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444072556 Beer Vector Dynamics Ism Ch12
동역학 1 전필 인필 핵심 (Inha University)
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CHAPTER 12
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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 139 lb when they were on the moon, determine (a) the
weight of the rocks on the earth, (b) the mass of the rocks in slugs. The acceleration due to gravity on the
moon is 5.30 ft/s2.
SOLUTION
Since the rocks weighed 139 lb on the moon, their mass is
=
m
(a)
Wmoon
139 lb
=
= 26.226 lb ⋅ s 2 /ft
g moon 5.30 ft/s 2
On the earth, Wearth = mg earth
=
w (26.226 lb ⋅ s 2 /ft)(32.2 ft/s 2 )
(b)
Since 1 slug
= 1 lb ⋅ s 2 /ft,
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w = 844 lb 
m = 26.2 slugs 
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PROBLEM 12.2
The value of g at any latitude φ may be obtained from the formula
g 32.09(1 + 0.0053 sin 2φ ) ft/s 2
=
which takes into account the effect of the rotation of the earth, as well as the fact that the earth is not truly
spherical. Knowing that the weight of a silver bar has been officially designated as 5 lb, determine to four
significant figures (a) the mass is slugs, (b) the weight in pounds at the latitudes of 0°, 45°, and 60°.
SOLUTION
g 32.09(1 + 0.0053 sin 2φ ) ft/s 2
=
φ = 0° :
g = 32.09 ft/s 2
φ= 45°:
g = 32.175 ft/s 2
φ= 90°:
g = 32.26 ft/s 2
m=
(a) Mass at all latitudes:
(b)
Weight:
5.00 lb
32.175 ft/s 2
=
m 0.1554 lb ⋅ s 2 /ft 
W = mg
φ = 0° :
W=
(0.1554 lb ⋅ s 2 /ft)(32.09 ft/s 2 ) =
4.987 lb

φ= 45°:
W=
(0.1554 lb ⋅ s 2 /ft)(32.175 ft/s 2 ) =
5.000 lb

φ= 90°:
W=
(0.1554 lb ⋅ s 2 /ft)(32.26 ft/s 2 ) =
5.013 lb

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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 gravity at 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 
=
(25.6 × 106 m/h) 
7.111 × 103 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) 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
ΣF =ma : Fs − W =−
W
a
g

a
W 1 −  =
Fs
g

or
=
W
Fs
60
=
1
a
1−
1−
9.81
g
W = 66.8 N 
(b)
=
ΣF ma : Fs=
−W
W
a
g

a
Fs W 1 + 
=
g

1 

= 66.811 +

 9.81 
For the balance system B,
ΣM
=
0: bFw − bF
=
0
0
p
Fw = Fp
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Fs = 73.6 N 
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PROBLEM 12.4 (Continued)
But
a

=
Fw Ww 1 + 
g

and

a
=
Fp W p 1 + 
g

so that
Ww = W p
and
m
=
w
W p 66.81
=
g
9.81
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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 3 ft/s 2 while still on a level
section of the highway. Knowing that the speed of the bus is 60 mi/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 50 mi/h.
SOLUTION
First consider when the bus is on the level section of the highway.
alevel = 3 ft/s 2
W
alevel
g
Σ=
Fx ma: =
P
We have
Now consider when the bus is on the upgrade.
We have
=
ΣFx ma: P − W =
sin 7°
Substituting for P
W
a′
g
W
W
alevel − W sin 7° = a′
g
g
a′ =alevel − g sin 7°
or
=
(3 − 32.2 sin 7°) ft/s 2
= −0.92419 ft/s 2
For the uniformly decelerated motion
2
v 2 =(v0 ) upgrade
+ 2a′( xupgrade − 0)
 5 
=
mi/h 
v0  , we have
when v 50
Noting that 60 mi/h = 88 ft/s, then =
 6 
2
5

2
2
=
 6 × 88 ft/s
 (88 ft/s) + 2(−0.92419 ft/s ) xupgrade


or
xupgrade = 1280.16 ft
xupgrade = 0.242 mi 
or
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PROBLEM 12.6
A 0.2-lb model rocket is launched vertically from rest at time t = 0 with a constant thrust
of 2 lb 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
W = 0.2 lbs
For 0 ≤ t ≤ 1 s Ft =
2 lbs
Given:
Free Body Diagram:
For t > 1 s Ft =
0 lbs
For the thrust phase, 0 ≤ t ≤ 1 s :
ΣF = ma : Ft − W = ma =
F

a g  t − 1=
=

W


=
v at
=
At t = 1 s :
=
y
For free flight phase, t > 1 s :
( 32.2 ) 
2

2
− 1=
 289.8 ft/s
0.2


( 289.8)(1=
)
1 2
=
at
2
W
a
g
289.8 ft/s
1
2
=
1)
144.9 ft
( 289.8)(
2
a =− g =− 32.2 ft/s
v = v1 + a ( t − 1) = 289.8 + ( − 32.2 )( t − 1)
(a) At maximum height:
=
v 0,
t −=
1
289.8
= 9.00 s, =
t 10.00 s
32.2
v 2 − v12 =
2a ( y − y1 ) =
−2 g ( y − y1 )
0 − ( 289.8 )
v 2 − v12
y − y1 =
1304.1 ft
−
=
−
=
2g
( 2 )( 32.2 )
2
ymax= h= 1304.1 + 144.9
h = 1449 ft 
t = 10.00 s 
(b) Time to reach max height:
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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
W = 3600 kN
Given:
Free Body Diagram:
W
⇒ m = 366972.5 kg
g
m/s, v2 2.5 m/s
=
v1 1.0
=
m=
81000
v =
81000 - 27000v N
3
From Graph:
Fnet =
81000 −
From FBD:
ΣFx = max ⇒ Fnet = max ⇒ ax =
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
ln ( 3 − v ) |1
t =
−
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
v1
0
x
=
m
2.5
3 − v − 3ln 3 − v ) |1
(
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
Σ=
Fy 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 = µ=
g 0.80(9.81)
m
m
a = 7.848 m/s 2
=
a
=
v 2 2=
ax 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
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 2=
ax 2(4.709 m/s 2 )(60 =
m) 565.1 m 2 /s 2
v = 23.77 m/s
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PROBLEM 12.8 (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 2=
ax 2(3.139 m/s 2 )(60 =
m) 376.7 m 2 /s 2
v = 19.41 m/s
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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 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,
vf 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
= −0.70789W
(a)
Going up a 5° incline.
ΣF =
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
v 2f − v02
x f − x0 =
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
=
b 1.71835°
100
W
− Fb + W sin b = a
g
−(0.70789 − sin b )(9.81)
a=
=
tan b
= −6.65028 m/s
0 − (25)2
x=
x
=
0
f
(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.
ΣF =
ma :
x-direction:
mg sin 5° − µk N − T cos15° = ma
y-direction:
N − mg cos 5° + T sin15° = 0
From y-direction,
=
N mg cos 5° − T sin15
=
° (20 kg)(9.81 m/s 2 ) cos 5° − (7 N)sin15°
= 193.64 N
From x-direction,
µk N
T cos15°
m
m
(0.1)(193.64 N) (7 N) cos 15°
2
= (9.81 m/s )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
vf =
v0 + at
(a)
(b)
x0 = 0
v f − v0
−1 m/s
2.2159 s
t= =
=
a
−0.45128 m/s 2
t = 2.22 s 
Time elapsed.
Corresponding distance.
x = x0 + v0t +
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
trailer shown are µ s = 0.40 and µk = 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
= µs N
Deceleration of load is same as deceleration of trailer, which is the maximum allowable deceleration a max .
ΣF=
0: N − W
= 0 N
= W
y
=
Fm µ=
0.40 W
sN
=
ΣFx ma :=
Fm mamax
=
0.40 W
W
=
amax
amax 3.924 m/s 2
g
a max = 3.92 m/s 2
Uniformly accelerated motion.
=
v0 72
=
km/h 20 m/s
v2 =
v02 + 2ax with v =
0
a=
−amax =
3.924 m/s 2
=
0 (20) 2 + 2(−3.924) x
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x = 51.0 m 
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PROBLEM 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 the train before it
comes to a 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:
a = 1.333 m/s 2
ΣFx =
Σma: 60 × 103 N =
(45 × 103 kg)a
v 2 = v10 + 2ax: 0 = (25) 2 + 2(−1.333) x
x = 234 m 
Stopping distance:
(b)
Car A:
ΣFx = 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 incline, determine (a) the
acceleration of each block, (b) the tension in the cable.
SOLUTION
(a)
We note that aB =
1
aA .
2
Block A
=
ΣFx mA a A : T − (200 lb)sin
=
30°
200
aA
32.2
(1)
Block B
=
ΣFy mB aB : 350
=
lb − 2T
(a)
(2)
Multiply Eq. (1) by 2 and add Eq. (2) in order to eliminate T:
−2(200)sin 30=
° + 350 2
(b)
350  1

aA 
32.2  2

From Eq. (1), T − (200)sin 30° =
200
350  1 
aA +
aA
32.2
32.2  2 
150 =
575
aA
32.2
=
aB
1
1
=
aA
(8.40 ft/s 2 ),
2
2
a A = 8.40 ft/s 2
200
(8.40)
32.2
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30° 
a B = 4.20 ft/s 2 
T = 152.2 lb 
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PROBLEM 12.14
Solve Problem 12.13, assuming that the coefficients of
friction between block A and the incline are µ s = 0.25 and
µk = 0.20.
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 incline,
determine (a) the acceleration of each block, (b) the tension
in the cable.
SOLUTION
We first determine whether the blocks move by computing the friction force required to maintain block A in
equilibrium. T = 175 lb. When B in equilibrium,
=
ΣFx 0: 175 − 200sin 30°=
− Freq 0
Freq = 75.0 lb
=
ΣFy 0: N − 200 cos=
30° 0 =
N 173.2 lb
=
FM µ=
0.25(173.2=
lb) 43.3 lb
sN
Since Freq > Fm , blocks will move (A up and B down).
We note that aB =
1
aA .
2
Block A
=
F µ=
(0.20)(173.2)
= 34.64 lb.
kN
=
ΣFx mA 0 A : − 200sin 30° − 34.64
=
+T
200
aA
32.2
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PROBLEM 12.14 (Continued)
Block B
=
ΣFy mB aB : 350lb
=
− 2T
(a)
(2)
Multiply Eq. (1) by 2 and add Eq. (2) in order to eliminate T:
−2(200)sin 30° − 2(34.64)
=
+ 350 2
81.32 =
=
aB
(b)
350  1 
aA
32.2  2 
200
350  1 
aA +
aA
32.2
32.2  2 
575
aA
32.2
a A = 4.55 ft/s 2
1
1
(4.52 ft/s 2 ),
=
aA
2
2
From Eq. (1), T − (200)sin 30° − 34.64 =
30° 
aB = 2.28 ft/s 2 
200
(4.52)
32.2
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T = 162.9 lb 
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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 10 ft, (c) the time required for block
A to reach a velocity of 20 ft/s.
SOLUTION
Let y be positive downward for both blocks.
constant
Constraint of cable: y A + yB =
a A + aB =
0
For blocks A and B,
Block A:
Block B:
aB = − a A
ΣF =
ma :
W
WA − T = A a A
g
or
=
T WA −
WA
aA
g
W
W
P + WB − T = B aB =
− B aA
g
g
P + WB − WA +
Solving for aA,
or
WA
W
aA =
− B aA
g
g
aA =
WA − WB − P
g
WA + WB
(1)
v A2 − (v A )02 = 2a A [ y A − ( y A )0 ] with (v A )0 = 0
=
vA
2a A [ y A − ( y A )0 ]
(2)
=
v A − (v A )0 a At with
=
(v A )0 0
t=
vA
aA
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(3)
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PROBLEM 12.15 (Continued)
(a)
Acceleration of block A.
=
WA 200=
lb, WB 100
=
lb, P 0
System (1):
(a A )1 =
System (2):
=
WA 200 lb, =
WB 0,=
P 50 lb
By formula (1),
(a A )2 =
By formula (1),
(c)
200 − 100
(32.2)
200
(a A )1 = 10.73 ft/s 2 
(a A )2 = 16.10 ft/s 2 
=
WA 2200
=
lb, WB 2100
=
lb, P 0
System (3):
(b)
200 − 100
(32.2)
200 + 100
By formula (1),
( a A )3 =
2200 − 2100
(32.2)
2200 + 2100
(a A )3 = 0.749 ft/s 2 
v A at y A − ( y A )0 =
10 ft. Use formula (2).
System (1):
(v A )1 = (2)(10.73)(10)
(v A )1 = 14.65 ft/s 
System (2):
(v A ) 2 = (2)(16.10)(10)
(v A )2 = 17.94 ft/s 
System (3):
(v A )3 = (2)(0.749)(10)
(v A )3 = 3.87 ft/s 
Time at v A = 20 ft/s. Use formula (3).
System (1):
t1 =
20
10.73
t1 = 1.864 s 
System (2):
t2 =
20
16.10
t2 = 1.242 s 
System (3):
t3 =
20
0.749
t3 = 26.7 s 
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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 ( µk ) A = 0.30 and ( µk ) B = 0.32, determine the
initial acceleration of each box.
SOLUTION
Assume that aB > a A so that the normal force NAB between the boxes is zero.
A:
=
ΣFy 0: NA − WA cos
=
15° 0
A:
=
NA WA cos 15°
or
FA = ( µk ) A NA
Slipping:
= 0.3WA cos 15°
=
ΣFx mA a A : FA − =
WA sin 15° mA a A
0.3WA cos 15° − WA sin 15° =
or
WA
aA
g
=
a A (32.2 ft/s 2 )(0.3 cos 15° − sin 15°)
or
= 0.997 ft/s 2
B:
B:
=
ΣFy 0: N B − WB cos
=
15° 0
=
N B WB cos 15°
or
Slipping:
FB = ( µk ) B N B
= 0.32WB cos 15°
=
ΣFx mB aB : FB − W
=
B sin 15° mB aB
or
or
0.32WB cos 15° − WB sin 15° =
WB
aB
g
=
aB (32.2 ft/s 2 )(0.32
=
cos 15° − sin 15°) 1.619 ft/s 2
aB > a A ⇒ assumption is correct
a A = 0.997 ft/s 2
15° 
a B = 1.619 ft/s 2
15° 
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PROBLEM 12.16 (Continued)
Note: If it is assumed that the boxes remain in contact ( NAB ≠ 0), then assuming NAB to be compression,
=
a A aB
and=
find (ΣFx ma) for each box.
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 = 1.273 ft/s 2 and NAB = −0.859 lb, which contradicts the assumption.
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PROBLEM 12.17
A 5000-lb truck is being used to lift a 1000 lb boulder B that
is on a 200 lb pallet A. Knowing the acceleration of the truck
is
1 ft/s2, determine (a) the horizontal force between the tires
and the ground, (b) the force between the boulder and the
pallet.
SOLUTION
aT = 1 m/s 2
Kinematics:
a=
a=
0.5 m/s 2
A
B
5000
= 155.28 slugs
32.2
200
=
mA = 6.211 slugs
32.2
1000
=
mB = 31.056 slugs
32.2
=
mT
Masses:
Let T be the tension in the cable. Apply Newton’s second law to the lower pulley, pallet and boulder.
Vertical components
:
2T − (mA + mB ) g = (mA + mB )a A
2T − (37.267)(32.2) =
(37.267)(0.5)
T = 609.32 lb
Apply Newton’s second law to the truck.
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PROBLEM 12.17 (Continued)
Horizontal components
(a)
: F −T =
mT aT
Horizontal friction force between tires and ground.
F=
T + mT aT =
609.32 + (155.28)(1.0)
F = 765 lb 
Apply Newton’s second law to the boulder.
Vertical components + :
FAB − mB g =
mB aB
FAB= mB ( g + a=
) 31.056(32.2 + 0.5)
(b)
Contact force between boulder and pallet:
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FAB = 1016 lb 
lOMoARcPSD|16598870
PROBLEM 12.18
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
µ s = 0.20 and µk = 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=
a A + a B/A
B
Then
aB= a A + (−2a A )
or
aB = − a A
(1)
First we determine if the blocks will move for the given value of θ . Thus, we seek the value of θ for
which the blocks are in impending motion, with the impending motion of A down the incline.
B:
=
ΣFy 0: N AB − WB cos
=
θ 0
or
N AB = mB g cos θ
Now
FAB = µ s N AB
B:
= 0.2mB g cos θ
ΣFx= 0: − T + FAB + WB sin θ= 0
or
=
T mB g (0.2cos θ + sin θ )
A:
Σ=
Fy 0: N A − N AB − WA cos
=
θ 0
or
N
=
(mA + mB ) g cos θ
A
Now
FA = µ s N A
A:
= 0.2(mA + mB ) g cos θ
ΣFx= 0: − T − FA − FAB + WA sin θ= 0
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PROBLEM 12.18 (Continued)
T= mA g sin θ − 0.2(mA + mB ) g cos θ − 0.2mB g cos θ
or
= g[mA sin θ − 0.2(mA + 2mB ) cos θ ]
Equating the two expressions for T
mB g (0.2cos θ + sin=
θ ) g[mA sin θ − 0.2(mA + 2mB ) cos θ ]
8(0.2 + tan
=
θ ) [40 tan θ − 0.2(40 + 2 × 8)]
or
tan θ = 0.4
or
or=
θ 21.8° for impending motion. Since θ < 25°, the blocks will move. Now consider the
motion of the blocks.
(a)
=
ΣFy 0: N AB − WB =
cos 25° 0
B:
=
N AB mB g cos 25°
or
Sliding:
=
FAB µ=
0.15mB g cos 25°
k N AB
=
ΣFx 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)
=
ΣFy 0: N A − N AB − WA cos
=
25° 0
or
NA =
(mA + mB ) g cos 25°
Sliding:
FA =
µk N A =
0.15(mA + mB ) g cos 25°
=
ΣFx mA a A : − T − FA − FAB + WA sin
=
25° mA a A
Substituting and using Eq. (1)
T mA g sin 25° − 0.15(mA + mB ) g cos 25°
=
− 0.15mB g cos 25° − mA (−aB )
= g[mA sin 25° − 0.15(mA + 2mB ) cos 25°] + mA 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
µ s = 0.20 and µk = 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=
a A + a B/A
B
Then
aB= a A + (−2a A )
or
aB = − 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:
=
ΣFy 0: N AB − WB cos
=
25° 0
B:
=
N AB mB g cos 25°
or
FAB = µ s N AB
Now
= 0.2 mB g cos 25°
ΣF
=
0: − T + FAB + WB sin 25
=
° 0
x
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
=
ΣFy 0: N A − N AB − WA cos 25° + P sin
=
25° 0
N=
(mA + mB ) g cos 25° − P sin 25°
A
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PROBLEM 12.19 (Continued)
FA = µ s N A
Now
=
FA 0.2[(mA + mB ) g cos 25° − P sin 25°]
or
ΣF
=
0: − T − FA − FAB + WA sin 25° + P cos 25
=
° 0
x
−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[(mA + 2mB ) g cos 25°] − mA g sin 25°
Then
P(0.2 sin 25°=
+ cos 25°) 47.39249 N + 9.81 m/s 2 {0.2[(40 + 2 × 8) cos 25° − 40 sin 25°] kg}
P = −19.04 N for impending motion.
or
Since P, < 40 N, the blocks will move. Now consider the motion of the blocks.
(a)
=
ΣFy 0: N AB − WB cos
=
25° 0
B:
=
N AB mB g cos 25°
or
FAB = µk N AB
Sliding:
= 0.15 mB g cos 25°
=
ΣFx mB aB : − T + FAB + WB sin
=
25° mB aB
or
T mB [ g (0.15 cos 25° + sin 25°) − aB ]
=
= 8[9.81(0.15 cos 25° + sin 25°) − aB ]
= 8(5.47952 − aB )
(N)
A:
=
ΣFy 0: N A − N AB − WA cos 25° + P sin
=
25° 0
or
N=
(mA + mB ) g cos 25° − P sin 25°
A
Sliding:
FA = µk N A
= 0.15[(mA + mB ) g cos 25° − P sin 25°]
=
ΣFx mA a A : − T − FA − FAB + WA sin 25° + P cos
=
25° mA a A
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PROBLEM 12.19 (Continued)
Substituting and using Eq. (1)
=
T mA g sin 25° − 0.15[(mA + mB ) g cos 25° − P sin 25°]
− 0.15 mB g cos 25° + P cos 25° − mA (−aB )
= g[mA sin 25° − 0.15(mA + 2mB ) cos 25°]
+ P(0.15 sin 25° + cos 25°) + mA 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/s 2
(b)
We have
25° 
=
T 8[5.47952 − (−1.79383)]
T = 58.2 N 
or
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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.
Σ=
Fy 0: N1 − =
W 0
For the upper beam,
N=
W
= mg
1
Σ=
Fy 0: N 2 − N1 −=
W 0
For the lower beam,
ΣF=
ma : 0.25 N1 + 0.30 N=
x
2
=
a 0.85
=
g
( 0.85)( 9.81)
For the upper beam,
T = 0.25mg + ma =
a = 8.34 m/s 2
or
=
N 2 2=
W 2mg
( 0.25 + 0.60 ) mg=
ma

Σ
=
Fx ma : T − 0.25=
N1 ma
( 0.25)(1500 )( 9.81) + (1500 )(8.34 ) =
16.19 × 103 N
(b) Maximum deceleration of trailer.
Case 1: Assume that only the top beam has impending motion. As in Part (a) N1 = mg.
=
ΣF ma =
: 0.25mg ma
=
a 0.25
=
g 2.45 m/s 2
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T = 16.19 kN 
lOMoARcPSD|16598870
PROBLEM 12.20 (Continued)
Case 2: Assume that both beams have impending slip. As before N 2 = 2mg.
=
ΣF
( 2m=
) a : ( 0.30 )( 2mg ) ( 2m ) a
=
a 0.30
=
g 2.94 m/s 2
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 suitcase B.
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 suitcase B, 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 the bag A – suitcase B combination treated as a single particle.
=
ΣFy ma y : − (mB + mA ) g cos 20=
°+ N 0
=
N (mA + mB ) g cos 30
=
° (30)(9.81) cos 20
=
° 276.55 N
µ B N (0.3)(276.55)
=
= 82.965 N
Fx max : µ B N + (mA + mB ) g sin 20
Σ=
=
° (mA + mB )a
µB N
82.965
9.81sin 20° −
a g sin 20° + =
=
30
mA mB
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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PROBLEM 12.21 (Continued)
=
ΣFy 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=
mA a
A g sin 20° − FAB
=
FAB mA ( g sin
=
20° − a) (10)(9.81sin 20° − 0.58972)
= 27.655 N
Since bag A does not slide on suitcase B,
µs >
FAB 27.655
=
= 0.300
N AB 92.184
µ 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 µ s = 0.40 and µk = 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 µ=
0.40 N1
s N1
ΣFy = mP a y : N1 − WP cos 20° = −mP aT sin 20°
=
N1 mP ( g cos 20° − aT sin 20°)
sin 20° mP aT cos 20°
=
ΣFx max : F1 − WP=
=
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/s 2
(b)
xP/T = ( xP/T )o + (vP /T )t +
=
a P /T
1
1
aP / T t 2 = 0 + 0 + aP / T t 2
2
2
2 xP /T (2)(2)
=
= 4.94 m/s 2
t2
(0.9) 2
a P / T = 4.94 m/s 2
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 (Continued)
ΣFx = Σmax : 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 µ=
0.30 N 2
k N2
=
mP ( g sin 20° + aT cos
20° − aP /T ) 0.30mP ( g cos 20° + aT sin 20°)
aT =
(0.30 cos 20° − sin 20°) g + aP /T
cos 20° + 0.30sin 20°
=
(−0.05767)(9.81) + (0.9594)(4.94)
= 4.17
aT = 4.17 m/s 2
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PROBLEM 12.23
To transport a series of bundles of shingles A
to a roof, a contractor uses a motor-driven 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 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
F1 µ=
0.30 N1
Acceleration a1: Impending slip. =
s N1
N1=
− WA mA a1 sin 65°
ΣFy =
mA a y :
N1 =
WA + mA a1 sin 65°
=
mA ( g + a1 sin 65°)
=
ΣFx mA a=
mA a1 cos 65°
x : F1
F1 = µ 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 =
= 19.53 m/s 2
a1 = 19.53 m/s 2
F2 µ=
0.30 N 2
Deceleration a 2 : Impending slip. =
s N2
ΣFy =
ma y : N1 − WA =
−mA a2 sin 65°
N1 =
WA − mA a2 sin 65°
ΣFx =
max :=
F2 mA a2 cos 65°
F2 = µ s N 2
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65° 
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PROBLEM 12.23 (Continued)
or
mA a=
2 cos 65° 0.30 m A ( g − a2 cos 65°)
a2 =
0.30 g
cos 65° + 0.30 sin 65°
= (0.432)(9.81)
= 4.24 m/s 2
a 2 = 4.24 m/s 2
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65° 
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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 − 2.25v 2 =
(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
=
=
km/h 66.67 m/s
For v1 240
=
x1
25 × 103
40 × 103
=
ln
5.556 ln1.333
4.5
40 × 103 − 2.25(66.67) 2
= 1.5982 × 103 m
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x1 = 1.598 km 
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PROBLEM 12.25
A 4-kg projectile is fired vertically with an initial velocity of 90 m/s, reaches a maximum height and falls to
the ground. The aerodynamic drag D has a magnitude D = 0.0024 v 2 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.
ΣFy =
ma :
(a) Upward motion.
− D − mg =
ma

D
kv 2 

a=
− g +  =
−  g +

m
m 



dv
kv 2 
k
mg 
=
−  g +
−  v2 +
v
 =

dy
m 
m
k 

v dv
k
= − dy
mg
m
v2 +
k
∫
0
v dv
k
= −
mg
v0 2
m
v +
k
∫
h
dy
0
0
1  2 mg 
kh
ln  v +
−
 =
2 
k v
m
0
mg

1
1  kv 2
kh
k
ln
=
− ln  0 + 1 =
−
2 v 2 + mg
2  mg
m

0
k
=
h

m  kv02
+ 1
ln  =
2k  mg

= 335.36 m
 ( 0.0024 )( 90 )2

4
+ 1
ln 
( 2 )( 0.0024 )  ( 4 )( 9.81)

h = 335 m 
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PROBLEM 12.25 (Continued)
ΣF =
ma :
(b) Downward motion.
D − mg =
ma
D
kv 2
−g =
−g
m
m
dv
kv 2
k  mg

= − g =− 
− v2 
v
dy
m
m k

a =
v dv
k
= − dy
mg
m
− v2
k
∫
vf
0
v dv
k
= −
mg
m
∫
0
dy
h
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
e− 2 kh/m
=
(
mg
vf =
±
1 − e −2 kh/m
k
vf
=
)
( 4 )( 9.81) 1 − e−( 2)( 0.0024)(335.36)/4 
0.0024 
= 73.6 m/s
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
v f = 73.6 m/s 
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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
=
ΣF 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
− ln
=
−
t=
or
ln
k
P
m
m
P − kv
P
= e− kt/m
v
or =
(1 − e− kt/m )
m
k
∫
∫
∫
x=
=
=
x
Pt
v dt =
0
k
t
t
0
t
P k

−  − e − kt/m 
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 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 + x 2
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 θ =
By geometry,
x
 + x2
2
Σ=
Fx max : − Fs cos=
θ ma
− k (  2 + x 2 − )
k
x
a=
− x−
m 
2 + x 2
∫
v
0
v dv =
x
 + x2
2
=
ma




0
∫ 0 a dx
x
v
x
1 2
k 0
−
v =
x−

2
x
2 0
m 0
 + x2
∫
0

k 1 2
2
2 
−  x −  + x 
 dx =

m2
x

0
k
1 2
1

−  0 −  2 − x02 +   2 + x02 
v =
2
2
m

k
=
2 2 + x02 − 2  2 + x02
v2
m
k  2
=
 + x02 − 2 2 + x02 + 2 

m 
)
(
(
)
answer: v =
k
m
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(
)
2 + x02 −  
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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. Neglect
the masses of the pulleys and axle friction.
SOLUTION
Let the position coordinate y be positive downward.
y A + yD =
constant
Constraint of cord AD:
v A +=
vD 0,
( yB − yD ) + ( yC − yD ) =
constant
Constraint of cord BC:
vB + vC − 2=
vD 0,
Eliminate aD .
a A +=
aD 0
aB + aC − 2=
aD 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
Pulley D:
1
a B t 2 , ( vB ) 0 = 0
2
2[ yB − ( yB )0 ] (2)(3)
= =
1.5 m/s 2
t2
(2)2
=
ΣFy 0: 2TBC −=
TAD 0
TAD = 2TBC
Block A:
or
=
ΣFy ma y : WA=
− TAD mA a A
aA
=
WA − TAD WA − 2TBC
=
mA
mA
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PROBLEM 12.28 (Continued)
Block C:
=
ΣFy ma y : WC =
− TBC mC aC
aC =
or
WC − TBC
mC
(3)
Substituting the value for aB and Eqs. (2) and (3) into Eq. (1), and solving
for TBC ,
 WC − TBC 
 W − 2TBC 
2 A
0
=
 + aB + 
mA


 mC

 m g − 2TBC
2 A
mA

 mC g − TBC

 + aB + 
mC



0
=

 4
1 
+
3 g + aB

 TBC =
 mA mC 
 4 1
 10 + 5  TBC = 3(9.81) + 1.5 or TBC = 51.55 N


Block B:
(a)
=
ΣFy ma y : P + WB −=
TBC mB aB
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 2=
TBC (2)(51.55)
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PROBLEM 12.29
A 40-lb sliding panel is supported by rollers at B and C. A 25-lb 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:
40
T −F = a
g
ΣFx =
ma :
(1)
Counterweight A: Its acceleration has two components
a A = a P + a A /P = a → + a
=
ΣFx max=
: F
25
a
g
=
ΣFg mag : 25
=
−T
(2)
25
a
g
(3)
Adding (1), (2), and (3):
40 + 25 + 25
a
g
25
25
=
a =
g
(32.2)
90
90
T − F + F + 25 − T =
a = 8.94 ft/s 2

Substituting for a into (3):
25 − T =
25  25 
g
g  90 
T = 25 −
625
90
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T = 18.06 lb 
lOMoARcPSD|16598870
PROBLEM 12.29 (Continued)
(b)
Panel:
40
a
g
(1)
25
25 − T = a
g
(2)
T=
ΣFy =
ma :
Counterweight A:
ΣFy =
ma :
40 + 25
a
g
25
a=
g
65
T + 25 − T =
Adding (1) and (2):
a = 12.38 ft/s 2

Substituting for a into (1):
T
=
(c)
40  25  1000
g
=
g  65 
65
T = 15.38 lb 
Since panel is accelerated to the left, there is no force exerted by panel on counterweight and vice
versa.
Panel:
ΣFx =
ma :
T=
40
a
g
(1)
Counterweight A: Same free body as in Part (b):
ΣFy =
ma :
25
25 − T = a
g
(2)
Since Eqs. (1) and (2) are the same as in (b), we get the same answers:
a = 12.38 ft/s 2
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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:
constant
x A + 4 yB =
v A + 4vB =
0
a A + 4 aB =
0
(1)
Uniform Acceleration of handle A:
( v A )o2 +2a A ( xA − ( xA )o )
32 =
02 + 2a A ( 0.3 − 0 )
=
v A2
a A = 15 m/s 2
From (1):
From FBD:
=
aB 3.75 m/s 2 ↑
∑F
y
Free Body Diagram:
= mB aB
mB aB
4 P − mB g =
mB =
=
4P
g + aB
4 (100 )
kg
9.81 + 3.75
mB = 29.50 kg 
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PROBLEM 12.31
A 10-lb block B rests as shown on a 20-lb bracket A. The
coefficients of friction are µ s = 0.30 and µk = 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 weight of block C if block B is not to slide on
bracket A. (b) If the weight 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 xB 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 = ( xB − x A ) + ( xP − x A ) + yC + constant
Differentiating and noting that xP = 0,
vB − 2v A + vC =
0
−2a A + aB + aC =
0
Here, a A 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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PROBLEM 12.31 (Continued)
Block B:
Block C:
=
ΣFx max : 2T =
− FAB
Bracket A:
WB
aB
g
=
ΣFx max : F
=
AB − T
Block B:
WA
aA
g
(2)
(3)
+ ΣFy ma y : N AB
=
=
− WB 0
N AB = WB
or
=
ΣFy ma y : m
=
C −T
Block C:
WC
aC
g
(4)
Adding Eqs. (2), (3), and (4), and transposing,
W
WA
W
a A + B aB + C aC =
WC
g
g
g
(5)
Subtracting Eq. (4) from Eq. (3) and transposing,
W
WB
aB − C aC =
FAB − WC
g
g
(a)
No slip between A and B.
aB = a A
From Eq. (1),
a=
a=
a=
a
A
B
C
From Eq. (5),
For impending slip,
a=
WC g
WA + WB + WC
=
FAB µ=
µ sWB
s N AB
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PROBLEM 12.31 (Continued)
Substituting into Eq. (6),
(WB − WC )(WC g )
= µ sWB − WC
WA + WB + WC
Solving for WC ,
WC =
=
µ sWB (WA + WB )
WA + 2WB − µ sWB
(0.30)(10)(20 + 10)
20 + (2)(10) − (0.30)(10)
WC = 2.43 lbs 
(b)
WC = 2.6757 lbs
WC increased by 10%.
=
FAB µ=
µkWB
k N AB
Since slip is occurring,
W
WB
aB − C aC =µkWB − WC
g
g
Eq. (6) becomes
or
10aB − 2.6757aC = [(0.25)(10) − 2.6757](32.2)
(7)
With numerical data, Eq. (5) becomes
20a A + 10aB + 2.6757aC =
(2.6757)(32.2)
(8)
Solving Eqs. (1), (7), and (8) gives
=
a A 3.144
=
ft/s 2 , aB 0.881
=
ft/s 2 , aC 5.407 ft/s 2
a A = 3.14 ft/s 2

a B = 0.881 ft/s 2

aC = 5.41 ft/s 2
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PROBLEM 12.32
Knowing that µ = 0.30, determine the acceleration of each block when
m A = m B = m C.
SOLUTION
Given:
µ = 0.30
m=
m=
m=
m
A
B
c
Kinematics:
y A + 2 yB + xC =
constant
v A + 2vB + vC =
0
(1)
aC + a A + 2aB =
0
Free Body Diagram of Block A:
Equation of Motion:
∑F
y
= ma A
mg − T =
ma A
a A= 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
maC
0.3mg − T =
=
aC 0.3 g −
T
m
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Problem 12.32 (Continued)
Substitute (2), (3) and (4) → (1)
3.3 g −
T=
6T
0
=
m
1.1
mg
2
(5)
Substitute (5) → (2):
a A= g −
1.1
g
2
=
a A 0.45g ↓ 
Substitute (5) → (3):
aB= g −
2.2
g
2
=
aB 0.10g ↑ 
Substitute (5) → (4):
=
aC 0.3 g −
1.1
g
2
=
aC 0.25g ← 
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PROBLEM 12.33
Knowing that µ = 0.30, determine the acceleration of each block
when mA = 5 kg, mB = 30 kg, and mC = 15 kg.
SOLUTION
Given:
Kinematics:
=
µ 0.30,
=
g 9.81 m/s 2
=
mA 5=
kg, mB 30=
kg, mC 15 kg
y A + 2 yB + 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
mA g − T =
mA a A
a A= 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
mA mB mC
Substitute (5) → (2):
a A= g −
8.25
g
5
=
a A 6.377 m/s 2 ↑ 
Substitute (5) → (3):
aB= g −
16.5
g
30
=
aB 4.415 m/s 2 ↓ 
Substitute (5) → (4):
=
aC 0.3 g −
8.25
g
15
aC =
−2.453 m/s 2 ← 
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PROBLEM 12.34
A 25-kg block A rests on an inclined surface, and a 15-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 aB / A = −a A
(1)
20° ) + ( aB / A
First note that a B =a A + a B / A =( a A
Block B:
=
ΣFx
mB ( aB ) x : =
mB g sin 20° − N AB
mB a A + N=
AB
20° )
mB a A
mB g sin 20°
15 a A + N AB =
50.328
=
ΣFy
mB ( aB )=
: mB g cos 20° − T
y
mB aB / A=
+T
(2)
mB aB / A
mB g cos 20°
15 aB / A + T =
138.276
Block A:
=
ΣFx
mAa A : mA g sin=
20° + N AB + T
mAa A − N AB=
+T
(3)
m Aa A
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 a A =
−0.0740 m/s 2 , a A =
0.0740 m/s 2
20° 
From Eq. (1), aB / A = 0.0740 m/s 2
From Eq. (3), T = 137.2 N
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T = 137.2 N 
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PROBLEM 12.35
Block B of mass 10-kg rests as shown on the upper surface of a 22-kg
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:
=
ΣFx mA a A : WA sin 30=
° + N AB cos 40° mA a A
(a)
NAB =
or
22 ( a A − 12 g )
cos 40°
a A + a B /A , where a B/A is directed along the top surface of A.
Now we note: a=
B
B:
ΣFy′ = 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°
aA
or =
(9.81)(1.1 + cos 20° cos 40°)
= 6.4061 m/s 2
2.2 + cos 40° sin 50°
=
ΣFx′ mB ax′ : WB sin
=
20° mB aB/A − mB a A cos50°
or
=
aB/A g sin 20° + a A cos 50°
= (9.81sin 20° + 6.4061cos 50°) m/s 2
= 7.4730 m/s 2
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PROBLEM 12.35 (Continued)
Finally
a=
a A + a B/ A
B
We have
aB2 = 6.40612 + 7.47302 − 2(6.4061 × 7.4730) cos 50°
or
aB = 5.9447 m/s 2
and
or
7.4730 5.9447
=
sin α
sin 50°
=
α 74.4°
a B = 5.94 m/s 2
(b)
75.6° 
Note: We have uniformly accelerated motion, so that
v= 0 + at
Now
At t = 0.5 s:
or
v B/A = v B − v A = a B t − a At = a B/At
=
vB/A 7.4730 m/s 2 × 0.5 s
v B/A = 3.74 m/s
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20° 
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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 θ that the cord
forms with pole BC, (b) the tension in the cord.
SOLUTION
a=
a=
A
n
First we note
ρ
ρ = l AB sin θ
where
(a)
v A2
=
ΣFy 0: TAB cos
=
θ − WA 0
TAB =
or
mA g
cos θ
=
ΣFx mA a=
mA
A : TAB sin θ
v A2
ρ
Substituting for TAB and ρ
mA g
v A2
sin θ = mA
l AB sin θ
cos θ
sin 2 θ = 1 − cos 2 θ
(4 m/s)2
1 − cos 2 θ =
cos θ
1.8 m × 9.81 m/s 2
or
cos 2 θ + 0.906105cos θ − 1 =0
cos θ = 0.64479
Solving
=
θ 49.9° 
or
(b)
From above
=
TAB
mA g 0.450 kg × 9.81 m/s 2
=
cos θ
0.64479
TAB = 6.85 N 
or
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PROBLEM 12.37
During a hammer thrower’s practice swings, the 7.1-kg head
A of the hammer revolves at a constant speed v in a horizontal
circle as shown. If ρ = 0.93 m and θ= 60°, determine
(a) the tension in wire BC, (b) the speed of the hammer’s
head.
SOLUTION
First we note
v A2
a=
a=
A
n
ρ
(a)=
ΣFy 0: TBC =
sin 60° − WA 0
or
7.1 kg × 9.81 m/s 2
sin 60°
= 80.426 N
TBC =
TBC = 80.4 N 
a A : TBC cos 60° mA
=
ΣFx mA=
(b)
or
v A2 =
v A2
ρ
(80.426 N) cos 60° × 0.93 m
7.1 kg
v A = 2.30 m/s 
or
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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. Knowing that the
centrifuge starts from rest and has a constant angular acceleration of
1.5 RPM per second until the pilot is pulling 9g’s 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 θ of the normal force once the pilot
reaches 9 g’s. Assume that the force parallel to the seat is zero.
SOLUTION
Given:
RPM/s
=
α 1.5
=
ω0 = 0
0.157 rad/s 2
N = 9 mg
R=7 m
Free Body Diagram of Pilot:
Equations of Motion:
∑F
y
= ma y
n
N sin θ − mg =
m ( 0)
N sin θ = mg
∑F
(1)
= man
N cos θ = mRω 2
ω=
N cos θ
mR
(2)
Substitute N=9mg into (1): 9mg sin θ = mg
1
 
θ 6.379°
=
θ = sin −1  
9
Substitute N=9mg and θ into (2):
9*9.81cos 6.379°
7
ω = 3.540 rad/s
ω=
For constant angular acceleration:
ω
= ω0 + α t
3.540= 0 + 0.157 * t
t = 22.55 s 
(a) Solving for t:
=
θ 6.379° 
From earlier:
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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
=
ΣFx ma: T (sin 30=
° + sin 45°)
mv 2
ρ
(1)
=
ΣFy 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
=
cos 30° + cos 45° ρ g
v 2 0.76733
m/s 2 ) 12.044 m 2 /s 2
=
=
ρ g 0.76733 (1.6 m)(9.81
=
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v = 3.47 m/s 
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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
ρ
(1)
TAC cos30° + TBC cos 45° = mg
(2)
Subtract Eq. (1) from Eq. (2). Since sin 45° = cos 45°, we obtain
°) mg −
TAC (cos 30° − sin 30=
mv 2
(3)
ρ
Multiply Eq. (1) by cos 30°, Eq. (2) by sin 30°, and subtract:
° sin 30°)
TBC (sin 45° cos 30° − cos 45=
=
sin15°
TBC
mv 2
ρ
mv 2
ρ
cos 30° − mg sin 30°
cos 30° − mg sin 30°
(4)
Making TAC = 60 N, m = 5=
kg, ρ 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
0.32
v12 = 8.668,
We have TAC ≤ 60 N for v ≥ v1 , that is, for
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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, ρ 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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PROBLEM 12.41
A 1-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=
dx
2
Given:
y=
Free Body Diagram of Sphere:
Equations of Motion:
tan θ
θ= 45°
or
=
+ ↑ ΣFy
0: N cos
=
θ − mg 0
mg
cosθ
N =
=
ΣFn
man=
: N sin θ
man
mg tan θ = m
v2
r
v 2 = gr tan θ
(a) v 2
=
(b) N
=
9.81)(1)(1.0000 )
(=
mg
=
cos 45°
9.81 m 2 /s 2
(1)( 9.81)
cos 45°
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v = 3.13 m/s 
N = 13.87 N 
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PROBLEM 12.42*
As part of an outdoor display, a 12-lb 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 26 lb.
SOLUTION
a=
a=
C
n
First note
vC2
ρ
ρ = 3 ft
where
=
ΣFx mC aC : TCA sin =
40° + TCB sin15°
WC vC2
g ρ
=
ΣFy 0: TCA cos 40° − TCB cos15
=
° − WC 0
Note that Eq. (2) implies that
(a)
when
=
TCB (T
=
(TCA )max
CB ) max , TCA
(b)
when
=
TCB (T
=
TCA (TCA )min
CB ) min ,
Case 1: TCA is maximum.
Let
Eq. (2)
or
TCA = 26 lb
(26 lb) cos 40° − TCB cos15° − (12 lb) = 0
TCB = 8.1964 lb
(TCB )(TCA )max < 26 lb
OK
[(TCB ) max = 8.1964 lb]
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(2)
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PROBLEM 12.42* (Continued)
Eq. (1)
=
(vC2 )(TCA )max
(32.2 ft/s 2 )(3 ft)
(26sin 40° + 8.1964 sin15°) lb
12 lb
(vC )(TCA )max = 12.31 ft/s
or
(cos15°)(Eq. 1) + (sin15°)(Eq. 2)
Now we form
40° sin15°
TCA sin 40° cos15° + TCA cos=
WC vC2
cos15° + WC sin15°
g ρ
=
TCA sin 55°
WC vC2
cos15° + WC sin15°
g ρ
or
(3)
(vc )max occurs when TCA = (TCA )max
(vC )max = 12.31 ft/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).
TCA cos 40° − (12 lb) = 0
Eq. (2)
TCA = 15.6649 lb, 26 lb OK
or
Note: Eq. (3) implies that when TCA = (TCA ) min , vC = (vC )min . Then
Eq. (1)
=
(vC2 ) min
(32.2 ft/s 2 )(3 ft)
(15.6649 lb)sin 40°
12 lb
(vC )min = 9.00 ft/s
or
0 < TCA ≤ TCB < 6 lb when
9.00 ft/s < vC < 12.31 ft/s 
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PROBLEM 12.43*
The 1.2-lb flyballs of a centrifugal governor revolve at a constant speed v
in the horizontal circle of 6-in. 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 17 lb.
SOLUTION
v2
First note
=
a a=
n
where
ρ = 0.5 ft
ρ
W v2
g ρ
(1)
=
ΣFy 0: TDA cos 20° − TDE=
cos 30° − W 0
(2)
=
ΣFx ma : TDA sin 20°=
+ TDE sin 30°
Note that Eq. (2) implies that
(a)
when
=
TDE (T
=
TDA (TDA )max
DE ) max ,
(b)
when
=
TDE (T
=
TDA (TDA )min
DE ) min ,
Case 1: TDA is maximum.
Let
Eq. (2)
or
Now let
Eq. (2)
or
TDA = 17 lb
(17 lb) cos 20° − TDE cos30° − (1.2 lb) = 0
=
TDE 17.06 lb unacceptable (> 17 lb)
TDE = 17 lb
TDA cos 20° − (17 lb) cos30° − (1.2 lb) = 0
=
TDA 16.9443 lb OK ( ≤ 17 lb)
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PROBLEM 12.43* (Continued)
(TDA ) max = 16.9443 lb
(TDE )max = 17 lb
(v 2 )(TDA )max
Eq. (1) =
(32.2 ft/s 2 )(0.5 ft)
(16.9443sin 20° + 17 sin 30°) lb
1.2 lb
v(TDA )max = 13.85 ft/s
or
(cos 30°) × [Eq. (1)] + (sin 30°) × [Eq. (2)]
Now form
20° sin 30°
TDA sin 20° cos 30° + TDA cos =
W v2
cos 30° + W sin 30°
g ρ
T=
DA sin 50°
W v2
cos 30° + W sin 30°
g ρ
or
(3)
vmax occurs when TDA = (TDA )max
vmax = 13.85 ft/s
Case 2: TDA is minimum.
Because (TDA ) min occurs when TDE = (TDE )min ,
let TDE = 0.
Eq. (2)
TDA cos 20° − (1.2 lb) = 0
or
TDA = 1.27701 lb, 17 lb
OK
Note: Eq. (3) implies that when TDA = (TDA ) min , v = vmin . Then
=
(v 2 )min
Eq.
(1)
(32.2 ft/s 2 ) (0.5 ft)
(1.27701 lb)sin 20°
1.2 lb
vmin = 2.42 ft/s
or
0 < TAB , TBC , TAD , TDE < 17 lb
2.42 ft/s < v < 13.85 ft/s 
when
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PROBLEM 12.44
A 130-lb wrecking ball B is attached to a 45-ft-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 13.2 ft/s.
SOLUTION
(a)
At C, the top of the swing, vB = 0; thus
=
an
vB2
=
0
LAB
=
ΣFn 0: TBA − WB cos
=
20° 0
TBA= (130 lb) × cos 20°
or
TBA = 122.2 lb 
or
(b)
=
ΣFn man : TBA=
− WB mB
or
(vB ) 2D
LAB
 130 lb
TBA (130 lb) + 
=
2
 32.2 ft/s
  (13.2 ft/s)
  45 ft

2

 
 
TBA = 145.6 lb 
or
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PROBLEM 12.45
During a high-speed chase, a 2400-lb sports car traveling at a
speed of 100 mi/h just loses contact with the road as it reaches
the crest A of a hill. (a) Determine the radius of curvature ρ of
the vertical profile of the road at A. (b) Using the value of ρ
found in part a, determine the force exerted on a 160-lb driver
by the seat of his 3100-lb car as the car, traveling at a constant
speed of 50 mi/h, passes through A.
SOLUTION
(a)
Note:
100 mi/h = 146.667 ft/s
Wcar v A2
g ρ
=
ΣFn man=
: Wcar
or
(146.667 ft/s)2
32.2 ft/s 2
= 668.05 ft
ρ=
ρ = 668 ft 
or
(b)
⇒ at 0; 50=
mi/h 73.333 ft/s
Note: v is constant=
=
ΣFn man : W=
−N
or
W v A2
g ρ


(73.333 ft/s)2
=
N (160 lb) 1 −

2
 (32.2 ft/s )(668.05 ft) 
N = 120.0 lb 
or
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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-kg passenger as the airplane
passes point B.
SOLUTION
8°
π
=
0.13963 rad
180°
Angle change over arc AB.
θ
=
=
( 6000 )( 0.13963
)
ρθ
=
Length of arc: s=
AB
837.76 m
sBC= 800 m, s AC= 837.76 + 800= 1637.76 m
160
=
∫180 v dv
1637.76
at
0
∫
1602 1802
or
=
−
2
2
ds
at (1637.76 )
at = −2.076 m/s 2
837.76
vB
∫180 v dv =
∫0 at ds
vB2
=
vB2 1802
−
=
( −2.076 )(837.76 )
2
2
or
28922
m 2 /s 2
vB 170.06 m/s
=
90 )( 9.81)
(=
Weight of passenger:
=
mg
882.9 N
Just before
point B. v 170.06
=
=
m/s, ρ
6000 m
an
=
Free Body Diagram of Pilot:
v2
=
ρ
(170.06=
)2
6000
4.820 m/s 2
ΣFn =N1 − W =
−m ( an )1 : N1 =
882.9 − ( 90 )( 4.820 ) =
449.1 N
ΣFt =Ft =mat :
Just after point B.
( Ft )1
=( 90 )( −2.076 ) =−186.8 N
v=
170.06 m/s,
Σ=
Fy
0 : N 2 −=
W
0
ρ =
∞, an =
0
N=
W
= 882.9 N
2
ΣFx =
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 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 ( µk = 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.
SOLUTION
=
ΣFn man : N =
− mg m
(a)
v2
ρ

v2 
N m  g + 
=
ρ


v2
µk m  g +
F µ=
=
kN
ρ




=
ΣFt mat=
: F mat
a=
t
Given data:

F
v2 
= µk  g + 
ρ
m

=
µk 0.20,
=
v 72=
km/h 20 m/s
=
g 9.81
=
m/s 2 , ρ 30 m

(20) 2 
=
at 0.20 9.81 +

30 

(b)
an = 0
at = 4.63 m/s 2 
ΣFn= man= 0: N − mg= 0
N = mg
=
F µ=
µk mg
k N
=
ΣFt mat=
: F mat
a=
t
F
g 0.20(9.81)
= µk =
m
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at = 1.962 m/s 2 
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PROBLEM 12.47 (Continued)
(c)
=
ΣFn man : mg
=
−N
mv 2
ρ

v2 
=
N m  g − 
ρ


v2 
µ
=
−
F µ=
N
m
g

k
k 

ρ 

=
ΣFt mat=
: F mat

F
v2
at = =µk  g −
m
ρ



(20)2 
 =0.20 9.81 −

45 


at = 0.1842 m/s 2 
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PROBLEM 12.48
A spherical-cap governor is fixed to a vertical shaft that rotates with angular
velocity ω. When the string-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:
ωmin 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 θ =
60° and α =
30°
R = 0.3*cos α
Free Body Diagram of Clapper:
Equations of Motion:
∑F
y
= ma y
n
FA cos θ − mg =
m ( 0)
FA =
Substitute (1)into (2):
ωwin =
mg
cos θ
∑F
(1)
= man
2
FA sin θ = mRωmin
ωwin =
FA sin θ
mR
(2)
m g tan θ
mR
9.81tan 60°
0.3cos 30°
= 8.087 rad/s
=
ωwin =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 θ defining the
Point B where the packages first slip relative to the belt.
SOLUTION
Assume package does not slip.
=
at 0, F f ≤ µ s N
On the curved portion of the belt
a=
n
v 2 (1 m/s)2
=
= 4 m/s 2
ρ 0.250 m
For any angle θ
mv 2
ΣFy =
−man =
−
ma y : N − mg cos θ =
ρ
=
N mg cos θ −
mv 2
ρ
(1)
ΣFx = max : −F f + mg sin θ = mat = 0
F f = mg sin θ
(a)
At Point A,
θ = 0°
=
N (0.5)(9.81)(1.000) − (0.5)(4)
(b)
At Point B,
(2)
Ff = µs N
mg sin θ µ s (mg cos θ − man )
=

an 
4 

sin
=
θ µ s  cos θ − =
 0.40 cos θ −
9.81 
g 


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PROBLEM 12.49 (Continued)
Squaring and using trigonometic identities,
θ 0.16cos 2 θ − 0.130479cos θ + 0.026601
1 − cos 2=
1.16cos 2 θ − 0.130479cos θ − 0.97340 =
0
cos θ = 0.97402
=
θ 13.09° 
Check that package does not separate from the belt.
N
=
F f mg sin θ
=
µs
µs
N > 0.
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PROBLEM 12.50
A 54-kg pilot flies a jet trainer in a half vertical loop of 1200-m
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 exerted 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.
ΣFn man : N=
m
At A: =
A −W
or
v A2
ρ
 1680 N

=
v A2 (1200 m) 
− 9.81 m/s 2 
 54 kg

= 25,561.3 m 2 /s 2
ΣFn man : N=
m
At C: =
C +W
or
vC2
ρ
 350 N

=
vC2 (1200 m) 
+ 9.81 m/s 2 
 54 kg

= 19,549.8 m 2 /s 2
Since at = constant, we have from A to C
vC2 =v A2 + 2at ∆s AC
or
or
19,549.8
=
m 2/s 2 25,561.3 m 2 /s 2 + 2at (π × 1200 m)
at = −0.79730 m/s 2
Then from A to B
vB2 =v A2 + 2at ∆s AB
π

= 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:
=
ΣFn man=
: NB m
vB2
ρ
22,555 m 2 /s 2
1200 m
or
N B = 54 kg
or
N B = 1014.98 N
=
ΣFt mat : W =
+ PB m | at |
or
or
Finally,
=
PB (54 kg)(0.79730 − 9.81) m/s 2
PB = 486.69 N
( Fpilot ) B =
N B2 + PB2 =
(1014.98) 2 + (486.69) 2
= 1126 N
(Fpilot ) B = 1126 N
or
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25.6° 
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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 v0 at 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
ΣF =
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° 


v0 = 12.19 m/s 
20°.
mv02
cos 20°.
R
Normal=
force:
N (80)(9.81)sin 20° +
80(12.1898) 2
=
cos 20° 2294 N
5.513
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PROBLEM 12.52
A curve in a speed track has a radius of 1000 ft and a rated speed of 120 mi/h.
(See Sample Problem 12.7 for the definition of rated speed). Knowing that a
racing car starts skidding on the curve when traveling at a speed of 180 mi/h,
determine (a) the banking angle θ , (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
a=
Acceleration
v2
ρ
=
ΣFx max : F + W sin
=
θ ma cos θ
=
F
mv 2
ρ
cos θ − mg sin θ
(1)
=
ΣFy ma y : N − W cos
=
θ ma sin θ
=
N
(a)
mv 2
ρ
sin θ + mg cos θ
(2)
Banking angle. Rated =
speed v 120
=
mi/h 176 ft/s. F = 0 at rated speed.
=
0
mv 2
ρ
cos θ − mg sin θ
v2
(176) 2
=
= 0.96199
ρ g (1000) (32.2)
=
θ 43.89°
tan=
θ
(b)
Slipping outward.
=
θ 43.9° 
=
v 180
=
mi/h 264 ft/s
F
= µN
µ=
µ
=
F v 2 cos θ − ρ g sin θ
=
N v 2 sin θ + ρ g cos θ
(264) 2 cos 43.89° − (1000) (32.2)sin 43.89°
(264) 2 sin 43.89° + (1000) (32.2) cos 43.89°
= 0.39009
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PROBLEM 12.52 (Continued)
(c)
Minimum speed.
F = −µ N
v 2 cos θ − ρ g sin θ
−µ =2
v sin θ + ρ g cos θ
ρ g (sin θ − µ cos θ )
v2 =
cos θ + µ sin θ
=
(1000) (32.2) (sin 43.89° − 0.39009 cos 43.89°)
cos 43.89° + 0.39009 sin 43.89°
= 13,369 ft 2 /s 2
v = 115.62 ft/s
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v = 78.8 mi/h 
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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 100 mi/h on a curved section of track banked
through an angle θ = 6° and with a rated speed of 60 mi/h,
determine (a) the magnitude of the side force felt by a passenger
of weight W in a standard car with no tilt (φ = 0), (b) the
required angle of tilt φ if the passenger is to feel no side force.
(See Sample Problem 12.7 for the definition of rated speed.)
SOLUTION
Rated speed:
=
vR 60
=
mi/h 88 ft/s, 100
=
mi/h 146.67 ft/s
From Sample Problem 12.6,
vR2 = g ρ tan θ
=
ρ
or
vR2
(88) 2
=
= 2288 ft
g tan θ 32.2 tan 6°
Let the x-axis be parallel to the floor of the car.
=
ΣFx max : Fs + W sin (θ=
+ φ ) man cos (θ + φ )
=
(a)
mv 2
ρ
cos (θ + φ )
φ = 0.
 v2

cos (θ + φ ) − sin (θ + φ ) 
Fs W 
=
 gρ

 (146.67)2

cos 6° − sin 6°
= W
 (32.2)(2288)

= 0.1858W
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PROBLEM 12.53 (Continued)
(b)
For Fs = 0,
v2
cos (θ + φ ) − sin (θ + φ ) =
0
gρ
v2
(146.67)2
=
= 0.29199
g ρ (32.2)(2288)
θ +=
φ 16.28°
=
φ 16.28° − 6°
tan (θ + φ ) =
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=
φ 10.28° 
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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 φ if passengers are to feel side forces
equal to 10% of their weights.
SOLUTION
=
vR 60
=
mi/h 88 ft/s, 100
=
mi/h 146.67 ft/s
Rated speed:
From Sample Problem 12.6,
vR2 = g ρ tan θ
or
=
ρ
vR2
(88) 2
=
= 2288 ft
g tan θ 32.2 tan 6°
Let the x-axis be parallel to the floor of the car.
=
ΣFx max : Fs + W sin (θ=
+ φ ) man cos (θ + φ )
=
Solving for Fs,
Now
So that
Let
Then
mv 2
ρ
cos (θ + φ )
 v2

cos (θ + φ ) − sin (θ + φ ) 
=
Fs W 
 gρ

v2
(146.67)2
= = 0.29199
=
and Fs 0.10W
g ρ (32.2)(2288)
=
0.10W W [0.29199 cos (θ + φ ) − sin (θ + φ )]
=
u sin (θ + φ )
cos (θ + φ ) = 1 − u 2
0.10 = 0.29199 1 − u 2 − u or 0.29199 1 − u 2 = 0.10 + u
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PROBLEM 12.54 (Continued)
Squaring both sides,
0.08526(1 − u 2 ) = 0.01 + 0.2u + u 2
or
1.08526u 2 + 0.2u − 0.07526 =
0
The positive root of the quadratic equation is u = 0.18685
Then,
θ +=
φ sin −1 =
u 10.77°
=
φ 10.77° − 6°
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=
φ 4.77° 
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PROBLEM 12.55
A 3-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 m, determine the maximum
allowable velocity v of the block.
SOLUTION
Let b be the slope angle of the dish. tan=
b
At =
r 2 m, tan b= 1
or
dy
=
dr
1
r
2
b= 45°
Draw free body sketches of the sphere.
=
ΣFy 0: N cos b − µ S N sin =
b − mg 0
N =
mg
cos b − µ S sin b
=
ΣFn man: N sin b + µ
=
S N cos b
mv 2
ρ
mg (sin b + µ S N cos b ) mv 2
=
cos b − µ S sin b
ρ
sin b + µ S cos b
sin 45° + 0.5cos 45°
=
(2)(9.81) = 58.86 m 2 /s 2
v 2 ρ=
g
cos b − µ S sin b
cos 45° − 0.5sin 45°
v = 7.67 m/s 
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PROBLEM 12.56
A polisher is started so that the fleece along the circumference
undergoes a constant tangential acceleration of 4 m/s 2 . Three
seconds after a polisher is started from rest, small tufts of
fleece from along the circumference of the 225-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/s 2 )(3 s)
v = 12.00 m/s 
or
(b) =
ΣFt mat=
: Ft mat
or
=
Ft (1.6 × 10−6 kg)(4 m/s 2 )
= 6.4 × 10−6 N
: Fn m
=
ΣFn man=
At t = 3 s:
=
Fn (1.6 × 10−6 kg)
v2
ρ
(12 m/s)2
m)
( 0.225
2
= 2.048 × 10−3 N
Finally,
F
=
tuft
Ft 2 + Fn2
= (6.4 × 10−6 N) 2 + (2.048 × 10−3 N) 2
F=
2.05 × 10−3 N 
tuft
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 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.
vB= 0 + ( a B ) t t
At t = 10 s:
=
vB (0.24
=
m/s 2 )(10 s) 2.4 m/s
In the plane of the turntable
=
ΣF mB a B=
: F mB (a B )t + mB (a B )n
Then
=
F mB (aB )t2 + (aB )n2
= mB (aB )t2 +
( )
vB2
2
ρ
+ ΣF=
0: N − W
= 0
y
or
At t = 10 s:
Then
N = mB g
=
F µ=
µ s mB g
sN
 v2
=
µ s mB g mB (aB )t2 +  ρB




2
1/ 2
or
1
=
µs
9.81 m/s 2
2

 (2.4 m/s)2  

2 2
+
(0.24
m/s
)


 
 2.5 m  

µ s = 0.236 
or
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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
=
m 80
=
kg, v0 13 m/s
Given:
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=
=
13
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
9.81tan 70° ( 5 + h cos 70° )
h = 3.714 m 
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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
=
m 80 kg,
=
µ s 0.2,
=
h 1.5 m
Given:
Geometry from figure:
5 + 1.5cos 70°
R=
= 5.513 m
Minimum and Maximum speeds occur if slipping is impending:
f = µs 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° + µ s N sin 70° − mg = 0
mg
cos 70° + µ s sin 70°
N=
N = 1480.9 N
∑F
n
= man
2
mvmin
R
2
mvmin
N sin 70° − µ s N cos 70° =
R
N sin 70° − f cos 70° =
vmin =
N ( sin 70° − µ 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° − µ s N sin 70° − mg = 0
mg
cos 70° − µ s sin 70°
N=
N = 5093.4 N
∑F
n
= man
2
mvmin
R
2
mvmin
N sin 70° + µ s N cos 70° =
R
N sin 70° + f cos 70° =
vmin =
N ( sin 70° + µ s cos 70° ) R
m
vmin = 18.81 m/s 
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PROBLEM 12.60
A semicircular slot of 10-in. radius is cut in a flat plate which rotates about
the vertical AD at a constant rate of 14 rad/s. A small, 0.8-lb block E is
designed to slide in the slot as the plate rotates. Knowing that the
coefficients of friction are µ s = 0.35 and µk = 0.25, determine whether
the block will slide in the slot if it is released in the position corresponding to
(a) θ= 80°, (b) θ= 40°. Also determine the magnitude and the direction
of the friction force exerted on the block immediately after it is released.
SOLUTION
First note
Then
1
(26 − 10 sin θ ) ft
12
vE = ρφABCD
=
ρ
a=
n
vE2
= ρ (φABCD ) 2
ρ
1

=  (26 − 10 sin θ ) ft  (14 rad/s) 2
12

98
=
(13 − 5 sin θ ) ft/s 2
3
Assume that the block is at rest with respect to the plate.
=
ΣFx max : N + W cos
=
θ m
or
vE2
ρ


v2
N=
W  − cos θ + E sin θ 
gρ


ΣFy =ma y : −F + W sin θ =−m
or
sin θ
vE2
ρ
cos θ


v2
F W  sin θ + E cos θ 
=


gρ


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PROBLEM 12.60 (Continued)
(a)
We have
θ= 80°
Then
1
98


=
N (0.8 lb)  − cos80° +
× (13 − 5sin 80°) ft/s 2 × sin 80°
2
3
32.2 ft/s


= 6.3159 lb
1
98


=
F (0.8 lb) sin 80° +
× (13 − 5sin 80°) ft/s 2 × cos80°
2
3
32.2
ft/s


= 1.92601 lb
Now
F=
µ=
0.35(6.3159=
lb) 2.2106 lb
max
sN
The block does not slide in the slot, and
F = 1.926 lb
(b)
We have
80° 
θ= 40°
Then
1
98


=
N (0.8 lb)  − cos 40° +
× (13 − 5sin 40°) ft/s 2 × sin 40°
2
3
32.2 ft/s


= 4.4924 lb
1
98


=
F (0.8 lb) sin 40° +
× (13 − 5sin 40°) ft/s 2 × cos 40°
2
3
32.2 ft/s


= 6.5984 lb
Now
Fmax = µ s N ,
from which it follows that
F > Fmax
Block E will slide in the slot
and
a=
a n + a E/plate
E
=
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.
=
ΣFx max : N + W =
cos 40° m
or
vE2
ρ
sin 40°


v2
N W  − cos 40° + E sin 40°  (as above)
=
gρ


= 4.4924 lb
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PROBLEM 12.60 (Continued)
F = µk N
Sliding:
= 0.25(4.4924 lb)
= 1.123 lb
Noting that F and a E/plane must be directed as shown (if their directions are reversed, then ΣFx is
while ma x is
), we have
the block slides downward in the slot and
F = 1.123 lb
40° 
Alternative solutions.
(a)
Assume that the block is at rest with respect to the plate.
=
ΣF ma : W=
+ R ma n
Then
W
W
g
=
=
2

man W vE ρ (φ ABCD )2
g ρ
tan (φ − 10=
°)
=
32.2 ft/s 2
98
(13 − 5sin 80°) ft/s 2
3
or
φ −=
10° 6.9588°
and
=
φ 16.9588°
Now
=
tan φs µ=
µ s 0.35
s
so that
(from above)
=
φs 19.29°
0 < φ < φs ⇒ Block does not slide and R is directed as shown.
Now
Then
F R=
=
sin φ and R
W
sin (φ − 10°)
sin16.9588°
sin 6.9588°
= 1.926 lb
F = (0.8 lb)
The block does not slide in the slot and
F = 1.926 lb
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PROBLEM 12.60 (Continued)
(b)
Assume that the block is at rest with respect to the plate.
=
ΣF ma : W=
+ R ma n
From Part a (above), it then follows that
g
=

ρ (φ ABCD )2
tan =
(φ − 50°)
or
φ −=
50° 5.752°
and
=
φ 55.752°
Now
=
φs 19.29°
so that
32.2 ft/s 2
98
(13 − 5sin 40°) ft/s 2
3
φ > φs
The block will slide in the slot and then
or
φ = φk , where =
tan φk µ=
µk 0.25
k
=
φk 14.0362°
To determine in which direction the block will slide, consider the free-body diagrams for the two
possible cases.
Now
ΣF= ma : W + R= ma n + ma E/plate
From the diagrams it can be concluded that this equation can be satisfied only if the block is sliding
downward. Then
R cos φk m
=
ΣFx max : W cos 40° +=
Now
Then
or
vE2
ρ
sin 40°
F = R sin φk
F
W vE2
sin 40°
W cos 40=
°+
tan φk
g ρ


v2
F µkW  − cos 40° + E sin 40° 
=


gρ


(see the first solution)
= 1.123 lb
The block slides downward in the slot and
F = 1.123 lb
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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 ≤ θ ≤ 150°. Knowing that the block begins to slide when
=
θ 150°, determine 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 =
θ 150°.
=
v a=
0,
t
=
g 9.81 m/s 2
=
ΣFt mat : − mg sin 30° =
+N 0
=
N mg sin 30°
=
ΣFn man : mg
=
cos 30° − F m
=
F mg cos30° −
Form the ratio
µ=
s
v2
ρ
mv 2
ρ
F
, and set it equal to µ s for impending slip.
N
F
=
N
g cos 30° − v 2 /ρ
9.81 cos 30° − (1.4) 2 /0.3
=
g sin 30°
9.81 sin 30°
µ s = 0.400 
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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 θ = 0 and depositing it at the next station when
=
θ 180°. 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 = 2.2 ft/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 θ for which sliding is impending.
SOLUTION
=
ΣFx max=
: F
W vB2
cos θ
g ρ
W vB2
−
sin θ
ma y : N − W =
+ ΣFy =
g ρ
or


v2
N W 1 − B sin θ 
=


gρ


Now


v2
µ=
µ s W 1 − B sin θ 
F=
max
sN
gρ


and for the component not to slide
F < Fmax
or
or


v2
W vB2
cos θ < µ s W 1 − B sin θ 


g ρ
gρ


µs >
cos θ
gρ
vB2
− sin θ
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PROBLEM 12.62 (Continued)
We must determine the values of θ which maximize the above expression. Thus
(
)
gρ
− sin θ − (cos θ )(− cos θ )
d  cos θ  − sin θ vB2
 gρ
 =
0
=
2
gρ
dθ  vB2 − sin θ 
− sin θ
v2
or
Now
B
)
=
sin θ
vB2
=
for
µ s ( µ s )min
gρ
=
sin θ
(2.2 ft/s) 2
=
0.180373
(32.2 ft/s 2 ) ( 10
ft )
12
θ=
10.3915° and θ =
169.609°
or
(a)
(
From above,
( µ s ) min
=
vB2
cos θ
where
sin
θ
=
gρ
gρ
− sin θ
v2
B
( µ s ) min
=
cos θ
cos θ sin θ
tan θ
=
=
1
− sin θ
1 − sin 2 θ
sin θ
= tan10.3915°
( µ s )min = 0.1834 
or
(b)
We have impending motion
to the left for
=
θ 10.39° 
to the right for
=
θ 169.6° 
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PROBLEM 12.63
Knowing that the coefficients of friction between the component I and member BC of the mechanism of
Problem 12.62 are µ s = 0.35 and µk = 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 θ for which sliding is impending.
SOLUTION
: F
=
ΣFx max=
W vB2
cos θ
g ρ
W vB2
−
sin θ
ma y : N − W =
+ ΣFy =
g ρ


v2
N W 1 − B sin θ 
=


gρ


or
Fmax = µ s N
Now


v2
= µ s W 1 − B sin θ 
gρ


and for the component not to slide
F < Fmax
or
or


v2
W vB2
cos θ < µ s W 1 − B sin θ 


g ρ
gρ


vB2 < µ s
gρ
cos θ + µ s sin θ
( )
To ensure that this inequality is satisfied, vB2
must be less than or equal to the minimum value
max
of µ s g ρ /(cos θ + µ s sin θ ), which occurs when (cos θ + µ s sin θ ) is maximum. Thus
d
(cos θ + µ s sin θ ) =
− sin θ + µ s cos θ =
0
dθ
or
tan θ = µ s
µ s = 0.35
or
=
θ 19.2900°
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PROBLEM 12.63 (Continued)
(a)
The maximum allowed value of vB is then
gρ
vB2
=
where tan θ µ s
µ=
s
max
cos θ + µ s sin θ
( )
tan θ
g ρ sin θ
= g=
ρ
cos θ + (tan θ ) sin θ
 10 
= (32.2 ft/s 2 )  ft  sin 19.2900°
 12 
(vB )max = 2.98 ft/s 
or
(b)
First note that for 90° < θ < 180°, Eq. (1) becomes
vB2 < µ s
gρ
cos α + µ s sin α
where α
= 180° − θ . It then follows that the second value of θ for which motion is impending is
=
θ 180° − 19.2900°
= 160.7100°
we have impending motion
to the left for
=
θ 19.29° 
to the right for
=
θ 160.7° 
Alternative solution.
=
ΣF ma : W=
+ R ma n
For impending motion, φ = φs . Also, as shown above, the values of θ for which motion is impending
(
minimize the value of vB, and thus the value of an is an =
vB2
ρ
). From the above diagram, it can be concluded
that an is minimum when ma n and R are perpendicular.
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PROBLEM 12.63 (Continued)
Therefore, from the diagram
θ= φ=
tan −1 µ s
s
and
or
(as above)
man = W sin φs
m
vB2
or
ρ
= mg sin θ
vB2 = g ρ sin θ
(as above)
α
= 180° − θ
(as above)
For 90° ≤ θ ≤ 180°, we have
from the diagram
α = φs
and
or
man = W sin φs
vB2 = g ρ sin θ
(as above)
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PROBLEM 12.64
A small 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 θ 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 ρ = r sin θ . v = ρω
an
Normal acceleration. =
v 2 ρ 2ω 2
=
= (r sin θ ) ω 2
ρ
ρ
∑ Fy =
ma y :
N cosθ − mg =
0
N =
mg
cosθ
ΣFx =
max :
N sin θ = ma
mg sin θ
= (m r sin θ )ω 2
cos θ
(1)
To satisfy (1), either:
sin θ = 0
or
cos θ =
g
rω 2
θ=
0° or 180°
or =
cosθ
9.81
=
0.3488
(0.5) (7.5) 2
θ =
0°, 180°, and 69.6° 
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PROBLEM 12.65
A small 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. Knowing that the coefficients of
friction are µs = 0.25 and µk = 0.20, indicate whether the collar will slide on the rod if
it is released in the position corresponding to (a) θ = 75°, (b) θ = 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 ρ = r sin θ . v = ρω
=
r 500
=
mm 0.500
=
m, ω 7.5 rad/s,
Given:
=
m 250
=
g 0.250 kg.
Normal acceleration:
=
an
v 2 ρ 2ω 2
=
= (r sin θ )ω 2
ρ
ρ
ΣF = ma:
F − mg sin θ =
− m (r sin θ ) ω 2 cos θ
F m ( g − rω 2 cos θ )sin θ
=
ΣF = ma :
N − mg cos θ =
m(r sin θ )ω 2 sin θ
N m( g cos θ + r ω 2 sin 2 θ )
=
(a) θ= 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
=
µ s N (0.25)(7.1950)
= 1.7987 N
Since F < µ s N , the collar does not slide.
F = 0.611 N
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PROBLEM 12.65 (Continued)
F = (0.25) 9.81 − (0.500 cos 40°) (7.5) 2  sin 40°
(b) θ= 40°.
= − 1.8858 N
=
N (0.25) 9.81cos 40° + (0.500sin 2 40°)(7.5) 2 
= 4.7839 N
=
µ s N (0.25)
=
(4.7839) 1.1960 N
Since F > µ s N , the collar slides.
Since the collar is sliding, F = µ k N .
∑ Fn =
+
ma :
N − mg cosθ =
man sin θ
=
N mg cos θ + m (r sin θ )ω 2 sin θ
= m  g cos θ + (r sin 2 θ )ω 2 
= (0.25) 9.81cos 40° + (0.500sin 2 40°)(7.5) 2 
= 4.7839 N
=
F µ=
(0.20) (4.7839)
= 0.957 N
kN
F = 0.957 N
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PROBLEM 12.66
An advanced spatial disorientation trainer allows the cab to rotate around
multiple axes as well as extend inwards and outwards. It can be used to
simulate driving, fixed wing aircraft flying, and helicopter maneuvering. In
one training scenario, the trainer rotates and translates in the horizontal plane
where the location of the pilot is defined by the relationships
π 
and θ 0.1( 2t 2 − t ) , where r, θ, and t are expressed
=
r 10 + 2 cos  t  =
3 
in feet, radians, and seconds, respectively. Knowing that the pilot has a mass
of 175 lbs, (a) find the magnitude of the resulting force acting on the pilot at
t= 5 s (b) plot the magnitudes of the radial and transverse components of the
force exerted on the pilot from 0 to 10 seconds.
SOLUTION
π 
=
r 10 + 2 cos  t  ft
3 
Given:
=
θ 0.1( 2t 2 − t ) rad
=
m
175 lbs
=
5.435 slugs
32.2 ft/s 2
Using Radial and Transverse Coordinates:
2π
π 
sin  t  ft/s
3
3 
2π 2
π 

r= −
cos  t  ft/s 2
9
3 
r = −
=
θ 0.1( 4t − 1) rad/s
θ = 0.4 rad/s 2
Free Body Diagram of pilot (top and side view) showing net forces in the radial and transverse directions:
Equations of Motion:
(
F
∑=
ma
=
m 
r − rθ 2
r
r
)
 2π 2
2
π  
 π 
Fr = m  −
cos  t  − 10 + 2 cos  t   0.12 ( 4t − 1) 
3  
 3 
 9

Fθ
∑=
(
=
ma
m rθ + 2rθ
θ
)


0.4π
 π 
π 
Fθ =
m 10 + 2 cos  t   ( 0.4 ) −
sin  t  ( 4t − 1) 
3
 3 
3 


∑F
z
=0 ⇒ Fz =mg
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PROBLEM 12.66 (Continued)
(a) Evaluate forces at t=5 s:
Fr = −221.78 lbs
Fθ = 61.37 lbs
Fz = 175 lbs
F=
F = 289.1 lb 
Fr2 + Fθ2 + Fz2
(b) Plot of Fr and Fθ from t=0 to t=10 s:
Radial and Transverse Components of Force for t=0 to 10 seconds
100
0
-100
Force, (lbs)
-200
-300
-400
-500
-600
-700
-800
Radial Component
Transverse Component
0
1
2
3
4
5
6
Time, (sec)
7
8
9
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
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PROBLEM 12.67
An advanced spatial disorientation trainer is programmed to only rotate
and translate in the horizontal plane. The pilot’s location is defined by the
(
relationships =
r 8 1 − e−t
)


and θ = 2 / π  sin
π 
t  , where r, θ, and t
2 
are expressed in feet, radians, and seconds, respectively. Determine the
radial and transverse components of the force exerted on the 175 lb pilot at
t = 3 s.
SOLUTION
=
r 8 (1 − e − t ) ft
Given:
 π 
t  rad
2 

175 lbs
=
m =
5.435 slugs
32.2 ft/s 2
θ = 2 / π  sin
Using Radial and Transverse Coordinates:
−t
r = 8e ft/s

r = −8e − t ft/s 2
π
θ = cos t rad/s
2
π
π
θ = − sin t rad/s 2
2
2
Free Body Diagram of pilot (top view) showing net forces in the radial and transverse directions:
Equations of Motion:
∑F
r
= mar
(
=
Fr m 
r − rθ 2
)
π 

Fr = m  −8e − t − 8 (1 − e − t ) cos 2 t 
2 

∑ Fθ = maθ
(
Fθ m rθ + 2rθ
=
)
π 

 π π
Fθ = m 8 (1 − e − t )  −  sin t − 16e − t cos t 
2
2 
 2

Evaluate Fr and Fθ at t=3 s:
Fr = −2.165 lb 
Fθ = 64.90 lb 
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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 θ = 0.75t , where θ and t are expressed in rad/s and seconds,
respectively. As the arm-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
θ = (0.75t ) rad/s
θ = 0.75 rad/s 2
Now

0 − [(0.5t ) m][(0.75t ) rad/s]2 =
ar =
r − rθ 2 =
−(0.28125t 3 ) m/s 2
and
a=
rθ + 2rθ
θ
= [(0.5t ) m][0.75 rad/s 2 ] + 2(0.5 m/s)[(0.75t ) rad/s]
= (1.125t ) m/s 2
Kinetics
Σ=
Fr mar : −=
T (3 kg)(−0.28125t 3 ) m/s 2
or
T = (0.84375t 3 ) N
2
=
ΣFθ mB a=
θ : Q (3 kg)(1.125t ) m/s
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
A 0.5-kg block B slides without friction inside a slot cut in arm OA which rotates in
a vertical plane. The rod has a constant angular acceleration θ = 10 rad/s2.
Knowing that when θ = 45º and r = 0.8 m 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
Given:
r =0.8 m, θ =
10 rad/s 2
=
m 0.5
=
kg, W mg
θ =45°,
Using Radial and Transverse Components:
vr= r= 0,
FBD of Block B
vθ= rθ= 0
θ = 0
(a)
(
=
ΣFθ
maθ : N − W cos=
45° m rθ + 2rθ
(
=
N mg cos 45° + m rθ + 2rθ
)
)
= (0.5)(9.81) cos 45° + 0.5 ( 0.8 )(10 ) + 0 
= 7.468
(b)
N = 7.47 N
(
Σ=
Fr mar : mg sin 45
=
° m 
r − rθ 2
45° 
)
mg
sin 45° + rθ 2
m
= g sin 45° + rθ 2
= ( 9.81) sin 45° + 0
= 6.937 m/s 2

=
r
a B / rod = 6.94 m/s 2
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PROBLEM 12.70
Pin B weighs 4 oz and is free to slide in a horizontal plane along the
rotating arm OC and along the circular slot DE of radius b = 20 in.
Neglecting friction and assuming that θ = 15 rad/s and
θ = 250 rad/s 2 for the position θ= 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 geometry of the system, we have
r = − (2b sin θ )θ
r = 2b cos θ
Then

r=
−2b(θsin θ + θ 2 cos θ )
and

ar =
r − rθ 2 =
−2b(θsin θ + θ 2 cos θ ) − (2b cos θ )θ 2 =
− 2b(θsin θ + 2θ 2 cos θ )
Now
 20 
= − 2
ft  [(250 rad/s 2 )sin 20° + 2(15 rad/s) 2 cos 20°] = −1694.56 ft/s 2
12


aθ = rθ + 2rθ = (2b cos θ )θ + 2(−2bθ sin θ )θ = 2b(θ cos θ − 2θ 2 sin θ )
and
 20 
ft  [(250 rad/s 2 ) cos=
20° − 2(15 rad/s) 2 sin 20°] 270.05 ft/s 2
= 2
 12 
Kinetics.
(a)
1
4
lb
We have
Fr =mar =
and
1
lb
4
× (270.05 ft/s 2 ) =
Fθ =
maθ =
2.0967 lb
32.2 ft/s 2
32.2 ft/s 2
Fr = −13.16 lb 
× (−1694.56 ft/s 2 ) =−13.1565 lb
Fθ = 2.10 lb 
ΣFr : −Fr =
−Q cos 20°
(b)
or
=
Q
1
=
(13.1565 lb) 14.0009 lb
cos 20°
ΣFθ : Fθ =
P − Q sin 20°
or
=
P (2.0967 + 14.0009sin 20°)=
lb 6.89 lb
P = 6.89 lb
70° 
Q = 14.00 lb
40° 
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PROBLEM 12.71
The two blocks are released from rest when r = 0.8 m and θ= 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 θ 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, 
r + aB =
0

r =
−aB
(1)
ΣFx mAa A : =
T cos θ mAa=
mAa A sec θ
For block A, =
A or T
(2)
For block B,
or
=
ΣFy mB aB : =
mB g − T mB aB
Adding Eq. (1) to Eq. (2) to eliminate=
T, mB g mAa A secθ + 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.

a A ⋅ er =
−a A cos θ
r − rθ 2 =
ar =
(5)
Noting that initially θ = 0, using Eq. (1) to eliminate r, and changing
signs gives
aB = a A cosθ
(6)
Substituting Eq. (6) into Eq. (4) and solving for a A ,
=
aA
mB g
=
mA sec θ + mB cos θ
(25) (9.81)
= 5.48 m/s 2
20sec30° + 25cos 30°
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/s 2
a B = 4.75 m/s 2 
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
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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 θ= 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 θ 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 to note the positive directions of the components.
r =
vr =
v A ⋅ er =
−v A cos30°
= −2cos30° = −1.73205 m/s
rθ =
vθ =
v A ⋅ eθ =
−v A sin 30°
= 2sin=
30° 1.000 m/s 2
=
θ
vθ
1.000
=
= 1.25 rad/s
r
0.8
constant,
Constraint of cable: r + yB =
r + vB =
0, 
r + aB =
0 or

r =
−aB
(1)
ΣFx mAa A :=
T cosθ mAa=
mAa A secθ
For block A, =
A or T
(2)
=
ΣFy mB aB: m=
mB aB
Bg − T
(3)
For block B,
Adding Eq. (1) to Eq. (2) to eliminate=
T, mB g mAa A secθ + mB aB
(4)
Radial and transverse components of a A.
Use a method similar to that used for the components of velocity.

a A ⋅ er =
−a A cos θ
r − rθ 2 =
ar =
(5)
Using Eq. (1) to eliminate r and changing signs gives
=
aB a A cosθ − rθ 2
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PROBLEM 12.72 (Continued)
Substituting Eq. (6) into Eq. (4) and solving for a A ,
=
aA
(
)
mB g + rθ 2
=
mA sec θ + mB cos θ
(25)[9.81 + (0.8)(1.25) 2 ]
= 6.18 m/s 2
20sec30° + 25cos 30°
2
From Eq. =
(6), aB 6.18cos 30° − (0.8)(1.25)
=
4.10 m/s 2
(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/s 2

a B = 4.10 m/s 2 
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PROBLEM 12.73*
Slider C has a weight of 0.5 lb and may move in a slot cut in arm
AB, which rotates at the constant rate θ0 = 10 rad/s in a horizontal
plane. The slider is attached to a spring of constant k = 2.5 lb/ft,
which is unstretched when r = 0. Knowing that the slider is
released from rest with no radial velocity in the position r = 18 in.
and neglecting friction, determine for the position r = 12 in. (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 )
Σ Fr = mar : − k (r − l0 ) = m(
r − rθ 2 )
kl
k


r = θ 2 −  r + 0
m
m

(1)
But
d
dr dr
dr
=
(r) = r
dt
dr dt
dr

kl 
k
  =
rdr
rdr = θ 2 −  r + 0  dr
m
m


=
r
Integrate using the condition r = r0 when r = r0 .
1 2 r  1   2 k  2 kl0
=  θ − r +
r
2 r0  2 
m
m
(
 r
r
 r0
)
kl
1 2 1 2 1  2 k  2
θ −  r − r02 + 0 (r − r0 )
r − r0=
2
2
2 
m
m
2kl0
k

(r − r0 )
r 2 =r02 +  θ 2 −  r 2 − r02 +
m
m

(
Data:
=
m
)
W
0.5 lb
=
= 0.01553 lb ⋅ s 2 /ft
g 32.2 ft/s 2
=
θ 10
=
=
rad/s, k 2.5
lb/ft, l0 0
=
in. 1.5 ft,=
in. 1.0 ft
r0 (=
vr )0 0,=
r0 18=
r 12=
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PROBLEM 12.73* (Continued)
(a)
Components of velocity when r = 12 in.
2.5 

r 2 =
0 + 102 −
(1.02 − 1.52 ) + 0

0.01553 

= 76.223 ft 2 /s 2
vr = r = ±8.7306 ft/s
Since r is decreasing, vr is negative
(b)
r = −8.7306 ft/s
vr = −8.73 ft/s 
v=
r=
θ (1.0)(10)
θ
vθ = 10.00 ft/s 
Components of acceleration.
Fr =−kr + kl0 =−(2.5)(1.0) + 0 =−2.5 lb
ar =
Fr
2.5
= −
m
0.01553
ar = 161.0 ft/s 2 
aθ =
rθ + 2rθ =+
0 (2)(−8.7306)(10)
aθ = −174.6 ft/s 2 
(c)
Transverse component of force.
=
Fθ ma
=
(0.01553)(−174.6)
θ
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Fθ = −2.71 lb 
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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 2θ and using Eq. (12.25), express
the radial and transverse components of the velocity v of the particle as
functions of θ .
SOLUTION
Since the particle moves under a central force, h = constant.
 h= r v
=
h r 2θ=
0
0 0
Using Eq. (12.27),
r0 v0 r0 v0 cos 2θ v0
=
=
θ =
cos 2θ
r0
r2
r0 2
or
Radial component of velocity.
vr= r=
= r0
r0
dr  d 
θ=

dθ
dθ  cos 2θ

sin 2θ
θ
θ= r0
3/ 2
(cos
2
)
θ

sin 2θ v0
cos 2θ
(cos 2θ )3/ 2 r
vr = v0
sin 2θ
cos 2θ

Transverse component of velocity.
vθ=
h r0 v0
cos 2θ
=
r
r0
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vθ = v0 cos 2θ 
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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 2θ and using
Eq. (12.25), express the radial and transverse components of the velocity
v of the particle as functions of θ.
SOLUTION
Since the particle moves under a central force, h = constant.
Using Eq. (12.27),
 h= r v
=
h r 2θ=
0
0 0
r0 v0 r0 v0 cos 2θ v0
cos 2θ
=
θ =
or =
r0
r2
r0 2
Differentiating the expression for r with respect to time,
=
r

r0
dr  d 
sin 2θ
sin 2θ v0
sin 2θ
=
=
=
=
θ
θ r0
cos
2θ v0

θ r0
3/ 2
3/ 2
dθ
dθ  cos 2θ 
r
(cos 2θ )
(cos 2θ )
cos 2θ
0
Differentiating again,

r
=
(a)
sin 2θ  
2 cos 2 2θ + sin 2 2θ  v0 2 2 cos 2 2θ + sin 2 2θ
dr  d 
θ
θ
=
=
=
 v0
 θ v0
dθ
dθ 
r0
(cos 2θ )3/ 2
cos 2θ 
cos 2θ
vr= r= v0
sin 2θ
cos 2θ
=
v0 r
sin 2θ
r0
v=
v=
r=
θ
θ
(vr )2 + (vθ )2 =
v0 r
cos 2θ
r0
v0 r
sin 2 2θ + cos 2 2θ
r0
v=
v0 r

r0
v 2 2 cos 2 2θ + sin 2 2θ
r0
v0 2

−
cos 2 2θ
ar =−
r rθ 2 =0
r0
cos 2θ
cos 2θ r0 2
=
v0 2 cos 2 2θ + sin 2 2θ
v0
v0 2 r
= =
r0
r0 2
r0 cos 2θ
cos 2θ
=
Fr ma
=
r
mv02 r
r02
:
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Fr =
mv02 r
r02

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PROBLEM 12.75 (Continued)
Since the particle moves under a central force, aθ = 0.
Magnitude of acceleration.
a=
ar 2 + aθ 2 =
v0 2 r
r0 2
Tangential component of acceleration.
v0 2 r
dv d  v0 r  v0

=
=
=
sin 2θ
r


dt dt  r0  r0
r0 2
=
at
Normal component of acceleration.
at =
r 
cos 2θ =  0 
r
But
an =
Hence,
(b)
But a=
n
a 2 − at 2 =
v2
ρ
or
v0 2 r
r0 2
1 − sin 2 2θ =
v0 2 r cos 2θ
r0 2
2
v0 2
r
=
ρ
v 2 v0 2 r 2 r
=
⋅
an
r0 2 v0 2
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ρ=
r3

r02
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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. (12.25), show that the speed of the particle is v = v0 /cos 2 θ .
SOLUTION
Since the particle moves under a central force, h = constant.
 h= r v
=
h r 2θ=
0
0 0
Using Eq. (12.27),
r0 v0
r0 v0
v0
=
θ =
=
2
2
2
r
r0 cos θ r0 cos 2 θ
or
Radial component of velocity.
vr = r =
d
(r0 cos θ ) = −(r0 sin θ )θ
dt
Transverse component of velocity.
θ (r0 cos θ )θ
v=
r=
θ
Speed.
v = vr 2 + vθ 2 = r0θ =
r0 v0
r0 cos θ
2
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v=
v0
cos 2 θ

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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. (12.25), show that the speed of the particle is v = v0 /cos 2 θ .
SOLUTION
Since the particle moves under a central force, h = constant
Using Eq. (12.27),
 h= r v
=
h r 2θ=
0
0 0
r0 v0
r0 v0
v0
=
=
θ =
2
2
2
r
r0 cos θ r0 cos 2 θ
Radial component of velocity.
vr = r =
d
(r0 cos θ ) = −(r0 sin θ )θ
dt
Transverse component of velocity.
θ (r0 cos θ )θ
v=
r=
θ
Speed.
v = vr 2 + vθ 2 = r0θ =
r0 v0
r0 cos θ
2
=
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 ma
t : Ft
2mv0 2 sin θ
r0 cos5 θ
θ 0,=
Ft 0
(a)=
(b)
θ =°
45 ,
Ft = 0 
2mv sin 45°
Ft = 0 5
cos 45°
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Ft =
8mv0 2

r0
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PROBLEM 12.78
Determine the mass of the earth knowing that the mean radius of the moon’s orbit about the earth is 238,910 mi
and that the moon requires 27.32 days to complete one full revolution about the earth.
SOLUTION
Mm
[Eq. (12.28)]
r2
We have
F =G
and
F
= F=
ma=
m
n
n
Then
or
Now
G
Mm
v2
=
m
r
r2
M=
v=
r 2
v
G
2π r
τ
2
so that
Noting that
and
=
M
2
r  2π r 
1  2π  3
=
r


G τ 
G  τ 
=
τ 27.32
=
days 2.3604 × 106 s
=
r 238,910
=
mi 1.26144 × 109 ft
we have
=
M
or
v2
r
2


2π
(1.26144 × 109 ft)3

4 
6
−9 4


34.4 × 10 ft /lb ⋅ s  2.3604 × 10 s 
1
M=
413 × 1021 lb ⋅ s 2 /ft 
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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, giving the answer in both SI and U.S. customary units.
SOLUTION
Mm
r2
We have
F =G
and
F
= F=
ma=
m
n
n
Then
G
GM
r
GM = gR 2
Now
v2 =
so that
=
τ
For one orbit,
[Eq. (12.30)]
g
gR 2
or v = R
r
r
2π r 2π r
=
v
R gr
 gτ 2 R 2
r = 
2
 4π
or
v2
r
Mm
v2
=
m
r
r2
v2 =
or
[Eq. (12.28)]
1/ 3




Q.E.D.
=
τ 27.3
=
days 2.35872 × 106 s
Now
=
=
R 3960
mi 20.9088 × 106 ft
1/3
SI:
 9.81 m/s 2 × (2.35872 × 106 s) 2 × (6.37 × 106 m) 2 
=
r 
=
382.81 × 106 m

2
4π


=
r 383 × 103 km 
or
U.S. customary units:
1/3
 32.2 ft/s 2 × (2.35872 × 106 s)2 × (20.9088 × 106 ft) 2 
r 
1256.52 × 106 ft
=
=

2
4π


=
r 238 × 103 mi 
or
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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. Give the answers in both SI and U.S. customary units.
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.
=
vτ 2π r
But
Then
=
v 2τ 2
GM τ 2
r
r
or
=
=
4π 2
3
GM τ 2
= 4π 2 r 2
r
 GM τ 2

2
 4π
1/3



=
=
τ 23.934
h 86.1624 × 103 s
Data:
(a)
or
=
g 9.81 m/s 2 ,=
R 6.37 × 106 m
In SI units:
GM =gR 2 =(9.81)(6.37 × 106 ) 2 =398.06 × 1012 m3 /s 2
1/3
 (398.06 × 1012 )(86.1624 × 103 ) 2 
r 
42.145 × 106 m
=
=

2
4π


h = 35,800 km 
altitude h =r − R =35.775 × 106 m
In U.S. units:
2
=
g 32.2 ft/s=
, R 3960
=
mi 20.909 × 106 ft
GM =gR 2 =(32.2)(20.909 × 106 ) 2 =14.077 × 1015 ft 3 /s 2
1/3
 (14.077 × 1015 )(86.1624 × 103 ) 2 
6
r 
=
=
 138.334 × 10 ft
2
4π


altitude h =r − R =117.425 × 106 ft
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PROBLEM 12.80 (Continued)
(b)
In SI units:
=
v
GM
=
r
398.06 × 1012
= 3.07 × 103 m/s
42.145 × 106
=
v
GM
=
r
14.077 × 1015
=
10.09 × 103 ft/s
6
138.334 × 10
v = 3.07 km/s 
In U.S. units:
=
v 10.09 × 103 ft/s 
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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 τ 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 τ = 3.551 days and r = 670.9 × 103 km for its moon Europa.
SOLUTION
Mm
r2
We have
F =G
and
F
= F=
ma=
m
n
n
Then
or
Now
so that
For one orbit,
G
[Eq. (12.28)]
v2
r
Mm
v2
=
m
r
r2
v2 =
GM
r
GM = gR 2
=
v2
τ
=
[Eq. (12.30)]
gR 2
g
=
or
v R
r
r
2π r 2π r
=
v
R gr
or
 gτ 2 R 2
r = 
2
 4π
Solving for g,
g = 4π 2
1/3




Q.E.D.
r3
τ 2 R2
and noting that τ = 3.551 days = 306,806 s, then
g Jupiter = 4π 2
= 4π 2
3
rEur
2
τ 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 ≈ 2.53g Earth
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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 /s 2
=
M
(a) Mass of sun.
(b) At the surface of the sun,
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
=
R2
1.3335 × 1020
(695.5 × 106 ) 2
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PROBLEM 12.83
A satellite is placed into a circular orbit about the planet Saturn at an altitude of 2100 mi. The satellite
describes its orbit with a velocity of 54.7 × 103 mi/h. Knowing that the radius of the orbit about Saturn and the
periodic time of Atlas, one of Saturn’s moons, are 85.54 × 103 mi 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
2π rA
Velocity of Atlas.
vA =
where
v A =85.54 × 103 mi =451.651 × 106 ft
τA
=
τ A 0.6017
=
days 51,987s
and
Gravitational force.
vA
=
(2π ) (451.651 × 106 )
= 54.587 × 103 ft/s
51,987
=
F
GMm mv 2
=
r
r2
from which
2
GM
= rv=
constant
For the satellite,
rs vs2 = rA v A2
rs =
rA v A2
vs 2
vs =
54.7 × 103 mi/h =
80.227 × 103 ft/s
where
(451.651 × 106 )(54.587 × 103 ) 2
= 209.09 × 106 ft
(80.227 × 103 ) 2
rs = 39, 600 mi
=
rs
(a)
Radius of Saturn.
R=
rs − (altitude) =
39,600 − 2100
(b)
R = 37,500 mi 
Mass of Saturn.
=
M
rA v A2 (451.651 × 106 )(54.587 × 103 ) 2
=
G
34.4 × 10−9
M=
39.1 × 1024 lb ⋅ s 2 /ft 
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PROBLEM 12.84
The periodic time (see Prob. 12.83) of an earth satellite in a circular polar
orbit is 120 minutes. 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
mv 2
r
GMm
=
r2
v
or =
GM
r
Let τ be the periodic time to complete one orbit.
GM
=
vτ 2=
πr
or
τ
2π r
r
1/ 3
Solving for r,
 GM τ 2 
r = 
2 

 4π 
For earth, R =3960 mi =20.909 × 106 ft, g =32.2 ft/s 2
(
GM =gR 2 =( 32.2 ) 20.909 × 106
)
2
=14.077 × 1015 ft 3/s 2
Data:
=
τ 120
=
min 7200 s
(
)
1/ 3
 14.077 × 1015 ( 7200 )2 

=
=
r 
26.441 × 106 ft


4π 2


(a) altitude h = r − R = 5.532 × 106 ft
(b) cos=
θ
R
=
r
h = 1048 mi 
20.909 × 106
= 0.79078
26.441 × 106
=
θ 37.74°
=
t AB
2θ
τ
=
360°
7200 )
( 75.48)( =
360
1509.6 s
t AB = 25.2 min 
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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
=
RE 6.37 × 106 m
First note that
rE = RE + hE = (6.37 × 106 + 4.5 × 106 ) m
Then
(a)
= 10.87 × 106 m
F=
We have
GMm
r2
[Eq. (12.28)]
GM = gR 2
and
[Eq. (12.29)]
R
2 m
=
=
F gR
W 
r2
r
Then
For the earth orbit,
2
 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
τ=
1  2π  3
r
G  τ 
2π r 3/ 2
GM
τE =
τM ⇒
2π rE3/ 2
GM E
2π r 3/ 2
=M
GM M
(1)
1/3
or
or
 MM 
1/3
6
=
rM =
 rE (0.01230) (10.87 × 10 m)
M
 E 
=
rM 2.509 × 106 m
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PROBLEM 12.85 (Continued)
(c)
GM = gR 2
We have
[Eq.(12.29)]
Substituting into Eq. (1)
2π rE3/ 2
RE g E
=
2π rM3/ 2
RM g M
2
or
3
2
 RE   rM 
 RE   M M
=
g M =
   gE 
 
 RM   rE 
 RM   M E

 gE

using the results of Part (b). Then
2
 6370 km 
2
gM = 
 (0.01230)(9.81 m/s )
 1737 km 
g moon = 1.62 m/s 2 
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,
=
Σ Fn man =
: F m
F =G
Eq. (12.28):
Then
G
or
v2
r
Mm
r2
Mm
v2
=m
2
r
r
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=
(v A )circ + ∆v A
A )TR
= (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
2200 km
× 1466.7 m/s
2080 km
= 1551.3 m/s
=
(vB )TR
(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 m3 /s 2
For circular orbit of earth, rE =
149.6 × 106 km =
149.6 × 109 m/s
( GM )sun
=
vE
For transfer orbit AB,
=
rE
=
rA rE ,
132.474 × 1018
= 29.758 × 103 m/s
149.6 × 109
=
rB r=
227.8 × 109 m
M
v A= vE + ( ∆v ) A= 29.758 × 103 + 2.94 × 103= 32.698 × 103 m/s
mrAv A = mrBvB
=
vB
rAv A
=
rB
(149.6 × 10 )(32.698 × 10=)
9
3
227.8 × 10
9
21.473 × 103 m/s
For circular orbit of Mars,
=
vM
( GM )sun
=
rM
132.474 × 1018
24.115 × 103 m/s
=
9
227.8 × 10
Speed increase at B.
( ∆v )B
= vM − vB = 24.115 × 103 − 21.473 × 103 = 2.643 × 103 m/s
( ∆v )B
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=
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 22,240 mi 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 185 mi, and then is propelled by an upperstage 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 4810 ft/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 3960
=
mi 20.909 × 106 ft
GM =gR 2 =(32.2)(20.909 × 106 ) 2 =14.077 × 1015 ft 3 /s 2
rA = 3960 + 185 = 4145 mi = 21.8856 × 106 ft
rB =
3960 + 22, 240 =
26, 200 mi =
138.336 × 106 ft
Speed on circular orbit through A.
(v A )circ =
=
GM
rA
14.077 × 1015
21.8856 × 106
= 25.362 × 103 ft/s
Speed on circular orbit through B.
(vB )circ =
=
GM
rB
14.077 × 1015
138.336 × 106
= 10.088 × 103 ft/s
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PROBLEM 12.88 (Continued)
(a)
Speed on transfer trajectory at B.
(vB ) tr= 10.088 × 103 − 4810
= 5.278 × 103
Conservation of angular momentum for transfer trajectory.
5280 ft/s 
rA (v A ) tr = rB (vB ) tr
(v A ) tr =
rB (vB ) tr
rA
(138.336 × 106 )(5278)
21.8856 × 106
= 33.362 × 103 ft/s
=
(b)
Change in speed at A.
∆v=
(v A ) tr − (v A )circ
A
= 33.362 × 103 − 25.362 × 103
= 8.000 × 103
∆v A =
8000 ft/s 
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PROBLEM 12.89
A space vehicle is in a circular orbit of 1400-mi radius around the moon.
To transfer to a smaller orbit of 1300-mi radius, the vehicle is first placed
in an elliptic path AB by reducing its speed by 86 ft/s as it passes through
A. Knowing that the mass of the moon is 5.03 × 1021 lb ⋅ s 2 /ft, 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
Circular orbits: v =
GM
r
=
rA 1400
=
mi 7.392 × 106 ft
=
( vA )1
10 )
(34.4 × 10 )(5.03 ×=
−9
21
7.392 × 106
4.8382 × 103 ft/s
=
rB 1300
=
mi 6.864 × 106 ft
=
( vB ) 2
10 )
(34.4 × 10 )(5.03 × =
−9
21
6.864 × 106
5.0208 × 103 ft/s
(a) Transfer orbit AB.
v)A
( v=
( vA )1 + ( ∆=
A )2
4.8382 × 103 −=
86 4.7522 × 103 ft/s
mrA ( v A )2 = mrB ( vB )1
=
( vB )1
rA ( v A )2
=
rB
10 )
( 7.392 × 10 )( 4.7522 ×=
6
6.864 × 103
3
5.1178 × 103 ft/s
( v=
B )1
5.12 × 103 ft/s 
(b) Speed change at B.
5.0208 × 103 ft/s − 5.1178 × 103 ft/s =
−97.0 ft/s
( ∆vB ) =
( vB )2 − ( vB )1 =
∆vB =
97.0 ft/s 
Speed reduction at B.
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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 θ = 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 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
F=
k (r − rA )
sp
First note
(a)
Fθ = 0 and at A,
Fr =
− Fsp =
0
(a A )r = 0 
(a A )θ = 0 
ΣFr =
mar :
(b)
Noting that
r − rθ 2 )
−Fsp = m(
acollar/rod = 
r , we have at A
=
0 m[acollar/rod − (150 mm)(16 rad/s) 2 ]
acollar/rod = 38400 mm/s 2
(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 )θ rB=
m (vB )θ where (v A )θ rA θ0
Then
=
(vB )θ
150 mm
=
[(150 mm)(16 rad/s)] 800 mm/s
450 mm
(vB )θ = 0.800 m/s 
or
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PROBLEM 12.91
A 1-lb ball A and a 2-lb 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 = 8 ft/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 θ be polar coordinates with the origin lying at the shaft.
Constraint of rod: θ B =
θ A + π radians; θB =
θA =
θ; θB =
θA =
θ.
(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 
Transverse components of acceleration.
(a A )θ =rAθ + 2rAθ =raθ
(aB=
)θ rBθ + 2rBθ
(1)
Since the rod is massless, it must be in equilibrium. Draw its free body
diagram, applying Newton’s 3rd Law.
=
ΣM 0 0: rA ( FA )θ + rB (=
FB )θ rAmA (a A )θ + rB mB =
(aB )θ 0
rAmArAθ + rB mB (rBθ + 2rBθ) =
0
θ =
−2mB rBθ
mArA2 + mB rB 2
t = 0, rB 0=
At
=
so that θ 0.
From Eq. (1),
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PROBLEM 12.91 (Continued)
(b)
Acceleration of B relative to the rod.
At=
= 96 in./s, =
θ
t 0, (v A=
)θ 8 ft/s
(v A )θ
96
= = 9.6 rad/s
rA
10

rB − rBθ 2 = (aB ) r = 0
2
 2 (8) (9.6)

=
rB r=
=
737.28 in./s 2
Bθ

rB = 61.4 ft/s 2 
(c)
Speed of A.
Substituting
d
(mr 2θ) for rFθ in each term of the moment equation
dt
gives
(
)
(
)
d
d
0
mArA2θ +
mB rB 2θ =
dt
dt
Integrating with respect to time,
mArA2θ + mB rB 2θ=
( m r θ ) + ( m r θ )
2
A A
2
0
B B
0
Applying to the final state with ball B moved to the stop at C,
 WA 2 WB 2  
 WA 2 WB

rA +
rC  θ f =
rA +
(rB )02  θ0


g
g
 g

 g

WArA2 + WB (rB )02 
θ f
θ0
=
=
WArA2 + WB rC 2
(1)(10) 2 + (2)(8) 2
=
(9.6) 3.5765 rad/s
(1)(10) 2 + (2)(16) 2

(v=
r=
(10)(3.5765)
= 35.765 in./s
A) f
Aθ f
(v A ) f = 2.98 ft/s 
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PROBLEM 12.92
Two 2.6-lb 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 = 0.6 ft 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 = 1.2 ft,
(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: m=
m=
A
B
2.6
= 0.08075 lb ⋅ s 2 /ft
32.2
(a) Conservation of angular momentum of collar A: ( H 0 ) 2 = ( H 0 )1
mAr1(vθ )1 = mAr2 (vθ )2
r1(vθ )1 r12θ1 (0.6) 2 (12)
= =
= 3.6
1.2
r2
r2
(=
vθ ) 2
(vθ )2 = 3.60 ft/s 
=
θ2
(vθ ) 2
3.6
= = 3.00 rad/s
1.2
rA
(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=
Collar B:
Collar A:

y= 
r
and
(a A ) r= 
r − rθ 2
ΣFy= mB aB : T − WB= mB 
y= mB 
r
(1)
mA (a A ) r : − T= mA (
r − rθ 2 )
ΣF=
r
(2)
Adding (1) and (2) to eliminate T,
−WB = (mA + mB )
r + mArθ 2
a A/rod = 
r=
mArθ 2 − WB (0.08075)(1.2)(3.00) 2 − (2.6)
=
= −10.70 ft/s 2
0.08075 + 0.08075
mA + mB
T= mB (
r + g )= (0.08075)(−10.70 + 32.2)
T = 1.736 lb 
a A / rod = 10.70 ft/s 2 radially inward. 
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PROBLEM 12.93
A small ball swings in a horizontal circle at the end of a cord of length l1 ,
which forms an angle θ1 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 , θ1 , and θ 2 . (b) If the ball is set in motion so that
initially l1 = 0.8 m and θ=
1 35°, determine the angle θ 2 when l2 = 0.6 m.
SOLUTION
(a)
For state 1 or 2, neglecting the vertical component of acceleration,
=
ΣFy 0: T cos=
θ −W 0
T = W cos θ
=
ΣFx man :=
=
θ cos θ
T sin θ W sin
But ρ =  sin θ
mv 2
ρ
so that
=
v2
ρW
=
sin 2 θ cos θ  g sin θ tan θ
m
v1 = 1 g sin θ1 tan θ1
v2 =  2 g sin θ 2 tan θ 2
and
0: H=
constant
ΣM=
y
y
r1mv1 = r2 mv2
or
v11 sin θ1 = v2  2 sin θ 2
3/2
3/2
1 g sin θ1 sin θ1 tan θ1 =  2 sin θ 2 sin θ 2 tan θ 2
31 sin 3 θ1 tan θ1 = 32 sin 3 θ 2 tan θ 2 
(b)
35°,
With θ1 =
1 =0.8 m, and  2 =0.6 m
(0.8)3 sin 3 35° tan 35° =(0.6)3 sin 3 θ 2 tan θ 2
sin 3 θ 2 tan θ 2 − 0.31320 =
0
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=
θ 2 43.6° 
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PROBLEM 12.94
A particle of mass m is projected from Point A with an initial velocity
v 0 perpendicular to OA and moves under a central force F along an
r r0 /(2 − cos θ ). Using Eq.
elliptic path defined by the equation=
(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 θ
,
=
r
r0
d 2u
2
F
+u =
=
2
r0 mh 2u 2
dθ
Solving for F,
F
=
du sin θ
,
=
dθ
r0
d 2u cos θ
=
r0
dθ 2
by Eq. (12.37).
2mh 2u 2 2mh 2
=
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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PROBLEM 12.95
A particle of mass m describes the logarithmic spiral r = r0 ebθ 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 −bθ
=
e
r r0
du
b
= − e−bθ
dθ
r0
d 2u b 2 −bθ
e
=
r0
dθ 2
d 2u
b 2 + 1 −bθ
F
=
+
u
=
e
r0
dθ 2
mh 2u 2
F =
=
(b 2 + 1)mh 2u 2 −bθ
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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PROBLEM 12.96
A particle of mass m describes the path defined by the equation
=
r r0 (6 cosθ − 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
6 cos θ − 5
r0
6sin θ
du
= −
dθ
r0
2
6 cos θ
d u
= −
dθ
r0
u=
1
=
r
d 2u
5
F
+ u =− = 2 2
r0
dθ 2
mh u
F = −
Solving for F,
Substitute u =
by Eq. (12.35).
5mh 2u 2
r0
1
r
F = −
5mh 2

r0r 2
1
, or inversely proportional to r 2. The minus sign
r2
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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PROBLEM 12.97
A particle of mass m describes the parabola y = x 2 4r0 under a central
force F directed toward the center of force C. Using Eq. (12.35) and
Eq. (12.37 ') with ε = 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
y=
x2
4r0
=
x r sin θ ,
y= r0 − r cosθ
or r0 − r cosθ =
(
)
r 2 sin 2 θ
=
4r0
r 2 1 − cos 2 θ
 1 − cos 2 θ

4r0

 2
0
 r + ( cosθ ) r − r0 =

4r0
Solving the quadratic equation for r,
r
=
2r0
1 − cos 2 θ

2

 − cosθ ± cos 2 θ − 4  1 − cos θ

4r0



2r0
θ ± 1)
( − cos=
1 − cos 2 θ
since r > 0. Simplifying gives
=
r
=
2r0
1 + cosθ


 ( −r0 ) 



2r0 (1 − cosθ )
1 − cos 2 θ
1 1 + cosθ
or
= = u
r
2r0
(1)
sin θ
cosθ
du
d 2u
=
−
=
−
and
dθ
2r0
2r0
dθ 2
d 2u
+ u=
dθ 2
Solving for F,
1
=
2r0
mh 2u 2
2r0
F =
Since m, h, and r0 are constants, F is proportional to
F
mh 2u 2
F =
mh 2

2r0r 2
1
, or inversely proportional to r 2.
r2
1 GM
= 2 (1 + ε cosθ ) =u
r
h
GM
1
Comparing with (1) shows that ε = 1 and 2 =
2r0
h
By Eq. (12.37 ') ,
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h=
2GMr0 
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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 /s 2
1
GM
=
(1 + ε )
r0
h2
ε
1 +=
h2
=
r0GM
(94.09 × 109 ) 2
= 3.33
(6.673 × 106 )(398.06 × 1012 )
=
ε 3.33 − 1
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PROBLEM 12.99
It was observed that during the Galileo spacecraft’s first flyby of the earth, its maximum altitude was 600 mi
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
=
R 3960
=
mi 20.909 × 106 ft
For the earth:
GM = gR 2 =(32.2) (20.909 × 106 ) 2 =14.077 × 1015 ft 3 /s 2
For a parabolic trajectory, ε = 1.
1 GM
Eq. (12.39′) :=
(1 + cos θ )
r
h2
1
At θ 0, =
=
r0
2GM
=
h2
2GM
r0 2v0 2
or =
v0
2GM
r0
At r0 = 3960 + 600 = 4560 mi = 24.077 × 106 ft,
=
v0
(2)(14.077 × 1015 )
= 34.196 × 103 ft/s
24.077 × 106
v0 = 6.48 mi/s 
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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,
(v A ) par
 2 × 66.73 × 10−12 m3 /kg ⋅ s 2 × 4.87 × 1024 kg 
=

6332 × 103 m


= 10,131.4 m/s
(v A ) par = 10.13 km/s 
or
(b)
We have
(v=
(v A ) par + ∆v A
A )circ
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
| ∆v A | =
2.97 km/s 
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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,
v=
Eq. (12.44)
GM
r
For the orbit of Tethys,
GM = rT vT2
For Voyager’s trajectory, we have
1 GM
(1 + ε cos θ )
=
r
h2
where h = r0 v0
At O,
=
r r0=
,θ 0
Then
1
GM
(1 + ε )
=
r0 (r0 v0 )2
or
r0 v02
r v2
−=
1 0 02 − 1
GM
rT vT
=
ε
2
185 × 103 km  21.0 km/s 
=
×
 −1
295 × 103 km  11.35 km/s 
ε = 1.147 
or
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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
=
+ C cos θ A
rA
h2
and
1 GM
=
+ C cos θ B .
rB
h2
But
so that
Adding,
θ B =θ A + 180°,
cos θ A = − cos θ B .
1 1 1 1 2GM
+ = + =
rA rB r0 r1
h2
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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 α 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 b v0 , where b < 1, by firing its engine for a short interval of
time. Determine the smallest permissible value of b if the probe is not to crash on the surface of the planet.
SOLUTION
GM
rA
For the circular orbit,
v0 =
where
rA =R + α R =R(1 + α )
Eq. (12.44),
=
GM v02 R(1 + α )
Then
From the solution to Problem 12.102, we have for the elliptic orbit,
1
1 2GM
+ =2
rA rB
h
=
h h=
rA (v A ) AB
A
Now
= [ R(1 + α )]( b v0 )
Then
2v 2 R(1 + α )
1
1
+ =0
R(1 + α ) rB [ R(1 + α ) b v0 ]2
=
2
b R(1 + α )
2
Now b min corresponds to rB → R.
Then
1
1
2
+ =2
R(1 + α ) R b min
R(1 + α )
b min =
or
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2

2 +α
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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 ε of the elliptic orbit.
SOLUTION
(
)
For earth, GM = gR 2 = ( 9.81) 6.37 × 106 = 398.06 × 1012 m3/s 2
rA =6370 + 19110 =25480 km =25.48 × 106 m
=
vcirc
GM
=
rA
=
vesc
398.06 × 1012
=
3.9525 × 103 m/s
6
25.48 × 10
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
GM
1
=
+ C cosθ B .
rB
h2
θB =
θ A + 180°, so that cosθ A =
− cosθ B
Adding,
h
=
GM
1
=
+ C cosθ A
rA
h2
1
1
+=
rA rB
2GMrArB
=
rA + rB
rA + rB
=
rArB
2GM
h2
( 2 ) ( 398.06 × 1012 )( 25.48 × 106 )(12.74 × 106 )
38.22 × 106
= 82.230 × 109 m 2 /s
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PROBLEM 12.104 (Continued)
v=
A
h
82.230 × 109
=
=
3.2272 × 103 m/s
6
rA
25.48 × 10
∆v =725 m/s 
(b) Decrease in speed. ∆v= vcirc − v A= 725 m/s
(c)
1
1
−=
rB rA
rA − rB
= C cosθ B − C cos
=
2C
A
rArB
=
C
rA − rB
=
2rArB
Ch 2
By Eq. (12.40), =
ε
=
GM
12.74 × 106
( 2 ) ( 25.48 × 10
6
=
19.623 × 10−9 m −1
6
12.74 × 10
)(
)
(19.623 × 10 )(82.230 × 10 )
−9
9
2
398.06 × 1012
ε = 0.333 
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PROBLEM 12.105
A space probe is to be placed in a circular orbit of 5600 mi 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 Δv 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 ΔvB . Finally, as the probe reaches Point C, it is
inserted in the desired circular orbit by reducing its speed by ΔvC .
Knowing that the mass of Venus is 0.82 times the mass of the earth,
that =
rA 9.3 × 103 mi and =
rB 190 × 103 mi, 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 =3690 mi =20.9088 × 106 ft, g =32.2 ft/s 2
For Earth,
GM earth =gR 2 =(32.2)(20.9088 × 106 ) 2 =14.077 × 1015 ft 3 /s 2
=
GM 0.82GM
=
11.543 × 1015 ft 3 /s 2
earth
For Venus,
For a parabolic trajectory with
rA =
9.3 × 103 mi =
49.104 × 106 ft
(v A=
)1 v=
esc
First transfer orbit AB.
2GM
=
rA
(2)(11.543 × 1015 )
= 21.683 × 103 ft/s
49.104 × 106
rB =
190 × 103 mi =
1003.2 × 106 ft
At Point A, where =
θ 180°
1 GM
GM
=
+ C cos 180
=
°
−C
2
2
rA
hAB
hAB
(1)
1 GM
GM
=2 + C cos 0 =2 + C
rB
hAB
hAB
(2)
At Point B, where θ = 0°
Adding,
r + rA 2GM
1
1
+= B =
2
rA rB
rA rB
hAB
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PROBLEM 12.105 (Continued)
Solving for hAB ,
(2)(11.543 × 1015 )(49.104 × 106 )(1003.2 × 106 )
= 1.039575 × 1012 ft 2 /s
6
1052.3 × 10
2GMrA rB
=
rB + rA
hAB
=
(v A=
)2
hAB 1.039575 × 1012
=
= 21.174 × 103 ft/s
rA
49.104 × 106
(v=
B )1
hAB 1.039575 × 1012
1.03626 × 103 ft/s
=
=
6
rB
1003.2 × 10
=
rC 5600
=
mi 29.568 × 106 ft
Second transfer orbit BC.
At Point B, where θ = 0
GM
1 GM
=2 + C cos 0 =2 + C
rB
hBC
hBC
At Point C, where =
θ 180°
GM
1 GM
=
+ C cos 180
=
°
−C
2
2
rC
hBC
hBC
Adding,
1
1 r + rC 2GM
+= B =
2
rB rC
rB rC
hBC
=
hBC
2GMrB rC
=
rB + rC
(2)(11.543 × 1015 )(1003.2 × 106 )(29.568 × 106 )
= 814.278 × 109 ft 2 /s
1032.768 × 106
(vB=
)2
hBC 814.278 × 109
=
= 811.69 ft/s
rB
1003.2 × 106
(v=
C )1
hBC 814.278 × 109
=
= 27.539 × 103 ft/s
rC
29.568 × 106
Final circular orbit.
=
rC 29.568 × 106 ft
(vC ) 2
=
GM
=
rC
11.543 × 1015
= 19.758 × 103 ft/s
29.568 × 106
Speed reductions.
(a)
At A:
(v A )1 − (v A )=
21.683 × 103 − 21.174 × 103
2
∆v A =
509 ft/s 
(b)
At B:
(vB )1 − (vB ) 2 = 1.036 × 103 − 811.69
∆vB =
224 ft/s 
(c)
At C:
(vC )1 − (vC )=
27.539 × 103 − 19.758 × 103
2
∆vC = 7.78 × 103 ft/s 
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PROBLEM 12.106
For the space probe of Problem 12.105, it is known that r=
9.3 × 103 mi and that the velocity of the probe is
A
reduced to 20,000 ft/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
Data from Problem 12.105:
rC =
29.568 × 106 ft, M =
0.82 M earth
For Earth,
R =3960 mi =20.9088 × 106 ft, g =32.2 ft/s 2
GM earth =gR 2 =(32.2)(20.9088 × 106 ) 2 =14.077 × 1015 m3 /s 2
=
GM 0.82GM
=
11.543 × 1015 ft 3 /s 2
earth
For Venus,
vA =
20, 000 ft/s, rA =
9.3 × 103 mi =
49.104 × 106 ft
Transfer orbit AB:
hAB =
rA v A =×
(49.104 106 )(20, 000) =
982.08 × 109 ft 2 /s
At Point A, where =
θ 180°
1 GM
GM
=
+ C cos 180
=
°
−C
2
2
rA
hAB
hAB
At Point B, where θ = 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)(11.543 × 1015 )
1
−
9 2
(982.08 × 10 )
49.104 × 106
= 3.57125 × 10−9 ft −1
(a)
Radial coordinate rB . =
rB 280.01 × 106 ft
(v=
B )1
=
rB 53.0 × 103 mi 
hAB 982.08 × 109
3.5073 × 103 ft/s
=
=
6
rB
280.01 × 10
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PROBLEM 12.106 (Continued)
Second transfer orbit BC. =
rC 5600
=
mi 29.568 × 106 ft
At Point B, where θ = 0
1 GM
GM
=2 + C cos 0 =2 + C
rB
hBC
hBC
At Point C, where =
θ 180°
GM
1 GM
=
+ C cos 180
=
°
−C
2
2
rC
hBC
hBC
r + rC 2GM
1
1
+= B =
2
rB rC
rB rC
hBC
Adding,
hBC
=
2GMrB rC
=
rB + rC
(2)(11.543 × 1015 )(280.01 × 106 )(29.568 × 106 )
309.578 × 106
= 785.755 × 109 ft 2 /s
Circular orbit with
(vB=
)2
hBC 785.755 × 109
=
= 2.8062 × 103 ft/s
rB
280.01 × 106
(v=
C )1
hBC 785.755 × 109
=
= 26.575 × 103 ft/s
rC
29.568 × 106
=
rC 29.568 × 106 ft
=
(vC ) 2
(b)
GM
=
rC
11.543 × 1015
= 19.758 × 103 ft/s
29.568 × 106
Speed reductions at B and C.
At B:
(vB )1 − (vB )=
3.5073 × 103 − 2.8062 × 103
2
∆vB =
701 ft/s 
At C:
(vC )1 − (vC )=
26.575 × 103 − 19.758 × 103
2
∆vC = 6.82 × 103 ft/s 
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PROBLEM 12.107
As it describes an elliptic orbit about the sun, a spacecraft
reaches a maximum distance of 202 × 106 mi from the center
of the sun at Point A (called the aphelion) and a minimum
distance of 92 × 106 mi 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
164.5 × 106 mi and 85.5 × 106 mi, 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 3960
mi 20.9088 × 106 ft
=
=
202 × 106 mi =
1066.56 × 109 ft
rA =
rB =
92 × 106 mi =
485.76 × 109 ft
From the solution to Problem 12.102, we have for any elliptic orbit about the sun
1 1 2GM sun
+ = 2
r1 r2
h
(a)
For the elliptic orbit AB, we have
=
r1 rA , r=
rB , =
h h=
rA v A
2
A
Also,
=
GM sun G[(332.8 × 103 ) M earth ]
2
= gRearth
(332.8 × 103 )
Then
using Eq. (12.30).
1
1 2 gR 2 (332.8 × 103 )
+ = earth
rA rB
(rA v A ) 2
1/ 2
or
R
v A = earth
rA
 665.6 g × 103 


1
1


+
rA
rB


3960 mi  665.6 × 103 × 32.2 ft/s 2
=
202 × 106 mi  1066.561× 109 ft + 485.761× 109 ft

= 52, 431 ft/s
1/ 2




=
v A 52.4 × 103 ft/s 
or
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PROBLEM 12.107 (Continued)
(b)
From Part (a), we have
1
1
=
2GM sun (rA v A ) 2  + 
 rA rB 
Then, for any other elliptic orbit about the sun, we have
(
2 1
1 1 (rA v A ) rA +
+ =
r1 r2
h2
1
rB
)
For the elliptic transfer orbit AB′, we have
=
r1 rA , r=
rB′ , =
h h=
rA (v A ) tr
2
tr
Then
(
2 1
1
1
1 (rA v A ) rA + rB
+
=
rA rB′
[rA (v A ) tr ]2
)
1/ 2
or
 1 + rA
 r1 + r1 
rB
A
B



(v A ) tr v=
v
=
A
A
 r1 + r1 
 1 + rA
rB′
 A B′ 

1/ 2




1/ 2
 1 + 202

92
= (52, 431 ft/s) 
 1 + 202 
85.5 

= 51,113 ft/s
Now
Then
=
htr (=
hA ) tr (hB′ ) tr : rA (=
v A ) tr rB′ (vB′ ) tr
(vB′ ) tr=
202 × 106 mi
× 51,113 ft/s= 120,758 ft/s
85.5 × 106 mi
For the elliptic orbit A′B′, we have
=
r1 rA=
rB=
′ , r2
′ , h rB′ vB′
Then
or
(
2 1
1
1 (rA v A ) rA +
+
=
rA′ rB′
(rB′ vB′ ) 2
r 
vB ′ = v A A 
rB′ 

1
rB
1
rA
+
1
rB
1
rA′
+
1
rB′
)
1/ 2




1
1
202 × 106 mi  202 × 106 + 92 × 106
= (52, 431 ft/s)
85.5 × 106 mi  164.51× 106 + 85.51× 106

= 116,862 ft/s
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1/ 2




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PROBLEM 12.107 (Continued)
Finally,
or
(v A ) tr= v A + ∆v A
∆=
v A (51,113 − 52, 431) ft/s
|∆v A | =
1318 ft/s 
or
and
=
vB′ (vB′ ) tr + ∆vB
or
=
∆vB′ (116,862 − 120, 758) ft/s
= −3896 ft/s
|∆vB | =
3900 ft/s 
or
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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
τH 
 aH 

 =

 τE 
 aE 
Thus
3
τ 
aH = aE  H 
 τE 
2/3
 76 years 
= rE 

 1 year 
= 17.94rE
But
2/3
1
(rmin + rmax )
2
11

=
rE
rE + rmax 
17.94

2 2

=
aH
1
rE
2
= (35.88 − 0.5)rE
=
rmax 2(17.94 rE ) −
= 35.38 rE
=
rmax (35.38)(150 × 106 km)
rmax
= 5.31 × 109 km 
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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 ε = 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,
2π RE 2π RE 3/ 2
GM
=
, τ0 =
RE
v0
GM
=
v0
or
=
GM
2π RE3/ 2
τ0
(1)
For the comet Hyakutake,
1 GM
=
=
(1 + ε ),
r0
h2
a=
=
h
τ
=
=
r0
1
(r0 + r1 )=
, b=
2
1− ε
r0 r1 =
1+ ε
r0
1− ε
GMr0 (1 + ε )
2π r02 (1 + ε )1/ 2
2π ab
=
h
(1 − ε )3/ 2 GMr0 (1 + ε )
2π r03/ 2
2π r03/ 2τ 0
=
3
3/ 2
GM (1 − ε )3/ 2 2π RE (1 − ε )
 r 
= 0 
 RE 
3/ 2
1
(1 − ε )3/ 2
= (0.230)3/ 2
Since
1 GM
1+ ε
=
(1 + ε ), r1 = r0
2
1− ε
r1
h
τ0
1
τ 0 91.8 × 103τ 0
=
3/ 2
(1 − 0.999887)
=
τ 0 1 yr,
=
τ (91.8 × 103 )(1.000)
=
τ 91.8 × 103 yr 
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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 /s 2
For Mars, GM = (0.1074)(398.43 × 1012 ) = 42.792 × 1012 m3 /s 2
=
rA 9000 km
= 9.0 × 106 m
=
rB 180000 km
= 180 × 106 m
For the parabolic approach trajectory at A,
2GM
=
rA
=
(v A )1
(2)(42.792 × 1012 )
= 3.0837 × 103 m/s
9.0 × 106
First elliptic transfer orbit AB.
1
GM
=
+ C cosθ A
2
rA
hAB
Using Eq. (12.39),
But
θB =
θ A + 180°,
Adding,
1
1
+=
rA rB
=
hAB
1
GM
=
+ C cosθ B .
2
rB
hAB
cosθ A =
− cosθ B .
so that
rA + rB
=
rArB
and
2GM
2
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 × 106 )(180 × 106 ) = 40.249 × 106 m
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PROBLEM 12.110 (Continued)
Periodic time for full ellipse:
For half ellipse AB,
τ=
AB
=
τ AB
τ =
2π ab
h
1
π ab
τ
=
2
h
π (94.5 × 106 )(40.249 × 106 )
27.085 × 109
= 444.81 × 103 s
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τ AB = 122.6 h 
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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/s 2
For earth,=
R 6.37 × 106 m ,
(
GM =gR 2 =( 9.81) 6.37 × 106
)
2
=398.06 × 1012 m3/s 2
For circular orbit of satellite,
r0 = 6370 + 500 = 6870 km = 6.87 × 106 m/s
=
v0
=
τ0
398.06 × 1012
= 7.6119 × 103 m/s
6.87 × 106
GM
=
r0
2π r0
=
v0
( 2π ) ( 6.87 × 106 )
= 5.6708 × 103 s
7.6119 × 103
For elliptic orbit of spacecraft it is given that
3
=
τ 0 8.5062 × 103 s
2
=
τ
1
a=
( rA + rB ) ,
2
Using Eq. (12.39),
But
θB =
θ A + 180°,
Adding,
so that
GM τ 2
a
=
=
4π 2
2π ab
=
h
2π ab a
=
GMb 2
or
3
4π 2
h=
GMb 2
a
2π a3 / 2
GM
10 )
(398.06 × 10 )(8.5062 ×=
12
1
GM
=
+ C cosθ B .
rB
h2
cosθ A =
− cosθ B .
1
1
r + rB
2a
2GM
+
= A
= 2 =
rA rB
rArB
b
h2
Periodic time:
τ
=
3
1
GM
and
=
+ C cosθ A
rA
h2
b=
rArB
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 × 106 m,
h
=
2π ab
=
τ
(
)(
b=
rArB = 8.7461 × 106 m
)
2π 9.0023 × 106 8.7461 × 106
= 58.159 × 109 m 2 /s
3
8.5062 × 10
v=
A
h
58.159 × 109
=
= 8.4656 × 103 m/s
6
rA
6.87 × 10
(a) Increase in speed at A.
∆v A = v A − v0 = 8.4656 × 103 − 7.6119 × 103
∆v A =
854 m/s 
(b) Periodic time for elliptic orbit.
τ = 141.8 min 
As calculated above τ = 8.5062 s
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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 /s 2
For moon,
GM = (0.01230)(398.06 × 1012 ) = 4.896 × 1012 m3 /s 2
rA = 1737 + 400 = 2137 km = 2.137 × 106 m
rB =1737 + 2940 = 4677 km = 4.677 × 106 m
Using Eq. (12.39),
1 GM
=
+ C cos θ A
rA
h2
But
θ B =+
θ A 180°, so that cos θ A =
− cos θ B .
Adding,
and
1 GM
=
+ C cos θ B .
rB
h2
1 1 rA + rB 2GM
+=
=
2
rA rB
rA rB
hAB
=
hAB
2GMrA rB
=
rA + rB
(2)(4.896 × 1012 )(2.137 × 106 )(4.677 × 106 )
6.814 × 106
= 3.78983 × 109 m 2 /s
1
(rA + rB )= 3.402 × 106 m
2
=
b =
rA rB 3.16145 × 106 m
a=
Periodic time.
τ
=
2π ab 2π (3.402 × 106 )(3.16145 × 106 )
=
= 17.831 × 103 s
9
hAB
3.78983 × 10
τ = 4.95 h 
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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
(v A ) par = 10,131.4 m/s
1
=
(v A ) par 7164.0 m/s
2
=
(v A )circ
and
rA =
(6052 + 280) km =
6332 km
Also,
For the parabolic trajectory BA, we have
1 GM v
(1 + ε cos θ )
=
2
r
hBA
[Eq. (12.39′)]
where ε = 1. Now
1 GM v
=
(1 + 1)
2
rA
hBA
at A, θ = 0:
rA =
or
at B, θ =
−90°:
2
hBA
2GM v
1 GM v
=
(1 + 0)
2
rB
hBA
rB =
or
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)=
A=
BA
BA
Now
2
4 2
rA=
rB
rA
3
3
dA 1
= h
dt 2
where h = constant
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PROBLEM 12.113 (Continued)
Then
=
A
2 ABA
1
=
ht or t BA =
hBA rA v A
2
hBA
=
t BA
2 × 43 rA2 8 rA
=
3 vA
rA v A
8 6332 × 103 m
3 10,131.4 m/s
= 1666.63 s
=
For the circular trajectory AC,
t AC
=
Finally,
π r
π 6332 × 103 m
2 A
=
= 1388.37 s
(v A )circ 2 7164.0 m/s
t BC
= t BA + t AC
= (1666.63 + 1388.37) s
=3055.0 s
t BC = 50 min 55 s 
or
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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 b v0, where b < 1, to place the probe on a
crash trajectory. Express in terms of n and b the angle AOB, where B denotes
the point of impact of the probe on the planet.
SOLUTION
r=
r=
nR
0
A
For the circular orbit,
v0
=
GM
=
r0
GM
nR
The crash trajectory is elliptic.
=
v A b=
v0
b 2GM
nR
=
=
h r=
nRv
AvA
A
b 2 nGMR
GM
1
= 2
2
h
b nR
1 GM
1 + ε cos θ
= 2 (1 + ε cos θ ) = 2
r
b nR
h
At Point A, =
θ 180°
1
1
1− ε
=
=
or b 2 =
1 − ε or ε =
1− b 2
2
rA nR b nR
At impact Point B, θ= π − φ
1 1
=
rB R
1 1 + ε cos (π − φ ) 1 − ε cos φ
= =
R
b 2 nR
b 2 nR
1 − nb 2 1 − nb 2
ε cos φ =
=
1 − nb 2 or cos φ =
ε
1− b 2
φ = cos −1[(1 − nb 2 ) /(1 − b 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 Rφ of the trajectory is 6000 km, determine (a) the velocity
of the projectile at C, (b) the eccentricity ε 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 /s 2
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
ϕ
=
s AB
=
R
6.00 × 106
rad 53.968°
= 0.94192 =
6.37 × 106
1 GM
For an elliptic trajectory,=
(1 + ε cos θ )
r
h2
At A,
ϕ
θ= 180° − = 153.016°,
At C,
1
=
θ 180° , =
rC
2
1
GM
=
(1 + ε cos153.016°)
rA
h2
GM
(1 − ε )
h2
Dividing Eq. (1) by Eq. (2),
rC
1 + ε cos153.016°
=
= 1.23548
1−ε
rA
ε
=
1.23548 − 1
= 0.68384
1.23548 + cos153.016°
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(1)
(2)
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PROBLEM 12.115 (Continued)
From Eq. =
(2), h
h=
GM (1 − ε )rC
(398.06 × 1012 )(0.31616)(7.87 × 106 ) = 31.471 × 109 m 2 /s
v=
C
(a) Velocity at C.
(b)
h
31.471 × 109
=
= 4.00 × 103 m/s
6
rC
7.87 × 10
vC = 4 km/s 
ε = 0.684 
Eccentricity of trajectory.
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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 /s 2
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,
398.06 × 1012
= 7.5773 × 103 m/s
6.933 × 106
GM
=
rA
=
vcirc
For the descent trajectory,
=
v A 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
=
r
At Point A,
=
θ 180°,
GM
(1 + ε cos θ )
h2
r = rA
1
GM
=
(1 − ε )
rA
h2
1=
−ε
h2
=
GM rA
(51.4795 × 109 ) 2
= 0.96028
(398.06 × 1012 )(6.933 × 106 )
ε = 0.03972
GM
1
=
(1 + ε cos θ B )
rB
h2
1 + ε cos θ B =
h2
(51.4795 × 109 ) 2
=
= 1.02567
GM rB
(398.06 × 1012 )(6.491 × 106 )
=
cosθ B
=
θ B 49.7°
1.02567 − 1
= 0.6463
ε
AOB
= 180° − θ =
130.3°
B
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AOB
= 130.3° 
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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 280 mi above the
surface of the planet. The trajectory of the probe is a hyperbola of eccentricity
ε = 1.031. Knowing that the radius and the mass of Jupiter are 44423 mi and
1.30 × 1026 slug, 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 vB of the probe at B.
SOLUTION
r=
(44.423 × 103 + 280) mi
= 44.703 × 103 mi
B
First we note
(a)
We have
1 GM j
=
(1 + ε cos θ )
r
h2
At A, θ = 0:
1 GM j
(1 + ε )
=
rA
h2
h2
= rA (1 + ε )
GM j
or
θ θ=
 AOB :
At B, =
B
or
Then
or
[Eq. (12.39′)]
1 GM j
=
(1 + ε cos θ B )
rB
h2
h2
= rB (1 + ε cos θ B )
GM j
rA (1 + ε ) = rB (1 + ε cos θ B )

1  rA
 (1 + ε ) − 1
ε  rB

cos=
θB

1  44.0 × 103 mi
(1 + 1.031) − 1

3
1.031  44.703 × 10 mi

=
= 0.96902
or
=
θ B 14.2988°
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 AOB
= 14.30° 
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PROBLEM 12.117 (Continued)
(b)
From above
where
Then
=
h 2 GM j rB (1 + ε cos θ B )
1
| rB × mv B | = rB vB sin φ
m
) 97.1988°
φ= (θ B + 82.9°=
h=
(rB vB=
sin φ ) 2 GM j rB (1 + ε cos θ B )
or
1/2

1  GM j
(1 + ε cos θ B ) 
vB
=

sin φ  rB

1/2
 34.4 × 10−9 ft 4 /lb ⋅ s 4 × (1.30 × 1026 slug)

1
× [1 + (1.031)(0.96902)]

6
sin 97.1988° 
236.03 × 10 ft

vB = 196.2 ft/s 
or
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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 
=
 + 
ρ 2  r0 r1 
SOLUTION
Using Eq. (12.39),
1 GM
=
+ C cos θ A
rA
h2
and
1 GM
=
+ C cos θ B .
rB
h2
But
so that
Adding,
θ B =θ A + 180°,
cos θ A = − cos θ B
1 1 2GM
+ =2
rA rB
h
At Points A and B, the radial direction is normal to the path.
a=
n
But
=
Fn
v2
=
ρ
h2
r 2ρ
GMm
mh 2
=
=
ma
n
r2
r 2ρ
1 GM 1  1
1
= =
 + 
2
ρ
2  rA rB 
h
1 1 1 1 
=
 +  
ρ 2  r0 r1 
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PROBLEM 12.119
(a) Express the eccentricity ε 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
RE 149.6 × 106 km, to determine the approximate
12.109, where =
maximum distance from the sun reached by comet Hyakutake.
SOLUTION
(a)
We have
1 GM
=
(1 + ε cos θ )
r
h2
At A, θ = 0:
1 GM
=
(1 + ε )
r0
h2
h2
= r0 (1 + ε )
GM
or
At B, =
θ 180°:
or
Then
Eq. (12.39′)
1 GM
(1 − ε )
=
r1
h2
h2
= r1 (1 − ε )
GM
r0 (1 + ε ) = r1 (1 − ε )
ε=
or
(b)
r1 − r0

r1 + r0
1+ ε
r0
1− ε
From above,
r1 =
where
r0 = 0.230 RE
Then
r1 =
1 + 0.999887
× 0.230(149.6 × 109 m)
1 − 0.999887
=
r1 609 × 1012 m 
or
=
=
RE or r1 0.064 lightyears.
Note: r1 4070
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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 and b =
rA rB
Using Eq. (12.37),
1 GM
=
+ C cos θ A
rA
h2
and
1 GM
=
+ C cos θ B .
rB
h2
But
so that
Adding,
θ B =θ A + 180°,
cos θ A = − cos θ B .
1
1 r +r
2a 2GM
+ = A B = 2 =
rA rB
rA rB
b
h2
h=b
GM
a
By Eq. (12.43),
2π ab 2π ab a 2π a 3/ 2
=
=
h
b GM
GM
=
τ
τ2 =
4π 2 a3
GM
For Orbits 1 and 2 about the same large mass,
and
τ12 =
4π 2 a13
GM
τ 22 =
4π 2 a23
GM
2
3
 τ1   a1 
  =  
 τ 2   a2 
Forming the ratio,
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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 ε about a planet of mass M can be expressed as
=
h
GMa (1 − ε 2 )
SOLUTION
By Eq. (12.39′),
1 GM
=
(1 + ε cos θ )
r
h2
At A, θ = 0°:
1 GM
=
=
(1 + ε )
or
rA
h2
h2
rA =
GM (1 + ε )
θ 180°:
At B, =
1 GM
(1 − ε )
or
=
=
rB
h2
h2
rB =
GM (1 − ε )
h2
1
 1
= 
+
GM  1 + ε 1 − ε
Adding,
rA + rB=
But for an ellipse,
rA + rB =
2a
2a =
2
2h

=
2
 GM (1 − ε )
2h 2
GM (1 − ε 2 )
=
h
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GMa (1 − ε 2 ) 
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PROBLEM 12.122
In the braking test of a sports car its velocity is reduced from 70 mi/h to zero in a distance of 170 ft with
slipping impending. Knowing that the coefficient of kinetic friction is 80 percent of the coefficient of static
friction, determine (a) the coefficient of static friction, (b) the stopping distance for the same initial velocity if
the car skids. Ignore air resistance and rolling resistance.
SOLUTION
(a)
Coefficient of static friction.
=
ΣFy 0:
N=
−W 0
N =W
=
v0 70
=
mi/h 102.667 ft/s
v 2 v02
−
= at (s − s0 )
2
2
at =
v 2 − v02
0 − (102.667)2
=
= − 31.001 ft/s 2
2(s − s0 )
(2)(170)
=
m | at |
For braking without
skidding µ µ=
s , so that µ s N
Σ=
Ft mat : − µ s=
N mat
µs =
−
(b)
mat
a
31.001
=
− t =
W
g
32.2
µ s = 0.963 
Stopping distance with skidding.
µ µ=
(0.80)(0.963)
= 0.770
Use =
k
ΣF =mat : µk N =−mat
µ N
− k =
−µk g =
− 24.801 ft/s 2
at =
m
Since acceleration is constant,
(s − =
s0 )
v 2 − v02
=
2at
0 − (102.667)2
(2)(− 24.801)
s − s0 =
212 ft 
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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 θ= 30°.
SOLUTION
Initial velocity of drop = velocity of bucket
=
ΣFy 0: =
T cos 30° mg
(1)
=
ΣFx max=
:
T sin 30° man
(2)
Divide (2) by (1):
tan 30=
°
an
=
g
v2
pg
v 2 ρ g tan 30°
Thus =
But
=
ρ L sin
=
30° (1.2 m) sin=
30° 0.6 m
Thus
=
v 2 0.6(9.81)
=
tan 30° 3.398 m 2 /s 2
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
y=
2L − L cos30=
° 2(1.2) − 1.2cos30=
° 1.361 m
0
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 12-lb block B rests as shown on the upper surface of a 30-lb 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=
a A + a B/ A
B
Block B:
=
ΣFx max : mB aB /A − m
=
B a A cos30° 0
=
aB/A a A cos30°
(1)
ΣFy =
ma y : N AB − WB =
−mB a A sin 30°
N AB =
WB − (WB sin 30°)
Block A:
aA
g
(2)
=
ΣF ma : WA sin 30° =
+ N AB sin 30° WA
WA sin 30° + WB sin 30° − (WB sin 2 30°)
=
aA
aA
g
aA
a
=WA A
g
g
(WA + WB )sin 30°
(30 + 12)sin 30°
=
=
(32.2) 20.49 ft/s 2
g
2
30 + 12sin 2 30°
WA + WB sin 30°
a A = 20.49 ft/s 2
(a)
30° 
=
aB/ A (20.49)
=
cos 30° 17.75 ft/s 2
a B/A = 17.75 ft/s 2
(b)
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PROBLEM 12.125
A 500-lb crate B is suspended from a cable attached to a 40-lb trolley A
which rides on an inclined I-beam as shown. Knowing that at the instant
shown the trolley has an acceleration of 1.2 ft/s2 up and to the right,
determine (a) the acceleration of B relative to A, (b) the tension in cable CD.
SOLUTION
(a)
a A + a B/A , where a B/A is directed perpendicular to cable AB.
First we note: a=
B
ΣFx =
mB ax : 0 =
−mB ax + mB a A cos 25°
B:
=
aB/A (1.2 ft/s 2 ) cos 25°
or
a B/A = 1.088 ft/s 2
or
(b)

For crate B
=
ΣFy mB a y : T=
AB − WB
WB
a A sin 25°
g
A:
 (1.2 ft/s 2 )sin 25° 
=
TAB (500 lb) 1 +

32.2 ft/s 2


= 507.87 lb
or
For trolley A
=
ΣFx mA a A : TCD − TAB sin 25° −=
WA sin 25°
or
WA
aA
g

1.2 ft/s 2
=
TCD (507.87 lb)sin 25° + (40 lb)  sin 25° +
32.2 ft/s 2




TCD = 233 lb 
or
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PROBLEM 12.126
The roller-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 (µk = 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.
ρ = 30 m
a=
n
v 2 (20)2
=
= 13.333 m/s 2
30
ρ
=
ΣFy ma y : N R + N F=
− mg man
N R + N F = m( g + an )
F= µk ( N R + N F )= µk m( g + an )
Σ=
Fx max : −=
F mat
F
− =
− µk ( g + an )
at =
m
| at |= µk ( g + an )= 0.25(9.81 + 13.33)
(b)
Between A and B.
| at | = 5.79 m/s 2 
ρ= ∞
an = 0
=
| at | µ=
(0.25)(9.81)
kg
(c)
Just passed Point B.
ρ = 45 m
a=
n
v 2 (20) 2
=
= 8.8889 m/s 2
45
ρ
ΣFy =
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= µk ( N R + N F )= µk m( g − an )
Σ=
Fx max : −=
F mat
F
− =
− µk ( g − an )
at =
m
| at |= µk ( g − an )= (0.25)(9.81 − 8.8889)
| at | = 0.230 m/s 2 
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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 θ is between 20º and 40º, (where t
= 0 at θ = 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 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
30°, θ =°
2 /s=
0.0349 rad/s, θ=0
θ0 =20°, θ =
Given:
=
=
T 18
kN, m 75 kg
θ= θ 0 + θt
30 = 20 + 2t ⇒ t = 5 s
Kinematics:
a=
aB + aP/ B
P
aP =
0 + ar e r + aθ eθ
Using Radial and Transverse Coordinates:
Free Body Diagram of parasailor (side view):
1 12
r = − t m/s
2
1 − 12

r = − t m/s 2
4
Equations of Motion:
∑F
r
= mar
(
FP cos b − T − mg sin θ= m 
r − rθ 2
FP cos b = T + m  g sin θ + 
r − rθ 2 
∑ Fθ = maθ
)
(
FP sin b − mg cos θ =
m rθ + 2rθ
(1)
FP=
sin b m  g cos θ + rθ + 2rθ 
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(2)
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PROBLEM 12.127 (Continued)
r=
121.27 m, r =
−1.118 m/s, 
r=
−0.1118 m/s 2
Evaluate at t=5 s:
(1)
2
FP cos
b 18000 + 75 9.81sin 30° − 0.1118 − 121.27 ( 0.0349 )  ⇒ FP cos
b 18348.4 N
=
=


(2)
=
FP sin b 75 9.81cos 30° + 0 + 2 ( −1.118 )( 0.0349 )  =
⇒ FP sin b 631.33 N
FP sin b
FP cos b
FP cos1.97° =18348.4 N
=
631.33
⇒ tan b = 0.03441 ⇒ b = 1.97°
18348.4
FP = 18.4 kN
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31.97° 
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PROBLEM 12.128
A small 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) θ= 90°, (b) θ= 75°, (c) θ= 45°. Indicate in each case the direction
of the impending motion.
SOLUTION
vC = (r sin θ )φAB
First note
= (0.6 m)(6 rad/s)sin θ
= (3.6 m/s)sin θ
(a)
With θ= 90°,
vC = 3.6 m/s
ΣF=
0: F − W=
0
y
C
or
F = mC g
Now
F = µs N
or
N=
1
µs
=
ΣFn mC a=
mC
n: N
1
or
or
µs
mC g = mC
µ=
s
mC g
vC2
r
vC2
r
gr (9.81 m/s 2 )(0.6 m)
=
vC2
(3.6 m/s)2
( µ 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 θ, it is not known whether the impending motion will be upward or
downward. To consider both possibilities for each value of θ, let Fdown correspond to impending motion
downward, Fup correspond to impending motion upward, then with the “top sign” corresponding to Fdown,
we have
=
ΣFy 0: N cos θ ± F sin=
θ − WC 0
F = µs N
Now
N cos θ ± µ s N sin θ − mC g =
0
Then
or
N=
mC g
cos θ ± µ s sin θ
and
F=
µ s mC g
cos θ ± µ s sin θ
vC2
=
ΣFn mC an : N sin θ =
 F cos θ mC =
ρ r sin θ
ρ
Substituting for N and F
mC g
µ s mC g
v2
sin θ 
cos θ = mC C
cos θ ± µ s sin θ
cos θ ± µ s sin θ
r sin θ
vC2
µs
tan θ

=
1 ± µ s tan θ 1 ± µ s tan θ gr sin θ
or
µs = ±
or
1+
vC2
gr sin θ
vC2
gr sin θ
tan θ
vC2
[(3.6 m/s)sin θ ]2
=
=
2.2018sin θ
gr sin θ (9.81 m/s 2 )(0.6 m)sin θ
Now
µs = ±
Then
(b)
tan θ −
tan θ − 2.2018 sin θ
1 + 2.2018sin θ tan θ
θ= 75°
tan 75° − 2.2018sin 75°
1 + 2.2018sin 75° tan 75°
±
=
± 0.1796
µs =
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PROBLEM 12.128 (Continued)
Then
downward:
µ s = + 0.1796
upward:
µs < 0
not possible
( µ s )min = 0.1796 
The direction of the impending motion is downward. 
(c)
θ= 45°
µ s =±
tan 45° − 2.2018sin 45°
=± (− 0.218)
1 + 2.2018sin 45° tan 45°
Then
downward:
µs < 0
upward:
µ s = 0.218
not possible
( µ s )min = 0.218 
The direction of the impending motion is upward. 
Note: When
or
tan θ − 2.2018sin θ =
0
=
θ 62.988°,
µ s = 0. Thus, for this value of θ , friction is not necessary to prevent the collar from sliding on the rod.
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PROBLEM 12.129
Telemetry technology is used to quantify kinematic values of a
200-kg roller coaster cart as it passes overhead. According to
r =
−2 m/s 2 , θ =
90°,
the system, r = 25 m, r = −10 m/s, 
θ = −0.4 rad/s, θ = −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
−10 m/s
vθ =
rθ =
(25 m)( − 0.4 rad/s) =
So the tangential direction is
45° and v = 10 2 m/s.

ar =
r − rθ 2 =
−2 m/s − (25 m)( − 0.4 rad/s) 2
= −6 m/s 2
a=
rθ + 2rθ
θ
= (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 =
ρ
=
v2
ρ
v2
(10 2)2
=
an
6cos 45°
ρ = 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 ρ the mean density of the planet, show that the time required by the moon to complete one full revolution
about the planet is (24 π /G ρ )1/2 , where G is the constant of gravitation.
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.
GM
=
vτ 2=
πr
or
τ
2π r
r
τ=
Solving for τ,
But
=
M
τ=
Then
2π r 3/2
GM
4
π
=
GM 2
G ρ R3/2
π R3 ρ , hence,
3
3
3π  r 
G ρ  R 
3/2
Using r = 2R as a given leads to
3π
3/2
=
τ 2=
Gρ
24π
Gρ
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τ = (24π /G ρ )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 40 mi 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 170 mi, 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
R=
3960 mi =
20.909 × 106 ft, g =
32.2 ft/s 2
For Earth,
GM =gR 2 =(32.2)(20.909 × 106 ) 2 =14.077 × 1015 ft 3 /s 2
(a)
For the elliptic orbit,
rA= 3960 + 40= 4000 mi= 21.12 × 106 ft
rB = 3960 + 170 = 4130 mi = 21.8064 × 106 ft
1
(rA + rB )= 21.5032 × 106 ft
2
=
b =
rA rB 21.4605 × 106 ft
a=
Using Eq. 12.39,
1 GM
=
+ C cos θ A
rA
h2
and
1 GM
=
+ C cos θ B
rB
rB2
But θ B =θ A + 180°, so that cos θ A = − cos θ B
1
1 r +r
2a 2GM
+ = A B = 2 =
rA rB
rA rB
b
h2
Adding,
h=
or
Periodic time.
=
τ
=
τ
GMb 2
a
2π ab 2π ab a 2π a 3/ 2
=
=
h
GM
GMb 2
2π (21.5032 × 106 )3/ 2
= 5280.6
=
s 1.4668 h
14.077 × 1015
The time to travel from A to B is one half the periodic time
τ AB = 0.7334 h
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PROBLEM 12.131 (Continued)
(b)
For the circular orbit,
a= b= rB= 21.8064 × 106 ft
=
τ circ
2π a3/ 2 2π (21.8064 × 106 )3/ 2
=
= 5393 s
GM
14.077 × 1015
τ circ = 1.498 h
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τ circ = 89.9 min 
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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
3
sun is 332.8 × 10 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 φ , 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 tprobe for the probe to travel from the earth to Venus:
1
tprobe = τ tr
2
where τ 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
τ tr2
=
2
τ earth
atr3
3
aearth
1
(rE + rv )
2
1
=
(93 × 106 + 67.2 × 106 ) mi
2
= 80.1 × 106 mi
where
=
atr
and
aearth ≈ rE
Then
tprobe =
=
( Note: ε earth =
0.0167)
1  atr 
 
2  rE 
3/ 2
τ earth
1  80.1 × 106 mi 


2  93.0 × 106 mi 
3/ 2
(365.25 days)
= 145.977 days
= 12.6124 × 106 s
In time tprobe , Venus travels through the angle θv given by
vv
=
θv θ=
tprobe
v tprobe
rv
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PROBLEM 12.132 (Continued)
Assuming that the orbit of Venus is circular (note: ε venus = 0.0068), then, for a circular orbit
vv =
Now
GM sun
rv
[Eq. (12.44)]
=
GM sun G (332.8 × 103 M earth )
(
2
= 332.8 × 103 gRearth
Then
(
using Eq. (12.30)
3
2
tprobe  332.8 × 10 gRearth

θv =
rv 
rv

= tprobe Rearth
where
)
) 
1/ 2


(332.8 g × 103 )1/ 2
rv3/ 2
=
Rearth 3960
=
mi 20.9088 × 106 ft
and
rv =
67.2 × 106 mi =
354.816 × 109 ft
Then
(12.6124 × 106 s)(20.9088 × 106 ft)
θv =
(332.8 × 103 × 32.2 ft/s 2 )1/ 2
(354.816 × 109 ft)1/2
= 4.0845 rad
= 234.02°
Finally,
φ =θv − 180° =234.02° − 180°
=
ϕ 54.0° 
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PROBLEM 12.133*
Disk A rotates in a horizontal plane about a vertical axis at the
constant rate θ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
r=
0, xsp =⇒
0 Fsp =
kr
when
=
r0 500
=
mm 0.5 m
and
 θ= 12 rad/s
θ=
0
then
θ = 0
mB ar : − Fsp= mB (
r − rθ02 )
ΣF=
r
 k


0
− θ02  r =
r +
 mB

ΣFθ= mB aθ : FA= mB (0 + 2rθ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
t 0,=
r 0 :
= 
r= 0 and at=
dt
∫
r
0
dr =
∫
0.1
0
(0) dt
r = 0
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(2)
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PROBLEM 12.133* (Continued)
dr
=
t 0,=
r0 0.5 m
= r= 0 and at
dt
and
∫
r
dr =
r0
∫
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

=
r
Now
d
(r) =
r vr
dt
dvr
dr

r = vr
Then
so that
dvr
0
+ 100r =
dr
vr
∫
t 0,=
vr 0,=
r r0 :
At=
d dr d
d
= = vr
dt dt dr
dr
vr
0
vr d vr = −100
∫
r
r dr
r0
vr2 =
−100(r 2 − r02 )
=
vr 10 r02 − r 2
v=
r
Now
t 0,=
r r0 :
At=
∫
r
r0
dr
=
r02 − r 2
dr
= 10 r02 − r 2
dt
t
10 dt
∫=
0
10t
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PROBLEM 12.133* (Continued)
Let
=
r r0=
sin φ ,
dr r0 cos φ dφ
sin −1 ( r/r0 )
∫π
Then
/2
r0 cos φ dφ
r02 − r02 sin 2 φ
sin −1 ( r/r0 )
∫π
/2
r
sin −1 
 r0
= 10t
dφ = 10t
 π
10 t
− =
 2
π

=
=
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)
F=
1 kg × 2 × [−(5 ft/s)sin (10 × 0.1)] (10 rad/s)
H
FH = −84.1 N 
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PROBLEM 12.CQ1
A 1000 lb boulder B is resting on a 200 lb 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 lb
(b) The tension in the cord connected to the truck is 1200 lb
(c) The tension in the cord connected to the truck is greater
than 1200 lb
(d ) The normal force between A and B is 1000 lb
(e) The normal force between A and B is 1200 lb
( 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 lb and (d) the normal force between A and B is 1000 lb.
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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 40 lb weights are connected
by an inextensible cord, and on the right,
a constant 40 lb 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 40 lb, 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) Since B 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 µk. 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 µk, 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 µk, 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 v in
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 5 in. and a constant k. Knowing that
the collar has a speed v at Point C, draw the FBD and KD of
the collar at this point.
SOLUTION
where x = 2/12 ft and r = 5/12 ft.
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PROBLEM 12.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 ω. 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 the constant 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, 
−0.08 rad/s, θ =
r=
0, θ =
0
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PROBLEM 12.F11
Disk A rotates in a horizontal plane about a vertical axis at the constant
rate θ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 and slides along the slot in the rotating arm
OC and along the slot DE which is cut in a fixed horizontal
plate. Neglecting friction and knowing that rod OC rotates at
the constant rate θ0 , draw a FBD and KD that can be used to
determine the forces P and Q exerted on pin B by rod OC and
the wall of slot DE, respectively.
SOLUTION
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