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Pulley Problem - Please give me hand here!

  1. Apr 16, 2006 #1
    Here's a pulley problem that has got me stopped; I'd appreciate any help that's offered.

    There are two blocks, A (44 N) and B (22 N) connected via a rope that stretches over a pulley. Block A is resting on a table, and block B is hanging over the edge. Block C is positioned on top of Block A.

    The question reads: "Given that the coefficient of static friction between Block A and the tabel is 0.20, what is the minimum weight of Block C to keep Block A from sliding off the table?"


    (The answer that is given is 66 N, but I can't seem to get this)


    I tried starting off with:
    Block A=M=44N Block B=m=22N
    (let F stand for the friction force)
    T-F=Ma and mg-T=ma
    Solving for Tension and setting them equal gives:
    mg-ma=F+Ma
    Rearranging some more and solving for a gives:
    a= (mg-F)/(m + M)
    The Tension on Block A then becomes M(mg-F)/(m + M) = 8.8 N
    So then in order for Block A to resist motion, the Friction between it and the table must be equal to T, so F=T=8.8 .20 * (4.489 + C) * 9.8 = 8.8 but then the mass of C ends up being zero. Where did I go wrong?
     
    Last edited: Apr 16, 2006
  2. jcsd
  3. Apr 16, 2006 #2

    Doc Al

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    Staff: Mentor

    Several problems here. One, you are trying to prevent block A from sliding, so the acceleration is zero. Two, you are given the weights of the blocks, not their masses. (22N is a force, not a mass.)

    Hint: Since the entire affair is in equilibrium, what must be the tension in the rope?
     
  4. Apr 16, 2006 #3
    Well, in order for Block A *not* to move, the the frictional force on Block A must be equal to the tension right?
    I redid the work for the acceleration and tension.

    T = (Mmg + mF) / (M + m) M=4.489 kg m=2.245 kg

    The tension (not taking Block C into account yet) is 17.6 N.
    Set this equal to u(M + C) g (the frictional force) and solve for C, and I get a mass of 8.97 kg, so a weight of 88 N, which is still off.
     
    Last edited: Apr 16, 2006
  5. Apr 16, 2006 #4

    Doc Al

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    Staff: Mentor

    This is true.

    Now answer my question: What's the tension in the rope? If you are looking at the situation correctly, you should be able to answer without doing any calculations.

    Hint: The hanging mass is in equilibrium.
     
  6. Apr 17, 2006 #5
    Yes, I think I got it. The tension is equal to the weight of the hanging block (22 N), since hte other block is resting on the table.
    It's in equilibrium, so a=0, so the whole term on that side of the equal sign becomes 0. Then T=mu * m (of Block A and C) * 9.8, then solve for the mass of Block C.
     
  7. Apr 17, 2006 #6

    Doc Al

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    Staff: Mentor

    Exactly. But don't waste time finding the masses; you are given the weights (w = mg) and that's all you need.
     
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