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Atwood Machine Help

  1. Sep 22, 2016 #1
    1. The problem statement, all variables and given/known data
    In the Atwood’s machine, what should M be, in terms of m1 and m2, so that it doesn’t move?
    atwood.JPG

    2. Relevant equations
    F=ma

    3. The attempt at a solution
    I've set T1=Mg as T1 is the tension of the rope attached to M. m1 and m2 are both connected together by the same rope so I assumed T1=2T2 .
    I set up the equations
    T2-m1g=m1a
    T2-m2g=-m2a (negative due to m1 and m2 moving in opposite directions).
    I then added the two together to get
    T2=[m1(g+a)+m2(g-a)]/2.
    Which, since T1=2T2=Mg, it can be expressed as
    M=[m1(g+a)+m2(g-a)]/g (I think).
    Which is where I'm stuck now. I'm unsure of how to get rid of g and a in order to have my answer in terms of just m1 and m2. It's probably pretty simple, but I can't quite get it.. It seems as g can be canceled out from the current equation, but I'm not sure about a.
    I've tried canceling out the g and then solving for a which gives me
    a=M/m1-m2
    But upon plugging that in as a, I am unsure of how to proceed.
     
    Last edited: Sep 22, 2016
  2. jcsd
  3. Sep 22, 2016 #2

    haruspex

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    Please either describe the set-up or post a diagram. Not all Atwood's machines are the same.
     
  4. Sep 22, 2016 #3
    Right, apologies. I've added a photo.
     
  5. Sep 22, 2016 #4
    You have 2 linear algebraic equations in two unknowns, T2 and a. Solve for them separately.
     
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