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(c)
For Block A, the equation of motion is
M
¨
x
=2
T
±,²1
Cable
Tension
−
k
(
x
+
x
eq
)
±
,²
1
Spring
Force
For Block B, the equation of motion is
m
¨
y
=
mg
±,²1
Weight
−
T
±,²1
Cable
Tension
and
y
x
Therefore, the equations of motion for Blocks A and B are
M
¨
x
T
−
k
(
x
+
x
eq
)a
n
d2
m
¨
x
=
mg
−
T
(d)
So, we can use Block B’s equation of motion to specify the cabletension, viz.,
T
=
mg
−
2
m
¨
x
Substituting this expression for
T
into Block A’s equation of motion tells us that
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This note was uploaded on 05/12/2010 for the course AME 301 taught by Professor Shiflett during the Spring '06 term at USC.
 Spring '06
 Shiflett

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