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A question about masses on a pully

  1. Oct 16, 2005 #1
    Three block, connecting ropes, and a light frictionless pully comprise a system, as shown. An external force P is applied downward on block A. The system accelerates at the rate of 2.5m/s2. The tension in the rope connecting block B and block C equals 60 N. I dont have a picture but ill discribe it.

    Ok there are two masses on one side of the pully B and C. B is on top and C is on bottom, the Force exerted on B by C is 60N. Mass of B is 18kg. Then there is a mass on the other side which has a mass of 12Kg.

    Ok i tired this problem but i keep getting the wrong answer.

    The total force on B is

    T - (60 N + wB) = maa

    and then the total force on B is

    T - wA = -maa

    Since A is accelerating downwards it has a negative acceleration and B has a positive accleration.

    The answer should be 190N but i cant seem to get the answer. What am i doing wrong.

  2. jcsd
  3. Oct 16, 2005 #2

    Doc Al

    User Avatar

    Staff: Mentor

    OK, except that it should be mba.
    I assume you mean this to be the total force on A. You forgot the applied force P (or -P, with your sign convention.)
    Right: the acceleration of B is +a; of A, -a.

    Correct your two equations and combine them to solve for P. (Which I'm guessing is the question.)
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