Adams, Russell – Homework 12 – Due: Dec 1 2006, 11:00 pm – Inst: Opyrchal, H
1
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printout
should
have
10
questions.
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before answering.
The due time is Central
time.
001
(part 1 of 1) 10 points
A solid sphere of radius
R
and mass
m
is
held against a wall by a string being pulled at
an angle
θ
.
W
P
F
θ
R
Determine the torque equation about the
point P.
1.
m g R
=
F
(1

sin
θ
)
R
2.
m g R
=
F
(1 + sin
θ
)
R
correct
3.
m g R
=
F
(1

cos
θ
)
R
4.
m g R
=
F
(1 + cos
θ
)
R
5.
m g R
=
F R
6.
m g R
= 2
F R
Explanation:
W
P
F
θ
R
R
(1 + sin
θ
)
A
θ
The clockwise torque about point P is
τ
cw
=
AP
×
F ,
where
AP
=
R
(1 + sin
θ
) and the counter
clockwise torque is
τ
ccw
=
m g R .
From equilibrium, we obtain the equation
τ
ccw
=
τ
cw
=
⇒
m g R
=
R
(1 + sin
θ
)
F
,
which is our answer.
keywords:
002
(part 1 of 1) 10 points
The system shown in the figure is in equilib
rium. A 15 kg mass is on the table. A string
attached to the knot and the ceiling makes an
angle of 45
◦
with the horizontal. The coeffi
cient of the static friction between the 15 kg
mass and the surface on which it rests is 0
.
38.
The acceleration of gravity is 9
.
8 m
/
s
2
.
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Adams, Russell – Homework 12 – Due: Dec 1 2006, 11:00 pm – Inst: Opyrchal, H
2
15 kg
m
45
◦
What is the largest mass
m
can have and
still preserve the equilibrium?
Correct answer: 5
.
7 kg.
Explanation:
Let :
M
= 15 kg
,
m
= 5
.
7 kg
,
and
θ
= 45
◦
.
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
.
38) (15 kg) (9
.
8 m
/
s
2
)
cos(45
◦
)
= 78
.
998 N
.
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 Spring '08
 moro
 Physics, Force, Work, Correct Answer, kg

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