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Table and frictionless pulleys

  1. Sep 21, 2010 #1
    Figure 5-58 shows three blocks attached by cords that loop over frictionless pulleys. Block B lies on a frictionless table; the masses are mA = 4.40 kg, mB = 9.80 kg, and mC = 12.0 kg. When the blocks are released, what is the tension in the cord at the right?

    all i could do on this one is make the two outside free body diagrams

    for block a i got T - 43.1 = 4.40a

    and for block c i got T-118 = 12.0a

    how would i get the net force for block b since the net force is just the two tensions?
     
  2. jcsd
  3. Sep 21, 2010 #2

    rl.bhat

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    Tension in the two segments of the rope are not the same.
    All the masses have the same acceleration.

    For block A .....> T2 = mA*a...(1)
    For block B.......> T1 - T2 = mB*a...(2)
    For block C.......> mC*g - T1 = mC*a...(3)
    Solve these equations and find T1 and T2.
     
  4. Sep 22, 2010 #3
    why isnt the weight taken into account for block a?
    and
    for block b why isnt it the tension minus the weight?
     
  5. Sep 22, 2010 #4

    rl.bhat

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    for block b why isnt it the tension minus the weight?


    Forces acting on block b are T1 and T2 only.
     
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