978-0073380308 Chapter 3 Solution Manual Part 24

subject Type Homework Help
subject Pages 9
subject Words 3928
subject Authors Francesco Costanzo, Gary Gray, Michael Plesha

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page-pf1
636 Solutions Manual
Problem 3.142
Two particles
A
and
B
, with masses
mA
and
mB
, respectively, are a distance
r0
apart. Particle
B
is
fixed in space, and
A
is initially at rest. Using Eq. (1.5) on p. 3 and assuming that the diameters of the
masses are negligible, determine the time it takes for the two particles to come into contact if
mAD1kg
,
mBD2kg
, and
r0D1
m. Assume that the two masses are infinitely far from any other mass. Hint:
Rr0
0pr0r=.r0r/dr D1
2r3=2
0.
Solution
We model
A
as a particle subject only to the gravitational attraction due to its interaction with
B
.
The motion of Awill only occur along the line connecting Aand B. Particle Bdoes not move.
page-pf2
Dynamics 2e 637
of McGraw-Hill, and must be surrendered upon request of McGraw-Hill. Any duplication or distribution, either in print or electronic form, without the
permission of McGraw-Hill, is prohibited.
page-pf3
638 Solutions Manual
Problem 3.143
The system shown is initially at rest when the bent bar starts to rotate about
the vertical axis
AB
with constant angular acceleration
˛0D3rad=s2
. The
coefficient of static friction between the collar of mass
mD2kg
and the bent
bar is
sD0:6
, the angle of the bend in the bar is
D30ı
, and the collar is
initially at dD70 cm from the spin axis AB.
Assuming the motion starts at
tD0
, determine the time at which the
collar starts to slip relative to the bent bar.
Solution
The figure at the right shows a side and a top view of the collar’s FBD. Until the
collar slides, it moves along a circle with center
O
and radius
dcos
. We use
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Dynamics 2e 639
of McGraw-Hill, and must be surrendered upon request of McGraw-Hill. Any duplication or distribution, either in print or electronic form, without the
permission of McGraw-Hill, is prohibited.
page-pf5
640 Solutions Manual
Problem 3.144
The system shown is initially at rest when the bent bar starts to rotate about
the vertical axis
AB
with constant angular acceleration
˛0D3rad=s2
. The
coefficient of static friction between the collar of mass
mD2kg
and the bent
bar is
sD0:6
, the angle of the bend in the bar is
D30ı
, and the collar is
initially at dD70 cm from the spin axis AB.
Determine the number of rotations undergone by the bent bar when the
collar starts to slip relative to it.
Solution
of McGraw-Hill, and must be surrendered upon request of McGraw-Hill. Any duplication or distribution, either in print or electronic form, without the
permission of McGraw-Hill, is prohibited.
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Dynamics 2e 641
of McGraw-Hill, and must be surrendered upon request of McGraw-Hill. Any duplication or distribution, either in print or electronic form, without the
permission of McGraw-Hill, is prohibited.
page-pf7
642 Solutions Manual
Problem 3.145
The centers of two spheres
A
and
B
, with weights
WAD3lb
and
WBD7lb
,
respectively, are a distance
r0D5ft
apart when they are released from rest.
Using Eq. (1.5) on p. 3, determine the speed with which they collide if the
diameters of spheres
A
and
B
are
dAD2:5 in:
and
dBD4in:
, respectively.
Assume that the two masses are only influenced by their mutual gravitational
attraction.
Solution
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Dynamics 2e 643
of McGraw-Hill, and must be surrendered upon request of McGraw-Hill. Any duplication or distribution, either in print or electronic form, without the
permission of McGraw-Hill, is prohibited.
page-pf9
644 Solutions Manual
Problem 3.146
A roller coaster goes over the top
A
of the track shown with a speed
vD135 km=h
. If the radius of
curvature at
A
is
D60
m, determine the minimum force that a restraint must apply to a person with a
mass of 85 kg to keep the person on his or her seat.
Solution
of McGraw-Hill, and must be surrendered upon request of McGraw-Hill. Any duplication or distribution, either in print or electronic form, without the
permission of McGraw-Hill, is prohibited.
page-pfa
Dynamics 2e 645
Problem 3.147
A
50;000 lb
aircraft is flying along a rectilinear path at a constant altitude with a
speed
vD720 mph
when the pilot initiates a turn by banking the plane
20ı
to the
right. Assuming that the initial rate of change of speed is negligible, determine
the components of the acceleration of the aircraft right at the beginning of the
turn if the pilot does not adjust the attitude of the aircraft so that the magnitude of
the lift remains the same as when the plane is flying straight and the aerodynamic
drag remains in the horizontal plane. Also determine the radius of curvature at
the beginning of the turn.
3
3
Solution
3
3
First we determine the components of the acceleration of the aircraft using a convenient
component system. Then we will determine the radius of curvature of the trajectory at the
start of the maneuver. Referring to the figure at the right, we consider the FBD of the aircraft
at the start of the maneuver. We sketched a view from the rear and a view from the top in
which we have indicated a Cartesian component system with base vectors
O{
,
O|
,
O
k
, such that
O|
is oriented as the velocity of the plane at the instant shown, and
O
k
is oriented opposite
to gravity. The coordinate system corresponding to these base vectors is fixed in space and
has its origin coinciding with the aircraft at the instant shown. We model the aircraft as a
particle subject to its own weight
mg
, the thrust
T
, the drag force
D
, and the lift force
FL
.
The lift force is shown forming an angle
with the vertical direction, where
D20ı
is the
bank angle.
Balance Principles. Applying Newton’s second law, we have
XFxWFLsin Dmax;(1)
XFyWTDDmay;(2)
XF´WFLcos mg Dma´;(3)
where ax,ay, and a´are the desired acceleration components.
Force Laws.
As indicated in the problem statement, at the instant shown, the magnitudes of the lift and
drag forces are the same as right before the start of the turn. Since the aircraft was traveling horizontally at
constant velocity, we must have
FLDmg and DDT: (4)
Kinematic Equations.
No kinematic relations are needed since the components of acceleration are the
primary unknowns.
Computation. Substituting Eq. (4) into Eqs. (1)–(3),after simplification we have
axDgsin ; ayD0; a´Dg.cos 1/: (5)
Recalling that gD32:2 ft=s2and D20ı, we have
axD11:01 ft=s2;a
yD0; a´D1:942 ft=s2:(6)
This solutions manual, in any print or electronic form, remains the property of McGraw-Hill, Inc. It may be used and/or possessed only by permission
of McGraw-Hill, and must be surrendered upon request of McGraw-Hill. Any duplication or distribution, either in print or electronic form, without the
permission of McGraw-Hill, is prohibited.
July 17, 2012

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