Now that we have determined
.Fm/to
, we can add lift to our model so that the Newton-Euler equations
Solving, we obtain
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Dynamics 2e 1447
Problem 7.17
A file cabinet weighing
230 lb
is being pushed to the right with a horizontal
force
P
applied a distance
h
from the floor. The width of the file cabinet is
wD15 in:
, its mass center
G
is a distance
dD2ft
above the floor, and
static friction is insufficient to prevent slipping between the cabinet and
the floor.
If
PD70 lb
and the coefficient of kinetic friction between the cabinet
and the floor is
kD0:28
, determine the maximum height
h
at which the
cabinet can be pushed so that it does not tip over, and find the corresponding
acceleration of the cabinet.
Solution
An FBD of the cabinet is shown on the right, where
F
is the friction force between
the cabinet and the floor. Since we want to find the maximum height
h
for no
tipping (i.e., tip is impending), we have placed the normal force at the corner of
the cabinet.
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which are three equations for the unknowns N,aGx , and h. Solving, we obtain
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Dynamics 2e 1449
Problem 7.18
A file cabinet weighing
230 lb
is being pushed to the right with a horizontal
force
P
applied a distance
h
from the floor. The width of the file cabinet is
wD15 in:
, its mass center
G
is a distance
dD2ft
above the floor, and
static friction is insufficient to prevent slipping between the cabinet and
the floor.
If the coefficient of kinetic friction between the file cabinet and the floor
is
kD0:15
and the horizontal force
P
is applied at a height
hD42 in:
,
determine the maximum value of
P
that can be applied so that the cabinet
does not tip over, and find the corresponding acceleration of the cabinet.
Solution
An FBD of the cabinet is shown on the right, where
F
is the friction force between
the cabinet and the floor. Since we want to find the maximum force
P
for no
tipping (i.e., tip is impending), we have placed the normal force at the corner of
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permission of McGraw-Hill, is prohibited.
which are three equations for the unknowns N,aGx , and P. Solving, we obtain
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permission of McGraw-Hill, is prohibited.
Dynamics 2e 1451
Problem 7.19
A conveyor belt must accelerate the cans from rest to
vD18 ft=s
as quickly as possible. Treating each can as a uniform circular
cylinder weighing
1:1 lb
, find the minimum possible time to reach
v
so that the cans do not tip or slip on the conveyer. Assume that
the acceleration is uniform, and use
wD2:71 in:
,
hD5in:
, and
sD0:5.
Solution
The FBD shown at the right assumes that each can does not tip. We will first find
the solution under the assumption that the cans do not tip, but that slip is impending.
We will then verify whether or not that assumption can be satisfied with the given
physical parameters. If our assumption is invalid, then we will solve the problem with
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permission of McGraw-Hill, is prohibited.
Problem 7.20
The uniform slender bar
AB
, with mass
mAB D75 kg
and length
LD4:5
m, is pin-connected at
A
to a trolley accelerating with
a0D3m=s2
along a horizontal rail. A crate with uniformly dis-
tributed mass
mCD250 kg
, height
hD1:5
m, and width
wD2
m
is contained in a cage with negligible mass that is pin-connected
to
AB
at
B
. The distance between
B
and the top of the crate is
dD0:75
m. Determine the angles
and
so that the bar and the
crate translate with the trolley.
Solution
Splitting up the bar
AB
and the crate
C
, their FBDs are shown at the right.
Note that the mass center
D
of the bar
AB
is at the geometric center of the
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Dynamics 2e 1453
Computation.
Substituting the kinematic equations, Eqs. (7) and (8), into the Newton-Euler equations,
Eqs. (1)–(6), yields the following system of six equations in the six unknowns Ax,Ay,Bx,By,, and :
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permission of McGraw-Hill, is prohibited.
Problem 7.21
The system shown lies in the vertical plane. The trolley
A
is
moving to the right with a constant acceleration aA. Attached to
the trolley is a rope
AB
of negligible mass. Attached to the end of
the rope is a thin uniform bar
BC
of length
L
and mass
m
. When
the trolley
A
is accelerating at constant
aA
, the angles
and
are both constant. Determine these two constant angles in this
steady state as functions of one or more of the given quantities
(i.e., L,m, and aA). Note that cos.
2x/ Dsin x.
Solution
The FBD of the bar
BC
is shown on the right, where
TAB
is the tension in the
rope AB.
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Dynamics 2e 1455
where we have used the facts that
L¤0
and
TAB ¤0
. Now, solving the first Newton-Euler equation for
TAB and substituting it into the second, we obtain
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