Problem 2/207 Page 103, Engineering Mechanics - Dynamics, Meriam and Kraige, 4th Edition

Given:  Pulley system shown and the known velocity of block A.

Find: Velocity of block B.

Parameter Values:
Velocity of block A = 3.6 ft/s to the right.

The length of the rope remains constant.  The rope can be seen to consist of 7 segments.  One segment reaches down to B and is of variable length.  Three segments stretch horizontally to A and are of variable length.  Three segments wrap around portions of pulley and are of fixed length.  Denoting the length to block B as YB and the length to block A as XA and the fixed pulley wrap length as L0, one obtains:

Constant rope length = YB + 3 XA + L0

Differentiating the above with respect to time:

0 = dYB/dt + 3 dXA/dt

But dXA/dt is the X component of the velocity of block A.  As block A is moving to the right, the distance XA is decreasing.  Thus:

dXA/dt = -3.6 ft/s

Plugging into the above equation:

dYB/dt = -3 dXA/dt = 10.8 ft/s

Thus the length YB is increasing at a rate of 10.8 ft/s.  However dYB/dt is the Y component of the velocity of block B.  Thus block B is moving downward (YB is increasing) at a rate of 10.8 ft/s.