; D 10.392 ft. The distance CD is given by 20 10.392 D
9.608 ft.
D
β
α
β
20.784+9.608
= 30.392
α
28.284
The distance AD is given by
A
AD D AC C CD D 20.784 C 9.608 D 30.392
B
The new angles are given by the cosine law
AB2
D
382
D
β
A
α
Reduce and solve:
cos ˛ D
C 30.3922
28.2842
23830.392
cos ˇ D
382
28.2842 C 382 30.3922
228.28438
y
D 0.6787, ˛ D 47.23&deg; .
D 0.6142, ˇ D 52.1&deg; .
Isolate the juncture A. The angle between the cable AD and the positive
x axis is ˛. The tension is:
Solve:
jTAB j D
and
The angle between x and the cable AB is 180&deg; ˇ. The tension is
TAB D jTAB ji cos ˇ C j sin ˇ.
The weight is W D 0i jWjj
F D 0 D W C TAB C TAD D 0.
cos ˛
cos ˇ
jWj cos ˇ
sin˛ C ˇ
.
For jWj D 1000 lb, and ˛ D 51.2&deg; , ˇ D 47.2&deg;
The equilibrium conditions are
x
W
0.6142
0.989
jTAB j D 622.3
0.6787
0.6142
D 621.03 lb,
D 687.9 lb
Substitute and collect like terms,
108
Fx D jTAD j cos ˛ jTAB j cos ˇi D 0,
Fy D jTAB j sin ˇ C jTAD j sin ˛ jWjj D 0.
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Problem 3.39 While working on another exhibit, a
curator at the Smithsonian Institution pulls the suspended
Voyager aircraft to one side by attaching three horizontal
cables as shown. The mass of the aircraft is 1250 kg.
Determine the tensions in the cable segments AB, BC,
and CD.
D
C
B
30&deg;
50&deg;
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&deg; ; the tension in cable AB
is TAB D jTAB ji cos ˛ C j sin ˛. The weight is W D 0i jWjj. The
tension in cable AT is T D jTji C 0j. The equilibrium conditions are
A
70&deg;
F D W C T C TAB D 0.
Substitute and collect like terms
Fx jTAB j cos ˛ jTji D 0,
Fy D jTAB j sin ˛ jWjj D 0.
y
Solve: the tension in cable AB is jTAB j D
For jWj D 1250 kg 9.81
jTAB j D
12262.5
0.94
m
s2
jWj
.
sin ˛
D 12262.5 N and ˛ D 70&deg;
B
x
α
A
T
D 13049.5 N
W
Isolate juncture B. The angles are ˛ D 50&deg; , ˇ D 70&deg; , and the tension
cable BC is TBC D jTBC ji cos ˛ C j sin ˛. The angle between the
cable BA and the positive x axis is 180 C ˇ; the tension is
y
C
x
TBA D jTBA ji cos180 C ˇ C j sin180 C ˇ
The tension in the left horizontal cable is T D jTji C 0j. The equilibrium conditions are
β
A
F D TBA C TBC C T D 0.
Substitute and collect like terms
α
B
T
D jTBA ji cos ˇ j sin ˇ
y
T
Fy D jTBC j sin ˛ jTBA j sin ˇj D 0.
Solve: jTBC j D
sin ˇ
sin ˛
D
x
Fx D jTBC j cos ˛ jTBA j cos ˇ jTji D 0
α
C
β
jTBA j.
B
For jTBA j D 13049.5 N, and ˛ D 50&deg; , ˇ D 70&deg; ,
jTBC j D 13049.5
0.9397
0.7660
D 16007.6 N
Isolate the cable juncture C. The angles are ˛ D 30&deg; , ˇ D 50&deg; . By
symmetry with the cable juncture B above, the tension in cable CD is
jTCD j D
sin ˇ
sin ˛
jTCB j.
Substitute: jTCD j D 16007.6
0.7660
0.5
D 24525.0 N.
This completes the problem solution.
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109
Problem 3.40 A truck dealer wants to suspend a 4000kg 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?
40 m
b
A
C
B
Solution: Determine the dimensions and angles of the cables. Iso-
15 m
late the cable juncture B, and solve the equilibrium conditions. The
dimensions of the triangles formed by the cables:
L D 25 m,
L
b
A
b D 15 m,
25 m
β
AB C BC D S D 42 m.
C
α
Subdivide into two right triangles with a common side of unknown
length. Let the unknown length of this common side be d, then by the
Pythagorean Theorem b2 C d2 D AB2 , L2 C d2 D BC2 .
B
y
Subtract the first equation from the second to eliminate the unknown d,
L 2 b2 D BC2 AB2 .
A
α
B
β
C
Note that BC2 AB2 D BC ABBC C AB.
W
Substitute and reduce to the pair of simultaneous equations in the
unknowns
x
BC AB D
L 2 b2
S
Solve:
2
1
L b2
CS
2
S
BC D
D
,
BC C AB D S
Substitute and collect like terms
2
1
25 152
C 42 D 25.762 m
2
42
Fx D jTBC j cos ˛ jTBA j cos ˇi D 0,
Fy D jTBC j sin ˛ C jTBA j sin ˇ jWjj D 0
and AB D S BC D 42 25.762 D 16.238 m.
Solve:
jTBC j D
The interior angles are found from the cosine law:
cos ˛ D
cos ˇ D
L
C b2
C BC2
2L C bBC
AB2
L C b2 C AB2 BC2
2L C bAB
and jTBA j D
D 0.9704
˛ D 13.97&deg;
cos ˇ
cos ˛
jTBA j,
jWj cos ˛
sin˛ C ˇ
.
For jWj D 40009.81 D 39240 N,
D 0.9238
ˇ D 22.52&deg;
Isolate cable juncture B. The angle between BC and the positive x axis
is ˛; the tension is
and ˛ D 13.97&deg; , ˇ D 22.52&deg; ,
jTBA j D 64033 D 64 kN,
jTBC j D 60953 D 61 kN
TBC D jTBC ji cos ˛ C j sin ˛
The angle between BA and the positive x axis is 180&deg; ˇ; the
tension is
TBA D jTBA ji cos180 ˇ C j sin180 ˇ
D jTBA ji cos ˇ C j sin ˇ.
The weight is W D 0i jWjj.
The equilibrium conditions are
110
F D W C TBA C TBC D 0.
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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?
B
12 in.
A
D
C
12 in.
h
8 in.
12 in.
Solution: Isolated the cable juncture. From the sketch, the angles
are found from
tan ˛ D
tan ˇ D
8
12
4
12
D 0.667
8 in.
y
B
12 in
α
˛ D 33.7&deg;
8 in
A
D
D 0.333
ˇ D 18.4&deg;
β
4 in
C
The angle between the cable AB and the positive x axis is 180&deg; ˛,
the tension in AB is:
x
TAB D jTAB ji cos180 ˛ C j sin180 ˛
TAB D jTAB ji cos ˛ C j sin ˛.
The angle between AC and the positive x axis is 180 C ˇ. The
tension is
TAC D jTAC ji cos180 C ˇ C j sin180 C ˇ
TAC D jTAC ji cos ˇ j sin ˇ.
The tension in the cable AD is
The equilibrium conditions are
F D TAC C TAB C TAD D 0.
Substitute and collect like terms,
Fx D jTAB j cos ˛ jTAC j cos ˇ C jTAD ji D 0
Fy D jTAB j sin ˛ jTAC j sin ˇj D 0.
Solve:
jTAB j D
and jTAC j D
sin ˇ
sin ˛
jTAC j,
sin ˛
sin˛ C ˇ
For jTAD j D 200 lb, ˛ D 33.7&deg; , ˇ D 18.4&deg;
jTAC j D 140.6 lb,
jTAB j D 80.1 lb
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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 ˛ &frac12; 45&deg; .
B
A
Strategy: Draw a diagram of the sum of the forces
exerted by the three cables at A.
W
Solution: Draw the free body diagram of the knot at point A. Then
y
TAC
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 ˛ &lt; 45&deg; .
TAB
and
B
α
x
A
W
In this case, we solved the problem without writing the equations of
equilibrium. For reference, these equations are:
C
β
α
B
Possible locations
for C lie on line
C?
C?
α
TAB
Fx D TAB cos ˛ C TAC cos ˇ D 0
Fy D TAB sin ˛ C TAC sin ˇ W D 0.
Candidate β
W
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?
1m
A
C
B
0.75 m
15⬚
Solution: Examine the geometry
h2 C 0.75 m2 C
tan ˛ D
)
0.75 m
2 kN
0.25 m
β
α
h2 C 0.25 m2 D 1.4 m
h
h
h
, tan ˇ D
0.75 m
0.25 m
h D 0.458 m, ˛ D 31.39&deg; , ˇ D 61.35&deg;
Now draw a FBD and solve for the tension. We can use either of the
equilibrium equations
Fx : T cos ˛ C T cos ˇ C 2 kN sin 15&deg; D 0
T
T
β
α
T D 1.38 kN
2 kN
15&deg;
112
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Problem 3.44 The masses m1 D 12 kg and m2 D 6 kg
are suspended by the cable system shown. The cable BC
is horizontal. Determine the angle ˛ and the tensions in
the cables AB, BC, and CD.
TAB
A
D
α
TBC
α
B
B
C
70⬚
m1
m2
117.7 N
Solution: We have 4 unknowns and 4 equations
TCD
FBx : TAB cos ˛ C TBC D 0
FBy : TAB sin ˛ 117.7 N D 0
FCx : TBC C TCD cos 70&deg; D 0
70&deg;
TBC
C
FCy : TCD sin 70&deg; 58.86 N D 0
Solving we find
˛ D 79.7&deg; , TAB D 119.7 N, TBC D 21.4 N, TCD D 62.6 N
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.
58.86 N
30 in
30 in
30 in
A
D
16 in
20 in
C
B
W2
W1
Solution: We have 4 unknowns and 4 equilibrium equations to use
3
15
TBC D 0
FBx : p TAB C p
229
13
TAB
2
15
2
3
2
2
TBC 50 lb D 0
FBy : p TAB C p
229
13
TBC
B
15
15
TBC C
TCD D 0
FCx : p
17
229
50 lb
2
8
TBC C
TCD W2 D 0
FCy : p
17
229
TCD
W2 D 25 lb, TAB D 75.1 lb
8
)
C
TBC D 63.1 lb, TCD D 70.8 lb
TBC
15
15
2
W2
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113
Problem 3.46 In the system shown in Problem 3.45,
assume that W2 D W1 /2. If you don’t want the tension
anywhere in the supporting cable to exceed 200 lb, what
is the largest acceptable value of W1 ?
TAB
Solution:
3
15
TBC D 0
FBx : p TAB C p
229
13
2
15
2
3
2
2
TBC W1 D 0
FBy : p TAB C p
229
13
TBC
B
15
15
TBC C
TCD D 0
FCx : p
17
229
W1
2
8
W1
TCD D0
TBC C
FCy : p
17
2
229
TCD
TAB D 1.502W1 , TBC D 1.262W1 , TCD D 1.417W1
8
C
AB is the critical cable
200 lb D 1.502W1 ) W1 D 133.2 lb
TBC
15
2
15
W2 = W1/2
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)
(b)
Draw the free-body diagram of the cylinder. (The
cylinder’s weight is negligible).
Determine the magnitudes of the forces exerted by
Solution: From the figure, if C is at the origin, then points A, B,
and D are located at
1m
D
C
Hydraulic
cylinder
1m
0.6 m
B
A
0.15 m
0.6 m
Scoop
y
A0.15, 0.6
B0.75, 0.6
D
FCD
D1.00, 0.4
and forces FCA , FBC , and FCD are parallel to CA, BC, and CD, respectively.
C
x
FBC
We need to write unit vectors in the three force directions and express
the forces in terms of magnitudes and unit vectors. The unit vectors
are given by
eCA D
rCA
D 0.243i 0.970j
jrCA j
eCB D
rCB
D 0.781i 0.625j
jrCB j
eCD D
rCD
D 0.928i C 0.371j
jrCD j
FCA
A
B
Now we write the forces in terms of magnitudes and unit vectors. We
can write FBC as FCB D 8eCB kN or as FCB D 8eCB kN (because
we were told it was directed from B toward C and had a magnitude
of 8 kN. Either way, we must end up with
FCB D 6.25i C 5.00j kN
Similarly,
FCA D 0.243FCA i 0.970FCA j
FCD D 0.928FCD i C 0.371FCD j
For equilibrium, FCA C FCB C FCD D 0
In component form, this gives


Fx D 0.243FCA C 0.928FCD 6.25 (kN) D 0
 F D 0.970F C 0.371F C 5.00 (kN) D 0
y
CA
CD
Solving, we get
FCA D 7.02 kN, FCD D 4.89 kN
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115
Problem 3.48 The 50-lb cylinder rests on two smooth
surfaces.
(a)
(b)
Draw the free-body diagram of the cylinder.
If ˛ D 30&deg; , what are the magnitudes of the forces
exerted on the cylinder by the left and right
surfaces?
α
Solution: Isolate the cylinder. (a) The free body diagram of the
45&deg;
y
isolated cylinder is shown. (b) The forces acting are the weight and the
normal forces exerted by the surfaces. The angle between the normal
force on the right and the x axis is 90 C ˇ. The normal force is
β
α
NL
NR D jNR ji cos90 C ˇ C j sin90 C ˇ
NR
W
x
NR D jNR ji sin ˇ C j cos ˇ.
The angle between the positive x axis and the left hand force is normal
90 ˛; the normal force is NL D jNL ji sin ˛ C j cos ˛. The weight
is W D 0i jWjj. The equilibrium conditions are
jNR j D
F D W C NR C NL D 0.
and jNL j D
Substitute and collect like terms,
Solve:
sin ˛
sin ˇ
jNL j,
jWj sin ˇ
sin˛ C ˇ
.
For jWj D 50 lb, and ˛ D 30&deg; , ˇ D 45&deg; , the normal forces are
Fx D jNR j sin ˇ C jNL j sin ˛i D 0,
jNL j D 36.6 lb,
jNR j D 25.9 lb
Fy D jNR j cos ˇ C jNL j cos ˛ jWjj D 0.
Problem 3.49 For the 50-lb cylinder in Problem 3.48,
obtain an equation for the force exerted on the cylinder
by the left surface in terms of the angle ˛ in two ways:
(a) using a coordinate system with the y axis vertical,
(b) using a coordinate system with the y axis parallel to
the right surface.
y
Solution: The solution for Part (a) is given in Problem 3.48 (see
free body diagram).
jNR j D
sin ˛
sin ˇ
jNL j jNL j D
jWj sin ˇ
sin˛ C ˇ
.
Part (b): The isolated cylinder with the coordinate system is shown.
The angle between the right hand normal force and the positive x axis
is 180&deg; . The normal force: NR D jNR ji C 0j. The angle between the
left hand normal force and the positive x is 180 ˛ C ˇ. The normal
force is NL D jNL ji cos˛ C ˇ C j sin˛ C ˇ.
The angle between the weight vector and the positive x axis is ˇ.
The weight vector is W D jWji cos ˇ j sin ˇ.
The equilibrium conditions are
NR
NL
W
x
Substitute and collect like terms,
Fx D jNR j jNL j cos˛ C ˇ C jWj cos ˇi D 0,
Fy D jNL j sin˛ C ˇ jWj sin ˇj D 0.
F D W C NR C NL D 0.
Solve:
116
β
α
jNL j D
jWj sin ˇ
sin˛ C ˇ
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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?
0.6 m
A
C
k
k
B
Solution:
F
F
F D k0.6 m 0.4 m
Fy : 2F sin 60&deg; 490.5 N D 0
60&deg;
60&deg;
k D 1416 N/m
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?
0.7 m
A
C
k
B
Solution: The Geometry
0.7 m
Law of Cosines and Law of Sines
φ
θ
0.72 D 0.52 C 0.452 20.50.45 cos ˇ
sin sin sin ˇ
D
D
0.45 m
0.5 m
0.7 m
0.5 m
0.45 m
β
ˇ D 94.8&deg; , D 39.8&deg; D 45.4&deg;
Now do the statics
TAB
F
F D k0.45 m 0.4 m
Fx : TAB cos C F cos D 0
θ
φ
Fy : TAB sin C F sin 490.5 N D 0
Solving: k D 7560 N/m
490.5 N
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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.
L
h
m
R
Solution: From the geometry we have
cos D
x
RChy
, sin D
L
L
cos D
x
y
, sin D
R
R
Thus the equilibrium equations can be written
x
x
Fx : T C N D 0
L
R
Fy :
RChy
y
T C N mg D 0
L
R
Solving, we find T D
mgL
RCh
Problem 3.53 The inclined surface is smooth. Determine the force T that must be exerted on the cable to
hold the 100-kg crate in equilibrium and compare your
T
60⬚
3T
Solution:
981 N
F- : 3 T 981 N sin 60&deg; D 0
T D 283 N
N
60&deg;
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 0.275mA g.
D
C
B
T
A
Solution: From the free-body diagram of pulley C
TD
TD 2T mC g D 0 ) TD D 2T C mC g
Then from the free-body diagram of pulley B
D
C
mg
T C T C 2T C mC g mB g mA g D 0
T
Thus
TD
TD
C
1
mA C mB mC g
4
T D 0.275mA g
TD ⫽ 2T ⫹ mg
B
1
mA C 0.3mA 0.2mA g D 0.275mA g
4
T
(a)
T
T
T
A
B
mg
mAg
(b)
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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.
Solution: Draw free body diagrams of each pulley and the object
A. Each pulley and the object A must be in equilibrium. The weights
of the pulleys and object A are W D mg and WA D mA g. The equilibrium equations for the weight A, the lower pulley, second pulley, third
pulley, and the top pulley are, respectively, B WA D 0, 2C B W D 0, 2D C W D 0, 2T D W D 0, and FS 2T W D 0.
Begin with the first equation and solve for B, substitute for B in the
second equation and solve for C, substitute for C in the third equation
and solve for D, and substitute for D in the fourth equation and solve
for T, to get T in terms of W and WA . The result is
T
A
Fs
W
T
W
B D WA ,
DD
CD
WA
W
C ,
2
2
3W
WA
7W
WA
C
, and T D
C
,
4
4
8
8
or in terms of the masses,
TD
120
g
mA C 7m.
8
T
T
T
W
D
D
D
C
W
C
C
B
B
WA
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Problem 3.56 The suspended mass m1 D 50 kg. Neglecting the masses of the pulleys, determine the value
of the mass m2 necessary for the system to be in
equilibrium.
A
B
C
m2
m1
Solution:
T1
T
T
FC : T1 C 2m2 g m1 g D 0
FB : T1 2m2 g D 0
C
m2 D
m1
D 12.5 kg
4
T1
m1 g
B
T = m2 g
T
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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
(a) One pulley
W
(b) Two pulleys
W
(c) Three pulleys
Solution:
(a)
(b)
(b) For two pulleys
T
Fy : 2T W D 0 ) T D
Fupper : 2T T1 D 0
Flower : 2T1 W D 0
T1
W
TD
4
(c)
T
W
2
T1
Fupper : 2T T1 D 0
Fmiddle : 2T1 T2 D 0
W
Flower : 2T2 W D 0
(c) For three pulleys
TD
T
W
8
T
(a) For one pulleys
T
T
T1
T1
W
T2
T2
T2
W
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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.
Solution: By extrapolation of the previous problem
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.
(a)
TD
W
2N
(b)
TD
W
1024
(a)
Problem 3.60 A 14,000-kg airplane is in steady flight
in the vertical plane. The flight path angle is D 10&deg; ,
the angle of attack is ˛ D 4&deg; , 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.
y
L
W D mg D 14,0009.81 N
The equilibrium equations are
x


Fx D T cos ˛ D W sin D 0

Fy D T sin ˛ C L W cos D 0
α
α = 4&deg;
γ = 10&deg;
T
γ
D
W
T D 60 kN D 60000 N
Solving, we get
γ
D D 36.0 kN, L D 131.1 kN
y
Path
x
T
γ
L
α
D
Horizon
W
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 jTji cos ˛ C j sin ˛
The lift vector: L D 0i C jLjj
The drag: D D jDji C 0j. The angle between the weight vector and
the positive x axis is 270 ;
W D jWji sin j cos .
The equilibrium conditions are
F D T C L C D C W D 0.
Substitute and collect like terms
and
Fx D jTj cos ˛ jDj jWj sin i D 0,
Fy D jTj sin ˛ C jLj jWj cos j D 0
Solve the equations for the terms in :
jWj sin D jTj cos ˛ jDj,
and jWj cos D jTj sin ˛ C jLj.
Take the ratio of the two equations
tan D
jTj cos ˛ jDj
jTj sin ˛ C jLj
.
Divide top and bottom on the right by jDj.
For ˛ D 0,
jLj
jTj
D 2,
D 4,
jDj
jDj
tan D
21
4
D
1
or D 14&deg;
4
y
Path
T
L
α
D
x
γ
Horizontal
W
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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 of 1000 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
y
and the positive x axis is ˛:
Path
x
T
T D jTji cos ˛ C j sin ˛.
L
α
The lift vector: L D 0i C jLjj.
The drag: D D jDji C 0j. The angle between the weight vector and
the positive x axis is 270 :
γ
Horizontal
D
W
W D jWji sin j cos .
The equilibrium conditions are
F D T C L C D C W D 0.
γ
Substitute and collect like terms:
1 km
Fx D jTj cos ˛ jDj jWj sin i D 0
γ
Fy D jTj sin ˛ C jLj jWj cos j D 0
h
Solve the equations for the terms in ,
jWj sin D jTj cos ˛ jDj,
and jWj cos D jTj sin ˛ C jLj
Part (a): Take the ratio of the two equilibrium equations:
tan D
jTj cos ˛ jDj
jTj sin ˛ C jLj
.
Divide top and bottom on the right by jDj.
For ˛ D 0, jTj D 0,
jLj
D 4,
jDj
tan D
1
4
D 14&deg;
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
126
1
,
h
hD
1
1
D D 4 km
1
tan 4
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 3.63 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?
C
(⫺2, 0, ⫺2) m
y
x
B
(4, 0, 2) m
(⫺3, 0, 3) m
D
Solution: The position vector from point A to point B can be used
to write the force TAB .
A (0, ⫺4, 0) m
z
rAB D 5i C 4j m
TAB D TAB
rAB
D TAB 0.781i C 0.625j
jrAB j
100 kg
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
y
Problem 3.64 The force F D 800i C 200j (lb) acts at
point A where the cables AB, AC, and AD are joined.
What are the tensions in the three cables?
F
D (0, 6, 0) ft
A
(12, 4, 2) ft
C
B
(0, 4, 6) ft
(6, 0, 0) ft
x
z
Solution: We first write the position vectors
rAB D 6i 4j 2k ft
rAC D 12i C 6k ft
rAD D 12i C 2j 2k ft
Now we can use these vectors to define the force vectors
TAB D TAB
rAB
D TAB 0.802i 0.535j 0.267k
jrAB j
TAC D TAC
rAC
D TAC 0.949i C 0.316k
jrAC j
D TAD 0.973i C 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
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
127
y
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.
F
D (0, 6, 0) ft
A
(12, 4, 2) ft
C
B
(0, 4, 6) ft
(6, 0, 0) ft
x
z
Solution: We first write the position vectors
rAB D 6i 4j 2k ft
rAC D 12i C 6k ft
rAD D 12i C 2j 2k ft
Now we can use these vectors to define the force vectors
TAB D T
rAB
D T0.802i 0.535j 0.267k
jrAB j
TAC D T
rAC
D T0.949i C 0.316k
jrAC j
D T0.973i C 0.162j 0.162k
The force F can be written F D Fx i C Fy j C Fz k
The equilibrium equations are then
Fx : 0.802T 0.949T 0.973T C Fx D 0 ) Fx D 2.72T
Fy : 0.535T C 0.162T C Fy D 0 ) Fy D 0.732T
Fz : 0.267T C 0.316T 0.162T C Fz D 0 ) Fz D 0.113T
We also have the constraint equation
) T D 363 lb
Fx 2 C Fy 2 C Fz 2 D 1000 lb
Solving, we find Fx D 990 lb, Fy D 135 lb, Fz D 41.2 lb
128
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 3.66 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?
y
B
(0, 6, 0) ft
C
(8, 4, 0) ft
Solution: The position vectors are
2 ft
rAB D 5i C 5j 4k ft
A
x
8 ft
rAC D 3i C 3j 4k ft
z
10 ft
The angle ˛ between the inclined surface the horizontal is
˛ D tan1 2/8 D 14.0&deg;
We identify the following force:
TAB D TAB
rAB
D TAB 0.615i C 0.615j 0.492k
jrAB j
TAC D TAC
rAC
D TAC 0.514i C 0.514j 0.686k
jrAC j
N D Ncos ˛j C sin ˛k D N0.970j C 0.243k
W D 10 lbj
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 F D 2i (kip)
at A. What are the tensions in cables AB, AC, and AD?
y
6 ft
C
8 ft
2 ft
Solution: Isolate the cable juncture. Express the tensions in terms
of unit vectors. Solve the equilibrium equations. The coordinates of
points A, B, C, D are:
A8, 0, 0,
B0, 3, 8,
C0, 2, 6,
B
A
3 ft
D0, 4, 0.
D
z
The radius vectors for these points are
4 ft
rA D 8i C 0j C 0k,
rB D 0i C 3j C 8k,
rC D 0i C 2j 6k,
rD D 0i C 4j C 0k.
8 ft
x
By definition, the unit vector parallel to the tension in cable AB is
eAB D
rB r A
.
jrB rA j
Carrying out the operations for each of the cables, the results are:
eAB D 0.6835i C 0.2563j C 0.6835k,
eAC D 0.7845i C 0.1961j 0.5883k,
eAD D 0.8944i 0.4472j C 0k.
The tensions in the cables are expressed in terms of the unit vectors,
TAB D jTAB jeAB ,
TAC D jTAC jeAC ,
The external force acting on the juncture is F D 2000i C 0j C 0k. The
equilibrium conditions are
F D 0 D TAB C TAC C TAD C F D 0.
Substitute the vectors into the equilibrium conditions:
Fx D 0.6835jTAB j 0.7845jTAC j 0.8944jTAD jC2000i D 0
Fy D 0.2563jTAB j C 0.1961jTAC j 0.4472jTAD jj D 0
Fz D 0.6835jTAB j 0.5883jTAC j C 0jTAD jk D 0
The commercial program TK Solver Plus was used to solve these
equations. The results are
jTAB j D 780.31 lb ,
130
jTAC j D 906.49 lb ,
jTAD j D 844.74 lb .
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y
Problem 3.68 Prior to its launch, a balloon carrying a
set of experiments to high altitude is held in place by
groups of student volunteers holding the tethers at B, C,
and D. The mass of the balloon, experiments package,
and the gas it contains is 90 kg, and the buoyancy force
on the balloon is 1000 N. The supervising professor
conservatively estimates that each student can exert at
least a 40-N tension on the tether for the necessary length
of time. Based on this estimate, what minimum numbers
of students are needed at B, C, and D?
A (0, 8, 0) m
C (10,0, –12) m
D
(–16, 0, 4) m
x
B (16, 0, 16) m
z
Solution:
1000 N
Fy D 1000 909.81 T D 0
T D 117.1 N
(90) g
A0, 8, 0
B16, 0, 16
T
C10, 0, 12
D16, 0, 4
We need to write unit vectors eAB , eAC , and eAD .
y
T
eAB D 0.667i 0.333j C 0.667k
(0, 8, 0)
eAC D 0.570i 0.456j 0.684k
A
FAC
eAD D 0.873i 0.436j C 0.218k
We now write the forces in terms of magnitudes and unit vectors

FAB D 0.667FAB i 0.333FAB j C 0.667FAB k




 FAC D 0.570FAC i 0.456FAC j 0.684FAC k





T D 117.1j (N)
C (10, 0, −12) m
D
x
(−16, 0, 4)
z
B (16, 0, 16) m
The equations of equilibrium are
Fx D 0.667FAB C 0.570FAC 0.873FAD D 0
Fy D 0.333FAB 0.456FAC 0.436FAC C 117.1 D 0
Fz D 0.667FAB 0.684FAC C 0.218FAC D 0
Solving, we get
FAB D 64.8 N &frac34; 2 students
FAC D 99.8 N &frac34; 3 students
FAD D 114.6 N &frac34; 3 students
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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,
y
C
B
D
A
1m
1m
2m
0.3 m
x
z
Solution: Points A, B, C, and D are located at
A0, 1.2, 0,
B0.3, 2, 1,
C0, 2, 1,
D2, 2, 0
y
C
FAC
B
FAB
Write the unit vectors eAB , eAC , eAD
D
A
eAB D 0.228i C 0.608j C 0.760k
W
eAC D 0i C 0.625j 0.781k
eAD D 0.928i C 0.371j C 0k
z
(20) (9.81) N
x
The forces are
FAB D 0.228FAB i C 0.608FAB j C 0.760FAB k
FAC D 0FAC i C 0.625FAC j 0.781FAC k
W D 209.81j
The equations of equilibrium are


Fx D 0.228FAB C 0 C 0.928FAD D 0




Fy D 0.608FAB C 0.625FAC C 0.371FAD 209.81 D 0





Fz D 0.760FAB 0.781FAC C 0 D 0
We have 3 eqns in 3 unknowns solving, we get
FAB D 150.0 N
FAC D 146.1 N
132
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Problem 3.70 The weight of the horizontal wall section is W D 20,000 lb. Determine the tensions in the
Solution: Set the coordinate origin at A with axes as shown. The
upward force, T, at point A will be equal to the weight, W, since the
cable at A supports the entire wall. The upward force at A is T D W
k. From the figure, the coordinates of the points in feet are
A
D
10 ft
A4, 6, 10,
B0, 0, 0,
C12, 0, 0,
and
7 ft
D4, 14, 0.
The three unit vectors are of the form
6 ft
C
B
4 ft
xI xA i C yI yA j C zI zA k
,
eAI D xI xA 2 C yI yA 2 C zI zA 2
14 ft
8 ft
W
where I takes on the values B, C, and D. The denominators of the unit
vectors are the distances AB, AC, and AD, respectively. Substitution
of the coordinates of the points yields the following unit vectors:
T
z
eAB D 0.324i 0.487j 0.811k,
A
y
TD
eAC D 0.566i 0.424j 0.707k,
10 ft
TB
D
7 ft
TC
and eAD D 0i C 0.625j 0.781k.
6 ft
The forces are
TAB D TAB eAB ,
14 ft
4 ft
TAC D TAC eAC ,
and
C
B
X
8 ft
W
The equilibrium equation for the knot at point A is
T C TAB C TAC C TAD D 0.
From the vector equilibrium equation, write the scalar equilibrium
equations in the x, y, and z directions. We get three linear equations
in three unknowns. Solving these equations simultaneously, we get
TAB D 9393 lb, TAC D 5387 lb, and
A
D
10 ft
6 ft
C
B
4 ft
8 ft
7 ft
14 ft
W
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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.
y
A
(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 10j C 5k ft
rAC D 6i 10j 5k ft
E
C
rAD D 8i 10j 4k ft
z
rAE D 6i 10j C 5k ft
B
TAB
rAB
D TAB 0.408i 0.816j C 0.408k
D TAB
jrAB j
x
(a)
TAC D TAC
rAC
D TAC 0.473i 0.788j 0.394k
jrAC j
TAE
rAE
D TAE 0.473i 0.788j C 0.394k
D TAE
jrAE j
8 ft
6 ft
C
D
5 ft
4 ft
x
The equilibrium equations are
5 ft
Fx : 0.408TAB C 0.473TAC 0.596TAD 0.473TAE D 0
E
Fy : 0.816TAB 0.788TAC 0.745TAD 0.788TAE C 3000 lb D 0
B
6 ft
5 ft
z
Fz : 0.408TAB 0.394TAC 0.298TAD C 0.394TAE D 0
(b)
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&deg;
O
x
B
C
D
y
Solution:
TOA
Fx D TOA sin 10&deg; D D 0,
Fy D TOA cos 10&deg; 6809.81 D 0.
Solving, we obtain D D 1176 N, TOA D 6774 N.
10&deg;
D
x
(680) (9.81) N
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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 5.3j C k (m),
rOD D 1.6i 5.4j k (m).
Dividing by the magnitudes, we obtain the unit vectors
eOB D 0.591i 0.806j,
eOC D 0.200i 0.963j C 0.182k,
eOD D 0.280i 0.944j 0.175k.
Using these unit vectors, we obtain the equilibrium equations
Fx D TOA sin 10&deg; 0.591TOB C 0.200TOC C 0.280TOD D 0,
Fy D TOA cos 10&deg; 0.806TOB 0.963TOC 0.944TOD D 0,
Fz D 0.182TOC 0.175TOD D 0.
From the solution of Problem 3.72, TOA D 6774 N. Solving these
equations, we obtain
TOB D 3.60 kN,
TOC D 1.94 kN,
TOD D 2.02 kN.
y
TOA
10&deg;
x
TOB
TOC
136
TOD
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 3.74 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.
y
(0, 4, ⫺3) m
C
B
(4, 3, 1) m
D
(0, 5, 5) m
Strategy: Draw a free-body diagram of the ball at B.
(The weight of the ball is negligible.)
x
A
E
z
Solution:
FAB D FAB
4i 3j k
p
26
, TBC D TBC
4i C j 4k
p
33
,
The forces
TBD D TBD
4i C 2j C 4k
6
, W D 200 kg9.81 m/s2 j
The equilibrium equations
4
4
4
Fx : p FAB p TBC TBD D 0
6
26
33
3
1
2
Fy : p FAB C p TBC C TBD 1962 N D 0
6
26
33
1
4
4
Fz : p FAB p TBC C TBD D 0
6
26
33
TBC D 1610 N
)
TBD D 1009 N
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
137
Problem 3.75* The 3400-lb car is at rest on the plane
surface. The unit vector en D 0.456i C 0.570j C 0.684k
is perpendicular to the surface. Determine the magnitudes of the total normal force N and the total friction
force f exerted on the surface by the car’s wheels.
y
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 N0.456i C 0.570j C 0.684k
The equilibrium equation is
x
z
Nen C f 3400 lbj 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 lbj &ETH; en D N 3400 lbj &ETH; en D 0
N D 3400 lb0.57 D 1940 lb
Now we can solve for the friction force f.
f D 3400 lbj Nen D 3400 lbj 1940 lb0.456i C 0.570j C 0.684k
f D 884i C 2300j 1330k lb
jfj D
884 lb2 C 2300 lb2 C 1330 lb2 D 2790 lb
jNj D 1940 lb, jfj D 2790 lb
138
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
y
Problem 3.76 The system shown anchors a stanchion
of a cable-suspended roof. If the tension in cable AB is
900 kN, what are the tensions in cables EF and EG?
G
(0, 1.4, ⫺1.2) m
Solution: Using the coordinates for the points we find
E
F
rBA D [3.4 2i C 1 1j C 0 0k] m
(2, 1, 0) m
(1, 1.2, 0) m
rBA D 1.4 mi eBA
rBA
Di
D
jrBA j
(3.4, 1, 0) m
B
(0, 1.4, 1.2) m
A
(2.2, 0, ⫺1) m
Using the same procedure we find the other unit vectors that we need.
D
eBC D 0.140i 0.700j C 0.700k
z
x
(2.2, 0, 1) m
C
eBD D 0.140i 0.700j C 0.700k
eBE D 0.981i C 0.196j
eEG D 0.635i C 0.127j 0.762k
eEF D 0.635i C 0.127j C 0.762k
We can now write the equilibrium equations for the connections at B
and E.
900 kNeBA C TBC eBC C TBD eBD C TBE eBE D 0, TBE eBE C TEF eEF C TEG eEG 0
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
y
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?
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
1500 kN
900 kN TAB D 1410 kN
TAB D
956 kN
G
(0, 1.4, ⫺1.2) m
F
E
(2, 1, 0) m
(1, 1.2, 0) m
(3.4, 1, 0) m
B
(0, 1.4, 1.2) m
A
(2.2, 0, ⫺1) m
D
z
C
x
(2.2, 0, 1) m
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
139
y
Problem 3.78 The 200-kg slider at A is held in place
on the smooth vertical bar by the cable AB.
2m
(a) Determine the tension in the cable.
(b) Determine the force exerted on the slider by
the bar.
B
A
5m
2m
x
2m
z
Solution: The coordinates of the points A, B are A2, 2, 0,
B0, 5, 2. The vector positions
rA D 2i C 2j C 0k,
T
rB D 0i C 5j C 2k
N
The equilibrium conditions are:
F D T C N C W D 0.
W
Eliminate the slider bar normal force as follows: The bar is parallel to
the y axis, hence the unit vector parallel to the bar is eB D 0i C 1j C
0k. The dot product of the unit vector and the normal force vanishes:
eB &ETH; N D 0. Take the dot product of eB with the equilibrium conditions:
eB &ETH; N D 0.
eB &ETH; F D eB &ETH; T C eB &ETH; W D 0.
The weight is
eB &ETH; W D 1j &ETH; jjWj D jWj D 2009.81 D 1962 N.
Note: For this specific configuration, the problem can be solved without eliminating the slider bar normal force, since it does not appear in
the y-component of the equilibrium equation (the slider bar is parallel
to the y-axis). However, in the general case, the slider bar will not be
parallel to an axis, and the unknown normal force will be projected
onto all components of the equilibrium equations (see Problem 3.79
below). In this general situation, it will be necessary to eliminate the
slider bar normal force by some procedure equivalent to that used
above. End Note.
The unit vector parallel to the cable is by definition,
eAB D
rB r A
.
jrB rA j
Substitute the vectors and carry out the operation:
eAB D 0.4851i C 0.7278j C 0.4851k.
(a)
The tension in the cable is T D jTjeAB . Substitute into the modified equilibrium condition
eB F D 0.7276jTj 1962 D 0.
Solve: jTj D 2696.5 N from which the tension vector is
T D jTjeAB D 1308i C 1962j C 1308k.
(b)
The equilibrium conditions are
F D 0 D T C N C W D 1308i C 1308k C N D 0.
Solve for the normal force: N D 1308i 1308k. The magnitude
is jNj D 1850 N.
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. Determine the tension in the cable.
Solution: The vector from B to C is now
y
B
4 ft
A
6 ft
rBC D 8 ft eBD
rBC D 8 ft
7
4
4
i jC k
9
9
9
7 ft
C
rBC D 3.56i 6.22j C 3.56k ft
O
x
We can now find the unit vector form C to A.
rCA D rOA rOB C rBC D [7j C 4k
z
4 ft
D
4 ft
f7j C 3.56i 6.22j C 3.56kg] ft
rCA D 3.56i C 6.22j C 0.444k ft
eCA D
rCA
D 0.495i C 0.867j C 0.0619k
jrCA j
Using N to stand for the normal force between the bar and the slider,
we can write the equilibrium equation:
TeCA C N 100 lbj D 0
We can use the dot product to eliminate N from the equation
[TeCA C N 100 lbj] &ETH; eBD D TeCA &ETH; eBD 100 lbj &ETH; eBD D 0
T
7
4
4
[0.495] C [0.867] C
[0.0619] 100 lb0.778 D 0
9
9
9
T0.867 C 77.8 lb D 0 ) T D 89.8 lb
T D 89.8 lb
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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?
y
0.15 m
0.4 m
B
Solution: We develop the following position vectors and unit
C
vectors
rCD D 0.2i 0.3j C 0.25k m
0.2 m 0.3 m
A
0.5 m
eCD D
rCD
D 0.456i 0.684j C 0.570k
jrCD j
O
x
0.25 m
D
0.2 m
rCA D 0.2 meCD D 0.091i 0.137j C 0.114k m
z
rAB D rOB rOC C rCA rAB D [0.5j C 0.15k f0.4i C 0.3jg C f0.091i 0.137j C 0.114kg] m
rAB D 0.309i C 0.337j C 0.036k m
eAB D
rAB
D 0.674i C 0.735j C 0.079k
jrAB j
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 kg9.81 m/s2 j D 0
To eliminate the normal force N we take a dot product with eCD .
[TeAB C N 8 kg9.81 m/s2 j] &ETH; eCD D 0
TeAB &ETH; eCD 78.5 Nj &ETH; eCD D 0
T[0.674][0.456] C [0.735][0.684] C [0.079][0.570] 78.5 N0.684 D 0
T0.150 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.
y
0.15 m
Solution: From Problem 3.81 we have
0.4 m
B
C
eAB D 0.674i C 0.735j C 0.079k
T D 357 N
0.2 m 0.3 m
A
0.5 m
The equilibrium equation is
O
TeAB C N 78.5 Nj D 0
D
We can now solve for the normal force N.
x
0.25 m
0.2 m
z
N D 78.5 Nj 357 N0.674i C 0.735j C 0.079k
N D 240i 184j 28.1k N
The magnitude of N is
jNj D
240 N2 C 184 N2 C 28.1 N2
jNj D 304 N
142
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y
Problem 3.82* The 10-kg collar A and 20-kg collar B
are held in place on the smooth bars by the 3-m cable
from A to B and the force F acting on A. The force F
is parallel to the bar. Determine F.
(0, 5, 0) m
(0, 3, 0) m
F
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
A
3m
B
(4, 0, 0) m
eCD D erCDx i C eCDy j C eCDz k,
(0, 0, 4) m
z
and eEG D erEGx i C eEGy j C eEGz k.
y
Using the coordinates of points C, D, E, and G from the picture, the
unit vectors are
D (0, 5, 0)
m
eCD D 0.625i C 0.781j C 0k,
yA D yC C CAeCDy ,
B
mBg
z
yB D yA C ABeABy ,
A
TAB
NB
and zA D zC C CAeCDz ,
xB D xA C ABeABx ,
NA
TAB
The location of point A is given by
where CA D 3 m. From these equations, we find that the location of
point A is given by A (2.13, 2.34, 0) m. Once we know the location
of point A, we can proceed to find the location of point B. We have
two ways to determine the location of B. First, B is 3 m from point A
along the line AB (which we do not know). Also, B lies on the line
EG. The equations for the location of point B based on line AB are:
F
G (0, 3, 0)
m
and eEG D 0i C 0.6j C 0.8k.
xA D xC C CAeCDx ,
mAg
3m
C (4, 0, 0) m
We now have two fewer equation than unknowns. Fortunately, there
are two conditions we have not yet invoked. The bars at A and B
are smooth. This means that the normal force on each bar can have
no component along that bar. This can be expressed by using the dot
product of the normal force and the unit vector along the bar. The two
conditions are
NA &ETH; eCD D NAx eCDx C NAy eCDy C NAz eCDz D 0
The equations based on line EG are:
for slider A and
yB D yE C EBeEGy ,
and zB D zE C EBeEGz .
We have six new equations in the three coordinates of B and the
distance EB. Some of the information in the equations is redundant.
However, we can solve for EB (and the coordinates of B). We get that
the length EB is 2.56 m and that point B is located at (0, 1.53, 1.96) m.
We next write equilibrium equations for bodies A and B. From the free
body diagram for A, we get
NAx C TAB eABx C FeCDx D 0,
x
E (0, 0, 4) m
and zB D zA C ABeABz .
xB D xE C EBeEGx ,
x
NB &ETH; eEG D NBx eEGx C NBy eEGy C NBz eEGz D 0.
Solving the eight equations in the eight unknowns, we obtain
F D 36.6 N .
Other values obtained in the solution are EB D 2.56 m,
NAx D 145 N,
NBx D 122 N,
NAy D 116 N,
NBy D 150 N,
NAz D 112 N,
and
NBz D 112 N.
NAy C TAB eABy C FeCDy mA g D 0,
and NAz C TAB eABz C FeCDz D 0.
From the free body diagram for B, we get
NBx TAB eABx D 0,
Nby TAB eABy mB g D 0,
and NBz TAB eABz D 0.
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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.
A
45⬚
B
30⬚
Solution: The free-body diagram is sketched. The equilibrium
equations are
Fx - : T cos45&deg; 30&deg; 100 lb sin 30&deg; D 0
F% : T sin45&deg; 30&deg; 100 lb cos 30&deg; C N 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
60⬚
20⬚
25⬚
B
A
W
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&deg; C W sin 60&deg; 32.2 lb D 0
Thus
WD
32.2 lb
sin 25&deg; C sin 60&deg;
W D 25.0 lb
144
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 3.85 The 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 ˛?
y
TAB
B
TAC
C
a
A
a
A
x
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 lb
2 sin ˛
If we limit the tensions to 400 lb, we have
400 lb D
400 lb
1
) sin ˛ D ) ˛ D 30&deg;
2 sin ˛
2
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&deg; 100 lb D 0
FxB : NB sin 40&deg; T D 0
FxA : T NA sin 20&deg; D 0
FyA : NA cos 20&deg; 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?
10⬚
4⬚
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&deg; C NL cos 4&deg; NR cos 3&deg; D 0
Fy : T cos 10&deg; 170 lb C NL sin 40&deg; C NR sin 3&deg; 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. Determine the force T necessary for the system to be in
equilibrium.
T
A
Solution: Break the system into four free body diagrams as shown.
Carefully label the forces to ensure that the tension in any single cord
is uniform. The equations of equilibrium for the four objects, starting
with the leftmost pulley and moving clockwise, are:
S 3T D 0,
R 3S D 0,
F
F 3R D 0,
R
and 2T C 2S C 2R mA g D 0.
We want to eliminate S, R, and F from our result and find T in
terms of mA and g. From the first two equations, we get S D 3T,
and R D 3S D 9T. Substituting these into the last equilibrium equation
results in 2T C 23T C 29T D mA g.
R
R
R
S
S
S
Solving, we get T D mA g/26 .
S
T
T
S
S
R
R
T
T
T
A
mAg
Note: We did not have to solve for F to find the appropriate value of T.
The final equation would give us the value of F in terms of mA and g.
We would get F D 27mA g/26. If we then drew a free body diagram
of the entire assembly, the equation of equilibrium would be F T mA g D 0. Substituting in the known values for T and F, we see that
this equation is also satisfied. Checking the equilibrium solution by
using the “extra” free body diagram is often a good procedure.
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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
F
60 lb.
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
F
45⬚
A
20⬚
Solution: Isolate the top block. Solve the equilibrium equations.
The weight is. The angle between the normal force N1 and the positive x axis is. The normal force is. The force N2 is. The equilibrium
conditions are
from which
Solve:
y
B
N2
F D N1 C N2 C W D 0
N1
Fx D 0.7071jN1 j jN2 ji D 0
Fy D 0.7071jN1 j 490.5j D 0.
N1 D 693.7 N,
W
α
x
y
N1
jN2 j D 490.5 N
Isolate the bottom block. The weight is
F
β
α
A
W D 0i jWjj D 0i 429.81j D 0i 412.02j (N).
The angle between the normal force N1 and the positive x axis is
270&deg; 45&deg; D 225&deg; .
x
N3
W
The normal force:
N1 D jN1 ji cos 225&deg; C j sin 225&deg; D jN1 j0.7071i 0.7071j.
The angle between the normal force N3 and the positive x-axis is
90&deg; 20&deg; D 70&deg; .
The normal force is
N1 D jN3 ji cos 70&deg; C j sin 70&deg; D jN3 j0.3420i C 0.9397j.
The force is . . . F D jFji C 0j. The equilibrium conditions are
F D W C N1 C N3 C F D 0,
from which:
Fx D 0.7071jN1 j C 0.3420jN3 j C jFji D 0
Fy D 0.7071jN1 j C 0.9397jN3 j 412j D 0
For jN1 j D 693.7 N from above:
jFj D 162 N
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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?
y
B
(2, 2, 0) m
B
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, A3, 0, 4, B2, 2, 0,
C5, 2, 1.
rB D 2i C 2j C 0k,
C
A
The vector locations of the points A, B, C are:
rA D 3i C 0j C 4k,
x
A
(3, 0, 4) m
z
Solution: Get the unit vectors parallel to the ropes using the coor-
C
(5, 2, ⫺1) m
W
N
rC D 5i C 2j 1k.
Substitute and collect like terms,
The unit vector parallel to the tension acting between the points A, B
in the direction of B is
eAB D
rB r A
jrB rA j
The unit vectors are
eAB D 0.2182i C 0.4364j 0.8729k,
Fx D 0.2182jTAB j C 0.3482jTAC j 0.286jNji D 0
Fy D 0.4364jTAB j C 0.3482jTAC j C 0.429jNj 784.8j D 0
Fz D 0.8729jTAB j C 0.8704jTAC j 0.857jNjk D 0
We have three linear equations in three unknowns. The solution is:
eAC D 0.3482i C 0.3482j 0.8704k,
jTAB j D 100.7 N ,
jTAC j D 889.0 N ,
jNj D 1005.5 N .
and eN D 0.286i C 0.429j C 0.857k.
where the last was given by the problem statement. The forces are
expressed in terms of the unit vectors,
TAB D jTAB jeAB ,
TAC D jTAC jeAC ,
N D jNjeN .
The weight is
W D 0i jWjj C 0k D 0i 809.81j C 0k 0i 784.8j C 0k.
The equilibrium conditions are
150
F D 0 D TAB C TAC C N C W D 0.
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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 coordinates of the end points. Express the tensions in terms of these
unit vectors, and solve the equilibrium conditions. The coordinates
of points A, B, C are A3, 0, 4, B4, 2, 0, C5, 2, 1. The vector
locations of the points A, B, C are:
rA D 3i C 0j C 4k,
rB D 4i C 2j C 0k,
rC D 5i C 2j 1k.
The unit vectors are
eAB D C0.2182i C 0.4364j 0.8729k,
eAC D C0.3482i C 0.3482j 0.8704k,
eN D 0.286i C 0.429j C 0.857k.
where the last was given by the problem statement. The forces are
expressed in terms of the unit vectors,
TAB D jTAB jeAB ,
TAC D jTAC jeAC ,
N D jNjeN .
The weight is
W D 0i jWjj C 0k D 0i 809.81j C 0k 0i 784.8j C 0k.
The equilibrium conditions are
F D 0 D TAB C TAC C N C W D 0.
Substitute and collect like terms,
Fx D C0.281jTAB j C 0.3482jTAC j 0.286jNji D 0
Fy D 0.4364jTAB j C 0.3482jTAC j C 0.429jNj 784.8j D 0
Fz D 0.8729jTAB j C 0.8704jTAC j 0.857jNjk D 0
The HP-28S hand held calculator was used to solve these simultaneous
equations. The solution is:
jTAB j D 420.5 N ,
jTAC j D 532.7 N ,
jNj D 969.3 N .
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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.
A
20⬚
B
C
40⬚
Solution: Isolate the box. The weight vector is
75
W2 D 229.81j D 215.8j (N).
D
The angle between the normal force and the positive x axis is 90&deg; 60&deg; D 30&deg; .
60⬚
The normal force is NB D jNB j0.866i 0.5j.
The angle between the rope CD and the positive x axis is 180&deg; 75&deg; D 105&deg; ; the tension is:
T
y
β
T2 D jT2 ji cos 105&deg; C j sin 105&deg; D jT2 j0.2588i C 0.9659j
The equilibrium conditions are
N
Fx D 0.866jNB j C 0.2588jT2 ji D 0,
α
x
W
Fy D 0.5jNB j C 0.9659jT2 j 215.8j D 0.
Solve:
NB D 57.8 N,
jT2 j D 193.5 N.
y
T1
Isolate the friend. The weight is
20&deg;
W D 909.81j D 882.9j (N).
The angle between the normal force and the positive x axis is
40&deg; D 50&deg; . The normal force is:
90&deg;
40&deg;
N
75&deg;
T2
W
x
NF D jNF j0.6428i C 0.7660j.
The angle between the lower rope and the x axis is 75&deg; ; the tension is
T2 D jT2 j0.2588i C 0.9659j.
The angle between the tension in the upper rope and the positive x
axis is 180&deg; 20&deg; D 160&deg; , the tension is
T1 D jT1 j0.9397i C 0.3420j.
The equilibrium conditions are
F D W C T1 C T2 C NF D 0.
From which:
Fx D 0.6428jNF j C 0.2588jT2 j 0.9397jT1 ji D 0
Fy D 0.7660jNF j 0.9659jT2 j C 0.3420jT1 j 882.9j D 0
Solve, for jT2 j D 193.5 N. The result:
jNF j D 1051.6 N ,
152
jT1 j D 772.6 N .
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 3.94 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 perpendicular to the road. Determine the magnitudes of the total
normal and friction forces exerted on the car by the road.
L
D
15⬚
Solution: The free-body diagram is shown. If we write the equilibrium equations parallel and perpendicular to the road, we have:
F- : N 2800 lb cos 15&deg; C 120 lb D 0
F% : f 270 lb 2800 lb sin 15&deg; 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 equally spaced 40-m towers. The receiving platform has
a mass of 20 Mg (megagrams) and is 10 m below the
tops of the towers. What tension would the cables be
subjected to?
TOP VIEW
Solution: Isolate the platform. Choose a coordinate system with
the origin at the center of the platform, with the z axis vertical, and
the x,y axes as shown. Express the tensions in terms of unit vectors,
and solve the equilibrium conditions. The cable connections at the
platform are labeled a, b, c, and the cable connections at the towers
are labeled A, B, C. The horizontal distance from the origin (center of
the platform) to any tower is given by
LD
20 m
65 m
C
65
D 37.5 m.
2 sin60
z
c
The coordinates of points A, B, C are
b
a
A37.5, 0, 10,
B
A
y
x
B37.5 cos120&deg; , 37.5 sin120&deg; .10,
C37.5 cos240&deg; , 37.5 sin240&deg; , 10,
The vector locations are:
The tensions in the cables are expressed in terms of the unit vectors,
rA D 37.5i C 0j C 10k,
rB D 18.764i C 32.5j C 10k,
rC D 18.764i C 32.5j C 10k.
The distance from the origin to any cable connection on the platform is
dD
20
D 11.547 m.
2 sin60&deg; The coordinates of the cable connections are
a11.547, 0, 0,
b11.547 cos120&deg; , 11547 sin120&deg; , 0,
TaA D jTaA jeaA ,
TcC D jTcC jecC .
The weight is W D 0i 0j 200009.81k D 0i C 0j 196200k.
The equilibrium conditions are
F D 0 D TaA C TbB C TcC C W D 0,
from which:
c11.547 cos240&deg; , 11.547 sin240&deg; , 0.
The vector locations of these points are,
ra D 11.547i C 0j C 0k,
TbB D jTbB jebB ,
Fx D 0.9333jTaA j 0.4666jTbB j 0.4666jTcC ji D 0
Fy D 0jTaA j C 0.8082jTbB j 0.8082jTcC jj D 0
Fz D 0.3592jTaA j 0.3592jTbB j
rb D 5.774i C 10j C 0k,
C 0.3592jTcC 196200jk D 0
rc D 5.774i C 10j C 0k.
The unit vector parallel to the tension acting between the points A, a
in the direction of A is by definition
eaA
r A ra
D
.
jrA ra
Perform this operation for each of the unit vectors to obtain
eaA D C0.9333i C 0j 0.3592k
The commercial package TK Solver Plus was used to solve these
equations. The results:
jTaA j D 182.1 kN ,
Fz D 3jTj sin 196200 D 0,
where D tan1
10
37.5 11.547
D 21.07&deg; ,
from which each tension is jTj D 182.1 kN.
154
jTcC j D 182.1 kN .
Check: For this geometry, where from symmetry all cable tensions may
be assumed to be the same, only the z-component of the equilibrium
equations is required:
ebB D 0.4667i C 0.8082j 0.3592k
ecC D 0.4667i C 0.8082j C 0.3592k
jTbB j D 182.1 kN ,
check.
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 3.96 To support the tent, the tension in the
rope AB must be 35 lb. What are the tensions in the
y
Solution: We develop the following position vectors
rAB D 2i ft
rAC D 6i C j 3k ft
(0, 5, 0) ft
rAD D 6i C 2j C 3k ft
C
(0, 6, 6) ft
rAE D 3i 4j ft
(6, 4, 3) ft
A
B
(8, 4, 3) ft
D
x
If we divide by the respective magnitudes we can develop the unit
vectors that are parallel to these position vectors.
E
(3, 0, 3) ft
eAB D 1.00i
eAC D 0.885i C 0.147j 0.442k
z
eAD D 0.857i C 0.286j C 0.429k
eAE D 6.00i 0.800j
The equilibrium equation is
TAB eAB C TAC eAC C TAD eAD C TAE eAE 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
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
155
y
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?
5m
5m
C
D
Solution: Get the unit vectors parallel to the cables using the
coordinates of the end points. Express the tensions in terms of these
unit vectors, and solve the equilibrium conditions. The coordinates of
points A, B, C, D, O are found from the problem sketch: The coordinates of the points are A6, 2, 0, B12, 3, 0, C0, 8, 5, D0, 4, 5,
O0, 0, 0.
4m
8m
(6, 2, 0) m
The vector locations of these points are:
rA D 6i C 2j C 0k,
rB D 12i C 3j C 0k,
O
rC D 0i C 8j C 5k,
B
A
z
3m
12 m
rD D 0i C 4j 5k,
rO D 0i C 0j C 0k.
x
The unit vector parallel to the tension acting between the points A, B
in the direction of B is by definition
eAB D
y
rB r A
.
jrB rA j
5m
5m
Perform this for each of the unit vectors
D
4m
C
eAB D C0.9864i C 0.1644j C 0k
eAC D 0.6092i C 0.6092j C 0.5077k
8m
O
(6, 2, 0) m
eAD D 0.7442i C 0.2481j 0.6202k
A
eAO D 0.9487i 0.3162j C 0k
The tensions in the cables are expressed in terms of the unit vectors,
TAB D jTAB jeAB D 50eAB ,
TAC D jTAC jeAC ,
TAO D jTAO jeAO .
The equilibrium conditions are
F D 0 D TAB C TAC C TAD C TAO D 0.
Substitute and collect like terms,
Fx D 0.986450 0.6092jTAC j 0.7422jTAD j
0.9487jTAO ji D 0
Fy D 0.164450 C 0.6092jTAC j C 0.2481jTAD j
0.3162jTAO jj D 0
Fz D C0.5077jTAC j 0.6202jTAD jk D 0.
This set of simultaneous equations in the unknown forces may be
solved using any of several standard algorithms. The results are:
jTAO j D 43.3 kN,
156
jTAC j D 6.8 kN,
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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 0.231i C 0.923j C 0.308k is perpendicular to the
surface. The y axis points upward. The direction cosines
of the cable from A to B are cos x D 0.816, cos y D
0.408, cos z D 0.408, and the tension in the cable is
1.2 kN. Determine the magnitudes of the normal and
friction forces the car’s wheels exert on the surface.
y
en
B
ep
x
z
Solution: Assume that all forces act at the center of mass of the
car. The vector equation of equilibrium for the car is
y
en
&quot;car&quot;
FN
TAB
FS C TAB C W D 0.
Writing these forces in terms of components, we have
x
W D mgj D 13509.81 D 13240j N,
FS D FSx i C FSy j C FSz k,
FS
F
W
z
and TAB D TAB eAB ,
where
eAB D cos x i C cos y j C cos z k D 0.816i C 0.408j 0.408k.
Substituting these values into the equations of equilibrium and solving
for the unknown components of FS , we get three scalar equations of
equilibrium. These are:
FSx TABx D 0,
FSy TABy W D 0,
and FSz TABz D 0.
Substituting in the numbers and solving, we get
FSx D 979.2 N,
FSy D 12, 754 N,
and FSz D 489.6 N.
The next step is to find the component of FS normal to the surface.
This component is given by
FN D FN &ETH; en D FSx eny C FSx eny C FSz enz .
Substitution yields
FN D 12149 N .
From its components, the magnitude of FS is FS D 12800 N. Using
the Pythagorean theorem, the friction force is
fD
F2S F2N D 4033 N.
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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 0.941i C 0.131j C 0.314k is parallel to the plane
surface and aligned with the car’s longitudinal axis.
What is the tension in the cable?
Solution: Only the cable and the car’s weight exert forces in the
direction parallel to ep . Therefore
ep &ETH; T mgj D 0:
0.941i C 0.131j C 0.314k
&ETH; [T0.816i C 0.408j 0.408k mgj] D 0,
0.9410.816T
C 0.1310.408T mg C 0.3140.408T D 0.
Solving, we obtain T D 2.50 kN.
158
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Problem 4.1 In Active Example 4.1, the 40-kN force
points 30&deg; above the horizontal. Suppose that the force
points 30&deg; 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?
40 kN
30&deg;
A
6m
Solution: The perpendicular distance from A to the line of action
of the force is unchanged
D D 6 m sin 30&deg; D 3 m
The magnitude of the moment is therefore unchanged
M D 3 m40 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
0.35 m
0.35 m
A
B
m1
m2
Solution: The total moment about B is
MB D m2 9.81 m/s2 0.35 m C 20 kg9.81 m/s2 0.7 m
D 170 N-m
Solving, we find m2 D 9.51 kg
The moment about A is then
jMA j D 20 kg9.81 m/s2 0.35 m C 9.51 kg9.81 m/s2 0.7 m
jMA j 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.
10 ft
25 ft
B
A
15 ft
C
D
Solution: Use 2-dimensional moment strategy: determine normal
distance to line of action D; calculate magnitude DF; determine sign.
(a)
The normal distances from A to the lines of action are DAB D
10 ft, and DAC D 35 ft. The moments are clockwise (negative).
Hence,
(b)
10 ft
25 ft
15 ft
A
D
B
C
MA D 1040 3544 D 1940 ft-kip .
The normal distances from D to the lines of action are DDB D
40 ft, and DDC D 15 ft. The actions are positive; hence
MD D C4040 C 1544 D 2260 ft-kip
Problem 4.4 What force F applied to the pliers is
required to exert a 4 N-m moment about the center of
the bolt at P?
Solution:
MP D 4 N-m D F0.165 m sin 42&deg; ) F D
4 N-m
0.165 m sin 42&deg;
D 36.2 N
P
F
165
mm
42⬚
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?
Solution:
300 mm
380 mm
F
Mnut
center
D F cos 30&deg; 0.3 m C F cos 20&deg; 0.38 m
F
30⬚
D 50 N-m
50 N-m
FD
D 81.1 N
0.3 m cos 30&deg; C 0.38 m cos 20&deg;
Problem 4.6 The force F D 8 kN. What is the
moment of the force about point P?
20⬚
F
F
y
(3, 7) m
Solution: The angle between the force F and the x axis is
F
˛ D tan1 5/4 D 51.3&deg;
Q
(8, 5) m
The force can then be written
P
(3, 2) m
F D 8 kNcos ˛i sin ˛j D 5.00i 6.25j kN
(7, 2) m
x
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 m5.00 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?
y
(3, 7) m
F
Solution: The angle between the force F and the x axis is
Q
(8, 5) m
˛ D tan1 5/4 D 51.3&deg;
The force can then be written
F D Fcos ˛i sin ˛j D F0.625i 0.781j
P
(3, 2) m
(7, 2) m
x
Treating counterclockwise moment as positive, the total moment about
point Q is
MQ D 0.781F5 m 0.625F2 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?
F
30&deg;
B
A
25&deg;
Solution:
MA D L &ETH; F sin 30&deg; D L
15
2
30&deg;
F = 15 kN
30&deg;
MA D 7.5 L kN &ETH; m
L
set MA D MAmax D 18 kN &ETH; m D 7.5 Lmax
Lmax D 2.4 m
25&deg;
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
45⬚
A
30⬚
T
T
Solution:
(a)
MA D 50 N0.12 m 50 N0.12 m D 0
MA D 0
(b)
P
45⬚
MP D 50 N0.12 m
50 N cos 30&deg; 0.65 m sin 45&deg; C 0.12 m cos 30&deg; 50 N sin 30&deg; 0.65 m cos 45&deg; C 0.12 m sin 30&deg; MP D 31.4 N-m
162
or
MP D 31.4 N-m CW
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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&deg; .
F
α
1m
P
2m
Solution: We have the moment about P
12 kN
MP D 12 kN sin ˛2 m 12 kN cos ˛1 m
α
MP D 122 sin ˛ cos ˛ kN-m
The moment must not exceed 5 kN-m
1m
Thus
5 kN-m &frac12; j122 sin ˛ cos ˛jkN-m
P
The limits occur when
122 sin ˛ cos ˛ D 5 ) ˛ D 37.3
2m
122 sin ˛ cos ˛ D 5 ) ˛ D 15.83&deg;
So we must have 15.83&deg; ˛ 37.3&deg;
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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
30&deg;
C
20&deg;
A
Solution: We have
1.75 kN-m D F0.35 m sin 30&deg; ) F D 10 kN
F
4.20 kN-m D FLAC cos 20&deg; ) LAC D 0.447 m
In summary F D 10 kN, LAC D 447 mm
B
C
30&deg;
20&deg;
F
d1
30&deg;
50
0.3
m
0.450
m
20&deg;
d2
164
600 N
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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
4 kN
30⬚
2 kN
300 mm
O 3 kN
5 kN
300 mm
400 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 counterclockwise moments as positive, the moment due to the 2-kN force is
0.3 m2 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 1.039 C 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
point with coordinates x D 7 m, y D 5 m.
y
40 N
P
30⬚
40 N
2m
Solution:
30⬚
Q
2m
y
40 N
40 N
MP D 40 N cos 30&deg; 2 m C 40 N cos 30&deg; 4 m
x
(a)
D 69.3 N-m CCW
30&deg;
30&deg;
(b)
MQ D 40 N cos 30&deg; 2 m D 69.3 N-m CCW
M D 40 N sin 30&deg; 5 m C 40 N cos 30&deg; 5 m
(c)
x
P
2m
2m
Q
40 N sin 30&deg; 5 m 40 N cos 30&deg; 3 m
D 69.3 N-m CCW
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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
S
13
in.
30&deg;
12
E
40&deg;
in.
Solution: The key is the geometry
From trigonometry,
cos 40&deg; D
Thus
d1 D 12 in cos 30&deg;
d1 D
and
d2
d1
, cos 30&deg; D
13 in
12 in
d1
S
30&deg;
13 in
12 i
10.3900
n
W
40&deg;
d2 D 13 in cos 40&deg;
d2
E
d2 D 9.9600
We are given that
299 in-lb D d2 W D 9.96 W
W D 30.0 lb
Now,
Ms D d1 C d2 W
Ms D 20.3530.0
Ms D 611 in-lb clockwise
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 ?
D
FC
D
FA
FB
6m
A
Solution: The angles between the three cables and the pillar are
˛A D tan1 4/6 D 33.7&deg;
B
C
E
4m
4m
4m
˛B D tan1 8/6 D 53.1&deg;
˛C D tan1 12/6 D 63.4&deg;
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. Determine the sum of the moments of the three forces about
point P.
2 kN
Solution:
MP D 4 kN0.2 m C 2 kN0.6 m 2 kN cos 20&deg; 0.2 m
C 2 kN sin 20&deg; 0.4 m D 10.18 kN-m
20⬚
MP D 0.298 kN-m CCW
2 kN
4 kN
0.2 m
2 kN
20&deg;
4 kN
P
0.2 m
0.2 m
0.2 m
P
0.2 m
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
Solution: The moment about point A due to F1 is zero. Treating
counterclockwise moments as positive the sum of the moments is
2 kN
0.2 m
0.2 m
0.2 m
y
F3
A
C
30⬚
F1
2m
MA D F3 sin 30&deg; 8 m C F2 cos 45&deg; 2 m
B
45⬚
F2
MA D 273 N-m counterclockwise
8m
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?
x
y
F3
A
30⬚
C
F1
2m
Solution: The sums of the forces in the x and y directions equal
zero:
Fx : F1 C F2 cos 45&deg; F3 cos 30&deg; D 0
B
45⬚
F2
8m
Fy :
F2 sin 45&deg;
C F3 sin 30&deg;
x
D0
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&deg; 8 m C F2 cos 45&deg; 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?
y
FD
30⬚
FA
FB
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
6 ft
x
FC
4 ft
D 40 lb6 ft C 20 lb10 ft D 40 ft-lb
M D 40 ft-lb clockwise
y
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)
(b)
FD
30⬚
FA
FB
Determine the forces FB , FC , and FD .
Determine the sum of the moments of the forces
about the right end of the beam.
6 ft
x
FC
4 ft
Solution:
(a)
The sum of the forces and the sum of the moments equals zero
Fx : FA cos 30&deg; FD D 0
Fy : FA sin 30&deg; FB C FC D 0
Morigin : FB 6 ft C FC 10 ft D 0
Setting FA D 30 lb and solving yields
(b)
FB D 37.5 lb, FC D 22.5 lb, FD D 26.0 lb
The sum of the moments about the right end is
MRight
End
: FB 4 ft FA sin 30&deg; 10 ft
D 37.5 lb4 ft 30 lb10 ft
D0
168
MRight
End
D0
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y
Problem 4.21 Three forces act on the car. The sum of
the forces is zero and the sum of the moments of the
forces about point P is zero.
(a)
(b)
3 ft
6 ft
Determine the forces A and B.
Determine the sum of the moments of the forces
x
B
P
Q
2800 lb
A
Solution:
6 ft
(a)
Fy : A C B 2800 lb D 0
MP : 2800 lb6 ft C A9 ft D 0
Solving we find
A D 1867 lb, B D 933 lb
(b)
3 ft
Q
P
2800 lb
MQ D 2800 lb3 ft B9 ft D 0
B
A
MQ D 0
80 lb
Problem 4.22 Five forces act on the piping. The vector
sum of the forces is zero and the sum of the moments
of the forces about point P is zero.
(a)
(b)
45⬚
y
Determine the forces A, B, and C.
Determine the sum of the moments of the forces
2 ft
20 lb
Q
x
A
P
C
2 ft
B
2 ft
2 ft
Solution:
80 lb
The conditions given in the problem are:
Fx : A C 80 lb cos 45&deg; D 0
45&deg;
y
20 lb
2 ft
(a)
Fy : B C 20 lb C 80 lb sin 45&deg; D 0
P
MP : 20 lb2 ft C6 ft 80 lb cos 45&deg; 2 ft
Q
x
A
2 ft
C 80 lb sin 45&deg; 4 ft D 0
Solving we have
B
2 ft
2 ft
C
A D 56.6 lb, B D 24.4 lb, C D 12.19 lb
MQ : 80 lb cos 45&deg; 2 ft 80 lb sin 45&deg; 2 ft
(b)
C20 lb4 ft C B6 ft D 0
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169
Problem 4.23 In Example 4.3, suppose that the attachment point B is moved upward and the cable is lengthened 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?
B
A
7 ft
W
4 ft
C
2 ft
2 ft
Solution: The angle ˛ between the cable AB and the horizontal is.
˛ D tan1 5/4 D 51.3&deg;
The sum of the moments about C is
MC : W2 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?
Solution: Let T be the tension in the cable. The sum of the
MA : T30 in C T sin 30&deg; 90 in 800 lb60in D 0
A
B
30
30 in
30 in
C
800 lb
30 in
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
150
600
mm
mm
40 N
C
20&deg;
B
m
0m
Solution: The strategy is to find the perpendicular distance from
40&deg;
60
160 N
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:
A
160 N
dAB D 0.300 cos 40&deg; D 0.2298 m
The distance from A to the action line of the weight of the arm BC is
dBC D 0.600cos 40&deg; C 0.300cos 20&deg; D 0.7415 m.
The distance from A to the line of action of the force is
dF D 0.600cos 40&deg; C 0.600cos 20&deg; C 0.150cos 20&deg; 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?
2.2 m
2.2 m
F
F
G
5&deg;
18 m
Solution: The key to this problem is getting the geometry correct.
The simplest way to do this is to break each force into components
parallel and perpendicular to the axis of the shuttle and then to sum
the moments of the components. (This will become much easier in the
next section)
6&deg;
12 m
F sin 6&deg;
F sin 5&deg;
5˚
6&deg;c
2.2 m
18 m
F cos 5&deg;
FRONT
2.2 m
12 m
F cos 6&deg;
REAR
CMFRONT&yacute; D 18F sin 5&deg; 2.2F cos 5&deg;
CMREAR&yacute; D 2.2F cos 6&deg; 12F sin 6&deg;
CMTOTAL D MFRONT C MREAR
CMTOTAL D 4.80 C 7.19 N-m
CMTOTAL D 2.39 N-m
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171
Problem 4.27 The force F exerts a 200 ft-lb counterclockwise moment about A and a 100 ft-lb
clockwise moment about B. What are F and ?
y
A
(–5, 5) ft
F
θ
(4, 3) ft
x
B
(3, – 4) ft
Solution: The strategy is to resolve F into x- and y-components,
and compute the perpendicular distance to each component from A
and B. The components of F are: F D iFX C jFY . The vector from A
to the point of application is:
rAF D 4 5i C 3 5j D 9i 2j.
The perpendicular distances are dAX D 9 ft, and dAY D 2 ft, and the
actions are positive. The moment about A is MA D 9FY C 2FX D
200 ft-lb. The vector from B to the point of application is rBF D
4 3i C 3 4j D 1i C 7j; the distances dBX D 1 ft and dBY D
7 ft, the action of FY is positive and the action of FX is negative. The moment about B is MB D 1FY 7FX D 100 ft-lb. The
two simultaneous equations have solution: FY D 18.46 lb and FX D
16.92 lb. Take the ratio to find the angle:
D tan1
FY
FX
D tan1
18.46
16.92
y
A
(–5, 5) ft
F
θ
(4, 3) ft
x
B
(3, –4) ft
D tan1 1.091 D 47.5&deg; .
From the Pythagorean theorem
jFj D
172
F2Y C F2X D
p
18.462 C 16.922 D 25.04 lb
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 4.28 Five forces act on a link in the gearshifting mechanism of a lawn mower. The vector sum
of the five forces on the bar is zero. The sum of their
moments about the point where the forces Ax and Ay act
is zero.
(a) Determine the forces Ax , Ay , and B.
(b) Determine the sum of the moments of the forces
about the point where the force B acts.
Ay
Ax
25 kN
20&deg;
650 mm
450 mm
30 kN
45&deg;
B
650 mm
350 mm
Solution: The strategy is to resolve the forces into x- and
y-components, determine the perpendicular distances from B to the line
of action, determine the sign of the action, and compute the moments.
The angles are measured counterclockwise from the x axis. The
forces are
F2 D 30i cos 135&deg; C j sin 135&deg; D 21.21i C 21.21j
F1 D 25i cos 20&deg; C j sin 20&deg; D 23.50i C 8.55j.
(a)
The sum of the forces is
F D A C B C F1 C F2 D 0.
Substituting:
and
FX
D AX C BX C 23.5 21.2i D 0,
FY D AY C 21.2 C 8.55j D 0.
Solve the second equation: AY D 29.76 kN. The distances of
the forces from A are: the triangle has equal base and altitude,
hence the angle is 45&deg; , so that the line of action of F1 passes
through A. The distance to the line of action of B is 0.65 m,
with a positive action. The distance to the line of action of the
y-component of F2 is 0.650 C 0.350 D 1 m, and the action is
positive. The distance to the line of action of the x-component
of F2 is 0.650 0.450 D 0.200 m, and the action is positive.
MA D 8.551 C 23.50.2 C BX 0.65 D 0.
Solve: BX D 20.38 kN. Substitute into the force equation to
obtain AX D 18.09 kN
(b)
The distance from B to the line of action of the y-component of
F1 is 0.350 m, and the action is negative. The distance from B
to the line of action of AX is 0.650 m and the action is negative.
The distance from B to the line of action of AY is 1 m and the
action is positive. The distance from B to the line of action of
the x-component of F2 is 0.450 m and the action is negative. The
sum of the moments about B:
MB D 0.35021.21 0.65018.09
C 129.76 0.45023.5 D 0
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
173
F
Problem 4.29 Five forces act on a model truss built by
a civil engineering student as part of a design project.
The dimensions are b D 300 mm and h D 400 mm; F D
100 N. The sum of the moments of the forces about the
point where Ax and Ay act is zero. If the weight of the
truss is negligible, what is the force B?
F
60&deg;
60&deg;
h
Ax
Ay
b
b
b
b
b
b
B
Solution: The x- and y-components of the force F are
F D jFji cos 60&deg; C j sin 60&deg; D jFj0.5i C 0.866j.
The distance from A to the x-component is h and the action is positive.
The distances to the y-component are 3b and 5b. The distance to B is
6b. The sum of the moments about A is
MA D 2jFj0.5h 3bjFj0.866 5bjFj0.866 C 6bB D 0.
Substitute and solve: B D
1.6784jFj
D 93.2 N
1.8
Problem 4.30 Consider the truss shown in Problem
4.29. The dimensions are b D 3 ft and h D 4 ft; F D
300 lb. The vector sum of the forces acting on the truss
is zero, and the sum of the moments of the forces about
the point where Ax and Ay act is zero.
(a)
(b)
Determine the forces Ax , Ay , and B.
Determine the sum of the moments of the forces
about the point where the force B acts.
Solution: The forces are resolved into x- and y-components:
Solve the first: Ax D 300 lb. The distance from point A to the
x-components of the forces is h, and the action is positive. The
distances between the point A and the lines of action of the ycomponents of the forces are 3b and 5b. The actions are negative.
The distance to the line of action of the force B is 6b. The action
is positive. The sum of moments about point A is
F D 300i cos 60&deg; C j sin 60&deg; D 150i 259.8j.
(a)
The sum of the forces:
F D 2F C A C B D 0.
MA D 2150 h 3b259.8 5b259.8 C 6b B D 0.
The x- and y-components:
Substitute and solve: B D 279.7 lb. Substitute this value into the
force equation and solve: Ax D 519.6 279.7 D 239.9 lb
Fx D Ax 300i D 0,
(b)
Fy D 519.6 C Ay C Bj D 0.
The distances from B and the line of action of AY is 6b and the
action is negative. The distance between B and the x-component
of the forces is h and the action is positive. The distance between
B and the y-components of the forces is b and 3b, and the action
is positive. The sum of the moments about B:
174
MB D 6b239.9 C 2150 h C b259.8 C 3b259.8 D 0
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 4.31 The mass m D 70 kg. What is the moment about A due to the force exerted on the beam at B
by the cable?
B
A
45&deg;
30&deg;
3m
m
Solution: The strategy is to resolve the force at B into components
parallel to and normal to the beam, and solve for the moment using
the normal component of the force. The force at B is to be determined
from the equilibrium conditions on the cable juncture O. Angles are
measured from the positive x axis. The forces at the cable juncture are:
FOB
FOC
O
FOB D jFOB ji cos 150&deg; C j sin 150&deg; D jFOB j0.866i C 0.5j
W
FOC D jFOC ji cos 45&deg; C j sin 45&deg; D jFOC j0.707i C 0.707j.
W D 709.810i 1j D 686.7j (N).
The equilibrium conditions are:
Fx D 0.866jFOB j C 0.7070jFOC ji D 0
FY D 0.500jFOB j C .707jFOC j 686.7j D 0.
Solve: jFOB j D 502.70 N. This is used to resolve the cable tension at B:
FB D 502.7i cos 330&deg; C j sin 330&deg; D 435.4i 251.4j. The distance
from A to the action line of the y-component at B is 3 m, and the
action is negative. The x-component at passes through A, so that the
action line distance is zero. The moment at A is MA D 3251.4 D
754.0 N-m
Problem 4.32 The weights W1 and W2 are suspended
by the cable system shown. The weight W1 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
50⬚
W1
B
6 ft
C
W2
P
Solution: Isolate part of the cable system near point B. The equilibrium equations are
Fx : TBC TAC cos 50&deg; 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 lb6 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
4m
A
1m
O
1m
3m
Solution: The strategy is to resolve the force at B into components
FB
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.
B
By inspection, the bar forms a 3, 4, 5 triangle. The angle the bar makes
with the horizontal is cos D 35 D 0.600, and sin D 45 D 0.800. The
force at B is FB D jFB j0.600i C 0.800j. The perpendicular distance
from O to the line of action of the x-component is 4 C 1 D 5 m, and
the action is positive. The distance from O to the line of action of
the y-component is 1 m, and the action is positive. The moment about
O is
MO D 50.600jFB j C 10.800jFB j D 3.8jFB j D 38 kN, from
which jFB j D 10 kN
4m
A
θ
O
1m
1m
3m
Problem 4.34 A contestant in a fly-casting contest
snags his line in some grass. If the tension in the line is
5 lb, what moment does the force exerted on the rod by
the line exert about point H, where he holds the rod?
H
Solution: The strategy is to resolve the line tension into a compo-
6 ft
nent normal to the rod; use the length from H to tip as the perpendicular distance; determine the sign of the action, and compute the
moment.
4 ft
The line and rod form two right triangles, as shown in the sketch. The
angles are:
˛ D tan1
2
D 15.95&deg;
7
ˇ D tan1
7 ft
6
15
α
D 21.8&deg; .
β
α
2 ft
7 ft
The angle between the perpendicular distance line and the fishing line
is D ˛ C ˇ D 37.7&deg; . The force normal to the
p distance line is F D
5sin 37.7&deg; D 3.061 lb. The distance is d D 22 C 72 D 7.28 ft, and
the action is negative. The moment about H is MH D 7.283.061 D
22.3 ft-lb Check: The tension can be resolved into x and y components,
15 ft
6 ft
15 ft
β
Fx D F cos ˇ D 4.642 lb, Fy D F sin ˇ D 1.857 lb.
The moment is
M D 2Fx C 7Fy D 22.28 D 22.3 ft-lb. check.
176
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A
Problem 4.35 The cables AB and AC help support the
tower. The tension in cable AB is 5 kN. The points A,
B, C, and O are contained in the same vertical plane.
(a)
What is the moment about O due to the force exerted
on the tower by cable AB?
(b) If the sum of the moments about O due to the forces
exerted on the tower by the two cables is zero, what
is the tension in cable AC?
Solution: The strategy is to resolve the cable tensions into components normal to the vertical line through OA; use the height of the
tower as the perpendicular distance; determine the sign of the action,
and compute the moments.
(a)
(b)
20 m
60&deg;
45&deg;
C
O
A
B
FN
60&deg;
FN
A
45&deg;
The component normal to the line OA is FBN D 5cos 60&deg; D
2.5 kN. The action is negative. The moment about O is MOA D
2.520 D 50 kN-m
By a similar process, the normal component of the tension in
the cable AC is FCN D jFC j cos 45&deg; D 0.707jFC j. The action is
positive. If the sum of the moments is zero,
MO D 0.70720jFC j 50 D 0,
from which
jFC j D
50 kN m
D 3.54 kN
0.70720 m
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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 backstay?
Solution: Triangle ABP
tan ˛ D
Triangle BCQ
tan ˇ D
y
4
, ˛ D 18.73&deg;
11.8
5
, ˇ D 22.62&deg;
12
CMO D 13230 sin ˛ 13660 sin ˇ
B (4,13) m
CMO D 2340 N-m
B (4,13)
230 N
660 N
β
α
A
(0,1.2) m
C
(9,1) m
x
P
A (0,1.2)
C (9,1)
O (4,0)
Q
660 sin β
230 sin α
α
β
13 m
O
178
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.
A
B
8m
C
G
Boom
P
16 m
38 m
56 m
Solution:
A
56
8
290
8
8
290 kN56 m p
148 kN16 m
MP D p
3200
320
kN
8
kN
16
14
40 m
B
D 3.36 MNm
8
C
16 m
P
MP D 3.36 MN-m CW
Problem 4.38 The mass of the building crane’s boom
in Problem 4.37 is 9000 kg. Its weight acts at G. The
sum of the moments about P due to the boom’s weight,
the force exerted at B by the cable AB, and the force
exerted at C by the cable AC is zero. Assume that the
tensions in cables AB and AC are equal. Determine the
tension in the cables.
Solution:
A
56
8
8
8
8
TAB 56 m p
TAC 16 m
MP D p
3200
320
C 9000 kg9.81 m/s2 38 m D 0
using TAB D TAC we solve and find
T AB
B
18 m
C
16
TA
22 m
C
16 m
P
9000 kg (9.81 m/s2)
TAB D TAC D 223 kN
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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&deg; ; (b) if
˛ D 50&deg; .
F
y
0.28 m
Solution: O is the origin of the coordinate system
MO D F1.2 m cos ˛
12 kg9.81 m/s2 0.28 cos ˛ 0.14 sin ˛ D 0
Solving we find
x
FD
12 kg9.81 m/s2 0.28 cos ˛ 0.14 sin ˛
1.2 m cos ˛
(a)
For ˛ D 30&deg; We find
F D 19.54 N
(b)
For ˛ D 50&deg; We find
F D 11.10 N
0.14 m
1.2 m
A
a
C
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
Solution: The strategy is to resolve the force exerted by the hydraulic cylinder into the normal component about the crane; determine the
distance; determine the sign of the action, and compute the moment.
Two right triangles are constructed: The angle formed by the hydraulic
cylinder with the horizontal is
ˇ D tan1
2.4
1.2
D 63.43&deg; .
The angle formed by the crane with the horizontal is
C
˛ D tan1
1.4
3
D 25.02&deg; .
A
2.4 m
1m
B
1.8 m
1.2 m
7m
The angle between the hydraulic cylinder and the crane is D ˇ ˛ D
38.42&deg; . The normal component of the force is: Fp
N D 300sin 38.42&deg; D 186.42 kN. The distance from point A is d D 1.42 C 32 D 3.31 m.
The action is positive. The moment about A is MO D C3.31186.42 D
617.15 kN-m Check: The force exerted by the actuator can be resolved
into x- and y-components, Fx D F cos ˇ D 134.16 kN, Fy D F sin ˇ
D 268.33 kN. The moment about the point A is M D 1.4Fx C 3.0
Fy D 617.15 kN m. check.
β
α
1.2 m
α
3m
2.4 m
β
1.4 m
180
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 4.41 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
6 ft
W
Solution: The angle between the piston AB and the horizontal is
3 ft
˛ D tan1 3/6 D 26.6&deg;
A
B
C
The sum of the counterclockwise moment about C is
MC : W6 ft 400 lb cos ˛3 ft 400 lb sin ˛3 ft D 0
6 ft
3 ft
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.
1m
D
C
Hydraulic
cylinder
1m
0.6 m
B
A
0.6 m
Scoop
0.15 m
Solution: Use x, y coords with origin A. We need the unit vector
from C to B, eCB . From the geometry,
eCB D 0.780i 0.625j
5.00 kN
6.25 kN
C (−0.15, + 0.6)
The force FCB is given by
FCB D 0.7808i 0.6258j kN
0.6 m
FCB D 6.25i 5.00j kN
For the moments about A and D, treat the components of FCB as two
separate forces.
0.15 m
CMA D 5, 000.15 0.66.25 kN &ETH; m
A (0 , 0)
CMA D 3.00 kN &ETH; m
5.0 kN
m
D
C
0,4 m
MD D 5 kN1 m C 6.25 kN0.4 m
C
6.25 kN
CMD D 7.5 kN &ETH; m
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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
320
mm
C
Shaft
100
mm
Scoop
260
mm
H
B
180
260 mm
mm
J
D
160
mm
L
K
380
mm
1040
mm 1120
mm
Solution:
320
320
700 N0.52 m D 353 Nm
MK D p
108800
80
700 N
520 mm
MK D 353 Nm CW
K
Problem 4.44 In the structure shown in Problem 4.43,
the bar BC exerts a force at C that points from C
toward B. The hydraulic cylinder DH exerts a 1550-N
force at D that points from D toward H. The sum of the
moments of these two forces about K is zero. What is
the magnitude of the force that bar BC exerts at C?
Solution:
320
80
260 mm
320
1120
1550 N0.26 m p
F0.52 D 0
MK D p
1264400
108800
Solving we find
BC
1120
F D 796 N
100
1550 N
260 mm
K
182
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
y
Problem 4.45 In Active Example 4.4, what is the
moment of F about the origin of the coordinate system?
C (7, 7, 0) ft
A
(0, 6, 5) ft
Solution: The vector from the origin to point B is
F
x
r D 11i C 4k ft
B (11, 0, 4) ft
From Active Example 4.4 we know that the force F is
z
F D 40i C 70j 40k lb
The moment of F about the origin is
i
j
k M D r &eth; F D 11
0
4 D 280i C 280j C 770k ft-lb
40 70 40 M D 280i C 280j C 770k ft-lb
Problem 4.46 Use Eq. (4.2) to determine the moment
of the 80-N force about the origin O letting r be the
vector (a) from O to A; (b) from O to B.
Solution:
(a)
MO D rOA &eth; F
y
D 6i &eth; 80j D 480k N-m.
80j (N)
B
MO D rOB &eth; F
(b)
(6, 4, 0) m
D 6i C 4j &eth; 80j
D 480k N-m.
O
x
A (6, 0, 0) m
Problem 4.47 A bioengineer studying an injury sustained in throwing the javelin estimates that the magnitude of the maximum force exerted was jFj D 360 N and
the perpendicular distance from O to the line of action of
F was 550 mm. The vector F and point O are contained
in the xy plane. Express the moment of F about the
shoulder joint at O as a vector.
Solution: The magnitude of the moment is jFj0.55 m D 360 N
0.55 m D 198 N-m. The moment vector is perpendicular to the xy
plane, and the right-hand rule indicates it points in the positive z direction. Therefore MO D 198k N-m.
y
F
y
550 mm
F
O
O
x
x
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
183
y
Problem 4.48 Use Eq. (4.2) to determine the moment
A
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 0i C
0 6j D 8i 6j. The force is F D 100j (kN). The cross product is
i
rAF &eth; F D 8
0
100j (kN)
6m
j
k 6 0 D 800k (kN-m)
100 0 B
x
8m
12 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:
i
rBF &eth; F D 4
0
j
k 0
0 D 400k (kN-m)
100 0 Problem 4.49 The cable AB exerts a 200-N force on
the support at A that points from A toward B. Use
Eq. (4.2) to determine the moment of this force about
point P in two ways: (a) letting r be the vector from P
to A; (b) letting r be the vector from P to B.
y
P (0.9, 0.8) m
(0.3, 0.5) m
Solution: 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 mi C 0.2 0.5 mj D 0.7i 0.3j m
jABj D
p
0.7 m2 C 0.3 m2 D 0.58 m
A
B
(1, 0.2) m
x
200 N
0.7i 0.3j
FD p
0.58
(a)
200 N
0.7i 0.3j
MP D PA &eth; F D 0.6 mi 0.3 mj &eth; p
0.58
Carrying out the cross product we find
MP D 102.4 N-mk
(b)
200 N
0.7i 0.3j
MP D PB &eth; F D 0.1 mi 0.6 mj &eth; p
0.58
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 xy plane. The moment of F about O is 140k (Nm), and the moment of F about A is 280k (N-m). What
are the components of F?
y
A (0, 7, 0) m
F
(5, 3, 0) m
Solution: The strategy is to find the moments in terms of the
components of F and solve the resulting simultaneous equations. The
position vector from O to the point of application is rOF D 5i C 3j.
The position vector from A to the point of application is rAF D 5 0i C 3 7j D 5i 4j. The cross products:
i
rOF &eth; F D 5
FX
j
3
FY
k 0 D 5FY 3FX k D 140k, and
0
i
rAF &eth; F D 5
FX
j
4
FY
k 0 D 5FY C 4FX k D 280k.
0
Take the dot product of both sides with k to eliminate k. The simultaneous equations are:
x
O
5FY 3FX D 140, 5FY C 4FX D 280.
Solving: FY D 40, FX D 20, from which F D 20i C 40j (N)
y
A
(0,7,0)
F
(5,3,0)
x
O
Problem 4.51 Use Eq. (4.2) to determine the sum of
the moments of the three forces (a) about A, (b) about B.
y
6 kN
3 kN
Solution:
(a)
3 kN
B
A
x
MA D 0.2i &eth; 3j C 0.4i &eth; 6j C 0.6i &eth; 3j
0.2 m
0.2 m
0.2 m
0.2 m
D O.
(b)
MB D 0.2i &eth; 3j C 0.4i &eth; 6j C 0.6i &eth; 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.
Solution: The position vectors from O to the points of application of the forces are: rO1 D 3j, F1 D 200i; rO2 D 10i, F2 D 500j;
rO3 D 6i C 6j, F3 D 200i.
y
200 lb
3 ft
200 lb
3 ft
The sum of the moments about O is
i
MO D 0
200
j k i
3 0 C 10
0 0 0
j
k i
0
0 C 6
500 0 200
j k 6 0 lb
0 0
O
x
6 ft
4 ft
500 lb
D 600k 5000k 1200k D 5600k ft-lb
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
185
Problem 4.53 Three forces act on the plate. Use
Eq. (4.2) to determine the sum of the moments of the
y
4 kN
45⬚
Solution:
3 kN
r1 D 0.12i C 0.08j m, F1 D 4 cos 45&deg; i C 4 sin 45&deg; j kN
30⬚
0.18 m
P
0.10 m
r2 D 0.16i m, F2 D 3 cos 30&deg; i C 3 sin 30&deg; j kN
r3 D 0.16i 0.1j m, F3 D 12 cos 20&deg; i 12 sin 20&deg; j kN
20⬚
0.12 m
12 kN
0.28 m
MP D r1 &eth; F1 C r2 &eth; F2 C r3 &eth; F3
x
MP D 0.145 kN-mk D 145 N-mk
Problem 4.54 (a) Determine the magnitude of the
moment of the 150-N force about A by calculating the
perpendicular distance from A to the line of action of
the force.
y
(0, 6, 0) m
150k (N)
(b) Use Eq. (4.2) to determine the magnitude of the
moment of the 150-N force about A.
A
Solution:
(a)
x
(6, 0, 0) m
The perpendicular from A to the line of action of the force lies
in the xy plane
d
p
D
z
62 C 62 D 8.485 m
jMj D dF D 8.485150 D 1270 N-m
(b)
M D 6i C 6j &eth; 150k D 900j C 900i N-m
p
jMj D
186
9002 C 9002 D 1270 N-m
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y
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.
A
(0.6, 0.5, 0.4) m
(b) Use Eq. (4.2) to determine the magnitude of the
moment of the 600-N force about A.
Solution:
(a)
x
0.8 m
Choose some point Px, 0, 0.8 m. on the line of action of the
force. The distance from A to P is then
d D x 0.6 m2 C 0 0.5 m2 C 0.8 m 0.4 m2
600i (N)
z
The perpendicular distance is the shortest distance d which occurs
when x D 0.6 m. We have d D 0.6403 m. Thus the magnitude of
the moment is
M D 600 N0.6403 m D 384 N-m
(b)
Define the point on the end of the rod to be B. Then AB D
0.6i 0.5j C 0.4k m we have
M D AB &eth; F D 0.6i 0.5j C 0.4k m &eth; 600 Ni
M D 240j C 300k N-m
Thus the magnitude is
MD
240 Nm2 C 300 Nm2 D 384 N-m
y
Problem 4.56 what is the magnitude of the moment of
A
(4, 4, 2) ft
Solution: The position vector from B to A is
B (8, 1, ⫺2) ft
x
rBA D [4 8i C 4 1j C 2 2k] ft
z
rBA D 4i C 3j C 4k ft
The moment of F about B
i
MB D rBA &eth; F D 4
20
F ⫽ 20i ⫹ 10j ⫺ 10k (lb)
is
j
3
10
k 4 D 70i C 40j 100k ft-lb
10 Its magnitude is
jMB j D
70 ft-lb2 C 40 ft-lb2 C 100 ft-lb2 D 128 ft-lb
jMB j D 128 ft-lb
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
187
Problem 4.57 In Example 4.5, suppose that the attachment 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 excerted on
the attachment point A by the two cables?
Solution: The position vector from A to C is
y
C
(6, 3, 0) m
B
(0, 4, 8) m
O
x
A
(4, 0, 6) m
rAC D [8 4i C 2 0j C 0 6k] m
z
rAC D 4i C 2j 6k
The force exerted at A by cable AC can be written
FAC D 25 kN
rAC
D 13.4i C 6.68j 20.0k kN
jrAC j
The total force exerted at A by the two cables is
F D FAB C FAC D 6.70i C 13.3j 16.7k kN
i
j
0
MO D rAB &eth; F D 4
6.70 13.3
k 6 D 80.1i C 107j C 53.4k kN-m
16.7 MO D 80.1i C 107j C 53.4k kN-m
y
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.
Solution: The position vector from B to C is
B (5, 6, 1) ft
F
rBC D [3 5i C 0 6j C 4 1k] ft
A
x
rBC D 2i 6j C 3k ft
C (3, 0, 4) ft
The force F can be written
F D 200 lb
rBC
D 57.1i 171j C 85.7k lb
jrBC j
The moment of F about A is
i
j
MA D rAB &eth; F D 5
6
57.1 171
z
k 1 85.7 D 686i 486j 514k ft-lb
Its magnitude is
jMA j D
686 ft-lb2 C 486 ft-lb2 C 514 ft-lb2 D 985 ft-lb
jMA j D 985 ft-lb
188
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y
Problem 4.59 The force F D 30i C 20j 10k (N).
(a)
Determine the magnitude of the moment of F
(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?
Solution: The vector from A to the point of application of F is
F
A
(8, 2, – 4) m
(4, 3, 3) m
x
z
r D 4i 1j 7k m
and
p
jrj D
(a)
42 C 12 C 72 D 8.12 m
The moment of F about A is
MA
i
j
D r &eth; F D 4 1
30 20
jMA j D
(b)
k 7 D 150i 170j C 110k N-m
10 p
1502 C 1702 C 1102 D 252 N-m
The maximum moment occurs when r ? F. In this case
jMAmax j D jrjjFj
Hence, we need jFj.
p
jFj D
302 C 202 C 102 D 37.4 N
Thus,
jMAmax j D 8.1237.4 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.
y
z
O
F
3m
Solution: The strategy is to determine the perpendicular distance
x
from O to the action line of F, and to calculate the largest magnitude of
F from MO D DjFj. The position vector from O to the point of application of F is rOF D 3i (m). Resolve the position vector into components parallel and normal to F. The component parallel to F is rP D
rOF &ETH; eF eF , where the unit vector eF parallel to F is eF D i cos X C
j cos Y C k cos Z D 0.818i C 0.182j 0.545k. The dot product is
rOF &ETH; eF D 2.454. The parallel component is rP D 2.007i C 0.4466j 1.3374k. The component normal to F is rN D rOF rP D 3 2i 0.4466j C 1.3374k. The magnitude
p of the normal component is the
perpendicular distance: D D 12 C 0.44662 C 1.3372 D 1.7283 m.
The maximum moment allowed is MO D 1.7283jFj D 100 kN-m,
from which
jFj D
100 kN-m
D 57.86 &frac34;
D 58 kN
1.7283 m
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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 j C 13 k and its magnitude is 120 N. Determine
the magnitude of the moment of F about the origin O.
150 mm
y
F
Solution: The vector from O to the point of application of the
O
200 mm
force is
z
r D 0.25i C 0.2j 0.15k m
250 mm
x
and the force is F D jFje
or
F D 80i 80j C 40k N.
The moment of F about O is
i
j
MO D r &eth; F D 0.25 0.2
80 80
k 0.15 N-m
40 or
MO D 4i 22j 36k N-m
and
p
jMO j D
42 C 222 C 362 N-m
jMO j 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 j C 13 k. The
support at O will safely support a moment of 560 N-m
magnitude.
(a)
Based on this criterion, what is the largest safe
magnitude of F?
(b)
If the force F may be exerted in any direction, what
is its largest safe magnitude?
Solution: See the figure of Problem 4.61.
The moment in Problem 4.61 can be written as
i
j
MO D 0.25 0.2
2F 2F
3
3
k 0.15 where F D jFj
C 31 F MO D 0.0333i 0.1833j 0.3kF
And the magnitude of MO is
p
jMO j D 0.03332 C 0.18332 C 0.32 F
jMO j D 0.353 F
190
If we set jMO j D 560 N-m, we can solve for jFmax j
560 D 0.353jFmax j
jFmax j D 1586 N
(b)
If F can be in any direction, then the worst case is when r ? F.
The moment in this case is jMO j D jrjjFworst j
jrj D
p
0.252 C 0.22 C 0.152 D 0.3536 m
560 D 0.3536jFWORST j
jFworst j D 1584 N
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 4.63 A civil engineer in Boulder, Colorado
estimates that under the severest expected Chinook
winds, the total force on the highway sign will be
F D 2.8i 1.8j (kN). Let MO be the moment due to F
about the base O of the cylindrical column supporting
the sign. The y component of MO is called the torsion
exerted on the cylindrical column at the base, and the
component of MO parallel to the xz plane is called
the bending moment. Determine the magnitudes of the
torsion and bending moment.
y
F
8m
8m
O
x
Solution: The total moment is
z
M D 8j C 8k m &eth; 2.8i 1.8j kN
D 14.4i C 22.4j 22.4k kN-m
We now identify
Torsion D My D 22.4 kN-m
Bending moment D Mx 2 C Mz 2
D 14.4 kNm2 C 22.4 kNm2 D 26.6 kN-m
Problem 4.64 The weights of the arms OA and AB of
the robotic manipulator act at their midpoints. The direction cosines of the centerline of arm OA are cos x D
0.500, cos y D 0.866, and cos z D 0, and the direction
cosines of the centerline of arm AB are cos x D 0.707,
cos y D 0.619, and cos z D 0.342. What is the sum
of the moments about O due to the two forces?
Solution: By definition, the direction cosines are the scalar components of the unit vectors. Thus the unit vectors are e1 D 0.5i C 0.866j,
and e2 D 0.707i C 0.619j 0.342k. The position vectors of the midpoints of the arms are
r1 D 0.3e1 D 0.30.5i C 0.866j D 0.15i C 0.2598j
r2 D 0.6e1 C 0.3e2 D 0.512i C 0.7053j 0.1026k.
The sum of moments is
y
0
60
mm
B
160 N
M D r1 &eth; W 1 C r 2 &eth; W 2
i
j
k i
D 0.15 0.2598 0 C 0.512
0
200 0 0
j
0.7053
160
k
0.1026 0
D 16.42i 111.92k (N-m)
600 mm
A
200 N
O
z
x
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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 ?
Solution: We have the forces
100 lb
TAC
8j C 10k, F2 D p
14i 8j C 14k
F1 D p
164
456
Thus the total moment is
M D 8 ftj &eth; F1 C F2 D 625 ft lb C 5.24 ft TAC i
y
5.24 ftTAC K
The magnitude squared is then
625 ft lb C 5.24 ft TAC 2 C 5.24 ft TAC 2 D 1500 ft lb2
Solving we find
TAC D 134 lb
(0, 8, 0) ft
A
x
O
B
(0, 0, 10) ft
(14, 0, 14) ft
C
z
Problem 4.66* A force F acts at the top end A of the
pole. Its magnitude is jFj D 6 kN and its x component
is Fx D 4 kN. The coordinates of point A are shown.
Determine the components of F so that the magnitude
of the moment due to F about the base P of the pole is
as large as possible. (There are two answers.)
Solution: The force is given by F D 4 kNi C Fy j C Fz k.
Since the magnitude is constrained we must have
4 kN2 C Fy 2 C Fz 2 D 6 kN2 ) Fz D
20 kN2 Fy 2
Thus we will use (suppressing the units)
y
FD
F
4i C Fy j C
20 Fy 2 k
A
(4, 3, ⫺2) m
The moment is now given by
M D 4i C 3j 2k &eth; F
M D 2Fy C 3 20 Fy 2 i 8 C 4 20 Fy 2 j C 12 C 4Fy k
The magnitude is
P
x
M2 D 708 5Fy 2 C 64 20 Fy 2 C 12Fy 8 C 20 Fy 2
To maximize this quantity we solve
z
dM2
D 0 for the critical values
dFy
of Fy .
There are three solutions Fy D 4.00, 3.72, 3.38.
The first and third solutions produce the same maximum moment.
The second answer corresponds to a local minimum and is therefore discarded.
So the force that produces the largest moment is
F D 4i 4j C 2k
192
or
F D 4i 3.38j C 2.92k
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y
Problem 4.67 The force F D 5i (kN) acts on the ring
A where the cables AB, AC, and AD are joined. What is
the sum of the moments about point D due to the force
F and the three forces exerted on the ring by the cables?
D (0, 6, 0) m
A
Strategy: The ring is in equilibrium. Use what you
know about the four forces acting on it.
C
B
Solution: The vector from D to A is
rDA D 12i 2j C 2k m.
The sum of the moments about point D is given by
F
(12, 4, 2) m
(6, 0, 0) m
x
(0, 4, 6) m
z
A
F
MD D rDA &eth; FAD C rDA &eth; FAC C rDA &eth; FAB C rDA &eth; F
MD D rDA &eth; FAD C FAC C FAB C F
FAC
FAB
However, we are given that ring A is in equilibrium and this
implies that
FAD C FAC C FAB C F D O D 0
Thus,
MD D rDA &eth; O D 0
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193
Problem 4.68 In Problem 4.67, determine the moment about point D due to the force exerted on the ring
A by the cable AB.
Solution: We need to write the forces as magnitudes times the
D(0, 6, 0)
appropriate unit vectors, write the equilibrium equations for A in component form, and then solve the resulting three equations for the three
unknown magnitudes. The unit vectors are of the form
FAC
eAP D
A(12, 4, 2) m
F = 5i (kN)
xP xA i C yP yA j C zP zA k
jrAP j
C(0, 4, 6) m
Where P takes on values B, C, and D
B(6, 0, 0) m
Calculating the unit vectors, we get

e D 0.802i 0.535j 0.267k

 AB
eAC D 0.949i C 0j C 0.316k


eAD D 0.973i C 0.162j 0.162k
From equilibrium, we have
FAB eAB C FAC eAC C FAD eAD C 5i kN D 0
In component form, we get

i: 0.802FAB 0.949FAC 0.973FAD C 5 D 0


j: 0.535FAB C 0FAC C 0.162FAD D 0


k: 0.267FAB C 0.316FAC 0.162FAD D 0
Solving, we get
FAB D 779.5 N, FAC D 1976 N
The vector from D to A is
rDA D 12i 2j C 2k m
The force FAB is given by
FAB D FAB eAB
FAB D 0.625i 0.417j 0.208k kN
The moment about D is given by
MD D rDA &eth; FAB
i
D 12
0.625
j
2
0.417
k 2 0.208 MD D 1.25i C 1.25j 6.25k kN-m
194
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 4.69 The tower is 70 m tall. The tensions
in cables AB, AC, and AD are 4 kN, 2 kN, and 2 kN,
respectively. Determine the sum of the moments about
the origin O due to the forces exerted by the cables at
point A.
y
A
D
35 m
B
35 m
40 m
C
O
x
40 m
40 m
z
Solution: The coordinates of the points are A (0, 70, 0), B (40, 0,
0), C (40, 0, 40) D(35, 0, 35). The position vectors corresponding
to the cables are:
rAD D 35 0i C 0 70j C 35 0k
rAC D 40 0i C 0 70j C 40 0k
rAC D 40i 70j C 40k
The sum of the forces acting at A are
TA D 0.2792i 6.6615j C 0.07239k (kN-m)
The position vector of A is rOA D 70j. The moment about O is M D
rOA &eth; TA
i
M D 0
0.2792
j
70
6.6615
k
0
0.07239 D 700.07239i j0 k700.2792 D 5.067i 19.54k
rAB D 40 0i C 0 70j C 0 0k
rAB D 40i 70j C 0k
The unit vectors corresponding to these position vectors are:
35
70
35
D
i
j
85.73
85.73
85.73
D 0.4082i 0.8165j 0.4082k
eAC D
rAC
40
70
40
D i
jC
k
jrAC j
90
90
90
D 0.4444i 0.7778j C 0.4444k
eAB D
40
70
rAB
D
i
j C 0k D 0.4962i 0.8682j C 0k
jrAB j
80.6
80.6
The forces at point A are
TAB D 4eAB D 1.9846i 3.4729j C 0k
TAC D 2eAB D 0.8889i 1.5556j C 0.8889k
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
195
Problem 4.70 Consider the 70-m tower in Problem 4.69. Suppose that the tension in cable AB is 4 kN,
and you want to adjust the tensions in cables AC and
AD so that the sum of the moments about the origin O
due to the forces exerted by the cables at point A is zero.
Determine the tensions.
Solution: From Varignon’s theorem, the moment is zero only if
the resultant of the forces normal to the vector rOA is zero. From
Problem 4.69 the unit vectors are:
35
70
35
D
i
j
D
85.73
85.73
85.73
The tensions are TAB D 4eAB , TAC D jTAC jeAC , and TAD D jTAD jeAD .
The components normal to rOA are
D 0.4082i 0.8165j 0.4082k
eAC D
40
70
40
rAC
D i
jC
k
jrAC j
90
90
90
D 0.4444i 0.7778j C 0.4444k
eAB D
FX D 0.4082jTAD j 0.4444jTAC j C 1.9846i D 0
FZ D 0.4082jTAD j C 0.4444jTAC jk D 0.
The HP-28S calculator was used to solve these equations:
jTAC j D 2.23 kN, jTAD j D 2.43 kN
rAB
40
70
D
i
j C 0k D 0.4963i 0.8685j C 0k
jrAB j
80.6
80.6
Problem 4.71 The tension in cable AB is 150 N. The
tension in cable AC is 100 N. Determine the sum of the
moments about D due to the forces exerted on the wall
by the cables.
y
5m
5m
B
C
Solution: The coordinates of the points A, B, C are A (8, 0, 0),
B (0, 4, 5), C (0, 8, 5), D(0, 0, 5). The point A is the intersection of
the lines of action of the forces. The position vector DA is
4m
8m
8m
D
rDA D 8i C 0j 5k.
A
z
x
The position vectors AB and AC are
rAB D 8i C 4j 5k,
rAB D
rAC D 8i C 8j C 5k,
rAC D
p
82 C 42 C 52 D 10.247 m.
p
82 C 82 C 52 D 12.369 m.
The unit vectors parallel to the cables are:
eAB D 0.7807i C 0.3904j 0.4879k,
eAC D 0.6468i C 0.6468j C 0.4042k.
i
MD D 8
181.79
j
0
123.24
k 5 D 123.245i
C32.77 8C32.77 5181.79j C 8123.24k
MD D 616.2i 117.11j 985.9k (N-m)
(Note: An alternate method of solution is to express the moment in
terms of the sum: MD D rDC &eth; TC C rDB &eth; TB .
The tensions are
y
TAB D 150eAB D 117.11i C 58.56j 73.19k,
5m
TAC D 100eAC D 64.68i C 64.68j C 40.42k.
The sum of the forces exerted by the wall on A is
5m
B
4m
C
TA D 181.79i C 123.24j 32.77k.
The force exerted on the wall by the cables is TA . The moment about
D is MD D rDA &eth; TA ,
196
A
8m
z D
8m
F
x
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 4.72 Consider the wall shown in Problem 4.71. The total force exerted by the two cables in
the direction perpendicular to the wall is 2 kN. The
magnitude of the sum of the moments about D due to
the forces exerted on the wall by the cables is 18 kN-m.
What are the tensions in the cables?
Solution: From the solution of Problem 4.71, we have rDA D 8i C
0j 5k. Forces in both cables pass through point A and we can use
this vector to determine moments of both forces about D. The position
vectors AB and AC are
p
rAB D 8i C 4j 5k,
jrAB j D
rAC D 8i C 8j C 5k,
jrAC j D
82 C 42 C 52 D 10.247 m.
p
82 C 82 C 52 D 12.369 m.
The unit vectors parallel to the cables are:
eAB D 0.7807i C 0.3904j 0.4879k,
eAC D 0.6468i C 0.6468j C 0.4042k.
The tensions are
TBA D TBA eAB D TBA 0.7807i C 0.3904j 0.4879k, and
TCA D TCA eAC D TCA 0.6468i C 0.6468j C 0.4042k.
The sum of the forces exerted by the cables perpendicular to the wall
is given by
TPerpendicular D TAB 0.7807 C TAC 0.6468 D 2 kN.
The moments of these two forces about D are given by
MD D rDA &eth; TCA C rDA &eth; TBA D rDA &eth; TCA C TBA .
The sum of the two forces is given by
i
8
MD D TCA C TCB X
j
0
TCA C TCB Y
k
.
5
TCA C TCB Z This expression can be expanded to yield
MD D 5TCA C TCB Y i C [8TCA C TCB Z 5TCA C TCB X ]j
C 8TCA C TCB Y k.
The magnitude of this vector is given as 18 kN-m. Thus, we obtain
the relation
jMD j D
25TCA C TCB 2Y C [8TCA C TCB Z
D 18 kN-m.
5TCA C TCB X ]2 C 64TCA C TCB 2Y
We now have two equations in the two tensions in the cables. Either
algebraic substitution or a numerical solver can be used to give
TBA D 1.596 kN, and TCA D 1.166 kN.
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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.
Solution: We have the force and position vectors
FD
1 kN
4i C 2j C 4k, r D AB D 4i C 3j C k m
6
The moment is then
y
M D r &eth; F D 1.667i 3.33j C 3.33k kN-m
C
(0, 4, ⫺3) m
B
(4, 3, 1) m
D
(0, 5, 5) m
x
A
E
z
Problem 4.74* Suppose that the mass of the suspended object E in Problem 4.73 is 100 kg and the mass
of the bar AB is 20 kg. Assume that the weight of the
bar acts at its midpoint. By using the fact that the sum
of the moments about point A due to the weight of the
bar and the forces exerted on the “ball” at B by the three
cables BC, BD, and BE is zero, determine the tensions
in the cables BC and BD.
Solution: We have the following forces applied at point B.
F1 D 100 kg9.81 m/s2 j,
F3 D
TBC
F2 D p 4i C j 4k,
33
TBD
4i C 2j C 4k
6
In addition we have the weight of the bar F4 D 20 kg9.81 m/s2 j
The moment around point A is
MA D 4i C 3j C k m &eth; F1 C F2 C F3 C 2i C 1.5j C 0.5k m &eth; F4 D 0
Carrying out the cross products and breaking into components we find
Mx D 1079 2.26TBC C 1.667TBD D 0
My D 2.089TBC 3.333TBD D 0
Mz D 4316 C 2.785TBC C 3.333TBD D 0
Only two of these three equations are independent. Solving we find
TBC D 886 N, TBD D 555 N
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.
y
2m
B
A
5m
2m
2m
C
x
z
Solution: The slider is in equilibrium. The smooth bar exerts no
vertical forces on the slider. Hence, the vertical component of FAB
supports the weight of the slider.
FAB
The unit vector from A to B is determined from the coordinates of
points A and B A2, 2, 0, B0, 5, 2 m
Thus,
H
rAB D 2i C 3j C 2k m
eAB D 0.485i C 0.728j C 0.485k
and
−mg j
FAB D FAB eAB
The horizontal force exerted by the bar on the slider is
H D Hx i C H z k
Equilibrium requires H C FAB mgj D 0
i: Hx 0.485FAB D 0
m D 200 kg
j: 0.728FAB mg D 0
g D 9.81 m/s2
k: Hz C 0.485FAB D 0
Solving, we get
FAB D 2697N D 2, 70 kN
Hx D 1308N D 1.31 kN
Hz D 1308N D 1.31 kN
rCA D 2j m
FAB D FAB eAB
FAB D 1308i C 1962j C 1308k N
i
Mc D 0
1308
j
2
1962
k 0 1308 Mc D 2616i C 0j C 2616k N-m
Mc D 2.62i C 2.62i kN-m
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199
y
Problem 4.76 To evaluate the adequacy of the design
of the vertical steel post, you must determine the moment
about the bottom of the post due to the force exerted on
the post at B by the cable AB. A calibrated strain gauge
mounted on cable AC indicates that the tension in cable
AC is 22 kN. What is the moment?
5m
5m
C
D
4m
8m
(6, 2, 0) m
B
A
O
z
3m
12 m
x
D(0, 4, −5) 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
xi xA i C yi yA j C zi zA k
D xi xA 2 C yi yA 2 C zi zA 2
where i takes on the values B, C, D, and O. We get
eAB D 0.986i C 0.164j C 0k
eAC D 0.609i C 0.609j C 0.508k
eAD D 0.744i C 0.248j 0.620k
C (0, 8, 5) m
TAC
A (6, 2, 0) m
TAO
B(12, 3, 0) m
TAB
O(0, 0, 0) m
In component form,

T e

 AB ABx
TAB eABy C TAC eACy C TAD eADy C TAO eAOy D 0


TAB eABz C TAC eACz C TAD eADz C TAO eAOz D 0
We know TAC D 22 kN. Substituting this in, we have 3 eqns in 3
unknowns. Solving, we get
eAO D 0.949i 0.316j C 0k
We now write the forces as
TAB D 163.05 kN,
TAO D 141.28 kN
We now know that TAB is given as
TAB D TAB eAB
TAB D TAB eAB D 160.8i C 26.8j kN
TAC D TAC eAC
and that the force acting at B is TAB .
The moment about the bottom of the post is given by
TAO D TAO eAO
MBOTTOM D r &eth; TAB D 3j &eth; TAB We then sum the forces and write the equilibrium equations in component form.
For equilibrium at A,
200
FA D 0
Solving, we get
MBOTTOM D 482k kN-m
FA D TAB C TAC C TAD C TAO D 0.
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Problem 4.77 The force F D 20i C 40j 10k (N). Use
both of the procedures described in Example 4.7 to determine the moment due to F about the z axis.
y
F
Solution: First Method: We can use Eqs. (4.5) and (4.6)
r D 8i m
x
(8, 0, 0) m
F D 20i C 40j 10k N
Mzaxis D [k &ETH; r &eth; F]k
0
jMzaxis j D k &ETH; r &eth; F D 8 m
20 N
01
0
40 N
0 D 320 N-m
10 N z
Mzaxis D 320 N-mk
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
jMzaxis j D 40 N8 m D 320 N-m
Using the right-hand rule we see that the moment is about the Cz axis.
Thus
Mzaxis D 320 N-mk
Problem 4.78 Use Eqs. (4.5) and (4.6) to determine
the moment of the 20-N force about (a) the x axis,
(b) the y axis, (c) the z axis. (First see if you can write
down the results without using the equations.)
Solution: The force is parallel to the z axis. The perpendicular
distance from the x axis to the line of action of the force is 4 m. The
perpendicular distance frompthe y axis p
is 7 m and the perpendicular
distance from the z axis is 42 C 72 D 65 m.
By inspection, the moment about the x axis is
y
Mx D 420i (N-m)
(7, 4, 0) m
Mx D 80i N-m
By inspection, the moment about the y axis is My D 720j N-m
20 k (N)
x
My D 140j (N-m)
z
By inspection, the moment about the z axis is zero since F is parallel
to the z axis.
Mz D 0 N-m
Now for the calculations using (4.5) and (4.6)
ML D [e &ETH; r &eth; F]e
1
Mx D 7
0
0
4
0
0 0 i D 80i N-m
20 0
My D 7
0
1
4
0
0 0 j D 140j N-m
20 0
Mz D 7
0
0
4
0
1 0 k D 0k N-m
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201
y
Problem 4.79 Three forces parallel to the y axis act
on the rectangular plate. Use Eqs. (4.5) and (4.6) to
determine the sum of the moments of the forces about
the x axis. (First see if you can write down the result
without using the equations.)
3 kN
x
2 kN
6 kN
600 mm
900 mm
z
Solution: By inspection, the 3 kN force has no moment about the
M6
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 [20.6 60.6]i kN
Mx D 2.4i kN
kN
1
D 0
0
Mx D M3
0
0
6
kN
0 .6 i D 3.6i kN
0
C M2
kN
C M6
kN
Mx D 0 C 1.2i 3.6i kN
Mx D 2.4i kN
Calculating the result:
M3
M2
kN
kN
1
D 0
0
0
0
3
0 0 i D 0i kN
0
1
D 0
0
0
0
2
0 .6 i D 1.2i kN
0
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
Solution: (a) The magnitude of the moments about the y axis is
M D eY &ETH; r &eth; F. The position vectors of the three forces are given
in the solution to Problem 4.79. The magnitude for each force is:
0
1 0 eY &ETH; r &eth; F D 0.9
0 0 D 0,
0
3 0 0
1
eY &ETH; r &eth; F D 0.9 0
0
6
0 0.6 D 0,
0 0
eY &ETH; r &eth; F D 0
0
0 0.6 D 0
0 1
0
2
Thus the moment about the y axis is zero, since the magnitude of each
moment is zero.
202
(b) The magnitude of each moment about the z axis is
0
1
eZ &ETH; r &eth; F D 0.9 0
0
3
0 0 D 2.7,
0
0
0
eZ &ETH; r &eth; F D 0.9 0
0 C 6
1 0.6 D 5.4,
0 0
eZ &ETH; r &eth; F D 0
0
0
0
2
1 0.6 D 0.
0 Thus the moment about the z axis is
MZ D 2.7eZ C 5.4eZ D 2.7k (kN-m)
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Problem 4.81 The person exerts a force F D 0.2i 0.4j C 1.2k (lb) on the gate at C. Point C lies in the xy
plane. What moment does the person exert about the
gate’s hinge axis, which is coincident with the y axis?
y
A
C
Solution:
3.5 ft
M D [e &ETH; r &eth; F]e
e D j,
0
MY D 2
.2
r D 2i ft,
F is given
x
1
0 0
0 j D 2.4j ft-lb
.4 1.2 B
2 ft
y
Problem 4.82 Four forces act on the plate. Their
components are
FB
FA D 2i C 4j C 2k (kN),
FA
FB D 3j 3k (kN),
x
FD
FC
FC D 2j C 3k (kN),
FD D 2i C 6j C 4k (kN).
Determine the sum of the moments of the forces
z
2m
3m
Solution: Note that FA acts at the origin so no moment is generated
about the origin. For the other forces we have
i
j
k j
k i
0
2m
0
0 C3 m
MO D 3 m
0
2 kN 3 kN 3 kN 3 kN 0
i
C 0
2 kN
j
0
6 kN
k 2m
4 kN MO D 16i C 4j C 15k kN-m
Now we find
Mx D MO &ETH; i D 16 kN-m, Mz D MO &ETH; k D 15 kN-m
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203
y
Problem 4.83 The force F D 30i C 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?
F
(4, 2, 2) ft
Solution:
x
z
(a)
My D j &ETH; [4i C 2j C 2k ft &eth; 30i C 20j 10k lb]
0
My D 4 ft
30 lb
2 ft
2 ft D 100 ft lb
20 lb 10 lb 1
0
) My D 100 ft-lbj
(b)
p
p
Mymax D Fd D 302 C 202 C 102 lb 42 C 22 ft
D 167.3 ft-lb
Note that d is the distance from the y axis, not the distance from
the origin.
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:
eX
e ž r &eth; F D rX
FX
eY
rY
FY
eZ rZ ,
FZ 0
eZ &ETH; r &eth; F D 4
FX
0
2
80
1 2 D 320 2FX D 160
FZ Solve: FX D 80 lb, FZ D 40 lb, from which the force vector is F D
80i C 80j C 40k
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?
m
y
0m
C
60
160 N
600 mm
B
Solution: The unit vectors along AB and AC are of the form
200 N
e D cos x i C cos y j C cos z k.
The unit vectors are
A
eAB D 0.500i C 0.866j C 0k and eBC D 0.707i C 0.619j 0.342k.
z
The vector to point G at the center of arm AB is
x
rAG D 3000.500i C 0.866j C 0k D 150i C 259.8j C 0k mm,
and the vector from A to the point H at the center of arm BC is
given by
rAH D rAB C rBH D 600eAB C 300eBC
D 512.1i C 705.3j 102.6k mm.
The weight vectors acting at G and H are WG D 200j N, and WH D
160j N. The moment vectors of these forces about the z axis are of
the form
eX
e ž r &eth; F D rX
FX
ey
rY
FY
ez rZ .
FZ Here, WG and WH take on the role of F, and e D k.
Substituting into the form for the moment of the force at G, we get
0
e ž r &eth; F D 0.150
0
0
1 0.260 0 D 30 N-m.
200 0 Similarly, for the moment of the force at H, we get
0
0
e ž r &eth; F D 0.512 0.705
0
160
1 0.103 D 81.9 N-m.
0 The total moment about the z axis is the sum of the two moments.
Hence, Mz axis D 111.9 N-m
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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
1
0
0 e &ETH; r &eth; F D 0.150 0.260 0 D 0 N-m.
0
200 0 Similarly, for the moment of the force at H, we get
1
0
e &ETH; r &eth; F D 0.512 0.705
0
160
0 0.103 D 16.4 N-m.
0 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 2i C 3j C 6k (kN). Determine
the magnitude of the moment of the force about the axis
of the bar BC.
y
C
(0, 4, 0) m
F ⫽ ⫺2i ⫹ 6j ⫹ 3k (kN)
A (4, 2, 2) m
Solution: We have the following vectors
rBA D 4i C 2j 1k m
x
B
z
(0, 0, 3) m
F D 2i C 3j C 6k kN
rBC D 4j 3k m
eBC D
rBC
D 0.8j 0.6k
jrBC j
The moment of F about the axis of the bar is
0 0.8 0.6 jMBC j D eBC &ETH; r &eth; F D 4
2
1 D 27.2 kN-m
2 3
6 Thus MBC D 27.2 kN-meBC , jMBC j 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.
y
A (0, 5, 0) m
(7, 4, 0) m
20k (N)
B
(– 4, 0, 0) m
x
z
Solution: First, we need the unit vector
Using eAB
xB xA i C yB yA j C zB zA k
eAB D xB xA 2 C yB yA 2 C zB zA 2
0.625
ML D 7
0
0.781 0 1
0 0.625i 0.781j
0
20 eAB D 0.625i 0.781j D eBA
ML D 76.1i 95.1j N-m
Now, the moment of the 20k (N) force about AB is given as
ex
ML D rx
Fx
ey
ry
Fy
ez rz e
Fz Using eBA
0.625
ML D 7
0
where e is eAB or eBA
0.781 0 1
0 0.625i C 0.781j
0
20 For this problem, r must go from line AB to the point of application
of the force. Let us use point A.
ML D 76.1i 95.1j N-m
r D 7 0i C 4 5j C 0 0k m
Ł Results
are the same
r D 7i 1j C 0k m
Problem 4.89 The force F D 10i C 5j 5k (kip).
Determine the moment of F about the line AB. Draw
a sketch to indicate the sense of the moment.
y
B
(6, 6, 0) ft
Solution: The moment of F about pt. A is
MA D 6i &eth; F
i
j
D 6 0
10 5
F
k 0 5 A
x
(6, 0, 0) ft
z
D 30j 30k ft-kip.
y
The unit vector j is parallel to line AB, so the moment about AB is
(6, 6, 0) ft
MAB D j &ETH; MA j
B
D 30j ft-kip.
F
y
B
z
x
A (6, 0, 0) ft
−30j (ft-kip)
Direction of moment
x
A
z
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207
y
Problem 4.90 The force F D 10i C 12j 6k (N).
What is the moment of F about the line OA? Draw a
sketch to indicate the sense of the moment.
Solution: The strategy is to determine a unit vector parallel to OA
and to use this to determine the moment about OA. The vector parallel
to OA is rOA D 6j C 4k. The magnitude: F. The unit vector parallel
to OA is eOA D 0.8321j C 0.5547k. The vector from O to the point
of application of F is rOF D 8i C 6k. The magnitude of the moment
0
jMO j D eOA &ETH; rOF &eth; F D 8
10
A
(0, 6, 4) m
x
(8, 0, 6) m
0.5547 6 6 0.8321
0
12
F
O
z
D 89.8614 C 53.251 D 143.1 N-m.
The moment about OA is MOA D jMOA jeOA D 119.1j C 79.4k (N-m).
The sense of the moment is in the direction of the curled fingers of
the right hand when the thumb is parallel to OA, pointing to A.
Problem 4.91 The tension in the cable AB is 1 kN.
Determine the moment about the x axis due to the force
exerted on the hatch by the cable at point B. Draw a
sketch to indicate the sense of the moment.
Solution: The vector parallel to BA is
y
A
(400, 300, 0) mm
x
rBA D 0.4 1i C 0.3j 0.6k D 0.6i C 0.3j 0.6k.
600 mm
B
The unit vector parallel to BA is
1000 mm
z
eBA D 0.6667i C 0.3333j 0.6667k.
i
MO D rOB &eth; T D 1
0.6667
j
0
0.3333
k
0.6
0.66667 MO D 0.2i C 0.2667j C 0.3333k.
The magnitude is
jMX j D eX &ETH; MO D 0.2 kN-m.
The moment is MX D 0.2i kN-m. The sense is clockwise when
viewed along the x axis toward the origin.
208
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y
Problem 4.92 Determine the moment of the force applied at D about the straight line through the hinges A
and B. (The line through A and B lies in the yz plane.)
6 ft
20i – 60j (lb)
Solution: From the figure, we see that the unit vector along the
line from A toward B is given by eAB D sin 20&deg; j C cos 20&deg; k. The
position vector is rAD D 4i ft, and the force vector is as shown in the
figure. The moment vector of a force about an axis is of the form
eX
e ž r &eth; F D rX
FX
ey
rY
FY
E
A
D
4 ft
2 ft
ez rZ .
FZ x
B
z
20&deg;
C
4 ft
For this case,
0 sin 20&deg;
e ž r &eth; F D 4
0
20
60
cos 20&deg; 0 D 240 cos 20&deg; ft-lb
0 D 225.5 ft-lb.
The negative sign is because the moment is opposite in direction to
the unit vector from A to B.
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&deg; j C cos 20&deg; k. The position
vector is rBC D 4i ft. The coordinates of point C are (4, 4 sin 20&deg; ,
4 cos 20&deg; ). The unit vector along CE is 0.703i C 0.592j C 0.394k
and the force vector is as shown acting at point D.
The moment vector is a force about an axis is of the form
eX
e ž r &eth; F D rX
FX
ey
rY
FY
ez rZ .
FZ For this case,
rCE D 4i C 3.368j C 2.242k
TCE D 160eCE D 112.488i C 94.715j C 63.049k
0
sin 20&deg;
e ž r &eth; F D 4
0
112.488 94.715
cos 20&deg; 0 D 240 cos 20&deg; ft-lb
63.049 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
B
Solution: The position vectors:
rOA D 2.4i 0.6k (m), rOB D 2.2i C 0.7j 1.2k (m),
A
rBA D 2.4 C 2.2i C 0 0.7j C 0.6 C 1.2k (m)
z
D 0.2i 0.7j C 0.6k (m).
The magnitude is jrBA j D 0.9434 m.
The unit vector parallel to BA is
eBA D 0.2120i 0.7420j C 0.6360k.
The tension is TBA D 130eBA .
The moment of TBA about the origin is
MO D rOB &eth; TBA
or
i
D 2.2
27.56
j
0.7
96.46
k 1.2 ,
82.68 MO D 57.88i C 214.97j C 231.5k.
The magnitude of the moment about the y axis is
jMY j D eY &ETH; MO 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.
y
A
(2, 5, –2) ft
x
z
B (10, –2, 3) ft
y
Solution: The position vector from B to A is
443 ft-lb
rBA D 2 10i C [5 2]j C 2 3k
D 8i C 7j 5k ft,
187 ft-lb
So the force exerted on B is
F D 200
x
rBA
D 136.2i C 119.2j 85.1k lb.
jrBA j
The moment of F about the origin O is
i
rOB &eth; F D 10
136.2
j
2
119.2
k 3 85.1 919 ft-b
z
D 187i C 443j C 919k ft-lb.
The moments about the x, y, and z axes are
[rOB &eth; F &ETH; i]i D 187i ft-lb,
[rOB &eth; F &ETH; j]j D 443j ft-lb,
[rOB &eth; F &ETH; k]k D 919k ft-lb.
Problem 4.96 The total force exerted on the blades
of the turbine by the steam nozzle is F D 20i 120j C
100k (N), and it effectively acts at the point (100, 80,
300) mm. What moment is exerted about the axis of the
turbine (the x axis)?
y
Fixed
Rotating
x
Solution: The moment about the origin is
i
MO D 0.1
20
j
0.08
120
k 0.3 100 D 44.0i 4.0j 13.6k N-m.
The moment about the x axis is
z
MO &ETH; ii D 44.0i N-m.
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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.
Solution: The vector from A to B is
rAB D [60 480i C 100 40j C 30 40k] mm
rAB D 420i C 140j 70k mm
y
The 35-N force can be written
F D 35 N
rAB
D 32.8i C 10.9j 5.47k N
jrAB j
The moment about point O is
i
j
MO D rOB &eth; F D 60
100
32.8 10.9
B (60, 100, ⫺30) mm
k 30 5.47 O
D 219i C 1310j C 3940k N-mm
The magnitude of the moment about the z axis is
z
Mz D MO &ETH; k D 3940 N-mm D 3.94 N-m
x
Mz D 3.94 N-m
212
A
(480, ⫺40, 40) mm
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 4.98 The tension in cable AB is 80 lb. What
is the moment about the line CD due to the force exerted
by the cable on the wall at B?
y
8 ft
3 ft
Solution: The strategy is to find the moment about the point C
exerted by the force at B, and then to find the component of that
moment acting along the line CD. The coordinates of the points B,
C, D are B (8, 6, 0), C (3, 6, 0), D(3, 0, 0). The position vectors
are: rOB D 8i C 6j, rOC D 3i C 6j, rOD D 3i. The vector parallel to
CD is rCD D rOD rOC D 6j. The unit vector parallel to CD is
eCD D 1j. The vector from point C to B is rCB D rOB rOC D 5i.
B
The position vector of A is rOA D 6i C 10k. The vector parallel to
BA is rBA D rOA rOB D 2i 6j C 10k. The magnitude is jrBA j D
11.832 ft. The unit vector parallel to BA is
C
6 ft
eBA D 0.1690i 0.5071j C 0.8452k.
The tension acting at B is
x
D
TBA D 80eBA D 13.52i 40.57j C 67.62k.
The magnitude of the moment about CD due to the tension acting at
B is
z
A (6, 0, 10) ft
0
jMCD j D eCD &ETH; rCB &eth; TBA D 5
13.52
1
0
40.57
0 0 67.62 D 338.1 (ft lb).
The moment about CD is MCD D 338.1eCD D 338.1j (ft lb). The
sense of the moment is along the curled fingers of the right hand when
the thumb is parallel to CD, pointing toward D.
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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.
Solution: We have
rAB D 0.3i 0.1j 0.4k m,
rAB D
0.32 C 0.12 C 0.42 m D
p
0.26 m
y
1
0.3i 0.1j 0.4k
eAB D p
0.26
B
F D 0.2 N0.727i 0.364j C 0.582k
(200, 400, 0) mm
rAP D 0.26i 0.025j 0.11k m
(160, 475, 290) mm
Now the magnitude of the moment about the spool axis AB is
P
A
F
(⫺100, 500, 400) mm
x
MAB
0.2 N
D p
0.26
0.3
0.26 m
0.727
0.4 0.11 m D 0.0146 N-m
0.582 0.1
0.025 m
0.364
z
Problem 4.100 A motorist applies the two forces
shown to loosen a lug nut. The direction cosines of
4
3
F are cos x D 13
, cos y D 12
, and cos z D 13
. If the
13
magnitude of the moment about the x axis must be 32 ftlb to loosen the nut, what is the magnitude of the forces
the motorist must apply?
y
Solution: The unit vectors for the forces are the direction cosines.
The position vector of the force F is rOF D 1.333k ft. The magnitude
of the moment due to F is
1
0
jMOF j D eX &ETH; rOF &eth; F D 0.3077F
0
0
0.9231F
0
1.333 0.2308F jMOF j D 1.230F ft lb.
The magnitude of the moment due to F is
–F
F
jMOF j D eX &ETH; rOF &eth; F
1
D 0
.3077F
z
0
0
0.9231F
0
1.333 D 1.230F ft lb.
0.2308F The total moment about the x axis is
16 in
16 in
x
MX D 1.230Fi C 1.230Fi D 2.46Fi,
from which, for a total magnitude of 32 ft lb, the force to be applied is
FD
214
32
D 13 lb
2.46
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 4.101 The tension in cable AB is 2 kN. What
is the magnitude of the moment about the shaft CD due
to the force exerted by the cable at A? Draw a sketch to
indicate the sense of the moment about the shaft.
2m
C
A
Solution: The strategy is to determine the moment about C due
to A, and determine the component parallel to CD. The moment is
determined from the distance CA and the components of the tension,
which is to be found from the magnitude of the tension and the unit
vector parallel to AB. The coordinates of the points A, B, C, and
D are: A (2, 2, 0), B (3, 0, 1), C (0, 2, 0), and D (0,0,0). The unit
vector parallel to CD is by inspection eCD D 1j. The position vectors
parallel to DC, DA, and DB:
rDC D 2j, rDA D 2i C 2j, rDB D 3i C 1k.
2m
The vector parallel to CA is rCA D 2i. The vector parallel to AB is
rAB D rDB rDA D 1i 2j C 1k.
D
B
1m
The magnitude: jrAB j D 2.4495 m. The unit vector parallel to AB is
3m
eAB D 0.4082i 0.8165j C 0.4082k.
The tension is
TAB D 2eAB D 0.8165i 1.633j C 0.8165k.
The magnitude of the moment about CD is
0
jMCD j D eCD &ETH; rCA &eth; TAB D 2
0.8164
1
0
1.633
0 0 0.8165 D 1.633 kN-m.
MCD D eCD jMCD j D 1.633j (kN-m).
The sense is in the direction of the curled fingers of the right hand
when the thumb is parallel to DC, pointed toward D.
Problem 4.102 The axis of the car’s wheel passes
through the origin of the coordinate system and
its direction cosines are cos x D 0.940, cos y D 0,
cos z D 0.342. The force exerted on the tire by the road
effectively acts at the point x D 0, y D 0.36 m, z D 0
and has components F D 720i C 3660j C 1240k (N).
What is the moment of F about the wheel’s axis?
Solution: We have to determine the moment about the axle where
a unit vector along the axle is
e D cos x i C cos y j C cos z k
e D 0.940i C 0j C 0.342k
The vector from the origin to the point of contact with the road is
r D 0i 0.36j C 0k m
The force exerted at the point of contact is
F D 720i C 3660j C 1240k N
y
The moment of the force F about the axle is
MAXLE D [e &ETH; r &eth; F]e
x
MAXLE
z
0.940
D 0
720
0
0.36
C3660
0.342 0 0.940i C 0.342k N-m
C1240 MAXLE D 508.260.940i C 0.342k N-m
MAXLE D 478i 174k N-m
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215
Problem 4.103 The direction cosines of the centerline
OA are cos x D 0.500, cos y D 0.866, and cos z D 0,
and the direction cosines of the line AG are cos x D
0.707, cos y D 0.619, and cos z D 0.342. What is the
moment about OA due to the 250-N weight? Draw a
sketch to indicate the sense of the moment about the
shaft.
Solution: By definition, the direction cosines are the scalar components of the unit vectors. Thus the unit vectors are
e1 D 0.5i C 0.866j, and e2 D 0.707i C 0.619j 0.341k.
The force is W D 250j (N). The position vector of the 250 N weight is
rW D 0.600e1 C 0.750e2 D 0.8303i C 0.9839j 0.2565k
G
m
MOA D eOA eOA &ETH; rW &eth; W
m
50
y
0.5
D 0.8303
0
7
250 N
0.866
0.9839
250
0
0.2565 e1 D 32.06e1
0
D 16i 27.77j (N-m)
A
600 mm
The moment is anti parallel to the unit vector parallel to OA, with the
sense of the moment in the direction of the curled fingers when the
thumb of the right hand is directed oppositely to the direction of
the unit vector.
O
z
x
Problem 4.104 The radius of the steering wheel is
200 mm. The distance from O to C is 1 m. The center C
of the steering wheel lies in the x y plane. The driver
exerts a force F D 10i C 10j 5k (N) on the wheel at A.
If the angle ˛ D 0, what is the magnitude of the moment
about the shaft OC? Draw a sketch to indicate the sense
of the moment about the shaft.
y
Solution: The strategy is to determine the moment about C, and
to OC and CA are:
rOC D 1i cos 20&deg; C j sin 20&deg; C 0k D 0.9397i C 0.3420j.
The line from C to the x axis is perpendicular to OC since it lies in
the plane of the steering wheel. The unit vector from C to the x axis is
eCX D i cos20 90 C j sin20 90 D 0.3420i 0.9397j,
where the angle is measured positive counterclockwise from the x axis.
The vector parallel to CA is
F
C
A
O
z
rCA D 0.2eCX D C0.0684i 0.1879j (m).
The magnitude of the moment about OC
20&deg;
α
0.9397
jMOC j D eOC &ETH; rCA &eth; F D 0.0684
10
0.3420
0 0.1879 0 10
5 x
D 0.9998 D 1 N-m.
The sense of the moment is in the direction of the curled fingers of
the right hand if the thumb is parallel to OC, pointing from O to C.
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.)
y
Solution: The moment of the general force F D Fx i C Fy j C Fz k
about the line is developed by
eBA D
1
3i C 6j 6k
D i C 2j 2k,
9
3
rBP D 12i C 2j 2k m,
MBA D eBA &ETH; rBP &eth; F
A (3, 8, 0) m
This expression simplifies to MBA D We also have the constraint that 10 N2 D Fx 2 C Fy 2 C Fz 2
L
F
Since Fx does not contribute to the moment we set it equal to zero.
Solving the constraint equation for Fz and substituting this into the
expression for the moment we find
B
(0, 2, 6) m
P
(12, 4, 4) m
MBA D 22
Fy š
3
100 Fy 2 . )
x
z
22 m
Fy C Fz 3
dMBA
D0
dFy
p
p
) Fy D š5 2N ) Fz D š5 2
F D 7.07j C 7.07k N or
F D 7.07j C 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?
y
(12, 10, 0) ft
(0, 3, 0) ft
W
Solution: The strategy is to use the unit vector parallel to the bar
to locate point C relative to the origin, and then use this location to
find the unit vector parallel to the cable CD. With the tension resolved
into components about the origin, the moment about the origin can be
resolved into components along the z axis. Denote the top of the bar
by T and the bottom of the bar by B. The position vectors of the ends
of the bar are:
D
C
d
x
z
(3, 0, 10) ft
rOB D 3i C 0j C 10k, rOT D 12i C 10j C 0k.
The vector from the bottom to the top of the bar is
rBT D rOT rOB D 9i C 10j 10k.
The magnitude:
p
jrBT j D
92 C 102 C 102 D 16.763 ft.
The unit vector parallel to the bar, pointing toward the top, is
eBT D 0.5369i C 0.5965j 0.5965k.
The position vector of the point C relative to the bottom of the bar is
rBC D 2eBT D 1.074i C 1.193j 1.193k.
The position vector of point C relative to the origin is
rOC D rOB C rBC D 4.074i C 1.193j C 8.807k.
The position vector of point D is
rOD D 0i C 3j C 0k.
The vector parallel to CD is
rCD D rOD rOC D 4.074i C 1.807j 8.807k.
The magnitude is
jrCD j D
p
4.0742 C 1.8072 C 8.8072 D 9.87 ft.
The unit vector parallel to CD is
eCD D 0.4127i C 0.1831j 0.8923k.
The tension is
TCD D 100eCD D 41.27i C 18.31j 89.23k lb.
The magnitude of the moment about the z axis is
0
jMO j D eZ &ETH; rOC &eth; TCD D 4.074
41.27
0
1.193
18.31
1 8.807 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?
Solution: Note
(150, 152.5, 195).
that
the
coordinates
of
point
G
are
We calculate the moment about the line BA due to the two forces
as follows.
eBA D
0.1i C 0.07j 0.36k
p
0.1445
y
r1 D 0.2i 0.125j C 0.03k m,
A
(100, 500, 700) mm
(100, 250, 0) mm
F1 D TCD
D
0.1i C 0.445j C 0.31k
p
0.304125
r2 D 0.15i 0.0275j 0.165k m,
F2 D 4 kg9.81 m/s2 j
G
x
B
MBA D eBA &ETH; r1 &eth; F1 C r2 &eth; F2 The moment reduces to
(0, 180, 360) mm
C
MBA D 3.871 N-m 0.17793 mTCD D 0 ) TCD D 21.8 N
(200, 55, 390) mm
z
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
ⴚF
(6, 6, 0) m
(8, 3, 0) m
F
x
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 8i C 6 8j] m D 2i 2j m
The sum of the moments of the two forces is
i
j k M D r &eth; F D 2 2 0 D 28k N-m
10 4 0 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 xy
plane.
Solution: The right hand force is
F D [1000 lb]cos 60&deg; i sin 60&deg; j
(a)
(b)
Determine the moment of the couple and represent
it as shown in Fig. 4.28c.
What is the sum of the moments of the two forces
about the point (10, 40, 20) ft?
y
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 r &eth; F
1000 lb
1000 lb
60&deg;
MC D 40i &eth; 500i 867j ft-lb
60&deg;
MC D 34700 ft-lb k
x
20 ft
20 ft
or
Problem 4.110 The moment of the couple is 600 k
(N-m). What is the angle ˛?
MC D 34700 ft-lb) clockwise
Solution:
M D 100 N cos ˛4 m C 100 N sin ˛5 m D 600 N-m
y
˛ D 30.9&deg;
a
or
˛ D 71.8&deg;
(0, 4) m
100 N
100 N
a
x
(5, 0) m
Problem 4.111 Point P is contained in the xy
plane, jFj D 100 N, and the moment of the couple is
500k (N-m). What are the coordinates of P?
Solution: The force is
F D 100i cos30&deg; C j sin30&deg; D 86.6i 50j.
Let r be the distance OP. The vector parallel to OP is
y
P
r D ri cos 70&deg; C j sin 70&deg; D r0.3420i C 0.9397j.
30&deg;
F
The moment is
–F
70&deg;
x
i
M D r &eth; F D 0.3420r
86.6
From which, r D
j
0.9397r
50.0
k 0 D 98.48rk.
0
500
D 5.077 m. From above,
98.48
r D 5.0770.3420i C 0.9397j.
The coordinates of P are
x D 5.0770.3420 D 1.74 m, y D 5.0770.9397 D 4.77 m
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.
F
L
F
Solution:
(a)
(b)
F
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.
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&deg; , so the magnitude of
the moment due to the three force is
M D FL cos 30&deg;
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?
y
200 ft-lb
A
B
4 ft
4 ft
x
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 A C 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?
y
50 ft-lb
P
x
(– 4, 0, 0) ft
10 ft-lb
Solution: The moment of a couple is the same anywhere in the
plane. Hence the sum about the point P is
M D 50k C 10k D 40k ft lb
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221
Problem 4.115 Determine the sum of the moments
exerted on the plate by the two couples.
y
Solution: The moment due to the 30 lb couple, which acts in a
clockwise direction is
M30 D 330k D 90k ft lb.
30 lb
The moment due to the 20 lb couple, which acts in a counterclockwise
direction, is
3 ft
30 lb
M20 D 920k D 180k ft lb.
2 ft
The sum of the moments is
x
20 lb
20 lb
5 ft
4 ft
M D 90k C 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
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
400 lb
M100 D 3i &eth; 100j D 300k (ft-lb),
900 ft-lb
and
A
3 ft
B
4 ft
3 ft
M400 D 7i &eth; 400j D 2800k (ft-lb).
The moment of the couple is MC D 900k (ft-lb). Summing the
moments, we get
4 ft
MTotal D 2200k (ft-lb)
Problem 4.117 Determine the sum of the moments
exerted about A by the couple and the two forces.
Solution:
MA D 0.2i &eth; 200j C 0.4i C 0.2j
100 N
30&deg;
&eth; 86.7i C 50j C 300k N-m
200 N
0.2 m
A
MA D 40k C 2.66k C 300k N-m
MA D 262.7k N-m ' 263k N-m
300 N-m
0.2 m
0.2 m
0.2 m
Problem 4.118 The sum of the moments about point
A due to the forces and couples acting on the bar is zero.
Solution:
(a)
MA D 20 kN-m 2 kN5 m 4 kN3 m
(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 kN8 C C D 0
C D 26 kN-m
B
4 kN
3m
20 kN-m
MB D 3 kN3 m 4 kN3 m 5 kN5 m
(b)
C 20 kN-m C 26 kN-m D 0
C
A
4 kN
2 kN
5 kN
5m
222
3 kN
3m
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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.
y
20 N
30 N
30 N
2m
4m
4m
x
60⬚
20 N
60⬚
z
Solution: The magnitude of the moment of the 20-N couple is
unchanged,
2 m20 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 k C 30 N cos 60&deg; 4 m j 30 N sin 60&deg; 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.
y
(0, 4, 0) m
⫺2i ⫹ 2j ⫹ k (kN)
Solution:
2i ⫺ 2j ⫺ k (kN)
x
M D 4j 5k m &eth; 2i 2j k kN
(a)
D 14i 10j 8k kN-m
(b)
MD
FD
(0, 0, 5) m
142
C 102
C
82
kN-m D 18.97 kN-m
z
22 C 22 C 12 kN D 3 kN
M D Fd ) d D
M
18.97 kN-m
D
D 6.32 m
F
3 kN
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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 xz plane.)
y
3 ft
20 lb
3 ft
20 lb
40 lb
x
8 ft
40 lb
z
60&deg;
60&deg;
80 lb
Solution: The moments of two of the couples can be determined
The moment is
from inspection:
i
j
M3 D r3 &eth; F3 D 6
0
69.282 0
M1 D 320k D 60k ft lb.
M2 D 840j D 320j ft lb
The forces in the 3rd couple are resolved:
80 lb
k 0 D 240j.
40 The sum of the moments due to the couples:
M D 60k C 320j 240j D 80j 60k ft lb
F D 80i sin 60&deg; C k cos 60&deg; D 69.282i C 40k
The two forces in the third couple are separated by the vector
r3 D 6i C 8k 8k D 6i
Problem 4.122 What is the magnitude of the sum of
the moments exerted on the T-shaped structure by the
two couples?
y
3 ft
50i + 20j – 10k (lb)
3 ft
50j (lb)
3 ft
z
–50j (lb)
3 ft
x
–50i – 20j + 10k (lb)
Solution: The moment of the 50 lb couple can be determined by
y
3 ft
inspection:
F
3 ft
50j (lb)
M1 D 503k D 150k ft lb.
The vector separating the other two force is r D 6k. The moment is
i
M2 D r &eth; F D 0
50
j
0
20
k 6 D 120i C 300j.
10 3 ft
x
–F
–50j (lb)
3 ft
z
The sum of the moments is
M D 120i C 300j 150k.
The magnitude is
p
jMj D
224
1202 C 3002 C 1502 D 356.23 ft lb
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y
Problem 4.123 The tension in cables AB and CD is
500 N.
(a)
(b)
A (0, 2, 0) m
Show that the two forces exerted by the cables on
the rectangular hatch at B and C form a couple.
What is the moment exerted on the plate by the
cables?
3m
B
z
x
3m
C
D
Solution: One condition for a couple is that the sum of a pair
of forces vanish; another is for a non-zero moment to be the same
anywhere. The first condition is demonstrated by determining the unit
vectors parallel to the action lines of the forces. The vector position of
point B is rB D 3i m. The vector position of point A is rA D 2j. The
vector parallel to cable AB is
rBA D rA rB D 3i C 2j.
(6, –2, 3) m
The moment about the origin is
MO D rB rC &eth; TAB D rCB &eth; 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 &eth; TAB D rCB &eth; TAB which
is identical to the above expression. Thus the moment is the same
everywhere on the plate, and the forces form a couple.
The magnitude is:
jrAB j D
p
32 C 22 D 3.606 m.
The unit vector:
eAB D
rAB
D 0.8321i C 0.5547j.
jrAB j
The tension is
TAB D jTAB jeAB D 416.05i C 277.35j.
The vector position of points C and D are:
rC D 3i C 3k, rD D 6i 2j C 3k.
The vector parallel to the cable CD is rCD D rD rC D 3i 2j. The
magnitude is jrCD j D 3.606 m, and the unit vector parallel to the cable
CD is eCD D C0.8321i 0.5547j. The magnitude of the tension in
the two cables is the same, and eBA D eCD , hence the sum of the
tensions vanish on the plate. The second condition is demonstrated by
determining the moment at any point on the plate. By inspection, the
distance between the action lines of the forces is
rCB D rB rC D 3i 3i 3k D 3k.
The moment is
M D rCB &eth; TAB
i
D 0
416.05
j
k 0
3 277.35 0 D 832.05i 1248.15j (N-m).
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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.
(⫺1.6, 2.2, ⫺1.2) m
y
D
C
(0.2, 1.6, ⫺0.2) m
Solution:
FAB D 8 kN
A
1.4i 0.6j C 1.0k
p
, rDB D 3.2i 2.2j C 2.4k m
3.32
(0.2, 0.6, 0.2) m
M D rBD &eth; FAB D 3.34i C 0.702j C 5.09k kN-m
x
) M D 6.13 kN-m
B
z
Problem 4.125 The bar is loaded by the forces
y
FB
FB D 2i C 6j C 3k (kN),
A
FC D i 2j C 2k (kN),
and the couple
(1.6, 0, 1.2) m
B
MC
C
x
1m
z
1m
FC
MC D 2i C j 2k (kN-m).
Determine the sum of the moments of the two forces
Solution: The moments of the two forces about A are given by
MFB D 1i &eth; 2i C 6j C 3k (kN-m) D 0i 3j C 6k (kN-m) and
MFC D 2i &eth; 1i 2j C 2k (kN-m) D 0i 4j 4k (kN-m).
MC D 2i C 1j 2k (kN-m),
we get MTOTAL D 2i 6j C 0k (kN-m)
Problem 4.126 In Problem 4.125, the forces
Solution: From the solution to Problem 4.125, the sum of the
moments of the two forces about A is
FB D 2i C 6j C 3k (kN),
MForces D 0i 7j C 2k (kN-m).
FC D i 2j C 2k (kN),
The required moment, MC , must be the negative of this sum.
and the couple
Thus
MCy D 7 (kN-m), and MCz D 2 (kN-m).
MC D MCy j C MCz k (kN-m).
Determine the values for MCy and MCz , so that the sum
of the moments of the two forces and the couple about
A is zero.
226
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y
Problem 4.127 Two wrenches are used to tighten an
elbow fitting. The force F D 10k (lb) on the right
wrench is applied at (6, 5, 3) in, and the force F
on the left wrench is applied at (4, 5, 3) in.
(a)
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.
z
x
F
–F
Solution: The position vector of the force on the right wrench is
rR D 6i 5j 3k. The magnitude of the moment about the x axis is
1
jMR j D eX &ETH; rR &eth; F D 6
0
(a)
0
0 5 3 D 50 in lb
0
10 from which MXL D 50i in lb, which is opposite in direction and
equal in magnitude to the moment exerted on the x axis by the
right wrench. The left wrench force is applied 2 in nearer the
origin than the right wrench force, hence the moment must be
absorbed by the space between, where it is wanted.
The moment about the x axis is
MR D jMR jeX D 50i (in lb).
(b)
The moment of the couple is
i
MC D rR rL &eth; FR D 2
0
(c)
j
0
0
k 6 D 20j in lb
10 The objective is to apply a moment to the elbow relative to
connecting pipe, and zero resultant moment to the pipe itself.
A resultant moment about the x axis will affect the joint at the
origin. However the use of two wrenches results in a net zero
moment about the x axis the moment is absorbed at the juncture
of the elbow and the pipe. This is demonstrated by calculating
the moment about the x axis due to the left wrench:
1
jMX j D eX &ETH; rL &eth; FL D 4
0
0
5
0
0 3 D 50 in lb
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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.
System 1
y
100 N
x
50 N
1m
1m
System 2
y
50 N
x
2m
Solution: The strategy is to check the two conditions for equivalence: (a) the sums of the forces must be equal and (b) the sums of
the moments about an arbitrary point must be equal. The sums of the
forces of the two systems:
FX D 0, (both systems) and
FY1 D 100j C 50j D 50j (N)
FY2 D 50j (N).
The sums of the forces are equal. The sums of the moments about the
left end are:
M1 D 1100k D 100k (N-m)
M2 D 250k D 100k (N-m).
The sums of the moments about the left end are equal. Choose any
point P at the same distance r D xi from the left end on each beam.
The sums of the moments about the point P are
M1 D 50x C 100x 1k D 50x 100k (N-m)
M2 D 502 xk D 50x 100k (N-m).
Thus the sums of the moments about any point on the beam are equal
for the two sets of forces; the systems are equivalent. Yes
228
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Problem 4.129 Two systems of forces and moments
act on the beam. Are they equivalent?
System 1
y
20 lb
50 ft-lb
10 lb
x
Solution: The sums of the forces are:
2 ft
FX D 0 (both systems)
FY1 D 10j 20j D 10j (lb)
System 2
y
FY2 D 20j C 10j D 10j (lb)
Thus the sums of the forces are equal. The sums of the moments about
the left end are:
2 ft
20 lb
30 ft-lb
10 lb
x
M1 D 204k C 50k D 30k (ft lb)
2 ft
2 ft
M2 D C102k 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?
System 1
the moments about some point (the left end will be used) must be
the same.
System 2
10 kN
8m
Solution: For equivalence, the sum of the forces and the sum of
F kN
ML kN-m
10 kN
80 kN-m
8m
System 1
10j
80k
System 2
10j
80k
Systems 1, 2, and 4 are equivalent.
System 3
System 4
20 kN
10 kN
8m
System 3
10j
160k
System 4
10j
80k
y
x
20 kN
10 kN
80 kN-m
4m
4m
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229
Problem 4.131 The four systems shown in Problem
couple to one of the systems. Which system is it, and
Solution: From the solution to 4.130, all systems have
F D 10j kN
and systems 1, 2, and 4 have
ML D 80k kN-m
system 3 has
ML D 160k kN-m.
Thus, we need to add a couple M D 80k kN-m to system 3 (clockwise moment).
Problem 4.132 System 1 is a force F acting at a point
O. System 2 is the force F acting at a different point O0
along the same line of action. Explain why these systems
are equivalent. (This simple result is called the principle
of transmissibility.)
System 2
System 1
F
F
O'
O
O
Solution: The sum of forces is obviously equal for both systems.
Let P be any point on the dashed line. The moment about P is the
cross product of the distance from P to the line of action of a force
times the force, that is, M D rPL &eth; 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.
A
12 m
B
16 m
C
12 m
D
6m
Solution: The angle formed by the single cable with the positive
E
20 m
Solve:
x axis is
D 180&deg; tan1
12
16
jTL j D 0.3353jT1 j, and
D 143.13&deg; .
jTR j D 0.7160jT1 j.
The single cable tension is
T1 D jTji cos 143.13&deg; C j sin 143.13&deg; D jTj0.8i C 0.6j.
The position vector to the center of the log from the left end is rc D 10i.
The moment about the end of the log is
i
M D r &eth; T1 D jT1 j 10
0.8
j k 0 0 D jTj6k (N-m).
0.6 0 The tension in the right hand cable is TR D jT1 j0.71600.9079i C
0.4191j D jT1 j0.6500i C 0.3000. The position vector of the right
end of the log is rR D 20i m relative to the left end. The moments
about the left end of the log for the second system are
i
M2 D rR &eth; TR D jT1 j 20
0.6500
j
k 0
0 D jT1 j6k (N-m).
0.3000 0 This is equal to the moment about the left end of the log for System
1, hence the systems are equivalent.
For the two cables, the angles relative to the positive x axis are
1 D 180&deg; tan1
2 D 180 tan1
12
6
12
26
D 116.56&deg; , and
D 155.22&deg; .
The two cable vectors are
TL D jTL ji cos 116.56&deg; C j sin 116.56&deg; D jTL j0.4472i C 0.8945j,
TR D jTR ji cos 155.22&deg; C j sin 155.22&deg; D jTR j0.9079i C 0.4191j.
Since the vector sum of the forces in the two systems is equal, two
simultaneous equations are obtained:
0.4472jTL j C 0.9079jTR j D 0.8jT1 j, and
0.8945jTL j C 0.4191jTR j D 0.6jT1 j
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231
Problem 4.134 Systems 1 and 2 each consist of a
couple. If they are equivalent, what is F?
System 1
y
System 2
y
200 N
F
20&deg;
30&deg;
Solution: For couples, the sum of the forces vanish for both sys-
(5, 4, 0) m
5m
200 N
tems. For System 1, the two forces are located at r11 D 4i, and r12 D
C5j. The forces are F1 D 200i cos 30&deg; C j sin 30&deg; D 173.21i C 100j.
The moment due to the couple in System 1 is
i
M1 D r11 r12 &eth; F1 D 4
173.21
30&deg;
2m
x
j
k 5 0 D 1266.05k (N-m).
100 0 x
20&deg;
F
4m
For System 2, the positions of the forces are r21 D 2i, and r22 D
5i C 4j. The forces are
F2 D Fi cos20&deg; C j sin20&deg; D F0.9397i 0.3420j.
The moment of the couple in System 2 is
i
M2 D r21 r22 &eth; F2 D F 3
0.9397
j
k 4
0 D 4.7848Fk,
0.3420 0 from which, if the systems are to be equivalent,
FD
1266
D 264.6 N
4.7848
Problem 4.135 Two equivalent systems of forces and
moments act on the L-shaped bar. Determine the forces
FA and FB and the couple M.
System 2
System 1
120 N-m
FB
40 N
60 N
3m
FA
M
3m
50 N
3m
Solution: The sums of the forces for System 1 are
FX D 50, and
The sums of the forces for System 2 are
6m
The sum of the moments about the left end for
System 2 is
FY D FA C 60.
3m
M2 D 3FB C M D 150 C M N-m.
Equating the sums of the moments, M D 150 180 D 30 N-m
FX D FB , and
FY D 40.
For equivalent systems: FB D 50 N, and FA D 60 40 D 20 N.
The sum of the moments about the left end for
System 1 is
232
M1 D 3FA 120 D 180 N-m.
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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.
System 1
30 lb
System 2
10 lb
30 lb
100 in-lb
5 in
5 in
8 in
8 in
M
50 lb
30 lb
F
Solution: The sums of the forces for System 1 are
FX D 30 lb,
FY D 50 10 D 40 lb.
The sums of the forces for System 2 are
FX D 30 lb,
FY D F 30 lb.
For equivalent forces, F D 30 C 40 D 70 lb. The sum of the moments
about the lower left corner for System 1 is
M1 D 530 810 C M D 230 C M in lb.
The sum of the moments about the lower left corner for System 2 is
M2 D 100 in lb.
Equating the sum of moments, M D 230 100 D 130 in lb
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233
System 1
Problem 4.137 In Example 4.13, suppose that the 30kN 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
30j (kN)
20i ⫹ 20j (kN)
x
O
3m
2m
210 kN-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 forces in system 1:
F2 D F1
F D 230 kN j C 20i C 20j kN
F D 20i C 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:
M2 D M1
M D 230 kN3 m C 20 kN5 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
M3 D M2
1000 kN-m D 250 kND
DD
1000 kN-m
D 4m
250 kN
DD4m
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?
If you represent system 1 by the force F (system 3),
what is the distance D?
(b)
System 1
y
30 lb
40 lb
20 lb
30 ft-lb
x
A
2 ft
2 ft
System 2
y
F
M
x
A
System 3
y
F
x
A
D
Solution: The sum of the forces in System 1 is
FX D 0i,
FY D 20 C 40 30j D 10j lb.
The sum of the moments about the left end for System 1 is
M1 D 240 430 C 30k D 10k ft lb.
(a)
For System 2, the force at A is F D 10j lb
The moment at A is M2 D 10k ft lb
(b)
For System 3 the force at D is F D 10j lb. The distance D is
the ratio of the magnitude of the moment to the magnitude of the
force, where the magnitudes are those in System 1:
DD
10
D 1 ft
10
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235
y
Problem 4.139 Represent the two forces and couple
acting on the beam by a force F. Determine F and determine where its line of action intersects the x axis.
60i + 60j (N)
280 N-m
x
– 40 j (N)
3m
3m
y
Solution: We first represent the system by an equivalent system
consisting of a force F at the origin and a couple M:
F
This system is equivalent if
M
F D 40j C 60i C 60j
x
D 60i C 20j N,
M D 280 C 660
y
D 80 N-m.
F
We then represent this system by an equivalent system consisting of
F alone:
x
For equivalence, M D dFy , so
dD
d
M
80
D
D 4 m.
Fy
20
y
Problem 4.140 The bracket is subjected to three forces
and a couple. If you represent this system by a force F,
what is F, and where does its line of action intersect the
x axis?
400 N
180 N
0.4 m
140 N-m
200 N
0.2 m
x
0.65 m
Solution: We locate a single equivalent force along the x axis a
distance d to the right of the origin. We must satisfy the following
three equations:
Fx D 400 N 200 N D Rx
Fy D 180 N D Ry
MO D 400 N0.6 m C 200 N0.2 m C 180 N0.65 m
C 140 Nm D Ry d
Solving we find
Rx D 200 N, Ry D 180 N, d D 0.317 m
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.
Solution: (a) The sum of the forces is
(a)
(b)
(c)
Determine the forces Ax and Ay , and the couple
MA .
Determine the sum of the moments about the right
end of the beam.
If you represent the 600-N force, the 200-N force,
and the 30 N-m couple by a force F acting at the
left end of the beam and a couple M, what are F
and M?
FY D AY 600 C 200j D 0,
from which AY D 400 N. The sum of the moments is
ML D MA 0.38600 30 C 0.560200k D 0,
from which MA D 146 N-m. (b) The sum of the moments about the
right end of the beam is
y
FX D AX i D 0 and
MR D 0.18600 30 C 146 0.56400 D 0.
600 N
MA
(c) The sum of the forces for the new system is
x
Ax
30 N-m
Ay
200 N
380 mm
FY D AY C Fj D 0,
from F D AY D 400 N, or F D 400j N. The sum of the moments
for the new system is
180 mm
M D MA C M 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.
Solution: (a) The sum of the forces is
(a)
(b)
from which AX D 12 kip
(c)
Determine the forces Ax , Ay , and B.
If you represent the 2-kip, 4-kip, and 6-kip forces
by a force F, what is F, and where does its line of
action intersect the y axis?
If you replace the 2-kip, 4-kip, and 6-kip forces by
the force you determined in (b), what are the vector
sum of the forces acting on the truss and the sum
FX D AX 2 4 6i D 0,
FY D AY C Bj D 0.
The sum of the moments about the origin is
MO D 36 C 64 C 92 C 6B D 0,
from which B D 10j kip. (b) Substitute into the force balance equation to obtain AY D B D 10 kip. (b) The force in the new system
will replace the 2, 4, and 6 kip forces, F D 2 4 6i D 12i kip.
The force must match the moment due to these forces: FD D 36 C
60
D 5 ft, or the action
64 C 92 D 60 kip ft, from which D D
12
line intersects the y axis 5 ft above the origin. (c) The new system is
equivalent to the old one, hence the sum of the forces vanish and the
sum of the moments about O are zero.
2 kip
y
3 ft
4 kip
3 ft
6 kip
3 ft
Ax
O
x
B
Ay
6 ft
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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?
y
Solution: The equivalent force must equal the sum of the forces
x
exerted by the soil:
80 kN
3m
F D 80 C 35 C 30 C 40 C 85j D 270j kN
35 kN
30 kN
40 kN
3m
3m
3m
85 kN
The sum of the moments about any point must be equal for the two
systems. The sum of the moments are
M D 335 C 630 C 940 C 1285 D 1665 kN-m.
Equating the moments for the two systems FD D 1665 kN-m from
which
DD
1665 kN-m
D 6.167 m.
270 kN
Thus the action line intersects the x axis at a distance D D 6.167 m to
the right of the origin.
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?
Solution:
y
88 kN
16 kN
x
G
5m
22 kN-m
5.7 m
9m
Fy D 88 kN C 16 kN D Ry
MG D 88 kN0.7 m C 16 kN3.3 m C 22 kN-m D M
Solving we find
Ry D 104 kN, M D 13.2 kN-m
Problem 4.145 If you represent the two forces and
couple acting on the airplane in Problem 4.144 by a
force F, what is F, and where does its line of action
intersect the x axis?
Solution:
Fy D 88 kN C 16 kN D Ry
MOrigin D 88 kN5 m C 16 kN9 m C 22 kN-m D Ry x
Solving we find
F D Ry j D 104 kNj, x D 5.83 m
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?
y
60&deg;
B
40&deg;
FAC
FAB
C
A
A
100 lb
100 lb
x
Solution: The sum of the forces acting at A is in opposition to the
weight, or F D jWjj D 100j lb.
The moment about point A is zero.
Problem 4.147 Three forces act on a beam.
(a)
(b)
y
Represent the system by a force F acting at the
origin O and a couple M.
Represent the system by a single force. Where does
the line of action of the force intersect the x axis?
30 N
5m
x
O
Solution: (a) The sum of the forces is
30 N
6m
4m
50 N
FX D 30i N, and
FY D 30 C 50j D 80j N.
The equivalent at O is F D 30i C 80j (N). The sum of the moments
M D 530 C 1050 D 350 N-m
(b) The solution of Part (a) is the single force. The intersection is the
350
D 4.375 m
moment divided by the y-component of force: D D
80
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239
y
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?
A
400 mm
B
C
300 mm
D
O
Solution: From the right triangle, the angle between the positive
x axis and the cable AB is
D tan1
400
800
D 26.6&deg; .
x
800 mm
Check. (b) The equivalent single force retains the same scalar components, but must act at a point that duplicates the sum of the moments.
The distance on the y axis is the ratio of the sum of the moments to
the x-component of the equivalent force. Thus
419
D 0.456 m
919.6
The tension in AB is
DD
TAB D 400i cos26.6&deg; Cj sin26.6&deg; D 357.77i 178.89j (N).
Check: The moment is
The angle between the positive x axis and the cable CD is
i
M D rF &eth; F D 0
919.6
˛ D tan1
300
800
D 20.6&deg; .
from which D D
The tension in CD is
300 mm
j
k D
0 D 919.6Dk D 419k,
389.6 0 419
D 0.456 m, Check.
919.6
TCD D 600i cos20.6&deg; C j sin20.6&deg; D 561.8i 210.67j.
The equivalent force acting at the origin O is the sum of the forces
acting on the left post:
F D 357.77 C 561.8i C 178.89 210.67j
D 919.6i 389.6j (N).
The sum of the moments acting on the left post is the product of the
moment arm and the x-component of the tensions:
M D 0.7357.77k 0.3561.8k D 419k N-m
Check: The position vectors at the point of application are rAB D 0.7j,
and rCD D 0.3j. The sum of the moments is
M D rAB &eth; TAB C rCD &eth; TCD i
D 0
357.77
j
k i
0.7
0 C 0
178.89 0 561.8
j
k 0.3
0 210.67 0 D 0.7357.77k 0.3561.8k D 419k
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 400i cos26.6&deg; Cj sin26.6&deg; D 357.77i C 178.89j,
and
TCD D 400i cos20.6&deg; Cj sin20.6&deg; D 374.42i C 140.74j.
The equivalent force is equal to the sum of these forces:
F D 357.77 374.42i C 178.77 C 140.74j
D 732.19i C 319.5j (N).
The sum of the moments about O is
M D 0.3357.77 C 0.8140.74 C 178.89k D 363k (N-m).
The intersection is D D
363
D 0.496 m on the positive y axis.
732.19
Problem 4.150 If you represent the three forces acting
on the beam cross section by a force F, what is F, and
where does its line of action intersect the x axis?
y
500 lb
800 lb
6 in
Solution: The sum of the forces is
FX D 500 500i D 0.
x
6 in
z
FY D 800j.
Thus a force and a couple with moment M D 500k ft lb act on the
cross section. The equivalent force is F D 800j which acts at a positive
500
D 0.625 ft D 7.5 in to the right of the
x axis location of D D
800
origin.
500 lb
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241
Problem 4.151 In Active Example 4.12, suppose that
the force FB is changed to FB D 20i 15j C 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.
System 1
y
(4, 3, ⫺2) m
P
FB
FA
Solution: From Active Example 4.12 we know that
x
(6, 0, 0) m
FA D 10i C 10j 15k kN
MC D 90i C 150j C 60k kN-m
z
MC
The force F must equal the sum of the forces in system 1:
F2 D F1 :
F D FA C FB D 10i 5j C 15k kN
In system 2, the sum of the moments about P is M. Therefore equivalence 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 :

i
M D  4
10
j
3
10
k i
2 C 2
15 20
j
3
15

k 2 C 90i C 150j C 60k kN-m
30 M D 125i C 50j C 20k kN-m
Thus F D 10i 5j C 15k kN, M D 125i C 50j C 20k kN-m
Problem 4.152 The wall bracket is subjected to the
force shown.
Determine the moment exerted by the force about
the z axis.
(b) Determine the moment exerted by the force about
the y axis.
(c) If you represent the force by a force F acting at O
and a couple M, what are F and M?
y
O
(a)
z
10i – 30j + 3k (lb)
12 in
x
Solution:
(a)
The moment about the z axis is negative,
MZ D 130 D 30 ft lb,
(b)
The moment about the y axis is negative,
MY D 13 D 3 ft lb
(c)
The equivalent force at O must be equal to the force at x D 12 in,
thus FEQ D 10i 30j C 3k (lb)
The couple moment must equal the moment exerted by the force at x D
12 in. This moment is the product of the moment arm and the y- and zcomponents of the force: M D 130k 13j D 3j 30k (ft lb).
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 21 in, and the angle ˛ D 120&deg; .
the forces applied: FEQ D 200j. The position vectors of the points of
application of the forces are r1 D h C ri, and r2 D ih C r cos ˛ kr sin ˛. The moments about the origin are
(a)
M D r1 &eth; F1 C r2 &eth; F2 D r1 C r2 &eth; F
(b)
If you represent the forces he exerts by a force F
acting at O and a couple M, what are F and M?
The glass backboard will shatter if jMj &gt; 4000 inlb. Does it break?
Solution: The equivalent force at the origin must equal the sum of
i
j
D 2h C r1 C cos ˛
0
0
100
k
r sin ˛ 0
D 100r sin ˛i 1002h C r1 C cos ˛k.
y
–100j (lb)
For the values of h, r, and ˛ given, the moment is M D 822.72i 3375k in lb. This is the p
couple moment required. (b) The magnitude
of the moment is jMj D 822.722 C 33752 D 3473.8 in lb. The backboard does not break.
O
α
r
–100j (lb)
h
x
z
Problem 4.154 In Example 4.14, suppose that the 30lb 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 xz plane?
System 1
y
20j (lb)
30j (lb)
(⫺3, 0, –2) ft
(6, 0, 2) ft
(2, 0, 4) ft
of the forces in system 1:
z
F2 D F1
x
O
Solution: The sum of the forces in system 2 must equal the sum
⫺10j (lb)
System 2
y
F D 20 C 25 10j 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
F
M2 D M1
i
x
0
j k i
y z D 6
35 0 0
j k i
0 2 C 2
25 0 0
j
j
k i
0
4 C 3 0
10 0 0 20
k 2 0 x
O
P
z
Expanding the determinants results in the equations
35z D 50 C 40 C 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
C
A
0.8 m
NA D 5104j (N),
x
NB D 5027j (N),
0.8 m
NC D 3613j (N),
D
1.4 m
1.4 m
B
z
ND D 3559j (N).
y
If you represent these forces by a single equivalent force
N, what is N, and where does its line of action intersect
the xz plane?
Solution: We must satisfy the following three equations
Fy :5104 N C 5027 N C 3613 N C 3559 N D Ry
x
Mx :5104 N C 3613 N0.8 m
5027 N C 3559 N0.8 m D Ry z
Mz :5104 N C 5027 N1.4 m
3613 N C 3559 N1.4 m D Ry x
Solving we find
Ry D 17303 N, x D 0.239 m, z D 0.00606 m
Problem 4.156 Two forces act on the beam. If you
represent them by a force F acting at C and a couple M,
what are F and M?
Solution: The equivalent force must equal the sum of forces: F D
100j C 80k. The equivalent couple is equal to the moment about C:
M D 380j 3100k D 240j 300k
y
100 N
80 N
z
C
x
3m
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?
b
Pi
z
Solution: The equivalent force at the origin is equal to the applied
force F D Pi. The position vector of the applied force is r D hj C bk.
The moment is
i
M D r &eth; P D 0
P
j
h
0
h
O
k Cb D bPj C hPk.
0 x
This is the couple at the origin.
y
(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)
(b)
If you represent the forces acting on the brace by
a force F acting at the origin O and a couple M,
what are F and M?
If you represent the forces acting on the brace by
a force F0 acting at a point P with coordinates
xP , yP , zP and a couple M0 , what are F0 and M0 ?
Solution: (a) Equivalent force at the origin O has the same value
as the sum of forces,
y
h
r
h
B
z
O
1
A
2
A
B
x
1
A
2
FX D B Bi D 0,
FY D A C 12 A C 12 A j D 0,
thus F D 0. The equivalent couple moment has the same value as the
moment exerted on the brace by the forces,
MO D rAi.
Thus the couple at O has the moment M D rAi. (b) The equivalent
force at xP , yP , zP has the same value as the sum of forces on the
brace, and the equivalent couple at xP , yP , zP has the same moment
as the moment exerted on the brace by the forces: F D 0, M D rAi.
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245
y
Problem 4.159 Two forces and a couple act on the
cube. If you represent them by a force F acting at point
P and a couple M, what are F and M?
P
FB =
2i – j (kN)
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).
FA =
– i + j + k (kN)
x
MC =
4i – 4j + 4k (kN-m)
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:
1m
z
rA D i j C k, and rB D Ck. The moment of the couple:
M D rA &eth; FA C rB &eth; FB C MC
i
D 1
1
j k i
1 1 C 0
1 1 2
j k 0 1 C MC
1 0 D 3i 2j C 2k (kN-m).
Problem 4.160 The two shafts are subjected to the
torques (couples) shown.
If you represent the two couples by a force F acting
at the origin O and a couple M, what are F and M?
(b) What is the magnitude of the total moment exerted
by the two couples?
y
(a)
6 kN-m
4 kN-m
40&deg;
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:
30&deg;
x
z
M D M1 C M2 .
The sense of M1 is (see sketch) negative with respect to both y and
z, and the sense of M2 is positive with respect to both x and y.
M1 D 4j sin 30&deg; k cos 30&deg; D 2j 3.464k,
M2 D 6i cos 40&deg; C j sin 40&deg; D 4.5963i C 3.8567j.
Thus the couple at the origin is MO D 4.6i C 1.86j 3.46k (kN-m)
(b) The magnitude
of the total moment exerted by the two couples is
p
jMO j D 4.62 C 1.862 C 3.462 D 6.05 (kN-m)
246
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Problem 4.161 The two systems of forces and
moments acting on the bar are equivalent. If
y
FA D 30i C 30j 20k (kN),
FA
FB D 40i 20j C 25k (kN),
z
A
2m
MB D 10i C 40j 10k (kN-m),
MB
2m
B
what are F and M?
x
FB
System 1
y
Solution:
F
F D FA C FB D 70i C 10j C 5k kN
z
M D 2 mi &eth; FA C 4 mi &eth; FB C MB
x
M
D 10i 20j 30k kNm
System 2
Problem 4.162 Point G is at the center of the block.
The forces are
FA D 20i C 10j C 20k (lb),
Solution: The equivalent force is the sum of the forces:
F D 20i C 10 C 10j C 20 10k
D 20i C 20j C 10k (lb).
FB D 10j 10k (lb).
The equivalent couple is the sum of the moments about G. The position
vectors are:
If you represent the two forces by a force F acting at G
and a couple M, what are F and M?
rA D 15i C 5j C 10k (in),
rB D 15i C 5j 10k.
y
The sum of the moments:
MG D rA &eth; FA C rB &eth; FB FB
FA
10 in
x
G
i
D 15
20
j
k i
j
5 10 C 15 5
10 20 0 10
k 10 10 D 50i C 250j C 100k (in lb)
20 in
z
30 in
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247
Problem 4.163 The engine above the airplane’s fuselage exerts a thrust T0 D 16 kip, and each of the engines
under the wings exerts a thrust TU D 12 kip. The dimensions are h D 8 ft, c D 12 ft, and b D 16 ft. If you represent the three thrust forces by a force F acting at the
origin O and a couple M, what are F and M?
y
T0
c
O
z
h
2 TU
y
Solution: The equivalent thrust at the point G is equal to the sum
of the thrusts:
T D 16 C 12 C 12 D 40 kip
b
The sum of the moments about the point G is
x
O
b
M D r1U &eth; TU C r2U &eth; TU C rO &eth; TO D r1U C r2U &eth; TU C rO &eth; TO .
The position vectors are r1U D Cbi hj, r2U D bi hj, and rO D
Ccj. For h D 8 ft, c D 12 ft, and b D 16 ft, the sum of the moments
is
i
M D 0
0
j
k i
16 0 C 0
0
12 0
j
12
0
k 0 D 192 C 192i D 0.
16 Thus the equivalent couple is M D 0
Problem 4.164 Consider the airplane described in
Problem 4.163 and suppose that the engine under the
wing to the pilot’s right loses thrust.
Solution: The sum of the forces is now
(a)
The sum of the moments is now:
If you represent the two remaining thrust forces by
a force F acting at the origin O and a couple M,
what are F and M?
(b) If you represent the two remaining thrust forces
by the force F alone, where does its line of action
intersect the xy plane?
F D 12 C 16 D 28k (kip).
M D r2U &eth; TU C rO &eth; TO .
For h D 8 ft, c D 12 ft, and b D 16 ft, using the position vectors for
the engines given in Problem 4.163, the equivalent couple is
i
M D 16
0
j
k i
8 0 C 0
0 12 0
j
k 12 0 D 96i 192j (ft kip)
0 16 (b) The moment of the single force is
i j
M D x y
0 0
k z D 28yi 28xj D 96i 192j.
28 From which
xD
96
192
D 6.86 ft, and y D
D 3.43 ft.
28
28
As to be expected, z can have any value, corresponding to any point
on the line of action. Arbitrarily choose z D 0, so that the coordinates
of the point of action are (6.86, 3.43, 0).
248
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Problem 4.165 The tension in cable AB is 100 lb, and
the tension in cable CD is 60 lb. Suppose that you want
to replace these two cables by a single cable EF so that
the force exerted on the wall at E is equivalent to the
two forces exerted by cables AB and CD on the walls
at A and C. What is the tension in cable EF, and what
are the coordinates of points E and F?
y
y
C
(4, 6, 0) ft
(0, 6, 6) ft
E
x
A
x
D
(7, 0, 2) ft
B
F
(3, 0, 8) ft
z
z
Solution: The position vectors of the points A, B, C, and D are
For the systems to be equivalent, the moments about the origin must
be the same. The moments about the origin are
rA D 0i C 6j C 6k,
rB D 3i C 0j C 8k,
MO D rA &eth; FA C rC &eth; FC i
D 0
42.86
rC D 4i C 6j C 0k, and
j
6
85.71
k i
6 C 4
28.57 25.72
j
6
51.43
k 0 17.14 rD D 7i C 0j C 2k.
D 788.57i C 188.57j 617.14k.
The unit vectors parallel to the cables are obtained as follows:
rAB D rB rA D 3i 6j C 2k,
jrAB j D
p
32 C 62 C 22 D 7,
This result is used to establish the coordinates of the point E. For the
one cable system, the end E is located at x D 0. The moment is
M1 D r &eth; FEF
from which
i
D 0
68.58
j
y
137.14
k z 45.71 eAB D 0.4286i 0.8571j C 0.2857k.
D 45.71y C 137.14zi C 68.58zj 68.58yk
rCD D rD rC D 3i 6j C 2k,
D 788.57i C 188.57j 617.14k,
jrCD j D
p
32 C 62 C 22 D 7,
from which
eCD D 0.4286i 0.8571j C 0.2857k.
Since eAB D eCD , the cables are parallel . To duplicate the force, the
single cable EF must have the same unit vector.
The force on the wall at point A is
FA D 100eAB D 42.86i 85.71j C 28.57k (lb).
The force on the wall at point C is
from above. From which
yD
617.14
D 8.999 . . . D 9 ft
68.58
zD
188.57
D 2.75 ft.
68.58
Thus the coordinates of point E are E (0, 9, 2.75) ft. The coordinates
of the point F are found as follows: Let L be the length of cable
EF. Thus, from the definition of the unit vector, yF yE D Ley with
9
D 10.5 ft. The other coordithe condition that yF D 0, L D
0.8571
nates are xF xE D LeX , from which xF D 0 C 10.50.4286 D 4.5 ft
zF zE D LeZ , from which zF D 2.75 C 10.50.2857 D 5.75 ft The
coordinates of F are F (4.5, 0, 5.75) ft
FC D 60eCD D 25.72i 51.43j C 17.14k (lb).
The total force is
FEF D 68.58i 137.14j C 45.71k (lb),
jFEF j D 160 lb.
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249
y
Problem 4.166 The distance s D 4 m. If you represent
the force and the 200-N-m couple by a force F acting at
origin O and a couple M, what are F and M?
(2, 6, 0) m
s
Solution: The equivalent force at the origin is
F D 100i C 20j 20k.
100i ⫹ 20j ⫺ 20k (N)
O
The strategy is to establish the position vector of the action point of
the force relative to the origin O for the purpose of determining the
moment exerted by the force about the origin. The position of the top
of the bar is
200 N-m
x
(4, 0, 3) m
z
rT D 2i C 6j C 0k. The vector parallel to the bar, pointing toward the
base, is rTB D 2i 6j C 3k, with a magnitude of jrTB j D 7. The unit
vector parallel to the bar is
eTB D 0.2857i 0.8571j C 0.4286k.
The vector from the top of the bar to the action point of the force is
rTF D seTB D 4eTB D 1.1429i 3.4286j C 1.7143k.
The position vector of the action point from the origin is
rF D rT C rTF D 3.1429i C 2.5714j C 1.7143k.
The moment of the force about the origin is
i
MF D r &eth; F D 3.1429
100
j
2.5714
20
k 1.7143 20 D 85.71i C 234.20j 194.3k.
The couple is obtained from the unit vector and the magnitude. The
sense of the moment is directed positively toward the top of the bar.
MC D 200eTB D 57.14i C 171.42j 85.72k.
The sum of the moments is
M D MF C MC D 142.86i C 405.72j 280k.
This is the moment of the equivalent couple at the origin.
250
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Problem 4.167 The force F and couple M in system
1 are
System 1
System 2
y
F D 12i C 4j 3k (lb),
y
M
M D 4i C 7j C 4k (ft-lb).
F
Suppose you want to represent system 1 by a
wrench (system 2). Determine the couple Mp and the
coordinates x and z where the line of action of the force
intersects the xz plane.
O
z
O
x
z
Mp
F
x
(x, 0, z)
Solution: The component of M that is parallel to F is found as
follows: The unit vector parallel to F is
eF D
F
D 0.9231i C 0.3077j 0.2308k.
jFj
The component of M parallel to F is
MP D eF &ETH; MeF D 4.5444i C 1.5148j 1.1361k (ft-lb).
The component of M normal to F is
MN D M MP D 0.5444i C 5.4858j C 5.1361k (ft-lb).
The moment of F must produce a moment equal to the normal component of M. The moment is
i
MF D r &eth; F D x
12
k z D 4zi C 3x C 12zj C 4xk,
3 j
0
4
from which
zD
0.5444
D 0.1361 ft
4
xD
5.1362
D 1.2840 ft
4
Problem 4.168 A system consists of a force F acting
at the origin O and a couple M, where
F D 10i (lb),
M D 20j (ft-lb).
If you represent the system by a wrench consisting of
the force F and a parallel couple Mp , what is Mp , and
where does the line of action F intersect the yz plane?
Solution: The component of M parallel to F is zero, since MP D
eF &ETH; MeF D 0. The normal component is equal to M. The equivalent
force must produce the same moment as the normal component
i j
M D r &eth; F D 0 y
10 0
from which z D
k z D 10zj 10yk D 20j,
0
20
D 2 ft and y D 0
10
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251
Problem 4.169 A system consists of a force F acting
at the origin O and a couple M, where
F D i C 2j C 5k (N),
M D 10i C 8j 4k (N-m).
If you represent it by a wrench consisting of the force F
and a parallel couple Mp , (a) determine Mp , and determine where the line of action of F intersects (b) the xz
plane, (c) the yz plane.
Solution: The unit vector parallel to F is
eF D
F
D 0.1826i C 0.3651j C 0.9129k.
jFj
from which
zD
9.8
5
D 4.9 m, and x D
D 2.5 m
2
2
(a) The component of M parallel to F is
(c) The intersection with the yz plane is
MP D eF &ETH; MeF D 0.2i C 0.4j C 1.0k (N-m).
i
MN D r &eth; F D 0
1
The normal component is
j
y
2
k z D 5y 2zi C zj yk
5
D 9.8i C 7.6j 5k,
MN D M MP D 9.8i C 7.6j 5k.
The moment of the force about the origin must be equal to the normal
component of the moment. (b) The intersection with the xz plane:
i j
MN D r &eth; F D x 0
1 2
k z D 2zi 5x zj C 2xk
5
from which
y D 5 m and z D 7.6 m
D 9.8i C 7.6j 5k,
Problem 4.170 Consider the force F acting at the origin
O and the couple M given in Example 4.15. If you represent this system by a wrench, where does the line of action
of the force intersect the xy plane?
6j C 2k (N), and M D 12i C 4j C 6k (N-m).
The normal component of the moment is
MN D 7.592i 4.816j C 3.061k (N-m).
y
The moment produced by the force must equal the normal component:
F
i j
MN D r &eth; F D x y
3 6
M
O
Solution: From Example 4.15 the force and moment are F D 3i C
k 0 2
x
D 2yi 2xj C 6x 3yk D 7.592i 4.816j C 3.061k,
z
from which
xD
252
4.816
7.592
D 2.408 m and y D
D 3.796 m
2
2
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 4.171 Consider the force F acting at the origin O and the couple M given in Example 4.15. If you
represent this system by a wrench, where does the line
of action of the force intersect the plane y D 3 m?
is F D 3i C 6j C 2k, and the normal component of the moment is
MN D 7.592i 4.816j C 3.061k.
The moment produced by the force must be equal to the normal component:
i
MN D r &eth; F D x
3
j k 3 z D 6 6zi 2x 3zj C 6x 9k
6 2
D 7.592i 4.816j C 3.061k,
from which
xD
Problem 4.172 A wrench consists of a force of magnitude 100 N acting at the origin O and a couple of magnitude 60 N-m. The force and couple point in the direction
from O to the point (1, 1, 2) m. If you represent the
wrench by a force F acting at point (5, 3, 1) m and a
couple M, what are F and M?
9 C 3.061
6 7.592
D 2.01 m and z D
D 0.2653 m
6
6
Solution: The vector parallel to the force is rF D i C j C 2k, from
which the unit vector parallel to the force is eF D 0.4082i C 0.4082j C
0.8165k. The force and moment at the origin are
F D jFjeOF D 40.82i C 40.82j C 81.65k (N), and
M D 24.492i C 24.492j C 48.99k (N-m).
The force and moment are parallel. At the point (5, 3, 1) m the equivalent force is equal to the force at the origin, given above. The moment
of this force about the origin is
i
MF D r &eth; F D 5
40.82
j
3
40.82
k 1 81.65 D 204.13i 367.43j C 81.64k.
For the moments to be equal in the two systems, the added equivalent
couple must be
MC D M MF D 176.94i C 391.92j 32.65k (N-m)
Problem 4.173 System 1 consists of two forces and
a couple. Suppose that you want to represent it by a
wrench (system 2). Determine the force F, the couple
Mp , and the coordinates x and z where the line of action
of F intersects the xz plane.
Solution: The sum of the forces in System 1 is F D 300j C
600k (N). The equivalent force in System 2 must have this value.
The unit vector parallel to the force is eF D 0.4472j C 0.8944k. The
sum of the moments in System 1 is
M D 6003i C 3004k C 1000i C 600j
System 1
y
System 2
D 2800i C 600j C 1200k (kN m).
y
1000i + 600j (kN-m)
The component parallel to the force is
MP D 599.963j C 1199.93k (kN-m) D 600j C 1200k (kN-m).
600k (kN)
3m
300j (kN)
Mp
x
x
4m
z
F
z
(x, 0, z)
The normal component is MN D M MP D 2800i. The moment of
the force
i
MN D x
0
j
0
300
k z D 300zi 600xj C 300xk D 2800i,
600 from which
x D 0, z D
2800
D 9.333 m
300
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
253
y
Problem 4.174 A plumber exerts the two forces shown
to loosen a pipe.
What total moment does he exert about the axis of
the pipe?
(b) If you represent the two forces by a force F acting
at O and a couple M, what are F and M?
(c) If you represent the two forces by a wrench
consisting of the force F and a parallel couple Mp ,
what is Mp , and where does the line of action of F
intersect the xy plane?
12 in
6 in
O
(a)
z
x
16 in
16 in
50 k (lb)
–70 k (lb)
Solution: The sum of the forces is
(a)
F D 50k 70k D 20k (lb).
The total moment exerted on the pipe is
M D 1620i D 320i (ft lb).
(b)
The equivalent force at O is F D 20k. The sum of the moments
MO D r1 &eth; F1 C r2 &eth; F2 i
D 12
0
j
16
0
k i
0 C 18
50 0
j
k 16
0 0
70 D 320i C 660j.
(c)
The unit vector parallel to the force is eF D k, hence the moment
parallel to the force is MP D eF &ETH; MeF D 0, and the moment
normal to the force is MN D M MP D 320i C 660j. The force
at the location of the wrench must produce this moment for the
wrench to be equivalent.
i j
MN D x y
0 0
k 0 D 20yi C 20xj D 320i C 660j,
20 from which x D
254
660
320
D 33 in, y D
D 16 in
20
20
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Problem 4.175 The Leaning Tower of Pisa is approximately 55 m tall and 7 m in diamteter. The horizontal
displacement of the top of the tower from the vertical
is approximately 5 m. Its mass is approximately 3.2 &eth;
106 kg. If you model the tower as a cylinder and assume
that its weight acts at the center, what is the magnitude
of the moment exerted by the weight about the point at
the center of the tower’s base?
5m
Solution: The unstretched length of spring is 1 m and the spring
N
. Assume that the bar is a quarter circle, with
m
a radius of 4 m. The stretched length of the spring is found from
the Pythagorean Theorem: The height of the attachment point is h D
4 sin ˛ m, and the distance from the center is 4 cos ˛. The stretched
length of the spring is
constant is k D 20
LD
4m
B
k
3m
α
3 h2 C 4 cos ˛2 m.
A
The spring force is F D 20L 1 N. The angle that the spring
makes with a vertical line parallel to A is
ˇ D tan1
3h
4 cos ˛
Moment at B as function of alpha
120
.
The horizontal component of the spring force is FX D F cos ˇ N. The
vertical component of the force is FY D F sin ˇ N. The displacement
of the attachment point to the left of point A is d D 41 cos ˛ m,
hence the action of the vertical component is negative, and the action
of the horizontal component is positive. The moment about A is
MA D dFY C hFX .
M
o
m
e
n
t
,
100
80
60
40
Collecting terms and equations,
N
m
20
h D 4 sin ˛ m,
0
FY D F sin ˇ N,
0
10
20
30
40
50
60
70
80
90
Alpha, deg
FX D F cos ˇ N,
F D 20L 1 N,
LD
3 h2 C 4 cos ˛2 m,
ˇ D tan1
3h
4 cos ˛
.
A programmable calculator or a commercial package such as TK
Solver or Mathcad is almost essential to the solution of this and
the following problems. The commercial package TK Solver PLUS
was used here to plot the graph of M against ˛. Using the graph as a
guide, the following tabular values were taken about the maximum:
˛, deg
Moment, N-m
41.5
42.0
42.5
101.463
101.483
101.472
The maximum value of the moment is estimated at MB D 101.49 N-m,
which occurs at approximately ˛ D 42.2&deg;
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255
y
Problem 4.176 The cable AB exerts a 300-N force on
the support A that points from A toward B. Determine the
magnitude of the moment the force exerts about point P.
B
(0.3, 0.6) m
A
(⫺0.4, 0.3) m
Solution:
F D 300 N
x
P
(0.5, ⫺0.2) m
0.7i C 0.3j
p
,
0.58
rPA D 0.9i C 0.5j m
MP D rPA &eth; F D 244 Nmk )
MP D 244 Nm
Problem 4.177 Three forces act on the structure. The
sum of the moments due to the forces about A is zero.
Determine the magnitude of the force F.
30⬚
45⬚
2 kN
4 kN
Solution:
b
p
MA D 4 kN 2b 2 kN cos 30&deg; 3b
A
F
C 2 kN sin 30&deg; b C F4b D 0
2b
b
b
Solving we find
F D 2.463 kN
Problem 4.178 Determine the moment of the 400-N
30&deg;
400 N
220 mm
A
Solution: Use the two dimensional description of the moment. The
vertical and horizontal components of the 200 N force are
260 mm
FY D 400 sin 30&deg; D 200 N,
FX D C400 cos 30&deg; D 346.41 N.
(a)
B
500 mm
The moment arm from A to the line of action of the horizontal
component is 0.22 m. The moment arm from A to the vertical
component is zero. The moment about A is negative,
MA D 0.22346.41 D 76.21 N-m
(b)
256
The perpendicular distances to the lines of action of the vertical
and horizontal components of the force from B are d1 D 0.5 m,
and d2 D 0.48 m. The action of the vertical component is positive, and the action of the horizontal component is negative.
The sum of the moments: MB D C0.5200 0.48346.41 D
66.28 N-m
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Problem 4.179 Determine the sum of the moments
exerted about A by the three forces and the couple.
A
5 ft
300 lb
800 ft-lb
Solution: Establish coordinates with origin at A, x horizontal, and
y vertical with respect to the page. The moment exerted by the couple
is the same about any point. The moment of the 300 lb force about A
is M300 D 6i 5j &eth; 300j D 1800k ft-lb.
200 lb
200 lb
6 ft
The moment of the downward 200 lb force about A is zero since the
line of action of the force passes through A. The moment of the 200 lb
force which pulls to the right is
3 ft
M200 D 3i 5j &eth; 200i D 1000k (ft-lb).
The moment of the couple is MC D 800k (ft-lb). Summing the four
moments, we get
MA D 1800 C 0 C 1000 800k D 1600k (ft-lb)
Problem 4.180 In Problem 4.179, if you represent the
three forces and the couple by an equivalent system
consisting of a force F acting at A and a couple M,
what are the magnitudes of F and M?
Solution: The equivalent force will be equal to the sum of the
forces and the equivalent couple will be equal to the sum of the
moments about A. From the solution to Problem 4.189, the equivalent couple will be C D MA D 1600k (ft-lb). The equivalent force
will be FEQUIV. D 200i 200j C 300j D 200i C 100j (lb)
Problem 4.181 The vector sum of the forces acting on
the beam is zero, and the sum of the moments about A
is zero.
(a)
(b)
30&deg;
220 mm
What are the forces Ax , Ay , and B?
What is the sum of the moments about B?
400 N
Ay
Ax
260 mm
Solution: The vertical and horizontal components of the 400 N
force are:
FX D 400 cos 30&deg; D 346.41 N,
500 mm
B
FY D 400 sin 30&deg; D 200 N.
The sum of the forces is
FX D AX C 346.41 D 0,
from which AX D 346.41 N
FY D AY C B 200 D 0.
The sum of the moments about A is
MA D 0.5B 0.22346.41 D 0,
from which B D 152.42 N. Substitute into the force equation to get
AY D 200 B D 47.58 N
(b) The moments about B are
MB D 0.5AY 0.48346.41 0.26AX C 0.5200 D 0
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257
Problem 4.182 The hydraulic piston BC exerts a 970lb force on the boom at C in the direction parallel to the
piston. The angle ˛ D 40&deg; . The sum of the moment about
A due to the force exerted on the boom by the piston
and the weight of the suspended load is zero. What is
the weight of the suspended load?
t
6f
t
9f
Solution: The horizontal (x) and vertical (y) coordinates of point
C relative to point B are
C
a
A
x D 9 ft cos ˛ 6 ft D 0.894 ft
B
6 ft
y D 9 ft sin ˛ D 5.79 ft
The angle between the piston BC and the horizontal is
ˇ D tan1 y/x D 81.2&deg;
The horizontal and vertical components of the force exerted by the
piston at C are
Cx D 970 lb cos ˇ D 148 lb
Cy D 970 lb sin ˇ D 959 lb
The sum of the moments about A due to the pistion force and the
suspended weight W is
MA D W15 ft cos ˛ C Cy 9 ft cos a Cx 9 ft sin ˛ D 0
Solving, yields W D 501 lb
y
Problem 4.183 The force F D 60i C 60j (lb).
F
(a) Determine the moment of F about point A.
(b) What is the perpendicular distance from point A to
the line of action of F?
(4, – 4, 2) ft
x
A
(8, 2, 12) ft
z
Solution: The position vector of A and the point of action are
(b)
The magnitude of the moment is
p
6002 C 6002 C 6002 D 1039.3 ft lb.
rA D 8i C 2j C 12k (ft), and rF D 4i 4j C 2k.
jMA j D
The vector from A to F is
p
The magnitude of the force is jFj D 602 C 602 D 84.8528 lb.
The perpendicular distance from A to the line of action is
rAF D rF rA D 4 8i C 4 2j C 2 12k
DD
D 4i 6j 10k.
(a)
1039.3
D 12.25 ft
84.8528
i
MA D rAF &eth; F D 4
60
j
k 6 10 60
0 D 600i C 600j 600k (ft lb)
258
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y
Problem 4.184 The 20-kg mass is suspended by cables
attached to three vertical 2-m posts. Point A is at (0, 1.2,
0) m. Determine the moment about the base E due to the
force exerted on the post BE by the cable AB.
C
B
D
A
1m
1m
E
2m
0.3 m
x
z
Solution: The strategy is to develop the simultaneous equations in
the unknown tensions in the cables, and use the tension in AB to find
the moment about E. This strategy requires the unit vectors parallel to
the cables. The position vectors of the points are:
The equilibrium conditions are TAB C TAC C TAD D W. Collect like
terms in i, j, k:
rOA D 1.2j,
FX D 0.2281TAB C 0TAC C 0.9284TAD i D 0
FY D C0.6082 &ETH; TAB C 0.6247 &ETH; TAC
rOB D 0.3i C 2j C 1k,
rOC D 2j 1k,
rOD D 2i C 2j,
rOE D 0.3i C 1k.
C 0.3714 &ETH; TAD 196.2j D 0
FZ D C0.7603 &ETH; TAB 0.7809 &ETH; TAC C 0 &ETH; TAD k D 0
Solve:
TAB D 150.04 N,
The vectors parallel to the cables are:
TAC D 146.08 N,
rAB D rOB rOA D 0.3i C 0.8j C 1k,
rAC D rOC rOA D C0.8j 1k,
rAD D rOD rOA D C2i C 0.8j.
The unit vectors parallel to the cables are:
eAB D
rAB
D 0.2281i C 0.6082j C 0.7603k :
jrAB j
eAC D 0i C 0.6247j 0.7809k,
ME D rEB &eth; TAB eAB D TAB rEB &eth; eAB i
D 150 0
0.2281
j
2
C0.6082
k
0
C0.7603 D 228i 68.43k (N-m)
eAD D C0.9284i C 0.3714j C 0k.
The tensions in the cables are
TAB D TAB eAB ,
TAC D TAC eAC , and
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259
Problem 4.185 What is the total moment due to the
two couples?
(a) Express the answer by giving the magnitude and
stating whether the moment is clockwise or counterclockwise.
(b) Express the answer as a vector.
y
100 N
4m
100 N
2m
x
2m
100 N
100 N
Solution:
(a)
(b)
4m
The couple in which the forces are 4 m apart exerts a counterclockwise moment of magnitude 100 N4 m D 400 N-m.
The couple in which the forces are 8 m apart exerts a clockwise moment of magnitude 100 N8 m D 800 N-m. The sum
of their moments is a clockwise moment of 400 N-m.
The vector representation of the clockwise moment of 400 N-m
magnitude is 400k (N-m). This expression can also be obtained
by calculating the sum of the moments of the four forces about
any point. The sum of the moments about the origin is
M D 2 mi &eth; 100 Nj C 2 mi &eth; 100 Nj
C 4 mj &eth; 100 Ni C 4 mj &eth; 100 Ni
D 400 N-mk
(a) 400 N-m clockwise
(b) 400k N-m
Problem 4.186 The bar AB supporting the lid of the
grand piano exerts a force F D 6i C 35j 12k (lb) at
B. The coordinates of B are (3, 4, 3) ft. What is the
moment of the force about the hinge line of the lid (the
x axis)?
y
Solution: The position vector of point B is rOB D 3i C 4j C 3k.
The moment about the x axis due to the force is
MX D eX &ETH; rOB &eth; F D i &ETH; rOB &eth; F
1
0
MX D 3
4
6 35
0 3 D 153 ft lb
12 B
x
A
z
260
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Problem 4.187 Determine the moment of the vertical
y
800 lb
A (4, 3, 4) ft
B
D (6, 0, 0) ft
Solution: The force vector acting at A is F D 800j (lb) and the
x
position vector from C to A is
rCA D xA xC i C yA yC j C zA zC k
z
C (5, 0, 6) ft
D 4 5i C 3 0j C 4 6k D 1i C 3j 2k (ft).
i
MC D 1
0
j
3
800
k 2 D 1600i C 0j C 800k (ft-lb)
0 Problem 4.188 In Problem 4.187, determine the moment of the vertical 800-lb force about the straight line
through points C and D.
Solution: In Problem 4.197, we found the moment of the 800 lb
force about point C to be given by
MC D 1600i C 0j C 800j (ft-lb).
The vector from C to D is given by
rCD D xD xC i C yD yC j C zD zC k
D 6 5i C 0 0j C 0 6k
D 1i C 0j 6j (ft),
and its magnitude is
jrCD j D
p
p
12 C 62 D 37 (ft).
The unit vector from C to D is given by
6
1
eCD D p i p k.
37
37
The moment of the 800 lb vertical force about line CD is given by
MCD D
D
6
1
p i p k &ETH; 1600i C 0j C 800j (ft-lb)
37
37
1600 4800
p
37
(ft-lb).
Carrying out the calculations, we get MCD D 1052 (ft-lb)
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
261
Problem 4.189 The system of cables and pulleys supports the 300-lb weight of the work platform. If you
represent the upward force exerted at E by cable EF
and the upward force exerted at G by cable GH by a
single equivalent force F, what is F, and where does its
line of action intersect the x axis?
Solution: The cable-pulley combination does not produce a moment. Hence the equivalent force does not. The equivalent force is
600
j D 300j (lb). The
equal to the total supported weight, or F D C
2
8
force occurs at midpoint of the platform width, x D D 4 ft
2
H
F
E
B
y
A
G
60&deg;
D
60&deg;
C
x
8 ft
Problem 4.190 Consider the system in Problem 4.189.
Solution: The vertical component of the tension is each cable must
equal half the weight supported.
(a) What are the tensions in cables AB and CD?
(b) If you represent the forces exerted by the cables at
A and C by a single equivalent force F, what is
F, and where does its line of action intersect the
x axis?
262
TAB sin 60&deg; D 150 lb, from which TAB D
symmetry, the tension TCD D 173.2 lb.
150
D 173.2 lb. By
sin 60&deg;
The single force must equal the sum of the vertical components; since
there is no resultant moment produced by the cables, the force is
F D 300j lb and it acts at the platform width midpoint x D 4 ft.
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Problem 4.191 The two systems are equivalent. Determine the forces Ax and Ay , and the couple MA .
System 1
y
20 N
400 mm
Solution: The sum of the forces for System 1 is
Ax
FX D AX C 20i,
x
30 N
Ay
FY D AY C 30j.
600 mm
400 mm
The sum of forces for System 2 is
System 2
y
FX D 20i and
8 N-m
FY D 80 10j.
Equating the two systems:
400 mm
AX C 20 D 20 from which AX D 40 N
20 N
MA
10 N
x
AY C 30 D 80 10 from which AY D 40 N
80 N
The sum of the moments about the left end for System 1 is
600 mm
400 mm
M1 D 0.420 C 301 D 22 N-m.
The sum of moments about the left end for System 2 is
M2 D MA 101 8 D MA 18.
Equating the moments for the two systems:
MA D 18 C 22 D 40 N-m
Problem 4.192 If you represent the equivalent systems
in Problem 4.191 by a force F acting at the origin and
a couple M, what are F and M?
Solution: Summing the forces in System 1, F D AX C 20i C
AY C 30j. Substituting from the solution in Problem 4.201,
F D 20i C 70j. The moment is M D 200.4k C 30k D 22k (N-m)
Problem 4.193 If you represent the equivalent systems
in Problem 4.191 by a force F, what is F, and where does
its line of action intersect the x axis?
Solution: The force is F D 20i C 70j. The moment to be represented is
i
M D r &eth; F D 22k D x
20
from which x D
j k 0 0 D 70xk,
70 0 22
D 0.3143 m
70
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263
Problem 4.194 The two systems are equivalent. If
Solution: The sum of forces in the two systems must be equal,
thus F0 D F D 100i C 40j C 30k (lb).
F D 100i C 40j C 30k (lb),
The moment for the unprimed system is MT D r &eth; F C M.
0
M D 80i C 120j C 40k (in-lb),
The moment for the primed system is M0T D r0 &eth; F C M0 .
determine F0 and M.
The position vectors are r D 0i C 6j C 6k, and r0 D 4i C 6j C 6k.
Equating the moments and solving for the unknown moment
System 1
y
System 2
y
M
MD
F
4 in
M0
C r0
i
r &eth; F D 80i C 120j C 40k C 4
100
j
0
40
k 0 30 4 in
F'
D 80i C 200k (in-lb)
M'
6 in
D 80i C 120j C 40k 120j C 160k
6 in
x
x
6 in
6 in
z
z
Problem 4.195 The tugboats A and B exert forces
FA D 1 kN and FB D 1.2 kN on the ship. The angle
D 30&deg; . If you represent the two forces by a force F
acting at the origin O and a couple M, what are F
and M?
y
A
FA
Solution: The sums of the forces are:
FX D 1 C 1.2 cos 30&deg; i D 2.0392i (kN)
60 m
O
x
FY D 1.2 sin 30&deg; j D 0.6j (kN).
60 m
FB
The equivalent force at the origin is
FEQ D 2.04i C 0.6j
θ
B
The moment about O is MO D rA &eth; FA C rB &eth; FB . The vector positions are
25 m
rA D 25i C 60j (m), and
rB D 25i 60j (m).
The moment:
i
MO D 25
1
j k i
60 0 C 25
0 0 1.0392
j
k 60 0 0.6 0 D 12.648k D 12.6k (kN-m)
Check: Use a two dimensional description: The moment is
MO D 25FB sin 30&deg; C 60FB cos 30&deg; 60FA D 39.46FB 60FA D 12.6 kN-m
264
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 4.196 The tugboats A and B in Problem 4.195
exert forces FA D 600 N and FB D 800 N on the ship.
The angle D 45&deg; . If you represent the two forces by
a force F, what is F, and where does its line of action
intersect the y axis?
Solution: The equivalent force is
Check: Use a two dimensional description:
F D 0.6 C 0.8 cos 45&deg; i C 0.8 sin 45&deg; j D 1.1656i C 0.5656j (kN).
MO D 25FB sin 45&deg; C 60FB cos 45&deg; 60FA
The moment produced by the two forces is
D 24.75FB 60FA D 16.20 kN-m.
The single force must produce this moment.
MO D rA &eth; FA C rB &eth; FB .
rA D 25i C 60j (m), and rB D 25i 60j (m).
i
MO D 0
1.1656
The moment:
from which
The vector positions are
j k i
60 0 C 25
0 0 0.5656
i
MO D 25
0.6
j
k 60
0 D 16.20k (kN-m)
0.5656 0 yD
j
k y
0 D 1.1656yk D 16.20k,
0.5656 0 16.20
D 13.90 m
1.1656
Problem 4.197 The tugboats A and B in Problem 4.195
want to exert two forces on the ship that are equivalent
to a force F acting at the origin O of 2-kN magnitude. If
FA D 800 N, determine the necessary values of FB and
angle .
Solution: The equivalent force at the origin is FA C FB cos 2 C
sin 2
D
FB
must be zero:
20002 .
The moment about the origin due to FA and FB
MO D 60FA C 60FB cos 25FB sin D 0.
These are two equations in two unknowns FB sin and FB cos . For
brevity write x D FB cos , y D FB sin , so that the two equations
become x2 C 2FA x C F2A C y 2 D 20002 and 60x 25y 60FA D 0.
Eliminate y by solving each equation for y2 and equating the results:
y 2 D 20002 x 2 2FA x F2A D
60
60
FA C
x
25
25
Reduce to obtain the quadratic in x:
1C
60
25
2 C 1C
x 2 C 2FA 1 60
25
60
25
Substitute FA D 800 N to obtain 6.76x2 7616x C 326400 D 0. In
canonical form: x2 C 2bx C c D 0, where
p b D 563.31, and c D
48284.0, with the solutions x D b š b2 c D 1082.0, D 44.62.
From the second equation, y D 1812.9, D 676.81. The force FB has
two solutions: Solve for FB and : (1)
FB D
p
44.62 C 1812.92 D 1813.4 N
at the angle
D tan1
2
.
FB D
1812.9
44.6
D 88.6&deg; , and 2
p
676.82 C 1082.02 D 1276.2 N,
at the angle
2 x
D tan1
676.8
1082.0
D 32.0&deg;
2 F2A 20002 D 0.
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
265
y
Problem 4.198 If you represent the forces exerted by
the floor on the table legs by a force F acting at the
origin O and a couple M, what are F and M?
2m
1m
Solution: The sum of the forces is the equivalent force at the
origin. F D 50 C 48 C 50 C 42j D 190j (N). The position vectors of
the legs are, numbering the legs counterclockwise from the lower left
in the sketch:
50 N
x
r1 D C1k,
z
42 N
48 N
r2 D 2i C 1k,
50 N
r3 D 2i,
r4 D 0.
The sum of the moments about the origin is
i
MO D 0
0
j
0
48
k i
1 C 2
0 0
j
0
50
k i
1 C 2
0 0
j k 0 0 42 0 D 98i C 184k (N-m).
This is the couple that acts at the origin.
Problem 4.199 If you represent the forces exerted by
the floor on the table legs in Problem 4.198 by a force
F, what is F, and where does its line of action intersect
the xz plane?
Solution: From the solution to Problem 4.198 the equivalent force
is F D 190j. This force must produce the moment M D 98i C 184k
obtained in Problem 4.198.
i
M D x
0
j
k 0
z D 190zi C 190xk D 98i C 184k,
190 0 from which
266
xD
184
D 0.9684 m and
190
zD
98
D 0.5158 m
190
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Problem 4.200 Two forces are exerted on the crankshaft by the connecting rods. The direction cosines of
FA are cos x D 0.182, cos y D 0.818, and cos z D
0.545, and its magnitude is 4 kN. The direction cosines
of FB are cos x D 0.182, cos y D 0.818, and cos z D
0.545, and its magnitude is 2 kN. If you represent the
two forces by a force F acting at the origin O and a
couple M, what are F and M?
Solution: The equivalent force is the sum of the forces:
FA D 40.182i C 0.818j C 0.545k
D 0.728i C 3.272j C 2.18k (kN)
FB D 20.182i C 0.818j 0.545k D 0.364iC1.636j1.09k (kN).
The sum: FA C FB D 0.364i C 4.908j C 1.09k (kN)
y
FB
The equivalent couple is the sum of the moments. M D rA &eth; FA C
rB &eth; FB . The position vectors are:
FA
rA D 0.16i C 0.08k,
360 mm
O
rB D 0.36i 0.08k.
The sum of the moments:
z
160 mm
80 mm
80 mm x
i
M D 0.16
0.728
j
0
3.272
k i
0.08 C 0.36
2.180 0.364
j
0
1.636
k 0.08 1.090 M D 0.1309i 0.0438j C 1.1125k (kN-m)
Problem 4.201 If you represent the two forces exerted
on the crankshaft in Problem 4.200 by a wrench consisting of a force F and a parallel couple Mp , what are F
and Mp , and where does the line of action of F intersect
the xz plane?
Solution: From the solution to Problem 4.200,
F D 0.364i C 4.908j C 1.09k (kN) and
M D 0.1309i 0.0438j C 1.1125k (kN-m).
The unit vector parallel to F is
eF D
F
D 0.0722i C 0.9737j C 0.2162k.
jFj
The moment parallel to the force is
MP D eF &ETH; MeF .
Carrying out the operations:
MP D 0.2073eF D 0.01497i C 0.2019j C 0.0448k (kN-m).
This is the equivalent couple parallel to F.
The component of the moment perpendicular to F is
MN D M MP D 0.1159i 0.2457j C 1.0688k.
The force exerts this moment about the origin.
i
MN D x
0.364
j
0
4.908
k z 1.09 D 4.908zi 1.09x C 0.364zj C 4.908xk
D 0.1159i 0.2457j C 1.06884k.
From which
xD
1.0688
D 0.2178 m,
4.908
zD
0.1159
D 0.0236 m
4.908
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267
Problem 5.1 In Active Example 5.1, suppose that the
beam is subjected to a 6kN-m counterclockwise couple
at the right end in addition to the 4-kN downward force.
Draw the free-body diagram of the beam and apply the
equilibrium equations to determine the reactions at A.
4 kN
A
2m
Solution: The equilibrium equations are
Fx : Ax D 0
Fy : Ay 4 kN D 0
MA : MA 4 kN2 m
C 6 kN-m D 0
Solving yields
Ax D 0
Ay D 4 kN
MA D 2 kN-m
Problem 5.2 The beam has a fixed support at A and is
loaded by two forces and a couple. Draw the free-body
diagram of the beam and apply equilibrium to determine
the reactions at A.
4 kN
A
2 kN
6 kN-m
60⬚
1m
1.5 m
1.5 m
Solution: The free-body diagram is drawn.
The equilibrium equations are
Fx : Ax C 2 kN cos 60&deg; D 0
Fy : Ay C 4 kN C 2 kN sin 60&deg; D 0
MA : MA C 6 kN-m C 4 kN2.5 m C 2 kN sin 60&deg; 4 m D 0
We obtain: Ax D 1 kN, Ay D 5.73 kN, MA D 22.9 kN-m
268
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Problem 5.3 The beam is subjected to a load F D
400 N and is supported by the rope and the smooth
surfaces at A and B.
(a)
(b)
Draw the free-body diagram of the beam.
What are the magnitudes of the reactions at A
and B?
F
A
B
30&deg;
45&deg;
1.2 m
y
Solution:
C
FX D 0:
A cos 45&deg; B sin 30&deg; D 0
FY D 0:
A sin 45&deg; C B cos 30&deg; T 400 N D 0
MA D 0:
1.2T 2.7400 C 3.7B cos 30&deg; D 0
1.5 m
1m
A
F
45&deg;
x
B
1.5 m
1.2 m
1m
30&deg;
T
Solving, we get
A D 271 N
B D 383 N
T D 124 N
Problem 5.4 (a) Draw the free-body diagram of the
beam. (b) Determine the tension in the rope and the
reactions at B.
30⬚
30⬚ 600 lb
B
A
5 ft
9 ft
Solution: Let T be the tension in the rope.
The equilibrium equations are:
Fx : T sin 30&deg; 600 lb sin 30&deg; C Bx D 0
Fy : T cos 30&deg; 600 lb cos 30&deg; C By D 0
MB : 600 lb cos 30&deg; 9 ft T cos 30&deg; 14 ft D 0
Solving yields T D 368 lb, Bx D 493 lb, By D 186 lb
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
269
Problem 5.5 (a) Draw the free-body diagram of the
60-lb drill press, assuming that the surfaces at A and B
are smooth.
(b) Determine the reactions at A and B.
Solution: The system is in equilibrium.
(a)
(b)
The free body diagram is shown.
The sum of the forces:
FX D 0,
FY D FA C FB 60 D 0
The sum of the moments about point A:
MA D 1060 C 24FB D 0,
from which FB D
60 lb
600
D 25 lb
24
Substitute into the force balance equation:
FA D 60 FB D 35 lb
A
B
10 in
14 in
60 lb
A
B
10 in
FA
270
14 in
FB
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 5.6 The masses of the person and the diving
board are 54 kg and 36 kg, respectively. Assume that
they are in equilibrium.
(a)
(b)
Draw the free-body diagram of the diving board.
Determine the reactions at the supports A and B.
Solution:
(a)
(b)
A
FX D 0:
AX D 0
FY D 0:
AY C BY 549.81 369.81 D 0
MA D 0:
1.2BY 2.4369.81
B
4.6549.81 D 0
WD
WP
AX D 0 N
Solving:
1.2 m
AY D 1.85 kN
2.4 m
BY D 2.74 kN
4.6 m
4.6 m
2.4 m
1.2 m
AX
Draw the free-body diagram of the ironing board.
Determine the reactions at A and B.
A
B
x
3 lb
10 lb
12 in
10 in
20 in
Solution: The system is in equilibrium.
Substitute into the force balance equation:
(a)
(b)
FA D 13 FB D C15.833 lb
The free-body diagram is shown.
The sums of the forces are:
FX D 0,
WP
y
Problem 5.7 The ironing board has supports at A and
B that can be modeled as roller supports.
(a)
(b)
WD
BY
AY
y
A
B
x
FY D FA C FB 10 3 D 0.
12 in
10 in
10 lb
20 in
3 lb
FB
10 lb
3 lb
The sum of the moments about A is
MA D 12FB 2210 423 D 0,
from which FB D
FA
346
D 28.833 in.
12
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271
Problem 5.8 The distance x D 9 m.
(a) Draw the free-body diagram of the beam.
(b) Determine the reactions at the supports.
10 kN
A
B
6m
x
Solution:
10 kN
(a) The FBD
(b) The equilibrium equations
x= 9 m
Ax
Fx : Ax D 0
6m
Fy : Ay C By 10 kN D 0
Ay
By
MA : By 6 m 10 kN9 m D 0
Solving we find
Ax D 0, Ay D 5 kN, By D 15 kN
Problem 5.9 In Example 5.2, suppose that the 200-lb
downward force and the 300 ft-lb counterclockwise
couple change places; the 200-lb downward force acts
at the right end of the horizontal bar, and the 300 ft-lb
counterclockwise couple acts on the horizontal bar 2 ft
to the right of the support A. Draw a sketch of the object
of the object and apply the equilibrium equations to
determine the reactions at A.
100 lb
30
2 ft
200 lb
A
300 ft-lb
2 ft
2 ft
2 ft
Solution: The sketch and free-body diagram are shown.
The equilibrium equations are
Fx : Ax C 100 lb cos 30&deg; D 0
Fy : Ay C 100 lb sin 30&deg; 200 lb D 0
MA : MA C 300 ft-lb
C 100 lb sin 30&deg; 4 ft
100 lb cos 30&deg; 2 ft
200 lb6 ft D 0
We obtain
Ax D 86.6 lb
Ay D 150 lb
MA D 873 ft-lb
272
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Problem 5.10 (a) Draw the free-body diagram of the
beam.
(b) Determine the reactions at the supports.
100 lb
400 lb
900 ft-lb
A
B
3 ft
Solution: (a) Both supports are roller supports. The free body
diagram is shown. (b) The sum of the forces:
and
4 ft
3 ft
100 lb
4 ft
400 lb
3 ft
4 ft
3 ft
4 ft
FX D 0,
A
FY D FA C FB C 100 400 D 0.
900 ft lb
100 lb
3 ft
4 ft
3 ft
B
400 lb
4 ft
The sum of the moments about A is
900 ft lb
MA D 3100 C 900 7400 C 11FB D 0.
From which FB D
FA
FB
2200
D 200 lb
11
Substitute into the force balance equation to obtain
FA D 300 FB D 100 lb
Problem 5.11 The person exerts 20-N forces on the
pliers. The free-body diagram of one part of the pliers
is shown. Notice that the pin at C connecting the two
parts of the pliers behaves like a pin support. Determine
the reactions at C and the force B exerted on the pliers
by the bolt.
Solution: The equilibrium equations
MC :B25 mm 20 N cos 45&deg; 80 mm
20 N sin 45&deg; 50 mm D 0
25
mm
B
C
Fx :Cx 20 N sin 45&deg; D 0
Cx
Fy :Cy B 20 N
cos 45&deg;
80
mm
D0
Solving:
Cy
50 mm
45⬚
20 N
C
B D 73.5 N, Cx D 14.14 N, Cy D 87.7 N
20 N
20 N
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
273
Problem 5.12 (a) Draw the free-body diagram of the
beam.
8 kN
8 kN
2 kN-m
A
B
30⬚
(b) Determine the reactions at the pin support A.
600
mm
Solution:
500
mm
8 kN
600
mm
8 kN
(a) The FBD
(b) The equilibrium equations
MA : 8 kN0.6 m C 8 kN1.1 m 2 kNm
600
mm
30&deg;
2 kN-m
B
Ax
B cos 30&deg; 2.3 m D 0
Ay
Fx :Ax B sin 30&deg; D 0
Fy :Ay 8 kN C 8 kN B cos 30&deg; D 0
Solving
Ax D 0.502 kN, Ay D 0.870 kN, B D 1.004 kN
Problem 5.13 (a) Draw the free-body diagram of the
beam.
y
A
(b) Determine the reactions at the supports.
6m
40 kN
B
x
8m
12 m
Solution:
A
(a)
(b)
The FBD
The equilibrium equations
MB : 40 kN4 m C A6 m D 0
40 kN
Bx
Fx : A C Bx D 0
Fy : 40 kN C By D 0
Solving we find
By
A D Bx D 26.7 kN, By D 40 kN
274
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Problem 5.14 (a) Draw the free-body diagram of the
beam.
A
(b) If F D 4 kN, what are the reactions at A and B?
2 kN-m
F 0.2 m
0.3 m
0.2 m
0.3 m
0.4 m
B
2 kN-m
Solution:
Ax
(a)
(b)
The free-body diagram
The equilibrium equations
F = 4 kN
MA : 2 kN-m 4 kN0.2 m C B1.0 m D 0
Ay
Fx : Ax 4 kN D 0
Fy : Ay C B D 0
Solving:
B
Ax D 4 kN, Ay D 2.8 kN, B D 2.8 kN
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
275
Problem 5.15 In Example 5.3, suppose that the attachment point for the suspended mass is moved toward
point B such that the horizontal distance from A to
the attachment point increases from 2 m to 3 m. Draw
a sketch of the beam AB showing the new geometry.
Draw the free-body diagram of the beam and apply the
equilibrium equations to determine the reactions at A
to B.
2m
2m
B
3m
A
Solution: From Example 5.3, we know that the mass of the
suspended object is 2-Mg. The sketch and free-body diagram are
shown.
The equilibrium equations are
Fx : Ax C Bx D 0
Fy : By 2000 kg9.81 m/s2 D 0
MB : Ax 3 m
C 2000 kg9.81 m/s2 1 m D 0
We obtain
Ax D 6.54 kN
Bx D 6.54 kN
By D 19.6 kN
276
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 5.16 A person doing push-ups pauses in the
position shown. His 180-lb weight W acts at the point
shown. The dimensions a = 15 in, b = 42 in, and c = 16
in. Determine the normal force exerted by the floor on
each of his hands and on each of his feet.
c
W
a
b
Solution: The free-body diagram is shown. The equilibrium
equations are
Fy : 2H C 2F 180 lb D 0
MH : Wa C 2Fa C b 0
We find that
H D 66.3 lb, F D 23.7 lb
Thus
66.3 lb on each hand
23.7 lb on each foot
Problem 5.17 The hydraulic piston AB exerts a 400-lb
force on the ladder at B in the direction parallel to the
piston. Determine the weight of the ladder and the reactions at C.
6 ft
W
3 ft
A
B
C
6 ft
3 ft
Solution: The free-body diagram of the ladder is shown.
The angle between the piston AB and the horizontal is
˛ D tan1 3/6 D 26.6&deg;
The equilibrium equations are
Fx : Cx C 400 lb cos ˛ D 0
Fy : Cy C 400 lb sin ˛ W D 0
MC : W6 ft 400 lb cos ˛3 ft
400 lb sin ˛3 ft D 0
Solving yields
Cx D 358 lb, Cy D 89.4 lb, W D 268 lb
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
277
Problem 5.18 Draw the free-body diagram of the
structure by isolating it from its supports at A and E.
Determine the reactions at A and E.
D
400 lb
2 ft
200 ft-lb
A
1 ft
B
C
100 lb
1 ft
E
2 ft
2 ft
2 ft
Solution: The free-body diagram is shown.
The equilibrium equations are
Fx : Ax C 100 lb D 0
Fy : Ay 400 lb C Ey D 0
MA : 100 lb1 ft 400 lb6 ft
200 ft-lb C Ey 4 ft D 0
Solving yields
Ax D 100 lb
Ay D 225 lb
Ey D 625 lb
Problem 5.19 (a) Draw the free-body diagram of the
beam.
(b) Determine the tension in the cable and the reactions
at A.
A
B
30&deg;
30 in
30 in
Solution:
800 lb
30 in
T
(a) The FBD
(b) The equilibrium equations
C
T
30&deg;
Ax
MA : 800 lb60 in C T30 in
C T sin 30&deg; 90 in D 0
Fx :Ax T cos 30&deg; D 0
Ay
800 lb
Fy :Ay C T C T sin 30&deg; 800 lb D 0
Solving:
Ax D 554 lb, Ay D 160 lb, T D 640 lb
278
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 5.20 The unstretched length of the spring CD
is 350 mm. Suppose that you want the lever ABC to
exert a 120-N normal force on the smooth surface at A.
Determine the necessary value of the spring constant k
and the resulting reactions at B.
C
k
230
mm
D
450
mm
20⬚
180
mm
B
A
Solution: We have
F D k
0.23
m2
C 0.3
m2 0.35
330
mm
m
300
mm
A D 120 N
30
F0.45 m C A cos 20&deg; 0.18 m
MB : p
1429
F
23
30
C A sin 20&deg; 0.33 m D 0
Fx :A cos 20&deg; C Bx C p
30
1429
Fy : A sin 20&deg; C By p
FD0
23
1429
Bx
FD0
A
By
Solving we find:
20&deg;
k D 3380 N/m, Bx D 188 N, By D 98.7 N
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
279
Problem 5.21 The mobile is in equilibrium. The fish
B weighs 27 oz. Determine the weights of the fish A, C,
and D. (The weights of the crossbars are negligible.)
12 in
3 in
A
6 in
2 in
B
7 in
2 in
C
D
Solution: Denote the reactions at the supports by FAB , FCD , and
and D. The sum of the forces is
FCD
D
C
7 in
2 in
FY D C D C FCD D 0,
FBCD
from which FCD D C C D.
6 in
For the cross bar supporting the weight B, the sum of the forces is
FY D B C FBCD FCD D 0,
from which, substituting, FBCD D B C C C D.
B
FCD
2 in
FAB
FBCD
A
12 in
3 in
For the crossbar supporting C and D, the sum of the moments about
the support is
MCD D 7D C 2C D 0,
from which D D
2C
.
7
For the crossbar supporting B, the sum of the moments is
MBCD D 6FCD 2B D 0,
from which, substituting from above
FCD D
2C
9C
2B
DCCD DCC
D
,
6
7
7
or C D 7B/27 D 7 oz,
and D D 2C/7 D 2 oz.
The sum of the moments about the crossbar supporting A is
MAB D 12A 3FBCD D 0,
from which, substituting from above,
280
27 C 7 C 2
3B C C C D
D
D 9 oz
12
4
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
y
Problem 5.22 The car’s wheelbase (the distance
between the wheels) is 2.82 m. The mass of the car is
1760 kg and its weight acts at the point x D 2.00 m,
y D 0.68 m. If the angle ˛ D 15&deg; , what is the total
normal force exerted on the two rear tires by the sloped
ramp?
x
α
W
Solution: Split W into components:
y
x
α
W cos ˛ acts ? to the incline
W sin ˛ acts parallel to the incline
R
FX : f W sin ˛ D 0
FY :
NR C NF W cos ˛ D 0
0.68
NF
α = 15&deg;
W = (1760X9.81) N
α
m
f
MR :
2m
m
2.82
NR
2W cos ˛ C 0.68W sin ˛ C 2.82NF D 0
Solving: NR D 5930 N, NF D 10750 N
Problem 5.23 The link AB exerts a force on the bucket
of the excavator at A that is parallel to the link. The
weight W = 1500 lb. Draw the free-body diagram of the
bucket and determine the reactions at C. (The connection
at C is equivalent to a pin support of the bucket.)
14 in
16 in
B
A
4 in
C
D
W
8 in
8 in
Solution: The free-body diagram is shown.
The angle between the link AB and the horizontal is
˛ D tan1 12/14 D 40.6&deg;
The equilibrium equations are
Fx : Cx C TAB cos ˛ D 0
Fy : Cy C TAB sin ˛ 1500 lb D 0
MC : 1500 lb8 in TAB cos ˛4 in
TAB sin ˛16 in D 0
Solving yields
TAB D 892 lb, Cx D 677 lb, Cy D 919 lb
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281
Problem 5.24 The 14.5-lb chain saw is subjected to
the loads at A by the log it cuts. Determine the reactions
R, Bx , and By that must be applied by the person using
the saw to hold it in equilibrium.
Solution: The sum of the forces are
y
FX D 5 C BX R cos 60&deg; D 0.
FY D 10 14.5 C BY R sin 60&deg; D 0.
R
The sum of the moments about the origin is
60&deg;
By
1.5 in
7 in
x
A
Bx
5 lb
MO D 7R cos 60&deg; C 8BY 214.5 1310 51.5 D 0.
From which 7R cos 60&deg; C 8BY 166.5 D 0. Collecting equations and
reducing to 3 equations in 3 unknowns:
14.5 lb
10 lb
BX C 0BY 0.5R D 5
13 in
6 in
2 in
0BX C BY 0.866R D 4.5
0BX C 8BY C 3.5R D 166.5.
Solving:
BX D 11.257 lb,
BY D 15.337 lb,
and R D 12.514 lb
Problem 5.25 The mass of the trailer is 2.2 Mg (megagrams). The distances a D 2.5 m and b D 5.5 m. The
truck is stationary, and the wheels of the trailer can turn
freely, which means that the road exerts no horizontal
force on them. The hitch at B can be modeled as a pin
support.
(a) Draw the free-body diagram of the trailer.
(b) Determine the total normal force exerted on the rear
tires at A and the reactions exerted on the trailer at
the pin support B.
B
Solution:
(a)
(b)
The free body diagram is shown.
The sum of forces:
FX D BX D 0.
FY D FA W C FB D 0.
The sum of the moments about A:
MA D aW C a C bFB D 0,
from which
FB D
W
A
2.52.2 &eth; 103 9.81
aW
D
D 6.744 kN
aCb
2.5 C 5.5
Substitute into the force equation:
a
b
FA D W FB D 14.838 kN
B
BX
FB
W
A
FA
282
a
b
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 5.26 The total weight of the wheelbarrow
and its load is W = 100 lb. (a) What is the magnitude of
the upward force F necessary to lift the support at A off
the ground? (b) What is the magnitude of the downward
force necessary to raise the wheel off the ground?
F
W
B
A
40 in
12 in
14 in
Solution: The free-body diagram is shown. The equilibrium
equations are
Fy : A C B C F W D 0
MA : B26 in W12 in
F40 in D 0
(a)
Set A D 0 and solve. We find that
F D 21.2 lb
(b)
Set B D 0 and solve. We find that
F D 30 lb
So we have a 21.2 lb, b 30 lb.
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
283
Problem 5.27 The airplane’s weight is W D 2400 lb.
Its brakes keep the rear wheels locked. The front (nose)
wheel can turn freely, and so the ground exerts no horizontal force on it. The force T exerted by the airplane’s
propeller is horizontal.
(a)
Draw the free-body diagram of the airplane. Determine the reaction exerted on the nose wheel and
the total normal reaction on the rear wheels
(b) when T D 0,
(c) when T D 250 lb.
T
4 ft
W
A
5 ft
B
2 ft
Solution: (a) The free body diagram is shown. (b) The sum of the
forces:
FX D BX D 0
FY D AY W C BY D 0.
The sum of the moments about A is
MA D 5W C 7BY D 0,
from which BY D
5W
D 1714.3 lb
7
Substitute from the force balance equation:
AY D W BY D 685.7 lb
(c) The sum of the forces:
FX D 250 C BX D 0,
from which BX D 250 lb
FY D AY W C BY D 0.
The sum of the moments about A:
MA D 2504 5W C 7BY D 0,
from which BY D 1571.4 lb. Substitute into the force balance equation
to obtain: AY D 828.6 lb
4 ft
AY
284
W
A
5 ft
B
2 ft
BX
BY
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 5.28 A safety engineer establishing limits on
the load that can be carried by a forklift analyzes the
situation shown. The dimensions are a = 32 in, b = 30
in, and c = 26 in. The combined weight of the forklift and
operator is WF = 1200 lb. As the weight WL supported
by the forklift increases, the normal force exerted on the
floor by the rear wheels at B decreases. The forklift is
on the verge of tipping forward when the normal force
at B is zero. Determine the value of WL that will cause
this condition.
WL
WF
A
a
B
b
c
Solution: The equilibrium equations and the special condition for
this problem are
Fy : A C B WL 1200 lb D 0
MA : WL a WF b C Bc D 0
BD0
We obtain
WL D 1125 lb
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285
Problem 5.29 Paleontologists speculate that the stegosaur could stand on its hind limbs for short periods
to feed. Based on the free-body diagram shown and
assuming that m D 2000 kg, determine the magnitudes
of the forces B and C exerted by the ligament–muscle
brace and vertebral column, and determine the angle ˛.
580
mm
160
mm
mg
C
B
22&deg;
Solution: Take the origin to be at the point of application of the
force C. The position vectors of the points of application of the forces
B and W are:
α
415
mm
790
mm
rB D 415i C 160j (mm),
rW D 790i C 580j (mm).
The forces are
C D Ci cos90&deg; ˛ C j sin90&deg; ˛
D Ci sin ˛ C j cos ˛.
B D Bi cos270&deg; 22&deg; C j sin270&deg; 22&deg; D B0.3746i 0.9272j.
W D 29.81j D 19.62j (kN).
i
MC D 415
0.3746B
i
C 790
0
k 160
0
0.9272B 0 j
k 580
0 D 0
19.62 0 j
D 444.72B 15499.8 D 0,
from which
BD
15499.8
D 34.85 kN.
444.72
The sums of the forces:
FX D C sin ˛ 0.3746Bi D 0,
from which C sin ˛ D 13.06 kN.
FY D C cos ˛ 0.9272B 19.62j D 0,
from which C cos ˛ D 51.93 kN.
The angle ˛ is
˛ D tan1
13.06
51.93
D 14.1&deg; .
The magnitude of C,
CD
286
p
13.062 C 51.932 D 53.55 kN
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Problem 5.30 The weight of the fan is W D 20 lb. Its
base has four equally spaced legs of length b D 12 in.
Each leg has a pad near the end that contacts the floor
and supports the fan. The height h D 32 in. If the fan’s
blade exerts a thrust T D 2 lb, what total normal force
is exerted on the two legs at A?
T
b
W
h
T
A
B
Side View
Top View
Solution: The free-body diagram is shown.
The equilibrium equations are
Fy : A C B W D 0
b
2b
MB : W p A p Th D 0
2
2
We obtain A D 6.23 lb
Problem 5.31 The weight of the fan is W D 20 lb. Its
base has four equally spaced legs of length b D 12 in.
Each leg has a pad near the end that contacts the floor
and supports the fan. The height h D 32 in. As the thrust
T of the fan increases, the normal force supported by the
two legs at A decreases. When the normal force at A is
zero, the fan is on the verge of tipping over. Determine
the value of T that will cause this condition.
T
b
W
h
T
A
B
Side View
Top View
Solution: The free-body diagram is shown.
The equilibrium equations are
Fy : A C B W D 0
2b
b
MB : W p A p Th D 0
2
2
We set A D 0 and solve to obtain
T D 5.30 lb
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287
Problem 5.32 In a measure to decrease costs, the
manufacturer of the fan described in Problem 5.31
proposes to support the fan with three equally spaced
legs instead of four. An engineer is assigned to analyze
the safety implications of the change. The weight of the
fan decreases to W D 19.6 lb. The dimensions b and h
are unchanged. What thrust T will cause the fan to be
on the verge of tipping over in this case? Compare your
b
T
Solution: The free-body diagram is shown.
The equilibrium equations are
Fy : A C B W D 0
MB : Wb cos 60&deg; Ab C b cos 60&deg; Th D 0
We set A D 0 and solve to obtain
T D 3.68 lb
This configuration is less stable than the one in Problem 5.31 using
four legs.
Problem 5.33 A force F D 400 N acts on the bracket.
What are the reactions at A and B?
F
A
80 mm
B
320 mm
Solution: The joint A is a pinned joint; B is a roller joint. The
pinned joint has two reaction forces AX , AY . The roller joint has one
reaction force BX . The sum of the forces is
FX D AX C BX D 0,
F
AY
AX
80
mm
BX
320
mm
FY D AY F D 0,
from which
AY D F D 400 N.
The sum of the moments about A is
MA D 0.08BX 0.320F D 0,
from which
BX D
0.320400
D 1600 N.
0.08
Substitute into the sum of forces equation to obtain:
AX D BX D 1600 N
288
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Problem 5.34 The sign’s weight WS D 32 lb acts at
the point shown. The 10-lb weight of bar AD acts at the
midpoint of the bar. Determine the tension in the cable
AE and the reactions at D.
11 in
30 in
11 in
E
30⬚
15⬚
A
B
C
D
Ws
33 in
Solution: Treat the bar AD and sign as one single object. Let TAE
be the tension in the cable. The equilibium equations are
Fx : TAE cos 15&deg; C Dx D 0
Fy : TAE sin 15&deg; C Dy Ws 10 lb D 0
MD : TAE cos 15&deg; 52 in tan 30&deg;
TAE sin 15&deg; 52 in
C 32 lb33 in C 10 lb26 in D 0
Solving yields
TAE D 31.0 lb
Dx D 29.9 lb
Dy D 34.0 lb
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289
Problem 5.35 The device shown, called a swape or
kind were used in Egypt at least as early as 1550 B.C.
and are still in use in various parts of the world.) The
dimensions a D 3.6 m and b D 1.2 m. The mass of the
bar and counterweight is 90 kg, and their weight W acts
at the point shown. The mass of the load being lifted
is 45 kg. Determine the vertical force the person must
is just above the ground (the position shown); (b) when
the load is 1 m above the ground. Assume that the rope
remains vertical.
Solution:
MO : 441 N F3.6 m cos 883 N1.2 m cos D 0
Solving we find
F D 147.2 N
Notice that the angle is not a part of this answer therefore
(a)
F D 147.2 N
(b)
F D 147.2 N
a
Oy
b
θ
25&deg;
Ox
F
W
883 N
441 N
Problem 5.36 This structure, called a truss, has a pin
support at A and a roller support at B and is loaded by
two forces. Determine the reactions at the supports.
Strategy: Draw a free-body diagram, treating the entire
truss as a single object.
30&deg; 2 kN
45&deg;
4 kN
Solution:
p
MA : 4 kN 2b 2 kN cos 30&deg; 3 b
C 2 kN sin 30&deg; b C B4 b D 0
Fx : Ax C 4 kN sin 45&deg; 2 kN sin 30&deg; D 0
Fy : Ay 4 kN cos 45&deg; 2 kN cos 30&deg; C B D 0
Solving:
b
Ax D 1.828 kN, Ay D 2.10 kN, B D 2.46 kN
B
A
b
b
b
b
45&deg;
4 kN
30&deg;
2 kN
Ax
Ay
290
B
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Problem 5.37 An Olympic gymnast is stationary in
the “iron cross” position. The weight of his left arm
and the weight of his body not including his arms are
shown. The distances are a D b D 9 in and c D 13 in.
Treat his shoulder S as a fixed support, and determine
the magnitudes of the reactions at his shoulder. That
is, determine the force and couple his shoulder must
support.
S
8 lb
144 lb
a
b
c
Solution: The shoulder as a built-in joint has two-force and couple
reactions. The left hand must support the weight of the left arm and
half the weight of the body:
FH D
FH
8 lb
144
C 8 D 80 lb.
2
The sum of the forces on the left arm is the weight of his left arm and
the vertical reaction at the shoulder and hand:
FH
8 lb
144 lb
FH
SX
M
FX D SX D 0.
8 lb
SY
b
FY D FH SY 8 D 0,
c
from which SY D FH 8 D 72 lb. The sum of the moments about the
shoulder is
MS D M C b C cFH b8 D 0,
where M is the couple reaction at the shoulder. Thus
M D b8 b C cFH D 1688 in lb D 1688 (in lb)
1 ft
12 in
D 140.67 ft lb
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291
Problem 5.38 Determine the reactions at A.
A
5 ft
300 lb
800 ft-lb
200 lb
200 lb
6 ft
3 ft
Solution: The built-in support at A is a two-force and couple reaction support. The sum of the forces for the system is
FX D AX C 200 D 0,
from which
AX D 200 lb
FY D AY C 300 200 D 0,
from which AY D 100 lb
The sum of the moments about A:
M D 6300 C 5200 800 C MA D 0,
from which MA D 1600 ft lb which is the couple at A.
AY
MA
300 lb
AX
5 ft
800 ft-lb
200 lb
200 lb
6 ft
292
3 ft
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 5.39 The car’s brakes keep the rear wheels
locked, and the front wheels are free to turn. Determine
the forces exerted on the front and rear wheels by the
road when the car is parked (a) on an up slope with
˛ D 15&deg; ; (b) on a down slope with ˛ D 15&deg; .
n
70 i
n
36 i
n
20 i
y
α
3300 lb
Solution: The rear wheels are two force reaction support, and the
front wheels are a one force reaction support. Denote the rear wheels
by A and the front wheels by B, and define the reactions as being
parallel to and normal to the road. The sum of forces:
AX D 854.1 lb.
20 in.
α
FX D AX 3300 sin 15&deg; D 0,
from which
x
BY
3300 lb
AX
AY
36
in.
70
in.
FY D AY 3300 cos 15&deg; C BY D 0.
Since the mass center of the vehicle is displaced above the point A,
a component of the weight (20W sin ˛) produces a positive moment
about A, whereas the other component (36W cos ˛) produces a negative
MA D 363300 cos 15&deg; C 203300 sin 15&deg; C BY 106 D 0,
from which
BY D
C97669
D 921.4 lb.
106
Substitute into the sum of forces equation to obtain AY D 2266.1 lb
(b) For the car parked down-slope the sum of the forces is
FX D AX C 3300 sin 15&deg; D 0,
from which AX D 854 lb
FY D AY 3300 cos 15&deg; C BY D 0.
The component (20W sin ˛) now produces a negative moment about
A. The sum of the moments about A is
MA D 330036 cos 15&deg; 330020 sin 15&deg; C 106BY D 0,
from which
BY D
131834
D 1243.7 lb.
106
Substitute into the sum of forces equation to obtain AY D 1943.8 lb
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293
Problem 5.40 The length of the bar is L D 4 ft. Its
weight W D 6 lb acts at the midpoint of the bar. The
floor and wall are smooth. The spring is unstretched
when the angle ˛ = 0. If the bar is in equilibrium when
˛ D 40&deg; , what is the spring constant k?
k
α
L
Solution: The free-body diagram is shown.
The stretch in the spring is L L cos ˛, so the upward force exerted on
the bar by the spring is F D kL 1 cos ˛. Let N and R be the normal
forces exerted by the floor and the wall, respectively. The equilibrium
equations for the bar are
Fx : R D 0
Fy : F C N W D 0
Mbottom : W
L
sin ˛ RL cos ˛
2
FL sin ˛ D 0
Because R D 0, the moment equation can be solved for the force
exerted by the spring.
F D 0.5W D 3 lb D kL 1 cos ˛
Solving yields k D 3.21 lb/ft
294
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 5.41 The weight W of the bar acts at its
midpoint. The floor and wall are smooth. The spring is
unstretched when the angle ˛ D 0. Determine the angle
˛ at which the bar is in equilibrium in terms of W, k,
and L.
k
α
L
Solution: The free-body diagram is shown.
The stretch in the spring is L L cos ˛, so the upward force exerted on
the bar by the spring is F D kL1 cos ˛. Let N and R be the normal
forces exerted by the floor and the wall, respectively. The equilibrium
equations for the bar are
Fx : R D 0
Fy : F C N W D 0
Mbottom : W
L
sin ˛ RL cos ˛
2
FL sin ˛ D 0
Because R D 0, the moment equation can be solved for the force
exerted by the spring.
FD
W
D kL1 cos ˛
2
W
Solving yields ˛ D cos1 1 2L
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295
Problem 5.42 The plate is supported by a pin in a
smooth slot at B. What are the reactions at the supports?
2 kN-m
6 kN-m
A
B
60&deg;
2m
Solution: The pinned support is a two force reaction support. The
smooth pin is a roller support, with a one force reaction. The reaction
at B forms an angle of 90&deg; C 60&deg; D 150&deg; with the positive x axis. The
sum of the forces:
6 kN-m
2 kN-m
FX D AX C B cos 150&deg; D 0
FY D AY C B sin 150&deg; D 0
A
60&deg;
B
The sum of the moments about B is
2m
MB D 2AY C 2 6 D 0,
from which
AY D 4
D 2 kN.
2
2 kN-m
6 kN-m
Substitute into the force equations to obtain
BD
AY
D 4 kN,
sin 150&deg;
and AX D B cos 150&deg; D 3.464 kN.
AX
150&deg;
AY
B
2m
The horizontal and vertical reactions at B are
BX D 4 cos 150&deg; D 3.464 kN,
and BY D 4 sin 150&deg; D 2 kN.
296
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 5.43 Determine the reactions at the fixed
support A.
Solution: The free-body diagram is shown.
The equilibrium equations are
y
30 lb
40 lb
150 ft-lb
45⬚
A
x
Fx : Ax C 40 lb cos 45&deg; D 0
3 ft
3 ft
6 ft
Fy : Ay C 30 lb
C 40 lb sin 45&deg; D 0
MA : MA C 30 lb3 ft
C 40 lb sin 45&deg; 6 ft
C 150 ft-lb D 0
Solving yields Ax D 28.3 lb, Ay D 58.3 lb, MA D 410 ft-lb.
Problem 5.44 Suppose the you want to represent the
two forces and couple acting on the beam in Problem
5.43 by an equivalent force F as shown. (a) Determine F
and the distance D at which its line of action crosses the
x axis. (b) Assume that F is the only load acting on the
beam and determine the reactions at the fixed support A.
y
F
A
x
D
Solution: The free-body diagram is shown.
(a)
To be equivalnet, F must equal the sum of the two forces:
F D 30 lbj C 40 lbcos 45&deg; i C sin 45&deg; j
F D 28.3i C 58.3j lb
The force F must be placed so that the moment about a point due to
F is equal to the moment about the same point due to the two forces
and couple. Evaluating the moments about the origin,
58.3 lbD D 30 lb3 ft C 40 lb sin 45&deg; 6 ft C 150 ft-lb
The distance D D 7.03 ft
(b)
The equilibrium equations are
Fx : Ax C 28.3 lb D 0
Fy : Ay C 58.3 lb D 0
MA : MA C 58.3 lb7.03 ft D 0
Solving yields Ax D 28.3 lb, Ay D 58.3 lb, MA D 410 ft-lb.
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
297
Problem 5.45 The bicycle brake on the right is pinned
to the bicycle’s frame at A. Determine the force exerted
by the brake pad on the wheel rim at B in terms of the
cable tension T.
T
35&deg;
40 mm
B
Wheel rim
45 mm
A
40 mm
Solution: From the force balance equation for the cables: the force
TB
on the brake mechanism TB in terms of the cable tension T is
35&deg;
T 2TB sin 35&deg; D 0,
from which TB D
T
D 0.8717T.
2 sin 35&deg;
40
mm
B
Take the origin of the system to be at A. The position vector of the
point of attachment of B is rB D 45j (mm). The position vector of the
point of attachment of the cable is rC D 40i C 85j (mm).
The force exerted by the brake pad is B D Bi. The force vector due
the cable tension is
AY
45
mm
AX
40
mm
TB D TB i cos 145&deg; C j sin 145&deg; D TB 0.8192i C 0.5736j.
MA D rB &eth; B C rC &eth; TB D 0
i
MA D 0
B
j
45
0
i
k 45 C 40
0 0.8192
k 85
85 TB D 0
0.5736 0 j
MA D 45B C 92.576TB k D 0,
from which B D
92.576TB
D 2.057TB .
45
Substitute the expression for the cable tension:
B D 2.0570.8717T D 1.793T
298
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 5.46 The mass of each of the suspended
weights is 80 kg. Determine the reactions at the supports
at A and E.
A
B
C
300 mm
D
E
200 mm
Solution: From the free body diagram, the equations of
equilibrium for the rigid body are
and
Fx D AX C EX D 0,
200 mm
y
AY
0.2 m
0.2 m
AX A
x
mg
0.3 m
Fy D AY 2809.81 D 0,
mg
E
EX
MA D 0.3EX 0.2809.81 0.4809.81 D 0.
We have three equations in the three components of the support
reactions. Solving for the unknowns, we get the values
AX D 1570 N,
AY D 1570 N,
and EX D 1570 N.
Problem 5.47 The suspended weights in Problem 5.46
are each of mass m. The supports at A and E will each
safely support a force of 6 kN magnitude. Based on this
criterion, what is the largest safe value of m?
Solution: Written with the mass value of 80 kg replaced by the
symbol m, the equations of equilibrium from Problem 5.46 are
and
Fx D AX C EX D 0,
Fy D AY 2 m9.81 D 0,
MA D 0.3EX 0.2 m9.81 0.4 m9.81 D 0.
We also need the relation
jAj D
A2X C A2Y D 6000 N.
We have four equations in the three components of the support
reactions plus the magnitude of A. This is four equations in four
unknowns. Solving for the unknowns, we get the values
AX D 4243 N,
AY D 4243 N,
EX D 4243 N,
and m D 216.5 kg.
Note: We could have gotten this result by a linear scaling of all of the
numbers in Problem 5.46.
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299
Problem 5.48 The tension in cable BC is 100 lb.
Determine the reactions at the built-in support.
C
6 ft
A
B
300 ft-lb
200 lb
3 ft
3 ft
6 ft
Solution: The cable does not exert an external force on the system,
and can be ignored in determining reactions. The built-in support is a
two-force and couple reaction support. The sum of forces:
FX D AX D 0.
MA
AY
300 ft-lb
AX
200 lb
3 ft
FY D AY 200 D 0,
from which AY D 200 lb.
The sum of the moments about A is
M D MA 3200 300 D 0,
from which MA D 900 ft lb
Problem 5.49 The tension in cable AB is 2 kN. What
are the reactions at C in the two cases?
60&deg;
A
B
2m
C
A
1m
2m
(a)
Solution: First Case: The sum of the forces:
FX D CX T cos 60&deg; D 0,
from which CX D 20.5 D 1 kN
T
T
C
1m
(b)
CY
2m
1m
Case (a)
MC
CX
CY
from which CY D 1.8662 D 3.732 kN.
B
60&deg;
FY D CY C T sin 60&deg; C T D 0,
The sum of the moments about C is
60&deg;
Case (b)
MC
CX
M D MC T sin 60&deg; 3T D 0,
from which MC D 3.8662 D 7.732 kN
Second Case: The weight of the beam is ignored, hence there are no
external forces on the beam, and the reactions at C are zero.
300
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 5.50 Determine the reactions at the supports.
6 in
5 in
50 lb
A
3 in
100 in-lb
Solution: The reaction at A is a two-force reaction. The reaction
3 in
at B is one-force, normal to the surface.
B
The sum of the forces:
30&deg;
FX D AX B cos 60&deg; 50 D 0.
AX
FY D AY C B sin 60&deg; D 0.
50 lb
AY
The sum of the moments about A is
6 in.
100
MA D 100 C 11B sin 60&deg; 6B cos 60&deg; D 0,
B
11 in.
from which
60&deg;
100
D 15.3 lb.
BD
11 sin 60&deg; 6 cos 60&deg; Substitute into the force equations to obtain
AY D B sin 60&deg; D 13.3 lb
and AX D B cos 60&deg; C 50 D 57.7 lb
Problem 5.51 The weight W D 2 kN. Determine the
tension in the cable and the reactions at A.
30&deg;
A
W
0.6 m
AY
Solution: Equilibrium Eqns:
C
FX D 0:
0.6 m
T
AX C T cos 30&deg; D 0
AX
FY D 0:
AY C T C T sin 30&deg;
MA D 0:
0, 6W C 0.6T sin 30&deg; WD0
T
30&deg;
0.6 m
0.6 m
W = 2 kN = 2000 N
C 1, 2T D 0
Solving, we get
AX D 693 N,
AY D 800 N,
T D 800 N
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301
Problem 5.52 The cable shown in Problem 5.51 will
safely support a tension of 6 kN. Based on this criterion,
what is the largest safe value of the weight W?
Solution: The equilibrium equations in the solution of problem are
C
FX D 0:
AX C T cos 30&deg; D 0
FY D 0:
AY C T C T sin 30&deg; W D 0
MA D 0:
0, 6W C 0, 6T sin 30&deg; C 1, 2T D 0
We previously had 3 equations in the 3 unknowns AX , AY and T (we
knew W). In the current problem, we know T but don’t know W.
We again have three equations in three unknowns (AX , AY , and W).
Setting T D 6 kN, we solve to get
AX D 5.2 kN
AY D 6.0 kN
W D 15.0 kN
Problem 5.53 The blocks being compressed by the
clamp exert a 200-N force on the pin at D that points
from A toward D. The threaded shaft BE exerts a force
on the pin at E that points from B toward E.
50 mm
(a)
50 mm
Draw a free-body diagram of the arm DCE of the
clamp, assuming that the pin at C behaves like a
pin support.
(b) Determine the reactions at C.
E
A
C
50 mm
D
FBE
(a) The free-body diagram
(b) The equilibrium equations
125 mm
B
Solution:
125 mm
125 mm
Cy
MC : 200 N0.25 m FBE 0.1 m D 0
200 N
Fx : Cx C FBE D 0
Cx
Fy : Cy 200 N D 0
Solving
Cx D 500 N, Cy D 200 N
302
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Problem 5.54 Consider the clamp in Problem 5.53.
The blocks being compressed by the clamp exert a 200N force on the pin at A that points from D toward A.
The threaded shaft BE exerts a force on the pin at B that
points from E toward B.
(a)
Draw a free-body diagram of the arm ABC of the
clamp, assuming that the pin at C behaves like a
pin support.
Determine the reactions at C.
(b)
FBE
Solution:
(a)
(b)
The free-body diagram
The equilibrium equations
Cx
MC : 200 N0.25 m C FBE 0.1 m D 0
Fx : FBE C Cx D 0
Cy
200 N
Fy : 200 N C Cy D 0
Solving we find
Cx D 500 N, Cy D 200 N
Problem 5.55 Suppose that you want to design the
safety valve to open when the difference between the
pressure p in the circular pipe diameter D 150 mm
and the atmospheric pressure is 10 MPa (megapascals;
a pascal is 1 N/m2 ). The spring is compressed 20 mm
when the valve is closed. What should the value of the
spring constant be?
250 mm
150 mm
k
A
p
150 mm
Solution: The area of the valve is
0.15
2
150
mm
2
250
mm
D 17.671 &eth; 103 m2 .
k
The force at opening is
A
F D 10a &eth; 106 D 1.7671 &eth; 105 N.
150 mm
The force on the spring is found from the sum of the moments about
A,
MA D 0.15F 0.4kL D 0.
k∆L
A
Solving,
F
0.15F
0.151.7671 &eth; 105 kD
D
0.4L
0.40.02
D 3.313 &eth; 106
0.15
m
0.25
m
N
m
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303
Problem 5.56 The 10-lb weight of the bar AB acts
at the midpoint of the bar. The length of the bar is 3 ft.
Determine the tension in the string BC and the reactions
at A.
C
B
3 ft
A
30⬚
1 ft
Solution: Geometry:
tan D
3 ft 3 ft sin 30&deg;
D 0.4169 ) D 22.63&deg;
1 ft C 3 ft cos 30&deg;
The equilibrium equations
MA : TBC cos 3 ft sin 30&deg; C TBC sin 3 ft cos 30&deg; 10 lb1.5 ft cos 30&deg; D 0
Fx : TBC cos C Ax D 0
Fy : TBC sin 10 lb C Ay D 0
Solving:
Ax D 5.03 lb, Ay D 7.90 lb, T D 5.45 lb
TBC
θ
10 lb
Ax
Ay
304
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 5.57 The crane’s arm has a pin support at A.
The hydraulic cylinder BC exerts a force on the arm at
C in the direction parallel to BC. The crane’s arm has a
mass of 200 kg, and its weight can be assumed to act at a
point 2 m to the right of A. If the mass of the suspended
box is 800 kg and the system is in equilibrium, what
is the magnitude of the force exerted by the hydraulic
cylinder?
C
A
2.4 m
1m
B
1.8 m
1.2 m
7m
Solution: The free-body
diagram of the arm is shown.
2.4
D 63.4&deg;
1.2
The equilibrium equations are
The angle D tan1
Fx : Ax C FH cos D 0
Fy : Ay C FH sin 200 kg9.81 m/s2 800 kg9.81 m/s2 D 0
MA : FH sin 3 m FH cos 1.4 m
200 kg9.81 m/s2 2 m
800 kg9.81 m/s2 7 m D 0
We obtain Ax D 12.8 kN, Ay D 15.8 kN, FH D 28.6 kN
Thus FH D 28.6 kN
Problem 5.58 In Problem 5.57, what is the magnitude
of the force exerted on the crane’s arm by the pin support
at A?
Solution: See the solution to Problem 5.57.
Ax D 12.8 kN, Ay D 15.8 kN, FH D 28.6 kN
jAj D
12.8 kN2 C 15.8 kN2 D 20.3 kN
C
A
2.4 m
1m
B
Thus jAj D 20.3 kN
1.8 m
1.2 m
7m
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305
Problem 5.59 A speaker system is suspended by the
cables attached at D and E. The mass of the speaker
system is 130 kg, and its weight acts at G. Determine
the tensions in the cables and the reactions at A and C.
0.5 m 0.5 m 0.5 m
0.5 m
1m
E
C
A
1m
B
D
G
Solution: The weight of the speaker is W D mg D 1275 N. The
1m
AY
equations of equilibrium for the entire assembly are
1.5 m
CY
E
A
Fx D CX D 0,
Fy D AY C CY mg D 0
C CX
B
D
(where the mass m D 130 kg), and
mg
MC D 1AY 1.5mg D 0.
Solving these equations, we get
1.5 m
CX D 0,
1m
T2
CY D 3188 N,
T1
and AY D 1913 N.
From the free body diagram of the speaker alone, we get
and
Mleft
mg
Fy D T1 C T2 mg D 0,
support
D 1mg C 1.5T2 D 0.
Solving these equations, we get
T1 D 425. N
and T2 D 850 N
306
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 5.60 The weight W1 D 1000 lb. Neglect the
weight of the bar AB. The cable goes over a pulley at
C. Determine the weight W2 and the reactions at the pin
support A.
B
50&deg;
35&deg;
W1
A
C
W2
Solution: The strategy is to resolve the tensions at the end of bar
AB into x- and y-components, and then set the moment about A to
zero. The angle between the cable and the positive x axis is 35&deg; .
The tension vector in the cable is
T2 D W2 i cos35&deg; C j sin35&deg; .
35&deg;
T2
rB
T1
50&deg;
AY
D W2 0.8192i 0.5736jlb.
AX
Assume a unit length for the bar. The angle between the bar and the
positive x axis is 180&deg; 50&deg; D 130&deg; . The position vector of the tip of
the bar relative to A is
rB D i cos130&deg; C j sin130&deg; , D 0.6428i C 0.7660j.
The tension exerted by W1 is T1 D 1000j. The sum of the moments
MA D rB &eth; T1 C rB &eth; T2 D rB &eth; T1 C T2 i
D L 0.6428
0.8191W2
j
0.7660
0.5736W2 1000 MA D 0.2587W2 C 642.8k D 0,
from which W2 D 2483.5 lb
The sum of the forces:
FX D AX C W2 0.8192i D 0,
from which AX D 2034.4 lb
FY D AY W2 0.5736 1000j D 0,
from which AY D 2424.5 lb
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
307
Problem 5.61 The dimensions a D 2 m and b D 1 m.
The couple M D 2400 N-m. The spring constant is k D
6000 N/m, and the spring would be unstretched if h D 0.
The system is in equilibrium when h D 2 m and the
beam is horizontal. Determine the force F and the reactions at A.
k
h
A
M
F
a
Solution: We need to know the unstretched length of the spring, l0
b
Unstretched
l0 D a C b D 3 m
(a + b)
We also need the stretched length
l2 D h2 C a C b2
AY
θ
M
l D 3.61 m
AX
a
FS D kl l0 tan D
F
b
h
a C b
D 33.69&deg;
Equilibrium eqns:
C
FX :
AX FS cos D 0
FY :
AY C FS sin F D 0
MA :
M aF C a C bFS sin D 0
a D 2 m,
b D 1 m,
M D 2400 N-m,
h D 2 m,
k D 6000 N/m.
Substituting in and solving, we get
FS D 6000l l0 D 3633 N
and the equilibrium equations yield
AX D 3023 N
AY D 192 N
F D 1823 N
308
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 5.62 The bar is 1 m long, and its weight W
acts at its midpoint. The distance b D 0.75 m, and the
angle ˛ D 30&deg; . The spring constant is k D 100 N/m, and
the spring is unstretched when the bar is vertical. Determine W and the reactions at A.
k
α
W
A
b
p
Solution: The unstretched length of the spring is L D b2 C 12 D
1.25 m. The obtuse angle is 90 C ˛, so the stretched length can be
determined from the cosine law:
L22
D
12
C 0.752
20.75 cos90 C ˛ D 2.3125
β
m2
α
W
from which L2 D 1.5207 m. The force exerted by the spring is
A
T D kL D 1001.5207 1.25 D 27.1 N.
b
The angle between the spring and the bar can be determined from the
sine law:
1.5207
b
D
,
sin ˇ
sin90 C ˛
T
β
from which sin ˇ D 0.4271,
α
ˇ D 25.28&deg; .
W
The angle the spring makes with the horizontal is 180 25.28 90 ˛ D 34.72&deg; . The sum of the forces:
AX
FX D
AX T cos 34.72&deg;
D 0,
AY
from which AX D 22.25 N.
FY D AY W T sin 34.72&deg; D 0.
The sum of the moments about A is
MA D T sin 25.28&deg; W
2
sin ˛ D 0,
from which
WD
2T sin 25.28&deg;
D 46.25 N.
sin ˛
Substitute into the force equation to obtain: AY D W C T sin 34.72&deg; D
61.66 N
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309
B
Problem 5.63 The boom derrick supports a suspended
15-kip load. The booms BC and DE are each 20 ft long.
The distances are a D 15 ft and b D 2 ft, and the angle
D 30&deg; . Determine the tension in cable AB and the reactions at the pin supports C and D.
E
θ
Solution: Choose a coordinate system with origin at point C, with
the y axis parallel to CB. The position vectors of the labeled points
are:
C
A
D
rD D 2i
a
b
rE D rD C 20i sin 30&deg; C j cos 30&deg; D 12i C 17.3j,
The components:
rB D 20j,
Dx D 0.6jDj D 7.67 kip,
rA D 15i.
The unit vectors are:
eDE D
rE r D
D 0.5i C 0.866j,
jrE rD j
eEB D
rB r E
D 0.976i C 0.2179j.
jrB rE j
eCB D
rB r C
D 1j,
jrB rC j
eAB D
rA r B
D 0.6i 0.8j.
jrA rB j
Dy D 0.866jDj D 13.287 kip,
and Cy D 1jCj D 11.94 kip
B
E
θ
A
C
Fx D 0.5jDj 0.976jTEB j D 0,
TAB
Fy D 0.866jDj C 0.2179jTEB j 15 D 0,
from which
b
a
Isolate the juncture at E: The equilibrium conditions are
D
TEB
TEB
15 kip
C
D
Juncture B Juncture E
jDj D 15.34 kip
and jTEB j D 7.86 kip.
Isolate the juncture at B: The equilibrium conditions are:
and
Fx D 0jCj 0.6jTAB j C 0.976jTEB j,
Fy D 1jCj 0.6jTAB j 0.2179jTEB j D 0,
from which
jTAB j D 12.79 kip,
and jCj D 11.94 kip.
310
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Problem 5.64 The arrangement shown controls the
elevators of an airplane. (The elevators are the horizontal
control surfaces in the airplane’s tail.) The elevators are
attached to member EDG. Aerodynamic pressures on the
elevators exert a clockwise couple of 120 in-lb. Cable
BG is slack, and its tension can be neglected. Determine
the force F and the reactions at pin support A.
A
D
2.5 in
C
F
3.5 in
Elevator
E
B
6 in
120 in-lb
G
2 in
2.5 in
2.5 in
1.5 in
120 in
(Not to scale)
Solution: Begin at the elevator. The moment arms at E and G are
6 in. The angle of the cable EC with the horizontal is
˛ D tan1
12
D 5.734&deg; .
119.5
Denote the horizontal and vertical components of the force on point
E by FX and FY . The sum of the moments about the pinned support
on the member EG is
2 in
FX
α
C
A
TEC
2.5 in
F
3.5 in
E
FY D
TEC
α
C
6 in
120 in-lb
2.5
in
MEG D 2.5FY C 6FX 120 D 0.
This is the tension in the cable EC. Noting that
FX D TEC cos ˛,
and FY D TEC sin ˛,
then TEC D
120
.
2.5 sin ˛ C 6 cos ˛
The sum of the moments about the pinned support BC is
MBC D 2TEC sin ˛ C 6TEC cos ˛ 2.5F D 0.
Substituting:
FD
120
2.5
6 cos ˛ 2 sin ˛
6 cos ˛ C 2.5 sin ˛
D 480.9277 D 44.53 lb.
The sum of the forces about the pinned joint A:
Fx D Ax F C TEC cos ˛ D 0
from which Ax D 25.33 lb,
Fy D Ay C TEC sin ˛ D 0
from which Ay D 1.93 lb
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
311
Problem 5.65 In Example 5.4 suppose that ˛ D 40&deg; ,
d D 1 m, a D 200 mm, b D 500 mm, R D 75 mm, and
the mass of the luggage is 40 kg. Determine F and N.
Solution: (See Example 5.4.)
The sum of the moments about the center of the wheel:
MC D dF cos ˛ C aW sin ˛ bW cos ˛ D 0,
from which F D
b a tan ˛W
D 130.35 N.
d
The sum of the forces:
FY D N W C F D 0,
F
from which N D 262.1 N
d
A
d
b
h
a
a W
α
R
C
W
b
F
R
h
a
C
N
Problem 5.66 In Example 5.4 suppose that ˛ D 35&deg; ,
d D 46 in, a D 10 in, b D 14 in, R D 3 in, and you don’t
want the user to have to exert a force F larger than 20 lb.
What is the largest luggage weight that can be placed on
the carrier?
N
Solution: (See Example 5.4.) From the solution to Problem 5.65,
the force is
FD
b a tan ˛W
.
d
Solve for W:
WD
Fd
.
b a tan ˛
For F D 20 lb,
W D 131.47 D 131.5 lb
Problem 5.67 One of the difficulties in making design
decisions is that you don’t know how the user will place
the luggage on the carrier in Example 5.4. Suppose you
assume that the point where the weight acts may be
anywhere within the “envelope” R a 0.75c and 0 b 0.75d. If ˛ D 30&deg; , c D 14 in, d D 48 in, R D 3 in,
and W D 80 lb, what is the largest force F the user will
have to exert for any luggage placement?
Solution: (See Example 5.4.) From the solution to Problem 5.65,
the force is
FD
b a tan ˛W
.
d
The force is maximized as
b ! 0.75d,
and a ! R.
Thus
FMAX D
312
0.75d R tan ˛W
D 57.11 lb
d
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 5.68 In our design of the luggage carrier
in Example 5.4, we assumed a user that would hold
the carrier’s handle at h D 36 in above the floor. We
assumed that R D 3 in, a D 6 in, and b D 12 in, and we
chose the dimension d D 4 ft. The resulting ratio of the
force the user must exert to the weight of the luggage
is F/W D 0.132. Suppose that people with a range of
heights use this carrier. Obtain a graph of F/W as a
function of h for 24 h 36 in.
Solution: (See Example 5.4.) From the solution to Problem 5.67,
the force that must be exerted is
FD
b a tan ˛W
,
d
b a tan ˛
F
D
.
W
d
from which
The angle a is given by
˛ D sin1
hR
d
.
F
/
W
,
d
i
m
e
n
s
i
o
n
l
e
x
e
F/W versus height
.2
.19
.18
.17
.16
.15
.14
.13
24
26
28
30
32
height h, in
34
36
The commercial package TK Solver Plus was used to plot a graph of
F
as a function of h.
W
Problem 5.69 (a) Draw the free-body diagram of the
beam and show that it is statically indeterminate. (See
Active Example 5.5.)
(b) Determine as many of the reactions as possible.
20 N-m
A
800 mm
Solution: (a) The free body diagram shows that there are four
unknowns, whereas only three equilibrium equations can be written.
(b) The sum of moments about A is
MA D M C 1.1BY D 0,
from which BY D 20
D 18.18 N.
1.1
The sum of forces in the vertical direction is
300 mm
20 N-m
A
800 mm
B
B
300 mm
BX
AX
AY
800 mm
300 mm
BY
FY D AY C BY D 0,
from which AY D BY D 18.18 N.
The sum of forces in the horizontal direction is
FX D AX C BX D 0,
from which the values of AX and BX are indeterminate.
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
313
Problem 5.70 Consider the beam in Problem 5.69.
Choose supports at A and B so that it is not statically
indeterminate. Determine the reactions at the supports.
Solution: One possibility is shown: the pinned support at B is
replaced by a roller support. The equilibrium conditions are:
20 N-m
A
B
FX D AX D 0.
800 mm
The sum of moments about A is
AX
MA D M C 1.1BY D 0,
from which BY D 300 mm
20 N-m
AY
800 mm
300 mm
BY
20
D 18.18 N.
1.1
The sum of forces in the vertical direction is
FY D AY C BY D 0,
from which AY D BY D 18.18 N.
Problem 5.71 (a) Draw the free-body diagram of the
beam and show that it is statically indeterminate. (The
external couple M0 is known.)
M0
A
B
(b) By an analysis of the beam’s deflection, it is determined that the vertical reaction B exerted by the roller
support is related to the couple M0 by B D 2M0 /L. What
are the reactions at A?
Solution:
(a)
C
L
Eqn (3) and Eqn (4) yield
FX :
AX D 0
(1)
MA D MO 2MO
FY :
AY C B D 0
(2)
MA D MO
MA :
MA MO C BL D 0
(3)
MA was assumed counterclockwise
Unknowns: MA , AX , AY , B.
3 Eqns in 4 unknowns
MA D jMO j clockwise
AX D 0
AY D 2MO /L
∴ Statistically indeterminate
(b)
Given B D 2MO /L
(4)
We now have 4 eqns in 4 unknowns and can solve.
Eqn (1) yields AX D 0
AY
MO
MA
AX
L
B
Eqn (2) and Eqn (4) yield
AY D 2MO /L
314
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 5.72 Consider the beam in Problem 5.71.
Choose supports at A and B so that it is not statically
indeterminate. Determine the reactions at the supports.
Solution: This result is not unique. There are several possible
FX :
A
MO
L
AX D 0
FY :
AY C BY D 0
MA :
Mo C BL D 0
O
MO
AX
AX D 0
L
B
AY
B D MO /L
AY D MO /L
Problem 5.73 Draw the free-body diagram of the
L-shaped pipe assembly and show that it is statically
indeterminate. Determine as many of the reactions as
possible.
B
80 N
A
Strategy: Place the coordinate system so that the x
axis passes through points A and B.
300 mm
Solution: The free body diagram shows that there are four reactions, hence the system is statically indeterminate. The sum of the
forces:
and
FX D AX C BX D 0,
FY D AY C BY C F D 0.
300
mm
100 N-m
700 mm
The moment about the point A is
MA D LBN 0.3F C M D 0,
from which BN D
76
M C 0.3F
D
D 99.79 N,
L
0.76157
from which
A strategy for solving some statically indeterminate problems is to
select a coordinate system such that the indeterminate reactions vanish
from the sum of the moment equations. The choice here is to locate
the x axis on a line passing through both A and B, with the origin at
A. Denote the reactions at A and B by AN , AP , BN , and BP , where the
subscripts indicate the reactions are normal to and parallel to the new
x axis. Denote
The sum of the forces normal to the new axis is
F D 80 N,
The reactions parallel to the new axis are indeterminate.
FN D AN C BN C F cos D 0,
from which
AN D BN F cos D 26.26 lb
M D 100 N-m.
The length from A to B is
LD
p
0.32 C 0.72 D 0.76157 m.
The angle between the new axis and the horizontal is
D tan1
0.3
0.7
BN
80 N
AN
D 23.2&deg; .
300
mm
BP
300
mm
AP
100 N-m
700
mm
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
315
Problem 5.74 Consider the pipe assembly in Problem
5.73. Choose supports at A and B so that it is not
statically indeterminate. Determine the reactions at the
supports.
Solution: This problem has no unique solution.
Problem 5.75 State whether each of the L-shaped bars
shown is properly or improperly supported. If a bar
is properly supported, determine the reactions at its
supports. (See Active Example 5.6.)
F
C
–12 L
F
L
–1 L
2
Solution:
(1)
is properly constrained. The sum of the forces
FX D F C BX D 0,
A
B
L
(1)
(2)
FY D BY C Ay D 0,
–12 L
F
–12 L
A
MB D LAY C LF D 0,
B
45&deg;
from which AY D F, and By D F
(2)
is improperly constrained. The reactions intersect at B, while the
force produces a moment about B.
(3)
is properly constrained. The forces are neither concurrent nor
parallel. The sum of the forces:
45&deg;
C
45&deg;
from which By D Ay . The sum of the moments about B:
B
L
from which BX D F.
A
L
(3)
FX D C cos 45&deg; B A cos 45&deg; C F D 0.
FY D C sin 45&deg; A sin 45&deg; D 0
from which A D C. The sum of the moments about A:
MA D 21 LF C LC cos 45&deg; C LC sin 45&deg; D 0,
F
F
from which C D p . Substituting and combining: A D p ,
2 2
2 2
F
BD
2
316
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 5.76 State whether each of the L-shaped bars
shown is properly or improperly supported. If a bar
is properly supported, determine the reactions at its
supports. (See Active Example 5.6.)
C
C
–12 L
F
F
–12 L
–12 L
Solution:
(1)
(2)
(3)
is improperly constrained. The reactions intersect at a point P,
and the force exerts a moment about that point.
is improperly constrained. The reactions intersect at a point P
and the force exerts a moment about that point.
is properly constrained. The sum of the forces:
A
L
45&deg;
(2)
(1)
C
FX D C F D 0,
–12 L
F
–12 L
from which C D F.
A
B
FY D A C B D 0,
from which A D B. The sum of the moments about B: LA C
L
1
1
F LC D 0, from which A D F, and B D F
2
2
2
B
A
B
L
–12 L
L
(3)
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
317
Problem 5.77 The bar AB has a built-in support at A
and is loaded by the forces
y
A
FB D 2i C 6j C 3k (kN),
FB
z
FC D i 2j C 2k (kN).
B
1m
(a) Draw the free-body diagram of the bar.
(b) Determine the reactions at A.
C
1m
Strategy: (a) Draw a diagram of the bar isolated from
its supports. Complete the free-body diagram of the bar
by adding the two external forces and the reactions
due to the built-in support (see Table 5.2). (b) Use the
scalar equilibrium equations (5.16)–(5.21) to determine
the reactions.
FC
Solution:
x
AY
MA = MAX i + MAY j + MAZ K
AX
MA D MAX i C MAY j C MAZ k
FB
(b)
Equilibrium Eqns (Forces)
1m
FX :
AX C FBX C FCX D 0
FY : AY C FBY C FCY D 0
FZ :
AZ
B
C
1m
x
FC
AZ C FBZ C FCZ D 0
Equilibrium Equations (Moments) Sum moments about A
rAB &eth; FB D 1i &eth; 2i C 6j C 3k kN-m
rAB &eth; FB D 3j C 6k (kN-m)
rAC &eth; FC D 2i &eth; 1i 2j C 2k kN-m
rAC &eth; FC D 4j 4k (kN-m)
x:
y:
z:
MA :
M AX D 0
MA :
M AY 3 4 D 0
MA :
M AZ C 6 4 D 0
Solving, we get
AX D 3 kN,
AY D 4 kN,
AZ D 5 kN
MAx D 0,
MAy D 7 kN-m,
MAz D 2 kN-m
318
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
y
Problem 5.78 The bar AB has a built-in support at A.
The tension in cable BC is 8 kN. Determine the reactions
at A.
A
z
C
(3,0.5,–0.5)m
2m
B
x
Solution:
AY
MA = MAX i + MAY j + MAZ K
AX
MA D MAx i C MAy j C MAz k
AZ
We need the unit vector eBC
eBC
2m
xC xB i C yC yB j C zC zB k
D xC xB 2 C yC yB 2 C zC zB 2
TBC
C (3, 0.5, −0.5)
B (2, 0, 0)
x
eBC D 0.816i C 0.408j 0.408k
TBC D 8 kNeBC
TBC D 6.53i C 3.27j 3.27k (kN)
The moment of TBC about A is
MBC D rAB &eth; TBC
i
D 2
6.53
j
0
3.27
k 0 3.27 MBC D rAB &eth; TBC D 0i C 6.53j C 6.53k (kN-m)
Equilibrium Eqns.
FX :
AX C TBCX D 0
FY :
AY C TBCY D 0
FZ :
AZ C TBCZ D 0
MX :
MAX C MBCX D 0
MY :
MAY C MBCY D 0
MZ :
MAZ C MBCZ D 0
Solving, we get
AX D 6.53 (kN),
AY D 3.27 (kN),
AZ D 3.27 (kN)
MAx D 0,
MAy D 6.53 (kN-m),
MAz D 6.53 (kN-m)
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
319
y
Problem 5.79 The bar AB has a fixed support at A.
The collar at B is fixed to the bar. The tension in the
rope BC is 300 lb. (a) Draw the free-body diagram of
the bar. (b) Determine the reactions at A.
B (6, 6, 2) ft
A
x
C (8, 0, 3) ft
z
Solution:
(a) The free-body diagram is shown.
(b) We need to express the force exerted by the rope in terms of its
components.
The vector from B to C is
rBC D [8 6i C 0 6j C 3 2k] ft
D 2i 6j C k ft
The force in the rope can now be written
T D TBC
rBC
D TBC 0.312i 0.937j C 0.156k
jrBC j
The equilibrium equations for the bar are
Fx : Ax C 0.312TBC D 0
Fy : Ay 0.937TBC D 0
Fz : Az C 0.156TBC D 0
i
MA : MAx i C MAy j C MAz k C 6 ft
0.312TBC
j
6 ft
0.937TBC
k
2 ft D 0
0.156TBC These last equations can be written as
MAx D 0.218 ftTBC , MAy D 0.312 ftTBC , MAz D 7.50 ftTBC
Setting TBC D 300 lb, expanding and solving we have
Ax D 93.7 lb, Ay D 281 lb, Az D 46.9 lb
MAx D 843 ft-lb, MAy D 93.7 ft-lb, MAz D 2250 ft-lb
320
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
y
Problem 5.80 The bar AB has a fixed support at A.
The collar at B is fixed to the bar. Suppose that you
don’t want the support at A to be subjected to a couple
of magnitude greater than 3000 ft-lb. What is the largest
allowable tension in the rope BC?
B (6, 6, 2) ft
Solution: See the solution to Problem 5.79. The magnitude of the
couple at A can be expressed in terms of the tension in the rope as
A
2
MAx C MAy C MAz
jMA j D
D
2
2
x
2.81 ft2 C 0.312 ft2 C 7.50 ft2 TBC
Setting jMA j D 3000 ft-lb and solving for TBC yields TBC D 374 lb
C (8, 0, 3) ft
z
Problem 5.81 The total force exerted on the highway
sign by its weight and the most severe anticipated winds
is F D 2.8i 1.8j (kN). Determine the reactions at the
fixed support.
y
F
8m
Solution: The applied load is F D 2.8i 1.8j kN applied at r D
8j C 8k m
The force reaction at the base is
8m
O
x
R D Ox i C Oy j C Oz k
The moment reaction at the base is
MO D MOx i C MOy j C MOz k
z
For equilibrium we need


Fx : 2.8 kN C Ox D 0



FDFCRD0)
Fy : 1.8 kN C Oy D 0




Fz : 0 C Oz D 0
Ox D 2.8 kN
)
Oy D 1.8 kN
Oz D 0


Mx : 14.4 kN-m C MOx D 0



M D r &eth; F C MO D 0 )
My : 22.4 kN-m C MOy D 0




Mz : 22.4 kN-m C MOz D 0
MOx D 14.4 kN-m
)
MOy D 22.4 kN-m
MOz D 22.4 kN-m
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
321
Problem 5.82 The tension in cable AB is 800 lb.
Determine the reactions at the fixed support C.
y
4 ft
Solution: The force in the cable is
F D 800 lb
2i 4j k
p
21
C
5 ft
We also have the position vector
4 ft
rCA D 4i C 5k ft
A
x
The force reaction at the base is
R D Cx i C Cy j C Cz k
B
The moment reaction at the base is
z
(6, 0, 4) ft
MC D MCx i C MCy j C MCz k
For equilibrium we need


Fx : Cx C 349 lb D 0



FDFCRD0)
Fy : Cy 698 lb D 0




Fz : Cz 175 lb D 0
Cx D 349 lb
)
Cy D 698 lb
Cz D 175 lb


Mx : MCx C 3490 ft-lb D 0



MDrCFCRD0)
My : MCy C 2440 ft-lb D 0




Mz : MCz 2790 ft-lb D 0
MCx D 3490 ft-lb
)
MCy D 2440 ft-lb
MCz D 2790 ft-lb
322
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 5.83 The tension in cable AB is 24 kN.
Determine the reactions in the built-in support D.
2m
C
A
2m
Solution: The force acting on the device is
D
B
F D FX i C FY j C FZ k D 24 kNeAB ,
3m
and the unit vector from A toward B is given by
eAB D
1m
1i 2j C 1k
p
.
6
The force, then, is given by
F D 9.80i 19.60j C 9.80k kN.
The position from D to A is
r D 2i C 2j C 0k m.
The force equations of equilibrium are
DX C FX D 0,
DY C FY D 0,
and DZ C FZ D 0.
The moment equation, in vector form, is
M D MD C r &eth; F.
Expanded, we get
i
M D MDX i C MDY j C MDZ k C 2
9.80
j
2
19.60
k 0 D 0.
9.80 The corresponding scalar equations are
MDX C 29.80 D 0,
MDY 29.80 D 0,
and MDZ C 219.60 29.80 D 0.
Solving for the support reactions, we get
DX D 9.80 kN,
OY D 19.60 kN,
OZ D 9.80 kN.
MDX D 19.6 kN-m,
MDY D 19.6 kN-m,
and MDZ D 58.8 kN-m.
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
323
Problem 5.84 The robotic manipulator is stationary
and the y axis is vertical. The weights of the arms AB and
BC act at their midpoints. The direction cosines of the
centerline of arm AB are cos x D 0.174, cos y D 0.985,
cos z D 0, and the direction cosines of the centerline
of arm BC are cos x D 0.743, cos y D 0.557, cos z D
0.371. The support at A behaves like a built-in support.
y
600 mm
C
160 N
B
(a)
What is the sum of the moments about A due to
the weights of the two arms?
(b) What are the reactions at A?
600 mm
200 N
A
z
x
Solution: Denote the center of mass of arm AB as D1 and that of
BC as D2 . We need
(a)
rAB ,
and rBD2 .
We now have the geometry determined and are ready to determine the moments of the weights about A.
where
i
0
To get these, use the direction cosines to get the unit vectors eAB and
eBC . Use the relation
e D cos X i C cos Y j C cos Z k
j
k 0.2955 0 200 0 rAD1 &eth; W1 D 10.44k N-m
eAB D 0.174i C 0.985j C 0k
and
i
0
eBC D 0.743i C 0.557j 0.371k
j
0.7581
160
k
0.1113 0
rAD2 &eth; W2 D 17.81i 52.37k
rAB D 0.6eAB m
Thus,
MW D 17.81i 62.81k (N-m)
rBC D 0.6eBC m
rBD2 D 0.3eBC m
(b)
WAB D 200j N
Equilibrium Eqns
FX : AX D 0
WBC D 160j N
Thus rAD1 D 0.0522i C 0.2955j m
rAB D 0.1044i C 0.5910j m
rBD2 D 0.2229i C 0.1671j 0.1113k m
rBC D 0.4458i C 0.3342j 0.2226k m
and rAD2 D rAB C rBD2
rAD2 D 0.3273i C 0.7581j 0.1113k m
324
FY :
AY W1 W2 D 0
FZ :
AZ D 0
A : MA C
MW D 0
MX :
MAx 17.81 D 0 (N-m)
MY :
MAy C 0 D 0
MZ :
MZ 62.81 D 0 (N-m)
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
5.84 (Continued )
Thus:
AX D 0, AY D 360 (N), AZ D 0,
MAx D 17.81 (N-m), MAy D 0, MAz D 62.81 (N-m)
C
W2
W1
D2
B
D1
MA
MA = MAXi + MAYj + MAZk
W1 = 200 N
W2 = 160 N
AX
AZ
AY
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
325
Problem 5.85 The force exerted on the grip of the
exercise machine is F D 260i 130j (N). What are the
reactions at the built-in support at O?
150
mm
y
F
O
200
mm
z
250
mm
Solution:
MO D MOx i C MOy j C MOz k
OX
rOP D 0.25i C 0.2j 0.15k
z
Equilibrium (Forces)
0.15 P
m
y
MO
OZ
x
OY
0.2
F = 260 i – 130 j (N)
5m
0.2 m
x
FX :
OX C FX D OX C 260 D 0 (N)
FY :
OY C FY D OY 130 D 0 (N)
FZ :
OZ C FZ D OZ D 0 (N)
Thus, OX D 260 N, OY D 130 N, OZ D 0
MX :
MOX C MFX D 0
MY :
MOY C MFY D 0
MZ :
MOZ C MFZ D 0
where
i
MF D rOP &eth; F D 0.25
260
j
0.2
130
k 0.15 0 MF D 19.5i 39j 84.5k (N-m)
and from the moment equilibrium eqns,
MOX D 19.5 (N-m)
MOY D 39.0 (N-m)
MOZ D 84.5 (N-m)
326
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
y
Problem 5.86 In Active Example 5.7, suppose that
cable BD is lengthened and the attachment point D
moved form (0, 600, 400) mm to (0, 600, 600) mm.
(The end B of bar AB remains where it is.) Draw a
sketch of the bar and its supports showing cable BD in
its new position. Draw the free-body diagram of the bar
and apply equilibrium to determine the tensions in the
cables and the reactions at A.
400
mm
1000 mm
C
B
D
600 mm
600
mm
A
x
⫺200j (N)
z
Solution: The sketch and free-body diagram are shown.
We must express the force exerted on the bar by cable BD in terms of
its components. The vector from B to D is
rBD D [0 1000i C 600 600j
C 600 400k] mm
D 1000i C 200j mm
The force exerted by cable BD can be expressed as
TBD
rBD
D TBD 0.981i C 0.196k
jrBD j
The equilibrium equations are
Fx : Ax 0.981TBD D 0
Fy : Ay 200 N D 0
Fz : Az C 0.196TBD TBC D 0
i
1
MA : 0.981TBD
i
j
k
C 0.5
0.6
0.4
0 0.196TBD TBC 0
j
0.3
200
k 0.2 D 0
0 Expanding and solving these equations, we find
Ax D 166.7 N, Ay D 200 N, Az D 66.7 N, TBC D 100 N,
TBD D 170 N
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
327
Problem 5.87 The force F acting on the boom ABC
at C points in the direction of the unit vector 0.512i 0.384j C 0.768k and its magnitude is 8 kN. The boom
is supported by a ball and socket at A and the cables BD
and BE. The collar at B is fixed to the boom.
y
1.5 m
2m
D
E
(a) Draw the free-body diagram of the boom.
(b) Determine the tensions in the cables and the reactions at A.
1m
2m
A
B
z
2m
C
2m
x
Solution:
F
(a) The free-body diagram
(b) We identify the following forces, position vectors, and reactions
rAC D 4 mi, F D 8 kN0.512i 0.384j C 0.768k

2i C 2j C 1.5k


p
T
D
T

BD
BD

10.25
D 2 mi,

2i
C
j 2k


 TBE D TBE
3
rAB
Az
Ax
TBE
B
Ay
TBD
C
R D Ax i C Ay j C Az k
Force equilibrium requires:
F D R C TBD C TBE C F D 0.
F
In component form we have
2
2
TBD TBE D 0
Fx : Ax C 8 kN0.512 p
3
10.25
2
1
TBD C TBE D 0
Fy : Ay 8 kN0.384 C p
3
10.25
1.5
2
TBD TBE D 0
Fz : Az C 8 kN0.768 C p
3
10.25
Moment equilibrium requires:
MA D rAB &eth; TBD C TBE C rAC &eth; F D 0.
In components:
Mx : 0 D 0
My : 8 kN0.7684 m p
C
TBD 2 m
2
TBE 2 m D 0
3
Mz : 8 kN0.3844 m C p
C
1.5
10.25
2
10.25
TBD 2 m
1
TBE 2 m D 0
3
Solving five equations for the five unknowns we find
Ax D 8.19 kN, Ay D 3.07 kN, Az D 6.14 kN,
TBD D 0, TBE D 18.43 kN
328
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 5.88 The cables BD and BE in Problem 5.87
will each safely support a tension of 25 kN. Based on
this criterion, what is the largest acceptable magnitude
of the force F?
Solution: We have the force and distances:
rAC D 4 mi, F D F0.512i 0.384j C 0.768k
rAB

2i C 2j C 1.5k


p

 TBD D TBD
10.25
D 2 mi,

2i
C
j 2k


 TBE D TBE
3
The moment equations are
1.5
2
TBD 2 m C TBE 2 m D 0
My : F0.7684 m p
3
10.25
2
1
TBD 2 m C TBE 2 m D 0
Mz : F0.3844 m C p
3
10.25
Solving we find
TBE D 2.304F, TBD D 0
Thus:
25 kN D 2.304F ) F D 10.85 kN
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
329
Problem 5.89 The suspended load exerts a force F D
600 lb at A, and the weight of the bar OA is negligible.
Determine the tensions in the cables and the reactions at
the ball and socket support O.
y
C
(0, 6, –10) ft
A
(8, 6, 0) ft
B
(0, 10, 4) ft
–Fj
x
O
Solution: From the diagram, the important points in this problem
z
are A (8, 6, 0), B (0, 10, 4), C (0, 6, 10), and the origin O (0, 0, 0)
with all dimensions in ft. We need unit vectors in the directions A to
B and A to C. Both vectors are of the form
eAP D xP xA i C yP yA j C zP zA k,
If we carry through these operations in the sequence described, we get
the following vectors:
where P can be either A or B. The forces in cables AB and AC are
TAB D TAB eAB D TABX i C TABY j C TABZ k,
and TAC D TAC eAB D TACX i C TACY j C TACZ k.
eAB D 0.816i C 0.408j C 0.408k,
eAC D 0.625i C 0j 0.781k,
TAB D 387.1i C 193.5j C 193.5k lb,
The weight force is
jTAB j D 474.1 lb,
F D 0i 600j C 0k,
and the support force at the ball joint is
TAC D 154.8i C 0j 193.5k lb,
jTAC j D 247.9 lb,
S D SX i C SY j C SZ k.
The vector form of the force equilibrium equation (which gives three
scalar equations) for the bar is
MAB D rOA &eth; TAB D 1161i 1548j C 3871k ft-lb,
MAC D rOA &eth; TAC D 1161i C 1548j C 929k ft-lb,
TAB C TAC C F C S D 0.
Let us take moments about the origin. The moment equation, in vector
form, is given by
and S D 541.9i C 406.5j C 0k lb
MO D rOA &eth; TAB C rOA &eth; TAC
C rOA &eth; F D 0,
where rOA D 8i C 6j C 0k.
The cross products are evaluated using the form
i
M D r &eth; H D 8
HX
j
6
HY
k 0 ,
HZ where H can be any of the three forces acting at point A. The vector
moment equation provides another three equations of equilibrium.
Once we have evaluated and applied the unit vectors, we have six
vector equations of equilibrium in the five unknowns TAB , TAC , SX , SY ,
and SZ (there is one redundant equation since all forces pass through
the line OA). Solving these equations yields the required values for
the support reactions at the origin.
330
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 5.90 In Problem 5.89, suppose that the
suspended load exerts a force F D 600 lb at A and bar
OA weighs 200 lb. Assume that the bar’s weight acts at
its midpoint. Determine the tensions in the cables and
the reactions at the ball and socket support O.
Solution: Point G is located at (4, 3, 0) and the position vector of
G with respect to the origin is
rOG D 4i C 3j C 0k ft.
The weight of the bar is
WB D 0i 200j C 0k lb,
and its moment around the origin is
MWB D 0i C 0j 800k ft-lb.
The mathematical representation for all other forces and moments from
Problem 5.89 remain the same (the numbers change!). Each equation
of equilibrium has a new term reflecting the addition of the weight of
the bar. The new force equilibrium equation is
TAB C TAC C F C S C WB D 0.
The new moment equilibrium equation is
MO D rOA &eth; TAB C rOA &eth; TAC
C rOA &eth; F C rOG &eth; WB D 0.
As in Problem 5.89, the vector equilibrium conditions can be reduced
to six scalar equations of equilibrium. Once we have evaluated and
applied the unit vectors, we have six vector equations of equilibrium
in the five unknowns TAB , TAC , SX , SY , and SZ (As before, there is one
redundant equation since all forces pass through the line OA). Solving
these equations yields the required values for the support reactions at
the origin.
If we carry through these operations in the sequence described, we get
the following vectors:
eAB D 0.816i C 0.408j C 0.408k,
eAC D 0.625i C 0j 0.781k,
TAB D 451.6i C 225.8j C 225.8k lb,
jTAB j D 553.1 lb,
TAC D 180.6i C 0j 225.8k lb,
jTAC j D 289.2 lb,
MAB D rOA &eth; TAB D 1355i 1806j C 4516k ft-lb,
MAC D rOA &eth; TAC D 1354i C 1806j C 1084k ft-lb,
and S D 632.3i C 574.2j C 0k lb
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
331
Problem 5.91 The 158,000-kg airplane is at rest on
the ground (z D 0 is ground level). The landing gear
carriages are at A, B, and C. The coordinates of the point
G at which the weight of the plane acts are (3, 0.5, 5) m.
What are the magnitudes of the normal reactions exerted
on the landing gear by the ground?
21 m
6m
B
G
A
x
C
6m
y
Solution:
mg
3m
FY D NL C NR C NF W D 0
MR D 3 mg C 21NF D 0
21 m
Side
View
x
R
Solving,
NF D 221.4 kN
F
(NL + NR)
(1)
NF
Z
NL C NR D 1328.6 kN
(2)
0.5 m
W
FY D NR C NL C NF W D 0
(same equation as before)
C
MO D 0.5 W 6NR C 6NL D 0
(3)
Front
View
y
Solving (1), (2), and (3), we get
6
6
NF D 221.4 kN
NR D 728.9 kN
NF
NR
NL
z
NL D 599.7 kN
332
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 5.92 The horizontal triangular plate is
suspended by the three vertical cables A, B, and C.
The tension in each cable is 80 N. Determine the x
and z coordinates of the point where the plate’s weight
effectively acts.
y
A
B
C
0.3 m
0.4 m
(x, 0, z)
x
z
Solution:
80 N
80 N
Mx : 240 Nz 80 N0.4 m D 0
Mz : 80 N0.3 m 240 Nx D 0
z
80 N
x
X
Solving
x D 0.1 m, z D 0.1333 m
240 N
Z
Problem 5.93 The 800-kg horizontal wall section is
supported by the three vertical cables, A, B, and C. What
are the tensions in the cables?
B
7m
Solution: All dimensions are in m and all forces are in N. Forces
A, B, C, and W act on the wall at (0, 0, 0), (5, 14, 0), (12, 7, 0),
and (4, 6, 0), respectively. All forces are in the z direction. The force
equilibrium equation in the z direction is A C B C C W D 0. The
moments are calculated from
MB D rOB &eth; Bk,
C
A
7m
7m
6m
4m
8m
mg
MC D rOC &eth; Ck,
and MG D rOG &eth; Wk.
The moment equilibrium equation is
MO D MB C MC C MG D 0.
Carrying out these operations, we get
A D 3717 N,
B D 2596 N,
C D 1534 N,
and W D 7848 N.
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
333
Problem 5.94 The bar AC is supported by the cable
BD and a bearing at A that can rotate about the z axis.
The person exerts a force F D 10j (lb) at C. Determine
the tension in the cable and the reactions at A.
y
A
x
B
C
8 in
14 in
z
(18, ⫺8, 7) in
D
Solution: The force in the cable is
TBD D TBD
10i 8j C 7k
p
213
Ax, MAx
Az
F = 10 lbj
We have the following six equilibrium equations
10
TBD D 0
Fx : Ax C p
213
8
TBD C 10 lb D 0
Fy : Ay p
213
Ay
MAy
TBD
7
TBD D 0
Fz : Az C p
213
Mx : MAx D 0
My : MAy p
7
213
TBD 8 in D 0
8
TBD 8 in C 10 lb22 in D 0
Mz : p
213
Solving we find
Ax D 34.4 lb, Ay D 17.5 lb, Az D 24.1 lb
MAx D 0, MAy D 192.5 lb in, TBD D 50.2 lb
334
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
y
Problem 5.95 The L-shaped bar is supported by a
bearing at A and rests on a smooth horizontal surface
at B. The vertical force F D 4 kN and the distance
b D 0.15 m. Determine the reactions at A and B.
F
b
A
x
Solution: Equilibrium Eqns:
B
FX :
0.2 m
OD0
0.3 m
z
FY :
AY C B F D 0
FZ :
AZ D 0
y
Sum moments around A
F
b
x:
Fb 0.3B D 40.15 0.3B D 0
y:
M AY D 0
z:
MAZ C 0.2F 0.2B D 0
MZ
0.2 m
A
B
3
0.
m
x
AY
b = 0.15 m
F = 4 kN
AZ
B
z
(MAX ≡ 0)
AX ≡ 0
MY
Solving,
AX D 0,
AY D 2 (kN),
AZ D 0
MAX D 0,
MAY D 0,
MAZ D 0.4 (kN-m)
Problem 5.96 In Problem 5.95, the vertical force F D
4 kN and the distance b D 0.15 m. If you represent the
reactions at A and B by an equivalent system consisting
of a single force, what is the force and where does its
line of action intersect the xz plane?
Solution: We want to represent the forces at A &amp; B by a single
zR 4 D C0.32
force. From Prob. 5.95
zR D C0.15 m
A D C2j (kN),
xR 4 D 0.22
B D C2j (kN)
xR D 0.1 m
MA D 0.4k (kN-m)
y
We want a single equivalent force, R that has the same resultant force
and moment about A as does the set A, B, and MA .
R
F
b
A
R D A C B D 4j (kN)
x
Let R pierce the xz plane at xR , zR B
MX :
zR R D 0.3B
z
MZ :
0.2 m
0.3 m
xR R D 0.2AY
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
335
Problem 5.97 In Problem 5.95, the vertical force F D
4 kN. The bearing at A will safely support a force of
2.5-kN magnitude and a couple of 0.5 kN-m magnitude.
Based on these criteria, what is the allowable range of
the distance b?
Solution: The solution to Prob. 5.95 produced the relations
AY C B F D 0
F D 4 kN
Fb 0.3B D 0
MAZ C 0.2F 0.2B D 0
AX D AZ D MAX D MAY D 0
Set the force at A to its limit of 2.5 kN and solve for b. In this case,
MAZ D 0.5 (kN-m) which is at the moment limit. The value for b is
b D 0.1125 m
We make AY unknown, b unknown, and B unknown F D 4 kN,
MAY D C0.5 (kN-m), and solve we get AY D 2.5 at b D 0.4875 m
However, 0.3 is the physical limit of the device.
Thus, 0.1125 m b 0.3 m
y
F
b
A
x
B
z
336
0.2 m
0.3 m
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 5.98 The 1.1-m bar is supported by a ball
and socket support at A and the two smooth walls. The
tension in the vertical cable CD is 1 kN.
(a)
(b)
y
B
Draw the free-body diagram of the bar.
Determine the reactions at A and B.
400 mm
D
x
C
(a)
(b)
600 mm
700 mm
z
Solution:
A
From which,
The ball and socket cannot support a couple reaction, but can
support a three force reaction. The smooth surface supports oneforce normal to the surface. The cable supports one force parallel
to the cable.
The strategy is to determine the moments about A, which will
contain only the unknown reaction at B. This will require the
position vectors of B and D relative to A, which in turn will
require the unit vector parallel to the rod. The angle formed by the
bar with the horizontal is required to determine the coordinates
of B:
˛ D cos1
p
0.62 C 0.72
1.1
BZ D
0.3819
D 0.6365 kN,
0.6
BX D
0.4455
D 0.7425 kN.
0.6
The reactions at A are determined from the sums of the forces:
D 33.1&deg; .
FX D BX C AX i D 0, from which AX D 0.7425 kN.
FY D AY 1j D 0, from which AY D 1 kN.
FZ D BZ C AZ k D 0, from which AZ D 0.6365 kN
The coordinates of the points are: A (0.7, 0, 0.6), B 0, 1.1
sin 33.1&deg; , 0 D 0, 0.6, 0, from which the vector parallel to the
bar is
FB
rAB D rB rA D 0.7i C 0.6j 0.6k (m).
The unit vector parallel to the bar is
eAB
T
FY
rAB
D 0.6364i C 0.5455j 0.5455k.
D
1.1
FZ
The vector location of the point D relative to A is
FX
rAD D 1.1 0.4eAB D 0.7eAB
D 0.4455i C 0.3819j 0.3819k.
The reaction at B is horizontal, with unknown x-component and
z-components. The sum of the moments about A is
i
MA D rAB &eth; B C rAD &eth; D D 0 D 0.7
BX
i
C 0.4455
0
j
0.3819
1
j
0.6
0
k 0.6 BZ k
0.3819 D 0
0
Expand and collect like terms:
MA D 0.6BZ 0.3819i 0.6BX 0.7BZ j
C0.6BX C 0.4455k D 0.
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
337
Problem 5.99 The 8-ft bar is supported by a ball and
socket at A, the cable BD, and a roller support at C. The
collar at B is fixed to the bar at its midpoint. The force
F D 50k (lb). Determine the tension in the cable BD
and the reactions at A and C.
y
A
3 ft
Solution: The strategy is to determine the sum of the moments
B
F
about A, which will involve the unknown reactions at B and C. This
will require the unit vectors parallel to the rod and parallel to the cable.
z
The angle formed by the rod is
4 ft
D
2 ft
3
D 22&deg; .
˛ D sin1
8
C
x
The vector positions are:
rA D 3j,
from which jTj D
rD D 4i C 2k
CY D
and rC D 8 cos 22&deg; i D 7.4162i.
75
D 62.92 lb
1.192
2.036jTj
D 17.27 lb.
7.4162
The reaction at A is determined from the sums of forces:
The vector parallel to the rod is
FX D AX C 0.1160jTji D 0,
rAC D rC rA D 7.4162i 3j.
from which AX D 7.29 lb,
The unit vector parallel to the rod is
eAC D 0.9270i 0.375j.
FY D AY 0.5960jTj C CY j D 0,
from which AY D 20.23 lb
The location of B is
rAB D 4eAC D 3.7081i 1.5j.
FZ D AZ C 0.7946jTj 50k D 0,
from which AZ D 0 lb
The vector parallel to the cable is
rBD D rD rA C rAB D 0.2919i 1.5j C 2k.
y
The unit vector parallel to the cable is
A
eBD D 0.1160i 0.5960j C 0.7946k.
F
B
3 ft
The tension in the cable is T D jTjeBD . The reaction at the roller
support C is normal to the xz plane. The sum of the moments about
A
C
D
z
4 ft
MA D rAB &eth; F C rAB &eth; T C rAC &eth; C D 0
i
D 3.7081
0
j
k 1.5
0 0
50 i
C jTj 3.7081
0.1160
i
C 7.4162
0
2 ft
j
1.5
0.5960
x
y
k 0 0.7946 j k 3 0 D 0
CY 0 AY
AZ
AX
T
F
z
D
D 75i C 185.4j C jTj1.192i 2.9466j 2.036k
CY
x
C 7.4162CY k D 0,
338
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 5.100 Consider the 8-ft bar in Problem 5.99.
The force F D Fy j 50k (lb). What is the largest value
of Fy for which the roller support at C will remain on
the floor?
Solution: From the solution to Problem 5.99, the sum of the
j
k 1.5
0 FY 50 i
MA D 3.7081
0
i
C jTj 3.7081
0.1160
i
C 7.4162
0
j
1.5
0.5960
k 0 0.7946 j k 3 0 D 0
CY 0 D 75i C 185.4j C 3.7081FY k
C jTj1.192i 2.9466j 2.036k
C 7.4162CY k D 0,
from which, jTj D
75
D 62.92 lb.
1.192
Collecting terms in k, 3.7081FY C 2.384jTj 7.4162CY D 0.
For CY D 0, FY D
128.11
D 34.54 lb
3.708
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
339
Problem 5.101 The tower is 70 m tall. The tension in
each cable is 2 kN. Treat the base of the tower A as a
built-in support. What are the reactions at A?
y
B
Solution: The strategy is to determine moments about A due to
the cables. This requires the unit vectors parallel to the cables.
C
40 m
The coordinates of the points are:
A0, 0, 0, B0, 70, 0, C50, 0, 0,
50 m
E
A
D20, 0, 50, E40, 0, 40.
40 m
D
The unit vectors parallel to the cables, directed from B to the points
E, D, and C
50 m
z
20 m
x
rBE D 40i 70j 40k,
The force reactions at A are determined from the sums of forces. (Note
that the sums of the cable forces have already been calculated and used
above.)
rBD D 20i 70j C 50k,
rBC D 50i 70j.
FX D AX C 0.17932i D 0,
The unit vectors parallel to the cables, pointing from B, are:
from which AX D 0.179 kN,
eBE D 0.4444i 0.7778j 0.4444k,
FY D AY 4.7682j D 0,
eBD D 0.2265i 0.7926j C 0.5661k,
from which AY D 4.768 kN,
eBC D 0.5812i 0.8137j C 0k.
The tensions in the cables are:
TBD D 2eBD D 0.4529i 1.5852j C 1.1323k (kN),
FZ D AZ C 0.2434k D 0,
from which AZ D 0.2434 kN
TBE D 2eBE D 0.8889i 1.5556j 0.8889k (kN),
y
B
TBC D 2eBC D 1.1625i 1.6275j 0k.
TBC
The sum of the moments about A is
TBD
MA D MA C rAB &eth; TBE
AY ,
C
A
MY
rAB &eth; TBD C rAB &eth; TBC D 0
C
D MA C rAB &eth; TBE C TBC C TBD i
A
MA D M C 0
0.1793
TBE
j
70
4.7682
k 0 D 0
0.2434 AZ ,MA
EA
AX , MA
Z
z
D
X
x
D MAX C 17.038i C MAY C 0j
C MAZ 12.551k D 0
from which
MAX D 17.038 kN-m,
MAY D 0,
MAZ D 12.551 kN-m.
340
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 5.102 Consider the tower in Problem 5.101.
If the tension in cable BC is 2 kN, what must the tensions in cables BD and BE be if you want the couple
exerted on the tower by the built-in support at A to be
zero? What are the resulting reactions at A?
Solution: From the solution to Problem 5.101, the sum of the
moments about A is given by
MA D MA C rAB &eth; TBE C TBC C TBD D 0.
If the couple MA D 0, then the cross product is zero, which is possible
only if the vector sum of the cable tensions is zero in the x and z
directions. Thus, from Problem 5.101,
ex &ETH; TBC C jTBE jeBE C jTBD jeBD D 0,
and ez &ETH; TBC C jTBE jeBE C jTBD jeBD D 0.
Two simultaneous equations in two unknowns result;
0.4444jTBE j C 0.2265jTBD j D 1.1625
0.4444jTBE j C 0.5661jTBD j D 0.
Solve:
jTBE j D 1.868 kN,
jTBD j D 1.467 kN.
The reactions at A oppose the sum of the cable tensions in the x-, y-,
and z-directions.
AX D 0,
AY D 4.243 kN,
AZ D 0.
(These results are to be expected if there is no moment about A.)
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
341
Problem 5.103 The space truss has roller supports at
B, C, and D and is subjected to a vertical force F D
20 kN at A. What are the reactions at the roller supports?
y
F
A (4, 3, 4) m
B
Solution: The key to this solution is expressing the forces in terms
D (6, 0, 0) m
of unit vectors and magnitudes-then using the method of joints in three
dimensions. The points A, B, C, and D are located at
x
A4, 3, 4 m, B0, 0, 0 m,
C5, 0, 6 m, D6, 0, 0 m
z
C (5, 0, 6) m
we need eAB , eAC , eAD , eBC , eBD , and eCD . Use the form
ePQ D
xQ xP i C yQ yP j C zQ zP k
[xQ xP 2 C yQ yP 2 C zQ zP 2 ]1/2
F
Joint A :
eAB D 0.625i 0.469j 0.625k
TAB
eAC D 0.267i 0.802j C 0.535k
TAC
D
B
eBC D 0.640i C 0j C 0.768k
C
eBD D 1i C 0j C 0k
eCD D 0.164i C 0j 0.986k
Joint B :
–TAB
We will write each force as a magnitude times the appropriate unit
vector.
TAB D TAB eAB , TAC D TAC eAC
TBD
NBJ
TBC
Joint C :
TBD D TBD eBD , TCD D TCD eCD
–TAC
Each force will be written in component form, i.e.

TABX D TAB eABX 


TABY D TAB eABY
etc.



TABZ D TAB eABZ
Joint A:
TAB C TAC C TAD C F D 0
TABX C TACX C TADX D 0
TABY C TACY C TADY 20 D 0
TCD
–TBC
NCJ
Joint D :
–TBD
–TCD
NDJ
TABZ C TACZ C TADZ D 0
Joint B:
TAB C TBC C TBD C NB j D 0
Joint C:
TAC TBC C TCD C NC j D 0
Joint D:
TAD TBD TCD C ND j D 0
Solving for all the unknowns, we get
NB D 4.44 kN
NC D 2.22 kN
ND D 13.33 kN
Also, TAB D 9.49 kN, TAC D 16.63 kN
TAD D 3.99 kN, TBC D 7.71 kN
TBD D 0.99 kN, TCD D 3.00 kN
342
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
y
Problem 5.104 In Example 5.8, suppose that the cable
BD is lengthened and the attachment point B is moved to
the end of the bar at C. The positions of the attachment
point D and the bar are unchanged. Draw a sketch of
the bar showing cable BD in its new position. Draw the
free-body diagram of the bar and apply equilibrium to
determine the tension in the cable and the reactions at A.
D
(2, 2, ⫺1) ft
x
A
30⬚
B
z
C
⫺100j (lb)
Solution: The sketch and free-body diagram are shown.
We must express the force exerted on the bar by cable BD in terms of
its components. The bar AC is 4 ft long. The vector from C to D is
rCD D [2 4 cos 30&deg; i C 2 f4 sin 30&deg; gj
C 1 0j] ft
rCD D 1.46i C 4j k ft
The force exerted by the cable CD can be expressed
T
rCD
D T0.335i C 0.914j 0.229k
jrCD j
The equilibrium equations are
Fx : Ax 0.335T D 0
Fy : Ay C 0.914T 100 lb D 0
Fz : Az 0.229T D 0
MA : MAx i C MAy j
i
C 3.464
0.335T
j
2
0.914T 100
k
D0
0
0.229T Expanding the determinant and solving the six equations, we obtain
T D 139 lb, Ax D 46.4 lb, Ay D 26.8 lb, Az D 31.7 lb
MAx D 63.4 ft-lb, MAy D 110 ft-lb
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
343
Problem 5.105 The 40-lb door is supported by hinges
at A and B. The y axis is vertical. The hinges do not
exert couples on the door, and the hinge at B does not
exert a force parallel to the hinge axis. The weight of
the door acts at its midpoint. What are the reactions at
A and B?
y
4 ft
1 ft
B
Solution: The position vector of the midpoint of the door:
5 ft
rCM D 2 cos 50&deg; i C 3.5j C 2 cos 40&deg; k
A
D 1.2856i C 3.5j C 1.532k.
1 ft
x
The position vectors of the hinges:
40
rA D j,
rB D 6j.
z
The forces are: W D 40j,
y
A D AX i C AY j C AZ k,
BX
B D BX i C BZ k.
BZ
The position vectors relative to A are
AY
rACM D rCM rA D 1.2856i C 2.5j C 1.532k,
rAB D rB rA D 5j.
AZ
The sum of the moments about A
AX
x
z
MA D rACM &eth; W C rAB &eth; B
i
D 1.2856
0
W
j
2.5
40
k i
1.532 C 0
0 BX
j k 5 0 D 0
0 BZ MA D 5BZ C 401.532i C 5BX 401.285k D 0,
from which BZ D
401.532
D 12.256 lb
5
and
401.285
D 10.28 lb.
5
BX D
The reactions at A are determined from the sums of forces:
FX D AX C BX i D 0,
from which AX D 10.28 lb,
FY D AY 40j D 0,
from which AY D 40 lb,
FZ D AZ C BZ k D 0,
from which AZ D 12.256 lb
344
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 5.106 The vertical cable is attached at A.
Determine the tension in the cable and the reactions at
the bearing B due to the force F D 10i 30j 10k (N).
y
200 mm
100 mm
100 mm
Solution: The position vector of the point of application of the
force is
B
200 mm
rF D 0.2i 0.2k.
F
z
The position vector of the bearing is
x
A
rB D 0.1i.
The position vector of the cable attachment to the wheel is
rC D 0.1k.
he position vectors relative to B are:
rBC D rC rB D 0.1i C 0.1k,
rBF D rF rB D 0.1i 0.2k.
The sum of the moments about the bearing B is
or
MB D MB C rBF &eth; F C rBC &eth; C D 0,
i
MB D MB C 0.1
10
j
j
k i
0
0.2 C 0.1 0
T
30 10 0
k 0.1 0 D 6 C 0.1Ti C MBY 1j
C MBZ 3 C 0.1Tk D 0,
from which T D
6
D 60 N,
0.1
MBY D C1 N-m,
MBZ D 0.1T C 3 D 3 N-m.
The force reactions at the bearing are determined from the sums of
forces:
FX D BX C 10i D 0,
from which BX D 10 N.
FY D BY 30 60j D 0,
from which BY D 90 N.
FZ D BZ 10j D 0,
from which BZ D 10 N.
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
345
Problem 5.107 In Problem 5.106, suppose that the z
component of the force F is zero, but otherwise F is
unknown. If the couple exerted on the shaft by the
bearing at B is MB D 6j 6k N-m, what are the force
F and the tension in the cable?
Solution: From the diagram of Problem 5.106, the force equilibrium equation components are
and
Fx D BX C FX D 0,
Fy D BY C FY D 0,
Fz D BZ C FZ D 0,
where FZ D 0 is given in the problem statement. The moment
equations can be developed by inspection of the figure also. They are
and
Mx D MBX C MAX C MFX D 0,
MY D MBY C MAY C MFY D 0,
MZ D MBZ C MAZ C MFZ D 0,
where MB D 6j 6k N-m. Note that MBX D 0 can be inferred. The
moments which need to be substituted into the moment equations are
MA D 0.1Ai C 0j C 0.1Ak N-m,
and MF D 0.2FY i 0.2FX j C 0.1FY k N-m.
Substituting these values into the equilibrium equations, we get F D
30i 60j C 0k N, and A D 120 N.
346
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
y
Problem 5.108 The device in Problem 5.106 is badly
designed because of the couples that must be supported
by the bearing at B, which would cause the bearing to
“bind”. (Imagine trying to open a door supported by
only one hinge.) In this improved design, the bearings
at B and C support no couples, and the bearing at C
does not exert a force in the x direction. If the force
F D 10i 30j 10k (N), what are the tension in the
vertical cable and the reactions at the bearings B and C?
200 mm
50 mm
100 mm
50 mm
200 mm
B
C
F
z
x
A
Solution: The position vectors relative to the bearing B are: the
position vector of the cable attachment to the wheel is
y
rBT D 0.05i C 0.1k.
BY
The position vector of the bearing C is:
BX CY
BZ
F
CZ
z
rBC D 0.1i.
T
x
The position vector of the point of application of the force is:
rBF D 0.15i 0.2k.
The sum of the moments about B is
MB D rBT &eth; T C rBC &eth; C C rBF &eth; F D 0
i
j
MB D 0.05 0
0
T
i
C 0.15
10
j
0
30
j
k i
0.1 C 0.1 0
0 0 CY
k 0 CZ k 0.2 D 0
10 MB D 0.1T 6i C 0.1CZ C 1.5 2j
C 0.05T C 0.1CY 4.5k D 0.
From which: T D 60 N,
CZ D
0.5
D 5 N,
0.1
CY D
4.5 0.05T
D 15 N.
0.1
The reactions at B are found from the sums of forces:
FX D BX C 10i D 0,
from which BX D 10 N.
FY D BY C CY T 30j D 0,
from which BY D 75 N.
FZ D BZ C CZ 10k D 0,
from which BZ D 15 N
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
347
Problem 5.109 The rocket launcher is supported by
the hydraulic jack DE and the bearings A and B. The
bearings lie on the x axis and support shafts parallel to
the x axis. The hydraulic cylinder DE exerts a force on
the launcher that points along the line from D to E. The
coordinates of D are (7, 0, 7) ft, and the coordinates of
E are (9, 6, 4) ft. The weight W D 30 kip acts at (4.5,
5, 2) ft. What is the magnitude of the reaction on the
launcher at E?
y
W
E
A
B
x
D
3 ft 3 ft
Solution: The position vectors of the points D, E and W are
rD D 7i C 7k,
rE D 9i C 6j C 4k (ft),
rW D 4.5i C 5j C 2k (ft).
The vector parallel to DE is
rDE D rE rD D 2i C 6j 3k.
The unit vector parallel to DE is
eDE D 0.2857i C 0.8571j 0.4286k.
Since the bearings cannot exert a moment about the x axis, the sum of
the moments due to the weight and the jack force must be zero about
the x axis. The sum of the moments about the x axis is:
1
1
0
0 5
2 CjFDE j 9
MX D 4.5
0 30 0 0.2857
0
0
D0
6
4
0.8571 0.4286 D 60 6jFDE j D 0.
From which
jFDE j D
60
D 10 kip
6
y
30 kip
BY
AY
AX
AZ
BZ
x
FDE
z
348
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 5.110 Consider the rocket launcher described
in Problem 5.109. The bearings at A and B do not exert
couples, and the bearing B does not exert a force in the
x direction. Determine the reactions at A and B.
Solution: See the solution of Problem 5.109. The force FDE can
be written
FDE D FDE 0.2857i C 0.8571j 0.4286k.
The equilibrium equations are
FX D AX C 0.2857FDE D 0,
FY D AY C BY C 0.8571FDE 30 D 0,
FZ D AZ C BZ 0.4286FDE D 0,
Morigin
i
D 3
AX
j
0
AY
k i
0 C 6
AZ 0
i
C FDE 7
0.2857
j
0
BY
j
0
0.8571
k 0 BZ k
7
0.4286 i
j
k C 4.5
5
2 D 0
0 30 0 The components of the moment eq. are
5.9997FDE C 60 D 0,
3AZ 6BZ C 5.0001FDE D 0,
3AY C 6BY C 5.9997FDE 135 D 0.
Solving, we obtain
FDE D 10.00 kip, AX D 2.86 kip,
AY D 17.86 kip, AZ D 8.09 kip,
BY D 3.57 kip, BZ D 12.38 kip.
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
349
Problem 5.111 The crane’s cable CD is attached to a
stationary object at D. The crane is supported by the
bearings E and F and the horizontal cable AB. The
tension in cable AB is 8 kN. Determine the tension in
the cable CD.
Strategy: Since the reactions exerted on the crane by
the bearings do not exert moments about the z axis, the
sum of the moments about the z axis due to the forces
exerted on the crane by the cables AB and CD equals
zero. (See the discussion at the end of Example 5.9.)
y
Solution: The position vector from C to D is
C
rCD D 3i 6j 3k (m),
A
so we can write the force exerted at C by cable CD as
B
TCD
rCD
D TCD
D TCD 0.408i 0.816j 0.408k.
jrCD j
The coordinates of pt. B are x D
4
3 D 2 m, y D 4 m.
6
F
E
z
The moment about the origin due to the forces exerted by the two
cables is
i
MO D 2
8
i
j k 3
4 0 C 0 0 0.408TCD
j
6
0.816TCD
2m
k
0
0.408TCD 2m
D
3m
y
x
C
D 32k 2.448TCD i C 1.224TCD j 4.896TCD k.
A
The moment about the z axis is
B
6m
k &ETH; MO D 32 4.896TCD D 0,
so
TCD D 6.54 kN.
4m
D
3m
350
x
3m
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
y
Problem 5.112 In Example 5.9, suppose that the cable
CE is shortened and its attachment point E is moved to
the point (0, 80, 0) mm. The plate remains in the same
position. Draw a sketch of the plate and its supports
showing the new position of cable CE. Draw the freebody diagram of the plate and apply equilibrium to determine the reactions at the hinges and the tension in the
cable.
100
mm
E
80 mm
A
B
z
C
200 mm
D
200 mm
x
⫺400j (N)
Solution: The sketch and free-body diagram are shown.
The vector from C to E is
rCE D [0 200i C 80 0j C 0 0k] mm
D 200i C 80j mm
The force exerted by cable CE can be expresses as
T
rCE
D T0.928i C 0.371j
jrCE j
The equilibrium equations for the plate are
Fx : Ax C Bx 0.928T D 0
Fy : Ay C By C 0.371T 400 N D 0
Fz : Az C Bz D 0
i
MB : 0.2
0.928T
i
C 0
Ax
j
0
Ay
j
k 0
0 0.371T 0 j
k i
0
0.2 C 0.2
0 0 400
k 0.2 D 0
0 Expanding and solving we find
Ax D 0, Ay D 400 N, T D 1080 N
Bx D 1000 N, By D 400 N, Bz D 0,
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
351
Problem 5.113 The plate is supported by hinges at A
and B and the cable CE, and it is loaded by the force at
D. The edge of the plate to which the hinges are attached
lies in the yz plane, and the axes of the hinges are
parallel to the line through points A and B. The hinges
do not exert couples on the plate. What is the tension in
cable CE?
y
3m
2i – 6j (kN)
E
A
D
2m
1m
Solution:
B
z
C
20&deg;
2m
F D A C B C FD C TCE D 0
However, we just want tension in CE. This quantity is the only
unknown in the moment equation about the line AB. To get this, we
need the unit vector along CE.
y
3m
Point C is at (2, 2 sin 20&deg; , 2 cos 20&deg; ) Point E is at (0, 1, 3)
eCE
AX
A
1m
z
eCE D 0.703i C 0.592j C 0.394k
BZ
BY
AZ
B
20&deg;
FD = 2i – 6j
AY
E
rCE
D
jrCE j
We also need the unit vector eAB . A(0, 0, 0), B0, 2 sin 20&deg; , 2 cos 20&deg; x
D
x
TCE
2m
BX
2m
C
eAB D 0i 0.342j C 0.940k
The moment of FD about A (a point on AB) is
MFD D rAD &eth; FD1 D 2i &eth; 2i 6j
MFD D 12k
The moment of TCE about B (another point on line CE) is
MTCE D rBC &eth; TCE eCE D 2i &eth; TCE eCE ,
where eCE is given above.
The moment of FD about line AB is
MFDAB D MFD &ETH; eAB
MFDAB D 11.27 N-m
The moment of TCE about line AB is
MCEAB D TCE 2i &eth; eCE &ETH; eAB
MCEAB D TCE 0.788j C 1.184k &ETH; eAB
MCEAB D 1.382TCE
The sum of the moments about line AB is zero. Hence
MFDAB C MCEAB D 0
11.27 C 1.382TCE D 0
TCE D 8.15 kN
352
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 5.114 In Problem 5.113, the hinge at B does
not exert a force on the plate in the direction of the hinge
axis. What are the magnitudes of the forces exerted on
the plate by the hinges at A and B?
Solution: From the solution to Problem 5.113, TCE D 8.15 kN
y
Also, from that solution,
F = + 2i – 6j (kN)
AY
eAB D 0i 0.342j C 0.940k
AX
We are given that the force at force at hinge B does not exert a force
parallel to AB at B. This implies
θ
z
BY
B &ETH; eAB D 0.
B &ETH; eAB D 0.342BY C 0.940BZ D 0
D
AZ
x
TCE
2
BX
C (2, –2sinθ , + 2cosθ)
(1)
BZ
We also had, in the solution to Problem 5.113
eCE D 0.703i C 0.592j C 0.394k
and
TCE D TCE eCE (kN)
For Equilibrium,
F D A C B C TCE C F D 0
FX :
AX C BX C TCE eCEX C 2 D 0 (kN)
(2)
FY :
AY C BY C TCE eCEY 6 D 0 (kN)
(3)
FZ :
AZ C BZ C TCE eCEZ D 0 (kN)
(4)
Summing Moments about A, we have
rAD &eth; F C rAC &eth; TCE C rAB &eth; B D 0
rAD &eth; F D 2i &eth; 2i 6j D 12k (kN)
rAC &eth; TCE D 2 sin TZ 2 cos TY i
C 2 cos TX 2TZ j
C 2TY C 2TX sin k
rCE &eth; B D 2BZ sin 2BY cos i
C 2BX cos j C C2BX sin k
MA D 0,
Hence
x:
2 sin TZ 2 cos TY 2BZ sin 2BY cos D 0
(5)
y:
2 cos TX 2TZ C 2BX cos D 0
(6)
z:
12 C 2TY C 2TX sin C 2BX sin D 0
(7)
Solving Eqns (1)–(7), we get
jAj D 8.53 (kN), jBj D 10.75 (kN)
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
353
Problem 5.115 The bar ABC is supported by ball and
socket supports at A and C and the cable BD. The
suspended mass is 1800 kg. Determine the tension in
the cable.
(⫺2, 2, ⫺1) m
y
D
2m
4m
B
A
C
x
4m
z
Solution: We take moments about the line AC to eliminate the
TBD
reactions at A and C.
We have
rAB D 4 mk, TBD D TBD
2i C 2j k
3
Az
rAW D 2i 4k m, W D 1800 kg9.81 m/s2 j
eCA D
1
6i 4k
p
D p 3i 2k
52
13
Ax
Cx
Ay
17.66 kN
The one equilibrium equation we need is
MAC D eCA &ETH; rAB &eth; TAB C rAW &eth; W D 0
Cz
Cy
This equation reduces to the scalar equation
1
p 3i 2k &ETH;
13
8
8
m TBD i C
m TBD j [4 m][17.66 kN]i
3
3
[2 m][17.66 kN]k D 0
8
1
p 3
m TBD [4 m][17.66 kN] C 2 f[2 m][17.66 kN]g D 0
3
2
Solving we find
TBD D 17.66 kN
354
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 5.116* In Problem 5.115, assume that the
ball and socket support at A is designed so that it
exerts no force parallel to the straight line from A to
C. Determine the reactions at A and C.
Solution: We have
TBD D TBD
2i C 2j k
3
There are 7 unknowns. We have the following 6 equilibrium equations
Fx : Ax C Cx 2
TBD D 0
3
Fy : Ay C Cy C
2
TBD 17.66 kN D 0
3
Fz : Az C Cz 1
TBD D 0
3
Mx : Cy 4 m D 0
My : Cx 4 m Az 6 m D 0
Mz : Ay 6 m 17.66 kN2 m D 0
The last equation comes from the fact that the ball and socket at A
exerts no force in the direction of the line from A to C
6i C 4k
1
p
Ax i C Ay j C Az k &ETH;
D p 6Ax C 4Az D 0
52
52
Solving these 7 equations we find
Ax D 3.62 kN, Ay D 5.89 kN, Az D 5.43 kN
Cx D 8.15 kN, Cy D 0, Cz D 0.453 kN
TBD D 17.66 kN
TBD
Az
Ax
Cx
Ay
17.66 kN
Cz
Cy
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
355
y
Problem 5.117 The bearings at A, B, and C do not
exert couples on the bar and do not exert forces in the
direction of the axis of the bar. Determine the reactions
at the bearings due to the two forces on the bar.
200 i (N)
300 mm
x
Solution: The strategy is to take the moments about A and solve
C
180 mm
the resulting simultaneous equations. The position vectors of the bearings relative to A are:
B
z
rAB D 0.15i C 0.15j,
A
rAC D 0.15i C 0.33j C 0.3k.
100 k (N)
Denote the lower force by subscript 1, and the upper by subscript 2:
150 mm
rA1 D 0.15i,
rA2 D 0.15i C 0.33j.
y
CY
The sum of the moments about A is:
i
C 0.15
200
k i
0 C 0.15
100 BX
j
k 0.15 0 0
BZ j
k i
0.33 0 C 0.15
0
0 CX
j
0.33
CY
k 0.3 D 0
0 200 N
x
MA D rA1 &eth; F1 C rAB &eth; B C rA2 &eth; F2 C rAC &eth; C D 0
i
j
MA D 0.15 0
0
0
150 mm
z
CX
BX
BZ
100 N
AY
AZ
MA D 0.15BZ 0.3CY i C 15 C 0.15BZ C 0.3CX j
C 0.15BX 66 0.15CY 0.33CX k D 0.
This results in three equations in four unknowns; an additional equation
is provided by the sum of the forces in the x-direction (which cannot
have a reaction term due to A)
FX D BX C CX C 200i D 0.
The four equations in four unknowns:
0BX C 0.15BZ C 0CX 0.3CZ D 0
0BX C 0.15BZ C 0.3CX C 0CY D 15
0.15BX C 0BZ 0.33CX 0.15CY D 66
BX C 0BZ C CX C 0CZ D 200.
(The HP-28S hand held calculator was used to solve these equations.)
The solution:
BX D 750 N,
BZ D 1800 N,
CX D 950 N,
CY D 900 N.
The reactions at A are determined by the sums of forces:
356
FY D AY C CY j D 0, from which AY D CY D 900 N
FZ D AZ C BZ C 100k D 0, from which AZ D 1900 N
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
y
Problem 5.118 The support that attaches the sailboat’s
mast to the deck behaves like a ball and socket support.
The line that attaches the spinnaker (the sail) to the top
of the mast exerts a 200-lb force on the mast. The force
is in the horizontal plane at 15&deg; from the centerline of the
boat. (See the top view.) The spinnaker pole exerts a 50lb force on the mast at P. The force is in the horizontal
plane at 45&deg; from the centerline. (See the top view.)
The mast is supported by two cables, the back stay AB
and the port shroud ACD. (The forestay AE and the
starboard shroud AFG are slack, and their tensions can
be neglected.) Determine the tensions in the cables AB
and CD and the reactions at the bottom of the mast.
A
A
Spinnaker
50 ft
C
C
F
P
P
6 ft
E
x
B
D
Side View
G
D
15 ft
21 ft
Aft View
x
z (Spinnaker not shown)
Top View
F
200 lb
G
15&deg;
A
B
E
P
C
50 lb
45&deg;
D
Solution: Although the dimensions are not given in the sketch,
assume that the point C is at the midpoint of the mast (25 ft above
the deck). The position vectors for the points A, B, C, D, and P are:
(5) The force due to the spinnaker pole:
rA D 50j,
The sum of the moments about the base of the mast is
FP D 500.707i C 0.707k D 35.35i C 35.35k.
MQ D rA &eth; FA C rA &eth; TAB C rA &eth; TAC C rC &eth; TCE
rB D 21i,
rP D 6j,
C rP &eth; FP D 0
MQ D rA &eth; FA C TAB C TAC C rC &eth; TCE C rP &eth; FP D 0.
rC D 25j 7.5k.
The vector parallel to the backstay AB is
From above,
FA C TAB C TAC D FTX i C FTY j C FTZ k
rAB D rB rA D 21i 50j.
D 193.2 0.3872jTAB ji C 0.922jTAB j
The unit vector parallel to backstay AB is
0.9578jTAC jj C 51.76 0.2873jTAC jk
eAB D 0.3872i 0.9220j.
The vector parallel to AC is
i
MQ D 0
FTX
i
j
C 0
6
35.35 0
rAC D rC rA D 25j 7.5k.
The forces acting on the mast are: (1) The force due to the spinnaker
at the top of the mast:
j
25
0
k
7.5
0.2873jTAC j k 6 D 0
35.35 C 50FTX C 212.1k D 0.
(2) The reaction due to the backstay:
(3) The reaction due to the shroud:
k i
50 C 0
FTZ 0
D 50FTZ C 250.2873jTAC j C 212.1i
FA D 200i cos 15&deg; C k cos 75&deg; D 193.19i C 51.76k.
TAB D jTAB jeAB
j
50
FTY
Substituting and collecting terms:
2800 7.1829jTAC ji C 9447.9 C 19.36jTAB jk D 0,
from which
TAC D jTAC jeAC
(4) The force acting on the cross spar CE:
TCE D k &ETH; TAC k D 0.2873jTAC jk.
jTAC j D
2800
D 389.81 lb,
7.1829
jTAB j D 488.0 lb.
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
357
5.118 (Continued )
The tension in cable CD is the vertical component of the tension
in AC,
jTCD j D jTAC jj &ETH; eAC D jTAC j0.9578 D 373.37 lb.
The reaction at the base is found from the sums of the forces:
FX D QX C 193.19 35.35 jTAB j0.3872 D 0,
from which QX D 31.11 lb
FY D QY 0.922jTAB j 0.9578jTAC jj D 0,
from which QY D 823.24 lb
FZ D QZ C 51.76 C 0.2873jTAC j
0.2873jTAC j C 35.35k D 0,
from which QZ D 87.11 lb
Collecting the terms, the reaction is
Q D 31.14i C 823.26j 87.12k (lb)
y
y
FA
TAC
FA
TCD TCE
TAB
z
FP
QX
QY
TCD
FP
QY
QB
z
SIDE VIEW z AFT VIEW
FA
QB
x
QX
FP TAB
TOP VIEW
358
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
y
Problem 5.119* The bar AC is supported by the cable
BD and a bearing at A that can rotate about the axis AE.
The person exerts a force F D 50j (N) at C. Determine
the tension in the cable.
(0.3, 0.5, 0) m
E
Strategy: Use the fact that the sum of the moments
about the axis AE due to the forces acting on the freebody diagram of the bar must equal zero.
C
(0.82, 0.60, 0.40) m
(0.3, 0.4, 0.3) m
A
B (0.46, 0.46, 0.33) m
x
Solution: We will take moments about the line AE in order to
eliminate all of the reactions at the bearing A. We have:
eAE D
z
0.1j 0.3k
p
D 0.316j 0.949k
0.1
D
(0.7, 0, 0.5) m
rAB D 0.16i C 0.06j C 0.03km,
TBD D TBD
0.24i 0.46j C 0.17k
p
0.2981
rAC D 0.52i C 0.2j C 0.1km,
F D 50jN
Then the equilibrium equation is
MAE D eAE &ETH; rAB &eth; TBD C rAC &eth; F D 0
This reduces to the single scalar equation
TBD D 174.5 N
Problem 5.120* In Problem 5.119, determine the
reactions at the bearing A.
Solution: See the previous problem for setup. We add the reactions
Strategy: Write the couple exerted on the free-body
diagram of the bar by the bearing as MA D MAx i C
MAy j C MAz k. Then, in addition to the equilibrium
equations, obtain an equation by requiring the
component of MA parallel to the axis AE to equal zero.
A D Ax i C Ay j C Az k, MA D MAx i C MAy j C MAz k
(force, moment)
This gives us too many reaction moments. We will add the constrain
that
MA &ETH; eAE D 0
We have the following 6 equilibrium equations:


Fx : Ax C 0.440TBD D 0



F D A C TBD C F D 0 )
Fy : Ay 0.843TBD C 50 N D 0




Fx : Az C 0.311TBD D 0
MA D MA C rAB &eth; TBD C rAC &eth; F D 0


MAx : MAx 5 N-m C 0.0440 mTBD D 0



)
MAy : MAy 0.0366 mTBD D 0




MAz : MAz C 26 N-m 0.161 mTBD D 0
eAE &ETH; MA D 0
) 0.316MAy 0.949MAz D 0
Solving these 7 equations we find
Ax D 76.7 N, Ay D 97.0 N, Az D 54.3 N
MAx D 2.67 N-m, MAy D 6.39 N-m, MAz D 2.13 N-m
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
359
Problem 5.121 In Active Example 5.10, suppose that
the support at A is moved so that the angle between the
bar AB and the vertical decreases from 45&deg; to 30&deg; . The
position of the rectangular plate does not change. Draw
the free-body diagram of the plate showing the point
P where the lines of action of the three forces acting
on the plate intersect. Determine the magnitudes of the
reactions on the plate at B and C.
45
A
C
B
4 ft
Solution: The equilibrium equations are
Fx : B sin 30&deg; C sin 30&deg; D 0
Fy : B cos 30&deg; C C cos 30&deg; 100 lb D 0
Solving yields
B D C D 57.7 lb
360
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Problem 5.122 The magnitude of the reaction exerted
on the L-shaped bar at B is 60 lb. (See Example 5.11.)
(a) What is the magnitude of the reaction exerted on the
bar by the support at A?
(b) What are the x and y components of the reaction
exerted on the bar by the support at A?
y
14 in
B
17 in
A
x
Solution: The angle between the line AB and the x axis is
D tan1 17/14 D 50.5&deg;
(a)
The bar is a two-force member, so the magnitude of the reaction
at A is
A D 60 lb
(b)
The reaction at A must be parallel to the line from A to B, but it
may point either from A toward B or from B toward A.
In the first case, the components are
Ax D 60 lb cos , Ay D 60 lb sin In the second case the components are
Ax D 60 lb cos , Ay D 60 lb sin Thus Ax D 38.1 lb, Ay D 46.3 lb or Ax D 38.1 lb, Ay D 46.3 lb
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361
Problem 5.123 The suspended load weighs 1000 lb.
The structure is a three-force member if its weight is
neglected. Use this fact to determine the magnitudes of
the reactions at A and B.
A
5 ft
B
10 ft
Solution: The pin support at A is a two-force reaction, and the
roller support at B is a one force reaction. The moment about A is
MA D 5B 101000 D 0, from which the magnitude at B is B D
2000 lb. The sums of the forces:
A
5 ft
FX D AX C B D AX C 2000 D 0, from which AX D 2000 lb.
B
10 ft
FY D AY 1000 D 0, from which AY D 1000 lb.
1000 lb
p
The magnitude at A is A D 20002 C 10002 D 2236 lb
AX
AY
5 ft
B
1000 lb
10 ft
Problem 5.124 The weight W D 50 lb acts at the
center of the disk. Use the fact that the disk is a threeforce member to determine the tension in the cable and
the magnitude of the reaction at the pin support.
60&deg;
Solution: Denote the magnitude of the reaction at the pinned joint
by B. The sums of the forces are:
and
FX D BX T sin 60&deg; D 0,
W
FY D BY C T cos 60&deg; W D 0.
The perpendicular distance to the action line of the tension from the
center of the disk is the radius R. The sum of the moments about
the center of the disk is MC D RBY C RT D 0, from which BY D T.
Substitute into the sum of the forces to obtain: T C T0.5 W D 0,
from which
TD
60&deg;
W
2
W D 33.33 lb.
3
Substitute into the sum of forces to obtain
60&deg;
BX D T sin 60&deg; D 28.86 lb.
The magnitude of the reaction at the pinned joint is
BD
362
p
T
R
BX
W
BY
33.332 C 28.862 D 44.1 lb
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Problem 5.125 The weight W D 40 N acts at the
center of the disk. The surfaces are rough. What force
F is necessary to lift the disk off the floor?
F
150 mm
W
50 mm
Solution: The reaction at the obstacle acts through the center of
the disk (see sketch) Denote the contact point by B. When the moment
is zero about the point B, the disk is at the verge of leaving the floor,
hence the force at this condition is the force required to lift the disk.
The perpendicular distance from B to the action line of the weight is
d D R cos ˛, where ˛ is given by (see sketch)
˛ D sin1
Rh
R
D sin1
150 50
150
F
150 mm
50 mm
W
D 41.81&deg; .
α
The perpendicular distance to the action line of the force is
F
D D 2R h D 300 50 D 250 mm.
The sum of the moments about the contact point is
W
MB D R cos ˛W C 2R hF D 0,
from which F D
150 cos 41.81&deg; W
D 0.4472W D 17.88 N
250
h
b
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363
Problem 5.126 Use the fact that the horizontal bar is
a three-force member to determine the angle ˛ and the
magnitudes of the reactions at A and B. Assume that
0 ˛ 90&deg; .
2m
Solution: The forces at A and B are parallel to the respective bars
since these bars are 2-force members. Since the horizontal bar is a
3-force member, all of the forces must intersect at a point. Thus we
have the following picture:
a
3 kN
60⬚ B
A
1m
30⬚
From geometry we see that
d D 1 m cos 30&deg;
d sin 30&deg; D e sin ˛
d cos 30&deg; C e cos ˛ D 3 m
Solving we find
˛ D 10.89&deg;
To find the other forces we look at the force triangle
FB D 3 kN cos 40.89&deg; D 2.27 kN
FA D 3 kN sin 40.89&deg; D 1.964 kN
FA
e
d
α
60&deg;
30&deg;
α
30&deg;
2m
3 kN
1m
FB
3 kN
40.89&deg;
FA
FB
364
90&deg;
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Problem 5.127 The suspended load weighs 600 lb.
Use the fact that ABC is a three-force member to
determine the magnitudes of the reactions at A and B.
3 ft
B
4.5 ft
30⬚
C
45⬚
A
Solution: All of the forces must intersect at a point.
From geometry
tan D
3 ft
D 0.435
3 C 4.5 cos 30&deg; ft
)
D 23.5&deg;
Now using the force triangle we find
FB D 600 lb cot D 1379 lb
FA D 600 lb csc D 1504 lb
FB
θ
FA
600 lb
FB
θ
600 lb
FA
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365
2 kN
Problem 5.128 (a) Is the L-shaped bar a three-force
member?
(b) Determine the magnitudes of the reactions at A and B.
(c) Are the three forces acting on the L-shaped bar
concurrent?
3 kN-m
B
300 mm
150 mm
700 mm
A
250
mm
500
mm
Solution: (a) No. The reaction at B is one-force, and the reaction
at A is two-force. The couple keeps the L-shaped bar from being a three
force member.(b) The angle of the member at B with the horizontal is
˛ D tan1
150
250
D 30.96&deg; .
The sum of the moments about A is
MA D 3 0.52 C 0.7B cos ˛ D 0,
from which B D 6.6637 kN. The sum of forces:
FX D AX C B cos ˛ D 0,
from which AX D 5.7143 kN.
FY D AY B sin ˛ 2 D 0,
from which AY D 5.4281 kN. The magnitude at A:
p
5.712 C 5.432 D 7.88 kN (c) No, by inspection.
0.5
m
3 kN-m
α
2 kN
0.3 m
B
0.7 m
Ay
Ax
366
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Problem 5.129 The hydraulic piston exerts a horizontal force at B to support the weight W D 1500 lb
of the bucket of the excavator. Determine the magnitude
of the force the hydraulic piston must exert. (The vector
sum of the forces exerted at B by the hydraulic piston,
the two-forces member AB, and the two-force member
BD must equal zero.)
Solution: See the solution to Problem 5.23.
14 in
16 in
B
A
4 in
C
The angle between the two-force member AB and the horizontal is
D
˛ D tan1 12/14 D 40.6&deg;
and the magnitude of the force exerted at B by member AB is
TAB D 892 lb
W
in the direction from B toward A. Let F be the force exerted by
the piston, and let TBD be the force exerted at B by member BD
in the direction from B toward D. The angle between member BD and
the horizontal is
ˇ D tan1 16/12 D 53.1&deg;
8 in
8 in
The sum of the forces at B is
Fx : F C TBD cos ˇ TAB cos ˛ D 0
Fy : TBD sin ˇ TAB sin ˛ D 0
Solving yields TBD D 726 lb, F D 1110 lb
Thus F D 1110 lb
Problem 5.130 The member ACG of the front-end
loader is subjected to a load W D 2 kN and is supported
by a pin support at A and the hydraulic cylinder BC.
Treat the hydraulic cylinder as a two-force member.
A
0.75 m
B
(a)
(b)
Draw a free-body diagrams of the hydraulic
cylinder and the member ACG.
Determine the reactions on the member ACG.
C
1m
G
0.5 m
W
1.5 m
Solution: This is a very simple Problem. The free body diagrams
are shown at the right. From the free body diagram of the hydraulic
cylinder, we get the equation BX C CX D 0. This will enable us to find
BX once the loads on member ACG are known. From the diagram of
ACG, the equilibrium equations are
and
CX
BX
B
AX
Fx D AX C CX D 0,
0.75 m
Fy D AY W D 0,
1m
MA D 0.75CX 3W D 0.
1.5 m
AY
CX
0.5 m
1.5 m
1.5 m
W
Using the given value for W and solving these equations, we get
AX D 8 kN,
AY D 2 kN,
CX D 8 kN,
and BX D 8 kN.
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367
Problem 5.131 In Problem 5.130, determine the reactions of the member ACG by using the fact that it is a
three-force member.
Solution: The easiest way to do this is take advantage of the fact
that for a three force member, the three forces must be concurrent. The
fact that the force at C is horizontal and the weight is vertical make
it very easy to find the point of concurrency. We then use this point
to determine the direction of the force through A. We can even know
which direction this force must take along its line — it must have an
upward component to support the weight — which is down. From the
geometry, we can determine the angle between the force A and the
horizontal.
y
A
A
0.75 m
1m
θ
CX C
1.5 m
x
1.5 m
G
W = 2 kN
tan D 0.75/3,
or D 14.04&deg; .
Using this, we can write force equilibrium equations in the form
Fx D A cos C CX D 0, and
Fy D A sin W D 0.
Solving these equations, we get A D 8.246 kN, and CX D 8 kN. The
components of A are as calculated in Problem 5.130.
368
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Problem 5.132 A rectangular plate is subjected to
two forces A and B (Fig. a). In Fig. b, the two forces
are resolved into components. By writing equilibrium
equations in terms of the components Ax , Ay , Bx , and By ,
show that the two forces A and B are equal in magnitude,
opposite in direction, and directed along the line between
their points of application.
B
B
A
h
A
b
(a)
y
By
Bx
B
h
Ay
Ax
A
x
b
(b)
Solution: The sum of forces:
b
B
FX D AX C BX D 0,
h
A
from which AX D BX
By
y
FY D AY C BY D 0,
Ay
from which AY D By . These last two equations show that A and B
are equal and opposite in direction, (if the components are equal and
opposite, the vectors are equal and opposite). To show that the two
vectors act along the line connecting the two points, determine the
angle of the vectors relative to the positive x axis. The sum of the
Fig a
Ax
Bx
Fig b
x
MA D Bx h bBy D 0,
from which the angle of direction of B is
tan1
BY
BX
D tan1
h
D ˛B .
b
or 180 C ˛B . Similarly, by substituting A:
tan1
AY
AX
D tan1
h
D ˛A ,
b
or 180 C ˛A . But
˛ D tan1
h
b
describes direction of the line from A to B. The two vectors are opposite
in direction, therefore the angles of direction of the vectors is one of
two possibilities: B is directed along the line from A to B, and A is
directed along the same line, oppositely to B.
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369
Problem 5.133 An object in equilibrium is subjected
to three forces whose points of application lie on a
straight line. Prove that the forces are coplanar.
F2
F3
Solution: The strategy is to show that for a system in equilibrium
under the action of forces alone, any two of the forces must lie in
the same plane, hence all three must be in the same plane, since the
choice of the two was arbitrary. Let P be a point in a plane containing
the straight line and one of the forces, say F2 . Let L also be a line,
not parallel to the straight line, lying in the same plane as F2 , passing
through P. Let e be a vector parallel to this line L. First we show
that the sum of the moments about any point in the plane is equal to
the sum of the moments about one of the points of application of the
forces. The sum of the moments about the point P:
F1
F2
F3
F1
P
L
M D r1 &eth; F1 C r2 &eth; F2 C r3 &eth; F3 D 0,
where the vectors are the position vectors of the points of the application of the forces relative to the point P. (The position vectors lie in
the plane.) Define
d12 D r2 r1 ,
and d13 D r3 r1 .
Then the sum of the moments can be rewritten,
M D r1 &eth; F1 C F2 C F3 C d12 &eth; F2 C d13 &eth; F3 D 0.
Since the system is in equilibrium,
F1 C F2 C F3 D 0,
and the sum of moments reduces to
M D d12 &eth; F2 C d13 &eth; F3 D 0,
which is the moment about the point of application of F1 . (The vectors
d12 , d13 are parallel to the line L.) The component of the moment
parallel to the line L is
e &ETH; d12 &eth; F2 e C e &ETH; d13 &eth; F3 e D 0,
or F2 &ETH; d12 &eth; ee C F3 &ETH; d13 &eth; ee D 0.
But by definition, F2 lies in the same plane as the line L, hence it is
normal to the cross product d12 &eth; e 6D 0, and the term
F2 &ETH; d12 &eth; e D 0.
But this means that
F3 &ETH; d13 &eth; ee D 0,
which implies that F3 also lies in the same plane as F2 , since
d13 &eth; e 6D 0.
Thus the two forces lie in the same plane. Since the choice of the point
about which to sum the moments was arbitrary, this process can be
repeated to show that F1 lies in the same plane as F2 . Thus all forces
lie in the same plane.
370
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Problem 5.134 The suspended cable weighs 12 lb.
(a)
B
Draw the free-body diagram of the cable. (The
tensions in the cable at A and B are not equal.)
Determine the tensions in the cable at A and B.
What is the tension in the cable at its lowest point?
(b)
(c)
50⬚
A
32⬚
Solution:
(a)
(b)
The FBD
The equilibrium equations
Fx : TA cos 32&deg; C TB cos 50&deg; D 0
Fy : TA sin 32&deg; C TB sin 50 12 lb D 0
Solving we find
TA D 7.79 lb, TB D 10.28 lb
(c)
Consider the FBD where W represents only a portion of the total
weight. We have
Fx : TA cos 32&deg; C T D 0
Solving
T D 6.61 lb
TB
50&deg;
TA
32&deg;
12 lb
TA
32&deg;
T
W
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371
Problem 5.135 Determine the reactions at the fixed
support.
4 kN
3m
A
20 kN-m
2 kN
5m
3 kN
3m
Solution: The equilibrium equations
4 kN
Fx : Ax C 4 kN D 0
Fy : Ay 2 kN 3 kN D 0
Ax
20 kN-m
MA : MA 2 kN5 m 4 kN3 m
3 kN8 m C 20 kN-m D 0
MA
Ay
Solving
2 kN
3 kN
Ax D 4 kN, Ay D 5 kN, MA D 26 kN-m
Problem 5.136 (a) Draw the free-body diagram of the
50-lb plate, and explain why it is statically indeterminate.
y
(b) Determine as many of the reactions at A and B as
possible.
A
12 in
Solution:
(a)
(b)
The pin supports at A and B are two-force supports, thus there
are four unknown reactions AX , AY , BX , and BY , but only three
equilibrium equations can be written, two for the forces, and one
for the moment. Thus there are four unknowns and only three
equations, so the system is indeterminate.
B
x
20 in
50 lb
Sums the forces:
A
FX D AX C BX D 0,
or AX D BX , and
8 in
12 in
8 in
B
20 in
FY D AY C BY 50 D 0.
The sum of the moments about B
50 lb
x
AY
AX
MB D 20AX 5020 D 0,
from which AX D 50 lb,
and from the sum of forces BX D 50 lb.
372
12 in.
BY
8 in.
BX
50 lb
x
20 in.
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Problem 5.137 The mass of the truck is 4 Mg. Its
wheels are locked, and the tension in its cable is T D
10 kN.
(003) 676-5942
(a)
(b)
Draw the free-body diagram of the truck.
Determine the normal forces exerted on the truck’s
30&deg;
AL's
To w i n g
2m
T
2.5 m
3m
2.2 m
mg
Solution: The weight is 40009.81 D 39.24 kN. The sum of the
MB D 3T sin 30&deg; 2.2T cos 30&deg; C 2.5W 4.5AN D 0
from which
AN D
D
2.5W T3 sin 30&deg; C 2.2 cos 30&deg; 4.5
64.047
D 14.23 N
4.5
The sum of the forces:
FY D AN W C BN T cos 30&deg; D 0,
from which BN D T cos 30&deg; AN C W D 33.67 N
30&deg;
AX
A
AN
W
BX
T
B
BN
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373
Problem 5.138 Assume that the force exerted on the
head of the nail by the hammer is vertical, and neglect
the hammer’s weight.
(a) Draw the free-body diagram of the hammer.
(b) If F D 10 lb, what are the magnitudes of the forces
exerted on the nail by the hammer and the normal
and friction forces exerted on the floor by the
hammer?
F
11 in.
Solution: Denote the point of contact with the floor by B. The
65&deg;
perpendicular distance from B to the line of action of the force is 11
in. The sum of the moments about B is MB D 11F 2FN D 0, from
11F
D 5.5F. The
which the force exerted by the nail head is FN D
2
sum of the forces:
FX D F cos 25 C Hx D 0,
2 in.
from which the friction force exerted on the hammer is HX D 0.9063F.
FY D NH FN C F sin 25&deg; D 0,
from which the normal force exerted by the floor on the hammer is
NH D 5.077F
If the force on the handle is
F D 10 lb,
then FN D 55 lb,
HX D 9.063 lb,
and NH D 50.77 lb
F
11 in.
65&deg;
2 in.
F
11 in.
65&deg;
HX
374
B NH
FN
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Problem 5.139 The spring constant is k D 9600 N/m
and the unstretched length of the spring is 30 mm. Treat
the bolt at A as a pin support and assume that the surface
at C is smooth. Determine the reactions at A and the
normal force at C.
A
24 mm
B
15 mm
30 mm
30&deg;
C
k
Solution: The length of the spring is
lD
p
302 C 302 mm D
p
1800 mm
30 mm
AY
l D 42.4 mm D 0.0424 m
The spring force is kυ where υ D l l0 . l0 is give as 30 mm. (We
must be careful because the units for k are given as N/m) We need to
use length units as all mm or all meters). k is given as 9600 N/m. Let
us use l0 D 0.0300 m and l D 0.0424 m
AX
24 mm
B
kδ
Equilibrium equations:
FX D 0:
AX kl l0 sin 45&deg;
NC cos 60&deg; D 0
FY D 0:
50 mm
15 mm
30 mm
45&deg;
60&deg;
50 mm
30 mm
30&deg;
C
NC
AY kl l0 cos 45&deg;
Solving, we get
C NC sin 60&deg; D 0
MB D 0:
0.024AX C 0.050NC sin 60&deg; 0.015NC cos 60&deg; D 0
AX D 126.7 N
AY D 10.5 N
NC D 85.1 N
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375
Problem 5.140 The engineer designing the release
mechanism shown in Problem 5.139 wants the normal
force exerted at C to be 120 N. If the unstretched length
of the spring is 30 mm, what is the necessary value of
the spring constant k?
Solution: Refer to the solution of Problem 5.139. The equilibrium
equations derived were
FX D 0:
AX kl l0 sin 45 NC cos 60&deg; D 0
FY D 0:
AY kl l0 cos 45 C NC sin 60&deg; D 0
MB D 0:
0.024AX C 0.050NC sin 60&deg;
0.015NC cos 60&deg; D 0
where l D 0.0424 m, l0 D 0.030 m, NC D 120 N, and AX , AY , and k
are unknowns.
Solving, we get
AX D 179.0 N,
AY D 15.1 N,
k D 13500 N/m
376
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Problem 5.141 The truss supports a 90-kg suspended
object. What are the reactions at the supports A and B?
400 mm
700 mm
300 mm
B
A
Solution: Treat the truss as a single element. The pin support at
A is a two force reaction support; the roller support at B is a single
force reaction. The sum of the moments about A is
MA D B400 W1100 D 0,
from which B D
1100W
D 2.75W
400
B D 2.75909.81 D 2427.975 D 2.43 kN.
The sum of the forces:
FX D AX D 0
FY D AY C B W D 0,
from which AY D W B D 882.9 2427.975 D 1.545 kN
AX
W
B
AY
400 mm
700 mm
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377
Problem 5.142 The trailer is parked on a 15&deg; slope.
Its wheels are free to turn. The hitch H behaves like a
pin support. Determine the reactions at A and H.
y
1.4 ft
H
x
870 lb
1.6 ft
A
8 ft
2.8 ft
15&deg;
Solution: The coordinate system has the x axis parallel to the road.
The wheels are a one force reaction normal to the road, the pin H is
a two force reaction. The position vectors of the points of the center
of mass and H are:
rW D 1.4i C 2.8j ft and
rH D 8i C 1.6j.
The angle of the weight vector realtive to the positive x axis is
˛ D 270&deg; 15&deg; D 255&deg; .
The weight has the components
W D Wi cos 255&deg; C j sin 255&deg; D 8700.2588i 0.9659j
D 225.173i 840.355j (lb).
The sum of the moments about H is
MH D rW rH &eth; W C rA rH &eth; A,
j
k i
1.2
0 C 8
840.355 0 0
i
MH D 6.6
225.355
j
k 1.6 0 D 0
AY
0
D 5816.55 8AY D 0,
from which AY D
5816.55
D 727.1 lb.
8
The sum of the forces is
FX D HX 225.173i D 0, from which HX D 225.2 lb,
FY D AY C HY 840.355j D 0, from which HY D 113.3 lb
1.4 ft
1.2 ft
W
15&deg;
AY
378
HX
HY
8 ft
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Problem 5.143 To determine the location of the point
where the weight of a car acts (the center of mass),
an engineer places the car on scales and measures the
normal reactions at the wheels for two values of ˛,
obtaining the following results:
y
h
x
B
W
Ay (kN)
˛
B (kN)
10&deg;
10.134
4.357
20&deg;
10.150
3.677
α
Ax
Ay
b
2.7 m
What are the distances b and h?
Solution: The position vectors of the cm and the point B are
These two simultaneous equations in two unknowns were solved using
the HP-28S hand held calculator.
rCM D 2.7 bi C hj,
b D 1.80 m,
rB D 2.7i.
h D 0.50 m
The angle between the weight and the positive x axis is ˇ D 270 ˛.
The weight vector at each of the two angles is
W10 D Wi cos 260&deg; C j sin 260&deg; W10 D W0.1736i 0.9848j
W20 D Wi cos 250&deg; C j sin 250&deg; or
W20 D W0.3420i 0.9397j
The weight W is found from the sum of forces:
FY D AY C BY C W sin ˇ D 0,
from which Wˇ D
AY C BY
.
sin ˇ
Taking the values from the table of measurements:
W10 D 10.134 C 4.357
D 14.714 kN,
sin 260&deg;
[check :W20 D 10.150 C 3.677
D 14.714 kN check ]
sin 250&deg;
MA D rCM &eth; W C rB &eth; B D 0.
Taking the values at the two angles:
M10
A
i
D 2.7 b
2.5551
j
k j
k i
0
0 D 0
h
0 C 2.7
14.4910 0 0 4.357 0 D 14.4903b C 2.5551h 27.3618 D 0
M20
A
i
D 2.7 b
5.0327
j
k j
k i
0
0 h
0 C 2.7
13.8272 0 0 3.677 0 D 013.8272b C 5.0327h 27.4054 D 0
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
379
Problem 5.144 The bar is attached by pin supports to
collars that slide on the two fixed bars. Its mass is 10 kg,
it is 1 m in length, and its weight acts at its midpoint.
Neglect friction and the masses of the collars. The spring
is unstretched when the bar is vertical (˛ D 0), and the
spring constant is k D 100 N/m. Determine the values
of ˛ in the range 0 ˛ 60&deg; at which the bar is in
equilibrium.
k
α
Solution: The force exerted by the spring is given by FS D kL L cos ˛. The equations of equilibrium, from the free body diagram, are
and
Spring Constant (K) in N/m vs Alpha (deg)
90000
80000
70000
Fx D NB D 0,
Fy D FS C NA mg D 0,
MB D L sin ˛NA C
L
sin ˛ mg D 0.
2
60000
K
= 50000
N
− 40000
m
30000
20000
10000
These equations can be solved directly with most numerical solvers and
the required plot can be developed. The plot over the given ˛ range is
shown at the left and a zoom-in is given at the right. The solution and
the plot were developed with the TK Solver Plus commercial software
package. From the plot, the required equilibrium value is ˛ &frac34;
D 59.4&deg; .
0
0
10
20
30
40
50
60
Alpha (deg)
Spring Constant (K) in N/m vs Alpha (deg)
116
114
FS
112
B
y
NB
α
mg
110
K
= 108
N
− 106
m
104
102
A
x
NA
100
98
55
55.5
56
56.5
57
57.5
58
58.5
59
59.5
60
Alpha (deg)
380
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 5.145 With each of the devices shown you
can support a load R by applying a force F. They are
called levers of the first, second, and third class.
(a)
(b)
R
F
A
A
The ration R/F is called the mechanical advantage.
Determine the mechanical advantage of each lever.
Determine the magnitude of the reaction at A for
each lever. (Express you answers in terms of F)
L
L
L
First-class lever
R
A
Solution:
R/F D 1
First Class MA : LF LR D 0 ) F D R
R/F D 2
Second Class MA : 2LF LR D 0 ) F D R/2
Third Class MA : LF 2LR D 0 ) F D 2R
(b)
L
Second-class lever
F
(a)
F
R
R/F D 1/2
First Class Fy : F C A R D 0 )
A D 2F
Second Class Fy : A R C F D 0 )
Third Class Fy : A C F R D 0 )
L
L
Third-class lever
A D F/2
Problem 5.146 The force exerted by the weight of the
horizontal rectangular plate is 800 N. The weight of the
plate acts at its midpoint. If you represent the reactions
exerted on the plate by the three cables by a single
equivalent force, what is the force, and where does its
line of action intersect the plate?
A
C
B
2m
0.5 m
1m
Solution: The equivalent force must equal the sum of the
reactions: FEQ D TA C TB C TC . FEQ D 300 C 100 C 400 D 800 N.
The moment due to the action of the equivalent force must equal the
moment due to the reactions: The moment about A is
i
MA D 2
0
j
j
k i
0
0
0 C 1.5
100 0 0 400
k i
1 D x
0 0
j
k 0
z 800 0 MA D 400i C 800k D 800zi C 800xk,
y
TA
TC
1m
z
TB
2m
0.5 m
800 N
x
from which z D 0.5 m, and x D 1 m, which corresponds to the
midpoint of the plate. Thus the equivalent force acts upward at the
midpoint of the plate.
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
381
y
Problem 5.147 The 20-kg mass is suspended by cables
attached to three vertical 2-m posts. Point A is at
(1, 1.2, 0) m. Determine the reactions at the built-in
support at E.
C
B
D
A
Solution: All distances will be in meters, all forces in Newtons,
and all moments in Newton-meters. To solve the three dimensional
point equilibrium problem at A, we will need unit vectors eAB , eAC ,
and eAD . To determine these, we need the coordinates of points A, B,
C, and D. The rest of the problem will require knowing where points
E, G (under C), and H(under D) are located. From the diagram, the
required point locations are A (0, 1.2, 0), B (0.3, 2, 1), C (0, 2, 1),
D (2, 2, 0), E (0.3, 0, 1), G(0, 0, 1), and H(2, 0, 0). The required
unit vectors are calculated from the coordinates of the points of the
ends of the lines defining the vector. These are
1m
2m
0.3 m
x
z
The couple ME is the couple exerted on the post by the built in support.
Solving these equations, we get
E D 34.2i C 91.3j C 114.1k N
eAB D 0.228i C 0.608j C 0.760k,
eAC D 0i C 0.625j 0.781k,
1m
E
and ME D 228.1i C 0j C 68.44k N-m.
and eAD D 0.928i C 0.371j C 0k.
Also,
The force TAB in cable AB can be written as
Using a procedure identical to that followed for post EB above, we
can find the built-in support forces and moments for posts CG and
DH. The results for CG are:
TAB D TABX i C TABY j C TABZ k,
where TABX D jTAB jeABX , etc. Similar equations can be written for the
forces in AC and AD. The free body diagram of point A yields the
following three equations of equilibrium.
and
jME j D 238.2 N-m.
G D 0i C 91.3j 114.1k N
and MG D 228.1i C 0j C 0k N-m.
Fx D TABX C TACX C TADX D 0,
Also,
jMG j D 228.1 N-m.
Fy D TABY C TACY C TADY W D 0,
The results for DH are:
H D 34.2i C 13.7j C 0 kN
Fz D TABZ C TACZ C TADZ D 0,
and MH D 0i C 0j C 68.4k N-m.
where W D mg D 209.81 D 196.2 N. Solving the equations above
after making the substitutions related to the force components yields
the tensions in the cables. They are
jTAB j D 150 N,
Also,
jMG j D 68.4 N-m
B
jTAC j D 146 N, and
Now that we know the tensions in the cables, we are ready to tackle
the reactions at E (also G and H). The first step is to draw the free
body diagram of the post EB and to write the equations of equilibrium
for the post. A key point is to note that the force on the post from
cable AB is opposite in direction to the force found in the first part of
the problem. The equations of equilibrium for post EB are
Z
C
−TAB
MG
ZM
ME
E
x
EY
D
−TAC
EX
H
GZ
HX
G
HZ
GY GX
HY
EZ
Fx D EX TABX D 0,
Fy D EY TABY D 0,
Fz D EZ TABZ D 0,
and, summing moments around the base point E,
382
M D ME C 2j &eth; TAB D 0.
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Problem 5.148 In Problem 5.147, the built-in support
of each vertical post will safely support a couple of
800 N-m magnitude. Based on this criterion, what is the
maximum safe value of the suspended mass?
Solution: We have all of the information necessary to solve this
problem in the solution to Problem 5.147 above. All of the force and
moment equations are linear and we know from the solution that a
20 kg mass produces a couple of magnitude 238.2 N-m at support
E and that the magnitudes of the couples at the other two supports
are smaller than this. All we need to do is scale the Problem. The
scale factor is f D 800/238.2 D 3.358 and the maximum value for
the suspended mass is mmax D 20f D 67.16 kg
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
383
y
Problem 5.149 The 80-lb bar is supported by a ball
and socket support at A, the smooth wall it leans against,
and the cable BC. The weight of the bar acts at its
midpoint.
5 ft
3 ft
(a) Draw the free-body diagram of the bar.
(b) Determine the tension in cable BC and the reactions
at A.
B
C
4 ft
3 ft
3 ft
x
A
z
Solution: (a) The ball and socket is a three reaction force support;
the cable and the smooth wall are each one force reaction supports.
(b) The coordinates of the points A, B and C are A (3, 0, 3), B (5, 4,
0), and C(0, 4, 3).
The vector parallel to the bar is
rAB D rB rA D 2i C 4j 3k.
Solve:
jTj D
80
D 23.32 lb
3.43
jBj D
120 2.058jTj
D 18.00 lb.
4
The reactions at A are found from the sums of forces:
The length of the bar is
p
jrAB j D 22 C 42 C 32 D 5.3852.
The unit vector parallel to the bar is
eAB D 0.3714i C 0.7428j 0.5571k.
FX D AX jTj0.8575 D 0 from which AX D 20 lb
FY D AY 80 D 0, from which AY D 80 lb
FZ D AZ C jTj0.5145 C jBj D 0, from which AZ D 30 lb
The vector parallel to the cable is
rBC D rC rB D 5i C 3k.
B
The unit vector parallel to the cable is
eBC D 0.8575i C 0.5145k.
The cable tension is T D jTjeBC . The point of application of the weight
relative to A is
AY
AX
rAW D 2.6936eAB
AZ
rAW D 1.000i C 2.000j 1.500k.
The reaction at B is B D jBjk, since it is normal to a wall in the yz
plane. The sum of the moments about A is
MA D rAW &eth; W C rAB &eth; B C rAB &eth; T D 0
i
MA D 1
0
j
2
80
k i
1.5 C 2
0 0
i
j
C 2
4
0.8575 0
j
4
0
k 3 jBj k 3 jTj D 0
0.5145 MA D 120 C 4jBj C 2.058jTji 2jBj 1.544jTjj
C 3.43jTj 80k D 0.
384
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Problem 5.150 The horizontal bar of weight W is
supported by a roller support at A and the cable BC. Use
the fact that the bar is a three-force member to determine
the angle ˛, the tension in the cable, and the magnitude
of the reaction at A.
C
A
B
α
W
L/2
L/2
Solution: The sum of the moments about B is
MB D LAY C
L
W D 0,
2
from which AY D
W
. The sum of the forces:
2
FX D T cos ˛ D 0,
from which T D 0 or cos ˛ D 0. The choice is made from the sum of
forces in the y-direction:
FY D AY W C T sin ˛ D 0,
W
. This equation cannot be satisfied
2
W
if T D 0, hence cos ˛ D 0, or ˛ D 90&deg; , and T D
2
from which T sin ˛ D W AY D
T
α
AY
W
L/ 2
L/ 2
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385
Problem 6.1 In Active Example 6.1, suppose that in
addition to the 2-kN downward force acting at point D,
a 2-kN downward force acts at point C. Draw a sketch of
forces in members AB and AC of the truss.
A
3m
C
3m
B
D
5m
5m
2 kN
Solution: The new sketch, a free-body diagram of the entire truss
and a free-body diagram of the joint at A are shown. The angle ˛
between CD and BD is
˛ D tan1 6/10 D 31.0&deg;
Using the entire truss, the equilibrium equations are
Fx : Ax C B D 0
Fy : Ay 2 kN 2 kN D 0
MA : 2 kN5 m 2 kN10 m
C B6 m D 0
Solving yields
Ax D 5 kN,
Ay D 4 kN,
B D 5 kN
Using the free-body diagram of joint A, the equilibrium equations are:
Fx : Ax C TAC cos ˛ D 0
Fy : Ay TAB TAC sin ˛ D 0
Solving yields TAB D 1 kN, TAC D 5.83 kN
Because both values are positive, we know that both are in tension
AB : 1 kN (T),
386
AC : 5.83 kN (T)
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Problem 6.2 Determine the axial forces in the
members of the truss and indicate whether they are in
tension (T) or compression (C).
20⬚
800 N
A
0.4 m
C
B
0.7 m
0.7 m
Solution: We start at joint A
Next we move to joint C
7
7
Fx : p FAB C p FAC 800 N sin 20&deg; D 0
65
65
4
4
Fy : p FAB p FAC 800 N cos 20&deg; D 0
65
65
Solving we have
7
Fx : p FAC FBC D 0 ) FBC D 521 N
65
FAC
7
4
FAB D 915 N, FAC D 600 N
C
20&deg;
800 N
FCB
Cy
A
4
In summary we have
4
7
7
FAB
FAB D 915 NC, FAC D 600 NC, FBC D 521 NT
FAC
A
Problem 6.3 Member AB of the truss is subjected to a
1000-lb tensile force. Determine the weight W and the
axial force in member AC.
60 in
W
B
C
60 in
60 in
Solution: Using joint A
1
1
2
Fx : p 1000 lb p FAC D 0
5
2
1
1000 lb
1
1
Fy : p 1000 lb p FAC W D 0
5
2
Solving we have
FAC D 1265 lb, W D 447 lb
In summary we have
A
2
1
FAC
W
W D 447 lb, FAC D 1265 lbC
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387
Problem 6.4 Determine the axial forces in members
BC and CD of the truss.
600 lb
E
3 ft
C
D
3 ft
A
B
3 ft
3 ft
Solution: The free-body diagrams for joints E, D, and C are
shown. The angle ˛ is
˛ D tan1 3/4 D 36.9&deg;
Using Joint E, we have
Fx : 600 lb TCE sin ˛ D 0
Fy : TCE cos ˛ TDE D 0
Using Joint D, we have
Fx : TCD TBD sin ˛ D 0
Fy : TDE TBD cos ˛ D 0
Finally, using Joint C, we have
Fx : TCD C TCD sin ˛ TAC sin ˛ D 0
Fy : TCE cos ˛ TAC cos ˛ TBC D 0
Solving these six equations yields
TCE D 1000 lb, TDE D 800 lb
TCD D 600 lb, TAC D 2000 lb
TBC D 800 lb, TBD D 1000 lb
A positive value means tension and a negative value means compression
Thus
388
BC : 800 lb (T),
CD : 600 lb (C)
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Problem 6.5 Each suspended weight has mass m D
20 kg. Determine the axial forces in the members of
the truss and indicate whether they are in tension (T) or
compression (C).
A
0.4 m
C
B
D
m
0.32 m
0.16 m 0.16 m
Finally work with joint A
5
Fy : p TAD 196.2 N D 0
61
6
Fx : p TAD TCD D 0
61
5
5
Fy : p TAB C TAC p TAD D 0
29
61
) TAB D 423 N
T
A
TAD D 306 N, TCD D 235 N
Solving:
m
6
2
5
5
6
5
5
2
D
TCD
TAB
TAC
In summary:
TAB D 423 NC
196.2 N
TAC D 211 NT
Now work with joint C
TBC D 314 NC
5
Fy : p TAC 196.2 N D 0
29
TCD D 235 NC
2
Fx : p TAC TBC C TCD D 0
29
Solving:
TAC D 211 N, TBC D 313 N
TAC
5
2
C
TBC
TCD
196.2 N
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389
Problem 6.6 Determine the largest tensile and compressive forces that occur in the members of the truss,
and indicate the members in which they occur if
(a)
(b)
the dimension h D 0.1 m;
the dimension h D 0.5 m.
Observe how a simple change in design affects the
B
A
h
D
0.7 m
1 kN
0.4 m
C
0.6 m
Solution: To get the force components we use equations of the
y
form TPQ D TPQ ePQ D TPQX i C TPQY j where P and Q take on the
designations A, B, C, and D as needed.
Equilibrium yields
At joint A:
and
Fx D TABX C TACX D 0,
h
0.4
m
1.2 m
B
−TAB
TAB
−TBD
TBD T
BC
TAC
DX D
−TBC
T
−TCD CD
−TAC
DY
C
CY
0.6 m
A
1 kN
0.7 m
x
1.2 m
Fy D TABY C TACY 1 kN D 0.
At joint B:
and
Fx D TABX C TBCX C TBDX D 0,
Fy D TABY C TBCY C TBDY D 0.
At joint C:
and
and
eAB D 0.986i C 0.164j,
eAC D 0.864i 0.504j,
Fx D TBCX TACX C TCDX D 0,
eBC D 0i 1j,
Fy D TBCY TACY C TCDY C CY D 0.
eBD D 0.768i 0.640j,
At joint D:
(b) For this part of the problem, we set h D 0.5 m. The unit vectors
change because h is involved in the coordinates of point B. The new
unit vectors are
Fx D TCDX TBDX C DX D 0,
and eCD D 0.832i C 0.555j.
We get the force components as above, and the equilibrium forces at
the joints remain the same. Solving the equilibrium equations simultaneously for this situation yields
Fy D TCDY TBDY C DY D 0.
TAB D 1.35 kN,
Solve simultaneously to get
TAC D 1.54 kN,
TAB D TBD D 2.43 kN,
TBC D 1.33,
TAC D 2.78 kN,
TBD D 1.74 kN,
TBC D 0, TCD D 2.88 kN.
and TCD D 1.60 kN.
Note that with appropriate changes in the designation of points, the
forces here are the same as those in Problem 6.4. This can be explained
by noting from the unit vectors that AB and BC are parallel. Also note
that in this configuration, BC carries no load. This geometry is the
same as in Problem 6.4 except for the joint at B and member BC
which carries no load. Remember member BC in this geometry — we
will encounter things like it again, will give it a special name, and will
learn to recognize it on sight.
390
These numbers differ significantly from (a). Most significantly,
member BD is now carrying a compressive load and this has reduced
the loads in all members except member BD. “Sharing the load” among
more members seems to have worked in this case.
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 6.7 This steel truss bridge is in the Gallatin
National Forest south of Bozeman, Montana. Suppose
that one of the tandem trusses supporting the bridge is
loaded as shown. Determine the axial forces in members
AB, BC, BD, and BE.
B
D
F
A
C
E
MH : 10 kip17 ft C 10 kip34 ft C 10 kip51 ft
G
10 kip
17 ft
17 ft
Solution: We start with the entire structure in order to find the
reaction at A. We have to assume that either A or H is really a roller
10 kip
17 ft
10 kip
17 ft
Finally work with joint B
A68 ft D 0 ) A D 15 kip
17
17
FAB C p
FBE C FBD D 0
Fx : p
353
353
8
8
FAB p
FBE FBC D 0
Fy : p
353
353
Solving we find
FBD D 42.5 kip, FBE D 11.74 kip
B
17 ft
17 ft
8
8
A
10 kip
10 kip
17
10 kip
H
FBC
8
FAB C A D 0 ) FAB D 35.2 kip
Fy : p
353
17
FBE
FAB
Now we examine joint A
FBD
17
17 ft
17 ft
8 ft
H
FAB
In Summary we have
FAB D 35.2 kipC, FBC D 10 kipT,
8
FBD D 42.5 kipC, FBE D 11.74 kipT
FAC
A
Now work with joint C
Fy : FBC 10 kip D 0 ) FBC D 10 kip
FBC
FCE
FAC
C
10 kip
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
391
Problem 6.8 For the bridge truss in Problem 6.7,
determine the largest tensile and compressive forces that
occur in the members, and indicate the members in
which they occur.
Solution: Continuing the solution to Problem 6.7 will show the
largest tensile and compressive forces that occur in the structure.
Examining joint A we have
17
FAB C FAC D 0 ) FAC D 31.9 kip
Fx : p
353
Examining joint C
Fx : FAC C FCE D 0 ) FCE D 31.9 kip
Examining joint D
Fy : FDE D 0 ) FDE D 0
D
FBD
FDF
FDE
The forces in the rest of the members are found by symmetry. We have
FAB D FFH D 35.2 kipC
FAC D FGH D 31.9 kipT
FBC D FFG D 10 kipT
FBD D FDF D 42.5 kipC
FBE D FEF D 11.74 kipT
FCE D FEG D 31.9 kipT
FDE D 0
The largest tension and compression members are then
FAC D FEG D FCE D FGH D 31.9 kipT
FBD D FDH D 42.5 kipC
392
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Problem 6.9 The trusses supporting the bridge in
Problems 6.7 and 6.8 are called Pratt trusses. Suppose
that the bridge designers had decided to use the truss
shown instead, which is called a Howe truss. Determine
the largest tensile and compressive forces that occur
in the members, and indicate the members in which
Problem 6.8.
Solution: We start with the entire structure in order to find the
reaction at A. We have to assume that either A or H is really a roller
MH : 10 kip17 ft C 10 kip34 ft C 10 kip51 ft
B
D
F
A
H
C
17 ft
E
8 ft
G
10 kip
17 ft
10 kip
17 ft
10 kip
17 ft
Next work with joint C
A68 ft D 0 ) A D 15 kip
Fy : FBC C p
8
353
FCD 10 kip D 0 ) FCD D 11.74 kip
17
FCD FAC D 0 ) FCE D 42.5 kip
Fx : FCE C p
353
FBC
FCD
17
8
FAC
FCE
C
A
10
kips
10
kips
10
kips
H
10 kip
Now we examine joint A
Finally from joint E we find
8
FAB C A D 0 ) FAB D 35.2 kip
Fy : p
353
17
FAB C FAC D 0 ) FAC D 31.9 kip
Fx : p
353
FAB
17
Fy : FDE 10 kip D 0 ) FDE D 10 kip
FDE
FCE
FEG
E
8
FAC
10 kip
The forces in the rest of the members are found by symmetry. We have
A
FAB D FFH D 35.2 kipC
Now work with joint B
FAC D FGH D 31.9 kipT
FBD D FDF D 31.9 kipC
17
FAB C FBD D 0 ) FBD D 31.9 kip
Fx : p
353
8
FAB FBC D 0 ) FBC D 15 kip
Fy : p
353
B
FBD
17
FBC D FFG D 15 kipT
FCD D FDG D 11.74 kipC
FCE D FEG D 42.5 kipT
8
FDE D 10 kipT
FAB
The largest tension and compression members are then
FBC
FCE D FEG D 42.5 kipT
FAB D FFH D 35.2 kipC
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
393
Problem 6.10 Determine the axial forces in members
BD, CD, and CE of the truss.
G
300 mm
E
C
F
Solution: The free-body diagrams of the entire truss and of joints
A, B, and C are shown. The angle
300 mm
A
˛ D tan1 3/4 D 36.9&deg;
From the free-body diagram of the entire truss
B
400
mm
D
400
mm
6 kN
400
mm
Fy : Ay 6 kN D 0
MG : 6 kN400 mm C Ax 600 mm
Ay 1200 mm D 0
Solving, Ax D 8 kN, Ay D 6 kN
Using joint A,
Fx : Ax C TAB C TAC cos ˛ D 0
Fy : Ay C TAC sin ˛ D 0
Solving we find
TAB D 0,
TAC D 10 kN
Because joint B consists of three members, two of which are parallel,
and is subjected to no external load, we can recognize that
TBD D TAB D 0
and
TBD D 0
Finally we examine joint C
Fx : TCE C TCD cos ˛ TAC cos ˛ D 0
Fy : TAC sin ˛ TCD sin ˛ TBC D 0
In summary
394
BD : 0,
CD : 10 kN (T),
) TCD D 10 kN, TCE D 16 kN
CE : 16 kN (C)
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 6.11 The loads F1 D F2 D 8 kN. Determine
the axial forces in members BD, BE, and BG.
F1
D
3m
F2
B
E
3m
Solution: First find the external support loads and then use the
method of joints to solve for the required unknown forces. (Assume
all unknown forces in members are tensions).
G
A
C
4m
F1 = 8 kN
D
y
3m
B
E
C
8m
DE D 6 kN C
F2 = 8 kN
Joint E :
y
x
C
AY
BD D 10 kN T
3m
G
A
AX
Solving,
4m
DE
GY
F2 = 8 kN
BE
Fx :
Ax C F1 C F2 D 0 (kN)
x
Fy :
Ay C Gy D 0
MA :
8Gy 3F2 6F1 D 0
EG
Solving for the external loads, we get
DE D 6 kN
Ax D 16 kN to the left
Ay D 9 kN downward
Gy D 9 kN upward
Now use the method of joints to determine BD, BE, and BG.
Fx D DE EG D 0
Fy D BE C F2 D 0
Solving:
EG D 6 kN C
BE D 8 kN T
Joint D:
y
D
F1 = 8 kN
DE
θ
x
BD
cos D 0.8
sin D 0.6
D 36.87&deg;
Fx :
F1 BD cos D 0
Fy :
BD sin DE D 0
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
395
6.11 (Continued )
Joint G:
Fx :
CG BG cos D 0
y
EG
BG
Fy : BG sin C EG C Gy D 0
Solving, we get
BG D 5 kN C
θ
x
CG D 4 kN T
CG
GY
Thus, we have
BD D 10 kN T
BE D 8 kN T
BG D 5 kN C
EG D 6 kN C
Gy D 9 kN
Problem 6.12 Determine the largest tensile and
compressive forces that occur in the members of the
truss, and indicate the members in which they occur if
(a)
(b)
B
D
h
C
E
the dimension h D 5 in;
the dimension h D 10 in.
20 in
A
20 in
20 in
30⬚
800 lb
Observe how a simple change in design affects the
Solution: Starting at joint A
Finally joint C
20
FAB FAC C 800 lb sin 30&deg; D 0
Fx : p
h2 C 202
h
FAB 800 lb cos 30&deg; D 0
Fy : p
h2 C 202
20
20
FCD C p
FBC FCE C FAC D 0
Fx : p
h2 C 202
h2 C 202
h
h
FCD C p
FBC D 0
Fy : p
h2 C 202
h2 C 202
FBC
FCD
FAB
20
h
20
h
20
h
A
FAC
C
FCE
FAC
(a) Using h D 5 in we find:
FAB D 2860 lbT, FAC D 2370 lbC, FBD D 5540 lbT
800 lb
FBC D 2860 lbC, FCD D 2860 lbT, FCE D 7910 lbC
Next joint B
20
20
FBC C p
FAB D 0
Fx : FBD p
h2 C 202
h2 C 202
FCE D 7910 lbC
h
h
FBC p
FAB D 0
Fy : p
h2 C 202
h2 C 202
FBD
B
h
20
FBC
396
FBD D 5540 lbT
)
(b) Using h D 10 in we find:
FAB D 1549 lbT, FAC D 986 lbC, FBD D 2770 lbT
FBC D 1549 lbC, FCD D 1549 lbT, FCE D 3760 lbC
h
20
FBD D 2770 lbT
FAB
)
FCE D 3760 lbC
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 6.13 The truss supports loads at C and E.
If F D 3 kN, what are the axial forces in members BC
and BE?
1m
1m
A
1m
B
D
1m
G
C
E
F
2F
Solution: The moment about A is
AY
AX
MA D 1F 4F C 3G D 0,
1m
from which G D
5
F D 5 kN. The sums of forces:
3
F
1m
FY D AY 3F C G D 0,
1m
2F
1m
DG
from which AY D
45&deg;
4
F D 4 kN.
3
BD
EG
G
Joint G
DG
DE 45&deg;
Joint D
BE
45&deg;
CE
DE
EG
Joint E
FX D AX D 0,
AY
from which AX D 0. The interior angles GDE, EBC are
45&deg; ,
AC
AB
45&deg;
45&deg; BC
AC
CE
F
Joint C
1
from which sin ˛ D cos ˛ D p .
2
Joint A
Denote the axial force in a member joining I, K by IK.
from which
(1)
5
BD D F D 5 kN C.
3
Joint G:
DG
Fy D p C G D 0,
2
from which
p
p
p
5 2
F D 5 2 kN C.
DG D 2G D 3
DG
Fx D p EG D 0,
2
from which
5
DG
EG D p D F D 5kN T.
3
2
(2)
G
Joint D:
(3)
Joint E :
BE
Fy D p 2F C DE D 0,
2
p
p
from which BE D 2 2F 2DE D
p
p
2
F D 2 kN T.
3
BE
Fx D CE p C EG D 0,
2
from which
4
BE
CE D EG p D F D 4 kN T.
3
2
DG
Fy D DE p D 0,
2
from which
DE D
5
F D 5 kN T.
3
DG
Fx D BD C p D 0,
2
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
397
6.13 (Continued )
(4)
Joint A:
AC
Fy D Ay p D 0,
2
from which AC D
p
p
4 2
F D 4 2 kN T.
3
AC
Fx D AB C p D 0,
2
4
from which AB D F D 4 kN C.
3
(5)
Joint C :
AC
Fy D BC C p F D 0,
2
AC
1
from which BC D F p D F D 1 kN C.
3
2
Problem 6.14 If you don’t want the members of the
truss to be subjected to an axial load (tension or compression) greater than 20 kN, what is the largest acceptable
magnitude of the downward force F?
A
12 m
F
B
4m
Fx : FAB cos 36.9&deg; FAC sin 30.5&deg; D 0
C
D
Fy : FAB sin 36.9&deg; FAC cos 30.5&deg; F D 0
3m
A
Finally examine joint D
36.9&deg;
Fy : FBD D 0
30.5&deg;
FBD
FAB
F
FAC
Dx
Now work with joint C
Fx :
FCD FBC sin 36.9&deg;
FAC
36.9&deg;
FCD
398
Solving we find
C FAC sin 30.5&deg;
Fy : FBC cos 36.9&deg; C FAC cos 30.5&deg; D 0
FBC
30.5&deg;
FCD
D
D0
FAB D 1.32F, FAC D 2.08F, FCD D 2.4F,
FBC D 2.24F, FBD D 0
The critical member is CD. Thus
2.4F D 20 kN ) F D 8.33 kN
C
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 6.15 The truss is a preliminary design for a
structure to attach one end of a stretcher to a rescue
helicopter. Based on dynamic simulations, the design
engineer estimates that the downward forces the stretcher
will exert will be no greater than 1.6 kN at A and at B.
What are the resulting axial forces in members CF, DF,
and FG?
G
300
mm
290
mm
390
mm
150 mm
F
480 mm
C
E
D
200 mm
B
A
48
FCF 1.6 kN D 0 ) FCF D 2.06 kN
Fy : p
3825
FCF
39
48
C
FCD
1.6 kN
Now use joint F
59
29
39
FFG p
FDF C p
FCF D 0
Fx : p
3706
3145
3825
15
48
48
FFG p
FDF p
FCF D 0
Fy : p
3706
3145
3825
Solving we find
FDF D 1.286 kN, FCF D 2.03 kN
59
FFG
15
F
39
48
48
29
FDF
FCF
In Summary
FCF D 2.06 kNT, FDF D 1.29 kNC, FCF D 2.03 kNT
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
399
Problem 6.16 Upon learning of an upgrade in the helicopter’s engine, the engineer designing the truss does
new simulations and concludes that the downward forces
the stretcher will exert at A and at B may be as large as
1.8 kN. What are the resulting axial forces in members
DE, DF, and DG?
Solution: Assume all bars are in tension.
Next work with joint B
Start at joint C
16
TCF 1.8 kN D 0 ) TCF D 2.32 kN
Fy : p
425
13
TCF TCD D 0 ) TCD D 1.463 kN
Fx : p
425
TBE
3
Fx : p TBE D 0 ) TBE D 0
13
2
Fy : p TBE C TBD 1.8 kN D 0 ) TBD D 1.8 kN
13
TBD
TCF
2
13
B
3
16
C
1.8 kN
TCD
Finally work with joint D
1.8 kN
Next work with joint F
59
29
13
TFG p
TDF C p
TCF D 0
Fx : p
3706
3145
425
10
29
TDG C p
TDF C TCD D 0
Fx : TDE p
541
3145
21
48
TDG C p
TDF TBD D 0
Fy : p
541
3145
Solving:
TDG D 6.82 kN, TDE D 7.03 kN
TDF
TDG
15
48
48
TFG p
TDF p
TCF D 0
Fy : p
3706
3145
425
21
Solving
TDF D 5.09 kN, TFG D 4.23 kN
48
29
10
59
TFG
15
TDE
F
D
TCD
13
48
TBD
16
29
TDF
TCF
In summary:
TDE D 7.03 kNC, TDF D 5.09 kNC, TDG D 6.82 kNT
400
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 6.17 Determine the axial forces in the
members in terms of the weight W.
B
E
1m
A
D
W
1m
C
0.8 m
0.8 m
0.8 m
Solution: Denote the axial force in a member joining two points
I, K by IK. The angle between member DE and the positive x axis
is ˛ D tan1 0.8 D 38.66&deg; . The angle formed by member DB with the
positive x axis is 90&deg; C ˛. The angle formed by member AB with the
positive x axis is ˛.
Joint E :
Fy D DE cos ˛ W D 0,
from which DE D 1.28W C .
Fy D BE DE sin ˛ D 0,
from which BE D 0.8W T
Joint D:
Fx D DE cos ˛ C BD cos ˛ CD cos ˛ D 0,
from which BD CD D DE.
Fy D BD sin ˛ C DE sin ˛ CD sin ˛ D 0,
from which BD C CD D DE.
Solving these two equations in two unknowns:
CD D DE D 1.28W C , BD D 0
Joint B :
Fx D BE AB sin ˛ BD sin ˛ D 0,
from which AB D
BE
D 1.28WT
sin ˛
Fy D AB cos ˛ BC D 0,
from which BC D AB cos ˛ D WC
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
401
Problem 6.18 The lengths of the members of the truss
are shown. The mass of the suspended crate is 900 kg.
Determine the axial forces in the members.
A
12 m
B
13 m
5m
C
13 m
12 m
D
40⬚
Finally work with joint B
Fx : FAB cos 40&deg; FAC sin 27.4&deg; D 0
Fy : FAB cos 50&deg; FBC sin 50&deg; FBD cos 27.4&deg; D 0
Fy : FAB sin 40&deg; FAC cos 27.4&deg; 900 kg9.81 m/s2 D 0
FAB
50&deg;
A
40&deg;
T
B
50&deg;
27.4&deg;
FAB
27.4&deg;
FBC
FAC
8829 N
Solving we find
Next work with joint C
FBD
Fx : FCD cos 40&deg; FBC cos 50&deg; C FAC sin 27.4&deg; D 0
FAB D 10.56 kN D 10.56 kNT
FAC D 17.58 kN D 17.58 kNC
Fy : FCD sin 40&deg; C FBC sin 50&deg; C FAC cos 27.4&deg; D 0
FAC
FBC
27.4&deg;
FCD D 16.23 kN D 16.23 kNC
FBC D 6.76 kN D 6.76 kNT
FBD D 1.807 kN D 1.807 kNT
50&deg;
C
40&deg;
FCD
402
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 6.19 The loads F1 D 600 lb and F2 D
300 lb. Determine the axial forces in members AE, BD,
and CD.
F1
G
D
F2
B
6 ft
C
3 ft
E
A
4 ft
4 ft
Solution: The reaction at E is determined by the sum of the
F1
GX
MG D C6E 4F1 8F2 D 0,
F2
GY
6 ft
from which
ED
4F1 C 8F2
D 800 lb.
6
E
EG
The interior angle EAG is
˛D
tan1
6
D 36.87&deg; .
8
E
AC
AE
Joint E
From similar triangles this is also the value of the interior angles ACB,
CBD, and CGD. Method of joints: Denote the axial force in a member
joining two points I, K by IK.
from which BD D
Joint E :
4 ft
α
AE
AB
4 ft
BD
α
BC
Joint A
F2
F1
DG
AB
Joint B
CD
α
BD
Joint D
300
F2 C AB
D
D 500 lbC .
0.6
0.6
Fx D BC BD cos ˛ D 0,
Fy D E C AE D 0,
from which BC D BD0.8 D 400 lbT.
from which AE D E D 800 lb C .
Joint D:
Fy D EG D 0,
from which EG D 0.
Fy D BD sin ˛ CD F1 D 0,
from which CD D F1 BD0.6 D 300 lbC
Joint A:
Fy D AE AC cos ˛ D 0,
from which AC D AE
D 1000 lbT.
0.8
Fy D AC sin ˛ C AB D 0,
from which AB D AC0.6 D 600 lbC.
Joint B :
Fy D BD sin ˛ AB F1 D 0,
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
403
Problem 6.20 Consider the truss in Problem 6.19. The
loads F1 D 450 lb and F2 D 150 lb. Determine the axial
forces in members AB, AC, and BC.
Solution: From the solution to Problem 6.19 the angle ˛ D 36.87&deg;
4F1 C 8F2
D 500 lb. Denote the axial
6
force in a member joining two points I, K by IK.
EG
and the reaction at E is E D
E
Joint E :
AE
Joint E
AC
AB
α
AE
Joint A
BD
α
BC
F2
AB
Joint B
Fy D EG D 0.
Fx D AE C E D 0,
from which AE D E D 500 lbC.
Joint A:
Fx D AE AC cos ˛ D 0,
from which AC D AE
D 625 lbT .
0.8
Fy D AC sin ˛ C AB D 0,
from which AB D AC0.6 D 375 lbC
Joint B:
Fy D BD sin ˛ F2 AB D 0,
from which BD D
F2 C AB
D 375 lbC
0.6
Fx D BC BD cos ˛ D 0,
from which BC D BD0.8 D 300 lbT
404
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
C
Problem 6.21 Determine the axial forces in members
BD, CD, and CE of the truss.
E
4 ft
G
B
D
F
12 kip
4 ft
H
A
4 ft
4 ft
4 ft
Solution: The free-body diagrams for the entire truss as well as
for joints A, B and C are shown.
From the entire truss:
Fx : Ax D 0
FH : 12 kip8 ft Ay 12 ft D 0
Solving, yields Ax D 0, Ay D 8 kip
From joint A:
Fx : Ax C TAD cos 45&deg; D 0
Fy : Ay C TAB C TAD sin 45&deg; D 0
Solving yields TAB D 8 kip, TAD D 0
From joint B:
Fx : TBD C TBC cos 45&deg; D 0
Fy : TBC C sin 45&deg; TAB D 0
Solving yields TBD D 8 kip, TBC D 11.3 kip
From joint C:
Fx : TCE TBC cos 45&deg; D 0
Fy : TBC sin 45&deg; TCD D 0
Solving yields TCD D 8 kip, TCE D 8 kip
Thus we have
BC : 11.3 kip (C),
CD : 8 kip (T),
CE : 8 kip (C)
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
405
Problem 6.22 The Warren truss supporting the
walkway is designed to support vertical 50-kN loads at
B, D, F, and H. If the truss is subjected to these loads,
what are the resulting axial forces in members BC, CD,
and CE?
B
D
F
H
2m
A
C
6m
Solution: Assume vertical loads at A and I Find the external loads
at A and I, then use the method of joints to work through the structure
to the members needed.
50 kN
50 kN
6m
50 kN
6m
3m
AY
3m
6m
I
6m
D 33.69&deg;
6m
6m
G
AB D 180.3 kN
50 kN
E
Fx : BC cos C BD AB cos D 0
Fy :
50 AB sin BC sin D 0
x
IY
Solving,
BC D 90.1 kN T
BD D 225 kN C
Fy : Ay C Iy 450 D 0 (kN)
MA :
Solving
Joint C :
350 950 1550 2150 C 24 Iy D 0
y
Ay D 100 kN
BC
Iy D 100 kN
θ
CD
θ
Joint A:
AC
y
C
CE
x
D 33.69&deg;
AB
AC D 150 kN T
BC D 90.1 kN T
θ
A
x
AC
AY
CE D 300 kN T
D 33.69&deg;
Fy : CD sin C BC sin D 0
Solving,
2
3
tan D
Fx : CE AC C CD cos BC cos D 0
CD D 90.1 kN C
Fx :
AB cos C AC D 0
Fy :
AB sin C Ay D 0
Hence
Solving,
BC D 90.1 kN T
CD D 90.1 kN C
CE D 300 kN T
AB D 180.3 kN C
AC D 150 kN T
Joint B :
y
50 kN
B
θ
BD
x
θ
BC
AB
406
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 6.23 For the Warren truss in Problem 6.22,
determine the axial forces in members DF, EF, and FG.
Solution: In the solution to Problem 6.22, we solved for the forces
Solving, we get
in AB, AC, BC, BD, CD, and CE. Let us continue the process. We
ended with Joint C. Let us continue with Joint D.
EF D 0
Joint D:
EG D 300 kN T
y
Note: The results are symmetric to this point!
50 kN
Joint F :
D
BD
DF
y
DF
DE
CD
50 kN
x
θ
θ
x
θ
θ
D 33.69&deg;
FH
F
BD D 225 kN C
EF
CD D 90.1 kN C
D 33.69&deg;
Fx :
DF BD C DE cos CD cos D 0
Fy :
50 CD sin DE sin D 0
Solving,
FG
DF D 300 kN C
EF D 0
DF D 300 kN C
DE D 0
At this point, we have solved half of a symmetric truss with a
the remaining members. We will continue, and use symmetry as a
check.
Joint E :
FH DF C FG cos EF cos D 0
Fy :
50 EF sin FG sin D 0
Solving:
FH D 225 kN C
FG D 90.1 kN C
Thus, we have
DF D 300 kN C
EF D 0
FG D 90.1 kN C
y
EF
DE
Fx :
Note-symmetry holds!
θ
θ
CE
E
EG
x
D 33.69&deg;
CE D 300 kN T
DE D 0
Fx :
EG CE C EF cos DE cos D 0
Fy :
DE sin C EF sin D 0
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
407
Problem 6.24 The Pratt bridge truss supports five
forces (F D 300 kN). The dimension L D 8 m. Determine the axial forces in members BC, BI, and BJ.
L
L
L
L
L
L
B
C
D
E
G
I
J
K
L
M
L
A
H
F
Solution: Find support reactions at A and H. From the free body
and
Fx D AX D 0,
L
8
MA D 68HY 3008 C 16 C 24 C 32 C 40 D 0.
L
8
L
K
L
8
L
8
F
L
8
L
H
HY
Joint I
y
TAB
TBI
θ
TAI
F
F
F
F = 300 kN
y
A
M
L
8
Joint A
From the geometry, the angle D 45&deg;
J
F
F
L=8m
Fy D AY C HY 5300 D 0,
From these equations, AY D HY D 750 kN.
F
G
θ I
A
AY
Joint A: From the free body diagram,
F
B
diagram,
F
I
x
TIJ
TAI
x
F
Fx D AX C TAB cos C TAI D 0,
AY
Joint B
y
Fy D TAB sin C AY D 0.
TBC
From these equations,
TAB D 1061 kN
and TAI D 750 kN.
θ
θ
x
TBJ
TAB
TBI
Joint I: From the free body diagram,
Fx D TIJ TAI D 0,
Fy D TBI 300 D 0.
From these equations,
TBI D 300 kN
and TIJ D 750 kN.
Joint B: From the free body diagram,
Fx D TBC C TBJ cos TAB cos D 0,
Fy D TBI TBJ sin TAB sin D 0.
From these equations,
TBC D 1200 kN
and TBJ D 636 kN.
408
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 6.25 For the roof truss shown, determine the
axial forces in members AD, BD, DE, and DG. Model
the supports at A and I as roller supports.
10 kN
8 kN
E
6 kN
C
8 kN
F
B
6 kN
H
D
3m
3m
Solution: Use the whole structure to find the reaction at A.
Next go to joint C
MI : 6 kN3 m C 8 kN6 m C 10 kN9 m
C 8 kN12 m C 6 kN15 m
C A18 m D 0 ) A D 19 kN
G
3m
3m
3m
3m
Fy : 8 kN FCD C FCE FBC sin 21.8&deg; D 0
Fx : FCE FBC cos 21.8&deg; D 0
Solving:
FCD D 8 kN, FCE D 43.1 kN
10 kN
6 kN
3.6 m
I
A
8 kN
8 kN
8 kN
FCD
6 kN
C
FCD
FBC
A
I
Finally examine joint D
Now work with joint A
Fx : FAD C FDG FBD cos 21.8&deg; C FDE cos 50.19&deg; D 0
Fy : FAB sin 21.8&deg; C A D 0 ) FAB D 51.2 kN
Fx : FAD C FAB cos 21.8&deg; D 0 ) FAD D 47.5 kN
Solving:
Fy : FBD sin 21.8&deg; C FCD C FDE sin 50.19&deg; D 0
FDE D 14.3 kN, FDG D 30.8 kN
FCD
FAB
FDE
FBD
21.8&deg;
A
50.19&deg;
A
FDG
In Summary
Next use joint B
D
Fx : FAB C FBC C FBD cos 21.8&deg; D 0
FAD D 47.5 kNT, FBD D 8.08 kNC,
FDE D 14.32 kNT, FDG D 30.8 kNT
Fy :
Solving:
FAB C FBC FBD sin 21.8&deg;
6 kN D 0
FBC D 43.1 kN, FBD D 8.08 kN
6 kN
FBC
B
FAB
FBD
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
409
Problem 6.26 The Howe truss helps support a roof.
Model the supports at A and G as roller supports. Determine the axial forces in members AB, BC, and CD.
800 lb
600 lb
600 lb
D
400 lb
400 lb
C
E
8 ft
B
F
A
G
H
I
4 ft
4 ft
J
4 ft
Solution: The strategy is to proceed from end A, choosing joints
8
12
BH D 4
1400 lb
BI
α Pitch
from which the angle
˛HIB D tan1
CI D 8
2.6667
4
D 33.7&deg; .
from which the angle
˛IJC D tan1
5.333
4
CI
AH
HI
Joint H
600 lb
CD
400 lb
α Pitch
BC
α Pitch
AB
BH BI
Joint B
α Pitch
α IJC
BC CI
CJ
Joint C
Joint H :
BH
AH
HI
IJ
Joint I
8
D 5.3333 ft,
12
600 lb
400 lb
G
α Pitch
Joint A
D 2.6667 ft,
4 ft
4 ft 4 ft 4 ft 4 ft 4 ft 4 ft
D 33.7&deg; .
AB
8
12
4 ft
A
The length of the vertical members:
L
800 lb
The interior angles HIB and HJC differ. The pitch angle is
˛Pitch D tan1
4 ft
600 lb
400 lb
with only one unknown axial force in the x- and/or y-direction, if
possible, and if not, establish simultaneous conditions in the unknowns.
K
Fy D BH D 0, or, BH D 0.
Fx D AH C HI D 0,
D 53.1&deg; .
from which HI D 2100 lb T
MG D 4 C 20400 C 8 C 16600 C 12800 24A D 0,
33600
D 1400 lb. Check: The total load is 2800 lb.
24
From left-right symmetry each support A, G supports half the total
from which A D
Joint B :
Fx D AB cos ˛Pitch C BC cos ˛Pitch
C BI cos ˛Pitch D 0,
from which BC C BI D AB
The method of joints: Denote the axial force in a member joining two
points I, K by IK.
Joint A:
Fy D AB sin ˛P C 1400 D 0,
from which AB D 1400
D 2523.9 lb C
sin ˛p
Fx D AB cos ˛Pitch C AH D 0,
from which AH D 2523.90.8321 D 2100 lb T
410
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6.26 (Continued )
Fy D 400 AB sin ˛Pitch C BC sin ˛Pitch
BI sin ˛Pitch D 0,
from which BC BI D AB C
400
.
sin ˛Pitch
Solve the two simultaneous equations in unknowns BC, BI:
BI D and
400
D 360.56 lb C,
2 sin ˛Pitch
BC D AB BI D 2163.3 lb C
Joint I :
Fx D BI cos ˛Pitch HI C IJ D 0,
from which IJ D 1800 lb T
Fy D CBI sin ˛Pitch C CI D 0,
from which CI D 200 lb (T)
Joint C:
Fx D BC cos ˛Pitch C CD cos ˛Pitch C CJ cos ˛IJC D 0,
from which CD0.8321 C CJ0.6 D 1800
Fy D 600 CI BC sin ˛Pitch C CD sin ˛Pitch
CJ sin ˛IJC D 0,
from which CD0.5547 CJ0.8 D 400
Solve the two simultaneous equations to obtain CJ D 666.67 lb C,
and
CD D 1682.57 lb C
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
411
Problem 6.27 The plane truss forms part of the
supports of a crane on an offshore oil platform. The
crane exerts vertical 75-kN forces on the truss at B, C,
and D. You can model the support at A as a pin support
and model the support at E as a roller support that can
exert a force normal to the dashed line but cannot exert
a force parallel to it. The angle ˛ D 45&deg; . Determine the
axial forces in the members of the truss.
C
B
D
1.8 m
2.2 m
F
G
H
A
α
E
3.4 m
3.4 m
3.4 m
3.4 m
Solution: The included angles
D tan1
ˇ D tan1
D tan1
4
3.4
2.2
3.4
1.8
3.4
75 kN 75 kN 75 kN
D 49.64&deg; ,
D 32.91&deg; ,
AY 3.4
m
D 27.9&deg; .
AB
AX
Ex cos 45&deg;
C Ey cos 45&deg;
with this relation and the fact that
obtain Ex D 112.5 kN and Ey D 112.5 kN. From
D 0, we
FAy D Ay 375 C Ey D 0,
The method of joints: Denote the axial force in a member joining two
points I, K by IK.
Joint A:
3.4
m
EY
3.4
m
BF
DE
EX
AF
γ β
CD
θ
DG
EY
Joint E
75 kN
γ
DH
Joint D
DE
β
AF
DH
FG
Joint F
75 kN
BC
CD
GH
β
EH
Joint H
CG
Joint C
FAx D Ax C Ex D 0, AX D EX D 112.5 kN.
from which Ay D 112.5 kN. Thus the reactions at A and E are symmetrical about the truss center, which suggests that symmetrical truss
members have equal axial forces.
β γ
EH
AY
Joint A
75 kN
BC
θ
γ
BF BG
AB
Joint B
MA D 753.41 C 2 C 3 C 43.4Ey D 0.
3.4
m
The complete structure as a free body: The sum of the moments about
A is
EX
AX
Fy D AB sin C Ay C AF sin ˇ D 0,
from which two simultaneous equations are obtained.
AF D 44.67 kN C ,
and
and
EH D 44.67 kNC ,
DE D 115.8 kNC
Joint F :
Fx D AF cos ˇ C FG D 0,
from which FG D 37.5 kN C
Fx D AB cos C Ax C AF cos ˇ D 0,
Solve:
Solve:
Fy D AF sin ˇ C BF D 0,
from which BF D 24.26 kN C
Joint H:
Fx D EH cos ˇ GH D 0,
AB D 115.8 kN C
Joint E:
Fy D DE cos C Ex EH cos ˇ D 0.
Fy D DE sin C Ey C EH sin ˇ D 0,
from which two simultaneous equations are obtained.
412
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6.27 (Continued )
from which GH D 37.5 kN C
from which DG D 80.1 kN T
Fy D EH sin ˇ C DH D 0,
Fx D DE cos CD DG cos D 0,
from which DH D 24.26 kN C
from which CD D 145.8 kN C
Joint B:
Joint C :
Fy D AB sin BF C BG sin 75 D 0,
Fx D CD BC D 0,
from which BG D 80.1 kN T
from which CD D BC Check.
Fx D AB cos C BC C BG cos D 0,
from which BC D 145.8 kN C
Fy D CG 75 D 0,
from which CG D 75 kN C
Joint D:
Fy D DE sin DH DG sin 75 D 0,
Problem 6.28 (a) Design a truss attached to the
supports A and B that supports the loads applied at points
C and D.
(b) Determine the axial forces in the members of the
truss you designed in (a)
1000 lb
2000 lb
C
D
4 ft
A
5 ft
5 ft
5 ft
Obstacle
Problem 6.29 (a) Design a truss attached to the
supports A and B that goes over the obstacle and
supports the load applied at C.
(b) Determine the axial forces in the members of the
truss you designed in (a).
2 ft
B
C
4m
2m
B
A
6m
3.5 m
10 kN
4.5 m
1m
Solution: This is a design problem with many possible solutions.
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
413
Problem 6.30 Suppose that you want to design a truss
supported at A and B (Fig. a) to support a 3-kN downward load at C. The simplest design (Fig. b) subjects
member AC to 5-kN tensile force. Redesign the truss so
that the largest force is less than 3 kN.
A
A
1.2 m
C
C
B
B
3 kN
3 kN
1.6 m
(a)
Solution: There are many possible designs. To better understand
D 36.87&deg;
the problem, let us calculate the support forces in A and B and the
forces in the members in Fig. (b).
Ay
Fx :
(b)
BC AC cos D 0
Fy : AC sin 3 kN D 0
Solving: BC D 4 kN C AC D 5 kN C
Ax
A
Thus, AC is beyond the limit, but BC (in compression) is not,
Joint B :
1.2 m
θ
AB
C
B
Bx
x
1.6 m
BX
BC
3 kN
1.2
1.6
tan D
D 36.87&deg;
sin D 0.6
Fx :
Ax C Bx D 0
Fy :
Ay 3 kN D 0
MA :
1.2Bx 1.63 D 0
C
Fy : AB D 0
Solving, BC and Bx are both already known. We get AB D 0
cos D 0.8
Fx : Bx C BC D 0
Thus, we need to reduce the load in AC. Consider designs like that
shown below where D is inside triangle ABC. Move D around to adjust
A
Solving, we get
Ax D 4 kN
Bx D 4 kN
D
Ay D 3 kN
Note: These will be the external reactions for every design that we
produce (the supports and load do not change).
B
C
Reference Solution (Fig. (b))
Joint C :
However, the simplest solution is to place a second member parallel
to AC, reducing the load by half.
AC
θ
BC
3 kN
414
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 6.31 The bridge structure shown in Example
6.2 can be given a higher arch by increasing the 15&deg;
angles to 20&deg; . If this is done, what are the axial forces
in members AB, BC, CD, and DE?
F
F
b
F
b
F
b
F
b
2b
(1)
F
F
G
F
F
I
J
H
b
a
b
15⬚
b
15⬚
K
b
2b
C
D
B
F
a
A
E
(2)
Solution: Follow the solution method in Example 6.3. F is known
For joint C,
Joint B :
y
F
Fx :
TBC cos 20&deg; C TCD cos 20&deg; D 0
Fy :
F TBC sin 20&deg; TCD sin 20&deg; D 0
TBC
TBC D TCD D 1.46F C
For joint B.
20&deg;
x
α
TAB
Fx :
TBC cos 20 TAB cos ˛ D 0
Fy :
TBC sin 20&deg; F TAB sin ˛ D 0
Solving, we get ˛ D 47.5&deg; and TAB D 2.03F C
Joint C :
For the new truss (using symmetry)
F
C
20&deg;
TBC
20&deg;
TCD
Members
Forces
AG, BH, CI,
DJ, EK
F
AB, DE
2.03F (C)
BC, CD
1.46F (C)
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
415
Problem 6.32 In Active Example 6.3, use the method
of sections to determine the axial forces in members BC,
BI and HI.
A
B
C
D
E
F
G
H
I
J
K
L
1m
M
100 kN
Solution: The horizontal members of the truss are each 1 m in
length. We cut through the relevant members and draw a free-body
diagram of the section to the right of the cut.
We will use equilibrium equations for this section that are designed to
allow us to easily solve for the unknowns.
The equilibrium equations
MI : TBC 1 m 100 kN4 m D 0 ) TBC D 400 kN
MB : THI 1 m 100 kN5 m D 0 ) THI D 500 kN
Fy : TBI sin 45&deg; 100 kN D 0 ) TBI D 141 kN
In summary we have
BC : 400 kN (T),
BI : 141 kN (T),
HI : 500 kN (C)
Problem 6.33 In Example 6.4, obtain a section of the
truss by passing planes through members BE, CE, CG,
and DG. Using the fact that the axial forces in members
DG and BE have already been determined, use your
section to determine the axial forces in members CE
and CG.
G
D
J
L
C
I
L
A
B
E
F
L
H
2F
L
K
F
L
L
Solution: From Example 6.4 we know that
TDG D F, TBE D F
Ax D 0, Ay D 2F
We make the indicated cuts and isolate the section to the left of the
cuts. The equilibrium equations are
Fx : TDG C TBE C TCG cos 45&deg; C TCE cos 45&deg; D 0
Fy : Ay F C TCG sin 45&deg; TCE sin 45&deg; D 0
F
F
Solving yields TCE D p , TCG D p
2
2
We have
416
F
CE : p T,
2
F
CG : p C
2
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Problem 6.34 The truss supports a 100-kN load at J.
The horizontal members are each 1 m in length.
(a)
(b)
Use the method of joints to determine the axial
force in member DG.
Use the method of sections to determine the axial
force in member DG.
A
B
C
D
E
F
G
H
1m
J
100 kN
Solution:
(a)
We draw free-body diagrams of joints J, H, and D.
From joint J we have
Fy : TDJ sin 45&deg; 100 kN D 0
) TDJ D 141 kN
From joint H we have Fy : TDH D 0
From joint D we have
Fy : TDG sin 45&deg; TDH TDJ sin 45&deg; D 0
Solving yields TDG D 141 kN
(b)
We cut through CD, DG and GH. The free-body diagram of
the section to the right of the cut is shown. From this diagram
we have
Fy : TDG sin 45&deg; 100 kN D 0
) TDG D 141 kN
In summary
(a), (b) DG : 141 kN (C)
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
417
Problem 6.35 For the truss in Problem 6.34, use the
method of sections to determine the axial forces in
members BC, CF, and FG.
BC
Solution:
Fx :
C
D
45&deg;
BC CF cos 45 FG D 0
1m
CF
CF sin 45&deg; 100 D 0
Fy :
J
F
MC :
FG
G
1FG 2100 D 0
H
1m
1m
100 kN
Solving
BC D 300 kN T
CF D 141.4 kN C
FG D 200 kN C
Problem 6.36 Use the method of sections to determine
the axial forces in members AB, BC, and CE.
1m
1m
A
B
1m
D
1m
G
C
E
F
2F
Solution: First, determine the forces at the supports
Θ = 45&deg;
D
B
Method of Sections:
y
AY = 1. 33 F
AX
AY
C
1m
F
E
1m
1m
θ
1m
AX = 0
AX = 0
Fx :
Ax D 0
Fy :
Ay C Gy 3F D 0
MA :
1F 22F C 3Gy D 0
C
1m
AY
1m
C
CE
x
F
Fx :
CE C AB D 0
Fy :
BC C Ay F D 0
MB :
1Ay C 1CE D 0
Solving
B
BC
GY
2F
AB
Ax D 0
Gy D 1.67F
Ay D 1.33F
C
Solving, we get
AB D 1.33F C
CE D 1.33F T
BC D 0.33F C
418
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
18 kN
Problem 6.37 Use the method of sections to determine
the axial forces in members DF, EF, and EG.
24 kN
C
E
G
H
A
300 mm
B
F
D
400 mm
400 mm
400 mm
400 mm
Solution: We will first use the free-body diagram of the entire
structure to find the reaction at F.
MB : 18 kN 400 mm
24 kN 1200 mm
C F 800 mm D 0
) F D 27 kN
Next we cut through DF, EF, EG and look at the section to the right
of the cut. The angle ˛ is given by
˛ D tan1 3/4 D 36.9&deg;
The equilibrium equations are
MF : TEG 300 mm 24 kN 400 mm D 0
ME : TDF 300 mm 24 kN 800 mm
C F400 mm D 0
Fy : F 24 kN C TEF sin ˛ D 0
Solving yields TDF D 28 kN, TEF D 5 kN, TEG D 32 kN
Thus
DF : 28 kN (C), EF : 5 kN (C), EG : 32 kN (T)
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
419
Problem 6.38 The Pratt bridge truss is loaded as
shown. Use the method of sections to determine the axial
forces in members BD, BE, and CE.
B
D
F
A
H
C
17 ft
E
8 ft
G
10 kip
30 kip
20 kip
17 ft
17 ft
17 ft
Solution: Use the whole structure to find the reaction at A.
MH : 20 kip17 ft C 30 kip34 ft
C 10 kip51 ft A68 ft D 0
10
kip
A
20
kip
30
kip
H
) A D 27.5 kip
B
Now cut through BD, BE, CE and use the left section
FBD
8
MB : A17 ft C FCE 8 ft D 0 ) FCE D 58.4 kip
17
FBE
ME : 10 kip17 ft A34 ft FBD 8 ft D 0
C
A
FCE
) FBD D 95.6 kip
Fy : A 10 kip p
8
353
FBE D 0 ) FBE D 41.1 kip
10 kip
In Summary
A
FCE D 58.4 kipT, FBD D 95.6 kipC, FBE D 41.1 kipT
Problem 6.39 The Howe bridge truss is loaded as
shown. Use the method of sections to determine the axial
forces in members BD, CD, and CE.
B
D
F
A
H
C
17 ft
E
10 kip
17 ft
G
30 kip
17 ft
20 kip
17 ft
B
Solution: Use the whole structure to find the reaction at A (same
FBD
as 6.38) A D 27.5 kip
Now cut through BD, CD, and CE and use the left section.
MC : A17 ft FBD 8 ft D 0 ) FBD D 58.4 kip
17
FCD
8
C
A
FCE
MD : A34 ft C 10 kip17 ft C FCE 8 ft D 0
) FCE D 95.6 kip
8 ft
Fy : A 10 kip C p
10 kip
8
353
A
FCD D 0 ) FCD D 41.1 kip
In Summary
FBD D 58.4 kipC, FCE D 95.6 kipT, FCD D 41.1 kipC
420
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 6.40 For the Howe bridge truss in Problem
6.39, use the method of sections to determine the axial
forces in members DF, DG, and EG.
D
Solution: Same truss as 6.39.
17
FDG
MD : A34 ft C 10 kip17 ft C FEG 8 ft D 0
E
) FEG D 95.6 kip
FDF
8
Cut through DF, DG, and EG and use left section
FEG
MG : A51 ft C 10 kip34 ft C 30 kip17 ft FDF 8 ft
A
30 kip
10 kip
D 0 ) FDF D 69.1 kip
8
FDG D 0 ) FDG D 29.4 kip
Fy : A 10 kip 30 kip p
353
In summary
FEG D 95.6 kipT, FDF D 69.1 kipC, FDG D 29.4 kipC
Problem 6.41 The Pratt bridge truss supports five
forces F D 340 kN. The dimension L D 8 m. Use the
method of sections to determine the axial force in
member JK.
L
L
L
L
L
L
B
C
D
E
G
I
J
K
L
M
L
A
H
Solution: First determine the external support forces.
F
AX
L
L
L
F
AY
F
L
F
L
F
HY
F = 340 kN, L = 8 M
C
Solving:
F
F
F
D 45&deg;
L
F
F
L D 8M
F D 340 kN
Ay D 850 kN
Fx : Ax D 0
Fy : Ay 5F C Hy D 0
MA : 6LHy LF 2LF 3LF 4LF 5LF D 0
Ax D 0,
C
Ay D 850 kN
Fx :
CD C JK C CK cos D 0
Fy :
Ay 2F CK sin D 0
MC :
LJK C LF 2LAy D 0
Solving,
JK D 1360 kN T
Also,
CK D 240.4 kN T
Hy D 850 kN
Note the symmetry:
Method of sections to find axial force in member JK.
C
B
θ
I
A
L
CD D 1530 kN C
CD
D
CK
J
K
L
AY
F
F
JK
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
421
Problem 6.42 For the Pratt bridge truss in Problem 6.41, use the method of sections to determine the
axial force in member EK.
Solution: From the solution to Problem 6.41, the support forces
L
are Ax D 0, Ay D Hy D 850 kN.
L
L
L
L
L
B
C
D
E
G
I
J
K
L
M
Method of Sections to find axial force in EK.
L
E
DE
A
G
H
θ
F
EK
EK D 240.4 kN T
Also,
KL D 1360 kN T
F
F
F
F
HY
DE D 1530 kN C
Solution:
F
L
KL
F
Fx :
DE EK cos KL D 0
Fy : Hy 2F EK sin D 0
ME :
LKL LF C 2LHy D 0
Problem 6.43 The walkway exerts vertical 50-kN
loads on the Warren truss at B, D, F, and H. Use
the method of sections to determine the axial force in
member CE.
B
D
F
H
2m
A
C
6m
Solution: First, find the external support forces. By symmetry,
Solving:
Ay D Iy D 100 kN (we solved this problem earlier by the method of
joints).
Also,
B
y
50 kN
BD
6m
C
6m
G
6m
I
6m
CE D 300 kN T
BD D 225 kN C
CD D 90.1 kN C
D
2m
A
E
CD
θ
CE
x
AY
tan D
2
3
D 33.69&deg;
422
Fx :
BD C CD cos C CE D 0
Fy :
Ay 50 C CD sin D 0
MC :
6Ay C 350 2BD D 0
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 6.44 Use the method of sections to determine
the axial forces in members AC, BC, and BD.
600 lb
E
4 ft
C
D
4 ft
A
B
3 ft
3 ft
Solution: Obtain a section by passing a plane through members
AC, BC, and BD, isolating the part of the truss above the planes. The
angle between member AC and the horizontal is
˛ D tan1 4/3 D 53.3&deg;
The equilibrium equations are
MC : 600 lb 4 ft TBD cos ˛ 3 ft D 0
MB : 600 lb 8 ft C TAC sin ˛ 4 ft D 0
Fy : TBC TAC cos ˛ TBD cos ˛ D 0
Solving yields
TBD D 1000 lb,
TAC D 2000 lb,
TBC D 800 lb
BD : 100 lb (T),
AC : 2000 lb (C),
BC : 800 lb (T)
Thus
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
423
Problem 6.45 Use the method of sections to determine
the axial forces in member FH, GH, and GI.
I
300 mm
C
E
G
H
300 mm
A
B
D
F
6 kN
400 mm
400 mm
400 mm
4 kN
400 mm
Solution: The free-body diagram of the entire truss is used to find
the force I.
MA : I600 mm 4 kN 1200 mm
6 kN 800 mm D 0
) I D 16 kN
Obtain a section by passing a plane through members FH, GH, and
GI, isolating the part of the truss to the right of the planes. The angle
˛ is
˛ D tan1 3/4 D 36.9&deg;
The equilibrium equations for the section are
MH : TGI cos ˛ 300 mm C I300 mm D 0
MG : I300 mm TFH cos ˛ 400 mm D 0
Fx : TGH TGI sin ˛ TFH sin ˛ D 0
Solving yields TGI D 20 kN, TFH D 20 kN, TGH D 16 kN
Thus
424
GI : 20 kN (C),
FH : 20 kN (T),
GH : 16 kN (C)
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 6.46 Use the method of sections to determine
the axial forces in member DF, DG, and EG.
I
300 mm
C
E
G
H
300 mm
A
B
D
F
6 kN
400 mm
400 mm
400 mm
4 kN
400 mm
Solution: The free-body diagram of the entire truss is used to find
the force I.
MA : I600 mm 4 kN 1200 mm
6 kN 800 mm D 0
) I D 16 kN
Obtain a section by passing a plane through members DF, DG, and
EG, isolating the part of the truss to the right of the planes. The angle
˛ is
˛ D tan1 3/4 D 36.9&deg;
The equilibrium equations for the section are
MG : I 300 mm TDF 300 mm D 0
MD : TEG 300 mm C I600 mm
4 kN400 mm D 0
Fy : TDG sin ˛ 4 kN D 0
Solving yields TDF D 16 kN, TEG D 26.7 kN, TDG D 6.67 kN
Thus
DF : 16 kN (T),
EG : 26.7 kN (C),
DG : 6.67 kN (C)
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
425
Problem 6.47 The Howe truss helps support a roof.
Model the supports at A and G as roller supports.
2 kN
2 kN
2 kN
(a)
(b)
Use the method of joints to determine the axial
force in member BI.
Use the method of sections to determine the axial
force in member BI.
D
2 kN
2 kN
C
E
4m
B
F
A
G
Solution: The pitch of the roof is
˛ D tan1
H
4
D 33.69&deg; .
6
2m
I
2m
J
K
2m
L
2m
2m
2m
F
F
This is also the value of interior angles HAB and HIB. The complete
structure as a free body: The sum of the moments about A is
F
F = 2 kN
F
MA D 221 C 2 C 3 C 4 C 5 C 62G D 0,
G
A
30
from which G D
D 5 kN. The sum of the forces:
6
FY D A 52 C G D 0,
2m 2m 2m 2m 2m 2m
BH
AB
from which A D 10 5 D 5 kN.
The method of joints: Denote the axial force in a member joining I, K
by IK.
(a)
(a)
A
α
AH
Joint A
AH
HI
2 kN
BH
Joint B
Joint H
Joint A:
F
B
Fy D A C AB sin ˛ D 0,
α
(b)
5
A
D
D 9.01 kN (C).
from which AB D
sin ˛
0.5547
Fx D AB cos ˛ C AH D 0,
BC
α
BI
α
AB
A
BC
α
α
BI
HI
2m
from which AH D AB cos ˛ D 7.5 kN (T).
Joint H :
Fy D BH D 0.
Joint B :
Fx D AB cos ˛ C BI cos ˛ C BC cos ˛ D 0,
Fy D 2 AB sin ˛ BI sin ˛ C BC sin ˛ D 0.
Solve: BI D 1.803 kN C , BC D 7.195 kN C
(b)
Make the cut through BC, BI and HI. The section as a free body:
The sum of the moments about B:
MB D A2 C HI2 tan ˛ D 0,
from which HI D
3
A D 7.5 kNT. The sum of the forces:
2
Fx D BC cos ˛ C BI cos ˛ C HI D 0,
Fy D A F C BC sin ˛ BI sin ˛ D 0.
Solve: BI D 1.803 kN C .
426
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 6.48 Consider the truss in Problem 6.47. Use
the method of sections to determine the axial force in
member EJ.
Solution: From the solution to Problem 6.47, the pitch angle is
˛ D 36.69&deg; , and the reaction G D 5 kN. The length of member EK is
LEK D 4 tan ˛ D
16
D 2.6667 m.
6
The interior angle KJE is
ˇ D tan1
LEK
2
DE
β
F
E
F
EJ
α
JK
2m
G
2m
D 53.13&deg; .
Make the cut through ED, EJ, and JK. Denote the axial force in a
member joining I, K by IK. The section as a free body: The sum of
ME D C4G 2F JK2.6667 D 0,
from which JK D
20 4
D 6 kN T.
2.6667
The sum of the forces:
Fx D DE cos ˛ EJ cos ˇ JK D 0.
Fy D DE sin ˛ EJ sin ˇ 2F C G D 0,
from which the two simultaneous equations:
0.8321DE C 0.6EJ D 6,
0.5547DE 0.8EJ D 1.
Solve: EJ D 2.5 kN C .
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
427
C
Problem 6.49 Use the method of sections to determine
the axial forces in member CE, DE, and DF.
E
4 ft
G
B
D
F
12 kip
4 ft
H
A
4 ft
4 ft
4 ft
Solution: The free-body diagrams for the entire structure and the
section to the right of the cut are shown.
From the entire structure:
MA : 12 kip 4 ft H 12 ft D 0
) H D 4 kip
Using the section to the right of the cut we have
ME : H4 ft TDF 4 ft D 0
MD : H8 ft C TCE 4 ft D 0
Fy : H TDE sin 45&deg; D 0
Solving yields
TDF D 4 kip,
TCE D 8 kip,
TDE D 5.66 kip
Thus we have
DF : 4 kip (T)
CE : 8 kip (C)
DE : 5.66 kip (T)
428
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 6.50 For the bridge truss shown, use the
method of sections to determine the axial forces in
members CE, CF, and DF.
200 kN
200 kN
200 kN
D
B
200 kN
200 kN
F
H
J
E
G
C
3m
7m
4m
I
A
5m
5m
5m
5m
Solution: From the entire structure we find the reactions at A
Now we cut through DF, CF, and CE and use the left section.
Fx : Ax D 0
) FDF D 375 kN
MI : 200 kN5 m C 200 kN10 m C 200 kN15 m
C 200 kN20 m Ay 20 m D 0 ) Ay D 500 kN
200 kN
200 kN
200 kN
200 kN
MC : 200 kN5 m Ay 5 m C Ax 3 m FDF 4 m D 0
200 kN
MF : 200 kN10 m C 200 kN5 m Ay 10 m C Ax 7 m
5
1
C p FCE 4 m p FCE 5 m D 0 ) FCE D 680 kN
26
26
5
5
Fx : Ax C FDF C p FCE C p FCF D 0
26
41
) FCF D 374 kN
200 kN
200 kN
Ax
D
I
FDF
Ay
FCF
5
4
FCE
1
C
5
Ax
Ay
Summary:
FDF D 375 kNC, FCE D 680 kNT, FCF D 374 kNC
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
429
Problem 6.51 The load F D 20 kN and the dimension
L D 2 m. Use the method of sections to determine the
axial force in member HK.
L
L
A
B
C
F
Strategy: Obtain a section by cutting members HK,
HI, IJ, and JM. You can determine the axial forces in
members HK and JM even though the resulting freebody diagram is statically indeterminate.
L
E
D
F
G
L
I
H
J
K
M
L
Solution: The complete structure as a free body: The sum of the
2L
F
moments about K is MK D FL2 C 3 C ML2 D 0, from which
5F
D 50 kN. The sum of forces:
MD
2
FY D KY C M D 0,
L
F
2L
from which KY D M D 50 kN.
FX D KX C 2F D 0,
KX
M
KY
from which KX D 2F D 40 kN.
The section as a free body: Denote the axial force in a member joining
I, K by IK. The sum of the forces:
from which HI IJ D Kx . Sum moments about K to get MK D
ML2 C JML2 IJL C HIL D 0.
HK
KX
Fx D Kx HI C IJ D 0,
Substitute HI IJ D Kx , to obtain JM D M HI
IJ
JM
L
M
KY
2L
Kx
D 30 kN C.
2
Fy D Ky C M C JM C HK D 0,
from which HK D JM D 30 kNT
430
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 6.52 The weight of the bucket is W D
1000 lb. The cable passes over pulleys at A and D.
D
A
C
(a) Determine the axial forces in member FG and HI.
(b) By drawing free-body diagrams of sections, explain
why the axial forces in members FG and HI are
equal.
F
B
H
3 ft 6 in
J
3 ft
E
3 ft 3 in L
G
I
35&deg;
3 ft
W
K
Solution: The truss is at angle ˛ D 35&deg; relative to the horizontal.
The angles of the members FG and HI relative to the horizontal are
ˇ D 45&deg; C 35&deg; D 80&deg; . (a) Make the cut through FH, FG, and EG,
and consider the upper section. Denote the axial force in a member
joining, ˛, ˇ by ˛ˇ.
W
FH
β
3.25 ft
The section p
as a free body: The perpendicular distance from point F
is LFW D 3 2 sin ˇ C 3.5 D 7.678 ft.
3 ft
The sum of the moments about F is MF D WLFW C W3.25 jEGj3 D 0, from which EG D 1476.1 lb C.
W
α
FG
EG
W
3.5 ft
W
The sum of the forces:
FY D FG sin ˇ FH sin ˛ EG sin ˛ W sin ˛ W D 0,
JH
HI
GI
FX D FG cos ˇ FH cos ˛ EG cos ˛ W cos ˛ D 0,
from which the two simultaneous equations:
0.9848FG 0.5736FH D 726.9, and 0.1736FG 0.8192FH D
389.97.
Solve: FG D 1158.5 lb C , and FH D 721.64 lb T. Make the
cut through JH, HI, and GI, and consider the upper section.
(b) Choose a coordinate system with the y axis parallel to JH. Isolate
a section by making cuts through FH, FG, and EG, and through HJ,
HI, and GI. The free section of the truss is shown. The sum of the
forces in the x- and y-direction are each zero; since the only external
x-components of axial force are those contributed by FG and HI, the
two axial forces must be equal:
The section as a free body: The perpendicular distance from
point
p
H to the line of action of the weight is LHW D 3 cos ˛ C 3 2 sin ˇ C
3.5 D 10.135 ft. The sum of the moments about H is MH D WL jGIj3 C W3.25 D 0, from which jGIj D 2295 lb C.
Fx D HI cos 45&deg; FG cos 45&deg; D 0,
from which HI D FG
FY D HI sin ˇ JH sin ˛ GI sin ˛ W sin ˛ W D 0,
FX D HI cos ˇ JH cos ˛ GI cos ˛ W cos ˛ D 0,
from which the two simultaneous equations:
0.9848HI 0.5736JH D 257.22,
and
0.1736HI 0.8192JH D 1060.8.
Solve:
and
HI D 1158.5 lbC ,
JH D 1540.6 lbT .
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
431
Problem 6.53 Consider the truss in Problem 6.52. The
weight of the bucket is W D 1000 lb. The cable passes
over pulleys at A and D. Determine the axial forces in
members IK and JL.
Solution: Make a cut through JL, JK, and IK, and consider the
upper section. Denote the axial force in a member joining, ˛, ˇ by
˛ˇ. The section as a free body: The perpendicular distance p
from point
J to the line of action of the weight is L D 6 cos ˛ C 3 2 sin ˇ C
3.5 D 12.593 ft. The sum of the moments about J is MJ D WL C
W3.25 IK3 D 0, from which IK D 3114.4 lbC.
The sum of the forces:
W
W
β
JL
α
3.5 ft
3.25 ft
3 ft
Fx D JL cos ˛ IK cos ˛
JK
IK
W cos ˛ JK cos ˇ D 0,
and
Fy D JL sin ˛ IK sin ˛
W sin ˛ W JK sin ˇ D 0,
from which two simultaneous equations:
0.8192JL C 0.1736JK D 1732
and
0.5736JL C 0.9848JK D 212.75.
Solve:
and
JL D 2360 lbT ,
JK D 1158.5 lbC .
Problem 6.54 The truss supports loads at N, P, and R.
Determine the axial forces in members IL and KM.
2m
2m
2m
2m
2m
K
M
O
Q
I
L
N
P
1 kN
2 kN
1m
J
R
2m
Solution: The strategy is to make a cut through KM, IM, and
IL, and consider only the outer section. Denote the axial force in a
member joining, ˛, ˇ by ˛ˇ.
H
G
1 kN
2m
F
E
2m
The section as a free body: The moment about M is
D
MM D IL 21 42 61 D 0,
from which
C
2m
A
B
IL D 16 kN C .
6m
The angle of member IM is ˛ D tan1 0.5 D 26.57&deg; .
The sums of the forces:
KM
α
Fy D IM sin ˛ 4 D 0,
1m
IM
IL
4
D 8.944 kN (C).
from which IM D sin ˛
1 kN
2m
2 kN
2m
1 kN
2m
Fx D KM IM cos ˛ IL D 0,
from which KM D 24 kNT
432
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 6.55 Consider the truss in Problem 6.54.
Determine the axial forces in members HJ and GI.
Solution: The strategy is to make a cut through the four members
AJ, HJ, HI, and GI, and consider the upper section. The axial force
in AJ can be found by taking the moment of the structure about B.
1m
AJ
The complete structureas a free body: The angle formed by AJ with the
4
D 26.57&deg; . The moment about B is MB D
vertical is ˛ D tan1
8
6AJ cos ˛ 24 D 0, from which AJ D 4.47 kN (T).
I
HJ
αβ
γ
HI
GI
2m
2m
1 kN
2m
2 kN
2m
1 kN
2m
The section as a free body: The
of members HJ and HI
angles
relative
2
1.5
to the vertical are ˇ D tan1
D 14.0&deg; , and D tan1
D
8
2
&deg;
36.87 respectively. Make a cut through the four members AJ, HJ,
HI, and GI, and consider the upper section. The moment about
the point I is MI D 24 C 2AJ cos ˛ C 2HJ cos ˇ D 0. From which
HJ D 8.25 kN T . The sums of the forces:
Fx D AJ sin ˛ C HJ sin ˇ HI sin D 0,
from which HI D
22
AJ sin ˛ HJ sin ˇ
D
D 0.
sin sin FY D AJ cos ˛ HJ cos ˇ HI cos GI 4 D 0,
from which GI D 16 kN C
Problem 6.56 Consider the truss in Problem 6.54. By
drawing free-body diagrams of sections, explain why the
axial forces in members DE, FG, and HI are zero.
Solution: Define ˛, ˇ to be the interior angles BAJ and ABJ
respectively. The sum of the forces in the x-direction at the base yields
AX C BX D 0, from which Ax D Bx . Make a cut through AJ, BD and
BC, from which the sum of forces in the x-direction, Ax BD sin ˇ D
0. Since Ax D AJ sin ˛, then AJ sin ˛ BD sin ˇ D 0. A repeat of the
solution to Problem 6.55 shows that this result holds for each section,
where BD is to be replaced by the member parallel to BD. For example:
make a cut through AJ, FD, DE, and CE. Eliminate the axial force
in member AJ as an unknown by taking the moment about A. Repeat
the solution process in Problem 6.55, obtaining the result that
DE D
AJ sin ˛ DF sin ˇ
D0
cos DE
where DE is the angle of the member DE with the vertical. Similarly,
a cut through AJ, FH, FG, and EG leads to
FG D
AJ sin ˛ FH sin ˇ
D 0,
cos FG
and so on. Thus the explanation is that each member BD, DF, FH and
HJ has equal tension, and that this tension balances the x-component
in member AJ
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
433
y
Problem 6.57 In Active Example 6.5, draw the freebody diagram of joint B of the space truss and use it to
determine the axial forces in members AB, BC, and BD.
1200 lb
A (5, 3, 2) ft
B
D (10, 0, 0) ft
x
z
C (6, 0, 6) ft
Solution: From Active Example 6.5 we know that the vertical
reaction force at B is 440 lb.
The free-body diagram of joint B is shown. We have the following
position vectors.
rBA D 5i C 3j C 2k ft
rBC D 6i C 6k ft
rBD D 10i ft
The axial forces in the rods can then be written as
TAB
rBA
D TAB 0.811i C 0.487j C 0.324k
jrBA j
TBC
rBC
D TBC 0.707i C 0.707k
jrBC j
TBD
rBD
D TBD i
jrBD j
The components of the equilibrium equations are
Fx : 0.811TAB C 0.707TBC C TBD D 0
Fy : 0.487TAB C 440 lb D 0
Fz : 0.324TAB C 0.707TBC D 0
Solving yields TAB D 904 lb, TBC D 415 lb, TBD D 440 lb
Thus
434
AB : 904 lb (C),
BC : 415 lb (T),
BD : 440 lb (T)
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 6.58 The space truss supports a vertical 10kN load at D. The reactions at the supports at joints A,
B, and C are shown. What are the axial forces in the
y
10 kN
D (4, 3, 1) m
Ay
Ax A
Cy
C (6, 0, 0) m
Az
By
z
Solution: Consider the joint D only. The position vectors parallel
to the members from D are
Cz
B (5, 0, 3) m
x
10 kN
rDA D 4i 3j k,
rDB D i 3j C 2k,
TDA
TDC
TDB
rDC D 2i 3j k.
The unit vectors parallel to the members from D are:
eDA D
rDA
D 0.7845i 0.5883j 0.1961k
jrDA j
eDB D
rDB
D 0.2673i 0.8018j C 0.5345k
jrDB j
eDC D
rDC
D 0.5345i 0.8018j 0.2673k
jrDC j
The equilibrium conditions for the joint D are
F D TDA eDA C TDB eDB C TDC eDC FD D 0,
from which
Fx D 0.7845TDA C 0.2673TDB C 0.5345TDC D 0
Fy D 0.5883TDA 0.8018TDB 0.8108TDC 10 D 0
Fz D 0.1961TDA C 0.5345TDB 0.2673TDC D 0.
Solve:
TDA D 4.721 kN C , TDB D 4.157 kN C
TDC D 4.850 kN C
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
435
Problem 6.59 Consider the space truss in Problem 6.58. The reactions at the supports at joints A, B,
and C are shown. What are the axial forces in members
Solution: The reactions at A are required for a determination of
the equilibrium conditions at A.
The complete structure as a free body: The position vectors are rAB D
5i C 3k, rAC D 6i, rAD D 4i C 3j C k. The sum of the forces:
and
Ay
Ax
TAC
Az
Fx D Ax D 0,
TAB
Fy D Ay C Cy C By 10 D 0,
Fz D Az C Cz D 0.
The moments due to the reactions:
M D rAB &eth; FB C rAC &eth; FC C rAD &eth; FD D 0
i
M D 5
0
j
0
By
k i
3 C 6
0 0
j
0
Cy
k i
0 C 4
Cz 0
j
3
10
k 1 D 0
0
D 3By C 10i 6Cz j C 5By C 6Cy 40k D 0.
These equations for the forces and moments are to be solved for the
unknown reactions. The solution:
Ax D Cz D 0,
Ay D 2.778 kN,
By D 3.333 kN,
and Cy D 3.889 kN
The method of joints: Joint A: The position vectors are given above.
The unit vectors are:
eAB D 0.8575i C 0.5145k,
eAC D i,
eAD D 0.7845i C 0.5883j C 0.1961k.
The equilibrium conditions are:
F D TAB eAB C TAC C eAC C TAD eAD C A D 0,
from which
Fx D 0.8575TAB C TAC C 0.7845TAD D 0
Fy D 0TAB C 0TAC C 0.5883TAD C 2.778 D 0
Fz D 0.5145jTAB j C 0jTAC j C 0.1961jTAD j D 0.
Solve:
TAB D 1.8 kN T , TAC D 2.16 kN T
436
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 6.60 The space truss supports a vertical load
F at A. Each member is of length L, and the truss rests on
the horizontal surface on roller supports at B, C, and D.
Determine the axial forces in members AB, AC, and AD.
F
A
B
D
C
Solution: By symmetry, the axial forces in members AB, AC, and
AD are equal. We just need to determine the angle between each of
these members and the vertical:
we see that
b
D tan 30&deg;
L
2
F
and
A
TAB
from which we obtain
TAC = TAB
1
Ltan 60&deg; tan 30&deg; .
2
c
Then D arcsin
L
cD
θ
θ
θ
D 35.26&deg;
F C 3TAB cos D 0,
so
TAB D TAC D TAD D F
.
3 cos and
From the top view,
L
bCc
D tan 60&deg; ,
L
2
TAB D TAC D TAD D F
3 cos 35.26&deg;
D 0.408F.
C
60&deg;
b
30&deg;
L /2
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
437
Problem 6.61 For the truss in Problem 6.60, determine the axial forces in members AB, BC, and BD.
Solution: See the solution of Problem 6.60. The axial force in
From the equilibrium equation
member AB is TAB D 0.408F, and the angle between AB and the
vertical is D 35.26&deg; . The free-body diagram of joint B is
TAB sin C 2TBC cos 30&deg; D 0,
we obtain
TAB
TBC D TBD D 0.136F.
θ
TBD = TBC
30&deg;
30&deg;
TBC
Problem 6.62 The space truss has roller supports at B,
C, and D and supports a vertical 800-lb load at A. What
are the axial forces in members AB, AC, and AD?
y
800 lb
A (4, 3, 4) ft
B
D (6, 0, 0) ft
x
z
Solution: The position vectors of the points A, B, C, and D are
rA D 4i C 3j C 4k,
rC D 5i C 6k,
rD D 6i.
C (5, 0, 6) ft
The equilibrium conditions at point A:
The position vectors from joint A to the vertices are:
Fx D 0.6247TAB C 0.2673TAC C 0.3714TAD D 0
Fy D 0.4685TAB 0.8018TAB 0.5570TAD 800 D 0
Fz D 0.6247TAB C 0.5345TAC 0.7428TAD D 0.
rAB D rB rA D 4i 3j 4k,
800 lb
rAC D rC rA D 1i 3j C 2k,
rAD D rD rA D 2i 3j 4k
Joint A: The unit vectors parallel to members AB, AC, and AD are
eAB D
rAB
D 0.6247i 0.4685j 0.6247k,
jrAB j
eAC D
rAC
D 0.2673i 0.8018j C 0.5345k,
jrAC j
D 0.3714i 0.5570j 0.7428k.
TAB
TAC
Solve:
and
438
TAB D 379.4 lb C , TAC D 665.2 lb C ,
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 6.63 The space truss shown models an
airplane’s landing gear. It has ball and socket supports
at C, D, and E. If the force exerted at A by the wheel is
F D 40j (kN), what are the axial forces in members AB,
y
E (0, 0.8, 0) m
D
0.4 m
B
x
(1, 0, 0) m
0.6 m
A
(1.1, –0.4, 0) m
C
z
F
Solution: The important points in this problem are A (1.1, 0.4,
0), B (1, 0, 0), C (0, 0, 6), and D (0, 0, 0.4). We do not need point
E as all of the needed unknowns converge at A and none involve the
location of point E. The unit vectors along AB, AC, and AD are
y
E
(0, 0.8, 0) m
uAB D 0.243i C 0.970j C 0k,
uAC D 0.836i C 0.304j C 0.456k,
and uAD D 0.889i C 0.323j 0.323k.
D
0.4
m
0.6
m
C
The forces can be written as
z
B
x
(1, 0, 0) m
TAB
TAC
F
A
(1.1, −0.4, 0) m
TRS D TRS uRS D TRSX i C TRSY j C TRSZ k,
where RS takes on the values AB, AC, and AD. We now have three
forces written in terms of unknown magnitudes and known directions.
The equations of equilibrium for point A are
Fx D TAB uABX C TAC uACX C TAD uADX C FX D 0,
and
Fy D TAB uABY C TAC uACY C TAD uADY C FY D 0,
Fz D TAB uABZ C TAC uACZ C TAD uADZ C FZ D 0,
where F D FX i C FY j C FZ k D 40j kN. Solving these equations for
the three unknowns, we obtain TAB D 45.4 kN (compression),
TAC D 5.26 kN (tension), and TAD D 7.42 kN (tension).
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
439
Problem 6.64 If the force exerted at point A of
the truss in Problem 6.63 is F D 10i C 60j C 20k (kN),
what are the axial forces in members BC, BD and BE?
Solution: The important points in this problem are A (1.1, 0.4,
0), B (1, 0, 0), C (0, 0, 0.6), D (0, 0, 0.4), and E (0, 0.8, 0). The
unit vectors along AB, AC, AD, BC, BD, and BE are
y
E
(0, 0.8, 0) m
uAB D 0.243i C 0.970j C 0k,
uAC D 0.836i C 0.304j C 0.456k,
0.4
m
0.6
m
uAD D 0.889i C 0.323j 0.323k,
TBC
C
z
uBC D 0.857i C 0j C 0.514k,
DT TDE
F
B
x
(1, 0, 0) m
TAB
A
(1.1, −0.4, 0) m
uBD D 0.928i C 0j 0.371k,
and uBE D 0.781i C 0.625j C 0k.
The forces can be written as TRS D TRS uRS D TRSX i C TRSY j C
TRSZ k, where RS takes on the values AB, AC, and AD when dealing
with joint A and AB, BC, BD, and BD when dealing with joint B. We
now have three forces written in terms of unknown magnitudes and
known directions.
Joint A: The equations of equilibrium for point A are,
and
Fx D TAB uABX C TAC uACX C TAD uADX C FX D 0,
Fy D TAB uABY C TAC uACY C TAD uADY C FY D 0,
Fz D TAB uABZ C TAC uACZ C TAD uADZ C FZ D 0,
where F D FX i C FY j C FZ k D 10i C 60j C 20k kN. Solving these
equations for the three unknowns at A, we obtain TAB D 72.2 kN
(compression), TAC D 13.2 kN (compression), and TAD D 43.3 kN
(tension).
Joint B: The equations of equilibrium at B are
and
Fx D TAB uABX C TBC uBCX C TBD uBDX C TBE uBEX D 0,
Fy D TAB uABY C TBC uBCY C TBD uBDY C TBE uBEY D 0,
Fz D TAB uABZ C TBC uBCZ C TBD uBDZ C TBE uBEZ D 0.
Since we know the axial force in AB, we have three equations in the
three axial forces in BC, BD, and BE. Solving these, we get TBC D
32.7 kN (tension), TBD D 45.2 kN (tension), and TBE D 112.1 kN
(compression).
440
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 6.65 The space truss is supported by roller
supports on the horizontal surface at C and D and a ball
and socket support at E. The y axis points upward. The
mass of the suspended object is 120 kg. The coordinates
of the joints of the truss are A: (1.6, 0.4, 0) m, B: (1.0,
1.0, 0.2) m, C: (0.9, 0, 0.9) m, D: (0.9, 0, 0.6) m,
and E: (0, 0.8, 0) m. Determine the axial forces in
y
B
E
D
A
C
x
z
Solution: The important points in this problem are A: (1.6, 0.4,
0) m, B: (1, 1, 0.2) m, C: (0.9, 0, 0.9) m, and D: (0.9, 0, 0.6) m.
We do not need point E as all of the needed unknowns converge at A
and none involve the location of point E. The unit vectors along AB,
y
E
TAB
D
uAB D 0.688i C 0.688j 0.229k,
uAC D 0.579i 0.331j C 0.745k,
C
and uAD D 0.697i 0.398j 0.597k.
The forces can be written as TRS D TRS uRS D TRSX i C TRSY j C
TRSZ k, where RS takes on the values AB, AC, and AD. We now
have three forces written in terms of unknown magnitudes and known
directions. The equations of equilibrium for point A are
and
B
z
A
mg
x
TAC
L
Fx D TAB uABX C TAC uACX C TAD uADX C FX D 0,
Fy D TAB uABY C TAC uACY C TAD uADY C FY D 0,
Fz D TAB uABZ C TAC uACZ C TAD uADZ C FZ D 0,
where F D FX i C FY j C FZ k D mgj D 1177j N. Solving these
equations for the three unknowns, we obtain TAB D 1088 N (tension),
TAC D 316 N (compression), and TAD D 813 N (compression).
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
441
Problem 6.66 The free-body diagram of the part of the
construction crane to the left of the plane is shown. The
coordinates (in meters) of the joints A, B, and C are (1.5,
1.5, 0), (0, 0, 1), and (0, 0, 1), respectively. The axial
forces P1 , P2 , and P3 are parallel to the x axis. The axial
forces P4 , P5 , and P6 point in the directions of the unit
vectors
e4 D 0.640i 0.640j 0.426k,
e5 D 0.640i 0.640j 0.426k,
e6 D 0.832i 0.555k.
The total force exerted on the free-body diagram by the
weight of the crane and the load it supports is Fj D
44j (kN) acting at the point (20, 0, 0) m. What is
the axial force P3 ?
Strategy: Use the fact that the moment about the line
that passes through joints A and B equals zero.
y
A
F
z
B
P6
P2
C
P5
P1
P4
P3
x
Solution: The axial force P3 and F are the only forces that exert
moments about the line through A and B. The moment they exert about
pt B is

i
j
MB D  20
0
0
44
 
k
i
1  C  0
0
P3
j
0
0

k
2 
0
D 44i 2P3 j C 880k (kN-m).
The position vector from B to A is
rBA D 1.5i C 1.5j k (m),
and the unit vector that points from B toward A is
eBA D
rBA
D 0.640i C 0.640j 0.426k.
jrBA j
From the condition that
eBA &ETH; MB D 0.64044 C 0.6402P3 0.426880 D 0,
we obtain P3 D 315 kN.
442
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 6.67 In Problem 6.66, what are the axial
forces P1 , P4 , and P5 ?
Strategy: Write the equilibrium equations for the
entire free-body diagram.
Solution: The equilibrium equations are
Fx D P1 C P2 C P3 C 0.64P4 C 0.64P5 C 0.832P6 D 0,
Fy D 0.64P4 0.64P5 44 D 0,
Fz D 0.426P4 C 0.426P5 0.555P6 D 0,

i
MB D  20
0
j
0
44

i
C  1.5
P1
j
1.5
0

i
C  1.5
0.64P4

i
C  1.5
0.64P5
 
k
i
1  C  0
0
P3
j
0
0

k
2 
0

k
1 
0
j
1.5
0.64P4

k

1
0.426P4
j
1.5
0.64P5

k
1  D 0.
0.426P5
The components of the moment equation are
MBx D 44 1.279P4 0.001P5 D 0,
MBy D 2P3 P1 0.001P4 1.279P5 D 0,
MBz D 880 1.5P1 1.92P4 1.92P5 D 0.
Solving these equations, we obtain
P1 D 674.7 kN,
P2 D P3 D 315.3 kN,
P4 D P5 D 34.4 kN,
and P6 D 0.
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
443
Problem 6.68 The mirror housing of the telescope is
supported by a 6-bar space truss. The mass of the
housing is 3 Mg (megagrams), and its weight acts at G.
The distance from the axis of the telescope to points A,
B, and C is 1 m, and the distance from the axis to points
D, E, and F is 2.5 m. If the telescope axis is vertical
(˛ D 90&deg; ), what are the axial forces in the members of
the truss?
A
G
F
C
B
60&deg;
4m
60&deg;
G
B
60&deg;
Solution: A cut through the 6-bar space truss leads to six equations
60&deg;
C
E
x
60&deg;
y
60&deg;
60&deg;
D
A
F
G
60&deg;
E
F
A
60&deg;
D
α
1m
END VIEW y
D
in the unknowns (see Problem 6.59). However for this problem an
alternate strategy based on reasonable assumptions about the equality
of the tensions is used to get the reactions. Assume that each support
carries one-third of the weight, which is equally divided between the
two bars at the support.
y
Mirror housing
z
60&deg;
60&deg;
Mirror housing
y
C
B
E
x
z
60&deg;
A
F
G
B
The coordinate system has its origin in the upper platform, with the
x axis passing though the point C. The coordinates of the points are:
C
D
α
1m
A cos 60&deg; , sin 60&deg; , 0 D 0.5, 0.866, 0,
E
4m
B cos 60&deg; , sin 60&deg; , 0 D 0.5, 0.866, 0,
r=1m
A
C1, 0, 0,
y
C
B
x
D2.5, 0, 4,
4m
E2.5 cos 60&deg; , 2.5 sin 60&deg; , 4 D 1.25, 2.165, 4,
R = 2.5 m
F2.5 cos 60&deg; , 2.5 sin 60&deg; , 4 D 1.25, 2.165, 4.
Consider joint B in the upper housing. The position vectors of the
points E and D relative to B are
D
E
rBD D 2i C 0.866j 4k,
rBE D 1.75i 1.299j 4k.
The unit vectors are
eBD D 0.4391i C 0.1901j 0.8781k,
and eBE D 0.3842i 0.2852j 0.8781k.
The weight is balanced by the z components:
Fz D W
0.8781TBD 0.8781TBE D 0.
3
Assume that the magnitude of the axial force is the same in both
members BD and BE, TBE D TBD . The weight is W D 39.81 D
29.43 kN. Thus the result: TBE D TBD D 5.5858 kN C . From
symmetry (and the assumptions made above) the axial force is the
same in all members.
444
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 6.69 Consider the telescope described in
Problem 6.68. Determine the axial forces in the members
of the truss if the angle ˛ between the horizontal and the
telescope axis is 20&deg; .
Solution: The coordinates of the points are,
y
F
A cos 60&deg; , sin 60&deg; , 0 D 0.5, 0.866, 0 m,
A
B cos 60&deg; , sin 60&deg; , 0 D 0.5, 0.866, 0 m,
x
C
D
B
C1, 0, 0 m,
E
D2.5, 0, 4 m,
E2.5 cos 60&deg; , 2.5 sin 60&deg; , 4 D 1.25, 2.165, 4 m,
F2.5 cos 60&deg; , 2.5 sin 60&deg; , 4 D 1.25, 2.165, 4 m.
20000
The coordinates of the center of gravity are G (0, 0, 1) (m). Make a
cut through the members just below the upper platform supports, such
that the cut members have the same radial distance from the axis as
the supports. Consider the upper section.
A
x
i
a
l
The section as a free body: The strategy is to sum the forces and
moments to obtain six equations in the six unknown axial forces. The
axial forces and moments are expressed in terms of unit vectors. The
position vectors of the points E, D, and F relative to the points A, B,
and C are required to obtain the unit vectors parallel to the members.
The unit vectors are obtained from these vectors. The vectors and their
associated unit vectors are given in Table I. Note: While numerical
values are shown below to four significant figures, the calculations
were done with the full precision permitted (15 digits for TK Solver
Plus.)
Vector
rAF
rBD
rBE
rCE
rCF
x
y
z
2
1.75
2
1.75
0.25
0.25
0.866
1.299
0.866
1.299
2.165
2.165
4
4
4
4
4
4
Table I
Unit
Vector
eAF
eBD
eBE
eCE
eCF
Axial Forces in Bars
25000
|AF| &amp; |CF|
15000
10000
5000
0
F −5000
, −10000
N −15000
−20000
|CE| &amp; |BD|
−25000
−100
−50
0
50
100
alpha, deg
x
y
z
0.4391
0.3842
0.4391
0.3842
0.0549
0.0549
0.1901
0.2852
0.1901
0.2852
0.4753
0.4753
0.8781
0.8781
0.8781
0.8781
0.8781
0.8781
The equilibrium condition for the forces is
jTAB jeAD C jTAF jeAF C jTBD jeBD C jTBE jeBE C jTCE jeCE
C jTCF jeCF C W D 0.
This is three equations in six unknowns. The unit vectors are given in
Table I. The weight vector is W D jWjj cos ˛ k sin ˛, where ˛ is
the angle from the horizontal of the telescope housing. The remaining
three equations in six unknowns are obtained from the moments:
rA &eth; TAD C TAF C rB &eth; TBD C TBE C rC &eth; TCE
C TCF C rG &eth; W D 0.
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
445
6.69 (Continued )
Carry out the indicated operations on the moments to obtain the vectors
defining the moments:
i
0.4391
k
0
0.8781 j
0.866
0.1901
The six equations in six unknowns are:
jTAD jeADx C jTAF jeAFx C jTBD jeBDx C jTBE jeBEx C jTCE jeCEx
C jTCF jeCFx C Wx D 0
jTAD jeADy C jTAF jeAFy C jTBD jeBDy C jTBE jeBEy C jTCE jeCEy
D jTAD j0.7605i 0.4391j C 0.4753
C jTCF jeCFy C Wy D 0
i
D jTAF j 0.5
0.3842
j
0.866
0.2852
k
0
0.8781 jTAD jeADz C jTAF jeAFz C jTBD jeBDz C jTBE jeBEz C jTCE jeCEz
C jTCF jeCFz C Wz D 0
jTAD juADx C jTAF juAFx C jTBD juBDx C jTBE juBEx C jTCE juCEx
D jTAF j0.7605i 0.4391j 0.4753k
D jTAF jiuAFx C juAFy C kuAFz rB &eth; TBD
i
D jTBD j 0.5
0.4391
j
0.866
0.1901
k
0
0.8781 D jTBD j0.7605i 0.4391j 0.4753k
D jTBD jiuBDx C juBDy C kuBDz rB &eth; TBE
rC &eth; TCE
i
D jTBE j 0.5
0.3842
j
0.866
0.2852
k
0
0.8781 jTAD juADy C jTAF juAFy C jTBD juBDy C jTBE juBEy C jTCE juCEy
C jTCF juCFy D 0,
jTAD juADz C jTAF juAFz C jTBD juBDz C jTBE juBEz C jTCE juCEz
C jTCF juCFz D 0
This set of equations was solved by iteration using TK Solver 2. For
˛ D 20&deg; the results are:
jTAD j D jTBD j D 1910.5 N C ,
D jTBE j0.7605i 0.4391j 0.4753k
jTAF j D jTCF j D 16272.5 N T ,
D jTBE jiuBEx C juBEy C kuBEz jTBE j D jTCE j D 19707 N C .
i
D jTCE j 1
0.0549
j
0
0.4753
k
0
0.8781 D jTCE j0i C 0.8781j 0.4753k
D jTCE jiuCEx C juCEy C kuCEz rC &eth; TCF
C jTCF juCFx C MWx D 0
i
D jTCF j 1
0.0549
j
0
0.4753
k
0
0.8781 D jTCF j0i C 0.8781j C 0.4753k
D jTCF jiuCFx C juCFy C kuCFz i
rG &eth; W D jWj 0
0
j
0
cos ˛
k 1 sin ˛ Check: For ˛ D 90&deg; , the solution is jTAD j D jTAF j D jTBD j D jTBE j D
jTCE j D jTCF j D 5585.8 N C, which agrees with the solution to
Problem 6.68, obtained by another method. check.
Check: The solution of a six-by-six system by iteration has risks, since
the matrix of coefficients may be ill-conditioned. As a reasonableness
test for the solution process, TK Solver Plus was used to graph the
axial forces in the supporting bars over the range 90&deg; &lt; ˛ &lt; 90&deg; .
The graph is shown. The negative values are compression, and the
positive values are tension. When ˛ D 90&deg; , the telescope platform is
pointing straight down, and the bars are in equal tension, as expected.
When ˛ D 90&deg; the telescope mount is upright and the supporting bars
are in equal compression, as expected. The values of compression
and tension at the two extremes are equal and opposite in value,
and the values agree with those obtained by another method (see
Problem 6.58), as expected. Since the axial forces go from tension
to compression over this range of angles, all axial forces must pass
through zero in the interval. check.
D jWji cos ˛ j0 C k0 D iMWx 446
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 6.70 In Active Example 6.6, suppose that in
frame is subjected to a 400-N force at C that is horizontal and points toward the left. Draw a sketch of the
and couples acting on members AB of the frame.
A
B
200 N-m
400 mm
C
Solution: The sketch of the frame with the new loading is shown.
We break the frame into separate bars and draw the free-body diagram
of each bar.
600 mm
400 mm
Starting with bar BC, we have the equilibrium equations
MB : C400 mm
400 N400 mm
200 N-m D 0
Fy : C By D 0
Fx : Bx 400 N D 0
Now using bar AB we have the equilibrium equations
Fx : Ax C Bx D 0
Fy : Ay C By D 0
MA : MA C By 600 mm D 0
Solving these six equations yields C D 900 N and
Ax D 400 N, Ay D 900 N
Bx D 400 N, By D 900 N
MA D 540 N-m
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
447
Problem 6.71 The object suspended at E weighs
200 lb. Determine the reactions on member ACD at A
and C.
D
3 ft
E
C
B
5 ft
A
4 ft
6 ft
We have the equilibrium equations:
Fx : Ax D 0
Fy : Ay 200 lb D 0
MA : MA 200 lb 6 ft D 0
Next we use the free-body diagram of the post ACD. Notice that BD
is a two-force body and the angle ˛ is
˛ D tan1 3/4 D 36.9&deg;
The equilibrium equations are
MC : MA C Ax 5 ft C TBD cos ˛ 3 ft D 0
Fx : Ax C Cx TBD cos ˛ D 0
Fy : Ay C Cy TBD sin ˛ D 0
Solving these six equations we find TBD D 500 lb and
Ax D 0,
Ay D 200 lb
Cx D 400 lb, Cy D 500 lb
MA D 1200 ft-lb
448
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 6.72 The mass of the object suspended at G
is 100 kg. Determine the reactions on member CDE at
C and E.
E
B
F
G
800 mm
D
200 mm
A
C
400 mm
400 mm
800 mm
400 mm
Solution: The free-body diagram of the entire frame and of member CDE are shown. The angle ˛ is
˛ D tan1 4/8 D 26.6&deg;
The equilibrium equations are
MC : TAB cos ˛ 400 mm
C TAB sin ˛ 800 mm
981 N1200 mm D 0
Fx : Cx TAB sin ˛ D 0
Fy : Cy TAB cos ˛ 981 N D 0
The free-body diagram for bar CDE is shown. Note that DF is a
two-force member. The angle ˇ is
ˇ D tan1 3/4 D 36.9&deg;
The equilibrium equations are
ME : TDF cos ˇ600 mm C Cx 800 mm D 0
Fx : TDF cos ˇ C Ex C Cx D 0
Fy : TDF sin ˇ C Ey C Cy D 0
Solving these six equations, we find
TAB D 1650 N,
Cx D 736 N,
Ex D 245 N,
TDF D 1230 N and
Cy D 2450 N
Ey D 1720 N
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
449
Problem 6.73 The force F D 10 kN. Determine the
shown in Fig. 6.25.
F
E
D
A
B
C
1m
1m
Solution: The complete structure as a free body: The sum of the
MG D C3F 5A D 0,
2m
1m
F
G
GY
GX
A
2m
3F
D 6 kN which is the reaction of the floor. The
5
sum of the forces:
3m
from which A D
F
Fy D Gy F C A D 0,
D
1m
A
Fx D Gx D 0.
1m
Element DEG: The sum of the moments about D
M D F C 3E C 4Gy D 0,
from which E D
10 16
F 4Gy
D
D 2 kN.
3
3
2m
E
1m
F
from which Gy D F A D 10 6 D 4 kN.
GY
A
1m
C = −E
B = −D
8 kN
2 kN
B
C
3m
6 kN
The sum of the forces:
Fy D Gy F C E C D D 0,
from which D D F E Gy D 10 C 2 4 D 8 kN.
Element ABC : Noting that the reactions are equal and opposite:
B D D D 8 kN ,
and
C D E D 2 kN .
The sum of the forces:
Fy D A C B C C D 0,
from which A D 8 2 D 6 kN. Check
450
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 6.74 In Example 6.7, suppose that the frame
is redesigned so that the distance from point C to the
attachment point E of the two-force member BE is
increased from 8 in to 10 in. Determine the forces acting
at C on member ABCD.
D
6 in
3 in
G
E
C
6 in
B
W
6 in
A
8 in
8 in
Solution: The analysis of the free-body diagram of the entire structure as presented in Example 6.7 is unchanged.
From the example we know that
Ax D 42.2 lb,
Ay D 40 lb,
D D 42.4 lb
The free-body diagram for ABCD is shown. Note that BE is a two-force
body. The angle ˛ is now
˛ D tan1 6/10 D 31.0&deg;
The equilibrium equations are
MC : TBE cos ˛ 6 in C D 6 in C Ax 6 in D 0
Fx : TBE cos ˛ C Cx C Ax D D 0
Fy : TBE sin ˛ C Cy C Ay D 0
Solving yields TBE D 124 lb and
Cx D 66.7 lb,
Cy D 24 lb
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
451
Problem 6.75 The tension in cable BD is 500 lb.
Determine the reactions at A for cases (1) and (2).
E
G
6 in
D
6 in
A
B
C
300 lb
8 in
8 in
(1)
G
E
6 in
D
6 in
A
B
C
300 lb
8 in
8 in
Solution: Case (a) The complete structure as a free body: The sum
(2)
Gy
MG D 16300 C 12Ax D 0,
(a) 12 in
from which Ax D 400 lb . The sum of the forces:
Ey
Gy
Ay
Ax
Gx
16 in
Ex
300 lb
Fx D Ax C Gx D 0,
Ay
from which Gx D 400 lb.
Gx
(b)
B
α
Ax
Fy D Ay 300 C Gy D 0,
8 in
8 in
Cy
Cx
300 lb
from which Ay D 300 Gy . Element GE : The sum of the moments
ME D 16Gy D 0,
from which Gy D 0, and from above Ay D 300 lb.
Case (b) The complete structure as a free body: The free body diagram,
except for the position of the internal pin, is the same as for case (a).
The sum of the moments about G is
MG D 16300 C 12Ax D 0,
from which Ax D 400 lb .
Element ABC : The tension at the lower end of the cable is up and to
the right, so that the moment exerted by the cable tension about point
C is negative. The sum of the moments about C:
MC D 8B sin ˛ 16Ay D 0,
noting that B D 500 lb and ˛ D tan1
then
452
6
D 36.87&deg; ,
8
Ay D 150 lb.
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 6.76 Determine the reactions on member
ABCD at A, C, and D.
B
A
0.4 m
E
C
600 N
0.4 m
D
0.6 m
0.4 m
0.4 m
Solution: Consider the entire structure first
MA : Dy 0.6 m 600 N1.0 m D 0 ) Dy D 1000 N
Fx : Ax D 0
Fy : Ay C Dy 600 N D 0 ) Ay D 400 N
Ax
Ay
E
C
600 N
Dy
Now examine bar CE. Note that the reactions on ABD are opposite to
those on CE.
ME : 600 N0.4 m C Cy 0.8 m D 0 ) Cy D 300 N
MB : Cx 0.4 m 600 N0.4 m D 0 ) Cx D 600 N
T
Cy
Cx
E
600 N
In Summary we have
Ax D 0, Ay D 400 N
Cx D 600 N, Cy D 300 N
Dx D 0, Dy D 1000 N
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
453
Problem 6.77 Determine the forces exerted on member
ABC at A and C.
D
400 lb
2 ft
1 ft
A
B
C
100 lb
1 ft
E
2 ft
2 ft
2 ft
Two of the equilibrium equations for the whole frame are
Fx : Ax C 100 lb D 0
ME : Ax 2 ft Ay 4 ft
100 lb 1 ft
400 lb 2 ft D 0
Next we examine the free-body diagram of bar ABC. Note that BD
is a two-force body and that the angle ˛ D 45&deg; . The equilibrium
equations are
MC : Ay 4 ft TBD sin ˛ 2 ft
400 lb 2 ft D 0
Fx : Ax C TBD cos ˛ C Cx D 0
Fy : Ay C TBD sin ˛ C Cy 400 lb D 0
Solving, we find that TBD D 70.7 lb and
Ax D 100 lb, Ay D 175 lb
Cx D 150 lb, Cy D 625 lb
454
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 6.78 An athlete works out with a squat thrust
machine. To rotate the bar ABD, she must exert a vertical
force at A that causes the magnitude of the axial force in
the two-force member BC to be 1800 N. When the bar
ABD is on the verge of rotating, what are the reactions
on the vertical bar CDE at D and E?
0.6 m
0.6 m
C
A
0.42 m
B
D
1.65 m
E
Solution: Member BC is a two force member. The force in BC is
along the line from B to C.
C
y
FBC
Ay
0.6 m
0.6 m
tan Θ =
Dy 0.42 m
θ
D
0.42
0.6
Dx
x (FBC = 1800 N)
Θ = 34.990
FBC D 1800 N
tan D
0.42
D 34.99&deg; .
0.6
C
Fx :
Dx FBC cos D 0
Fy :
Ay FBC sin C Dy D 0
MD :
Solving, we get
1.2Ay C 0.6FBC sin D 0
Dx D 1475 N
Dy D 516 N
Ay D 516 N
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
455
Problem 6.79 The frame supports a 6-kN load at C.
Determine the reactions on the frame at A and D.
6 kN
0.4 m
A
B
1.0 m
C
0.5 m
D
E
F
Solution: Note that members BE and CF are two force members.
Consider the 6 kN load as being applied to member ABC.
Ay
Ax
0.4 m
0.4 m
6 kN
1.0 m
B
0.8 m
C
FCF
FBE
φ
θ
tan D
0.5
0.4
D 51.34&deg;
tan D
0.5
0.2
D 68.20&deg;
Member DEF
FBE
θ
Dx
0.8 m
FCF
E
F
φ
0.4 m
Dy
Equations of equilibrium:
Member ABC:
Fx :
Ax C FBE cos FCF cos D 0
Fy :
Ay FBE sin FCF sin 6 D 0
MA :
0.4FBE sin 1.4FCF sin 1.46 D 0
C
Member DEF:
C
Fx :
Dx FBE cos C FCF cos D 0
Fy :
Dy C FBE sin C FCF sin D 0
MD :
0.8FBE sin C 1.2FCF sin D 0
Unknowns Ax , Ay , Dx , Dy , FBE , FCF we have 6 eqns in 6
unknowns.
Solving, we get
Also,
Ax
Ay
Dx
Dy
D 16.8 kN
D 11.25 kN
D 16.3 kN
D 5.25 kN
FBE D 20.2 kN T
FCF D 11.3 kN C
456
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 6.80 The mass m D 120 kg. Determine the
shown in Fig. 6.25.
A
B
C
300 m m
D
E
m
200 mm
Solution: The equations of equilibrium for the entire frame are
AX
AY
CX
CX
DY
FY D AY 2mg D 0,
CY
DY
EX
MA D 0.3EX 0.2mg 0.4mg D 0.
Solving yields AX D 2354 N, AY D 2354 N, and EX D 2354 N.
Member ABC: The equilibrium equations are
and
CY
BY
and summing moments at A,
200 mm
BY
FX D AX C EX D 0,
m
2354 N
B
2354 N A
4708 N
4708 N 2354 N
C
2354 N
B
FX D AX C CX D 0,
2354 N
2354 N
4708 N
FY D AY BY C CY D 0,
MA D 0.2BY C 0.4CY D 0.
We have three equations in the three unknowns BY , CX , and CY .
Solving, we get BY D 4708 N, CX D 2354 N, and CY D 2354 N. This
gives all of the forces on member ABC. A similar analysis can be made
for each of the other members in the frame. The results of solving for
all of the forces in the frame is shown in the figure.
4708 N
E
2354 N
C
D
1177 N
1177 N
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
457
30 lb
Problem 6.81 Determine the reactions on member
BCD.
F
D
G
8 in.
E
40 lb
8 in.
C
8 in.
B
A
18 in.
12 in.
8 in.
Solution: We will use frame ADG, bar DFG and bar BCD. The
free-body diagrams ares shown.
The angle ˛ D tan1 18/24 D 36.9&deg;
MD : Bx 24 in C 40 lb8 in
30 lb 20 in D 0
MA : By 18 in 40 lb16 in
30 lb38 in D 0
Fy : By 30 lb TAD cos ˛ D 0
From DFG we have
MF : Dy C TAD cos ˛ 12 in
30 lb8 in D 0
And finally from BCD we have
Fy : By C Cy C Dy D 0
MD : Bx 24 in C Cx 16 in D 0
Fx : Dx C Cx C Bx D 0
Solving these seven equations, we find TAD D 86.1 lb and
Bx D 11.7 lb, By D 98.9 lb
Cx D 17.5 lb, Cy D 50 lb
Dx D 5.83 lb,
458
Dy D 48.9 lb
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 6.82 The weight of the suspended object is
W D 50 lb. Determine the tension in the spring and the
reactions at F. (The slotted member DE is vertical.)
A
B
4 in
E
6 in
W
C
10 in
F
D
8 in
8 in
10 in
Finally examine DCE
1
MA : 50 lb8 in C p FB 16 in D 0 ) FB D 35.4 lb
2
10 in
MD : T16 in C FC 10 in D 0 ) T D 62.5 lb
FB
T
1
1
Ax
FC
Ay
50 lb
Now examine BCF
p
MF : FB 20 2 in FC 10 in D 0 ) FC D 100 lb
Dx
1
Fx : p FB C FC C Fx D 0 ) Fx D 75 lb
2
1
Fy : p FB C Fy D 0 ) Fy D 25 lb
2
Dy
Summary
Tension in Spring D 62.5 lb
Fx D 25 lb, Fy D 75 lb
1
1
FB
FC
Fx
Fy
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
459
Problem 6.83 The mass m D 50 kg. Bar DE is
horizontal. Determine the forces on member ABCD,
1m
1m
D
E
1m
C
m
1m
B
1m
A
F
Solution: The weight of the mass hanging is W D mg D
509.81 D 490.5 N The complete structure as a free body: The sum
MA D 2W C Fy D 0,
from which Cy D 981 N.
Fy D Ey C Cy D 0,
Fx D Ex C Cx D 0,
from which Fy D 981 N. The sum of the forces:
from which Cx D 981 N, and
Fy D Ay C Fy W D 0,
from which Ay D 490.5 N,
Element ABCD: All reactions on ABCD have been determined
above. The components at B and C have the magnitudes
p
B D C D 9812 C 9812 D 1387 N , at angles of 45&deg; .
Fx D Ax C Fx D 0,
from which Ax D Fx . Element BF: The sum of the moments
MF D Bx By D 0,
from which By D Bx . The sum of the forces:
Dy
Dy
Cy
Cx
Cy
Bx
By
Fy D By C Fy D 0,
Ey
Ey
W
Ex
By
Cx
Bx
Fy
Ay
Ax
from which By D 981 N, and Bx D 981 N.
Ex
Dx
Dx
Fx
Fx D Bx C Fx D 0,
490.5 N
from which Fx D 981 N, and from above, Ax D 981 N ,
Element DE: The sum of the moments about D:
MD D Ey 2W D 0,
D
C
981 N
1387 N
45&deg;
B
from which Ey D 981 N. The sum of the forces:
Fy D Dy Ey W D 0,
A
45&deg;
1387 N
981 N
490.5 N
from which Dy D 490.5 N .
Fx D Dx Ex D 0,
from which Dx D Ex . Element CE : The sum of the moments
MC D Ey Ex D 0,
from which Ex D 981 N, and from above Dx D 981 N .
460
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 6.84 Determine the forces on member BCD.
400 lb
6 ft
B
A
4 ft
C
4 ft
D
E
8 ft
Solution: The following is based on free body diagrams of the
elements: The complete structure as a free body: The sum of the
MD D 6400 C 8Ey D 0,
from which Ey D 300 lb. The sum of the forces:
Fx D Dx D 0.
MA D 8By 6400 D 0,
from which By D 300 lb. The sum of forces:
The reactions are now known:
By D 300 lb , Bx D 400 lb , Cy D 200 lb ,
Dx D 0 , Dy D 100 lb ,
where negative sign means that the force is reversed from the direction
shown on the free body diagram.
Fy D Ey C Dy 400 D 0,
from which Dy D 100 lb. Element AB: The sum of the moments
Element BCD:
Fy D By Ay 400 D 0,
400 lb
Ay
Ax
Ax
By
Bx
Cy
Cx E
Ay
Bx
Cy
Dy
Cx
By
Dx
from which Ay D 100 lb.
Fx D Ax Bx D 0,
from which (1) Ax C Bx D 0 Element ACE: The sum of the moments
ME D 8Ax C 4Cx 8Ay C 4Cy D 0,
from which (2) 2Ax C Cx 2Ay C Cy D 0. The sum of the forces:
Fy D Ay C Ey Cy D 0,
from which Cy D 200 lb .
Fx D Ax Cx D 0,
from which (3) Ax D Cx . The three numbered equations are solved:
Ax D 400 lb, Cx D 400 lb , and Bx D 400 lb .
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
461
Problem 6.85 Determine the forces on member ABC.
E
6 kN
1m
D
C
1m
A
B
2m
Solution: The frame as a whole: The equations of equilibrium are
EX
FX D AX C EX D 0,
FY D AY C EY 6000 N D 0,
EY
E DY
DX D
DX D
DY
BX B
BY BX
B
AX A
BY
AY
2m
CY
1m
6 kN
C
CY
C
ME D 2AX 56000 D 0.
Solving for the support reactions, we get AX D 15,000 N and EX D
15,000 N. We cannot yet solve for the forces in the y direction at A
and E.
Member ABC: The equations of equilibrium are
FX D AX BX D 0,
FY D AY BY CY D 0,
MA D 2BY 4CY D 0.
Member BDE: The equations of equilibrium are
FX D EX C DX C BX D 0,
FY D EY C DY C BY D 0,
ME D 1DY C 1DX C 2BY C 2BX D 0.
Member CD: The equations of equilibrium are
FX D DX D 0,
FY D DY C CY 6000 D 0,
MD D 46000 C 3CY D 0.
Solving these equations simultaneously gives values for all of the
forces in the frame. The values are AX D 15,000 N, AY D 8,000 N,
BX D 15,000 N, BY D 16,000 N, CY D 8,000 N, DX D 0, and
DY D 2,000 N.
462
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 6.86 Determine the forces on member ABD.
8 in
8 in
8 in
A
60 lb
8 in
60 lb
B
E
8 in
C
Solution: The equations of equilibrium for the frame as a
AY
whole are
AX
FX D AX C CX D 0,
and
BX
DX
FY D AY 60 60 D 0,
D
BY
BX
B
60 lb
EX
EY
DY
DY
CX
MA D 16CX 1660 2460 D 0.
60 lb
BY
EY
EX
DX
Solving these three equations yields
AX D 150 lb,
AY D 120 lb,
and CX D 150 lb.
Member ABD: The equilibrium equations for this member are:
FX D AX BX DX D 0,
and
FY D AY BY DY D 0,
MA D 8BY 8DY 8BX 16DX D 0.
Member BE: The equilibrium equations for this member are:
FX D BX C EX D 0,
and
FY D BY C EY 60 60 D 0,
MB D 860 1660 C 16EY D 0.
Member CDE: The equilibrium equations for this member are:
and
FX D CX C DX EX D 0,
FY D DY EY D 0,
MD D 8EX 16EY D 0.
Solving these equations, we get BX D 180 lb, BY D 30 lb, DX D
30 lb, DY D 90 lb, EX D 180 lb, and EY D 90 lb. Note that we have
12 equations in 9 unknowns. The extra equations provide a check.
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
463
Problem 6.87 The mass m D 12 kg. Determine the
forces on member CDE.
A
200 mm
100 mm
E
B
200 mm
C
D
200 mm
m
400 mm
Ax
Eq. for entire frame:
C
!
C &quot;
C
Ay
Fx D 0:
Ax C Cx D 0 ) Ax D Cx
1
B
Fy D 0:
Mc D 0:
Ay 117.7 D 0 ) Ay D 117.7 N
Ax 0.4 117.70.7 D 0
Ax D 206 N
Cx
W = (9.81) (12)
W = 117.7
D
∴ Ax D Cx D 206 N.
Now look at free-body diagram ABD.
Ax = 206
Eq. for ABD:
C
MB D 0:
Ay = 117.7
T = 117.7
By
Dx 0.2 117.70.2 C 2060.2 117.70.1 D 0
Bx
Dx D 29.45 N
C
!
Fx D 0:
Dx
206 C 117.7 C Bx 29.45 D 0
C &quot;
Bx D 117.75 N
Dy
Fy D 0:
117.7 By C Dy D 0
Draw free-body diagram of CDE
Ey
Ex
Eq. for CDE:
C
!
Fx D 0:
206 C 29.45 C Ex D 0
C
Dx = 29.45
Ex D 235.45 or Ex D 235.45
Cx = 206
MD D 0:
Ex 0.2 C Ey 0.4 D 0
Ey D
Dy
235.450.2
Ex 0.2
D
0.4
0.4
Ey D 117.7 or Ey D 117.7 N
C &quot;
Fy D 0:
Ey Dy D 0 or Dy D Ey D 117.7
464
Dy D 117.7 N#
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 6.88 The weight W D 80 lb. Determine the
forces on member ABCD.
11 in
5 in
12 in
3 in
B
A
D
C
8 in
W
E
Solution: The complete structure as a free body: The sum of the
Ay
Fx D Ex C Ax D 0,
Cx
By
Cx
Ey
F
Dy
Dx
Cy
Bx
Ax
MA D 31W C 8Ex D 0,
from which Ex D 310 lb. The sum of the forces:
F
Cy
W
Ex
from which Ax D 310 lb .
Fy D Ey C Ay W D 0,
from which (1) Ey C Ay D W.
Element CFE: The sum of the forces parallel to x:
Fx D Ex Cx D 0,
from which Cx D 310 lb . The sum of the moments about E:
ME D 8F 16Cy C 8Cx D 0.
For frictionless pulleys, F D W, and thus Cy D 195 lb . The sum of
forces parallel to y:
Fy D Ey Cy C F D 0,
from which Ey D 115 lb .
Equation (1) above is now solvable: Ay D 35 lb .
Element ABCD: The forces exerted by the pulleys on element ABCD
are, by inspection: Bx D W D 80 lb , By D 80 lb , Dx D 80 lb ,
and Dy D 80 lb , where the negative sign means that the force is
reversed from the direction of the arrows shown on the free body
diagram.
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
465
Problem 6.89 The woman using the exercise machine
is holding the 80-lb weight stationary in the position
shown. What are the reactions at the built-in support E
and the pin support F? (A and C are pinned connections.)
2 ft 2 in
B
A
9 in
1 ft 6 in
2 ft
D
C
60⬚
6 ft
80 lb
E
Solution: The complete structure as a free body: The sum of the
26
in
F
42
in
60&deg;
M D 26W 68W sin 60&deg; C 50Fy 81W cos 60&deg; C ME D 0
W
from which (1) 50Fy C ME D 10031. The sum of the forces:
81 in
Fx D Fx C W cos 60&deg; C Ex D 0,
ME
from which (2) Fx C Ex D 40.
W
Ey
Ex
Fx
50 in
Fy D W W sin 60&deg; C Ey C Fy D 0,
from which (3) Ey C Fy D 149.28
Ay
Cy
Element CF: The sum of the moments about F:
Fy
Ax
Cx
M D 72Cx D 0,
from which Cx D 0. The sum of the forces:
ME
Fx D Cx C Fx D 0,
Fx
from which Fx D 0 . From (2) above, Ex D 40 lb
Element AE: The sum of the moments about E:
Fy
Ex
Ey
M D ME 72Ax D 0, .
from which (4) ME D 72Ax . The sum of the forces:
Fy D Ey C Ay D 0,
from which (5) Ey C Ay D 0.
Fx D Ax C Ex D 0;
from which Ax D 40 lb, and from (4) ME D 2880 in lb D 240 ft lb .
From (1) Fy D 143.0 lb , and from (2) Ey D 6.258 lb . This
completes the determination of the 5 reactions on E and F.
466
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 6.90 Determine the reactions on member
ABC at A and B.
80 lb
E
9 in
B
C
8 in
D
A
13 in
Solution: We first examine the entire structure.
Next examine body ABC
MD : Ay 13 in C 80 lb21 in D 0
Solving:
Ay D 129.2 lb
80 lb
4 in
MB : Ax 8 in Ay 13 in C 80 lb 4 in D 0
Fx : Ax C Bx D 0
Fy : Ay C By C 80 lb D 0
80 lb
Bx
By
Dx
Ax
Ax
Ay
Dy
Ay
Solving and summarizing we have
Ax D 170 lb, Ay D 129.2 lb
Bx D 170 lb, By D 209 lb
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
467
Problem 6.91 The mass of the suspended object is
m D 50 kg. Determine the reactions on member ABC.
0.2 m
A
B
0.6 m
E
D
0.8 m
C 0.2 m
0.6 m
m
Solution: Begin with an examination of the pulley at B.
Finally look at member ABC
1
Fx : Bx C p 490.5 N D 0 ) Bx D 347 N
2
1
Fy : By 490.5 N p 490.5 N D 0 ) By D 837 N
2
1
By
MC : Ax 0.6 m Ay 1.4 m Bx 0.6 m By 0.6 m D 0
) Ay D 771 N
1
Fx : Ax C Bx C Cx D 0 ) Cx D 961 N
Fy : Ay C By C Cy D 0 ) Cy D 66.6 N
By
Ay
Bx
490.5 N
Ax
Bx
490.5 N
Cy
Now examine the entire structure
MD : 490.5 N1.6 m Ax 0.6 m D 0 ) Ax D 1308 N
Cx
In Summary
Ay
Ax D 1308 N, Ay D 771 N
Ax
Bx D 347 N, By D 837 N
Cx D 961 N, Cy D 66.6 N
Dy
Dx
490.5 N
468
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 6.92 The unstretched length of the string is
LO . Show that when the system is in equilibrium the
angle ˛ satisfies the relation sin ˛ D 2LO 2F/kL.
F
1– L
4
1– L
4
k
1– L
2
α
α
Solution: Since the action lines of the force F and the reaction E
are co-parallel and coincident, the moment on the system is zero, and
the system is always in equilibrium, for a non-zero force F. The object
is to find an expression for the angle ˛ for any non-zero force F.
The complete structure as a free body:
The solution for angle ˛: The spring force is
Cy D T D k
L
sin ˛ LO ,
2
L
sin ˛ LO D 2F.
2
2F
2 LO k
Solve: sin ˛ D
L
from which k
The sum of the moments about A
MA D FL sin ˛ C EL sin ˛ D 0,
from which E D F. The sum of forces:
F
Fx D Ax D 0,
L
from which Ax D 0.
α
Fy D Ay C E F D 0,
from which Ay D 0, which completes a demonstration that F does not
exert a moment on the system. The spring C: The elongation of the
L
spring is s D 2 sin ˛ LO , from which the force in the spring is
4
TDk
L
sin ˛ LO
2
Ax
Ay
E
By
Cy
Bx
L
4
α
E
L
4
Element BE: The strategy is to determine Cy , which is the spring force
on BE. The moment about E is
L
L
L
ME D Cy cos ˛ By cos ˛ Bx cos ˛ D 0,
4
2
2
from which
Cy
C By D Bx . The sum of forces:
2
Fx D Bx D 0,
from which Bx D 0.
Fy D Cy C By C E D 0,
from which Cy C By D E D F. The two simultaneous equations
are solved: Cy D 2F, and By D F.
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
469
Problem 6.93 The pin support B will safely support a
force of 24-kN magnitude. Based on this criterion, what
is the largest mass m that the frame will safely support?
C
500 mm
100 mm
E
D
B
300 mm
m
A
300 mm
Solution: The weight is given by W D mg D 9.81 g
Sum the forces in the x-direction:
Fx D Ax D 0,
Cx
Cx
W
By
W
Bx
By Bx
W
Ay
from which Ax D 0
Element ABC: The sum of the moments about A:
400 mm
400 mm
Cy
Cy
The complete structure as a free body:
F
Ey
Ex
Ey
Ex
F
Ax
MA D C0.3Bx C 0.9Cx 0.4W D 0,
from which (1) 0.3Bx C 0.9Cx D 0.4W. The sum of the forces:
Fx D Bx Cx C W C Ax D 0,
from which (2) Bx C Cx D W. Solve the simultaneous equations (1)
5
and (2) to obtain Bx D W
6
Element BE : The sum of the moments about E:
ME D 0.4W 0.7By D 0,
from which By D
jBj D W
4
W. The magnitude of the reaction at B is
7
2 2
4
5
C
D 1.0104W.
6
7
24
D 23.752 kN is the
1.0104
maximum load that can be carried. Thus, the largest mass that can be
supported is m D W/g D 23752 N/9.81 m/s2 D 2421 kg.
For a safe value of jBj D 24 kN, W D
470
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 6.94 Determine the reactions at A and C.
C
A
3 ft
72 ft-lb
36 lb
3 ft
B
18 lb
4 ft
Solution: The complete structure as a free body:
The sum of the moments about A:
8 ft
Cy
Ay
Ax
Cx 3 ft
72 ft-lb
36 lb
MA D 418 C 336 C 12Cy 72 D 0,
18 lb
from which Cy D 3 lb. The sum of the forces:
8 ft
4 ft
Ay
Fy D Ay C Cy 18 D 0,
Ax
from which Ay D 15 lb.
Fx D Ax C Cx C 36 D 0,
72 ft-lb
By 6 ft
Bx
18 lb
from which (1) Cx D Ax 36
Element AB: The sum of the forces:
Fy D Ay By 18 D 0,
from which By D 3 lb. The sum of the moments:
MA D 6Bx 418 4By 72 D 0,
from which Bx D 22 lb. The sum of the forces:
Fx D Ax C Bx D 0,
from which Ax D 22 lb From equation (1) Cx D 14 lb
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471
Problem 6.95 Determine the forces on member AD.
200 N
130 mm
D
400 mm
C
A
B
400 mm
Solution: Denote the reactions of the support by Rx and Ry . The
400 mm
Dy
200 N
complete structure as a free body:
from which Rx D 400 N. The sum of moments:
MA D 800C 400930 C 400530 400200 D 0,
Ay
Ax
Ry
Dy
400 N
Ay
Fx D Rx 400 D 0,
400 N
By
400 N
Ax
Rx
Dx
Dx
Bx
By
Bx
C
from which C D 300 N.
Fy D C C Ry 400 200 D 0,
from which Ry D 300 N. Element ABC : The sum of the moments:
MA D 4By C 8C D 0,
from which By D 600 N. Element BD: The sum of the forces:
Fy D By Dy 400 D 0,
from which Dy D 200 N.
Element AD: The sum of the forces:
Fy D Ay C Dy 200 D 0,
from which Ay D 0: Element AD: The sum of the forces:
and
Fx D Ax C Dx D 0
MA D 400200 C 800Dy 400Dx D 0
Ax D 200 N, and Dx D 200 N.
Element BD: The sum of forces:
Fx D Bx Dx 400 D 0
from which Bx D 600 N. This completes the solution of the nine
equations in nine unknowns, of which Ax , Ay , Dx , and Dy are the
values required by the Problem.
472
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 6.96 The frame shown is used to support
high-tension wires. If b D 3 ft, ˛ D 30&deg; , and W D
200 lb, what is the axial force in member HJ?
A
B
α
C
D
α
E
G
F
W
H
α
I
J
α
W
W
b
Solution: Joints B and E are sliding joints, so that the reactions
are normal to AC and BF, respectively. Member HJ is supported by
pins at each end, so that the reaction is an axial force. The distance
h D b tan ˛ D 1.732 ft
Member ABC. The sum of the forces:
B
b
b
Ay
h
Ax
B
Dy
Dx E
b G
y
W
Fx D Ax C B sin ˛ D 0,
b
Gx
H
W
W
Fy D Ay W B cos ˛ D 0.
The sum of the moments about B:
MB D bAy hAx C bW D 0.
These three equations have the solution: Ax D 173.21 lb, Ay D
100 lb, and B D 346.4 lb.
Member BDEF: The sum of the forces:
Fx D Dx B sin ˛ E sin ˛ D 0,
Fy D Dy W C B cos ˛ E cos ˛ D 0.
The sum of the moments about D:
MD D 2bW bE cos ˛ hE sin ˛ bB cos ˛ C hB sin ˛ D 0.
These three equations have the solution: Dx D 259.8 lb, Dy D
350 lb, E D 173.2 lb.
Member EGHI: The sum of the forces:
Fx D Gx C E sin ˛ H cos ˛ D 0,
Fy D Gy W C E cos ˛ C H sin ˛ D 0.
The sum of the moments about H:
MH D bGy hGx C bW C 2bE cos ˛ 2hE sin ˛ D 0.
These three equations have the solution: Gx D 346.4 lb, Gy D 200 lb,
and H D 300 lb. This is the axial force in HJ.
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
473
Problem 6.97 Determine the force exerted on the ball
by the bolt cutters and the magnitude of the axial force
in the two-force member AB.
20 lb
A
20 in
B
3 in
6 in
4 in
20 lb
Solution: Free-body diagrams of the top head and the top handle
are shown.
From the head we learn that
Fx : Cx D 0
From the handle we have
MD : 20 lb20 in
C Cy 4 in D 0
) Cy D 100 lb
MA : Cy 6 in F3 in D 0
Fy : F TAB C Cy D 0
Solving yields
Force on the ball D F D 200 lb,
474
Axial force D TAB D 300 lb
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 6.98 The woman exerts 20-N forces to the
pliers as shown.
(a)
What is the magnitude of the forces the pliers exert
on the bolt at B?
(b) Determine the magnitude of the force the members
of the pliers exert on each other at the pinned
connection C.
25 mm
80 mm
B
C
50 mm
45⬚
20 N
20 N
Solution: Look at the piece that has the lower jaw of the pliers
(a)
B D 73.5 N
(b)
Cy
MC : B25 mm 20 N cos 45&deg; 80 mm
20 N sin 45&deg; 50 mm D 0
B
Cx
45&deg;
Fx : Cx 20 N sin 45&deg; D 0 ) Cx D 14.14 N
20 N
Fy : Cy B 20 N cos 45&deg; D 0 ) Cy D 87.7 N
Thus the magnitude is
CD
Cx 2 C Cy 2 D 88.8 N
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
475
Problem 6.99 Figure a is a diagram of the bones and
biceps muscle of a person’s arm supporting a mass.
Tension in the biceps muscle holds the forearm in the
horizontal position, as illustrated in the simple mechanical model in Fig. b. The weight of the forearm is 9 N,
and the mass m D 2 kg.
(a)
(b)
Determine the tension in the biceps muscle AB.
Determine the magnitude of the force exerted on
the upper arm by the forearm at the elbow joint C.
B
290
mm
(a)
A
50
mm
C
9N
m
200 mm
150 mm
(b)
Solution: Make a cut through AB and BC just above the elbow
joint C. The angle formed
by the biceps muscle with respect to the
290
D 80.2&deg; . The weight of the mass is W D
forearm is ˛ D tan1
50
29.81 D 19.62 N.
T
W
9N
200
mm
α Cy
50
150 mm
mm
Cx
The section as a free body: The sum of the moments about C is
MC D 50T sin ˛ C 1509 C 350W D 0,
from which T D 166.76 N is the tension exerted by the biceps muscle
AB. The sum of the forces on the section is
FX D Cx C T cos ˛ D 0,
from which Cx D 28.33 N.
FY D Cy C T sin ˛ 9 W D 0,
from which Cy D 135.72. The magnitude of the force exerted by the
forearm on the upper arm at joint C is
FD
476
C2x C C2y D 138.65 N
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 6.100 The bones and the tendons in a horse’s
rear leg are shown in Fig. a. A biomechanical model of
the leg is shown in Fig. b. If the horse is stationary and
the normal force everted on its leg by the ground is
N D 1200 N, determine the tension s in the superficial
digital flexor BC and the patellar ligament DF.
6 cm
6 cm
6 cm
C
D
F
E
40 cm
A
B
3 cm
72 cm
N
(a)
8 10 8
cm cm cm
(b)
Solution: The free-body diagrams for AB and AE are shown. The
angle
˛ D tan1 18/58 D 17.2&deg;
The equilibrium equations for AB are
MA : 1200 N10 cm
C TBC cos ˛8 cm
C TBC sin ˛3 cm D 0
Fx : Ax TBC sin ˛ D 0
Fy : Ay C TBC cos ˛ C 1200 N D 0
One of the equilibrium equations for AE is
ME : TDF 8 cm Ax 49 cm
Ay 10 cm D 0
Solving these four equations yields
Ax D 417 N, Ay D 2540 N
TBC D 1410 N,
TDF D 625 N
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
477
Problem 6.101 The pressure force exerted on the
piston is 2 kN toward the left. Determine the couple
M necessary to keep the system in equilibrium.
B
300 mm
350 mm
45&deg;
A
C
M
400 mm
Solution: From the diagram, the coordinates of point B are d, d
b can be determined from the
where d D 0.3 cos45&deg; . The distance
Pythagorean Theorem as b D 0.352 d2 . From the diagram, the
angle D 37.3&deg; . From these calculations, the coordinates of points B
and C are B (0.212, 0.212), and C (0.491, 0) with all distances being
measured in meters. All forces will be measured in Newtons.
B
0.3 m
45&deg;
A
0.35 m
d
d
θ
b
C
The unit vector from C toward B is uCB D 0.795i C 0.606j.
y
The equations of force equilibrium at C are
and
FX D FBC cos 2000 D 0,
FBC
FBCY
c
2000 N
x
FBCX
N
FY D N FBC sin D 0.
Solving these equations, we get N D 1524 Newtons(N), and FBC D
2514 N.
The force acting at B due to member BC is FBC uBC D 2000i C
1524j N.
B
y
FBC uCB
M
A
rAB
x
The position vector from A to B is rAB D 0.212i C 0.212j m, and the
moment of the force acting at B about A, calculated from the cross
product, is given by MFBC D 747.6k N-m (counter - clockwise). The
moment M about A which is necessary to hold the system in equilibrium, is equal and opposite to the moment just calculated. Thus,
M D 747.6k N-m (clockwise).
478
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Problem 6.102 In Problem 6.101, determine the forces
on member AB at A and B.
Solution: In the solution of Problem 6.101, we found that the force
FBC uCB
acting at point B of member AB was FBC uBC D 2000i C 1524j N,
and that the moment acting on member BC about point A was given by
M D 747.6k N-m (clockwise). Member AB must be in equilibrium,
and we ensured moment equilibrium in solving Problem 6.101.
y
B
M
From the free body diagram, the equations for force equilibrium are
and
AX
A
x
FX D AX C FBC uBCX D AX 2000 N D 0,
AY
FY D AY C FBC uBCY D AY C 1524 N D 0.
Thus, AX D 2000 N, and AY D 1524 N.
Problem 6.103 In Example 6.8, suppose that the
object being held by the plier’s is moved to the left
so that the horizontal distance from D to the object at
E decreases from 30 mm to 20 mm. Draw a sketch of
the pliers showing the new position of the object. What
forces are exerted on the object at E as a result of the
150-N forces on the pliers?
150 N
A
C
E
B
150 N
30 mm
70 mm
30 mm
D
30 mm
30 mm
Solution: The analysis of the bottom grip of the pliers (member 3)
is unchanged.
The reactions Dx D 1517 N, Dy D 500 N.
From the free-body diagram of the lower jaw (member 2) we obtain
MC : E20 mm Dx 30 mm D 0
Therefore E D 2280 N
2280 N
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479
Problem 6.104 The shovel of the excavator is
supported by a pin support at E and the two-force
member BC. The 300-lb weight W of the shovel acts at
the point shown. Determine the reactions on the shovel
at E and the magnitude of the axial force in the two-force
member BC.
Hydraulic
cylinder
Shovel
15 in
A
B
3 in
E
C
12 in
D
W
12 in 7 in
20 in
Solution: The angle
˛ D tan1 3/15 D 11.3&deg;
The equilibrium equations for the shovel are
Fx : Ex TBC cos ˛ D 0
Fy : Ey C TBC sin ˛ 300 lb D 0
MC : 300 lb20 in C Ex 12 in
Ey 7 in D 0
Solving yields
Ex D 604 lb,
Thus
480
Ex D 604 lb,
Ey D 179 lb,
Ey D 179 lb,
TBC D 616 lb
axial force D 616 lb
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 6.105 The shovel of the excavator has a pin
support at E. The position of the shovel is controlled by
the horizontal hydraulic piston AB, which is attached to
the shovel through a linkage of the two-force members
BC and BD. The 300-lb weight W of the shovel acts
at the point shown. What is the magnitude of the force
the hydraulic piston must exert to hold the shovel in
equilibrium?
Hydraulic
cylinder
Shovel
15 in
A
B
3 in
E
C
12 in
D
W
12 in 7 in
20 in
Solution: From the solution to Problem 6.104 we know that the
force in member BC is
TBC D 616 lb
We draw a free-body diagram of joint B and note that AB is horizontal.
The angles are
˛ D tan1 3/15 D 11.3&deg;
ˇ D tan1 4/15 D 14.9&deg;
The equilibrium equations for joint B are
Fx : TBC cos ˛ TAB TBD sin ˇ D 0
Fy : TBD cos ˇ TBC sin ˛ D 0
Solving yields TAB D 637 lb, TBD D 125 lb
637 lb
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
481
Problem 6.106 The woman exerts 20-N forces on the
handles of the shears. Determine the magnitude of the
forces exerted on the branch at A.
20 N
D
C
B
A
E
36 mm 25 mm 25 mm
65 mm
Solution: Assume that the shears are symmetrical.
Consider the 2 pieces CD and CE
20 N
Now examine CD by itself
MC D 20 N90 mm C Dy 25 mm D 0 ) Dy D 72 N
20 N
Fx D 0 ) Dx D Ex
Dy
Fy D 0 ) Dy D Ey
Cy
MC D 0 ) Dx D Ex D 0
Dx = 0
20 N
Dy
Cx
Finally examine DBA
Dx
MB : A36 mm Dy 50 mm D 0
A
C
Dx = 0
By
Ex
Ey
Bx
20 N
Dy
Solving we find
482
A D 100 N
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 6.107 The person exerts 40-N forces on the
handles of the locking wrench. Determine the magnitude
of the forces the wrench exerts on the bolt at A.
A
B
40 N
8 mm 40 mm
E
C
D
50 mm
30 mm
75 mm
Solution: Recognize that DE is a 2-force member. Examine
FDE
75
part CD
40 N
8
75
D
Cx
FDE C Cx D 0
Fx : p
5689
Cy
8
FDE C Cy C 40 N D 0
Fy : p
5689
MC : p
8
5689
40 N
FDE 30 mm C 40 N105 mm D 0
By
Solving we find
Cx D 1312.5 N, Cy D 100 N, FDE D 1320 N
A
Bx
Now examine ABC
MB : A50 mm Cx 40 mm D 0
Fx : Bx Cx D 0
Cx
Fy : By Cy A D 0
Solving:
Cy
A D 1050 N, Bx D 1312.5 N
By D 1150 N
A D 1050 N
Problem 6.108 In Problem 6.107, determine the
magnitude of the force the members of the wrench exert
on each other at B and the axial force in the two-force
member DE.
Solution: From the previous problem we have
BD
Bx 2 C By 2 D 1745 N
FDE D 1320 NC
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
483
Problem 6.109 The device is designed to exert a large
force on the horizontal bar at A for a stamping operation.
If the hydraulic cylinder DE exerts an axial force of
800 N and ˛ D 80&deg; , what horizontal force is exerted on
the horizontal bar at A?
90&deg;
D
m
m
α
25
0m
B
0m
25
25
0m
m
A
E
C
400 mm
Solution: Define the x-y coordinate system with origin at C. The
projection of the point D on the coordinate system is
Fy
B
Ry D 250 sin ˛ D 246.2 mm,
and Rx D 250 cos ˛ D 43.4 mm.
Py
D
Px
Fx
Cx
Cy
The angle formed
by member
DE with the positive x axis is D
Ry
180 tan1
D 145.38&deg; . The components of the force
400 Rx
produced by DE are Fx D F cos D 658.3 N, and Fy D F sin D
454.5 N. The angle of the element AB with the positive x axis is ˇ D
180 90 ˛ D 10&deg; , and the components of the force for this member
are Px D P cos ˇ and Py D P sin ˇ, where P is to be determined. The
angle of the arm BC with the positive x axis is D 90 C ˛ D 170&deg; .
The projection of point B is Lx D 250 cos D 246.2 mm, and Ly D
250 sin D 43.4 mm. Sum the moments about C:
MC D Rx Fy Ry Fx C Lx Py Ly Px D 0.
Substitute and solve: P D 2126.36 N, and Px D P cos ˇ D 2094 N is
the horizontal force exerted at A.
484
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 6.110 This device raises a load W by
extending the hydraulic actuator DE. The bars AD and
BC are 4 ft long, and the distances b D 2.5 ft and h D
1.5 ft. If W D 300 lb, what force must the actuator exert
to hold the load in equilibrium?
b
W
A
B
h
D
C
Solution: The angle ADC is ˛ D sin1
h
D 22.02&deg; . The
4
(1)
dCx hCy dB D 0. The sum of the forces:
distance CD is d D 4 cos ˛.
E
Fx D Cx Ex D 0, from which
The complete structure as a free body: The sum of the forces:
(2)
Fy D W C Cy C Dy D 0.
Ex Cx D 0,
Fx D Cx C Dx D 0.
Fy D Cy Ey C B D 0,
from which
The sum of the moments about C:
W
MC D bW C dDy D 0.
A
These have the solution:
B
Ex
B
A
Cy
Cy D 97.7 lb,
Ey
Cx E
Ex
Dy D 202.3 lb,
Dx
and Cx D Dx .
Divide the system into three elements: the platform carrying the
weight, the member AB, and the member BC.
(3)
Dy
Cy Ey C B D 0
The Platform: (See Free body diagram) The moments about the
point A:
MA D bW dB D 0.
The sum of the forces:
Fy D A C B C W D 0.
ME D
d
h
d
Dy C
Dx A D 0,
2
2
2
from which
(4)
dDy C hDx dA D 0.
These are four equations in the four unknowns: EX , EY , Dx , CX and
DX
These have the solution:
Solving, we obtain Dx D 742 lb.
B D 202.3 lb,
and A D 97.7 lb.
Element BC: The sum of the moments about E is
MC D h
d
d
Cy C
Cx C
B D 0, from which
2
2
2
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
485
Problem 6.111 The four-bar linkage operates the forks
of a fork lift truck. The force supported by the forks is
W D 8 kN. Determine the reactions on member CDE.
0.7 m
0.15 m
0.2 m
W
C
0.15 m
B
D E
Forks
0.2 m
0.3 m
A
F
0.2 m
Solution: Consider body BC. Note that AB is a 2-force body.
W = 8 kN
Fx : Cx D 0
MB : Cy 0.2 m 8 kN0.9 m D 0
Cx
) Cx D 0, Cy D 36 kN
Now examine CDE. Note that DF is a 2-force body.
Cy
3
ME : Cy 0.15 m Cx 0.15 m C p FDF 0.15 m D 0
13
2
Fx : Cx C Ex C p FDF D 0
13
FAB
Cy
3
Fy : Cy C Ey p FDF D 0
13
Solving we find
Note that
Cx
FDF D 43.3 kN, Ex D 24 kN, Ey D 0
2
3
Dx D p FDF , Dy D p FDF
13
13
Summary:
Cx D 0, Cy D 36 kN
Dx D 24 kN, Dy D 36 kN
Ex D 24 kN, Ey D 0
Ey
Ex
D
3
2
FDF
486
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 6.112 If the horizontal force on the scoop is
F D 2000 lb, what is the magnitude of the axial force
in the hydraulic actuator AC?
C
38 in
B
28 in
Scoop
D
10 in
A
F
10 in
20 in
12 in
that BC is a two-force body. The angle
˛ D tan1 38/32 D 49.9&deg;
We have the following equilibrium equation
MA : 2000 lb 10 in
C TBC cos ˛28 in
C TBC sin ˛12 in D 0
) TBC D 735 lb
Now we work with the free-body diagram of joint C. The angles
ˇ D tan1 20/66 D 16.9&deg;
D tan1 10/38 D 14.7&deg;
The equilibrium equations are
Fx : TBC cos ˛ C TAC sin ˇ TCD sin D 0
Fy : TBC sin ˛ TAC cos ˇ TCD cos D 0
Solving yields TAC D 1150 lb, TCD D 553 lb
Thus
1150 lb
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
487
Problem 6.113 A 10-kip horizontal force acts on the
bucket of the excavator. Determine the reactions on
member ACF at A and F.
9 ft
2 ft
E
D
1 ft 4 in
F
C
2 ft
4 ft 4 in
A
1 ft 8 in
B
5 ft 6 in
Bucket
2 ft
3 ft
10 kip
Solution: We start with the free-body diagram of the entire structure. The angle
˛ D tan1 36/72 D 26.6&deg;
The equilibrium equations are
Fx : Ax TBC sin ˛ C 10 kip D 0
Fy : Ay TBC cos ˛ D 0
MA : 10 kip 66 in
C TBC sin ˛ 52 in
TBC cos ˛ 60 in D 0
Next we examine the free-body diagram of the member on the right.
The angle
ˇ D tan1 84/4 D 87.3&deg;
The equilibrium equations are
Fx : Fx C 10 kip TDE sin ˇ D 0
Fy : Fy TDE cos ˇ D 0
MF : TDE sin ˇ 24 in C 10 kip 120 in D 0
Solving these six equations we find
TBC D 21.7 kip, TDE D 49.2 kip
Ax D 0.294 kip,
Fx D 59.2 kip,
488
Ay D 19.4 kip
Fy D 2.34 kip
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 6.114 The structure shown in the diagram
(one of the two identical structures that support the
scoop of the excavator) supports a downward force F D
1800 N at G. Members BC and DH can be treated as
two-force members. Determine the reactions on member
CDK at K.
320
mm
C
Shaft
100
mm
Scoop
260
mm
H
D
J
160
mm
L
Now examine CDK
1800 N0.2 m D 0
) FBC D 873 N
FBC
1
380
mm
1120
mm
200
mm
56
4
FDH 0.26 m p FBC 0.52 m D 0
MK : p
3161
17
56
4
FDH C p FBC C Kx D 0
Fx : p
3161
17
5
1
FDH p FBC C Ky D 0
Fy : p
3161
17
Solving we find
4
G
F
K
1040
mm
4
1
MJ : p FBC 0.44 m p FBC 0.06 m
17
17
260
mm
B
180
mm
Kx D 847 N, Ky D 363 N
4
1800 N
1
FBC
Jx
56
5
Jy
FDH
Kx
Ky
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
489
Problem 6.115 The loads F1 D 440 N and F2 D 160 N.
Determine the axial forces in the members. Indicate
whether they are in tension (T) or compression (C).
F1
A
F2
400 mm
C
200 mm
B
700 mm
Solution: The sum of the moments about C is
MC D 0.7BY C 0.7F1 C 0.4F2 D 0,
from which By D
0.7F1 C 0.4F2
D 531.43 N .
0.7
The axial loads at joint B are
AB D By D 531.4 N C ,
and
BC D 0 .
Similarly, the sum of the forces at the joint A is
FAx D F2 C AC cos ˇ D 0,
from which
AC D
490
F2
D 184.3 N T
cos ˇ
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
B
Problem 6.116 The truss supports a load F D 10 kN.
Determine the axial forces in the members AB, AC,
and BC.
3m
C
A
D
4m
3m
F
Solution: Find the support reactions at A and D.
Fx : Ax D 0
3m
C
Fy : Ay C Dy 10 D 0
AX
4m
MA : 410 C 7Dy D 0
3m
DY
AY
Solving,
10 kN
Ax D 0,
Ay D 4.29 kN
FAB
Dy D 5.71 kN
y
θ
Joint A:
tan D
3
4
FAC
D 36.87&deg;
x
AY
Ay D 4.29 kN
Fx :
FAB cos C FAC D 0
Fy :
Ay C FAB sin D 0
Solving,
FAB D 7.14 kN C
FBC
FAC
FCD
10 kN
FAC D 5.71 kN T
Joint C:
Fx :
FCD FAC D 0
Fy :
FBC 10 kN D 0
Solving
FBC D 10 kN T
FCD D C5.71 kN T
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
491
Problem 6.117 Each member of the truss shown in
Problem 6.116 will safely support a tensile force of
40 kN and a compressive force of 32 kN. Based on this
criterion, what is the largest downward load F that can
safely be applied at C?
Solution: Assume a unit load F and find the magnitudes of the
tensile and compressive loads in the truss. Then scale the load F
up (along with the other loads) until either the tensile limit or the
compressive limit is reached.
B
Fx :
Ax D 0
(1)
Fy :
Ay C Dy F D 0
(2)
2
5
1
C
A
MA :
4F C 7Dy D 0
D
3
4
4m
3m
3m
F
(3)
Joint A:
3
2
tan D
D 36.87&deg;
Fx :
AX
FAC C FAB cos D 0
(4)
F
Fy :
FAB sin C Ay D 0
(5)
y
Joint C
DY
AY
FAB
Fx :
FCD FAC D 0
(6)
Fy :
FBC F D 0
(7)
θ
x
FAC
Joint D
tan D
3
3
AY
D 45&deg;
y
Fx :
FCD FBD cos D 0
(8)
Fy :
FBD sin C Dy D 0
(9)
FBD
φ
Setting F D 1 and solving, we get the largest tensile load of 0.571 in
AC and CD. The largest compressive load is 0.808 in member BD.
x
FCD
Largest Tensile is in member BC. BC D F D 1
DY
The compressive load will be the limit
32
Fmax
D
1
0.808
y
FBC
Fmax D 40 kN
FCD
FAC
x
F
492
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 6.118 The Pratt bridge truss supports loads
at F, G, and H. Determine the axial forces in members
BC, BG, and FG.
B
C
D
4m
E
A
F
G
60 kN
80 kN
20 kN
4m
4m
4m
4m
Solution: The angles of the cross-members are ˛ D 45&deg; .
H
4m
4m
4m
4m
The complete structure as a free body:
Ay
MA D 604 808 2012 C 16E D 0,
from which E D 70 kN. The sum of the forces:
Fx D Ax D 0.
4m
Ax
The sum of the moments about A:
Ay
AB
α
AF
Joint A
60 kN 80 kN 20 kN
BF
AF FG
60 kN
Joint F
E
α
α BC
AB BF BG
Joint B
Fy D Ay 60 80 20 C E D 0,
from which Ay D 90 kN
The method of joints: Joint A:
FY D Ay C AB sin ˛ D 0,
from which AB D 127.3 kN C,
Fx D AB cos ˛ C AF D 0,
from which AF D 90 kN T. Joint F:
Fx D AF C FG D 0,
from which FG D 90 kN T .
Fy D BF 60 D 0,
from which BF D 60 kN C. Joint B:
and
Fx D AB cos ˛ C BC C BG cos ˛ D 0,
Fy D AB sin ˛ BF BG sin ˛ D 0,
from which:
AB sin ˛ BF BG sin ˛ D 0.
Solve: BG D 42.43 kN T ,
and AB cos ˛ C BC C BG cos ˛ D 0,
from which BC D 120 kN C
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
493
Problem 6.119 Consider the truss in Problem 6.118.
Determine the axial forces in members CD, GD,
and GH.
Solution: Use the results of the solution of Problem 6.130:
BC D 120 kN C,
BG D 42.43 kN T,
and FG D 90 kN T.
BC
CD
CG
Joint C
BG
α
CG
GD
α
GH
80 kN
Joint G
The angle of the cross-members with the horizontal is ˛ D 45&deg; .
Joint C:
Fx D BC C CD D 0,
from which CD D 120 kN C
FY D CG D 0,
from which CG D 0.
Joint G:
Fy D BG sin ˛ C GD sin ˛ C CG 80 D 0,
from which GD D 70.71 kN T .
Fy D BG cos ˛ C GD cos ˛ FG C GH D 0,
from which GH D 70 kN T
494
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 6.120 The truss supports loads at F and H.
Determine the axial forces in members AB, AC, BC, BD,
CD, and CE.
200 lb
F
100 lb
4 in
D
4 in
H
B
E
C
4 in
J
G
A
I
6 in
6 in
6 in
Solution: The complete structure as a free body: The sum of the
6 in
200 lb
100 lb
MA D 1006 C 20012 24AY D 0,
from which AY D 125 lb. The sum of forces:
Ax
Fx D Ax D 0.
The method of joints: The angles of the inclined members with the
horizontal are
12
in
6
in
6
in
CD
AB
BD
α
BC
AC
α
˛ D tan1 0.6667 D 33.69&deg;
Joint A:
I
Ay
Ay
Joint A
AB
Joint B
CE
BC
α
AC
Joint C
Fx D AC cos ˛ D 0,
from which AC D 0.
Fy D Ay C AB C AC sin ˛ D 0,
from which AB D 125 lb C
Joint B :
Fyt D AB C BD sin ˛ D 0,
from which BD D 225.3 lb C .
Fx D BD cos ˛ C BC D 0,
from which BC D 187.5 lb T
Joint C :
Fx D BC AC cos ˛ C CE cos ˛ D 0,
from which CE D 225.3 lb T
Fy D AC sin ˛ C CD C CE sin ˛ D 0,
from which CD D 125 lb C
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
495
Problem 6.121 Consider the truss in Problem 6.120.
Determine the axial forces in members EH and FH.
Solution: Use the results from the solution to Problem 6.132:
DF
α
DE
CE D 225.3 lb T,
BD
CD D 125 lb C,
200 lb
α
α
FH
DF
CD
Joint D
EF
Joint F
EF
DE
α
CE
EH
α
EG
Joint E
BD D 225.3 lb C.
The method of joints: The angle of inclined members with the horizontal is ˛ D 33.69&deg; .
Joint D:
Fy D BD sin ˛ CD C DF sin ˛ D 0,
from which DF D 450.7 lb C.
Fx D DF cos ˛ C DE BD cos ˛ D 0,
from which DE D 187.5 lb T
Joint F :
Fx D DF cos ˛ C FH cos ˛ D 0,
from which FH D 450.7 lb C
Fy D 200 DF sin ˛ FH sin ˛ EF D 0,
from which EF D 300 lb T
Joint E :
Fy D CE sin ˛ C EF EG sin ˛ D 0,
from which EG D 315 lb T
Fx D DE C EH CE cos ˛ C EG cos ˛ D 0,
from which EH D 112.5 lb T
496
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 6.122 Determine the axial forces in members
BD, CD, and CE.
10 kN
A
2m
14 kN
C
B
2m
D
E
2m
G
F
2m
I
H
6m
Solution: Use the method of sections
y
10 kN
A
2m
Θ
B
1.5 m
Θ
14 kN
FBD
C
x
Θ
FCD
FCE
D
tan D
2
1.5
D 53.13&deg;
Fx : FCE cos FCD cos C 24 D 0
Fy :
FBD FCD sin FCE sin D 0
MB :
210 1.5FCD sin 1.5FCE sin D 0
3 eqns-3 unknowns.
Solving
FBD D 13.3 kN,
FCD D 11.7 kN,
FCE D 28.3 kN
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
497
Problem 6.123 For the truss in Problem 6.122, determine the axial forces in members DF, EF, and EG.
Solution: Use method of sections
A
10 kN
2m
A
14 kN
10 kN
C
B
2m
2m
14 kN
D
E
2m
2m
D
FDF
G
F
2m
3
E
I
H
φ
Θ
2
FEG
FEF
3
tan D
6m
1.5
2
1.5
D 53.13&deg;
tan D
2
3
D 33.69&deg;
Fx : 24 C FEG cos FEF cos D 0
Fy :
FDF FEF sin FEG sin D 0
ME : 3FDF 214 410 D 0
Solving,
FEG D 32.2 kN C
FDF D 22.67 kN T
FEF D 5.61 kN T
498
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 6.124 The truss supports a 400-N load at G.
Determine the axial forces in members AC, CD, and CF.
400 N
A
C
E
G
300 mm
600 mm
H
F
D
B
300 mm
Solution: The complete structure as a free body: The sum of the
400 N
Ax
MA D 900400 C 600B D 0,
600 mm
B
Fx D Ax C B D 0,
AB
B
from which Ax D 600 N.
300 mm
900 mm
Ay
from which B D 600 N. The sum of forces:
300 mm
from which Ay D 400 N.
The method of joints: The angle from the horizontal of element BD is
300
900
AC
AB
Joint A
AC
Joint D
CE
αCF
CF
CD
Joint C
D 18.43&deg; .
The angle from the horizontal of element AD is
DF
θ
BD
AY
AX
Joint B
Fy D Ay 400 D 0,
D tan1
θ
BD
tan1
300
600 300 tan Joint D:
D
59.04&deg; .
Fx D AD cos ˛AD BD cos C DF cos D 0,
from which DF D 505.96 N C
The angle from the horizontal of element CF is
˛CF D 90 tan1
300
6001 tan Fy D AD sin ˛AD C CD BD sin C DF sin D 0,
D 53.13&deg; .
Joint B:
Fx D B C BD cos D 0,
from which CD D 240 N C
Joint C :
Fy D CD CF sin ˛CF D 0,
from which BD D 632.5 N C
from which CF D 300 N T
Fy D AB C BD sin D 0,
from which AB D 200 N T
Joint A:
from which AD D 233.2 N T
from which AC D 480 N T
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
499
Problem 6.125 Consider the truss in Problem 6.124.
Determine the axial forces in members CE, EF,
and EH.
Solution: Use the results of the solution of Problem 6.124:
AC
CE
αCF
AC D 480 N T,
CD
CF D 300 N T,
CF
Joint C
CF
αCF
θ
DF
EF
EG
αEH
CE
FH
EF
Joint F
EH
Joint E
DF D 505.96 N C,
Joint F :
D 18.4&deg; ,
˛CF D 53.1&deg; .
The method of joints: The angle from the horizontal of element EH is
˛EH D 90 tan1
300
600 900 tan Fy D CF cos ˛CF DF cos C FH cos D 0,
from which FH D 316.2 N C
D 45&deg;
Fy D EF C CF sin ˛CF DF sin C FH sin D 0,
from which EF D 300 N C
Joint C:
Joint E :
Fx D AC C CE C CF cos ˛CF D 0,
Fy D EH sin ˛EH EF D 0,
from which CE D 300 N T
from which EH D 424.3 N T
Problem 6.126 Consider the truss in Problem 6.124.
Which members have the largest tensile and compressive
forces, and what are their values?
Solution: The axial forces for all members have been obtained in
Problems 6.124 and 6.125 except for members EG and GH. These are:
CE
Joint E:
EF
Fx D CE C EG C EH cos ˛EH D 0,
EG
αEH
Joint E
EH
400 N
EG
GH
Joint G
from which EG D 0
Joint G:
Fy D GH 400 D 0,
from which GH D 400 N C.
This completes the determination for all members. A comparison of
tensile forces shows that AC D 480 N T is the largest value, and a
comparison of compressive forces shows that BD D 632.5 N C
is the largest value.
500
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 6.127 The Howe truss helps support a roof.
Model the supports at A and G as roller supports. Use
the method of joints to determine the axial forces in
members BC, CD, CI, and CJ.
6 kN
4 kN
4 kN
D
2 kN
2 kN
C
E
4m
B
F
A
G
H
I
2m
Solution: The free body diagrams for the entire truss and the
required joints are shown.
The whole truss: The equations of equilibrium for the entire truss are:
FX D 0,
2m
J
K
2m
L
2m
2m
2m
6 kN
y
4 kN
D E 4 kN
C
2 kN
F 2 kN
4m
B
A
G
x
H I
J K L
AY
12 m
GY
FY D AY C GY 18 kN D 0.
y
Instead of using the moment equation here (it would work), we see
that the loading is symmetric. Thus, AY D GY D 9 kN.
A
TAH
We need unit vectors along AB, BC, CD, (note that these are the same),
and along BI, and CJ. We get
E
uAB D uBC D uCD D 0.832i C 0.555j,
TDF
F
uBI D 0.832i 0.555j,
y
TAB
TFH
TEF
TFG
TBX
THI
x
TAH H THI
y
4 kN
y
TCX
I
x
TBH
x
TIJ
TBC
C
TCD
x
TCJ
TCI
and uCJ D 0.6i 0.8j.
Joint C:
Joint A:
The equations of equilibrium are
and
FX D TAB uABX C TAH D 0
FY D TAB uABY C AY D 0.
Joint H: The equations of equilibrium are
and
FX D TAH C THI D 0,
FY D TBH D 0.
Joint B:
FX D TBC uBCX C TCJ uCJX C TCD uCDX D 0,
FY D TBC uBCY C TCJ uCJY C TCD uCDY TCI 4 D 0.
Solving these equations in sequence (we can solve at each joint before
going to the next), we get
TAB D 16.2 kN, TAH D 13.5 kN, TBH D 0 kN,
THI D 13.5 kN, TBC D 14.4 kN, TBI D 1.80 kN,
TIJ D 12.0 kN, TCI D 1.00 kN, TCJ D 4.17 kN,
and TCD D 11.4 kN.
FX D TAB uABX C TBC uBCX C TBI uBIX D 0,
FY D TAB uABY C TBC uBCY C TBI uBIY TBH 2 D 0,
Joint I:
and
FX D THI C TIJ TBI uBIX D 0,
FY D TCI TBI uBIY D 0,
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
501
Problem 6.128 For the roof truss in Problem 6.127,
use the method of sections to determine the axial forces
in members CD, CJ, and IJ.
Solution: The free body diagram of the section is shown at the
right. The support force at A is already known from the solution to
Problem 6.139. The equations of equilibrium for the section are
and
FX D TCD uCDX C TCJ uCJX C TIJ D 0,
FY D TCD uCDY C TCJ uCJY C AY D 0,
MC D yC TIJ 4AY D 0.
Solving, we get
TIJ D 12.0 kN,
TCJ D 4.17 kN,
and TCD D 11.4 kN.
Note that these values check with the values obtained in
Problem 6.139.
4 kN
TCD
2 kN
C
I
D
TCJ
TIJ
J
AY
502
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 6.129 A speaker system is suspended from
the truss by cables attached at D and E. The mass of
the speaker system is 130 kg, and its weight acts at G.
Determine the axial forces in members BC and CD.
0.5 m 0.5 m 0.5 m
0.5 m
1m
C
E
A
1m
B
D
G
Solution: The speaker as a free body: The weight of the speaker
Cy
1m
is W D 1309.81 D 1275.3 N. Make a cut through the suspension
cables D, E, the sum of the moments about cable D is
A
Cx
B
0.5 m
The structure as a free body: The sum of the moments about C is
D
D
W
Fy D D C E W D 0,
from which D D 425.1 N.
E
E
MD D 1W C 1.5E D 0,
from which E D 850.2 N. The sum of the forces:
2m
1 m 0.5 m
CY
CE
α
CD
E
DE
BD
MC D C1A 0.5D 2E D 0,
Joint E
β
α
DE
D
Joint D
AC
CE
β
β
BC
CD
Joint C
from which A D 1912.95 N. The sum of the forces:
Joint D:
Fy D A C Cy W D 0,
Fy D CD sin ˇ C DE sin ˛ D D 0,
from which Cy D 637.65 N and
from which CD D 1425.8 N T
Fx D Cx D 0.
Joint C :
The method of joints:
angle of member DE relative to the hori The
1
D 33.69&deg; . The angles of members AB, BC,
zontal is ˛ D tan1
1.5
1
and CD are ˇ D 90 tan 0.5 D 63.43&deg; .
Fy D CD sin ˇ BC sin ˇ C Cy D 0,
from which BC D 2138.7 N T
Joint E :
Fy D E DE sin ˛ D 0,
from which DE D 1532.72 N C.
Fx D CE DE cos ˛ D 0,
from which CE D 1275.3 N T
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
503
y
Problem 6.130 The mass of the suspended object is
900 kg. Determine the axial forces in the bars AB
and AC.
D (0, 4, 0) m
Strategy: Draw the free-body diagram of joint A.
A (3, 4, 4) m
B (0, 0, 3) m
C (4, 0, 0) m
x
z
Solution: The free-body diagram of joint A is.
TAB
TAC
(900) (9.81) N
The position vectors from pt A to pts B, C, and D are
rAB D 3i 4j k (m),
rAC D i 4j 4k (m),
Dividing these vectors by their magnitudes, we obtain the unit vectors
eAB D 0.588i 0.784j 0.196k,
eAC D 0.174i 0.696j 0.696k,
From the equilibrium equation
We obtain the equations
0.588TAB C 0.174TAC 0.6TAD D 0,
0.784TAB 0.696TAC 9009.81 D 0,
Solving, we obtain
TAB D 7200 N,
TAC D 4560 N,
504
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 6.131 Determine the forces on member ABC,
(a)
When you draw the free-body diagrams
individual members, place the 400-lb load
free-body diagram of member ABC.
(b) When you draw the free-body diagrams
individual members, place the 400-lb load
free-body diagram of member CD.
of the
on the
1 ft
1 ft
200 lb
C
D
400 lb
1 ft
of the
on the
B
1 ft
E
Solution: The angle of element BE relative to the horizontal is
˛ D tan1
1
2
1 ft
D 26.57&deg; .
A
The complete structure as a free body: The sum of the moments
Cy
MA D 3400 1200 C 2Fy D 0
from which Fy D 700 lb. The sum of forces:
F
400 lb
C
Cx y
B
200 lb
Dx
Dy
Dy
Dx
B
Fy D Ay C Fy 200 D 0,
Ax
from which Ay D 500 lb .
Cx
B
Ay
B
Fx
Fy
Fx D Ax C Fx C 400 D 0.
100 lb
(a)
400 lb
Element CD: The sum of the moments about D:
26.6&deg;
MD D 200 C 2Cy D 0,
1341
lb
from which Cy D 100 lb .
400 lb
400 lb
500 lb
Fy D Dy Cy 200 D 0,
from which Dy D 100.
from which Check:
Fx D Cx Dx D 0,
BD
from which Dx D Cx .
500 C 100
D 1341.6 lb.
sin ˛
Element DEF: The sum of the moments about F:
check. From above: Cx D Dx D 400 lb .
MF D 3Dx C B cos ˛ D 0,
cos ˛ .
from which Dx D B
3
Fy D Fy C B sin ˛ C Dy D 0,
from which
BD
700 C 100
D 1341.6 lb , and Dx D
sin ˛
Fx D 400 C Cx C B cos ˛ C Ax D 0,
from which Ax D 400 lb .
(b)
When the 400 lb load is applied to element CD instead, the
following changes to the equilibrium equations occur: Element
CD:
400 lb
Fx D Cx Dx C 400 D 0,
Element ABC:
from which Cx C Dx D 400. Element ABC:
MA D 2B cos ˛ 3400 3Cx D 0.
The sum of the forces
Fy D Cy B sin ˛ C Ay D 0,
Fx D Cx C Ax B cos ˛ D 0.
Element DEF : No changes. The changes in the solution for
Element ABC Cx D 800 lb when the external load is removed,
instead of Cx D 400 lb when the external load is applied, so that
the total load applied to point C is the same in both cases.
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
505
Problem 6.132 The mass m D 120 kg. Determine the
forces on member ABC.
A
B
C
300 mm
D
m
E
200 mm
Solution: The weight of the hanging mass is given by
m
W D mg D 120 kg 9.81 2 D 1177 N.
s
FX D AX C EX D 0,
Cx
B
W
B
Cx
B
Cy
B
Ex
and
Cy
Ay
Ax
The complete structure as a free body: The equilibrium equations are:
200 mm
FY D AY W D 0,
MA D 0.3EX 0.4W D 0.
Solving, we get
AX D 1570 N,
AY D 1177 N,
and EX D 1570 N.
Element ABC: The equilibrium equations are
and:
FX D Ax C CX D 0,
FY D AY C CY BY W D 0,
MA D 0.2BY C 0.4cY 0.4W D 0.
Solution gives BY D 2354 N (member BD is in tension),
CX D 1570 N,
and CY D 2354 N.
506
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Problem 6.133 Determine the reactions on member
ABC at B and C.
4 kN
A
0.2 m
D
B
2 kN-m
0.2 m
C
E
0.2 m
0.2 m
Solution: We draw free-body diagrams for the entire structure, and
for members BD and ABC.
From the entire structure:
Fx : Cx C 4 kN D 0
ME : Cy 0.4 m 4 kN0.4 m
2 kN-m D 0
From body ABC
MA : Bx 0.2 m C Cx 0.4 m D 0
And from body BD
MD : By 0.2 m 2 kN-m D 0
Solving these four equations yields
Bx D 8 kN, By D 10 kN
Cx D 4 kN, Cy D 9 kN
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
507
Problem 6.134 The truck and trailer are parked on a
10&deg; slope. The 14,000-lb weight of the truck and the
8000-lb weight of the trailer act at the points shown.
The truck’s brakes prevent its rear wheels at B from
turning. The truck’s front wheels at C and the trailer’s
wheels at A can turn freely, which means they do not
exert friction forces on the road. The trailer hitch at D
behaves like a pin support. Determine the forces exerted
on the truck at B, C, and D.
2 ft
y
9 ft
3 ft
14 ft
D
4 ft
3 ft
8 kip
6 ft
10⬚
B
14 kip
5 ft 6 in
x
C
A
Solution: We separate the two vehicles and draw a free-body
diagram of each. Starting with the trailer we have
MA : 8 kip cos 10&deg; 4 ft C 8 kip sin 10&deg; 6 ft
C Dx 5.5 ft Dy 16 ft D 0
Fx : 8 kip sin 10&deg; Dx D 0
Now we use the free-body diagram for the truck
MB : C11 ft C Dy 2 ft Dx 5.5 ft
14 kip cos 10&deg; 8 ft C 14 kip sin 10&deg; 3 ft D 0
Fx : Bx C Dx 14 kip sin 10&deg; D 0
Fy : By C Dy C C 14 kip cos 10&deg; D 0
Solving yields
Bx D 3820 lb, By D 6690 lb
Dx D 1390 lb, Dy D 1930 lb,
508
C D 9020 lb
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 6.135 The 600-lb weight of the scoop acts at
a point 1 ft 6 in to the right of the vertical line CE. The
line ADE is horizontal. The hydraulic actuator AB can
be treated as a two-force member. Determine the axial
force in the hydraulic actuator AB and the forces exerted
on the scoop at C and E.
B
C
2 ft
A
Solution: The free body diagrams are shown at the right. Place the
coordinate origin at A with the x axis horizontal. The coordinates (in
ft) of the points necessary to write the needed unit vectors are A (0,
0), B (6, 2), C (8.5, 1.5), and D (5, 0). The unit vectors needed for
this problem are
uBA D 0.949i 0.316j,
E
D
5 ft
TCB
1 ft 6 in
1 ft
2 ft 6 in
Scoop
C
1.5 ft
1.5 ft
G
EX
E
EY
uBC D 0.981i 0.196j,
600 lb
and uBD D 0.447i 0.894j.
y
The scoop: The equilibrium equations for the scoop are
and
FX D TCB uBCX C EX D 0,
TBA
x
TCB
FY D TCB uBCY C EY 600 D 0,
TBD
MC D 1.5EX 1.5600 lb D 0.
Solving, we get
EX D 600 lb,
EY D 480 lb,
and TCB D 611.9 lb.
Joint B: The equilibrium equations for the scoop are
and
FX D TBA uBAX C TBD uBDX C TCB uBCX D 0,
FY D TBA uBAY C TBD uBDY C TCB uBCY D 0.
Solving, we get
TBA D 835 lb,
and TBD D 429 lb.
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509
Problem 6.136 Determine the force exerted on the bolt
by the bolt cutters.
100 N
A
75
mm
40 mm
C 55 mm
B
D
90 mm
60 mm 65 mm
300 mm
100 N
Solution: The equations of equilibrium for each of the members
will be developed.
AY
AX
F
Member AB: The equations of equilibrium are:
and
100 N
λ
40 mm
B
55 mm
75 mm
A
FX D AX C BX D 0,
BX
BY
FY D AY C BY D 0,
90 mm
MB D 90F 75AX 425100 D 0
60 mm 65 mm
300 mm
Member BD: The equations are
AY
AX
CY
FX D BX C DX D 0,
40 mm
75 mm
and
FY D BY C DY C 100 D 0,
MB D 15DX C 60DY C 425100 D 0.
90 mm
Member AC: The equations are
and
FX D AX C CX D 0,
FY D AY C CY C F D 0,
60 mm 65 mm
300 mm
BY
DX
B
BX
D
DY
MA D 90F C 125CY C 40CX D 0.
Member CD: The equations are:
D
F
C
55 mm X
C
60 mm 65 mm
300 mm
100 N
FX D CX DX D 0,
CY
FY D CY DY D 0.
Solving the equations simultaneously (we have extra (but compatible)
equations, we get F D 1051 N, AX D 695 N, AY D 1586 N, BX D
695 N, BY D 435 N, CX D 695 N, CY D 535 N, DX D 695 N,
and Dy D 535 N
C
CX
DX
D
DY
Problem 6.137 For the bolt cutters in Problem 6.136,
determine the magnitude of the force the members exert
on each other at the pin connection B and the axial force
in the two-force member CD.
Solution: From the solution to 6.136, we know BX D 695 N,
and BY D 435 N. We also know that CX D 695 N, and CY D 535 N,
from which the axial load in member CD can be calculated. The load
in CD is given by TCD D C2X C C2Y D 877 N
510
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Problem 7.1 In Active Example 7.1, Suppose that the
triangular area is oriented as shown. Use integration to
determine the x and y coordinates of its centroid. (Notice
of Active Example 7.1.)
Solution:
Theheight of the vertical strip is h h/b x so the area
is dA D
h
h
x dx. Use this expression to evaluate Eq. (7.6).
b
The x coordinate of the centroid is
2
b
x3
x
h
h
x h x dx
b
2
3b 0
b
D 0 b x D A
D b D 3
2
h
x
dA
h x dx
h x
A
b
0
2b 0
b
xdA
y
1
h
The y coordinate of the midpoint of the vertical strip is h h x D
2
b
1
h
h C x . We let this value be the value of y in Eq. (7.7):
2
b
h
x
b
b
ydA
y D A
D
0
yD
2h
3
dA
A
b
,
3
xD
Problem 7.2 In Example 7.2, suppose that the area is
redefined as shown. Determine the x coordinate of the
centroid.
1
2
b
x3
h2
h
h
x 2
hC x
h x dx
2h
2
3b 0
b
b
D
D
b
2 b
3
h
x
h x dx
h x
b
0
2b 0
Solution: The height of the vertical strip is 1 x2 , so the area is
dA D 1 x2 dx. Use this expression to evaluate Eq. (7.6).
The x coordinate of the centroid is
1
x4
x2
3
2
4 0
x D A
D
D 01
D 3 1
8
x
2
dA
1 x dA
x
A
0
3 0
y
1
xdA
y⫽1
(1, 1)
x1 x2 x
3
8
xD
y ⫽ x2
x
Problem 7.3 In Example 7.2, suppose that the area is
redefined as shown. Determine the y coordinate of the
centroid.
y
y⫽1
Solution: The height of the vertical strip is 1 x2 , so the area is
dA D 1 x2 dx.
The y coordinate of the midpoint of the vertical strip is
1
1
1 1 x2 D 1 C x2 .
2
2
We let this be the value of y in Eq. (7.7):
(1, 1)
ydA
y D A
D
dA
A
y ⫽ x2
yD
0
1
1
x5
x
1
1 C x2 1 x2 x
3
2
10 0
2
D D
1
3 1
5
x
2
1 x dA
x
0
3 0
3
5
x
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511
Problem 7.4 Determine the centroid of the area.
Solution: The height of a vertical strip of width dx is x2 x C 1,
y
so the area is dA D x2 x C 1 dx. Use this expression to evaluate
Eq. (7.6).
The x coordinate of the centroid is
2
x3
x2
x4
C
4
3
2 0
x D A
D 01
D 2 D 1.25
3
2
x
x
2
dA
x x C 1 dA
C
x
A
0
3
2
0
1
xdA
y ⫽ x2 ⫺ x ⫹ 1
xx 2 x C 1 x
The y coordinate of the midpoint of the vertical strip is
We let this be the value of y in Eq. (7.7):
1
ydA
x
2
y D A
D 0
dA
A
1
1 2
x x C 1.
2
1 2
x x C 12 x
2
x 2 x C 1 dA
0
2
2x4
1 x5
C x3 x2 C x
2 5
4
0
D 0.825
D
2
3
x2
x
Cx
3
2
0
x D 1.25,
Problem 7.5 Determine the coordinates of the centroid
of the area.
y D 0.825
y
Solution: Use a vertical strip - The equation of the line is y D
8
2x
3
6
9
xy dx
3
xD
9
y dx
3
yD
9
3
y dx
3
9
2
x 8 x dx
11
3
D
D 3 9 2
2
8 x dx
3
3
1
yy dx
2
9
2
3
9
x
1
2 2
8 x dx
13
2
3
D 3 9 D
6
2
8 x dx
3
3
9
x D 5.5
y D 2.17
512
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 7.6 Determine the x coordinate of the
values given in Appendix B.
Solution:
0
0
y = cx
n
xD
a
x
cxn
dy dx D
y
0
b
1
A
b
cbnC1
provided that n &gt; 1
nC1
cxn
bn C 1
nC2
x dy dx D
0
0
Matches the appendix
Problem 7.7 Determine the y coordinate of the centroid
in Appendix B.
Solution: See solution to 7.6
yD
1
A
b
cxn
y dy dx D
0
0
bn cn C 1
4n C 2
Matches the appendix
Problem 7.8 Suppose that an art student wants to paint
a panel of wood as shown, with the horizontal and
vertical lines passing through the centroid of the painted
area, and asks you to determine the coordinates of the
centroid. What are they?
Solution: The area:
y
The x-coordinate:
1
x C x 3 dx D
0
1
xx C x 3 dx D
0
y = x + x3
x2
x4
C
2
4
x3
x5
C
3
5
Divide by the area: x D
1
D
0
1
D
0
3
.
4
8
.
15
32
D 0.711
45
The y-coordinate: The element of area is dA D 1 x dy. Note that
dy D 1 C 3x2 dx, hence dA D 1 x1 C 3x2 dx. Thus
yA D
y dA D
A
0
1 ft
x
1
x C x 3 1 x1 C 3x2 dx,
0
from which
1
x x 2 C 4x3 4x4 C 3x5 3x6 dx
0
D
1
4
4
3
3
1
C C D 0.4381.
2
3
4
5
6
7
Divide by A y D 0.5841
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513
Problem 7.9 Determine the value of the constant c so
that the y coordinate of the centroid of the area is y D 2.
What is the x coordinate of the centroid?
y
Solution: The height of a vertical strip of width dx is cx2 so the
area is dA D cx 2 dx.
The y coordinate of the midpoint of the vertical strip is 12 cx 2 . We let
this be the value of y in Eq. (7.7):
4
4
c x5
1 2
cx
cx2 dx
2 5
2
D 4 2 D 5.31c
4
x3
cx 2 dx
2
3 2
ydA
y D A
y ⫽ cx
2
D
dA
2
A
D 2 ) c D 0.376
0
2
4
x
The x coordinate of the centroid is
4
xdA
x D A
D
xcx 2 dx
2
4
dA
D 2
cx dx
A
2
x4
4
x3
3
4
2
4 D 3.21
2
Notice that the value of x does not depend on the value of c.
c D 0.376,
Problem 7.10 Determine the coordinates of the centroid of the metal plate’s cross-sectional area.
y
Solution: Let dA be a vertical strip:
The area dA D y dx D
where 4 1
y = 4 – – x 2 ft
4
Therefore
4
1 2
x
4
dx. The curve intersects the x axis
1 2
x D 0, or x D š4.
4
4 4
1 3
2 x
2x
x
4x
dx
x dA
16 4
4
D 44 D 0.
x D A
D 3 4
1
x
2
dA
4 x
dx
4x
A
4
4
12 4
x
x D 3.21
4
To determine y, let y in equation (7.7) be the height of the midpoint
of the vertical strip:
4
y dA
y D A
D
4
1
2
dA
A
1
1
4 x2
4 x 2 dx
4
4
4
1 2
4 x
dx
4
4
4
x5
x3
1 4
C
8x x
8 x2 C
dx
3
532 4
32
D 4 4 D
3 4
x2
x
4
dx
4x 4
4
12 4
D
4
34.1
D 1.6 ft.
21.3
y
dA
y
x
x
514
dx
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4
3
y, m
Problem 7.11 An architect wants to build a wall with
the profile shown. To estimate the effects of wind loads,
he must determine the wall’s area and the coordinates
of its centroid. What are they?
y = 2 + 0.02x2
2
1
0
0
2
4
6
8
10
x, m
Solution:
10
Area D
10
y dx D
0
2 C 0.02x2 dx
0
10
x3
Area D 2x C 0.02
D 26.67 m2
3 0
dA D y dx D 2 C 0.02x2 dx
Y
X
10
10
x dA
x D 0
10
2x C 0.02x3 dx
0
D
26.67
dA
0
2
10
x
x 4 2 C 0.02
2
4 0
m
xD
26.67
104
100 C 0.02
150
4
xD
D
26.67
26.67
x D 5.62 m
yD
0
10
y 2
y dx
D
10
dA
1
2
10
2 C 0.02x2 2 dx
0
26.67
0
yD
1
226.67
10
4 C 0.08x2 C 0.0004x4 dx
0
3
5 10
x
x
4x C 0.08
C 0.0004
3
5
0
yD
226.67
yD
74.67
53.34
y D 1.40 m
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
515
Problem 7.12 Determine the coordinates of the centroid of the area.
y
y⫽⫺
1 2
x ⫹ 4x ⫺ 7
4
x
Solution: Use a vertical strip. We first need to find the x intercepts.
1
y D x 2 C 4x 7 D 0 ) x D 2, 14
4
xy dx
2
xD
14
y dx
2
14
2
yD
1
x x 2 C 4x 7 dx
4
D 2 14 D8
1 2
x C 4x 7 dx
4
2
14
1
yy dx
2
14
y dx
2
14
2
1
1
x 2 C 4x 7 dx
18
2
4
D 2 14 D
5
1 2
x C 4x 7 dx
4
2
14
xD8
y D 3.6
Problem 7.13 Determine the coordinates of the centroid of the area.
y
y⫽⫺
1 2
x ⫹ 4x ⫺ 7
4
y⫽5
x
Solution: Use a vertical strip. We first need to find the x intercepts.
1
y D x 2 C 4x 7 D 5 ) x D 4, 12
4
xy dx
4
xD
12
y dx
4
1
x x 2 C 4x 7 5 dx
4
D 4 12 D8
1 2
x C 4x 7 5 dx
4
4
12
12
12
yc y dx
4
yD
12
y dx
4
D
12
4
1
2
1
1
x 2 C 4x 7 5 dx
x 2 C 4x 7 C 5
33
4
4
D
12 5
1 2
x C 4x 7 5 dx
4
4
x D 8, y D 6.6
516
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 7.14 Determine the x coordinate of the centroid of the area.
y
y = x3
y=x
x
Solution: Work this problem like Example 7.2
1
1
x dA
0
xD y = x3
0
D 1
1
dA
0
xx x 3 dx
y
y=x
x x 3 dx
0
1
1
1
2
3
5
15
0
D
D 0.533
xD 1 D 1
1
1
x4
x2
4
2
4
2
4 0
x5
x3
3
5
dA
x
x D 0.533
Problem 7.15 Determine the y coordinate of the centroid of the area shown in Problem 7.14.
Solution: Solve this problem like example 7.2.
1
y dA
y D A
D
dA
A
0
1
x C x 3 x x 3 dx
2
1
x x 3 dx
0
1
x3
x7
1
3
7 0
D y D 0 1
1
2
2
x
x4
3
x x dx
2
0
2
4 0
1
x 2 x 6 dx
1
1
4
8
3
7
21
D
D
D 0.381
yD 1
1
2
21
2
2
4
y D 0.381
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
517
Problem 7.16 Determine the x component of the centroid of the area.
y
Solution: The value of the function y D x2 x C 1 at x D 0 is
y D 1, and its value at x D 2 is y D 3. We need a function describing
a straight line that passes through those points. Let y D ax C b. Determining the constants a and b from the conditions that y D 1 when
x D 0 and y D 3 when x D 2, we obtain a D 1 and b D 1. The straight
line is described by the function y D x C 1.
The height of the vertical strip of width dx is x C 1 x2 x C 1 D
2x x2 , so the area is dA D 2x x3 dx. Using this expression to
evaluate Eq. (7.6).
2
xdA
x D A
D 0
x2x x3 dx
2
dA
D 3
2x x dx
A
0
x5
2x3
3
5
x4
2x2
2
4
y ⫽ x2 ⫺ x ⫹ 1
2
2
0
2 D 1
x
x D 1.
0
y
Problem 7.17 Determine the x coordinate of the centroid of the area.
y = x 2 – 20
y=x
x
Solution: The intercept of the straight line with the parabola
occurs at the roots of the simultaneous equations: y D x, and y D x2 20. This is equivalent to the solution of the quadratic x2 x 20 D 0,
x1 D 4, and x2 D 5. These establish the limits on the integration.
The area: Choose a vertical strip dx wide. The length of the strip
is x x 2 C 20, which is the distance between the straight line
y D x and the parabola y D x2 20. Thus the element of area is
dA D x x 2 C 20 dx and
C5
x x 2 C 20 dx D
4
x2
x3
C 20x
2
3
C5
D 121.5.
4
The x-coordinate:
xA D
A
D
xD
518
C5
x 2 x 3 C 20x dx
x dA D
4
x3
x4
C 10x2
3
4
C5
D 60.75.
4
60.75
D 0.5
121.5
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 7.18 Determine the y coordinate of the centroid of the area in Problem 7.17.
Solution: Use the results of the solution to Problem 7.17 in the
following.
The y-coordinate: The centroid of the area element occurs at the
midpoint of the strip enclosed by the parabola and the straight line,
and the y-coordinate is:
yDx
1
1
x x 2 C 20 D
x C x 2 20.
2
2
yA D
y dA D
A
yD
5
1
x C x 2 20x x2 C 20 dx
2
4
D
C5
1
x 4 C 41x2 400 dx
2
4
D
5
5
1
41x3
x
400x
D 923.4.
C
2
5
3
4
923.4
D 7.6
121.5
Problem 7.19 What is the x coordinate of the centroid
of the area?
y
y⫽⫺
1 2
x ⫹ 2x
6
2
Solution: Use vertical strips, do an integral for the parabola then
x
subtract the square
6
2
First find the intercepts
1
y D x 2 C 2x D 0 ) x D 0, 12
6
1
x x 2 C 2x dx 722
65
6
x D 0 12 D
11
1 2
x C 2x dx 22
6
0
12
x D 5.91
Problem 7.20 What is the y coordinate of the centroid
of the area in Problem 7.19?
Solution: Use vertical strips, do an integral for the parabola then
subtract the square
First find the intercepts
1
y D x 2 C 2x D 0 ) x D 0, 12
6
2
1
x 2 C 2x dx 122
139
6
y D 0 12 D
55
1 2
x C 2x dx 22
6
0
12
1
2
y D 2.53
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
519
Problem 7.21 An agronomist wants to measure the
rainfall at the centroid of a plowed field between two
roads. What are the coordinates of the point where the
rain gauge should be placed?
y
0.5 mi
Solution: The area: The element of area is the vertical strip yt yb long and dx wide, where yt D mt x C bt and yb D mb x C bb are
the two straight lines bounding the area, where
0.8 0.3
D 0.3846,
1.3 0
mt D
0.3 mi
0.3 mi
x
0.5 mi
0.6 mi
0.2 mi
and bt D 0.8 1.3 mt D 0.3.
Similarly:
0.3 0
D 0.2308,
1.3 0
mb D
and bb D 0.
The element of area is
dA D yt yb dx D mt mb x C bt bb dx
D 0.1538x C 0.3 dx,
from which
1.1
0.1538x C 0.3 dx
0.5
1.1
x2
D 0.1538 C 0.3x
D 0.2538 sq mile.
2
0.5
The x-coordinate:
1.1
0.1538x C 0.3x dx
x dA D
A
0.5
1.1
x2
x3
D 0.2058.
D 0.1538 C 0.3
3
2 0.5
x D 0.8109 mi
The y-coordinate: The y-coordinate of the centroid of the elemental
area is
y D yb C 12 yt yb D 12 yt C yb D 0.3077x C 0.15.
Thus,
yA D
y dA
A
1.1
0.3077x C 0.150.1538x C 0.3 dx
D
0.5
1.1
D
0.0473x2 C 0.1154x C 0.045 dx
0.5
1.1
x3
x2
D 0.0471 C 0.1153 C 0.045x
D 0.1014.
3
2
0.5
Divide by the area: y D
520
0.1014
D 0.3995 mi
0.2538
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 7.22 The cross section of an earth-fill dam is
shown. Determine the coefficients a and b so that the y
coordinate of the centroid of the cross section is 10 m.
y
y = ax – bx3
x
100 m
Solution: The area: The elemental area is a vertical strip of length
y and width dx, where y D ax bx3 . Note that y D 0 at x D 100, thus
b D a &eth; 104 . Thus
100
dA D a
x 104 x 3 dx
0
A
D 0.5a[x2 0.5 &eth; 104 x 4 ]100
0
D 0.5a &eth; 104 0.25b &eth; 108 ,
and the area is A D 0.25a &eth; 104 . The y-coordinate: The y-coordinate
of the centroid of the elemental area is
y D 0.5ax bx3 D 0.5ax 104 x 3 ,
from which
yA D
y dA
A
100
D 0.5a2
x 104 x 3 2 dx
0
100
D 0.5a2
x 2 2104 x 4 C 108 x 6 dx
0
D 0.5a2 x3
2x5
x7
104 C 108 3
5
7
100
0
D 3.81a2 &eth; 104 .
Divide by the area:
yD
3.810a2 &eth; 104
D 15.2381a.
0.25a &eth; 104
For y D 10, a D 0.6562 , and b D 6.562 &eth; 105 m2
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521
Problem 7.23 The Supermarine Spitfire used by Great
Britain in World War II had a wing with an elliptical
profile. Determine the coordinates of its centroid.
y
y2 = 1
x2 + —
—
a2
b2
Solution:
y
a2
x
2b
y2 = 1
x2 + —
—
b2
b
a
x
b
a
By symmetry, y D 0.
From the equation of the ellipse,
yD
bp 2
a x2
a
By symmetry, the x centroid of the wing is the same as the x centroid
of the upper half of the wing. Thus, we can avoid dealing with š
values for y.
y = ab a2 – x2
y
b
dA = y dx
0
a
x
dx
b a
x a2 x 2 dx
a 0
a
D
xD b
dA
a2 x 2 dx
a 0
x dA
Using integral tables
x
a2 x 2 dx D a2 x 2 dx D
a2 x 2 3/2
3
p
x
x a2 x 2
a2
C
sin1
2
2
a
Substituting, we get
a2 x 2
3/2
a
/3
0
xD p
x a
x a2 x 2
a2
C
sin1
2
2
a
0
0 C a3 /3
a3 /3
xD D 2
2
a /4
a
00
0C
2 2
xD
522
4a
3
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Problem 7.24 Determine the coordinates of the
centroid of the area.
y
Strategy: Write the equation for the circular boundary
in the form y D R2 x 2 1/2 and use a vertical “strip”
of width dx as the element of area dA.
R
Solution: The area: The equation of the circle is x2 Cpy 2 D R2 .
2
2
x
Take the elemental area to be a vertical strip of height
p y D R x
and width dx, hence the element of area is dA D R2 x 2 dx. The
Acircle
R2
area is A D
D
. The x-coordinate:
4
4
xA D
x
0
A
xD
R
R2 x 2 3/2
R3
R2 x 2 dx D D
:
3
3
0
R
x dA D
4R
3
The y-coordinate: The y-coordinate of the centroid of the element of
area is at the midpoint:
p
y D 12 R2 x 2 ,
hence yA D
y dA D
A
D
yD
R
1
R2 x 2 dx
2
0
R
x3
R3
1
R2 x D
2
3 0
3
4R
3
Problem 7.25* If R D 6 and b D 3, what is the y
coordinate of the centroid of the area?
Solution: We will use polar coordinates. First find the angle ˛
˛ D cos1
y
˛
b
3
D cos1
D 60&deg; D
R
6
3
R
/3 6
rdrd D 6
rdrd D
0
yD
1
A
0
/3
0
0
6
0
0
r 2 sin drd D
6
D 1.910
R
x
b
R
α
b
Problem 7.26* What is the x coordinate of the centroid of the area in Problem 7.25?
Solution: See the solution to 7.25
1
xD
A
/3
0
0
6
p
6 3
D 3.31
r cos drd D
2
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
523
Problem 7.27 In Active Example 7.3, suppose that the
area is placed as shown. Let the dimensions R D 6 in,
c D 14 in, and b D 18 in. Use Eq. (7.9) to determine the
x coordinate of the centroid.
y
R
x
Solution: Let the semicircular area be area 1, let the rectangular
b
c
area be area 2, and let the triangular area be area 3. The areas and the
x coordinates of their centroids are
A1 D
1 2
R ,
2
x1 D A2 D c 2R,
A3 D
x2 D
1
b 2R,
2
4R
,
3
1
c,
2
x3 D c C
1
b
3
The x coordinate of the centroid of the composite area is
xD
x 1 A1 C x 2 A2 C x 3 A3
A1 C A2 C A3
D
4R
3
1 2
R
2
C
1
c
2
1
2cR C c C b bR
3
1 2
R C 2cR C bR
2
Substituting the values for R, b, and c yields
x D 9.60 in
Problem 7.28 In Example 7.4, suppose that the area is
given a second semicircular cutout as shown. Determine
the x coordinate of the centroid.
y
Solution: Let the rectangular area without the cutouts be area 1,
let the left cutout be area 2, and let the right cutout be area 3. The
areas and the x coordinates of their centroids are
A1 D 200 280 mm2 ,
x 1 D 100 mm,
1
A2 D 1002 mm2 ,
2
x2 D
1
A3 D 502 mm2 ,
2
x 3 D 200 100 mm
x
50 mm
4100
mm,
3
450
mm.
3
140 mm
140 mm
200 mm
The x coordinated of the centroid of the composite area is
xD
x 1 A1 C x 2 A2 C x 3 A3
A1 C A2 C A3
450
1
1
1002 C 200 502
2
3
2
1
1
200 280 1002 502
2
2
100 [200 280] C
D
4100
3
D 116 mm.
x D 116 mm
524
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Problem 7.29 Determine the coordinates of the
centroids.
Solution: Break into a rectangle, a triangle and a circular hole
xD
y
yD
5[108] C 12
108 C
1
1
2 86 4[108] C 13 8
108 C
2 86
1
2 86
4[22 ]
22
3[22 ]
1
2
2 86 2
D 6.97 in
D 3.79 in
2 in
8 in
x D 6.97 in
y D 3.79 in
3 in
x
4 in
6 in
10 in
Problem 7.30 Determine the coordinates of the
centroids.
Solution: The strategy is to find the centroid for the half circle
area, and use the result in the composite algorithm. The area: The
element of area is a vertical
strip y high and dx wide. From the equation
p
of the circle, y D š R2 x 2 . The
p height of the strip will be twice
the positive value, so that dA D 2 R2 x 2 dx, from which
R
dA D 2
R2 x 2 1/2 dx
10 in
x
20 in
0
A
y
p
R
R2
R2
x R2 x 2
1 x
C
sin
D
D2
2
2
R
2
0
The x-coordinate:
R
x dA D 2
A
x
R2 x 2 dx
0
R
R2 x 2 3/2
2R3
.
D2 D
3
3
0
Divide by A: x D
4R
3
The y-coordinate: From symmetry, the y-coordinate is zero.
420
D 8.488 in. For
3
the inner half circle x2 D 4.244 in. The areas are
The composite: For a complete half circle x1 D
A1 D 628.32 in2
and A2 D 157.08 in2 .
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525
Problem 7.31 Determine the coordinates of the
centroids.
y
Solution: Use a big triangle and a triangular hole
xD
2
1
2
1
3 1.0 2 1.00.8 0.6 C 3 0.4
2 0.40.8
1
1
2 1.00.8 2 0.40.8
yD
1
1
3 0.8 2 1.00.8 1
2 1.00.8 0.8 m
D 0.533
1
1
3 0.8
2 0.40.8
1
2 0.40.8
D 0.267
x
0.6 m
x D 0.533 m
y D 0.267 m
1.0 m
y
Problem 7.32 Determine the coordinates of the
centroid.
Solution: Let the area be divided into parts as shown. The areas
2
and the coordinates are
A1 D 40 50 in2 ,
x 1 D 25 in,
y 1 D 20 in,
A2 D 20 30 in2 ,
x 2 D 10 in,
y 2 D 40 C 15 in,
A3 D
1
302 in2 ,
4
x 3 D 20 C
430
in.
3
y 3 D 40 C
3
30 in
1
40 in
430
in.
3
x
20 in
The x coordinate of the centroid of the composite area is
xD
x 1 A1 C x 2 A2 C x 3 A3
A1 C A2 C A3
430
1
25 [40 50] C [10] [20 30] C 20 C
302
3
4
D
1
40 50 C 20 30 C 302
4
D 23.9 in.
The y coordinate of the centroid of the composite area is
yD
y 1 A1 C y 2 A2 C y 3 A3
A1 C A2 C A3
430
1
302
20 [40 50] C [55] [20 30] C 40 C
3
4
D
D 33.3 in.
1
2
40 50 C 20 30 C 30
4
x D 23.9 in, y D 33.3 in.
526
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Problem 7.33 Determine the coordinates of the
centroids.
y
Solution: Break into 4 pieces (2 rectangles, a quarter circle, and
400
mm
a triangle)
4[0.4]
[0.4]2
0.2[0.40.3] C 0.4 3
4
1
C 0.550.30.7 C 0.8 0.30.7
2
xD
[0.4]2
1
0.40.3 C
C 0.30.7 C 0.30.7
4
2
300 mm
x
300
mm
300
mm
[0.4]2
4[0.4]
0.15[0.40.3] C 0.3 C
3
4 1
1
0.7
0.30.7
C 0.350.30.7 C
3
2
yD
2
1
[0.4]
C 0.30.7 C 0.30.7
0.40.3 C
4
2
x D 0.450 m, y D 0.312 m
y
Problem 7.34 Determine the coordinates of the centroid.
3
2
Solution: Let the area be divided into parts as shown. The areas
and the coordinates are
A1 D 4 3 ft2 ,
x 1 D 2 ft,
A2 D
1
4 4 ft2 ,
2
x2 D
A3 D
1
22 ft2 ,
2
x3 D 4 C
y 1 D 1.5 ft,
2
4 ft,
3
y2 D 3 C
3 ft
1
4 ft,
3
2 ft
1
x
4 ft
42
ft. y 3 D 3 C 2 ft.
3
The x coordinate of the centroid of the composite area is
xD
x 1 A1 C x 2 A2 C x 3 A3
A1 C A2 C A3
1
42
1
4 4 C 4 C
22
2
3
2
D 2.88 ft.
1
1
2
4 3 C 4 4 C 2
2
2
2 [4 3] C
D
2
4
3
The y coordinate of the centroid of the composite area is
yD
y 1 A1 C y 2 A2 C y 3 A3
A1 C A2 C A3
1
1
1
1.5 [4 3] C 3 C 4
4 4 C [3 C 2]
22
3
2
2
D
D 3.20 ft.
1
1
2
4 3 C 4 4 C 2
2
2
x D 2.88 ft,
y D 3.20 ft.
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
527
y
Problem 7.35 Determine the coordinates of the
centroids.
20 mm
30 mm
20 mm
10
mm
30 mm
x
90 mm
Solution: Determine this result by breaking the compound object
into parts
For the composite:
y
m
20 m
m
50 m
m
10 m
30 mm
A1
=
A2
+
+
A3
–
A4
xD
x1 A1 C x2 A2 C x3 A3 x4 A4
A1 C A2 C A3 A4 xD
155782
D 35.3 mm
4414.2
yD
y1 A1 C y2 A2 C y3 A3 y4 A4
A1 C A2 C A3 A4
yD
146675
D 33.2 mm
4414.2
40 mm
20 mm
30 mm
90 mm
A1 :
x
A1 D 3090 D 2700 mm2
x1 D 45 mm
The value for y is not the same as in the new problem statement. This
value seems correct. (The x value checks).
y1 D 15 mm
A2 :
(sits on top of A1 )
A2 D 4050 D 2000 mm2
x2 D 20 mm
y2 D 30 C 25 D 55 mm
A3 :
A3 D
1 2
r D 202 D 628.3 mm2
2 0
2
x3 D 20 mm
y3 D 80 mm C
A4 :
4r0
D 88.49 mm
3
A4 D 3020 C ri2
A4 D 600 C 102 D 914.2 mm2
x4 D 20 mm
y4 D 50 C 15 D 65 mm
Area (composite)
D A1 C A2 C A3 A4
D 4414.2 mm2
528
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 7.36 Determine the coordinates of the
centroids.
y
5 mm
15 mm
50 mm
5 mm
5 mm
15 mm
x
15 mm
10 15 15 10
mm mm mm mm
y
Solution: Comparison of the solution to Problem 7.29 and our
areas 1, 2, and 3, we see that in order to use the solution of
Problem 7.29, we must set a D 25 mm, R D 15 mm, and r D 5 mm.
If we do this, we find that for this shape, measuring from the y axis,
x D 18.04 mm. The corresponding areas for regions 1, 2, and 3 is
1025 mm2 . The centroids of the rectangular areas are at their geometric
centers. By inspection, we how have the following information for the
five areas
Area 1: Area1 D 1025 mm2 , x1 D 18.04 mm, and y1 D 50 mm.
Area 2: Area2 D 1025
mm2 ,
x2 D 18.04 mm, and y2 D 0 mm.
Area 3: Area3 D 1025 mm2 , x3 D 18.04 mm, and y3 D 0 mm.
1
5 mm
5
15
15 mm
50 mm
y 4
5 mm
3
2
15
15 mm
5 mm
x
15 mm
10 15 15 10
mm mm mm mm
Area 4: Area4 D 600 mm2 , x4 D 0 mm, and y4 D 25 mm.
Area 5: Area5 D 450 mm2 , x5 D 7.5 mm, and y5 D 50 mm.
Combining the properties of the five areas, we can calculate the
centroid of the composite area made up of the five regions shown.
AreaTOTAL D Area1 C Area2 C Area3 C Area4 C Area5
D 4125 mm2 .
Then, x D x1 Area1 C x2 Area2 C x3 Area3 C x4 Area4
C x5 Area5 /AreaTOTAL D 3.67 mm,
and y D y1 Area1 C y2 Area2 C y3 Area3 C y4 Area4
C y5 Area5 /AreaTOTAL D 21.52 mm.
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
529
y
Problem 7.37 The dimensions b D 42 mm and h D
22 mm. Determine the y coordinate of the centroid of
the beam’s cross section.
200 mm
h
120 mm
x
b
Solution: Work as a composite shape
y
100 mm
100 mm
A2
A
1
h
120 mm
x
b
b D 42 mm
h D 22 mm
A1 D 120 b mm2 D 5040 mm2
x1 D 0
y1 D 60 mm
by symmetry
A2 D 200 h D 4400 mm2
x2 D 0
y2 D 120 C
xD
h
D 131 mm
2
A1 x1 C A2 x2
0C0
D
A1 C A2
9440
xD0
yD
530
A1 y1 C A2 y2
D 93.1 mm
A1 C A2
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 7.38 If the cross-sectional area of the beam
shown in Problem 7.37 is 8400 mm2 and the y coordinate of the centroid of the area is y D 90 mm, what are
the dimensions b and h?
Solution: From the solution to Problem 7.37
A1 D 120 b, A2 D 200 h
and y D
y1 A1 C y2 A2
A1 C A2
h
60120 b C 120 C
200 h
2
yD
120 b C 200 h
where
y1 D 60 mm
y D 90 mm
A1 C A2 D 8400 mm2
Also,
y2 D 120 C h/2
Solving these equations simultaneously we get
h D 18.2 mm
b D 39.7 mm
200 mm
h
A2
A1
120 mm
b
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531
y
Problem 7.39 Determine the y coordinate of the centroid of the beam’s cross section.
5 in
2 in
8 in
x
3 in
5 in
3 in
5 in
Solution: Take advantage of the symmetry. Work with only one
half of the structure. Break into 2 rectangles, a quarter circle, and a
quarter circular hole.
[5]2
4[5]
438 C 11.533 C 8 C
3
4
4[2]
[2]2
8C
3
4
yD
[2]2
[5]2
38 C 33 C
4
4
y D 7.48 in
y
Problem 7.40 Determine the coordinates of the centroid of the airplane’s vertical stabilizer.
11 m
48&deg;
x
70&deg;
12.5 m
Solution: We work with a rectangle and two triangular holes
y
e
We have
d D 12.5 m C 11 m cot 70&deg; D 16.50 m
11 m
e D 11 m tan 48&deg; D 12.22 m
48&deg;
In the x direction
1
2 d[d11]
xD
yD
x
13 e 12 e11 12.5 C 23 d 12.5
1
2 d 12.511
70&deg;
12.5 m
d
d11 12 e11 12 d 12.511
1
2 11[d11]
23 11
1
2 e11
d11 12 e11
1
1
3 11
2 d 12.5
m11 m
12 d 12.511
Solving we find
x D 9.64 m, y D 4.60 m
532
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 7.41 The area has elliptical boundaries. If
a D 30 mm, b D 15 mm, and ε D 6 mm, what is the x
coordinate of the centroid of the area?
y
Solution: The equation of the outer ellipse is
x2
y2
C
D1
a C ε2
b C ε2
b
and for the inner ellipse
y2
x2
C 2 D1
2
a
b
x
a
We will handle the problem by considering two solid ellipses
For any ellipse
ˇ ˛
x ˛2 x 2 dx
x dA
˛ 0
D
xD ˇ
dA
˛2 x 2 dx
˛
y
ε
From integral tables
b
x
˛2 x 2 3/2
dx D 3
p
x
˛2
x ˛2 x 2
C
sin1
˛2 x 2 dx D
2
2
˛
˛2
x2
˛2 x 2
ε
a
3/2 ˛
A1
=
0
Substituting x D p
x ˛
˛2
x ˛2 x 2
C
sin
3
2
˛
–
x
A2
0
0 C ˛3 /3
˛3 /3
xD D 2
2
˛ /4
˛
00
0C
2 2
xD
dA D
ˇ
˛
˛2 C x 2 dx
0
˛2 2
2
x
For the composite
˛
0
ˇ
˛
β
α
p
x ˛
˛2
ˇ x ˛2 x 2
C
sin1
D
˛
2
2
˛
Area D
β
y = α √α 2 – x2
dA = y dx
4˛
3
Also Area D
y
D ˛ˇ/4
(The area of a full ellipse is ˛ˇ so this checks.
Now for the composite area.
xD
x1 A1 x2 A2
A1 A2
Substituting, we get
x1 D 15.28 mm
x2 D 12.73 mm
A1 D 2375 mm2
A2 D 1414 mm2
and x D 19.0 mm
For the outer ellipse, ˛ D a C ε ˇ D b C ε and for the inner ellipse
˛Da ˇDb
Outer ellipse
x1 D
4a C ε
3
A1 D
a C εb C ε
4
Inner Ellipse
x2 D
4a
3
A2 D
ab
4
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533
Problem 7.42 By determining the x coordinate of the
centroid of the area shown in Problem 7.41 in terms of
a, b, and ε, and evaluating its limit as ε ! 0, show that
the x coordinate of the centroid of a quarter-elliptical
line is
xD
4aa C 2b
.
3a C b
Solution: From the solution to 7.41, we have
x1 D
4a C ε
3
A1 D
4a
x2 D
3
so x1 A1 D
x2 A2 D
a C εb C ε
4
C a2 ε C 2aε2 C ε3 a2 b
x1 A1 x2 A2 D 13 2ab C a2 ε
ab
A2 D
4
a
x1 A1 x2 A2 D 13 a2 b C 2abε C bε2
C ε2 b C
C 2a C bε2 C ε3 ε
Finally x D
3
a2 b
3
A1 A2 D
ab C aε C bε C ε2 ab
4
A1 A2 D
aε C bε C ε2 4
x1 A1 x2 A2
A1 A2
1
2ab C a2 C 2a C bε C ε2 ε
3
xD
[a C b C ε] ε
4
xD
42a C bε
4 2
4aa C 2b
C
C
ε
3a C b
3
3
Taking the limit as ε ! 0
xD
Problem 7.43 Three sails of a New York pilot
schooner are shown. The coordinates of the points are
in feet. Determine the centroid of sail 1.
4aa C 2b
3a C b
Solution: Divide the object into three areas: (1) The triangle with
altitude 21 ft and base 20 ft. (2) The triangle with altitude 21 ft
and base 20 16 D 4ft, and (3) the composite sail. The areas and
coordinates are:
(1)
A1 D 210 ft2 ,
x1 D
y1 D
1
2
3
(2)
2
20 D 13.33 ft,
3
1
21 D 7 ft.
3
A2 D 42 ft2 ,
(a)
x2 D 16 C
y
y
y
y2 D 7 ft.
(14, 29)
(12.5, 23)
(20, 21)
(3, 20)
1
(16, 0)
(3)
(3.5, 21)
2
The composite area: A D A1 A2 D 168 ft2 .
The composite centroid:
3
x
x
x
(10, 0)
(23, 0)
2
4 D 18.67 ft,
3
xD
A1 x1 A2 x2
D 12 ft ,
A
yD
A1 y1 A2 y2
D 7 ft
A
(b)
534
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 7.44 Determine the centroid of sail 2 in
Problem 7.43.
Solution: Divide the object into five areas: (1) a triangle on the
(3)
A3 D
left with altitude 20 ft and base 3 ft, (2) a rectangle in the middle 23 ft
by 9.5 ft, (3) a triangle at the top with base of 9.5 ft and altitude of
3 ft. (4) a triangle on the right with altitude of 23 ft and base of 2.5 ft.
(5) the composite sail. The areas and centroids are:
(1)
A1 D
x1 D
y1 D
(2)
x3 D 3 C
320
D 30 ft2 ,
2
2
3 D 2 ft,
3
(4)
A4 D
y2 D
9.5
2
y4 D
D 7.75 ft,
2
3 D 22 ft
3
1
2.523 D 28.75 ft2 ,
2
x4 D 10 C
A2 D 239.5 D 218.5 ft2 ,
1
9.5 D 6.167 ft,
3
y3 D 20 C
1
20 D 6.67 ft.
3
x2 D 3 C
1
39.5 D 14.25 ft2 ,
2
2
2.5 D 11.67 ft,
3
1
23 D 7.66 ft
3
The composite area: A D A1 C A2 A3 A4 D 205.5 ft2 .
The composite centroid:
(5)
23
D 11.5 ft
2
xD
A1 x1 C A2 x2 A3 x3 A4 x4
D 6.472 ft ,
A
yD
A1 y1 C A2 y2 A3 y3 A4 y4
D 10.603 ft
A
Problem 7.45 Determine the centroid of sail 3 in
Problem 7.43.
Solution: Divide the object into six areas: (1) The triangle Oef,
with base 3.5 ft and altitude 21 ft. (2) The rectangle Oabc, 14 ft by
29 ft. (3) The triangle beg, with base 10.5 ft and altitude 8 ft. (4) The
triangle bcd, with base 9 ft and altitude 29 ft. (5) The rectangle agef
3.5 ft by 8 ft. (6) The composite, Oebd. The areas and centroids are:
(1)
a
f
g
b
e
A1 D 36.75 ft2 ,
x1 D 1.167 ft,
o
c
d
y1 D 14 ft.
(2)
A2 D 406 ft2 ,
(5)
A5 D 28 ft2 ,
x5 D 1.75 ft,
x2 D 7 ft,
y5 D 25 ft.
y2 D 14.5 ft.
(6)
(3)
The composite area:
A3 D 42 ft2 ,
A D A1 C A2 A3 C A4 A5 D 429.75 ft2 .
x3 D 7 ft,
The composite centroid:
y3 D 26.33 ft
(4)
A4 D 130.5 ft2 ,
xD
A1 x1 C A2 x2 A3 x3 C A4 x4 A5 x5
D 10.877 ft
A
yD
A1 y1 C A2 y2 A3 y3 C A4 y4 A5 y5
D 11.23 ft
A
x4 D 17 ft,
y4 D 9.67 ft.
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
535
Problem 7.46 In Active Example 7.5, suppose that the
distributed load is modified as shown. Determine the
reactions on the beam at A and B.
60 N/m
A
B
8m
Solution: We can treat the distributed load as two triangular distributed loads. Using the area analogy, the magnitude of the left one
is 12 8 m60 N/m D 240 N, and the magnitude of the right one is
1
2 4 m60 N/m D 120 N. They must be placed at the centroids of
the triangular distributions.
The equilibrium equations are
Fx : Ax D 0,
Fy : Ay C B 240 N 120 N D 0,
4m
MA : 12 m B [ 23 8 m]240 N [12 m 2
3
4 m]120 N D 0.
Solving yields
Ax D 0,
Problem 7.47 Determine the reactions at A and B.
Ay D 160 N,
B D 200 N.
6 ft
200 lb/ft
B
A
6 ft
4 ft
200 lb/ft
Solution: From the free-body diagram of the bar (with the
distributed loads represented by equivalent force), the equilibrium
equations are
Fx : Ax 600 lb D 0
Fy : Ay C B 800 lb D 0
MA : 800 lb 2 ft
600 lb 4 ft C B10 ft D 0
Solving yields
Ax D 600 lb,
536
Ay D 400 lb,
B D 400 lb.
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 7.48 In Example 7.6, suppose that the
distributed loads are modified as shown. Determine the
reactions on the beam at A and B.
400 N/m
6m
600 N/m
A
6m
400 N/m
B
6m
Solution: The distributed loads are represented by three equivalent
forces. The equilibrium equations are
Fx : 1200 N C Ax D 0
Fy : Ay C B 2400 N 600 N D 0
MA : 1200 N 4 m D 600 N 2 m
2400 N 3 m C B 6 m D 0
Solving yields
Ax D 1200 N,
Ay D 800 N,
B D 2200 N
Problem 7.49 In Example 7.7, suppose that the
distributed load acting on the beam from x D 0 to x D
10 ft is given by w D 350 C 0.3x3 lb/ft. (a) Determine
the downward force and the clockwise moment about
A exerted by the distributed load. (b) Determine the
reactions at the fixed support.
y
2000 lb
w
10,000 ft-lb
x
A
10 ft
10 ft
Solution:
(a)
The force and moment are
10
350 C 0.3x3 dx D 4250 lb
RD
0
R D 4250 lb, M D 23,500 ft-lb
10
3
x350 C 0.3x dx D 23500 ft-lb
MD
0
(b)
The equilibrium equations are
Fx : Ax D 0,
Fy : Ay 4250 lb C 2000 lb D 0,
MA : MA 23500 ft-lb
C 2000 lb20 ft
C 10,000 ft-lb D 0.
Solving yields
Ax D 0, Ay D 2250 lb, MA D 26,500 ft-lb
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
537
Problem 7.50 Determine the reactions at the fixed
support A.
y
= 3(1 – x 2/25) kN/m
x
A
5m
Solution: The free-body diagram of the beam is: The downward
= 3(1 – x 2/25) kN/m
force exerted by the distributed load is
5
w dx D
0
L
x2
3 1
dx
25
5
x3
D 10 kN.
D3 x
75 0
Ma
5m
Ax
x
Ay
The clockwise moment about the left end of the beam due to the
5
xw dx D
0
L
D3
x3
3 x
dx
25
x4
x2
2
100
5
D 18.75 kN-m.
0
From the equilibrium equations
Fx D Ax D 0,
Fy D Ay 10 D 0,
mleftend D Ma C 5Ay 18.75 D 0,
we obtain
Ax D 0,
Ay D 10 kN,
and Ma D 31.25 kN-m.
538
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 7.51
exerted by the
foundation and
(a)
(b)
An engineer measures the forces
soil on a 10-m section of a building
finds that they are described by the
w D 10x x2 C 0.2x 3 kN/m.
Solution:
(a)
The total force is
12
FD
Determine the magnitude of the total force exerted
on the foundation by the distributed load.
Determine the magnitude of the moment about A
10x C x2 0.2x3 dx
0
10
0.2 4
x3
C
x
D 5x2 3
4
0
y
jFj D 333.3 kN
2m
10 m
(b)
The moment about the origin is
A
x
10
MD
10x C x2 0.2x3 x dx
0
1
0.2 5 10
10
x
,
D x3 x4 C
3
4
5
0
jMj D 1833.33 kN.
The distance from the origin to the equivalent force is
dD
jMj
D 5.5 m,
F
from which
jMA j D d C 2F D 2500 kN m.
Problem 7.52 Determine the reactions on the beam at
A and B.
Solution: Replace the distributed load with three equivalent single
forces.
The equilibrium equations
3 kN/m
2 kN/m
A
B
4m
2m
Fx : Ax D 0
Fy : Ay C B 8 kN 2 kN 3 kN D 0
MA : B4 m 8 kN2 m 2 kN
2
34
m
3 kN 4 m C 13 2 m D 0
Ax D 0, Ay D 4.17 kN B D 8.83 kN
2 kN
8 kN
3 kN
Ax
Ay
B
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539
Problem 7.53 The aerodynamic lift of the wing is
y
p
w D 300 1 0.04x 2 N/m.
The mass of the wing is 27 kg, and its center of mass is
located 2 m from the wing root R.
x
R
2m
(a)
(b)
Determine the magnitudes of the force and the
moment about R exerted by the lift of the wing.
Determine the reactions on the wing at R.
5m
Solution:
w
(a)
The force due to the lift is
5
F D w D
MR
3001 0.04x2 1/2 dx,
mg
0
FD
300
5
F D 60
5
FR
2m
25 x2 1/2 dx
3m
0
5
p
25 1 x x 25 x2
C
sin
D 375 N,
2
2
5
0
jFj D 1178.1 N.
The moment about the root due to the lift is
5
M D 300
1 0.04x2 1/2 x dx,
0
M D 60
25 x2 3/2
3
5
D
0
60253/2
D 2500
3
jMj D 2500 Nm.
(b)
The sum of the moments about the root:
M D MR C 2500 27g2 D 0,
from which MR D 1970 N-m. The sum of forces
Fy D FR C 1178.1 27g D 0,
from which FR D 1178.1 C 27g D 913.2 N
540
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Problem 7.54 Determine the reactions on the bar at A
and B.
400 lb/ft
B
y
2 ft
600 lb/ft
400 lb/ft
2 ft
x
4 ft
A
4 ft
Solution: First replace the distributed loads with three equivalent forces.
The equilibrium equations
Fx : Bx C 800 lb D 0
MB : 800 lb1 ft A4 ft C 1600 lb6 ft
C 400 lb6.67 ft D 0
Fy : A C By 1600 lb 400 lb D 0
Solving:
A D 3267 lb, Bx D 800 lb, By D 1267 lb
Bx
800 lb
By
400 lb
1600 lb
A
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541
Problem 7.55 Determine the reactions on member AB
at A and B.
300 lb/ft
A
B
6 ft
6 ft
6 ft
C
300 lb/ft
Solution: From the free-body diagram of the entire structure (with
the distributed loads represented by equivalent forces), one of the equilibrium equations is
MC : Ax 6 ft
3600 lb 6 ft
900 lb 4 ft
1800 lb 9 ft D 0
From the free-body diagram of member AB we have the equilibrium
equations
Fx : Ax C Bx D 0
Fy : Ay C By 3600 lb D 0
MA : By 12 ft
3600 lb 6 ft D 0
Solving yields
Ax D 6900 lb,
Bx D 6900 lb,
542
Ay D 1800 lb,
By D 1800 lb.
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 7.56 Determine the axial forces in members
BD, CD, and CE of the truss and indicate whether they
are in tension (T) or compression (C).
2m
2m
B
A
2m
2m
H
F
D
2m
C
G
E
4 kN/m
8 kN/m
Solution: We start by analyzing the horizontal bar with the distributed load
MG : 16 kN0.667 m C 32 kN2 m
FC
FG
FC 4 m D 0 ) FC D 18.67 kN
Fy D FC C FG 32 kN 16 kN D 0
32 kN
) FG D 29.33 kN
Now work with the whole structure in order to find the reactions at A
Fx : Ax D 0 ) Ax D 0
16 kN
Ax
MH : FG 2 m C FC 6 m
Ay
Ay 8 m D 0 ) Ay D 21.3 kN
H
Finally, cut through the truss and look at the left section
MC : Ax 2 m Ay 2 m BD2 m D 0
FC
MD : Ay 4 m C FC 2 m C CE2 m D 0
Ax
FG
B
BD
1
Fy : Ay FC C p CD D 0
2
CD
Solving we find
BD D 21.3 kN, CD D 3.77 kN, CE D 24 kN
Ay
C
CE
In summary:
FC
BD D 21.3 kNC, CD D 3.77 kNC, CE D 24 kNT
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543
Problem 7.57 Determine the reactions on member
ABC at A and B.
400 N/m
200 N/m
C
160 mm
B
D
240 mm
E
A
400 N/m
160
mm
Solution: Work on the entire structure first to find the reactions
at A. Replace the distributed forces with equivalent concentrated forces
160
mm
160
mm
48 N
96 N
Fx : Ax C 160 N D 0
ME : 96 N0.08 m C 48 N0.16 m
160 N0.2 m Ay 0.32 m D 0
Solving:
Ax D 160 N, Ay D 52 N
Now look at body ABC. Take advantage of the two-force body CD.
160 N
MB : Ax 0.24 m C 160 N0.04 m
48 N0.16 m 96 N0.24 m
4
7
p CD0.32 m C p CD0.16 m D 0
65
65
Ax
E
Ay
4
Fx : Ax C Bx C 160 N p CD D 0
65
48 N
7
Fy : Ay C By 48 N 96 N p CD D 0
65
96 N
Solving:
Ax D 160 N, Ay D 52 N
Bx D 157 N, By D 78.4 N
C
7
By
4
Bx
CD
160 N
Ax
Ay
544
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Problem 7.58 Determine the forces on member ABC
of the frame.
A
1m
3 kN/m
B
1m
C
2m
Solution: The free body diagram of the member on which the
1m
(4 m)(3 kN/m) = 12 kN
BX
From the equilibrium equations
2m
2m
BY
Fx D Bx D 0,
1m
E
AX
Fy D By C E 12 D 0,
4 kN
AY
mleftend D 3E 212 D 0,
CX
we find that Bx D 0, By D 4 kN, and E D 8 kN. From the lower fbd,
writing the equilibrium equation
mleftend D 2Cy 48 D 0,
CX
DX
we obtain Cy D 16 kN. Then from the middle free body diagram,
we write the equilibrium equations
8 kN
CY
DY
CY
2m
2m
Fx D Ax C Cx D 0,
Fy D Ay 4 16 D 0,
mrightend D 2Ax 2Ay C 14 D 0
obtaining Ax D 18 kN, Ay D 20 kN, Cx D 18 kN.
Problem 7.59 Use the method described in Active
Example 7.8 to determine the centroid of the truncated
cone.
y
Solution: Just as in Active Example 7.8, the volume of the disk
element is
dV D R
x
h
2
z
dx
x
the x coordinate of the centroid is
2
R
x x
h
45
h/2
h
D h x D V
2 D
56
R
dV
x
x
V
h
h/2
xdV
h
2
h
x
R
45
h
xD
56
h
2
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545
Problem 7.60 A grain storage tank has the form of a
surface of revolution with the profile shown. The height
of the tank is 7 m and its diameter at ground level is
10 m. Determine the volume of the tank and the height
above ground level of the centroid of its volume.
y
y = ax1/2
7m
10 m
x
Solution:
O
y
y = ax1/2
y
dx
dV = π y2dx
x
dV D y 2 dx
7
x D 0
7
y 2 dx
7
D 0
y 2 dx
0
a2 x dx
7
a2 x dx
0
7
x 3 /3 0
xD 7 D 4.67 m
x2 /2 0
The height of the centroid above the ground is 7 m x
h D 2.33 m
The volume is
7
VD
a2 x dx D a2
0
49
2
m3
To determine a,
y D 5, m when x D 7 m.
p
y D ax1/2 , 5 D a 7
p
a D 5/ 7a2 D 25/7
VD
25
7
49
2
D 275 m3
V D 275 m3
546
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
y
Problem 7.61 The object shown, designed to serve
as a pedestal for a speaker, has a profile obtained by
revolving the curve y D 0.167x 2 about the x axis. What
is the x coordinate of the centroid of the object?
z
x
0.75 m
0.75 m
Solution:
y = 0.167 x2
dV = π y2dx
x
dv
1.50
xdV
x D V
D 0.75
1.50
dV
V
x0.167x2 2 dx
0.167x2 2 dx
0.75
1.5
0.1672
&ETH; 0.75
xD
1.5
0.1672
x 5 dx
x 4 dx
1.5
x 6 /6 0.75
D 1.5
x5 /5 0.75
0.75
x D 1.27 m
y
Problem 7.62 The volume of a nose cone is generated
by rotating the function y D x 0.2x2 about the x axis.
(a)
(b)
What is the volume of the nose cone?
What is the x coordinate of the centroid of
the volume?
z
x
2m
Solution:
(a)
2 m
VD
0
(b)
xD
2 m
2 m
y 2 dx D
x 0.2x2 2 dx D 4.16 m3
0
xy 2 dx
0
V
D
2
xx 0.2x2 2 dx
0
4.155 m3
D 1.411 m
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
547
Problem 7.63 Determine the centroid of the hemispherical volume.
y
R
z
x
Solution: The equation of the surface of a sphere is x2 C y 2 C
z 2 D R2 .
The volume: The element of volume is a disk of radius and thickness
dx. The radius of the disk at any point within the hemisphere is 2 D
y 2 C z2 . From the equation of the surface of the sphere, 2 D R2 x 2 . The area is 2 , and the element of volume is dV D R2 x 2 dx, from which
Vsphere
2 3
D
R .
2
3
VD
The x-coordinate is:
R
x dV D R2 x 2 x dx
0
V
D
D
x4
R2 x 2
2
4
R
0
4
R .
4
Divide by the volume:
xD
R4
4
3
2R3
D
3
R.
8
By symmetry, the y- and z-coordinates of the centroid are zero.
y
x
R
548
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 7.64 The volume consists of a segment of a
sphere of radius R. Determine its centroid.
y
x
R
R
2
z
Solution: The volume: The element of volume is a disk of radius
and thickness dx. The area of the disk is 2 , and the element of
volume is 2 dx. From the equation of the surface of a sphere (see
solution to Problem 7.63) 2 D R2 x 2 , from which the element of
volume is dV D R2 x 2 dx. Thus
VD
R
dV D V
R2 x 2 dx
R/2
R
x3
5
R3 .
D
D R2 x 3 R/2
24
The x-coordinate:
R
x dV D V
R2 x 2 x dx
R/2
D
R2 x 2
x4
2
4
R
D
R/2
9 4
R .
64
Divide by the volume:
xD
9R4
64
24
5R3
D
27
R D 0.675R.
40
By symmetry the y- and z-coordinates are zero.
y
R
–
2
x
R
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
549
Problem 7.65 A volume of revolution is obtained
x2
y2
by revolving the curve 2 C 2 D 1 about the x axis.
a
b
Determine its centroid.
y
y2
x 2 + ––
––
=1
a2 b2
z
x
Solution: The volume: The element of volume is a disk of radius
y and thickness dx. The area of the disk is y2 . From the equation for
the surface of the ellipse,
y
x2 + y2 = 1
a2 b2
x2
y 2 D b2 1 2
a
and dV D
y 2
dx D
b2
x2
1 2
a
x
dx,
from which
a
dV D b2
VD
0
V
x2
1 2 dx
a
a
x3
2b2 a
.
D b2 x 2 D
3a 0
3
The x-coordinate:
a
x dV D b2
0
V
D b2
x2
1 2 x dx
a
x4
x2
2
2
4a
a
D
0
b2 a2
.
4
Divide by volume:
xD
b2 a2
4
3
2b2 a
D
3
a.
8
By symmetry, the y- and z-coordinates of the centroid are zero.
550
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 7.66 In Example 7.9, determine the y coordinate of the centroid of the line.
y
(1, 1)
y ⫽ x2
L
x
Solution: The expression derived in Example 7.9 for the element
dL of the line in terms of x is
dL D
y
(1, 1)
1 C 4x2 dx
dL
The y coordinate of the centroid is
1
ydL
y D L
D
x2
D 0.410
1
dL
dy
1 C 4x2 dx
0
1 C 4x2 dx
L
x
x
dx
0
y D 0.410
y
Problem 7.67 Determine the coordinates of the centroid of the line.
y ⫽x2
Solution:
2
2
dy
x
1C
dx
xds
x 1 C 2x2 dx
dx
1
1
D
x D 12
D
2
2
2
dy
ds
1 C 2x2 dx
1
C
dx
1
1
dx
1
2
2
2
2
dy
y 1C
dx
x 2 1 C 2x2 dx
dx
1
1
D
y D 12
D
2
2
2
dy
ds
1 C 2x2 dx
dx
1
C
1
1
dx
1
2
2
yds
⫺1
2
x
x D 0.801
y D 1.482
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
551
Problem 7.68 Determine the x coordinate of the
centroid of the line.
Solution: The length: Noting that
of length is
y
1C
dL D
dy
dx
2
dx D
dy
D x 11/2 , the element
dx
p
x dx
from which
2
y = – (x – 1)3/2
3
LD
5
dL D
x1/2 dx D
1
L
2 3/2
x
3
5
D 6.7869.
1
The x-coordinate:
5
x 3/2 dx D
x dL D
0
5
0
L
x
Divide by the length: x D
Problem 7.69 Determine the x coordinate of the
centroid of the line.
2 5/2
x
5
5
D 21.961.
1
21.961
D 3.2357
6.7869
Solution: The length: Noting that
length is
y
1C
dL D
dy
dx
2
dx D
dy
D x1/2 the element of
dx
p
1 C x dx
from which
2
y = – x 3/2
3
2
dL D
LD
1 C x1/2 dx D
0
L
2
1 C x3/2
3
2
D 2.7974
0
The x-coordinate:
0
2
2
x dL D
x
L
x1 C x1/2 dx D 2
0
1 C x5/2
1 C x3/2
5
3
2
0
33/2
1
1
35/2
C
D 3.0379.
D2
5
3
5
3
Divide by the length: x D 1.086
Problem 7.70 Use the method described in Example 7.10 to determine the centroid of the circular arc.
y
Solution: The length of the differential line element of the circular
arc is dL D Rd. The coordinates of the centroid are
˛
R cos Rd
xdL
R sin ˛
D 0 ˛
D
x D L
˛
dL
Rd
L
0
ydL
y D L
D
dL
0
552
x
R
˛
R sin Rd
R 1 cos ˛
˛
D
˛
Rd
L
Thus
a
0
xD
R sin ˛
,
˛
yD
R1 cos ˛
˛
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
y
Problem 7.71 In Active Example 7.11, suppose that
the cylinder is hollow with inner radius R/2 as shown.
If the dimensions R D 6 in, h D 12 in, and b D 10 in,
what is the x coordinate of the centroid of the volume?
z
R
h
Solution: Let the cone be volume 1, let the solid cylinder be
volume 2, and let the cylindrical hole be volume 3. The volumes and
the x coordinates of the their centroids are
V1 D
1
3
R2 h,
V2 D R2 b,
x1 D
3
4
x2 D h C
b
1
2
x
R
2
h,
b,
V3 D 12 R2 b, x 3 D h C
1
2
b
The x coordinate of the centroid of the composite volume is
xD
x 1 V1 C x 2 V2 C x 3 V3
V1 C V2 C V3
3
h
4
D
2 1
1
1
1
R2 h C h C b R2 b C h C b
R b
3
2
2
2
2
1
1
R2 h C R2 b R b
3
2
Substituting the values for R, h, and b, we have
x D 14.2 in.
Problem 7.72 Use the procedure described in Example 7.12 to determine the x component of the centroid
of the volume.
y
y
25 mm
Solution: Let the rectangular part without the cutout be volume 1,
let the semicylindrical part be volume 2, and let the cylindrical hole
be volume 3. The volumes and the x coordinates of their centroids are
V1 D 60 50 20 mm3 ,
V2 D
1
252 20 mm3 ,
2
V3 D 102 20 mm3 ,
x z
10 mm
x 1 D 30 mm,
x 2 D 60 C
425
mm,
3
60 mm
20
mm
x 3 D 60 mm.
The x coordinate of the centroid of the composite volume is
xD
x 1 V1 C x 2 V2 C x 3 V3
V1 C V2 C V3
425
1
252 20 C 60 102 20
30 [60 50 20] C 60 C
3
2
D
1
60 50 20 C 252 20 102 20
2
D 38.3 mm.
x D 38.3 mm
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
553
Problem 7.73 Determine the centroids of the volumes.
y
Solution: The object will be divided into a cone and a hemisphere.
From symmetry y D z D 0
Using tables we have in the x direction
xD
3
4R
4
3R
1 2
2R3
R [4R] C 4R C
83R
3
8
3
D
24
1 2
2R3
R [4R] C
3
3
z
R
x
4R
In summary
xD
83R
, y D 0, z D 0
24
Problem 7.74 Determine the centroids of the volumes.
y
Solution: We have a hemisphere and a hemispherical hole. From
symmetry y D z D 0
200 mm
In the x direction we have
3[300 mm]
2[300 mm]3
8
3
2[200 mm]3
3[200 mm]
8
3
xD
2[200 mm]3
2[300 mm]3
3
3
300 mm
z
x
We have
x D 128 mm, y D 0, z D 0
y
Problem 7.75 Determine the centroids of the volumes.
Solution: This is a composite shape. Let us consider a solid
cylinder and then subtract the cone. Use information from the appendix
Volume
Cylinder
Cone
R2 L
1
2
3 r h
Volume (mm3 )
&eth; 107
1.1706
1.3572 &eth; 106
x
x (mm)
L/2
L-h/4
230
370
z
60 mm
90 mm
R D 90 mm
L D 460 mm
360 mm
x
460 mm
r D 60 mm
h D 360 mm
xD
XCyL VCyL XCONE VCONE
VCyL VCONE
x D 211.6 mm
y D z D 0 mm
554
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 7.76 Determine the centroids of the volumes.
y
20 mm
25 mm
75 mm
x
120 mm
25 mm
100 mm
z
Solution: Break the composite object into simple shapes, find the
volumes and centroids of each, and combine to find the required
centroid.
Object
Volume (V)
x
y
z
1
2
LWH
hWD
0
0
L/2
D/2
3
R2 D/2
0
4
r 2 D
0
H/2
H
C h/2 4R
HChC
3
H C h
y
x
(H)
25 mm
+
D/2
D/2
12
0m
(L)
m
100
mm (W)
y
where R D W/2. For the composite,
z
x1 V1 C x2 V2 C x3 V3 x4 V4
V1 C V2 C V3 V4
+
with similar eqns for y and z
–
50
xD
mm
1
3
x
The dimensions, from the figure, are
L D 120 mm
W D 100 mm
H D 25 mm
r D 20 mm
h D 75 mm
D D 25 mm
R D 50 mm
Object
V mm3
x (mm)
y (mm)
z (mm)
C1
C2
C3
4
300000
187500
98175
31416
0
0
0
0
12.5
62.5
121.2
100
60
12.5
12.5
12.5
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
555
7.76 (Continued )
y
Substituting into the formulas for the composite, we get
xD0
100
mm
25
mm
(D)
y D 43.7 mm
mm
75
z D 38.2 mm
x
(h)
2
H
y
z
r = 20 mm
x
4
z
Problem 7.77 Determine the centroids of the volumes.
y
1.75 in
1 in
5 in
z
4 in
1 in
x
Solution: Divide the object into six volumes: (1) A cylinder 5 in
long of radius 1.75 in, (2) a cylinder 5 in long of radius 1 in, (3) a
block 4 in long, 1 in thick, and 21.75 D 3.5 in wide. (4) Semicylinder 1 in long with a radius of 1.75 in, (5) a semi-cylinder 1 in
long with a radius of 1.75 in. (6) The composite object. The volumes
and centroids are:
1 in
x
1.75 in
z
z
x 5 in
Volume
V1
V2
V3
V4
V5
Vol, cu in
48.1
15.7
14
4.81
4.81
x, in
0
0
2
0.743
0
y, in
2.5
2.5
0.5
0.5
4.743
z, in
0
0
0
0
0
4 in
y
1 in
x
The composite volume is V D V1 V2 C V3 V4 C V5 D 46.4 in3 .
The composite centroid:
xD
V1 x1 V2 x2 C V3 x3 V4 x4 C V5 x5
D 1.02 in,
V
yD
V1 y1 V2 y2 C V3 y3 V4 y4 C V5 y5
D 1.9 in,
V
zD0
556
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
y
Problem 7.78 Determine the centroids of the volumes.
30 mm
60 mm
z
x
180
mm
Solution: Consider the composite volume as being made up of
three volumes, a cylinder, a large cone, and a smaller cone which is
removed
Cone 1
Cone 2
Cylinder
Cone 1
Cone 2
V
r 2 L/2
1 2
R L
3
1 2 L
r
3
2
y
x
2R
Object
Cylinder
180
mm
2r
L/4
3L/4
L/2
3(L/2)/4
(mm3 )
(mm)
5.089 &eth; 105
1.357 &eth; 106
1.696 &eth; 105
90
270
135
L/2
y
60 mm
Cylinder
L D 360 mm
1
x
r D 30 mm
R D 60 mm
y
For the composite shape
xCyl VCyL C x1 V1 x2 V2
VCyL C V1 V2
120 mm
cone
+
xD
360 mm
2
x
x D 229.5 mm
y
cone
–
60 mm
3
x
180 mm
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
557
Problem 7.79 The dimensions of the Gemini spacecraft (in meters) are a D 0.70, b D 0.88, c D 0.74,
d D 0.98, e D 1.82, f D 2.20, g D 2.24, and h D 2.98.
Determine the centroid of its volume.
y
g
e
b
a
Solution: The spacecraft volume consists of three truncated cones
and a cylinder. Consider the truncated cone of length L with radii at
the ends R1 and R2 , where R2 &gt; R1 . Choose the origin of the x –y
coordinate system at smaller end. The radius of the cone is a linear
function of the length; from geometry, the length of the cone before
truncations was
R2 L
with volume
R2 R1 (1)
HD
(2)
R22 H
. The length of the truncated portion is
3
(3)
(4)
D
R1 L
with volume
R2 R1 R12
. The volume of the truncated cone is the difference of the
3
two volumes,
L
(5) V D
3
cone is
R23 R13
R2 R1
. The centroid of the removed part of the
3
, and the centroid of the complete cone is
4
(6)
x D
(7)
3
H, measured from the pointed end. From the
4
composite theorem, the centroid of the truncated cone is
(8)
558
c
f
d
h
x
Beginning from the left, the volumes are (1) a truncated cone, (2) a
cylinder, (3) a truncated cone, and (4) a truncated cone. The algorithm
and the data for these volumes were entered into TK Solver Plus and
the volumes and centroids determined. The volumes and x-coordinates
of the centroids are:
Volume
V1
V2
V3
V4
Composite
Vol, cu m
0.4922
0.5582
3.7910
11.8907
16.732
x, m
0.4884
1.25
2.752
4.8716
3.999
The last row is the composite volume and x-coordinate of the centroid
of the composite volume.
The total length of the spacecraft is 5.68 m, so the centroid of the
volume lies at about 69% of the length as measured from the left
end of the spacecraft. Discussion: The algorithm for determining the
centroid of a system of truncated cones may be readily understood
if it is implemented for a cone of known dimensions divided into
sections, and the results compared with the known answer. Alternate
algorithms (e.g. a Pappus-Guldinus algorithm) are useful for checking
but arguably do not simplify the computations End discussion.
xh D
Vh xh V x
C x, where x is the x-coordinate of the left
V
hand edge of the truncated cone in the specific coordinate system.
These eight equations are the algorithm for the determination of
the volumes and centroids of the truncated cones forming the
spacecraft.
xD
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
y
Problem 7.80 Two views of a machine element are
shown. Determine the centroid of its volume.
y
24 mm
Solution: We divide the volume into six parts as shown. Parts 3
and 6 are the “holes”, which each have a radius of 8 mm. The volumes
are
8 mm
18 mm
60 mm
V1 D 604850 D 144,000 mm3 ,
V2 D 12 242 50 D 45, 239 mm3 ,
x
V3 D 82 50 D 10, 053 mm3 ,
V4 D 163620 D 11, 520
8 mm
z
20
mm
mm3 ,
16
mm
50 mm
V5 D 12 182 20 D 10, 179 mm3 ,
y
V6 D 82 20 D 4021 mm3 .
The coordinates of the centroids are
2
x1 D 25 mm,
3
y1 D 30 mm,
5
x2 D 25 mm,
y2 D 60 C
1
6
z1 D 0,
4
z
424
D 70.2 mm,
3
z5 D 24 C 16 C
z2 D 0,
x3 D 25 mm,
x6 D 10 mm,
y3 D 60 mm,
y6 D 18 mm,
z3 D 0,
418
D 47.6 mm,
3
z6 D 24 C 16 D 40 mm.
x4 D 10 mm,
The x coordinate of the centroid is
y4 D 18 mm,
xD
z4 D 24 C 8 D 32 mm,
x5 D 10 mm,
x1 V1 C x2 V2 x3 V3 C x4 V4 C x5 V5 x6 V6
D 23.65 mm.
V1 C V2 V3 C V4 C V5 V6
Calculating the y and z coordinates in the same way, we obtain y D
36.63 mm and z D 3.52 mm
y5 D 18 mm,
y
Problem 7.81 In Example 7.13, suppose that the
circular arc is replaced by a straight line as shown.
Determine the centroid of the three-segment line.
Solution: Let the new straight-line segment be line 1 and let the
segment in the x-z plane be line 2. Let the other line segment be line 3.
The centroid locations of the parts and their lengths are
x 1 D 0,
x 2 D 2 m,
y 1 D 1 m,
y 2 D 1 m,
z1 D 1 m,
z2 D 2 m,
L1 D 2.83 m,
L2 D 4 m,
x 3 D 2 m,
y 3 D 1 m,
z3 D 1 m,
L2 D 4.90 m.
(0, 2, 0) m
(0, 0, 2) m
z
x
(4, 0, 2) m
Applying Eqs. (7.18) yields
x D 1.52 m,
y D 0.659 m,
z D 1.34 m
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
559
Problem 7.82 Determine the centroids of the lines.
y
Solution: The object is divided into two lines and a composite.
3m
(1)
(2)
(3)
L1 D 6 m, x1 D 3 m, y1 D 0.
6
L2 D 3 m, x2 D 6 C m (Note: See Example 7.13) y2 D 3.
The composite length: L D 6 C 3 m. The composite centroid:
xD
L1 x1 C L2 x2
D 6 m,
L
x
6m
3
yD
D 1.83 m
2C
y
Problem 7.83 Determine the centroids of the lines.
Solution: Break the composite line into three parts (the quarter
circle and two straight line segments) (see Appendix B).
Part 1
Part 2
Part 3
xD
xi
yi
Li
2R/
3 m
0
2R/
0
3 m
R/2
2 m
2 m
x 1 L 1 C x2 L 2 C x3 L 3
D 1.4 m
L1 C L2 C L3
2m
(R D 2 m)
2m
x
2m
2m
y1 L1 C y2 L2 C y3 L3
yD
D 1.4 m
L1 C L2 C L3
560
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 7.84 The semicircular part of the line lies in
the x –z plane. Determine the centroid of the line.
y
100 mm
160 mm
x
120 mm
z
Solution: The bar is divided into three segments plus the
composite. The lengths and the centroids are given in the table: The
composite length is:
3
LD
y
100 mm
2
3
160 mm
Li .
iD1
z
The composite coordinates are:
3
xD
and y D
120 mm
x
Li xi
iD1
L
3
1
,
Li yi
iD1
L
Segment
Length, mm
x, mm
240
0
y, mm
L1
120
L2
100
L3
188.7
80
50
Composite
665.7
65.9
21.7
z, mm
0
120
50
0
0
68.0
Problem 7.85 Determine the centroid of the line.
y
Solution: Break into a straight line and an arc.
2/3
mm tan 60&deg; 2 cos 30&deg; C
0
1
200 mm2
C cos d
cos 60&deg;
xD
D 332 mm
2/3
200 mm tan 60&deg; C
200 mm d
1
2 200
200 mm
60⬚
x
0
1
&deg;
2 200mm tan 60 200
2/3
2
mm sin 60&deg; 200 mm sin d
D 118 mm
2/3
200 mm tan 60&deg; C
200 mm d
C
yD
0
0
x D 332 mm, y D 118 mm
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
561
y
Problem 7.86 Use the method described in Active
Example 7.14 to determine the area of the curved part
of the surface of the truncated cone.
z
R
x
h
2
h
2
Solution: Work with the solid line shown. The surface area is
given by
A D 2yL D 2
3R
4
2 2
h
R
C
2
2
3R p 2
h C R2
4
Problem 7.87 Use the second Pappus–Guldinus theorem to determine the volume of the truncated cone.
Solution: Work with the trapezoidal area
A D R/2h/2 C 1/2R/2h/2 D
3Rh
8
R
R/2
yD
7R
R/2h/2R/4 C 1/2R/2h/2[1/3R/2 C R/2]
D
A
18
V D 2yA D 2
VD
7R
18
3Rh
8
D
h/2
h/4
h/4
7R2 h
24
7R2 h
24
Problem 7.88 The area of the shaded semicircle is
1
R2 . The volume of a sphere is 43 R3 . Extend the
2
approach described in Example 7.15 to the second
Pappus–Guldinus theorem and determine the centroid
y S of the semicircular area.
Solution: The semicrcular area is A D
1
2
R2 ,
and y s is the y
coordinate of its centroid. Rotating the area about the x axis generates
the volume of a sphere. The second Pappus–Guldinus theorem states
that the volume of the sphere is
y
_
yS
x
R
V D 2y s A
4
3
R3 D 2y s
1
2
R2
Solving for ys yields
562
ys D
4R
3
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 7.89 Use the second Pappus–Guldinus
theorem to determine the volume generated by revolving
the curve about the y axis.
Solution: The x coordinate of the centroid:. The element of area
is the vertical strip of height 1 y and width dx. Thus
y
1
1
1 y dx D
0
1 x2 dx.
0
Integrating,
(1, 1)
1
2
x3
D .
3 0
3
y ⫽ x2
1
x dA D
A
x
x x 3 dx D
0
divide by the area: x D
x2
x4
2
4
1
D
0
1
,
4
3
. The volume is V D 2xA D
8
2
Problem 7.90 The length of the curve is L D 1.479,
and the area generated by rotating it about the x axis
is A D 3.810. Use the first Pappus–Guldinus theorem to
determine the y coordinate of the centroid of the curve.
Solution: The surface area is A D 2yL, from which
yD
A
D 0.41
2L
Problem 7.91 Use the first Pappus–Guldinus theorem
to determine the area of the surface generated by
revolving the curve about the y axis.
Solution: The length of the line is given in Problem 7.90.
L D 1.479. The elementary length of the curve is
dy
dx
2
dL D
1C
Noting
dy
D 2x, the element of line is dL D 1 C 4x2 1/2 .
dx
dx.
The x-coordinate:
1
x dL D
x1 C 4x2 1/2 dx
0
L
D
1
1 53/2 1
1 C 4x2 3/2 0 D
D 0.8484.
12
12
Divide by the length to obtain x D 0.5736. The surface area is
A D 2xL D 5.33
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
563
y
Problem 7.92 A nozzle for a large rocket engine is
designed by revolving the function y D 23 x 13/2 about
the y axis. Use the first Pappus–Guldinus theorem to
determine the surface area of the nozzle.
y = _2 (x – 1)3 / 2
3
x
5 ft
dy
D x 11/2 , the element
dx
Solution: The length: Noting that
of length is
1C
dL D
dy
dx
2
dx D
p
x dx
from which
5
dL D
LD
x1/2 dx D
1
L
2 3/2
x
3
5
D 6.7869 ft
1
The x-coordinate:
5
x dL D
x 3/2 dx D
1
L
2 5/2
x
5
5
D 21.961.
1
Divide by the length: x D 3.2357. The area
A D 2xL D 138 ft2
Problem 7.93 The coordinates of the centroid of the
line are x D 332 mm and y D 118 mm. Use the first
Pappus-Guldinus theorem to determine the area of the
surface of revolution obtained by revolving the line about
the x axis.
y
200 mm
60⬚
x
Solution:
L D 200 mm tan 60&deg; C 200 mm
120&deg;
180&deg;
D 765 mm
A D 2yL D 20.118 m0.765 m D 0.567 m2
564
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 7.94 The coordinates of the centroid of the
area between the x axis and the line in Problem 7.93 are
x D 355 mm and y D 78.4 mm. Use the second PappusGuldinus theorem to determine the volume obtained by
revolving the area about the x axis.
Solution: The area is
1
0.2 m0.2 m tan 60&deg; C
2
120&deg;
0.2 m2 D .0765 m2
360&deg;
V D 2yA D 20.0784 m0.0765 m2 D 0.0377 m3
Problem 7.95 The volume of revolution contains a
(a)
(b)
R+a
Use integration to determine its volume.
Use the second Pappus–Guldinus theorem to determine its volume.
R
h
Solution:
(a)
The element of volume is a disk of radius y and thickness dx.
The area of the disk is y 2 R2 . The radius is
yD
a
h
from which dV D Denote m D
a
h
x C R,
2
xCR
dx R2 dx.
a
, dV D m2 x 2 C 2mRx dx,
h
from which
h
dV D m
VD
mx 2 C 2Rx dx
0
V
3
h
x
mh
CR
D m m C Rx 2 D mh2
3
3
0
D ah
(b)
a
3
CR .
The area of the triangle is A D 12 ah. The y-coordinate of the
centroid is y D R C 13 a. The volume is
V D 2yA D ahR C 13 a
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
565
Problem 7.96 Determine the volume of the volume of
revolution.
Solution: The area of the semicircle is A D
y DRC
4r
. The volume is
3
V D 2
r 2
2
RC
4r
3
r 2
. The centroid is
2
4r
D 2 r 2 R C
.
3
For r D 40 mm and R D 140 mm, V D 2.48 &eth; 103 m3
140
mm
80
mm
Problem 7.97 Determine the surface area of the
volume of revolution in Problem 7.96.
Solution: The length and centroid of the semicircle is Lo D r,
yDRC
y D R.
2r
. The length and centroid of the inner line is Li D 2r, and
2r
A D 2r R C
C 22rR D 2rR C 2r C 2R.
For r D 40 mm and R D 140 mm, A D 0.201 m2
Problem 7.98 The volume of revolution has an
elliptical cross section. Determine its volume.
230 mm
130 mm
180 mm
Solution: Use the second theorem of Pappus-Guldinus. The
centroid of the ellipse is 180 mm from the axis of rotation. The area
of the ellipse is ab where a D 115 mm, b D 65 mm.
2b
The centroid moves through a distance jdj D 2R D 2 (180 mm) as
the ellipse is rotated about the axis.
2a
V D Ad D abd D 2.66
&eth; 107
mm3
v D 0.0266 m3
566
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 7.99 Suppose that the bar in Active
Example 7.16 is replace with this 100-kg homogeneous
bar. (a) What is the x coordinate of the bar’s center of
mass? (b) Determine the reactions at A and B.
y
0.5 m
B
1m
A
x
1m
Solution:
(a)
Let the new horizontal segment of the bar part 3. The x coordinate
of the centroid of the bar’s axis, which coincides with its center
of mass is
xD
x 1 L1 C x 2 L2 C x 3 L3
0.51 C 11 C 0.750.5
D 0.75 m
D
L1 C L2 C L3
0.5 C 1 C 0.5
x D 0.75 m
(b)
The equilibrium equations are
MA : B1 m 981 N 0.75 m D 0,
Fx : Ax B D 0,
Fy : Ay 981 N D 0.
Solving yields
Ax D 736 N,
Ay D 981 N,
B D 736 N.
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
567
Problem 7.100 The mass of the homogeneous flat
plate is 50 kg. Determine the reactions at the supports A
and B.
100 mm
400 mm
200 mm
A
B
600 mm
800 mm
600 mm
Solution: Divide the object into three areas and the composite.
Since the distance to the action line of the weight is the only item of
importance, and since there is no horizontal component of the weight,
it is unnecessary to determine any centroid coordinate other than the xcoordinate. The areas and the x-coordinate of the centroid are tabulated.
The last row is the composite area and x-coordinate of the centroid.
500 N
AX
X
B
AY
Area
A, sq mm
x
Rectangle
3.2
&eth; 105
400
Circle
3.14 &eth; 104
600
Triangle
1.2
&eth; 105
1000
Composite
4.09 &eth; 105
561
1400 mm
The composite area is A D Arect Acirc C Atriang . The composite xcoordinate of the centroid is
xD
Arect xrect Acirc xcirc C Atriang xtriang
.
A
The sum of the moments about A:
MA D 500561 C 1400B D 0,
from which B D 200 N. The sum of the forces:
Fy D Ay C B 500 D 0,
from which Ay D 300 N.
568
Fx D Ax D 0
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Problem 7.101 The suspended sign is a homogeneous
flat plate that has a mass of 130 kg. Determine the axial
forces in members AD and CE. (Notice that the y axis
is positive downward.)
A
2m
4m
C
1m
E
B
x
D
y
y = 1 + 0.0625x2
Solution: The strategy is to determine the distance to the action
line of the weight (x-coordinate of the centroid) from which to apply
the equilibrium conditions to the method of sections.
The area: The element of area is the vertical strip of length y and width
dx. The element of area dA D y dx D 1 C ax2 dx, where a D 0.0625.
Thus
4
dA D
0
A
4
ax 3
1 C ax2 dx D x C
D 5.3333 sq ft.
3 0
The x-coordinate:
4
x dA D
A
x1 C ax2 dx D
0
Divide A: x D
ax 4
x2
C
2
4
4
D 12.
0
12
D 2.25 ft.
5.3333
The equilibrium conditions: The angle of the member CE is
˛ D tan1 14 D 14.04&deg; .
The weight of the sign is W D 1309.81 D 1275.3 N. The sum of the
MD D 2.25W C 4CE sin ˛ D 0,
from which CE D 2957.7 N T .
Method of sections: Make a cut through members AC, AD and BD
and consider the section to the right. The angle of member AD is
ˇ D tan1 12 D 26.57&deg; .
The section as a free body: The sum of the vertical forces:
FY D AD sin ˇ W D 0
from which AD D 2851.7 N T
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569
Problem 7.102 The bar has a mass of 80 kg. What are
the reactions at A and B?
A
2m
2m
B
y
Solution: Break the bar into two parts and find the masses and
centers of masses of the two parts. The length of the bar is
X1
m1g
m2g
AX
L D L1 C L2 D 2 m C 2R/4R D 2 m
X2
L D2C m
AY
Lengthi (m)
Part
1
2
2
m1 D 31.12 kg
x1 D 1 m
m2 D 48.88 kg
x2 D 3.27 m
Fx :
Ax D 0
Fy :
Ay C By m1 g m2 g D 0
MA :
x1 m1 g x2 m2 g C 4By D 0
Massi (kg)
2
80
2C
80
2C
xi (m)
x
4m
BY
1
2C
2R
Solving
Ax D 0, Ay D 316 N, B D 469 N
570
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Problem 7.103 The mass of the bar per unit length is
2 kg/m. Choose the dimension b so that part BC of the
suspended bar is horizontal. What is the dimension b,
and what are the resulting reactions on the bar at A?
A
1m
30⬚
B
b
Solution: We must have
C
Ay
MA : g1.0 m0.5 m cos 30&deg; gb
b
1.0 m cos 30&deg;
2
Ax
D0
) b D 2.14 m
Then
Fx : Ax D 0
ρg(1.0 m)
ρ gb
Fy : Ay g1.0 m gb D 0
)
Ax D 0, Ay D 61.6 N, b D 2.14 m
Problem 7.104 The semicircular part of the homogeneous slender bar lies in the x –z plane. Determine the
center of mass of the bar.
Solution: The bar is divided into three segments plus the composite. The lengths and the centroids are given in the table: The
composite length is:
y
3
LD
Li .
iD1
The composite coordinates are:
10 in
3
16 in
12 in
xD
Li xi
iD1
L
,
z
x
3
and y D
Li yi
iD1
L
Segment
Length, in
0
12
10
x, in
24
0
L1
12
L2
y, in
z, in
5
0
L3
18.868
8
5
0
Composite
66.567
6.594
2.168
6.796
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571
y
Problem 7.105 The density of the cone is given by the
equation D 0 1 C x/h, where 0 is a constant. Use
the procedure described in Example 7.17 to show that
the mass of the cone is given by m D 7/40 V, where
V is the volume of the cone, and that the x coordinate
of the center of mass of the cone is x D 27/35h.
z
R
x
h
Solution: Consider an element of volume dV of the cone in the
form of a “disk” of width dx. The radius of such a disk at position x
is (R/h)x, so dV D [R/hx]2 dx.
The mass of the cone is
h
7
7
dV D
0 1 C x/h [R/hx]2 dx D
0 R2 h D 0 V.
mD
12
3
V
0
The x coordinate of the center of mass is
h
xdV
x0 1 C x/h [R/hx]2 dx
27
h.
D 0h
x D V
D
35
2
dV
0 1 C x/h [R/h x] dx
V
572
0
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y
Problem 7.106 A horizontal cone with 800-mm length
and 200-mm radius has a built-in support at A. Its density
is D 60001 C 0.4x 2 kg/m3 , where x is in meters.
What are the reactions at A?
200 mm
A
x
800 mm
Solution: The strategy is to determine the distance to the line of
action of the weight, from which to apply the equilibrium conditions.
The mass: The element of volume is a disk of radius y and thickness
dx. y varies linearly with x: y D 0.25x. Denote a D 0.4. The mass of
the disk is
dm D y 2 dx D 60001 C ax2 0.25x2 dx
y
200
mm
x
A
800 mm
D 3751 C ax2 x 2 dx,
from which
0.8
m D 375
1 C ax2 x 2 dx D 375
0
x5
x3
Ca
3
5
0.8
0
D 231.95 kg
The x-coordinate of the mass center:
x dm D 375
0.8
1 C ax2 x 3 dx D 375
0
m
x6
x4
Ca
4
6
0.8
0
D 141.23.
Divide by the mass: x D 0.6089 m
The equilibrium conditions: The sum of the moments about A:
M D MA mgx D 0,
from which
MA D mgx D 231.949.810.6089
D 1385.4 N-m .
The sum of the vertical forces:
FY D AY mg D 0
from which AY D 2275.4 N . The horizontal component of the
reaction is zero,
FX D 0.
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573
y
Problem 7.107 In Active Example 7.18, suppose that
bar 1 is replaced by a bar with the same dimensions
that consists of aluminum alloy with a density of
2600 kg/m3 . Determine the x coordinate of the center
of mass of the machine part.
240 mm
1
40 mm
2
80 mm
Solution: The mass of bar 1 is
4
m D 7.68 &eth; 10
3
80 mm
z
3
240 mm
m 2600 kg/m D 2.00 kg
x
The x coordinate of the center of mass is
xD
x 1 m1 C x 2 m2
40 mm 2.00 kg C 200 mm 5.99 kg
D 160 mm
D
m 1 C m2
2.00 kg C 5.99 kg
x D 160 mm
y
Problem 7.108 The cylindrical tube is made of
aluminum with mass density 2700 kg/m3 . The cylindrical plug is made of steel with mass density
7800 kg/m3 . Determine the coordinates of the center of
mass of the composite object.
z
x
y
y
Tube
A
Plug
20 mm
x
z
35 mm
100
mm
100
mm
A
Section A-A
Solution: The volume of the aluminum tube is
VAl D 0.0352 0.022 0.2 D 5.18 &eth; 104 m3 .
The mass of the aluminum tube is mAl D 2700VAl D 1.4 kg. The
centroid of the aluminum tube is xAL D 0.1 m, yAl D zAl D 0.
The volume of the steel plug is VFe D 0.022 0.1 D 1.26 &eth;
104 m3 . The mass of the steel plug is mFe D 7800VFe D
0.9802 kg. The centroid of the steel plug is xFe D 0.15 m, yFe D
zFe D 0.
The composite mass is m D 2.38 kg. The composite centroid is
xD
mAl 0.1 C mFe 0.15
D 0.121 m
m
yDzD0
574
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Problem 7.109 In Example 7.19, suppose that the
object is redesigned so that the radius of the hole in
the hollow cylinder is increased from 2 in to 3 in. What
is the x coordinate of the center of mass of the object?
Front View
Side View
y
y
2 in
x
z
4 in
10 in
5 in
5 in
12 in
Solution: The volume of the cylinder is
Vcylinder D 12 in [4 in2 3 in2 ] D 264 in3 D 0.153 ft3
Its weight is
Wcylinder D 0.153 ft3 530 lb/ft3 D 80.9 lb.
The x coordinate of the center of mass (same as center of weight) is
xD
D
x bar Wbar C x cylinder Wcylinder
Wbar C Wcylinder
1.86 in 15.6 lb C 10 in 80.9 lb
D 8.68 in.
15.6 lb C 80.9 lb
x D 8.68 in.
Problem 7.110 A machine consists of three parts. The
masses and the locations of the centers of mass of two
of the parts are:
Part
1
2
Mass (kg)
2.0
4.5
x (mm)
100
150
y (mm)
50
70
z (mm)
20
0
The mass of part 3 is 2.5 kg. The design engineer wants
to position part 3 so that the center of mass location
of the machine is x D 120 mm, y D 80 mm, and z D 0.
Determine the necessary position of the center of mass
of part 3.
Solution: The composite mass is m D 2.0 C 4.5 C 2.5 D 9 kg.
The location of the third part is
x3 D
1209 2100 4.5150
D 82 mm
2.5
y3 D
809 250 4.570
D 122 mm
2.5
z3 D
220
D 16 mm
2.5
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
575
Problem 7.111 Two views of a machine element
are shown. Part 1 is aluminum alloy with density
2800 kg/m3 , and part 2 is steel with density 7800 kg/m3 .
Determine the coordinates of its center of mass.
y
y
1
24 mm
2
8 mm
18 mm
x
20
mm
60 mm
8 mm
z
16
mm
50 mm
Solution: The volumes of the parts are
V1 D 6048 C 12 242 82 50
D 179, 186 mm3 D 17.92 &eth; 105 m3 ,
V2 D 1636 C 12 182 82 20 D 17, 678 mm3
D 1.77 &eth; 105 m3 ,
so their masses are
m1 D S1 V1 D 280017.92 &eth; 105 D 0.502 kg,
m2 D S2 V2 D 78001.77 &eth; 105 D 0.138 kg.
The x coordinates of the centers of mass of the parts are x1 D 25 mm,
x2 D 10 mm, so
xD
576
x1 m1 C x2 m2
D 21.8 mm
m1 C m 2
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Problem 7.112 The loads F1 D F2 D 25 kN. The
mass of the truss is 900 kg. The members of the truss are
homogeneous bars with the same uniform cross section.
(a) What is the x coordinate of the center of mass of the
truss? (b) Determine the reactions at A and G.
y
D
F1
3m
F2
B
E
3m
G
A
x
C
4m
4m
Solution:
(a)
The center of mass of the truss is located at the centroid of
the composite line of the axes of the members. The lengths of
the diagonal members are 4 m2 C 3 m2 D 5 m. The lengths
and x coordinates of the centroids of the axes of the members are
Member
AB
AC
BC
BD
BE
BG
CG
DE
EG
Length
5m
4m
3m
5m
4m
5m
4m
3m
3m
x coordinate
2m
2m
4m
6m
6m
6m
6m
8m
8m
The x coordinate of the centroid of the composite line, which is
coincident with the center of mass of the truss, is
xD
xi Li
2 5 C 4 C 4 3 C 6 5 C 4 C 5 C 4 C 8 3 C 3
D
D 5.17 m
Li
5C4C3C5C4C5C4C3C3
x D 5.17 m
(b)
The equilibrium equations for the truss are
Fx : Ax C 25 kN C 25 kN,
Fy : Ay C G 9009.81N D 0,
MA : 25 kN 3 m 25 kN 6 m 900 9.81 N5.17 m C G8 m D 0.
Solving yields
Axc D 50 kN, Ay D 25.0 kN, G D 33.8 kN.
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577
Problem 7.113 With its engine removed, the mass of
the car is 1100 kg and its center of mass is at C. The
mass of the engine is 220 kg.
E
Suppose that you want to place the center of mass
E of the engine so that the center of mass of the
car is midway between the front wheels A and the
rear wheels B. What is the distance b?
(b) If the car is parked on a 15&deg; slope facing up the
slope, what total normal force is exerted by the
road on the rear wheels B?
C
(a)
0.6 m
0.45 m
A
B
1.14 m
b
2.60 m
Solution:
(a)
The composite mass is m D mC C mE D 1320 kg. The xcoordinate of the composite center of mass is given:
xD
2.6
D 1.3 m,
2
from which the x-coordinate of the center of mass of the engine is
xE D b D
1.3 m 1.14 mC D 2.1 m.
mE
The y-coordinate of the composite center of mass is
yD
(b)
0.45 mC C 0.6 mE
D 0.475 m.
m
Assume that the engine has been placed in the new position, as
given in Part (a). The sum of the moments about B is
MA D 2.6A C ymg sin15&deg; 2.6 xmg cos15&deg; D 0,
from which A D 5641.7 N. This is the normal force exerted by
the road on A. The normal force exerted on B is obtained from;
FN D A mg cos15&deg; C B D 0,
from which B D 6866 N
578
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Problem 7.114 The airplane is parked with its landing
gear resting on scales. The weights measured at A, B, and
C are 30 kN, 140 kN, and 146 kN, respectively. After a
crate is loaded onto the plane, the weights measured at A,
B, and C are 31 kN, 142 kN, and 147 kN, respectively.
Determine the mass and the x and y coordinates of the
center of mass of the crate.
B
6m
A
x
Solution: The weight of the airplane is WA D 30 C 140 C 146 D
316 kN. The center of mass of the airplane:
10 m
Myaxis D 3010 xA WA D 0,
from which xA D 0.949 m.
C
6m
y
Mxaxis D 140 1466 C yA WA D 0,
from which yA D 0.114 m. The weight of the loaded plane:
W D 31 C 142 C 147 D 320 kN.
The center of mass of the loaded plane:
Myaxis D 3110 xW D 0,
from which x D 0.969 m.
Mxaxis D 142 1476 C yW D 0,
from which y D 0.0938 m. The weight of the crate is Wc D W WA D 4 kN. The center of mass of the crate:
xc D
Wx WA xA
D 2.5 m,
Wc
yc D
Wy WA yA
D 1.5 m.
Wc
The mass of the crate:
mc D
Wc &eth; 103
D 407.75 kg
9.81
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
579
Problem 7.115 A suitcase with a mass of 90 kg is
placed in the trunk of the car described in Example 7.20.
The position of the center of mass of the suitcase is
x s D 0.533 m, y s D 0.762 m, and z s D 0.305 m. If
the suitcase is regarded as part of the car, what is the
new position of the car’s center of mass?
Solution: In Example 7.20, the following results were obtained
The new center of mass is at
for the car without the suitcase
xN D
Wc D 17303 N
xc Wc C xs Ws
Wc C Ws with similar eqns for yN and zN
xc D 1.651 m
Solving, we get
yc D 0.584 m
xN D 1.545 m, yN D 0.593 m, zN D 0.717 m
zc D 0.769 m
For the suitcase
Ws D 90 g,
y D 0.762 m,
xs D 0.533 m,
z D 0.305 m.
Problem 7.116 A group of engineering students
constructs a miniature device of the kind described in
Example 7.20 and uses it to determine the center of mass
of a miniature vehicle. The data they obtain are shown
in the following table:
Wheelbase = 36 in
Track = 30 in
Left front wheel, NLF
Right front wheel, NRF
Left rear wheel, NLR
Right rear wheel, NRR
˛D0
˛ D 10&deg;
35
32
36
33
27
34
29
30
Determine the center of mass of the vehicle. Use the
same coordinate system as in Example 7.20.
Solution: The weight of the go-cart: W D 35 C 36 C 27 C 29 D
127 lb. The sum of the moments about the z axis
With the go-cart in the tilted position, the sum of the moments about
the z axis
Mzaxis D WheelbaseNLF C NRF xW D 0,
Mzaxis D WheelbaseNLF C NRF C yW sin10&deg; xW cos10&deg; D 0,
from which
xD
3635 C 36
D 20.125 in.
W
The sum of the moments about the x axis:
Mxaxis D zW TrackNRF C NRR D 0,
from which
yD
xW cos10&deg; 3632 C 33
W sin10&deg; D 8.034 in.
from which
zD
580
3036 C 29
D 15.354 in.
W
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 7.117 Determine the centroid of the area by
letting dA be a vertical strip of width dx.
y
(1, 1)
y = x2
x
Solution: The area: The length of the vertical strip is 1 y, so
that the elemental area is dA D 1 y dx D 1 x2 dx. The area:
1
dA D
0
A
1
2
x3
1
1 x2 dx D x D1 D .
3 0
3
3
The x-coordinate:
xA D
1
x dA D
x1 x2 dx D
0
A
x4
x2
2
4
1
D
0
3
1
: xD
4
8
The y-coordinate: The y-coordinate of the centroid of each element of
area is located at the midpoint of the vertical dimension of the area
element.
y D y C 12 1 x2 .
Thus
1
y dA D
x2 C
0
A
D
yD
1
1 x2 1 x2 dx
2
1
1
2
x5
D .
x
2
5 0
5
3
5
Problem 7.118 Determine the centroid of the area in
Problem 7.117 by letting dA be a horizontal strip of
height dy.
Solution: The area: The length of the horizontal strip is x, hence
Divide by the area: x D
the element of area is
3
8
The y-coordinate:
dA D x dy D y 1/2 dy.
yA D
Thus
A
1
0
y 1/2 dy D
1
2y 3/2
3
0
D
2
3
1
y dA D
D
0
1
y y 1/2 dy
0
y 3/2 dy D
2y 5/2
5
Check:
The x-coordinate: The x-coordinate of the centroid of each element of
area is x D 12 x D 12 y 1/2 . Thus
Divide by the area: y D
1
D
0
2
.
5
3
5
1
2 1
1
1
1
1
y
y dy D
D .
y 1/2 dA D
2
2
2 0
4
A 2
0
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
581
y
Problem 7.119 Determine the centroid of the area.
60 cm
x
80 cm
Solution: The strategy is to develop useful general results for the
triangle and the rectangle.
Divide by the area: y D
The rectangle: The area of the rectangle of height h and width w is
xD
w
60 cm
a
20 cm. The composite:
3
xR AR C xT AT
404800 C 1001800
D
AR C AT
4800 C 1800
h dx D hw D 4800 cm2 .
0
D 56.36 cm
The x-coordinate:
w
x2
hx dx D h
2
0
w
0
1
D
hw2 .
2
yD
304800 C 201800
4800 C 1800
D 27.27 cm
Divide by the area: x D
w
D 40 cm
2
The y-coordinate:
w
1
1
h2 dx D
h2 w.
2
2
0
Divide by the area: y D
1
h D 30 cm
2
The triangle: The area of the triangle of altitude a and base b is
(assuming that the two sides a and b meet at the origin)
b
b
yx dx D
0
0
b
a
ax 2
ax
x C a dx D b
2b
0
ab
ab
C ab D
D 1800 cm2
D 2
2
Check: This is the familiar result. check.
The x-coordinate:
b
0
b
a
ax 3
ax2
ab2
C
.
D
x C a x dx D b
3b
2 0
6
Divide by the area: x D
b
D 20 cm
3
The y-coordinate:
y dA D
A
b 2
a
1
x C a dx
2
b
0
D
582
3 b
ba2
b a
.
D
xCa
6a
b
6
0
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
y
Problem 7.120 Determine the centroid of the area.
40 mm
20 mm
Solution: Divide the object into five areas:
40 mm
(1)
(2)
(3)
(4)
(5)
The rectangle 80 mm by 80 mm,
The rectangle 120 mm by 80 mm,
the semicircle of radius 40 mm,
The circle of 20 mm radius, and
the composite object. The areas and centroids:
80 mm
(1)
A1 D 6400 mm2 ,
x1 D 40 mm, y1 D 40 mm,
(2)
A2 D 9600 mm2 ,
x2 D 120 mm, y2 D 60 mm,
(3)
A3 D 2513.3 mm2 ,
x3 D 120 mm, y3 D 136.98 mm,
(4)
A4 D 1256.6 mm2 ,
x4 D 120 mm, y4 D 120 mm.
(5)
The composite area: A D A1 C A2 C A3 A4 D 17256.6 mm2 .
The composite centroid:
x
120 mm
160 mm
xD
A1 x1 C A2 x2 C A3 x3 A4 x4
D 90.3 mm .
A
yD
A1 y1 C A2 y2 C A3 y3 A4 y4
D 59.4 mm
A
y
Problem 7.121 The cantilever beam is subjected to a
triangular distributed load. What are the reactions at A?
Solution: The load distribution is a straight
line with intercept
w D 200 N/m at x D 0, and slope 200
10
D 20 N/m2 . The sum
200 N/m
of the moments is
x
A
10
20x C 200x dx D 0,
M D MA 10 m
0
from which
10
20
D 3333.3 N-m.
MA D x 3 C 100x2
3
0
The sum of the forces:
200 N/m
AX
MA
AY
10 m
10
20x C 200 dx D 0,
Fy D Ay 0
from which
10
Ay D 10x2 C 200x 0 D 1000 N,
and
Fx D Ax D 0
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
583
Problem 7.122 What is the axial load in member BD
of the frame?
C 100 N/m
5m
B
D
5m
A
E
10 m
Solution: The distributed load is two straight lines: Over the
Over the interval 5 y 10, the load is a constant w D 100 N/m.
5
ME D
10
100y dy,
20yy dy C
0
Cy
Cx
interval 0 y 5 the intercept is w D 0 at y D 0 and the slope is
100
D 20.
C
5
Cx
By
Bx
Ay
Cy
Bx
By
Dy
Dx
Dx Dy
Ey
Ex
5
from which
ME D
20 3
y
3
5
C
0
100 2
y
2
10
D 4583.33 N-m.
5
Check: The area of the triangle is
F1 D 12 5100 D 250 N.
The area of the rectangle: F2 D 500 N. The centroid distance for the
triangle is
d1 D 23 5 D 3.333 m.
The centroid distance of the rectangle is d2 D 7.5 m. The moment
ME D d1 F1 C d2 F2 D 4583.33 Nm check.
The Complete Structure: The sum of the moments about E is
M D 10AR C ME D 0,
where AR is the reaction at A, from which AR D 458.33 N.
The element ABC : Element BD is a two force member, hence By D 0.
The sum of the moments about C:
MC D 5Bx 10Ay D 0,
where Ay is equal and opposite to the reaction of the support, from
which
Bx D 2Ay D 2AR D 916.67 N.
Since the reaction in element BD is equal and opposite, Bx D
916.67 N, which is a tension in BD.
584
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 7.123 An engineer estimates that the maximum wind load on the 40-m tower in Fig. a is described
by the distributed load in Fig. b. The tower is supported
by three cables A, B, and C from the top of the tower to
equally spaced points 15 m from the bottom of the tower
(Fig. c). If the wind blows from the west and cables B
and C are slack, what is the tension in cable A? (Model
the base of the tower as a ball and socket support.)
200 N/m
B
N
A
40 m
15 m
C
400 N/m
(a)
Solution: The load distribution is a straight line with the intercept
w D 400 N/m, and slope 5. The moment about the base of the tower
due to the wind load is
40
5y C 400y dy,
MW D
(b)
(c)
200 N/m
θ
40 m
TA
0
40
5
D 213.33 kN-m,
MW D y 3 C 200y 2
3
0
clockwise about the base, looking North. The angle formed by the
cable with the horizontal at the top of the tower is
D 90&deg; tan1
15
40
Fx
Fy
400 N/m
D 69.44&deg; .
The sum of the moments about the base of the tower is
M D MW C 40TA cos D 0,
from which
TA D
1
40 cos MW D 15.19 kN
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
585
Problem 7.124 Determine the reactions on member
ABCD at A and D.
2 kN/m
2 kN/m
D
E
1m
1m
C
1m
B
1m
A
F
1m
Solution: First, replace the distributed forces with equivalent
2 kN / m
concentrated forces, then solve for the loads. Note that BF and CE
are two force members.
y
By area analogy, concentrated load is applied at y D šm. The load is
1
2 (2)(3) kN
3m
F1 D 3 kN
By the area analogy,
F2 D 4 kN applied at x D 1 m
Assume FCE and FBF are tensions
For ABCD:
Fx :
Ax C FBF cos 45&deg; C FCE cos 45&deg; C Dx C 3 kN D 0
Fy :
Ay C Dy FBF sin 45&deg; C FCE sin 45&deg; D 0
x
y
2 kN / m
2m
MA :
1FBF cos 45&deg; 2FCE cos 45&deg; 23 3Dx D 0
For DE:
586
Fx :
Dx FCE cos 45&deg; D 0
Fy :
Dy FCE sin 45&deg; 4 D 0
ME : 1Dy D 0
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
7.124 (Continued )
Solving, we get
Ax D 7 kN
Ay D 6 kN
Dx D 4 kN
Dy D 0
also FBF D 14.14 kN(c)
FCE D 5.66 kN(c)
DY
DX
FCE
45&deg;
3 kN
C
B
2m
FBF
45&deg;
AX
AY
1
4 kN
1m
DX
E
DY
45&deg;
FCE
2
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
587
Problem 7.125 Estimate the centroid of the volume of
the Apollo lunar return configuration (not including its
rocket nozzle) by treating it as a cone and a cylinder.
y
x
12.8 ft
Nozzle
10 ft
14 ft
Solution: The volume of the cone is
V1 D
R2 h
D 428.93 ft3 .
3
The x-coordinate of the centroid from the nose of the cone is x1 D
3h
D 7.5 ft. The volume of the cylinder is V2 D R2 L D 1801.5 ft3 .
4
The x-coordinate of the centroid from the nose of the cone is x2 D
L
h C D 17 ft. The composite volume is V D V1 C V2 D 2230.4 ft3 .
2
The x-coordinate of the composite centroid is
xD
V1 x1 C V2 x2
D 15.2 ft .
V
The y- and z-coordinates are zero, from symmetry.
588
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 7.126 The shape of the rocket nozzle of the
Apollo lunar return configuration is approximated by
revolving the curve shown around the x axis. In terms
of the coordinate system shown, determine the centroid
of the volume of the nozzle.
y
y = 0.350 + 0.435x – 0.035x 2
x
Solution:
x dV
xD .
dV
2.83 m
Let dV be a disk of radius y and thickness dx
Thus, dV D y 2 dx, where
y 2 D 0.350 C 0.435x 0.035x2 2
y 2 D a C bx C cx 2 2
y 2 D a2 C 2abx C 2ac C b2 x 2 C 2bcx3 C c2 x 4
a D 0.350
b D 0.435
c D 0.035
2.83
a2 x C 2abx2 C 2ac C b2 x 3 C 2bcx4 C c2 x 5 dx
0
xD 2.83
a2 C 2abx C 2ac C b2 x 2 C 2bcx3 C c2 x 4 dx
0
4 2.83
3
x
x
C 2ab
C 2ac C b2 3
4 

6
5


x
x
2
C2bc
Cc
5 6
xD
3 2.830

x2
x
2
2
C 2ac C b a x C 2ab
2
3 

5
4


x
x
2
Cc
C2bc
4
5
0

a2
x2
2
Evaluating,
xD
4.43
D 1.87 m
3.37
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
589
y
Problem 7.127 Determine the coordinates of the centroid of the volume.
120 mm
40 mm
100
mm
z
30 mm
20 mm
x
Solution: From symmetry y D z D 0
xD
0.042 0.032 0.120.06 C 0.032 0.022 0.220.11
0.042 0.032 0.12 C 0.032 0.022 0.22
x D 0.0884 m D 88.4 mm, y D z D 0
Problem 7.128 Determine the surface area of the
volume of revolution in Problem 7.127.
5 in
9 in
6 in
Solution: The outer surface: The length of the line is L1 D
p
42 C 62 D 7.2111 in. The y-coordinate of the centroid is y D 5 C
2 D 7 in. The surface area is A1 D 2L1 y1 D 317.16 in2 .
The side surface: The length of the line is L2 D 4 in. The ycoordinate of the centroid is y2 D 5 C 2 D 7 in. The surface area is
A2 D 2L2 y2 D 87.96 in2 .
The inner surface: The length of the line is L3 D 6 in. The y-coordinate
is y3 D 5 in. The surface area is A3 D 2L3 y3 D 188.5 in2 . The total
surface: A D A1 C A2 C A3 D 681.6 in2
590
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 7.129 Determine the y coordinate of the
center of mass of the homogeneous steel plate.
y
20 mm
10 mm
20 mm
20 mm
x
80 mm
Solution: Divide the object into five areas: (1) The lower rectangle
20 by 80 mm, (2) an upper rectangle, 20 by 40 mm, (3) the semicircle
of radius 20 mm, (4) the circle of radius 10 mm, and (5) the composite
part. The areas and the centroids are tabulated. The last row is the
composite and the centroid of the composite. The composite area is
3
Ai A4 .
1
The centroid:
3
xD
,
A
3
and y D
Ai xi A4 x4
1
Ai yi A4 y4
1
A
.
The following relationships were used for the centroids: For a rectangle: the centroid is at half the side and half the base. For a semicircle,
4R
from the base. For a circle,
the centroid is on the centerline and at
3
the centroid is at the center.
Area
A, sq mm
x, mm
y, mm
A1
1600
40
10
A2
800
60
30
A3
628.3
60
48.5
A4
314.2
60
40
48.2
21.3
Composite
2714
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
591
Problem 7.130 Determine the x coordinate of the
center of mass of the homogeneous steel plate.
y
220 mm
x2
Solution: The quarter circle: The equation of the circle is C
y 2 DpR2 . Take the elemental area to be a vertical strip of height
ypD R2 x 2 and width dx, hence the element of area is dA D
R2 x 2 dx, and the area is
R
R2 x 2 dx D
0
150 mm
p
x R
R2
R2
x R2 x 2
C
sin1
D
2
2
R
4
x
0
50 mm
The x-coordinate:
xC A D
A
xC D
R
x dA D
x
0
R
R2 x 2 3/2
R3
R2 x 2 dx D D
3
3
0
4R
3
The rectangle: The area is A D 50150 D 7500 mm2 . The xcoordinate of the centroid is xR D 25 mm.
The composite: The area of the quarter circle is
AC D
2202
D 3.8013 &eth; 104 mm.
4
The area of the rectangle is AR D 50150 D 0.75 &eth; 104 mm2 . The
composite area is A D AC AR D 3.0513 &eth; 104 mm2 . The centroid:
AC xC AR xR
xD
D 110 mm
A
Problem 7.131 The area of the homogeneous plate is
10 ft2 . The vertical reactions on the plate at A and B are
80 lb and 84 lb, respectively. Suppose that you want to
equalize the reactions at A and B by drilling a
1-ft-diameter hole in the plate. What horizontal distance
from A should the center of the hole be? What are the
resulting reactions at A and B?
A
B
5 ft
Solution: The weight of the plate is W D 80 C 84 D 164 lb. From
the sum of moments about A, the centroid is
xD
845
D 2.56 ft.
W
A
The weight density is
wD
5 ft
W
D 16.4 lb/ft2 .
10
The weight of the cutout is WC D 0.52 w D 12.88 lb. The new
weight of the plate is W2 D W WC D 151.12 lb. The new centroid
must be at
x2 D
5
D 2.5 ft for the reactions to be equal.
2
B
X
A
X2
B
XC
Therefore the x-coordinate of the center of the circle will be
xC D
Wx W2 x2
D 3.26 ft.
WC
The reactions at A and B will be
592
W2
D 75.56 lb
2
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 7.132 The plate is of uniform thickness and
is made of homogeneous material whose mass per unit
area of the plate is 2 kg/m2 . The vertical reactions at
A and B are 6 N and 10 N, respectively. What is the x
coordinate of the centroid of the hole?
1m
A
B
2m
Solution: Choose an origin at A. The basic relation is WC xC D
Wx WH xH , where WC is the weight of the composite plate (the one
with the hole), W is the weight of the plate without the hole, WH is
the weight of the material removed from the hole, and xC , x, and xH
are the x-coordinates of the centroids of the composite plate, the plate
without the hole, and the hole, respectively.
The composite weight:
FY D A C B WC D 0,
WC
A
WH
B
XC
XH
2m
from which WC D 16 N. The x-coordinate of the centroid:
MA D WC xC C 2B D 0,
from which xC D 1.25 m. The weight of the plate without the hole
and the x-coordinate of the centroid:
W D Ag D 12 2129.81 D 19.62 N,
and x D 23 2 D 1.3333 m.
The weight of the material removed from the hole:
WH D W WC D 3.62 N.
Solve: xH D
Wx WC xC
D 1.702 m
WH
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
593
y
Problem 7.133 Determine the center of mass of the
homogeneous sheet of metal.
Solution: Divide the object into four parts: (1) The lower plate,
x
(2) the left hand plate, (3) the semicircular plate, and (4) the composite
plate. The areas and centroids are found by inspection:
(1)
Area: A1 D 912 D 108 in2 ,
x1 D 0.5 in, y1 D 8 in, z1 D 6 in.
(2)
A2 D 812 D 96 in2 ,
x2 D 4 in,
y2 D 4 in, z2 D 6 in.
(3)
4 in
8 in
z
12 in
9 in
A3 D 412 D 150.8 in2 ,
24
x3 D 0, y D
D 2.546 in, z D 6 in.
The composite area is
3
Ai D 354.796 in2 .
1
The centroid for the composite:
3
xD
A
3
yD
594
D 0.930 in
Ai yi
1
A
3
zD
Ai xi
1
D 2.435 in
Ai zi
1
A
D 6 in
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 7.134 Determine the center of mass of the
homogeneous object.
60 mm
z
10 mm
y
(1) A triangular solid 30 mm altitude, 60 mm base, and 10 mm thick.
(2) A rectangle 60 by 70 mm by 10 mm. (3) A semicircle with radius
20 mm and 10 mm thick.
The volumes and their centroids are determined by inspection:
(1)
V1 D
y1 D 10 C
z1 D
(2)
30
mm
z
x
x
y
1
306010 D 9000 mm3 ,
2
x1 D 5 mm,
60 mm
20 mm
Solution: Divide the object into three parts and the composite:
30 mm
z
10 mm
30
D 20 mm,
3
60
D 20 mm
3
V2 D 607010 D 42000 mm3 ,
x2 D 35 mm,
y2 D 5 mm,
z2 D 30 mm
(3)
V3 D
202
10 D 6283.2 mm3 ,
2
x3 D 70 420
D 61.51 mm,
3
y3 D 5 mm,
z3 D 30 mm.
The composite volume is V D V1 C V2 V3 D 44716.8 mm3 . The
centroid is
xD
V1 x1 C V2 x2 V3 x3
D 25.237 mm
V
yD
V1 y1 C V2 y2 V3 y3
D 8.019 mm
V
zD
V1 z1 C V2 z2 V3 z3
D 27.99 mm
V
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
595
Problem 7.135 Determine the center of mass of the
homogeneous object.
5 in
1.5 in
y
x
z
y
Solution: Divide the object into five parts plus the composite.
(1) A solid cylinder with 1.5 in radius, 3 in long. (2) A rectangle 3
by 5 by 1 in (3) A solid cylinder with radius 1.5 in, 2 in long. (4) A
semicircle with radius 1.5 in, 1 inch thick, (5) a semicircle with radius
1.5 in, 1 inch thick. The volumes and centroids are determined by
inspection. These are tabulated:
Part No
Vol, cu in
x, in
V1
21.205
0
V2
15
2.5
V3
14.137
5
V4
3.534
0.6366
3.534
V5
Composite
43.27
y, in
z, in
1
0
0
0
0.5
0
0
0
4.363
0
0
2.09
0.3267
0
Top View
z
1 in
x
3 in
2 in
x
Side View
The composite is
VD
3
Vi 5
1
Vi .
4
The centroid:
3
xD
Vi xi 1
5
4
V
Vi xi
,
with a corresponding expression for y. The z-coordinate is zero because
of symmetry.
596
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 7.136 The arrangement shown can be used to
determine the location of the center of mass of a person.
A horizontal board has a pin support at A and rests on
a scale that measures weight at B. The distance from A
to B is 2.3 m. When the person is not on the board, the
scale at B measures 90 N.
(a)
(b)
y
When a 63-kg person is in position (1), the scale
at B measures 496 N. What is the x coordinate of
the person’s center of mass?
When the same person is in position (2), the scale
measures 523 N. What is the x coordinate of his
center of mass?
B
x
A
(1)
y
B
x
A
(2)
Solution:
(a)
WB
1.15 m
W D mg D 63 g D 618 N
Fy :
Ay C By WB D 0
MA : 1.15WB C 2.3By D 0
X
Fy :
Ay C By WB W D 0
By = 90 N
W
WB
Solving, Ay D 90 N, WB D 180 N
1.15 m
(b)
2.3 m
Ay
Ay
2.3 m
By
MA : 2.3By 1.15WB xW D 0
W D 618 N, WB D 180 N
For (a), By D 496 N. Solving the equations for this case yields
x D 1.511 m
For (b), By D 523 N. Solving the equations for this case yields
x D 1.611 m
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
597
Problem 7.137 If a string is tied to the slender bar at
A and the bar is allowed to hang freely, what will be the
angle between AB and the vertical?
B
4 in
A
8 in
Solution: When the bar hangs freely, the action line of the weight
will pass through the mass center. With a homogenous, slender bar,
the mass center corresponds to the centroid of the lines making up
the bar. Choose the origin at A, with the x axis parallel to the lower
bar. Divide the bar into three segments plus the composite: (1) The
segment from A to the semi circle, (2) the segment AB, and (3) the
semicircle.
(1)
L1 D 8 in, x1 D 4 in, y1 D 0.
(2)
L2 D
(3)
L3 D 4 D 12.566 in, x3 D 8 C
p
82 C 82 D 11.314 in, x2 D 4, y2 D 4
24
D 10.546 in, y3 D 4.
The composite length
LD
3
Li D 31.88 in.
1
The composite centroid:
xD
L1 x1 C L2 x2 C L3 x3
D 6.58 in,
L
yD
L1 y1 C L2 y2 C L3 y3
D 2.996 in
L
The angle from the point A to the centroid relative to the lower bar is
˛ D tan1
y
x
D 24.48&deg; .
The angle between AB and the lower bar is 45&deg; , hence the angle
between the line from A to the centroid and AB is
ˇ D 45 ˛ D 20.52&deg;
Since the line from A to the centroid will be vertical, this is the angle
between AB and the vertical.
598
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
y
Problem 7.138 When the truck is unloaded, the total
reactions at the front and rear wheels are A D 54 kN
and B D 36 kN. The density of the load of gravel is
D 1600 kg/m3 . The dimension of the load in the z
direction is 3 m, and its surface profile, given by the
function shown, does not depend on z. What are the
total reactions at the front and rear wheels of the loaded
truck?
y = 1.5 – 0.45x + 0.062x2
x
A
B
2.8 m
3.6 m
5.2 m
Solution: First, find the location of the center of mass of the
unloaded truck (and its mass). Then find the center of mass and mass
Fx : no forces
Fy :
54000 C 36000 mT g D 0N
MA :
xT mT g C 5.236 D 0
Fx : no forces
Fy : Ay C By mT g mL g D 0
MA : 5.2By xT mT g dL mL g D 0
Solving Ay D 80.7 kN, By D 171.6 kN
Solving xT D 2.08 m, mT D 9174 kg
Next, find xL and mL (for the load)
x dm
Num
m
D
xL D L
mL
dm
XT
5.2 m
mL
dm, Num D
mL
0
mL g
y
3.6
3y dx D 3
mL D
x dm
mL
3.6
B
36 kN
54 kN
where mL D
mTg
2
1.5 0.45x C 0.062x dx
y = 1.5 − 0.45x + 0.062x2
dm = p(3)y dx
0
3 3.6
2
x
x
C 0.062
mL D 3 1.5x 0.45
2
3
0
XL
mL D 16551 kg
3.6
Num D 3
1.5x 0.45x2 C 0.062x3 dx
0
3.6 m
x
0
2
3
4 3.6
x
x
x
Num D 3 1.5
0.45
C 0.062
2
3
4
0
dL
XT
Num D 25560 kg &ETH; m
xL D
mL g
mT g
Num
D 1.544 m
mL
measured from the front of the load
AY
5.2 m
BY
The horizontal distance from A to the center of mass of the load is
dL D xL C 2.8 m D 4.344 m
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
599
Problem 7.139 The mass of the moon is 0.0123 times
the mass of the earth. If the moon’s center of mass is
383,000 km from the center of mass of the earth, what
is the distance from the center of mass of the earth to
the center of mass of the earth–moon system?
Solution:
mE
mM
xmE C mM D 383,000 mM so
xD
mM
383,000
mE C m M
X
mM /mE
383,000
D
1 C mM /mE
D
383,000 km
0.0123
383,000
1 C 0.0123
D 4650 km.
(The earth’s radius is 6370 km, so the center of mass of the earth-moon
system is within the earth.)
600
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 8.1 Use the method described in Active
Example 8.1 to determine IY and ky for the rectangular
area.
Solution: The height of the vertical strip of width dx is 0.6 m, so
the area is dA D (0.6 m) dx.
We can use this expression to determine Iy .
y
x 2 dx D (0.6 m)
Iy D
0.4 m
x 2 dx
0.2m
A
D (0.6 m)
x3
3
0.4
m
D 0.0416 m4
0.2 m
0.6 m
ky D
Iy
D
A
0.0416 m4
D 0.416 m
(0.4 m) (0.6 m)
x
0.2 m
Iy D 0.0416 m4 , ky D 0.416 m
0.4 m
Problem 8.2 Use the method described in Active
Example 8.1 to determine Ix and kx for the rectangular
area.
y
Solution: It was shown in Active Example 8.1 that the moment
of inertia about the x axis of a vertical strip of width dx and height
f(x) is
1
Ix strip D [fx]3 dx.
3
For the rectangular strip, fx D 0.6 m. Integrating to determine Ix
for the rectangular area.,
0.4m
Ix D
0.2m
1
1
4
0.6 m3 dx D 0.6 m3 [x]0.4m
0.2m D 0.0288 m
3
3
0.6 m
kx D
Ix
D
A
0.0288 m4
D 0.346 m
(0.4 m)(0.6 m)
x
0.2 m
Ix D 0.0288 m4 , ky D 0.346 m
0.4 m
Problem 8.3 In Active Example 8.1, suppose that the
triangular area is reoriented as shown. Use integration
to determine Iy and ky .
Solution: The height of a vertical strip of width dx is h h/bx,
so the area
dA D
h
h
x
b
y
h
dx.
We can use this expression to determine Iy :
b
x 2 dA D
Iy D
x2 h 0
A
h
x
b
dx D h
x3
3
x4
4b
x
b
D
0
1 3
hb
12
b
ky D
Iy D
Iy
D
A
1 3
hb
b
12
D p
1
6
hb
2
hb3
b
, ky D p .
12
6
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
601
Problem 8.4 (a) Determine the moment of inertia Iy
of the beam’s rectangular cross section about the y axis.
Solution:
(b) Determine the moment of inertia Iy 0 of the beam’s
values, show that Iy D Iy 0 C d2x A, where A is the area of
the cross section.
(a)
40 mm
0
x 2 dydx D 1.28 &eth; 106 mm4
0
(b)
60 mm
Iy D
20 mm
30 mm
Iy 0 D
20 mm
x 2 dydx D 3.2 &eth; 105 mm4
30 mm
y
Iy D Iy 0 C dx 2 A
y
dx
1.28 &eth; 106 mm4
D 3.2 &eth; 105 mm4 C 20 mm2 [40 mm60 mm]
60 mm
x
O
dy
x
O
40 mm
Problem 8.5 (a) Determine the polar moment of
inertia JO of the beam’s rectangular cross section about
the origin O.
(b) Determine the polar moment of inertia JO0 of the
beam’s cross section about the origin O0 . Using your
numerical values, show that JO D JO0 C d2x C d2y A,
where A is the area of the cross section.
Solution:
(a)
40 mm
0
(b)
60 mm
JO D
x 2 C y 2 dydx D 4.16 &eth; 106 mm4
0
20 mm
30 mm
JO 0 D
20 mm
x 2 C y 2 dydx D 1.04 &eth; 106 mm4
30 mm
JO D JO0 C dx 2 C dy 2 A
(c)
4.16 &eth; 106 mm4 D 1.04 &eth; 106 mm4 C [20 mm2
C 30 mm2 ][40 mm60 mm]
Problem 8.6 Determine Iy and ky .
y
Solution:
A D 0.3 m1 m C
1 m 0.3 mC0.3x
Iy D
0
0.6 m
0.3 m
ky D
x
1
0.3 m1 m D 0.45 m2
2
0
Iy
D
A
x 2 dydx D 0.175 m4
0.175 m4
D 0.624 m
0.45 m2
1m
602
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 8.7 Determine JO and kO .
Solution:
A D 0.3 m1 m C
1m
0.3 mC0.3x
JO D
0
1
0.3 m1 m D 0.45 m2
2
x 2 C y 2 dydx D 0.209 m4
0
0.209 m4
D 0.681 m
0.45 m2
kO D
Problem 8.8 Determine Ixy .
Solution:
1m
0.3 mC0.3x
Ixy D
0
xydydx D 0.0638 m4
0
Problem 8.9 Determine Iy .
y
Solution: The height of a vertical strip of width dx is 2 x2 , so
y 2 x2
the area is
dA D 2 x2 dx.
We can use this expression to determine Iy :
x 2 dA D
Iy D
1
x 2 2 x2 dx D
0
A
2x3
x5
3
5
1
0
D 0.467.
Iy D 0.467.
1
Problem 8.10 Determine Ix .
Solution: It was shown in Active Example 8.1 that the moment
of inertia about the x axis of a vertical strip of width dx and height
fx is
1
Ix strip D [fx]3 dx.
3
x
y
y 2 x2
In this problem fx D 2 x3 . Integrating to determine Ix for the area,
1
Ix D
0
D
D
1
3
1
3
1
2 x2 3 dx
3
1
8 12x2 C 6x4 x 6 dx
0
1
12x3
6x5
x7
8x C
D 1.69.
3
5
7 0
1
x
Ix D 1.69.
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
603
Problem 8.11 Determine JO .
y
y 2 x2
Solution: See the solutions to Problems 8.9 and 8.10. The polar
moment of inertia is
JO D Ix C Iy D 1.69 C 0.467 D 2.15.
JO D 2.15.
1
Problem 8.12 Determine Ixy .
x
y
y 2 x2
Solution: It was shown in Active Example 8.1 that the product of
inertia of a vertical strip of width dx and height f(x) is
Ixy strip D
1
[fx]2 xdx.
2
In this problem, fx D 2 x2 . Integrating to determine Ixy for the
area,
1
1
2 x2 2 xdx
2
Ixy D
0
D
D
1
2
1
4x 4x3 C x 5 x
0
1
4x4
x6
1 4x2
C
D 0.583.
2 2
4
6 0
1
x
Ixy D 0.583.
Problem 8.13 Determine Iy and ky .
y
y
1 2
x 4x 7
4
Solution: First we need to locate the points where the curve intersects the x axis.
4 š
1
x 2 C 4x 7 D 0 ) x D
4
Now
14
x
x2 /4C4x7
dydx D 72
2
0
14
x2 /4C4x7
Iy D
2
ky D
604
p
16 41/47
D 2,14
21/4
0
Iy
D
A
x 2 dydx D 5126
5126
D 8.44
72
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 8.14 Determine Ix and kx .
Solution: See Solution to Problem 8.13
14
x2 /4C4x7
Ix D
2
0
Ix
D
A
kx D
y 2 dydx D 1333
1333
D 4.30
72
Problem 8.15 Determine JO and kO .
Solution: See Solution to 8.13 and 8.14
JO D Ix C Iy D 1333 C 5126 D 6459
JO
6459
D
D 9.47
kO D
A
72
Problem 8.16 Determine Ixy .
Solution:
14
x2 /4C4x7
xydydx D 2074
Ixy D
2
0
Problem 8.17 Determine Iy and ky .
y
y
1 2
x 4x 7
4
y5
Solution: First we need to locate the points where the curve intersects the line.
4 š
1 2
x C 4x 7 D 5 ) x D
4
12
p
16 41/412
D 4,12
21/4
x
x2 /4C4x7
dydx D 21.33
4
5
12
x2 /4C4x7
Iy D
4
ky D
5
Iy
D
A
x 2 dydx D 1434
1434
D 8.20
21.33
Problem 8.18 Determine Ix and kx .
Solution: See Solution to Problem 8.17
12
x2 /4C4x7
Ix D
4
kx D
5
Ix
D
A
y 2 dydx D 953
953
D 6.68
21.33
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
605
Problem 8.19 (a) Determine Iy and ky by letting dA
be a vertical strip of width dx.
(b) The polar moment of inertia of a circular area with
its center at the origin is JO D 12 R4 . Explain how you
y
x
R
Solution:
The equation of the circle is x2 C y 2 D R2 , from which
p
y D š pR2 x 2 . The strip dx wide and y long has the elemental area
dA D 2 R2 x 2 dx. The area of the semicircle is
R2
Iy D
2
R
x 2 dA D 2
x2
R2 x 2 dx
0
A
R
R2 xR2 x 2 1/2
R4
x
xR2 x 2 3/2
C
C
sin1
D2 4
8
8
R 0
(b) If the integration were done for a circular area with the center at
the origin, the limits of integration for the variable x would be from
R to R, doubling the result. Hence, doubling the answer above,
Iy D
R4
.
4
By symmetry, Ix D Iy , and the polar moment would be
JO D 2Iy D
D
R4
8
ky D
R4
,
2
which is indeed the case. Also, since kx D ky by symmetry for the full
circular area,
R
Iy
D
A
2
kO D
Iy
Ix
C
D
A
A
2
Iy
D
A
JO
A
as required by the definition. Thus the result checks.
Problem 8.20 (a) Determine Ix and kx for the area
in Problem 8.19 by letting dA be a horizontal strip of
height dy.
(b) The polar moment of inertia of a circular area with
its center at the origin is JO D 12 R4 . Explain how you
Solution: Use the results of the solution to Problem 8.19, A D
R2
. The equation for the circle is x2 C y 2 D R2 , from which x D
2
š R2 y 2 . The horizontal strip is from 0 to R, hence the element of
area is
CR
y 2 dA D
Ix D
A
y2
D
kx D
606
and JO D 2Ix D
R2 y 2 dy
R
R4
R4
C
D
8 2
8 2
R4
.
4
R
y
R2 yR2 y 2 1/2
R4
yR2 y 2 3/2
C
C
sin1
D 4
8
8
R R
Ix D
By symmetry Iy D Ix ,
R2 y 2 dy.
dA D
(b) If the area were circular, the strip would be twice as long, and the
moment of inertia would be doubled:
R4
R4
,
2
which is indeed the result. Since kx D ky by symmetry for the full
circular area, the
kO D
Iy
Ix
C
D
A
A
2
Ix
D
A
JO
A
8
as required by the definition. This checks the answer.
R
Ix
D .
A
2
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 8.21 Use the procedure described in
Example 8.2 to determine the moment of inertia Ix
and Iy for the annular ring.
y
Ro
x
Ri
Solution: We first determine the polar moment of inertia JO by
integrating in terms of polar coordinates. Because of symmetry and
the relation JO D Ix C Iy , we know that Ix and Iy each equal 12 JO .
Integrating as in Example 8.2, the polar moment of inertia for the
annular ring is
r 2 dA D
JO D
A
Therefore Ix D Iy D
Ro
r 2 2rdr D
Ri
1
Ro4 Ri4 2
1
Ro4 Ri4 4
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
607
y
Problem 8.22 What are the values of Iy and ky for the
elliptical area of the airplane’s wing?
y2
x2
21
b
a2
Solution:
a
x 2 dA D
Iy D
A
0
a
y
y
x 2 dy dx
y
0
a
Iy D 2
0
5m
x2
b1 2 1/2
a
[x2 y]0
y
dx
0
1/2
x2
x2 b 1 2
dx
a
a
x
2m
x 2 dy dx
Iy D 2
0
a
Iy D 2
x2 + y2 = 1
a2 b2
2b
x
x2
Iy D 2b
1
0
x2
a2
dx
a
Rewriting
Iy D
2b
a
a
x2
y = b 1– x2
a2
a2 x 2 dx
0
p
xa2 x 2 3/2
a2 x a2 x 2
2b
Iy D
C
a
4
8
x
a4
C sin1
8
a
a
0
D
a
D

0
p 0

2b  aa2 a2 3/2 a3 a2 a2
Iy D
C

a 
4
8



a  
a4
sin1

8
a
x2
b 1
dx
a
2b
a
a
a2 x 2 1/2 dx
0
p
a
x
x a2 x 2
a2
C
sin1
2
2
a
0
(from the integral tables)
C
2b
a
0
p
a2
a
2b
a 0
C
sin1
D
a
2
2
a
0
0a2 3/2
p
a2
0
0 a
C
sin1
2
2
a
4

0p

a2 &ETH; 0 a2 a4 01 0 
C
C sin


8
8
a

ab
2b a2 D
a 2 2
2
Evaluating, we get
A D 7.85 m2
Iy D
a4
2b
a 8 2
Finally
Iy
D
A
2a3 b
Iy D
8
ky D
Evaluating, we get
ky D 2.5 m
49.09
7.85
Iy D 49.09 m4
The area of the ellipse (half ellipse) is
a
x2
b 1 a
0
608
1/2
dy dx
0
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 8.23 What are the values of Ix and kx for the
elliptical area of the airplane’s wing in Problem 8.22?
Solution:
y
y 2 dA D 2
Ix D
a
Ix D 2
0
b
3 a
y
3
p
b
yD a
y2
x2
—
+ —2 = 1
a2
b
p
a2 x2
y 2 dy dx
0
0
A
2b
a
y = b a2 – x2
a
dx
0
b3 2
a x 2 3/2 dx
3a3
Ix D 2
0
2b3
3a3
p
3a2 x a2 x 2
xa2 x 2 3/2
C
4
8
3 4 1 x
a sin
8
a
C
2b3
Ix D 3
3a
3
a4
8
2
Ix D
2b3
&ETH;
3a3
Ix D
3ab3 ab3 D
3.8
8
a
Evaluating (a D 5, b D 1)
0
p
3a3 0
3 a0
C
C a4
4
8
8 2
Ix D
p
3a2 &ETH; 0 a2
0a2 C0
4
8
5
D 1.96 m4
8
From Problem 8.22, the area of the wing is A D 7.85 m2
x
a2 x2
a
Ix D
a
kx D
Ix
D
A
1.96
7.85
Problem 8.24 Determine Iy and ky .
kx D 0.500 m
y
Solution: The straight line and curve intersect where x D x2 20.
Solving this equation for x, we obtain
xD
1š
y = x 2 – 20
p
1 C 80
D 4, 5.
2
y=x
If we use a vertical strip: the area
x
dA D [x x2 20] dx.
Therefore
A
D
5
x 2 dA D
Iy D
x 2 x x 2 C 20 dx
4
x4
x5
20x3
C
4
5
3
5
D 522.
The area is
y = x2 – 20
A
ky D
y=x
x x 2 C 20 dx
x
4
x2
x3
C 20x
2
3
So
5
dA D
D
y
4
Iy
D
A
dA
5
D 122.
4
dx
522
D 2.07.
122
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
609
Problem 8.25 Determine Ix and kx for the area in
Problem 8.24.
Solution: Let us determine the moment of inertia about the x axis
of a vertical strip holding x and dx fixed:
y 2 dAs D
Ix strip D
As
x
y
y 2 dx dy D dx
x2 20
y3
3
x
y=x
y=
x2 20
x2 –
20
dAs
x
dx
x 6 C 60x4 C x 3 1200x2 C 8000.
3
D
Integrating this value from x D 4 to x D 5 (see the solution to
Problem 8.24), we obtain Ix for the entire area:
5
x
1
x 6 C 60x4 C x 3 1200x2 C 8000 dx
3
Ix D
4
dx
7
5
x
x4
400x3
8000x
D D 10,900.
C 4x5 C
C
21
12
3
3
4
From the solution to Problem 8.24, A D 122 so
kx D
Ix
D
A
10,900
D 9.45.
122
Problem 8.26 A vertical plate of area A is beneath the
surface of a stationary body of water. The pressure of
the water subjects each element dA of the surface of the
plate to a force p0 C y dA, where p0 is the pressure
at the surface of the water and is the weight density
of the water. Show that the magnitude of the moment
about the x axis due to the pressure on the front face of
the plate is
Mx
x
A
y
D p0 yA C Ix ,
axis
where y is the y coordinate of the centroid of A and Ix
is the moment of inertia of A about the x axis.
Solution: The moment about the x axis is dM D yp0 C y dA
integrating over the surface of the plate:
p0 C yy dA.
MD
A
Noting that p0 and are constants over the area,
y dA C M D p0
y 2 dA.
A
By definition,
y dA
A
yD
A
y 2 dA,
and Ix D
A
then M D p0 yA C IX , which demonstrates the result.
610
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 8.27 Using the procedure described in Active
Example 8.3, determine Ix and kx for the composite area
by dividing it into rectangles 1 and 2 as shown.
Solution: Using results from Appendix B and applying the
parallel-axis theorem, the moment on inertia about the x axis for area
1 is
Ix 1 D Ix C d2y A D
y
1m
1
4m
2
1m
1
(1 m)(3 m)3 C (2.5 m)[(1 m)(3 m)]
12
x
3m
D 21.0 m4
The moment of inertia about the x axis for area 2 is
Ix 2 D
1
(3 m)(1 m)3 D 1 m4 .
3
The moment of inertia about the x axis for the composite area is
Ix D Ix 1 C Ix 2 D 22.0 m4 .
kx D
Ix
D
A
22.0 m4
D 1.91 m
6 m2
Ix D 22.0 m4 , kx D 1.91 m.
Problem 8.28 Determine Iy and ky for the composite
area by dividing it into rectangles 1 and 2 as shown.
Solution: Using results from Appendix B, the moment of inertia
about the y axis for area 1 is
Iy 1 D
1m
1
4m
1
(3 m)(1 m)3 D 1 m4 .
3
The moment of inertia about the y axis for area 2 is
Iy 2 D
y
2
1m
3m
x
1
(1 m)(3 m)3 D 9 m4 .
3
The moment of inertia about the y axis for the composite area is
Iy D Iy 1 C Iy 2 D 10 m4 .
ky D
Iy
D
A
10 m4
D 1.29 m
6 m2
Iy D 10 m4 , kx D 1.29 m.
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
611
Problem 8.29 Determine Ix and kx .
y
Solution: Break into 3 rectangles
Ix D
1
1
0.60.23 C
0.20.63 C 0.20.60.52
3
12
0.8 m
1
0.80.23 C 0.80.20.92 D 0.1653 m4
12
C
0.2 m
A D 0.20.6 C 0.60.2 C 0.80.2 D 0.4 m2
0.6 m
kx D
Ix
D
A
0.1653 m4
D 0.643 m
0.4 m2
0.2 m
x
Ix D 0.1653 m4
)
0.2 m
kx D 0.643 m
0.6 m
Problem 8.30 In Example 8.4, determine Ix and kx for
the composite area.
y
Solution: The area is divided into a rectangular area without the
20 mm
cutout (part 1), a semicircular areas without the cutout (part 2), and
the circular cutout (part 3).
x
Using the results from Appendix B, the moment of inertia of part 1
Ix 1 D
1
120 mm80 mm3 D 5.12 &eth; 106 mm4 ,
12
40 mm
120 mm
the moment of inertia of part 2 is
Ix 2 D
1
40 mm4 D 1.01 &eth; 106 mm4 ,
8
and the moment of inertia of part 3 is
Ix 3 D
1
20 mm4 D 1.26 &eth; 105 mm4 .
4
The moment of inertia of the composite area is
Ix D Ix 1 C Ix 2 C Ix 3 D 6.00 &eth; 106 mm4 .
From Example 8.4, the composite area is A D 1.086 &eth; 104 mm4 , so
kx D
Ix
D
A
6.00 &eth; 106 mm4
D 23.5 mm.
1.086 &eth; 104 mm2
Ix D 6.00 &eth; 106 mm4 , kx D 23.5 mm.
612
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 8.31 Determine Ix and kx .
y
0.8 m
0.2 m
x
0.6 m
0.2 m
0.2 m
0.6 m
Solution: Break into 3 rectangles — See 8.29
First locate the centroid
dD
0.60.20.1 C 0.20.60.5 C 0.80.20.9
D 0.54 m
0.60.2 C 0.20.6 C 0.80.2
1
0.20.63 C 0.20.6d 0.52
12
1
0.60.23 C 0.60.2d 0.12
C
12
1
0.80.23 C 0.80.20.9 d2 D 0.0487 m4
C
12
Ix
0.0487 m4
D
kx D
D 0.349 m
A
0.4 m2
Ix D
d
Problem 8.32 Determine Iy and ky .
Solution: Break into 3 rectangles — See 8.29
Iy D
1
1
1
0.60.23 C
0.20.63 C
0.20.83
12
12
12
D 0.01253 m4
Iy
0.01253 m4
ky D
D 0.1770 m
D
A
0.4 m2
Problem 8.33 Determine JO and kO .
Solution: See 8.29, 8.31 and 8.32
JO D Ix C Iy D 0.0612 m4
JO
0.0612 m4
D
KO D
D 0.391 m
A
0.4 m2
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
613
y
Problem 8.34 If you design the beam cross section so
that Ix D 6.4 &eth; 105 mm4 , what are the resulting values
of Iy and JO ?
h
Solution: The area moment of inertia for a triangle about the
x
base is
Ix D
1
12
bh3 ,
h
from which Ix D 2
1
12
60h3 D 10h3 mm4 ,
30
mm
30
mm
Ix D 10h3 D 6.4 &eth; 105 mm4 ,
from which h D 40 mm.
Iy D 2
1
12
2h303 D
from which Iy D
1
h303 3
1
40303 D 3.6 &eth; 105 mm4
3
and JO D Ix C Iy D 3.6 &eth; 105 C 6.4 &eth; 105 D 1 &eth; 106 mm4
614
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
y
Problem 8.35 Determine Iy and ky .
160
mm
Solution: Divide the area into three parts:
Part (1): The top rectangle.
A1 D 16040 D 6.4 &eth; 103 mm2 ,
dx1
160
D 80 mm,
D
2
Iyy1 D
1
12
40 mm
200
mm
40
mm
40 mm
x
120
mm
401603 D 1.3653 &eth; 107 mm4 .
From which
Iy1 D d2x1 A1 C Iyy1 D 5.4613 &eth; 107 mm4 .
Part (2): The middle rectangle:
A2 D 200 8040 D 4.8 &eth; 103 mm2 ,
dx2 D 20 mm,
Iyy2 D
1
12
120403 D 6.4 &eth; 105 mm4 .
From which,
Iy2 D d2x2 A2 C Iyy2 D 2.56 &eth; 106 mm4 .
Part (3) The bottom rectangle:
A3 D 12040 D 4.8 &eth; 103 mm2 ,
dx3 D
120
D 60 mm,
2
Iyy3 D
1
12
401203 D 5.76 &eth; 106 mm4
From which
Iy3 D d2X3 A3 C Iyy3 D 2.304 &eth; 107 mm4
The composite:
Iy D Iy1 C Iy2 C Iy3 D 8.0213 &eth; 107 mm4
ky D
Iy
D 70.8 mm.
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
615
Problem 8.36 Determine Ix and kx .
Solution: Use the solution to Problem 8.35. Divide the area into
from which
three parts:
Part (1): The top rectangle.
A1 D 6.4 &eth; 103 mm2 ,
Ix2 D d2y2 A2 C Ixx2 D 5.376 &eth; 107 mm4
Part (3) The bottom rectangle:
A3 D 4.8 &eth; 103 mm2 ,
dy1 D 200 20 D 180 mm,
Ixx1 D
1
12
dy3 D 20 mm,
160403 D 8.533 &eth; 105 mm4 .
Ixx3 D
From which
Ix1 D d2y1 A1 C Ixx1 D 2.082 &eth; 108 mm4
Part (2): The middle rectangle:
A2 D 4.8 &eth; 103 mm2 ,
dy2
120
C 40 D 100 mm,
D
2
Ixx2 D
1
12
1
12
120403 D 6.4 &eth; 105 mm4
and Ix3 D d2y3 A3 C Ixx3 D 2.56 &eth; 106 mm4 .
The composite:
Ix D Ix1 C Ix2 C Ix3 D 2.645 &eth; 108 mm4
kx D
Ix
D 128.6 mm
A1 C A2 C A3 401203 D 5.76 &eth; 106 mm4
Problem 8.37 Determine Ixy .
Solution: (See figure in Problem 8.35). Use the solutions in
from which
Problems 8.35 and 8.36. Divide the area into three parts:
Part (1): A1 D 16040 D 6.4 &eth; 103 mm2 ,
dx1
160
D 80 mm,
D
2
dy1 D 200 20 D 180 mm,
Ixy2 D dx2 dy2 A2 D 9.6 &eth; 106 mm4 .
Part (3): A3 D 12040 D 4.8 &eth; 103 mm2 ,
dx3 D
120
D 60 mm,
2
dy3 D 20 mm,
Ixxyy1 D 0,
from which
from which
Ixy1 D dx1 dy1 A1 C Ixxyy1 D 9.216 &eth; 107 mm4 .
Ixy3 D dx3 dy3 A3 D 5.76 &eth; 106 .
The composite:
Part (2) A2 D 200 8040 D 4.8 &eth; 103 mm2 ,
Ixy D Ixy1 C Ixy2 C Ixy3 D 1.0752 &eth; 108 mm4
dx2 D 20 mm,
dy2 D
616
120
C 40 D 100 mm,
2
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 8.38 Determine Ix and kx .
y
Solution: The strategy is to use the relationship Ix D d2 A C Ixc ,
160
mm
where Ixc is the area moment of inertia about the centroid. From this
Ixc D d2 A C Ix . Use the solutions to Problems 8.35, 8.36, and 8.37.
Divide the area into three parts and locate the centroid relative to the
coordinate system in the Problems 8.35, 8.36, and 8.37.
40 mm
x
200
mm
40
mm
Part (1) A1 D 6.4 &eth; 103 mm2 ,
40 mm
120
mm
dy1 D 200 20 D 180 mm.
Part (2) A2 D 200 8040 D 4.8 &eth; 103 mm2 ,
dx1 D
160
D 80 mm,
2
dy2 D
120
C 40 D 100 mm,
2
dx2 D 20 mm,
Part (3) A3 D 12040 D 4.8 &eth; 103 mm2 ,
dx3 D
120
D 60 mm,
2
dy3 D 20 mm.
The centroid coordinates are
xD
A1 dx1 C A2 dx2 C A3 dx3
D 56 mm,
A
yD
A1 dy1 C A2 dy2 C A3 dy3
D 108 mm
A
from which
Ixc D y2 A C Ix D 1.866 &eth; 108 C 2.645 &eth; 108
D 7.788 &eth; 107 mm4
The total area is
A D A1 C A2 C A3 D 1.6 &eth; 104 mm2 .
kxc D
Ixc
D 69.77 mm
A
Problem 8.39 Determine Iy and ky .
Solution: The strategy is to use the relationship Iy D d2 A C Iyc ,
where Iyc is the area moment of inertia about the centroid. From
this Iyc D d2 A C Iy . Use the solution to Problem 8.38. The centroid
coordinates are x D 56 mm, y D 108 mm, from which
Iyc D x2 A C Iy D 5.0176 &eth; 107 C 8.0213 &eth; 107
D 3.0 &eth; 107 mm4 ,
kyc D
Iyc
D 43.33 mm
A
Problem 8.40 Determine Ixy .
Solution: Use the solution to Problem 8.37. The centroid
coordinates are
x D 56 mm,
y D 108 mm,
from which Ixyc D xyA C Ixy D 9.6768 &eth; 107 C 1.0752 &eth; 108
D 1.0752 &eth; 107 mm4
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
617
Problem 8.41 Determine Ix and kx .
Solution: Divide the area into two parts:
y
3 ft
4 ft
Part (1): a triangle and Part (2): a rectangle. The area moment of inertia
for a triangle about the base is
Ix D
1
12
3 ft
x
bh3 .
The area moment of inertia about the base for a rectangle is
Ix D
1
bh3 .
3
Part (1) Ix1 D
Part (2) Ix2 D
1
12
433 D 9 ft2 .
1
333 D 27.
3
The composite: Ix D Ix1 C Ix2 D 36 ft4 . The area:
1
43 C 33 D 15 ft2 .
2
kx D
Ix
D 1.549 ft.
A
Problem 8.42 Determine JO and kO .
Solution: (See Figure in Problem 8.41.) Use the solution to
from which
Problem 8.41.
Part (1): The area moment of inertia about the centroidal axis parallel
to the base for a triangle is
Iyc D
1
36
bh3 D
1
36
343 D 5.3333 ft4 ,
from which
Iy1
2
8
D
A1 C Iyc D 48 ft4 .
3
Iy2 D 5.52 A2 C Iyc D 279 ft4 ,
where A2 D 9 ft2 .
The composite: Iy D Iy1 C Iy2 D 327 ft4 , from which, using a result
from Problem 8.41,
JO D Ix C Iy D 327 C 36 D 363 ft4
and kO D
JO
D 4.92 ft
A
where A1 D 6 ft2 .
Part (2): The area moment of inertia about a centroid parallel to the
base for a rectangle is
Iyc D
618
1
12
bh3 D
1
12
333 D 6.75 ft4 ,
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 8.43 Determine Ixy .
Solution: (See Figure in Problem 8.41.) Use the results of the
solutions to Problems 8.41 and 8.42. The area cross product of the
moment of inertia about centroidal axes parallel to the bases for a
1 2 2
b h , and for a rectangle it is zero. Therefore:
triangle is Ix0 y 0 D
72
Ixy1 D
1
72
42 32 C
8
3
A1 D 18 ft4
3
3
and Ixy2 D 1.55.5A2 D 74.25 ft4 ,
Ixy D Ix0 y 0 1 C Ixy2 D 92.25 ft4
y
Problem 8.44 Determine Ix and kx .
Solution: Use the results of Problems 8.41, 8.42, and 8.43. The
strategy is to use the parallel axis theorem and solve for the area
moment of inertia about the centroidal axis. The centroidal coordinate
yD
A1 1 C A2 1.5
D 1.3 ft.
A
4 ft
3 ft
3 ft
x
From which
Ixc D y2 A C Ix D 10.65 ft4
and kxc D
Ixc
D 0.843 ft
A
Problem 8.45 Determine JO and kO .
Solution: Use the results of Problems 8.41, 8.42, and 8.43. The
strategy is to use the parallel axis theorem and solve for the area
moment of inertia about the centroidal axis. The centroidal coordinate:
A1
xD
8
C A2 5.5
3
D 4.3667 ft,
A
from which
IYC D x2 A C IY D 40.98 ft4 .
Using a result from Problem 8.44,
JO D IXC C IYC D 10.65 C 40.98 D 51.63 ft4
and kO D
JO
D 1.855 ft
A
Problem 8.46 Determine Ixy .
Solution: Use the results of Problems 8.41–8.45. The strategy is
to use the parallel axis theorem and solve for the area moment of inertia
about the centroidal axis. Using the centroidal coordinates determined
in Problems 8.44 and 8.45,
Ixy D xyA C Ixy D 85.15 C 92.25 D 7.1 ft4
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
619
Problem 8.47 Determine Ix and kx .
y
120
mm
20
mm
80
mm
x
40 mm
80 mm
Solution: Let Part 1 be the entire rectangular solid without the
y
hole and let part 2 be the hole.
m
20 m
Area D hb
R2
D 80120 40 mm
R2
y′
Ix1 D 13 bh3
where b D 80 mm
h D 120 mm
Ix1 D 13 801203 D 4.608 &eth; 107 mm4
40 mm
120 mm
Part 2
x′
For Part 2,
Ix0 2 D 14 R4 D 14 204 mm4
Ix0 2 D 1.257
&eth; 105
dy = 80 mm
mm4
Ix2 D Ix0 2 C d2y A
Part 1
x
where A D R2 D 1257 mm2
80 mm
d D 80 mm
Ix2 D 1.257 &eth; 105 C 202 802
Ix2 D 0.126 &eth; 106 C 8.042 &eth; 106 mm4
D 8.168 &eth; 106 mm4 D 0.817 &eth; 107 mm4
Ix D Ix1 Ix2 D 3.79 &eth; 107 mm4
Area D 8343 mm2
kx D
620
Ix
D 67.4 mm
Area
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 8.48 Determine JO and kO .
Solution: For the rectangle,
y
40 mm
JO1 D Ix1 C Iy1 D 13 bh3 C 13 hb3
JO1 D 4.608 &eth; 107 C 2.048 &eth; 107 mm4
y′
R = 20 mm
JO1 D 6.656 &eth; 107 mm4
x′
A1 D bh D 9600 mm2
120 mm
A2
For the circular cutout about x0 y 0
J0O2 D Ix0 2 C Iy 0 D 14 R4 C 14 R4
2
(h) 80 mm
A1
J0O2 D 1.257 &eth; 105 C 1.257 &eth; 105 mm4
J0O2 D 2.513 &eth; 105 mm2
x
Using the parallel axis theorem to determine JO2 (about x, y)
80 mm
(b)
JO2 D J002 C d2x C d2y A2
A2 D R2 D 1257 mm2
JO2 D 1.030 &eth; 107 mm4
JO D 5.63 &eth; 107 mm4
kO D
JO D JO1 JO2
JO
D
Area
JO
A1 A2
kO D 82.1 mm
JO D 6.656 &eth; 107 1.030 &eth; 107 mm4
Problem 8.49 Determine Ixy .
Solution:
A1 D 80120 D 9600
y
80 mm
mm2
A2 D R2 D 202 D 1257 mm2
R = 20 mm
y′
For the rectangle A1 x′
Ixy1 D 14 b2 h2 D 14 802 1202
Ixy1 D 2.304 &eth; 107 mm2
For the cutout
A2
A1
120 mm
A2
dy = 80 mm
A1
Ix 0 y 0 2 D 0
and by the parallel axis theorem
x
dx = 40 mm
Ixy2 D Ix0 y 0 2 C A2 dx dy Ixy2 D 0 C 12574080
Ixy2 D 4.021 &eth; 106 mm4
Ixy D Ixy1 Ixy2
Ixy D 2.304 &eth; 107 0.402 &eth; 107 mm4
Ixy D 1.90 &eth; 107 mm4
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
621
y
Problem 8.50 Determine Ix and kx .
20
mm
120
mm
x
80
mm
40 mm
80 mm
Solution: We must first find the location of the centroid of the total
y
area. Let us use the coordinates XY to do this. Let A1 be the rectangle
and A2 be the circular cutout. Note that by symmetry Xc D 40 mm
Y
80 mm
Rectangle1
Circle2
Area
Xc
Yc
9600 mm2
40 mm
60 mm
mm2
40 mm
80 mm
1257
R = 20 mm
120 mm
A1 D 9600 mm2
A2 D 1257 mm2
x
80 mm
For the composite,
Xc D
A1 Xc1 A2 Xc2
D 40 mm
A1 A2
Yc D
A1 Yc1 A2 Yc2
D 57.0 mm
A1 A2
Now let us determine Ix and kx about the centroid of the
composite body.
Rectangle about its centroid (40, 60) mm
Ix1 D
1 3
1
bh D
801203
12
12
Ix1 D 1.152
&eth; 107
mm3 ,
X
40 mm
40 mm
Now to C ! dy2 D 80 57 D 23 mm
Ixc2 D Ix2 C dy2 2 A2
Ixc2 D 7.91 &eth; 105 mm4
For the composite about the centroid
Ix D Ixc1 Ixc2
Now to C
Ix D 1.08 &eth; 107 mm4
Ixc1 D Ix1 C 60
Yc 2 A1
The composite Area D 9600 1257 mm2
Ixc1 D 1.161 &eth; 107 mm4
D 8343 mm2
Circular cut out about its centroid
A2 D
R2
D
202 D 1257
mm2
kx D
Ix
D 36.0 mm
A
Ix2 D 14 R4 D 204 /4
Ix2 D 1.26 &eth; 105 mm4
622
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 8.51 Determine Iy and ky .
Solution: From the solution to Problem 8.50, the centroid of the
y
composite area is located at (40, 57.0) mm.
The area of the rectangle, A1 , is 9600 mm2 .
The area of the cutout, A2 , is 1257 mm2 .
The area of the composite is 8343 mm2 .
(1)
Rectangle about its centroid (40, 60) mm.
Iy1 D
1 3
1
hb D
120803
12
12
+
x
80 mm
Iy1 D 5.12 &eth; 106 mm4
dx1 D 0
(2)
40 mm
Circular cutout about its centroid (40, 80)
80 mm
Iy2 D R4 /4 D 1.26 &eth; 105 mm4
dx2 D 0
Since dx1 and dx2 are zero. (no translation of axes in the xdirection), we get
Iy D Iy1 Iy2
Iy D 4.99 &eth; 106 mm4
Finally,
ky D
Iy
D
A1 A2
4.99 &eth; 106
8343
ky D 24.5 mm
Problem 8.52 Determine JO and kO .
y
Solution: From the solutions to Problems 8.51 and 8.52,
Ix D 1.07 &eth; 107 mm4
Iy D 4.99 &eth; 106 mm4
and A D 8343 mm2
20
mm
JO D Ix C Iy D 1.57 &eth; 107 mm4
kO D
JO
D 43.4 mm
A
x
120 mm
80 mm
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
623
y
Problem 8.53 Determine Iy and ky .
12 in
x
20 in
Solution: Treat the area as a circular area with a half-circular
y
y
y
cutout: From Appendix B,
1
Iy 1 D 14 204 in4
and Iy 2 D 18 124 in4 ,
12
in.
20 in.
2
x
x
2 in.
x
20 in.
so Iy D 14 204 18 124 D 1.18 &eth; 105 in4 .
The area is A D 202 12 122 D 1030 in2
so,
ky D
Iy
D
A
1.18 &eth; 105
1.03 &eth; 103
D 10.7 in
Problem 8.54 Determine JO and kO .
Solution: Treating the area as a circular area with a half-circular
cutout as shown in the solution of Problem 8.53, from Appendix B,
JO 1 D Ix 1 C Iy 1 D 12 204 in4
and JO 2 D Ix 2 C Iy 2 D 14 124 in4 .
Therefore JO D 12 204 14 124
D 2.35 &eth; 105 in4 .
From the solution of Problem 8.53,
A D 1030 in2 Ro D
D
624
JO
A
2.35 &eth; 105
D 15.1 in.
1.03 &eth; 103
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 8.55 Determine Iy and ky if h D 3 m.
y
Solution: Break the composite into two parts, a rectangle and a
semi-circle.
1.2 m
For the semi-circle
Ix 0 c D
9
8
8
R4
h
Iy 0 c D
1 4
4R
R d D
8
3
y′
x
y
d
x′
1.2 m
AC
d = 4R
3π
AR
To get moments about the x and y axes, the dxc , dyc for the semicircle are
dxc D 0,
dyc D 3 m C
3m = h
4R
3
x
and Ac D R2 /2 D 2.26 m2
Iy 0 c D
To get moments of area about the x, y axes, dxR D 0, dyR D 1.5 m
1 4
R
8
and Iyc D Iy 0 c C d2xc A
0
IyR D Iy 0 R C dxR 2 bh
dx D 0
1
32, 43 m4
12
Iyc D Iy 0 c D 1.24 /8
IyR D Iy 0 R D
Iyc D 0.814 m4
IyR D 3.456 m2
For the Rectangle
Ix 0 R D
1 3
bh
12
Iy 0 R D
1 3
hb
12
AR D bh D 7.2 m2
Iy D Iyc C IyR
Iy D 4.27 m2
To find ky , we need the total area, A D AR C Ac
AR D bh
A D 7.20 C 2.26 m2
y′ y
A D 9.46 m2
2.4 m
ky D
h
3m
b
Iy
D 0.672 m
A
x′
x
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
625
Problem 8.56 Determine Ix and kx if h D 3 m.
Solution: Break the composite into two parts, the semi-circle and
the rectangle. From the solution to Problem 8.55,
Ix 0 c D
9
8
8
dyc D
3C
4R
3
y
Ac
R4
m
Ac D 2.26 m2
R = 1.2 m
h=3m
b = 2.4 m
AR
3m
h
Ixc D Ix0 c C Ac d2yc
Substituting in numbers, we get
dyc D 3.509 m
x
2.4 m
Ix0 c D 0.0717 m4
yc′
and Ixc D Ix0 c C Ac d2y
Ixc D 27.928 m2
xc′
R
For the Rectangle h D 3 m, b D 2.4 m
Area: AR D bh D 7.20 m2
Ix 0 R D
1 3
bh , dyR D 1.5 m
12
4R
3π
IxR D Ix0 R C d2yR AR
Substituting, we get
Ix0 R D 5.40 m4
IxR D 21.6 m4
For the composite,
Ix D IxR C Ixc
Ix D 49.5 m4
Also kx D
Ix
D 2.29 m
AR C Ac
kx D 2.29 m
626
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 8.57 If Iy D 5 m4 , what is the dimension
of h?
Solution: From the solution to Problem 8.55, we have:
y c′
y
xc′
For the semicircle
Iy 0 c D Iy D 1.24/8 D 0.814 m2
h
For the rectangle
Iy 0 R D IyR D
1.2
y ′R
m
h
x ′R
b
1
h2.43 m4
12
2.4 m
x
Also, we know IyR C Iyc D 5 m4 .
Hence 0.814 C
1
h2.43 D 5
12
Solving, h D 3.63 m
Problem 8.58 Determine Iy and ky .
Solution: Let the area be divided into parts as shown. The areas
and the coordinates of their centroids are
A1 D 4050 D 2000 in2 ,
x 1 D 25 in,
y 1 D 20 in,
A2 D 2030 D 600 in2 ,
x 2 D 10 in,
y 2 D 55 in,
A3 D
1
302 D 707 in2 ,
4
x 3 D 20 C
430
D 32.7 in,
3
y 3 D 40 C
430
D 52.7 in.
3
Using the results from Appendix B, the moments of inertia of the parts
Iy 1 D
1
40 in50 in3 D 167 &eth; 104 in4 ,
3
Iy 2 D
1
30 in20 in3 D 8.00 &eth; 104 in4 ,
3
Iy 3 D
4
16
9
&quot;
430 in 2 ! 30 in4 C 20 in C
30 in2
3
4
D 80.2 &eth; 104 in4 .
The moment of inertia of the composite area about the y axis is
Iy D Iy 1 C Iy 2 C Iy 3 D 2.55 &eth; 106 in4 .
2
The composite area is A D A1 C A2 C A3 D 3310
in .
Iy
D 27.8 in.
A
Iy D 2.55 &eth; 106 in4 , ky D 27.8 in.
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
627
Problem 8.59 Determine Ix and kx .
Solution: See the solution to Problem 8.58.
Let the area be divided into parts as shown. The areas and the coordinates of their centroids are
A1 D 4050 D 2000 in2 ,
x 1 D 25 in,
y 1 D 20 in,
A2 D 2030 D 600 in2 ,
x 2 D 10 in,
y 2 D 55 in,
x 3 D 20 C
430
D 32.7 in,
3
A3 D
1
302 D 707 in2 ,
4
y 3 D 40 C
430
D 52.7 in.
3
Using the results from Appendix B, the moments of inertia of the parts
Ix 1 D
1
50 in40 in3 D 1.07 &eth; 106 in4 ,
3
Ix 2 D
1
20 in30 in3 C 55 in2 600 in2 D 1.86 &eth; 106 in4 ,
12
Ix 3 D
4
4
9
&quot;
430 in 2 ! 30 in4 C 40 in C
30 in2 D 2.01 &eth; 106 in4 .
3
4
The moment of inertia of the composite area about the x axis is
Ix D Ix 1 C Ix 2 C Ix 3 D 4.94 &eth; 106 in4 .
2
The composite area is A D A1 C A2 C A3 D 3310
in .
Ix
D 38.6 in.
A
Ix D 4.94 &eth; 106 in4 , kx D 38.6 in
628
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 8.60 Determine Ixy .
Solution: See the solution to Problem 8.58.
Let the area be divided into parts as shown. The areas and the coordinates of their centroids are
A1 D 4050 D 2000 in2 ,
x 1 D 25 in,
y 1 D 20 in,
A2 D 2030 D 600 in2 ,
x 2 D 10 in,
y 2 D 55 in,
x 3 D 20 C
430
D 32.7 in,
3
A3 D
1
302 D 707 in2 ,
4
y 3 D 40 C
430
D 52.7 in.
3
Using the results from Appendix B, the products of inertia of the parts
Ixy 1 D
1
50 in2 40 in2 D 10.0 &eth; 105 in4 ,
4
Ixy 2 D 10 in55 in600 in2 D 3.30 &eth; 105 in4 ,
Ixy 3 D
4
1
8
9
430 in
430 in
30 in4 C 20 in C
40 in C
[707 in2 ]
3
3
D 12.1 &eth; 105 in4 .
The product of inertia of the composite area is
Ixy D Ixy 1 C Ixy 2 C Ixy 3 D 2.54 &eth; 106 in4 .
Ixy D 2.54 &eth; 106 in4 .
y
Problem 8.61 Determine Iy and ky .
Solution: See the solution to Problem 8.58.
In terms of the coordinate system used in Problem 8.58, the areas and
the coordinates of their centroids are
A1 D 4050 D 2000 in2 ,
x 1 D 25 in,
y 1 D 20 in,
A2 D 2030 D 600 in2 ,
x 2 D 10 in,
y 2 D 55 in,
x 3 D 20 C
430
D 32.7 in,
3
A3 D
1
302 D 707 in2 ,
4
30 in
40 in
x
430
D 52.7 in.
y 3 D 40 C
3
20 in
The composite area is A D A1 C A2 C A3 D 3310 in2 .
x 1 A1 C x 2 A2 C x 3 A3
The x coordinate of its centroid is x D
D 23.9 in
A
The moment of inertia about the y axis in terms of the coordinate
system used in Problem 8.58 is Iy D 2.55 &eth; 106 in4 . Applying the
parallel axis theorem, the moment of inertia about the y axis through
the centroid of the area is
Iy D 2.55 &eth; 106 in4 23.9 in2 3310 in2 D 6.55 &eth; 105 in4 .
Iy
D 14.1 in.
A
Iy D 6.55 &eth; 105 in4 , ky D 14.1 in.
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
629
y
Problem 8.62 Determine Ix and kx .
Solution: See the solution to Problem 8.59.
In terms of the coordinate system used in Problem 8.59, the areas and
the coordinates of their centroids are
A1 D 4050 D 2000 in2 ,
A2 D 2030 D 600 in2 ,
A3 D
1
302 D 707 in2 ,
4
x 1 D 25 in,
y 1 D 20 in,
x 2 D 10 in,
y 2 D 55 in,
x 3 D 20 C
430
D 32.7 in,
3
30 in
40 in
x
430
D 52.7 in.
3
y 3 D 40 C
20 in
2
The composite area is A D A1 C A2 C A3 D 3310 in .
y A1 C y 2 A2 C y 3 A3
The x coordinate of its centroid is y D 1
D 33.3 in
A
The moment of inertia about the x axis in terms of the coordinate
system used in Problem 8.59 is Ix D 4.94 &eth; 106 in4 . Applying the
parallel axis theorem, the moment of inertia about the x axis through
the centroid of the area is
Ix D 4.94 &eth; 106 in4 33.3 in2 3310 in2 D 1.26 &eth; 106 in4 .
Ix
D 19.5 in.
A
Ix D 1.26 &eth; 106 in4 , kx D 19.5 in.
Problem 8.63 Determine Ixy .
Solution: See the solution to Problem 8.60.
Ixy D [0 C 1.0 m0.8 mdx 0.5 mdy 0.4 m]
[0 C 0.2 m2 dx 0.4 mdy 0.3 m]
C
1
1
0.8 m2 0.6 m2 0.8 m0.6 m0.2 m
24
2
C
Solving:
0.8 m
3
0.8 m
1
0.8 m0.6 mdx 1.2 m dy 2
3
Ixy D 0.0230 m4
Check using the noncentroidal product of inertia from Problem 8.60
we have
Ixy D Ixy 0 Adx dy D 0.2185 m4 0.914 m2 0.697 m0.379 m
D 0.0230 m4
630
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 8.64 Determine Iy and ky .
y
18 in
x
6 in
6 in
Solution: Divide the area into three parts:
The composite area:
Part (1) The rectangle 18 by 18 inches; Part (2) The triangle with base
6 in and altitude 18 in; Part (3) The semicircle of 9 in radius.
3
#
6 in
Ai D 397.23 in2 .
1
Part (1): A1 D 1818 D 324 in2 ,
x1 D 9 in,
Iy D x21 A1 C Iyy1 x22 A2 Iyy2 C x23 A3 C Iyy3 ,
y1 D 9 in,
Ixx1 D
Iyy1 D
1
12
1
12
The area moment of inertia:
Iy D 4.347 &eth; 104 in4 ,
18183 D 8748 in4 ,
ky D
Iy
D 10.461 in
A
18183 D 8748 in4 .
1
186 D 54 in2 ,
2
Part (2): A2 D
x2 D 9 in,
y2 D
1
18 D 6 in,
3
Ixx2 D
1
36
6183 D 972 in4 ,
Iyy2 D 1/181833 D 27 in4 .
Part (3) A3 D
92 D 127.23 in2 ,
2
x3 D 9 in,
y3 D 18 C
Ixx3 D
Iyy3 D
49
3
D 21.82 in,
1
49 2
A3 D 720.1 in4 ,
94 8
3
1
94 D 2576.5 in4 .
8
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
631
Problem 8.65 Determine Ix and kx .
Solution: Use the results of the solution to Problem 8.64.
IX D y21 A1 C IXX1 y22 A2 IXX2 C y23 A3 C IXX3 ,
Ix D 9.338 &eth; 104 in4 ,
kx D
Ix
D 15.33 in
A
Problem 8.66 Determine Ixy .
Solution: Use the results of the solutions to Problems 8.63
and 8.64.
Ixy D x1 y1 A1 x2 y2 A2 C x3 y3 A3
Ixy D 4.8313 &eth; 104 in4
Problem 8.67 Determine Iy and ky .
y
Solution: We divide the composite area into a triangle (1), rect-
6 in
angle (2), half-circle (3), and circular cutout (4):
2 in
Triangle:
Iy 1 D 14 1283 D 1536 in4
x
Rectangle:
Iy 2 D
8 in
1
1283 C 122 812 D 14,336 in4 .
12
8 in
y
Half-Circle:
Iy 3 D
8
8
9
2
46 2 1
62 D 19,593 in4
64 C 16 C
3
2
3
4
1
x
Circular cutout:
Iy 4 D 14 24 C 162 22 D 3230 in4 .
y
8 in.
y
Therefore
Iy D Iy 1 C Iy 2 C Iy 3 Iy 4 D 3.224 &eth; 104 in4 .
12 in.
2
1
12 in.
The area is
x
A D A1 C A2 C A3 A4
D
1
1
128 C 812 C 62 22 D 188 in2 ,
2
2
so ky D
Iy
D
A
x
8 in.
12 in.
y
y
3
6 in.
4
3.224 &eth; 104
D 13.1 in.
188
6 in.
x
16 + 4(6)
in.
3π
632
2 in.
x
16 in.
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 8.68 Determine JO and kO .
Solution: Iy is determined in the solution to Problem 8.67. We
will determine Ix and use the relation JO D Ix C Iy . Using the figures
in the solution to Problem 8.67,
Triangle:
Ix 1 D
1
8123 D 1152 in4 .
12
Rectangle:
Ix 2 D 13 8123 D 4608 in4 .
Half Circle:
Ix 3 D 18 64 C 62 21 62 D 2545 in4 .
Circular Cutout:
Ix 4 D 14 24 C 62 22 D 465 in4 .
Therefore
Ix D Ix 1 C Ix 2 C Ix 3 Ix 4 D 7840 in4 .
Using the solution of Problem 8.67,
JO D Ix C Iy D 0.784 &eth; 104 C 3.224 &eth; 104 D 4.01 &eth; 104 in4 .
From the solution of Problem 8.67, A D 188 in2 , so
R0 D
JO
D
A
4.01 &eth; 104
D 14.6 in.
188
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
633
Problem 8.69 Determine Iy and ky .
y
4 in
Solution: Divide the area into four parts: Part (1) The rectangle
8 in by 16 in. Part (2): The rectangle 4 in by 8 in. Part (3) The semicircle of radius 4 in, and Part (4) The circle of radius 2 in.
Part (1): A1 D 168 D 128 in2 ,
x1 D 8 in,
2 in
4 in
8 in
y1 D 4 in,
Ixx1 D
Iyy1 D
1
12
1
12
x
1683 D 682.67 in4 ,
16 in
8163 D 2730.7 in4 .
Part (2): A2 D 48 D 32 in2 ,
Iyy2 D
1
12
1
12
843 D 42.667 in4 ,
42 D 25.133 in2 ,
2
Ixx3 D
24 D 12.566 in4 ,
3
#
Ai A4 D 172.566 in2 .
1
The area moment of inertia:
Iy D x21 A1 C Iyy1 C x22 A2 C Iyy2 C x23 A3 C Iyy3 x24 A4 Iyy4
44
3
D 13.698 in.
The area moments of inertia about the centroid of the semicircle are
Iyy3 D
4
Iyy4 D Ixx4 D 12.566 in4 .
483 D 170.667 in4 .
x3 D 12 in.
\$1%
The composite area:
Part (3): A3 D
y3 D 12 C
x4 D 12 in,
Ixx4 D
y2 D 10 in,
Part (4): A4 D 22 D 12.566 in2 ,
y4 D 12 in,
x2 D 12 in,
Ixx2 D
12 in
1
44 D 100.53 in4 ,
8
Iy D 1.76 &eth; 104 in4 ,
ky D
Iy
D 10.1 in
A
1
44 2
A3 D 28.1 in4 .
44 8
3
Check:
Ixx3 D 0.1098R4 D 28.1 in4 .
check.
Problem 8.70 Determine Ix and kx .
Solution: Use the results in the solution to Problem 8.69.
Ix D y21 A1 C Ixx1 C y22 A2 C Ixx2 C y23 A3 C Ixx3 y24 A4 Ixx4
Ix D 8.89 &eth; 103 in4
kx D
634
Ix
D 7.18 in
A
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 8.71 Determine Ixy .
Solution: Use the results in the solution to Problem 8.69.
Ixy D x1 y1 A1 C x2 y2 A2 C x3 y3 A3 x4 y4 A4 ,
Ixy D 1.0257 &eth; 104 in4
Problem 8.72 Determine Iy and ky .
y
4 in
2 in
4 in
Solution: Use the results in the solutions to Problems 8.69 to 8.71.
The centroid is
xD
x1 A1 C x2 A2 C x3 A3 x4 A4
A
1024 C 384 C 301.6 150.8
D 9.033 in,
D
172.567
x
8 in
12 in
16 in
from which
Iyc D x2 A C Iy D 1.408 &eth; 104 C 1.7598 &eth; 104 D 3518.2 in4
kyc D
Iyc
D 4.52 in
A
Problem 8.73 Determine Ix and kx .
Solution: Use the results in the solutions to Problems 8.69 to 8.71.
The centroid is
yD
y1 A1 C y2 A2 C y3 A3 y4 A4
D 5.942 in,
A
from which
Ixc D y2 A C Ix D 6092.9 C 8894 D 2801 in4
kxc D
Ixc
D 4.03 in
A
Problem 8.74 Determine Ixy .
Solution: Use the results in the solutions to Problems 8.69–8.71.
Ixyc D xyA C Ixy D 9.263 &eth; 103 C 1.0257 &eth; 104 D 994.5 in4
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
635
Problem 8.75 Determine Iy and ky .
y
Solution: We divide the area into parts as shown:
Iy 1 D
5 mm
1
50 C 15 C 15303 D 180,000 mm4
12
Iy 2 D Iy 3 D Iy 4 D
15 mm
50 mm
1
30103 C 202 1030
12
5 mm
5 mm
D 122,500 mm4
Iy 5 D Iy 6 D Iy 7 D
8
8
9
x
15 mm
15 mm
154
10 15 15 10
mm mm mm mm
415 2 1
152 D 353,274 mm4
C 25 C
3
2
y
1
Iy 8 D Iy 9 D Iy 10
2 5
8
1
D 54 C 252 52 D 49,578 mm4 .
4
Therefore,
50 mm
7 4
10
Iy D Iy 1 C 3Iy 2 C 3Iy 5 3Iy 8 D 1.46
&eth; 106
3 6
9
mm4 .
x
10
15
mm mm
The area is
A D A1 C 3A2 C 3A5 3A8
D 3080 C 31030 C 3
1
152 352
2
D 4125 mm2
so ky D
Iy
D
A
1.46 &eth; 106
D 18.8 mm
4125
Problem 8.76 Determine JO and kO .
Solution: Iy is determined in the solution to Problem 8.75. We
will determine Ix and use the relation JO D Ix C Iy . Dividing the area
as shown in the solution to Problem 8.75, we obtain
Ix 1 D
1
30803 C 252 3080 D 2, 780, 000 mm4
12
Ix 2 D
1
10303 C 502 1030 D 772, 500 mm4
12
Ix 3 D Ix 4 D
Ix 5 D
1
10303 D 22, 500 mm4
12
Therefore
Ix D Ix 1 C Ix 2 C 2Ix 3 C Ix 5 C 2Ix 6 Ix 8 2Ix 9
D 4.34 &eth; 106 mm4
and JO D Ix C Iy D 5.80 &eth; 106 mm4 .
From the solution to Problem 8.75, A D 4125 mm2
so kO D
1
1
154 C 502 152 D 903,453 mm4
8
2
D
Ix 6 D Ix 7 D
Ix 8 D
5.80 &eth; 106
4125
D 37.5 mm.
1
54 C 52 502 ,
4
Ix 9 D Ix 10 D
636
1
154 D 19,880 mm4 ,
8
JO
A
1
54 D 491 mm4 .
4
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
y
Problem 8.77 Determine Ix and Iy for the beam’s
cross section.
5 in
2 in
Solution: Use the symmetry of the object
1
1
Ix
D 3 in.8 in3 C
3 in3 in3 C 3 in2 11.5 in2
2
3
12
C
5 in4
5 in2
16
4
C
4[5 in]
3
2
5 in2
4
8 in
4[5 in] 2
8 in C
3
x
3 in
in4
2
16
in2
2
4
C
4[2 in]
3
5 in
3 in
5 in
2
2 in2
4
4[2 in] 2
8 in C
3
Solving we find
Ix D 7016 in4
Iy
1
1
D 3 in3 in3 C
8 in3 in3 C 8 in3 in6.5 in2
2
3
12
C
5 in4
5 in2
16
4
4[5 in]
3
2
5 in2
C
4
2 in2
2 in4
16
4
C
4[2 in]
3
4[5 in]
3 in C
3
2 2
2 in2
4
3 in C
4[2 in]
3
2 Solving we find
Iy D 3122 in4
y
Problem 8.78 Determine Ix and Iy for the beam’s
cross section.
5 in
2 in
x
Solution: Use Solution 8.77 and 7.39. From Problem 7.39 we
know that
y D 7.48 in, A D 98.987 in2
8 in
Ix D 7016 in4 Ay 2 D 1471 in4
Iy D 3122 in4 A02 D 3122 in4
3 in
5 in
5 in
3 in
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
637
Problem 8.79 The area A D 2 &eth; 104 mm2 . Its moment
of inertia about the y axis is Iy D 3.2 &eth; 108 mm4 . Determine its moment of inertia about the yO axis.
ŷ
y
Solution: Use the parallel axis theorem. The moment of inertia
A
about the centroid of the figure is
Iyc D x 2 A C Iy D 1202 2 &eth; 104 C 3.2 &eth; 108
D 3.20 &eth; 107 mm4 .
x, x̂
The moment of inertia about the yO axis is
100 mm
120 mm
IyO D x2 A C Iyc
IyO D 2202 2 &eth; 104 C 3.2 &eth; 107
D 1 &eth; 109 mm4
Problem 8.80 The area A D 100 in2 and it is
symmetric about the x 0 axis. The moments of inertia
Ix0 D 420 in4 , Iy 0 D 580 in4 , JO D 11000 in4 , and Ixy D
4800 in4 . What are Ix and Iy ?
y
y'
A
x'
O'
O
x
Solution: The basic relationships:
(1)
(2)
(3)
(4)
(5)
(6)
Ix D y 2 A C Ixc ,
Iy D x2 A C Iyc ,
JO D Ar 2 C Jc ,
JO D Ix C Iy ,
Jc D Ixc C Iyc , and
Ixy D Axy C Ixyc ,
where the subscript c applies to the primed axes, and the others to the
unprimed axes. The x, y values are the displacement of the primed
axes from the unprimed axes. The steps in the demonstration are:
From symmetry about the xc axis, the product of inertia Ixyc D 0.
JO J c
D 100 in2 , from which r 2 D x2 C
From (3): r 2 D
A
y 2 D 100 in2
Ixy
(iii) From (6) and Ixyc D 0, y D
, from which x2 r 2 D x4 C
Ax
2
Ixy
. From which: x4 100x2 C 2304 D 0.
A
(iv) The roots: x12 D 64, and x22 D 36. The corresponding values of y
p
are found from y D r 2 x 2 from which x1 , y1 D 8, 6, and
x2 , y2 D 6, 8.
(v) Substitute these pairs to obtain the possible values of the area
moments of inertia:
(i)
(ii)
Ix1 D Ay12 C Ixc D 4020 in4 ,
Iy1 D Ax12 C Iyc D 6980 in4
Ix2 D Ay22 C Ixc D 6820 in4 ,
Iy2 D Ax22 C Iyc D 4180 in4
638
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 8.81 Determine the moment of inertia of the
with the moment of inertia of a solid square cross section
of equal area. (See Example 8.5.)
y
20 mm
x
Solution: We first need to find the location of the centroid of the
composite. Break the area into two parts. Use X, Y coords.
160 mm
y
2
100
20
A1 = 2000 mm2
XC = 0
160 mm
1
A2 = 3200 mm2
XC2 = 0
20 mm
YC2 = 80 mm
100 mm
1
YC1 = 170 mm
x
100 mm
y Y
20 mm
Xc D
Xc1 A1 C Xc2 A2
D0
A1 C A2
Yc D
Yc1 A1 C Yc2 A2
A1 C A2
C1
1
For the composite
20 mm
x
160 mm
2
C2
Substituting, we get
Xc D 0 mm
X
Yc D 114.6 mm
We now find Ix for each part about its center and use the parallel axis
theorem to find Ix about C.
20 mm
Part (1): b1 D 100 mm, h1 D 20 mm
Finally, Ix D Ix1 C Ix2
1
1
b1 h13 D
100203 mm4
12
12
Ix D 1.686 &eth; 107 mm4
Ix 0 1 D
for our composite shape.
Ix0 1 D 6.667 &eth; 104 mm4
dy1 D Yc1 Yc D 55.38 mm
Ix1 D Ix0 1 C dy1 2 A1 Now for the comparison. For the solid square with the same total area
A1 C A2 D 5200 mm2 , we get a side of length
l2 D 5200:
l D 72.11 mm
And for this solid section
Ix1 D 6.20 &eth; 106 mm4
Part (2) b2 D 20 mm, h2 D 160 mm
Ix 0 2 D
1
1
b2 h2 3 D
201603 mm4
12
12
IxSQ D
1 3
1 4
bh D
l
12
12
IxSQ D 2.253 &eth; 106 mm4
1.686 &eth; 107
2.253 &eth; 106
Ix0 2 D 6.827 &eth; 106 mm4
Ratio D Ix /IxSQ D
dy2 D Yc2 Yc D 34.61 mm
Ratio D 7.48
Ix2 D Ix0 2 C dy2 A2
This matches the value in Example 8.5.
Ix2 D 1.066 &eth; 107 mm4
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
639
Problem 8.82 The area of the beam cross section is
5200 mm2 . Determine the moment of inertia of the beam
the moment of inertia of a solid square cross section of
equal area. (See Example 8.5.)
y
x
20 mm
Solution: Let the outside dimension be b mm, then the inside
dimension is b 40 mm. The cross section is A D b2 b 402 D
5200 mm2 . Solve: b D 85 mm. Divide the beam cross section into two
parts: the inner and outer squares. Part (1)
A1 D 852 D 7225 mm2 ,
Ixx1 D
1
12
85853 D 4.35 &eth; 106 .
Part (2)
A2 D 452 D 2025 mm2 .
Ixx2 D
1
12
45453 D 3.417 &eth; 105 .
The composite moment of inertia about the centroid is
Ix D Ixx1 Ixx2 D 4.008 &eth; 106 mm4 .
For a square cross section of the same area, h D
p
5200 D 72.111 mm.
The area moment of inertia is
Ixb D
1
12
72.11172.1113 D 2.253 &eth; 106 in4 .
The ratio:
RD
4.008 &eth; 106
D 1.7788 D 1.78
2.253 &eth; 106
which confirms the value given in Example 8.5.
640
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 8.83 If the beam in Fig. a is subjected to
couples of magnitude M about the x axis (Fig. b), the
beam’s longitudinal axis bends into a circular are whose
y
y
z
x
EIx
RD
,
M
where Ix is the moment of inertia of the beam’s cross
section about the x axis. The value of the term E, which
is called the modulus of elasticity, depends on the material of which the beam is constructed. Suppose that a
beam with the cross section shown in Fig. c is subjected
to couples of magnitude M D 180 N-m. As a result,
the beam’s axis bends into a circular arc with radius
R D 3 m. What is the modulus of elasticity of the beam’s
material? (See Example 8.5.)
Solution: The moment of inertia of the beam’s cross section about
the x axis is
&amp;
'
1
1
mm4
393 C 2
933 C 62 93
Ix D
12
12
D 2170 mm4 D 2.17 &eth; 109 m4 .
The modulus of elasticity is
ED
RM
3 m180 N-m
D
D 2.49 &eth; 1011 N/m2
Ix
2.17 &eth; 109 m4
M
R
M
(b) Subjected to couples at the ends.
y
3 mm
x
9 mm
3 mm
3
mm
9 mm
(c) Beam cross section.
E D 2.49 &eth; 1011 N/m2 .
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
641
Problem 8.84 Suppose that you want to design a beam
made of material whose density is 8000 kg/m3 . The
beam is to be 4 m in length and have a mass of
320 kg. Design a cross section for the beam so that
Ix D 3 &eth; 105 m4 . (See Example 8.5.)
Solution: The strategy is to determine the cross sectional area,
and then use the ratios given in Figure 8.14 to design a beam. The
volume of the beam is V D AL D 4A m3 . The mass of the beam is m D
V8000 D 32000A D 320 kg, from which A D 0.01 m2 . The moment
of inertia for a beam of square cross section with this area is
Ixxb D
1
12
h
0.10.13 D 8.333 &eth; 106 m4 .
The ratio is R D
b
&eth; 105
3
D 3.6.
8.333 &eth; 106
From Figure 8.6, this ratio suggests an I-beam of the form shown in
the sketch. Choose an I-beam made up of three equal area rectangles,
of dimensions b by hm in section. The moment of inertia about the
centroid is Ix D y21 A1 C Ixx1 C y22 A2 C Ixx2 C y23 A3 C Ixx3 .
Since all areas are equal, A1 D A2 D A3 D bh, and y1 D
0, and y3 D y1 , this reduces to
Ix D
bCh
, y2 D
2
1
bCh 2
1
bh3 C 2
hb3 .
hb C
6
2
12
A
, where A is the known total cross section area. These
3
are two equations in two unknowns. Plot the function
Note that bh D
fb D
1
bCh 2
1
bh C
bh3 C 2
hb3 Ix
6
2
12
Problem 8.85 The area in Fig. (a) is a C230&eth;30
American Standard Channel beam cross section. Its cross
sectional area is A D 3790 mm2 and its moments of
inertia about the x and y axes are Ix D 25.3 &eth; 106 mm4
and Iy D 1 &eth; 106 mm4 . Suppose that two beams with
C230&eth;30 cross sections are riveted together to obtain a
composite beam with the cross section shown in Fig. (b).
What are the moments of inertia about the x and y axes
of the composite beam?
f
(
b
)
*
E
+
5
I - beam Flange dimension
1
.8
.6
.4
.2
0
–.2
–.4
–.6
–.8
–1
.03 .04 .05 .06 .07 .08 .09 .1
b
A
.The function was graphed using
3
TK Solver Plus. The graph crosses the zero axis at approximately
b D 0.0395 m. and b D 0.09 m. The lower value is an allowable value
for h and the greater value corresponds to an allowable value of b.
Thus the I beam design has the flange dimensions, b D 90 mm and
h D 39.5 mm.
subject to the condition that hb D
y
y
x
x
14.8 mm
(a)
(b)
Solution:
Ix D 225.3 &eth; 106 mm4 D 50.6 &eth; 106 mm4
Iy D 2106 mm4 C [3790 mm2 ][14.8 mm]2 D 3.66 &eth; 106 mm4
642
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 8.86 The area in Fig. (a) is an L152&eth;102&eth;
12.7 Angle beam cross section. Its cross sectional
area is A D 3060 mm2 and its moments of inertia
about the x and y axes are Ix D 7.24 &eth; 106 mm4 and
Iy D 2.61 &eth; 106 mm4 . Suppose that four beams with
L152&eth;102&eth;12.7 cross sections are riveted together to
obtain a composite beam with the cross section shown
in Fig. (b). What are the moments of inertia about the x
and y axes of the composite beam?
y
y
24.9
mm
x
Solution:
x
50.2 mm
(a)
Ix D 47.24 &eth; 106 mm4 C [3060 mm2 ][50.2 mm]2 D 59.8 &eth; 106 mm4
Iy D 42.61 &eth; 106 mm4 C [3060 mm2 ][24.9 mm]2 D 18.0 &eth; 106 mm4
(b)
Problem 8.87 In Active Example 8.6, suppose that the
vertical 3-m dimension of the triangular area is increased
to 4 m. Determine a set of principal axes and the corresponding principal moments of inertia.
y
Solution: From Appendix B, the moments and products of inertia
of the area are
Ix D
1
4 m4 m3 D 21.3 m4 ,
12
Iy D
1
4 m3 4 m D 64 m4 ,
4
Ixy D
1
4 m2 4 m2 D 32 m4 .
8
3m
x
4m
From Eq. (8.26),
tan2p D
2Ixy
232
D 1.50 ) p D 28.2&deg; .
D
Iy Ix
64 21.3
From Eqs. (8.23) and (8.24), the principal moments of inertia are
Ix ‘ D
Ix C Iy
Ix Iy
C
cos 2p Ixy sin 2p
2
2
D
21.3 C 64
2
C
21.3 64
2
cos2[28.2&deg; ] 32 sin2[28.2&deg; ]
D 4.21 m4
Iy ‘ D
Ix Iy
Ix C Iy
cos 2p C Ixy sin 2p
2
2
D
21.3 C 64
2
21.3 64
2
cos2[28.2&deg; ] C 32 sin2[28.2&deg; ]
D 81.8 m4
p D 28.2&deg; , principal moments of inertia are
4.21 m4 , 81.1 m4
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
643
y
Problem 8.88 In Example 8.7, suppose that the area is
reoriented as shown. Determine the moments of inertia
Ix0 , Iy 0 and Ix0 y 0 if D 30o .
Solution: Based on Example 8.7, the moments and product of
1 ft
3 ft
inertia of the reoriented area are
1 ft
Ix D 10 ft4 , Iy D 22 ft4 , Ixy D 6 ft4 .
x
4 ft
Applying Eqs. (8.23)–(8.25),
Ix 0 D
D
Iy 0 D
D
Ix 0 y 0 D
D
Ix Iy
Ix C Iy
C
cos 2 Ixy sin 2
2
2
10 C 22
10 22
C
cos 60&deg; 6 sin 60&deg; D 7.80 ft4 ,
2
2
Ix Iy
Ix C Iy
C
cos 2 C Ixy sin 2
2
2
10 C 22
10 22
cos 60&deg; C 6 sin 60&deg; D 24.2 ft4 ,
2
2
Ix Iy
sin 2 C Ixy cos 2
2
12 22
sin 60&deg; C 6 cos 60&deg; D 2.20 ft4
2
Ix0 D 7.80 ft4 , Iy 0 D 24.2 ft4 , Ix0 y 0 D 2.20 ft4 .
y
Problem 8.89 In Example 8.7, suppose that the area
is reoriented as shown. Determine a set of principal
axes and the corresponding principal moments of inertia.
Based on the result of Example 8.7, can you predict a
value of p without using Eq. (8.26)?
Solution: Based on Example 8.7, the moments and product of
1 ft
3 ft
inertia of the reoriented area are
1 ft
Ix D 10 ft4 , Iy D 22 ft4 , Ixy D 6 ft4 .
x
2Ixy
26
D
D 1 ) p D 22.5&deg;
Ix Iy
22 10
This value could have been anticipated from Example 8.7 by reorienting the axes.
Substituting the angle into Eqs. (8.23) and (8.24), the principal
moments of inertia are
From Eq. (8.26), tan 2p D
Ix 0 D
D
Iy 0 D
D
4 ft
Ix Iy
Ix C Iy
C
cos 2p Ixy sin 2p
2
2
10 C 22
10 22
C
cos 45&deg; 6 sin 45&deg; D 7.51 ft4 ,
2
2
Ix Iy
Ix C Iy
cos 2p C Ixy sin 2p
2
2
10 22
10 C 22
cos 45&deg; C 6 sin 45&deg; D 24.5 ft4 ,
2
2
p D 22.5&deg; , principal moments of inertia are
7.51 ft4 , 24.5 ft4 .
644
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 8.90 The moment of inertia of the area are
y
y
Ix D 1.26 &eth; 106 in4 ,
Iy D 6.55 &eth; 105 in4 ,
Ixy D 1.02 &eth; 105 in4
Determine the moments of inertia of the area Ix0 , Iy 0 and
Ix0 y 0 if D 30&deg; .
x
u
x
Solution:
Applying Eqs. (8.23)–(8.25),
Ix 0 D
Ix Iy
Ix C Iy
C
cos 2 Ixy sin 2&ETH;
2
2
D
1.26 C 0.655
1.26 0.655
C
cos 60&deg; 0.102 sin 60&deg; &eth; 106 in4
2
2
D 1.20 &eth; 106 in4
Iy 0 D
Ix Iy
Ix C Iy
cos 2 C Ixy sin 2
2
2
D
1.26 0.655
1.26 C 0.655
cos 60&deg; C 0.102 sin 60&deg; &eth; 106 in4
2
2
D 7.18 &eth; 105 in4
Ix 0 y 0 D
Ix Iy
sin 2 C Ixy cos 2
2
D
1.26 0.655
sin 60&deg; C 0.102 cos 60&deg; &eth; 106 in4
2
D 2.11 &eth; 105 in4
Ix0 D 1.20 &eth; 106 in4 , Iy 0 D 7.18 &eth; 105 in4 , Ix0 y 0 D 2.11 &eth; 105 in4 .
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
645
Problem 8.91 The moment of inertia of the area are
y
y
Ix D 1.26 &eth; 106 in4 ,
Iy D 6.55 &eth; 105 in4 ,
Ixy D 1.02 &eth; 105 in4
x
u
Determine a set of principal axes and the corresponding
principal moments of inertia.
x
Solution: From Eq. (8.26),
tan 2p D
2Ixy
2.102
D
D 0.337
Iy Ix
0.655 1.26
) p D 9.32&deg;
Substituting this angle into Eqs. (8.23) and (8.24), the principal
moments of inertia are
Ix 0 D
Ix Iy
Ix C Iy
C
cos 2p Ixy sin 2p
2
2
D
1.26 0.655
1.26 C 0.655
C
cos 18.63&deg; 0.102 sin 18.63&deg;
2
2
&eth; 106 in4 D 1.28 &eth; 106 in4
Iy 0 D
Ix C Iy
Ix Iy
cos 2p C Ixy sin 2p
2
2
D
1.26 0.655
1.26 C 0.655
cos 18.63&deg; C 0.102 sin 18.63&deg;
2
2
&eth; 106 in4 D 6.38 &eth; 105 in4
p D 9.32&deg; , principal moments of inertia are
1.28 &eth; 106 in4 , 6.38 &eth; 105 in4 .
646
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
y
Problem 8.92* Determine a set of principal axes and
the corresponding principal moments of inertia.
160 mm
Solution: We divide the area into 3 rectangles as shown: In terms
of the xO , yO coordinate system, the position of the centroid is
x̂ D
40 mm
x̂1 A1 C x̂2 A1 C x̂3 A3
A1 C A2 C A3
x
D
ŷ D
2040200 C 10012040 C 808040
D 56 mm,
40200 C 12040 C 8040
200 mm
40 mm
ŷ1 A1 C ŷ2 A1 C ŷ3 A3
A1 C A2 C A3
40 mm
10040200 C 18012040 C 208040
D
D 108 mm.
40200 C 12040 C 8040
120 mm
The moments and products of inertia in terms of the xO , yO system are
y
y^
(Ix D (Ix 1 C (Ix 2 C (Ix 3
D
160 mm
2
1
1
402003 C
120403 C 1802 12040
3
12
C
1
80403 D 26.5 &eth; 107 mm4 ,
3
200
mm
x
40
mm
(Iy D (Iy 1 C (Iy 2 C (Iy 3
40 mm
3
120 mm
D
1
1
200403 C
401203 C 1002 12040
3
12
C
40 mm
1
x^
y y′
1
40803 C 802 8040 D 8.02 &eth; 107 mm4 ,
12
(Ixy D (Ixy 1 C (Ixy 2 C (Ixy 3
x
12.1&deg;
x′
D 2010040200 C 10018040120
C 20804080 D 10.75 &eth; 107 mm.
The moments and product of inertia in terms of the xO , yO system are
Ix D (Ix yO 2 A D 77.91 &eth; 106 mm4 ,
Iy D (Iy xO 2 A D 30.04 &eth; 106 mm4 ,
Ixy D (Ixy x̂ŷA D 10.75 &eth; 106 mm4 ,
from Equation (8.26),
tan 2p D
2Ixy
210.75 &eth; 106 ,
D
Iy Ix
30.04 &eth; 106 77.91 &eth; 106 we obtain p D 12.1&deg; . We can orient the principal axes as shown:
Substituting the values of Ix , Iy and Ixy into Equations (8.23)
and (8.24) and setting D 12.1&deg; , we obtain
Ix1 D 80.2 &eth; 106 mm4
Iy 1 D 27.7 &eth; 106 mm4 .
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
647
Problem 8.93 Solve Problem 8.87 by using Mohr’s
Circle.
y
3m
x
4m
Solution: The vertical 3-m dimension is increased to 4 m. From
Problem 8.87, the moments and product of inertia for the unrotated
system are
Ix D
1
4 m4 m3 D 21.3 m4 ,
12
Iy D
1
4 m3 4 m D 64 m4 ,
4
Ixy D
1
4 m2 4 m2 D 32 m4 .
8
Mohr’s circle (shown) has a center and radius given by
CD
21.3 C 64
D 42.7 m4
2
RD
21.3 64
2
2
C 322 D 38.5 m4
The angle and principal moments are now
tan2p D
32
) p D 28.2&deg; ,
64 42.7
I1 D C C R D 81.1 m4 , I2 D C R D 4.21 m4 .
p D 28.2&deg; , principal moments of inertia are 4.21 m4 , 81.1 m4
648
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
y
Problem 8.94 Solve Problem 8.88 by using Mohr’s Circle.
1 ft
3 ft
1 ft
x
4 ft
Solution: Based on Example 8.7, the moments and product of
inertia of the reoriented area are
Ix D 10 ft4 , Iy D 22 ft4 , Ixy D 6 ft4 .
For Mohr’s circle we have the center, radius, and angle
CD
10 C 22
D 16 ft4 ,
2
RD
p D
22 10
2
1
tan1
2
2
C 62 D 8.49 ft4 ,
6
22 16
D 22.5&deg;
Now we can calculate the new inertias
Ix D C R cos 15&deg; D 7.80 ft4
Iy D C C R cos 15&deg; D 24.2 ft4
Ixy D R sin 15&deg; D 2.20 ft4
Ix0 D 7.80 ft4 , Iy 0 D 24.2 ft4 , Ix‘y‘ D 2.20 ft4 .
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
649
y
Problem 8.95 Solve Problem 8.89 by using Mohr’s Circle.
1 ft
3 ft
1 ft
x
4 ft
Solution: Based on Example 8.7, the moments and product of
inertia of the reoriented area are
Ix D 10 ft4 , Iy D 22 ft4 , Ixy D 6 ft4 .
For Mohr’s circle we have the center, radius, and angle
CD
10 C 22
D 16 ft4 ,
2
RD
p D
22 10
2
1
tan1
2
2
C 62 D 8.49 ft4 ,
6
22 16
D 22.5&deg;
Now we can calculate the principal moments of inertias
I1 D C C R D 24.5 ft4
I2 D C R D 7.51 ft4
p D 22.5&deg; , principal moments of inertia are 7.51 ft4 , 24.5 ft4 .
650
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 8.96 Solve Problem 8.90 by using Mohr’s Circle. y
y
x
u
x
Solution: For Mohr’s circle we have the center, radius, and angle
CD
12.6 C 6.55
2
RD
p D
12.6 6.55
2
1
tan1
2
&eth; 105 D 9.58 &eth; 105 in4 ,
2
C 1.022 &eth; 105 D 3.19 in4 ,
1.02
12.6 9.58
D 9.32&deg;
Now we can calculate the new inertias
Ix0 D C C R cos 22.7&deg; D 12.0 &eth; 105 in4 ,
Iy 0 D C R cos 22.7&deg; D 7.18 &eth; 105 in4 ,
Ix0 y 0 D R sin 22.7&deg; D 2.11 &eth; 105 in4 .
Ix0 D 1.20 &eth; 106 in4 , Iy 0 D 7.18 &eth; 105 in4 , Ix0 y 0 D 2.11 &eth; 105 in4 .
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
651
Problem 8.97 Solve Problem 8.91 by using Mohr’s Circle. y
y
x
u
x
Solution: For Mohr’s circle we have the center, radius, and angle
CD
12.6 C 6.55
2
RD
p D
12.6 6.55
2
1
tan1
2
&eth; 105 D 9.58 &eth; 105 in4 ,
2
C 1.022 &eth; 105 D 3.19 in4 ,
1.02
12.6 9.58
D 9.32&deg;
Now we can calculate the principal inertias
I1 D C C R D 12.8 &eth; 105 in4 ,
Iy 0 D C R D 6.38 &eth; 105 in4 ,
p D 9.32&deg; , principal moments of inertia are
1.28 &eth; 106 in4 , 6.38 &eth; 105 in4 .
652
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 8.98* Solve Problem 8.92 by using Mohr’s
circle.
Solution: The moments and product of inertia are derived in terms
of the xy coordinate system in the solution of Problem 8.92:
50
◊106
(+)
Ix D 77.91 &eth; 106 mm4
Iy D 30.04 &eth; 106 mm4
1
2θp
Ixy D 10.75 &eth; 106 mm4 .
The Mohr’s circle is: Measuring the 2p, angle we estimate that p D
12&deg; , and the principle moments of inertia are approximately 81 &eth;
106 mm4 and 28 &eth; 106 mm4 the orientation of the principal axes is
shown in the solution of Problem 8.92.
2
(30.0, –10.8)
(77.9,10.8)
(+)
100
6
◊10
- 50
◊106
Problem 8.99 Derive Eq. (8.22) for the product of
inertia by using the same procedure we used to derive
Eqs. (8.20) and (8.21).
Solution: Suppose that the area moments of inertia of the area A
x′
are known in the coordinate system x, y,
y
2
y dA,
Ix D
y′
A
A
0
x 2 dA,
Iy D
x
A
and Ixy D
xyA.
A
The objective is to find the product of inertia in the new coordinate
system x 0 , y 0 in terms of the known moments of inertia. The new
x 0 , y 0 system is formed from the old x, y system by rotation about
the origin through a counterclockwise angle .
By definition,
Substitute into the definition:
Ix0 y 0 D cos2 sin2 xy dA
A
A
0 0
x y dA.
Ix 0 y 0 D
x 2 dA ,
y 2 dA C cos sin A
from which
A
From geometry,
x 0 D x cos C y sin ,
Ix0 y 0 D cos2 sin2 Ixy C Ix Iy sin cos ,
which is the expression required.
and y 0 D x sin C y cos .
The product is
x0 y 0 D xy cos2 xy sin2 C y 2 cos sin x2 cos sin .
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
653
Problem 8.100 The axis LO is perpendicular to both
segments of the L-shaped slender bar. The mass of the
bar is 6 kg and the material is homogeneous. Use the
method described in Example 8.10 to determine is moment
Solution: Use Example 8.10 as a model for this solution.
1m
LO
2m
Introduce the coordinate system shown and divide the bar into two
parts as shown
y=1
y
dy
2
r
dx
1
0
2
r 2 dm D
I0 1 D
Ax 2 dx D A
0
I0 1 D
x
2
x
x3
3
2
0
8
A
3
However m1 D Al1 D 2A.
Since part 1 is 2/3 of the length, its mass is 2/36 kg D 4 kg. Part 2
has mass 2 kg.
For part 2, dm D A dy and
22 C y 2
rD
1
r 2 dm D
I0 2 D
I0 2 D A4y C
I0TOTAL D
A22 C y 2 dy
0
m2
y3
3
1
D A
0
13
3
13
8
21
A C A D
A
3
3
3
I0 TOTAL D 7A
The total mass D 3A D 6 kg
I0TOTAL D
654
7
6 kg &ETH; m2 D 7 kg m2
6
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 8.101 Two homogenous slender bars, each
of mass m and length l, are welded together to form
the T-shaped object. Use integration to determine the
moment of inertia of the object about the axis through
point O that is perpendicular to the bars.
l
O
l
Solution: Divide the object into two pieces, each corresponding
to a slender bar of mass m; the first parallel to the y axis, the second
to the x axis. By definition
l
r 2 dm C
ID
0
r 2 dm.
m
For the first bar, the differential mass is dm D A dr. Assume that
the second bar is very slender, so that the mass is concentrated at a
distance l from O. Thus dm D A dx, where x lies between the limits
l
l
x .
2
2
The distance to a differential dx is r D
becomes
l
I D A
r 2 dr C A
l
2
0
D A
r3
3
D ml2
l
2
p
l2 C x 2 . Thus the definition
l2 C x 2 dx I
l/2
x3
C A l2 x C
3 l/2
0
l
1
1
C1C
3
12
D
17 2
ml
12
Problem 8.102 The slender bar lies in the x –y plane.
Its mass is 6 kg and the material is homogeneous. Use
integration to determine its moment of inertia about the
z axis.
y
2m
50
x
1m
Solution:
1
m
2
Iz D
6 kg
D 2 kg/m
The density is D
3m
x 2 dx
y
0
C
m
2m
[1 m C s cos 50&deg; 2 C s sin 50&deg; 2 ]ds
0
50&deg;
Iz D 15.14 kg m2
Problem 8.103 Use integration to determine the
moment of inertia of the slender bar in Problem 8.102
x
1m
Solution: See solution for 8.102
Iy D
0
1
m
2
x 2 dx C
m
1 m C s cos 50&deg; 2 ds D 12.01 kg m2
0
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
655
y
Problem 8.104 The homogeneous thin plate has mass
m D 12 kg and dimensions b D 2 m and h D 1 m. Use
the procedure described in Active Example 8.9 to
determine the moments of inertia of the plate about the
x and y axes.
h
x
Solution: From Appendix B, the moments of inertia about the x
and y axes are
1 3
1 3
bh , Iy D
hb .
Ix D
36
36
b
Therefore the moments of inertia of the plate about the x and y axes
are
1 3
m
m
1
1
b D
mh2 D 12 kg1 m2 D 0.667 kg-m2
Ixaxis D Ix 1
A
36
18
8
bh
2
Iyaxis D
m
m
Iy D
1
A
bh
2
1 3
hb
36
D
1
1
mb2 D
12 kg2 m2 D 2.67 kg-m2
18
18
Ixaxis D 0.667 kg-m2 , Iyaxis D 2.67 kg-m2 .
y
Problem 8.105 The homogenous thin plate is of
uniform thickness and mass m.
(a)
(b)
Determine its moments of inertia about the x and z
axes.
Let Ri D 0, and compare your results with the
values given in Appendix C for a thin circular
plate.
Ro
Ri
x
Solution:
(a)
(b)
The area moments of inertia for a circular area are Ix D Iy D
R4
. For the plate with a circular cutout, Ix D Ro4 Ri4 . The
4
4
m
area mass density is , thus for the plate with a circular cut,
A
m
m
,
from
which the moments of inertia
D
2
A
Ro2 Ri Ri
Ro
m
mRo4 Ri4 D Ro2 C Ri2 4
4Ro2 Ri2 Ix
axis
D
Iz
axis
D 2Ix
axis
D
m 2
R C Ri2 .
2 o
Let Ri D 0, to obtain
Ix
axis
D
m 2
R ,
4 o
Iz
axis
D
m 2
R ,
2 o
which agrees with table entries.
656
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
y
Problem 8.106 The homogenous thin plate is of
uniform thickness and weighs 20 lb. Determine its
moment of inertia about the y axis.
1
y = 4 – – x 2 ft
4
Solution:
y D4
1 2
x ft
4
x
y
The plate’s area is
4
4
4
1 2
x
4
dx D 21.3 ft2 .
y=4–
1 2
x ft
4
The plate’s density per unit area is
υ D 20/32.2/21.3 D 0.0291 slug/ft2 .
x
–4 ft
x
The moment of inertia about the y axis is
Iy
axis
D
4 ft
dx
1
x 2 υ 4 x 2 dx
4
4
4
D 1.99 slug-ft2 .
Problem 8.107 Determine the moment of inertia of the
plate in Problem 8.106 about the x axis.
Solution: See the solution of Problem 8.106. The mass of the strip
so the moment of inertia of the plate about the x axis is
element is
mstrip
1
D υ 4 x2
4
dx.
Ix
axis
4
D
4
3
1
1
dx D 2.27 slug-ft2 .
υ 4 x2
3
4
The moment of inertia of the strip about the x axis is
Istrip D
D
2
1
1
mstrip 4 x 2
3
4
3
1
1
υ 4 x2
dx,
3
4
Problem 8.108 The mass of the object is 10 kg. Its
moment of inertia about L1 is 10 kg-m2 . What is its
moment of inertia about L2 ? (The three axes lie in the
same plane.)
Solution: The strategy is to use the data to find the moment of
inertia about L, from which the moment of inertia about L2 can be
determined.
IL D 0.62 10 C 10 D 6.4 m2 ,
0.6 m
L
from which IL2 D 1.22 10 C 6.4 D 20.8 m2
0.6 m
L1
L2
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657
Problem 8.109 An engineer gathering data for the
design of a maneuvering unit determines that the
astronaut’s center of mass is at x D 1.01 m, y D 0.16 m
and that her moment of inertia about the z axis is
105.6 kg-m2 . Her mass is 81.6 kg. What is her moment
of inertia about the z0 axis through her center of mass?
y
y
x
x
Solution: The distance d from the z axis to the z0 axis is
1.012 C 0.162
dD
D 1.0226 m.
From the parallel-axis theorem,
D Iz0
axis
C d2 m :
105.6 D Iz0
axis
C 1.02262 81.6.
Iz
axis
Solving, we obtain
Iz0
D 20.27 kg-m2 .
axis
Problem 8.110 Two homogenous slender bars, each of
mass m and length l, are welded together to form the Tshaped object. Use the parallel axis theorem to determine
the moment of inertia of the object about the axis through
point O that is perpendicular to the bars.
l
O
l
Solution: Divide the object into two pieces, each corresponding
to a bar of mass m. By definition
l
ID
r 2 dm.
0
For the first bar, the differential mass is dm D A dr, from which the
moment of inertia about one end is
l
I1 D A
r 2 dr D A
0
r3
3
l
D
0
ml2
.
3
For the second bar
I2 D A
l
2
l
2
r3
r dr D A
3
2
2l
l
2
D
ml2
12
is the moment of inertia about the center of the bar. From the parallel
axis theorem, the moment of inertia about O is
Io D
658
17 2
ml2
ml2
C l2 m C
D
ml .
3
12
12
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 8.111 Use the parallel-axis theorem to determine the moment of inertia of the T-shaped object in
Problem 8.110 about the axis through the center of mass
of the object that is perpendicular to the two bars. (See
Active Example 8.11.)
Solution:
The location of the center of mass of the object is x D
\$ %
l
2
C lm
3
D l. Use the results of Problem 8.110 for the moment
2m
4
of inertia of a bar about its center. For the first bar,
m
I1 D
2
l
ml2
7 2
mC
D
ml .
4
12
48
For the second bar,
I2 D
2
l
ml2
7 2
mC
D
ml .
4
12
48
The composite:
Ic D I1 C I2 D
7 2
ml
24
y
Problem 8.112 The mass of the homogenous slender
bar is 20 kg. Determine its moment of inertia about the
z axis.
y'
x'
Solution: Divide the object into three segments. Part (1) is the 1 m
bar on the left, Part (2) is the 1.5 m horizontal segment, and Part (3)
is the segment on the far right. The mass density per unit length is
1m
x
20
m
p D 5.11 kg/m.
D
D
L
1 C 1.5 C 2
1.5 m
1m
The moments of inertia about the centers of mass and the distances to
the centers of mass from the z axis are:
Part (1)
I1 D l31
12
D m1
l21
D 0.426 kg-m2 ,
12
m1 D 5.11 kg,
d1 D 0.5 m,
Part (2),
I2 D l2
l32
D m2 2 D 1.437 kg- m2 ,
12
12
m2 D 7.66 kg,
d2 D
Part (3)
p
0.752 C 12 D 1.25 m
I3 D p
22
l33
D m3
D 1.204 kg-m2 ,
12
12
m3 D 7.23 kg,
d3 D
p
22 C 0.52 D 2.062 m.
The composite:
I D d21 m1 C I1 C d22 m2 C I2 C d23 m3 C I3 D 47.02 kg-m2
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
659
Problem 8.113 Determine the moment of inertia of the
bar in Problem 8.112 about the z0 axis through its center
of mass.
Solution: The center of mass:
xD
D
yD
D
x 1 m 1 C x2 m 2 C x3 m 3
20
0 C 0.757.66 C 27.23
D 1.01 m.
20
0.5m1 C 1m2 C 0.5m3
20
0.55.11 C 17.66 C 0.57.23
D 0.692 m.
20
The distance from the z axis to the center of mass is d D x2 C y2 D
1.224 m. The moment of inertia about the center o mass:
Ic D d2 20 C Io
D 17.1 kg-m2
Problem 8.114 The homogeneous slender bar weighs
5 lb. Determine its moment of inertia about the z axis.
y
y'
Solution: The Bar’s mass is m D 5/32.2 slugs. Its length is
L D L1 C L 2 C L 3 D 8 C
4 in
p
82 C 82 C 4 D 31.9 in.
The masses of the parts are therefore,
L1
m1 D
mD
L
m2 D
L2
mD
L
m3 D
L3
mD
L
8
31.9
5
32.2
x'
D 0.0390 slugs,
p
264
5
D 0.0551 slugs,
31.9
32.2
4
31.9
5
32.2
x
8 in
D 0.0612 slugs.
y′
y
The center of mass of part 3 is located to the right of its center C a
distance 2R/ D 24/ D 2.55 in. The moment of inertia of part 3
2
C
r 2 dm D m3 r 2 D 0.061242 D 0.979 slug-in2 .
m3
1
The moment of inertia of part 3 about the center of mass of part 3 is
therefore
4 in
x′
3
x
8 in
I3 D 0.979 m3 2.552 D 0.582 slug-in2 .
The moment of inertia of the bar about the z axis is
Iz
axis
D 13 m1 L12 C 13 m2 L22 C I3 C m3 [8 C 2.552 C 42 ]
D 11.6 slug-in2 D 0.0802 slug-ft2 .
660
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Problem 8.115 Determine the moment of inertia of the
bar in Problem 8.114 about the z0 axis through its center
of mass.
Solution: In the solution of Problem 8.114, it is shown that the
moment of inertia of the bar about the z axis is Iz axis D 11.6 slug-in2 .
The x and y coordinates of the center of mass coincide with the
centroid of the axis:
xD
x1 L1 C x2 L2 C x3 L3
L1 C L2 C L3
p
24
4
48 C 4 82 C 82 C 8 C
p
D
D 6.58 in,
8 C 82 C 82 C 4
yD
D
y1 L1 C y2 L2 C y3 L3
L1 C L2 C L3
p
0 C 4 82 C 82 C 44
p
D 3.00 in.
8 C 82 C 82 C 4
The moment of inertia about the z axis is
Iz0
axis
D Iz
axis
x2 C y2 5
32.2
D 3.43 slug-in2 .
y
Problem 8.116 The rocket is used for atmospheric
research. Its weight and its moment of inertia about the
z axis through its center of mass (including its fuel) are
10 kip and 10,200 slug-ft2 , respectively. The rocket’s
fuel weighs 6000 lb, its center of mass is located at
x D 3 ft, y D 0, z D 0, and the moment of inertia of
the fuel about the axis through the fuel’s center of mass
parallel to z is 2200 slug-ft2 . When the fuel is exhausted,
what is the rocket’s moment of inertia about the axis
through its new center of mass parallel to z?
Solution: Denote the moment of inertia of the empty rocket as IE
about a center of mass xE , and the moment of inertia of the fuel as IF
about a mass center xF . Using the parallel axis theorem, the moment
of inertia of the filled rocket is
x
From which
xE D 186.335
124.224
3 D 4.5 ft
2m ,
IR D IE C xE2 mE C IF C xF
F
is the new location of the center of mass. Substitute:
about a mass center at the origin (xR D 0). Solve:
2m
IE D IR xE2 mE IF xF
F
2m .
IE D IR xE2 mE IF xF
F
D 10200 2515.5 2200 1677.01
The objective is to determine values for the terms on the right from
the data given. Since the filled rocket has a mass center at the origin,
the mass center of the empty rocket is found from 0 D mE xE C mF xF ,
from which
D 3807.5 slug-ft2
xE D mF
mE
xF .
Using a value of g D 32.2 ft/s2 ,
mF D
WF
6000
D
D 186.34 slug,
g
32.2
mE D
10000 6000
WR WF D
D 124.23 slug.
g
32.2
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661
Problem 8.117 The mass of the homogenous thin plate
is 36 kg. Determine its moment of inertia about the x
axis.
y
0.4 m
0.4 m
Solution: Divide the plate into two areas: the rectangle 0.4 m by
0.6 m on the left, and the rectangle 0.4 m by 0.3 m on the right. The
m
mass density is D . The area is
A
0.3 m
A D 0.40.6 C 0.40.3 D 0.36 m2 ,
0.3 m
x
from which
D
36
D 100 kg/m2 .
0.36
The moment of inertia about the x axis is
Ix
axis
D
\$1%
3
0.40.63 C \$1%
3
0.40.33 D 3.24 kg-m2
Problem 8.118 Determine the moment of inertia of the
plate in Problem 8.117 about the z axis.
Solution: The basic relation to use is
Iz
axis
D Ix
axis
C Iy
axis .
The value of Ix axis is given in the solution of Problem 8.117. The
moment of inertia about the y axis using the same divisions as in
Problem 8.117 and the parallel axis theorem is
Iy
axis
D
1
1
0.60.43 C 0.30.43
3
12
C 0.62 0.30.4 D 5.76 kg-m2 ,
from which
Iz
axis
D Ix
axis
C Iy
axis
D 3.24 C 5.76 D 9 kg-m2.
y
Problem 8.119 The homogenous thin plate weighs
10 lb. Determine its moment of inertia about the x axis.
5 in
5 in
Solution: Divide the area into two parts: the lower rectangle 5 in
by 10 in and the upper triangle 5 in base and 5 in altitude. The mass
W
. The area is
density is D
gA
A D 510 C
\$1%
2
Using g D 32 ft/s2 , the mass density is
D
10 in
55 D 62.5 in2 .
W
D 0.005 slug/in2 .
gA
5 in
x
Using the parallel axis theorem, the moment of inertia about the x axis
is
Ix
axis
D
1
1
1053 C 553 3
36
5 2 1
C 5C
55 D 4.948 slug-in2
3
2
Ix
662
axis
D 0.03436 slug-ft2
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Problem 8.120 Determine the moment of inertia of the
plate in Problem 8.119 about the y axis.
Solution: Use the results of the solution in Problem 8.119 for the
area and the mass density.
Iy
axis
D
1
1
5103 C 553 3
36
10 2 1
C 5C
55
3
2
D 12.76 slug-in2 D 0.0886 slug-ft2
Problem 8.121 The thermal radiator (used to eliminate excess heat from a satellite) can be modeled as a
homogenous, thin rectangular plate. Its mass is 5 slugs.
Determine its moment of inertia about the x, y, and
z axes.
y
3 ft
6 ft
3 ft
Solution: The area is A D 93 D 27 ft2 . The mass density is
D
5
m
D
D 0.1852 slugs/ft2 .
A
27
2 ft
x
The moment of inertia about the centroid of the rectangle is
Ixc D Iyc D 1
12
1
12
933 D 3.75 slug-ft2 ,
393 D 33.75 slug-ft2 .
Use the parallel axis theorem:
Ix
axis
D A2 C 1.52 C Ixc D 65 slug-ft2 ,
Iy
axis
D A4.5 32 C Iyc D 45 slug-ft2 .
Iz
axis
D Ix
axis
C Iy
axis
D 110 slug-ft2
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
663
Problem 8.122 The homogeneous cylinder has mass m,
length l, and radius R. Use integration as described in
Example 8.13 to determine its moment of inertia about
the x axis.
y
x
Solution: The volume of the disk element is R2 dz and its mass is
dm D R2 dz, where is the density of the cylinder. From Appendix
C, the moment of inertia of the disk element about the x0 axis is
dIx0 axis D
1
1
dmR2 D R2 dzR2 .
4
4
R
l
Applying the parallel-axis theorem, the moment of inertia of the disk
element about the x axis is
dIxaxis
z
1
D dIx0 axis C z2 dm D R2 dzR2 C z2 R2 dz
4
Integrating this expression from z D 0 to z D l gives the moment of
inertia of the cylinder about the x axis.
l
Ixaxis D
0
1
1
1
R4 C R2 z2 dz D R4 l C R2 l3 .
4
4
3
In terms of the mass of the cylinder m D R2 l,
Ixaxis D
1 2 1 2
mR C ml
4
3
Problem 8.123 The homogenous cone is of mass m.
Determine its moment of inertia about the z axis, and
compare your result with the value given in Appendix C.
(See Example 8.13.)
y
x
Solution: The differential mass
dm D
R
3m
m
r 2 dz D 2 r 2 dz.
V
R h
The moment of inertia of this disk about the z axis is 12 mr 2 . The radius
varies with z,
rD
h
z
R
z,
h
from which
Iz
664
axis
D
3mR2
2h5
h
0
z4 dz D
3mR2
2h5
z5
5
h
D
0
3mR2
10
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Problem 8.124 Determine the moments of inertia of
the homogenous cone in Problem 8.123 about the x and
y axes, and compare your results with the values given
in Appendix C.
3m
m
D
. The differential
V
R2 h
element of mass is dm D r 2 dz. The moment of inertia of this
elemental disk about an axis through its center of mass, parallel to
the x- and y-axes, is
Solution: The mass density is D
\$1%
dIx D
4
r 2 dm.
Use the parallel axis theorem,
\$1%
Ix D
m
4
r 2 dm C
z2 dm.
m
Noting that r D
R
z, then
h
r 2 dm D and z2 dm D R4
h4
R2
h2
z4 dz,
z4 dz.
Substitute:
Ix D R4
4h4
h
z4 dz C 0
R2
h2
h
z4 dz.
0
Integrating and collecting terms:
Ix D
3mR2
3m
C 3
4h5
h
z5
5
h
D m
0
3 2 3 2
R C h .
20
5
By symmetry, Iy D Ix
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665
Problem 8.125 The mass of the homogeneous wedge
is m. Use integration as described in Example 8.13 to
determine its moment of inertia about the z axis. (Your
answer should be in terms of m, a, b, and h.)
y
x
Solution: Consider a triangular element of the wedge of thickness
dz. The mass of the element is the product of the density of the
material and the volume of the element, dm D 12 bhdz. The moments
of inertia of the triangular element about the x’ and y’ axes are given
by Eqs. (8.30) and (8.31) in terms of the mass of the element, its
triangular area, and the moments of inertia of the triangular area:
dIx0 axis
1
bhdz 1
dm 0
1
2
I D
rbh3 dz,
bh3 D
D
1
A x
12
12
bh
2
dIy 0 axis
1
bhdz 1
dm 0
1
I D 2
hb3 D rhb3 dz,
D
1
A y
4
4
bh
2
h
a
z
b
The moment of inertia of this thin plate about the z axis is
dIzaxis D dIx0 axis C dIy 0 axis D
1
1
bh3 dz C hb3 dz.
12
4
Integrating this expression from z D 0 to z D a gives the moment of
inertia of the wedge about the z axis:
a
Izaxis D
0
1
1
1
1
bh3 C hb3 dz D
bh3 a C hb3 a.
12
4
12
4
In terms of the mass m D 12 bha,
Izaxis D
666
1 2 1 2
mh C mb .
6
2
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 8.126 The mass of the homogeneous wedge
is m. Use integration as described in Example 8.13 to
determine its moment of inertia about the x axis. (Your
answer should be in terms of m, a, b, and h.)
y
x
Solution: Consider a triangular element of the wedge of thickness
dz. The mass of the element is the product of the density of the
material and the volume of the element, dm D 12 bhdz. The moments
of inertia of the triangular element about the x’ axis is given by Eq.
(8.30) in terms of the mass of the element, its triangular area, and the
moments of inertia of the triangular area:
dIx0 axis
h
a
z
b
1
bhdz 1
dm
1
Ix 0 D 2
rbh3 dz,
bh3 D
D
1
A
36
36
bh
2
Applying the parallel-axis theorem, the moment of inertia of the
triangular element about the x axis is
1
dIxaxis D dIx0 axis C z2 C h2 dm
3
D
1
1
1
1
1
bh3 dz C [z2 C h2 ] bhdz D
bh3 dz C bhz2 dz
36
3
2
12
2
Integrating this expression from z D 0 to z D a gives the moment of
inertia of the wedge about the x axis:
a
Ixaxis D
0
1
1
1
1
bh3 C bhz2 dz D
bh3 a C bha3 .
12
2
12
6
In terms of the mass m D 12 bha,
Ixaxis D
1 2 1 2
mh C ma .
6
3
Problem 8.127 In Example 8.12, suppose that part of
the 3-kg bar is sawed off so that the bar is 0.4 m long
and its mass is 2 kg. Determine the moment of inertia
of the composite object about the perpendicular axis L
through the center of mass of the modified object.
Solution: The mass of the disk is 2 kg. Measuring from the left
0.2 m
0.6 m
L
end of the rod, we locate the center of mass
xD
2 kg0.2 m C 2 kg0.6 m
D 0.4 m.
2 kg C 2 kg
The center of mass is located at the point where the rod and disk are
connected. The moment of inertia is
'
&amp;
1
1
I D 2 kg0.4 m2 C
2 kg0.2 m2 C 2 kg0.2 m2
3
2
I D 0.227 kg-m2 .
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
667
y
Problem 8.128 The L-shaped machine part is composed of two homogeneous bars. Bar 1 is tungsten alloy
with mass density 14,000 kg/m3 , and bar 2 is steel
with mass density 7800 kg/m3 . Determine its moment
of inertia about the x axis.
240 mm
1
40 mm
2
80 mm
Solution: The masses of the bars are
m1 D 14,0000.240.080.04 D 10.75 kg
80 mm
z
240 mm
m2 D 78000.240.080.04 D 5.99 kg.
Using Appendix C and the parallel axis theorem the moments of inertia
of the parts about the x axis are
Ix
axis1
D
1
m1 [0.042 C 0.242 ] C m1 0.122 D 0.2079 kg-m2 ,
12
Ix
axis2
D
1
m2 [0.042 C 0.082 ] C m2 0.042 D 0.0136 kg-m2 .
12
x
Therefore
Ix
668
axis
D Ix
axis1
C Ix
axis2
D 0.221 kg-m2
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 8.129 The homogeneous object is a cone
with a conical hole. The dimensions R1 D 2 in, R2 D
1 in, h1 D 6 in, and h2 D 3 in. It consists of an
aluminum alloy with a density of 5 slug/ft3 . Determine
its moment of inertia about the x axis.
y
x
R1
Solution: The density of the material is
D 5 slug/ft3
1 ft
12 in
3
D 0.00289 slug/in3 .
The volume of the conical object without the conical hole is
V1 D
R2
h1
z
h2
1 2
1
R h1 D 2 in2 6 in D 25.1 in3 .
3 1
3
The mass of the conical object without the conical hole is m1 D V1 D
0.0727 slug. From Appendix C, the moment of inertia of the conical
object without the conical hole about the x axis is
Ix 1 D m1
3 2
3 2
h C
R
5 1 20 1
D 0.0727 slug
3
3
6 in2 C
2 in2 D 1.61 slug-in2
5
20
The volume of the conical hole is
V2 D
1 2
1
R h2 D 1 in2 3 in D 3.14 in3 .
3 2
3
The mass of the material that would occupy the conical hole is m2 D
V2 D 0.00909 slug. The z coordinate of the center of mass of the
material that would occupy the conical hole is
z D h1 h2 C
3
3
h2 D 6 in 3 in C 3 in D 5.25 in.
4
4
Using Appendix C and applying the parallel-axis theorem, the moment
of inertia about the x axis of the material that would occupy the conical
hole is
Ix 2 D m2 3 2
3 2
h C
R C z2 m2 D 0.255 slug-in2 .
80 2 20 2
The moment of inertia of the conical object with the conical hole is
Ix D Ix 1 Ix 2 D 1.36 slug-in2 .
Ix D 1.36 slug-in2 .
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
669
Problem 8.130 The circular cylinder is made of
aluminum (Al) with density 2700 kg/m3 and iron (Fe)
with density 7860 kg/m3 . Determine its moments of
inertia about the x 0 and y 0 axes.
y
y
Al
z
Fe
600 mm
200 mm
z
Solution: We have Al D 2700 kg/m3 , Fe D 7860 kg/m3
600 mm
x, x
We first locate the center of mass
xD
Al [0.1 m]2 [0.6 m]0.3 m C Fe [0.1 m]2 [0.6 m]0.9 m
Al [0.1 m]2 [0.6 m] C Fe [0.1 m]2 [0.6 m]
D 0.747 m
We also have the masses
mAl D Al 0.1 m2 0.6 m, mFe D Fe 0.1 m2 0.6 m
Now find the moments of inertia
1
1
mAl 0.1 m2 C mFe 0.1 m2 D 0.995 kg m2
2
2
[0.1 m]2
[0.6 m]2
C
D mAl
C mAl x 0.3 m2
12
4
[0.1 m]2
[0.6 m]2
C mFe
C
C mFe 0.9 m x2
12
4
D 20.1 kg m2
Ix 0 D
Iy 0
Problem 8.131 The homogenous half-cylinder is of
mass m. Determine its moment of inertia about the axis
L through its center of mass.
Solution:
The centroid of the half cylinder is located a distance
R
L
T
4R
from the edge diameter. The strategy is to use the parallel
3
axis theorem to treat the moment of inertia of a complete cylinder as
the sum of the moments of inertia for the two half cylinders. From
Problem 8.118, the moment of inertia about the geometric axis for a
cylinder is IcL D mR2 , where m is one half the mass of the cylinder.
of
By the parallel axis theorem,
IcL D 2
4R
3
2
m C IhL
.
Solve
IhL D
IcL
2
D mR2
D mR2
670
4R
3
2 2 16
mR
mR2
m D
2
2
9
16
1
2
92
16
1
2
92
D 0.31987 mR2 D 0.32 mR2
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 8.132 The homogeneous machine part is
made of aluminum alloy with density D 2800 kg/m3 .
Determine its moment of inertia about the z axis.
y
y
20 mm
x
z
40 mm
120 mm
Solution: We divide the machine part into the 3 parts shown: (The
dimension into the page is 0.04 m). The masses of the parts are
y
m1 D 28000.120.080.04 D 1.075 kg,
m2 D 2800 12 0.042 0.04 D 0.281 kg,
40
mm
0.12
m
1
0.08
m
x
m3 D 28000.022 0.04 D 0.141 kg.
y
Iz
axis1
0.12 m
C 2
0.12 m
1
m1 [0.082 C 0.122 ] C m1 0.062
D
12
The moment of inertia of part 2 about the axis through the center C
that is parallel to the z axis is
x
0.02 m
Iz
axis2
D 0.000144 C m2 0.12 C 0.0172 D 0.00543 kg-m2 .
The moment of inertia of the material that would occupy the hole 3
D 12 m2 0.042 .
The distance along the x axis from C to the center of mass of part 2 is
40.04/3 D 0.0170 m.
Iz
axis3
Therefore Iz
Therefore, the moment of inertia of part 2 about the z axis through its
center of mass that is parallel to the axis is
1
2
2 m2 0.04
3
x –
0.04 m
Using this result, the moment of inertia of part 2 about the z axis is
D 0.00573 kg-m2 .
1
2
2 m2 R
y
+
Using Appendix C and the parallel axis theorem the moment of inertia
of part 1 about the z axis is
axis
D 12 m3 0.022 C m3 0.122 D 0.00205 kg-m2 .
D Iz
axis1
C Iz
axis2
Iz
axis3
D 0.00911 kg-m2 .
m2 0.01702 D 0.000144 kg-m2 .
Problem 8.133 Determine the moment of inertia of the
machine part in Problem 8.132 about the x axis.
Solution: We divide the machine part into the 3 parts shown in
the solution to Problem 8.132. Using Appendix C and the parallel axis
theorem, the moments of inertia of the parts about the x axis are:
1
m1 [0.082 C 0.042 ] D 0.0007168 kg-m2
12
Ix
axis1
D
Ix
axis2
D m2
1
1
0.042 C 0.042
12
4
D 0.0001501 kg-m2
Ix
axis3
D m3
1
1
0.042 C 0.022
12
4
D 0.0000328 kg-m2 .
Therefore, Ix
axis
D Ix
axis1
C Ix
axis2
Ix
axis3
D 0.000834 kg-m2 .
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
671
Problem 8.134 The object consists of steel of density
D 7800 kg/m3 . Determine its moment of inertia about
the axis LO .
20 mm
O
100 mm
Solution: Divide the object into four parts: Part (1) The semicylinder of radius R D 0.02 m, height h1 D 0.01 m.
Part (2): The rectangular solid L D 0.1 m by h2 D 0.01 m by w D
0.04 m. Part (3): The semi-cylinder of radius R D 0.02 m, h1 D
0.01 m. Part (4) The cylinder of radius R D 0.02 m, height h D
0.03 m.
10 mm
30 mm
LO
Part (1)
m1 D
I1 D
R2 h1
D 0.049 kg,
2
m1 R 2
D 4.9 &eth; 106 kg-m2 ,
4
Part (2):
m2 D wLh2 D 0.312 kg,
I2 D
1
12
m2 L 2 C w2 C m2
2
L
D 0.00108 kg-m2 .
2
Part (3)
m3 D m1 D 0.049 kg,
I3 D 4R
3
2
4R 2
m2 C I 1 C m 3 L D 0.00041179 kg m2 .
3
Part (4)
m4 D R2 h D 0.294 kg,
I4 D
\$1%
2
m4 R2 C m4 L 2 D 0.003 kg m2 .
The composite:
ILo D I1 C I2 I3 C I4 D 0.003674 kg m2
Problem 8.135 Determine the moment of inertia of
the object in Problem 8.134 about the axis through the
center of mass of the object parallel to LO .
Solution: The center of mass is located relative to LO
4R
4R
m1 C m2 0.05 m3 0.1 C m4 0.1
3
3
xD
m1 C m 2 m 3 C m 4
D 0.066 m,
Ic D x2 m C ILo D 0.00265 C 0.00367 D 0.00102 kg m2
672
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 8.136 The thick plate consists of steel of
density D 15 slug/ft3 . Determine its moment of inertia
y
y
4 in
2 in
2 in
x
Solution: Divide the object into three parts: Part (1) the rectangle
8 in by 16 in, Parts (2) &amp; (3) the cylindrical cut outs. Part (1):
m1 D 8164 D 4.444 slugs.
I1 D
1
12
z
4 in
4 in
8 in
4 in
4 in
m1 162 C 82 D 118.52 slug in2 .
Part (2):
m2 D 22 4 D 0.4363 slug,
I2 D
m2 22 C m2 42 D 7.854 slug in2 .
2
Part (3):
m3 D m2 D 0.4363 slugs,
I3 D I2 D 7.854 slug-in2 .
The composite:
Iz
axis
D I1 2I2 D 102.81 slug-in2
Iz
axis
D 0.714 slug-ft2
Problem 8.137 Determine the moment of inertia of the
plate in Problem 8.136 about the x axis.
Solution: Use
Problem 8.136.
the
same
divisions
of
the
object
as
in
Part (1):
I1x
axis
D
1
12
m1 82 C 42 D 29.63 slug-in2 ,
Part (2):
I2x
axis
D
1
12
m2 322 C 42 D 1.018 slug-in2 .
The composite:
Ix
axis
D I1x
axis
2I2x
axis
D 27.59 slug in2
D 0.1916 slug ft2
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
673
Problem 8.138 Determine Iy and ky .
y
(1, 1)
Solution:
1
dA D dx dyA D
dx
Iy
D
A
1
x 2 dx D .
3
x2
x 2 dx
1
dy D
0
0
A
ky D
0
1
x 2 dA D
Iy D
1
dy D
0
0
y = x2
x2
x 4 dx D
0
x5
5
1
D
0
1
5
x
3
5
Problem 8.139 Determine Ix and kx .
Solution: (See figure in Problem 8.138.) dA D dx dy,
kx D
[x7 ]10 D
y 2 dy D
0
0
1
21
x2
dx
A
D
1
y 2 dA D
Ix D
1
3
1
x 6 dx
0
1
21
Ix
1
D p
A
7
Problem 8.140 Determine JO and kO .
Solution: (See figure in Problem 8.138.)
JO D Ix C Iy D
kO D
1
26
1
C
D
,
5
21
105
kx2 C ky2 D
1
3
C D
5
7
26
35
Problem 8.141 Determine Ixy .
Solution: (See figure in Problem 8.138.) dA D dx dy
D
674
1
2
0
1
x2
x dx
0
A
1
xy dA D
Ixy D
x 5 dx D
y dy
0
1 6 1
1
[x ]0 D
12
12
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 8.142 Determine Iy and ky .
Solution: By definition,
y
y = x – 1– x 2
4
x 2 dA.
Iy D
A
The element of area is dA D dx dy. The limits on the variable x are
0 x 4. The area is
4
dy D
0
0
4
Iy D
x3
x2
2
12
xx2 /4
x 2 dx
4
0
x5
x4
4
20
4
D 2.6667
0
x
dy D
0
D
xx2 /4
dx
x
0
x2
4
x 2 dx
4
D 12.8
0
from which
ky D
Iy
D 2.19
A
Problem 8.143 Determine Ix and kx .
Solution: By definition,
y 2 dA,
Ix D
A
from which
4
0
Ix D
xx2 /4
dx
Ix D
y 2 dy D
0
4 3
1
x2
dx
x
3
4
0
4
4
1
3 6
x7
x
3 5
D 0.6095.
x C
x 3
4
20
96
448 0
From Problem 8.142,
A D 2.667, kx D
Ix
D 0.4781
A
Problem 8.144 Determine Ixy .
Solution:
Ixy D
xy dA,
A
4
x dx
D
0
xx2 /4
y dy
0
D
4 2
1
x2
x dx
x
2
4
0
D
4
4
x
x5
x6
1
C
D 2.1333
2
4
10
96 0
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
675
Problem 8.145 Determine Iy 0 and ky 0 .
Solution: The limits on the variable x are 0 x 4. By definition,
y dA D
Ay D
4
y dy
0
4 2
1
x2
D
dx
x
2
4
0
D
y'
y = x – 1– x 2
4
xx2 /4
dx
0
A
y
x'
x
3
4
x4
x5
1
x
C
D 1.06667.
2
3
8
80 0
From Problem 8.142 the area is A D 2.667, from which y D 0.3999 D
0.4. Similarly,
4
XX2 /4
x dx
Ax D
0
dy
0
4
D
0
3
4
x2
x
x4
x x
D 5.3333,
dx D
4
3
16 0
from which x D 1.9999 D 2. The area moment of inertia is Iyy D
x2 A C Iy . Using the result of Problem 8.142, Iy D 12.8, from which
the area moment of inertia about the centroid is
Iy 0 D 10.6666 C 12.8 D 2.133
and ky 0 D
Iy 0
D 0.8944
A
Problem 8.146 Determine Ix0 and kx0 .
Solution: Using the results of Problems 8.143 and 8.145, Ix D
0.6095 and y D 0.4. The area moment of inertia about the centroid is
Ix0 D y2 A C Ix D 0.1828
and k D
x0
Ix 0
D 0.2618
A
Problem 8.147 Determine Ix0 y 0 .
Solution: From Problems 8.143 and 8.144, Ixy D 2.133 and x D
2, y D 0.4. The product of the moment of inertia about the centroid is
Ix0 y 0 D xyA C Ixy D 2.133 C 2.133 D 0
676
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
y
Problem 8.148 Determine Iy and ky .
Solution: Divide the section into two parts: Part (1) is the upper
rectangle 40 mm by 200 mm, Part (2) is the lower rectangle, 160 mm
by 40 mm.
Part (1) A1 D 0.0400.200 D 0.008 m2 ,
y1 D 0.180 m
40
mm
160
mm
x1 D 0,
Iy1 D
1
12
0.040.23 D 2.6667 &eth; 105 m4 .
x
80
mm
40
mm
80
mm
Part (2): A2 D 0.040.16 D 0.0064 m2 ,
y2 D 0.08 m,
x2 D 0,
Iy2 D
1
12
0.160.043 D 8.5 &eth; 107 m4 .
The composite:
A D A1 C A2 D 0.0144 m2 ,
Iy D Iy1 C Iy2 ,
Iy D 2.752 &eth; 105 m4 D 2.752 &eth; 107 mm4 ,
and ky D
Iy
D 0.0437 m D 43.7 mm
A
Problem 8.149 Determine Ix and kx for the area in
Problem 8.148.
Solution: Use the results in the solution to Problem 8.148. Part (1)
A1 D 0.0400.200 D 0.008 m2 ,
y1 D 0.180 m,
Ix1 D
1
12
0.20.043 C 0.182 A1 D 2.603 &eth; 104 m4 .
Part (2):
A2 D 0.040.16 D 0.0064 m2 ,
y2 D 0.08 m,
Ix2 D
1
12
0.040.163 C 0.082 A2 D 5.461 &eth; 105 m4 .
The composite: A D A1 C A2 D 0.0144 m2 , The area moment of
inertia about the x axis is
Ix D Ix1 C Ix2 D 3.15 &eth; 104 m4 D 3.15 &eth; 108 mm4 ,
and kx D
Ix
D 0.1479 m D 147.9 mm
A
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
677
y
Problem 8.150 Determine Ix and kx .
Solution: Use the results of the solutions to Problems 8.148–
40
mm
8.149. The centroid is located relative to the base at
xc D
x1 A1 C x2 A2
D 0,
A
yc D
y1 A1 C y2 A2
D 0.1356 m.
A
x
160
mm
The moment of inertia about the x axis is
80
mm
Ixc D y2C A C IX D 5.028 &eth; 107 mm4
and kxc D
40
mm
80
mm
Ixc
D 59.1 mm
A
Problem 8.151 Determine JO and kO for the area in
Problem 8.150.
Solution: Use the results of the solutions to Problems 8.148–
8.149. The area moments of inertia about the centroid are
Ixc D 5.028 &eth; 105 m4
and Iyc D Iy D 2.752 &eth; 105 m4 ,
from which
JO D Ixc C Iyc D 7.78 &eth; 105 m4 D 7.78 &eth; 107 mm4
and kO D
JO
D 0.0735 m
A
D 73.5 mm
Problem 8.152 Determine Iy and ky .
y
Solution: For a semicircle about a diameter:
Iyy D Ixx D
Iy D
1
1
44 24 D 44 24 D 94.25 ft4 ,
8
8
8
ky D
678
1
R4 ,
8
2 ft
x
4 ft
2Iy
D 2.236 ft
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 8.153 Determine JO and kO . for the area in
Problem 8.152.
Solution: For a semicircle:
Iyy D Ixx D
1
R4 .
8
4
4 24 D 94.248 ft4 .
8
Ix D
kx D
2Ix
D 2.236 ft.
42 22 Also use the solution to Problem 8.152.
JO D Ix C Iy D 294.248 D 188.5 ft4
kO D
2JO
D 3.16 ft
42 22 Problem 8.154 Determine Ix and kx .
Solution: Break the area into three parts: Part (1) The rectangle
with base 2a and altitude h; Part (2) The triangle on the right with
base b a and altitude h, and Part (3) The triangle on the left with
base b a and altitude h. Part (1) The area is
y
3 ft
3 ft
6 ft
A1 D 2ah D 24 ft2 .
The centroid is
x
x1 D 0
2 ft
h
and y1 D D 3 ft.
2
2 ft
y
b
The area moment of inertia about the centroid is
Ixc1 D
1
12
2ah3 D
Part (2): A2 D
x2 D a C
y2 D
ba
D 2.3333 ft,
3
1
36
h
1
hb a D 3 ft2 ,
2
2
h D 4 ft,
3
Ixc2 D
1
ah3 D 72 ft4 .
6
a
x
The composite moment of inertia
Ix D y1 2 A1 C Ixc1 C y2 2 A2 C Ixc2 C y3 2 A3 C Ixc3 ,
Ix D 396 ft4
b ah3 D 6 ft4 .
kx D
Ix
D
A
396
D 3.633 ft
30
Part (3): A3 D A2 ,
x3 D x2 , y3 D y2 , Ixc3 D Ixc2 .
The composite area is
A D A1 C A2 C A2 D 30 ft2 .
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
679
Problem 8.155 Determine Iy and ky for the area in
Problem 8.154.
Solution: Divide the area as in the solution to Problem 8.154.
Part (1) The area is A1 D 2ah D 24 ft2 . The centroid is x1 D 0 and
h
y1 D D 3 ft. The area moment of inertia about the centroid is
2
Iyc1 D
1
12
h2a3 D
Part (2): A2 D
x2 D a C
y2 D
2
ha3 D 32 ft4
3
1
hb a D 3 ft2 ,
2
ba
D 2.3333 ft,
3
2
h D 4 ft,
3
Iyc2 D
1
36
hb a3 D 0.1667 ft4 .
Part (3): A3 D A2 ,
x3 D x2 , y3 D y2 , Iyc3 D Iyc2 .
The composite area is
A D A1 C A2 C A2 D 30 ft2 .
The composite moment of inertia,
Iy D x21 A1 C Iyc1 C x22 A2 C Iyc2 C x23 A3 C Iyc3 ,
Iy D 65 ft4
and ky D
Iy
D 1.472 ft
A
Problem 8.156 The moments of inertia of the area are
Ix D 36 m4 , Iy D 145 m4 , and Ixy D 44.25 m4 . Determine a set of principal axes and the principal moment
of inertia.
Solution: The principal angle is
D
1
2Ixy
tan1
D 19.54&deg; .
2
Iy Ix
y
3m
4m
3m
x
The principal moments of inertia are
IxP D Ix cos2 2Ixy sin cos C Iy sin2 D 20.298 D 20.3 m4
IyP D Ix sin2 C 2Ixy sin cos C Iy cos2 D 160.70 m4
680
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 8.157 The moment of inertia of the 31-oz bat
about a perpendicular axis through point B is 0.093 slugft2 . What is the bat’s moment of inertia about a perpendicular axis through point A? (Point A is the bat’s “instantaneous center,” or center of rotation, at the instant shown.)
31
D 0.06023 slugs.
1632.17
Use the parallel axis theorem to obtain the moment of inertia about the
center of mass C, and then use the parallel axis theorem to translate
to the point A.
Solution: The mass of the bat is m D
IC D IA D
12
12
2
12 C 14
12
m C 0.093 D 0.0328 slug-ft2
2
m C 0.0328 D 0.3155 slug-ft2
C
12 in
B
14 in
A
Problem 8.158 The mass of the thin homogenous plate
is 4 kg. Determine its moment of inertia about the y axis.
Solution: Divide the object into two parts: Part (1) is the semicircle of radius 100 mm, and Part (2) is the rectangle 200 mm by
280 mm. The area of Part (1)
A1 D
y
100 mm
140 mm
x
R2
D 15708 mm2 .
2
140 mm
The area of Part (2) is
200 mm
A2 D 280200 D 56000 mm2 .
The composite area is A D A2 A1 D 40292 mm2 . The area mass
density is
D
4
D 9.9275 &eth; 105 kg/mm2 .
A
For Part (1) x1 D y1 D 0,
Iy1 D 1
R4 D 3898.5 kg-mm2 .
8
For Part (2) x2 D 100 mm.
Iy2 D x22 A2 C 1
12
2802003 D 74125.5 kg-mm2 .
The composite:
Iy D Iy2 Iy1 D 70226 kg-mm2 D 0.070226 kg-m2
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
681
Problem 8.159 Determine the moment of inertia of the
plate in Problem 8.158 about the z axis.
Solution: Use the same division of the parts and the results of the
solution to Problem 8.158. For Part (1),
Ix1 D 1
R4 D 3898.5 kg-mm2 .
8
For Part (2)
Ix2 D 1
12
2002803 D 36321.5 kg-mm2 .
The composite: Ix D Ix2 Ix1 D 32423 kg-mm2 ,
from which, using the result of the solution to Problem 8.158
Iz D Ix C Iy D 32422 C 70226 D 102649 kg-mm2
D 0.10265 kg-m2
y
Problem 8.160 The homogenous pyramid is of mass m.
Determine its moment of inertia about the z axis.
Solution: The mass density is
D
x
3m
m
D 2 .
V
w h
h
The differential mass is dm D ω2 dz. The moment of inertia of this
element about the z axis is
z
1
dIZ D
ω2 dm.
6
Noting that ω D
dIz D w4
6h4
z4 dz D
y
y
wz
, then
h
mw2 4
z dz.
2h5
Integrating:
Iz
axis
D
x
w
z
w
h
2
mw
2h5
h
z4 dz D
0
1
mw2
10
Problem 8.161 Determine the moment of inertia of the
homogenous pyramid in Problem 8.160 about the x and
y axes.
Solution: Use the results of the solution of Problem 8.160 for the
Noting that ω D
mass density. The elemental disk is dm D ω2 dz. The moment of
inertia about an axis through its center of mass parallel to the x axis is
dIX D
1
12
Ix
axis
D
w4
12h4
ω2 dm.
Ix
682
axis
D
1
12
h
z4 dz C
0
w2
h2
h
z4 dz.
0
Integrating and collecting terms
Use the parallel axis theorem:
w
z, the integral is
h
Ix
ω2 dm C
m
axis
Dm
1 2 3 2
w C h .
20
5
z2 dm.
m
By symmetry, Iy
axis
D Ix
axis
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 8.162 The homogenous object weighs
400 lb. Determine its moment of inertia about the x axis.
y
y
Solution: The volumes are
Vcyl D
4692
z
6 in
9 in
3
D 11,706 in ,
Vcone D 13 62 36 D 1357in3 ,
36 in
x
36 in
46 in
x
46 in
so V D Vcyl Vcone D 10,348 in3 .
The masses of the solid cylinder and the material that would occupy
the conical hole are
mcyl D
mcone D
Vcyl
V
Vcone
V
400
32.2
400
32.2
D 14.052 slug,
D 1.629 slug.
Using results from Appendix C,
Ix
axis
D
1
3
mcyl 92 mcone 62
2
10
D 551 slug-in2
D 3.83 slug-ft2
Problem 8.163 Determine the moments of inertia of
the object in Problem 8.162 about the y and z axes.
Solution: See the solution of Problem 8.162. The position of the
center of mass of the material that would occupy the conical hole is
3
36 D 37 in.
4
x D 46 36 C
6 in
x , x′
From Appendix C,
X
Iy 0
axiscone
y′
y
D mcone
3
3
362 C
62
80
20
36 in
D 87.97 slug-in2 .
The moment of inertia about the y axis for the composite object is
Iy
axis
D mcyl
)1
2
3 46
\$
Iy 0
C 14 92
axiscone
*
C x2 mcone
%
D 7877 slug-in2
D 54.7 slug-ft2 .
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
683
Problem 8.164 Determine the moment of inertia of the
14-kg flywheel about the axis L.
50 mm
70 mm
120 mm
L
100 mm
440 mm
500 mm
150 mm
Solution: The flywheel can be treated as a composite of the objects
shown:
The volumes are
=
–
+
–
+
4
V1 D 1502502 D 294.5 &eth; 105 mm3 ,
1
2
–
5
6
3
V2 D 1502202 D 228.08 &eth; 105 mm3 ,
V3 D 502202 D 76.03 &eth; 105 mm3 ,
V4 D 50602 D 5.65 &eth; 105 mm3 ,
V5 D 100602 D 11.31 &eth; 105 mm3 ,
V6 D 100352 D 3.85 &eth; 105 mm3 .
The volume
V D V1 V2 C V3 V4 C V5 V6
D 144.3 &eth; 105 mm3 ,
so the density is
υD
14
D 9.704 &eth; 107 kg/mm3 .
V
The moment of inertia is
IL D 12 υV1 2502 12 υV2 2202
C 12 υV3 2202 12 υV4 602
C 12 υV5 602 12 υV6 352
D 536,800 kg-mm2
D 0.5368 kg-m2 .
684
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 9.1 In Active Example 9.1, suppose that the
coefficient of static friction between the 180-lb crate
and the ramp is s D 0.3. What is the magnitude of
the smallest horizontal force the rope must exert on the
crate to prevent it from sliding down the ramp?
20⬚
Solution: The free-body diagram is shown.
Assume that the crate is on the verge of slipping down the ramp. Then
the friction force is f D s N and points up the ramp. The equilibrium
equations are
y
x
T
20⬚
Fx : s N C T cos 20&deg; W sin 20&deg; D 0,
Fy : N T sin 20&deg; W cos 20&deg; D 0.
Setting W D 180 lb and s D 0.3 and solving yields
W
f
N
N D 173 lb, T D 10.4 lb.
10.4 lb.
Problem 9.2 A person places a 2-lb book on a table
that is tilted at 5&deg; relative to the horizontal. She finds that
if she exerts a very small force on the book as shown,
the book remains in equilibrium, but if she removes the
force, the book slides down the table. What force would
she need to exert on the book (in the direction parallel
to the table) to cause it to slide up the table?
15⬚
Solution: If the person can hold the book in equilibrium with
a very small force, but the book slips when she removes the force,
the maximum friction force must be nearly equal to the value necessary to maintain the book in equilibrium. Assume that the book is in
equilibrium and f D s N (Fig. a). The equilibrium equations are
Fx : f 2 lb sin 15&deg; D 0,
Fy : N 2 lb cos 15&deg; D 0.
The coefficient of friction is then
s D
f
D tan 15&deg; D 0.268.
N
Now assume that the person exerts a force F on the book that is
parallel to the table and slip up the table is impending (Fig. b). Then
the friction force f D s N opposes the impending motion and the
equilibrium equations are
Fx : F f 2 lb sin 15&deg; D 0,
Fy : N 2 lb cos 15&deg; D 0.
Solving yields F D 1.04 lb .
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685
Problem 9.3 A student pushes a 200-lb box of books
across the floor. The coefficient of kinetic friction
between the carpet and the box is k D 0.15.
(a)
(b)
If he exerts the force F at angle ˛ D 25&deg; , what is
the magnitude of the force he must exert to slide
the box across the floor?
If he bends his knees more and exerts the force
F at angle ˛ D 10&deg; , what is the magnitude of the
force he must exert to slide the box?
F
F
Solution:
a
200 lb
Fx : F cos ˛ f D 0
α
Fy : N 200 lb F sin ˛ D 0
f D 0.15 N
(a)
˛ D 25&deg; ) F D 35.6 lb
f
(b)
˛D
10&deg;
) F D 31.3 lb
N
Problem 9.4 The 2975-lb car is parked on a sloped
street. The brakes are applied to both its front and rear
wheels.
(a)
(b)
If the coefficient of static friction between the car’s
tires and the road is s D 0.8, what is the steepest
slope (in degrees relative to the horizontal) on
which the car could remain in equilibrium?
If the street were icy and the coefficient of static
friction between the car’s tires and the road was
s D 0.2, what is the steepest slope on which the
car could remain in equilibrium?
Solution: Let ˛ be the slope of the street in degrees. The equilibrium equations and impending slip friction equation are
Fx : W sin ˛ f D 0,
Fy : N W cos ˛ D 0,
f D s N
Solving, we find that
f D W sin ˛, N D W cos ˛, s D tan ˛.
˛ D tan1 s .
a ˛ D tan1 0.8 D 38.7&deg; .
b ˛ D tan1 0.2 D 11.3&deg; .
a ˛ D 38.7&deg; , b ˛ D 11.3&deg; .
686
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Problem 9.5 The truck’s winch exerts a horizontal
force on the 200-kg crate in an effort to pull it down
the ramp. The coefficient of static friction between the
crate and the ramp is s D 0.6.
20⬚
(a)
If the winch exerts a 200-N horizontal force on the
crate, what is the magnitude of the friction force
exerted on the crate by the ramp?
(b) What is the magnitude of the horizontal force the
winch must exert on the crate to cause it to start
moving down the ramp?
Solution: Assume the crate doesn’t slip
1962 N
F% : N 1962 N cos 20&deg; C 200 N sin 20&deg; D 0
F- : f 1962 N sin 20&deg; 200 N cos 20&deg; D 0
20&deg;
fmax D 0.6 N
F = 200 N
f
(a)
Solving
N
N D 1775 N, f D 859 N, fmax D 1065 N
Sincef &lt; fmax ,
(b)
f D 859 N
F% : N 1962 N cos 20&deg; C F sin 20&deg; D 0
F- : f 1962 N sin 20&deg; F cos 20&deg; D 0 ) F D 380 N
f D 0.6 N
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687
Problem 9.6 The device shown is designed to position
pieces of luggage on a ramp. It exerts a force parallel
to the ramp. The suitcase weighs 40 lb. The coefficients
of friction between the suitcase and ramp are s D 0.20
and k D 0.18.
20⬚
(a)
Will the suitcase remain stationary on the ramp
when the device exerts no force on it?
(b) What force must the device exert to push the
suitcase up the ramp at a constant speed?
Solution:
(a)
Assume that the suitcase is in equilibrium with no external force
exerted on it (Fig. a). From the equilibrium equations
Fx : f W sin 20&deg; D 0,
Fy : N W cos 20&deg; D 0,
we obtain
f D W sin 20&deg; D 13.7 lb,
N D W cos 20&deg; D 37.6 lb.
The maximum friction force
fmax D s N D 0.237.6 lb D 7.25 lb
is less than the friction force necessary for equilibrium, so the
suitcase will not remain in equilibrium with no force exerted on
it.
(b) Now assume that the device exerts a force F on the suitcase and
is pushing it up the ramp at a constant speed (Fig. b). Then the
friction force f D k N opposes the motion and the equilibrium
equations are
Fx : F k N W sin 20&deg; D 0,
Fy : N W cos 20&deg; D 0.
Solving yields N D 37.6 lb, F D 20.4 lb. a No, b 20.4 lb.
688
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Problem 9.7 The coefficient of static friction between
the 50-kg crate and the ramp is s D 0.35. The
unstretched length of the spring is 800 mm, and the
spring constant is k D 660 N/m.
What is the minimum value of x at which the crate can
remain stationary on the ramp?
x
k
50⬚
Solution:
490.5 N
Fs D 660 N/m0.8 m x
F% : Fs 490.5 N sin 50&deg; C f D 0
50&deg;
F- : N 490.5 N cos 50&deg; D 0
N
f D 0.35 N
f
Solving:
x D 0.398 m D 398 mm
Fs
Problem 9.8 The coefficient of kinetic friction between the 40-kg crate and the slanting floor is k D 0.3.
If the angle ˛ D 20&deg; , what tension must the person exert
on the rope to move the crate at constant speed?
a
10⬚
Solution:
392.4 N
T
˛ D 20&deg; , k D 0.3
α
F% : T cos ˛ f 392.4 N sin 10&deg; D 0
F- : T sin ˛ C N 392.4 N cos 10&deg; D 0
f
f D 0.3 N
10&deg;
Solving
T D 177 N
N
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689
Problem 9.9 In Problem 9.8, for what angle ˛ is the
tension necessary to move the crate at constant speed a
minimum? What is the necessary tension?
Solution: See figure for 9.8
k D 0.3
F% : T cos ˛ f 392.4 N sin 10&deg; D 0
F- : T sin ˛ C N 392.4 N cos 10&deg; D 0
f D 0.3 N
) TD
184.1 N
cos ˛ C 0.3 sin ˛
To find the angle for minimum T
184.1 Nsin ˛ 0.3 cos ˛
dT
D
D 0 ) tan ˛ D 0.3
d˛
cos ˛ 0.3 sin ˛2
˛ D 16.7&deg;
)
T D 176.3 N
Problem 9.10 Box A weighs 100 lb, and box B weighs
30 lb. The coefficients of friction between box A and the
ramp are s D 0.30 and k D 0.28. What is the magnitude of the friction force exerted on box A by the ramp?
A
B
30&deg;
Solution: The sum of the forces parallel to the inclined surface is
B
F D A sin ˛ C B C f D 0,
α
A
from which f D A sin ˛ B D 100 sin 30&deg; 30 D 20 lb
f
N
690
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Problem 9.11 In Problem 9.10, box A weighs 100 lb,
and the coefficients of friction between box A and the
ramp are s D 0.30 and k D 0.28. For what range
of the weights of the box B will the system remain
stationary?
Solution: The upper and lower limits on the range are determined
by the weight required to move the box up the ramp, and the weight
that will allow the box to slip down the ramp. Assume impending
slip. The friction force opposes the impending motion. For impending
motion up the ramp the sum of forces parallel to the ramp are
BMAX
α
A
N
&micro; sN
F D A sin ˛ BMAX C S A cos ˛ D 0,
BMIN
from which
α
A
BMAX D Asin ˛ C s cos ˛
&micro; sN
D 100sin 30&deg; C 0.3 cos 30&deg; D 75.98 lb
N
For impending motion down the ramp:
F D A sin ˛ BMIN s A cos ˛ D 0,
from which
B D Asin ˛ s cos ˛
D 100sin 30&deg; 0.3 cos 30&deg; D 24.02 lb
Problem 9.12 The mass of the box on the left is 30 kg,
and the mass of the box on the right is 40 kg. The coefficient of static friction between each box and the inclined
surface is s D 0.2. Determine the minimum angle ˛ for
with the boxes will remain stationary.
a
Solution: If the boxes slip when ˛ is decreased, they will slip
toward the right. Assume that slip toward the right impends, the free
body diagrams are as shown.
y
y
T
α
x
T
30&deg;
(40)(9.81)
(30)(9.81)
The equilibrium equations are
30⬚
0.2 NB
Fx D T 0.2 NA 309.81 sin ˛ D 0,
(1)
Fy D NA 309.81 cos ˛ D 0,
(2)
Fx D T 0.2 NB C 409.81 sin 30&deg; D 0,
(3)
Fy D NB 409.81 cos 30&deg; D 0,
(4)
0.2 NA
NA
NB
x
Summing Equations (1) and (3), we obtain 0.2 NA 0.2 NB 309.81 sin ˛ C 409.81 sin 30&deg; D 0. Solving Equation (2) for
NA and Equation (4) for NB and substituting the results into
&deg;
Equation (5) gives
15 sin ˛ C 3 cos ˛ D 10 4 cos 30 . (6) Using the
identity cos ˛ D 1 sin2 ˛ and solving Equation (6) for sin ˛, we
obtain sin ˛ D 0.242, so ˛ D 14.0&deg;
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691
Problem 9.13 The coefficient of kinetic friction between
the 100-kg box and the inclined surface is 0.35. Determine the tension T necessary to pull the box up the
surface at a constant rate.
T
60⬚
3T
Solution:
F- : 3T 981 N sin 60&deg; f D 0
981 N
F% : N 981 N cos 60&deg; D 0
60&deg;
f D 0.35 N
Solving:
T D 340 N
N
f
692
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Problem 9.14 The box is stationary on the inclined
surface. The coefficient of static friction between the
box and the surface is s .
(a)
(b)
If the mass of the box is 10 kg, ˛ D 20&deg; , ˇ D 30&deg; ,
and s D 0.24, what force T is necessary to start
the box sliding up the surface?
Show that the force T necessary to start the box
sliding up the surface is a minimum when
tan ˇ D s .
T
β
α
Solution:
T
mg
y
α
β
α
f
x
N
˛ D 20&deg;
s D 0.24
m D 10 kg
g D 9.81 m/s2
(a)
Fx :
T cos ˇ C f C mg sin ˛ D 0
Fy :
N C T sin ˇ mg cos ˛ D 0
ˇ D 30&deg; , f D s N
Substituting the known values and solving, we get
T D 56.5 N,
N D 64.0 N,
f D 15.3 N
Solving the 2nd equilibrium eqn for N and substituting for
f f D s N in the first eqn, we get
T cos ˇ C s mg cos ˛ s T sin ˇ C mg sin ˛ D 0
Differentiating with respect to ˇ, we get
Tsin ˇ s cos ˇ
dT
D
dˇ
cos ˇ C s sin ˇ
Setting
dT
D 0, we get
dˇ
tan ˇ D s
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693
Problem 9.15 To explain observations of ship launchings at the port of Rochefort in 1779, Coulomb analyzed
the system shown in Problem 9.14 to determine the
minimum force T necessary to hold the box stationary
on the inclined surface. Show that the result is
TD
sin ˛ s cos ˛mg
.
cos ˇ s sin ˇ
Solution:
mg
y
T
β
α
α
x
N
f = &micro;sN
Fx :
T cos ˇ C mg sin ˛ s N D 0
Fy :
N C T sin ˇ mg cos ˛ D 0
˛ is fixed, ˇ is variable. Solve the second eqn for N and substitute
into the first. We get
0 D Ts sin ˇ cos ˇ D mgsin ˛ s cos ˛
or T D
mgsin ˛ s cos ˛
cos ˇ s sin ˇ
To get the conditions for the minimum, set
dT
D0
dˇ
Tsin ˇ C s cos ˇ
dT
D
D0
dˇ
cos ˇ s sin ˇ
For the min.
tan ˇ D s .
Note ˇ is negative!
694
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Problem 9.16 Two sheets of plywood A and B lie on
the bed of a truck. They have the same weight W, and
the coefficient of static friction between the two sheets
of wood and between sheet B and the truck bed is s .
(a)
(b)
If you apply a horizontal force to sheet A and apply
no force to sheet B, can you slide sheet A off the
truck without causing sheet B to move? What force
is necessary to cause sheet A to start moving?
If you prevent sheet A from moving by applying
a horizontal force on it, what horizontal force on
sheet B is necessary to start it moving?
A
B
Solution:
(a)
The friction force exerted by sheet A on B at impending motion
is fAB D s W. The friction force exerted by sheet B on the bed
of the truck is fBT D s 2W, since the normal force is due to
the weight of both sheets. Since fBT &gt; fAB , the top sheet will
begin moving before the bottom sheet. Yes
fAB
(a)
fAB
W
fBT
fAB
The force required to start sheet A to move is
(b)
2W
W
FB
F D fAB D s W.
(b)
F
W
fBT
2W
The force on B is the friction between A and B and the friction
between B and the truck bed. Thus the force required to start B
in motion is
FB D fAB C fBT D 3s W.
Problem 9.17 The weights of the two boxes are W1 D
100 lb and W2 D 50 lb. The coefficients of kinetic friction between the left box and the inclined surface are
s D 0.12 and k D 0.10. Determine the tension the
man must exert on the rope to pull the boxes upward
at a constant rate.
30⬚
W1
30⬚
W2
Solution:
F- : T 100 lb sin 30&deg; f 50 lb D 0
100 lb
T
F% : N 100 lb cos 30&deg; D 0
f D 0.10 N
Solving:
30&deg;
50 lb
T D 109 lb
f
N
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695
Problem 9.18 In Problem 9.17, for what range of
tensions exerted on the rope by the man will the boxes
remain stationary?
Solution: See the figure in 9.17.
First solve for the largest force Tmax
F- : Tmax 100 lb sin 30&deg; f 50 lb D 0
F% : N 100 lb cos 30&deg; D 0
) Tmax D 110.4 lb
f D 0.12 N
Next solve for the smallest force Tmin . We need to turn the friction
force in the opposite direction.
F- : Tmin 100 lb sin 30&deg; C f 50 lb D 0
F% : N 100 lb cos 30&deg; D 0
) Tmin D 89.6 lb
f D 0.12 N
Thus for the boxes to remain stationary we must have
89.6 lb &lt; T &lt; 110.4 lb
Problem 9.19 Each box weighs 10 lb. The coefficient
of static friction between box A and box B is 0.24, and
the coefficient of static friction between box B and the
inclined surface is 0.3. What is the largest angle ˛ for
which box B will not slip?
A
B
Strategy: Draw individual free-body diagrams of
the two boxes and write their equilibrium equations
assuming that slip of box B is impending.
a
Solution: We have 6 unknowns, 4 equilibrium equations and 2
10 lb
T
friction equations
FA% : T 10 lb sin ˛ f2 D 0
FA- : N2 10 lb cos ˛ D 0
A
f2
FB% : f2 C f1 10 lb sin ˛ D 0
N2
10 lb
FB- : N1 N2 10 lb cos ˛ D 0
f1
B
f1 D 0.3N1 , f2 D 0.24N2
Solving we find
˛ D 40.0&deg;
α
N1
696
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Problem 9.20 The masses of the boxes are mA D
15 kg and mB D 60 kg. The coefficient of static friction
between boxes A and B and between box B and the
inclined surface is 0.12. What is the largest force F for
which the boxes will not slip?
A
F
B
20⬚
Solution: We have 6 unknowns, 4 equilibrium equations and 2
friction equations.
147.15 N
T
FA% : T F 147.15 N sin 20&deg; C f2 D 0
FA- : N2 147.15 N cos 20&deg; D 0
A
f2
FB% : T 588.6 N sin 20&deg; f1 f2 D 0
T
F
N2
FB- : N1 N2 588.6 N cos 20&deg; D 0
B
f1 D 0.12N1 , f2 D 0.12N2
Solving we find
588.6 N
f1
F D 267 N
20&deg;
N1
Problem 9.21 In Problem 9.20, what is the smallest
force F for which the boxes will not slip?
Solution: See the solution for 9.20 — change the directions of all
of the friction forces.
FA% : T F 147.15 N sin 20&deg; f2 D 0
FA- : N2 147.15 N cos 20&deg; D 0
FB% : T 588.6 N sin 20&deg; C f1 C f2 D 0
) F D 34.8 N
FB- : N1 N2 588.6 N cos 20&deg; D 0
f1 D 0.12N1 , f2 D 0.12N2
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697
Problem 9.22 In Example 9.2, what clockwise couple
M would need to be applied to the disk to cause it to
rotate at a constant rate in the clockwise direction?
M
C
1
h
2
1
h
2
r
E
D
B
A
b
Solution: Assume that the disk is rotating in the clockwise direction. From the free-body diagram of the disk,
MD : M C R sin k r D 0.
From the free-body diagram of the brake,
MA : F 12 h C R cos k h C R sin k b D 0.
Solving these two equations yields
MD
1
2 hrFk
.
h C bk
Problem 9.23 The homogeneous horizontal bar AB
weighs 20 lb. The homogeneous disk weighs 30 lb. The
coefficient of kinetic friction between the disk and the
sloping surface is k D 0.24. What is the magnitude of
the couple that would need to be applied to the disk
to cause it to rotate at a constant rate in the clockwise
direction?
5 ft
1 ft
A
B
20⬚
Solution: From the free-body diagram of the bar,
MB : 20 lb2.5 ft C Ay 5 ft D 0
) Ay D 10 lb.
From the free-body diagram of the disk.
Fx : Ax C N sin 20&deg; C k N cos 20&deg; D 0,
Fy : N cos 20&deg; k N sin 20&deg; 30 lb D 0,
MA : M C k N1 ft D 0.
Solving yields Ax D 26.5 lb, N D 46.6 lb, M D 11.2 ft-lb.
M D 11.2 ft-lb.
698