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System of masses over a pulley

  1. Sep 28, 2008 #1
    1. The problem statement, all variables and given/known data
    [​IMG]

    Two blocks are arranged as shown. The pulley can be considered to be massless, and friction is negligible. M1 is four times more massive than M2.

    If the system is released from rest, how far will M1 travel in 0.809 s?


    2. Relevant equations

    F=ma



    3. The attempt at a solution

    Okay, basically I've tried multiple times to try and solve this and I keep getting it wrong...I'm not quite sure what else to try. I think I'm looking at the forces wrong somehow. When I input a wrong answer, I get the message "One force acts on the hanging mass, but the net force acts on both masses. Once you have the acceleration, you can calculate the distance." Wouldn't there be two forces, weight and tension?

    My free body diagram for the mass on the table is its weight, its normal force, and its tension. It is four times more massive than the hanging mass, so T=4mg. For the hanging mass, I have only weight and tension. T-mg=ma. I tried solving for T in the second sum and plugging it into the first to find a, but I keep getting it wrong. I tried assuming that the mass of the hanging mass is added to the tension in the first, and that didn't work. Again, I assumed that M2=M and M1=4M. I assumed that the acceleration has to be the same for both blocks. To get the distance at the end, that would obviouslt be x=0.5st^2

    ANY pointer or help would be greatly appreciated, thanks!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 28, 2008 #2
    If the pulley is massless and frictionless, then the tension will be the same in both cases. Does that help?
     
  4. Sep 28, 2008 #3
    Yes, I realize that the ension will be equal in both cases...that's what I was trying to account for but I'm not quite sure how to account for it. Like I said, I solved for T in the hanging mass and used that in for T in the first mass...wouldn't that be assuming they are equal?
     
  5. Sep 28, 2008 #4
    What answers have you plugged in so far?
     
  6. Sep 28, 2008 #5
    Numerically? 1.07 m, 0.802 m, and 2.14 m.
     
  7. Sep 28, 2008 #6
    Ah! I just got it right. I was using the wrong directiond for my sums. For the x sum on M1, I assumed right to be positive and for the y sum on Mw I assumed up to be positive. This doesn't work because I need to assume that the pulley only changes the direction, and therefore the downward direction would have to be positive for the second sum.

    Thanks for the assistance! =)
     
  8. Sep 28, 2008 #7
    No problem ;) Congrats on getting it!
     
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