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Physics 40A  Final Exam Solutions  Fall 2005
Problem 1
[25 points]
A box of mass
m
1
= 10 kg rests on a surface inclined at 30° to the horizontal. It is connected by a
massless cord, which passes over a massless and frictionless pulley, to a second box of mass
m
2
,
as shown in the figure. (a) If the coefficient of static friction is
μ
s
= 0.4, what is the maximum
value that
m
2
can have if the boxes are to remain at rest? (b) If the coefficient of kinetic friction
is
k
= 0.3, and
m
2
= 10 kg, determine the acceleration of the system.
Solution:
(a) We apply Newton's second law to
m
2
T
m
2
g
=
m
2
0
⇒
T
=
m
2
g
1
If
m
2
is very large
m
1
would tend to move up the plane so the frictional force is directed down the
plane. Thus, Newton's second law applied to
m
1
gives
N
m
1
g
cos
=
m
1
0
N
=
m
1
g
cos
(perpendicular to plane)
T
m
1
g
sin
f
s
=
m
1
0
(up the plane)
Combining with Eq. (1) gives
m
2
g
=
m
1
g
sin
f
s
Since the maximum value of
f
s
is
f
s,
max
=
s
N
s
m
1
g
cos
, the maximum value of
m
2
is
given by
m
2
g
=
m
1
g
sin
s
m
1
g
cos
Therefore,
m
2
=
m
1
[
sin
s
cos
]
⇒
m
2
=
10 kg
[
sin 30
°
0.4
cos30
°
]
=
8.46 kg
(b) Since
m
2
is now greater than 8.46 kg,
m
1
will move up the plane. Newton's second law applied
to
m
1
and
m
2
gives
m
1
a
=
T
m
1
g
sin
f
k
m
2
a
=
T
m
2
g
⇒
T
=
m
2
g
a
Setting
f
k
k
N
k
m
1
g
cos
we get
m
1
a
=
m
2
g
a
m
1
g
sin
k
m
1
g
cos
Solving for
a
gives
a
=
m
2
g
m
1
g
sin
k
m
1
g
cos
m
1
m
2
a
=
9.8 m/s
2
10 kg
1
sin 30
°
0.3cos30
°
20 kg
=
1.18 m/s
2
1
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View Full Document Physics 40A  Final Exam Solutions  Fall 2005
Problem 2
[20 points]
A child stands 1.2m from the center (axis of rotation) of a playground merrygoround. When the
merrygoround rotates at a rate just exceeding 0.345 rev/s, the child slips off. What is the
coefficient of static friction between the child and the merrygoround?
Solution:
In the radial direction, outward from the center of the merrygoround, the only force acting is the
frictional force. It acts toward the center of the merrygoround. Newton's second law applied to
the child gives
F
r
=
ma
r
⇒
f
s
=
m
v
2
r
where
v
is the velocity of the child and
r
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This note was uploaded on 04/16/2008 for the course PHYS 40 taught by Professor Ellison during the Winter '08 term at UC Riverside.
 Winter '08
 Ellison
 Friction, Mass

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