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Problem 1.1 The value of is 3.14159265... If Cis
Solution: CD2r )r
CD1
2D0.159154943.
Problem 1.2 The base of natural logarithms is eD
2.718281828 ...
(a) Express eto five significant digits.
Solution: The value of e is: eD2.718281828
(a) To five significant figures eD2.7183
Problem 1.3 A machinist drills a circular hole in a
Solution:
Problem 1.4 The opening in the soccer goal is 24 ft
wide and 8 ft high, so its area is 24 ft ð8ftD192 ft2.
What is its area in m2to three significant digits?
Problem 1.5 The Burj Dubai, scheduled for comple-
tion in 2008, will be the world’s tallest building with a
Solution:
Problem 1.6 Suppose that you have just purchased
a Ferrari F355 coupe and you want to know whether
with dimensions nD5 mm, 10 mm, 15 mm, 20 mm,
and 25 mm. Defining a wrench to fitifwis no more
than 2% larger than n, which of your wrenches can you
use?
n
Solution: Convert the metric size nto inches, and compute the
percentage difference between the metric sized nut and the SAE
25.4mm
0.19685 100
D27.0%
25.4mm
D0.3937.. in,0.3937 0.5
0.3937 100 D27.0%
25.4mm
0.5905 100 DC15.3%
25.4mm
0.7874 100 DC4.7%
25.4mm
0.9843 100 D1.6%
A negative percentage implies that the metric nut is smaller than the
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) h1D29032 ft
h2D29034 ft
b) In meters we have
Problem 1.8 The maglev (magnetic levitation) train
Solution:
Problem 1.9 In the 2006 Winter Olympics, the men’s
15-km cross-country skiing race was won by Andrus
Solution:
a) vD15 km
38 C1.3
1h D23.7 km/h vD23.7 km/h
2
Problem 1.10 The Porsche’s engine exerts 229 ft-lb
Solution:
Problem 1.11 The kinetic energy of the man in Active
Example 1.1 is defined by 1
Solution:
Problem 1.12 The acceleration due to gravity at sea
Solution: Use Table 1.2. The result is:
Problem 1.13 Afurlong per fortnight is a facetious
engineers must deal with. A furlong is 660 ft (1/8 mile).
Solution:
0.3048 m 1 furlong
660 ft 3600 s
hr 24 hr
1 day 14 day
1 fortnight
Problem 1.14 Determine the cross-sectional area of
the beam (a) in m2; (b) in in2.
y
40 mm
200 mm
Solution:
AD200 mm2280 mm120 mmD20800 mm2
25.4 mm 2
Problem 1.15 The cross-sectional area of the C12ð30
American Standard Channel steel beam is AD8.81 in2.
What is its cross-sectional area in mm2?
y
A
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.
Problem 1.17 A horsepower is 550 ft-lb/s. A watt is
1 N-m/s. Determine how many watts are generated by
Solution:
Problem 1.18 Chapter 7 discusses distributed loads
Solution:
Problem 1.19 The moment of inertia of the rectan-
gular area about the xaxis is given by the equation
Solution:
mis in kilograms and the velocity of light cis in meters
4
Problem 1.21 The equation
(m), and Iis in meters to the fourth power (m4).
What are the SI units of ?
(b) If MD2000 N-m, yD0.1 m, and ID7ð
105m4, what is the value of in U.S. Customary
base units?
Solution:
Problem 1.22 The acceleration due to gravity on the
Solution:
Problem 1.23 The 1 ft ð1ft ð1 ft cube of iron
weighs 490 lb at sea level. Determine the weight in
newtons of a 1 m ð1m ð1 m cube of the same
material at sea level. 1 ft
Solution: The weight density is D490 lb
1ft
3
The weight of the 1 m3cube is:
Problem 1.24 The area of the Pacific Ocean is
64,186,000 square miles and its average depth is 12,925 ft.
Solution: The volume of the ocean is
32.2 ft/s22.312 ð1019 ft3D4.60 ð1019 slugs
GD6.67 ð1011 N-m2/kg2. Use this information to
determine the mass of the earth.
mEDgR2
GD
9.81 m/s26370 km2103m
km 2
6.671011N-m2
kg2
Problem 1.26 A person weighs 180 lb at sea level. The
radius of the earth is 3960 mi. What force is exerted on
Solution: Use Eq (1.5).
the moon’s surface.
Problem 1.28 If an object is near the surface of the
earth, the variation of its weight with distance from the
center of the earth can often be neglected. The acceler-
Solution: Use a variation of Eq (1.5).
Problem 1.29 The planet Neptune has an equatorial
Solution:
Problem 1.30 At a point between the earth and the
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
Solution: Let rEp be the distance from the Earth to the point where
The fact that the gravitational attractions by the Earth and the Moon
at this point are equal leads to the equation
9.81 m
D1.62 m
,
6
