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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.
Dynamics 2e 1577
Problem 7.84
A uniform thin rod is slightly nudged at
B
from the
D0
position so that it falls to the
right. The coefficient of static friction between the rod and the floor is s.
(a)
Determine as a function of
the normal force (
N
) and the frictional force (
F
)
exerted by the ground on the rod as the rod falls over.
(b)
Knowing that the rod will slip when
jF=N j
exceeds
s
, determine whether the rod
will slip as it falls.
(c)
Plot
F=.mg/
,
N=.mg/
, and
jF=N j
as a function of
for
0=2 rad
. Use
those plots to show that for smaller values of
s
, end
A
of the rod slips to the left,
and for larger values of s, it slips to the right.
Solution
Part (a)
The FBD of the rod as it falls is shown on the right.
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permission of McGraw-Hill, is prohibited.
Computation.
Substituting Eqs. (4) and (5) into the Newton-Euler equations and using
˛rD R
, we
obtain the followin system of three equations in the three unknowns F,N, and P
Part (b)
For a given value of
s
, the rod will slip if there is a value of
0ı< < 90ı
such that
jF=N j> s
. This
exists if there is a value of , say Q
, such that
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permission of McGraw-Hill, is prohibited.
Dynamics 2e 1579
0.5
1.0
1.5
Θ
"1.5
"1.0
"0.5
F!"mg#
0.0
0.5
1.0
1.5
Θ
0.2
0.4
0.6
0.8
1.0
N!"mg#
0.0
0.5
1.0
1.5
Θ
0.5
1.0
1.5
2.0
!F"N!
The above plots were obtained using the following Mathematica code:
FF !
3
4
g m !"2#3 Cos"Θ#$Sin"Θ#;NN !
1
4
g m !1"3 Cos"Θ#$2;
Plot%FF
m g
,&Θ, 0, 90 Degree', AxesLabel %&"Θ", "F(!mg$"')
Plot%NN
m g
,&Θ, 0, 90 Degree', AxesLabel %&"Θ", "N(!mg$"')
Plot%Abs%FF
NN ),&Θ, 0, 90 Degree', PlotRange %&Automatic, &0, 2'', AxesLabel %&"Θ", "*F(N*"')
Referring to the plot of
jF=N j
, the value of
jF=N j
for a given value of
can be interpreted as the value that
s
would have to have for slip to occur at the angle
. With this in mind, observe that there is a value of
near
0:65 rad
for which the function
jF=N j
achieves a local maximum. Calling
cr
the value in question, and
referring to the plot of
F
, we see that if the given value of
s< cr
, then slip would occur while
F > 0
and
point
A
on the bar would move to the left. However, if the given value of
s> cr
, then slip would occur
only for value of Fthat are negative thus implying that Awould move to the right.
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.
Problem 7.85
The roadster weighs
2570 lb
, and its mass is evenly distributed
between its front and rear wheels. It can accelerate from 0 to
60 mph
in
6:98
s. The rear wheel, shown in the blowup above the
roadster, weighs
47 lb
, its mass center is at its geometric center,
and its mass moment of inertia
IB
is
0:989 slugft2
. With this in
mind, we want to determine the forces on the rear wheel shown in
Fig. 2 of Example 7.9.
(a)
Assuming that its acceleration is uniform, determine the
forces on the front and rear wheels due to the pavement.
(b)
Now that you have the normal and friction forces between
the rear wheels and the pavement, isolate one of the rear
wheels and determine the forces and moments exerted by
the axle on that rear wheel.
Assume that the mass is evenly distributed between the right and
left sides of the car and that friction is sufficient to prevent slipping
of the wheels.
Solution
Part (a)
Force Laws. All forces have been accounted for on the FBD.
Kinematic Equations. Since the car is translating in the horizontal direction, we have
f
60 ft=s
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permission of McGraw-Hill, is prohibited.
Dynamics 2e 1581
Computation.
Substituting the kinematic equations into the Newton-Euler equations, we obtain the follow-
ing system of three equations in the three unknowns Nr,N
f, and Fr
f
`
`
Part (b)
The FBD of one of the rear wheels is shown on the right, where
Ax
,
Ay
, and
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permission of McGraw-Hill, is prohibited.
Computation.
Substituting the kinematic equations into the Newton-Euler equations yields the following
gt
f
2FrMAD 2IBv
dt
f
the solutions of which is
f
f.Wc2Ww/
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permission of McGraw-Hill, is prohibited.
Dynamics 2e 1583
Problem 7.86
A spool of mass
m
, radius
r
, and radius of gyration
kG
rolls without slipping
on the incline, whose angle with respect to the horizontal is
. A linear elastic
spring with constant
k
and unstretched length
L0
connects the center of the
spool to a fixed wall. Determine the equation(s) of motion of the spool, using
the xcoordinate shown.
mg sin k.x L0/mk2
r2RxDmRx;
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.
which can be rearranged to obtain the following equation of motion
G
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permission of McGraw-Hill, is prohibited.
Dynamics 2e 1585
Problem 7.87
A spool of mass
mD200 kg
, radius
rD0:8
m, and radius of gyration
kGD
0:65
m rolls without slipping on the incline, whose angle with respect to the
horizontal is
D38ı
. A linear elastic spring with constant
kD500 N=m
and unstretched length
L0D1:5
m connects the center of the spool to a fixed
wall. Determine the equation(s) of motion of the spool, using the
x
coordinate
shown; solve them for
15
s, using the initial conditions
x.0/ D2:5
m and
Px.0/ D0m=s
; and then plot
x
versus
t
. What is the approximate period of
oscillation of the spool?
mg sin k.x L0/mk2
r2RxDmRx;
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