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Pulley Problem

  1. Apr 4, 2004 #1
    Two blocks A and B are connected via a pulley.

    Diagram

    They are initially at height h = 2 m from the ground. When block A hits the ground, it is moving at 3 m/s. For part a), I already determined the amount of frictional energy dissipated, with E = 45.7 J. This is correct, according to the textbook. Mass of A = 11 kg, Mass of B = 5 kg.

    Now I'm asked to calculate the tension in the cables.

    Isolating Block A, we have the TA, mg and the frictional force acting on it. Using the principle of work-energy:

    0 + (11)(9.81)(2 m) - (2 m)TA - 45.7 = 1/2(11)(3)^2
    Solving, I get TA = 60.3 N

    For Block B:

    0 - (5)(9.81)(2 m) + (2 m)TB - 45.7 = 1/2(5)(3)^2
    Solving, I get TB = 83.2 N.

    However, the back of the textbook has the answers listed as TA = 83.2 N and TB = 60.3 N. I must be on the right track since I have the right numbers, but can anyone point out where I went wrong to get the exact reversal of answers? It makes sense that TA > TB, since A is heavier than B, but I can't find the mistake in my equations.

    Any help would be appreciated!
     
  2. jcsd
  3. Apr 4, 2004 #2
    Are you sure the answers in the book aren't just wrong? (They could have mixed up the symbols.)
     
  4. Apr 4, 2004 #3

    Doc Al

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    Staff: Mentor

    The book's answers are correct. Your "work-energy" relations are not. (The friction does not act directly on the blocks--its effect is transmitted via the tensions in the cables.)
    For block A, the correct equation would be:
    MAg(2) - TA(2) = ΔKEA

    For B:
    -MBg(2) + TB2 = ΔKEB
     
  5. Apr 5, 2004 #4
    Just to confirm that I understand this correctly:

    For an ideal system with friction = 0, we would be able to calculate TA and TB and they'd be equal. But once friction is taken into account, this increases the tension in each part, making them unequal. (And since it's included in the tension, once each block is isolated, the friction can be "ignored.") So, it's only when the entire system is taken into account that the friction needs to be added, because when we use the entire system we're assuming TA (without friction) = TB (without friction) = internal forces, while friction is external?

    Thanks for your help!
     
  6. Apr 5, 2004 #5

    Doc Al

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    Staff: Mentor

    Right!
    Right, the friction changes the tension in the cable.
    Since you chose to isolate each block, you need only consider the forces on the blocks. The friction is a force on the pulley/cable, not on the blocks.
    If you take as your system the blocks plus cable plus pulley then the work done by the tension in the cable will always cancel. You can think of the tension as a passive internal force: The cable--assumed massless--merely transmits force from one end to the other. But you are correct that the friction is an external force doing work on the system and dissipating energy.
     
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