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I have a question about a pulley and tension-related problem

  1. Mar 3, 2016 #1
    1. The problem statement, all variables and given/known data
    The figure shows three blocks attached by cords that loop over frictionless pulleys. Block B lies on a frictionless table; the masses are mA = 4.80 kg, mB = 7.10 kg, and mC = 13.0 kg. When the blocks are released, what is the tension in the cord at the right?

    Figure A is on the left, figure C on the right, C being the largest mass.

    2. Relevant equations
    For part A I got T-m1g=ma, (Because the tension will be more used when it gets pulled up)
    For Part B, This is where i am confused, I originally used T1-T2=M2A, because I thought that since the Box B will move to the right, then its tension would be less because Box C is pulling more with M3g, but I checked the answer and it said T2-T1=M2A, but why is that the right answer
    For Part C I got, M3g-T=m3a (since gravity will do most of the work

    3. The attempt at a solution
    I got the answer of -7.22 on my first attempt and 3.22 for the acceleration on the second, which 3.22 is the correct answer. Help!! I am just confused in part B!!
     
  2. jcsd
  3. Mar 3, 2016 #2

    Merlin3189

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    Your picture doesn't seem to have uploaded and I think it is needed to understand the problem. Perhaps you can try again.
     
  4. Mar 3, 2016 #3

    haruspex

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    I think I can guess the diagram. Mass 1 hangs from a pulley on the left. The cord passes up over the pulley and extends horizontally to the right to mass 2. The second cord continues from mass 2 horizontally to the right, over a second pulley and straight down to mass 3.
    Your working would be easier to follow if you were to use suffixes consistently.

    You correctly presumed mass 1 would rise and mass 3 would fall, and got the right equations for those.
    And, as you say, mass 2 will move to the right. But I cannot understand your explanation for why you chose T1-T2 there.
    If the acceleration is to the right, which tension must be greater?
     
  5. Mar 3, 2016 #4
    I see to understand now!! thanks sorry for not posting a picture!!
     
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