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Question

# Blocks B and C are connected by a single inextensible cable, with this cable being wrapped around pulleys at D and E. In addition, the cable is wrapped around a pulley attached to block A as shown. Assume the radii of the pulleys to be small. Blocks B and C move downward with speeds of VB=6 ft/s and VC=18 ft/s, respectively. Determine the velocity of block A when SA=4 ft.

A
6 ft/s
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B
15 ft/s
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C
18 ft/s
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D
20 ft/s
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Solution

## The correct option is B 15 ft/sSince the cable is inextensible, the length of the cable remains constant. ⇒l1+l2+l3+l4=constant ---- (1) ⇒dl1dt+dl2dt+dl3dt+dl4dt=0 --- (2) From the question diagram, l2=l3=√S2A+32 Substituting in eqn (1), SC+√S2A+32+√S2A+32+SB=constant Equation (2) becomes, ⇒dScdt−2ddt√S2A+32+dSBat=0 (negative sign because SA is decreasing as A is moving up) ⇒VC−2.SA√S2A+32dSAdt+VB=0 [∵dScdt=Vc dSBdt=VB, & dSAdt=VA] Putting values VC=6 ft/s,VB=18 ft/s SA=4ft, we get 6−85VA+18=0 VA=15 ft/s

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