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Atwood Machine Problem

  1. Dec 26, 2007 #1
    1. The problem statement, all variables and given/known data

    In the three figures given in the attachment consisting of three atwood machines with, the blocks A, B and C of mass m have accelerations a1, a2 and a3 respectively.F1 and F2 are external forces of magnitude 2mg and mg acting on the first and third diagrams respectively.

    How do the accelerations of the block differ and why is it so?

    2. Relevant equations

    Basic Newton's Laws of Motion equations.

    3. The attempt at a solution

    Acceleration of masses in atwood machine is given by:

    [tex] a = (\frac{m_{2}-m_{1}}{m_{2}+m_{1}})g [/tex]

    I really don't get it how the acceleration are not the same in each case as the forces add to the same?

    Attached Files:

  2. jcsd
  3. Dec 26, 2007 #2

    Doc Al

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    Staff: Mentor

    Don't use a "canned" formula for Atwood's machine--that's only good in certain situations. Instead, derive the acceleration yourself using Newton's 2nd law. Hint: In the cases with two masses, analyze each mass separately. Then combine the resulting equations to solve for the acceleration.
  4. Dec 26, 2007 #3
    For the second pulley we can straight forward apply the equation and with that we get the acceleration a2 = g/3

    Yes, you were right. The acceleration have to be calculated individually.

    For the first pulley let the tension in the string be T1 which will be equal to the force pulling F1=2mg.

    for the block we can write: ma=T-mg or ma=mg or a=g...(this must be right) or a1=g

    For the third pulley I can't figure out.
    Last edited: Dec 26, 2007
  5. Dec 26, 2007 #4
    Lets see...F2=mg,

    [tex] F_{2}+mg-T=ma ...(1) [/tex]
    [tex] T-mg=ma ...(2) [/tex]

    Adding both sides, we get

    [tex]mg+mg-mg+T-T=2ma [/tex]
    then [tex]a_{3}=\frac{g}{2}[/tex]

    Therefore, the correct option would be no. (b)
    Last edited: Dec 26, 2007
  6. Dec 26, 2007 #5

    Doc Al

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    Staff: Mentor

    For the third pulley: What forces act on the second mass? Apply Newton!

    (Looks like you did it while I was typing. Good!)
    Last edited: Dec 26, 2007
  7. Dec 26, 2007 #6
    Is this correct?
  8. Dec 26, 2007 #7

    Doc Al

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    Staff: Mentor

    Yes. You got it.
  9. Dec 26, 2007 #8
    Thank you very much for your help and support!
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