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Double Atwood's machine

  1. Jan 2, 2016 #1
    1. The problem statement, all variables and given/known data
    Masses m1 and m2 are connected by a light during A over a light frictionless pulley B. The axel of pulley B is connected by a light string C over a second light frictionless pulley D to a mass m3. Pulley D is attached o the ceiling. The system is released from rest.
    In terms of m1, m2, m3 and g what are
    a) the acceleration of the block m3
    b) the acceleration of pulley B
    c) the acceleration of block m1 and m2
    d) The tension in the string A
    e) The tension in the string C

    2. Relevant equations
    F = ma

    3. The attempt at a solution
    a) As the strings are weightless the tension either side of the pulley will be the same.
    I came up with equations
    TA - m1g = m1a1
    TA - m2g = -m2a1
    TB - m3g = m3a2
    TB - (m1 + m2)g = -(m1 + m2)a2
    TB = 2TA

    Rearranging the first 2 equations i got
    TA = (2m1m2g)/(m1 + m2)

    I then substituted this into equation 3 to get
    (4m1m2g)/(m1 + m2) - m3g = m3a
    which when i rearrange goes
    (4m1m2g - m1m3g - m2m3g)/(m1m3 + m2m3) = a

    However this is not the answer stated in my textbook and I'm not sure where I've gone wrong.
    Any help would be much appreciated!
     
  2. jcsd
  3. Jan 2, 2016 #2
    Your first equations would be right only if the pulley B were fixed.
     
  4. Jan 2, 2016 #3
    So would it be TA - m1g = m1(a1 + a2) as it is accelerating not only on its own pulley system but also has the acceleration of pulley B?
    This would also change equation 2 to T1 - m2g = -m2(a1 + a2)
     
  5. Jan 2, 2016 #4

    TSny

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    Homework Helper
    Gold Member

    If a1, a2, and a3 are accelerations measured relative to the earth, then equation 1 is OK. The next three need modification. You are right that you are going to need to think about relative accelerations.
     
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