TAM 212
Spring, 2004
1.61 Given:
The length of the string is 44
m
.
The system shown starts from rest in the position shown.
At
that instant,
2
/
2
s
m
x
B
=
.
(Also given is the information in the picture.)
Required: The speed of A just before it hits the pulley, that is, after it has traveled 5m up.
Solution: Let
( 29
t
y
A
be the distance of the object A from the pulley P, so
( 29
t
y
A
is the velocity of A at an
arbitrary time t (positive down, given that
( 29
t
y
A
is measured down from the pulley).
Let
( 29
t
x
B
be the distance of the object B from the wall, so
( 29
t
x
B
is the velocity of B at an
arbitrary time t (positive down, given that
( 29
t
x
B
is measured to the right of the wall).
Let
( 29
t
s
be the distance of the object B from the pulley P, so
( 29
t
s
is the speed at which this is
changing, that is, the lengthening of the distance of the rope extended between P and B…at an
arbitrary time t.
We have two constraints to work with:
(1)
The length of the rope is constant:
( 29 ( 29 ( 29( 29
K
t
y
t
d
t
x
m
L
A
B
+
+
+
=
=
44
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 Spring '08
 Staff
 Pulley, Velocity, The Wall, arbitrary time, pulley P. Equation

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