CHAPTER 12
PROBLEM 12.1
The value of g at any latitude φ may be obtained from the formula
g = 9.7807(1 + 0.0053 sin2f) m/s2
which takes into account the effect of the rotation of the earth, as well as the fact that the earth is not truly
spherical. Determine to four significant figures (a) the weight in newtons, (b) the mass in kilograms, at the
latitudes of 0°, 45°, and 90°, of a silver bar, the mass of which has been officially designated as 5 kg.
SOLUTION
g = 9.7807(1 + 0.0053 sin2f) m/s2
(a)
(b)
Weight:
φ = 0° :
g = 9.7807 m/s2
φ = 45°:
g = 9.8066 m/s2
φ = 90°:
g = 9.8325 m/s2
W = mg
φ = 0° :
W = (5 kg)(9.7807 m/s2) = 48.90 N
φ = 45°:
W = (5 kg)(9.8066 m/s2) = 49.03 N
φ = 90°:
W = (5 kg)(9.8325 m/s2) = 49.16 N
m = 5.000 kg
Mass: At all latitudes:
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
No part
part of
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced or
or distributed
distributed in
in any
any form
form or
reproduced
distribution to
toteachers
teachersand
andeducators
educatorspermitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
If you
you are
are aastudent
studentusing
usingthis
thisManual,
Manual,
distribution
you are
are using
using itit without
without permission.
permission.
you
275
275
PROBLEM 12.2
The acceleration due to gravity on the moon is 1.62 m/ss 2 . Determine (a) the weight in newtons, (b) the mass
in kilograms, on the moon, of a gold bar, the mass
ss of which has been officially designated as 2 kg.
SOLUTION
(a)
Weight:
(b)
Mass: Same as on earth:
W = mg
mg = (2 kg)(1.62 m/ss 2 )
W = 3.24 N �
m = 2.00 kg � ��
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
276
PROBLEM 12.3
A 200-kg satellite is in a circular orbit 1500 km above the surface of Venus. The acceleration due to the
gravitational attraction of Venus at this altitude is 5.52 m/ss 2 . Determine the magnitude of the linear momentum
of the satellite knowing that its orbital speed is 23.4 × 103 km/h
km .
SOLUTION
First note
v = 23.4
23.4 × 1
10
03 km/h
h=6
6500 m/s
Now
L = mv
mv = 200 kg × 6500 m/s
L = 1.300
1.300 × 1
1006 kg ⋅ m
m/s ��
or
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
277
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 1.2 m/s2 the spring
scale indicates a load of 3 kg, 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 1.2 m/s2.
SOLUTION
Assume g = 9.81 m/s2
(a)
m=
ΣF = ma : W − Fs =
W 1−
W
g
W
a
g
a
= Fs
g
W=
or
Fs
1−
a
g
=
3 × 9.81
1.2
1−
9.81
W = 33.53 N
(b)
ΣF = ma : Fs − W =
W
a
g
Fs = W 1 +
a
g
= 33.53 1 +
1.2
9.81
Fs = 37.63 N
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
part of
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
No part
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced or
or distributed
distributed in
in any
any form
form or
reproduced
distribution to
to teachers
teachers and
andeducators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
If you
you are
are aa student
student using
using this
thisManual,
Manual,
distribution
you are
are using
using itit without
without permission.
permission.
you
278
278
PROBLEM 12.4 (Continued)
For the balance system B,
ΣM 0 = 0: bFw − bbF
Fp = 0
Fw = Fp
But
Fw = Ww 1 +
a
g
and
Fp = W p 1 +
a
g
so that
Ww = W p
and
mw =
Wp
g
=
33.53
9.81
mw = 3.42 kg
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
279
PROBLEM 12.5
A hockey player hits a puck so that it comes to rest in 9 s after sliding 30 m on the ice. Determine (a) the
initial velocity of the puck, (b) the coefficient of friction between the puck and the ice.
SOLUTION
(a)
(b)
Assume uniformly decelerated motion.
Then
v = v0 + at
At t = 9 s:
0 = v0 + a(9)
or
v
a=− 0
9
Also
v 2 = v02 + 2a( x − 0)
At t = 9 s:
0 = v02 + 2a (30)
Substituting for a
� v �
0 = v02 + 2 � − 0 � (30) = 0
� 9�
or
v0 = 6.6667 m/s
and
a=−
or v0 = 6.67 m/s �
6.6667
= −0.74074 m/s 2
9
We have
+ ΣFy = 0: N − W = 0
Sliding:
or
F = µ k N = µ k mg
or − µ k mg = ma
ΣFx = ma : −F = ma
or
N = W = mg
µk = −
a
−0.74074 m/s 2
=−
g
9.81 m/s 2
or µk = 0.0755 �
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
All rights
rights reserved.
reserved. No
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
Inc. All
No part
part of
of this
reproduced or
or distributed
distributed in
in any
any form
form or
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced
are aa student
student using
using this
this Manual,
Manual,
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
individual course
distribution
for their
their individual
course preparation.
preparation. If
If you
you are
you are
are using
using it
it without
without permission.
you
permission.
280
280
PROBLEM 12.6
Determine the maximum theoretical speed that an automobile starting from rest can reach after traveling
400 m. Assume that the coefficient of static friction is 0.80 between the tires and the pavement and that (a) the
automobile has front-wheel drive and the front wheels support 62 percent of the automobile’s weight, (b) the
automobile has rear-wheel drive and the rear wheels support 43 percent of the automobile’s weight.
SOLUTION
(a)
For maximum acceleration
FF = Fmax = µ s N F = 0.8(0.62 W )
= 0.496 W = 0.496 mg
Now
or
ΣFx = ma: FF = ma
0.496 mg = ma
a = 0.496(9.81 m/s 2 ) = 4.86576 m/s 2
Then
Since a is constant, we have
v 2 = 0 + 2a( x − 0)
When
2
x = 400 m: vmax
= 2(4.86576 m/s 2 )(400 m)
vmax = 62.391 m/s
or
vmax = 225 km/h ��
or
(b)
For maximum acceleration
FR = Fmax = µ s N R = 0.8(0.43 W )
= 0.344 W = 0.344 mg
Now
or
ΣFx = ma: FR = ma
0.344 mg = ma
a = 0.344(9.81 m/s 2 ) = 3.37464 m/s 2
Then
Since a is constant, we have
v 2 = 0 + 2a( x − 0)
When
or
2
x = 400 m: vmax
= 2(3.37464 m/s 2 )(400 m)
vmax = 51.959 m/s
vmax = 187.1 km/h �
or
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
No part
part of
reproduced or
or distributed
distributed in
in any
any form
form or
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced
you are
are aa student
student using
using this
this Manual,
Manual,
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
distribution
If you
you are
are using
using it
it without
without permission.
permission.
you
281
281
PROBLEM 12.7
In anticipation of a long 7° upgrade, a bus driver accelerates at a constant rate of 1 m/s2 while still on a level
section of the highway. Knowing that the speed of the bus is 90 km/h as it begins to climb the grade and that
the driver does not change the setting of his throttle or shift gears, determine the distance traveled by the bus
up the grade when its speed has decreased to 80 km/h.
SOLUTION
First consider when the bus is on the level section of the highway.
alevel = 1 m/s2
We have
ΣFx = ma: P =
W
alevel
g
Now consider when the bus is on the upgrade.
We have
Substituting for P
ΣFx = ma: P − W sin 7° =
W
a′
g
W
W
alevel − W sin 7° = a′
g
g
a′ = alevel − g sin 7°
or
= (1 − 9.81 sin 7°) m/s2
= −0.1955 m/s2
For the uniformly decelerated motion
v 2 = (v0 ) 2upgrade + 2a′( xupgrade − 0)
Noting that 90 km/h = 25 m/s, then when v = 80 km/h =
8
v0 , we have
9
2
8
× 25 m/s = (25 m/s)2 + 2(−0.1955 m/s2) xupgrade
9
or
xupgrade = 335.5 m
xupgrade = 335.5 m
or
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
No part
part of
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced or
or distributed
distributed in
in any
any form
form or
reproduced
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
you are
are aa student
student using
using this
this Manual,
Manual,
distribution
If you
you are
are using
using it
it without
without permission.
permission.
you
282
282
PROBLEM 12.8
If an automobile’s braking distance from 96 km/h is 45 m on level pavement, determine the automobile’s
braking distance from 96 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,
v 0 = 96 km/h = 26.7 m/s, v f = 0
xf − x0 = 45 m
v 2f = v02 + 2a( x f − x0 )
a=
=
v 2f − v02
2( x f − x0 )
0 − (26.7)2
(2)(45)
= −7.92 m/s2
Braking force.
Fb = ma
W
= a
g
7.92
W
=−
9.81
= −0.8073W
(a)
Going up a 5° incline.
ΣF = ma
W
a
g
F + W sin 5°
a=− b
g
W
= −(0.8073 + sin 5°)(9.81)
− Fb − W sin 5° =
= −8.775 m/s2
x f − x0 =
v 2f − v02
2a
0 − (26.7)2
=
(2)(−8.775)
x f − x0 = 40.6 m
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
No part
part of
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced or
or distributed
distributed in
in any
any form
form or
reproduced
distribution to
toteachers
teachersand
andeducators
educatorspermitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
If you
you are
are aastudent
studentusing
usingthis
thisManual,
Manual,
distribution
you are
are using
using itit without
without permission.
permission.
you
283
283
PROBLEM 12.8 (Continued)
(b)
Going down a 3 percent incline.
3
β = 1.71835°
100
W
− Fb + W sin β = a
g
a = −(0.8073 − sin b )(9.81)
= −7.625 m/s2
0 − (26.7)2
x f = x0 =
(2)(−7.625)
tan β =
x f − x0 = 46.7 m
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
No part
part of
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced or
or distributed
distributed in
in any
any form
form or
reproduced
distribution
If you
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
you are
are aa student
student using
using this
this Manual,
Manual,
you
you are
are using
using it
it without
without permission.
permission.
284
284
PROBLEM 12.9
A 20-kg package is at rest on an incline when a force P is applied to it.
Determine the magnitude of P if 10 s is required for the package to
travel 5 m up the incline. The static and kinetic coefficients of friction
between the package and the incline are both equal to 0.3.
SOLUTION
Kinematics: Uniformly accelerated motion. ( x0 = 0, v0 = 0)
x = x0 + v0 t +
or
a=
1 2
at ,
2
2 x (2)(5)
= 0.100 m/s 2
=
2
2
t
(10)
ΣFy = 0: N − P sin 50° − mg cos 20° = 0
N = P sin 50° + mg cos 20°
ΣFx = ma : P cos 50° − mg sin 20° − µ N = ma
or
P cos 50° − mg sin 20° − µ ( P sin 50° + mg cos 20°) = ma
P=
ma + mg (sin 20° + µ cos 20°)
cos 50° − µ sin 50°
For motion impending, set a = 0 and µ = µs = 0.4
(20)(0) + (20)(9.81)(sin 20° + 0.4 cos 20°)
cos 50° − 0.4 sin 50°
= 419 N
P=
�
For motion with a = 0.100 m/s 2 , use µ = µ k = 0.3.
P=
(20)(0.100) + (20)(9.81)(sin 20° + 0.3 cos 20°)
cos 50° − 0.3 sin 50°
P = 301 N ��
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
No part
part of
reproduced or
or distributed
distributed in
in any
any form
form or
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced
you are
are aa student
student using
using this
this Manual,
Manual,
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
distribution
If you
you are
are using
using it
it without
without permission.
permission.
you
285
285
PROBLEM 12.10
The acceleration of a package sliding at Point A is 3 m/s2.
Assuming that the coefficient of kinetic friction is the same
for each section, determine the acceleration of the package
at Point B.
SOLUTION
For any angle θ .
Use x and y coordinates as shown.
ay = 0
ΣFy = ma y : N − mg cos θ = 0
N = mg cos θ
ΣFx = max : mg sin θ − µk N = max
ax = g (sin θ − µ k cos θ )
At Point A.
θ = 30°, ax = m/s 2
µk =
g sin 30° − ax
g cos 30°
9.81 sin 30° − 3
9.81 cos 30°
= 0.22423
=
At Point B.
θ = 15°, µk = 0.22423
ax = 9.81(sin 15° − 0.22423 cos 15°)
a = 0.414 m/s 2
= 0.414 m/s
15° �
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
All rights
rights reserved.
reserved. No
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
Inc. All
No part
part of
of this
reproduced or
or distributed
distributed in
in any
any form
form or
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced
are aa student
student using
using this
this Manual,
Manual,
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
individual course
distribution
for their
their individual
course preparation.
preparation. If
If you
you are
you are
are using
using it
it without
without permission.
you
permission.
286
286
PROBLEM 12.11
The two blocks shown are originally at rest. Neglecting the masses of
the pulleys and the effect of friction in the pulleys and between block A
and the horizontal surface, determine (a) the acceleration of each block,
(b) the tension in the cable.
SOLUTION
From the diagram
x A + 3 yB = constant
Then
v A + 3vB = 0
and
a A + 3aB = 0
a A = −3aB
or
(a)
A:
ΣFx = m A a A :
(1)
−T = mA aB
T = 3m A aB
Using Eq. (1)
B:
�
ΣFy = mB aB : WB − 3T = mB aB
Substituting for T
A:
mB g − 3(3mA aB ) = mB aB
aB =
or
�
B:�
�
(b)
mA
mB
9.81 m/s 2
= 0.83136 m/s 2
30 kg
1+ 9
25 kg
��
a B = 0.831 m/s 2 �
and
�
1+ 9
=
a A = 2.49 m/s 2
Then
�
g
T = 3 × 30 kg × 0.83136 m/s 2
We have
T = 74.8 N ��
or
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
No part
part of
reproduced or
or distributed
distributed in
in any
any form
form or
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced
you are
are aa student
student using
using this
this Manual,
Manual,
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
distribution
If you
you are
are using
using it
it without
without permission.
permission.
you
287
287
PROBLEM 12.12
The two blocks shown are originally at rest. Neglecting the masses
of the pulleys and the effect of friction in the pulleys and assuming
that the coefficients of friction between block A and the horizontal
surface are µs = 0.25 and µk = 0.20, determine (a) the acceleration
of each block, (b) the tension in the cable.
SOLUTION
From the diagram
x A + 3 yB = constant
Then
v A + 3vB = 0
and
a A + 3aB = 0
a A = −3aB
or
(1)
First determine if the blocks will move with a A = aB = 0. We have
A:
A:
1
ΣFy = 0: WB − 3T = 0 or T = mB g
3
B:
ΣFx = 0:
Then
FA − T = 0
1
FA = × 25 kg × 9.81 m/s 2 = 81.75 N
3
ΣFy = 0: WA − N A = 0 or
Also,
B:
N A = mA g
( FA ) max = ( µ s ) A N A = ( µ s ) A m A g
= 0.25 × 30 kg × 9.81 m/s 2
= 73.575 N
FA � ( FA ) max , which implies that the blocks will move.�
�
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
All rights
rights reserved.
reserved. No
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
Inc. All
No part
part of
of this
reproduced or
or distributed
distributed in
in any
any form
form or
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced
are aa student
student using
using this
this Manual,
Manual,
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
individual course
distribution
for their
their individual
course preparation.
preparation. If
If you
you are
you are
are using
using it
it without
without permission.
you
permission.
288
288
PROBLEM 12.12 (Continued)
(a)
A:
ΣFy = 0: WA − N A = 0 or
Sliding:
FA = ( µk ) A N A = 0.20 m A g
N A = mA g
ΣFx = m A a A : FA − T = m A a A
Using Eq. (1)
B:
or
T = 0.20 mA g + 3m A aB
ΣFy = mB aB : WB − 3T = mB aB
mB g − 3(0.20 m A g + 3mA aB ) = mB aB
or
�
m �
g �1 − 0.6 A �
mB �
aB = �
mA
1+ 9
mB
�
30 kg �
(9.81 m/s 2 ) �1 − 0.6
�
25 kg �
�
=
kg
1 + 9 30
25 kg
= 0.23278 m/s 2
a A = 0.698 m/s 2
Then
a B = 0.233 m/s 2 ��
and
(b)
We have
�
T = (30 kg)(0.20 × 9.81 + 3 × 0.23278) m/s 2
T = 79.8 N ��
or
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
No part
part of
reproduced or
or distributed
distributed in
in any
any form
form or
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced
you are
are aa student
student using
using this
this Manual,
Manual,
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
distribution
If you
you are
are using
using it
it without
without permission.
permission.
you
289
289
PROBLEM 12.13
The coefficients of friction between the load and the flat-bed
trailer shown are µs = 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 .
ΣFy = 0: N − W = 0 N = W
Fm = µ s N = 0.40 W
ΣFx = ma : Fm = mamax
0.40 W =
W
amax
g
amax = 3.924 m/s2
a max = 3.9 m/s2
Uniformly accelerated motion.
v 2 = v02 + 2ax with v = 0
v0 = 72 km/h = 20 m/s
a = − amax = −3.924 m/s2
0 = (20)2 + 2(−3.924)x
x = 51 m
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
part of
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
No part
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced or
or distributed
distributed in
in any
any form
form or
reproduced
distribution to
to teachers
teachers and
andeducators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
If you
you are
are aa student
student using
using this
thisManual,
Manual,
distribution
you are
are using
using itit without
without permission.
permission.
you
290
290
PROBLEM 12.14
A tractor-trailer is traveling at 96 km/h when the driver
applies his brakes. Knowing that the braking forces of the
tractor and the trailer are 18 kN and 68 kN, respectively,
determine (a) the distance traveled by the tractor-trailer
before it comes to a stop, (b) the horizontal component of the
force in the hitch between the tractor and the trailer while
they are slowing down.
SOLUTION
ΣFx = ma: − ( Fbr ) trac − ( Fbr ) trl = mtotal a
(a)
a=−
or
(18 + 68)103 N
(7500 + 8700)kg
= −5.31 m/s2
For uniformly decelerated motion
v2 = v02 + 2a(x − x0)
When v = 0:
v0 = 96 km/h = 26.67 m/s
0 = (26.67 m/s)2 + 2(−5.31 m/s2)(∆x)
∆x = 67 m
or
ΣFx = mtrl a : − ( Fbr ) trl + Phitch = mtrl a
(b)
Then
Phitch = 68000 N + 8700 kg × (−5.31 m/s2)
Phitch = 21.8 kN (tension)
or
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
No part
part of
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced or
or distributed
distributed in
in any
any form
form or
reproduced
distributionto
toteachers
teachersand
andeducators
educatorspermitted
permitted by
byMcGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
If you
you are
areaastudent
studentusing
usingthis
thisManual,
Manual,
distribution
you are
are using
using itit without
without permission.
permission.
you
291
291
PROBLEM 12.15
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 diagram
2 x A + xB/A = constant
Then
2v A + vB/A = 0
and
2a A + aB/A = 0
Now
a B = a A + a B/A
Then
aB = a A + ( −2a A )
or
aB = − a A
(1)
First we determine if the blocks will move for the given value of 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
T = mB g (0.2cos θ + sin θ )
or
A:
A:
ΣFy = 0: N A − N AB − WA cos θ = 0
or
N A = (m A + mB ) g cos θ
Now
FA = µ s N A
= 0.2( mA + mB ) g cos θ
ΣFx = 0: − T − FA − FAB + WA sin θ = 0
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
All rights
rights reserved.
reserved. No
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
Inc. All
No part
part of
of this
reproduced or
or distributed
distributed in
in any
any form
form or
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced
are aa student
student using
using this
this Manual,
Manual,
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
individual course
distribution
for their
their individual
course preparation.
preparation. If
If you
you are
you are
are using
using it
it without
without permission.
you
permission.
292
292
PROBLEM 12.15 (Continued)
T = m A g sin θ − 0.2(m A + mB ) g cos θ − 0.2mB g cos θ
or
= g[ mA sin θ − 0.2(m A + 2mB ) cos θ ]
Equating the two expressions for T
mB g (0.2cos θ + sin θ ) = g[m A 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.
ΣFy = 0: N AB − WB cos 25° = 0
(a)
B:
or
N AB = mB g cos 25°
Sliding:
FAB = µk N AB = 0.15mB g cos 25°
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
N A = (m A + mB ) g cos 25°
Sliding:
FA = µk N A = 0.15(m A + mB ) g cos 25°
ΣFx = m A a A : − T − FA − FAB + WA sin 25° = m A a A
Substituting and using Eq. (1)
T = m A g sin 25° − 0.15( mA + mB ) g cos 25°
− 0.15mB g cos 25° − m A (− aB )
= g[ mA sin 25° − 0.15(m A + 2mB ) cos 25°] + m A aB
= 9.81[40 sin 25° − 0.15(40 + 2 × 8) cos 25°] + 40aB
= 91.15202 + 40aB
(N)
Equating the two expressions for T
8(5.47952 − aB ) = 91.15202 + 40aB
or
aB = −0.98575 m/s 2
a B = 0.986 m/s 2
(b)
We have
or
25° �
T = 8[5.47952 − (−0.98575)]
T = 51.7 N �
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
No part
part of
reproduced or
or distributed
distributed in
in any
any form
form or
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced
you are
are aa student
student using
using this
this Manual,
Manual,
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
distribution
If you
you are
are using
using it
it without
without permission.
permission.
you
293
293
PROBLEM 12.16
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 diagram
2 x A + xB/A = constant
Then
2v A + vB/A = 0
and
2a A + aB/A = 0
Now
a B = a A + a B/A
Then
aB = a A + ( −2a A )
or
aB = − a A
(1)
First we determine if the blocks will move for the given value of 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
or
N AB = mB g cos 25°
Now
FAB = µ s N AB
B:
= 0.2 mB g cos 25°
ΣFx = 0: − T + FAB + WB sin 25° = 0
A:
T = 0.2 mB g cos 25° + mB g sin 25°
or
= (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 A = (m A + mB ) g cos 25° − P sin 25°
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
All rights
rights reserved.
reserved. No
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
Inc. All
No part
part of
of this
reproduced or
or distributed
distributed in
in any
any form
form or
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced
are aa student
student using
using this
this Manual,
Manual,
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
individual course
distribution
for their
their individual
course preparation.
preparation. If
If you
you are
you are
are using
using it
it without
without permission.
you
permission.
294
294
PROBLEM 12.16 (Continued)
Now
FA = µ s N A
or
FA = 0.2[(mA + mB ) g cos 25° − P sin 25°]
ΣFx = 0: − T − FA − FAB + WA sin 25° + P cos 25° = 0
or
−T − 0.2[(m A + mB ) g cos 25° − P sin 25°] − 0.2mB g cos 25° + mA g sin 25° + P cos 25° = 0
or
P(0.2 sin 25° + cos 25°) = T + 0.2[(m A + 2mB ) g cos 25°] − m A g sin 25°
Then
P(0.2 sin 25° + cos 25°) = 47.39249 N + 9.81 m/s 2 {0.2[(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.
ΣFy = 0: N AB − WB cos 25° = 0
(a)
B:
or
N AB = mB g cos 25°
Sliding:
FAB = µk N AB
= 0.15 mB g cos 25°
ΣFx = mB aB : − T + FAB + WB sin 25° = mB aB
T = mB [ g (0.15 cos 25° + sin 25°) − aB ]
or
= 8[9.81(0.15 cos 25° + sin 25°) − aB ]
= 8(5.47952 − aB )
(N)
A:
ΣFy = 0: N A − N AB − WA cos 25° + P sin 25° = 0
or
N A = (m A + mB ) g cos 25° − P sin 25°
Sliding:
FA = µk N A
= 0.15[(m A + mB ) g cos 25° − P sin 25°]
ΣFx = m A a A : − T − FA − FAB + WA sin 25° + P cos 25° = m A a A
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
No part
part of
reproduced or
or distributed
distributed in
in any
any form
form or
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced
you are
are aa student
student using
using this
this Manual,
Manual,
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
distribution
If you
you are
are using
using it
it without
without permission.
permission.
you
295
295
PROBLEM 12.16 (Continued)
Substituting and using Eq. (1)
T = m A g sin 25° − 0.15[( mA + mB ) g cos 25° − P sin 25°]
− 0.15 mB g cos 25° + P cos 25° − m A (− aB )
= g[ mA sin 25° − 0.15(m A + 2mB ) cos 25°]
+ P(0.15 sin 25° + cos 25°) + m A aB
= 9.81[40 sin 25° − 0.15(40 + 2 × 8) cos 25°]
+ 40(0.15 sin 25° + cos 25°) + 40aB
= 129.94004 aB
(N)
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
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
All rights
rights reserved.
reserved. No
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
Inc. All
No part
part of
of this
reproduced or
or distributed
distributed in
in any
any form
form or
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced
are aa student
student using
using this
this Manual,
Manual,
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
individual course
distribution
for their
their individual
course preparation.
preparation. If
If you
you are
you are
are using
using it
it without
without permission.
you
permission.
296
296
PROBLEM 12.17
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 a B
a A so that the normal force NAB between the boxes is zero.
A:
ΣFy = 0: NA − WA cos 15° = 0
or
NA = WA cos 15°
Slipping:
FA = ( µk ) A NA
A:
= 0.3WA cos 15°
ΣFx = m A a A : FA − WA sin 15° = m A a A
0.3WA cos 15° − WA sin 15° =
or
WA
aA
g
a A = (9.81 m/s2)(0.3 cos 15° − sin 15°)
or
= 0.304 m/s2
B:
B:
ΣFy = 0: N B − WB cos 15° = 0
or
N B = WB cos 15°
Slipping:
FB = ( µk ) B N B
= 0.32WB cos 15°
ΣFx = mB aB : FB − WB sin 15° = mB aB
or
or
0.32WB cos 15° − WB sin 15° =
WB
aB
g
aB = (9.81 m/s2)(0.32 cos 15° − sin 15°) = 0.493 m/s2
aB
aA
assumption is correct
a A = 0.304 m/s2
15°
a B = 0.493 m/s2
15°
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
No part
part of
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced or
or distributed
distributed in
in any
any form
form or
reproduced
distributionto
toteachers
teachersand
andeducators
educatorspermitted
permitted by
byMcGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
If you
you are
areaastudent
studentusing
usingthis
thisManual,
Manual,
distribution
you are
are using
using itit without
without permission.
permission.
you
297
297
PROBLEM 12.17 (Continued)
Note: If it is assumed that the boxes remain in contact ( NAB ≠ 0), then
a A = aB
and find (ΣFx = ma )
A:
0.3WA cos 15° − WA sin 15° − N AB =
WA
a
g
B:
0.32WB cos 15° − WB sin 15° + N AB =
WB
a
g
Solving yields a = 0.386 m/s2 and NAB = −4.3 N, which contradicts the assumption.
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
No part
part of
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced or
or distributed
distributed in
in any
any form
form or
reproduced
distribution
If you
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
you are
are aa student
student using
using this
this Manual,
Manual,
you
you are
are using
using it
it without
without permission.
permission.
298
298
PROBLEM 12.18
Knowing that the system shown starts from rest,
find the velocity at t = 1.2 s of (a) collar A,
(b) collar B. Neglect the masses of the pulleys
and the effect of friction.
SOLUTION
Referring to solution of Problem 11.51 we note that
1
aB = − a A
2
(1)
where minus sign indicates that a A and a B have opposite sense.
Block B.
ΣF = ma : 25 − 2T = 7.5a A
(2)
Block A.
ΣF = ma : 2T − T = 10a A
T = 10a A
(a)
Substituting for T from (3) into (2):
25 − 2(10 a A ) = 7.5a A
25 = 27.5a A
v A = (v A )0 + a At = 0 + 0.909(1.2),
(b)
(3)
a A = 0.909 m/s 2 →
v A = 1.091 m/s → �
a B = 0.455 m/s 2 ←
Substituting a = 0.909 into Eq. (1):
vB = (vB )0 + aB t = 0 + 0.455(1.2)
v B = 0.545 m/s ← ��
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
No part
part of
reproduced or
or distributed
distributed in
in any
any form
form or
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced
you are
are aa student
student using
using this
this Manual,
Manual,
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
distribution
If you
you are
are using
using it
it without
without permission.
permission.
you
299
299
PROBLEM 12.19
Each of the systems shown is initially at rest.
Neglecting axle friction and the masses of the
pulleys, determine for each system (a) the
acceleration of block A, (b) the velocity of block
A after it has moved through 3 m, (c) the time
required for block A to reach a velocity of 6 m/s.
SOLUTION
Let y be positive downward for both blocks.
Constraint of cable: y A + yB = constant
a A + aB = 0
For blocks A and B,
aB = − a A
or
ΣF = ma :
WA
aA
g
Block A:
WA − T =
Block B:
P + WB − T =
WB
W
aB = − B a A
g
g
P + WB − WA +
WA
W
aA = − B aA
g
g
Solving for aA,
T = WA −
or
aA =
WA
aA
g
WA − WB − P
g
WA + WB
(1)
v A2 − (v A )02 = 2a A [ y A − ( y A )0 ] with (v A )0 = 0
v A = 2a A [ y A − ( y A ) 0 ]
v A − (v A )0 = a A t
t=
(2)
with (v A )0 = 0
vA
aA
(3)
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
part of
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
No part
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced or
or distributed
distributed in
in any
any form
form or
reproduced
distribution to
to teachers
teachers and
andeducators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
If you
you are
are aa student
student using
using this
thisManual,
Manual,
distribution
you are
are using
using itit without
without permission.
permission.
you
300
300
PROBLEM 12.19 (Continued)
(a)
Acceleration of block A.
System (1):
By formula (1),
System (2):
By formula (1),
System (3):
By formula (1),
(b)
(c)
WA = 981 N, WB = 490.5 N, P = 0
(a A )1 =
981 − 490.5
(9.81)
981 + 490.5
WA = 981 N,
(a A ) 2 =
(a A )1 = 3.27 m/s2
WB = 0 , P = 500 N
981 − 500
(9.81)
981
(a A ) 2 = 4.81 m/s2
WA = 10,791 N, WB = 10300.5 N, P = 0
( a A )3 =
10,791 − 10300.5
(9.81)
10,791 + 10300.5
(a A )3 = 0.23 m/s2
v A at y A − ( y A )0 = 3 m. Use formula (2).
System (1):
(v A )1 = (2)(3.27)(3)
(v A )1 = 4.43 m/s
System (2):
(v A ) 2 = (2)(4.81)(3)
( v A )2 = 5.37 m/s
System (3):
(v A )3 = (2)(0.23)(3)
(v A )3 = 1.18 m/s
Time at v A = 6 m/s. Use formula (3).
System (1):
t1 =
6
3.27
t1 = 1.835 s
System (2):
t2 =
6
4.81
t2 = 1.247 s
System (3):
t3 =
6
0.23
t3 = 26.087 s
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
No part
part of
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced or
or distributed
distributed in
in any
any form
form or
reproduced
distribution
distribution to
toteachers
teachersand
andeducators
educatorspermitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
If you
you are
are aastudent
studentusing
usingthis
thisManual,
Manual,
you
you are
are using
using itit without
without permission.
permission.
301
301
PROBLEM 12.20
A man standing in an elevator that is moving with a constant
acceleration holds a 3-kg block B between two other blocks in such a
way that the motion of B relative to A and C is impending. Knowing
that the coefficients of friction between all surfaces are µ s = 0.30
and µk = 0.25, determine (a) the acceleration of the elevator if it is
moving upward and each of the forces exerted by the man on blocks
A and C has a horizontal component equal to twice the weight of B,
(b) the horizontal components of the forces exerted by the man on
blocks A and C if the acceleration of the elevator is 2.0 m/s2
downward.
SOLUTION
First we observe that because B is not moving relative to A and to C that a B = a EL .
(a)
We have
F = µs N
= 0.30(2WB )
= 0.6WB = 0.6mB g
For a EL to be ↑, the net vertical force must be ↑, which requires that the frictional forces be acting as
shown. It then follows that the impending motion of B relative to A and C is downward. Then
ΣFy = mB aEL : 2 F − WB = mB aEL
or
or
2(0.6mB g ) − mB g = mB aEL
aEL = 0.2 × 9.81 m/s 2
a EL = 1.962 m/s 2 ↑ �
or
(b)
We have
F = µs N
= 0.30 N
Now we observe that because the direction of the impending motion is unknown, the directions of the
frictional forces is also unknown (although Fnet must be downward).
ΣFy = mB aEL : ± 2 F − WB = − mB |aEL |
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
All rights
rights reserved.
reserved. No
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
Inc. All
No part
part of
of this
reproduced or
or distributed
distributed in
in any
any form
form or
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced
are aa student
student using
using this
this Manual,
Manual,
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
individual course
distribution
for their
their individual
course preparation.
preparation. If
If you
you are
you are
are using
using it
it without
without permission.
you
permission.
302
302
PROBLEM 12.20 (Continued)
or
±2 F = mB ( g − |aEL |)
= 3 kg × (9.81 − 2) m/s 2
Since the magnitude of F must be positive, it then follows that F ↑, and that the impending motion of B
relative to A and C is downward. Finally
2(0.30 N) = 3 kg × (9.81 − 2) m/s 2
N AB = N BC = 39.1 N �
or
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
No part
part of
reproduced or
or distributed
distributed in
in any
any form
form or
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced
you are
are aa student
student using
using this
this Manual,
Manual,
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
distribution
If you
you are
are using
using it
it without
without permission.
permission.
you
303
303
PROBLEM 12.21
A package is at rest on a conveyor belt, which is initially at rest. The belt
is started and moves to the right for 1.3 s with a constant acceleration of
2 m/s2. The belt then moves with a constant deceleration a2 and comes
to a stop after a total displacement of 2.2 m. Knowing that the
coefficients of friction between the package and the belt are µ s = 0.35
and µk = 0.25, determine (a) the deceleration a2 of the belt, (b) the
displacement of the package relative to the belt as the belt comes to a
stop.
SOLUTION
(a)
Kinematics of the belt. vo = 0
1. Acceleration phase with a1 = 2 m/s 2
v1 = vo + a1t1 = 0 + (2)(1.3) = 2.6 m/s
x1 = xo + vo t1 +
1 2
1
a1t1 = 0 + 0 + (2)(1.3)2 = 1.69 m
2
2
2. Deceleration phase: v2 = 0 since the belt stops.
v22 − v12 = 2a2 ( x2 − x1 )
a2 =
t2 − t1 =
(b)
v22 − v12
0 − (2.6)2
=
= −6.63
2( x2 − x1 ) 2(2.2 − 1.69)
a 2 = 6.63 m/s 2
�
v2 − v1 0 − 2.6
=
= 0.3923 s
a2
−6.63
Motion of the package
1. Acceleration phase. Assume no slip. (a p )1 = 2 m/s 2
ΣFy = 0: N − W = 0 or N = W = mg
ΣFx = ma : F f = m(a p )1
The required friction force is Ff.
The available friction force is µ s N = 0.35W = 0.35mg
Ff
m
= (a p )1 �
µs N
m
= µ s g = (0.35)(9.81) = 3.43 m/s 2
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
All rights
rights reserved.
reserved. No
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
Inc. All
No part
part of
of this
reproduced or
or distributed
distributed in
in any
any form
form or
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced
are aa student
student using
using this
this Manual,
Manual,
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
individual course
distribution
for their
their individual
course preparation.
preparation. If
If you
you are
you are
are using
using it
it without
without permission.
you
permission.
304
304
PROBLEM 12.21 (Continued)
Since 2.0 m/s 2 � 3.43 m/s 2, the package does not slip.
(v p )1 = v1 = 2.6 m/s and (x p )1 = 1.69 m.
2. Deceleration phase. Assume no slip. (a p ) 2 = −11.52 m/s 2
ΣFx = ma : − F f = m( a p )2
Ff
m
µs N
m
=
µs mg
m
= (a p ) 2 = −6.63 m/s 2
= µ s g = 3.43 m/s 2 � 6.63 m/s 2
Since the available friction force µ s N is less than the required friction force Ff for no slip, the
package does slip.
(a p ) 2 � 6.63 m/s 2 ,
F f = µk N
ΣFx = m( a p )2 : − µk N = m(a p ) 2
µk N
= − µk g
m
= −(0.25)(9.81)
(a p ) 2 = −
= −2.4525 m/s 2
(v p ) 2 = (v p )1 + (a p ) 2 (t2 − t1 )
= 2.6 + (−2.4525)(0.3923)
= 1.638 m/s 2
( x p ) 2 = ( x p )1 + (v p )1 (t2 − t1 ) 2 +
= 1.69 + (2.6)(0.3923) +
1
(a p ) 2 (t2 − t1 ) 2
2
1
( −2.4525)(0.3923) 2
2
= 2.521 m
Position of package relative to the belt
( x p )2 − x2 = 2.521 − 2.2 = 0.321
x p/belt = 0.321 m
�
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
No part
part of
reproduced or
or distributed
distributed in
in any
any form
form or
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced
you are
are aa student
student using
using this
this Manual,
Manual,
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
distribution
If you
you are
are using
using it
it without
without permission.
permission.
you
305
305
PROBLEM 12.22
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
Acceleration a1: Impending slip.
ΣFy = m A a y :
F1 = µs N1 = 0.30 N1
N1 − WA = mA a1 sin 65°
N1 = WA + mA a1 sin 65°
= m A ( g + a1 sin 65°)
ΣFx = m A ax : F1 = m A a1 cos 65°
F1 = µs N
or
m A a1 cos 65° = 0.30m A ( g + a1 sin 65°)
a1 =
0.30 g
cos 65° − 0.30 sin 65°
= (1.990)(9.81)
= 19.53 m/s 2
Deceleration a 2 : Impending slip.
a1 = 19.53 m/s 2
65° �
F2 = µs N 2 = 0.30 N 2
ΣFy = ma y : N1 − WA = − mA a2 sin 65°
N1 = WA − mA a2 sin 65°
ΣFx = max :
F2 = mA a2 cos 65°
F2 = µs N 2
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
All rights
rights reserved.
reserved. No
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
Inc. All
No part
part of
of this
reproduced or
or distributed
distributed in
in any
any form
form or
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced
are aa student
student using
using this
this Manual,
Manual,
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
individual course
distribution
for their
their individual
course preparation.
preparation. If
If you
you are
you are
are using
using it
it without
without permission.
you
permission.
306
306
PROBLEM 12.22 (Continued)
or
m A a2 cos 65° = 0.30 m A ( g − a2 cos 65°)
0.30 g
cos 65° + 0.30 sin 65°
= (0.432)(9.81)
a2 =
= 4.24 m/s 2
a 2 = 4.24 m/s 2
65° ��
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
No part
part of
reproduced or
or distributed
distributed in
in any
any form
form or
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced
you are
are aa student
student using
using this
this Manual,
Manual,
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
distribution
If you
you are
are using
using it
it without
without permission.
permission.
you
307
307
PROBLEM 12.23
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 µs = 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 = µs N1 = 0.40 N1
ΣFy = mP a y : N1 − WP cos 20° = − mP aT sin 20°
N1 = mP ( g cos 20° − aT sin 20°)
ΣFx = max : F1 − WP sin 20° = mP aT cos 20°
F1 = mP ( g sin 20° + aT cos 20°)
mP ( g sin 20° + aT cos 20°) = 0.40 mP ( g cos 20° − aT sin 20°)
(0.40 cos 20° − sin 20°)
g
cos 20° + 0.40sin 20°
= (0.03145)(9.81)
= 0.309
aT =
aT = 0.309 m/s 2
(b)
xP / T = ( xP / T )o + (vP / T )t +
aP / T =
2 xP / T
t
2
=
�
1
1
aP / T t 2 = 0 + 0 + aP / T t 2
2
2
(2)(2)
= 4.94 m/s 2
(0.9)2
a P / T = 4.94 m/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°)
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
All rights
rights reserved.
reserved. No
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
Inc. All
No part
part of
of this
reproduced or
or distributed
distributed in
in any
any form
form or
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced
are aa student
student using
using this
this Manual,
Manual,
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
individual course
distribution
for their
their individual
course preparation.
preparation. If
If you
you are
you are
are using
using it
it without
without permission.
you
permission.
308
308
PROBLEM 12.23 (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 = µk N 2 = 0.30 N 2
mP ( g sin 20° + aT cos 20° − aP / T ) = 0.30mP ( g cos 20° + aT sin 20°)
aT =
(0.30 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
��
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
No part
part of
reproduced or
or distributed
distributed in
in any
any form
form or
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced
you are
are aa student
student using
using this
this Manual,
Manual,
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
distribution
If you
you are
are using
using it
it without
without permission.
permission.
you
309
309
PROBLEM 12.24
The propellers of a ship of weight W can produce a propulsive force F0; they produce a force of the same
magnitude but of opposite direction when the engines are reversed. Knowing that the ship was proceeding
forward at its maximum speed v0 when the engines were put into reverse, determine the distance the ship
travels before coming to a stop. Assume that the frictional resistance of the water varies directly with the
square of the velocity.
SOLUTION
F0 = kv02 = 0 k =
At maximum speed a = 0.
F0
v02
When the propellers are reversed, F0 is reversed.
ΣFx = ma : − F0 − kv 2 = ma
− F0 − F0
v2
= ma
v02
dx =
�
x
0
a−
F0
(v + v )
2
0
mv02
mv02 vdv
vdv
=
a
F0 v02 + v 2
dx = −
x=−
=−
(
mv02
F0
0
vdv
v0
2
0
� v +v
2
)
2
0
mv02 1
ln v02 + v 2
v0
F0 2
(
)
mv02 � 2
mv 2
ln v0 − ln 2v02 � = 0 ln 2
� 2 F0
2 F0 �
( )
x = 0.347
m0 v02
��
F0
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
All rights
rights reserved.
reserved. No
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
Inc. All
No part
part of
of this
reproduced or
or distributed
distributed in
in any
any form
form or
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced
are aa student
student using
using this
this Manual,
Manual,
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
individual course
distribution
for their
their individual
course preparation.
preparation. If
If you
you are
you are
are using
using it
it without
without permission.
you
permission.
310
310
PROBLEM 12.25
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 the 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
or
ln
=−
t = − ln
k
P
m
m
P − kv
P
or
= e− kt/m
v = (1 − e− kt/m )
m
k
�
�
t
t
t
Pt
P� k
�
− � − e− kt/m �
k 0 k� m
�0
x=
�
=
Pt P − kt/m
Pt P
+ (e
− 1) =
− (1 − e− kt/m )
k m
k m
x=
Pt kv
− , which is linear.
k
m
0
v dt =
�
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
No part
part of
reproduced or
or distributed
distributed in
in any
any form
form or
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced
you are
are aa student
student using
using this
this Manual,
Manual,
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
distribution
If you
you are
are using
using it
it without
without permission.
permission.
you
311
311
PROBLEM 12.26
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 − �)
x
cos θ =
By geometry,
� + x2
2
ΣFx = max : − Fs cos θ = ma
− k ( � 2 + x 2 − �)
a=−
�
v
0
v dv =
x
� + x2
2
= ma
k�
�x �
�x−
�
m ��
�2 + x 2 ��
0
� 0 a dx
x
v
1 2
k
v =−
2 0
m
�
0
�x �
k �1 2
2
2 �
�
�
−
=
−
−
+
x
dx
x
x
�
�
�
x0 �
m �� 2
�x
�2 + x 2 ��
�
0
0 �
1 2
1
k�
�
v = − � 0 − � 2 − x02 + � � 2 + x02 �
2
2
m�
�
k
2� 2 + x02 − 2� � 2 + x02
v2 =
m
k
= � � 2 + x02 − 2� � 2 + x02 + � 2 �
�
m�
)
(
(
)
answer: v =
k
m
( � + x − �) �
2
2
0
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
All rights
rights reserved.
reserved. No
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
Inc. All
No part
part of
of this
reproduced or
or distributed
distributed in
in any
any form
form or
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced
are aa student
student using
using this
this Manual,
Manual,
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
individual course
distribution
for their
their individual
course preparation.
preparation. If
If you
you are
you are
are using
using it
it without
without permission.
you
permission.
312
312
PROBLEM 12.27
Determine the maximum theoretical speed that a 1200-kg automobile starting from rest can reach after
traveling 400 m if air resistance is considered. Assume that the coefficient of static friction between the tires
and the pavement is 0.70, that the automobile has front-wheel drive, that the front wheels support 62 percent
of the automobile’s weight, and that the aerodynamic drag D has a magnitude D = 0.012v2, where D and v are
expressed in newtons and m/s, respectively.
SOLUTION
F = Fmax for v = vmax
F = µ s N F = 0.70(0.62 W)
= 0.434 W
W
ΣFx = ma : F − D = a
g
g
(0.434 W − 0.012 v 2 )
W
g
= 0.002 (217 W − 6v 2 )
W
a=
or
v
Now
At x = 0, v = 0:
or
or
or
0.002
g
W
0.002
x
0
g
dv
= a = 0.002 (217 W − 6v 2 )
dx
W
dx =
vdv
0 217 W − 6v 2
v
g
1
217 W − 6v 2
x = − ln
W
12
217 W
g
217 W − 6v 2
−0.024 W x
=e
217 W
v=
g
217
W 1 − e−0.024 W x
6
1/ 2
When x = 400 m:
217
400
(1200 × 9.81)(1 − e−0.024(1200 ))
6
= 58.24 m/s
1/2
vmax =
vmax = 209.7 km/h
or
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
No part
part of
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced or
or distributed
distributed in
in any
any form
form or
reproduced
distribution to
toteachers
teachersand
andeducators
educatorspermitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
If you
you are
are aastudent
studentusing
usingthis
thisManual,
Manual,
distribution
you are
are using
using itit without
without permission.
permission.
you
313
313
PROBLEM 12.28
The coefficients of friction between blocks A and C and the
horizontal surfaces are µs = 0.24 and µk = 0.20. Knowing
that m A = 5 kg, mB = 10 kg, and mC = 10 kg, determine
(a) the tension in the cord, (b) the acceleration of each
block.
SOLUTION
We first check that static equilibrium is not maintained:
( FA ) m + ( FC )m = µs ( mA + mC ) g
= 0.24(5 + 10) g
= 3.6 g
Since WB = mB g = 10g � 3.6g, equilibrium is not maintained.
Block A:
ΣFy : N A = m A g
FA = µk N A = 0.2m A g
ΣFλ = mA a A : T − 0.2mA g = mA a A
Block C:
(1)
ΣFy : NC = mC g
FC = µk NC = 0.2mC g
ΣFx = mC aC : T − 0.2mC g = mC aC
Block B:
ΣFy = mB aB
mB g − 2T = mB aB
aB =
From kinematics:
(a)
(2)
(3)
1
(a A + aC )
2
Tension in cord. Given data:
(4)
m A = 5 kg
mB = mC = 10 kg
Eq. (1): T − 0.2(5) g = 5a A
a A = 0.2T − 0.2 g
(5)
Eq. (2): T − 0.2(10) g = 10aC
aC = 0.1T − 0.2 g
(6)
Eq. (3): 10 g − 2T = 10aB
aB = g − 0.2T
(7)
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
All rights
rights reserved.
reserved. No
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
Inc. All
No part
part of
of this
reproduced or
or distributed
distributed in
in any
any form
form or
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced
are aa student
student using
using this
this Manual,
Manual,
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
individual course
distribution
for their
their individual
course preparation.
preparation. If
If you
you are
you are
are using
using it
it without
without permission.
you
permission.
314
314
PROBLEM 12.28 (Continued)
Substitute into (4):
1
(0.2T − 0.2 g + 0.1T − 0.2 g )
2
24
24
1.2 g = 0.35T
T=
g=
(9.81 m/s 2 )
7
7
g − 0.2T =
(b)
T = 33.6 N �
Substitute for T into (5), (7), and (6):
� 24 �
a A = 0.2 � g � − 0.2 g = 0.4857(9.81 m/s 2 )
� 7 �
a A = 4.76 m/s 2 → �
� 24 �
aB = g − 0.2 � g � = 0.3143(9.81 m/s 2 )
� 7 �
a B = 3.08 m/s 2 ↓ �
� 24 �
aC = 0.1� g � − 0.2 g = 0.14286(9.81 m/s 2 )
� 7 �
aC = 1.401 m/s 2 ← �
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
No part
part of
reproduced or
or distributed
distributed in
in any
any form
form or
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced
you are
are aa student
student using
using this
this Manual,
Manual,
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
distribution
If you
you are
are using
using it
it without
without permission.
permission.
you
315
315
PROBLEM 12.29
Solve Problem 12.28, assuming that m A = 5 kg, mB = 10 kg, and
mC = 20 kg.
PROBLEM 12.28 The coefficients of friction between blocks A
and C and the horizontal surfaces are µs = 0.24 and µk = 0.20.
Knowing that m A = 5 kg, mB = 10 kg, and mC = 10 kg, determine
(a) the tension in the cord, (b) the acceleration of each block.
SOLUTION
We first check that static equilibrium is not maintained:
( FA ) m + ( FC )m = µs ( mA + mC ) g = 0.24(5 + 20) g = 6 g
WB = mB g = 10 g � 6 g , equilibrium is not maintained.
Since
We shall assume that all 3 blocks move and use Eqs. (1), (2), (3), (4) derived in solution of Problem 12.28.
(a)
Tension in cord.
Given data:
m A = 5 kg, mB = 10 kg, mC = 20 kg
Eq. (1): T − 0.2(5) g = 5a A
a A = 0.2T − 0.2 g
(5′)
Eq. (2): T − 0.2(20) g = 20aC
aC = 0.05T − 0.2 g
(6′)
Eq. (3): 10 g − 2T = 10aB
aB = g − 0.2T
(7′)
Substituting into (4):
1
(0.2T − 0.2 g + 0.05T − 0.2 g )
2
48
48
T=
g=
1.2 g = 0.325T
(9.81 m/s 2 )
13
13
g − 0.2T =
Substituting into (6′):
T = 36.2 N �
� 48 �
aC = 0.05 � g � − 0.2 g = −0.0154 g (impossible)
� 13 �
This means that our assumption that block C moves was wrong.
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
All rights
rights reserved.
reserved. No
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
Inc. All
No part
part of
of this
reproduced or
or distributed
distributed in
in any
any form
form or
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced
are aa student
student using
using this
this Manual,
Manual,
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
individual course
distribution
for their
their individual
course preparation.
preparation. If
If you
you are
you are
are using
using it
it without
without permission.
you
permission.
316
316
PROBLEM 12.29 (Continued)
Assuming now that C does not move, we have
aC = 0 �
Substituting aC = 0 and from (5′) and (6′) into Eq. (4):
1
(0.2T − 0.2 g )
2
11
11
T = g = (9.81 m/s 2 ) = 35.97 N
1.1g = 0.3T
3
3
g − 0.2T =
T = 36.0 N �
Substituting for T into Eqs. (5′) and (7′):
� 11 �
a A = 0.2 � g � − 0.2 g = 0.5333(9.81 m/s 2 )
�3 �
� 11 �
aB = g − 0.2 � g � = 0.2667(9.81 m/s 2 )
�3 �
a A = 5.23 m/s 2
�
a B = 2.62 m/s 2 �
We also check that T = 113 g is less than ( FC ) m = 0.24(20 g ) = 4.8 g .
The value assumed for aC is thus correct:
aC = 0 �
as well as that obtained for T:
T = 36.0 N �
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
No part
part of
reproduced or
or distributed
distributed in
in any
any form
form or
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced
you are
are aa student
student using
using this
this Manual,
Manual,
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
distribution
If you
you are
are using
using it
it without
without permission.
permission.
you
317
317
PROBLEM 12.30
Blocks A and B weigh 20 kg each, block C weighs 14 kg, and block D
weighs 16 kg. Knowing that a downward force of magnitude 24 kg is
applied to block D, determine (a) the acceleration of each block, (b) the
tension in cord ABC. Neglect the weights of the pulleys and the effect
of friction.
SOLUTION
Note: As shown, the system is in equilibrium.
From the diagram:
A:
Cord 1:
2 y A + 2 yB + yC = constant
Then
2v A + 2vB + vC = 0
and
2a A + 2aB + aC = 0
Cord 2:
( yD − y A ) + ( yD − yB ) = constant
Then
2 vD − v A − v B = 0
and
2aD − a A − aB = 0
(1)
(2)
ΣFy = m A a A : WA − 2T1 + T2 = mAaA
(a)
20g − 2T1 + T2 = 20aA
or
B:
(3)
ΣFy = mB aB : WB − 2T1 + T2 = mBaB
20g − 2T1 + T2 = 20aB
or
Note: Eqs. (3) and (4)
Then
(4)
a A = aB
Eq. (1)
aC = −4a A
Eq. (2)
aD = a A
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
No part
part of
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced or
or distributed
distributed in
in any
any form
form or
reproduced
distributionto
toteachers
teachersand
andeducators
educatorspermitted
permitted by
byMcGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
If you
you are
areaastudent
studentusing
usingthis
thisManual,
Manual,
distribution
you are
are using
using itit without
without permission.
permission.
you
318
318
PROBLEM 12.30 (Continued)
C:
ΣFy = mC aC : WC − T1 =
or
D:
WC
aC
g
T1 = 14g 1−
aC
g
= 14g 1+
4a A
g
ΣFy = mD aD : WD − 2T2 + ( FD )ext =
T2 =
or
(5)
WD
aD
g
a
1
16 1 − D + 24
2
g
= 20 − 8
aA
g
g
Substituting for T1 [Eq. (5)] and T2 [Eq. (6)] in Eq. (3)
20 − 2 × 14 1 +
aA =
or
4a A
a
20
+ 20 − 8 A =
aA
g
g
g
3
3
g=
× 9.81 m/s2 = 0.841 m/s2
35
35
a A = a B = a D = 0.84 m/s2↓
aC = −4(0.841 m/s2)
and
(b)
or aC = 3.36 m/s2↑
Substituting into Eq. (5)
T1 = 14 × 9.81 1 +
4 × 0.841
9.81
or T1 = 184 N
PROPRIETARY
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
No part
part of
of this
this Manual
Manual may
may be
be displayed,
displayed,
reproduced
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced or
or distributed
distributed in
in any
any form
form or
distribution
usingthis
thisManual,
Manual,
distributionto
toteachers
teachersand
andeducators
educatorspermitted
permittedby
byMcGraw-Hill
McGraw-Hill for
for their
theirindividual
individual course
course preparation.
preparation. If
If you
you are
areaastudent
studentusing
you
youare
areusing
using itit without
without permission.
permission.
319
319
PROBLEM 12.31
Blocks A and B weigh 20 kg each, block C weighs 14 kg, and block D
weighs 16 kg. Knowing that a downward force of magnitude 10g N is
applied to block B and that the system starts from rest, determine at
t = 3 s the velocity (a) of D relative to A, (b) of C relative to D. Neglect
the weights of the pulleys and the effect of friction.
SOLUTION
Note: As shown, the system is in equilibrium.
From the diagram:
Cord 1:
2 y A + 2 yB + yC = constant
Then
2v A + 2vB + vC = 0
and
2a A + 2aB + aC = 0
Cord 2:
( yD − y A ) + ( yD − yB ) = constant
Then
2 vD − v A − vB = 0
and
2aD − a A − aB = 0
(1)
(2)
We determine the accelerations of blocks A, C, and D, using the
blocks as free bodies.
A:
ΣFy = m A a A : WA − 2T1 + T2 = mAaA
20g − 2T1 + T2 = 20 aA
or
(3)
ΣFy = mB aB : WB − 2T1 + T2 + ( FB )ext = mBaB
B:
20g − 2T1 + T2 + 10g = 20 aB
or
(3) − (4)
Forming
−10g = 20(aA − aB)
aB = a A +
or
(4)
1
g
2
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
No part
part of
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced or
or distributed
distributed in
in any
any form
form or
reproduced
distribution to
toteachers
teachersand
andeducators
educatorspermitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
If you
you are
are aastudent
studentusing
usingthis
thisManual,
Manual,
distribution
you are
are using
using itit without
without permission.
permission.
you
320
320
PROBLEM 12.31 (Continued)
Then
Eq. (1):
2a A + 2 a A +
1
g + aC = 0
2
aC = −4a A − g
or
2aD − a A − a A +
Eq. (2):
aD = a A +
or
C:
WC
aC
g
ΣFy = mC aC : WC − T1 =
T1 =14g 1−
or
D:
1
g =0
2
ΣFy = mD aD : WD − 2T2 =
T2 =
or
1
g
4
aC
1
a
= 14g 1 − (−4aA − g ) = 28g 1 + 2 A
g
g
g
(5)
WD
aD
g
a
1
1
1
× 16g 1 − D = 8g 1 − aA + g
g
2
g
4
= 8g
3 aA
−
4 g
(6)
Substituting for T1 [Eq. (5)] and T2 [Eq. (6)] in Eq. (3)
20 − 2 28 1 + 2
aA
g
+8
3 aA
20
−
=
aA
4 g
g
3
3
g = − (9.81 m/s2) = −2.102 m/s2
14
14
or
aA = −
Then
aC = −4(−2.102 m/s2) − 9.81 m/s2 = −1.402 m/s2
aD = −2.102 m/s2 +
1
(9.81 m/s2) = 0.351 m/s2
4
Note: We have uniformly accelerated motion, so that
v = 0 + at
(a)
We have
v D/A = v D − v A
or
v D/A = aD t − a A t = [0.351 − (−2.102)] m/s2 × 3s
v D/A = 7.36 m/s ↓
or
(b)
And
v C/ D = v C = v D
or
vC/D = aC t − aD t = (−1.402 − 0.351) m/s2 × 3s
v C/D = 5.26 m/s ↑
or
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Manual may
may be
be displayed,
displayed,
PROPRIETARY
Inc. All
All rights
rights reserved.
reserved. No
No part
part of
of this
this Manual
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced or
or distributed
distributed in
in any
any form
form or
reproduced
of the
distribution
for their
their individual
individual course
course preparation.
preparation. If
If you
you are
are aa student
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
student using
using this
this Manual,
Manual,
you
it without
without permission.
permission.
you are
are using
using it
321
321
PROBLEM 12.32
The 15-kg block B is supported by the 25-kg block A and is
attached to a cord to which a 225-N horizontal force is applied
as shown. Neglecting friction, determine (a) the acceleration of
block A, (b) the acceleration of block B relative to A.
SOLUTION
(a)
First we note a B = a A + a B/A , where a B/A is directed along the inclined surface of A.
B:
ΣFx = mB ax : P − WB sin 25° = mB a A cos 25° + mB aB/A
or
225 − 15 g sin 25° = 15( a A cos 25° + aB/A )
or
15 − g sin 25° = a A cos 25° + aB/A
B:
(1)
ΣFy = mB a y : N AB − WB cos 25° = −mB a A sin 25°
N AB = 15( g cos 25° − a A sin 25°)
or
A:
ΣFx′ = mA a A : P − P cos 25° + N AB sin 25° = m A a A
or
N AB = [25a A − 225(1 − cos 25°)]/ sin 25°
A:
Equating the two expressions for N AB
15( g cos 25° − a A sin 25°) =
or
25a A − 225(1 − cos 25°)
sin 25°
3(9.81) cos 25° sin 25° + 45(1 − cos 25°)
5 + 3sin 2 25°
= 2.7979 m/s 2
aA =
a A = 2.80 m/s 2 ← �
(b)
From Eq. (1)
aB/A = 15 − (9.81)sin 25° − 2.7979cos 25°
a B/A = 8.32 m/s 2
or
25° �
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
All rights
rights reserved.
reserved. No
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
Inc. All
No part
part of
of this
reproduced or
or distributed
distributed in
in any
any form
form or
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced
are aa student
student using
using this
this Manual,
Manual,
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
individual course
distribution
for their
their individual
course preparation.
preparation. If
If you
you are
you are
are using
using it
it without
without permission.
you
permission.
322
322
PROBLEM 12.33
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 = m A a A : WA sin 30° + N AB cos 40° = m A a A
(a)
1 �
�
22 � a A − g �
2 �
NAB = �
cos 40°
or
Now we note: a B = a A + a B/A , where a B/A is directed along the top surface of A.
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°
or
aA =
(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 cos 50°
or
aB/A = g sin 20° + a A cos 50°
= (9.81sin 20° + 6.4061cos 50°) m/s 2
= 7.4730 m/s 2
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
No part
part of
reproduced or
or distributed
distributed in
in any
any form
form or
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced
you are
are aa student
student using
using this
this Manual,
Manual,
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
distribution
If you
you are
are using
using it
it without
without permission.
permission.
you
323
323
PROBLEM 12.33 (Continued)
Finally
a B = a A + a B/ A
We have
aB2 = 6.40612 + 7.47302 − 2(6.4061 × 7.4730) cos 50°
or
aB = 5.9447 m/s 2
and
7.4730 5.9447
=
sin α
sin 50°
or
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
v B/A = v B − v A = a B t − a At = a B/At
At t = 0.5 s:
vB/A = 7.4730 m/s 2 × 0.5 s
v B/A = 3.74 m/s
or
20° �
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
All rights
rights reserved.
reserved. No
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
Inc. All
No part
part of
of this
reproduced or
or distributed
distributed in
in any
any form
form or
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced
are aa student
student using
using this
this Manual,
Manual,
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
individual course
distribution
for their
their individual
course preparation.
preparation. If
If you
you are
you are
are using
using it
it without
without permission.
you
permission.
324
324
PROBLEM 12.34
A 40-kg sliding panel is
supported by rollers at B and
C. A 25-kg counterweight A
is attached to a cable as
shown and, in cases a and c,
is initially in contact with a
vertical edge of the panel.
Neglecting friction, determine
in each case shown the
acceleration of the panel and
the tension in the cord
immediately after the system
is released from rest.
SOLUTION
(a)
F = Force exerted by counterweight
Panel:
ΣFx = ma :
T − F = 40a
(1)
Counterweight A: Its acceleration has two components
a A = a P + a A/P = a → + a ↓
ΣFx = max : F = 25a
(2)
ΣFg = mag : 25g − T = 25a
(3)
Adding (1), (2), and (3):
T − F + F + 25g − T = (40 + 25 + 25)a
a=
25
25
(9.81)
g=
90
90
a = 2.73 m/s2 ←
Substituting for a into (3):
25g − T = 25(2.73)
T = 177 N
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
No part
part of
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced or
or distributed
distributed in
in any
any form
form or
reproduced
distribution to
toteachers
teachersand
andeducators
educatorspermitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
If you
you are
are aastudent
studentusing
usingthis
thisManual,
Manual,
distribution
you are
are using
using itit without
without permission.
permission.
you
325
325
PROBLEM 12.34 (Continued)
(b)
Panel:
ΣFy = ma :
T = 40a
(1)
25g − T = 25a
(2)
Counterweight A:
ΣFy = ma :
Adding (1) and (2):
T + 25g − T = (40 + 25)a
a=
25
g
65
a = 3.77 m/s2 ←
Substituting for a into (1):
T = 40(3.77)
(c)
T = 151 N
Since panel is accelerated to the left, there is no force exerted by panel on counterweight and vice
versa.
Panel:
ΣFx = ma :
T = 40a
(1)
Counterweight A: Same free body as in Part (b):
ΣFy = ma :
25g − T = 25a
(2)
Since Eqs. (1) and (2) are the same as in (b), we get the same answers:
a = 3.77 m/s2 ←; T = 151 N
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
No part
part of
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced or
or distributed
distributed in
in any
any form
form or
reproduced
distribution
If you
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
you are
are aa student
student using
using this
this Manual,
Manual,
you
you are
are using
using it
it without
without permission.
permission.
326
326
PROBLEM 12.35
A 500-kg crate B is suspended from a cable attached to a 40-kg 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 m/s2 up and to the
right, determine (a) the acceleration of B relative to A, (b) the tension in
cable CD.
SOLUTION
(a)
First we note: a B = a A + a B/A , where a B/A is directed perpendicular to cable AB
ΣFx = mB ax : 0 = −mB ax + mB a A cos 25°
B:
aB/A = (1.2 m/s2) cos 25°
or
a B/A = 1.088 m /s2
or
(b)
For crate B
ΣFy = mB a y : TAB − WB =
or
WB
a A sin 25°
g
TAB = (500g N) 1 +
A:
(1.2 m/s2)sin 25°
9.81 m/s2
= 5158.6 N
For trolley A
ΣFx = m A a A : TCD − TAB sin 25° − WA sin 25° +
or
TCD = (5158.6 N)sin25° + (40g N) sin25° +
WA
aA
g
1.2 m/s2
9.81 m/s2
TCD = 2.4 kN
or
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
No part
part of
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced or
or distributed
distributed in
in any
any form
form or
reproduced
distribution to
toteachers
teachersand
andeducators
educatorspermitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
If you
you are
are aastudent
studentusing
usingthis
thisManual,
Manual,
distribution
you are
are using
using itit without
without permission.
permission.
you
327
327
PROBLEM 12.36
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
a A = an =
v 2A
ρ
ΣFy = 0: TBC sin 60° − WA = 0
(a)
or
7.1 kg × 9.81 m/s 2
sin 60°
= 80.426 N
TBC =
TBC = 80.4 N �
ΣFx = m A a A : TBC cos 60° = 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
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
All rights
rights reserved.
reserved. No
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
Inc. All
No part
part of
of this
reproduced or
or distributed
distributed in
in any
any form
form or
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced
are aa student
student using
using this
this Manual,
Manual,
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
individual course
distribution
for their
their individual
course preparation.
preparation. If
If you
you are
you are
are using
using it
it without
without permission.
you
permission.
328
328
PROBLEM 12.37
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 = an =
First we note
v 2A
ρ
ρ = l AB sin θ
where
ΣFy = 0: TAB cos θ − WA = 0
(a)
TAB =
or
mA g
cos θ
ΣFx = m A a A : TAB sin θ = m A
v A2
ρ
Substituting for TAB and ρ
mA g
v A2
sin θ = m A
l AB sin θ
cos θ
1 − cos 2 θ =
or
sin 2 θ = 1 − cos 2 θ
(4 m/s) 2
cos θ
1.8 m × 9.81 m/s 2
cos 2 θ + 0.906105cos θ − 1 = 0
cos θ = 0.64479
Solving
θ = 49.9° ��
or
(b)
From above
TAB =
m A g 0.450 kg × 9.81 m/s 2
=
cos θ
0.64479
TAB = 6.85 N �
or
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
No part
part of
reproduced or
or distributed
distributed in
in any
any form
form or
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced
you are
are aa student
student using
using this
this Manual,
Manual,
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
distribution
If you
you are
are using
using it
it without
without permission.
permission.
you
329
329
PROBLEM 12.38
A single wire ACB of length 80 cm passes through a ring at C that is
attached to a sphere which revolves at a constant speed v in the
horizontal circle shown. Knowing that θ1 = 60° and θ 2 = 30° and that
the tension is the same in both portions of the wire, determine the speed v.
SOLUTION
vC2
First we note
aC = aN =
where
ρ = LAC sin 30° = LBC sin 60°
LAC + LBC = LABC
Now
or
or
ρ
ρ
1
1
+
= 80 cm
sin 30° sin 60°
ρ = 25.359 cm
ΣFy = 0: TCA cos 30° + TCB cos 60° − WC = 0
or
T=
mC g
= 0.73205 mC g
cos 30° + cos 60°
ΣFx = mC aC : TCA sin 30° + TCB sin 60° = mC
or
or
0.73205 mC g (sin 30° + sin 60°) = mC
vC2
ρ
vC2
ρ
vC2 = 0.73205 (9.81 m/s2)(0.25359 m) × (sin30° + sin60°)
vC = 2.5 m
or
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
No part
part of
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced or
or distributed
distributed in
in any
any form
form or
reproduced
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
you are
are aa student
student using
using this
this Manual,
Manual,
distribution
If you
you are
are using
using it
it without
without permission.
permission.
you
330
330
PROBLEM 12.39
A single wire ACB passes through a ring at C that is attached to a 1-kg
sphere which revolves at a constant speed v in the horizontal circle
shown. Knowing that q 1 = 50° and d = 0.8 m and that the tension in
both portions of the wire is 6 N determine (a) the angle q 1, (b) the
speed v.
SOLUTION
(a)
ΣFy = 0: T cos θ1 + T cos θ 2 − W = 0
cos θ 2 =
=
W
− cos θ1
T
(1)(9.81)
− cos50° = 0.9922
6
θ 2 = 7.16°
(b)
q2 = 7.2°
θ3 = θ1 − θ 2 = 50° − 7.16° = 42.84°
1
sin θ 2
=
d
sin θ3
1 =
or
d sin θ 2
sin θ3
Radius of horizontal circle
ρ=
1 sin θ1 =
d sin θ 2 sin θ1
sin θ3
(0.8)(sin 7.16°)(sin 50°)
sin 42.84°
= 0.1123 m
=
ΣFx = man : T sin θ1 + T sin θ 2 =
v2 =
=
mv 2
ρ
ρT ( sin θ1 + sin θ 2 )
m
(0.1123)(6)(sin 50° + sin 7.16°)
1
v = 0.78 m/s
= 0.6 m2/s2
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
No part
part of
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced or
or distributed
distributed in
in any
any form
form or
reproduced
you are
are aa student
student using
using this
this Manual,
Manual,
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
distribution
If you
you are
are using
using it
it without
without permission.
permission.
you
331
331
PROBLEM 12.40
Two wires AC and BC are tied at C to a 7-kg sphere which revolves
at a constant speed v in the horizontal circle shown. Knowing that
θ 1 = 55° and θ 2 = 30° and that d = 1.4 m, determine the range of
values of v for which both wires remain taut.
SOLUTION
vC2
First we note
aC = an =
where
ρ = l tan 55° and
ρ
ρ = (d + l ) tan 30°
�
�
Then
ρ = �d +
or
ρ=
ρ
�
tan 30°
tan 55° ��
1.4 m
1
1
−
tan 30° tan 55°
= 1.35680 m
ΣFx = mC aC : TCA sin 30° + TCB sin 55° = mC
vC2
ρ
(1)
ΣFy = 0: TCA cos 30° + TCB cos 55° − WC = 0
or
TCA cos 30° + TCB cos 55° = mC g
Case 1: TCA = 0:
Eq. (2) � TCB =
Substituting into Eq. (1)
mC g
v2
sin 55° = mC C
cos 55°
ρ
(2)
mC g
cos 55°
or
(vC2 )TCA=0 = (1.35680 m)(9.81 m) tan 55°
or
(vC )TCA=0 = 4.36 m/s
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
All rights
rights reserved.
reserved. No
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
Inc. All
No part
part of
of this
reproduced or
or distributed
distributed in
in any
any form
form or
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced
are aa student
student using
using this
this Manual,
Manual,
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
individual course
distribution
for their
their individual
course preparation.
preparation. If
If you
you are
you are
are using
using it
it without
without permission.
you
permission.
332
332
PROBLEM 12.40 (Continued)
Now we form
(cos30°)(1) − (sin 30°)(2)
TCB sin 55° cos 30° − TCB cos 55° sin 30° = mC
ρ
vC2
TCB sin 25° = mC
or
vC2
ρ
cos 30° − mC g sin 30°
cos 30° − mC g sin 30°
(vC)max occurs when TCB = (TCB )max , which occurs when TCA = 0.
(vC)max = 4.36 m/s and wire AC will be taut if vC � 4.36 m/s.
mC g
cos 30°
Case 2: TCB = 0:
Eq. (2) � TCA =
Substituting into Eq. (1)
mC g
v2
sin 30° = mC C
cos 30°
ρ
or
( vC2 )T g =0 = (1.35680 m)(9.81 m/s2 ) tan 30°
or
(vC )TC g =0 = 2.77 m/s
C
Now we form
(cos 55°) (1) − (sin 55°) (2)
TCA sin 30° cos 55° − TCA cos 30° sin 55° = mC
or
−TCA sin 25° = mC
vC2
ρ
vC2
ρ
cos 55° − mC g sin 55°
cos 55° − mC g sin 55°
(vC)min occurs when TCA = (TCA ) max , which occurs when TCB = 0.
(vC)min = 2.77 m/s and wire BC will be taut if vC > 2.77 m/s.
2.77 m/s � v � 4.36 m/s �
Both wires are taut when
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
No part
part of
reproduced or
or distributed
distributed in
in any
any form
form or
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced
you are
are aa student
student using
using this
this Manual,
Manual,
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
distribution
If you
you are
are using
using it
it without
without permission.
permission.
you
333
333
PROBLEM 12.41
A 100-g sphere D is at rest relative to drum ABC, which rotates
at a constant rate. Neglecting friction, determine the range of
the allowable values of the velocity v of the sphere if neither of
the normal forces exerted by the sphere on the inclined
surfaces of the drum is to exceed 1.1 N.
SOLUTION
vD2
First we note
aD = an =
where
ρ = 0.2 m
ρ
ΣFx = mD aD : N1 cos 60° + N 2 cos 20° = mD
vD2
ρ
(1)
ΣFy = 0: N1 sin 60° + N 2 sin 20° − wD = 0
N1 sin 60° + N 2 sin 20° = mD g
or
(2)
Case 1: N1 is maximum.
N1 = 1.1 N
Let
Eq. (2)
(1.1 N) sin 60° + N 2 sin 20° = (0.1 kg) (9.81 m/s 2 )
N 2 = 0.082954 N
or
( N 2 )( N1 )max � 1.1 N.
O.K.
0.2 m
(1.1cos 60° + 0.082954 cos 20°) N
0.1 kg
Eq. (1)
(vD2 )( N1 )max =
or
(vD )( N1 )max = 1.121 m/s
Now we form
(sin 20°) (1) − (cos 20°) (2)
N1 cos 60° sin 20° − N1 sin 60° cos 20° = mD
or
− N1 sin 40° = mD
vD2
ρ
vD2
ρ
sin 20° − mD g cos 20°
sin 20° − mD g cos 20°
(vD )min occurs when N1 = ( N1 ) max
(vD )min = 1.121 m/s
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
All rights
rights reserved.
reserved. No
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
Inc. All
No part
part of
of this
reproduced or
or distributed
distributed in
in any
any form
form or
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced
are aa student
student using
using this
this Manual,
Manual,
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
individual course
distribution
for their
their individual
course preparation.
preparation. If
If you
you are
you are
are using
using it
it without
without permission.
you
permission.
334
334
PROBLEM 12.41 (Continued)
Case 2: N 2 is maximum.
N 2 = 1.1 N
Let
Eq. (2)
N1 sin 60° + (1.1 N)sin 20° = (0.1 kg)(9.81 m/s 2 )
N1 = 0.69834 N
or
( N1 )( N 2 )max � 1.1 N.
O.K.
0.2 m
(0.69834cos 60° + 1.1cos 20°) N
0.1 kg
Eq. (1)
(vD2 )( N 2 )max =
or
(vD )( N 2 )max = 1.663 m/s
Now we form
(sin 60°) (1) − (cos 60°) (2)
N 2 cos 20° sin 60° − N 2 sin 20° cos 60° = mD
or
N 2 cos 40° = mD
vD2
ρ
vD2
ρ
sin 60° − mD g cos 60°
sin 60° − mD g cos 60°
(vD ) max occurs when N 2 = ( N 2 )max
(vD ) max = 1.663 m/s
For N1 , N 2 � 1.1 N
1.121 m/s � vD � 1.663 m/s �
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
No part
part of
reproduced or
or distributed
distributed in
in any
any form
form or
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced
you are
are aa student
student using
using this
this Manual,
Manual,
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
distribution
If you
you are
are using
using it
it without
without permission.
permission.
you
335
335
PROBLEM 12.42*
As part of an outdoor display, a 6-kg model C of the earth is attached
to wires AC and BC and revolves at a constant speed v in the horizontal
circle shown. Determine the range of the allowable values of v if both
wires are to remain taut and if the tension in either of the wires is not to
exceed 120 N.
SOLUTION
vC2
First note
aC = an =
where
ρ = 1.5 m
ρ
ΣFx = mC aC : TCA sin 40° + TCB sin15° = m C
ΣFy = 0: TCA cos 40° − TCB cos15° − WC = 0
vC2
ρ
(1)
(2)
Note that Eq. (2) implies that
(a)
when
TCB = (TCB ) max , TCA = (TCA ) max
(b)
when
TCB = (TCB ) min ,
TCA = (TCA )min
Case 1: TCA is maximum.
TCA = 120 N
Let
Eq. (2)
(120 N) cos 40° − TCB cos15° − (6 kg)(9.81 m/s2) = 0
TCB = 33.07 N
or
(TCB )(TCA )max
120 N
O.K.
[(TCB )max = 33.07 N ]
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
No part
part of
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced or
or distributed
distributed in
in any
any form
form or
reproduced
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
you are
are aa student
student using
using this
this Manual,
Manual,
distribution
If you
you are
are using
using it
it without
without permission.
permission.
you
336
336
PROBLEM 12.42* (Continued)
Eq. (1)
(vC2 )(TCA )max =
(1.5 m)
(120 sin 40° + 33.07 sin 15°)N
(6 kg)
(vC )(TCA )max = 4.64 m/s
or
Now we form
(cos 15°) + [Eq. (2)](sin15°)
[Eq. (1)]
TCA sin 40° cos15° + TCA cos 40° sin15° = mC
TCA sin 55° = mC
or
v C2
ρ
v C2
ρ
cos15° + WC sin15°
cos15° + WC sin15°
(3)
(vc ) max occurs when TCA = (TCA ) max
(vC )max = 4.64 m/s
Case 2: TCA is minimum.
Because (TCA ) min occurs when TCB = (TCB ) min ,
let TCB = 0 (note that wire BC will not be taut).
TCA cos 40° − (6 × 9.81 N) = 0
Eq. (2)
TCA = 76.84 N
or
120 N O.K.
Note: Eq. (3) implies that when TCA = (TCA )min , vC = (vC ) min . Then
(1.5 m)
(76.84 N) sin40°
(6 kg)
Eq. (1)
(vC2 )min =
or
(vC )min = 3.51 m/s
0
TCA , TCB
120 N when
3.51 m/s
vC
4.64 m/s
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
No part
part of
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced or
or distributed
distributed in
in any
any form
form or
reproduced
distribution
distribution to
toteachers
teachersand
andeducators
educatorspermitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
If you
you are
are aastudent
studentusing
usingthis
thisManual,
Manual,
you
you are
are using
using itit without
without permission.
permission.
337
337
PROBLEM 12.43*
The 0.5-kg flyballs of a centrifugal governor revolve at a constant speed
v in the horizontal circle of 0.15 m 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 75 N.
SOLUTION
v2
First note
a = an =
where
ρ = 0.15 m
ρ
ΣFx = ma : TDA sin 20° + TDE sin 30° = m
v2
ρ
(1)
ΣFy = 0: TDA cos 20° − TDE cos 30° − W = 0
(2)
Note that Eq. (2) implies that
(a)
when
TDE = (TDE ) max ,
TDA = (TDA ) max
(b)
when
TDE = (TDE ) min ,
TDA = (TDA ) min
Case 1: TDA is maximum.
Let
Eq. (2)
TDA = 75 N
(75 N) cos 20° − TDE cos 30° − (0.5 × 9.81 N) = 0
or
TDE = 75.72 N unacceptable ( 75 N)
Now let
TDE = 75 N
Eq. (2)
or
TDA cos 20° − (75 N) cos30° − (0.5 × 9.81 N) = 0
TDA = 74.34 N
O.K. (< 75 N)
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
No part
part of
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced or
or distributed
distributed in
in any
any form
form or
reproduced
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
you are
are aa student
student using
using this
this Manual,
Manual,
distribution
If you
you are
are using
using it
it without
without permission.
permission.
you
338
338
PROBLEM 12.43* (Continued)
(TDA ) max = 74.34 N
(TDE ) max = 75 N
(0.15 m)
(74.34 sin 20°+ 75sin 30°) N
0.5 kg
Eq. (1)
(v 2 )(TDA )max =
or
v(TDA )max = 4.34 m/s
Now form
[Eq. (1)]
(cos 30°) + [Eq. (2)](sin30°)
TDA sin 20° cos 30° + TDA cos 20° sin 30° = m
TDA sin 50° = m
or
v2
ρ
v2
ρ
cos 30° + W sin 30°
cos30° + W sin 30°
(3)
vmax occurs when TDA = (TDA ) max
vmax = 4.34 m/s
Case 2: TDA is minimum.
Because (TDA )min occurs when TDE = (TDE )min ,
let TDE = 0.
Eq. (2)
TDA cos 20° − (0.5 × 9.81 N) = 0
or
TDA = 5.22 N
75 N
O.K.
Note: Eq. (3) implies that when TDA = (TDA ) min , v = vmin . Then
Eq. (1)
(0.15 m)
(v 2 )min =
(5.22 N) sin 20°
0.5 kg
vmin = 0.73 m/s
or
0
TAB , TBC , TAD , TDE
75 N
when
0.73 m/s
v
4.34 m/s
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
No part
part of
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced or
or distributed
distributed in
in any
any form
form or
reproduced
distribution
distribution to
toteachers
teachersand
andeducators
educatorspermitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
If you
you are
are aastudent
studentusing
usingthis
thisManual,
Manual,
you
you are
are using
using itit without
without permission.
permission.
339
339
PROBLEM 12.44
A child having a mass of 22 kg sits on a swing and is held in the
position shown by a second child. Neglecting the mass of the
swing, determine the tension in rope AB (a) while the second child
holds the swing with his arms outstretched horizontally,
(b) immediately after the swing is released.
SOLUTION
Note: The factors of “ 12 ” are included in the following free-body diagrams because there are two ropes and
only one is considered.
(a)
For the swing at rest
1
ΣFy = 0: TBA cos 35° − W = 0
2
TBA =
or
22 kg × 9.81 m/s 2
2 cos 35°
TBA = 131.7 N �
or
(b)
At t = 0, v = 0, so that
an =
v2
ρ
=0
1
ΣFn = 0: TBA − W cos 35° = 0
2
or
TBA =
1
(22 kg)(9.81 m/s 2 ) cos 35°
2
TBA = 88.4 N �
or
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
All rights
rights reserved.
reserved. No
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
Inc. All
No part
part of
of this
reproduced or
or distributed
distributed in
in any
any form
form or
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced
are aa student
student using
using this
this Manual,
Manual,
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
individual course
distribution
for their
their individual
course preparation.
preparation. If
If you
you are
you are
are using
using it
it without
without permission.
you
permission.
340
340
PROBLEM 12.45
A 60-kg wrecking ball B is attached to a 15-m-long steel cable
AB and swings in the vertical arc shown. Determine the tension
in the cable (a) at the top C of the swing, (b) at the bottom D of
the swing, where the speed of B is 4.2 m/s.
SOLUTION
(a)
At C, the top of the swing, vB = 0; thus
an =
vB2
=0
LAB
ΣFn = 0: TBA − WB cos 20° = 0
TBA = (60 kg) (9.81 m/s 2 ) × cos 20°
or
TBA = 553 N �
or
ΣFn = man : TBA − WB = mB
(b)
or
(vB ) 2D
LAB
�
(4.2 m/s)2 �
TBA = (60 kg) �9.81 m/s 2 +
�
15 m �
�
TBA = 659 N �
or
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
No part
part of
reproduced or
or distributed
distributed in
in any
any form
form or
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced
you are
are aa student
student using
using this
this Manual,
Manual,
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
distribution
If you
you are
are using
using it
it without
without permission.
permission.
you
341
341
PROBLEM 12.46
During a high-speed chase, a 1200-kg sports car traveling at a
speed of 160 km/h just loses contact with the road as it
reaches the crest A of a hill. (a) Determine the radius of
curvature 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 an 80-kg driver by the seat of his 1500-kg car as the car,
traveling at a constant speed of 80 km/h, passes through A
SOLUTION
(a)
Note:
160 km/h = 44.44 m/s
ΣFn = man : Wcar =
Wcar v A2
g ρ
(44.44 m/s)2
9.81 m/s2
= 201.3 m
ρ=
or
ρ = 201 m
or
(b)
Note: v is constant
a t = 0; 80 km/h = 22.22 m/s
ΣFn = man : W − N =
or
W v 2A
g ρ
N = (80 × 9.81 N) 1 −
(22.22 m/s)2
(9.81 m/s2)(201.3 m)
N = 589 N ↑
or
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
No part
part of
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced or
or distributed
distributed in
in any
any form
form or
reproduced
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
you are
are aa student
student using
using this
this Manual,
Manual,
distribution
If you
you are
are using
using it
it without
without permission.
permission.
you
342
342
PROBLEM 12.47
The portion of a toboggan run shown is contained in a vertical
plane. Sections AB and CD have radii of curvature as indicated, and
section BC is straight and forms an angle of 20° with the horizontal.
Knowing that the coefficient of kinetic friction between a sled and
the run is 0.10 and that the speed of the sled is 7 m/s at B,
determine the tangential component of the acceleration of the sled
(a) just before it reaches B, (b) just after it passes C.
SOLUTION
(a)
Note: Just before B, ρ B = 18 m
ΣFn = man : N − W cos 20° =
W vB2
g ρB
or
N = W cos 20° +
Sliding:
F = µk N
vB2
g ρB
= µkW cos 20° +
ΣFt = mat : W sin 20° − F =
vB2
g ρB
W
at
g
vB2
or
at = g (sin 20° − µt cos 20°) − µk
Then
at = (9.81 m/s2)(sin 20° − 0.1cos 20°) − 0.1
ρB
(7 m/s)2
18 m
at = 2.16 m/s2
or
20°
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
No part
part of
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced or
or distributed
distributed in
in any
any form
form or
reproduced
distributionto
toteachers
teachersand
andeducators
educatorspermitted
permitted by
byMcGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
If you
you are
areaastudent
studentusing
usingthis
thisManual,
Manual,
distribution
you are
are using
using itit without
without permission.
permission.
you
343
343
PROBLEM 12.47 (Continued)
(b)
It is first necessary to determine vC .
For section BC
ΣFy = 0: N BC − W cos 20° = 0
or
N BC = W cos 20°
Sliding:
FBC = µk N BC = µ kW cos 20°
ΣFx = maBC : W sin 20° − FBC =
W
aBC
g
aBC = g (sin 20° − µ k cos 20°)
or
= (9.81 m/s2)(sin 20° − 0.1cos 20°)
= 2.43 m/s2
For this uniformly accelerated motion we have
vC2 = vB2 + 2aBC ∆ xBC
= (7 m/s)2 + 2(2.43 m/s2)(12 m)
vC = 13.81 m/s2
or
Now, just after C, ρC = 40 m
ΣFn = man : W cos 20° − N =
W vC2
g ρC
or
N = W cos 20° −
Sliding:
F = µk N
vC2
g ρC
= µ kW cos 20° −
ΣFt = mat : W sin 20° − F =
or
Note:
Then
vC2
g ρC
W
at
g
at = g (sin 20° − µk cos 20°) + µk
vC2
ρC
g (sin 20° − µ k cos 20°) = aBC
at = 2.43 m/s2 + 0.1
(13.81 m/s)2
40 m
at = 2.91 m/s2
or
20°
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
No part
part of
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced or
or distributed
distributed in
in any
any form
form or
reproduced
distribution
If you
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
you are
are aa student
student using
using this
this Manual,
Manual,
you
you are
are using
using it
it without
without permission.
permission.
344
344
PROBLEM 12.48
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 0.40, determine
(a) the force exerted by the belt on a package just after it has passed
Point A, (b) the angle θ defining 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
an =
v2
ρ
=
(1 m/s) 2
= 4 m/s 2
0.250 m
For any angle θ
ΣFy = ma y : N − mg cos θ = − man = −
mv 2
N = mg cos θ −
ρ
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)
N = 2.905 N �
F f = µs N
mg sin θ = µ s ( mg cos θ − man )
�
a �
4 �
�
sin θ = µ s � cos θ − n � = 0.40 �cos θ −
g �
9.81 ��
�
�
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
No part
part of
reproduced or
or distributed
distributed in
in any
any form
form or
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced
you are
are aa student
student using
using this
this Manual,
Manual,
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
distribution
If you
you are
are using
using it
it without
without permission.
permission.
you
345
345
PROBLEM 12.48 (Continued)
Squaring and using trigonometic identities,
1 − cos 2 θ = 0.16 cos 2 θ − 0.130479cos θ + 0.026601
1.16 cos 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=
Ff
µs
=
mg sin θ
N � 0. �
µs
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
All rights
rights reserved.
reserved. No
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
Inc. All
No part
part of
of this
reproduced or
or distributed
distributed in
in any
any form
form or
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced
are aa student
student using
using this
this Manual,
Manual,
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
individual course
distribution
for their
their individual
course preparation.
preparation. If
If you
you are
you are
are using
using it
it without
without permission.
you
permission.
346
346
PROBLEM 12.49
A 54-kg pilot files 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.
At A:
v2
ΣFn = man : N A − W = m A
ρ
� 1680 N
�
v A2 = (1200 m) �
− 9.81 m/s 2 �
� 54 kg
�
or
= 25,561.3 m 2 /s 2
At C:
v2
ΣFn = man : N C + W = m C
or
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 2A + 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
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
No part
part of
reproduced or
or distributed
distributed in
in any
any form
form or
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced
you are
are aa student
student using
using this
this Manual,
Manual,
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
distribution
If you
you are
are using
using it
it without
without permission.
permission.
you
347
347
PROBLEM 12.49 (Continued)
At B:
v2
ΣFn = man : N B = m B
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
PB = (54 kg)(0.79730 − 9.81) m/s 2
or
PB = 486.69 N ↑
Finally,
( Fpilot ) B = N B2 + PB2 = (1014.98) 2 + (486.69) 2
= 1126 N
(Fpilot ) B = 1126 N
or
25.6° ��
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
All rights
rights reserved.
reserved. No
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
Inc. All
No part
part of
of this
reproduced or
or distributed
distributed in
in any
any form
form or
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced
are aa student
student using
using this
this Manual,
Manual,
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
individual course
distribution
for their
their individual
course preparation.
preparation. If
If you
you are
you are
are using
using it
it without
without permission.
you
permission.
348
348
PROBLEM 12.50
A 250-g block B fits inside a small cavity cut in arm OA, which
rotates in the vertical plane at a constant rate such that v = 3 m/s.
Knowing that the spring exerts on block B a force of magnitude
P = 1.5 N and neglecting the effect of friction, determine the range of
values of θ for which block B is in contact with the face of the cavity
closest to the axis of rotation O.
SOLUTION
ΣFn = man : P + mg sin θ − Q = m
v2
ρ
To have contact with the specified surface, we need Q � 0,
or
Q = P + mg sin θ −
sin θ �
Data:
mv 2
ρ
�0
1 � v2 P �
� − �
g �� ρ m ��
(1)
m = 0.250 kg, v = 3 m/s, P = 1.5 N, ρ = 0.9 m
Substituting into (1):
sin θ �
1 � (3) 2 1.5 �
−
�
�
9.81 � 0.9 0.25 �
sin θ � 0.40775
24.1° � θ � 155.9° �
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
No part
part of
reproduced or
or distributed
distributed in
in any
any form
form or
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced
you are
are aa student
student using
using this
this Manual,
Manual,
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
distribution
If you
you are
are using
using it
it without
without permission.
permission.
you
349
349
PROBLEM 12.51
A curve in a speed track has a radius of 300 m and a rated speed of 192 km/h.
(See sample Problem 12.6 for the definition of rated speed). Knowing that a
racing car starts skidding on the curve when traveling at a speed of 288 km/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
Weight
W = mg
Acceleration
a=
v2
ρ
Fx = max : F + W sin θ = ma cos θ
F=
mv 2
ρ
cos θ − mg sin θ
(1)
Fy = 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 = 192 km/h = 53.33 m/s. F = 0 at rated speed.
0=
mv 2
ρ
cos θ − mg sin θ
v2
(53.33)2
=
= 0.96639
ρ g (300)(9.81)
θ = 44.02°
tan θ =
(b)
Slipping outward.
θ = 44°
v = 288 km/h = 80 m/s
F = µN
µ=
µ=
F v 2 cos θ − ρ g sin θ
=
N v 2 sin θ + ρ g cos θ
(80)2 cos 44.02° − (300)(9.81) sin 44.02°
(80)2 sin 44.02° + (300)(9.81) cos 44.02°
= 0.3896
µ = 0.390
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
No part
part of
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced or
or distributed
distributed in
in any
any form
form or
reproduced
distribution to
toteachers
teachersand
andeducators
educatorspermitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
If you
you are
are aastudent
studentusing
usingthis
thisManual,
Manual,
distribution
you are
are using
using itit without
without permission.
permission.
you
350
350
PROBLEM 12.51 (Continued)
(c)
Minimum speed.
F = −µ N
v 2 cos θ − ρ g sin θ
v 2 sin θ + ρ g cos θ
ρ g (sin θ − µ cos θ )
v2 =
cos θ + µ sin θ
−µ =
(300)(9.81)(sin 44.02° − 0.3896 cos 44.02°)
=
cos 44.02° + 0.3896 sin 44.02°
v = 35.12 m/s
v = 126 km/h
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
No part
part of
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced or
or distributed
distributed in
in any
any form
form or
reproduced
distribution
distribution to
toteachers
teachersand
andeducators
educatorspermitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
If you
you are
are aastudent
studentusing
usingthis
thisManual,
Manual,
you
you are
are using
using itit without
without permission.
permission.
351
351
PROBLEM 12.52
A car is traveling on a banked road at a constant speed v. Determine the range of values of v for which the car
does not skid. Express your answer in terms of the radius r of the curve, the banking angle θ , and the angle of
static friction φs between the tires and the pavement.
SOLUTION
Case 1: v = vmax
Note: R = F + N
ΣFn = man :
v2
R sin (θ + φs ) = m max
r
R cos (θ + φs ) − W = 0
+ ΣFy = 0:
R cos (θ + φs ) = mg
or
(1)
Forming (2)
(1)
(2)
v2
m max
R sin (θ + φs )
r
=
R cos (θ + φs )
mg
vmax = gr tan (θ + φs )
or
Case 2: v = vmin
Note: R = F + N
ΣFn = man :
v2
R sin (θ − φs ) = m min
r
R cos (θ − φs ) − W = 0
+ ΣFy = 0:
R cos (θ − φs ) = mg
or
(3)
Forming (4)
or
(3)
(4)
v2
m min
R sin (θ − φs )
r
=
R cos (θ − φs )
mg
vmin = gr tan (θ − φs )
For the car not to skid
gr tan (θ − φs ) � v � gr tan (θ + φs ) ��
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
All rights
rights reserved.
reserved. No
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
Inc. All
No part
part of
of this
reproduced or
or distributed
distributed in
in any
any form
form or
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced
are aa student
student using
using this
this Manual,
Manual,
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
individual course
distribution
for their
their individual
course preparation.
preparation. If
If you
you are
you are
are using
using it
it without
without permission.
you
permission.
352
352
PROBLEM 12.53
Tilting trains, such as the American Flyer, which will run from
Washington to New York and Boston, are designed to travel safely
at high speeds on curved sections of track, which were built for
slower, conventional trains. As it enters a curve, each car is tilted
by hydraulic actuators mounted on its trucks. The tilting feature of
the cars also increases passenger comfort by eliminating or greatly
reducing the side force Fs (parallel to the floor of the car) to which
passengers feel subjected. For a train traveling at 160 km/h on a
curved section of track banked through an angle θ = 6° and with a
rated speed of 96 km/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 12.6 for the definition of
rated speed.)
SOLUTION
Rated speed:
vR = 96 km/h = 26.67 m/s, 160 km/h = 44.44 m/s
From Sample Problem 12.6,
vR2 = g ρ tan θ
ρ=
or
vR2
(26.67)2
=
= 690 m
g tan θ 9.81 tan6°
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.
Fs = W
v2
cos (θ + φ ) − sin (θ + φ )
gρ
=W
(44.44)2
cos 6° − sin 6°
(9.81)(690)
Fs = 0.186W
= 0.1856W
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
No part
part of
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced or
or distributed
distributed in
in any
any form
form or
reproduced
distribution to
toteachers
teachersand
andeducators
educatorspermitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
If you
you are
are aastudent
studentusing
usingthis
thisManual,
Manual,
distribution
you are
are using
using itit without
without permission.
permission.
you
353
353
PROBLEM 12.53 (Continued)
(b)
For Fs = 0,
v2
cos (θ + φ ) − sin (θ + φ ) = 0
gρ
(44.44)2
v2
=
= 0.29176
g ρ (9.81)(690)
θ + φ = 16.27°
φ = 16.27° − 6°
tan (θ + φ ) =
φ = 10.3°
PROPRIETARY MATERIAL.
MATERIAL. ©
2009 The
The McGraw-Hill
McGraw-Hill Companies,
be displayed,
displayed,
PROPRIETARY
© 2010
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
No part
part of
of this
this Manual
Manual may
may be
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
or used
used beyond
beyond the
the limited
limited
reproduced or
or distributed
distributed in
in any
any form
form or
reproduced
of the
the publisher,
publisher, or
distribution
to teachers
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. IfIf you
you are
are aa student
distribution to
teachers and
and educators
educators permitted
permitted by
student using
using this
this Manual,
Manual,
you
are using
using itit without
without permission.
permission.
you are
354
354
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.
PROBLEM 12.53 Tilting trains, such as the American Flyer,
which will run from Washington to New York and Boston, are
designed to travel safely at high speeds on curved sections of track,
which were built for slower, conventional trains. As it enters a
curve, each car is tilted by hydraulic actuators mounted on its
trucks. The tilting feature of the cars also increases passenger
comfort by eliminating or greatly reducing the side force Fs
(parallel to the floor of the car) to which passengers feel subjected.
For a train traveling at 160 km/h on a curved section of track
banked through an angle θ = 6° and with a rated speed of 96 km/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 12.6 for the definition of rated speed.)
SOLUTION
vR = 96 km/h = 26.67 m/s, 160 km/h = 44.44 m/s
Rated speed:
From Sample Problem 12.6,
vR2 = g ρ tan θ
or
ρ=
vR2
(26.67)2
=
= 690 m
g tan θ 9.81 tan6°
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,
Fs = W
mv 2
ρ
cos (θ + φ )
v2
cos (θ + φ ) − sin (θ + φ )
gρ
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
All rights
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
Inc. All
rights reserved.
reserved. No
No part
part of
of this
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced or
or distributed
distributed in
in any
any form
form or
reproduced
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
student using
using this
this Manual,
Manual,
distribution
for their
their individual
individual course
course preparation.
preparation. If
If you
you are
are aa student
you are
are using
using it
it without
without permission.
you
permission.
355
355
PROBLEM 12.54 (Continued)
Now
So that
Let
Then
(44.44)2
v2
=
= 0.29176
g ρ (9.81)(690)
and Fs = 0.10W
0.10W = W [0.29176 cos (θ + φ ) − sin (θ + φ )]
u = sin (θ + φ )
cos (θ + φ ) = 1 − u 2
0.10 = 0.29176 1 − u 2 − u or 0.29176 1 − u 2 = 0.10 + u
Squaring both sides,
0.08512(1 − u 2 ) = 0.01 + 0.2u + u 2
or
1.08512u 2 + 0.2u − 0.07512 = 0
The positive root of the quadratic equation is u = 0.18663
Then,
θ + φ = sin −1 u = 10.76°
φ = 10.76° − 6°
φ = 4.76°
PROPRIETARY MATERIAL.
MATERIAL. ©
2009 The
The McGraw-Hill
McGraw-Hill Companies,
be displayed,
displayed,
PROPRIETARY
© 2010
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
No part
part of
of this
this Manual
Manual may
may be
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
or used
used beyond
beyond the
the limited
limited
reproduced or
or distributed
distributed in
in any
any form
form or
reproduced
of the
the publisher,
publisher, or
distribution
to teachers
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. IfIf you
you are
are aa student
distribution to
teachers and
and educators
educators permitted
permitted by
student using
using this
this Manual,
Manual,
you
are using
using itit without
without permission.
permission.
you are
356
356
PROBLEM 12.55
A small, 300-g collar D can slide on portion AB of a rod which is bent as shown.
Knowing that α = 40° and that the rod rotates about the vertical AC at a constant rate of
5 rad/s, determine the value of r for which the collar will not slide on the rod if the effect
of friction between the rod and the collar is neglected.
SOLUTION
First note
vD = rθ�ABC
+ ΣFy = 0: N sin 40° − W = 0
or
N=
mg
sin 40°
v2
ΣFn = man : N cos 40° = m D
r
or
( rθ�ABC ) 2
mg
cos 40° = m
sin 40°
r
or
r=
g
2
θ ABC
1
tan 40°
9.81 m/s 2
1
2
(5 rad/s) tan 40°
= 0.468 m
=
r = 468 mm �
or
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
No part
part of
reproduced or
or distributed
distributed in
in any
any form
form or
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced
you are
are aa student
student using
using this
this Manual,
Manual,
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
distribution
If you
you are
are using
using it
it without
without permission.
permission.
you
357
357
PROBLEM 12.56
A small, 200-g collar D can slide on portion AB of a rod which is bent as
shown. Knowing that the rod rotates about the vertical AC at a constant
rate and that α = 30° and r = 600 mm, determine the range of values of
the speed v for which the collar will not slide on the rod if the coefficient
of static friction between the rod and the collar is 0.30.
SOLUTION
Case 1: v = vmin , impending motion downward
ΣFx = max : N − W sin 30° = m
or
v2
cos 30°
r
�
�
v2
N = m �� g sin 30° + cos 30° ��
r
�
�
ΣFy = ma y : F − W cos 30° = −m
v2
sin 30°
r
or
�
�
v2
F = m �� g cos 30° − sin 30° ��
r
�
�
Now
F = µs N
Then
�
�
�
�
v2
v2
m �� g cos 30° − sin 30° �� = µs × m �� g sin 30° + cos 30° ��
r
r
�
�
�
�
or
v 2 = gr
1 − µ s tan 30°
µ s + tan 30°
= (9.81 m/s 2 )(0.6 m)
or
1 − 0.3tan 30°
0.3 + tan 30°
vmin = 2.36 m/s
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
All rights
rights reserved.
reserved. No
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
Inc. All
No part
part of
of this
reproduced or
or distributed
distributed in
in any
any form
form or
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced
are aa student
student using
using this
this Manual,
Manual,
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
individual course
distribution
for their
their individual
course preparation.
preparation. If
If you
you are
you are
are using
using it
it without
without permission.
you
permission.
358
358
PROBLEM 12.56 (Continued)
Case 2: v = vmax , impending motion upward
ΣFx = max : N − W sin 30° = m
v2
cos 30°
r
�
�
v2
N = m �� g sin 30° + cos 30° ��
r
�
�
or
ΣFy = ma y : F + W cos 30° = m
v2
sin 30°
r
or
�
�
v2
F = m �� − g cos 30° + sin 30° ��
r
�
�
Now
F = µs N
Then
�
�
�
�
v2
v2
m �� − g cos 30° + sin 30° �� = µs × m �� g sin 30° + cos30° ��
r
r
�
�
�
�
or
v 2 = gr
1 + µs tan 30°
tan 30° − µ s
= (9.81 m/s 2 )(0.6 m)
or
1 + 0.3tan 30°
tan 30° − 0.3
vmax = 4.99 m/s
For the collar not to slide
2.36 m/s � v � 4.99 m/s ��
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
No part
part of
reproduced or
or distributed
distributed in
in any
any form
form or
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced
you are
are aa student
student using
using this
this Manual,
Manual,
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
distribution
If you
you are
are using
using it
it without
without permission.
permission.
you
359
359
PROBLEM 12.57
A small, 300-g collar D can slide on portion AB of a rod which is bent as shown.
Knowing that r = 200 mm and that the rod rotates about the vertical AC at
a constant rate of 10 rad/s, determine the smallest allowable value of the
coefficient of static friction between the collar and the rod if the collar is not to
slide when (a) α = 15°, (b) α = 45°. Indicate for each case the direction of the
impending motion.
SOLUTION
First we note that v = rθ ABC = (0.2 m)(10 rad/s) = 2 m/s, and that requiring µs = ( µs )min implies that sliding of
collar D is impending.
µs = tan φs
Also,
Now we consider the two possible cases of impending motion.
Case 1: Impending motion downward.
ΣFx = max : N − W sin α =
W v2
cos α
g r
N = W sin α −
or
ΣFy = ma y : F − W cos α = −
W v2
sin α
g r
or
F = W cos α −
Now
F = µs N
Then
or
W cos α −
v2
cos α
gr
v2
sin α
gr
v2
v2
sin α = µs × W sin α +
cos α
gr
gr
v 2 1 − µs tan α 1 − tan φs tan α
=
=
gr tan α + µ s
tan α + tan φs
=
1
tan (α + φs )
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
No part
part of
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced or
or distributed
distributed in
in any
any form
form or
reproduced
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
you are
are aa student
student using
using this
this Manual,
Manual,
distribution
If you
you are
are using
using it
it without
without permission.
permission.
you
360
360
PROBLEM 12.57 (Continued)
Case 2: Impending motion upward.
ΣFx = max : N − W sin α =
W v2
cos α
g r
N = W sin α +
or
ΣFy = ma y : F + W cos α =
v2
cos α
gr
W v2
sin α
g r
or
F = W − cos α +
Now
F = µs N
W − cos α +
Then
v2
sin α
gr
v2
v2
sin α = µ s × W sin α + cos α
gr
gr
v 2 1 + µ s tan α 1 + tan α tan φs
=
=
gr
tan α − µ s
tan α − tan φs
or
=
1
tan (α − φs )
gr (9.81 m/s2)(0.2 m)
=
= 0.4905
(2 m/s)2
v2
Now
tan (α ± φs ) = 0.4905
Then
a ± fs = 26.13°,
or
fs 0
and where the “+” corresponds to impending motion downward and the “–” to impending motion upward.
(a)
α = 15°:
15° ± φs = 26.13°
We have
φs
0
“+”
so that
φs = 11.13°
Then
( µ s )min = tan11.13°
or
( µ s ) min = 0.1967
motion impending downward
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
No part
part of
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced or
or distributed
distributed in
in any
any form
form or
reproduced
distribution
distribution to
toteachers
teachersand
andeducators
educatorspermitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
If you
you are
are aastudent
studentusing
usingthis
thisManual,
Manual,
you
you are
are using
using itit without
without permission.
permission.
361
361
PROBLEM 12.57 (Continued)
(b)
α = 45° :
45° ± φs = 26.13°
We have
φs
Then
or
0
“–”
so that
φs = 18.87°
( µs )min = tan18.87°
µs = 0.342 motion impending upward
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
No part
part of
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced or
or distributed
distributed in
in any
any form
form or
reproduced
distribution
If you
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
you are
are aa student
student using
using this
this Manual,
Manual,
you
you are
are using
using it
it without
without permission.
permission.
362
362
PROBLEM 12.58
A semicircular slot of 250-mm radius is cut in a flat plate which rotates about
the vertical AD at a constant rate of 14 rad/s. A small, 0.4-kg 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
r = (0.65 − 0.25 sinq)m
vE = ρφ ABCD
Then
an =
vE2
ρ
= ρ (φ ABCD ) 2
= [(0.65 − 0.25 sinq)m](14 rad/s)2
= 9.8(13 − 5sin q ) m/s2
Assume that the block is at rest with respect to the plate.
v2
ΣFx = max : N + W cos θ = m E sin θ
ρ
or
N = W − cos θ +
vE2
sin θ
gρ
v2
ΣFy = ma y : −F + W sin θ = − m E cos θ
ρ
or
F = W sin θ +
vE2
cos θ
gρ
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
No part
part of
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced or
or distributed
distributed in
in any
any form
form or
reproduced
distributionto
toteachers
teachersand
andeducators
educatorspermitted
permittedby
byMcGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
If you
you are
areaastudent
studentusing
usingthis
thisManual,
Manual,
distribution
you are
are using
using itit without
without permission.
permission.
you
363
363
PROBLEM 12.58 (Continued)
(a)
We have
θ = 80°
Then
1
× 9.8(13 − 5sin80°) m/s2 × sin80°
9.81 m/s2
N = (0.4 × 9.81 N) −cos80°+
= 30.5 N
F = (0.4 × 9.81 N) sin80°+
1
× 9.8(13 − 5sin80°) m/s2 × cos80°
2
9.81 m/s
= 9.36 N
Fmax = µs N = 0.35(30.5 N) = 10.68 N
Now
The block does not slide in the slot, and
F = 9.36 N
(b)
We have
80°
θ = 40°
Then
N = (0.4 × 9.81 N) −cos40° +
1
× 9.8(13 − 5sin40°) m/s2 × sin40°
9.81 m/s2
= 21.65 N
F = (0.4 × 9.81 N) sin40° +
1
× 9.8(13 − 5sin40°) m/s2 × cos40°
9.81 m/s2
= 31.91 N
Now
Fmax = µs N ,
F
from which it follows that
Fmax
Block E will slide in the slot
and
a E = a n + a E/plate
= a n + (a E/plate )t + (a E/plate ) n
At t = 0, the block is at rest relative to the plate, thus (a E/plate )n = 0 at t = 0, so that a E/plate must be
directed tangentially to the slot.
v2
ΣFx = max : N + W cos 40° = m E sin 40°
ρ
or
N = W − cos 40° +
vE2
sin 40°
gρ
(as above)
= 21.65 N
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
No part
part of
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced or
or distributed
distributed in
in any
any form
form or
reproduced
distribution
If you
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
you are
are aa student
student using
using this
this Manual,
Manual,
you
you are
are using
using it
it without
without permission.
permission.
364
364
PROBLEM 12.58 (Continued)
F = µk N
Sliding:
= 0.25(21.65 N)
= 5.41 N
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
F = 5.41 N
40°
the block slides downward in the slot and
Alternative solutions.
(a)
Assume that the block is at rest with respect to the plate.
ΣF = ma : W + R = ma n
tan (φ − 10°) =
Then
=
W
W
g
=
=
2
man W vE ρ (φ ABCD ) 2
g ρ
9.81 m/s2
9.8(13 − 5sin80°) m/s2
or
φ − 10° = 7.066°
and
φ = 17.066°
tan φs = µ s
Now
φ
µ s = 0.35
φs = 19.29°
so that
0
(from above)
φs
Block does not slide and R is directed as shown.
Now
F = R sin φ
Then
F = (0.4 × 9.81 N)
and R =
W
sin (φ − 10°)
sin17.066°
sin 7.066°
= 9.36 N
F = 9.36 N
The block does not slide in the slot and
80°
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
No part
part of
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced or
or distributed
distributed in
in any
any form
form or
reproduced
distribution
distribution to
toteachers
teachersand
andeducators
educatorspermitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
If you
you are
are aastudent
studentusing
usingthis
thisManual,
Manual,
you
you are
are using
using itit without
without permission.
permission.
365
365
PROBLEM 12.58 (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
tan (φ − 50°) =
×°
g
ρ (φ ABCD )
or
φ − 50° = 5.84°
and
φ = 55.84°
Now
φs = 19.29°
so that
φ
=
2
9.81 m/s2
9.8(13 − 5 sin 40°) m/s2
φs
The block will slide in the slot and then
φ = φk , where
φk = 14.0362°
or
tan φk = µ k
µk = 0.25
To determine in which direction the block will slide, consider the free-body diagrams for the two
possible cases.
ΣF = ma : W + R = ma n + ma E/plate
Now
From the diagrams it can be concluded that this equation can be satisfied only if the block is sliding
downward. Then
v2
ΣFx = max : W cos 40° + R cos φk = m E sin 40°
ρ
F = R sin φk
Now
Then
or
W cos 40° +
F
W vE2
=
sin 40°
g ρ
tan φk
F = µkW − cos 40° +
= 5.41 N
vE2
sin 40°
gρ
(see the first solution)
F = 5.41 N
The block slides downward in the slot and
40°
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
No part
part of
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced or
or distributed
distributed in
in any
any form
form or
reproduced
distribution
If you
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
you are
are aa student
student using
using this
this Manual,
Manual,
you
you are
are using
using it
it without
without permission.
permission.
366
366
PROBLEM 12.59
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. If the polisher is
started so that the fleece along the circumference undergoes a
constant tangential acceleration of 4 m/s 2 , 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
v2
ΣFn = man : Fn = m
At t = 3 s:
Fn = (1.6 × 10−6 kg)
ρ
(12 m/s)2
� 0.225 �
� 2 m�
�
�
= 2.048 × 10−3 N
Finally,
Ftuft = Ft 2 + Fn2
= (6.4 × 10−6 N) 2 + (2.048 × 10−3 N)2
Ftuft = 2.05 × 10−3 N �
or
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
No part
part of
reproduced or
or distributed
distributed in
in any
any form
form or
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced
you are
are aa student
student using
using this
this Manual,
Manual,
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
distribution
If you
you are
are using
using it
it without
without permission.
permission.
you
367
367
PROBLEM 12.60
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
vB = (0.24 m/s 2 )(10 s) = 2.4 m/s
At t = 10 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 ) 2n
= mB
� v2 �
(aB )t2 + � B �
� �
2
� ρ �
+ ΣFy = 0: N − W = 0
or
N = mB g
At t = 10 s:
F = µ s N = µ s mB g
Then
µs mB g = mB
� v2 �
( aB )t2 + �� B ��
2
� ρ �
1/2
or
2
�
� (2.4 m/s)2 � ��
1
�
2 2
µs =
�(0.24 m/s ) + �
� �
9.81 m/s 2 �
� 2.5 m � ��
�
µs = 0.236 �
or
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
All rights
rights reserved.
reserved. No
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
Inc. All
No part
part of
of this
reproduced or
or distributed
distributed in
in any
any form
form or
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced
are aa student
student using
using this
this Manual,
Manual,
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
individual course
distribution
for their
their individual
course preparation.
preparation. If
If you
you are
you are
are using
using it
it without
without permission.
you
permission.
368
368
PROBLEM 12.61
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 = 0.66 m/s, determine
(a) the minimum value of the coefficient
of static friction between the component
and BC if the component is not to slide
on BC while being transferred, (b) the
values of θ for which sliding is impending.
SOLUTION
ΣFx = max : F =
W vB2
cos θ
g ρ
+ ΣFy = ma y : N − W = −
or
Now
W vB2
sin θ
g ρ
N =W 1−
vB2
sin θ
gρ
Fmax = µ s N = µ s W 1 −
vB2
sin θ
gρ
and for the component not to slide
F
or
W vB2
cos θ
g ρ
or
µs
Fmax
µs W 1 −
vB2
sin θ
gρ
cos θ
gρ
− sin θ
vB2
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
No part
part of
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced or
or distributed
distributed in
in any
any form
form or
reproduced
distributionto
toteachers
teachersand
andeducators
educatorspermitted
permitted by
byMcGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
If you
you are
areaastudent
studentusing
usingthis
thisManual,
Manual,
distribution
you are
are using
using itit without
without permission.
permission.
you
369
369
PROBLEM 12.61 (Continued)
We must determine the values of θ which maximize the above expression. Thus
d
dθ
− sin θ
cos θ
gρ
=
− sin θ
vB2
gρ
− sin θ − (cos θ )(− cos θ )
vB2
gρ
− sin θ
vB2
2
or
sin θ =
vB2
gρ
Now
sin θ =
(0.66 m/s)2
= 0.1776
(9.81 m/s2)(0.25 m)
(a)
µ s = (µ s )min
for
θ = 10.07°
or
=0
and θ = 169.93°
From above,
( µs ) min =
( µs ) min =
cos θ
gρ
− sin θ
vB2
where sin θ =
vB2
gρ
cos θ
cos θ sin θ
= tan θ
=
1
1 − sin 2 θ
− sin θ
sin θ
= tan10.07°
( µs )min = 0.1776
or
(b)
We have impending motion
to the left for
θ = 10.1°
to the right for
θ = 169.9°
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
No part
part of
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced or
or distributed
distributed in
in any
any form
form or
reproduced
distribution
If you
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
you are
are aa student
student using
using this
this Manual,
Manual,
you
you are
are using
using it
it without
without permission.
permission.
370
370
PROBLEM 12.62
Knowing that the coefficients of friction between the component I and member BC of the mechanism of
Problem 12.61 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
ΣFx = max : F =
W vB2
cos θ
g ρ
+ ΣFy = ma y : N − W = −
W vB2
sin θ
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
(1)
( )
must be less than or equal to the minimum value
To ensure that this inequality is satisfied, vB2
max
of µ s g ρ /(cos θ + µ s sin θ ), which occurs when (cosθ + 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°
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
No part
part of
reproduced or
or distributed
distributed in
in any
any form
form or
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced
you are
are aa student
student using
using this
this Manual,
Manual,
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
distribution
If you
you are
are using
using it
it without
without permission.
permission.
you
371
371
PROBLEM 12.62 (Continued)
(a)
(0.66 m/s)2
(9.81 m/s2)(0.25 m)
The maximum allowed value of vB is then
(v )
2
B
max
= µs
gρ
cos θ + µ s sin θ
= gρ
tan θ
= g ρ sin θ
cos θ + (tan θ ) sin θ
where tan θ = µ s
= (9.81 m/s2)(0.25 m) sin 19.2900°
(vB ) max = 0.81 m
or
(b)
First note that for 90°
θ
180°, Eq. (1) becomes
vB2
µs
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 = man
Then
For impending motion, φ = φs . Also, as shown above, the values of θ for which motion is impending
(
v2
)
minimize the value of vB, and thus the value of an is an = ρB . From the above diagram, it can be concluded
that an is minimum when ma n and R are perpendicular.
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
No part
part of
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced or
or distributed
distributed in
in any
any form
form or
reproduced
distribution
If you
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
you are
are aa student
student using
using this
this Manual,
Manual,
you
you are
are using
using it
it without
without permission.
permission.
372
372
PROBLEM 12.62 (Continued)
Therefore,
from the diagram
θ = φs = tan −1 µ s
and
man = W sin φs
or
v2
m B = mg sin θ
or
vB2 = g ρ sin θ
(as above)
ρ
(as above)
For 90° � θ � 180°, we have
from the diagram
α = 180° − θ
(as above)
α = φs
and
man = W sin φs
or
vB2 = g ρ sin θ
(as above)�
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
No part
part of
reproduced or
or distributed
distributed in
in any
any form
form or
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced
you are
are aa student
student using
using this
this Manual,
Manual,
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
distribution
If you
you are
are using
using it
it without
without permission.
permission.
you
373
373
PROBLEM 12.63
In the cathode-ray tube shown, electrons emitted by the cathode and
attracted by the anode pass through a small hole in the anode and
then travel in a straight line with a speed v0 until they strike the
screen at A. However, if a difference of potential V is established
between the two parallel plates, the electrons will be subjected to a
force F perpendicular to the plates while they travel between the
plates and will strike the screen at Point B, which is at a distance δ
from A. The magnitude of the force F is F = eV /d , where −e is the
charge of an electron and d is the distance between the plates.
Derive an expression for the deflection d in terms of V, v0 , the
charge −e and the mass m of an electron, and the dimensions d, �,
and L.
SOLUTION
Consider the motion of one electron. For the horizontal motion, let x = 0 at the left edge of the plate
and x = � at the right edge of the plate. At the screen,
x=
�
2
+L
Horizontal motion: There are no horizontal forces acting on the electron so that ax = 0.
Let t1 = 0 when the electron passes the left edge of the plate, t = t1 when it passes the right edge, and t = t2
when it impacts on the screen. For uniform horizontal motion,
x = v0t ,
so that
t1 =
and
t2 =
�
v0
�
2v0
+
L
.
v0
Vertical motion: The gravity force acting on the electron is neglected since we are interested in the deflection
produced by the electric force. While the electron is between plates (0 � t � t1 ), the vertical force on the
electron is Fy = eV /d . After it passes the plates (t1 � t � t2 ), it is zero.
For 0 � t � t1 ,
ΣFy = ma y : a y =
Fy
m
=
eV
md
v y = (v y ) 0 + a y t = 0 +
y = y0 + ( v y ) 0 t +
eVt
md
1 2
eVt 2
ayt = 0 + 0 +
2
2md
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
All rights
rights reserved.
reserved. No
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
Inc. All
No part
part of
of this
reproduced or
or distributed
distributed in
in any
any form
form or
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced
are aa student
student using
using this
this Manual,
Manual,
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
individual course
distribution
for their
their individual
course preparation.
preparation. If
If you
you are
you are
are using
using it
it without
without permission.
you
permission.
374
374
PROBLEM 12.63 (Continued)
At t = t1 ,
(v y )1 =
eVt1
md
and
y1 =
eVt12
2md
For t1 � t � t2 , a y = 0
y = y1 + (v y )1 (t − t1 )
At t = t2 ,
y2 = δ = y1 + (v y )1 (t2 − t1 )
δ=
=
eVt12 eVt1
eVt
1
+
( t2 − t1 ) = 1 �� t2 − t1 ��
2md md
2 �
md �
eV � � �
L 1 ��
+ −
�
�
mdv0 � 2v0 v0 2 v0 �
or
δ=
eV �L
��
mdv02
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
No part
part of
reproduced or
or distributed
distributed in
in any
any form
form or
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced
you are
are aa student
student using
using this
this Manual,
Manual,
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
distribution
If you
you are
are using
using it
it without
without permission.
permission.
you
375
375
PROBLEM 12.64
In Problem 12.63, determine the smallest allowable value of the
ratio d /� in terms of e, m, v0, and V if at x = �, the minimum
permissible distance between the path of the electrons and the
positive plate is 0.05d .
PROBLEM 12.63 In the cathode-ray tube shown, electrons emitted
by the cathode and attracted by the anode pass through a small hole
in the anode and then travel in a straight line with a speed v0 until
they strike the screen at A. However, if a difference of potential V is
established between the two parallel plates, the electrons will be
subjected to a force F perpendicular to the plates while they travel
between the plates and will strike the screen at Point B, which is at a
distance δ from A. The magnitude of the force F is F = eV /d ,
where –e is the charge of an electron and d is the distance between
the plates. Neglecting the effects of gravity, derive an expression for
the deflection δ in terms of V, v0, the charge –e and the mass m of
an electron, and the dimensions d, �, and L.
SOLUTION
Consider the motion of one electron. For the horizontal motion, let x = 0 at the left edge of the plate
and x = � at the right edge of the plate. At the screen,
x=
�
2
+L
Horizontal motion: There are no horizontal forces acting on the electron so that ax = 0.
Let t1 = 0 when the electron passes the left edge of the plate, t = t1 when it passes the right edge, and t = t2
when it impacts on the screen. For uniform horizontal motion,
x = v0t ,
so that
t1 =
and
t2 =
�
v0
�
2v0
+
L
.
v0
Vertical motion: The gravity force acting on the electron is neglected since we are interested in the deflection
produced by the electric force. While the electron is between the plates (0 � t � t1 ), the vertical force on the
electron is Fy = eV/d . After it passes the plates (t1 � t � t2 ), it is zero.
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
All rights
rights reserved.
reserved. No
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
Inc. All
No part
part of
of this
reproduced or
or distributed
distributed in
in any
any form
form or
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced
are aa student
student using
using this
this Manual,
Manual,
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
individual course
distribution
for their
their individual
course preparation.
preparation. If
If you
you are
you are
are using
using it
it without
without permission.
you
permission.
376
376
PROBLEM 12.64 (Continued)
For 0 � t � t1 ,
ΣFy = ma y : a y =
Fy
m
=
v y = (v y ) 0 + a y t = 0 +
y = y0 + ( v y ) 0 t +
At t = t1 ,
�
v0
, y=
eV
md
eVt
md
eVt 2
1 2
ayt = 0 + 0 +
2
2md
eV � 2
2mdv02
d
− 0.05d = 0.450d
2
But
y�
so that
eV � 2
� 0.450 d
2mdv02
d2
eV
eV
1
= 1.111 2
�
2
2
0.450 2mv0
mv0
�
d
eV
� 1.054
��
�
mv02
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
No part
part of
reproduced or
or distributed
distributed in
in any
any form
form or
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced
you are
are aa student
student using
using this
this Manual,
Manual,
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
distribution
If you
you are
are using
using it
it without
without permission.
permission.
you
377
377
PROBLEM 12.65
The current model of a cathode-ray tube is to be modified so that the
length of the tube and the spacing between the plates are reduced by
40 percent and 20 percent, respectively. If the size of the screen is to
remain the same, determine the new length �′ of the plates, assuming
that all of the other characteristics of the tube are to remain
unchanged. (See Problem 12.63 for a description of a cathode-ray
tube.)
PROBLEM 12.63 In the cathode-ray tube shown, electrons emitted
by the cathode and attracted by the anode pass through a small hole
in the anode and then travel in a straight line with a speed v0 until
they strike the screen at A. However, if a difference of potential V is
established between the two parallel plates, the electrons will be
subjected to a force F perpendicular to the plates while they travel
between the plates and will strike the screen at Point B, which is at a
distance δ from A. The magnitude of the force F is F = eV /d , where
–e is the charge of an electron and d is the distance between the
plates. Neglecting the effects of gravity, derive an expression for
the deflection δ in terms of V, v0, the charge –e and the mass m of an
electron, and the dimensions d, �, and L.
SOLUTION
Consider the motion of one electron. For the horizontal motion, let x = 0 at the left edge of the plate
and x = � at the right edge of the plate. At the screen,
x=
�
2
+L
Horizontal motion: There are no horizontal forces acting on the electron so that ax = 0.
Let t1 = 0 when the electron passes the left edge of the plate, t = t1 when it passes the right edge, and t = t2
when it impacts on the screen. For uniform horizontal motion,
x = v0t ,
so that
t1 =
and
t2 =
�
v0
�
2v0
+
L
.
v0
Vertical motion: The gravity force acting on the electron is neglected since we are interested in the deflection
produced by the electric force. While the electron is between the plates (0 � t � t1 ), the vertical force on the
electron is Fy = eV /d . After it passes the plates (t1 � t � t2 ), it is zero.
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
All rights
rights reserved.
reserved. No
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
Inc. All
No part
part of
of this
reproduced or
or distributed
distributed in
in any
any form
form or
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced
are aa student
student using
using this
this Manual,
Manual,
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
individual course
distribution
for their
their individual
course preparation.
preparation. If
If you
you are
you are
are using
using it
it without
without permission.
you
permission.
378
378
PROBLEM 12.65 (Continued)
For 0 � t � t1 ,
Fy
eV
md
eVt
v y = (v y )0 + a y t = 0 +
md
eVt 2
1
y = y0 + (v y )0 t + a y t 2 = 0 + 0 +
2
2md
ΣFy = ma y : a y =
m
=
Let δ, �, L, and d be dimensions taken from the current model and δ ′, �′, L′, and d′ be those of the modified
model.
eV �L
mdv02
eV �′L′
δ′ =
md ′v02
δ=
Since e, V, m, and v0 are constants
δ ′ � �′ �� L′ �� d �
=
�� �
δ �� � ��
�� L �� d ′ �
But
δ ′ = δ , L′ = L − 0.40 L = 0.6 L,
d ′ = d − 0.20 L = 0.8 d
Then
δ ′ � �′ �� 0.6 L �� d �
=
��
�
δ �� � ��
�� L �� 0.8d �
L = 0.75
�′
�
�′ = 1.333 � ��
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
No part
part of
reproduced or
or distributed
distributed in
in any
any form
form or
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced
you are
are aa student
student using
using this
this Manual,
Manual,
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
distribution
If you
you are
are using
using it
it without
without permission.
permission.
you
379
379
PROBLEM 12.66
Rod OA rotates about O in a horizontal plane. The motion of the
300 g collar B is defined by the relations r = 300 + 100 cos (0.5π t)
and θ = π (t2 − 3t), where r is expressed in millimeters, t in seconds,
and θ in radians. Determine the radial and transverse components of
the force exerted on the collar when (a ) t = 0, (b) t = 0.5 s.
SOLUTION
Polar coordinates and their derivatives.
(a)
For t = 0.
r = 0.3 + 0.1 cos (0.5π t ) m
θ = π (t 2 − 3t ) rad
r� = −0.05π sin (0.5π t ) m/s
θ� = π (2t − 3) rad/s
r�� = −.025π 2 cos (0.5π t ) m/s 2
θ�� = 2π rad/s 2
r = 0.4 m
θ =0
r� = 0
θ� = −3π rad/s
r�� = −0.025π 2 m/s 2
θ�� = 2π rad/s 2
Components of acceleration.
ar = ��
r − rθ� 2
= −0.025π 2 − 0.4(−3π ) 2
= −35.777 m/s 2
aθ = rθ�� + 2r�θ�
= (0.4)(2π ) + 2(−3π )(0)
= 2.513 m/s 2
Components of force.
�
(b)
For t = 0.5 s.
Fr = mar = (0.3)(−35.777)
Fr = −10.73 N ��
Fθ = maθ = (0.3)(2.513) �
Fθ = 0.754 N �
r = 0.37071 m
θ = −3.927 rad
r� = −0.11107 m/s
θ� = −2π rad/s
r�� = −0.17447 m/s 2
θ�� = 2π rad/s 2
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
All rights
rights reserved.
reserved. No
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
Inc. All
No part
part of
of this
reproduced or
or distributed
distributed in
in any
any form
form or
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced
are aa student
student using
using this
this Manual,
Manual,
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
individual course
distribution
for their
their individual
course preparation.
preparation. If
If you
you are
you are
are using
using it
it without
without permission.
you
permission.
380
380
PROBLEM 12.66 (Continued)
Components of acceleration.
ar = ��
r − rθ� 2
= −14.809 m/s 2
aθ = rθ�� + 2r�θ�
= 3.725 m/s 2
Components of force.
Fr = mar = (0.3)(−14.809)
Fr = −4.44 N ��
Fθ = maθ = (0.3)(3.725)
Fθ = 1.118 N �
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
No part
part of
reproduced or
or distributed
distributed in
in any
any form
form or
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced
you are
are aa student
student using
using this
this Manual,
Manual,
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
distribution
If you
you are
are using
using it
it without
without permission.
permission.
you
381
381
PROBLEM 12.67
For the motion defined in Problem 12.66, determine the radial
and transverse components of the force exerted on the collar
when t = 1.5 s.
PROBLEM 12.66 Rod OA rotates about O in a horizontal
plane. The motion of the 300 g collar B is defined by the
relations r = 300 + 100 cos (0.5πt) and θ = π (t2 − 3t), where r
is expressed in millimeters, t in seconds, and θ in radians.
Determine the radial and transverse components of the force
exerted on the collar when (a ) t = 0, (b) t = 0.5 s.
SOLUTION
Polar coordinates and their derivatives.
For t = 1.5 s.
r = 0.3 + 0.1 cos (0.5π t ) m
θ = π (t 2 − 3t ) rad
r� = −0.05π sin (0.5π t ) m/s
θ� = π (2t − 3) rad/s
r�� = −.025π 2 cos (0.5π t ) m/s 2
θ�� = 2π rad/s 2
r = 0.22929 m
θ = −7.068 rad
r� = −0.11107 m/s
θ� = 0
r�� = 0.17447 m/s 2
θ�� = 2π rad/s 2
Components of acceleration.
ar = ��
r − rθ� 2 = 0.17447 m/s 2
a = rθ�� + 2r�θ� = 1.441 m/s 2
θ
Components of force.
Fr = mar = (0.3)(0.17447)
Fr = 0.0523 N �
Fθ = maθ = (0.3)(1.44107)
Fθ = 0.432 N �
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
All rights
rights reserved.
reserved. No
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
Inc. All
No part
part of
of this
reproduced or
or distributed
distributed in
in any
any form
form or
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced
are aa student
student using
using this
this Manual,
Manual,
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
individual course
distribution
for their
their individual
course preparation.
preparation. If
If you
you are
you are
are using
using it
it without
without permission.
you
permission.
382
382
PROBLEM 12.68
Rod OA oscillates about O in a horizontal plane. The motion of the 2-kg
collar B is defined by the relations r = 3/(t + 4) and q = (2/ p ) sin p t,
where r is expressed in meters, t in seconds, and q in radians. Determine the
radial and transverse components of the force exerted on the collar when
(a) t = 1 s, (b) t = 6 s.
SOLUTION
θ=
3
m/s
(t + 4) 2
θ = (2cos π t ) rad/s
6
m/s2
(t + 4)3
θ = −(2π sin π t ) rad/s 2
r=
Then
r=−
and
r=
(a)
At t = 1 s:
Now
2
3
m
t+4
We have
π
sin π t rad
r = 0.6 m
r = −0.12 m/s
θ = −2 rad/s
r = 0.048 m/s2
θ =0
ar = r − rθ 2
= (0.048 m/s2) − (0.6 m)(−2 rad/s)2
= 2.352 m/s2
and
aθ = r + 2rθ
= 0 + 2(−0.12 m/s)(−2 rad/s)
= 0.48 m/s2
Finally
Fr = mB ar
= 2 kg(−2.352 m/s2)
Fr = −4.7 N
or
Fθ = mB aθ
= 2 kg(0.48 m/s2)
Fθ = 0.96 N
or
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
No part
part of
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced or
or distributed
distributed in
in any
any form
form or
reproduced
distributionto
toteachers
teachersand
andeducators
educatorspermitted
permitted by
byMcGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
If you
you are
areaastudent
studentusing
usingthis
thisManual,
Manual,
distribution
you are
are using
using itit without
without permission.
permission.
you
383
383
PROBLEM 12.68 (Continued)
(b)
At t = 6 s:
r = 0.3 m
r = −0.03 m/s
θ = 2 rad/s
r = 0.006 m/s2
θ =0
Now
ar = r − rθ 2 = (0.006 m/s2) − (0.3 m)(2 rad/s)2 = −1.194 m/s2
and
aθ = rθ + 2rθ = 0 + 2(−0.03 m/s)(2 rad/s) = −0.12 m/s2
Finally
Fr = mB ar
= 2 kg(−1.194 m/s2)
Fr = −2.4 N
or
Fθ = mB aθ
= 2 kg(−0.12 m/s2)
Fθ = −0.24 N
or
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
No part
part of
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced or
or distributed
distributed in
in any
any form
form or
reproduced
distribution
If you
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
you are
are aa student
student using
using this
this Manual,
Manual,
you
you are
are using
using it
it without
without permission.
permission.
384
384
PROBLEM 12.69
A collar B of mass m 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� = ct , where c is a constant. As the arm-drum assembly
rotates, a mechanism within the drum winds in the cord so that
the collar moves toward O with a constant speed k. Knowing that
at t = 0, r = r0, express as a function of m, c, k, r0, and t,
(a) the tension T in the cord, (b) the magnitude of the horizontal
force Q exerted on B by arm AA′.
SOLUTION
Kinematics
dr
= r� = −k
dt
We have
At t = 0, r = r0 :
r
t
r0
0
� dr = � −kdt
or
r = r0 − kt
Also,
��
r =0
θ� = ct
� = c
ar = ��
r − rθ� 2
Now
= 0 − ( r0 − kt )(ct ) 2
= −c 2 (r0 − kt )t 2
aθ = rθ�� + 2r�θ�
and
= (r0 − kt )(c) + 2(−k )(ct )
= c(r0 − 3kt )
Kinetics
ΣFr = mar :
(a)
−T = m[−c 2 (r0 − kt )t 2 ]
T = mc 2 (r0 − kt )t 2 �
or
ΣFθ = maθ :
(b)
Q = m[c( r0 − 3kt )]
Q = mc( r0 − 3kt ) ��
or
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
No part
part of
reproduced or
or distributed
distributed in
in any
any form
form or
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced
you are
are aa student
student using
using this
this Manual,
Manual,
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
distribution
If you
you are
are using
using it
it without
without permission.
permission.
you
385
385
PROBLEM 12.70
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
t
0
0
� dr = � 0.5 dt
or
r = (0.5t ) m
Also,
��
r =0
θ� = (0.75t ) rad/s
� = 0.75 rad/s 2
Now
ar = ��
r − rθ� 2 = 0 − [(0.5t ) m][(0.75t ) rad/s]2 = −(0.28125t 3 ) m/s 2
and
aθ = rθ�� + 2r�θ�
= [(0.5t ) m][0.75 rad/s 2 ] + 2(0.5 m/s)[(0.75t ) rad/s]
= (1.125t ) m/s 2
Kinetics
ΣFr = mar : − T = (3 kg)( −0.28125t 3 ) m/s 2
or
T = (0.84375t 3 ) N
ΣFθ = mB aθ : Q = (3 kg)(1.125t ) m/s 2
or
Q = (3.375t ) N
Now require that
T =Q
or
or
(0.84375t 3 ) N = (3.375t ) N
t 2 = 4.000
t = 2.00 s �
or
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
All rights
rights reserved.
reserved. No
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
Inc. All
No part
part of
of this
reproduced or
or distributed
distributed in
in any
any form
form or
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced
are aa student
student using
using this
this Manual,
Manual,
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
individual course
distribution
for their
their individual
course preparation.
preparation. If
If you
you are
you are
are using
using it
it without
without permission.
you
permission.
386
386
PROBLEM 12.71
The 100-g pin B 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 = 12 rad/s, determine for any given value of θ
(a) the radial and transverse components of the resultant force F
exerted on pin B, (b) the forces P and Q exerted on pin B by
rod OC and the wall of slot DE, respectively.
SOLUTION
Kinematics
From the drawing of the system, we have
r=
Then
0.2
m
cos θ
sin θ � �
�
r� = � 0.2
θ � m/s
2
cos
θ �
�
θ� = 12 rad/s
� = 0
and
cos θ (cos θ ) − sin θ (−2cos θ sin θ ) � 2
θ
cos 4 θ
� 1 + sin 2θ � 2 �
θ �� m/s 2
= �� 0.2
3
θ
cos
�
�
2
��
r = 0.2
Substituting for θ�
Now
r� = 0.2
�
sin θ
sin θ �
(12) = �� 2.4
� m/s
2
cos θ
cos 2 θ ��
�
r�� = 0.2
�
1 + sin 2 θ
1 + sin 2 θ �
2
(12)2 = �� 28.8
�� m/s
3
3
cos θ
cos θ �
�
ar = ��
r − rθ� 2
�
1 + sin 2 θ � � 0.2 �
2
= �� 28.8
�−�
� (12)
3
�
θ
cos
cos θ � �
�
�
2
�
sin θ �
2
= �� 57.6
3 �
� m/s
cos
θ
�
�
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
No part
part of
reproduced or
or distributed
distributed in
in any
any form
form or
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced
you are
are aa student
student using
using this
this Manual,
Manual,
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
distribution
If you
you are
are using
using it
it without
without permission.
permission.
you
387
387
PROBLEM 12.71 (Continued)
aθ = rθ + 2rθ
and
= 0 + 2 2.4
= 57.6
sin θ
(12)
cos 2 θ
sin θ
m/s 2
cos 2 θ
Kinetics
(a)
We have
Fr = mB ar = (0.1 kg)
57.6
sin 2 θ
m/s 2
cos3 θ
Fr = (5.76 N) tan 2 θ sec θ
or
and
Fθ = mB aθ = (0.1 kg)
57.6
sin θ
m/s 2
2
cos θ
Fθ = (5.76 N) tan θ sec θ
or
(b)
Now
ΣFy : Fθ cos θ + Fr sin θ = P cos θ
or
P = 5.76 tan θ sec θ + (5.76 tan 2 θ sec θ ) tan θ
P = (5.76 N) tan θ sec3 θ
or
θ
ΣFr : Fr = Q cos θ
or
Q = (5.76 tan 2 θ sec θ )
1
cos θ
Q = (5.76 N) tan 2 θ sec 2 θ →
or
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
No part
part of
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced or
or distributed
distributed in
in any
any form
form or
reproduced
distribution
If you
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
you are
are aa student
student using
using this
this Manual,
Manual,
you
you are
are using
using it
it without
without permission.
permission.
388
388
PROBLEM 12.72*
Slider C has a mass of 230 g and may move in a slot cut in arm
AB, which rotates at the constant rate θ = 10 rad/s in a horizontal
plane. The slider is attached to a spring of constant k = 36 N/m,
which is unstretched when r = 0. Knowing that the slider is
released from rest with no radial velocity in the position r = 450 mm
and neglecting friction, determine for the position r = 300 mm (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 )
ΣFr = mar : − k ( r − l0 ) = m( r − rθ 2 )
r = θ2 −
kl
k
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 = r0 when r = r0 .
r
1 2 r
1 2 k 2 kl0
θ −
r
r +
=
r
2 r0
2
m
m
r0
1 2 1 2 1 2 k
r − r0 = θ −
2
2
m
2
r 2 = r02 + θ 2 −
Data:
( r − r ) + klm (r − r )
2
k
m
2
0
0
0
( r − r ) + 2mkl (r − r )
2
2
0
0
0
m = 0.23 kg
θ = 10 rad/s, k = 36 N/m,
l0 = 0
r0 = (vr )0 = 0, r0 = 0.45 m,
r = 0.3 m
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
No part
part of
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced or
or distributed
distributed in
in any
any form
form or
reproduced
distribution to
toteachers
teachersand
andeducators
educatorspermitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
If you
you are
are aastudent
studentusing
usingthis
thisManual,
Manual,
distribution
you are
are using
using itit without
without permission.
permission.
you
389
389
PROBLEM 12.72* (Continued)
(a)
Components of velocity when r = 0.3 m
r 2 = 0 + 10 2 −
36
(0.32 − 0.452) + 0
0.23
= 6.359 m 2/s 2
vr = r = ±2.52 m/s
Since r is decreasing, vr is negative
(b)
r = −2.52 m/s
vr = −2.52 m/s
vθ = rθ = (0.3)(10)
vθ = 3 m/s
Components of acceleration.
Fr = −kr + kl0 = −(3.6)(0.3) + 0 = −1.08 N
ar =
1.08
Fr
=−
m
0.23
ar = 4.7 m/s2
aθ = rθ + 2rθ = 0 + (2)( −2.52)(10)
aθ = −50.4 m/s2
(c)
Transverse component of force.
Fθ = maθ = (0.23)(−50.4)
Fθ = −11.6 N
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
No part
part of
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced or
or distributed
distributed in
in any
any form
form or
reproduced
distribution
If you
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
you are
are aa student
student using
using this
this Manual,
Manual,
you
you are
are using
using it
it without
without permission.
permission.
390
390
PROBLEM 12.73*
Solve Problem 12.72, assuming that the spring is unstretched when
slider C is located 50 mm to the left of the midpoint O of arm AB
(r = −50 mm).
PROBLEM 12.72 Slider C has a weight of 0.23 kg 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 = 36 N/m, which is unstretched when r = 0. Knowing that the
slider is released with no radial velocity in the position r = 450 mm
and neglecting friction, determine for the position r = 300 mm, (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 )
ΣFr = mar : − k ( r − l0 ) = m( r − rθ 2 )
r = θ2 −
kl
k
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 = r0 when r = r0 .
r
1 2 r
1 2 k 2 kl0
θ −
r
r +
=
r
2 r0
2
m
m
r0
1 2 1 2 1 2 k
r − r0 = θ −
2
2
m
2
r 2 = r02 + θ 2 −
( r − r ) + klm (r − r )
2
k
m
2
0
0
0
( r − r ) + 2mkl (r − r )
2
2
0
0
0
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
No part
part of
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced or
or distributed
distributed in
in any
any form
form or
reproduced
distribution to
toteachers
teachersand
andeducators
educatorspermitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
If you
you are
are aastudent
studentusing
usingthis
thisManual,
Manual,
distribution
you are
are using
using itit without
without permission.
permission.
you
391
391
PROBLEM 12.73* (Continued)
Data:
m = 0.23 kg
θ = 10 rad/s, k = 36 N/m, l0 = 0 − 0.05 m = −0.05 m
r0 = (vr )0 = 0, r0 = 0.45 m,
(a)
r = 0.3 m
Components of velocity when r = 0.3 m
r 2 = 0 + 102 −
36
(2)(36)(−0.05)
(0.32 − 0.452) +
(0.3 − 0.45)
0.23
0.23
= 8.707 m2/s2
vr = r = ±2.95 m/s
Since r is decreasing, vr is negative.
(b)
r = −2.95 m/s
vr = −2.95 m/s
vθ = rθ = (0.3)(10)
vθ = 3 m/s
Components of acceleration.
Fr = −kr + kl0 = −(36)(0.3) + (36)(−0.05) = −12.6 N
ar =
Fr
12.6
=−
m
0.23
ar = −54.8 m/s2
aθ = rθ + 2rθ = 0 + (2)(−2.95)(10)
aθ = −59 m/s2
(c)
Transverse component of force.
Fθ = maθ = (0.23)(59)
Fθ = −13.6 N
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
No part
part of
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced or
or distributed
distributed in
in any
any form
form or
reproduced
distribution
If you
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
you are
are aa student
student using
using this
this Manual,
Manual,
you
you are
are using
using it
it without
without permission.
permission.
392
392
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 along a
semicircular path of diameter OA. Observing that r = r0 cos θ and using
Eq. (12.27), show that the speed of the particle is v = v0 /cos 2 θ .
SOLUTION
Since the particle moves under a central force, h = constant.
Using Eq. (12.27),
h = r 2θ� = h0 = r0 v0
or
θ� = 0 20 =
rv
r
2
r0 v0
2
r0 cos θ
=
v0
r0 cos 2 θ
Radial component of velocity.
vr = r� =
d
(r0 cos θ ) = −( r0 sin θ )θ�
dt
Transverse component of velocity.
vθ = rθ� = ( r0 cos θ )θ�
Speed.
v = vr 2 + vθ 2 = r0θ� =
r0 v0
2
r0 cos θ
v=
v0
cos 2 θ
��
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
No part
part of
reproduced or
or distributed
distributed in
in any
any form
form or
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced
you are
are aa student
student using
using this
this Manual,
Manual,
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
distribution
If you
you are
are using
using it
it without
without permission.
permission.
you
393
393
PROBLEM 12.75
For the particle of Problem 12.74, 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°.
SOLUTION
Since the particle moves under a central force, h = constant
Using Eq. (12.27),
h = r 2θ� = h0 = r0 v0
rv
θ� = 0 20 =
r
2
r0 v0
2
r0 cos θ
=
v0
r0 cos 2 θ
Radial component of velocity.
vr = r� =
d
(r0 cos θ ) = −( r0 sin θ )θ�
dt
Transverse component of velocity.
vθ = rθ� = ( r0 cos θ )θ�
Speed.
v = vr 2 + vθ 2 = r0θ� =
r0 v0
2
r0 cos θ
=
v0
cos 2 θ
Tangential component of acceleration.
at =
v0
dv
(−2)( − sin θ )θ� 2v0 sin θ
= v0
=
⋅
3
3
dt
cos θ
cos θ r0 cos 2 θ
=
2v0 2 sin θ
r0 cos5 θ
Tangential component of force.
Ft = mat : Ft =
(a)
θ = 0,
Ft = 0
(b)
θ = 45°,
Ft =
2mv0 2 sin θ
r0 cos5 θ
Ft = 0 �
2mv0 sin 45°
Ft =
cos5 45°
8mv0 2
��
r0
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
All rights
rights reserved.
reserved. No
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
Inc. All
No part
part of
of this
reproduced or
or distributed
distributed in
in any
any form
form or
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced
are aa student
student using
using this
this Manual,
Manual,
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
individual course
distribution
for their
their individual
course preparation.
preparation. If
If you
you are
you are
are using
using it
it without
without permission.
you
permission.
394
394
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 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. (12.27), 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 2θ� = h0 = r0 v0
or
θ� = 0 20 = 0 0
rv
r v cos 2θ
r
2
r0
=
v0
cos 2θ
r0
Radial component of velocity.
vr = r� =
= r0
��
r0
dr � d �
sin 2θ
θ=
θ�
�
�θ = r0
dθ
dθ � cos 2θ �
(cos 2θ )3/2
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
vθ = v0 cos 2θ ��
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
No part
part of
reproduced or
or distributed
distributed in
in any
any form
form or
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced
you are
are aa student
student using
using this
this Manual,
Manual,
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
distribution
If you
you are
are using
using it
it without
without permission.
permission.
you
395
395
PROBLEM 12.77
For the particle of Problem 12.76, 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 12.76 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 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. (12.27), 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 2θ� = h0 = r0 v0
or
rv
r v cos 2θ
r
2
θ� = 0 2n = 0 0
r0
=
v0
cos 2θ
r0
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 �
(cos 2θ )
(cos 2θ ) r0
cos 2θ
Differentiating again,
r�� =
(a)
dr� � d �
sin 2θ � �
2 cos 2 2θ + sin 2 2θ � v0 2 2 cos 2 2θ + sin 2 2θ
θ=
v
θ
=
v
θ=
� 0
�
0
dθ
dθ �
r0
(cos 2θ )3/2
cos 2θ �
cos 2θ
vr = r� = v0
sin 2θ
cos 2θ
=
v0 r
sin 2θ
r0
vθ = rθ� =
v = (vr ) 2 + (vθ ) 2 =
ar = ��
r − rθ� 2 =
=
v0 r
cos 2θ
r0
v0 r
sin 2 2θ + cos 2 2θ
r0
v=
v0 r
�
r0
v0 2 2cos 2 2θ + sin 2 2θ
r0
v0 2
−
cos 2 2θ
2
r0
cos 2θ
cos 2θ r0
v0 2 cos 2 2θ + sin 2 2θ
v0
v 2r
=
= 02
r0
r0
r0 cos 2θ
cos 2θ
Fr = mar =
mv02 r
r02
Fr =
:
mv02 r
r02
�
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
All rights
rights reserved.
reserved. No
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
Inc. All
No part
part of
of this
reproduced or
or distributed
distributed in
in any
any form
form or
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced
are aa student
student using
using this
this Manual,
Manual,
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
individual course
distribution
for their
their individual
course preparation.
preparation. If
If you
you are
you are
are using
using it
it without
without permission.
you
permission.
396
396
PROBLEM 12.77 (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.
at =
v0 2 r
dv d � v0 r � v0
�
= �
=
=
sin 2θ
r
�
dt dt � r0 � r0
r0 2
Normal component of acceleration.
at = a 2 − at 2 =
(b)
But
�r �
cos 2θ = � 0 �
�r �
Hence,
an =
But an =
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
= 2 ⋅ 2
an
r0
v0
ρ=
r3
��
r02
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
No part
part of
reproduced or
or distributed
distributed in
in any
any form
form or
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced
you are
are aa student
student using
using this
this Manual,
Manual,
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
distribution
If you
you are
are using
using it
it without
without permission.
permission.
you
397
397
PROBLEM 12.78
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
or
v=
GM
r
Let τ be the periodic time to complete one orbit.
vτ = 2π r
τ
GM
= 2π r
r
hence,
GM = 2
or
2π r 3/2
Solving for τ,
τ=
But
4
M = π R3 ρ ,
3
Then
τ=
GM
3π � r �
G ρ �� R ��
π
3
G ρ R 3/2
3/2
Using r = 2R as a given leads to
τ = 23/2
3π
=
Gρ
24π
Gρ
τ = (24π /G ρ )1/2 ��
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
All rights
rights reserved.
reserved. No
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
Inc. All
No part
part of
of this
reproduced or
or distributed
distributed in
in any
any form
form or
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced
are aa student
student using
using this
this Manual,
Manual,
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
individual course
distribution
for their
their individual
course preparation.
preparation. If
If you
you are
you are
are using
using it
it without
without permission.
you
permission.
398
398
PROBLEM 12.79
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 = Fn = man = m
Then
G
[Eq. (12.28)]
v2
r
Mm
v2
=
m
r
r2
GM
r
or
v2 =
Now
GM = gR 2
so that
v2 =
For one orbit,
τ=
or
� gτ 2 R 2 �
r = ��
2 �
�
� 4π �
Solving for g,
g = 4π 2
[Eq. (12.30)]
gR 2
r
v=R
or
g
r
2π r
2π r
=
v
g
R
r
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 �
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
No part
part of
reproduced or
or distributed
distributed in
in any
any form
form or
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced
you are
are aa student
student using
using this
this Manual,
Manual,
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
distribution
If you
you are
are using
using it
it without
without permission.
permission.
you
399
399
PROBLEM 12.80
Communication satellites are placed in a geosynchronous orbit, i.e., in a circular orbit such that they complete
one full revolution about the earth in one sidereal day (23.934 h), and thus appear stationary with respect to
the ground. Determine (a) the altitude of these satellites above the surface of the earth, (b) the velocity with
which they describe their orbit.
SOLUTION
For gravitational force and a circular orbit,
Fr =
GMm mv 2
=
r
r2
or
v=
GM
r
Let τ be the period time to complete one orbit.
vτ = 2π r
Then
GM τ 2
r =
4π 2
Data:
τ = 23.934 h = 86.1624 × 103 s
(a)
3
In SI units:
v 2τ 2 =
GM τ 2
= 4π 2 r 2
r
But
or
or
GM τ 2
r=
4π 2
1/ 3
g = 9.81 m/s 2 , R = 6.37 × 106 m
GM = gR 2 = (9.81)(6.37 × 106 ) 2 = 398.06 × 1012 m3 /s 2
(398.06 × 1012 )(86.1624 × 103 ) 2
r=
4π 2
altitude h = r − R = 35.775 × 106 m
1/ 3
= 42.145 × 106 m
h = 35,800 km
PROPRIETARY
No part
part of
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
of this
this Manual
Manual may
may be
be displayed,
displayed,
reproduced
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced or
or distributed
distributed in
in any
any form
form or
distribution
If you
you are
are aa student
student using
using this
this Manual,
Manual,
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
you
you are
are using
using it
it without
without permission.
permission.
400
400
PROBLEM 12.80 (Continued)
(b)
v=
398.06 × 1012
GM
=
= 3.07 × 103 m/s
r
42.145 × 106
v = 3.07 km/s
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
No part
part of
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced or
or distributed
distributed in
in any
any form
form or
reproduced
distribution
distribution to
toteachers
teachersand
andeducators
educatorspermitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
If you
you are
are aastudent
studentusing
usingthis
thisManual,
Manual,
you
you are
are using
using itit without
without permission.
permission.
401
401
PROBLEM 12.81
Determine the mass of the earth, knowing that the mean radius of the moon’s orbit about the earth is 382,250 km
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 = Fn = man = m
Then
G
v2
r
Mm
v2
=
m
r
r2
r 2
v
G
or
M=
Now
v=
so that
M=
Noting that
τ = 27.32 days = 2.3604 × 106 s
and
r = 382.25 × 106 m
we have
M=
2π r
τ
r 2π r
G τ
2
=
1 2π
G τ
2
r3
2
1
2p
(382.25 × 106 m)3
6.224 × 10−11 m3/kg⋅ s2 2.3604 × 106 s
M = 6.36 × 1024 kg
or
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
No part
part of
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced or
or distributed
distributed in
in any
any form
form or
reproduced
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
you are
are aa student
student using
using this
this Manual,
Manual,
distribution
If you
you are
are using
using it
it without
without permission.
permission.
you
402
402
PROBLEM 12.82
A spacecraft is placed into a polar orbit about the planet Mars at an altitude of 380 km. Knowing that the
mean density of Mars is 3.94 Mg/m3 and that the radius of Mars is 3397 km, determine (a) the time τ required
for the spacecraft to complete one full revolution about Mars, (b) the velocity with which the spacecraft
describes its orbit.
SOLUTION
(a)
From the solution to Problem 12.78, we have
3π � r �
τ=
G ρ �� R ��
where
3/ 2
r = R + h = (3397 + 380) km
= 3777 km
1/2
Then
�
� � 3777 km �
3π
τ =�
�
3
2
3
3 � �
−12
� (66.73 × 10 m /kg ⋅ s )(3.94 × 10 kg/m ) � � 3397 km �
= 7019.5 s
τ = 1 h 57 min �
or
(b)
We have
3/2
v=
2π r
τ
2π (3777 × 103 m)
=
7019.5 s
v = 3380 m/s ��
or
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
No part
part of
reproduced or
or distributed
distributed in
in any
any form
form or
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced
you are
are aa student
student using
using this
this Manual,
Manual,
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
distribution
If you
you are
are using
using it
it without
without permission.
permission.
you
403
403
PROBLEM 12.83
A satellite is placed into a circular orbit about the planet Saturn at an altitude of 3360 km. The satellite
describes its orbit with a velocity of 87.5 × 103 km/h. Knowing that the radius of the orbit about Saturn and the
periodic time of Atlas, one of Saturn’s moons, are 136.9 × 103 km and 0.6017 days, respectively, determine
(a) the radius of Saturn, (b) the mass of Saturn. (The periodic time of a satellite is the time it requires to
complete one full revolution about the planet.)
SOLUTION
2π rA
Velocity of Atlas.
vA =
where
v A = 136.9 × 103 km = 136.9 × 106 m
and
τ A = 0.6017 days = 51,987 s
Gravitational force.
τA
vA =
(2π )(136.9 × 106 )
= 16.546 × 103 m/s
51,987
F=
GMm mv 2
=
r
r2
from which
GM = rv 2 = constant
For the satellite,
rs vs2 = rAv A2
r v2
rs = A 2A
vs
vs = 87.5 × 103 km/h = 24.31 × 103 m/s
where
rs =
(136.9 × 106 )(16.546 × 103)2
= 63.42 × 106 m
3 2
(24.31 × 10 )
rs = 63.42 × 103 km
(a)
Radius of Saturn.
R = rs − (altitude) = 63420 − 3360
(b)
R = 60060 km
Mass of Saturn.
r v 2 (136.9 × 106 )(16.546 × 103)2
M= A A =
G
6.224 × 10−11
M = 602.2 × 1024 kg
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
No part
part of
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced or
or distributed
distributed in
in any
any form
form or
reproduced
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
you are
are aa student
student using
using this
this Manual,
Manual,
distribution
If you
you are
are using
using it
it without
without permission.
permission.
you
404
404
PROBLEM 12.84
The periodic times (see Problem 12.83) of the planet Uranus’ moons Juliet and Titania have been observed to
be 0.4931 days and 8.706 days, respectively. Knowing that the radius of Juliet’s orbit is 64,360 km, determine
(a) the mass of Uranus, (b) the radius of Titania’s orbit.
SOLUTION
2π rJ
Velocity of Juliet.
vJ =
where
rJ = 64,360 km = 64.36 × 106 m
and
τ J = 0.4931 days = 42, 604 s
Gravitational force.
from which
(a)
Mass of Uranus.
τJ
vJ =
(2π )(64.36 × 106 )
= 9.4917 × 103 m/s
42, 604
F=
GMm mv 2
=
r
r2
GM = rv 2 = constant
r v2
M= J J
G
(64.36 × 106 )(9.4917 × 103 )2
66.73 × 10−12
= 86.893 × 1024 kg
M=
M = 86.9 × 1024 kg �
(b)
Radius of Titania’s orbit.
GM = rT vT2 =
4π 2 rT3
�τ �
rT3 = rJ3 � T �
�τJ �
τT 2
2
=
4π 2 rJ3
τJ2
2
� 8.706 �
3
24
= (64.36 × 106 )3 �
� = 81.103 × 10 m
� 0.4931 �
rT = 436.39 × 106 m
rT = 436, 000 km �
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
No part
part of
reproduced or
or distributed
distributed in
in any
any form
form or
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced
you are
are aa student
student using
using this
this Manual,
Manual,
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
distribution
If you
you are
are using
using it
it without
without permission.
permission.
you
405
405
PROBLEM 12.85
A 600-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 1700 km, determine (a) the gravitational force exerted on the
spacecraft as it was orbiting the earth, (b) the required radius of the orbit of the spacecraft about the moon if
the periodic times (see Problem 12.83) of the two orbits are to be equal, (c) the acceleration of gravity at the
surface of the moon.
SOLUTION
First note that
rE = RE + hE = (6370 + 4500) km
Then
(a)
RE = 6370 km
= 10870 km
We have
and
F =G
MM
r2
[Eq. (12.28)]
GM = gR 2
[Eq. (12.29)]
Then
For the earth orbit,
6370 km
F = (600 × 9.81 N)
10870 km
r
R
r
2
m
F = gR 2 2 = W
2
F = 2.02 kN
or
(b)
From the solution to Problem 12.81, we have
M=
1 2π
G τ
Then
τ=
2π r 3/2
GM
Now
τE =τM
or
MM
rM =
ME
2
r3
2π rE3/ 2
GM E
=
2π rM3/2
GM M
(1)
1/3
rE = (0.01230)1/3 (10870 km)
rM = 2509 km
or
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
No part
part of
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced or
or distributed
distributed in
in any
any form
form or
reproduced
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
you are
are aa student
student using
using this
this Manual,
Manual,
distribution
If you
you are
are using
using it
it without
without permission.
permission.
you
406
406
PROBLEM 12.85 (Continued)
(c)
GM = gR 2
We have
[Eq.(12.29)]
Substituting into Eq. (1)
2π rE3/2
RE g E
or
=
gM =
2π rM3/ 2
RM g M
2
RE
RM
rM
rE
3
gE =
RE
RM
2
MM
gE
ME
using the results of Part (b). Then
2
6370 km
gM =
(0.01230)(9.81 m/s2)
1700 km
g moon = 1.694 m/s2
or
Note:
g moon ≈
1
g earth
6
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
No part
part of
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced or
or distributed
distributed in
in any
any form
form or
reproduced
distribution
distribution to
toteachers
teachersand
andeducators
educatorspermitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
If you
you are
are aastudent
studentusing
usingthis
thisManual,
Manual,
you
you are
are using
using itit without
without permission.
permission.
407
407
PROBLEM 12.86
To place a communications satellite into a geosynchronous
orbit (see Problem 12.80) at an altitude of 35580 km above
the surface of the earth, the satellite first is released from a
space shuttle, which is in a circular orbit at an altitude of
296 km, and then is propelled by an upper-stage booster to
its final altitude. As the satellite passes through A, the
booster’s motor is fired to insert the satellite into an elliptic
transfer orbit. The booster is again fired at B to insert the
satellite into a geosynchronous orbit. Knowing that
the second firing increases the speed of the satellite by
1400 m/s, determine (a) the speed of the satellite as it
approaches B on the elliptic transfer orbit, (b) the increase
in speed resulting from the first firing at A.
SOLUTION
For earth,
R = 6370 km = 6.37 × 106 m
GM = gR2 = (9.81)(6.37 × 106)2 = 0.39806 × 1015 m3/s2
rA = 6370 + 296 = 6666 km = 6.666 × 106 m
rB = 6370 + 35580 = 41950 km = 41.95 × 106 m
Speed on circular orbit through A.
(v A )circ =
=
GM
rA
0.39806 × 1015
6.666 × 106
= 7.728 × 103 m/s
Speed on circular orbit through B.
(vB )circ =
=
GM
rB
0.39806 × 1015
41.95 × 106
= 3.08 × 103 m/s
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
part of
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
No part
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced or
or distributed
distributed in
in any
any form
form or
reproduced
distribution to
to teachers
teachers and
andeducators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
If you
you are
are aa student
student using
using this
thisManual,
Manual,
distribution
you are
are using
using itit without
without permission.
permission.
you
408
408
PROBLEM 12.86 (Continued)
(a)
Speed on transfer trajectory at B.
(vB ) tr = 3.08 × 103 − 1400
= 1.68 × 103 m/s
Conservation of angular momentum for transfer trajectory.
1680 m/s
rA (v A ) tr = rB (vB ) tr
r (v )
(v A ) tr = B B tr
rA
(41.95 × 106)(1680)
6.666 × 106
= 10.572 × 103 m/s
=
(b)
Change in speed at A.
∆v A = (v A ) tr − (v A )circ
= 10.572 × 103 − 1.68 × 103
= 8.892 × 103 m/s
∆v A = 8892 m/s
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
No part
part of
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced or
or distributed
distributed in
in any
any form
form or
reproduced
distribution
distribution to
toteachers
teachersand
andeducators
educatorspermitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
If you
you are
are aastudent
studentusing
usingthis
thisManual,
Manual,
you
you are
are using
using itit without
without permission.
permission.
409
409
PROBLEM 12.87
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,
ΣFn = man : F = m
v2
r
F =G
Mm
r2
Eq. (12.28):
G
Then
Mm
v2
m
=
r
r2
v2 =
or
GM
r
66.73 × 10−12 m3 /kg ⋅ s 2 × 73.49 × 1021 kg
2200 × 103 m
Then
2
(v A )circ
=
or
(v A )circ = 1493.0 m/s
and
2
=
(vB )circ
or
(vB )circ = 1535.5 m/s
(a)
We have
66.73 × 10−12 m3 /kg ⋅ s 2 × 73.49 × 1021 kg
2080 × 103 m
(v A )TR = (v A )circ + ∆v A
= (1493.0 − 26.3) m/s
= 1466.7 m/s
Conservation of angular momentum requires that
rA m(v A )TR = rB m(vB )TR
or
2200 km
× 1466.7 m/s
2080 km
= 1551.3 m/s
(vB )TR =
(vB )TR = 1551 m/s
or
(b)
Now
or
or
(v B )circ = (vB )TR + ∆vB
∆vB = (1535.5 − 1551.3) m/s
= 15.8 m/s
Amount to reduce = 15.8 m/s
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
No part
part of
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced or
or distributed
distributed in
in any
any form
form or
reproduced
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
you are
are aa student
student using
using this
this Manual,
Manual,
distribution
If you
you are
are using
using it
it without
without permission.
permission.
you
410
410
PROBLEM 12.88
Plans for an unmanned landing mission on the planet Mars called
for the earth-return vehicle to first describe a circular orbit at an
altitude dA = 2200 km above the surface of the planet with a
velocity of 2771 m/s. As it passed through point A, the vehicle
was to be inserted into an elliptic transfer orbit by firing its
engine and increasing its speed by ∆v A = 1046 m/s. As it passed
through point B, at an altitude dB = 100,000 km, the vehicle was to
be inserted into a second transfer orbit located in a slightly
different plane, by changing the direction of its velocity and
reducing its speed by ∆vB = −22.0 m/s. Finally, as the vehicle
passed through Point C, at an altitude dC = 1000 km, its speed
was to be increased by ∆vC = 660 m/s to insert it into its return
trajectory. Knowing that the radius of the planet Mars is
R = 3400 km, determine the velocity of the vehicle after
completion of the last maneuver.
SOLUTION
τ A = 3400 + 2200 = 5600 km = 5.60 × 106 m
τ B = 3400 + 100, 000 = 103, 400 km = 103.4 × 106 m
τ C = 3400 + 1000 = 4400 km = 4.40 × 106 m
First transfer orbit.
v A = 2771 m/s + 1046 m/s = 3817 m/s
Conservation of angular momentum:
τ A m v A = τ B m vB
(5.60 × 106 )(3817) = (103.4 × 106 )vB
vB = 206.7 m/s
Second transfer orbit.
vB′ = vB + ∆vB
= 206.7 − 22.0 = 184.7 m/s
Conservation of angular momentum:
τ B mvB′ = τ C mvC
(103.4 × 106 )(184.7) = (4.40 × 106 )vC
vC = 4340 m/s
After last maneuver.
v = vC + ∆vC = 4340 + 660 = 5000 m/s
��
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
No part
part of
reproduced or
or distributed
distributed in
in any
any form
form or
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced
you are
are aa student
student using
using this
this Manual,
Manual,
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
distribution
If you
you are
are using
using it
it without
without permission.
permission.
you
411
411
PROBLEM 12.89
A space shuttle S and a satellite A are in the circular orbits shown. In
order for the shuttle to recover the satellite, the shuttle is first placed
in an elliptic path BC by increasing its speed by ∆vB = 84 m/s as it
passes through B. As the shuttle approaches C, its speed is increased
by ∆vC = 78 m/s to insert it into a second elliptic transfer orbit CD.
Knowing that the distance from O to C is 6860 km, determine the
amount by which the speed of the shuttle should be increased as it
approaches D to insert it into the circular orbit of the satellite.
SOLUTION
R = 6370 km = 6.37 × 106 m
First note
rA = (6370 + 608) km = 6978 km = 6.978 × 106 m
rB = (6370 + 288) km = 6658 km = 6.658 × 106 m
ΣFn = man :
For a circular orbit,
F =G
Eq. (12.28):
Then
or
G
F =m
v2
r
Mm
r2
Mm
v2
=
m
r
r2
v2 =
GM gR 2
=
r
r
using Eq. (12.29).
9.81 m/s2 × (6.37 × 106 m)2
6.978 × 106 m
Then
2
(v A )circ
=
or
(v A )circ = 7553 m/s
and
2
(vB )circ
=
or
(vB )circ = 7732 m/s
We have
(vB )TRBC = (vB )circ + ∆vB = (7732 + 84) m/s
9.81 m/s2 × (6.37 × 106 m)2
6.658 × 106 m
= 7816 m/s
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
No part
part of
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced or
or distributed
distributed in
in any
any form
form or
reproduced
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
you are
are aa student
student using
using this
this Manual,
Manual,
distribution
If you
you are
are using
using it
it without
without permission.
permission.
you
412
412
PROBLEM 12.89 (Continued)
Conservation of angular momentum requires that
From Eq. (1)
BC : rB m (vB )TRBC = rC m (vC )TRBC
(1)
CD : rC m (vC )TRCD = rA m (vD )TRCD
(2)
r
6658 km
(vC )TRBC = B (vB )TRBC =
× 7816 m/s
rC
6860 km
= 7586 m/s
Now
(vC )TRCD = (vC )TRBC + ∆vC = (7586 + 78) m/s
= 7664 m/s
From Eq. (2)
6860 km
r
(vD )TRCD = C (vC )TRCD =
× 7664 m/s
6978 km
rA
= 7534 m/s
Finally,
or
(v A )circ = (vD )TRCD + ∆vD
∆vD = (7553 − 7534) m/s
∆vD = 19 m/s
or
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
No part
part of
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced or
or distributed
distributed in
in any
any form
form or
reproduced
distribution
distribution to
toteachers
teachersand
andeducators
educatorspermitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
If you
you are
are aastudent
studentusing
usingthis
thisManual,
Manual,
you
you are
are using
using itit without
without permission.
permission.
413
413
PROBLEM 12.90
A 1.5-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
Fsp = k ( r − rA )
First note
(a)
Fθ = 0 and at A,
Fr = − Fsp = 0
(a A ) r = 0
(a A )θ = 0
ΣFr = mar :
(b)
Noting that
−Fsp = m( r − rθ 2 )
acollar/rod = r , we have at A
0 = m[acollar/rod − (0.15 m)(16 rad/s)2]
or
(c)
(acollar/rod ) A = 38.4 m/s2
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
Then
(vB )θ =
where (v A )θ = rA θ 0
0.15 m
[(0.15 m)(16 rad/s)2]
0.45 m
(vB )θ = 0.8 m/s
or
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
No part
part of
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced or
or distributed
distributed in
in any
any form
form or
reproduced
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
you are
are aa student
student using
using this
this Manual,
Manual,
distribution
If you
you are
are using
using it
it without
without permission.
permission.
you
414
414
PROBLEM 12.91
For the collar of Problem 12.90, assuming that the rod initially rotates at the rate θ = 12 rad/s, determine for
position B of the collar (a) the transverse component of the velocity of the collar, (b) the radial and transverse
components of its acceleration, (c) the acceleration of the collar relative to the rod.
SOLUTION
First we note
At B:
Fsp = k ( r − rA )
( Fsp ) B = 30 N/m (0.45 m − 0.15 m)
=9N
(a)
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 = rAθ0
Then
(vB )θ =
0.15 m
[(0.15 m)(12 rad/s)]
0.45 m
(vB )θ = 0.6 m/s
or
(b)
We have
Now
or
FB = 0
(aB )θ = 0
ΣFr = mar : − ( Fsp )θ = m (aB ) r
(aB ) r = −9 N × 1.5 kg
= −6 m/s2
(aB ) r = −6 m/s2
or
(c)
We have
Now
Then at B:
ar = r − rθ 2
acollar/rod = r
and
θB =
(vB )θ
rB
(acollar/rod )θ = −6 m/s2 + 0.45 m ×
0.6 m/s
0.45 m
2
(acollar/rod )θ = −5.2 m/s2
or
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
PROPRIETARY
MATERIAL.
2009orThe
McGraw-Hill
Companies,
Inc.written
All rights
reserved.ofNo
of this or
Manual
may bethe
displayed,
reproduced
or distributed
in any©form
by any
means, without
the prior
permission
thepart
publisher,
used beyond
limited
reproduced to
orteachers
distributed
any form
or by any
means, without
prior
writtencourse
permission
of the publisher,
used beyond
theManual,
limited
distribution
and in
educators
permitted
by McGraw-Hill
forthe
their
individual
preparation.
If you are aor
student
using this
distribution
educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you
are usingtoitteachers
without and
permission.
you are using it without permission.
415
415
PROBLEM 12.92
A 200-g ball A and a 400-g ball B are mounted on a
horizontal rod which rotates freely about a vertical shaft. The
balls are held in the positions shown by pins. The pin holding
B is suddenly removed and the ball moves to position C as the
rod rotates. Neglecting friction and the mass of the rod and
knowing that the initial speed of A is v A = 2.5 m/s, determine
(a) the radial and transverse components of the acceleration
of ball B immediately after the pin is removed, (b) the
acceleration of ball B relative to the rod at that instant, (c) the
speed of ball A after ball B has reached the stop at C.
SOLUTION
Let r and 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 Third Law.
ΣM 0 = 0: rA ( FA )θ + rB ( FB )θ = rA mA (a A )θ + rB mB ( aB )θ = 0
rA m A rAθ�� + rB mB ( rBθ�� + 2r�Bθ�) = 0
� =
At t = 0,
r�B = 0
−2rB r�Bθ�
mA rA2 + mB rB 2
so that � = 0.
(aB )θ = 0 �
From Eq. (1),
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
All rights
rights reserved.
reserved. No
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
Inc. All
No part
part of
of this
reproduced or
or distributed
distributed in
in any
any form
form or
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced
are aa student
student using
using this
this Manual,
Manual,
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
individual course
distribution
for their
their individual
course preparation.
preparation. If
If you
you are
you are
are using
using it
it without
without permission.
you
permission.
416
416
PROBLEM 12.92 (Continued)
(b)
Acceleration of B relative to the rod.
(v )
2.5
= 10 rad/s
(v A )θ = 2.5 m/s, θ� = A θ =
rA
0.25
At t = 0,
��
rB − rBθ� 2 = (aB ) r = 0
��
rB = rBθ� 2 = (0.2)(10)2 = 20 m/s 2
��
rB = 20.0 m/s 2 �
(c)
Speed of A.
Substituting
d
(mr 2θ�) for rFθ in each term of the moment equation gives
dt
d
d
m A rA2θ� +
mB rB2θ� = 0
dt
dt
(
)
(
)
Integrating with respect to time,
(
m A rA2θ� + mB rB2θ� = m A rA2θ�
) + ( m r θ� )
0
2
B B
0
Applying to the final state with ball B moved to the stop at C,
( m r + m r )θ� = ��m r + m (r ) �� θ�
2
A A
2
B C
2
A A
f
θ� f =
B
2
B 0
0
mA rA2 + mB rB2
mA rA2 + mB rC2
(0.2)(0.25) 2 + (0.4)(0.2)2
(10)
(0.2)(0.25) 2 + (0.4)(0.4)2
= 3.7255 rad/s
=
(v A ) f = rAθ� f = (0.25)(3.7255) = 0.93137 m/s
(v A ) f = 0.931 m/s �
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
PROPRIETARY MATERIAL. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
reproduced to
orteachers
distributed
any form
or by any
means, without
prior
writtencourse
permission
of the publisher,
used beyond
theManual,
limited
distribution
and in
educators
permitted
by McGraw-Hill
forthe
their
individual
preparation.
If you are aorstudent
using this
distribution
educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are usingtoitteachers
withoutand
permission.
you are using it without permission.
417
417
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 θ1 = 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 : T sin θ = W sin θ cos θ =
But ρ = � sin θ
mv 2
ρ
so that
v2 =
ρW
m
sin 2 θ cos θ = � g sin θ tan θ
v1 = �1 g sin θ1 tan θ1
and
v2 = � 2 g sin θ 2 tan θ 2
ΣM y = 0: H y = constant
r1mv1 = r2 mv2
v1�1 sin θ1 = v2 � 2 sin θ 2
or
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)
With θ1 = 35°, �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
θ 2 = 43.6° �
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
All rights
rights reserved.
reserved. No
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
Inc. All
No part
part of
of this
reproduced or
or distributed
distributed in
in any
any form
form or
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced
are aa student
student using
using this
this Manual,
Manual,
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
individual course
distribution
for their
their individual
course preparation.
preparation. If
If you
you are
you are
are using
using it
it without
without permission.
you
permission.
418
418
PROBLEM 12.94
A particle of mass m describes the cardioid r = r0 (1 + cos θ )/2 under a
central force F directed toward the center of force O. Using Eq. (12.37),
show that F is inversely proportional to the fourth power of the distance r
from the particle to O.
SOLUTION
We have
F
d 2u
+u =
dθ 2
mh 2u 2
where
u=
1
r
Eq. (12.37)
and mh 2 = constant
� d 2u
�
F × u 2 �� 2 + u ��
� dθ
�
Now
u=
1 2
1
=
r r0 1 + cos θ
Then
� 2
du
d �2
1
sin θ
=
�
�=
dθ dθ � r0 1 + cos θ � r0 (1 + cos θ ) 2
and
d 2 u 2 cos θ (1 + cos θ ) 2 − sin θ [2(1 + cos θ )](− sin θ )
=
dθ 2 r0
(1 + cos θ ) 4
Then
=
2 1 + cos θ + sin 2 θ 2 �
1
1 − cos 2 θ �
+
= �
�
3
2
r0 (1 + cos θ )
r0 � (1 + cos θ )
(1 + cos θ )3 �
=
2
2 2 − cos θ
2 � r0 � � � 2r � �
=
� 2 − � − 1� �
r0 (1 + cos θ ) 2 r0 �� 2r �� � � r0
��
=
r0 �
2r �
3− �
2 �
r0 �
2r �
2
2r � 1 � 3 r0
�1� � r �
F × � � � 02 � 3 − � + � =
r0 � r � 2 r 4
� r � � 2r �
F×
1
r4
Q.E.D. �
Note: F � 0 implies that F is attractive.
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
PROPRIETARY MATERIAL. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
reproduced to
orteachers
distributed
any form
or by any
means, without
prior
writtencourse
permission
of the publisher,
used beyond
theManual,
limited
distribution
and in
educators
permitted
by McGraw-Hill
forthe
their
individual
preparation.
If you are aorstudent
using this
distribution
educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are usingtoitteachers
withoutand
permission.
you are using it without permission.
419
419
PROBLEM 12.95
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 elliptic path defined by the equation r = r0 /(2 − cos θ ).
Using Eq. (12.37), 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 2 u cos θ
=
r0
dθ 2
by Eq. (12.37).
2mh 2 u 2 2mh 2
=
r0
r0 r 2
Since m, h, and r0 are constants, F is proportional to 12 , or inversely proportional to r 2 . �
r
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
All rights
rights reserved.
reserved. No
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
Inc. All
No part
part of
of this
reproduced or
or distributed
distributed in
in any
any form
form or
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced
are aa student
student using
using this
this Manual,
Manual,
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
individual course
distribution
for their
their individual
course preparation.
preparation. If
If you
you are
you are
are using
using it
it without
without permission.
you
permission.
420
420
PROBLEM 12.96
A particle of mass m describes the path defined by the equation r = r0 sin θ under a central force F directed
toward the center of force O. Using Eq. (12.37), show that F is inversely proportional to the fifth power of the
distance r from the particle to O.
SOLUTION
We have
d 2u
F
+u =
2
dθ
mh 2u 2
where
u=
1
r
Eq. (12.37)
and mh 2 = constant
� d 2u
�
F × u 2 �� 2 + u ��
� dθ
�
Now
u=
1
1
=
r r0 sin θ
Then
du
1 � 1 �
1 cos θ
=
�
�=−
dθ dθ � r0 sin θ �
r0 sin 2 θ
and
1 � − sin θ (sin 2 θ ) − cos θ (2 sin θ cos θ ) �
d 2u
=
−
�
�
r0 �
dθ 2
sin 4 θ
�
=
Then
1 1 + cos 2 θ
r0 sin 3 θ
2
2
1 �
� 1 � � 1 1 + cos θ
F × � � ��
+
�
3
r0 sin θ ��
� r � � r0 sin θ
=
1 1 � 1 + cos 2 θ sin 2 θ �
+
�
�
r0 r 2 �� sin 3 θ
sin 3 θ ��
=
2 1 1
r0 r 2 sin 3 θ
=
�r�
sin 3 θ = � �
� r0 �
3
2r02
r3
F×
1
r5
Q.E.D. �
Note: F � 0 implies that F is attractive.�
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
PROPRIETARY MATERIAL. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
reproduced to
orteachers
distributed
any form
or by any
means, without
prior
writtencourse
permission
of the publisher,
used beyond
theManual,
limited
distribution
and in
educators
permitted
by McGraw-Hill
forthe
their
individual
preparation.
If you are aorstudent
using this
distribution
educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are usingtoitteachers
withoutand
permission.
you are using it without permission.
421
421
PROBLEM 12.97
For the particle of Problem 12.76, and using Eq. (12.37), show that the
central force F is proportional to the distance r from the particle to the
center of force O.
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 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. (12.27), express the radial and transverse components of the velocity
v of the particle as functions of θ .
SOLUTION
We have
F
d 2u
+u =
2
dθ
mh 2u 2
where
u=
1
r
Eq. (12.37)
and mh 2 = constant
� d 2u
�
F × u 2 �� 2 + u ��
� dθ
�
1 1
cos 2 θ
=
r r0
Now
u=
Then
�
du
d �1
1 sin 2θ
cos 2θ � = −
=
�
dθ dθ � r0
r0 cos 2θ
�
and
d 2u
1 2 cos 2θ cos 2θ − sin 2θ (− sin 2θ / cos 2θ )
=−
2
r0
cos 2θ
dθ
3
4
1 � r � � � r0 � �
1 1 + cos 2 2θ
1
=−
=
−
+
�
�
�
�
r0 (cos 2θ )3/ 2
r0 � r0 � � �� r �� �
�
�
4
r3 � � r � �
= − 4 �1 + � 0 � �
r0 �� � r � ��
Then
4
2
3
r
� 1 � �� r � � r � � 1 ��
F × � � �− 4 �1 + � 0 � � + � = − 4
r0
� r � �� r0 �� � r � �� r ��
F × r Q.E.D. �
Note: F � 0 implies that F is repulsive.
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
All rights
rights reserved.
reserved. No
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
Inc. All
No part
part of
of this
reproduced or
or distributed
distributed in
in any
any form
form or
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced
are aa student
student using
using this
this Manual,
Manual,
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
individual course
distribution
for their
their individual
course preparation.
preparation. If
If you
you are
you are
are using
using it
it without
without permission.
you
permission.
422
422
PROBLEM 12.98
It was observed that during the Galileo spacecraft’s first flyby of the earth, its minimum altitude was 960 km
above the surface of the earth. Assuming that the trajectory of the spacecraft was parabolic, determine the
maximum velocity of Galileo during its first flyby of the earth.
SOLUTION
First we note
R = 6.37 × 106 m
so that
r0 = (6.37 × 106 + 960 × 103 ) m
= 7.33 × 106 m
Now
vmax = v0
and from page 709 of the text
v0 =
2GM
=
r0
2 gR 2
r0
using Eq. (12.30).
1/2
Then
� 2 × 9.81 m/s 2 × (6.37 × 106 m)2 �
vmax = �
�
7.33 × 106 m
�
�
= 10, 421.7 m/s
vmax = 10.42 km/s ��
or
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
PROPRIETARY MATERIAL. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
reproduced to
orteachers
distributed
any form
or by any
means, without
prior
writtencourse
permission
of the publisher,
used beyond
theManual,
limited
distribution
and in
educators
permitted
by McGraw-Hill
forthe
their
individual
preparation.
If you are aorstudent
using this
distribution
educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are usingtoitteachers
withoutand
permission.
you are using it without permission.
423
423
PROBLEM 12.99
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 page 709 of the text, the velocity at the point of closest approach on a parabolic trajectory is
given by
v0 =
2GM
r0
1/2
Thus,
� 2 × 66.73 × 10−12 m3 /kg ⋅ s 2 × 4.87 × 1024 kg �
(v A )par = �
�
6332 × 103 m
�
�
= 10,131.4 m/s
(v A ) par = 10.13 km/s �
or
(b)
We have
(v A )circ = (v A ) par + ∆v A
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 ��
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
All rights
rights reserved.
reserved. No
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
Inc. All
No part
part of
of this
reproduced or
or distributed
distributed in
in any
any form
form or
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced
are aa student
student using
using this
this Manual,
Manual,
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
individual course
distribution
for their
their individual
course preparation.
preparation. If
If you
you are
you are
are using
using it
it without
without permission.
you
permission.
424
424
PROBLEM 12.100
It was observed that during its second flyby of the earth, the Galileo spacecraft had a velocity of 14 × 103 m/s
as it reached its minimum altitude of 300 km above the surface of the earth. Determine the eccentricity of the
trajectory of the spacecraft during this portion of its flight.
SOLUTION
First we note
R = 6370 km = 6.37 × 106 m
and
r0 = (6370 + 300) km = 6670 km
= 6.67 × 106 m
We have
1 GM
= 2 (1 + ε cos θ )
r
h
At Point O,
r = r0 , θ = 0, h = h0 = r0 v0
Also,
GM = gR 2
Eq. (12.39)
Eq. (12.30)
Then
gR 2
1
=
(1 + ε )
r0 (r0 v0 ) 2
or
ε = 0 02 − 1
r v2
gR
=
(6.67 × 106 m)(14 × 103 m/s)2
−1
(9.81 m/s2)(6.37 × 106 m)2
ε = 2.28
or
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
PROPRIETARY
MATERIAL.
2009orThe
McGraw-Hill
Companies,
Inc.written
All rights
reserved.ofNo
of this or
Manual
may bethe
displayed,
reproduced
or distributed
in any©form
by any
means, without
the prior
permission
thepart
publisher,
used beyond
limited
reproduced to
orteachers
distributed
any form
or by any
means, without
prior
writtencourse
permission
of the publisher,
used beyond
theManual,
limited
distribution
and in
educators
permitted
by McGraw-Hill
forthe
their
individual
preparation.
If you are aor
student
using this
distribution
educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you
are usingtoitteachers
without and
permission.
you are using it without permission.
425
425
PROBLEM 12.101
It was observed that as the Galileo spacecraft reached the point on its trajectory closest to Io, a moon of the
planet Jupiter, it was at a distance of 2800 km from the center of Io and had a velocity of 15 × 10 3 m/s.
Knowing that the mass of Io is 0.01496 times the mass of the earth, determine the eccentricity of the trajectory
of the spacecraft as it approached Io.
SOLUTION
First note
r0 = 2800 km = 2.8 × 106 m
Rearth = 6370 km = 6.37 × 106 m
We have
1 GM
= 2 (1 + ε cos θ )
r
h
At Point O,
r = r0 , θ = 0, h = h0 = r0 v0
Also,
Eq. (12.39)
GM Io = G (0.01496 M earth )
2
= 0.01496gRearth
using Eq. (12.30).
Then
2
1 0.01496 gRearth
=
(1 + ε )
r0
(r0 v0 ) 2
or
ε=
=
r0 v02
2
0.01496 gRearth
−1
(2.8× 106 m)(15 × 103 m/s)2
−1
0.01496(9.81 m/s2)(6.37 × 106 m)2
ε = 104.8
or
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
No part
part of
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced or
or distributed
distributed in
in any
any form
form or
reproduced
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
you are
are aa student
student using
using this
this Manual,
Manual,
distribution
If you
you are
are using
using it
it without
without permission.
permission.
you
426
426
PROBLEM 12.102
A satellite describes an elliptic orbit about a planet of mass M.
Denoting by r0 and r1 , respectively, the minimum and maximum
values of the distance r from the satellite to the center of the planet,
derive the relation
1 1 2GM
+ = 2
r0 r1
h
where h is the angular momentum per unit mass of the satellite.
SOLUTION
Using Eq. (12.39),
1 GM
= 2 + C cos θ A
rA
h
and
1 GM
= 2 + C cos θ B .
rB
h
But
θ B = θ A + 180°,
so that
cos θ A = − cos θ B .
Adding,
1 1 1 1 2GM
+ = + = 2 �
rA rB r0 r1
h
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
PROPRIETARY MATERIAL. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
reproduced to
orteachers
distributed
any form
or by any
means, without
prior
writtencourse
permission
of the publisher,
used beyond
theManual,
limited
distribution
and in
educators
permitted
by McGraw-Hill
forthe
their
individual
preparation.
If you are aorstudent
using this
distribution
educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are usingtoitteachers
withoutand
permission.
you are using it without permission.
427
427
PROBLEM 12.103
At main engine cutoff of its thirteenth flight, the space
shuttle Discovery was in an elliptic orbit of minimum
altitude 64 km and maximum altitude 538 km above the
surface of the earth. Knowing that at Point A the shuttle had
a velocity v0 parallel to the surface of the earth and that the
shuttle was transferred to a circular orbit as it passed
through Point B, determine (a) the speed v0 of the shuttle
at A, (b) the increase in speed required at B to insert the
shuttle into the circular orbit.
SOLUTION
For earth, R = 6370 km = 6.37 × 10 6 m
GM = gR2 = (9.81)(6.37 × 106 )2 = 0.3980615 m3/s2
rA = 6370 + 64 = 6434 km = 6.434 × 106 m
rB = 6370 + 538 = 6908 km = 6.908 × 106 m
Elliptic trajectory.
Using Eq. (12.39),
1 GM
= 2 + C cosθ A
rA
h
But
θ B = θ A + 180°, so that cos θ A = − cos θ B
Adding,
and
1 GM
= 2 + C cos θ B .
rB
h
1 1 rA + rB 2GM
+ =
= 2
rA rB
rA rB
h
h=
=
2GMrA rB
rA + rB
(2)(0.39806 × 1015)(6.434 × 106)(6.908 × 106)
13.342 × 106
= 51.5 × 109 m2/s
(a)
Speed v0 at A.
v0 = v A =
(vB )1 =
=
h
51.5 × 109
=
rA 6.434 × 106
v0 = 8 × 103 m/s
h
rA
51.5 × 109
6.908 × 106
= 7.455 × 103 m/s
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
No part
part of
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced or
or distributed
distributed in
in any
any form
form or
reproduced
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
you are
are aa student
student using
using this
this Manual,
Manual,
distribution
If you
you are
are using
using it
it without
without permission.
permission.
you
428
428
PROBLEM 12.103 (Continued)
For a circular orbit through Point B,
(vB )circ =
=
GM
rB
0.39806 × 1015
6.908 × 106
= 7.59 × 103 m/s
(b)
Increase in speed at Point B.
∆vB = (vB )circ − (vB )1
∆vB = 135 m/s
= 135 m/s
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
PROPRIETARY
MATERIAL.
2009orThe
McGraw-Hill
Companies,
Inc.written
All rights
reserved.ofNo
of this or
Manual
may bethe
displayed,
reproduced
or distributed
in any©form
by any
means, without
the prior
permission
thepart
publisher,
used beyond
limited
or by any
means, without
prior
writtencourse
permission
of the publisher,
used beyond
theManual,
limited
reproduced to
orteachers
distributed
any form
distribution
and in
educators
permitted
by McGraw-Hill
forthe
their
individual
preparation.
If you are aor
student
using this
distribution
educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you
are usingtoitteachers
without and
permission.
you are using it without permission.
429
429
PROBLEM 12.104
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 β v0 , where β � 1, by firing its engine for a short interval of
time. Determine the smallest permissible value of β 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 = hA = rA (v A ) AB
Now
= [ R (1 + α )]( β v0 )
Then
2v02 R(1 + α )
1
1
+ =
R(1 + α ) rB [ R (1 + α ) β v0 ]2
=
2
β R(1 + α )
2
Now β min corresponds to rB → R.
Then
1
1
2
+ = 2
R(1 + α ) R β min
R(1 + α )
β min =
or
2
��
2 +α
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
All rights
rights reserved.
reserved. No
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
Inc. All
No part
part of
of this
reproduced or
or distributed
distributed in
in any
any form
form or
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced
are aa student
student using
using this
this Manual,
Manual,
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
individual course
distribution
for their
their individual
course preparation.
preparation. If
If you
you are
you are
are using
using it
it without
without permission.
you
permission.
430
430
PROBLEM 12.105
As it describes an elliptic orbit about the sun, a spacecraft reaches a
maximum distance of 323 × 106 km from the center of the sun at
Point A (called the aphelion) and a minimum distance of 147 × 106 km
at Point B (called the perihelion). To place the spacecraft in a
smaller elliptic orbit with aphelion at A′ and perihelion at B′,
where A′ and B′ are located 263 × 10 6 km and 137 × 10 6 km,
respectively, from the center of the sun, the speed of the spacecraft
is first reduced as it passes through A and then is further reduced as
it passes through B′. Knowing that the mass of the sun is
332.8 × 103 times the mass of the earth, determine (a) the speed of
the spacecraft at A, (b) the amounts by which the speed of the
spacecraft should be reduced at A and B′ to insert it into the
desired elliptic orbit.
SOLUTION
First note
Rearth = 6370 km = 6.37 × 106 m
rA = 323 × 106 km = 323 × 109 m
rB = 147 × 106 km = 147 × 109 m
From the solution to Problem 12.102, we have for any elliptic orbit about the sun
1 1 2GM sun
+ =
r1 r2
h2
(a)
For the elliptic orbit AB, we have
r1 = rA , r2 = rB , h = hA = rA v A
Also,
GM sun = G[(332.8 × 103 ) M earth ]
2
= gRearth
(332.8 × 103 )
Then
or
using Eq. (12.30).
2
(332.8 × 103 )
1
1 2 gRearth
+ =
rA rB
(rAv A )2
R
665.6 g × 103
v A = earth
1
rA
+ r1
r
A
6370 km
=
323 × 106 km
1/2
B
665.6 × 103 × 9.81 m/s2
1
1
+
9
9
323 × 10 m
1/ 2
147 × 10 m
= 16,017 m/s
v A = 16 × 103 m/s
or
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
PROPRIETARY
MATERIAL.
2009orThe
McGraw-Hill
Companies,
Inc.written
All rights
reserved.ofNo
of this or
Manual
may bethe
displayed,
reproduced
or distributed
in any©form
by any
means, without
the prior
permission
thepart
publisher,
used beyond
limited
reproduced to
orteachers
distributed
any form
or by any
means, without
prior
writtencourse
permission
of the publisher,
used beyond
theManual,
limited
distribution
and in
educators
permitted
by McGraw-Hill
forthe
their
individual
preparation.
If you are aor
student
using this
distribution
educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you
are usingtoitteachers
without and
permission.
you are using it without permission.
431
431
PROBLEM 12.105 (Continued)
(b)
From Part (a), we have
1
1
+
rA rB
2GM sun = (rA v A ) 2
Then, for any other elliptic orbit about the sun, we have
(
)
1 1 (rA v A ) rA + rB
+ =
r1 r2
h2
2
1
1
For the elliptic transfer orbit AB′, we have
r1 = rA , r2 = rB′ , h = htr = rA (v A ) tr
Then
or
(
)
2 1
1
1
1 ( rAv A ) rA + rB
+
=
rA rB′
[rA (v A ) tr ]2
(v A ) tr = v A
1
+ r1
rA
B
r
1/2
= vA
1 + 1
rA
rB′
1 + rA
1/ 2
B
r
1 + rA′
B
1 + 323
147
1 + 323
137
= (16017 m/s)
1/2
= 15630 m/s
htr = ( hA ) tr = (hB′ ) tr : rA (v A ) tr = rB′ (vB′ ) tr
Now
Then
(vB′ ) tr =
323 × 106 km
× 15630 m/s = 36850 m/s
137 × 106 km
For the elliptic orbit A′B′, we have
r1 = rA′ , r2 = rB′ , h = rB′vB′
2
Then
or
(
1
1 (rA v A ) rA + rB
+
=
rA′ rB′
(rB′vB′ )2
r
vB ′ = v A A
rB′
1
1
)
1 + 1
rA
rB
= (16017 m/s)
1/ 2
1
+ r1′
rA′
B
323 × 106 km
= (16017 m/s)
137 × 106 km
1
+ 1
323 × 106 147 × 106
1
+ 1
263 × 106 137 × 106
1/ 2
= 35658 m/s
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
No part
part of
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced or
or distributed
distributed in
in any
any form
form or
reproduced
distribution
If you
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
you are
are aa student
student using
using this
this Manual,
Manual,
you
you are
are using
using it
it without
without permission.
permission.
432
432
PROBLEM 12.105 (Continued)
Finally,
or
(v A ) tr = v A + ∆v A
∆vA = (15,630 − 16,017) m/s
or
= 387 m/s
and
vB′ = (vB′ ) tr + ∆vB
or
∆vB′ = (35,658 − 36,850) m/s
|∆v A | = 387 m/s
= −1192 m/s
|∆vB | = 1192 m/s
or
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
PROPRIETARY
MATERIAL.
2009orThe
McGraw-Hill
Companies,
Inc.written
All rights
reserved.ofNo
of this or
Manual
may bethe
displayed,
reproduced
or distributed
in any©form
by any
means, without
the prior
permission
thepart
publisher,
used beyond
limited
or by any
means, without
prior
writtencourse
permission
of the publisher,
used beyond
theManual,
limited
reproduced to
orteachers
distributed
any form
distribution
and in
educators
permitted
by McGraw-Hill
forthe
their
individual
preparation.
If you are aor
student
using this
distribution
educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you
are usingtoitteachers
without and
permission.
you are using it without permission.
433
433
PROBLEM 12.106
A space probe is to be placed in a circular orbit of 8960 km
radius about the planet Venus in a specified plane. As the
probe reaches A, the point of its original trajectory closest to
Venus, it is inserted in a first elliptic transfer orbit by
reducing its speed by v 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 vB . Finally, as the probe reaches Point C, it is
inserted in the desired circular orbit by reducing its speed
by vC . Knowing that the mass of Venus is 0.82 times the
mass of the earth, that rA = 14.9 × 103 km and rB = 304 × 103 km,
and that the probe approaches A on a parabolic trajectory,
determine by how much the velocity of the probe should be
reduced (a) at A, (b) at B, (c) at C.
SOLUTION
For Earth,
R = 6370 km = 6.37 ×106 m, g = 9.81 m/s2
GM earth = gR2 = (9.81)(6.37 ×106)2 = 0.39806 ×1015 m3/s2
For Venus,
For a parabolic trajectory with
GM = 0.82 GM earth = 0.3264 ×1015 m3/s2
rA = 14.9 ×103 km = 14.9 ×106 m
(v A )1 = vesc =
First transfer orbit AB.
2GM
rA
(2)(0.3264 ×1015)
= 6.619 ×103 m/s
14.9 ×106
rB = 304 ×103 km = 304 ×106 m
At Point A, where θ = 180°
1
GM
GM
= 2 + C cos 180° = 2 − C
rA
hAB
hAB
(1)
1
GM
GM
= 2 + C cos 0 = 2 + C
rB
hAB
hAB
(2)
At Point B, where θ = 0°
Adding,
r + rA 2GM
1
1
= 2
+
= B
rA rB
rA rB
hAB
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
part of
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
No part
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced or
or distributed
distributed in
in any
any form
form or
reproduced
distribution to
to teachers
teachers and
andeducators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
If you
you are
are aa student
student using
using this
thisManual,
Manual,
distribution
you are
are using
using itit without
without permission.
permission.
you
434
434
PROBLEM 12.106 (Continued)
Solving for hAB ,
hAB =
2GMrA rB
=
rB + rA
(2)(0.3264 × 1015)(14.9 ×106)(304 ×106)
318.9 × 106
= 9.629 ×1010 m2/s
Second transfer orbit BC.
(v A ) 2 =
hAB
9.629 ×1010
=
= 6.462 ×103 m/s
rA
14.9 ×106
(vB )1 =
hAB 9.629 × 1010
=
= 0.3167 × 103 m/s
rB
304 × 106
rC = 8960 km = 8.96 × 106 m
At Point B, where θ = 0
1
GM
GM
= 2 + C cos 0 = 2 + C
rB
hBC
hBC
At Point C, where θ = 180°
1
GM
GM
= 2 + C cos 180° = 2 − C
rC
hBC
hBC
Adding,
1
1 rB + rC 2GM
= 2
+
=
rB rC
rB rC
hBC
hBC =
=
2GMrB rC
rB + rC
(2)(0.3264 ×1015)(304 × 106)(8.96 ×106)
312.96 ×106
= 75.377 × 109 m2/s
Final circular orbit.
( vB ) 2 =
hBC 75.377 × 109
= 247.95 m/s
=
rB
304 × 106
(vC )1 =
hBC 75.377 × 109
=
= 8.413 ×103 m/s
rC
8.96 × 106
rC = 8.96 ×106 m
(vC )2 =
=
GM
rC
0.3264 × 1015
8.96 × 106
= 6.036 × 103 m/s
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
PROPRIETARY
MATERIAL.
2009orThe
McGraw-Hill
Companies,
Inc.written
All rights
reserved.ofNo
of this or
Manual
may bethe
displayed,
reproduced
or distributed
in any©form
by any
means, without
the prior
permission
thepart
publisher,
used beyond
limited
or by any
means, without
prior
writtencourse
permission
of the publisher,
used beyond
theManual,
limited
reproduced to
orteachers
distributed
any form
distribution
and in
educators
permitted
by McGraw-Hill
forthe
their
individual
preparation.
If you are aor
student
using this
distribution
educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you
are usingtoitteachers
without and
permission.
you are using it without permission.
435
435
PROBLEM 12.106 (Continued)
Speed reductions.
(a)
At A:
(v A )1 − (v A ) 2 = 6.619 × 103 − 6.462 × 103
∆v A = 157 m/s
(b)
At B:
(vB )1 − (vB ) 2 = 0.3167 × 103 − 247.95
∆vB = 69 m/s
(c)
At C:
(vC )1 − (vC )2 = 8.413 × 103 − 6.036 × 103
∆vC = 2.38 × 103 m/s
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
No part
part of
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced or
or distributed
distributed in
in any
any form
form or
reproduced
distribution
If you
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
you are
are aa student
student using
using this
this Manual,
Manual,
you
you are
are using
using it
it without
without permission.
permission.
436
436
PROBLEM 12.107
For the space probe of Problem 12.106, it is known that
rA = 14.9 × 103 km and that the velocity of the probe is reduced
to 6000 m/s, as it passes through A. Determine (a) the
distance from the center of Venus to Point B, (b) the amounts
by which the velocity of the probe should be reduced at B and
C, respectively.
SOLUTION
Data from Problem 12.106:
rC = 8.96 ×106 m,
M = 0.82 Mearth
For Earth,
R = 6370 km = 6.37 ×106 m,
g = 9.81 m/s2
GMearth = gR2 = (9.81)(6.37 × 106)2 = 0.39806 × 1015 m3/s2
GM = 0.82GM earth = 0.3264 × 1015 m3/s2
For Venus,
Transfer orbit AB:
vA = 6,000 m/s, rA = 14.9 × 103 km = 14.9 × 106m
hAB = rAv A (14.9 × 106)(6,000) = 89.4 × 109 m2/s
At Point A, where θ = 180°
1
GM
GM
= 2 + C cos 180° = 2 − C
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
=
1
(2)(0.3264 × 1015)
−
9 2
14.9 × 109
(89.4 × 10 )
= 14 × 10−9 m−1
rB = 68.77 × 103 km
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
PROPRIETARY
MATERIAL.
2009orThe
McGraw-Hill
Companies,
Inc.written
All rights
reserved.ofNo
of this or
Manual
may bethe
displayed,
reproduced
or distributed
in any©form
by any
means, without
the prior
permission
thepart
publisher,
used beyond
limited
reproduced to
orteachers
distributed
any form
or by any
means, without
prior
writtencourse
permission
of the publisher,
used using
beyond
theManual,
limited
distribution
and in
educators
permitted
by McGraw-Hill
forthe
their
individual
preparation.
If you are aor
student
this
distribution
educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you
are usingtoitteachers
without and
permission.
you are using it without permission.
437
437
PROBLEM 12.107 (Continued)
(a)
Radial coordinate rB .
rB = 68.77 × 106 m
(vB )1 =
Second transfer orbit BC.
hAB
89.4 × 109
= 1.3 × 103 m/s
=
6
rB
68.77 × 10
rC = 8960 km = 8.96 × 106 m
At Point B, where θ = 0
1
GM
GM
= 2 + C cos 0 = 2 + C
rB
hBC
hBC
At Point C, where θ = 180°
1
GM
GM
= 2 + C cos 180° = 2 − C
rC
hBC
hBC
Adding,
r + rC
1
1
2GM
= 2
+
= B
rB
rC
rB rC
hBC
hBC =
2GMrB rC
=
rB + rC
(2)(0.3264 × 1015)(68.77 × 106)(8.96× 106)
77.73 × 106
= 71.936 × 109 m2/s
( vB ) 2 =
hBC
71.936 × 109
=
= 1.046 × 103 m/s
rB
68.77 × 106
(vC )1 =
hBC
71.936 × 109
=
= 7.675 × 103 m/s
rC
8.96 × 106
rC = 8.96 × 106 m
Circular orbit with
(vC )2 =
(b)
0.3264 × 1015
GM
=
= 6.036 × 103 m/s
rC
8.96 × 106
Speed reductions at B and C.
At B:
(vB )1 − (vB ) 2 = 1.3 × 103 − 1.046 × 103
∆vB = 254 m/s
At C:
(vC )1 − (vC )2 = 7.675 × 103 − 6.036 × 103
∆vC = 1.64 × 103 m/s
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
No part
part of
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced or
or distributed
distributed in
in any
any form
form or
reproduced
distribution
If you
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
you are
are aa student
student using
using this
this Manual,
Manual,
you
you are
are using
using it
it without
without permission.
permission.
438
438
PROBLEM 12.108
Determine the time needed for the probe of 12.106 to travel from A to B on its first transfer orbit.
SOLUTION
From Problem 12.106 for the first transfer orbit
rA = 14.9 × 106 m
rB = 304 × 106 m
hAB = 9.629 × 1010 m2/s
1
(rA + rB )
2
1
= (14.9 × 106 + 304 × 106)
2
= 159.45 × 106 m
a=
b = rA rB
= (14.9 × 106)(304 × 106)
= 67.302 × 106 m
Periodic time for full elliptical orbit.
τ=
2π ab
hAB
(2p)(159.45 × 106)(67.302 × 106)
9.629 × 1010
= 700.25 × 103 s
=
Time to travel from A to B.
1
t AB = τ = 350.1 × 103 s
2
t AB = 97.25 h
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
PROPRIETARY
MATERIAL.
2009orThe
McGraw-Hill
Companies,
Inc.written
All rights
reserved.ofNo
of this or
Manual
may bethe
displayed,
reproduced
or distributed
in any©form
by any
means, without
the prior
permission
thepart
publisher,
used beyond
limited
reproduced to
orteachers
distributed
any form
or by any
means, without
prior
writtencourse
permission
of the publisher,
used beyond
theManual,
limited
distribution
and in
educators
permitted
by McGraw-Hill
forthe
their
individual
preparation.
If you are aor
student
using this
distribution
educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you
are usingtoitteachers
without and
permission.
you are using it without permission.
439
439
PROBLEM 12.109
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 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
= 2 + C cos θ A
rA
h
But
θ B = θ A + 180°, so that cos θ A = − cos θ B .
Adding,
and
1 GM
= 2 + C cos θ B .
rB
h
1
1 r +r
2GM
+ = A B = 2
rA rB
rA rB
hAB
hAB =
2GMrA rB
(2)(4.896 × 1012 )(2.137 × 106 )(4.677 × 106 )
=
rA + rB
6.804 × 106
= 3.78856 × 109 m 2/s
1
a = ( rA + rB ) = 3.402 × 106 m
2
b = rA rB = 3.16145 × 106 m
Periodic time.
τ=
2π ab 2π (3.402 × 106 )(3.16145 × 106 )
=
= 17.837 × 103 s
hAB
3.78856 × 109
τ = 4.95 h ��
�
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
All rights
rights reserved.
reserved. No
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
Inc. All
No part
part of
of this
reproduced or
or distributed
distributed in
in any
any form
form or
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced
are aa student
student using
using this
this Manual,
Manual,
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
individual course
distribution
for their
their individual
course preparation.
preparation. If
If you
you are
you are
are using
using it
it without
without permission.
you
permission.
440
440
PROBLEM 12.110
A space probe in a low earth orbit is inserted into an elliptic
transfer orbit to the planet Venus. Knowing that the mass of the
sun is 332.8 × 103 times the mass of the earth and assuming that
the probe is subjected only to the gravitational attraction of the
sun, determine the value of 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. Now
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
= (148.8 × 106 + 107.5 × 106) km
2
= 128.15 × 106 km
where
atr =
and
aearth ≈ rE
Then
1 atr
tprobe =
2 rE
=
( Note: ε earth = 0.0167)
3/ 2
τ earth
1 128.15 × 106 km
2 148.8 × 106 km
3/ 2
(365.25 days)
= 145.96 days
= 12.6109 × 106 s
In time tprobe , Venus travels through the angle θ v given by
v
rv
θv = θv tprobe = v tprobe
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
PROPRIETARY
MATERIAL.
2009orThe
McGraw-Hill
Companies,
Inc.written
All rights
reserved.ofNo
of this or
Manual
may bethe
displayed,
reproduced
or distributed
in any©form
by any
means, without
the prior
permission
thepart
publisher,
used beyond
limited
reproduced to
orteachers
distributed
any form
or by any
means, without
prior
writtencourse
permission
of the publisher,
used beyond
theManual,
limited
distribution
and in
educators
permitted
by McGraw-Hill
forthe
their
individual
preparation.
If you are aor
student
using this
distribution
educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you
are usingtoitteachers
without and
permission.
you are using it without permission.
441
441
PROBLEM 12.110 (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
θv =
(
using Eq. (12.30)
3
2
tprobe 332.8 × 10 gRearth
rv
)
1/ 2
rv
= tprobe Rearth
where
)
(332.8 g × 103 )1/ 2
rv3/ 2
Rearth = 6370 km = 6.37 × 106 m
and
rv = 107.5 × 106 km = 107.5 × 109 m
Then
θ v = (12.6109 × 106 s)(6.37 × 106 m)
(332.8 × 103 × 9.81 m/s2)1/2
(107.5 × 109 m)1/2
= 4.1181 rad
= 235.95°
Finally,
φ = θv − 180° = 235.95° − 180°
ϕ = 56°
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
No part
part of
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced or
or distributed
distributed in
in any
any form
form or
reproduced
distribution
If you
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
you are
are aa student
student using
using this
this Manual,
Manual,
you
you are
are using
using it
it without
without permission.
permission.
442
442
PROBLEM 12.111
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,
v0 =
2π RE 2π RE 3/ 2
GM
, τ0 =
=
RE
v0
GM
or
GM =
2π RE3/ 2
τ0
(1)
For the comet Hyakutake,
1 GM
= 2 = (1 + ε ),
r0
h
a=
1 GM
1+ ε
r0
= 2 (1 + ε ), r1 =
r1
1− ε
h
r
1
1+ ε
r0
(r0 + r1 ) = 0 , b = r0 r1 =
1−ε
2
1− ε
h = GMr0 (1 + ε )
τ=
=
2π r02 (1 + ε )1/ 2
2π ab
=
h
(1 − ε )3/ 2 GMr0 (1 + ε )
2π r03/ 2
GM (1 − ε )3/2
� r �
=� 0 �
� RE �
3/2
1
(1 − ε )3/2
= (0.230)3/2
Since
=
2π r03/ 2τ 0
2π RE3 (1 − ε )3/2
τ0
1
τ 0 = 91.8 × 103τ 0
(1 − 0.999887)3/ 2
τ 0 = 1 yr, τ = (91.8 × 103 )(1.000)
τ = 91.8 × 103 yr ��
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
PROPRIETARY MATERIAL. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
reproduced to
orteachers
distributed
any form
or by any
means, without
prior
writtencourse
permission
of the publisher,
used beyond
theManual,
limited
distribution
and in
educators
permitted
by McGraw-Hill
forthe
their
individual
preparation.
If you are aorstudent
using this
distribution
educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are usingtoitteachers
withoutand
permission.
you are using it without permission.
443
443
PROBLEM 12.112
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 �
=τE �
�
� 1 year �
= 17.94τ E
But
2/3
1
(τ min + τ max )
2
1�1
�
17.94τ E = � τ E + τ max �
2� 2
�
aH =
1
2
τ max = 2(17.94τ E ) − τ E
= (35.88 − 0.5)τ E
= 35.38τ E
τ max = 150 × 106 km = 5.31 × 109 km ��
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
All rights
rights reserved.
reserved. No
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
Inc. All
No part
part of
of this
reproduced or
or distributed
distributed in
in any
any form
form or
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced
are aa student
student using
using this
this Manual,
Manual,
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
individual course
distribution
for their
their individual
course preparation.
preparation. If
If you
you are
you are
are using
using it
it without
without permission.
you
permission.
444
444
PROBLEM 12.113
Determine the time needed for the space probe of Problem 12.99 to travel
from B to C.
SOLUTION
From the solution to Problem 12.99, we have
(v A )par = 10,131.4 m/s
and
(v A )circ =
Also,
1
2
(v A ) par = 7164.0 m/s
rA = (6052 + 280) km = 6332 km
For the parabolic trajectory BA, we have
1 GM v
= 2 (1 + ε cos θ )
r
hBA
[Eq. (12.39′)]
where ε = 1. Now
at A, θ = 0:
1 GM v
= 2 (1 + 1)
rA
hBA
or
rA =
at B, θ = −90°:
1 GM v
= 2 (1 + 0)
rB
hBA
or
rB =
2
hBA
2GM v
2
hBA
GM v
rB = 2rA
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
PROPRIETARY MATERIAL. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
reproduced to
orteachers
distributed
any form
or by any
means, without
prior
writtencourse
permission
of the publisher,
used beyond
theManual,
limited
distribution
and in
educators
permitted
by McGraw-Hill
forthe
their
individual
preparation.
If you are aorstudent
using this
distribution
educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are usingtoitteachers
withoutand
permission.
you are using it without permission.
445
445
PROBLEM 12.113 (Continued)
As the probe travels from B to A, the area swept out is the semiparabolic area defined by Vertex A and
Point B. Thus,
(Area swept out) BA = ABA =
Now
2
4
(rA )(rB ) = rA2
3
3
dA 1
= h
dt 2
where h = constant
Then
A=
2 ABA
1
ht or t BA =
2
hBA
=
2 × 43 rA2
rAv A
hBA = rAv A
=
8 rA
3 vA
8 6332 × 103 m
3 10,131.4 m/s
= 1666.63 s
=
For the circular trajectory AC,
π
rA
π 6332 × 103 m
=
t AC = 2
= 1388.37 s
(v A )circ 2 7164.0 m/s
Finally,
t BC = t BA + t AC
= (1666.63 + 1388.37) s
=3055.0 s
t BC = 50 min 55 s �
or
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
All rights
rights reserved.
reserved. No
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
Inc. All
No part
part of
of this
reproduced or
or distributed
distributed in
in any
any form
form or
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced
are aa student
student using
using this
this Manual,
Manual,
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
individual course
distribution
for their
their individual
course preparation.
preparation. If
If you
you are
you are
are using
using it
it without
without permission.
you
permission.
446
446
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 β v0, where β � 1, to place the probe on a
crash trajectory. Express in terms of n and β the angle AOB, where B denotes
the point of impact of the probe on the planet.
SOLUTION
r0 = rA = nR
For the circular orbit,
v0 =
GM
GM
=
r0
nR
The crash trajectory is elliptic.
v A = β v0 =
β 2 GM
nR
h = rAv A = nRv A = β 2 nGMR
GM
1
= 2
2
β nR
h
1 GM
1 + ε cos θ
= 2 (1 + ε cos θ ) =
r
h
β 2 nR
At Point A, θ = 180°
1
1
1− ε
=
= 2
rA nR β nR
or β 2 = 1 − ε
or ε = 1 − β 2
At impact Point B, θ = π − φ
1 1
=
rB R
1 1 + ε cos (π − φ ) 1 − ε cos φ
=
=
R
β 2 nR
β 2 nR
ε cos φ = 1 − nβ 2 or cos φ =
1 − nβ 2
ε
=
1 − nβ 2
1− β 2
φ = cos −1[(1 − nβ 2 ) /(1 − β 2 )] �
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
PROPRIETARY MATERIAL. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
reproduced to
orteachers
distributed
any form
or by any
means, without
prior
writtencourse
permission
of the publisher,
used beyond
theManual,
limited
distribution
and in
educators
permitted
by McGraw-Hill
forthe
their
individual
preparation.
If you are aorstudent
using this
distribution
educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are usingtoitteachers
withoutand
permission.
you are using it without permission.
447
447
PROBLEM 12.115
Prior to the Apollo missions to the moon, several Lunar Orbiter
spacecraft were used to photograph the lunar surface to obtain
information regarding possible landing sites. At the conclusion of
each mission, the trajectory of the spacecraft was adjusted so that the
spacecraft would crash on the moon to further study the
characteristics of the lunar surface. Shown is the elliptic orbit of
Lunar Orbiter 2. Knowing that the mass of the moon is 0.01230 times
the mass of the earth, determine the amount by which the speed of
the orbiter should be reduced at Point B so that it impacts the lunar
surface at Point C. (Hint: Point B is the apogee of the elliptic impact
trajectory.)
SOLUTION
From the solution to Problem 12.102, we have for the elliptic orbit AB
1
1 2GM moon
+ =
2
rA rB
hAB
where
and
hAB = (hB ) AB = rB (vB ) AB
GM moon = G (0.01230M earth )
2
= 0.01230gRearth
using Eq. (12.30)
Then
2
)
1
1 2(0.01230 gRearth
+ =
2
rA rB
[rB (vB ) AB ]
or
� 0.0246 g �
R
(vB ) AB = earth � 1 1 �
rB � r + r �
� A B �
1/2
1/2
6.37 × 106 m �� 0.0246 × 9.81 m/s 2 ��
=
3600 × 103 m � 1790 ×1103 m + 3600 ×1103 m �
�
�
= 950.43 m/s
For the elliptic impact trajectory, we have
1 GM moon
=
+ C cos θ
2
r
hBC
where
[Eq. (12.39)]
hBC = (hB ) BC = rB (vB ) BC
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
All rights
rights reserved.
reserved. No
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
Inc. All
No part
part of
of this
reproduced or
or distributed
distributed in
in any
any form
form or
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced
are aa student
student using
using this
this Manual,
Manual,
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
individual course
distribution
for their
their individual
course preparation.
preparation. If
If you
you are
you are
are using
using it
it without
without permission.
you
permission.
448
448
PROBLEM 12.115 (Continued)
Noting that Point B is the apogee of this trajectory, we have
at B, θ = 180°:
1 GM moon
=
−C
2
rB
hBC
or
C=
at C, θ = −70°:
1 GM moon
=
+ C cos ( −70°)
2
R
hBC
or
C=
Then
or
GM moon
2
hBC
−
GM moon
2
hBC
−
1
rB
1 � 1 GM moon �
� −
��
2
cos 70° �� R
hBC
�
1
1 � 1 GM moon �
=
� −
��
2
rB cos 70° �� R
hBC
�
2
hBC
=
GM moon (1 + cos 70°)
1
+ cosr 70°
R
B
1/2
or
� 0.012309(1 + cos 70°) �
R
�
(vB ) BC = earth �
1 + cos 70°
rB �
�
R
rB
�
�
1/2
�
�
6.37 × 106 m � 0.01230(9.81 m/s 2 )(1 + cos 70°) �
(vB ) BC =
1
�
3600 × 103 m �
+ cos 70°3
3600 × 10 m
1737 × 103 m
�
�
= 869.43 m/s
Finally,
or
(vB ) BC = (vB ) AB + ∆vB
∆vB = (869.43 − 950.43) m/s
|∆vB | = 81.0 m/s �
or
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
PROPRIETARY MATERIAL. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
reproduced to
orteachers
distributed
any form
or by any
means, without
prior
writtencourse
permission
of the publisher,
used beyond
theManual,
limited
distribution
and in
educators
permitted
by McGraw-Hill
forthe
their
individual
preparation.
If you are aorstudent
using this
distribution
educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are usingtoitteachers
withoutand
permission.
you are using it without permission.
449
449
PROBLEM 12.116
As a spacecraft approaches the planet Jupiter, it releases a probe which is to
enter the planet’s atmosphere at Point B at an altitude of 450 km above the
surface of the planet. The trajectory of the probe is a hyperbola of
eccentricity ε = 1.031. Knowing that the radius and the mass of Jupiter are
71.492 × 103 km and 1.9 × 1027 kg, respectively, and that the velocity vB of
the probe at B forms an angle of 82.9° with the direction of OA, determine
(a) the angle AOB, (b) the speed vB of the probe at B.
SOLUTION
First we note
rB = (71.492 × 103 + 450) km = 71.942 × 103 km
(a)
We have
1 GM j
= 2 (1 + ε cos θ )
r
h
At A, θ = 0:
1 GM j
= 2 (1 + ε )
rA
h
or
At B, θ = θ B = � AOB :
h2
= rA (1 + ε )
GM j
1 GM j
= 2 (1 + ε cos θ B )
rB
h
or
h2
= rB (1 + ε cos θ B )
GM j
Then
rA (1 + ε ) = rB (1 + ε cos θ B )
or
[Eq. (12.39′)]
cos θ B =
=
�
1 � rA
� (1 + ε ) − 1�
ε � rB
�
�
1 � 70.8 × 103 km
(1 + 1.031) − 1�
�
3
1.031 � 71.942 × 10 km
�
= 0.96873
or
θ B = 14.3661°
� AOB = 14.37° �
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
All rights
rights reserved.
reserved. No
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
Inc. All
No part
part of
of this
reproduced or
or distributed
distributed in
in any
any form
form or
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced
are aa student
student using
using this
this Manual,
Manual,
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
individual course
distribution
for their
their individual
course preparation.
preparation. If
If you
you are
you are
are using
using it
it without
without permission.
you
permission.
450
450
PROBLEM 12.116 (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
φ = (θ B + 82.9°) = 97.2661°
h=
(rB vB sin φ )2 = GM j rB (1 + ε cos θ B )
1/ 2
or
vB =
=
�
1 � GM j
(1 + ε cos θ B ) �
�
sin φ � rB
�
1
sin 97.2661°
1/ 2
−12
2
2
27
� 66.73 × 10 m /kg ⋅ s × 1.9 × 10 kg
�
× [1 + (1.031)(0.96873)]�
�
6
71.942 × 10 m
�
�
vB = 59.8 km/s �
or
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
PROPRIETARY MATERIAL. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
reproduced to
orteachers
distributed
any form
or by any
means, without
prior
writtencourse
permission
of the publisher,
used beyond
theManual,
limited
distribution
and in
educators
permitted
by McGraw-Hill
forthe
their
individual
preparation.
If you are aorstudent
using this
distribution
educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are usingtoitteachers
withoutand
permission.
you are using it without permission.
451
451
PROBLEM 12.117
A space shuttle is describing a circular orbit at an altitude of 560 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 150 m/s and
begin its descent toward the earth. Determine the angle AOB so that the
altitude of the shuttle at Point B is 120 km. (Hint: Point A is the apogee
of the elliptic descent trajectory.)
SOLUTION
First we note
R = 6370 km = 6.37 × 106 m
rA = (6370 + 560) km = 6930 km
= 6.93 × 106 m
rB = (6370 + 120) km = 6490 km
For the circular orbit, we have
vcirc =
gR 2
rA
[Eq. (12.44)]
6
= 6.37 × 10 m
9.81 m/s2
6.93 × 106 m
1/ 2
= 7579 m/s
Now
(v A ) AB = vcirc + ∆v A = (7579 − 150) m/s
= 7429 m/s
For the elliptic descent trajectory, we have
1 GM
= 2 + C cos θ
r
h
[Eq. (12.39)]
Noting that Point A is at the apogee of this trajectory, we have
at A, θ = 180°:
1 GM
= 2 −C
rA
h
or
C=
GM 1
−
rA
h2
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
No part
part of
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced or
or distributed
distributed in
in any
any form
form or
reproduced
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
you are
are aa student
student using
using this
this Manual,
Manual,
distribution
If you
you are
are using
using it
it without
without permission.
permission.
you
452
452
PROBLEM 12.117 (Continued)
at B, θ = θ B = 180° − AOB :
1 GM
= 2 + C cos θ B
rB
h
or
C=
Then
or
1
1 GM
− 2
cos θ B rB
h
1
1 GM
GM 1
− =
− 2
2
cos
r
r
θ
h
h
A
B
B
1
− GM2
rB
h
cos θ B = GM
− r1
h2
Now
A
h = ( hA ) AB = rA (v A ) AB
and
GM = gR 2
[Eq. (12.30)]
From above,
gR 2 = rA (vcirc )2
[Eq. (12.44)]
Then
rA (vcirc )2
vcirc
GM
1
=
=
2
2
rA (v A ) AB
h
[ rA (v A ) AB ]
so that
cos θ B =
=
1 − 1
rB
rA
1
rA
vcirc
(v A ) AB
vcirc
( v A ) AB
(
2
2
A
6930 km
7579 m/s
− 7429
m/s
6490 km
(
7579 m/s
7429 m/s
=
− r1
) −1
rA
−
rB
2
vcirc
(v A ) AB
vcirc
(v A ) AB
2
2
−1
)
2
2
= 0.6621
or
θ B = 48.54°
Finally,
AOB = 180° − 48.54°
AOB = 131.5°
or
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
PROPRIETARY
MATERIAL.
2009orThe
McGraw-Hill
Companies,
Inc.written
All rights
reserved.ofNo
of this or
Manual
may bethe
displayed,
reproduced
or distributed
in any©form
by any
means, without
the prior
permission
thepart
publisher,
used beyond
limited
or by any
means, without
prior
writtencourse
permission
of the publisher,
used beyond
theManual,
limited
reproduced to
orteachers
distributed
any form
distribution
and in
educators
permitted
by McGraw-Hill
forthe
their
individual
preparation.
If you are aor
student
using this
distribution
educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you
are usingtoitteachers
without and
permission.
you are using it without permission.
453
453
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
ρ
=
1� 1 1 �
� + �
2 � r0 r1 �
SOLUTION
Using Eq. (12.39),
1 GM
= 2 + C cos θ A
rA
h
and
1 GM
= 2 + C cos θ B .
rB
h
But
θ B = θ A + 180°,
so that
cos θ A = − cos θ B
Adding,
1 1 2GM
+ = 2
rA rB
h
At Points A and B, the radial direction is normal to the path.
an =
But
Fn =
1
ρ
=
v2
ρ
=
h2
r 2ρ
GMm
mh 2
ma
=
=
n
r2
r 2ρ
GM 1 � 1 1 �
= � + �
2 � rA rB �
h2
1
ρ
=
1� 1 1 �
� + � �
2 � r0 r1 �
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
All rights
rights reserved.
reserved. No
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
Inc. All
No part
part of
of this
reproduced or
or distributed
distributed in
in any
any form
form or
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced
are aa student
student using
using this
this Manual,
Manual,
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
individual course
distribution
for their
their individual
course preparation.
preparation. If
If you
you are
you are
are using
using it
it without
without permission.
you
permission.
454
454
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 12.111, where RE = 149.6 × 106 km, to determine the
approximate maximum distance from the sun reached by
comet Hyakutake.
SOLUTION
(a)
We have
1 GM
= 2 (1 + ε cos θ )
r
h
At A, θ = 0:
1 GM
= 2 (1 + ε )
r0
h
or
At B, θ = 180°:
Eq. (12.39′)
h2
= r0 (1 + ε )
GM
1 GM
= 2 (1 − ε )
r1
h
or
h2
= r1 (1 − ε )
GM
Then
r0 (1 + ε ) = r1 (1 − ε )
r − r0
�
r1 + r0
ε= 1
or
(b)
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
Note: r1 = 4070 RE
or r1 = 0.064 lightyears.
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
PROPRIETARY MATERIAL. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
reproduced to
orteachers
distributed
any form
or by any
means, without
prior
writtencourse
permission
of the publisher,
used beyond
theManual,
limited
distribution
and in
educators
permitted
by McGraw-Hill
forthe
their
individual
preparation.
If you are aorstudent
using this
distribution
educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are usingtoitteachers
withoutand
permission.
you are using it without permission.
455
455
PROBLEM 12.120
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
= 2 (1 + ε cos θ )
r
h
At A, θ = 0°:
1 GM
= 2 = (1 + ε )
rA
h
or
rA =
h2
GM (1 + ε )
At B, θ = 180°:
1 GM
= 2 = (1 − ε )
rB
h
or
rB =
h2
GM (1 − ε )
h2
1 �
2h 2
� 1
=�
+
=
�
GM � 1 + ε 1 − ε � GM (1 − ε 2 )
Adding,
rA + rB =
But for an ellipse,
rA + rB = 2a
2a =
2h2
GM (1 − ε 2 )
h = GMa(1 − ε 2 ) �
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
All rights
rights reserved.
reserved. No
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
Inc. All
No part
part of
of this
reproduced or
or distributed
distributed in
in any
any form
form or
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced
are aa student
student using
using this
this Manual,
Manual,
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
individual course
distribution
for their
their individual
course preparation.
preparation. If
If you
you are
you are
are using
using it
it without
without permission.
you
permission.
456
456
PROBLEM 12.121
Derive Kepler’s third law of planetary motion from Eqs. (12.39) and (12.45).
SOLUTION
For an ellipse,
2a = rA + rB
and b = rA rB
Using Eq. (12.39),
1 GM
= 2 + C cos θ A
rA
h
and
1 GM
= 2 + C cos θ B .
rB
h
But
θ B = θ A + 180°,
so that
cos θ A = − cos θ B .
Adding,
1 1 rA + rB 2a 2GM
+ =
= 2 = 2
rA rB
rA rB
b
h
h=b
GM
a
By Eq. (12.45),
τ=
2π ab 2π ab a 2π a3/ 2
=
=
h
b GM
GM
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,
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
PROPRIETARY MATERIAL. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
reproduced to
orteachers
distributed
any form
or by any
means, without
prior
writtencourse
permission
of the publisher,
used beyond
theManual,
limited
distribution
and in
educators
permitted
by McGraw-Hill
forthe
their
individual
preparation.
If you are aorstudent
using this
distribution
educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are usingtoitteachers
withoutand
permission.
you are using it without permission.
457
457
PROBLEM 12.122
A 1500-kg automobile is being driven down a 5° incline at a speed of 80 km/h when the brakes are applied,
causing a total braking force of 6 kN to be applied to the automobile. Determine the distance traveled by
the automobile before it comes to a stop.
SOLUTION
We have
where
ΣFx = ma : W sin 5° − ( FF + FR ) =
W
a
g
FF + FR = Fbrake
a = (9.81 m/s2) sin 5° −
Then
6000 N
1500 × 9.81 N
= −3.145 m/s2
For this uniformly decelerated motion, we have
v 2 = v02 + 2a( x − 0)
where
v0 = 80 km/h = 22.22 m/s
Then when v = 0,
0 = (22.22 m/s)2 + 2(−3.145 m/s2)x
x = 78.5 m
or
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
No part
part of
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced or
or distributed
distributed in
in any
any form
form or
reproduced
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
you are
are aa student
student using
using this
this Manual,
Manual,
distribution
If you
you are
are using
using it
it without
without permission.
permission.
you
458
458
PROBLEM 12.123
A 6-kg block B rests as shown on a 10-kg 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 mass of block C if block B is not to slide on bracket A.
(b) If the mass of block C is 10% larger than the answer found in
a, determine the accelerations of A, B and C.
SOLUTION
Kinematics. Let x A and 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
(1)
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:
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
PROPRIETARY MATERIAL. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
reproduced to
orteachers
distributed
any form
or by any
means, without
prior
writtencourse
permission
of the publisher,
used beyond
theManual,
limited
distribution
and in
educators
permitted
by McGraw-Hill
forthe
their
individual
preparation.
If you are aorstudent
using this
distribution
educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are usingtoitteachers
withoutand
permission.
you are using it without permission.
459
459
PROBLEM 12.123 (Continued)
Block B:
Block C:
Bracket A:
ΣFx = max : 2T − FAB = m A a A
(2)
Block B:
ΣFx = max : FAB − T = mB aB
(3)
+ ΣFy = ma y : N AB − m A g = 0
N AB = m A g
or
Block C:
ΣFy = ma y : mC g − T = maC
(4)
Adding Eqs. (2), (3), and (4), and transposing,
m A a A + mB aB + mC aC = mC g
(5)
Subtracting Eq. (4) from Eq. (3) and transposing,
mB aB − mC aC = FAB − mC g
(a)
No slip between A and B.
aB = a A
From Eq. (1),
a A = aB = aC = a
From Eq. (5),
a=
(6)
mC g
m A + mB + mC
FAB = µ s N AB = µ s mB g
For impending slip,
Substituting into Eq. (6),
( mB − mC )( mC g )
= µ s mB g − mC g
m A + mB + mC
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
All rights
rights reserved.
reserved. No
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
Inc. All
No part
part of
of this
reproduced or
or distributed
distributed in
in any
any form
form or
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced
are aa student
student using
using this
this Manual,
Manual,
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
individual course
distribution
for their
their individual
course preparation.
preparation. If
If you
you are
you are
are using
using it
it without
without permission.
you
permission.
460
460
PROBLEM 12.123 (Continued)
Solving for mC ,
µ s mB ( mA + mB )
mA + 2mB − µ s mC
mC =
=
(0.30)(6)(10 + 6)
10 + (2)(6) − (0.30)(6)
mC = 1.426 kg �
(b)
mC increased by 10%.
mC = 1.568 kg
Since slip is occurring,
FAB = µ k N AB = µ k mB g
Eq. (6) becomes
mB aB − mC aC = ( µ k mB − mC ) g
or
6aB − 1.568aC = [(0.25)(6) − 1.568](9.81)
(7)
With numerical data, Eq. (5) becomes
10a A + 6aB + 1.568aC = (1.568)(9.81)
(8)
Solving Eqs. (1), (7), and (8) gives
a A = 1.053 m/s 2 , aB = 0.348 m/s 2 , aC = 1.759 m/s 2
a A = 1.503 m/s 2
�
a B = 0.348 m/s 2
�
aC = 1.759 m/s 2 ��
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
PROPRIETARY MATERIAL. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
reproduced to
orteachers
distributed
any form
or by any
means, without
prior
writtencourse
permission
of the publisher,
used beyond
theManual,
limited
distribution
and in
educators
permitted
by McGraw-Hill
forthe
their
individual
preparation.
If you are aorstudent
using this
distribution
educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are usingtoitteachers
withoutand
permission.
you are using it without permission.
461
461
PROBLEM 12.124
Block A weighs 10 kg, and blocks B and C weigh 5 kg each. Knowing that
the blocks are initially at rest and that B moves 2.4 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.
Constraint of cord AD:
y A + yD = constant
v A + vD = 0,
Constraint of cord BC:
( yB − yD ) + ( yC − yD ) = constant
vB + vC − 2vD = 0,
Eliminate aD .
a A + aD = 0
aB + aC − 2aD = 0
2a A + aB + aC = 0
(1)
We have uniformly accelerated motion because all of the forces are
constant.
y B = ( y B ) 0 + ( vB ) 0 t +
1
a B t 2 , ( vB ) 0 = 0
2
2[ yB − ( yB )0 ]
(2)(2.4)
= 1.2 m/s2
(2)2
aB =
Pulley D:
t
2
=
ΣFy = 0: 2TBC − TAD = 0
TAD = 2TBC
Block A:
ΣFy = ma y : WA − TAD =
or
aA =
WA
aA
g
W − 2TBC
WA − TAD
g
g= A
WA
WA
(2)
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
All rights
rights reserved.
reserved. No
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
Inc. All
No part
part of
of this
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced or
or distributed
distributed in
in any
any form
form or
reproduced
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
individual course
are aa student
student using
using this
this Manual,
Manual,
distribution
for their
their individual
course preparation.
preparation. If
If you
you are
you are
are using
using it
it without
without permission.
you
permission.
462
462
PROBLEM 12.124 (Continued)
Block C:
ΣFy = ma y : WC − TBC =
or
aC =
WC
aC
g
WC − TBC
g
WC
(3)
Substituting the value for aB and Eqs. (2) and (3) into Eq. (1), and solving
for TBC ,
2
WA − 2TBC
W − TBC
g + aB + C
g =0
WA
WC
a
4
1
TBC = 3 + B
+
g
WA WC
4
1
+
10 5
Block B:
(a)
TBC
1.2
=3+
or
9.81
9.81
TBC = 51.05 N
ΣFy = ma y : P + WB − TBC =
WB
aB
g
Magnitude of P.
P = TBC − WB +
WB
a
g B
= 51.05 − (5)(9.81) + (5)(1.2)
(b)
P=8N
Tension in cord AD.
TAD = 2TBC = (2)(51.05)
TAD = 102 Ν
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
PROPRIETARY
MATERIAL.
2009orThe
Companies,
Inc. written
All rights
reserved.ofNo
of this or
Manual
may be the
displayed,
reproduced
or distributed
in any©form
by McGraw-Hill
any means, without
the prior
permission
thepart
publisher,
used beyond
limited
or by any
means, without
prior
writtencourse
permission
of the publisher,
used beyond
theManual,
limited
reproduced to
orteachers
distributed
any formpermitted
distribution
andin
educators
by McGraw-Hill
forthe
their
individual
preparation.
If you are aorstudent
using this
distribution
to itteachers
educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you
are using
withoutand
permission.
you are using it without permission.
463
463
PROBLEM 12.125
A 6-kg block B rests as shown on the upper surface of a 15-kg
wedge A. Neglecting friction, determine immediately after the
system is released from rest (a) the acceleration of A, (b) the
acceleration of B relative to A.
SOLUTION
Acceleration vectors:
a A = aA
30°, a B/A = aB/A
a B = a A + a B/ A
Block B:
ΣFx = max : mB aB/A − mB a A cos 30° = 0
aB/A = a A cos 30°
(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°
(10 + 6) sin 30°
g=
(9.81) = 6.82 m/s2
2
2
10 + 6 sin 30°
WA + WB sin 30°
a A = 6.8 m/s2
(a)
30°
aB/ A = (6.82) cos 30° = 5.91 m/s2
a B/A = 5.9 m/s2
(b)
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
No part
part of
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced or
or distributed
distributed in
in any
any form
form or
reproduced
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
you are
are aa student
student using
using this
this Manual,
Manual,
distribution
If you
you are
are using
using it
it without
without permission.
permission.
you
464
464
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 (µ k = 0.25).
Determine the initial deceleration of the car if the brakes are
applied as the car (a) has almost reached A, (b) is traveling
between A and B, (c) has just passed B.
SOLUTION
v = 72 km/h = 20 m/s
(a)
Almost reached Point A.
ρ = 30 m
v2
(20) 2
= 13.333 m/s 2 ↑
30
ρ
ΣFy = ma y : N R + N F − mg = man
an =
=
N R + N F = m( g + an )
F = µ k ( N R + N F ) = µ k m ( g + an )
ΣFx = max : − F = mat
at = −
F
= − µk ( g + an )
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 | = µk g = (0.25)(9.81)
(c)
Just passed Point B.
| at | = 2.45 m/s 2
ρ = 40 m
v2
(20) 2
= 8.8889 m/s 2 ↓
45
ρ
ΣFy = ma y : N R + N F − mg = − man
an =
=
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
PROPRIETARY
MATERIAL.
2009orThe
McGraw-Hill
Companies,
Inc.written
All rights
reserved.ofNo
of this or
Manual
may bethe
displayed,
reproduced
or distributed
in any©form
by any
means, without
the prior
permission
thepart
publisher,
used beyond
limited
reproduced to
orteachers
distributed
any form
or by any
means, without
prior
writtencourse
permission
of the publisher,
used beyond
theManual,
limited
distribution
and in
educators
permitted
by McGraw-Hill
forthe
their
individual
preparation.
If you are aor
student
using this
distribution
educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you
are usingtoitteachers
without and
permission.
you are using it without permission.
465
465
PROBLEM 12.126 (Continued)
or
N R + N F = m ( g − an )
F = µ k ( N R + N F ) = µ k m ( g − an )
ΣFx = max : − F = mat
at = −
F
= − µ k ( g − an )
m
| at | = µk ( g − an ) = (0.25)(9.81 − 8.8889)
| at | = 0.230 m/s 2 ��
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
All rights
rights reserved.
reserved. No
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
Inc. All
No part
part of
of this
reproduced or
or distributed
distributed in
in any
any form
form or
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced
are aa student
student using
using this
this Manual,
Manual,
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
individual course
distribution
for their
their individual
course preparation.
preparation. If
If you
you are
you are
are using
using it
it without
without permission.
you
permission.
466
466
PROBLEM 12.127
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
ΣFy = 0: F − WC = 0
or
F = mC g
Now
F = µs N
or
N=
1
µs
ΣFn = mC an : N = mC
or
or
1
µs
mC g = mC
µs =
mC g
vC2
r
vC2
r
gr (9.81 m/s 2 )(0.6 m)
=
(3.6 m/s) 2
vC2
( µ s ) min = 0.454 �
or
The direction of the impending motion is downward.
�
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
PROPRIETARY MATERIAL. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
reproduced to
orteachers
distributed
any form
or by any
means, without
prior
writtencourse
permission
of the publisher,
used beyond
theManual,
limited
distribution
and in
educators
permitted
by McGraw-Hill
forthe
their
individual
preparation.
If you are aorstudent
using this
distribution
educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are usingtoitteachers
withoutand
permission.
you are using it without permission.
467
467
PROBLEM 12.127 (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
ΣFn = mC an : N sin θ � F cos θ = mC
vC2
ρ
ρ = r sin θ
Substituting for N and F
mC g
v2
µs mC g
sin θ �
cos θ = mC C
r sin θ
cos θ ± µ s sin θ
cos θ ± µ s sin θ
vC2
µs
tan θ
=
�
1 ± µ s tan θ 1 ± µ s tan θ gr sin θ
or
v2
µs = ±
or
v2
C
1 + 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
Then
(b)
C
tan θ − gr sin
θ
µs = ±
tan θ − 2.2018 sin θ
1 + 2.2018sin θ tan θ
µs = ±
tan 75° − 2.2018sin 75°
= ± 0.1796
1 + 2.2018sin 75° tan 75°
θ = 75°
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
All rights
rights reserved.
reserved. No
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
Inc. All
No part
part of
of this
reproduced or
or distributed
distributed in
in any
any form
form or
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced
are aa student
student using
using this
this Manual,
Manual,
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
individual course
distribution
for their
their individual
course preparation.
preparation. If
If you
you are
you are
are using
using it
it without
without permission.
you
permission.
468
468
PROBLEM 12.127 (Continued)
Then
downward:
µs = + 0.1796
upward:
µs � 0
not possible
( µ s )min = 0.1796 �
The direction of the impending motion is downward.
(c)
�
θ = 45°
tan 45° − 2.2018sin 45°
= ± (− 0.218)
1 + 2.2018sin 45° tan 45°
µs = ±
Then
downward:
µs � 0
upward:
µs = 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°,
µs = 0. Thus, for this value of θ , friction is not necessary to prevent the collar from sliding on the rod.�
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
PROPRIETARY MATERIAL. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
reproduced to
orteachers
distributed
any form
or by any
means, without
prior
writtencourse
permission
of the publisher,
used beyond
theManual,
limited
distribution
and in
educators
permitted
by McGraw-Hill
forthe
their
individual
preparation.
If you are aorstudent
using this
distribution
educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are usingtoitteachers
withoutand
permission.
you are using it without permission.
469
469
PROBLEM 12.128
Pin B weighs 110 g and is free to slide in a horizontal plane along
the rotating arm OC and along the circular slot DE of radius
b = 500 mm. Neglecting friction and assuming that 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 drawing of the system, we have
r = 2b cos θ
Then
r = − (2b sin θ )θ
and
r = −2b(θ sin θ + θ 2 cos θ )
Now
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 )
= −2(0.5 m)[(250 rad/s 2 )sin 20° + 2(15 rad/s) 2 cos 20°]
= −508.4 m/s2
aθ = rθ + 2rθ = (2b cos θ )θ + 2(−2bθ sin θ )θ
and
= 2b(θ cos θ − 2θ 2 sin θ )
= 2(0.5 m)[(250 rad/s2) cos 20° − 2(15 rad/s2) sin 20°]
= 81 m/s2
Kinetics.
(a)
We have
Fr = mar = 0.11 kg × (−508.4 m/s2) = −55.924 N
Fr = −55.9 N
or
and
Fθ = maθ = 0.11 kg × (81 m/s2) = 8.91 N
Fθ = 8.9 N
or
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
part of
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
No part
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced or
or distributed
distributed in
in any
any form
form or
reproduced
distribution to
to teachers
teachers and
andeducators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
If you
you are
are aa student
student using
using this
thisManual,
Manual,
distribution
you are
are using
using itit without
without permission.
permission.
you
470
470
PROBLEM 12.128 (Continued)
(2.096714.0009sin20)lbP=°
ΣFr : −Fr = −Q cos 20°
(b)
or
Q=
1
(55.924 N)
cos 20°
= 59.51 N
ΣFθ : Fθ = P − Q sin 20°
or
P = (8.91 + 59.51 sin20°) N
= 29.26 N
F = 29.3 N
70°
Q = 59.5 N
40°
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
PROPRIETARY
MATERIAL.
2009orThe
McGraw-Hill
Companies,
Inc.written
All rights
reserved.ofNo
of this or
Manual
may bethe
displayed,
reproduced
or distributed
in any©form
by any
means, without
the prior
permission
thepart
publisher,
used beyond
limited
or by any
means, without
prior
writtencourse
permission
of the publisher,
used beyond
theManual,
limited
reproduced to
orteachers
distributed
any form
distribution
and in
educators
permitted
by McGraw-Hill
forthe
their
individual
preparation.
If you are aor
student
using this
distribution
educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you
are usingtoitteachers
without and
permission.
you are using it without permission.
471
471
PROBLEM 12.129
A particle of mass m is projected from Point A with an initial
velocity v0 perpendicular to OA and moves under a central force F
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. (12.27), express the radial and transverse components of the
velocity v of the particle as functions of the angle θ.
SOLUTION
We have
r=
r0
cos 2θ
Then
r� =
2r0 sin 2θ �
Now
v = r�er + rθ�eθ
so that at t = 0,
v0 = r0θ�0
From Eq. (12.27):
r 2θ� = r02θ�0 = r0 v0
θ
cos 2 2θ
2
or
� cos 2θ �
rv
v0
2
θ� = 0 20 = r0 v0 �
� = cos 2θ
r0
r
� r0 �
Then
r� =
Now
vr = r�
�
2r0 sin 2θ � v0
2
� cos 2θ � = 2v0 sin 2θ
2
cos 2θ � r0
�
vr = 2v0 sin 2θ ��
or
and
vθ = r θ� =
r0
v
= 0 cos 2 2θ
cos 2θ r0
vθ = v0 cos 2θ �
or
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
All rights
rights reserved.
reserved. No
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
Inc. All
No part
part of
of this
reproduced or
or distributed
distributed in
in any
any form
form or
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced
are aa student
student using
using this
this Manual,
Manual,
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
individual course
distribution
for their
their individual
course preparation.
preparation. If
If you
you are
you are
are using
using it
it without
without permission.
you
permission.
472
472
PROBLEM 12.130
Show that the radius r of the moon’s orbit can be determined from the radius R of the earth, the acceleration
of gravity g at the surface of the earth, and the time τ required for the moon to complete one full revolution
about the earth. Compute r knowing that τ = 27.3 days.
SOLUTION
Mm
r2
We have
F =G
and
F = Fn = man = m
Then
G
[Eq. (12.28)]
v2
r
Mm
v2
=
m
r
r2
GM
r
or
v2 =
Now
GM = gR 2
so that
v2 =
gR 2
g
or v = R
r
r
For one orbit,
τ=
2π r
2π r
=
v
R g
[Eq. (12.30)]
r
1/ 3
or
gτ 2 R 2
r=
4π 2
Now
τ = 27.3 days = 2.35872 × 106 s
Q.E.D.
R = 6370 km = 6.37 × 106 m
9.81 m/s 2 × (2.35872 × 106 s) 2 × (6.37 × 106 m) 2
r=
4π 2
1/ 3
= 382.81 × 106 m
r = 383 × 103 km
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
PROPRIETARY
MATERIAL.
2009orThe
McGraw-Hill
Companies,
Inc.written
All rights
reserved.ofNo
of this or
Manual
may bethe
displayed,
reproduced
or distributed
in any©form
by any
means, without
the prior
permission
thepart
publisher,
used beyond
limited
reproduced to
orteachers
distributed
any form
or by any
means, without
prior
writtencourse
permission
of the publisher,
used beyond
theManual,
limited
distribution
and in
educators
permitted
by McGraw-Hill
forthe
their
individual
preparation.
If you are aor
student
using this
distribution
educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you
are usingtoitteachers
without and
permission.
you are using it without permission.
473
473
PROBLEM 12.131*
Disk A rotates in a horizontal plane about a vertical axis at the
constant rate θ 0 = 12 rad/s. Slider B weighs 230 g 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 = 380 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 = 33 N/m, (b) k = 48 N/m.
SOLUTION
First we note
r = 0,
when
X sp = 0
Fsp = kr
r0 = 0.38 m
and
θ = θ0 = 12 rad/s
then
θ =0
ΣFr = mB ar : − Fsp = mB (r − rθ 02 )
r = (mB − θ02 ) r = 0
or
ΣFq = mB aq : FA = mB(0 + 2ṙ qׂ0)
(1)
(2)
k = 33 N/m
(a)
Substituting the given values into Eq. (1)
r+
33 N/m
− (12 rad/s)2 r = 0
0.23 kg
r =0
or
Then
dr
= r = 0 and at t = 0, r = 0 :
dt
r
0
or
0.1
dr =
0
(0) dt
r =0
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
No part
part of
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced or
or distributed
distributed in
in any
any form
form or
reproduced
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
you are
are aa student
student using
using this
this Manual,
Manual,
distribution
If you
you are
are using
using it
it without
without permission.
permission.
you
474
474
PROBLEM 12.131* (Continued)
dr
= r = 0 and at t = 0, r0 = 0.38 m
dt
and
r
dr =
r0
0.1
0
(0) dt
r = r0
or
r = 0.38 m
Note: r = 0 implies that the slider remains at its initial radial position.
With r = 0, Eq. (2) implies
FH = 0
k = 48 N/m
(b)
Substituting the given values into Eq. (1)
r+
48 N/m
− (12 rad/s)2 r = 0
0.23 kg
or
r + 64.7 r = 0
Now
r=
Then
r = vr
so that
d
(r ) r = vr
dt
dvr
dr
dvr
+ 64.7r = 0
dr
vr
vr
At t = 0, vr = 0, r = r0 :
0
r
vr d vr = −64.7 r dr
r0
or
vr2 = −64.7(r 2 − r02)
or
vr = 8.044 r02 − r 2
Now
vr =
At t = 0, r = r0 :
d dr d
d
=
= vr
dt dt dr
dr
r
dr
r0
r02 − r 2
=
dr
= 8.044 r02 − r 2
dt
t
0
8.044dt = 8.044t
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
PROPRIETARY
MATERIAL.
2009orThe
McGraw-Hill
Companies,
Inc.written
All rights
reserved.ofNo
of this or
Manual
may bethe
displayed,
reproduced
or distributed
in any©form
by any
means, without
the prior
permission
thepart
publisher,
used beyond
limited
or by any
means, without
prior
writtencourse
permission
of the publisher,
used beyond
theManual,
limited
reproduced to
orteachers
distributed
any form
distribution
and in
educators
permitted
by McGraw-Hill
forthe
their
individual
preparation.
If you are aor
student
using this
distribution
educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you
are usingtoitteachers
without and
permission.
you are using it without permission.
475
475
PROBLEM 12.131* (Continued)
r = r0 sin φ ,
Let
Then
dr = r0 cos φ dφ
sin −1 ( r/r0 )
r0 cos φ dφ
π /2
r02 − r02 sin 2 φ
sin −1 ( r/r0 )
or
π /2
sin −1
or
= 8.044t
dφ = 8.044t
π
r
− = 8.044t
r0
2
π
or
r = r0 sin 8.044t +
Then
r = −(3.057 m/s) sin8.044t
Finally,
2
= r0 cos 8.044t = (0.38 m)cos 8.044t
at t = 0.1 s:
r = (0.38 m)cos(8.044 × 0.1)
r = 0.26 m
or
Eq. (2)
FH = 0.23 kg × 2 × [−(3.057 m/s)sin(8.044 × 0.1)](12 rad/s)
FH = −12.2 N
or
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
No part
part of
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced or
or distributed
distributed in
in any
any form
form or
reproduced
distribution
If you
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
you are
are aa student
student using
using this
this Manual,
Manual,
you
you are
are using
using it
it without
without permission.
permission.
476
476
PROBLEM 12.132
It was observed that as the Voyager I spacecraft reached the point of its trajectory closest to the planet Saturn,
it was at a distance of 185 × 103 km from the center of the planet and had a velocity of 21.0 km/s. Knowing
that Tethys, one of Saturn’s moons, describes a circular orbit of radius 295 × 103 km at a speed of 11.35 km/s,
determine the eccentricity of the trajectory of Voyager I on its approach to Saturn.
SOLUTION
For a circular orbit,
Eq. (12.44)
v=
GM
r
For the orbit of Tethys,
GM = rT vT2
For Voyager’s trajectory, we have
1 GM
= 2 (1 + ε cos θ )
r
h
where h = r0 v0
At O,
r = r0 , θ = 0
Then
GM
1
=
(1 + ε )
r0 (r0 v0 ) 2
or
ε = 0 0 − 1 = 0 02 − 1
r v2
GM
=
r v2
rT vT
3
2
185 × 10 km � 21.0 km/s �
�
� −1
295 × 103 km � 11.35 km/s �
ε = 1.147 �
or
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
PROPRIETARY MATERIAL. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
reproduced to
orteachers
distributed
any form
or by any
means, without
prior
writtencourse
permission
of the publisher,
used beyond
theManual,
limited
distribution
and in
educators
permitted
by McGraw-Hill
forthe
their
individual
preparation.
If you are aorstudent
using this
distribution
educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are usingtoitteachers
withoutand
permission.
you are using it without permission.
477
477
PROBLEM 12.133
At engine burnout on a mission, a shuttle had reached Point A at an
altitude of 64 km above the surface of the earth and had a horizontal
velocity v0. Knowing that its first orbit was elliptic and that the shuttle
was transferred to a circular orbit as it passed through Point B at an
altitude of 270 km, determine (a) the time needed for the shuttle to
travel from A to B on its original elliptic orbit, (b) the periodic time of
the shuttle on its final circular orbit.
SOLUTION
For Earth,
R = 6370 km = 6.37 × 106 m, g = 9.81 m/s2
GM = gR2 = (9.81)(6.37 × 106 )2 = 0.39806 × 1015 m3/s2
(a)
For the elliptic orbit,
rA = 6370 + 64 = 6434 km = 6.434 × 106 m
rB = 6370 + 270 = 6640 km = 6.64 × 106 m
1
(rA + rB ) = 6.537 × 106 m
2
b = rA rB = 6.536 × 106 m
a=
Using Eq. 12.39,
1 GM
= 2 + C cos θ A
rA
h
and
1 GM
= 2 + C cos θ B
rB
rB
But θ B = θ A + 180°, so that cos θ A = − cos θ B
Adding,
1 1 rA + rB 2a 2GM
+ =
= 2 = 2
rA rB
rA rB
b
h
or
h=
Periodic time.
τ=
τ=
GMb 2
a
2π ab 2π ab a 2π a3/ 2
=
=
h
GM
GMb2
2p(6.537 × 106)3/2
0.39806 × 1015
= 5263.5 s = 1.4621 h
The time to travel from A to B is one half the periodic time
τ AB = 0.7311 h
τ AB = 44.0 min
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
The McGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
No part
part of
or by
by any
any means,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher, or
or used
used beyond
beyond the
the limited
limited
reproduced or
or distributed
distributed in
in any
any form
form or
reproduced
distribution to
to teachers
teachers and
and educators
educators permitted
permitted by
by McGraw-Hill
McGraw-Hill for
for their
their individual
individual course
course preparation.
preparation. If
you are
are aa student
student using
using this
this Manual,
Manual,
distribution
If you
you are
are using
using it
it without
without permission.
permission.
you
478
478
PROBLEM 12.133 (Continued)
(b)
For the circular orbit,
a = b = rB = 6.64 × 106 m
τ circ =
2π a3/2
2p(6.64 × 106)3/2
=
= 5388 s
GM
0.39806 × 1015
τ circ = 1.497 h
τ circ = 90 min
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
PROPRIETARY
MATERIAL.
2009orThe
McGraw-Hill
Companies,
Inc.written
All rights
reserved.ofNo
of this or
Manual
may bethe
displayed,
reproduced
or distributed
in any©form
by any
means, without
the prior
permission
thepart
publisher,
used beyond
limited
or by any
means, without
prior
writtencourse
permission
of the publisher,
used beyond
theManual,
limited
reproduced to
orteachers
distributed
any form
distribution
and in
educators
permitted
by McGraw-Hill
forthe
their
individual
preparation.
If you are aor
student
using this
distribution
educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you
are usingtoitteachers
without and
permission.
you are using it without permission.
479
479