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SOLUTION MANUAL FOR ENGINEERING MECHANICS DYNAMICS 6TH EDITION CHAPTER 12_21 BY ANTHONY BEDFORD, WALLACE FOWLER €27,46
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SOLUTION MANUAL FOR ENGINEERING MECHANICS DYNAMICS 6TH EDITION CHAPTER 12_21 BY ANTHONY BEDFORD, WALLACE FOWLER

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SOLUTION MANUAL FOR ENGINEERING MECHANICS DYNAMICS 6TH EDITION CHAPTER 12_21 BY ANTHONY BEDFORD, WALLACE FOWLER

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Engineering
Mechanics Dynamics
6e Anthony Bedford,
Wallace Fowler
(Solu�ons Manual All
Chapters, 100%
Original Verified, A+
Grade)
(Chapter 12-21)

, Chapter 12

Problem 12.1 In 1967, the International Committee Solution: The number of cycles in two seconds is
of Weights and Measures defined one second to be the 2(9,192,631, 770) = 18,385, 263, 540.
time required for 9,192,631,770 cycles of the transition Expressed to four significant digits, this is 18,390, 000, 000 or 1.839E10
between two quantum states of the cesium-133 atom. cycles.
Express the number of cycles in two seconds to four sig- 18,390,000,000 or 1.839E10 cycles.
nificant digits.


Problem 12.2 The base of natural logarithms is Solution:
e = 2.71828183... . (a) Express e to three significant
(a) The rounded-off value is e = 2.72.
digits. (b) Determine the value of e 2 to three significant
(b) e 2 = 7.38905610..., so to three significant digits it is e 2 = 7.39.
digits. (c) Use the value of e you obtained in part (a) to (c) Squaring the three-digit number we obtained in part (a) and
determine the value of e 2 to three significant digits. expressing it to three significant digits, we obtain e 2 = 7.40.

[Comparing the answers of parts (b) and (c) demonstrates (a) e = 2.72. (b) e 2 = 7.39. (c) e 2 = 7.40.
the hazard of using rounded-off values in calculations.]



Problem 12.3 The base of natural logarithms (see Solution: The exact value rounded off to five significant digits
Problem 12.2) is given by the infinite series is e = 2.7183. Let N be the number of terms summed. We obtain the
results
1 1 1
e = 2+ + + +.
2! 3! 4! N Sum
Its value can be approximated by summing the first few 1 2
terms of the series. How many terms are needed for the 2 2.5
approximate value rounded off to five digits to be equal 3 2.666...
to the exact value rounded off to five digits? 4 2.708333...
5 2.71666...
6 2.7180555...
7 2.7182539...

We see that summing seven terms gives the rounded-off value 2.7183.
Seven.




Problem 12.4 The opening in the soccer goal is 24 ft
wide and 8 ft high, so its area is 24 ft × 8 ft = 192 ft 2 .
What is its area in m 2 to three significant digits?
Solution:
2
A = 192 ft 2 ( 3.281
1m
ft
) = 17.8 m 2

A = 17.8 m 2 .




Problem 12.5 In 2020, teams from China and Nepal, Solution:
based on their independent measurements using GPS
(a) The height of the mountain in kilometers to three significant digits is
satellites, determined that the height of Mount Everest is
8848.86 meters. Determine the height of the mountain 8848.86 m = 8848.86 m ( 1000
1 km
m
) = 8.85 km.
to three significant digits (a) in kilometers; (b) in miles.
(b) Its height in miles to three significant digits is

8848.86 m = 8848.86 m ( 3.281
1m
ft
)( 5280
1 mi
ft
) = 5.50 mi.
(a) 8.85 km. (b) 5.50 mi.




Copyright © 2024, 2008 by Pearson Education, Inc. or its affiliates.
1


BandF_6e_ISM_C12.indd 1 15/09/2

, Problem 12.6 The distance D = 6 in (inches). Solution: The value of M P is

The magnitude of the moment of the force M P = FD
F = 12 lb (pounds) about point P is defined to be = (12 lb)(6 in)
the product M P = FD. What is the value of M P in = 72 lb-in.
N-m (newton-meters)? Converting units, the value of M P in N-m is


F
72 lb-in = 72 lb-in ( 0.2248
1N
)( 1 m ) = 8.14 N-m.
lb 39.37 in

8.14 N-m.


P

D




Problem 12.7 The length of this Boeing 737 is Solution: Because 1 ft = 12 in, the length in meters is
110 ft 4 in and its wingspan is 117 ft 5 in. Its maxi-
mum takeoff weight is 154,500 lb. Its maximum range is (110 + 124 ft )( 0.3048
1 ft
m
) = 33.6 m.
3365 nautical miles. Express each of these quantities in In the same way, the wingspan in meters is
SI units to three significant digits.
(117 + 125 ft )( 0.3048
1 ft
m
) = 35.8 m.
The weight in newtons is

( 154, 500 ( 4.448
lb )
1 lb
N
) = 687,000 N.
One nautical mile is 1852 meters. Therefore, the range in meters is

(3365 nautical miles) ( 1 nautical
1852 m
mile
) = 6.23E6 m.
Length = 33.6 m, wingspan = 35.8 m,
weight = 687 kN, range = 6230 km.




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) in ft/s.
Solution:
(a)
v = 430
h
(
km 0.6214 mi
1 km
)
= 267mi/h .

v = 267 mi/h.
(b)
v = 430
h
(
km 1000 m
1 km
)( 0.3048
1 ft
)( 1 h )
m 3600 s
= 392 ft/s.
v = 392 ft/s.
Source: Courtesy of Qilai Shen/EPA/Shutterstock.




Copyright © 2024, 2008 by Pearson Education, Inc. or its affiliates.
2


BandF_6e_ISM_C12.indd 2 15/09/2

, Problem 12.9 In the 2006 Winter Olympics, the men’s Solution:
15-km cross-country skiing race was won by Andrus 15 km  60 min 
Veerpalu of Estonia in a time of 38 minutes, 1.3 seconds. (a) v =  
Determine his average speed (the distance traveled (
38 +
1.3
60
)
min  1 h 

divided by the time required) to three significant digits = 23.7 km/h.
(a) in km/h; (b) in mi/h. v = 23.7 km/h.

 1 mi 
(b) v = (23.7 km/h)  = 14.7 mi/h.
 1.609 km 
v = 14.7 mi/h.




Problem 12.10 The Porsche’s engine exerts 229 ft-lb
(foot-pounds) of torque at 4600 rpm. Determine the
value of the torque inN-m (newton-meters).
Solution:

T = 229 ft-lb ( 0.2248
1N
)( 1 m ) = 310 N-m.
lb 3.281 ft
T = 310 N-m.



Problem 12.10


Problem 12.11 The kinetic energy of the man in Solution:
Practice Example 12.1 is defined by 12 mv 2, where m
 1 slug  2
is his mass and υ is his velocity. The man’s mass is T = 1224 kg-m 2 /s 2   (
1 ft
 14.59 kg  0.3048 m )
68 kg and he is moving at 6 m/s, so his kinetic energy
= 903 slug-ft 2 /s.
is 12 (68 kg)(6 m/s) 2 = 1224 kg-m 2 /s 2. What is his
kinetic energy in US customary units? T = 903 slug-ft 2 /s.




Problem 12.12 The acceleration due to gravity at sea Solution: Use Table 12.2. The result is:
level in SI units is g = 9.81 m/s 2 . By converting units,
use this value to determine the acceleration due to gravity g = 9.81 ( sm )( 0.3048m
2
1ft
) = 32.185...( sft ) = 32.2( sft ).
2 2

at sea level in US customary units.


Problem 12.13 The value of the universal gravitational Solution: Converting units,
constant in SI units is G = 6.67E−11 m 3 /kg-s 2 . Use 3
this value and convert units to determine the value of G 6.67E − 11 m 3 /kg-s 2 = 6.67E − 11
kg-s 2 1m
(
m 3 3.281 ft
)
in US customary units.
 1 kg 
 0.0685 slug 
= 3.44E − 8 ft 3 /slug-s 2 .

G = 3.44E − 8 ft 3 /slug-s 2 .




Copyright © 2024, 2008 by Pearson Education, Inc. or its affiliates.
3


BandF_6e_ISM_C12.indd 3 15/09/2

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