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Problem 9.15
Repeat Prob. 9.14 if the reservoir is on the left-hand side of the dam.
Solution
The FBD for the dam is shown at the right. The water pressure at the bottom of the
reservoir (i.e., point A) is
m3/1
Sliding analysis: Using the FBD shown above, the equilibrium equations are
Since the value of
s
given in the problem statement exceeds
.s/min
(i.e., since
0:8 > 0:5232
), the dam will
not slide.
Tipping analysis: Using the FBD shown above,
XMAD0WN ` Wc1
3.6 m/P1
3.8 m/D0(7)
Problem 9.16
The owner of a small concrete gravity dam is considering attaching steel plate to the
face of the dam so that a greater depth of water can be retained. The surface between
the dam and earth has coefficients of friction
sD0:6
and
kD0:55
. Determine
if the dam is safe from both overturning (tipping) and sliding along its base. The
specific weights of concrete and water are
cD150 lb=ft3
and
wD62:4 lb=ft3
,
respectively. Neglect the weight of the steel plate.
Solution
The FBD for the dam is shown at the right. The water pressure at the bottom of
For a
1ft
thick section of dam (into the plane of the figure), the resultant force
P
2.1 ft/.2 ft/.1 ft/D150 lb:(3)
Sliding analysis: Using the FBD shown above, the equilibrium equations are
XFyD0WNW1W2D0)ND450 lb;(4)
Note that the dam may also fail by tipping, thus we carry out the following analysis.
Tipping analysis: Using the FBD shown above,
XMAD0W P .1 ft/CW1.1:5 ft/CW22
3.1 ft/N ` D0(7)
Problem 9.17
A wedge is used to level a structure. All contact surfaces have coefficients
of static and kinetic friction of 0.3 and 0.25, respectively, and
WD500
N.
Assume the dimensions of the wedge are small. Determine the value of
P
to
cause impending motion of the wedge:
(a) To the left.
(b) To the right.
Problem 9.18
Blocks
A
and
B
each have
2kg
mass. All contact surfaces have the same coefficient
of friction. Determine the force Pneeded to cause impending motion of block Bto
the left if the coefficient of static friction is
0.4.
Problem 9.19
Blocks
A
and
B
each have
2kg
mass. All contact surfaces have the same coefficient
of friction. Determine the force Pneeded to cause impending motion of block Bto
the left if the coefficient of static friction is
0.3.
Problem 9.20
A roll of paper weighs
5
N with center of gravity at point
A
. The roll is
supported by a steel bar
AB
that has negligible weight, and the roll rests against
a vertical wall with equal coefficients of static and kinetic friction of 0.5. If the
paper tears when angle
reaches
20ı
, determine the strength of the sheet of
paper.
Problem 9.21
In Prob. 9.20, let the strength of the paper be
4
N. Determine the largest value
of angle
so that paper may be pulled off the roll without tearing. Hint: The
equilibrium equations are difficult to solve analytically. Thus, a solution using
software such as Mathematica or Maple is helpful, or an approximate solution
(i.e., graphical or by trial and error) to obtain
with two-digit accuracy is also
acceptable.
(6) may be rearranged to obtain, respectively,
Dcos11
4NhTAB
60
100 8Ni;(14)
80
Problem 9.22
The bicycle shown has a brake for each wheel. The rider and bicycle
weigh
120 lb
with center of gravity at point
C
, and the bicycle descends a
straight slope with
D15ı
. For each of the braking situations given below,
determine the minimum coefficient of kinetic friction between the tires and
pavement so that the bicycle will descend the slope at a uniform speed, and
determine if the bicycle is safe from tipping.
(a)
Only the rear brake is applied, causing the rear wheel to skid on the
pavement.
(b)
Only the front brake is applied, causing the front wheel to skid on the
pavement.
(c)
Both the rear and front brakes are applied, causing both wheels to
skid on the pavement.
Problem 9.23
The machine shown is used to move boxes. Bar
ABC
slides horizontally in the
bearing of the fixed machine housing. Points
B
,
C
, and
D
are pins, and point
C
has a frictionless roller. The flywheel
E
rotates clockwise under the action of
moment
ME
. The horizontal surface on which the box rests has coefficients of
friction
sD0:3
and
kD0:25
, and all other contact surfaces are frictionless.
If the box weighs
900
N, determine the moment
ME
that must be applied to
the flywheel to initiate motion of the box, and determine if the motion is sliding
or tipping.
Problem 9.24
An
8ft
long ladder has seven rungs. The rungs are spaced
1ft
apart, and the top and
bottom rungs are
1ft
from their respective ends of the ladder. The top of the ladder
has a roller. Neglect the weight of the ladder and assume the worker’s hand applies no
force to the ladder.
(a)
If the worker weighs
140 lb
and stands on the middle rung, determine the mini-
mum value of the coefficient of friction so that the ladder does not slide.
(b) If the worker weighs 140 lb and stands on a different rung, does your answer to
Part (a) change? Explain.
(c)
If the worker weighs more than
140 lb
and stands on the middle rung, does your
answer to Part (a) change? Explain.
Problem 9.25
In Prob. 9.24, the roller at
B
is removed and the surfaces at
A
and
B
both have the
same coefficient of friction. If the worker weighs
140 lb
and stands on the middle rung,
determine the minimum value of the coefficient of friction so that the ladder does not
slide. Hint: The use of mathematical software is helpful, but is not required.
2.tan 60ıC/ and F2DW
2.tan 60ıC/ :(8)
2.tan 60ıC/ and F1DW
2.tan 60ıC/ :(9)
1364 Solutions Manual
2.tan 60ıC/ i:(10)
The expressions for F2given in Eqs. (8) and (10) must be equal, thus
Wcancels in the above equation, and we multiply both sides by 2.tan 60ıC/ to obtain
Problem 9.26
The mechanism for a gumball machine is shown in Figs. P9.26 and P9.27. To
begin dispensing a gumball, as shown in Fig. P9.26, a horizontal force
P1
is
applied to block
A
to cause it to begin sliding to the right while member
CD
simultaneously begins to lift door
B
. After the door
B
is fully open, as shown in
Fig. P9.27, the gumball drops out of the machine. Blocks
A
and
B
weigh
0:3 lb
and
0:2 lb
, respectively, and the weights of member
CD
and the gumballs may
be neglected. The pins at points
C
and
D
are frictionless, blocks
A
and
B
are a
loose fit in their tracks, and all contact surfaces have a coefficient of friction of
0.3. The motion of the blocks is slow enough so that static equilibrium may be
assumed.
Determine the value of
P1
that will cause block
A
to begin moving to the
right.
Problem 9.27
The mechanism for a gumball machine is shown in Figs. P9.26 and P9.27. To
begin dispensing a gumball, as shown in Fig. P9.26, a horizontal force
P1
is
0.3. The motion of the blocks is slow enough so that static equilibrium may be
assumed.
Just before the door
B
is fully opened, as shown in Fig. P9.27, determine
the value of P2.
Problem 9.28
The mechanism for a gumball machine is shown in Figs. P9.26 and P9.27. To begin dispensing a gumball,
as shown in Fig. P9.26, a horizontal force
P1
is applied to block
A
to cause it to begin sliding to the right
while member
CD
simultaneously begins to lift door
B
. After the door
B
is fully open, as shown in
Fig. P9.27, the gumball drops out of the machine. Blocks
A
and
B
weigh
0:3 lb
and
0:2 lb
, respectively,
and the weights of member
CD
and the gumballs may be neglected. The pins at points
C
and
D
are
frictionless, blocks
A
and
B
are a loose fit in their tracks, and all contact surfaces have a coefficient of
friction of 0.3. The motion of the blocks is slow enough so that static equilibrium may be assumed.
In Fig. P9.27, where the door is fully open, replace the force
P2
with a force
Q
in the opposite direction.
Determine the value of Qso that block Awill begin moving to the left.
Problem 9.29
The tool shown is used by a person to carry a large sheet of plywood. The
tool consists of aluminum clamps
A
and
B
, and steel member
CDE
, with
frictionless pins at points
C
and
D
. The weights of all members are negligible,
except for the weight of the plywood. If the coefficient of friction between
aluminum and plywood is 0.2, determine the largest value of
h
so that the
plywood does not slip out of the tool. Assume that the person rests the plywood
on their hip so that the sheet of plywood is vertical.
h;(12)
and then Eq. (11) becomes
0:2 W 8in.
h0:2 W 8in.
hCWD0: (14)
Canceling Wand solving for hprovides
