Overton, Mays – Oldquiz 4 – Due: May 3 2005, 2:00 pm – Inst: Turner
1
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The due time is Central
time.
001
(part 1 of 1) 10 points
The system shown in the figure is in equilib
rium. A 17 kg mass is on the table. A string
attached to the knot and the ceiling makes an
angle of 68
◦
with the horizontal. The coeffi
cient of the static friction between the 17 kg
mass and the surface on which it rests is 0
.
41.
The acceleration of gravity is 9
.
8 m
/
s
2
.
17 kg
m
68
◦
What is the largest mass
m
can have and
still preserve the equilibrium?
Correct answer: 17
.
2514 kg.
Explanation:
Let :
M
= 17 kg
,
m
= 17
.
2514 kg
,
and
θ
= 68
◦
.
Basic Concepts:
The net force exerted
on an object in equilibrium equals zero.
Solution:
For the system to remain in equi
librium, the net forces on both
M
and
m
should be zero; and thus the tension in the
rope has an upper bound value
T
max
T
max
cos
θ
=
μ M g ,
so
(1)
T
max
=
μ M g
cos
θ
=
(0
.
41) (17 kg) (9
.
8 m
/
s
2
)
cos(68
◦
)
= 182
.
341 N
.
For
m
to remain in equilibrium
T
max
sin
θ
=
m
max
g ,
so
(2)
m
max
=
T
max
sin
θ
g
=
(182
.
341 N) sin(68
◦
)
(9
.
8 m
/
s
2
)
= 17
.
2514 kg
.
Alternative Solution:
Equations 1 and 2
come directly from the freebody diagram for
the knot. Dividing Eq. 2 by Eq. 1, we have
tan
θ
=
m
max
μ M
,
so
m
max
=
μ M
tan
θ
(3)
= (0
.
41) (17 kg) tan(68
◦
)
= 17
.
2514 kg
.
002
(part 1 of 1) 10 points
A uniform ladder of mass 20 kg and length
6 m leans against a smooth wall at an angle
72
◦
with respect to the ground. A person of
mass 88 kg climbs up the ladder to a point
twothirds of the way to the top.
Assume
that there is no friction between the top of
the ladder and the wall.
The acceleration of gravity is 9
.
8 m
/
s
2
.
Compute the minimum coefficient of static
friction between the ladder and the ground
that would prevent the ladder from slipping.
Correct answer: 0
.
206584 .
Explanation:
Consider the freebody diagram for the lad
der shown below.
f
W =mg
W =Mg
O
θ
1
2
x
y
w
f
f
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Overton, Mays – Oldquiz 4 – Due: May 3 2005, 2:00 pm – Inst: Turner
2
The three conditions of equilibrium are
0 =
X
F
x
=
f
x

f
w
,
0 =
X
F
y
=
f
y

Mg

mg,
0 =
X
τ
◦
=
f
w
×
L
sin
θ

Mg
×
1
2
L
cos
θ

mg
×
2
3
L
cos
θ,
where the torques are with respect to the
point where the ladder contacts the ground.
Solving these equations, we find
f
x
=
f
w
=
1
2
M g
+
2
3
m g
tan
θ
=
1
2
(20 kg) (9
.
8 m
/
s
2
)
tan(72
◦
)
+
2
3
(88 kg) (9
.
8 m
/
s
2
)
tan(72
◦
)
= 218
.
649 N
and
f
y
=
M g
+
m g
= (20 kg) (9
.
8 m
/
s
2
)
+ (88 kg) (9
.
8 m
/
s
2
)
= 1058
.
4 N
.
Since the friction force
f
x
cannot exceed
μ
s
times the normal force
f
y
, we must have
μ
s
≥
f
x
f
y
= 0
.
206584
.
003
(part 1 of 1) 10 points
When an object oscillating in simple harmonic
motion is at its maximum displacement from
the equilibrium position, which of the follow
ing is true of the values of its speed and the
magnitude of the restoring force?
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 Spring '08
 Turner
 Physics, Frequency, Wavelength, Correct Answer, Overton

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