arellano (aa39398) – Homework 15 – weathers – (17101)
1
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printout
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have
10
questions.
Multiplechoice questions may continue on
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before answering.
001
10.0 points
An Atwood machine is constructed using a
hoop with spokes of negligible mass.
The
2
.
2 kg mass of the pulley is concentrated on
its rim, which is a distance 22
.
2 cm from the
axle. The mass on the right is 1
.
33 kg and on
the left is 1
.
99 kg.
1
.
9 m
2
.
2 kg
22
.
2 cm
ω
1
.
99 kg
1
.
33 kg
What is the magnitude of the linear acceler
ation of the hanging masses? The acceleration
of gravity is 9
.
8 m
/
s
2
.
Correct answer: 1
.
17174 m
/
s
2
.
Explanation:
Let :
M
= 2
.
2 kg
,
R
= 22
.
2 cm
,
m
1
= 1
.
33 kg
,
m
2
= 1
.
99 kg
,
h
= 1
.
9 m
,
and
v
=
ω R .
Consider the free body diagrams
1
.
99 kg
1
.
33 kg
T
2
T
1
m
2
g
m
1
g
a
a
The net acceleration
a
=
r α
is in the direc
tion of the heavier mass
m
2
.
For the mass
m
1
,
F
net
=
m
1
a
=
m
1
g

T
1
T
1
=
m
1
g

m
1
a
and for the mass
m
2
,
F
net
=
m
2
a
=
T
2

m
2
g
T
2
=
m
2
a
+
m
2
g .
The pulley’s mass is concentrated on the
rim, so
I
=
M r
2
,
and
τ
net
=
summationdisplay
τ
ccw

summationdisplay
τ
cw
=
I α
T
1
r

T
2
r
= (
m r
2
)
parenleftBig
a
r
parenrightBig
=
m r a
m a
=
T
1

T
2
m a
= (
m
1

m
2
)
g

(
m
1
+
m
2
)
a
m a
+ (
m
1
+
m
2
)
a
= (
m
1

m
2
)
g
a
=
(
m
1

m
2
)
g
m
+
m
1
+
m
2
=
(1
.
33 kg

1
.
99 kg) (9
.
8 m
/
s
2
)
2
.
2 kg + 1
.
33 kg + 1
.
99 kg
=
1
.
17174 m
/
s
2
.
002 (part 1 of 2) 10.0 points
A block of mass 3 kg and one of mass 6 kg are
connected by a massless string over a pulley
that is in the shape of a disk having a radius
of 0
.
26 m, and a mass of 8 kg. In addition, the
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arellano (aa39398) – Homework 15 – weathers – (17101)
2
blocks are allowed to move on a fixed block
wedge of angle 43
◦
, as shown. The coefficient
of kinetic friction is 0
.
17 for both blocks.
3 kg
6 kg
43
◦
0
.
26 m
8 kg
What is the acceleration of the two blocks?
The acceleration of gravity is 9
.
8 m
/
s
2
.
As
sume the positive direction is to the right.
Correct answer: 2
.
13791 m
/
s
2
.
Explanation:
Let :
m
1
= 3 kg
,
m
2
= 6 kg
,
M
= 8 kg
,
and
R
= 0
.
26 m
.
m
1
m
2
θ
M
T
1
T
2
Applying Newton’s law to
m
1
,
N
1

m
1
g
=
m
1
a
y
= 0
N
1
=
m
1
g ,
where the force of friction on
m
1
is
f
1
=
μ N
1
=
μ m
1
g
= (0
.
17) (3 kg) (9
.
8 m
/
s
2
)
= 4
.
998 N
and
T
1

f
1
=
m
1
a
T
1
=
m
1
a
+
f
1
.
(1)
For the mass
m
2
, applying Newton’s law
perpendicular to the slanted surface
N
2

m
2
g
cos
θ
=
m
2
a
perp
= 0
N
2
=
m
2
g
cos
θ ,
so the force of friction is
f
2
=
μ N
2
=
μ m
2
g
cos
θ
= (0
.
17) (6 kg) (9
.
8 m
/
s
2
) cos 43
◦
= 7
.
31061 N
.
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 Fall '10
 Weathers
 Force, Friction, Mass, Work, Arellano

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