Ph1a Solutions 3
CheFung Chan ([email protected]), Fall 2007
Each homework problem is worth 5 points. Please disregard the point values listed on the problem itself.
Use these instead.
3.1
QP3 (5 points)
3.1.a
(0.5 points)
In terms of the lengths in the diagram, we have:
l
1
= (
p
1

x
1
) +
πR
+ (
p
1

p
2
) = 2
p
1

x
1

p
2
+
πR
l
2
=
p
2
+
πR
+ (
p
2

x
2
) = 2
p
2

x
2
+
πR
3.1.b
(1 point)
The free body diagrams are in Fig. 1.
M
M
F
F
T
T
g
g
2
1
1
2
Figure 1: QP3.b
In the vertical direction, we have Newton’s Laws:
m
1
a
1
=
T
1

m
1
g
m
2
a
2
=
T
2

m
2
g
3.1.c
(1 point)
Taking two time derivatives of the equations in part a gives:
d
2
l
1
dt
2
= 0 =

a
1

d
2
p
2
dt
2
d
2
l
2
dt
2
= 0 = 2
d
2
p
2
dt
2

a
2
From this, we get
a
2
=

2
a
1
, or
a
1
=
ka
2
where
k
=

1
2
.
3.1.d
(1 point)
We can get the relationship between the tensions by considering Newton’s Laws applied to the (massless)
2
nd
pulley. For massless objects, the condition is that net force must be zero to avoid the unphysical result
of in±nite acceleration.
The force diagram is Fig. 2
m
P
2
a
P
2
=
T
1

2
T
2

m
P
2
g
1
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T
T
1
T
2
Figure 2: QP3.d
T
1

2
T
2
= 0
So we get
T
1
= 2
T
2
, or
T
1
=
k
2
T
2
with
k
2
= 2.
3.1.e
(1 point)
Solving for
a
1
and
T
2
is straightforward:
T
2
=
m
2
p
a
1
k
P
+
m
2
g
=
m
2
p
a
1
k
+
g
P
m
1
a
1
= (
k
2
T
2
)

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

m
1
g
a
1
=
(
m
2
k
2

m
1
)
kg
km
1

k
2
m
2
T
2
=
m
1
m
2
g
(
k

1)
km
1

k
2
m
2
Using the correct values of
k
=

1
/
2 and
k
2
= 2, we get
a
1
=
2
m
2

m
1
m
1
+ 4
m
2
g
T
2
=
3
m
1
m
2
g
m
1
+ 4
m
2
Using the incorrect assumption of
k
=

1 and
k
2
= 1
/
3, we get
a
1
=
m
2

3
m
1
3
m
1
+
m
2
g
T
2
=
6
m
1
m
2
g
3
m
1
+
m
2
3.1.f
(0.5 point)
If
m
1
=
m
2
the formula for
a
2
reduces to:
a
2
=
a
1
k
=
(
k
2
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 Fall '07
 Goodstein
 Force, Work

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