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Double Atwood machine

  1. Feb 11, 2015 #1
    1. The problem statement, all variables and given/known data
    The double Atwood machine shown in the figure has frictionless, mass-less pulleys and cords.
    Determine the acceleration of mA, mB, mC.
    GIANCOLI.ch04.p56.jpg

    2. Relevant equations
    F = ma

    3. The attempt at a solution

    http://photo1.ask.fm/779/725/638/-259996976-1shhdoh-qms4ifsem3hthf/preview/IMG_4082.jpg [Broken]
    http://photo1.ask.fm/905/398/673/-59997000-1shhdp1-2726aartaqghh2f/preview/IMG_4083.jpg [Broken]
    http://photo1.ask.fm/916/959/786/-239996977-1shhdpk-78i75a0215n926k/preview/IMG_4084.jpg [Broken]
    http://photo1.ask.fm/333/391/062/-249996981-1shhdrb-eeemt2tkfheqfp0/preview/IMG_4085.jpg [Broken]

    What did I do wrong?
     
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. Feb 11, 2015 #2

    BvU

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    First I see aa = ab , later on it becomes aa = -ab ?
     
  4. Feb 11, 2015 #3

    ehild

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    The accelerations of A and B are equal in magnitude with respect to the hanging pulley, which accelerates, too.
     
  5. Feb 11, 2015 #4

    BvU

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    Oops o:)
     
  6. Feb 15, 2015 #5
    Here is the answer given in my solutions manual.

    Because the pulleys are massless, the net force on them must be 0. Because the cords are massless, the tension will be the same at both ends of the cords. Use the free-body diagrams to write Newton’s second law for each mass. We take the direction of acceleration to be positive in the direction of motion of the object. We assume that mC is falling, mB is falling relative to its pulley, and mA is rising relative to its pulley. Also note that if the acceleration of mA relative to the pulley above it is aR , then aA = aR + aC. Then, the acceleration of mB is aB = aR - aC since aC is in the opposite direction of aB .

    Even though mA and mB are moving in opposite directions, how can they both be considered moving in the positive direction?

    The acceleration of m_A relative to the free pulley should be the acceleration of the free pulley plus the acceleration of m_A. The free pulley is rising and mA is rising too. Therefore, aR = aC + aA. Why is it necessary to consider the free pulley in the first place? I don't see any problem with ignoring it (my original attempt doesn't seem to have an errors) but I somehow got the wrong answers.


    Finally, what's the error in my original attempt in the starting post in this thread?
     
  7. Feb 15, 2015 #6

    Nathanael

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    ma and mb are not necessarily moving in opposite directions. It's only with respect to the lower pulley that they move in opposite directions. So in reality, they can both be moving in the same direction.

    Your error was the equation aa = -ab... This is not true! Imagine aa = ab ... They should both be at rest with respect to the lower pulley, right? According to your equation, aa+ab=0... But this is clearly not true because both a and b will moving with some nonzero acceleration in the same direction (they will move with the same acceleration that the lower pulley is moving with.)
     
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