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Atwood Machine problem

  1. Nov 2, 2003 #1
    There is a problem in my physcs book that I have been trying to solve for some time now, and I just can't get trough it. If someone could help me a little bit, it would really be great.
    My question is about the Atwood machine.

    There are two blocks of same weight (M) suspended on a rope of each side of a pulley. There is a squared plaque (weight = m) placed on one of the blocks. When the block is dropped, is accelerates on a distance H until the squared plaque is stopped by a ring, but the block underneath continues it way at a constant speed. The distance D traveled by the block at constant speed, after the plaque is stopped by the ring, lasts "t" seconds. Prove that gravity acceleration, "g", is represented by the formula:

    g = ((2M + m)D^2) / (2mHt^2)

    Help for this question would be really, really apprecied...thank you !

  2. jcsd
  3. Nov 3, 2003 #2


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    The sentence "The distance D traveled by the block at constant speed, after the plaque is stopped by the ring, lasts "t" seconds." cofused me. It really should be "it took t seconds for the block to travel distance D ...." since there will be nothing to stop the block.

    The two M blocks balance so the total force, after m is placed on, is -mg. The total mass of the system is 2M+ m so the acceleration, from -mg= (2M+m)a is -mg/(2M+m).

    The distance H move in time t1 is: H= (-mg/(2(2M+m))t12 ((1/2)a t2) and the speed at time t is V= (-mg/(2M+m))t1.

    Solve the first equation for t1 and plug that into the equation for V so you know the speed after m is removed.

    Once you have V in terms of g, m, M, use

    D= Vt and solve for g.
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