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Faulty approach to solving a double Atwood's machine

  1. Jul 28, 2015 #1
    Hello. I'm wondering why my approach to solving this problem (Morin 3.2) is faulty.

    1. The problem statement, all variables and given/known data
    A double Atwood's machine is shown, with masses ##m_1##, ##m_2##, and ##m_3##. Find the accelerations of the masses.
    3.2.PNG

    2. Relevant equations
    ##F = ma##

    3. The attempt at a solution
    Let's say the string over the bottom pulley has tension ##T## and the string over the top pulley has tension ##T_2##. Assuming that the pulleys and ropes are massless and defining the upwards direction as positive, I have several equations:

    (1) For ##m_1##: ##T_2 - m_1g = m_1a_1##
    (2) For ##m_2##: ##T - m_2g = m_2a_2##
    (3) For ##m_3##: ##T - m_3g = m_3a_3##
    (4) For the lower pulley: ##T_2 - 2T = 0##
    (5) From conservation of string: ##a_2 = -a_3##

    If I substitute ##-a_3## into (2) and then solve (2) and (3) for ##T##, I find ##T = \frac{2m_2m_3g}{m_2 + m_3}##. From (4), ##T_2## is twice this value.

    Then, I substitute my expression for ##T_2## into (1) and solve for ##a_1##, which yields:
    [tex]a_1 = g\frac{4m_2m_3 - m_1(m_2 + m_3)}{m_1(m_2 + m_3)}[/tex]

    This answer is close to what Morin writes, but is missing a term in the denominator. After reading his solution, which uses the fact that ##a_1 = -\left(\frac{a_2 + a_3}{2}\right)##, I feel that I can understand why his approach works. However, I do not see where I made a mistake.
     
  2. jcsd
  3. Jul 28, 2015 #2
    This is where you have made a mistake . Why ?

    Conservation of string can be looked at in two ways .
    Since you use all accelerations in one direction , this would be an easier way to understand -

    Total power by string must be zero ( Why ? ) . Power = F.v , so total sum of of each individual power provided by string must be equal to zero .

    So 2T*v1 + T*v2 + T*v3 = 0 , or ,
    2v1 + v2 + v3 = 0 .

    So what relation do you get for acceleration ?

    I hope this helps .
     
  4. Jul 28, 2015 #3
    Oh, right. I just realized that ##a_2 = -a_3## only holds in the reference frame of the pulley, not the ground. Don't know why that point eluded me for so long. Thanks.
     
  5. Jul 28, 2015 #4

    SammyS

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    Your error is with the equation you get from the "conservation of string", as Qwertywerty points out.

    A simple way to see that this is erroneous is: Suppose that m2 = m3 . Then we must have a2 = a3 , but these won't be zero in general.
     
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