Beer Vector Mechanics for Engineers DYNAMICS 10th

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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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