Solution Manual For Engineering Mechanics: Statics, 5th Edition

Name: Solution Manual For Engineering Mechanics: Statics, 5th Edition
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Solution Manual For Engineering Mechanics: Statics, 5th Edition offers textbook solutions that are easy to follow, helping you ace your assignments.

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Problem 1.1 The value of is 3.14159265. . . If C is
the circumference of a circle and r is its radius, deter-
mine the value of r/C to four significant digits.
Solution: C D 2r ) r
C D 1
2 D 0.159154943.
To four significant digits we have r
C D 0.1592
Problem 1.2 The base of natural logarithms is e D
2.718281828 . . .
(a) Express e to five significant digits.
(b) Determine the value of e2 to five significant digits.
(c) Use the value of e you obtained in part (a) to deter-
mine the value of e2 to five significant digits.
[Part (c) demonstrates the hazard of using rounded-off
values in calculations.]
Solution: The value of e is: e D 2.718281828
(a) To five significant figures e D 2.7183
(b) e2 to five significant figures is e2 D 7.3891
(c) Using the value from part (a) we find e2 D 7.3892 which is
not correct in the fifth digit.
Problem 1.3 A machinist drills a circular hole in a
panel with a nominal radius r D 5 mm. The actual radius
of the hole is in the range r D 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 r12 D 78.226 m2
to A2 D r22 D 78.854 m2 . These numbers are equal only if rounded
to one significant digit:
One: A D 80 mm2
Problem 1.4 The opening in the soccer goal is 24 ft
wide and 8 ft high, so its area is 24 ft ð 8 ft D 192 ft 2 .
What is its area in m 2 to three significant digits?
Solution:
A D 192 ft2
( 1 m
3.281 ft
)2
D 17.8 m2
A D 17.8 m2
Problem 1.5 The Burj Dubai, scheduled for comple-
tion in 2008, will be the world’s tallest building with a
height of 705 m. The area of its ground footprint will be
8000 m2 . Convert its height and footprint area to U.S.
customary units to three significant digits.
Solution:
h D 705 m
( 3.281 ft
1 m
)
D 2.31 ð 103 ft
A D 8000 m2
( 3.218 ft
1 m
)2
D 8.61 ð 104 ft2
h D 2.31 ð 103 ft, A D 8.61 ð 104 ft2

Problem 1.6 Suppose that you have just purchased
a Ferrari F355 coupe and you want to know whether
you can use your set of SAE (U.S. Customary Units)
wrenches to work on it. You have wrenches with widths
w D 1/4 in, 1/2 in, 3/4 in, and 1 in, and the car has nuts
with dimensions n D 5 mm, 10 mm, 15 mm, 20 mm,
and 25 mm. Defining a wrench to fit if w is no more
than 2% larger than n, which of your wrenches can you
use?
n
Solution: Convert the metric size n to inches, and compute the
percentage difference between the metric sized nut and the SAE
wrench. The results are:
5 mm
( 1 inch
25.4 mm
)
D 0.19685.. in,
( 0.19685 0.25
0.19685
)
100
D 27.0%
10 mm
( 1 inch
25.4 mm
)
D 0.3937.. in,
( 0.3937 0.5
0.3937
)
100 D 27.0%
15 mm
( 1 inch
25.4 mm
)
D 0.5905.. in,
( 0.5905 0.5
0.5905
)
100 D C15.3%
20 mm
( 1 inch
25.4 mm
)
D 0.7874.. in,
( 0.7874 0.75
0.7874
)
100 D C4.7%
25 mm
( 1 inch
25.4 mm
)
D 0.9843.. in,
( 0.9843 1.0
0.9843
)
100 D 1.6%
A negative percentage implies that the metric nut is smaller than the
SAE wrench; a positive percentage means that the nut is larger then
the wrench. Thus within the definition of the 2% fit, the 1 in wrench
will fit the 25 mm nut. The other wrenches cannot be used.
Problem 1.7 Suppose that the height of Mt. Everest is
known to be between 29,032 ft and 29,034 ft. Based on
this information, to how many significant digits can you
express the height (a) in feet? (b) in meters?.
Solution:
a) h1 D 29032 ft
h2 D 29034 ft
The two heights are equal if rounded off to four significant digits.
The fifth digit is not meaningful.
Four: h D 29,030 ft
b) In meters we have
h1 D 29032 ft
( 1 m
3.281 ft
)
D 8848.52 m
h2 D 29034 ft
( 1 m
3.281 ft
)
D 8849.13 m
These two heights are equal if rounded off to three significant
digits. The fourth digit is not meaningful.
Three: h D 8850 m
Problem 1.8 The maglev (magnetic levitation) train
from Shanghai to the airport at Pudong reaches a speed
of 430 km/h. Determine its speed (a) in mi/h; (b) ft/s.
Solution:
a) v D 430 km
h
( 0.6214 mi
1 km
)
D 267 mi/h v D 267 mi/h
b) v D 430 km
h
( 1000 m
1 km
) ( 1 ft
0.3048 m
) ( 1 h
3600 s
)
D 392 ft/s
v D 392 ft/s
Problem 1.9 In the 2006 Winter Olympics, the men’s
15-km cross-country skiing race was won by Andrus
Veerpalu of Estonia in a time of 38 minutes, 1.3 seconds.
Determine his average speed (the distance traveled
divided by the time required) to three significant digits
(a) in km/h; (b) in mi/h.
Solution:
a) v D 15 km
(
38 C 1.3
60
)
min
( 60 min
1 h
)
D 23.7 km/h v D 23.7 km/h
b) v D 23.7 km/h
( 1 mi
1.609 km
)
D 14.7 mi/h v D 14.7 mi/h

Problem 1.10 The Porsche’s engine exerts 229 ft-lb
(foot-pounds) of torque at 4600 rpm. Determine the
value of the torque in N-m (Newton-meters).
Solution:
T D 229 ft-lb
( 1 N
0.2248 lb
) ( 1 m
3.281 ft
)
D 310 N-m T D 310 N-m
Problem 1.11 The kinetic energy of the man in Active
Example 1.1 is defined by 1
2 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 (68 kg)(6 m/s)2 D 1224 kg-m2 /s 2 . What is his kinetic
energy in U.S. Customary units?
Solution:
T D 1224 kg-m2/s 2
( 1 slug
14.59 kg
) ( 1 ft
0.3048 m
)2
D 903 slug-ft2/s
T D 903 slug-ft2/s
Problem 1.12 The acceleration due to gravity at sea
level in SI units is g D 9.81 m/s2 . By converting units,
use this value to determine the acceleration due to
gravity at sea level in U.S. Customary units.
Solution: Use Table 1.2. The result is:
g D 9.81
( m
s2
) ( 1 ft
0.3048 m
)
D 32.185 . . .
( ft
s2
)
D 32.2
( ft
s 2
)
Problem 1.13 A furlong per fortnight is a facetious
unit of velocity, perhaps made up by a student as a
satirical comment on the bewildering variety of units
engineers must deal with. A furlong is 660 ft (1/8 mile).
A fortnight is 2 weeks (14 nights). If you walk to class
at 2 m/s, what is your speed in furlongs per fortnight to
three significant digits?
Solution:
v D 2 m/s
( 1 ft
0.3048 m
) ( 1 furlong
660 ft
) ( 3600 s
hr
) ( 24 hr
1 day
) ( 14 day
1 fortnight
)
v D 12,000 furlongs
fortnight
Problem 1.14 Determine the cross-sectional area of
the beam (a) in m2 ; (b) in in2 .
120 mm x
y
40 mm
40 mm
40
mm
200 mm
Solution:
A D 200 mm2 280 mm120 mm D 20800 mm2
a) A D 20800 mm2
( 1 m
1000 mm
)2
D 0.0208 m2 A D 0.0208 m2
b) A D 20800 mm2
( 1 in
25.4 mm
)2
D 32.2 in2 A D 32.2 in2
Problem 1.15 The cross-sectional area of the C12ð30
American Standard Channel steel beam is A D 8.81 in 2 .
What is its cross-sectional area in mm 2 ?
x
y
A
Solution:
A D 8.81 in2
( 25.4 mm
1 in
)2
D 5680 mm2

Problem 1.16 A pressure transducer measures a value
of 300 lb/in2 . Determine the value of the pressure in
pascals. A pascal (Pa) is one newton per meter squared.
Solution: Convert the units using Table 1.2 and the definition of
the Pascal unit. The result:
300
( lb
in2
) ( 4.448 N
1 lb
) ( 12 in
1 ft
)2 ( 1 ft
0.3048 m
)2
D 2.0683 . . . 106
( N
m2
)
D 2.07106 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
) ( 1 m
3.281 ft
) ( 1 N
0.2248 lb
)
D 5.22 ð 106 W
P D 5.22 ð 106 W
Problem 1.18 Chapter 7 discusses distributed loads
that are expressed in units of force per unit length. If
the value of a distributed load is 400 N/m, what is its
value in lb/ft?.
Solution:
w D 400 N/m
( 0.2248 lb
1 N
) ( 1 m
3.281 ft
)
D 27.4 lb/ft w D 27.4 lb/ft
Problem 1.19 The moment of inertia of the rectan-
gular area about the x axis is given by the equation
I D 1
3 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) m 4 ; (c) in4 .
h
b
x
y
Solution:
(a) I D 1
3 200 mm100 mm3 D 66.7 ð 106 mm4
(b) I D 66.7 ð 106 mm 4
( 1 m
1000 mm
)4
D 66.7 ð 106 m4
(c) I D 66.7 ð 106 mm 4
( 1 in
25.4 mm
)4
D 160 in4
Problem 1.20 In Example 1.3, instead of Einstein’s
equation consider the equation L D mc, where the mass
m is in kilograms and the velocity of light c is in meters
per second. (a) What are the SI units of L? (b) If the
value of L in SI units is 12, what is its value in U.S.
Customary base units?
Solution:
a) L D mc ) Units L D kg-m/s
b) L D 12 kg-m/s
( 0.0685 slug
1 kg
) ( 3.281 ft
1 m
)
D 2.70 slug-ft/s
L D 2.70 slug-ft/s

Problem 1.21 The equation
D My
I
is used in the mechanics of materials to determine
normal stresses in beams.
(a) When this equation is expressed in terms of SI base
units, M is in newton-meters (N-m), y is in meters
(m), and I is in meters to the fourth power (m4 ).
What are the SI units of ?
(b) If M D 2000 N-m, y D 0.1 m, and I D 7 ð
105 m4 , what is the value of in U.S. Customary
base units?
Solution:
(a) D My
I D (N-m)m
m4 D N
m2
(b)
D My
I D 2000 N-m0.1 m
7 ð 105 m 4
( 1 lb
4.448 N
) ( 0.3048 m
ft
)2
D 59,700 lb
ft 2
Problem 1.22 The acceleration due to gravity on the
surface of the moon is 1.62 m/s 2 . (a) What would the
mass of the C-clamp in Active Example 1.4 be on the surface
of the moon? (b) What would the weight of the C-clamp
in newtons be on the surface of the moon?
Solution:
a) The mass does not depend on location. The mass in kg is
0.0272 slug
( 14.59 kg
1 slug
)
D 0.397 kg mass D 0.397 kg
b) The weight on the surface of the moon is
W D mg D 0.397 kg1.62 m/s2 D 0.643 N W D 0.643N
Problem 1.23 The 1 ft ð 1 ft ð 1 ft cube of iron
weighs 490 lb at sea level. Determine the weight in
newtons of a 1 m ð 1 m ð 1 m cube of the same
material at sea level. 1 ft
1 ft 1 ft
Solution: The weight density is D 490 lb
1 ft 3
The weight of the 1 m3 cube is:
W D V D
( 490 lb
1 ft 3
)

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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/s 2 . The moon’s radius is
RM D 1738 km.
(a) What is the weight in newtons on the surface of
the moon of an object that has a mass of 10 kg?
(b) Using the approach described in Example 1.5, deter-
mine the force exerted on the object by the gravity
of the moon if the object is located 1738 km above
the moon’s surface.
Solution:
a) W D mgM D 10 kg1.26 m/s2 D 12.6 N W D 12.6 N
b) Adapting equation 1.4 we have aM D gM
( RM
r
)2
. The force is
then
F D maM D 10 kg1.62 m/s2
( 1738 km
1738 km C 1738 km
)2
D 4.05 N
F D 4.05 N

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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.
VU
45
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°. Draw a sketch of the
new situation. The magnitudes of the forces are jFABj D
100 kN and jFACj D 60 kN. Graphically determine the
magnitude and direction of the sum of the forces exerted
on the pylon by the cables.
Solution: Accurately draw the vectors and measure the magnitude
and direction of the resultant
jFAB C FACj D 146 kN
˛ D 32°

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Problem 2.3 The magnitude jFAj D 80 lb and the
angle ˛ D 65°. The magnitude jFA C FBj D 120 lb.
Graphically determine the magnitude of FB.
FB
FA
a
FC

Solution: Accurately draw the vectors and measure the magnitude
of FB.
jFBj D 62 lb
Problem 2.4 The magnitudes jFAj D 40 N, jFBj D
50 N, and jFCj D 40 N. The angle ˛ D 50° and ˇ D 80°.
Graphically determine the magnitude of FA C FB C FC.
FB
FA
a
FC

Solution: Accurately draw the vectors and measure the magnitude
of FA C FB C FC.
R D jFA C FB C FCj D 83 N

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Problem 2.5 The magnitudes jFAj D jFBj D jFCj D
100 lb, and the angles ˛ D 30°. Graphically determine
the value of the angle ˇ for which the magnitude
jFA C FB C FCj is a minimum and the minimum value
of jFA C FB C FCj.
FB
FA
a
FC

Solution: For a minimum, the vector FC must point back to the
origin.
R D jFA C FB C FCj D 93.2 lb
ˇ D 165°
Problem 2.6 The angle D 50°. Graphically determine
the magnitude of the vector rAC. 60 mm 150 mm
A C
B
rAB rBC
rAC

Solution: Draw the vectors accurately and then measure jrACj.
jrACj D 181 mm

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Problem 2.7 The vectors FA and FB represent
the forces exerted on the pulley by the belt.
Their magnitudes are jFAj D 80 N and jFBj D 60 N.
Graphically determine the magnitude of the total force
the belt exerts on the pulley.
45
FA
FB
10
Solution: Draw the vectors accurately and then measure jFA C
FBj.
jFA C FBj D 134 N
Problem 2.8 The sum of the forces FA C FB C
FC D 0. The magnitude jFAj D 100 N and the angle ˛ D
60°. Graphically determine the magnitudes jFBj and jFCj.
30
FB
FA
FC
a
Solution: Draw the vectors so that they add to zero.
jFBj D 86.6 N, jFCj D 50.0 N
Problem 2.9 The sum of the forces FA C FB C
FC D 0. The magnitudes jFAj D 100 N and jFBj D
80 N. Graphically determine the magnitude jFCj and the
angle ˛. 30
FB
FA
FC
a
Solution: Draw the vectors so that they add to zero.
jFCj D 50.4 N, ˛ D 52.5°

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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. Graph-
ically determine the magnitudes of the lift and drag.
W
D
L
25
Solution: Draw the vectors so that they add to zero. Then measure
the unknown magnitudes.
jLj D 453 lb
jDj D 211 lb
Problem 2.11 A spherical storage tank is suspended
from cables. The tank is subjected to three forces, the
forces FA and FB exerted by the cables and its weight W.
The weight of the tank is jWj D 600 lb. The vector sum
of the forces acting on the tank equals zero. Graphically
determine the magnitudes of FA and FB.
40˚
FA
W
FB
20˚ 20˚
Solution: Draw the vectors so that they add to zero. Then measure
the unknown magnitudes.
jFAj D jFBj D 319 lb

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Problem 2.12 The rope ABC exerts forces FBA and
FBC of equal magnitude on the block at B. The
magnitude of the total force exerted on the block by
the two forces is 200 lb. Graphically determine jFBAj.
20
FBC
FBA
B
C
A
B
Solution: Draw the vectors accurately and then measure the
unknown magnitudes.
jFBAj D 174 lb
Problem 2.13 Two snowcats tow an emergency shelter
to a new location near McMurdo Station, Antarctica.
(The top view is shown. The cables are horizontal.)
The total force FA C FB exerted on the shelter is in
the direction parallel to the line L and its magnitude
is 400 lb. Graphically determine the magnitudes of F A
and FB.
L
Top View
FA
FB50 30
Solution: Draw the vectors accurately and then measure the
unknown magnitudes.
jFAj D 203 lb
jFBj D 311 lb
Problem 2.14 A surveyor determines that the horizon-
tal distance from A to B is 400 m and the horizontal
distance from A to C is 600 m. Graphically determine
the magnitude of the vector rBC and the angle ˛.
East
North
60
20
C
B
A
rBC
a
Solution: Draw the vectors accurately and then measure the
unknown magnitude and angle.
jrBCj D 390 m
˛ D 21.2°

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Problem 2.15 The vector r extends from point A to
the midpoint between points B and C. Prove that
r D 1
2 rAB C rAC .
A
C
B
rAC
r
rABrAB
Solution: The proof is straightforward:
r D rAB C rBM , and r D rAC C rCM .
Add the two equations and note that rBM C rCM D 0, since the two
vectors are equal and opposite in direction.
Thus 2r D rAC C rAB , or r D ( 1
2
) rAC C rAB
rAC
rAB
r
A B
M
C
Problem 2.16 By drawing sketches of the vectors,
explain why
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

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Problem 2.19 A support is subjected to a force F D
Fx i C 80j (N). If the support will safely support a force
of 100 N, what is the allowable range of values of the
component Fx?
Solution: Use the definition of magnitude in Eq. (2.8) and reduce
algebraically.
100 ½ √Fx 2 C 802 , from which 1002 802 ½ Fx 2 .
Thus jFx j p3600, or 60 Fx C60 (N)
Problem 2.20 If FA D 600i 800j (kip) and FB D
200i 200j (kip), what is the magnitude of the force
F D FA 2FB?
Solution: Take the scalar multiple of FB, add the components of
the two forces as in Eq. (2.9), and use the definition of the magnitude.
F D 600 2200i C 800 2200j D 200i 400j
jFj D √2002 C 4002 D 447.2 kip
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.
y
x
W
D
L
Solution:
L D W D D 500j 200i C 100j D 200i C 400jlb
jLj D √200 lb2 C 400 lb2 D 447 lb
L D 200i C 400jlb, jLj D 447 lb

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Problem 2.22 Two perpendicular vectors U and V lie
in the x-y plane. The vector U D 6i 8j and jVj D 20.
What are the components of V? (Notice that this problem
has two answers.)
Solution: The two possible values of V are shown in the sketch.
The strategy is to (a) determine the unit vector associated with U,
(b) express this vector in terms of an angle, (c) add š90° to this
angle, (d) determine the two unit vectors perpendicular to U, and
(e) calculate the components of the two possible values of V. The
unit vector parallel to U is
eU D 6i
√62 C 82 8j
√62 C 82 D 0.6i 0.8j
Expressed in terms of an angle,
eU D i cos ˛ j sin ˛ D i cos53.1° j sin53.1°
Add š90° to find the two unit vectors that are perpendicular to this
unit vector:
ep1 D i cos143.1° j sin143

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Solution Manual For Engineering Mechanics: Statics, 5th Edition

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