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Pulley System- equation derivation

  1. May 26, 2010 #1
    1. The problem statement, all variables and given/known data

    Determine a formula for acceleration of the system shown in Fig. 4-45 if the cord has a non-negligible mass mc. Specify in terms of lA and lB, the length of cord from the respective masses to the pulley ( The total cord length is l= lA + lB)

    [PLAIN]http://img706.imageshack.us/img706/8575/3333v.png [Broken]




    2. Relevant equations

    F = ma

    3. The attempt at a solution

    Fby = Fax

    Fby = mB *g

    acceleration of system: a = (mB *g)/ma

    acceleration of system with non-negligible cord

    a = (mB + (mC * (lB/ l)))*g / (ma + (mC * (lA/ l)))


    need an answer check, I can't tell if its right.
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. May 27, 2010 #2

    ehild

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    This is not right. Draw the free-body diagram, showing the forces both at A and B.
     
  4. May 27, 2010 #3
    Is my whole solution wrong, or just the forces I used.

    I understand now that I didn't use tension force, but is the way I derived the equation wrong with

    Fby = Fax
     
  5. May 27, 2010 #4

    ehild

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    If you do not specify what Fbx and Fax are the equation Fbx=Fax has no sense.

    ehild
     
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