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Can anybody check if I solved this correctly?(tension forces + pulley)

  1. Oct 19, 2013 #1
    1. The problem statement, all variables and given/known data
    pulley_zps70f8d0bc.png
    The pulley can rotate and has no friction.
    Block [itex]m_2[/itex] is twice as big as [itex]m_1[/itex]
    Find the tensions, [itex]T_1, T_2, T_3[/itex]

    2. Relevant equations
    Newton's second law


    3. The attempt at a solution
    Tension 1:
    By creating a free body diagram only for the block, I got with Newton's second law:
    [itex]T_1-mg=ma[/itex]
    [itex]T_1=m(a+g)[/itex]

    Tension 2:
    [itex]T_2-2mg=ma[/itex]
    [itex]T_2=m(a+2g)[/itex]

    Tension 3:
    [itex]T_3-T_1-T_2=ma[/itex]

    Inserting above values for the other tensions:
    [itex]T_3=ma+m(a+g)+m(a+2g)[/itex]
    [itex]T_3=3m(a+g)[/itex]

    Did I solve this correctly? What about the acceleration? Can I do anything about that?
     
    Last edited: Oct 19, 2013
  2. jcsd
  3. Oct 19, 2013 #2

    the pulley has no net force acting on it.
     
  4. Oct 19, 2013 #3
    What exactly do you mean?
    Are the 2 first tensions correct?
     
  5. Oct 19, 2013 #4
    So we have:

    [itex]T_1-mg=ma[/itex]
    [itex]T_1=m(a+g)[/itex]

    [itex]2mg-T_2=ma[/itex]
    [itex]T_2=2m(g-a)[/itex]

    Since the tension is uniform on the string:
    [itex]m(a+g)=2m(g-a)[/itex]
    [itex]a=\frac{1}{3}g[/itex]

    Substituting this into the first equation:
    [itex]T_1-mg=\frac{1}{3}mg[/itex]
    [itex]T=\frac{4}{3}mg[/itex]


    The tension of the pulley since it has no net force acting:
    [itex]T_3-T_1-T_2=0[/itex]

    Since [itex]T_1=T_2=T[/itex]
    [itex]T_3=2T[/itex]

    Inserting the tensions:
    [itex]T_3=2\frac{4}{3}mg[/itex]
     
    Last edited: Oct 19, 2013
  6. Oct 20, 2013 #5

    haruspex

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    Assuming the pulley has no moment of inertia, that all looks correct. I wouldn't write the final answer as [itex]T_3=2\frac{4}{3}mg[/itex] though. It looks like you mean "two and four thirds".
     
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