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Homework Help: Atwood's machine with 3 masses

  1. Sep 23, 2010 #1
    1. The problem statement, all variables and given/known data
    A pulley is massless and frictionless, 2kg, 3kg, and 6kg masses are suspended. The 2 and 3 kilogram masses are suspended on the left side of the pulley (the 2 kilogram mass on the bottom) and are connected by a cord of negligible mass that does not stretch(T1), the 6kg mass is suspended on the right side and is connected around the pulley via a cord of negligible mass that does not stretch (T2). Down is the positive direction on the right side, up is positive on the left. what is the tension on the string between mass 1 and mass 2.


    2. Relevant equations
    Fnet=ma



    3. The attempt at a solution
    Sorry there's not a picture. I did not have anywhere to host it.

    I drew free body diagrams for each mass and came up with the Fnet equations for each. I assume we're only working with y accelerations.
    m3g-T2=m3a
    T2-T1=m2a
    T1-m1g=m1a

    I added the equations together for m2 and m3 because they're connected by a cord and
    the acceleration for the whole system should be the same.

    m3g-T2+T2-T1=(m2+m3)a
    m3g-T1=(m2+m3)a

    m3=6kg m2=3kg g=9.8

    (6kg)(9.8m/s^2)-T1 = 9a

    i solved for a

    T1-m1g=m1a

    a = (T1-m1g)/(m1)

    58.8-T1=9((T1-m1g)/(m1))

    58.8-T1=9((T1-(2)(9.8))/(2))

    58.8-T1=(9T1-176.4)/(2)

    117.6-2(T1)=9(T1)-176.4

    294-2(T1)=9(T1)

    11(T1) = 294

    T1 = 26.7272 N

    I felt like i did this right, and it's tormenting me..
     
  2. jcsd
  3. Sep 23, 2010 #2

    PhanthomJay

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    In your second equation T2 -T1 = m2a, you forgot to include the weight, m2g, in that equation.
     
  4. Sep 23, 2010 #3

    ehild

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    This equation i wrong. Gravity acts on all masses!

    ehild
     
  5. Sep 24, 2010 #4
    Thanks guys, I got it right. That small oversight cost me an hour at least.

    m3g-T2 = m3a
    T2-T1-m2g=m2a
    T1-m1g=m1a

    m3g-T1-m2g=(m2+m3)a

    a=(T1-m1g)/(m1)

    (6)(9.8)-T1-(3)(9.8)=9a
    58.8-T1-29.4=9a
    29.4-T1=9a

    sub in a
    29.4-T1=9((T1-m1g)/(m1))

    29.4-T1=((9)(T1)-(18)g)/2
    58.8-2(T1)=9(T1)-176.4

    -(2)(T1)=(9)(T1)-235.2
    -(11)(T1)=-235.2
    T1= (-235.2)/(-11)

    T1= 21.3818 N
     
  6. Sep 24, 2010 #5

    ehild

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    It is an easier job if you add up all equations. That cancels all inner forces (tensions here) and you get the acceleration as the total external force divided by the total mass. Knowing the acceleration, it is easy to find the tensions.

    ehild
     
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