Engineering Mechanics Dynamics 6th Edition Anthony Bedford, Wallace Fowler (Solu�ons Manual All Chapters Original) (Chapter 12-21)
Chapter 12 Problem 12.1 In 1967, the International Committee of Weights and Measures defined one second to be the time required for 9,192,631,770 cycles of the transition between two quantum states of the cesium-133 atom. Express the number of cycles in two seconds to four significant digits.
Solution:
Problem 12.2 The base of natural logarithms is e = 2.71828183... . (a) Express e to three significant digits. (b) Determine the value of e 2 to three significant digits. (c) Use the value of e you obtained in part (a) to determine the value of e 2 to three significant digits.
Solution:
[Comparing the answers of parts (b) and (c) demonstrates the hazard of using rounded-off values in calculations.]
Problem 12.3 The base of natural logarithms (see Problem 12.2) is given by the infinite series 1 1 1 e = 2+ + + + . 2! 3! 4! Its value can be approximated by summing the first few terms of the series. How many terms are needed for the approximate value rounded off to five digits to be equal to the exact value rounded off to five digits?
The number of cycles 2(9,192,631, 770) = 18,385, 263, 540.
in
two
seconds
is
Expressed to four significant digits, this is 18,390, 000, 000 or 1.839E10 cycles. 18,390,000,000 or 1.839E10 cycles.
(a) The rounded-off value is e = 2.72. (b) e 2 = 7.38905610..., so to three significant digits it is e 2 = 7.39. (c) Squaring the three-digit number we obtained in part (a) and expressing it to three significant digits, we obtain e 2 = 7.40. (a) e = 2.72. (b) e 2 = 7.39. (c) e 2 = 7.40.
Solution: The exact value rounded off to five significant digits is e = 2.7183. Let N be the number of terms summed. We obtain the results N
Sum
1
2
2
2.5
3
2.666...
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: A = 192 ft 2
2
1m ( 3.281 ) = 17.8 m ft
2
A = 17.8 m 2 .
Problem 12.5 In 2020, teams from China and Nepal, based on their independent measurements using GPS satellites, determined that the height of Mount Everest is 8848.86 meters. Determine the height of the mountain to three significant digits (a) in kilometers; (b) in miles.
Solution: (a) The height of the mountain in kilometers to three significant digits is 8848.86 m = 8848.86 m
1 km ( 1000 ) = 8.85 km. m
(b) Its height in miles to three significant digits is 8848.86 m = 8848.86 m
ft 1 mi ( 3.281 )( 5280 ) = 5.50 mi. 1m ft
(a) 8.85 km. (b) 5.50 mi.
Copyright © 2024, 2008 by Pearson Education, Inc. or its affiliates.
BandF_6e_ISM_C12.indd 1
1
15/09/23 12:43 PM
Problem 12.6 The distance D = 6 in (inches). The magnitude of the moment of the force F = 12 lb (pounds) about point P is defined to be the product M P = FD. What is the value of M P in N-m (newton-meters)? F
Solution:
The value of M P is
M P = FD = (12 lb)(6 in) = 72 lb-in. Converting units, the value of M P in N-m is 72 lb-in = 72 lb-in
1N ( 0.2248 )( 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 110 ft 4 in and its wingspan is 117 ft 5 in. Its maximum takeoff weight is 154,500 lb. Its maximum range is 3365 nautical miles. Express each of these quantities in SI units to three significant digits.
Solution:
Because 1 ft = 12 in, the length in meters is
m (110 + 124 ft )( 0.3048 ) = 33.6 m. 1 ft In the same way, the wingspan in meters is m (117 + 125 ft )( 0.3048 ) = 35.8 m. 1 ft The weight in newtons is N ( 4.448 ) = 687,000 N. 1 lb
( 154, 500 lb )
One nautical mile is 1852 meters. Therefore, the range in meters is (3365 nautical miles)
1852 m ( 1 nautical ) = 6.23E6 m. mile
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)
(
)
km 0.6214 mi = 267mi/h . h 1 km v = 267 mi/h.
v = 430
(b)
(
km 1000 m h 1 km = 392 ft/s.
v = 430
1 ft )( 0.3048 )( 1 h ) m 3600 s
v = 392 ft/s. Source: Courtesy of Qilai Shen/EPA/Shutterstock.
2
Copyright © 2024, 2008 by Pearson Education, Inc. or its affiliates.
BandF_6e_ISM_C12.indd 2
15/09/23 12:44 PM
Problem 12.9 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: 60 min 15 km 1.3 min 1 h 38 + 60 = 23.7 km/h.
(a) v =
(
)
v = 23.7 km/h. 1 mi = 14.7 mi/h. (b) v = (23.7 km/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
1N ( 0.2248 )( 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 Practice Example 12.1 is defined by 12 mv 2, where m is his mass and υ is his velocity. The man’s mass is 68 kg and he is moving at 6 m/s, so his kinetic energy is 12 (68 kg)(6 m/s) 2 = 1224 kg-m 2 /s 2. What is his kinetic energy in US customary units?
Solution:
Problem 12.12 The acceleration due to gravity at sea level in SI units is g = 9.81 m/s 2 . By converting units, use this value to determine the acceleration due to gravity at sea level in US customary units.
Solution:
Problem 12.13 The value of the universal gravitational constant in SI units is G = 6.67E−11 m 3 /kg-s 2 . Use this value and convert units to determine the value of G in US customary units.
Solution:
2 1 slug 1 ft T = 1224 kg-m 2 /s 2 14.59 kg 0.3048 m
(
)
= 903 slug-ft 2 /s. T = 903 slug-ft 2 /s.
Use Table 12.2. The result is:
1ft ( sm )( 0.3048m ) = 32.185...( sft ) = 32.2( sft ).
g = 9.81
2
2
2
Converting units,
6.67E − 11 m 3 /kg-s 2 = 6.67E − 11
m 3 3.281 ft 3 kg-s 2 1m
(
)
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.
BandF_6e_ISM_C12.indd 3
3
15/09/23 12:44 PM
Problem 12.14 The density (mass per unit volume) of aluminum is 2700 kg/m 3 . Determine its density in slug/ft 3 .
Solution:
Converting units, the density is
0.0685 slug 1 m 3 2700 kg/m 3 = 2700 kg/m 3 3.281 ft 1 kg
(
=
)
5.24 slug/ft 3 .
5.24 slug/ft 3 .
y
Problem 12.15 The cross-sectional area of the C12×30 American Standard Channel steel beam is A = 8.81 in 2 . What is its cross-sectional area in mm 2 ?
A
Solution: A = 8.81 in 2
2
( 25.41 inmm ) = 5680 mm . 2
x
Problem 12.16 A pressure transducer measures a value of 300 lb/in 2. Determine the value of the pressure in pascals. A pascal (Pa) is one newton per square meter.
Solution: Convert the units using Table 12.2 and the definition of the Pascal unit. The result: 300
2
N 12 in 1ft ( inlb )( 4.448 )( 1 ft ) ( 0.3048 ) 1 lb m
2
2
= 2.0683...(10 6 )
( mN ) = 2.07(10 ) Pa. 2
6
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: 550 ft-lb/s 1 m P = 7000 hp 1 hp 3.28 1ft
(
1N )( 0.2248 ) lb
= 5.22 × 10 6 W. P = 5.22 × 10 6 W.
4
Copyright © 2024, 2008 by Pearson Education, Inc. or its affiliates.
BandF_6e_ISM_C12.indd 4
15/09/23 12:44 PM