Muraj, Hamza – Homework 14 – Due: Feb 24 2006, 4:00 am – Inst: Florin
1
This printout should have 12 questions.
Multiplechoice questions may continue on
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be±ore answering.
The due time is Central
time.
001
(part 1 o± 5) 10 points
Given:
The mass on the le±t is equal to the
sum o± the other two masses;
i.e.
,
m
1
=
m
2
+
m
3
, and the center mass is greater than the
mass on the right hand side;
i.e.
,
m
2
> m
3
.
A massless pulley is attached to the ceiling,
in a uni±orm gravitational feld
g
, and rotates
with no ±riction about its pivot.
Another
massless pulley is attached to the end o± the
cord on the right and rotates with no ±riction
about its pivot.
m
1
T
1
m
2
m
3
T
2
T
3
T
4
Select the correct expression.
1.
Cannot be determined.
2.
T
2
=
m
2
g
3.
T
2
> m
2
g
4.
T
2
< m
2
g
correct
Explanation:
One can immediately see that the mass
m
2
is accelerating downward, so
T
1
< m
2
g
, which
can be seen in Eq. 2 below.
However, the dynamic expressions ±or the
three masses are derived below ±or your inter
est.
Consider the ±ree body diagram ±or the
right hand pulley.
T
23
T
23
T
1
There±ore :
T
23
=
T
2
=
T
3
,
T
1
=
T
4
,
and
T
1
=
T
23
+
T
23
.
Consider the ±ree body diagrams ±or mass
m
1
and the right hand pulley where
m
23
=
m
2
+
m
3
.
m
1
m
23
T
1
m
g
23
a
Consider the ±ree body diagrams ±or masses
m
2
and
m
3
.
Note:
The right hand pulley system has an
acceleration
a
1
,
which is downward.
m
2
m
3
2
3

+
The equations o± motion using these ±ree
body diagram are
T
1

m
1
g
=

m
1
a
1
(1)
T
23

m
2
g
=

m
2
(
a
23

a
1
)
(2)
T
23

m
3
g
= +
m
3
(
a
23
+
a
1
)
.
(3)
Since
T
1
= 2
T
23
and
m
1
=
m
2
+
m
3
, Eq. 4 is
Eq. 1 and Eq. 5 is Eqs. 2 plus 3.
2
T
23

(
m
2
+
m
3
)
g
=

(
m
2
+
m
3
)
a
1
(4)
2
T
23

(
m
2
+
m
3
)
g
= +(
m
2
+
m
3
)
a
1
+ (
m
2

m
3
)
a
23
.
(5)
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View Full DocumentMuraj, Hamza – Homework 14 – Due: Feb 24 2006, 4:00 am – Inst: Florin
2
Subtracting Eq. 4 from Eq. 5 and solving for
a
1
, we have
0 = 2 (
m
2
+
m
3
)
a
1

(
m
2

m
3
)
a
23
2 (
m
2
+
m
3
)
a
1
= (
m
2

m
3
)
a
23
a
1
=
(
m
2

m
3
)
2 (
m
2
+
m
3
)
a
23
.
(6)
Rearranging Eqs. 2 and 3, we have
T
23
=
m
2
g

m
2
(
a
23

a
1
)
(7)
T
23
=
m
3
g
+
m
3
(
a
23
+
a
1
)
.
(8)
Eliminating
T
23
, we have
(
m
2

m
3
)
g
=
m
2
(
a
23

a
1
) +
m
3
(
a
23
+
a
1
)
(
m
2

m
3
)
g
= (
m
2
+
m
3
)
a
23

(
m
2

m
3
)
a
1
.
(9)
Substituting
a
1
from Eq. 6, we have
(
m
2

m
3
)
g
= (
m
2
+
m
3
)
a
23

(
m
2

m
3
)
2
2 (
m
2
+
m
3
)
a
23
(10)
=
2 (
m
2
+
m
3
)
2

(
m
2

m
3
)
2
2 (
m
2
+
m
3
)
a
23
.
Solving for
a
23
, we have
a
23
=
2 (
m
2
+
m
3
)(
m
2

m
3
)
2 (
m
2
+
m
3
)
2

(
m
2

m
3
)
2
g .
(11)
Use
a
23
from Eq. 11 to substitute into Eq. 6
to solve for
a
1
a
1
=
(
m
2

m
3
)
2 (
m
2
+
m
3
)
a
23
(6)
=
(
m
2

m
3
)
2 (
m
2
+
m
3
)
×
2 (
m
2
+
m
3
) (
m
2

m
3
)
2 (
m
2
+
m
3
)
2

(
m
2

m
3
)
2
g
=
(
m
2

m
3
)
2
2 (
m
2
+
m
3
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 Turner
 Force, Friction, Mass, General Relativity

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