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Kinematic Equations. Relating the acceleration of Dto that of A, we have
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 1657
and
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 7.121
The uniform slender bar
AB
has mass
mAB
and length
L
. The
crate has a uniformly distributed mass
mC
and dimensions
h
and
w
. Bar
AB
is pin-connected to the trolley at
A
and to the crate at
B
. The trolley is constrained to move along the horizontal guide
shown. Point
O
on the trolley’s guide is a fixed reference point.
Neglect the mass of the trolley and friction.
Let
mAB D75 kg
,
LD4:5
m,
mCD250 kg
,
dD0:5
m,
hD1:5
m, and
wD2
m. Finally, assume that the system is
released from rest when
xAD0
,
D30ı
, and
D45ı
. Plot
xA
,
, and as functions of time for 0t15 s.
Solution
The FBDs of the arm and crate are shown on the right.
Balance Principles.
Based on the FBD of the bar, its Newton-Euler equations
Force Laws. All forces have been accounted for on the FBD.
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permission of McGraw-Hill, is prohibited.
Dynamics 2e 1659
Kinematic Equations. Relating the acceleration of Dto that of A, we have
2R
2P
„ ƒ‚ …
aDx
2R
2P
„ ƒ‚ …
aDy
where we have used the facts that
!AB D P
,
˛AB D R
,
aAx D RxA
, and
aAy D0
. To determine the
2R
2P
„ ƒ‚ …
aEy
where !CD P
and ˛CD R
are the angular velocity and angular acceleration of the crate, respectively.
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permission of McGraw-Hill, is prohibited.
and
though, we need the initial conditions, which are
4(2d$h)L mC(2 Sin&Θ&t'' Cos&Φ&t'' #Sin&Φ&t'' Cos&Θ&t'') $Φ'&t'%2$
1
2$mAB $2mC%gLSin&Θ&t'' '0,
#
1
2(2d$h)mCCos&Φ&t'' xA''&t'$
1
2(2d$h)L mCCos&Θ&t'#Φ&t'' Θ''&t'$
1
12 *h2$w2+$
1
4(2d$h)2mCΦ''&t'#
1
2(2d$h)L mC(Sin&Θ&t'' Cos&Φ&t'' #Sin&Φ&t'' Cos&Θ&t'') $Θ'&t'%2$
1
2
g(2d$h)mCSin&Φ&t'' '0,;
Motion !
NDSolve&-EOMs,xA&0''0, xA'&0''0, Θ&0''30. Degree, Θ'&0''0, Φ&0''45.0 Degree,
Φ'&0''0. /.Parameters,-xA,Θ,Φ.,-t, 0, 15.'
Plot&xA&t' /.Motion&&1'',-t, 0, 15., AxesLabel "-"t", "xA".'
Plot&Θ&t' /.Motion&&1'',-t, 0, 15., AxesLabel "-"t", "Θ".'
Plot&Φ&t' /.Motion&&1'',-t, 0, 15., AxesLabel "-"t", "Φ".'
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 6, 2012
2
4
6
8
10
12
14
t
!5
!4
!3
!2
!1
xA
2
4
6
8
10
12
14
t
!0.4
!0.2
0.2
0.4
Θ
2
4
6
8
10
12
14
t
!1.0
!0.5
0.5
1.0
Φ
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permission of McGraw-Hill, is prohibited.
Problem 7.122
A drum of mass
md
, radius
R
, radius of gyration
kG
, and with
center of mass at
G
is placed on a cart of mass
mc
for transport. The
system is initially at rest, and the cart is pushed to the right with
the force
P
. The coefficient of static friction between the cart and
the drum is
s
. Neglecting the mass of the wheels, determine the
maximum force
P
that can be applied to the cart so that the drum
does not slip on the cart, and find the corresponding acceleration of
the cart and of point G.
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.
k2
G!; ˛dDsRg
k2
G
k2
G!#:
The result obtained for
P
corresponds to the required maximum value of
P
since we have solved the problem
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permission of McGraw-Hill, is prohibited.
Problem 7.123
The uniform bar
AB
of mass
m
and length
L
is leaning against the
corner with
0
. A small box is placed on top of the bar at
A
.
End
B
of the bar is given a slight nudge so that end
A
starts sliding
down the wall as
B
slides along the floor. Assuming that friction is
negligible between the bar and the two surfaces against which it is
sliding, and neglecting the weight of the box, determine the angle
at
which the box will lose contact with the bar.
2sin NA
2cos DIG˛AB ;(3)
where ˛AB is the angular acceleration of the bar and its mass moment of inertia is given by
IGD1
12 mL2:(4)
Force Laws. All forces have been accounted for on the FBD.
Kinematic Equations.
The kinematic constraints on the bar are that end
A
only moves vertically and that
end Bonly moves horizontally. These are embodied in the relation
!2
AB
L
2.sin O{Ccos O|/:
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permission of McGraw-Hill, is prohibited.
the solution of which is
The value of
at which the box separates from the bar is found by finding the value of
for which the
which means that we still need to find
!AB
as a function of
. We can do that by integrating the
˛AB
using
where we have used the fact that the system starts from rest at
D0
. Substituting this expression for
!2
AB
D3g 3
2cos2cos 1
2(12)
31p2and cos 2D1
31Cp2;(13)
which correspond to the following values of the angle :
It can be verified that for
D36:4ı
, the bar is still in contact with the vertical wall (see solution of
Problem 7.88).
of McGraw-Hill, and must be surrendered upon request of McGraw-Hill. Any duplication or distribution, either in print or electronic form, without the
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