Mechanical Engineering Chapter 12 Problem The Value Ifsolutionc The Circumference Circle And Its Radius Determine The

Problem 12.1 The value of πis 3.1415962654….. If

Cis the circumference of a circle and ris its radius,

Solution:

C=0.1592

Problem 12.2 The base of natural logarithms is e=

[Part (c) demonstrates the hazard of using rounded-off

values in calculations.]

Solution: The value of e is: e=2.718281828

Problem 12.3 A machinist drills a circular hole in a

panel with a nominal radius r=5 mm. The actual radius

Solution:

Problem 12.4 The opening in the soccer goal is 25 ft

wide and 8 ft high, so its area is 24 ft ×8ft=192 ft2.

What is its area in m2to three signiﬁcant digits?

Solution:

Problem 12.5 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:

Problem 12.6 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)

Solution: Convert the metric size nto inches, and compute the

percentage difference between the metric sized nut and the SAE

wrench. The results are:

A negative percentage implies that the metric nut is smaller than the

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Problem 12.7 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

Solution:

(a) h1=29032 ft

(b) In meters we have

Problem 12.8 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:

Problem 12.9 In the 2006 Winter Olympics, the men’s

15-km cross-country skiing race was won by Andrus

Solution:

Problem 12.10 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).

Problem 12.11 The kinetic energy of the man in Active

Example 12.1 is deﬁned by 1

Solution:

Problem 12.12 The acceleration due to gravity at sea

Solution: Use Table 1.2. The result is:

Problem 12.13 Afurlong per fortnight is a facetious

Solution:

Problem 12.14 Determine the cross-sectional area of

the beam (a) in m2; (b) in in2.

y

Solution:

A=(200 mm)2−2(80 mm)(120 mm)=20800 mm2

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Problem 12.15 The cross-sectional area of the

C12×30 American Standard Channel steel beam is A=

8.81 in2. What is its cross-sectional area in mm2?

y

A

Solution:

1in 2

Problem 12.16 A pressure transducer measures a value

of 300 lb/in2. Determine the value of the pressure in

Solution: Convert the units using Table 12.2 and the deﬁnition of

the Pascal unit. The result:

Problem 12.17 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:

Problem 12.18 Distributed loads on beams are expres-

Solution:

Problem 12.19 The moment of inertia of the rectan-

gular area about the xaxis is given by the equation

h

b

x

Solution:

25.4mm

Problem 12.20 In Example 12.3, instead of Einstein’s

equation consider the equation L=mc, where the mass

Solution:

Problem 12.21 The equation

What are the SI units of σ?

(b) If M=2000 N-m, y=0.1 m, and I=7×

10−5m4, what is the value of σin U.S. Customary

base units?

Solution:

Problem 12.22 The acceleration due to gravity on the

Solution:

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Problem 12.23 The 1 ft ×1ft ×1 ft cube of iron

weighs 490 lb at sea level. Determine the weight in new-

tons of a 1 m ×1m×1 m cube of the same material

at sea level. 1 ft

Solution: The weight density is γ=490 lb

1ft

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Problem 12.24 The area of the Paciﬁc Ocean is

Solution: The volume of the ocean is

sea level is g=9.81 m/s2. The radius of the earth

is 6370 km. The universal gravitational constant is

G=6.67 ×10−11 N-m2/kg2. Use this information to

R2. Solve for the mass,

(9.81 m/s2)(6370 km)2103m

km 2

Problem 12.26 A person weighs 180 lb at sea level.

Solution: Use Eq. (12.5).

Problem 12.27 The acceleration due to gravity on the

Solution:

Problem 12.28 If an object is near the surface of the

earth, the variation of its weight with distance from the

Solution: Use a variation of Eq. (12.5).

Problem 12.29 The planet Neptune has an equatorial

diameter of 49,532 km and its mass is 1.0247 ×1026 kg.

If the planet is modeled as a homogeneous sphere, what

is the acceleration due to gravity at its surface? (The uni-

versal gravitational constant is G=6.67 ×10−11 h-m2/

kg2.)

Solution:

We have: W=GmNm

Problem 12.30 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

Solution: Let rEp be the distance from the Earth to the point where

the gravitational accelerations are the same and let rMp be the distance

from the Moon to that point. Then, rEp +rMp =rEM =383,000 km.

The fact that the gravitational attractions by the Earth and the Moon

at this point are equal leads to the equation

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