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DYNAMICS - EXAMPLE


Pulley System
  Example

 

Two blocks are attached to a system with four pulleys. If both blocks are simultaneously released from rest, what is the displacement of block B travel when its velocity reaches 4 ft/s. Assume the pulleys are massless.

 

   
    Solution


Displacement of Objects

 

Since both displacement and velocity are parameters of the problem, energy can be used to solve the problem.

First, arbitrarily assume a downward displacement for block B, ΔsB, and relate it to the displacement of Block A, ΔsA. This can then be used to relate their velocities. The length of the cable from block B does not change.
     4sA + sB = L
     4 ΔsA = ΔsB

Time derivative gives,
     4vA + vB = 0 
     vA = -(4)/4 = - 1 ft/s

The system conserves energy, thus the energy before lease and after release must stay constant.
     T1 + V1 = T2 + V2
     0 - mAgsA - mBgsB
              = -mAg(sA + ΔsA) - mBg(sB - ΔsB)
                               + 0.5mA vA2 + 0.5mB vB2

Cancel terms,
     mAgΔsA - mBgΔsB = 0.5mA vA2 + 0.5mB vB2

Substitute ΔsA = ΔsB,  vB = 4 m/s, and vA = -1 m/s
     50 (ΔsB /4) - 8 ΔsB = 0.5(50/g)(-1)2 + 0.5(8/g) 42
     4.5 ΔsB = (25 + 64) / 32.2

ΔsB = 0.6142 ft

     
   
 
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