978-0073380308 Chapter 2 Solution Manual Part 5

subject Type Homework Help
subject Pages 9
subject Words 2275
subject Authors Francesco Costanzo, Gary Gray, Michael Plesha

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70 Solutions Manual
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-pf2
Dynamics 2e 71
Problem 2.35
In the mechanism shown, block Bis fixed and has a profile described by the following relation:
yDh"1C1
2x
d2
1
4x
d4#:
The follower moves with the shuttle A, and the tip Cof the follower remains in contact with B.
Assume that
hD2mm
,
dD20 mm
, and
A
is made to move from
xDd
to
xDd
with a constant
speed v0D0:1 m=s. Determine the acceleration of Cfor xD15 mm. Express your answer in m=s2.
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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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 73
Problem 2.36
The Center for Gravitational Biology Research at NASAs Ames Research Center runs a large centrifuge
capable of
20g
of acceleration, where
g
is the acceleration due to gravity (
12:5g
is the maximum for
human subjects). The distance from the axis of rotation to the cab at either
A
or
B
is
RD25 ft
. The
trajectory of
A
is described by
yADqR2x2
A
for
yA0
and by
yAD
qR2x2
A
for
yA<0
. If
A
moves at the constant speed
vAD120 ft=s
, determine the velocity and acceleration of
A
when
xAD20 ft
and yA>0.
Photo credit: NASA
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-pf5
74 Solutions Manual
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-pf6
Dynamics 2e 75
Problem 2.37
Point
C
is a point on the connecting rod of a mechanism called a slider-
crank. The
x
and
y
coordinates of
C
can be expressed as follows:
xCDRcos C1
2pL2R2sin2
and
yCD.R=2/ sin
, where
describes
the position of the crank. The crank rotates at a constant rate such that
D!t
,
where tis time.
Find expressions for the velocity, speed, and acceleration of
C
as functions
of the angle and the parameters, R,L, and !.
Solution
Using the coordinate system and expressions given in the problem statement, the position of point
C
can be
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
Problem 2.38
Point
C
is a point on the connecting rod of a mechanism called a slider-
crank. The
x
and
y
coordinates of
C
can be expressed as follows:
xCDRcos C1
2pL2R2sin2
and
yCD.R=2/ sin
, where
describes
the position of the crank. The crank rotates at a constant rate such that
D!t
,
where tis time.
Let
t
be expressed in seconds,
RD0:1
m,
LD0:25
m, and
!D250 rad=s
.
Plot the trajectory of point
C
for
0t0:025
s. For the same interval of time,
plot the speed as a function of time, as well as the components of the velocity
and acceleration of C.
Solution
The velocity of point
C
is the time derivative of the position of
C
. Using the coordinate system shown and
since D!t, can be written as
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-pf8
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
78 Solutions Manual
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 79
Problem 2.39
The following four problems refer to a car traveling between two
stop signs, in which the car’s velocity is assumed to be given by
v.t/ DŒ9 9cos.2t =5/çm=s for 0t5s.
Determine
vmax
, the maximum velocity reached by the car.
Furthermore, determine the position
svmax
and the time
tvmax
at
which vmax occurs.
STOPSTOP
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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