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Simple pulley problem with weights

  1. Jul 19, 2011 #1
    1. The problem statement, all variables and given/known data
    Two 0.25 kg blocks are connected by a thread that passes over a frictionless pulley. The thread has no mass, and no friction. The two blocks are in balance. If a third block that weighs 0.20 kg is put on one of the blocks, then what is the acceleration of the 0.20 kg block?


    2. Relevant equations



    3. The attempt at a solution

    I was thinking that since the 0.20 kg block just disrupts the equilibrium of the 0.25 kg blocks, then wouldn't the 0.20kg block just accelerate as if it were just dropped, at 9.8 m/s^2?? That seems like too obvious of an answer though...
     
  2. jcsd
  3. Jul 19, 2011 #2

    Doc Al

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    Staff: Mentor

    Something accelerates at 9.8 m/s^2 only if there are no forces on it other than gravity--but that's not the case here, since it's in contact with another block.

    How would you figure out the acceleration using Newton's laws?
     
  4. Jul 19, 2011 #3
    Well.. I could try using F=ma, and then solve for acceleration. But if I were to use this equation, how would I figure out what F was???
     
  5. Jul 19, 2011 #4

    Doc Al

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    Staff: Mentor

    Why not treat it as a standard pulley problem? Apply F=ma to both sides, then combine the equations. What forces are acting on each mass?
     
  6. Jul 19, 2011 #5
    Ok, so how about this. If we do T-m1g=ma, and then T-m2g=-ma, then combine both equations and solve for a?
     
  7. Jul 19, 2011 #6

    Doc Al

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    Great! But make sure that the masses in the 'ma' terms have the proper subscripts also. For example: T - m1g = m1a.
     
  8. Jul 19, 2011 #7
    Oh yes, that's important. Thanks for your help!
     
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