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PROBLEM 12.1
Astronauts who landed on the moon during the Apollo 15, 16 and 17 missions brought back a large collection
of rocks to the earth. Knowing the rocks weighed 700 N when they were on the moon, determine (a) the weight
of the rocks on the earth, (b) the mass of the rocks in kg. Theacceleration due to gravity on the moon is
1.625 m/s2.
SOLUTION
Since the rocks weighed 700 N on the moon, their mass is
m
(a)
Wmoon
700 N

 430.77 kg
g moon 1.625 m/s 2
On the earth, Wearth  mg earth
(430.77 kg)(9.81 m/s2)
(b)
w  4230 N 
m  431 kg 
mass stays the same
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PROBLEM 12.2
The value of g at any latitude f may be obtained from the formula
g = 9.78(1 + 0.0053 sin 2f ) m/s2
which takes into account the effect of the rotation of the earth, as well as the fact that the earth is not truly
spherical. Determine to four significant figures (a) the weight in N, (b) the mass in kg, at the latitudes of 0°, 45°,
and 60°, of a silver bar, the mass of which has been officially designated as 5 kg.
SOLutiOn
g = 9.78(1 + 0.0053 sin 2f ) m/s2
(a)
Weight:
f = 0∞ :
g = 9.78 m/s2
f = 45∞ :
g = 9.8059 m/s2
f = 60∞ :
g = 9.8189 m/s2
W = mg
f = 0∞ :
(b)
W = (5 kg)(9.78 m/s2 ) = 48.9 N
b
2
f = 45∞ :
W = (5 kg)(9.8059 m/s ) = 49.0 N
b
f = 60∞ :
W = (5 kg)(9.8189 m/s2 ) = 49.1 N
b
m = 5.000 kg b
Mass: At all latitudes:
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4
PROBLEM 12.3
A 400-kg satellite has been placed in a circular orbit 1500 km above the surface of the earth. The acceleration
of gravityat this elevation is 6.43 m/s2. Determine the linear momentum of the satellite, knowing that its
orbital speed is 25.6  103 km/h.
SOLUTION
Mass of satellite is independent of gravity:
m  400 kg
v  25.6  103 km/h
 1h 
3
 (25.6  106 m/h) 
  7.111  10 m/s
3600
s


L  mv  (400 kg)(7.111  103 m/s)
L  2.84  106 kg  m/s 
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PROBLEM 12.4
A spring scale A and a lever scale B having equal lever arms
are fastened to the roof of an elevator, and identical packages
are attached to the scales as shown. Knowing that when the
elevator moves downward with an acceleration of 1 m/s 2 the
spring scale indicates a load of 60 N, determine (a
a) the weight
of the packages, (b)) the load indicated by the spring scale and
the mass needed to balance the lever scale when the elevator
moves upward with an acceleration of 1 m/s2.
SOLUTION
Assume
g  9.81 m/s 2
m
W
g
F  ma : Fs  W  
W
a
g

a
W  1    Fs
g

or
W
Fs
1
a
g

60
1
1
9.81
W  66.8 N 
(b)
F  ma : Fs  W 
W
a
g

a
Fs  W 1  
g

1 

 66.81 1 

 9.81 
Fs  73.6 N 
For the balance system B,
M 0  0: bFw  bFp  0
Fw  Fp
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PROBLEM 12.4 (Continued)
But
a

Fw  Ww 1  
g

and

a
Fp  W p 1  
g

so that
Ww  Wp
and
mw 
Wp
g

66.81
9.81
mw  6.81 kg 
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PROBLEM 12.5
In anticipation of a long 7° upgrade, a bus driver accelerates at a constant rate of 0.9 m/s2 while still on a level
section of the highway. Knowing that the speed of the bus is 100 km/h as it begins to climb the grade and that
the driver does not change the setting of his throttle or shift gears, determine the distance traveled by the bus
up the grade when its speed has decreased to 80 km/h.
SOLutiOn
First consider when the bus is on the level section of the highway.
alevel = 0.9 m/s2
We have
+
SFx = ma: P =
W
alevel
g
Now consider when the bus is on the upgrade.
We have
+
SFx = ma: P - W sin 7∞ =
Substituting for P
W
a¢
g
W
W
alevel - W sin 7∞ = a ¢
g
g
or
a ¢ = alevel - g sin 7∞
= (0.9 - 9.81sin 7∞) m/s2
= - 0.29554 m/s2
For the uniformly decelerated motion
v 2 = ( v0 )2upgrade + 2a ¢( xupgrade - 0)
Noting that 100 km/h =
250
Ê 5 ˆ
m/s, then when v = 80 km/h Á =
v , we have
Ë 0.8 0 ˜¯
9
2
2
250
ˆ
Ê 250
Ê
ˆ
m/s˜ = Á
m/s˜ + 2( -0.29554 m/s2 ) xupgrade
ÁË 0.8 ¥
¯
Ë
¯
9
9
or
xupgrade = 469.949 m
xupgrade = 0.470 km b
or
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8
PROBLEM 12.6
A .09 kg model rocket is launched vertically from rest at time t= 0 with a
constant thrust of 4 N for one second and no thrust for t> 1 s. Neglecting
air resistance and the decrease in mass of the rocket, determine (a) the
maximum height h reached by the rocket, (b) the time required to reach
this maximum height.
SOLUTION
ΣF = ma : Ft − W = ma =
For the thrust phase,
W
a
g
4
F



a = g  t − 1 = ( 9.81) 
− 1 = 34.6 m/s 2
W

 .09 9.81

At t = 1 s,
v = at = (34.6) (1) = 34.6 m/s
y=
1 2 1
2
at = ( 34.6 ) (1) = 17.3 m
2
2
2
For the free flight phase, t > 1 s. a = − g = − 9.81 m/s
v = v1 + a (t − 1) = 34.6 + ( − 9.81)(t − 1)
At v = 0,
t −1 =
34.6
= 3.527 s, t = 4.527 s
9.81
v 2 − v12 = 2a ( y − y1 ) = −2 g ( y − y1 )
0 − ( 34.6 )
v 2 − v12
=−
2g
( 2 )( 9.81)
2
y − y1 = −
(a)
ymax = h = 61 + 17.3
(b) As already determined,
= 61 m
h = 78.3 m W
t = 4.527 s W
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9
PROBLEM 12.7
A tugboat pulls a small barge through a harbor. The
propeller thrust minus the drag produces a net thrust
that varies linearly with speed. Knowing that the
combined weight of the tug and barge is 3600 kN,
determine (a) the time required to increase the speed
from an initial value v1= 1.0 m/s to a final value v2 =
2.5 m/s, (b) the distance traveled during this time
interval.
SOLUTION
Given:
W  3600 kN
Free Body Diagram:
W
 m  366972.5 kg
m
g
v1  1.0 m/s, v2  2.5 m/s
From Graph:
Fnet  81000 
From FBD:
Fx  max  Fnet  max  ax 
81000
v  81000 - 27000v N
3
81000  27000v
m
27000

 3  v  m/s2
m
Fnet
m
ax 
(a) From Kinematics:
ax 
t
dv
dv
 dt 
dt
ax
m v2 dv

27000 v1 3  v
m
2.5
t 
ln  3  v  |1
27000
 dt 
0
t  18.84 s 
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PROBLEM 12.7 (Continued)
(b) From Kinematics:
ax dx  vdv  dx 
x
m
v2
vdv
ax
vdv
 dx  27000  3  v
0
v1
x
m
 3  v  3ln 3  v |12.5
27000
x  36.14 m 
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PROBLEM 12.8
Determine the maximum theoretical speed that may be achieved over a distance of 60 m by a car starting from
rest, knowing that the coefficient of static friction is 0.80 between the tires and the pavement and that 60 percent
of the weight of the car is distributed over its front wheels and 40 percent over its rear wheels. Assume
(a) four-wheel drive, (b) front-wheel drive, (c) rear-wheel drive.
SOLUTION
(a)
Four-wheel drive
Fy  0: N1  N 2  W  0
F1  F2   N1   N 2
  ( N1  N 2 )   w
Fx  ma : F1  F2  ma
 w  ma
W
mg
  g  0.80(9.81)
m
m
a  7.848 m/s 2
a

v 2  2ax  2(7.848 m/s 2 )(60 m)  941.76 m 2 /s 2
v  30.69 m/s
(b)
v  110.5 km/h 
Front-wheel drive
F2  ma
 (0.6 W )  ma
0.6W 0.6 mg

m
m
 0.6  g  0.6(0.80)(9.81)
a
a  4.709 m/s 2
v 2  2ax  2(4.709 m/s 2 )(60 m)  565.1 m 2 /s 2
v  23.77 m/s
v  85.6 km/h 
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PROBLEM 12.8 (Continued)
(c)
Rear-wheel drive
F1  ma
 (0.4 W )  ma
0.4 W 0.4 mg

m
m
 0.4  g  0.4(0.80)(9.81)
a
a  3.139 m/s 2
v 2  2ax  2(3.139 m/s2 )(60 m)  376.7 m2 /s2
v  19.41 m/s
v  69.9 km/h 
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PROBLEM 12.9
If an automobile’s braking distance from 90 km/h is 45 m on level pavement, determine the automobile’s
braking distance from 90 km/h when it is ((a) going up a 5° incline, (b) going down a 3-percent
percent incline.
Assume the braking force is independent of grade.
SOLUTION
Assume uniformly decelerated motion in all cases.
For braking on the level surface,
v0  90 km/h  25 m/s, v f  0
x f  x0  45 m
v 2f  v02  2a ( x f  x0 )
a

v 2f  v02
2( x f  x0 )
0  (25) 2
(2)(45)
 6.9444 m/s 2
Braking force.
Fb  ma
W
 a
g
6.944
W

9.81
 0.70789 W
(a)
Going up a 5° incline.
F  ma
W
a
g
F  W sin 5
a b
g
W
 (0.70789  sin 5)(9.81)
 Fb  W sin 5 
 7.79944 m/s 2
x f  x0 
v 2f  v02
2a
0  (25) 2

(2)( 7.79944)
x f  x0  40.1 m 
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PROBLEM 12.9 (Continued)
(b)
Going down a 3 percent incline.
3
  1.71835
100
W
 Fb  W sin   a
g
a  (0.70789  sin  )(9.81)
tan  

 6.65028 m/s
0  (25) 2
x f  x0 

(2)(6.65028)
x f  x0  47.0 m 
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PROBLEM 12.10
A mother and her child are skiing together, and the
mother is holding the end of a rope tied to the child’s
waist. They are moving at a speed of 7.2 km/h on a
gently sloping portion of the ski slope when the
mother observes that they are approaching a steep
descent. She pulls on the rope with an average force
of 7 N. Knowing the coefficient of friction between
the child and the ground is 0.1 and the angle of the
rope does not change, determine (a)the time required
for the child’s speed to be cut in half, (b)the distance
traveled in this time.
SOLUTION
Draw free body diagram of child.
F  ma :
x-direction:
mg sin 5  k N  T cos15  ma
y-direction:
N  mg cos 5  T sin15  0
From y-direction,
N  mg cos5  T sin15  (20 kg)(9.81 m/s 2 ) cos5  (7 N)sin15°  193.64 N
From x-direction,
 k N T cos15

m
m
(0.1)(193.64
N) (7 N) cos 15°
 (9.81 m/s 2 ) sin 5 

20 kg
20 kg
a  g sin 5 
 0.45128 m/s 2
(in x-direction.)
v0  7.2 km/h  2 m/s
vf 
1
v0  1 m/s
2
v f  v0  at
(a)
(b)
x0  0
t
v f  v0
a

1 m/s
 2.2159 s
0.45128 m/s 2
Time elapsed.
Corresponding distance.
x  x0  v0 t 
t  2.22 s 
1 2
at
2
 0  (2 m/s)(2.2159 s) 
1
( 0.45128 m/s 2 )(2.2159 s)2
2
x  3.32 m 
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PROBLEM 12.11
The coefficients of friction between the load and the flat-bed
flat
trailer shown are s  0.40 and k  0.30. Knowing that the
speed of the rig is 72 km/h, determine the shortest distance in
which the rig can be brought to a stop if the load is not to shift.
SOLUTION
Load: We assume that sliding of load relative to trailer is impending:
F  Fm
 s N
Deceleration of load is same as deceleration of trailer, which is the maximum allowable deceleration amax .
Fy  0: N  W  0 N  W
Fm   s N  0.40 W
Fx  ma : Fm  mamax
0.40 W 
W
amax
g
amax  3.924 m/s 2
a max  3.92 m/s2
Uniformly accelerated motion.
v 2  v02  2ax with v  0
v0  72 km/h  20 m/s
a  amax  3.924 m/s2
0  (20)2  2(3.924) x
x  51.0 m 
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PROBLEM 1
12.12
A light train made up of two cars is traveling at 90 km/h when the
brakes are applied to both cars. Knowing that car A has a mass of 25
Mg and car B a mass of 20 Mg, and that the braking force is 30 kN
on each car, determine ((a) the distance traveled by thee train before it
comes to a stop,(
stop,(b)) the force in the coupling between the cars while
the train is showing down.
SOLUTION
v0  90 km/h  90/3.6  25 m/s
(a)
Both cars:
Fx  ma: 60  103 N  (45  103 kg)a
a  1.333 m/s 2
v 2  v10  2ax: 0  (25)2  2(1.333) x 

Stopping distance:
(b)
Car A:


x  234 m 
Fx  ma: 30  103 + P  (25  103 )a
P  (25  103 )(1.333)  30  103


Coupling force:
P  3332 N
P  3.33 kN (tension) 
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PROBLEM 12.13
The two blocks shown are originally at rest. Neglecting the masses of
the pulleys and the effect of friction in the pulleys and between block A
and the horizontal surface, determine (a) the acceleration of each block,
(b) the tension in the cable.
SOLutiOn
From the diagram
x A + 3 y B = constant
Then
v A + 3v B = 0
and
aA + 3aB = 0
aA = -3aB
or
(a)
A:
+
SFx = mA a A :
(1)
-T = mA aB
T = 3mA aB
Using Eq. (1)
B:
+
SFy = mB aB : WB - 3T = mB aB
Substituting for T
A:
mB g - 3(3mA aB ) = mB aB
aB =
or
B:
(b)
g
1+ 9
mA
mB
=
9.81 m/s2
= 0.83136 m/s2
30 kg
1+ 9
25 kg
Then
a A = 2.49 m/s2 Æ b
and
a B = 0.831 m/s2Ø b
We have
T = 3 ¥ 30 kg ¥ 0.83136 m/s2
T = 74.8 N b
or
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19
PROBLEM 12.14
The two blocks shown are originally at rest. Neglecting the masses
of the pulleys and the effect of friction in the pulleys and assuming that
the coefficients of friction between block A and the horizontal surface
are m s = 0.25 and m k = 0.20, determine (a) the acceleration of each
block, (b) the tension in the cable.
SOLutiOn
From the diagram
x A + 3 y B = constant
Then
v A + 3v B = 0
and
aA + 3aB = 0
aA = -3aB
or
(1)
First determine if the blocks will move with aA = aB = 0. We have
+
1
SFy = 0 : WB - 3T = 0 or T = mB g
3
B:
+
SFx = 0:
FA =
Then
+
A:
A:
Also,
B:
FA - T = 0
1
¥ 25 kg ¥ 9.81 m/s2 = 81.75 N
3
SFy = 0 : WA - N A = 0 or
N A = mA g
( FA ) max = ( m s ) A N A = ( m s ) A mA g
= 0.25 ¥ 30 kg ¥ 9.81 m/s2
= 73.575 N
FA ( FA ) max, which implies that the blocks will move.
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20
PROBLEM 12.14 (continued)
(a)
A:
+
N A = mA g
FA = ( m k ) A N A = 0.20 mA g
+
Sliding:
+
B:
SFx = mA a A : FA - T = mA a A
T = 0.20 mA g + 3mA aB
Using Eq. (1)
or
SFy = 0 : WA - N A = 0 or
SFy = mB aB : WB - 3T = mB aB
mB g - 3(0.20 mA g + 3mA aB ) = mB aB
or
Ê
m ˆ
g Á1 - 0.6 A ˜
mB ¯
Ë
aB =
mA
1+ 9
mB
=
Ê
30 kg ˆ
(9.81 m/s2 ) Á1 - 0.6
Ë
25 kg ˜¯
30 kg
1 + 9 25 kg
= 0.23278 m/s2
a A = 0.698 m/s2 Æ b
Then
a B = 0.233 m/s2 Ø b
and
(b)
We have
T = (30 kg)(0.20 ¥ 9.81 + 3 ¥ 0.23278) m/s2
T = 79.8 N b
or
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21
PROBLEM 12.15
Each of the systems shown is initially at rest.
Neglecting axle friction and the masses of the
pulleys, determine for each system (a) the
acceleration of block A, (b) the velocity of block A
after it has moved through 3 m, (c) the time required
for block A to reach a velocity of 6 m/s.
SOLutiOn
Let y be positive downward for both blocks.
Constraint of cable: y A + y B = constant
a A + aB = 0
or
aB = - a A
WA
aA
g
or
T = WA -
For blocks A and B, + SF = ma :
Block A:
WA - T =
Block B:
P + WB - T =
WB
W
aB = - B a A
g
g
P + WB - W A +
WA
W
aA = - B aA
g
g
Solving for aA,
aA =
WA
aA
g
W A - WB - P
g
W A + WB
(1)
v 2A - ( v A )20 = 2a A [ y A - ( y A )0 ] with ( v A )0 = 0
v A = 2a A [ y A - ( y A )0 ]
v A - ( v A ) 0 = a At
t=
(2)
with (v A )0 = 0
vA
aA
(3)
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22
PROBLEM 12.15 (continued)
(a)
Acceleration of block A.
System (1):
By formula (1),
System (2):
By formula (1),
System (3):
By formula (1),
(b)
(c)
W A = 100 g, WB = 50 g, P = 0
( aA )1 =
100 - 50
(9.81)
100 + 50
(a A )1 = 3.27 m/s2Ø b
W A = 100 g, WB = 0, P = 490.5 N = 50 g
( a A )2 =
100 - 50
(9.81) = 4.905 m/s2
100
(a A )2 = 4.91 m/s2Ø b
W A = 1100 g, WB = 1050 g, P = 0
( a A )3 =
1100 - 1050
(9.81)
1100 + 1050
(a A )3 = 0.228 m/s2 Ø b
v A at y A - ( y A )0 = 3 m. Use formula (2).
System (1):
( v A )1 = (2)(3.27)(3)
( v A )1 = 4.43 m/s Ø b
System (2):
( v A )2 = (2)( 4.905)(3)
(v A )2 = 5.42 m/s Ø b
System (3):
( v A )3 = (2)(0.228)(3)
( v A )3 = 1.170 m/s Ø b
Time at v A = 6 m/s. Use formula (3).
System (1):
t1 =
6
3.27
t1 = 1.835 s b
System (2):
t2 =
6
4.905
t2 = 1.223 s b
System (3):
t3 =
6
0.228
t3 = 26.3 s b
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23
PROBLEM 12.16
Boxes A and B are at rest on a conveyor belt that is initially at
rest. The belt is suddenly started in an upward direction so that
slipping occurs between the belt and the boxes. Knowing that
the coefficients of kinetic friction between the belt and the
boxes are ( m k ) A = 0.30 and ( m k ) B = 0.32, determine the initial
acceleration of each box.
SOLUTION
Assume that a B a A so that the normal force NAB between the boxes is zero.
+
A:
SFy = 0 : NA - WA cos 15∞ = 0
or
NA = WA cos 15∞
Slipping:
FA = ( m k ) A NA
A:
= 0.3WA cos 15∞
+
SFx = mA a A : FA - WA sin 15∞ = mA a A
0.3WA cos 15∞ - WA sin 15∞ =
or
WA
aA
g
aA = (9.81 m/s2 )(0.3 cos 15∞ - sin 15∞)
or
= 0.3037 m/s2
B:
+
SFy = 0 : N B - WB cos 15∞ = 0
or
N B = WB cos 15∞
Slipping:
FB = ( m k ) B N B
B:
= 0.32WB cos 15∞
+
or
or
SFx = mB aB : FB - WB sin 15∞ = mB aB
0.32WB cos 15∞ - WB sin 15∞ =
WB
aB
g
aB = (9.81 m/s2 )(0.32 cos 15∞ - sin 15∞) = 0.49322 m/s2
aB aA fi assumption is correct
a A = 0.304 m/s2
15° b
a B = 0.493 m/s2
15° b
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24
PROBLEM 12.16 (Continued)
Note: If it is assumed that the boxes remain in contact ( NAB π 0), then
a A = aB
and find ( SFx = ma)
A:
0.3WA cos 15∞ - WA sin 15∞ - N AB =
WA
a
g
B:
0.32WB cos 15∞ - WB sin 15∞ + N AB =
WB
a
g
Solving yields a = 0.3878 m/s2 and NAB = - 4.2168 N, which contradicts the assumption.
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25
PROBLEM 12.17
A 2500-kg truck is being used to lift a 500 kg boulder B that
is on a 100 kg pallet A. Knowing the acceleration of the truck
is 0.3 m/s2, determine (a) the horizontal force between the
tires and the ground, (b) the force between the boulder and
the pallet.
SOLUTION
aT = 0.3 m/s2
Kinematics:
a A = a B = 0.3 m/s 2
mT = 2500 kg
Masses:
mA = 100 kg
mB = 500 kg
Let T be the tension in the cable. Apply Newton’s second law to the lower pulley, pallet and boulder.
Vertical components
:
T − (mA + mB ) g = (mA + mB )a A
T − (600)(9.81) = 600(0.3)
T = 6066 N
Apply Newton’s second law to the truck.
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26
PROBLEM 12.17 (Continued)
Horizontal components
(a)
: F − T = mT aT
Horizontal force between lines and ground.
F = T + mT aT = 6066 + (2500)(0.3)
F = 6816 N W
Apply Newton’s second law to the boulder.
Vertical components
:
FAB − mB g = mB aB
FAB = mB ( g + a) = 500(9.81 + 0.3) = 5055 N
(b)
FAB = 5060 N W
Contact force:
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27
PROBLEM 12.18
Block A has a mass of 40 kg, and block B has a mass of8 kg.
The coefficients of friction between all surfaces of contact are
s  0.20 and k  0.15. If P  0, determine (a)
(
the
acceleration of block B, (b) the tension in the cord.
SOLUTION
From the constraint of the cord:
2 x A  xB/A  constant
Then
2v A  vB/A  0
and
2a A  aB/A  0
Now
a B  a A  a B/A
Then
aB  a A  (2a A )
or
aB   a A
(1)
First we determine if the blocks will move for the given value of  . Thus, we seek the value of  for
which the blocks are in impending motion, with the impending motion of A down the incline.
B:
Fy  0: N AB  WB cos   0
or
N AB  mB g cos 
Now
FAB  s N AB
B:
 0.2mB g cos 
Fx  0:  T  FAB  WB sin   0
T  mB g (0.2cos   sin  )
or
A:
A:
Fy  0: N A  N AB  WA cos   0
or
N A  (mA  mB ) g cos 
Now
FA  s N A
 0.2(mA  mB ) g cos 
Fx  0:  T  FA  FAB  WA sin   0
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PROBLEM 12.18 (Continued)
T  mA g sin   0.2(mA  mB ) g cos   0.2mB g cos 
or
 g[mA sin   0.2(mA  2mB ) cos  ]
Equating the two expressions for T
mB g (0.2cos  sin  )  g[mA sin   0.2(mA  2mB )cos ]
8(0.2  tan  )  [40 tan   0.2(40  2  8)]
or
tan   0.4
or
or   21.8 for impending motion. Since   25, the blocks will move. Now consider the
motion of the blocks.
Fy  0: N AB  WB cos 25  0
(a)
B:
or
N AB  mB g cos 25
Sliding:
FAB  k N AB  0.15mB g cos 25
Fx  mB aB :  T  FAB  WB sin 25  mB aB
or
T  mB [ g (0.15cos 25  sin 25)  aB ]
 8[9.81(0.15cos 25  sin 25)  aB ]
 8(5.47952  aB )
A:
(N)
Fy  0: N A  N AB  WA cos 25  0
or
N A  (mA  mB ) g cos 25
Sliding:
FA  k N A  0.15(mA  mB ) g cos 25
Fx  mA aA :  T  FA  FAB  WA sin 25  mA aA
Substituting and using Eq. (1)
T  m A g sin 25  0.15(m A  mB ) g cos 25
 0.15mB g cos 25  mA (aB )
 g[mA sin 25  0.15(mA  2mB ) cos 25]  m A aB
 9.81[40 sin 25  0.15(40  2  8) cos 25]  40aB
 91.15202  40aB
(N)
Equating the two expressions for T
8(5.47952  aB )  91.15202  40aB
or
aB  0.98575 m/s 2
a B  0.986 m/s 2
(b)
We have
or
25 
T  8[5.47952  (0.98575)]
T  51.7 N 
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PROBLEM 12.19
Block A has a mass of 40 kg, and block B has a mass of 8 kg.
The coefficients of friction between all surfaces of contact are
s  0.20 and k  0.15. If P  40 N
, determine (a)
( the
acceleration of block B, (b) the tension in the cord.
SOLUTION
From the constraint of the cord.
2 x A  xB/A  constant
Then
2v A  vB/A  0
and
2a A  aB/A  0
Now
a B  a A  a B/A
Then
aB  a A  (2a A )
or
a B  a A
(1)
First we determine if the blocks will move for the given value of P.. Thus, we seek the value of P for which
the blocks are in impending motion, with the impending motion of a down the incline.
B:
Fy  0: N AB  WB cos 25  0
or
N AB  mB g cos 25
Now
FAB   s N AB
B:
 0.2 mB g cos 25
Fx  0:  T  FAB  WB sin 25  0
or
A:
T  0.2 mB g cos 25  mB g sin 25
 (8 kg)(9.81 m/s 2 ) (0.2 cos 25  sin 25)
 47.39249 N
A:
or
Fy  0: N A  N AB  WA cos 25  P sin 25  0
N A  (mA  mB ) g cos 25  P sin 25
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PROBLEM 12.19 (Continued)
Now
FA  s N A
or
FA  0.2[(mA  mB ) g cos 25  P sin 25]
Fx  0:  T  FA  FAB  WA sin 25  P cos 25  0
T  0.2[(mA  mB ) g cos 25  P sin 25]  0.2mB g cos 25  mA g sin 25  P cos 25  0
or
or
P(0.2 sin 25  cos 25)  T  0.2[(m A  2mB ) g cos 25]  m A g sin 25
Then
P(0.2 sin 25  cos 25)  47.39249 N  9.81 m/s 2{0.2[(4
{0.2[(40
0  2  8)cos
8) cos 25  40 sin 25] kg}
or
P  19.04 N for impending motion.
Since P,  40 N, the blocks will move. Now consider the motion of the blocks.
Fy  0: N AB  WB cos 25  0
(a)
or
N AB  mB g cos 25
Sliding:
FAB  k N AB
B:
 0.15 mB g cos 25
Fx  mB aB :  T  FAB  WB sin 25  mB aB
T  mB [ g (0.15 cos 25  sin 25)  aB ]
or
 8[9.81(0.15 cos 25  sin 25)  aB ]
 8(5.47952  aB )
(N)
A:
Fy  0: N A  N AB  WA cos 25  P sin 25  0
or
N A  (mA  mB ) g cos 25  P sin 25
Sliding:
FA  k N A
 0.15[(mA  mB ) g cos 25  P sin 25]
Fx  mA a A :  T  FA  FAB  WA sin 25  P cos 25  mA aA
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PROBLEM 12.19 (C
(Continued)
Substituting and using Eq. (1)
T  m A g sin 25  0.15[(m A  mB ) g cos 25  P sin 25]
 0.15 mB g cos 25  P cos 25  m A ( aB )
 g[mA sin 25  0.15(m A  2mB ) cos 25]
 P (0.15 sin 25  cos 25)  m A aB
 9.81[40 sin 25  0.15(40  2  8) cos 25]
 40(0.15 sin 25  cos 25)  40aB
 129.94004  40aB
Equating the two expressions for T
8(5.47952  aB )  129.94004  40aB
or
aB  1.79383 m/s 2
a B  1.794 m/s2
(b)
We have
25° 
T  8[5.47952  (1.79383)]
or
T  58.2 N 
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PROBLEM 12.20
The flat-bed trailer carries two 1500-kg beams with the upper
beam secured by a cable. The coefficients of static friction
between the two beams and between the lower beam and the
bed of the trailer are 0.25 and 0.30, respectively. Knowing that
the load does not shift, determine (a) the maximum acceleration
of the trailer and the corresponding tension in the cable, (b) the
maximum deceleration of the trailer.
SOLUTION
(a) Maximum acceleration. The cable secures the upper beam to the cab; the lower beam has impending slip.
For the upper beam,
Fy  0: N1  W  0
N1  W  mg
For the lower beam,
Fy  0: N 2  N1  W  0
or
N 2  2W  2mg
Fx  ma : 0.25 N1  0.30 N 2   0.25  0.60  mg  ma
a  0.85g   0.85  9.81 a  8.34 m/s 2
For the upper beam,

Fx  ma : T  0.25 N1  ma
T  0.25mg  ma   0.25 1500  9.81  1500 8.34   16.19  10 3 N
T  16.19 kN 
(b) Maximum deceleration of trailer.
Case 1: Assume that only the top beam has impending motion. As in Part (a) N1  mg.
F  ma : 0.25mg  ma
a  0.25g  2.45 m/s 2
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PROBLEM 12.20 (Continued)
Case 2: Assume that both beams have impending slip. As before N 2  2mg.
F   2m  a :
 0.30  2mg    2m  a
a  0.30 g  2.94 m/s2
The smaller of deceleration value of Case 1 governs.
a  2.45 m/s 2  
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PROBLEM 12.21
A baggage conveyor is used to unload luggage from an airplane.
The 10-kg duffel bag A is sitting on top of the 20-kg
20
suitcaseB.
The conveyor is moving the bags down at a constant speed of
0.5 m/s when the belt suddenly stops. Knowing that the
coefficient of friction between the belt and B is 0.3 and that bag A
does not slip on suitcaseB,, determine the smallest allowable
coefficient of static friction between the bags.
SOLUTION
Since bag A does not slide on suitcase B, both have the same acceleration.
aa
20°
Apply Newton’s second law to thebag A – suitcase B combination treated as asingle particle.
Fy  ma y :  (mB  mA ) g cos 20  N  0
N  (mA  mB ) g cos 30  (30)(9.81) cos 20  276.55 N
 B N  (0.3)(276.55)  82.965 N
Fx  ma x :  B N  ( mA  mB ) g sin 20  ( mA  mB ) a
a  g sin 20 
B N
82.965
 9.81sin 20 
m A mB
30
a  0.58972 m/s 2
a  0.58972 m/s 2
Apply Newton’s second law to bag A alone.
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20°
PROBLEM 12.
12.21 (Continued)
Fy  ma y : N AB  m A g cos 20  0
N AB  ma g sin 20  (10)(9.81) cos 20  92.184 N
ZFx  max : M A g sin 20  FAB  mA a
FAB  mA ( g sin 20  a)  (10)(9.81sin 20  0.58972)
 27.655 N
Since bag A does not slide on suitcase B,
s 
FAB 27.655

 0.300
N AB 92.184
s  0.300 
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PROBLEM 12.22
To unload a bound stack of plywood from a truck, the driver first tilts the
bed of the truck and then accelerates from rest. Knowing that the
coefficients of friction between the bottom sheet of plywood and the bed
are s  0.40 and k  0.30, determine (a)) the smallest acceleration of
the truck which will cause the stack of plywood to slide, (b)
( the
acceleration of the truck which causes corner A of the stack to reach the
end of the bed in 0.9 s.
SOLUTION
Let a P be the acceleration of the plywood, aT be the acceleration of the truck, and a P/T be the acceleration
of the plywood relative to the truck.
(a)
Find the value of aT so that the relative motion of the plywood with respect to the truck is impending.
aP  aT and F1  s N1  0.40 N1
Fy  mP a y : N1  WP cos 20   mP aT sin 20
N1  mP ( g cos 20  aT sin 20)
Fx  max : F1  WP sin 20  mP aT cos 20
F1  mP ( g sin 20  aT cos 20)
mP ( g sin 20  aT cos 20)  0.40 mP ( g cos 20  aT sin 20)
(0.40 cos 20  sin 20)
g
cos 20  0.40sin 20
 (0.03145)(9.81)
 0.309
aT 
aT  0.309 m/s2
(b)
xP/T  ( xP/T )o  (vP /T )t 
aP /T 
2 xP /T
t
2

1
1
aP /T t 2  0  0  aP / T t 2
2
2
(2)(2)
 4.94 m/s 2
(0.9)2
a P / T  4.94 m/s2
20
a P  aT  a P / T  (aT )  (4.94 m/s 2
20)
Fy  mP a y : N 2  WP cos 20  mP aT sin 20
N 2  mP ( g cos 20  aT sin 20)
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
PROBLEM 12.22 (C
(Continued)
Fx  max : F2  WP sin 20  mP aT cos 20  mP aP / T
F2  mP ( g sin 20  aT cos 20  aP / T )
For sliding with friction
F2  k N 2  0.30 N 2
mP ( g sin 20  aT cos 20  aP /T )  0.30mP ( g cos 20  aT sin 20)
aT 
(0.30cos 20  sin 20) g  aP /T
cos 20  0.30sin 20
 (0.05767)(9.81)  (0.9594)(4.94)
aT  4.17 m/s2
 4.17
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38

PROBLEM 12.23
To transport a series of bundles of shinglesAto
shingles
a roof, a contractor uses a motor-driven
motor
lift
consisting of a horizontal platform BC which
rides on rails attached to the sides of a ladder.
The lift starts from rest and initially moves
with a constant acceleration
accelera
a1 as shown. The
lift then decelerates at a constant rate a2 and
comes to rest at D,, near the top of the ladder.
Knowing that the coefficient of static friction
between a bundle of shingles and the
horizontal platform is 0.30, determine the
largest allowable acceleration a1 and the
largest allowable deceleration a2 if the bundle
is not to slide on the platform.
SOLUTION
Acceleration a1: Impending slip.
Fy  mA a y :
F1  s N1  0.30 N1
N1  WA  mA a1 sin 65
N1  WA  mA a1 sin 65
 mA ( g  a1 sin 65)
Fx  mA ax : F1  mA a1 cos 65
F1  s N
or
mA a1 cos 65  0.30mA ( g  a1 sin 65)
0.30 g
cos 65  0.30 sin 65
 (1.990)(9.81)
a1 
a1  19.53 m/s 2
 19.53 m/s 2
Deceleration a 2 : Impending slip. F2  s N 2  0.30 N 2
Fy  ma y : N1  WA  mA a2 sin 65
N1  WA  mA a2 sin 65
Fx  max :
F2  mA a2 cos 65
F2  s N 2
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65 
PROBLEM 12.23 (Continued)
or
mA a2 cos 65  0.30 mA ( g  a2 cos 65)
a2 
0.30 g
cos 65  0.30 sin 65
 (0.432)(9.81)
a 2  4.24 m/s 2
 4.24 m/s 2
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65 
PROBLEM 12.24
An airplane has a mass of 25 Mg and its engines develop a total thrust of 40 kN during take-off. If the drag
D exerted on the plane has a magnitude D  2.25v 2 , where v is expressed in meters per second and D in
newtons, and if the plane becomes airborne at a speed of 240 km/h, determine the length of runway required
for the plane to take off.
SOLUTION
F  ma: 40  103 N  2.25v2  (25  103 kg)a
av
Substituting

x1
0
dv
dv
: 40  103  2.25 v 2  (25  103 ) v
dx
dx
(25  103 )vdv
0 40  103  2.25v 2
25  103
x1  
[ln(40  103  2.25v 2 )]v01
2(2.25)
dx 


v1
25  103
40  103
ln
4.5
40  103  2.25v12
For v1  240 km/h  66.67 m/s
x1 
25  103
40  103
ln
 5.556 ln1.333
4.5
40  103  2.25(66.67) 2
 1.5982  103 m
x1  1.598 km 
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41
PROBLEM 12.25
A 4-kg
kg projectile is fired vertically with an initial velocity of 90 m/s, reaches a maximum height and falls to
2
the ground. The aerodynamic drag D has a magnitude D = 0.0024 v where D and v are expressed in newtons
and m/s, respectively. Knowing that the direction of the drag is always opposite to the direction of the
velocity, determine (a)) the maximum height of the trajectory, ((b)) the speed of the projectile when it reaches
the ground.
SOLUTION
Let y be the position coordinate of the projectile measured upward from
the ground. The velocity and acceleration a taken to positive upward.
D  kv 2.
Fy  ma :
(a)) Upward motion.
 D  mg  ma

D
kv 2 

a    g      g 

m
m 



dv
kv 2 
k  2 mg 
v
   g 
    v 
dy
m 
m
k 

v dv
k
  dy
mg
m
v2 
k

v dv
k

mg
v0 2
m
v 
k
0

h
0
dy
0
1  2 mg 
kh
ln  v 
  
2 
k v
m
0
mg

1
1  kv 2
kh
k
  ln  0  1  
ln
2 v 2  mg
2  mg
m

0
k
  0.0024  90 2


m  kv02
4
ln 
 1  
ln 
 1
2k  mg
 2  0.0024    4 9.81


 335.36 m
h  335 m 
h 
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PROBLEM 12.25 (Continued)
F  ma :
(b) Downward motion.
D  mg  ma
D
kv 2
g 
g
m
m
dv
kv 2
k  mg

v

g   
 v2 
dy
m
m k

a 
v dv
k
  dy
mg
m
 v2
k

vf
0
v dv
k

mg
m

0
h
dy
vf
1  mg
kh

ln
 v2 

2  k
m
0
 mg
 v 2f
1  k
 ln 
2  mg

k



kh
 
m



kv 2f 
2 kh
  
ln 1 

mg 
m

1
kv 2f
mg
vf  
 e  2 kh/m
mg
1  e 2 kh/m
k

vf 
 73.6 m/s

 4  9.81 1  e 2  0.0024335.36/4 
0.0024 

v f  73.6 m/s 
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43
PROBLEM 12.26
A constant force P is applied to a piston and rod of total mass m to
make them move in a cylinder filled with oil. As the piston moves, the
oil is forced through orifices in the piston and exerts on the piston a
force of magnitude kv in a direction opposite to the motion of the
piston. Knowing that the piston starts from rest at t  0 and x  0,
show that the equation relating x, v, and t, where x is the distance
traveled by the piston and v is the speed of the piston, is linear in each
of these variables.
SOLUTION
F  ma : P  kv  ma
dv
P  kv
a
dt
m
t
v m dv
dt 
0
0 P  kv
m
v
  ln ( P  kv) 0
k
m
  [ln ( P  kv)  ln P]
k
m P  kv
P  kv
kt
t   ln
or
ln

k
P
m
m
P  kv
P
 kt/m
 kt/m
e
or
v  (1  e
)
m
k


x

x

t
0
v dt 
Pt
k
t
t

0
P  k  kt/m 
 e

k  m
0
Pt P  kt/m
Pt P
 (e
 1) 
 (1  e  kt/m )
k m
k m
Pt kv
 , which is linear.
k
m

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PROBLEM 12.27
A spring AB of constant k is attached to a support at A and to a
collar of mass m.. The unstretched length of the spring is .
Knowing that the collar is released from rest at x  x0 and
neglecting friction between the collar and the horizontal rod,
determine the magnitude
itude of the velocity of the collar as it passes
through Point C.
SOLUTION
Choose the origin at Point C and let x be positive to the right. Then x is a position coordinate of the slider B
and x0 is its initial value. Let L be the stretched length of the spring. Then, from the right triangle
L  2  x2
The elongation of the spring is e  L  , and the magnitude of the force exerted by the spring is
Fs  ke  k (  2  x 2   )
cos  
By geometry,
x
  x2
2
Fx  max :  Fs cos   ma
k (  2  x 2  )
a

v
0
v dv 
k
x
x
2
m 
  x2
x
  x2
2
 ma




0
 0 a dx
x
v
1 2
k
v 
2 0
m


x
x

2
x0
  x2

0
0

k 1 2
2
2 
 dx    x     x 

m2
x

0
1 2
k
1

v    0  2  x02   2  x02 
2
m
2

k
v2 
2 2  x02  2 2  x02
m
k
   2  x02  2 2  x02  2 

m 




answer: v 
k
m

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
2  x02   
PROBLEM 12.28
Block A has a mass of 10 kg, and blocks B and C have masses of 5 kg each.
Knowing that the blocks are initially at rest and that B moves through 3 m in 2 s,
determine (a)) the magnitude of the force P, (b) the tension in the cord AD.
AD Neglect
the masses of the pulleys and axle friction.
SOLUTION
Let the position coordinate y be positive downward.
Constraint of cord AD:
y A  yD  constant
v A  vD  0,
Constraint of cord BC:
( yB  yD )  ( yC  yD )  constant
vB  vC  2vD  0,
Eliminate aD .
a A  aD  0
aB  aC  2aD  0
2a A  aB  aC  0
(1)
We have uniformly accelerated motion because all of the forces are
constant.
y B  ( y B ) 0  (vB ) 0 t 
aB 
2[ yB  ( yB )0 ]
t
2

1
aB t 2 , (vB )0  0
2
(2)(3)
 1.5 m/s 2
(2)2
Fy  0: 2TBC  TAD  0
Pulley D:
TAD  2TBC
Fy  ma y : WA  TAD  mA aA
Block A:
aA 
or
WA  TAD WA  2TBC

mA
mA
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(2)
PROBLEM 12.28 (Continued)
Fy  ma y : WC  TBC  mC aC
Block C:
aC 
or
WC  TBC
mC
(3)
Substituting the value for aB and Eqs. (2) and (3) into Eq. (1), and solving
for TBC ,
 W  2TBC
2 A
mA

 WC  TBC

  aB  

 mC

0

 mC g  TBC 
 m g  2TBC 
2 A
0
  aB  
mA
mC




 4
1 


 TBC  3g  aB
 mA mC 
 4 1
   TBC  3(9.81)  1.5 or TBC  51.55 N
 10 5 
Fy  ma y : P  WB  TBC  mB aB
Block B:
(a)
Magnitude of P.
P  TBC  WB  mB aB
 51.55  5(9.81)  5(1.5)
(b)
P  10.00 N 
Tension in cord AD.
TAD  2TBC  (2)(51.55)
TAD  103.1 N 
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47
PROBLEM 12.29
A 20-kg sliding panel is supported by rollers at B and C. A 12.5-kg counterweight A is attached to a cable as
shown and, in cases a and c, is initially in contact with a vertical edge of the panel. Neglecting friction, determine
in each case shown the acceleration of the panel and the tension in the cord immediately after the system is
released from rest.
SOLutiOn
(a)
F = Force exerted by counterweight
Panel:
+
SFx = ma :
T - F = 20a
(1)
Counterweight A: Its acceleration has two components
a A = a P + a A /P = a Æ + a Ø
+
+
SFx = max : F = 12.5a
(2)
SFy = ma y : 12.5 g - T = 12.5a
(3)
Adding (1), (2), and (3):
T - F + F + 12.5 g - T = (20 + 12.5 + 12.5)a
a=
12.5 g 12.5 ¥ 9.81
=
= 2.725 m/s2
45
45
a = 2.73 m/s2 b
Substituting for a into (3):
12.5 ¥ 9.81 - T = 12.5 ¥ (2.725) T = 12.5 ¥ (9.81 - 2.725)
T = 88.6 N b
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48
PROBLEM 12.29 (continued)
(b)
Panel:
+
SFx = ma :
T = 20a
(1)
Counterweight A:
+
SFy = ma :
Adding (1) and (2):
12.5 g - T = 12.5a
(2)
T + 12.5 g - T = 32.5a
a=
5
g
13
a = 3.77 m/s2 ¨ b
Substituting for a into (1):
Ê5 ˆ
T = 20 Á g ˜
Ë 13 ¯
(c)
T = 75.5 N b
Since panel is accelerated to the left, there is no force exerted by panel on counterweight and vice versa.
Panel:
+
SFx = ma :
T = 20a
(1)
+
Counterweight A: Same free body as in Part (b):
SFy = ma :
12.5 g - T = 12.5a
(2)
Since Eqs. (1) and (2) are the same as in (b), we get the same answers:
a = 3.77m/s2 ¨; T = 75.5 N b
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49
PROBLEM 12.30
An athlete pulls handle A to the left with a constant
force of P = 100 N. Knowing that after the handle
A has been pulled 30 cm its velocity is 3 m/s,
determine the mass of the weight stack B.
SOLUTION
Given:
P  100 N
x A  0.30 m
v A  3 m/s
Kinematics:
x A  4 yB  constant
v A  4vB  0
a A  4aB  0
(1)
Uniform Acceleration of handle A:

v A 2   v A o +2 a A x A   x A o
2

32  0 2  2 a A  0.3  0 
a A  15 m/s 2
From (1):
a B  3.75 m/s 2 
From FBD:
F
y
Free Body Diagram:
 mB aB
4 P  mB g  mB aB
mB 

4P
g  aB
4 100 
kg
9.81  3.75
mB  29.50 kg 
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50
PROBLEM 12.31
A 6-kg block B rests as shown on a 10-kg bracket A. The coefficients
of friction are m s = 0.30 and m k = 0.25 between block B and bracket
A, and there is no friction in the pulley or between the bracket and
the horizontal surface. (a) Determine the maximum mass of block
C if block B is not to slide on bracket A. (b) If the mass of block C is
10% larger than the answer found in a, determine the accelerations
of A, B and C.
SOLutiOn
Kinematics. Let x A and x B be horizontal coordinates of A and B measured from a fixed vertical line to the left
of A and B. Let yC be the distance that block C is below the pulley. Note that yC increases when C moves
downward. See figure.
The cable length L is fixed.
L = ( x B - x A ) + ( xP - x A ) + yC + constant
Differentiating and noting that x P = 0,
v B - 2v A + vC = 0
-2aA + aB + aC = 0
(1)
Here, aA and aB are positive to the right, and aC is positive downward.
Kinetics. Let T be the tension in the cable and FAB be the friction force between blocks A and B. The free body
diagrams are:
Bracket A:
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51
PROBLEM 12.31 (continued)
Block B:
Block C:
+
(2)
+
Block B:
SFx = max : 2T - FAB = mA a A
SFx = max : FAB - T = mB aB
(3)
+
Bracket A:
SFy = ma y : N AB - mA g = 0
N AB = mA g
or
+
Block C:
SFy = ma y : mC g - T = maC
(4)
Adding Eqs. (2), (3), and (4), and transposing,
mA aA + mB aB + mC aC = mC g
(5)
Subtracting Eq. (4) from Eq. (3) and transposing,
mB aB - mC aC = FAB - mC g
(a)
(6)
No slip between A and B. aB = aA
From Eq. (1),
From Eq. (5),
For impending slip,
aA = aB = aC = a
a=
mC g
mA + mB + mC
FAB = m s N AB = m s mB g
Substituting into Eq. (6),
( mB - mC )( mC g )
= m s mB g - mC g
mA + mB + mC
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52
PROBLEM 12.31 (continued)
Solving for mC ,
mC =
=
m s mB ( mA + mB )
mA + 2mB - m s mC
(0.30)(6)(10 + 6)
10 + (2)(6) - (0.30)(6)
mC = 1.426 kg b
(b)
mC increased by 10%.
mC = 1.568 kg
Since slip is occurring,
FAB = m k N AB = m k mB g
Eq. (6) becomes
mB aB - mC aC = ( m k mB - mC ) g
or
6aB - 1.568aC = [(0.25)(6) - 1.568](9.81)
(7)
With numerical data, Eq. (5) becomes
10a A + 6aB + 1.568aC = (1.568)(9.81)
(8)
Solving Eqs. (1), (7), and (8) gives
aA = 1.053 m/s2 , aB = 0.348 m/s2 , aC = 1.759 m/s2
a A = 1.503 m/s2
b
a B = 0.348 m/s2
b
aC = 1.759 m/s2 b
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53
PROBLEM 12.32
Knowing that  = 0.30, determine the acceleration of each block when
mA = mB= mC.
SOLUTION
Given:
  0.30
mA  mB  mc  m
Kinematics:
y A  2 y B  xC  constant
v A  2vB  vC  0
aC  a A  2aB  0
(1)
Free Body Diagram of Block A:
Equation of Motion:
F
y
 maA
mg  T  ma A
aA  g 
Free Body Diagram of Block B:
T
m
(2)
Equation of Motion:
F
y
 maB
mg  2T  maB
aB  g 
Free Body Diagram of Block B:
2T
m
(3)
Equations of Motion:
F
y
0
N  mg
F
x
 maC
0.3mg  T  maC
aC  0.3g 
T
m
(4)
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Problem 12.32 (Continued)
3.3g 
Substitute (2), (3) and (4)  (1)
T
6T
0
m
1.1
mg
2
(5)
Substitute (5)  (2):
aA  g 
1.1
g
2
a A  0.45g  
Substitute (5)  (3):
aB  g 
2.2
g
2
a B  0.10g  
Substitute (5)  (4):
aC  0.3g 
1.1
g
2
aC  0.25g  
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PROBLEM 12.33
Knowing that  = 0.30, determine the acceleration of each block
when mA = 5 kg, mB= 30 kg, and mC = 15 kg.
SOLUTION
Given:
  0.30, g  9.81 m/s 2
mA  5 kg, mB  30 kg, mC  15 kg
Kinematics:
y A  2 y B  xC  constant
v A  2vB  vC  0
a A  2aB  aC  0
Free Body Diagram of Block A:
(1)
Equation of Motion:
F
y
 mA a A
m A g  T  mA a A
aA  g 
Free Body Diagram of Block B:
T
mA
(2)
Equation of Motion:
F
y
 mB aB
mB g  2T  mB aB
aB  g 
Free Body Diagram of Block B:
2T
mB
(3)
Equations of Motion:
F
y
0
N  mC g
F
x
 mC aC
0.3mC g  T  mC aC
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Problem 12.33 (Continued)
aC  0.3 g 
Substitute (2), (3) and (4)  (1)
3.3 g 
T
mC
(4)
T
4T T


0
m A mB mC
Substitute (5)  (2):
aA  g 
8.25
g
5
a A  6.377 m/s 2  
Substitute (5)  (3):
aB  g 
16.5
g
30
a B  4.415 m/s 2  
Substitute (5)  (4):
aC  0.3g 
8.25
g
15
aC  2.453 m/s 2  
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57
PROBLEM 12.34
A 25--kg block A rests on an inclined surface, and a 15-kg
kg counterweight
B is attached to a cable as shown. Neglecting friction, determine the
acceleration of A and the tension in the cable immediately after the
system is released from rest.
SOLUTION
Let the positive direction of x and y be those shown in the sketch, and let
the origin lie at the cable anchor.
Constraint of cable: x A  yB / A  constant or a A  aB / A  0, where the
positive directions of a A and aB / A are respectively the x and the y
directions. Then
a B / A  a A
First note that a B  a A  a B / A   aA
Block B
B:
(1)
20    aB / A
20
20 
Fx  m B  aB  x : m B g sin 20  N AB  mB a A
mB a A  N AB  mB g sin 20
15 a A  N AB  50.328
(2)
Fy  mB  aB  y : mB g cos 20  T  mB aB / A
mB aB / A  T  mB g cos 20
15 aB / A  T  138.276
Block A:
(3)
Fx  mAa A : mA g sin 20  N AB  T  mAaA
mAa A  N AB  T  mA g sin 20
25 aB / A  N AB  T  83.880
(4)
Eliminate aB / A using Eq. (1), then add Eq. (4) to Eq. (2) and
subtract Eq. (3).
55 a A  4.068 or aA  0.0740 m/s 2 , a A  0.0740 m/s 2
20 
From Eq. (1), aB / A  0.0740 m/s2
From Eq. (3), T  137.2 N
T  137.2 N 
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58
PROBLEM 12.35
Block B of mass 10-kg
kg rests as shown on the upper surface of a 22-kg
22
wedge A. Knowing that the system is released from rest and neglecting
friction, determine (a) the acceleration of B, (b)) the velocity of B relative to
A at t  0.5 s.
SOLUTION
A:
Fx  mA a A : WA sin 30  N AB cos 40  mA a A
(a)
NAB 
or
22  a A  12 g 
cos 40
Now we note: a B  a A  a B /A , where a B/A is directed along the top surface of A.
A
B:
Fy  mB a y : NAB  WB cos 20  mB a A sin 50
NAB  10 ( g cos 20  a A sin 50)
or
Equating the two expressions for NAB
1 

22  a A  g 
2 

 10( g cos 20  a A sin 50)
cos 40
or
aA 
(9.81)(1.1  cos 20 cos 40)
 6.4061 m/s 2
2.2  cos 40 sin 50
Fx  mB ax : WB sin 20  mB aB/A  mB a A cos 50
or
aB/A  g sin 20  a A cos50
 (9.81sin 20  6.4061cos50) m/s 2
 7.4730 m/s 2
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PROBLEM 12.35 (C
(Continued)
Finally
a B  a A  a B/ A
We have
aB2  6.40612  7.47302  2(6.4061  7.4730) cos50
or
aB  5.9447 m/s 2
and
or
7.4730 5.9447

sin 
sin 50
50
  74.4
a B  5.94 m/s2
(b)
75.6 
Note: We have uniformly accelerated motion, so that
v  0  at
Now
v B/A  v B  v A  a B t  a A t  a B/A t
At t  0.5 s:
vB/A  7.4730 m/s 2  0.5 s
v B/A  3.74 m/s
or
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20 
PROBLEM 12.36
A 450-g tetherball A is moving along a horizontal circular path at a
constant speed of 4 m/s. Determine (a)) the angle  that the cord
forms with pole BC, (b)) the tension in the cord.
SOLUTION
a A  an 
First we note
v A2

  l AB sin 
where
Fy  0: TAB cos  WA  0
(a)
TAB 
or
mA g
cos 
Fx  mA a A : TAB sin   mA
v A2

Substituting for TAB and 
mA g
v A2
sin   mA
cos 
l AB sin 
1  cos 2  
or
sin 2   1  cos2 
(4 m/s) 2
cos 
1.8 m  9.81 m/s 2
cos 2   0.906105cos   1  0
cos   0.64479
Solving
  49.9 
or
(b)
From above
TAB 
mA g 0.450 kg  9.81 m/s 2

cos 
0.64479
TAB  6.85 N 
or
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61
PROBLEM 12.37
During a hammer thrower’s practice swings, the 7.1-kg
7.1
head
A of the hammer revolves at a constant speed v in a horizontal
circle as shown. If   0.93 m and   60, determine
(a) the tension in wire BC, (b)) the speed of the hammer’s
head.
SOLUTION
First we note
a A  an 
v A2

Fy  0: TBC sin 60  WA  0
(a)
or
7.1 kg  9.81 m/s 2
sin 60
 80.426 N
TBC 
TBC  80.4 N 
Fx  mA a A : TBC cos 60  mA
(b)
or
v A2 
v A2

(80.426 N)
N)cos
cos 60  0.93 m
7.1 kg
v A  2.30 m/s 
or
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62
PROBLEM 12.38
Human centrifuges are often used to simulate different acceleration
levels for pilots. When aerospace physiologists say that a pilot is
pulling 9g’s,, they mean that the resultant normal force on the pilot from
the bottom of the seat is nine times their weight.
weight Knowing that the
centrifuge starts from rest and has a constant angular acceleration
a
of
1.5 RPM per second until the pilot is pulling 9g’s
9
and then continues
with a constant angular velocity, determine (a)) how long it will take for
the pilot to reach 9g’s (b) the angle  of the normal force once the pilot
reaches 9 g’s. Assume that the force parallel to the seat is zero.
SOLUTION
Given:
  1.5 RPM/s  0.157 rad/s 2
0  0
N  9 mg
R7 m
Free Body Diagram of Pilot:
Equations of Motion:
F
y
 ma y
N sin   mg  m  0 
N sin   mg
(1)
F
n
 man
N cos   mR 2
=
N cos 
mR
(2)
Substitute N=9mg into (1): 9mg sin   mg
1
  sin 1  
9
  6.379
Substitute N=9mg and θ into (2):
9*9.81cos 6.379
7
  3.540 rad/s
=
For constant angular acceleration
acceleration:
  0   t
3.540  0  0.157 * t
(a) Solving for t:
t  22.55 s 
From earlier:
  6.379 
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63
PROBLEM 12.39
A single wire ACB passes through a ring at C attached to a sphere which
revolves at a constant speed v in the horizontal circle shown. Knowing that
the tension is the same in both portions of the wire, determine the speed v.
SOLUTION
Fx  ma: T (sin 30  sin 45) 
mv 2

(1)
Fy  0: T (cos 30  cos 45)  mg  0
T (cos 30  cos 45)  mg
Divide Eq. (1) by Eq. (2):
(2)
sin 30  sin 45 v 2

cos30  cos 45  g
v 2  0.76733 g  0.76733 (1.6 m)(9.81 m/s2 )  12.044 m2 //s2
v  3.47 m/s 
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64
PROBLEM 12.40*
Two wires AC and BC are tied at C to a sphere which revolves at a
constant speed v in the horizontal circle shown. Determine the range of
the allowable values of v if both wires are to remain taut and if the
tension in either of the wires is not to exceed 60 N.
SOLUTION
From the solution of Problem 12.39, we find that both wires remain taut for
3.01 m/s  v  3.96 m/s 
To determine the values of v for which the tension in either wire will not exceed 60 N, we recall
Eqs. (1) and (2) from Problem 12.39:
TAC sin 30  TBC sin 45 
mv 2

(1)
TAC cos 30  TBC cos 45  mg
(2)
Subtract Eq. (1) from Eq. (2). Since sin 45°  cos 45°, we obtain
TAC (cos30  sin 30)  mg 
mv 2

(3)
Multiply Eq. (1) by cos 30°, Eq. (2) by sin 30°, and subtract:
TBC (sin 45 cos30  cos 45 sin 30) 
TBC sin15 
mv 2
cos30  mg sin 30

mv 2
cos30  mg sin 30

(4)
Making TAC  60 N, m  5 kg,   1.6 m, g  9.81 m/s 2 in Eq. (3), we find the value v1 of v for which
TAC  60 N:
60(cos 30  sin 30)  5(9.81) 
21.962  49.05 
5v12
1.6
v12 2
v1  8.668,
0.32
We have TAC  60 N for v  v1 , that is, for
v1  2.94 m/s
v  2.94 m/s 
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PROBLEM 12.40* (Continued)
Making TBC  60 N, m  5 kg,   1.6 m, g  9.81 m/s 2 in Eq. (4), we find the value v2 of v for which
TBC  60 N:
60sin 15 
5v22
cos 30  5(9.81) sin 30
1.6
15.529  2.7063v22  24.523
v22  14.80,
v2  3.85 m/s
We have TBC  60 N for v  v2 , that is, for
v  3.85 m/s 
Combining the results obtained, we conclude that the range of allowable value is
3.01 m/s  v  3.85 m/s 
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66
PROBLEM 12.41
A 1-kg
kg sphere is at rest relative to a parabolic dish which rotates at a
constant rate about a vertical axis. Neglecting friction and knowing
that r 1 m, determine (a) the speed v of the sphere, (b)
( the magnitude
of the normal force exerted by the sphere on the inclined surface of
the dish.
SOLUTION
r 2 dy
,
 r  1  tan 
2
dx
Given:
y 
Free Body Diagram of Sphere:
Equations of Motion:
  45
or
  Fy  0: N cos   mg  0
N 
mg
cos
Fn  man : N sin   man
mg tan   m
v2
r
v 2  gr tan 
(a) v 2   9.8111.0000   9.81 m 2 /s 2
(b) N 
1 9.81
mg

cos 45
cos 45
v  3.13 m/s 
N  13.87 N 
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67
PRoBlEM 12.42*
As part of an outdoor display, a 6 kg model C of the earth is attached to wires AC and
BC and revolves at a constant speed v in the horizontal circle shown. Determine the
range of the allowable values of v if both wires are to remain taut and if the tension in
either of the wires is not to exceed 125 N.
Solution
aC = an =
First note
vC2
r
r = 0.9 m
where
+
+
vC2
r
(1)
SFy = 0 : TCA cos 40∞ - TCB cos15∞ - mC g = 0
(2)
SFx = mC aC : TCA sin 40∞ + TCB sin15∞ = mC
Note that Eq. (2) implies that
(a)
when
TCB = (TCB ) max,
TCA = (TCA ) max
(b)
when
TCB = (TCB ) min ,
TCA = (TCA ) min
Case 1: TCA is maximum.
TCA = 125 N
Let
Eq. (2)
or
(125 N ) cos 40∞ - TCB cos15∞ - (6 ¥ 9.81) = 0
TCB = 68.4732 N
(TCB )(TCA )max < 125 N
O.K.
[(TCB ) max = 68.4732 N ]
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68
PRoBlEM 12.42* (continued)
Eq. (1)
( vC2 )(TCA )max =
0.9
(125 sin 40∞ + 68.4732 sin 15∞) N
6
( vC )(TCA )max = 3.8354 m/s
or
Now we form
(cos15∞) (1) + (sin 15∞) (2)
TCA sin 40∞ cos15∞ + TCA cos 40∞ sin 15∞ = mC
vC2
cos15∞ + mC g sin 15∞
r
TCA sin 55∞ = mC
vC2
cos15∞ + mC g sin 15∞
r
or
(3)
( vc ) max occurs when TCA = (TCA ) max
( vC ) max = 3.84 m/s
Case 2: TCA is minimum.
Because (TCA ) min occurs when TCB = (TCB ) min,
let TCB = 0 (note that wire BC will not be taut).
Eq. (2)
TCA cos 40∞ - (6 g ) = 0
TCA = 76.8363 N 125 N
or
O.K.
Note: Eq. (3) implies that when TCA = (TCA ) min , vC = ( vC ) min . Then
0.9
(76.8363)sin 40∞
6
Eq. (1)
( vC2 ) min =
or
( vC ) min = 2.7218 m/s
0 TCA , TCB 125 N when
2.72 m/s vC 3.84 m/s b
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69
PRoBlEM 12.43*
The 600 g flyballs of a centrifugal governor revolve at a constant speed v
in the horizontal circle of 150 mm radius shown. Neglecting the weights
of links AB, BC, AD, and DE and requiring that the links support only
tensile forces, determine the range of the allowable values of v so that the
magnitudes of the forces in the links do not exceed 80 N.
Solution
v2
r
First note
a = an =
where
r = 0.15 m
+
+
SFx = ma : TDA sin 20∞ + TDE sin 30∞ = m
v2
r
FBD of flyball
SFy = 0 : TDA cos 20∞ - TDE cos 30∞ - mg = 0
(1)
(2)
Note that Eq. (2) implies that
(a)
when
TDE = (TDE ) max,
TDA = (TDA ) max
(b)
when
TDE = (TDE ) min,
TDA = (TDA ) min
Case 1: TDA is maximum.
TDA = 80 N
Let
Eq. (2)
(80 N ) cos 20∞ - TDE cos 30∞ - (0.6 ¥ 9.81) = 0
or
TDE = 80.01 N unacceptable ( 80 N )
Now let
TDE = 80 N
Eq. (2)
or
TDA cos 20∞ - (80 N ) cos 30∞ - (0.6 ¥ 9.81) = 0
TDA = 79.9921 N O.K. ( 80 N)
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70
PRoBlEM 12.43* (continued)
(TDA ) max = 79.9921 N
(TDE ) max = 80 N
( v 2 )(TDA )max =
Eq. (1)
0.15 m
(79.9921sin 20∞ + 80 sin 30∞) N
0.6 kg
v(TDA )max = 4.10362 m/s
or
(cos 30∞)(1) + (sin 30∞)(2)
Now form
TDA sin 20∞ cos 30∞ + TDA cos 20∞ sin 30∞ = m
v2
cos 30∞ + mg sin 30∞
r
TDA sin 50∞ = m
v2
cos 30∞ + mg sin 30∞
r
or
(3)
vmax occurs when TDA = (TDA ) max
vmax = 4.10362 m/s
Case 2: TDA is minimum.
Because (TDA ) min occurs when TDE = (TDE ) min ,
let TDE = 0.
Eq. (2)
TDA cos 20∞ - (0.6 ¥ 9.81) = 0
or
TDA = 6.26375 N 80 N
O.K.
Note: Eq. (3) implies that when TDA = (TDA ) min , v = vmin . Then
Eq. (1)
( v 2 ) min =
(0.15 m)
(6.26375 N )sin 20∞
(0.6 kg)
vmin = 0.53558 m/s
or
0 TAB , TBC , TAD , TDE 80 N
when
0.536 m/s v 4.10 m/s b
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71
PRoBlEM 12.44
A 60-kg wrecking ball B is attached to a 15-m-long steel cable AB
and swings in the vertical arc shown. Determine the tension in the
cable (a) at the top C of the swing, (b) at the bottom D of the swing,
where the speed of B is 4.2 m/s.
Solution
(a)
At C, the top of the swing, v B = 0; thus
an =
+
v B2
=0
LAB
SFn = 0 : TBA - WB cos 20∞ = 0
TBA = (60 kg)(9.81 m/s2 ) ¥ cos 20∞
or
TBA = 553 N b
or
+
(b)
or
SFn = man : TBA - WB = mB
( v B )2D
LAB
È
( 4.2 m/s)2 ˘
TBA = (60 kg) Í9.81 m/s2 +
˙
15 m ˚
Î
TBA = 659 N b
or
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72
PRoBlEM 12.45
During a high-speed chase, a 1000 kg sports car traveling at a
speed of 160 km/h just loses contact with the road as it reaches
the crest A of a hill. (a) Determine the radius of curvature r of
the vertical profile of the road at A. (b) Using the value of r
found in part a, determine the force exerted on a 80 kg driver
by the seat of his 1200 kg car as the car, traveling at a constant
speed of 75 km/h, passes through A.
Solution
(a)
Note:
+
160 km/h =
400
m/s
9
S Fn = man : Wcar =
Wcar v 2A
g r
r=
or
Ê 400
ˆ
m/s˜
ÁË
¯
9
2
9.81 m/s2
= 201.357 m
r = 201 m b
or
(b)
Note: v is constant fi at = 0; 75 km/h =
125
m/s
6
+
SFn = man : W - N =
W v 2A
g r
2
or
N =W -
mv A2
S
Ê 125 ˆ
80 ¥ Á
Ë 6 ˜¯
= 80 ¥ 9.81 = 612.359 N
201.357
N = 612 N ≠ b
or
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73
PROBLEM 12.46
An airline pilot climbs to a new flight level along the path shown.
Knowing that the speed of the airplane decreases at a constant rate from
180 m/s at point A to 160 m/s at point C,, determine the magnitude of the
abrupt change in the force exerted on a 90
90-kg
kg passenger as the airplane
passes point B.
SOLUTION
Angle change over arc AB.  
8
  0.13963 rad
180
Length of arc: s AB     6000  0.13963  837.76 m
sBC  800 m, s AC  837.76  800  1637.76 m
160
v
180

1637.76
dv  0
at ds
or
1602 1802

 at 1637.76 
2
2
at  2.076 m/s 2
vB
v
180

837.76
dv  0
at ds
or
vB2  28922 m 2/s 2
vB2 1802

  2.076  837.76 
2
2
vB  170.
170.06 m/s
Weight of passenger: mg   90  9.81  882.9 N
v  170.06 m/s,   6000 m
Just before point B.
2
170.06   4.820 m/s 2
v2
an 


6000
Free Body Diagram of Pilot:
Fn  N1  W  m  an 1 : N1  882.9   90  4.820   449.1 N
Ft  Ft  mat :
Just after point B.
 Ft 1
  90  2.076   186.8 N
v  170.06 m/s,   ,
an  0
Fy  0 : N 2  W  0 N 2  W  882.9 N
Fx  mat : Ft  mat   90  2.076   186.8 N
Ft does not change.
N increases by 433.8 N.
magnitude of change of force  434 N 
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PROBLEM 12.47
The roller-coaster
coaster track shown is contained in a vertical plane.
The portion of track between A and B is straight and
horizontal, while the portions to the left of A and to the right of
B have radii of curvature as indicated. A car is traveling at a
speed of72 km/h when the brakes are suddenly applied,
causing the wheels of the car to slide on the track ( k  0.20).
Determine the initial deceleration of the car if the brakes are
applied as the car (a)) has almost reached A, (b) is traveling
between A and B, (c) has just passed B.
B
SOLUTION
Fn  man : N  mg  m
(a)
v2


v2 
N  m  g  



v2 
F  k N  k m  g  


Ft  mat : F  mat
at 
Given data:

F
v2
 k  g 
m





k  0.20, v  72 km/h  20 m/s
g  9.81 m/s 2 ,
  30 m

(20)2 
at  0.20 9.81 

30 

(b)
an  0
at  4.63 m/s2 
Fn  man  0: N  mg  0
N  mg
F  k N  k mg
Ft  mat : F  mat
at 
F
 k g  0.20(9.81)
m
at  1.962 m/s2 
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PROBLEM 12.47 (C
(Continued)
(c)
Fn  man : mg  N 
mv 2


v2 
N  m  g  
 


v2 
F  k N  k m  g  


Ft  mat : F  mat
at 


F
v2 
(20)2 
 k  g    0.20 9.81 

m

45 


at  0.1842 m/s 2 
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76
PROBLEM 12.48
A spherical-cap
cap governor is fixed to a vertical shaft that rotates with angular
velocity . When the string-supported
supported clapper of mass m touches the cap, a
cutoff switch is operated electrically to reduce the speed of the shaft.
Knowing that the radius of the clapper is small relative to the cap, determine
the minimum angular speed at which the cutoff switch operates.
SOLUTION
Given:
min is the angular velocity when the clapper hits the cap
Geometry From Figure:
Distance AB is 300 mm, the length of the Clapper Arm
Distance BC and AC is 300 mm, the radius of the Spherical Cap.
Therefore  ABC is equilateral, so   60 and   30
R  0.3*cos 
Free Body Diagram of Clapper:
Equations of Motion:
F
y
 ma y
n
FA cos   mg  m  0 
FA 
Substitute (1)into (2):
win =
mg
cos 
F
(1)
 man
2
FA sin   mRmin
win =
FA sin 
mR
(2)
m g tan 
mR
9.81tan 60
0.3cos 30
 8.087 rad/s

win =77.23 rpm 
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PROBLEM 12.49
A series of small packages, each with a mass of 0.5 kg, are discharged from a
conveyor belt as shown. Knowing that the coefficient of static friction between
each package and the conveyor belt is 0.4, determine (a)) the force exerted by the
belt on a package just after it has passed Point A, (b) the angle  defining the
Point B where the packages first slip relative to the belt.
SOLUTION
Assume package does not slip.
at  0, F f  s N
On the curved portion of the belt
an 
v 2 (1 m/s)2

 4 m/s 2
 0.250 m
For any angle 
Fy  ma y : N  mg cos   man  
mv 2

N  mg cos  
mv 2

(1)
Fx  max : Ff  mg sin   mat  0
F f  mg sin 
(a)
At Point A,
  0
N  (0.5)(9.81)(1.000)  (0.5)(4)
(b)
At Point B,
(2)
N  2.905 N 
F f  s N
mg sin   s (mg cos   man )

a 
4 

sin   s  cos   n   0.40 cos  
g 
9.81 


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PROBLEM 12.49 (Continued)
Squaring and using trigonometic identities,
1  cos 2   0.16 cos2   0.130479cos   0.026601
1.16 cos 2   0.130479cos   0.97340  0
cos   0.97402
  13.09 
Check that package does not separate from the belt.
N
Ff
s

mg sin 
s
N  0. 
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79
PROBLEM 12.50
A 54
54-kg
kg pilot flies a jet trainer in a half vertical loop of 1200-m
1200
radius so that the speed of the trainer decreases at a constant rate.
Knowing that the pilot’s apparent weights at Points A and C are 1680
N and 350 N, respectively, determine the force ex
exerted
erted on her by the
seat of the trainer when the trainer is at Point B.
SOLUTION
First we note that the pilot’s apparent weight is equal to the vertical force that she exerts on the
seat of the jet trainer.
At A:
Fn  man : N A  W  m
v 2A

 1680 N

v A2  (1200 m) 
 9.81 m/s2 
 54 kg

or
 25,561.3 m 2 /s 2
At C:
Fn  man : NC  W  m
vC2

 350 N

vC2  (1200 m) 
 9.81 m/s 2 
 54 kg

or
 19,549.8 m 2 /s 2
Since at  constant, we have from A to C
vC2  vA2  2at s AC
19,549.8 m 2/s 2  25,561.3 m 2 /s 2  2at (  1200 m)
or
at  0.79730 m/s 2
or
Then from A to B
vB2  vA2  2at s AB


 25,561.3 m 2/s 2  2(0.79730 m/s 2 )   1200 m 
2


 22,555 m 2/s 2
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PROBLEM 12.50 (Continued)
At B:
Fn  man : N B  m
vB2

22,555 m 2 /s 2
1200 m
or
N B  54 kg
or
N B  1014.98 N
Ft  mat : W  PB  m | at |
or
PB  (54 kg)(0.79730  9.81) m/s 2
or
PB  486.69 N
Finally,
( Fpilot ) B  N B2  PB2  (1014.98) 2  (486.69)2
 1126 N
(Fpilot ) B  1126 N
or
25.6° 
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81
PROBLEM 12.51
A carnival ride is designed to allow the general public to experience high acceleration motion. The ride rotates
about Point O in a horizontal circle such that the rider has a speed v0. The rider reclines on a platform A which
rides on rollers such that friction is negligible. A mechanical stop prevents the platform from rolling down the
incline. Determine (a) the speed v0at which the platform A begins to roll upwards, (b)) the normal force
experienced by an 80-kg rider at this speed.
SOLUTION
Radius of circle:
R  5  1.5cos 70  5.513 m
F  ma:
Components up the incline,
70°:
 mA g cos 20  
mv02
sin 20
R
1
(a)
Components normal to the incline,
N  mg sin 20 
(b)
1
 g R  2  (9.81 m/s) (5.513 m  2
Speed v0 : v0  
 
  12.1898 m/s
tan 20
 tan 20 


Normal force:
v0  12.19 m/s 
20°.
mv02
cos 20.
R
N  (80)(9.81)sin 20 
80(12.1898) 2
cos 20  2294 N
5.513
N  2290 N 
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82
PROBLEM 12.52
A curve in a speed track has a radius of 300 m and a rated speed of 190 km/h. (See
sample Problem 12.6 for the definition of rated speed). Knowing that a racing
car starts skidding on the curve when traveling at a speed of 290 km/h, determine
(a) the banking angle q, (b) the coefficient of static friction between the tires and
the track under the prevailing conditions, (c) the minimum speed at which the
same car could negotiate that curve.
SOLutiOn
W = mg
Weight
Acceleration
a=
v2
r
SFx = max : F + W sin q = ma cos q
F=
mv 2
cos q - mg sin q
r
(1)
SFy = ma y : N - W cos q = ma sin q
N=
(a)
mv 2
sin q + mg cos q
r
Banking angle. Rated speed v = 190 km / h =
0=
(2)
475
m / s. F = 0 at rated speed.
9
mv 2
cos q - mg sin q
r
2
Ê 475 ˆ
ÁË
˜
v
9 ¯
tan q =
=
r g (300)(9.81)
q = 43.425∞
2
(b)
Slipping outward.
q = 43.5∞ b
Ê 725 ˆ
m/s
v = 290 km / h = Á
Ë 9 ˜¯
F = mN
m=
F v 2 cos q - r g sin q
=
N v 2 sin q + r g cos q
2
m=
Ê 725 ˆ
ÁË
˜ cos ( 43.425∞) - (300)(9.81)sin 43.425∞
9 ¯
2
Ê 725 ˆ
ÁË
˜ sin ( 43.425∞) + (300)(9.81) cos 43.425∞
9 ¯
= 0.40768
m = 0.408 b
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379
83
PROBLEM 12.52 (continued)
(c)
Minimum speed.
F = - mN
-m =
v 2 cos q - r g sin q
v 2 sin q + r g cos q
r g (sin q - m cos q )
v2 =
cos q + m siin q
=
(300)(9.81)(sin 43.425∞ - 0.40768 cos 43.425∞)
(cos 43.425∞ + 0.40768 sin 43.425∞)
= 1144 m2 /s2
v = 33.8258 m/s
v = 121.8 km/h b
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84
PROBLEM 12.53
Tilting trains, such as the American Flyer, which will run from
Washington to New York and Boston, are designed to travel safely
at high speeds on curved sections of track, which were built for
slower, conventional trains. As it enters a curve, each car is tilted
by hydraulic actuators mounted on its trucks. The tilting feature of
the cars also increases passenger comfort by eliminating or greatly
reducing the side force Fs (parallel to the floor of the car) to which
passengers feel subjected. For a train traveling at 160 km/h on a
curved section of track banked through an angle q = 6∞ and with a
rated speed of 100 km/h, determine (a) the magnitude of the side
force felt by a passenger of weight W in a standard car with no
tilt (f = 0), (b) the required angle of tilt f if the passenger is to feel
no side force. (See Sample Problem 12.6 for the definition of rated
speed.)
SOLutiOn
vR = 100 km/h =
Rated speed:
250
400
m/s, 160 km/h =
m/s
9
9
From Sample Problem 12.6,
vR2 = g r tan q
2
Ê 250 ˆ
ÁË
˜
vR2
9 ¯
r=
=
= 748.352 m
g tan q 9.81 tan 6∞
or
Let the x-axis be parallel to the floor of the car.
+
SFx = max : Fs + W sin (q + f ) = man cos (q + f )
=
(a)
f = 0.
mv 2
cos (q + f )
r
È v2
˘
Fs = W Í cos (q + f ) - sin (q + f ) ˙
Î gr
˚
2
˘
È
Ê 400 ˆ
˙
Í
ÁË
˜¯
9
cos 6∞ - sin 6∞˙
=W Í
˙
Í (9.81)(748.352)
˙
Í
˚
Î
= 0.16306W
Fs = 0.1631W b
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85
PROBLEM 12.53 (continued)
(b)
For Fs = 0,
v2
cos (q + f ) - sin (q + f ) = 0
gr
2
Ê 400 ˆ
ÁË
˜
2
v
9 ¯
tan (q + f ) =
=
= 0.26907
g r (9.81)(748.352)
q + f = 15.060∞∞
f = 15.06∞ - 6∞
f = 9.06∞ b
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86
PROBLEM 12.54
Tests carried out with the tilting trains described in Problem 12.53
revealed that passengers feel queasy when they see through the car
windows that the train is rounding a curve at high speed, yet do not
feel any side force. Designers, therefore, prefer to reduce, but not
eliminate that force. For the train of Problem 12.53, determine the
required angle of tilt f if passengers are to feel side forces equal to
10% of their weights.
PROBLEM 12.53 Tilting trains, such as the American Flyer, which
will run from Washington to New York and Boston, are designed to
travel safely at high speeds on curved sections of track, which were
built for slower, conventional trains. As it enters a curve, each car
is tilted by hydraulic actuators mounted on its trucks. The tilting
feature of the cars also increases passenger comfort by eliminating
or greatly reducing the side force Fs (parallel to the floor of the car)
to which passengers feel subjected. For a train traveling at 160 km/h
on a curved section of track banked through an angle q = 6∞ and
with a rated speed of 100 km/h, determine (a) the magnitude of the
side force felt by a passenger of weight W in a standard car with no
tilt (f = 0), (b) the required angle of tilt f if the passenger is to feel
no side force. (See Sample Problem 12.6 for the definition of rated
speed.)
SOLutiOn
vR = 100 km/h =
Rated speed:
250
400
m/s, 160 km/h =
m/s
9
9
From Sample Problem 12.6,
vR2 = g r tan q
2
Ê 250 ˆ
ÁË
˜
9 ¯
r=
=
= 748.352 m
g tan q 9.81 tan 6∞
vR2
or
Let the x-axis be parallel to the floor of the car.
+
SFx = max : Fs + W sin (q + f ) = man cos (q + f )
=
Solving for Fs,
mv 2
cos (q + f )
r
È v2
˘
Fs = W Í cos (q + f ) - sin (q + f ) ˙
Î gr
˚
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87
PROBLEM 12.54 (continued)
2
Ê 400 ˆ
ÁË
˜
v2
9 ¯
=
= 0.26907 and
g r (9.81)(748.352)
Now
So that
Fs = 0.10W
0.10W = W [0.26907 cos (q + f ) - sin (q + f )]
u = sin (q + f )
Let
cos (q + f ) = 1 - u 2
Then
0.10 = 0.26907 1 - u 2 - u or 0.26907 1 - u 2 = 0.10 + u
Squaring both sides,
or
0.072399(1 - u 2 ) = 0.01 + 0.2u + u 2
1.072399u 2 + 0.2u - 0.062399 = 0
The positive root of the quadratic equation is u = 0.16537
Then,
q + f = sin -1 u = 9.5187∞
f = 9.5187∞ - q
f = 3.52∞ b
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88
PROBLEM 12.55
A 3-kg
kg block is at rest relative to a parabolic dish which rotates at a
constant rate about a vertical axis. Knowing that the coefficient of
static friction is 0.5 and that r 2
2 m, determine the maximum
allowable velocity v of the block.
SOLUTION
Let  be the slope angle of the dish. tan  
At r  2 m, tan   1
or
dy
1
 r
dr
2
  45
Draw free body sketches of the sphere.
Fy  0: N cos    S N sin   mg  0
N 
mg
cos    S sin 
Fn  man: N sin    S N cos  
mv 2

mg (sin   S N cos  ) mv 2

cos    S sin 

v2   g
sin    S cos 
sin 45  0.5cos 45
 (2)(9.81)
 58.86 m2 /s2
cos   S sin 
cos 45  0.5sin 45
v  7.67 m/s 
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89
PROBLEM 12.56
A polisher is started so that the fleece along the circumference
2
undergoes a constant tangential acceleration of 4 m/s . Three
seconds after a polisher is started from rest, small tufts of
fleece from along the circumference of the 225-mm
mm-diameter
polishing pad are observed to fly free of the pad. At this
instant, determine (a) the speed v of a tuft as it leaves the pad,
(b) the magnitude of the force required to free a tuft if the
average mass of a tuft is 1.6 mg.
SOLUTION
(a)
at  constant  uniformly acceleration motion
Then
v  0  at t
At t  3 s:
v  (4 m/s2 )(3 s)
or
(b)
v  12.00 m/s 
Ft  mat : Ft  mat
or
Ft  (1.6  10 6 kg)(4 m/s 2 )
 6.4  10 6 N
Fn  man : Fn  m
At t  3 s:
Fn  (1.6  106 kg)
v2

(12 m/s) 2
m
 0.225
2
 2.048  103 N
Finally,
Ftuft  Ft 2  Fn2
 (6.4  106 N) 2  (2.048  10 3 N) 2
Ftuft  2.05  103 N 
or
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PROBLEM 12.57
A turntable A is built into a stage for use in a theatrical production.
It is observed during a rehearsal that a trunk B starts to slide on the
turntable 10 s after the turntable begins to rotate. Knowing that the
trunk undergoes a constant tangential
angential acceleration of 0.24 m/s 2 ,
determine the coefficient of static friction between the trunk and
the turntable.
SOLUTION
First we note that (aB )t  constant implies uniformly accelerated motion.
v B  0  ( a B )t t
vB  (0.24 m/s2 )(10 s)  2.4 m/s
At t  10 s:
In the plane of the turntable
F  mB a B : F  mB (a B )t  mB (a B )n
F  mB (aB )t2  (aB )n2
Then
 mB (aB )t2 
 
vB2

2
 Fy  0: N  W  0
or
N  mB g
At t  10 s:
F   s N   s mB g
Then
or
 s mB g  mB
(aB )t2
1
s 
9.81 m/s 2
 v2
  B




2
1/2
2

 (2.4 m/s) 2  

2 2
(0.24 m/s )  
 

 2.5 m  
s  0.236 
or
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91
PROBLEM 12.58
The carnival ride from Prob 12.51 is modified so that
the 80 kg riders can move up and down the inclined
wall as the speed of the ride increases. Assuming
that the friction between the wall and the carriage is
negligible, determine the position h of the rider if the
speed v0 = 13 m/s.
SOLUTION
Given:
m  80 kg, v0  13 m/s
Geometry from figure:
R  5  h cos 70
Free Body Diagram of Rider:
Equations of Motion:
F
y
 ma y
n
N cos 70  mg  m  0 
N
Substitute (1)into (2):
v
mg
cos 70
F
(1)
 man
mv 2
R
N sin 70R
v=
m
N sin 70 
(2)
m g tan 70 R
m
13  9.81tan 70  5  h cos 70 
h  3.714 m 
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92
PROBLEM 12.59
The carnival ride from Prob 12.51 is modified so that
the 80 kg riders can move up and down the inclined
wall as the speed of the ride increases. Knowing that
the coefficient of static friction between the wall and
the platform is 0.2, determine the range of values of
the constant speed v0 for which the platform will
remain in the position shown.
SOLUTION
Given:
m  80 kg, s  0.2, h  1.5 m
Geometry from figure:
R  5  1.5cos 70
 5.513 m
Minimum and Maximum speeds occur if slipping is impending:
f  s N
Finding the minimum speed so that rider does not slide down the wall:
Free Body Diagram of Rider:
Equations of Motion:
F
y
 ma y
N cos 70  f sin 70  mg  m  0 
N cos 70  s N sin 70  mg  0
mg
cos 70  s sin 70
N
N  1480.9 N
F
n
 man
2
mvmin
R
2
mvmin
N sin 70  s N cos 70 
R
N sin 70  f cos 70 
vmin =
N  sin 70   s cos 70  R
m
vmin  9.430 m/s 
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PROBLEM 12.59 (Continued)
Finding the maximum speed so that rider does not slide up the wall:
Free Body Diagram of Rider:
Equations of Motion:
F
y
 ma y
N cos 70  f sin 70  mg  m  0 
N cos 70  s N sin 70  mg  0
mg
cos 70  s sin 70
N
N  5093.4 N
F
n
 man
2
mvmin
R
mv 2
N sin 70  s N cos 70  min
R
N sin 70  f cos 70 
vmin =
N  sin 70  s cos 70 R
m
vmin  18.81 m/s 
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94
PROBLEM 12.60
A semicircular slot of 250-mm radius is cut in a flat plate which rotates about
the vertical AD at a constant rate of 14 rad/s. A small, 400-g block E is designed
to slide in the slot as the plate rotates. Knowing that the coefficients of friction
are m s = 0.35 and m k = 0.25, determine whether the block will slide in the slot
if it is released in the position corresponding to (a) q = 80∞, (b) q = 40∞. Also
determine the magnitude and the direction of the friction force exerted on the
block immediately after it is released.
SOLutiOn
r = (650 - 250 sin q ) mm
First note
=(0.65 - 0.25 sin q ) m
vE = rf ABCD
an =
Then
vE2
= r(f ABCD )2
r
= (0.65 - 0.25 sin q ) m (14 rad/s)2
= (127.4 - 49 sin q ) m/s2
Assume that the block is at rest with respect to the plate.
+
vE2
sin q
r
Ê
ˆ
v2
N = W Á - cos q + E sin q ˜
gr
Ë
¯
or
+
or
SFx = max : N + W cos q = m
SFy = ma y : -F + W sin q = - m
vE2
cos q
r
Ê
ˆ
v2
F = W Á sin q + E cos q ˜
gr
Ë
¯
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95
PROBLEM 12.60 (continued)
(a)
We have
Then
q = 80∞
È
˘
1
N = (0.4 ¥ 9.81) Í- cos 80∞ +
¥ (127.4 - 49 sin 80∞) ¥ sin 80∞˙
2
9.81 m/s
Î
˚
= 30.4954 N
È
˘
1
F = (0.4 ¥ 9.81) Ísin 80∞ +
¥ (127.4 - 49 sin 80∞) ¥ cos 80∞˙
2
9.81m/s
Î
˚
= 9.3617 N
Now
Fmax = m s N = 0.35(30.4954) = 10.6734 N
The block does not slide in the slot, and
F = 9.3617 N
(b)
We have
80° b
q = 40∞
Then
1
È
˘
N = (0.4 ¥ 9.81) Í- cos 40∞ +
¥ (127.4 - 49 sin 40∞) ¥ sin 40∞˙
9.81
Î
˚
= 21.652 N
1
È
˘
F = (0.4 ¥ 9.81) Ísin 40∞ +
¥ (127.4 - 49 sin 40∞) ¥ cos 40∞˙
9
.
81
Î
˚
= 31.9088 N
Now
Fmax = m s N , from which it follows that
F Fmax
Block E will slide in the slot
and
a E = a n + a E /plate
= a n + (a E /plate )t + (a E /plate ) n
+
At t = 0, the block is at rest relative to the plate, thus (a E /plate ) n = 0 at t = 0, so that a E /plate must be directed
tangentially to the slot.
v2
SFx = max : N + W cos 40∞ = m E sin 40∞
r
or
Ê
ˆ
v2
N = W Á - cos 40∞ + E sin 40∞˜
gr
Ë
¯
(as above)
= 21.652 N
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96
PROBLEM 12.60 (continued)
Sliding:
F = mk N
= 0.25(21.652 N)
= 5.413 N
Noting that F and a E /plane must be directed as shown (if their directions are reversed, then SFx is
while ma xis ), we have
the block slides downward in the slot and
F = 5.413 N
40° b
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97
PROBLEM 12.61
A small block B fits inside a lot cut in arm OA which rotates in a vertical plane at a
constant rate. The block remains in contact with the end of the slot closest to A and
its speed is 1.4 m/s for 0    150. Knowing that the block begins to slide when
  150, determine
ne the coefficient of static friction between the block and the slot.
SOLUTION
Draw the free body diagrams of the block B when the arm is at   150.
v  at  0,
g  9.81 m/s2
Ft  mat :  mg sin 30  N  0
N  mg sin 30
Fn  man : mg cos30  F  m
F  mg cos30 
Form the ratio
s 
v2

mv 2

F
, and set it equal to  s for impending slip.
N
F
g cos30  v 2 /
9.81 cos30
cos 30  (1.4) 2 /0.3


N
g sin 30
9.81 sin 30
 s  0.400 
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98
PROBLEM 12.62
The parallel-link mechanism ABCD is
used to transport a component I between
manufacturing processes at stations E, F,
and G by picking it up at a station when
q = 0 and depositing it at the next station
when q = 180º. Knowing that member
BC remains horizontal throughout its
motion and that links AB and CD rotate at
a constant rate in a vertical plane in such
a way that vB = 0.7 m/s, determine (a)
the minimum value of the coefficient of
static friction between the component and
BC if the component is not to slide on BC
while being transferred, (b) the values of
q for which sliding is impending.
SOLUTION
SFx = max : F =
+
W v B2
cos q
g r
+
SFy = ma y : N - W = -
W v B2
sin q
g r
Ê
ˆ
v2
N = W Á1 - B sin q ˜
gr
Ë
¯
or
Ê
ˆ
v2
Fmax = m s N = m s W Á1 - B sin q ˜
gr
Ë
¯
Now
and for the component not to slide
F £ Fmax
or
or
Ê
ˆ
v2
W v B2
cos q m s W Á1 - B sin q ˜
g r
Ë gr
¯
ms cos q
gr
- sin q
v B2
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99
PROBLEM 12.62 (continued)
We must determine the values of q which maximize the above expression. Thus
Ê gr
ˆ
Ê cos q ˆ - sin q Á 2 - sin q ˜ - (cos q ) ( - cos q )
v
Ë
¯
d Á
B
˜=
gr
=0
2
dq Á 2 - sin q ˜
g
r
Ê
ˆ
Ë vB
¯
ÁË v 2 - sin q ˜¯
B
or
sin q =
Now
sin q =
for
m s = ( m s ) min
(0.7 m/s)2
(9.81 m/s2 ) (0.25 m)
q = 11.525º
or
(a)
v B2
gr
and
= 0.199796
q = 168.475º
From above,
( m s ) min =
cos q
gr
- sin q
v B2
where sin q =
v B2
gr
cos q sin q
cos q
=
= tan q
2
1
1
sin
q
- sin q
sin q
= tan 11.525∞
( m s ) min =
(ms)min = 0.204 b
or
(b)
We have impending motion
to the left for
q = 11.53° b
to the right for
q = 168.5° b
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100
PROBLEM 12.63
Knowing that the coefficients of friction between the component I and member BC of the mechanism of
Problem 12.61 are ms = 0.35 and mk = 0.25, determine (a) the maximum allowable constant speed vB if the
component is not to slide on BC while being transferred, (b) the values of q for which sliding is impending.
SOLUTION
SFx = max : F =
+
W v B2
cos q
g r
+
SFy = ma y : N - W = -
W v B2
sin q
g r
Ê
ˆ
v2
N = W Á1 - B sin q ˜
Ë gr
¯
or
Fmax = m s N
Now
Ê
ˆ
v2
= m s W Á1 - B sin q ˜
Ë gr
¯
and for the component not to slide
F Fmax
or
or
Ê
ˆ
v2
W v B2
cos q m s W Á1 - B sin q ˜
g r
gr
Ë
¯
v B2 m s
( )
To ensure that this inequality is satisfied, v B2
gr
cos q + m s sin q
max
(1)
must be less than or equal to the minimum value of
m s g r / (cos q + m s sin q ), which occurs when (cos q + m s sin q ) is maximum. Thus
d
(cos q + m s sin q ) = - sin q + m s cos q = 0
dq
or
tan q = m s
m s = 0.35
or
q = 19.2900∞
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101
PROBLEM 12.63 (continued)
(a)
The maximum allowed value of v B is then
(v )
2
B
max
= ms
gr
cos q + m s sin q
= gr
tan q
= gr sin q
cos q + (tan q ) sin q
where tan q = m s
= (9.81)(0.25m ) sin 19.29∞
( v B ) max = 0.900 m/s b
or
(b)
First note that for 90∞ q 180∞, Eq. (1) becomes
v B2 m s
gr
cos a + m s sin a
where a = 180∞ - q . It then follows that the second value of q for which motion is impending is
q = 180∞ - 19.2900∞
= 160.7100∞
we have impending motion
to the left for
q = 19.29∞ b
to the right for
q = 160.7∞ b
Alternative solution.
SF = ma : W + R = ma n
Then
For impending motion, f = f s . Also, as shown above, the values of q for which motion is impending
(
minimize the value of vB, and thus the value of an is an =
that an is minimum when ma n and R are perpendicular.
v B2
r
). From the above diagram, it can be concluded
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102
PROBLEM 12.63 (continued)
Therefore,
from the diagram
q = f s = tan -1 m s
(as above)
man = W sin f s
and
or
m
v B2
= mg sin q
r
v B2 = g r sin q
or
(as above)
For 90∞ q 180∞, we have
from the diagram
a = 180∞ - q
(as above)
a = fs
and
or
man = W sin f s
v B2 = g r sin q
(as above)
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399
103
PROBLEM 12.64
A small 250
250-g collar C can slide on a semicircular rod which is made to rotate about
the vertical AB at a constant rate of 7.5 rad/s. Determine the three values of  for
which the collar will not slide on the rod, assuming no friction between the collar
and the rod.
SOLUTION
If the collar is not sliding, it moves at constant speed on a circle of radius   r sin  . v  
Normal acceleration. an 
v 2 2 2

 (r sin  )  2


 Fy  ma y :
N cos   mg  0
N 
mg
cos 
Fx  ma x :
N sin   ma
mg sin 
 (m r sin  ) 2
cos 
(1)
To satisfy (1), either:
sin   0
or
cos  
g
r 2
  0 or 180
or
cos 
9.81
 0.3488
(0.5) (7.5) 2
  0, 180, and 69.6 
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PROBLEM 12.65
A small 250-gg collar C can slide on a semicircular rod which is made to rotate about
the vertical AB at a constant rate of 7.5 rad/s. Knowing that the coefficients of
friction are s = 0.25 and k = 0.20, indicate whether the collar will slide on the rod if
it is released in the position corresponding to ((a)  = 75, (b)  = 40.
40 Also,
determine the magnitude and direction of the friction force exerted on the collar
immediately after release.
SOLUTION
If the collar is not sliding, it moves at constant speed on a circle of radius   r sin  . v  
r  500 mm  0.500 m,   7.5 rad/s,
Given:
m  250g  0.250 kg.
Normal acceleration:
an 
v 2 2 2

 (r sin  ) 2


F 
ma:
F  mg sin    m (r sin  )  2 cos 
F  m ( g  r 2 cos  )sin 
F 
ma :
N  mg cos   m(r sin  ) 2 sin 
N  m( g cos  r  2 sin 2  )
(a)   75.
F  (0.25) 9.81  (0.500 cos 75)(7.5) 2  sin 75
 0.61112 N
N  (0.25) 9.81cos 75  (0.500sin 2 75)(7.5) 2 
 7.1950 N
 s N  (0.25)(7.1950)  1.7987 N
Since F   s N , the collar does not slide.
F  0.611 N
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75
PROBLEM 12.65 (Continued)
F  (0.25) 9.81  (0.500 cos 40) (7.5) 2  sin 40
  1.8858 N
(b)   40.
N  (0.25) 9.81cos 40  (0.500
(0.500sin
sin 2 40)(7.5) 2 
 4.7839 N
 s N  (0.25)(4.7839)  1.1960 N
Since F   s N , the collar slides.
Since the collar is sliding, F  k N .
 Fn  
ma :
N  mg cos  man sin 
N  mg cos   m (r sin  ) 2 sin 
 m  g cos   (r sin 2  ) 2 
 (0.25) 9.81cos 40  (0.500sin 2 40)(7.5) 2 
 4.7839 N
F  k N  (0.20)
(0.20)(4.7839)
(4.7839)  0.957 N
F  0.957 N
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40 
PROBLEM 12.66
A 0.5-kg block B slides without friction inside a slot cut in arm OA which
rotates in a vertical plane at a constant rate, θ = 2 rad/s. At the instant
when θ = 30°, r = 0.6 m and the force exerted on the block by the arm is
zero. Determine, at this instant, (a) the relative velocity of the block with
respect to the arm, (b) the relative acceleration of the block with respect
to the arm.
SOLUTION
θ = 30°, θ = 2 rad/s, θ = 0
r = 0.6 m, W = mg
Block B: Only force is weight
Fr = W cos 30°,
(
Fθ = −W sin 30°
)
(a) Fθ = maθ = m rθ + 2rθ :
2rθ =
r = −
Fθ
mg sin 30°
− rθ = −
− rθ = − g sin 30° − rθ
m
m
(9.81) sin 30° + (0.6 )(0 ) = −1.226 m/s
g sin 30° + rθ
= −
2θ
( 2 )(2 )
v B / rod = 1.226 m/s
(
60° W
)
(b) Fr = mar = m r − rθ 2 :
r = rθ +
Fr
mg cos 30°
= rθ 2 +
= rθ 2 + g cos 30°
m
m
= (0.6 )( 2 ) + (9.81) cos 30° = 10.90 m/s 2
2
a B / rod = 10.90 m/s 2
60° t
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PROBLEM 12.67
A 0.5-kg block B slides without friction inside a slot cut in arm OA which
rotates in a vertical plane. The motion of the rod is defined by the relation
θ = 10 rad/s 2 , constant. At the instant when θ = 45°, r = 0.8 m and the
velocity of the block is zero. Determine, at this instant, (a) the force
exerted on the block by the arm, (b) the relative acceleration of the block
with respect to the arm.
SOLUTION
θ = 45°,
vr = r = 0,
(a)
θ = 10 rad/s2
r = 0.8 m,
vθ = rθ = 0,
W = mg
(
ΣFθ = maθ : N − W cos 45° = m rθ + 2rθ
(
N = mg cos 45° + m rθ + 2rθ
)
)
= (0.5)(9.81) cos 45° + 0.5 ( 0.8)(10 ) + 0
= 7.468
N = 7.47 N
(b)
(
ΣFr = mar : mg sin 45° = m r − rθ 2
r =
45° t
)
mg
sin 45° + rθ 2 = g sin 45° + rθ 2
m
= (9.81) sin 45° + 0 = 6.937 m/s 2
a B / rod = 6.94 m/s 2
45° t
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108
PROBLEM 12.68
The 3-kg collar B slides on the frictionless arm AA. The arm is
attached to drum D and rotates about O in a horizontal plane at the
rate   0.75t , where  and t are expressed in rad/s and seconds,
respectively. As the arm-drum
drum assembly rotates, a mechanism
within the drum releases cord so that the collar moves outward from
O with a constant speed of 0.5 m/s. Knowing that at t 0, r 0,
determine the time at which the tension in the cord is equal to the
magnitude of the horizontal force exerted on B by arm AA.
SOLUTION
Kinematics
dr
 r  0.5 m/s
dt
We have
At t  0, r  0:

r
0
dr 

t
0
0.5 dt
or
r  (0.5t ) m
Also,

r 0
  (0.75t ) rad/s
  0.75 rad/s2
Now
ar  
r  r2  0  [(0.5t ) m][(0.75t ) rad/s]2  (0.28125t 3 ) m/s2
and
a  r  2r
 [(0.5t ) m][0.75 rad/s 2 ]  2(0.5 m/s)[(0.75t ) rad/s]
 (1.125t ) m/s 2
Kinetics
Fr  mar :  T  (3 kg)(0.28125t 3 ) m/s2
T  (0.84375t 3 ) N
or
F  mB a : Q  (3 kg)(1.125t ) m/s 2
or
Q  (3.375t ) N
Now require that
T Q
or
or
(0.84375t 3 ) N  (3.375t ) N
t 2  4.000
t  2.00 s 
or
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PROBLEM 12.69
Rod OA rotates about O in a horizontal plane. The motion of the 300 g
collar B is defined by the relations r = 300 + 100 cos (0.5pt) and
q = p(t2 - 3t), where r is expressed in millimeters, t in seconds, and q
in radians. Determine the radial and transverse components of the force
exerted on the collar when ( a) t = 0, (b) t = 0.5 s.
SOLUTION
Polar coordinates and their derivatives.
(a)
For t = 0.
r = 0.3 + 0.1 cos (0.5p t ) m
q = p (t 2 - 3t ) rad
r = -0.05p sin (0.5p t ) m/s
q = p (2t - 3) rad/s

r = -.025p 2 cos (0.5p t ) m/s2
q = 2p rad/s2
r = 0.4 m
q =0
r = 0
q = -3p rad/s

r = -0.025p 2 m/s2
q = 2p rad/s2
Components of acceleration.
ar = 
r - rq 2
= -0.025p 2 - 0.4( -3p )2
= -35.777 m/s2
aq = rq + 2rq
= (0.4)(2p ) + 2( -3p )(0)
= 2.513 m/s2
Components of force.
Fr = mar = (0.3)( -35.777)
Fr = -10.73 N b
Fq = maq = (0.3)(2.513)
(b)
For t = 0.5 s.
Fq = 0.754 N b
r = 0.37071 m
q = -3.927 rad
r = -0.11107 m/s
q = -2p rad/s

r = -0.17447 m/s2
q = 2p rad/s2
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110
PROBLEM 12.69 (continued)
Components of acceleration.
ar = 
r - rq 2
= -14.809 m/s2
aq = rq + 2rq
= 3.725 m/s2
Components of force.
Fr = mar = (0.3)( -14.809)
Fr = -4.44 N b
Fq = maq = (0.3)(3.725)
Fq = 1.118 N b
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111
PROBLEM 12.70
Pin B has a mass of 100 g and is free to slide in a horizontal plane
along the rotating arm OC and along the circular slot DE of radius
b = 500 mm. Neglecting friction and assuming that q = 15 rad/s and
q = 250 rad/s2 for the position q = 20∞, determine for that position (a)
the radial and transverse components of the resultant force exerted
on pin B, (b) the forces P and Q exerted on pin B, respectively, by rod
OC and the wall of slot DE.
SOLutiOn
Kinematics.
From the drawing of the system, we have
r = 2b cosq
Then
r = - (2b sin q )q
and

r = -2b(q sin q + q 2 cos q )
ar = 
r - rq 2 = -2b(q sin q + q 2 cos q ) - (2b cos q )q 2
= - 2b(q sin q + 2q 2 cos q )
Now
= - 2(0.5 m)[(250 rad/s2 )sin 20∞ + 2(15 rad/s)2 cos 20∞]
= - 508.367 m/s2
aq = rq + 2rq = (2b cos q )q + 2( -2bq sin q )q
= 2b(q cos q - 2q 2 sin q )
and
= 2(0.5 m)[(250 rad/s2 ) cos 20∞ - 2(15 rad/s)2 sin 20∞]
= 81.014 m/s2
Kinetics.
(a)
We have
Fr = mar = 0.1 kg ¥ ( -508.367 m/s2 ) = -50.8367 N
Fr = -50.8 N b
or
and
Fq = maq = 0.1 kg ¥ (81.014 m/s2 ) = 8.1014 N
Fq = 8.10 N b
or
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112
PROBLEM 12.70 (continued)
+
(b)
or
+
or
SFr : -Fr = -Q cos 20∞
1
(50.8367 N)
Q=
cos 20∞
= 54.099 N
SFq : Fq = P - Q sin 20∞
P = (8.1014 + 54.099 sin 20∞) N
= 26.604 N
P = 26.6 N
70∞ b
Q = 54.1 N
40∞ b
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113
PROBLEM 12.71
The two blocks are released from rest when r 0.8 m and   30.
Neglecting the mass of the pulley and the effect of friction in the
pulley and between block A and the horizontal surface, determine
(a)) the initial tension in the cable, ((b)) the initial acceleration of block
A,, ((c) the initial acceleration of block B.
SOLUTION
Let r and  be polar coordinates of block A as shown, and let yB be the
position coordinate (positive downward, origin at the pulley) for the
rectilinear motion of block B.
Constraint of cable:
r  yB  constant,
r  vB  0,
For block A,
r  aB  0

r  aB
(1)
Fx  mAa A : T cos   mAaA or T  mAaA sec
(2)
or
Fy  mB aB : mB g  T  mB aB
For block B,
Adding Eq. (1) to Eq. (2) to eliminate T, mB g  mAa A sec  mB aB
(3)
(4)
Radial and transverse components of a A.
Use either the scalar product of vectors or the triangle construction shown,
being careful to note the positive directions of the components.
r  r2  ar  a A  er  a A cos 
(5)
Noting that initial
initially   0, using Eq. (1) to eliminate r, and changing
signs gives
aB  a A cos
(6)
Substituting Eq. (6) into Eq. (4) and solving for a A ,
aA 
mB g
(25) (9.81)

 5.48 m/s 2
mA sec  mB cos 
20sec30  25cos30
From Eq. (6), aB  5.48cos30  4.75 m/s 2
(a)
From Eq. (2), T  (20)(5.48)sec30  126.6
(b)
Acceleration of block A.
(c)
Acceleration of block B.
T  126.6 N 
a A  5.48 m/s2
a B  4.75 m/s2 
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
PROBLEM 12.72
The velocity of block A is 2 m/s to the right at the instant when
r  0.8 m and   30. Neglecting the mass of the pulley and the
effect of friction in the pulley and between block A and the horizontal
surface, determine, at this instant, (a)) the tension in the cable, (b)
( the
acceleration of block A, (c)) the acceleration of block B.
SOLUTION
Let r and  be polar coordinates of block A as shown, and let yB be the
position coordinate (positive downward, origin at the pulley) for the
rectilinear motion of block B.
Radial and transverse components of v A.
Use either the scalar product of vectors or the triangle construction shown,
being careful
areful to note the positive directions of the components.
r  vr  v A  e r  v A cos30
 2cos30  1.73205 m/s
r  v  v A  e  v A sin 30
 2sin 30  1.000 m/s 2
v
1.000
   
 1.25 rad/s
r
0.8
Constraint of cable: r  yB  constant,
r  vB  0,
For block A,

r  aB  0
or

r  aB
Fx  mAa A : T cos  mAa A or T  mAaA sec
Fy  mB aB: mB g  T  mB aB
For block B,
Adding Eq. (1) to Eq. (2) to eliminate T, mB g  mAa A sec  mB aB
(1)
(2)
(3)
(4)
Radial and transverse components of a A.
Use a method similar to that used for the components of velocity.
r  r2  ar  a A  er  a A cos 
(5)
Using Eq. (1) to eliminate r and changing signs gives
aB  a A cos  r 2
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(6)
PROBLEM 12.72 (Continued)
Substituting Eq. (6) into Eq. (4) and solving for a A ,
aA 

mB g  r2

mA sec  mB cos 

(25)[9.81  (0.8)(1.25)2 ]
 6.18 m/s2
20sec30  25cos30
From Eq. (6), aB  6.18cos30  (0.8)(1.25)2  4.10 m/s2
(a)
From Eq. (2), T  (20)(6.18) sec30  142.7
(b)
Acceleration of block A.
(c)
Acceleration of block B.
T  142.7 N 
a A  6.18 m/s2

a B  4.10 m/s2 
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PROBLEM 12.73*
Slider C has a mass of 250 g and may move in a slot cut in arm AB,
which rotates at the constant rate q0 = 10 rad/s in a horizontal plane.
The slider is attached to a spring of constant k = 50 N/m, which is
unstretched when r = 0. Knowing that the slider is released from rest
with no radial velocity in the position r = 500 mm and neglecting
friction, determine for the position r = 300mm (a) the radial and
transverse components of the velocity of the slider, (b) the radial
and transverse components of its acceleration, (c) the horizontal
force exerted on the slider by arm AB.
SOLUTION
Let l0 be the radial coordinate when the spring is unstretched. Force exerted by the spring.
Fr = - k ( r - l0 )
SFr = mar : - k ( r - l0 ) = m( 
r - rq 2 )
kl
kˆ
Ê

r = Á q 2 - ˜ r + 0
Ë
m
m¯
But
(1)
d
dr dr
dr
( r) =
= r
dt
dr dt
dr
kl ˘
kˆ
ÈÊ
  = 
rdr
rdr = ÍÁ q 2 - ˜ r + 0 ˙ dr
¯
Ë
m
m˚
Î

r=
Integrate using the condition r = r0 when r = r0 .
1 2 r È 1 Ê  2 k ˆ 2 kl0 ˘ r
r
= Áq - ˜ r +
r
m ˙˚ r0
2 r0 ÍÎ 2 Ë
m¯
(
)
kl
1 2 1 2 1 Ê 2 k ˆ 2
r - r0 = Á q - ˜ r - r02 + 0 ( r - r0 )
2
2
2Ë
m¯
m
2kl
kˆ
Ê
r2 = r02 + Á q 2 - ˜ r 2 - r02 + 0 ( r - r0 )
Ë
m
m¯
(
Data:
)
m = 0.25 kg
q = 10 rad/s, k = 50 N/m, l0 = 0
r0 = ( vr )0 = 0, r0 = 500 mm = 0.5 m, r = 300 mm = 0.3 m
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117
PROBLEM 12.73* (continued)
(a)
Components of velocity when r = 0.3 m
50 ˆ
Ê
2
2
r2 = 0 + Á102 ˜¯ (0.3 - 0.5 ) + 0
Ë
0.25
= 16 m2 /s2
vr = r = ±4.00 m/s
Since r is decreasing, vr is negative
vr = -4.00 m/s b
r = -4 m/s
vq = rq = (0.3)(10)
(b)
vq = 3.00 m/s b
Components of acceleration.
Fr = - kr + kl0 = -(50)(0.3) + 0 = -15 N
F
15
ar = r = 0.25
m
aq = rq + 2rq = 0 + (2)( -4)(10)
ar = -60.0 m/s2 b
aq = -80.0 m/s2 b
(c)
Transverse component of force.
Fq = maq = (0.25)( -80)
Fq = -20.0 N b
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118
PROBLEM 12.74
A particle of mass m is projected from Point A with an initial velocity v0
perpendicular to line OA and moves under a central force F directed
away from the center of force O. Knowing that the particle follows a path
defined by the equation r  r0 / cos 2 and using Eq. (12.25), express
the radial and transverse components of the velocity v of the particle as
functions of  .
SOLUTION
Since the particle moves under a central force, h  constant.
Using Eq. (12.27),
h  r 2  h0  r0 v0
or
rv
r v cos 2 v0
  0 20  0 0 2
 cos 2
r0
r
r0
Radial component of velocity.
vr  r 
 r0

dr  d  r0
sin 2 



  r0
d
d  cos 2 
(cos 2 )3/ 2
sin 2 v0
cos 2
(cos 2 )3/ 2 r
vr  v0
sin 2
cos 2

Transverse component of velocity.
v 
h r0 v0

cos 2
r
r0
v  v0 cos 2 
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PROBLEM 12.75
For the particle of Problem 12.74, show (a) that the velocity of the
particle and the central force F are proportional to the distance r from
the particle to the center of force O, (b) that the radius of curvature of
the path is proportional to r3.
PROBLEM 12.74 A particle of mass m is projected from Point A with
an initial velocity v0 perpendicular to line OA and moves under a central
force F directed away from the center of force O. Knowing that the
particle follows a path defined by the equation r  r0 / cos 2 and using
Eq. (12.25), express the radial and transverse components of the velocity
v of the particle as functions of .
SOLUTION
Since the particle moves under a central force, h  constant.
Using Eq. (12.27),
h  r 2  h0  r0 v0
or
rv
r v cos 2 v0
  0 20  0 0 2
 cos 2
r0
r
r0
Differentiating the expression for r with respect to time,
r 

dr  d  r0
sin 2 
sin 2 v0
sin 2

  r0
cos 2  v0

  r0
3/2
3/ 2
d
d  cos 2 
(cos 2 )
(cos 2 ) r0
cos 2
Differentiating again,
r 
(a)
dr  d 
sin 2  
2 cos 2 2  sin 2 2  v0 2 2 cos 2 2  sin 2 2

v

v


 0

0
d
d 
r0
(cos 2 )3/2
cos 2 
cos 2
vr  r  v0
sin 2
cos 2

v0 r
vr
sin 2 v  r  0 cos 2
r0
r0
v  (vr ) 2  (v ) 2 
v0 r
sin 2 2  cos 2 2
r0
v
v0 r

r0
v 2 2 cos 2 2  sin 2 2
r0
v0 2
ar  
r  r 2  0

cos2 2
2
r0
cos 2
cos 2 r0

v0 2 cos 2 2  sin 2 2
v0
v 2r

 02
r0
r0
cos 2
r0 cos 2
Fr  mar 
mv02 r
r02
Fr 
:
mv02 r
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120
r02

PROBLEM 12.75 (Continued)
Since the particle moves under a central force, a  0.
Magnitude of acceleration.
a  ar 2  a 2 
v0 2 r
r02
Tangential component of acceleration.
at 
v0 2 r
dv d  v0 r  v0
 
  r  2 sin 2
dt dt  r0  r0
r0
Normal component of acceleration.
at  a 2  at 2 
r 
cos 2   0 
r 
But
an 
Hence,
(b)
But an 
v2

or
v0 2 r
r02
1  sin 2 2 
v02 r cos 2
r0 2
2
v0 2
r

v 2 v02 r 2 r
 2  2
an
r0
v0

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121
r3

r02
PROBLEM 12.76
A particle of mass m is projected from Point A with an initial velocity v0
perpendicular to line OA and moves under a central force F along a
semicircular path of diameter OA. Observing that r  r0 cos  and using
Eq. (12.25), show that the speed of the particle is v  v0 /cos2  .
SOLUTION
Since the particle moves under a central force, h  constant.
Using Eq. (12.27),
h  r 2  h0  r0 v0
or
rv
rv
v0
  0 20  2 0 0 2 
r
r0 cos  r0 cos 2 
Radial component of velocity.
vr  r 
d
(r0 cos  )  (r0 sin  )
dt
Transverse component of velocity.
v  r  (r0 cos  )
Speed.
v  vr 2  v 2  r0 
r0 v0
2
r0 cos 
v
cos 2 
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v0

PROBLEM 12.77
For the particle of Problem 12.76, determine the tangential component Ft
of the central force F along the tangent to the path of the particle for
(a) θ = 0, (b) θ = 45°.
PROBLEM 12.76 A particle of mass m is projected from Point A with an
initial velocity v0 perpendicular to line OA and moves under a central force F
along a semicircular path of diameter OA. Observing that r = r0 cos θ and
using Eq. (12.27), show that the speed of the particle is v = v0 /cos2 θ .
SOLUTION
Since the particle moves under a central force, h = constant
Using Eq. (12.27),
h = r 2θ = h0 = r0 v0
θ =
r0 v0
r
2
=
2
r0 v0
2
r0 cos θ
=
v0
r0 cos 2 θ
Radial component of velocity.
vr = r =
d
(r0 cos θ ) = −(r0 sin θ )θ
dt
Transverse component of velocity.
vθ = rθ = (r0 cos θ )θ
Speed.
v = vr 2 + vθ 2 = r0θ =
r0 v0
2
r0 cos θ
=
v0
cos 2 θ
Tangential component of acceleration.
at =
v0
dv
(−2)(− sin θ )θ 2v0 sin θ
= v0
=
⋅
3
3
dt
cos θ
cos θ r0 cos 2 θ
=
2v0 2 sin θ
r0 cos5 θ
Tangential component of force.
Ft = mat : Ft =
(a)
θ = 0,
(b)
θ = 45°,
2mv0 2 sin θ
r0 cos5 θ
Ft = 0
Ft =
Ft = 0 W
2mv0 sin 45°
Ft =
cos5 45°
8mv0 2
W
r0
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123
PROBLEM 12.78
Determine the mass of the earth, knowing that the mean radius of the moon’s orbit about the earth is 384,000 km
and that the moon requires 27.32 days to complete one full revolution about the earth.
SOLUTION
Mm
[Eq. (12.28)]
r2
v2
F = Fn = man = m
r
F =G
We have
and
Then
G
or
v2
Mm
=
m
r
r2
r
M = v2
G
v=
Now
2p r
t
2
M=
so that
2
1 Ê 2p ˆ 3
r Ê 2p r ˆ
˜¯ = ÁË ˜¯ r
ÁË
G t
G t
Noting that
t = 27.32 days = 2.3604 ¥ 106 s
and
r = 384, 000 km = 3.84 ¥ 108 m
we have
M=
2
1
66.73 ¥ 10 -12
m3
kg.s2
Ê
ˆ
2p
(3.84 ¥ 108 )3
Á
6 ˜
¥
.
s
2
3604
10
Ë
¯
M = 6.01 ¥ 1024 kg b
or
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124
PROBLEM 12.79
Show that the radius r of the moon’s orbit can be determined from the radius R of the earth, the acceleration of
gravity g at the surface of the earth, and the time τ required for the moon to complete one full revolution about
the earth. Compute r knowing that τ = 27.3 days.
SOLUTION
Mm
r2
We have
F =G
and
F = Fn = man = m
G
Then
GM
r
GM = gR 2
Now
v2 =
so that
v2
r
Mm
v2
=
m
r
r2
v2 =
or
[Eq. (12.28)]
[Eq. (12.30)]
g
gR 2
or v = R
r
r
2π r 2π r
=
v
R gr
For one orbit,
τ=
or
⎛ gτ 2 R 2 ⎞
r = ⎜⎜
⎟
2 ⎟
⎝ 4π ⎠
Now
τ = 27.3 days = 2.35872 × 106 s
1/ 3
Q.E.D.
W
R = 3960 mi = 20.9088 × 106 ft
1/3
⎡ 9.81 m/s 2 × (2.35872 × 106 s) 2 × (6.37 × 106 m)2 ⎤
r=⎢
⎥
4π 2
⎣
⎦
= 382.81 × 106 m
r = 383 × 103 km W
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125
PROBLEM 12.80
Communication satellites are placed in a geosynchronous orbit, i.e., in a circular orbit such that they complete
one full revolution about the earth in one sidereal day (23.934 h), and thus appear stationary with respect to
the ground. Determine (a) the altitude of these satellites above the surface of the earth, (b) the velocity with
which they describe their orbit.
SOLUTION
For gravitational force and a circular orbit,
Fr =
GMm mv 2
=
r
r2
or
v=
GM
r
Let τ be the period time to complete one orbit.
But
vτ = 2π r
Then
GM τ 2
r =
4π 2
3
or
GM τ 2
= 4π 2 r 2
r
⎛ GM τ 2
r = ⎜⎜
2
⎝ 4π
1/3
⎞
⎟⎟
⎠
τ = 23.934 h = 86.1624 × 103 s
Data:
(a)
v 2τ 2 =
or
In SI units:
g = 9.81 m/s2 , R = 6.37 × 106 m
GM = gR 2 = (9.81)(6.37 × 106 ) 2 = 398.06 × 1012 m3 /s 2
1/3
⎡ (398.06 × 1012 )(86.1624 × 103 ) 2 ⎤
6
r=⎢
⎥ = 42.145 × 10 m
2
4π
⎣
⎦
altitude h = r − R = 35.775 × 106 m
(b)
h = 35,800 km W
In SI units:
v=
GM
398.06 × 1012
=
= 3.07 × 103 m/s
6
r
42.145 × 10
v = 3.07 km/s W
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126
PROBLEM 12.81
Show that the radius r of the orbit of a moon of a given planet can be determined from the radius R of the
planet, the acceleration of gravity at the surface of the planet, and the time  required by the moon to complete
one full revolution about the planet. Determine the acceleration of gravity at the surface of the planet Jupiter
knowing that R 71,492 km and that  3.551 days and r 670.9  103 km for its moon Europa.
SOLUTION
Mm
r2
We have
F G
and
F  Fn  man  m
G
Then
v2
r
Mm
v2

m
r
r2
v2 
or
[Eq. (12.28)]
GM
r
GM  gR 2
Now
v2 
so that
[Eq. (12.30)]
gR 2
r
or
vR
g
r
2 r 2 r

v
R gr
For one orbit,

or
 g 2 R 2 
r  
2 

 4 
Solving for g,
g  4 2
1/3
Q.E.D.

r3
 2 R2
and noting that   3.551 days  306,806 s, then
g Jupiter  4 2
 4 2
3
rEur
2
 Eur
RJup
(670.9  106 m)3
(306,806 s)2 (71.492  106 m) 2
g Jupiter  24.8 m/s 2 
or
Note:
g Jupiter  2.53gEarth 
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PROBLEM 12.82
The orbit of the planet Venus is nearly circular with an orbital velocity of 126.5  103 km/h. Knowing that
the mean distance from the center of the sun to the center of Venus is 108  106 km and that the radius of the
sun is 695  103 km, determine (a) the mass of the sun, (b) the acceleration of gravity at the surface of the
sun.
SOLUTION
Let M be the mass of the sun and m the mass of Venus.
For the circular orbit of Venus,
GMm
mv 2

ma

n
r
r2
GM  rv 2
where r is radius of the orbit.
r  108  106 km  108  109 m
Data:
v  126.5  103 km/hr  35.139  103 m/s 
GM  (108  109 )(35.139  103 )2  1.3335  1020 m3 /s2
(a) Mass of sun.
(b) At the surface of the sun,
M 
GM
1.3335  1020 m3 /s 2

G
66.73  1012
M  1.998  1030 kg 
R  695.5  103 km  695.5  106 m
GMm
 mg
R2
g 
GM
1.3335  1020

R2
(695.5  106 ) 2
g  276 m/s2 
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PROBLEM 12.83
A satellite is placed into a circular orbit about the planet Saturn at an altitude of 3400 km. The satellite describes
its orbit with a velocity of 87,500 km/h Knowing that the radius of the orbit about Saturn and the periodic time
of Atlas, one of Saturn’s moons, are 136.9 3 103 km and 0.6017 days, respectively, determine (a) the radius
of Saturn, (b) the mass of Saturn. (The periodic time of a satellite is the time it requires to complete one full
revolution about the planet.)
SOLUTION
where
2p rA
tA
rA = 136.9 ¥ 103 km = 136.9 ¥ 106 m
and
t A = 0.6017 days = 51, 987 s
Velocity of Atlas.
vA =
vA =
Gravitational force.
F=
(2p )(136.9 ¥ 106 )
= 16.54583 ¥ 103 m/s
51, 987
GMm
r2
=
mv 2
r
from which
GM = rv 2 = constant
For the satellite,
rs vs2 = rA v 2A
rs =
rA v 2A
vs 2
vs = 87, 500 km/h = 24.30556 ¥ 103 m/s
where
rs =
(136.9 ¥ 106 )(16.54583 ¥ 103 )2
(24.30556 ¥ 103 )2
rs = 63, 400 km
(a)
= 63.4409 ¥ 106 m
Radius of Saturn.
R = rs - (altitude) = 63,400 - 3400
(b)
R = 60, 000 km b
Mass of Saturn.
M=
rA v 2A (136.9 ¥ 106 )(16.54583 ¥ 103 )2
=
G
66.73 ¥ 10 -12
M = 5.62 ¥ 1024 kg b
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129
PROBLEM 12.84
The periodic time (see Prob. 12.84) of an earth satellite in a circular polar
orbit is 120 min. Determine (a) the altitude h of the satellite, (b) the time
during which the satellite is above the horizon for an observer located at
the north pole.
SOLUTION
For gravitational force and a circular orbit,
Fr =
GMm
mv 2
=
r
r2
v=
or
GM
r
Let τ be the periodic time to complete one orbit.
vτ = 2π r
or
GM
= 2π r
r
τ
1/ 3
 GM τ 2 
r = 
2 

 4π 
Solving for r,
For earth, R = 6370 km = 6.37 × 106 m, g = 9.81 m/s 2
(
GM = gR 2 = (9.81) 6.37 × 106
)
2
= 398.06 × 1012 m3/s 2
Data: τ = 120 min = 7200 s
(
)
1/ 3
 398.06 × 1012 (7200 )2 

r=


4π 2


(a) altitude
h = r − R = 1.685 × 106 m
(b) cosθ =
6.37 × 106
R
=
= 0.7908
r
8.055 × 106
= 8.055 × 106 m
h = 1685 km t
θ = 37.74°
t AB =
(75.48)(7200 ) = 1509.6 s
2θ
τ =
360°
360
t AB = 25.2 min t
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130
PROBLEM 12.85
A 500 kg spacecraft first is placed into a circular orbit about the earth at an altitude of 4500 km and then is
transferred to a circular orbit about the moon. Knowing that the mass of the moon is 0.01230 times the mass
of the earth and that the radius of the moon is 1737 km, determine (a) the gravitational force exerted on the
spacecraft as it was orbiting the earth, (b) the required radius of the orbit of the spacecraft about the moon if
the periodic times (see Problem 12.83) of the two orbits are to be equal, (c) the acceleration of gravity at the
surface of the moon.
SOLUTION
First note that
rE  RE  hE  (6.37  106  4.5  106 ) m
Then
(a)
RE  6.37  106 m
 10.87  106 m
We have
and
F
GMm
r2
[Eq. (12.28)]
GM  gR 2
[Eq. (12.29)]
m
R
W  
2
r
r
2
Then
F  gR 2
For the earth orbit,
 6.37  106 m 
F  (500 kg)(9.81 m/s ) 

6
 10.87  10 m 
2
2
F  1684 N 
or
(b)
From the solution to Problem 12.78, we have
2
M 
Then
Now

1  2  3
r
G   
2 r 3/2
GM
E M 
2 rE3/ 2
GM E

2 rM3/ 2
(1)
GM M
1/3
or
M 
rM   M  rE  (0.01230)1/3 (10.87  106 m)
 ME 
or
rM  2.509  106 m
rM  2510 km 
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PROBLEM 12.85 (Continued)
(c)
GM  gR 2
We have
[Eq.(12.29)]
Substituting into Eq. (1)
2 rE3/ 2
RE g E

2 rM3/ 2
RM
gM
2
or
3
2
gM
 R  r 
 R  M 
  E   M  g E   E   M  gE
 RM   rE 
 RM   M E 
gM
 6370 km 
2

 (0.01230)(9.81 m/s )
1737
km


using the results of Part (b). Then
2
g moon  1.62 m/s2 
or
Note:
g moon 
1
g earth
6
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PROBLEM 12.86
A space vehicle is in a circular orbit of 2200-km radius around the moon.
To transfer it to a smaller circular orbit of 2080-km radius, the vehicle is
first placed on an elliptic path AB by reducing its speed by 26.3 m/s as it
passes through A. Knowing that the mass of the moon is 73.49  1021 kg,
determine (a) the speed of the vehicle as it approaches B on the elliptic
path, (b) the amount by which its speed should be reduced as it
approaches B to insert it into the smaller circular orbit.
SOLUTION
For a circular orbit,
 Fn  man : F  m
F G
Eq. (12.28):
G
Then
or
v2
r
Mm
r2
Mm
v2

m
r
r2
GM
v2 
r
66.73  1012 m3 /kg  s 2  73.49  1021 kg
2200  103 m
Then
2
(v A )circ

or
(v A )circ  1493.0 m/s
and
2
(vB )circ

or
(vB )circ  1535.5 m/s
(a)
We have
66.73  1012 m3 /kg  s 2  73.49  1021 kg
2080  103 m
(v A )TR  (v A )circ  v A
 (1493.0  26.3) m/s
 1466.7 m/s
Conservation of angular momentum requires that
rA m(v A )TR  rB m(vB )TR
or
(vB )TR 
2200 km
 1466.7 m/s
2080 km
 1551.3 m/s
(vB )TR  1551 m/s 
or
(b)
Now
or
(v B )circ  (vB )TR  vB
vB  (1535.5  1551.3) m/s
vB  15.8 m/s 
or
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PROBLEM 12.87
As a first approximation to the analysis of a space flight from the
earth to Mars, assume the orbits of the earth and Mars are
circular and coplanar. The mean distances from the sun to the
earth and to Mars are 149.6  106 km and 227.8  106 km,
respectively. To place the spacecraft into an elliptical transfer
orbit at point A, its speed is increased over a short interval of
time to v A which is 2.94 km/s faster than the earth’s orbital
speed. When the spacecraft reaches point B on the elliptical
transfer orbit, its speed vB is increased to the orbital speed of
Mars. Knowing that the mass of the sun is 332.8  103 times the
mass of the earth, determine the increase in speed required at B.
SOLUTION

For earth,  GM earth  gR 2   9.81 6.37  106


2
 398.06  1012 m3/s 2

For sun,  GM sun  332.8  103  GM earth  132.474  1018 m 3 /s 2
For circular orbit of earth, rE  149.6  106 km  149.6  109 m/s
vE 
For transfer orbit AB,
rA  rE ,
 GM sun
rE

132.474  1018
 29.758  103 m/s
149.6  109
rB  rM  227.8  109 m
vA  vE   v  A  29.758  103  2.94  103  32.698  103 m/s
mrAv A  mrB vB
vB 



149.6  109 32.698  103
rAvA

 21.473  103 m/s
9
rB
227.8  10
For circular orbit of Mars,
vM 
 GM sun
rM

132.474  1018
 24.115  103 m/s
227.8  109
Speed increase at B.
 v B
 vM  vB  24.115  103  21.473  103  2.643  103 m/s
 v  B
 2.64 km/s 
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PROBLEM 12.88
To place a communications satellite into a geosynchronous
orbit (see Problem 12.80) at an altitude of 35800 km above
the surface of the earth, the satellite first is released from
a space shuttle, which is in a circular orbit at an altitude of
300 km, and then is propelled by an upper-stage booster
to its final altitude. As the satellite passes through A, the
booster’s motor is fired to insert the satellite into an elliptic
transfer orbit. The booster is again fired at B to insert
the satellite into a geosynchronous orbit. Knowing that
the second firing increases the speed of the satellite by
1440 m/s, determine (a) the speed of the satellite as it
approaches B on the elliptic transfer orbit, (b) the increase
in speed resulting from the first firing at A.
SOLUTION
For earth,
R = 6370 km = 6.37 ¥ 106 m
GM = gR2 = (9.81 m/s2 )(6.37 ¥ 106 )2 = 3.9806 ¥ 1014 m3 /s2
rA = 6370 + 300 = 6670 km = 6.67 ¥ 106 m
rB = 6370 + 35800 = 42,170 km = 42.17 ¥ 106 m
Speed on circular orbit through A.
( v A )circ =
=
GM
rA
3.9806 ¥ 1014
6.67 ¥ 106
= 7.7252 ¥ 103 m/s
Speed on circular orbit through B.
( v B )circ =
=
GM
rB
3.9806 ¥ 1014
42.17 ¥ 106
= 3.07236 ¥ 103 m/s
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135
PROBLEM 12.88 (continued)
(a)
Speed on transfer trajectory at B.
( v B ) tr = 3.07236 ¥ 103 - 1440
= 1632.36 ¥ m/s
1632 m/s b
Conservation of angular momentum for transfer trajectory.
rA ( v A )tr = rB ( v B ) tr
r (v )
( v A )tr = B B tr
rA
( 42.17 ¥ 106 )(1632.36)
(6.67 ¥ 106 )
= 10, 320.33 m/s
=
(b)
Change in speed at A.
Dv A = ( v A )tr - ( v A )circ
= 10, 320.33 - 7725.2
= 2595.13 m/s
Dv A = 2600 m/s b
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136
PROBLEM 12.89
Plans for an unmanned landing mission on the planet Mars called
for the earth-return vehicle to first describe a circular orbit at an
altitude dA = 2200 km above the surface of the planet with a
velocity of 2771 m/s. As it passed through Point A, the vehicle
was to be inserted into an elliptic transfer orbit by firing its
engine and increasing its speed by ΔvA = 1046 m/s. As it passed
through Point B, at an altitude dB = 100,000 km, the vehicle was to
be inserted into a second transfer orbit located in a slightly
different plane, by changing the direction of its velocity and
reducing its speed by ΔvB = −22.0 m/s. Finally, as the vehicle
passed through Point C, at an altitude dC = 1000 km, its speed
was to be increased by ΔvC = 660 m/s to insert it into its return
trajectory. Knowing that the radius of the planet Mars is
R = 3400 km, determine the velocity of the vehicle after
completion of the last maneuver.
B
SOLUTION
rA = 3400 + 2200 = 5600 km = 5.60 × 106 m
rB = 3400 + 100, 000 = 103, 400 km = 103.4 × 106 m
rC = 3400 + 1000 = 4400 km = 4.40 × 106 m
First transfer orbit.
v A = 2771 m/s + 1046 m/s = 3817 m/s
Conservation of angular momentum:
rA m v A = rB m vB
6
(5.60 × 10 )(3817) = (103.4 × 106 )vB
vB = 206.7 m/s
Second transfer orbit.
vB′ = vB + ΔvB
= 206.7 − 22.0 = 184.7 m/s
Conservation of angular momentum:
rB mvB′ = rC mvC
(103.4 × 106 )(184.7) = (4.40 × 106 )vC
vC = 4340 m/s
After last maneuver.
v = vC + ΔvC = 4340 + 660
v = 5000 m/s W
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137
PROBLEM 12.90
A 1 kg collar can slide on a horizontal rod, which is free to
rotate about a vertical shaft. The collar is initially held at A by a
cord attached to the shaft. A spring of constant 30 N/m is
attached to the collar and to the shaft and is undeformed when
the collar is at A.. As the rod rotates at the rate   16 rad/s, the
cord is cut and the collar moves out along the rod. Neglecting
friction and the mass of the rod, determine (a)
( the radial and
transverse components of the acceleration of the collar
c
at A,
(b)) the acceleration of the collar relative to the rod at A, (c) the
transverse component of the velocity of the collar at B.
SOLUTION
Fsp  k (r  rA )
First note
(a)
F  0 and at A,
Fr   Fsp  0
(a A ) r  0 
(a A )  0 
Fr  mar :
(b)
Noting that
Fsp  m ( 
r  r 2 )
acollar/rod  
r , we have at A
0  m[acollar/rod  (150 mm)(16 rad/s)2 ]
acollar/rod  38400 mm/s2
(acollar/rod ) A  38.4 m/s 2 
or
(c)
After the cord is cut, the only horizontal force acting on the collar is due to the spring. Thus, angular
momentum about the shaft is conserved.
rA m (v A )  rB m (vB )
Then
(vB ) 
where (vA )  rA 0
150 mm
[(150 mm)(16 rad/s)]  800 mm/s
450 mm
(vB )  0.800 m/s 
or
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PROBLEM 12.91
A 200-g ball A and a 400-g ball B are mounted on a horizontal rod
which rotates freely about a vertical shaft. The balls are held in the
positions shown by pins. The pin holding B is suddenly removed
and the ball moves to position C as the rod rotates. Neglecting
friction and the mass of the rod and knowing that the initial speed
of A is v A = 2.5 m/s, determine (a) the radial and transverse
components of the acceleration of ball B immediately after the
pin is removed, (b) the acceleration of ball B relative to the rod
at that instant, (c) the speed of ball A after ball B has reached the
stop at C.
SOLutiOn
Let r and q be polar coordinates with the origin lying at the shaft.
Constraint of rod: q B = q A + p radians; q B = q A = q; qB = qA = q.
(a)
Components of acceleration.
Sketch the free body diagrams of the balls showing the radial and
transverse components of the forces acting on them. Owing to
frictionless sliding of B along the rod, ( FB )r = 0.
Radial component of acceleration of B.
Fr = mB ( aB ) r :
( aB ) r = 0 b
Transverse components of acceleration.
( aA )q = rAq + 2rAq = raq
( aB )q = rBq + 2rBq
(1)
Since the rod is massless, it must be in equilibrium. Draw its freebody diagram, applying Newton’s Third Law.
+
SM 0 = 0 : rA ( FA )q + rB ( FB )q = rA mA ( aA )q + rB mB ( aB )q = 0
rA mA rAq + rB mB ( rBq + 2rBq ) = 0
q =
At t = 0,
rB = 0
From Eq. (1),
-2rB rBq
mA rA2 + mB rB 2
so that
q = 0.
( aB )q = 0 b
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139
PROBLEM 12.91 (continued)
(b)
Acceleration of B relative to the rod.
(v )
2.5
(v A )q = 2.5 m/s, q = A q =
= 10 rad/s
rA
0.25
At t = 0,

rB - rBq 2 = ( aB ) r = 0

rB = rBq 2 = (0.2)(10)2 = 20 m/s2

rB = 20.0 m/s2 b
(c)
Speed of A.
Substituting
d
(mr 2q ) for rFq in each term of the moment equation gives
dt
(
)
(
)
d
d
mA rA2q +
mB rB2q = 0
dt
dt
Integrating with respect to time,
(
mA rA2q + mB rB2q = mA rA2q
) + (m r q )
0
2
B B
0
Applying to the final state with ball B moved to the stop at C,
(m r
2
A A
)
+ mB rC2 q f = ÈÎmA rA2 + mB ( rB )20 ˘˚ q0
m r 2 + mB rB2
q f = A A2
mA rA + mB rC2
=
(0.2)(0.25)2 + (0.4)(0.2)2
(10)
(0.2)(0.25)2 + (0.4)(0.4)2
= 3.7255 rad/s
( v A ) f = rAq f = (0.25)(3.7255) = 0.93137 m/s
( v A ) f = 0.931 m/s b
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140
PROBLEM 12.92
Two 1.25 kg collars A and B can slide without friction on a frame, consisting
of the horizontal rod OE and the vertical rod CD, which is free to rotate about
CD. The two collars are connected by a cord running over a pulley that is
attached to the frame at O and a stop prevents collar B from moving. The
frame is rotating at the rate θ = 12 rad/s and r = 200 mm when the stop is
removed allowing collar A to move out along rod OE. Neglecting friction and
the mass of the frame, determine, for the position r = 400 mm, (a) the
transverse component of the velocity of collar A, (b) the tension in the cord
and the acceleration of collar A relative to the rod OE.
SOLUTION
Masses: mA = mB = 1.25 kg
(a) Conservation of angular momentum of collar A: ( H 0 )2 = ( H 0 )1
mAr1(vθ )1 = mAr2 (vθ )2
(vθ )2 =
r1(vθ )1 r12θ1 (0.2) 2 (12)
=
=
= 3.6
r2
r2
0.4
(vθ )2 = 1.200 m/s W
θ2 =
(vθ ) 2
1.2
=
= 3.00 rad/s
rA
0.4
(b) Let y be the position coordinate of B, positive upward with origin at O.
Constraint of the cord: r − y = constant
or
y = r
Kinematics:
(aB ) y = y = r
Collar B:
Collar A:
and
(a A ) r = r − rθ 2
ΣFy = mB aB : T − WB = mB y = mB
r
(1)
ΣFr = mA (a A )r : − T = mA (
r − rθ2 )
(2)
Adding (1) and (2) to eliminate T,
−WB = (mA + mB )
r + mArθ2
a A/rod = r=
m Ar θ 2 − WB (1.25)(0.4)(3)2 − (1.25)(9.81)
=
= −3.105 m/s 2
m A + mB
(1.25 + 1.25)
T = mB (
r + g ) = (1.25)(−3.105 + 9.81)
T = 8.38 N W
a A / rod = 3.11 m/s 2 radially inward. W
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141
PROBLEM 12.93
A small ball swings in a horizontal circle at the end of a cord of length l1 ,
which forms an angle 1 with the vertical. The cord is then slowly drawn
through the support at O until the length of the free end is l2 . (a) Derive a
relation among l1 , l2 , 1 , and  2 . (b)) If the ball is set in motion so that
initially l1  0.8 m and 1  35, determine the angle  2 when l2  0.6 m.
SOLUTION
(a)
For state 1 or 2, neglecting the vertical component of acceleration,
Fy  0: T cos  W  0
T  W cos 
Fx  man : T sin   W sin  cos  
But    sin 
mv 2

so that
v2 
W
sin 2  cos    g sin  tan 
m
v1  1 g sin 1 tan 1
and
v2   2 g sin 2 tan 2
M y  0: H y  constant
r1mv1  r2 mv2
or
v11 sin 1  v2  2 sin  2
3/2
3/2
1 g sin 1 sin 1 tan 1   2 sin  2 sin  2 tan  2
31 sin 3 1 tan 1  32 sin 3  2 tan  2 
(b)
With 1  35, 1  0.8 m, and  2  0.6 m
(0.8)3 sin3 35 tan 35  (0.6)3 sin3  2 tan  2
sin 3  2 tan  2  0.31320  0
 2  43.6 
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142
PROBLEM 12.94
A particle of mass m is projected from Point A with an initial velocity
v0 perpendicular to OA and moves under a central force F along an
elliptic path defined by the equation r  r0 /(2  cos  ). Using Eq.
(12.35), show that F is inversely proportional to the square of the
distance r from the particle to the center of force O.
SOLUTION
u
1 2  cos 
,

r
r0
d 2u
2
F
u  
2
r0 mh 2u 2
d
Solving for F,
F
du sin 
,

d
r0
d 2u cos 

r0
d 2
by Eq. (12.37).
2mh 2u 2 2mh2

r0
r0 r 2
Since m, h, and r0 are constants, F is proportional to
1
r2
2
, or inversely proportional to r . 
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143
PROBLEM 12.95
A particle of mass m describes the logarithmic spiral r  r0 eb under a central force F directed toward the
center of force O. Using Eq. (12.35) show that F is inversely proportional to the cube of the distance r from
the particle to O.
SOLUTION
u
1 1 b
 e
r r0
du
b
  e b
d
r0
d 2u b 2 b

e
r0
d 2
d 2u
b 2  1 b
F
u
e



2
r0
d
mh2u 2
F

(b 2  1)mh 2u 2 b
e
r0
(b 2  1)mh 2u 2 (b 2  1)mh 2

r
r3
Since b, m, and h are constants, F is proportional to
1
r3
, or inversely proportional to r 3.
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144
PROBLEM 12.96
A particle of mass m describes the path defined by the equation
r  r0 (6 cos  5) under a central force F directed away from the
center of force O. Using Eq. (12.35), show that F is inversely
proportional to the square of the distance r from the particle to O.
SOLUTION
1 6 cos   5

r
r0
du
6sin 

d
r0
d 2u
6 cos 

d
r0
u 
d 2u
5
F
u   
by Eq. (12.35).
2
r0
d
mh 2u 2
F 
Solving for F,
Substitute u 
5mh 2u 2
r0
1
r
F 
5mh 2

r0r 2
1
, or inversely proportional to r 2. The minus sign
2
r
indicates that the force is repulsive, as shown in Fig. P12.96.
Since m, h, and r0 are constants, F is proportional to
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145
PROBLEM 12.97
A particle of mass m describes the parabola y  x2 4r0 under a central
force F directed toward the center of force C. Using Eq. (12.35) and
Eq. 12.37' with   1, show that F is inversely proportional to the square
of the distance r from the particle to the center of force and that the angular
momentum per unit mass h  2GMr0 .
SOLUTION
From Fig. P12.97,
But
x2
y 
4r0
x  r sin  ,
y  r0  r cos

r 2 1  cos2 
r 2 sin 2 

or r0  r cos 
4r0
4r0
 1  cos 2 

4r0


 2
 r   cos  r  r0  0

Solving the quadratic equation for r,
r 

2r0
1  cos 2 

2

  cos  cos 2   4  1  cos 

4r0





  r0  



2r 1  cos 
2r0
 cos  1  0
2 
1  cos 
1  cos 2 
since r  0. Simplifying gives
r 
2r0
1  cos
1 1  cos

u
r
2r0
or
du
sin 

d
2r0
and
(1)
d 2u
cos

2
2r0
d
d 2u
1
F
u 

2
2r0
d
mh 2u 2
Solving for F,
F 
Since m, h, and r0 are constants, F is proportional to
mh 2u 2
2r0
F 
mh2

2r0r 2
1
, or inversely proportional to r 2.
r2
1 GM
 2 1   cos   u
r
h
GM
1
Comparing with (1) shows that   1 and 2 
2r0
h
By Eq. 12.37 ' ,
h 
2GMr0 
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146
PROBLEM 12.98
It was observed that during its second flyby of the earth, the Galileo spacecraft had a velocity of 14.1 km/s as
it reached its minimum altitude of 303 km above the surface of the earth. Determine the eccentricity of the
trajectory of the spacecraft during this portion of its flight.
SOLUTION
For earth, R  6.37  106 m
r0  6.37  106  303.  103  6.673  106 m
h  r0v0  (6.673  106 )(14.1  103 )  94.09  109 m 2 /s
GM  gR 2  (9.81)(6.37  106 )2  398.06  1012 m3 /s2
1
GM
 2 (1   )
r0
h
1 
h2
(94.09  109 )2

 3.33
r0GM
(6.673  106 )(398.06  1012 )
  3.33  1
  2.33 
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PROBLEM 12.99
It was observed that during the Galileo spacecraft’s first flyby of the earth, its minimum altitude was 960 km
above the surface of the earth. Assuming that the trajectory of the spacecraft was parabolic, determine the
maximum velocity of Galileo during its first flyby of the earth.
SOLutiOn
First we note
R = 6.37 ¥ 106 m
so that
r0 = (6.37 ¥ 106 + 960 ¥ 103 ) m
= 7.33 ¥ 106 m
vmax = v0
Now
and from page 709 of the text
v0 =
2GM
2 gR2
=
r0
r0
using Eq. (12.30).
1/ 2
Then
vmax
È 2 ¥ 9.81 m/s2 ¥ (6.37 ¥ 106 m)2 ˘
=Í
˙
7.33 ¥ 106 m
Î
˚
= 10, 421.7 m/s
vmax = 10.42 km/s b
or
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148
PROBLEM 12.100
As a space probe approaching the planet Venus on a parabolic trajectory reaches
Point A closest to the planet, its velocity is decreased to insert it into a circular
orbit. Knowing that the mass and the radius of Venus are 4.87  1024 kg and 6052
km, respectively, determine (a) the velocity of the probe as it approaches A,
(b) the decrease in velocity required to insert it into the circular orbit.
SOLUTION
First note
(a)
rA  (6052  280) km  6332 km
From the textbook, the velocity at the point of closest approach on a parabolic trajectory is given by
v0 
2GM
r0
1/2
Thus,
(vA )par
 2  66.73  1012 m3 /kg  s 2  4.87  1024 kg 


6332  103 m


 10,131.4 m/s
(vA )par  10.13 km/s 
or
(b)
We have
(vA )circ  (vA )par  vA
Now
(v A )circ 

Then
v A 
GM
r0
1
2
1
2
Eq. (12.44)
(v A ) par
(v A ) par  (v A ) par
 1


 1 (10.1314 km/s)
 2

 2.97 km/s
| vA |  2.97 km/s 
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149
PROBLEM 12.101
It was observed that as the Voyager I spacecraft reached the point of its trajectory closest to the planet Saturn,
it was at a distance of 185  103 km from the center of the planet and had a velocity of 21.0 km/s. Knowing
that Tethys, one of Saturn’s moons, describes a circular orbit of radius 295  103 km at a speed of 11.35 km/s,
determine the eccentricity of the trajectory of Voyager I on its approach to Saturn.
SOLUTION
For a circular orbit,
Eq. (12.44)
v
GM
r
For the orbit of Tethys,
GM  rT vT2
For Voyager’s trajectory, we have
1 GM
 2 (1   cos  )
r
h
where h  r0 v0
At O,
r  r0 ,   0
Then
1
GM

(1   )
r0 ( r0 v0 ) 2
or

r0 v02
r v2
 1  0 02  1
GM
rT vT
2

185  103 km  21.0 km/s 

 1
295  103 km  11.35 km/s 
  1.147 
or
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150
PROBLEM 12.102
A satellite describes an elliptic orbit about a planet of mass M.
Denoting by r0 and r1 , respectively, the minimum and maximum
values of the distance r from the satellite to the center of the planet,
derive the relation
1 1 2GM
  2
r0 r1
h
where h is the angular momentum per unit mass of the satellite.
SOLUTION
Using Eq. (12.39),
1 GM
 2  C cos  A
rA
h
and
1 GM
 2  C cos  B .
rB
h
But
 B   A  180,
cos  A   cos  B .
so that
Adding,
1
1 1 1 2GM
    2 
rA rB r0 r1
h
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151
PROBLEM 12.103
A space probe is describing a circular orbit about a planet of radius R.. The altitude of the probe above the
surface of the planet is  R and its speed is v0. To place the probe in an elliptic orbit which will bring it closer
to the planet, its speed is reduced from v0 to  v0 , where   1, by firing its engine for a short interval of
time. Determine the smallest permissible value of  if the probe is not to crash on the surface of the planet.
SOLUTION
GM
rA
For the circular orbit,
v0 
where
rA  R   R  R(1   )
Eq. (12.44),
GM  v02 R(1   )
Then
From the solution to Problem 12.102, we have for the elliptic orbit,
1 1 2GM
  2
rA rB
h
h  hA  rA (v A ) AB
Now
 [ R(1   )](  v0 )
Then
2v02 R (1   )
1
1


R (1   ) rB [ R (1   )  v0 ]2

2
 R (1   )
2
Now  min corresponds to rB  R.
Then
1
1
2
  2
R (1   ) R  min R(1   )
 min 
or
2

2 
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152
PROBLEM 12.104
A satellite describes a circular orbit at an altitude of 19 110 km above
the surface of the earth. Determine (a) the increase in speed required at
point A for the satellite to achieve the escape velocity and enter a
parabolic orbit, (b) the decrease in speed required at point A for the
satellite to enter an elliptic orbit of minimum altitude 6370 km, (c) the
eccentricity  of the elliptic orbit.
SOLUTION


For earth, GM  gR 2   9.81 6.37  106  398.06  1012 m 3/s 2
rA  6370  19110  25480 km  25.48  106 m
vcirc 
GM

rA
vesc 
398.06  1012
 3.9525  103 m/s
25.48  106
2GM

rA
2 vcirc  5.5897  103 m/s
(a) Increase in speed at A.
v  vesc  vcirc  1.637  103 m/s
v  1.637  103 m/s 
Elliptical orbit with rB  6370  6370  12740 km  12.74  106 m.
Using Eq. (12.39),
But
and
1
GM
 2  C cos B .
rB
h
 B   A  180, so that cos A   cos B
Adding,
h
1
GM
 2  C cos A
rA
h
1
1
r  rB
2GM

 A

rA rB
rArB
h2
2GMrArB

rA  rB
 2   398.06  1012  25.48  106 12.74  106 
38.22  106
 82.230  109 m 2 /s
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153
PROBLEM 12.104 (Continued)
vA 
h
82.230  109

 3.2272  103 m/s
rA
25.48  106
(b) Decrease in speed. v  vcirc  v A  725 m/s
(c)
v  725 m/s 
1
1
r  rB

 A
 C cos B  C cos A  2C
rB rA
rArB
C 
By Eq. (12.40),
rA  rB
12.74  106

 19.623  10 9 m 1
6
6
2rArB


2
25.48
10
12.74
10
 




19.623  10 9 82.230  109
Ch 2

 
GM
398.06  1012


2
  0.333 
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154
PROBLEM 12.105
A space probe is to be placed in a circular orbit of 9000 km
radius about the planet Venus in a specified plane. As the probe
reaches A, the point of its original trajectory closest to Venus, it
is inserted in a first elliptic transfer orbit by reducing its speed by
Dv A . This orbit brings it to Point B with a much reduced velocity.
There the probe is inserted in a second transfer orbit located in the
specified plane by changing the direction of its velocity and further
reducing its speed by Dv B . Finally, as the probe reaches Point C,
it is inserted in the desired circular orbit by reducing its speed
by DvC . Knowing that the mass of Venus is 0.82 times the mass of
the earth, that rA = 15 ¥ 103 km and rB = 300 ¥ 103 km, and that the
probe approaches A on a parabolic trajectory, determine by how
much the velocity of the probe should be reduced (a) at A, (b) at B,
(c) at C.
SOLutiOn
R = 6370 km = 6.37 ¥ 106 m, g = 9.81 m/s2
For Earth,
GM earth = gR2 = (9.81)(6.37 ¥ 106 )2 = 3.98059 ¥ 1014 m3 /s2
GM = 0.82 GM earth = 3.2641 ¥ 1014 m3 /s2
For Venus,
For a parabolic trajectory with
rA = 15 ¥ 103 km = 15 ¥ 106 m
( v A )1 = vesc =
First transfer orbit AB.
(2)(3.2641 ¥ 1014 )
2GM
=
= 6.5971 ¥ 103 m/s
rA
15 ¥ 106
rB = 300 ¥ 103 km = 300 ¥ 106 m
At Point A, where q = 180∞
1 GM
GM
= 2 + C cos 180∞ = 2 - C
rA hAB
hAB
(1)
1 GM
GM
= 2 + C cos 0 = 2 + C
rB
hAB
hAB
(2)
At Point B, where q = 0∞
Adding,
r + rA 2GM
1
1
+
= B
= 2
rA rB
rA rB
hAB
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155
PROBLEM 12.105
A space probe is to be placed in a circular orbit of 9000 km
radius about the planet Venus in a specified plane. As the probe
reaches A, the point of its original trajectory closest to Venus, it
is inserted in a first elliptic transfer orbit by reducing its speed by
Dv A . This orbit brings it to Point B with a much reduced velocity.
There the probe is inserted in a second transfer orbit located in the
specified plane by changing the direction of its velocity and further
reducing its speed by Dv B . Finally, as the probe reaches Point C,
it is inserted in the desired circular orbit by reducing its speed
by DvC . Knowing that the mass of Venus is 0.82 times the mass of
the earth, that rA = 15 ¥ 103 km and rB = 300 ¥ 103 km, and that the
probe approaches A on a parabolic trajectory, determine by how
much the velocity of the probe should be reduced (a) at A, (b) at B,
(c) at C.
SOLutiOn
R = 6370 km = 6.37 ¥ 106 m, g = 9.81 m/s2
For Earth,
GM earth = gR2 = (9.81)(6.37 ¥ 106 )2 = 3.98059 ¥ 1014 m3 /s2
GM = 0.82 GM earth = 3.2641 ¥ 1014 m3 /s2
For Venus,
For a parabolic trajectory with
rA = 15 ¥ 103 km = 15 ¥ 106 m
( v A )1 = vesc =
First transfer orbit AB.
(2)(3.2641 ¥ 1014 )
2GM
=
= 6.5971 ¥ 103 m/s
rA
15 ¥ 106
rB = 300 ¥ 103 km = 300 ¥ 106 m
At Point A, where q = 180∞
1 GM
GM
= 2 + C cos 180∞ = 2 - C
rA hAB
hAB
(1)
1 GM
GM
= 2 + C cos 0 = 2 + C
rB
hAB
hAB
(2)
At Point B, where q = 0∞
Adding,
r + rA 2GM
1
1
+
= B
= 2
rA rB
rA rB
hAB
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156
PROBLEM 12.105 (continued)
Solving for hAB ,
hAB =
2GMrA rB
(2)(3.2641 ¥ 1014 )(15 ¥ 106 )(300 ¥ 106 )
=
rB + rA
315 ¥ 106
(
)
= 9.65712 ¥ 1010 m2 /s
Second transfer orbit BC.
( v A )2 =
hAB 9.65712 ¥ 1010
=
= 6.43808 ¥ 103 m/s
6
rA
15 ¥ 10
(v B )1 =
hAB 9.65712 ¥ 1010
=
= 0.321904 ¥ 103 m/s
rB
300 ¥ 106
rC = 9000 km = 9 ¥ 106 m
At Point B, where q = 0
1 GM
GM
= 2 + C cos 0 = 2 + C
rB
hBC
hBC
At Point C, where q = 180∞
1 GM
GM
= 2 + C cos 180∞ = 2 - C
rC
hBC
hBC
Adding,
1 1 rB + rC 2GM
+ =
= 2
rB rC
rB rC
hBC
hBC =
=
2GMrB rC
rB + rC
(2)(3.2641 ¥ 1014 )((300 ¥ 106 )(9 ¥ 106 )
(309 ¥ 106 )
= 7.55265 ¥ 1010 m2 /s
Final circular orbit.
( v B )2 =
hBC 7.55265 ¥ 1010
=
= 251.755 m/s
rB
300 ¥ 106
( vC )1 =
hBC 7.55265 ¥ 1010
= 8391.83 m/s
=
rC
9 ¥ 106
rC = 9 ¥ 106 m
( vC )2 =
=
GM
rC
3.2641 ¥ 1014
9 ¥ 106
= 6.02227 ¥ 103 m/s
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157
PROBLEM 12.105 (continued)
Speed reductions.
(a)
At A:
( v A )1 - ( v A )2 = 6.5971 ¥ 103 - 6.43808 ¥ 103 = 159.02 m s
Dv A = 159.0 m/s b
(b)
At B:
( v B )1 - ( v B )2 = 0.321904 ¥ 103 - 251.755 = 70.149 m s
Dv B = 70.1 m/s b
(c)
At C:
( vC )1 - ( vC )2 = (8391.83 - 6.02227 ¥ 103 )
= 2369.13 m s
DvC = 2370 m/s b
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158
PROBLEM 12.106
For the space probe of Problem 12.105, it is known that
rA = 15 ¥ 103 km and that the velocity of the probe is reduced to
6000 m/s , as it passes through A. Determine (a) the distance from the
center of Venus to Point B, (b) the amounts by which the velocity of
the probe should be reduced at B and C, respectively.
SOLutiOn
For Earth,
R = 6370 km = 6.37 ¥ 106 m, g = 9.81 m/s2
GM earth = gR2 = (9.81)(6.37 ¥ 106 )2 = 3.98059 ¥ 1014 m3 /s2
For Venus,
Transfer orbit AB:
GM = 0.82 GM earth = 3.2641 ¥ 1014 m3 /s2
v A = 6000 m/s, rA = 15 ¥ 103 km = 15 ¥ 106 m
hAB = rA v A = (15 ¥ 106 )(6000) = 9 ¥ 1010 m2 /s
At Point A, where q = 180∞
1 GM
GM
= 2 + C cos 180∞ = 2 - C
rA hAB
hAB
At Point B, where q = 0∞
1 GM
GM
= 2 + C cos 0 = 2 + C
rB
hAB
hAB
Adding,
1
1
2GM
+
= 2
rA rB
hAB
1 2GM 1
= 2 rB
rA
hAB
=
(2)(3.2641 ¥ 1014 )
1
10 2
(9 ¥ 10 )
15 ¥ 106
= 1.39284 ¥ 10 -8 m -1
rB = 71.79578 ¥ 106 m
rB = 71.8 ¥ 103 km b
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159
PROBLEM 12.106 (continued)
(a)
rB = 71.79578 ¥ 106 m
Radial coordinate rB .
( v B )1 =
hAB
9 ¥ 1010
= 1.25356 ¥ 103 m/s
=
6
rB
71.79578 ¥ 10
rC = 9000 km = 9 ¥ 106 m
Second transfer orbit BC.
At Point B, where q = 0
1 GM
GM
= 2 + C cos 0 = 2 + C
rB
hBC
hBC
At Point C, where q = 180∞
1 GM
GM
= 2 + C cos 180∞ = 2 - C
rC
hBC
hBC
Adding,
r + rC 2GM
1
1
+
= B
= 2
rB rC
rB rC
hBC
hBC =
2GMrB rC
=
rB + rC
(2)(3.2641 ¥ 1014 )(71.79578 ¥ 106 )(9 ¥ 106 )
(80.779578 ¥ 106 )
= 7.22555 ¥ 1010 m2 /s
( v B )2 =
hBC 7.22555 ¥ 1010
=
= 1.0064 ¥ 103 m/s
rB
71.79578 ¥ 106
( vC )1 =
hBC 7.22555 ¥ 1010
=
= 8.0284 ¥ 103 m/s
rC
9 ¥ 106
rC = 9 ¥ 106 m
Circular orbit with
( vC )2 =
(b)
GM
3.2641 ¥ 1014
=
= 6.0223 ¥ 103 m/s
rC
9 ¥ 106
Speed reductions at B and C.
At B:
( v B )1 - ( v B )2 = 1.25356 ¥ 103 - 1.0064 ¥ 103 = 247.16 m s
Dv B = 247 m/s b
At C:
( vC )1 - ( vC )2 = 8.0284 ¥ 103 - 6.0223 ¥ 103
= 2006.1 m s
DvC = 2010 m s b
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160
PROBLEM 12.107
As it describes an elliptic orbit about the sun, a spacecraft reaches
a maximum distance of 320 ¥ 106 km from the center of the sun at
Point A (called the aphelion) and a minimum distance of 150 ¥ 106 km
at Point B (called the perihelion). To place the spacecraft in a smaller
elliptic orbit with aphelion at A¢ and perihelion at B ¢, where A¢ and
B ¢ are located 260 ¥ 106 and 135 ¥ 106 km, respectively, from the
center of the sun, the speed of the spacecraft is first reduced as it
passes through A and then is further reduced as it passes through
B ¢. Knowing that the mass of the sun is 332.8 ¥ 103 times the mass
of the earth, determine (a) the speed of the spacecraft at A, (b) the
amounts by which the speed of the spacecraft should be reduced at
A and B ¢ to insert it into the desired elliptic orbit.
SOLutiOn
First note
Rearth = 6370 km = 6.37 ¥ 106 m
rA = 320 ¥ 106 km = 320 ¥ 109 m
rB = 150 ¥ 106 km = 150 ¥ 109 m
From the solution to Problem 12.102, we have for any elliptic orbit about the sun
1 1 2GM sun
+ =
r1 r2
h2
(a)
For the elliptic orbit AB, we have
r1 = rA , r2 = rB , h = hA = rA v A
Also,
GM sun = G[(332.8 ¥ 103 ) M earth ]
2
= gRearth
(332.8 ¥ 103 )
Then
or
using Eq. (12.30).
2
(332.8 ¥ 103 )
1 1 2 gRearth
+ =
rA rB
(rA v A )2
R
v A = earth
rA
Ê 665.6 g ¥ 103 ˆ
Á
˜
1
1
Ë
¯
rA + rB
1/ 2
(6.37 ¥ 10 m) ÊÁ 665.6 ¥ 9.81 ¥ 10 ˆ˜
=
+
˜¯
(320 ¥ 10 m) ÁË
6
9
3
1
320 ¥ 109
1/ 2
1
150 ¥ 109
= 16, 255.6 m/s
v A = 16.3 ¥ 103 m/s b
or
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161
PROBLEM 12.107 (continued)
(b)
From Part (a), we have
Ê 1 1ˆ
2GM sun = ( rA v A )2 Á + ˜
Ë rA rB ¯
Then, for any other elliptic orbit about the sun, we have
(
2 1
1
1 1 ( rA v A ) rA + rB
+ =
r1 r2
h2
)
For the elliptic transfer orbit AB¢, we have
r1 = rA , r2 = rB ¢ , h = htr = rA ( v A ) tr
Then
or
(
2 1
1
1
1 ( rA v A ) rA + rB
+
=
rA rB ¢
[rA ( v A ) tr ]2
Ê r1 + r1 ˆ
( v A ) tr = v A Á 1A 1B ˜
Ë rA + rB¢ ¯
1/ 2
)
Ê 1 + rA ˆ
rB
˜
= vA Á
ÁË 1 + rrA ˜¯
B¢
320 ˆ
Ê 1 + 150
= (16, 255.6 m/s) Á
320 ˜
Ë 1 + 135 ¯
1/ 2
1/ 2
= 15, 673.6 m/s
htr = ( hA ) tr = ( hB ¢ ) tr : rA ( v A ) tr = rB ¢ ( v B ¢ ) tr
Now
Then
( v B¢ )tr =
320 ¥ 109 km
135 ¥ 109 km
¥ 15673.6 m/s = 37,152.2 m/s
For the elliptic orbit A¢ B ¢, we have
r1 = rA¢ , r2 = rB ¢ , h = rB ¢ v B ¢
2
Then
or
(
1
1 ( rA v A ) rA + rB
+
=
rA¢ rB ¢
( rB ¢ v B ¢ )2
vB¢
1
1
)
1
1
rA Ê rA + rB ˆ
= vA
Á
˜
rB ¢ Ë r1 + r1 ¯
A¢
B¢
1/ 2
1
1
Ê 320 ¥ 109 m ˆ Ê 320 ¥ 109 + 150 ¥ 109 ˆ
˜
= (16, 255.6 m/s) Á
˜Á
Ë 135 ¥ 109 m ¯ ÁË 260 1¥ 109 + 135 ¥1 109 ˜¯
1/ 2
= 35, 942.0 m/s
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465
162
PROBLEM 12.107 (continued)
Finally,
or
( v A )tr = v A + Dv A
Dv A = (15673.6 - 16, 255.6) m/s
|Dv A | = 582 m/s b
or
and
or
v B ¢ = ( v B ¢ )tr + Dv B
Dv B¢ = (35, 942 - 37,152.2) m/s
= -1210.2 m/s
|Dv B | = 1210 m/s b
or
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163
PROBLEM 12.108
Halley’s comet travels in an elongated elliptic orbit for which the minimum distance from the sun is
approximately 12 rE , where rE  150  106 km is the mean distance from the sun to the earth. Knowing that the
periodic time of Halley’s comet is about 76 years, determine the maximum distance from the sun reached by
the comet.
SOLUTION
We apply Kepler’s Third Law to the orbits and periodic times of earth and Halley’s comet:
2
3
H 
 aH 

 

 E 
 aE 
Thus
 
aH  aE  H 
 E 
2/3
 76 years 
 rE 

 1 year 
 17.94rE
But
2/3
1
(rmin  rmax )
2
11

17.94rE   rE  rmax 
22

aH 
1
rmax  2(17.94rE )  rE
2
 (35.88  0.5)rE
 35.38 rE
rmax  (35.38)(150  106 km)
rmax  5.31  109 km 
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164
PROBLEM 12.109
Based on observations made during the 1996 sighting of comet Hyakutake, it was concluded that the
trajectory of the comet is a highly elongated ellipse for which the eccentricity is approximately   0.999887.
Knowing that for the 1996 sighting the minimum distance between the comet and the sun was 0.230 RE ,
where RE is the mean distance from the sun to the earth, determine the periodic time of the comet.
SOLUTION
For Earth’s orbit about the sun,
v0 
2 RE 2 RE 3/ 2
GM
, 0 

RE
v0
GM
or
GM 
2 RE3/2
0
(1)
For the comet Hyakutake,
1 GM
 2  (1   ),
r0
h
a
1 GM
1
r0
 2 (1   ), r1 
r1
1 
h
r
1
1 
(r0  r1 )  0 , b  r0 r1 
r0
2
1 
1
h  GMr0 (1   )


2 r02 (1   )1/2
2 ab

h
(1   )3/ 2 GMr0 (1   )
2 r03/ 2
GM (1   )3/2
 r 
 0 
 RE 
3/2
2 r03/2 0
2 RE3 (1   )3/2
1
0
(1   )3/2
 (0.230)3/2
Since

1
 0  91.8  103 0
(1  0.999887)3/2
 0  1 yr,   (91.8  103 )(1.000)
  91.8  103 yr 
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165
PROBLEM 12.110
A space probe is to be placed in a circular orbit of radius 4000
km about the planet Mars. As the probe reaches A, the point of
its original trajectory closest to Mars, it is inserted into a first
elliptic transfer orbit by reducing its speed. This orbit brings it
to Point B with a much reduced velocity. There the probe is
inserted into a second transfer orbit by further reducing its
speed. Knowing that the mass of Mars is 0.1074 times the
mass of the earth, that rA 9000 km and rB 180,000 km, and
that the probe approaches A on a parabolic trajectory,
determine the time needed for the space probe to travel from A
to B on its first transfer orbit.
SOLUTION
For earth, R  6373 km  6.373  106 m
GM  gR 2  (9.81) (6.373  106 )2  398.43  1012 m3 /s2
For Mars, GM  (0.1074)(398.43  1012 )  42.792  1012 m3 /s2
rA  9000 km  9.0  106 m rB  180000 km  180  106 m
For the parabolic approach trajectory at A,
(vA )1 
2GM

rA
(2)(42.792  1012 )
 3.0837  103 m/s
9.0  106
First elliptic transfer orbit AB.
Using Eq. (12.39),
But
1
GM
 2  C cos A
rA
hAB
 B   A  180,
Adding,
so that
and
1
GM
 2  C cos B .
rB
hAB
cos A   cos B .
1
1
r  rB
2GM

 A
 2
rA rB
rArB
hAB
hAB 
2GMrArB

rA  rB
(2)(42.792  1012 )(9.0  106 )(180  106 )
189.0  106
hAB  27.085  109 m 2 /s
a 
b 
1
1
(rA  rB )  (9.0  106  180  106 )  94.5  106 m
2
2
rArB 
(9.0  10 6 )(180  10 6 )  40.249  106 m
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PROBLEM 12.110 (Continued)
Periodic time for full ellipse:
For half ellipse AB,
 
2 ab
h
 AB 
1
 ab
 
2
h
 AB 
 (94.5  106 )(40.249  106 )
 444.81  103 s
9
27.085  10
 AB  122.6 h 
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167
PROBLEM 12.111
A spacecraft and a satellite are at diametrically opposite positions
in the same circular orbit of altitude 500 km above the earth. As it
passes through point A, the spacecraft fires its engine for a short
interval of time to increase its speed and enter an elliptic orbit.
Knowing that the spacecraft returns to A at the same time the
satellite reaches A after completing one and a half orbits,
determine (a) the increase in speed required, (b) the periodic time
for the elliptic orbit.
SOLUTION
g  9.81 m/s2
For earth, R  6.37  106 m ,

GM  gR 2   9.81 6.37  106

2
 398.06  1012 m 3/s 2
For circular orbit of satellite,
r0  6370  500  6870 km  6.87  106 m/s
v0 
398.06  1012
 7.6119  103 m/s
6.87  106
GM

r0


 2  6.87  106
2 r0
0 

 5.6708  103 s
v0
7.6119  103
For elliptic orbit of spacecraft it is given that
3
 0  8.5062  103 s
2
 
a 
Using Eq. (12.39),
But
1
GM
 2  C cos A
rA
h
 B   A  180,
Adding,
1
 rA  rB  ,
2
so that

rArB
1
GM
 2  C cos B .
rB
h
and
cos A   cos B .
1
1
r  rB
2a
2GM

 A
 2 
rA rB
rArB
b
h2
Periodic time:  
b
h
or
GMb 2
a
2 ab
2 ab a
2 a 3 / 2


h
GM
GMb 2

398.06  1012 8.5062  103
GM  2
3

a 
4 2
4 2

2
 729.558  1018 m3
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PROBLEM 12.111 (Continued)
a  9.0023  106 m,
rA  r0  6.87  106 m
rB  2a  rA  11.1346  10 6 m,


b 
rArB  8.7461  10 6 m

2 9.0023  106 8.7461  106
2 ab

 58.159  109 m 2 /s
h

8.5062  103
vA 
h
58.159  109

 8.4656  103 m/s
rA
6.87  106
(a) Increase in speed at A.
vA  v A  v0  8.4656  103  7.6119  103
v A  854 m/s 
(b) Periodic time for elliptic orbit.
As calculated above   8.5062 s
  141.8 min 
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169
PROBLEM 12.112
The Clementine spacecraft described an elliptic orbit of
minimum altitude hA  400 km and a maximum altitude of
hB  2940 km above the surface of the moon. Knowing that the
radius of the moon is 1737 km and that the mass of the moon is
0.01230 times the mass of the earth, determine the periodic
time of the spacecraft.
SOLUTION
For earth,
R  6370 km  6.370  106 m
GM  gR 2  (9.81)(6.370  106 )2  398.06  1012 m3 /s2
For moon,
GM  (0.01230)(398.06  1012 )  4.896  1012 m3 /s2
rA  1737  400  2137 km  2.137  106 m
rB  1737  2940  4677 km  4.677  106 m
Using Eq. (12.39),
1 GM
 2  C cos  A
rA
h
But
 B   A  180, so that cos  A   cos  B .
Adding,
and
1 GM
 2  C cos  B .
rB
h
1 1 rA  rB 2GM
 
 2
rA rB
rA rB
hAB
hAB 
2GMrA rB
(2)(4.896  1012 )(2.137  106 )(4.677  106 )

rA  rB
6.814  106
 3.78983  109 m 2 /s
1
a  (rA  rB )  3.402  106 m
2
b  rA rB  3.16145  106 m
Periodic time.

2 ab 2 (3.402  106 )(3.16145  106 )

 17.831  103 s
hAB
3.78983  109
  4.95 h 

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170
PROBLEM 12.113
Determine the time needed for the space probe of Problem 12.100 to travel from B to C.
SOLUTION
From the solution to Problem 12.100, we have
(vA )par  10,131.4 m/s
(v A )circ 
and
1
2
(v A )par  7164.0 m/s
rA  (6052  280) km  6332 km
Also,
For the parabolic trajectory BA, we have
1 GM v
 2 (1   cos  )
r
hBA
[Eq. (12.39)]
where   1. Now
at A,   0:
1 GM v
 2 (1  1)
rA
hBA
or
rA 
at B,   90:
1 GM v
 2 (1  0)
rB
hBA
or
rB 
2
hBA
2GM v
2
hBA
GM v
rB  2rA
As the probe travels from B to A, the area swept out is the semiparabolic area defined by Vertex A and
Point B. Thus,
(Area swept out) BA  ABA 
2
4
rA rB  rA2
3
3
dA 1
 h
dt 2
Now
where h  constant
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PROBLEM 12.113 (Continued)
Then
A
2A
1
ht or t BA  BA
2
hBA
t BA 
2  43 rA2
rA v A
hBA  rA v A

8 rA
3 vA
8 6332  103 m
3 10,131.4 m/s
 1666.63 s

For the circular trajectory AC,
t AC 
Finally,
 r
2 A
(v A )circ

 6332  103 m
 1388.37 s
2 7164.0 m/s
tBC  t BA  t AC
 (1666.63  1388.37) s
=3055.0 s
tBC  50 min 55 s 
or
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172
PROBLEM 12.114
A space probe is describing a circular orbit of radius nR with a velocity v0
about a planet of radius R and center O. As the probe passes through Point
A, its velocity is reduced from v0 to  v0, where   1, to place the probe ona
crash trajectory. Express in terms of n and  the angle AOB, where B denotes
the point of impact of the probe on the planet.
SOLUTION
r0  rA  nR
For the circular orbit,
v0 
GM
GM

r0
nR
The crash trajectory is elliptic.
v A   v0 
 2 GM
nR
h  rA v A  nRv A   2 nGMR
GM
1
 2
2
h
 nR
1 GM
1   cos 
 2 (1   cos  ) 
r
 2 nR
h
At Point A,   180
1
1
1 

 2
rA nR  nR
or  2  1  
or   1   2
At impact Point B,     
1 1

rB R
1 1   cos (   ) 1   cos 


R
 2 nR
 2 nR
 cos   1  n 2
or cos  
1  n  2 1  n 2


1  2
  cos1[(1  n 2 ) /(1   2 )] 
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PROBLEM 12.115
A long-range ballistic trajectory between Points A and B on the earth’s
surface consists of a portion of an ellipse with the apogee at Point C.
Knowing that Point C is 1500 km above the surface of the earth and
the range R of the trajectory is 6000 km, determine (a) the velocity
of the projectile at C, (b) the eccentricity  of the trajectory.
SOLUTION
For earth, R  6370 km  6.37  106 m
GM  gR 2  (9.81)(6.37  106 )2  398.06  1012 m3 /s2
For the trajectory, rC  6370  1500  7870 km  7.87  106 m
rA  rB  R  6.37  106 m,
rC
7870

 1.23548
rA
6370
Range A to B: s AB  6000 km  6.00  106 m
s AB
6.00  106

 0.94192 rad  53.968
R
6.37  106
 
For an elliptic trajectory,
At A,
  180 
At C,
  180 ,
1 GM
 2 (1   cos  )
r
h

 153.016,
2
1
GM
 2 (1   cos153.016)
rA
h
1
GM
 2 (1   )
rC
h
(2)
Dividing Eq. (1) by Eq. (2),
rC
1   cos153.016

 1.23548
rA
1
 
(1)
1.23548  1
 0.68384
1.23548  cos153.016
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174
PROBLEM 12.115 (Continued)
From Eq. (2), h 
h
GM (1   )rC
(398.06  1012 )(0.31616)(7.87  106 )  31.471  109 m2 /s
(a) Velocity at C.
(b)
vC 
h
31.471  109

 4.00  103 m/s
rC
7.87  106
Eccentricity of trajectory.
vC  4 km/s 
  0.684 
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175
PROBLEM 12.116
A space shuttle is describing a circular orbit at an altitude of 563 km above
the surface of the earth. As it passes through Point A, it fires its engine for
a short interval of time to reduce its speed by 152 m/s and begin its descent
toward the earth. Determine the angle AOB so that the altitude of the
shuttle at Point B is 121 km. (Hint: Point A is the apogee of the elliptic
descent trajectory.)
SOLUTION
GM  gR 2  (9.81)(6.37  106 )2  398.06  1012 m3 /s2
rA  6370  563  6933 km  6.933  106 m
rB  6370  121  6491 km  6.491  106 m
For the circular orbit through Point A,
vcirc 
GM

rA
398.06  1012
 7.5773  103 m/s
6.933  106
For the descent trajectory,
vA  vcirc  v  7.5773  103  152  7.4253  103 m/s
h  rAv A  (6.933  106 )(7.4253  103 )  51.4795  109 m 2 /s
1 GM
 2 (1   cos  )
r
h
At Point A,
  180, r  rA
1
GM
 2 (1   )
rA
h
1 
h2
(51.4795  109 )2

 0.96028
GM rA
(398.06  1012 )(6.933  106 )
  0.03972
1
GM
 2 (1   cos  B )
rB
h
1   cos  B 
h2
(51.4795  109 ) 2

 1.02567
GM rB
(398.06  1012 )(6.491  106 )
cos B 
 B  49.7
1.02567  1
 0.6463

AOB  180   B  130.3
AOB  130.3 
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176
PROBLEM 12.117
As a spacecraft approaches the planet Jupiter, it releases a probe which is to
enter the planet’s atmosphere at Point B at an altitude of 450 km above the
surface of the planet. The trajectory of the probe is a hyperbola of eccentricity
e = 1.031. Knowing that the radius and the mass of Jupiter are 71.492 ¥ 103 km
and 1.9 ¥ 1027 kg, respectively, and that the velocity vB of the probe at B forms
an angle of 82.9° with the direction of OA, determine (a) the angle AOB, (b)
the speed v B of the probe at B.
SOLutiOn
rB = (71.492 ¥ 103 + 450) km = 71.942 ¥ 103 km
First we note
(a)
1 GM j
= 2 (1 + e cos q )
r
h
We have
1 GM j
= 2 (1 + e )
rA
h
At A, q = 0:
h2
= rA (1 + e )
GM j
or
At B, q = q B =  AOB :
or
Then
or
[Eq. (12.39¢)]
1 GM j
= 2 (1 + e cos q B )
rB
h
h2
= rB (1 + e cos q B )
GM j
rA (1 + e ) = rB (1 + e cos q B )
cos q B =
=
˘
1 È rA
Í (1 + e ) - 1˙
e Î rB
˚
˘
1 È 70.8 ¥ 103 km
(1 + 1.031) - 1˙
Í
3
1.031 Î 71.942 ¥ 10 km
˚
= 0.96873
or
q B = 14.3661∞
 AOB = 14.37∞ b
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177
PROBLEM 12.117 (continued)
(b)
From above
where
Then
h2 = GM j rB (1 + e cos q B )
1
|rB ¥ mv B | = rB v B sin f
m
f = (q B + 82.9∞) = 97.2661∞
h=
( rB v B sin f )2 = GM j rB (1 + e cos q B )
1/ 2
or
vB =
=
˘
1 È GM j
(1 + e cos q B )˙
Í
sin f Î rB
˚
1
sin 97.2661∞
1/ 2
-12
2
2
27
ÔÏ 66.73 ¥ 10 m /kg ◊ s ¥ 1.9 ¥ 10 kg
Ô¸
¥ [1 + (1.031)(0.96873)]˝
Ì
6
71.942 ¥ 10 m
ÔÓ
˛Ô
v B = 59.8 km/s b
or
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178
PROBLEM 12.118
A satellite describes an elliptic orbit about a planet. Denoting by r0
and r1 the distances corresponding, respectively, to the perigee and
apogee of the orbit, show that the curvature of the orbit at each of
these two points can be expressed as
1 1 1 1 
   
 2  r0 r1 
SOLUTION
Using Eq. (12.39),
1 GM
 2  C cos  A
rA
h
and
1 GM
 2  C cos  B .
rB
h
But
 B   A  180,
cos  A   cos  B
so that
Adding,
1 1 2GM
  2
rA rB
h
At Points A and B, the radial direction is normal to the path.
But
an 
v2
h2
 2
 r
Fn 
GMm
mh 2

ma

n
r2
r 2
1 GM 1  1
1
 2    

2  rA rB 
h
1 1 1 1 
   
 2  r0 r1 
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PROBLEM 12.119
(a) Express the eccentricity  of the elliptic orbit described by a
satellite about a planet in terms of the distances r0 and r1
corresponding, respectively, to the perigee and apogee of the orbit.
(b) Use the result obtained in Part a and the data given in Problem
12.109, where RE  149.6  106 km, to determine the approximate
maximum distance from the sun reached by comet Hyakutake.
SOLUTION
(a)
We have
1 GM
 2 (1   cos  ) Eq. (12.39)
r
h
At A,   0:
1 GM
 2 (1   )
r0
h
or
At B,   180 :
or
Then
h2
 r0 (1   )
GM
1 GM
 2 (1   )
r1
h
h2
 r1 (1   )
GM
r0 (1   )  r1 (1   )

or
(b)
1 
r0
1 
From above,
r1 
where
r0  0.230RE
Then
r1 
1  0.999887
 0.230(149.6  109 m)
1  0.999887
r1  609  1012 m 
or
Note: r1  4070RE
r1  r0

r1  r0
or r1  0.064 lightyears.
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180
PROBLEM 12.120
Derive Kepler’s third law of planetary motion from Eqs. (12.37) and (12.43).
SOLUTION
For an ellipse,
2a  rA  rB
Using Eq. (12.37),
1 GM
 2  C cos  A
rA
h
and
1 GM
 2  C cos  B .
rB
h
But
 B   A  180,
and b  rA rB
cos  A   cos  B .
so that
Adding,
1
1 r r
2a 2GM
  A B  2  2
rA rB
rA rB
b
h
hb
GM
a
By Eq. (12.43),

2 
2 ab 2 ab a 2 a 3/2


h
b GM
GM
4 2 a3
GM
For Orbits 1 and 2 about the same large mass,
and
 12 
4 2 a13
GM
 22 
4 2 a23
GM
2
3
 1   a1 
    
  2   a2 
Forming the ratio,
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181
PROBLEM 12.121
Show that the angular momentum per unit mass h of a satellite describing an elliptic orbit of semimajor axis a
and eccentricity  about a planet of mass M can be expressed as
h  GMa (1   2 )
SOLUTION
By Eq. (12.39),
1 GM
 2 (1   cos  )
r
h
At A,   0 :
1 GM
 2  (1   )
rA
h
or
rA 
h2
GM (1   )
At B,   180 :
1 GM
 2  (1   )
rB
h
or
rB 
h2
GM (1   )
h2
1
 1


GM  1   1  
Adding,
rA  rB 
But for an ellipse,
rA  rB  2a
2a 
2h 2
GM (1   2 )
2
2h


2
 GM (1   )
h  GMa (1   2 ) 
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182
PROBLEM 12.122
Determine the maximum theoretical speed that may be achieved over a distance of 60 m by a car starting from
rest, knowing that the coefficient of static friction is 0.80 between the tires and the pavement and that 60 percent
of the weight of the car is distributed over its front wheels and 40 percent over its rear wheels. Assume
(a) four-wheel drive, (b) front-wheel drive, (c) rear-wheel drive.
SOLUTION
(a)
Four-wheel drive
ΣFy = 0: N1 + N 2 − W = 0
F1 + F2 = μ N1 + μ N 2
= μ ( N1 + N 2 ) = μ w
ΣFx = ma : F1 + F2 = ma
μ w = ma
μW
mg
= μ g = 0.80(9.81)
m
m
a = 7.848 m/s 2
a=
=μ
v 2 = 2ax = 2(7.848 m/s2 )(60 m) = 941.76 m2 /s2
v = 30.69 m/s
(b)
v = 110.5 km/h W
Front-wheel drive
F2 = ma
μ (0.6 W ) = ma
0.6μW 0.6μ mg
=
m
m
= 0.6 μ g = 0.6(0.80)(9.81)
a=
a = 4.709 m/s 2
v 2 = 2ax = 2(4.709 m/s 2 )(60 m) = 565.1 m 2 /s 2
v = 23.77 m/s
v = 85.6 km/h W
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183
PROBLEM 12.122 (Continued)
(c)
Rear-wheel drive
F1 = ma
μ (0.4 W ) = ma
0.4μW 0.4μ mg
=
m
m
= 0.4 μ g = 0.4(0.80)(9.81)
a=
a = 3.139 m/s 2
v 2 = 2ax = 2(3.139 m/s 2 )(60 m) = 376.7 m 2 /s 2
v = 19.41 m/s
v = 69.9 km/h W
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184
PROBLEM 12.123
A bucket is attached to a rope of length L 1.2 m and is made to revolve in a
horizontal circle. Drops of water leaking from the bucket fall and strike the floor
along the perimeter of a circle of radius a.. Determine the radius a when   30.
SOLUTION
Initial velocity of drop  velocity of bucket
Fy  0:
T cos30  mg
Fx  max :
Divide (2) by (1):
(1)
T sin 30  man
tan 30 
(2)
an
v2

g
pg
Thus
v 2   g tan 30
But
  L sin 30  (1.2 m) sin 30°  0.6 m
Thus
v 2  0.6(9.81) tan 30  3.398 m2 /s2
v  1.843 m/s
Assuming the bucket to rotate clockwise (when viewed from
above), and using the axes shown, we find that the components
of the initial velocity of the drop are
(v0 ) x  0, (v0 ) y  0, (v0 )2  1.843 m/s
Free fall of drop
y  y0  (v0 ) y t 
1 2
gt
2
y  y0 
1 2
gt
2
When drop strikes floor:
y 0
y0 
1 2
gt  0
2
But
y0  2L  L cos30  2(1.2)  1.2cos30  1.361 m
Thus
1.361 
1
(9.81) t 2  0
2
t  0.5275
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PROBLEM 12.123 (Continued)
Projection on horizontal floor (uniform motion)
x  x0  (v0 ) z t  L sin 30  0,
x  0.6 m
z  z0  (v0 ) z t  0  1.843(0.527)  0.971 m
Radius of circle: a 
a
x2  z 2
(0.6) 2  (0.971)2
a  1.141 m 
Note: The drop travels in a vertical plane parallel to the yz plane.
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PROBLEM 12.124
A 6-kg block B rests as shown on the upper surface of a 15 kg wedge A.
Neglecting friction, determine immediately after the system is released from
rest (a) the acceleration of A, (b) the acceleration of B relative to A.
SOLutiOn
Acceleration vectors:
a A = aA
30∞, a B /A = aB /A
a B = a A + a B/ A
SFx = max : mB aB /A - mB a A cos 30∞ = 0
+
Block B:
aB /A = aA cos 30∞
(1)
+
SFy = ma y : N AB - WB = - mB aA sin 30∞
N AB = WB - (WB sin 30∞)
(2)
aA
g
a
a
W A sin 30∞ + WB sin 30∞ - (WB sin 2 30∞) A = WA A
g
g
+
Block A:
aA
g
SF = ma : WA sin 30∞ + N AB sin 30∞ = WA
aA =
(W A + WB )sin 30∞
2
g=
( mA + mB )sin 30∞
WA + WB sin 30∞
mA + mB sin 2 30∞
21sin 30∞
¥ (9.81)
=
15 + 6 sin 2 30∞
g
a A = 6.24 m/s2
(a)
30∞ b
aB/ A = (6.24) cos 30∞ = 5.40 m/s2
a B /A = 5.40 m/s2
(b)
b
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187
PROBLEM 12.125
A 250-kg crate B is suspended from a cable attached to a 20-kg trolley A which
rides on an inclined I-beam as shown. Knowing that at the instant shown the
trolley has an acceleration of 0.4 m/s2 up and to the right, determine (a) the
acceleration of B relative to A, (b) the tension in cable CD.
SOLutiOn
(a)
First we note: a B = a A + a B /A , where a B /A is directed perpendicular to cable AB
SFx = mB ax : 0 = - mB aB / A + mB aA cos 25∞
B:
+
aB /A = (0.4 m/s2 ) cos 25∞
or
a B /A = 0.363 m/s2 ¨ b
or
For crate B
+
(b)
SFy = mB a y : TAB - mB g = mB a A sin 25∞
A:
(Not that ( aB ) y = aA sin 25o )
TAB = 250( g + aA sin 25∞)
= 2494.76 N
or
For trolley A
+
or
SFx = mAaA : TCD - TAB sin 25∞ - mA g sin 25∞ = mAa A
TCD = (2494.76 N )sin 25∞ + 20(9.81 ¥ sin 25∞ + 0.4)
TCD = 1145 N b
or
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188
PROBLEM 12.126
The roller-coaster
coaster track shown is contained in a vertical plane.
The portion of track between A and B is straight and
horizontal, while the portions to the left of A and to the right
of B have radii of curvature as indicated. A car is traveling at a
speed of 72 km/h when the brakes are suddenly applied,
causing the wheels of the car to slide on the track (k  0.25).
Determine the initial deceleration of the car if the brakes are
applied as the car (a)) has almost reached A, (b) is traveling
between A and B, (c) has just passed B.
SOLUTION
v  72 km/h  20 m/s
(a)
Almost reached Point A..
  30 m
an 
v 2 (20)2

 13.333 m/s 2

30
Fy  ma y : N R  N F  mg  man
N R  N F  m( g  an )
F   k ( N R  N F )   k m ( g  an )
Fx  max :  F  mat
at  
F
  k ( g  an )
m
| at |  k ( g  an )  0.25(9.81  13.33)
(b)
Between A and B.
 
an  0
| at |  k g  (0.25)(9.81)
(c)
Just passed Point B.
| at |  5.79 m/s2 
| at |  2.45 m/s 2 
  45 m
an 
v 2 (20) 2

 8.8889 m/s 2

45
Fy  ma y : N R  N F  mg  man
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PROBLEM 12.126 (Continued)
or
N R  N F  m( g  an )
F  k ( N R  N F )  k m( g  an )
Fx  max :  F  mat
at  
F
   k ( g  an )
m
| at |  k ( g  an )  (0.25)(9.81  8.8889)
| at |  0.230 m/s2 
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PROBLEM 12.127
The parasailing system shown uses a
winch to pull the rider in towards the
boat, which is travelling with a
constant velocity. During the interval
when  is between 20º and 40º, (where t
= 0 at  = 20º) the angle increases at the
constant rate of 2 º/s. During this time,
the length of the rope is defined by the
1
3
relationship r  125  t
3/2
, where r
and t are expressed in meters and
seconds, respectively. At the instant
when the rope makes a 30 degree angle
with the water, the tension in the
th rope is
18 kN. At this instant, what is the
magnitude and direction of the force of
the parasail on the 75 kg parasailor?
SOLUTION
1 3
r  125  t 2 m
3
 0  20,   30,   2 / s  0.0349 rad/s, =0
Given:
T  18 kN, m  75 kg
  0  t
30  20  2t  t  5 s
Kinematics:
aP  aB  aP/ B
a P  0  ar er  a e
Using Radial and Transverse Coordinates:
Free Body Diagram of parasailor (side view):
1 1
r   t 2 m/s
2
1 1

r   t 2 m/s2
4
Equations of Motion:
F
r
 mar

FP cos   T  mg sin   m r  r 2
FP cos   T  m  g sin   
r  r2 
 F  ma


FP sin   mg cos   m r  2r
(1)
FP sin   m  g cos   r  2r 

(2)
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191
PROBLEM 12.127 (Continued)
Evaluate at t=5 s:
r  121.27 m, r  1.118 m/s, 
r  0.1118 m/s2
(1)
2
FP cos   18000  75 9.81sin 30  0.1118  121.27  0.0349    FP cos
cos   18348.4 N


(2)
FP sin   75 9.81cos 30  0  2  1.118  0.0349    FP sin   631.33 N
FP sin 
FP cos 

631.33
 tan   0.03441    1.97
18348.4
FP cos1.97  18348.4 N
FP  18.4 kN
31.97 
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PROBLEM 12.128
A small 200
200-g collar C can slide on a semicircular rod which is made to rotate about
the vertical AB at the constant rate of 6 rad/s. Determine the minimum required value
of the coefficient of static friction between the collar and the rod if the collar is not to
slide when ((a)   90, (b)   75, (c)   45. Indicate in each case the direction
of the impending motion.
SOLUTION
First note
vC  ( r sin  )AB
 (0.6 m)(6 rad/s)sin 
 (3.6 m/s) sin 
(a)
With   90,
vC  3.6 m/s
Fy  0: F  WC  0
or
F  mC g
Now
F  s N
or
N
1
mC g
s
Fn  mC an : N  mC
vC2
r
or
v2
1
mC g  mC C
s
r
or
s 
gr (9.81 m/s 2 )(0.6 m)

vC2
(3.6 m/s)2
( s )min  0.454 
or
The direction of the impending motion is downward. 
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PROBLEM 12.128 (Continued)
(b) and (c)
First observe that for an arbitrary value of , it is not known whether the impending motion will be upward or
downward. To consider both possibilities for each value of , let Fdown correspond to impending motion
downward, Fup correspond to impending motion upward, then with the “top sign” corresponding to Fdown,
we have
Fy  0: N cos  F sin   WC  0
F  s N
Now
N cos  s N sin   mC g  0
Then
or
N
mC g
cos   s sin 
and
F
s mC g
cos   s sin 
Fn  mC an : N sin   F cos   mC
vC2

  r sin 
Substituting for N and F
mC g
s mC g
v2
sin  
cos   mC C
cos    s sin 
cos   s sin 
r sin 
s
vC2
tan 


1  s tan  1  s tan  gr sin 
or
or
s  
1
vC2
gr sin 
vC2
gr sin 
tan 
vC2
[(3.6 m/s)sin  ]2

 2.2018sin 
gr sin  (9.81 m/s 2 )(0.6 m)sin 
Now
Then
(b)
tan  
s  
tan   2.2018 sin 
1  2.2018sin  tan 
s  
tan 75  2.2018sin 75
  0.1796
1  2.2018sin 75 tan 75
  75
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PROBLEM 12.128 (Continued)
Then
downward:
s   0.1796
upward:
s  0
not possible
( s )min  0.1796 
The direction of the impending motion is downward. 
(c)
  45
s  
tan 45  2.2018sin 45
  (  0.218)
1  2.2018sin 45 tan 45
Then
downward:
s  0
upward:
s  0.218
not possible
( s )min  0.218 
The direction of the impending motion is upward. 
Note: When
or
tan   2.2018sin   0
  62.988,
s  0. Thus, for this value of  , friction is not necessary to prevent the collar from sliding on the rod.
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195
PROBLEM 12.129
Telemetry technology is used to quantify kinematic values of a
200-kg roller coaster cart as it passes overhead. According to
the system, r  25 m, r  10 m/s, r  2 m/s2 ,   90,
  0.4 rad/s,   0.32 rad/s 2. At this instant, determine
(a) the normal force between the cart and the track, (b) the
radius of curvature of the track.
SOLUTION
Find the acceleration and velocity using polar coordinates.
vr  r  10 m/s
v  r  (25 m)( 0.4 rad/s)  10 m/s
So the tangential direction is
45° and v  10 2 m/s.
ar  r  r 2  2 m/s  (25 m)( 0.4 rad/s) 2
 6 m/s 2
a  r  2r
 (25 m)(0.32) rad/s 2 | 2(10 m/s)( 0.4 rad/s)
0
So the acceleration is vertical and downward.
(a)
To find the normal force use Newton’s second law.
y-direction
N  mg sin 45  ma cos 45
N  m( g sin 45  a cos 45)
 (200 kg)(9.81) m/s 2  6 m/s 2 )(0.70711)
 538.815 N
N  539 N 
(b)
Radius l curvature of the track.
an 
 
v2

v2
(10 2)2

an
6cos 45
  47.1 m 
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PROBLEM 12.130
The radius of the orbit of a moon of a given planet is equal to twice the radius of that planet. Denoting
by  the mean density of the planet, show that the time required by the moon to complete one full revolution
about the planet is (24  /G  )1/2 , where G is the constant of gravitation.
SOLUTION
For gravitational force and a circular orbit,
Fr 
GMm mv 2

r
r2
or
v
GM
r
Let be the periodic time to complete one orbit.
v  2 r
Solving for ,


Then
GM
 2 r
r
hence,
GM  2
2 r 3/2
GM
4
M   R3  ,
3
But

or
3  r 
G   R 

G  R3/2
3
3/2
Using r 2R as a given leads to
  23/2
3

G
24
G
  (24 /G  )1/2 
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PROBLEM 12.131
At engine burnout on a mission, a shuttle had reached Point A at an
altitude of 80 km above the surface of the earth and had a horizontal
velocity v0. Knowing that its first orbit was elliptic and that the shuttle
was transferred to a circular orbit as it passed through Point B at an
altitude of 270 km, determine (a) the time needed for the shuttle to travel
from A to B on its original elliptic orbit, (b) the periodic time of the
shuttle on its final circular orbit.
SOLutiOn
For Earth,
R = 6370 km = 6.37 ¥ 106 m, g = 9.81 m/s2
GM = gR2 = (9.81)(6.37 ¥ 106 )2 = 3.98059 ¥ 1014 m3 /s2
(a)
For the elliptic orbit,
rA = 6370 + 80 = 6450 km = 6.45 ¥ 106 m
rB = 6370 + 270 = 6640 km = 6.64 ¥ 106 m
1
a = ( rA + rB ) = 6.545 ¥ 106 m
2
b = rA rB =
Using Eq. 12.39,
and
(6.45 ¥ 10 ) (6.545 ¥ 10 ) = 6.49733 ¥ 10
6
6
6
m
1 GM
= 2 + C cosq A
rA
h
1 GM
= 2 + C cosq B
rB
rB
But q B = q A + 180∞, so that cos q A = - cos q B
Adding,
1 1 rA + rB 2a 2GM
+ =
= 2 = 2
rA rB
rA rB
b
h
or
h=
Periodic time.
t=
t=
GMb2
a
2p ab 2p ab a 2p a3/ 2
=
=
h
GM
GMb2
2p (6.545 ¥ 106 )3/ 2
3.98059 ¥ 1014
= 5273.156 s = 1 hr 27 min 53sec
The time to travel from A to B is one half the periodic time
t AB = 2636.58 s
t AB = 43 min 56.6 sec b
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198
PROBLEM 12.131 (continued)
(b)
For the circular orbit,
a = b = rB = 6.64 ¥ 106 m
t circ =
2p a3/ 2 2p (6.64 ¥ 106 )3/2
=
= 5388 s
GM
3.98059 ¥ 1014
t circ = 1 h 29 min 48sec
t circ = 1 h 29 min 48 sec b
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199
PROBLEM 12.132
A space probe in a low earth orbit is inserted into an elliptic
transfer orbit to the planet Venus. Knowing that the mass of the
sun is 332.8 ¥ 103 times the mass of the earth and assuming that
the probe is subjected only to the gravitational attraction of the
sun, determine the value of f, which defines the relative position of
Venus with respect to the earth at the time the probe is inserted into
the transfer orbit.
SOLUTION
First determine the time t probe for the probe to travel from the earth to Venus. Now
1
t probe = t tr
2
where t tr is the periodic time of the elliptic transfer orbit. Applying Kepler’s Third Law to the orbits about the
sun of the earth and the probe, we obtain
atr3
t tr2
=
3
2
aearth
t earth
where
and
Then
1
atr = ( rE + rv )
2
1
= (150 ¥ 106 + 108 ¥ 106 ) km
2
= 129 ¥ 106 km
aearth ª rE
t probe
( Note: e earth = 0.0167)
1Êa ˆ
= Á tr ˜
2 Ë rE ¯
3/ 2
t earth
1 Ê 129 ¥ 106 km ˆ
= Á
2 Ë 150 ¥ 106 km ˜¯
3/ 2
(365.25 days)
= 145.649 days
= 12.5841 ¥ 106 s
In time t probe , Venus travels through the angle q v given by
q v = q v t probe =
vv
t probe
rv
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200
PROBLEM 12.132 (Continued)
Assuming that the orbit of Venus is circular (note: e venus = 0.0068), then, for a circular orbit
vv =
GM sun
rv
[Eq. (12.44)]
GM sun = G(332.8 ¥ 103 M earth )
Now
(
) using Eq. (12.30)
È 332.8 ¥ 10 ( gR ) ˘
˙
Í
2
= 332.8 ¥ 103 gRearth
qv =
Then
t probe
3
rv Í
Î
2
earth
rv
= t probe Rearth
1/ 2
˙
˚
(332.8 g ¥ 103 )1/ 2
rv3/ 2
Rearth = 6370 km = 6.37 ¥ 106 m
where
and
rv = 108 ¥ 106 km = 108 ¥ 109 m
Then
q v = (12.5841 ¥ 106 s)(6.37 ¥ 106 m)
(332.8 ¥ 103 ¥ 9.81 m/s2 )1/ 2
(108 ¥ 109 )3/2
= 4.0809 rad
= 233.818∞
Finally,
f = q v - 180∞ = 233.818∞ - 180∞ = 53.818∞
j = 53.8∞ b
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201
PROBLEM 12.133*
Disk A rotates in a horizontal plane about a vertical axis at the
constant rate 0  10 rad/s. Slider B has mass 1 kg and moves in a
frictionless slot cut in the disk. The slider is attached to a spring of
constant k, which is undeformed when r  0. Knowing that the slider
is released with no radial velocity in the position r  500 mm,
determine the position of the slider and the horizontal force exerted
on it by the disk at t  0.1 s for (a) k  100 N/m, (b) k  200 N/m.
SOLUTION
First we note
when
r  0, xsp  0  Fsp  kr
r0  500 mm  0.5 m
and
  0  12 rad/s
then
  0
Fr  mB ar :  Fsp  mB (r  r02 )
 k


 02  r  0
r 
 mB

F  mB a : FA  mB (0  2r0 )
(a)
k  100 N/m
Substituting the given values into Eq. (1)
100 N/m

r  
 (10 rad/s)2  r  0
 1 kg


r 0
Then
dr
 
r  0 and at t  0, r  0 :
dt

r
0
dr 

0.1
0
(0) dt
r  0
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(1)
(2)
PROBLEM 12.133* (Continued)
and
dr
 r  0 and at t  0, r0  0.5 m
dt

r
r0
dr 

0.1
0
(0) dt
r  r0
r  0.5 m 
Note: r  0 implies that the slider remains at its initial radial position.
With r  0, Eq. (2) implies
FH  0 
(b)
k  200 N/m
Substituting the given values into Eq. (1)
 200


r
 (10 rad/s) 2  r  0
1 kg


r  100 r  0
d
( r)
dt
Now

r
Then

r  vr
so that
d dr d
d

 vr
dt dt dr
dr
dvr
dr
dvr
 100r  0
dr
vr

At t  0, vr  0, r  r0 :
r  vr
vr
0
vr d vr  100

r
r0
r dr
vr2  100(r 2  r02 )
vr  10 r02  r 2
Now
At t  0, r  r0 :
vr 

r
r0
dr
r02  r 2

dr
 10 r02  r 2
dt
t
 10 dt  10t
0
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PROBLEM 12.133* (Continued)
r  r0 sin  ,
Let
sin 1 ( r/r0 )

Then
/2
dr  r0 cos  d
r0 cos  d
r02  r02 sin 2 
sin 1 ( r/r0 )

/2
 10t
d  10t
r 
sin 1     10 t
 r0  2


r  r0 sin 10 t    r0 cos10 t  (0.5 ft) cos10 t
2

r  (5 m/s)sin10 t
Then
Finally,
at t  0.1 s:
r  (0.5 ft) cos (10  0.1)
r  0.270 m 
Eq. (2)
FH  1 kg  2  [(5 ft/s)sin (10  0.1)] (10 rad/s)
FH  84.1 N 
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PROBLEM 12.CQ1
A 1000 N boulder B is resting on a 200 N platform A when truck
C accelerates to the left with a constant acceleration. Which of
the following statements are true (more than one may be true)?
(a) The tension in the cord connected to the truck is 200 N
(b) The tension in the cord connected to the truck is 1200 N
(c) The tension in the cord connected to the truck is greater
than 1200 N
(d ) The normal force between A and B is 1000 N
(e) The normal force between A and B is 1200 N
( f ) None of the above
SOLUTION
The boulder and platform are accelerating upwards; therefore it will require a force larger than their weight to
achieve this acceleration:
Answers: (c) The tension will be greater than 1200 N and (d) the normal force will be greater than 1000 N.
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PROBLEM 12.CQ2
Marble A is placed in a hollow tube, and the tube is swung in a horizontal
plane causing the marble to be thrown out. As viewed from the top, which
of the following choices best describes the path of the marble after leaving
the tube?
(a) 1
(b) 2
(c) 3
(d ) 4
(e) 5
SOLUTION
Answer: (d ) The particle will have velocity components along the tube and perpendicular to the tube when it
leaves. After it leaves, it will not accelerate in the horizontal plane since there are no forces in that plane;
therefore, it will travel in a straight line.
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PROBLEM 12.CQ3
The two systems shown start from rest. On
the left, two 200 N weights are connected
by an inextensible cord, and on the right,
a constant 200 N force pulls on the cord.
Neglecting all frictional forces, which of
the following statements is true?
(a) Blocks A and C will have the same
acceleration
(b) Block C will have a
acceleration than block A
larger
(c) Block A will have a
acceleration than block C
larger
(d ) Block A will not move
(e) None of the above
SOLUTION
Answer: (b) If you draw a FBD of B, you will see that since it is accelerating downward, the tension in the
cable will be less than 200 N, so the acceleration of A will be less than the acceleration of C. Also, the system
on the left has more inertia, so it will require more force to have the same acceleration as the system on the
right. 
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PROBLEM 12.CQ4
The system shown is released from rest in the position shown. Neglecting
friction, the normal force between block A and the ground is
(a) less than the weight of A plus the weight of B
(b) equal to the weight of A plus the weight of B
(c) greater than the weight of A plus the weight of B
SOLUTION
Answer: (a) SinceB has an acceleration component downward the normal force between A and the ground
will be less than the sum of the weights. To see this, you can draw an FBD of both block A and B. From
block A you will find that the normal force on the ground is the sum of the weight of block A plus the vertical
component of the normal force of block B on A since block A does not accelerate in the vertical direction.
From the FBD of block B and using F=ma, you will see that the vertical component of the normal force of B
on A is less than the weight of block B since it is accelerating downwards.
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PROBLEM 12.CQ5
People sit on a Ferris wheel at Points A, B, C and D. The Ferris wheel
travels at a constant angular velocity. At the instant shown, which
person experiences the largest force from his or her chair (back and
seat)? Assume you can neglect the size of the chairs, that is, the people
are located the same distance from the axis of rotation.
(a) A
(b) B
(c) C
(d ) D
(e) The force is the same for all the passengers.
SOLUTION
Answer: (c) Draw a FBD and KD at each location and it will be clear that the maximum force will be
experienced by the person at Point C. At this point the normal force isdirectly opposite the force of gravity
and the acceleration is in the same direction as the normal force making the normal force the sum of mg and
ma.
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PROBLEM 12.CQ6
A uniform crate C with mass mC is being
transported to the left by a forklift with a
constant speed v1. What is the magnitude
of the angular momentum of the crate
about Point D, that is, the upper left
corner of the crate?
(a) 0
(b) mv1a
(c) mv1b
(d ) mv1 a 2  b 2
SOLUTION
Answers: (b) The angular momentum is the moment of the momentum, so simply take the linear momentum,
mv1, and multiply it by the perpendicular distance from the line of action of mv1 and Point D.
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PROBLEM 12.CQ7
A uniform crate C with mass mC is being
transported to the left by a forklift with
a constant speed v1. What is the
magnitude of the angular momentum of
the crate about Point A, that is, the point
of contact between the front tire of the
forklift and the ground?
(a) 0
(b) mv1d
(c) 3mv1
(d ) mv1 32  d 2
SOLUTION
Answer: (b) The angular momentum is the moment of the momentum, so simply take the linear momentum,
mv1, and multiply it by the perpendicular distance from the line of action of mv1 and Point A.
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PROBLEM 12.F1
Crate A is gently placed with zero initial velocity onto a moving
conveyor belt. The coefficient of kinetic friction between the crate and
the belt is k. Draw the FBD and KD for A immediately after it
contacts the belt.
SOLUTION
Answer:
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PROBLEM 12.F2
Two blocks weighing WA and WB are at rest on a conveyor that is initially at
rest. The belt is suddenly started in an upward direction so that slipping
occurs between the belt and the boxes. Assuming the coefficient of friction
between the boxes and the belt is k,draw the FBDs and KDs for blocks A
and B. How would you determine if A and B remain in contact?
SOLUTION
Answer:
Block A
Block B
To see if they remain in contact assume aAaB and then check to see if NAB is greater than zero.
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PROBLEM 12.F3
Objects A, B, and C have masses mA, mB, and mC respectively.
The coefficient of kinetic friction between A and B is k, and
the friction between A and the ground is negligible and the
pulleys are massless and frictionless. Assuming B slides on A
draw the FBD and KD for each of the three masses A, B and C.
SOLUTION
Answer:
Block A
Block B
Block C
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PROBLEM 12.F4
Blocks A and B have masses mA and mB respectively. Neglecting friction
between all surfaces, draw the FBD and KD for each mass.
SOLUTION
Block A
Block B
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PROBLEM 12.F5
Blocks A and B have masses mA and mB respectively.
Neglecting friction between all surfaces, draw the FBD
and KD for the two systems shown.
SOLUTION
System 1
System 2
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PROBLEM 12.F6
A pilot of mass m flies a jet in a half vertical loop of radius R so that
the speed of the jet, v, remains constant. Draw a FBD and KD of the
pilot at Points A, B and C.
SOLUTION
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PROBLEM 12.F7
Wires AC and BC are attached to a sphere which revolves at a constant speed vin
the horizontal circle of radius r as shown. Draw a FBD and KD of C.
SOLUTION
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PROBLEM 12.F8
A collar of mass m is attached to a spring and slides without
friction along a circular rod in a vertical plane. The spring has
an undeformed length of 125 mm a nd constant k. Knowing
that the collar has a speed v at Point C,
C draw the FBD and KD
of the collar at this point.
SOLUTION
where x  0.05 m and r  0.125 m
m.
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219
PROBLEM12.F9
Four pins slide in four separate slots cut in a horizontal circular plate as
shown. When the plate is at rest, each pin has a velocity directed as shown
and of the same constant magnitude u. Each pin has a mass m and
maintains the same velocity relative to the plate when the plate rotates
about O with a constant counterclockwise angular velocity . Draw the
FBDs and KDs to determine the forces on pins P1 and P2.
SOLUTION
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PROBLEM 12.F10
At the instant shown, the length of the boom AB is being decreased at the
constant rate of 0.2 m/s, and the boom is being lowered at theconstant rate of
0.08 rad/s. If the mass of the men and lift connected to the boom at Point B is m,
draw the FBD and KD that could be used to determine the horizontal and
vertical forces at B.
SOLUTION
Where r  6 m, r  0.2 m/s, 
r  0,   0.08 rad/s,   0
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PROBLEM 12.F11
Disk A rotates in a horizontal plane about a vertical axis at the constant
rate 0 . Slider B has a mass m and moves in a frictionless slot cut in
the disk. The slider is attached to a spring of constant k, which is
undeformed when r  0. Knowing that the slider is released with no
radial velocity in the position r r0, draw a FBD and KD at an arbitrary
distance r from O.
SOLUTION
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PROBLEM 12.F12
Pin B has a mass m andslides along the slot in the rotating arm
OC andalong the slot DE which is cut in a fixed horizontal
plate. Neglectingfriction and knowing that rod OC rotates at the
constant rate 0 , draw a FBD and KD that can be used to
determine the forcesP and Q exerted on pin B by rod OC and
the wall of slot DE, respectively.
SOLUTION
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223