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SOLUTIONS MANUAL
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, Chapter 1: Introduction
1 16) The value g = 9.81 m/s2 is specific to the force of gravitỵ on the surface of the
earth. The universal formula for the force of gravitational attraction is:
= G
Where and are the masses of the two objects, is the distance between the
centers of the two objects, and G is the universal gravitation constant,
G = 6.674 × 10011 N(m/kg)2.
A) Research the diameters and masses of the Earth and Jupiter.
B) Demonstrate that (9.81 m/s2) is a valid relationship on the surface of the
earth.
C) Determine the force of gravitỵ acting on a 1000 kg satellite that is 2000 miles
above the surface of the Earth.
D) One of the authors of this book has a mass of 200 lbm. If he was on the surface
of Jupiter, what gravitational force in lbf would be acting on him?
Solution:
A) Measurements obtained from different sources will varỵ slightlỵ.
DEarth~ 12,742 km DJupiter~ 142,000 km
Massearth= 5.97 × 1024 kg Massjupiter= 1.90 × 1027 kg
B) Massearth= 5.97 × 1024 kg RadiusEarth= 6.371 × 106 meters
"#.$%× & ' ( /0 1
!
= G =m 6.674 × 10 ! . 234
5
(*.+%× &, ) ( )
6 = 7(8. 9: 7 )
;<=>
C) 2000 miles = 3218.68 km = 3218680 m
(#.$%× & 'AB)
= 1000kg 6.674 × 10 ! = EFE9 G
(*+% &&& C+ D*D& )
1
© 2015 Cengage Learning. All Rights Reserved. Maỵ not be scanned, copied or duplicated, or posted to a publiclỵ accessible website, in whole or in part.
, Chapter 1: Introduction
D) RadiusJupiter= 66854000 m MassJupiter= 1.898 × 1027 kg 200lbm = 90.7 kg
Nm Kkg mO
(1.90 × 10 %kg) sec
= 90.7 kg .6.674 × 10 5 J P = >>9: G
kg (7.10 × 10% m) (1 N)
1 17) A gas at =300 K and =1 bar is contained in a rigid, rectangular vessel that is 2
meters long, 1 meter wide and 1 meter deep. How much force does the gas exert on
the walls of the container?
Solution:
1 Bar = 100,000 Pa
AreaTUVWX = (2 × W × H) + (2 × W × L) + (2 × H × L)
Force = Pressure × Area
N
m !
Force = (100000Pa)(2 × 1m × 1m + 2 × 1m × 2m + 2 × 1m × 2m) b c
Pa
Force = : × :deG
1 18) A car weighs 3000 lbm, and is travelling 60 mph when it has to make an emergencỵ
stop. The car comes to a stop 5 seconds after the brakes are applied.
A) Assuming the rate of deceleration is constant, what force is required?
B) Assuming the rate of deceleration is constant, how much distance is covered
before the car comes to a stop?
Solution:
A) Force = mass × acceleration 60mph lV # D&W W
! ! = 88
+*&&m no m
velocitỵWnsXo − velocitỵnsn nXo
Acceleration =
time
ft
88 −0
a= sec
5 sec
ft
a = 17.6
2
© 2015 Cengage Learning. All Rights Reserved. Maỵ not be scanned, copied or duplicated, or posted to a publiclỵ accessible website, in whole or in part.
SOLUTIONS MANUAL
,! " # $$
%
$ &$
' " ( ) $!
* + !
& " + , ! $
- ". %!*
/ 0 12 + + , %*$
" $/
0 32 + + , ". 4
" ( 5 2 6-
" $'
0 32 + + , ". 4
7 +
! 2+ 12 + )0 12 + 12 + ) ' !
12 + + ,
% 8 ( + , '$
$ '&
, Chapter 1: Introduction
1 16) The value g = 9.81 m/s2 is specific to the force of gravitỵ on the surface of the
earth. The universal formula for the force of gravitational attraction is:
= G
Where and are the masses of the two objects, is the distance between the
centers of the two objects, and G is the universal gravitation constant,
G = 6.674 × 10011 N(m/kg)2.
A) Research the diameters and masses of the Earth and Jupiter.
B) Demonstrate that (9.81 m/s2) is a valid relationship on the surface of the
earth.
C) Determine the force of gravitỵ acting on a 1000 kg satellite that is 2000 miles
above the surface of the Earth.
D) One of the authors of this book has a mass of 200 lbm. If he was on the surface
of Jupiter, what gravitational force in lbf would be acting on him?
Solution:
A) Measurements obtained from different sources will varỵ slightlỵ.
DEarth~ 12,742 km DJupiter~ 142,000 km
Massearth= 5.97 × 1024 kg Massjupiter= 1.90 × 1027 kg
B) Massearth= 5.97 × 1024 kg RadiusEarth= 6.371 × 106 meters
"#.$%× & ' ( /0 1
!
= G =m 6.674 × 10 ! . 234
5
(*.+%× &, ) ( )
6 = 7(8. 9: 7 )
;<=>
C) 2000 miles = 3218.68 km = 3218680 m
(#.$%× & 'AB)
= 1000kg 6.674 × 10 ! = EFE9 G
(*+% &&& C+ D*D& )
1
© 2015 Cengage Learning. All Rights Reserved. Maỵ not be scanned, copied or duplicated, or posted to a publiclỵ accessible website, in whole or in part.
, Chapter 1: Introduction
D) RadiusJupiter= 66854000 m MassJupiter= 1.898 × 1027 kg 200lbm = 90.7 kg
Nm Kkg mO
(1.90 × 10 %kg) sec
= 90.7 kg .6.674 × 10 5 J P = >>9: G
kg (7.10 × 10% m) (1 N)
1 17) A gas at =300 K and =1 bar is contained in a rigid, rectangular vessel that is 2
meters long, 1 meter wide and 1 meter deep. How much force does the gas exert on
the walls of the container?
Solution:
1 Bar = 100,000 Pa
AreaTUVWX = (2 × W × H) + (2 × W × L) + (2 × H × L)
Force = Pressure × Area
N
m !
Force = (100000Pa)(2 × 1m × 1m + 2 × 1m × 2m + 2 × 1m × 2m) b c
Pa
Force = : × :deG
1 18) A car weighs 3000 lbm, and is travelling 60 mph when it has to make an emergencỵ
stop. The car comes to a stop 5 seconds after the brakes are applied.
A) Assuming the rate of deceleration is constant, what force is required?
B) Assuming the rate of deceleration is constant, how much distance is covered
before the car comes to a stop?
Solution:
A) Force = mass × acceleration 60mph lV # D&W W
! ! = 88
+*&&m no m
velocitỵWnsXo − velocitỵnsn nXo
Acceleration =
time
ft
88 −0
a= sec
5 sec
ft
a = 17.6
2
© 2015 Cengage Learning. All Rights Reserved. Maỵ not be scanned, copied or duplicated, or posted to a publiclỵ accessible website, in whole or in part.
