ROBLEM 11.53
Collar A starts from rest and moves upward with a constant acceleration. Knowing
that after 8 s the relative velocity of collar B with respect to collar A is 24 in./s,
determine (a) the accelerations of A and B, (b) the velocity and the change in position
of B after 6 s.
From the diagram
2y +y + (y − y )= constant
A
B
B
A
Then
vA + 2vB = 0
(1)
and
aA + 2aB = 0
(2)
(a)
Eq. (1) and (v ) = 0 (v
A 0
) =0
B 0
Also, Eq. (2) and a A is constant and negative
positive.
Then
Now
vA = 0+aA t
aB
is constant and
vB = 0+aBt
vB/A =vB − vA= (aB − aA )t
1
From Eq. (2)
So that
aB =−
vB/A =−
2
3
2
a
A
aA t
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68
PROBLEM 11.53 (Continued)
At t = 8 s:
3
24 in./s =− a (8 s)
A
2
or
aA = 2.00 in./s
2
aB = 1.000 in./s
2
1
and then
a =− (− 2 in./s2 )
B
2
or
(b)
At t = 6 s:
v = (1 in./s 2)(6 s)
B
vB = 6.00 in./s
or
Now
At t = 6 s:
or
1
y = (y ) + 0+ a t
B
B 0
B
2
1
2
2
y − (y ) = (1 in./s )(6 s)
B
B 0
2
2
y − (y ) =18.00 in.
B
B 0