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Double Atwood's Machine

  1. Aug 19, 2016 #1
    • Moved from a technical forum, so homework template missing
    Masses m1 and m2 are connected by a light string (A) over a light, friction-less pulley (B). The Axle of Pulley B is connected to another light string (C), which goes over a light, friction-less pulley (D), to a mass m3. Pulley D is attached to the ceiling by an attachment to its axle. The system is released from rest.

    I'm trying to find the acceleration of m1, m2, m3, and pulley B, as well as the tension on string A and string C.

    I've split the whole system into two smaller systems, where [m1, m2, string A, and pulley B] is system 1 and [m1, m2, m3, pulley B and D, and strings A and C] (everything in the system) is system 2.

    relevant relationships/equations: F=ma, Tsys1 = (m1+m2)g - (m1+m2)asys1

    m1+m2+m3 = M

    and (I think):

    asys2 = [(m3 - (m1+m2)) * g] / M

    I'm not sure if I just can't do the algebra here, or if I'm missing something big. I have two pages of me trying to figure it out, but I need some insight.

    Thank you.

    20160819_172509.jpg
     
  2. jcsd
  3. Aug 19, 2016 #2

    Charles Link

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    I would suggest you write equations for the acceleration of each mass separately, with the tension of the attached string being included. The coordinates for masses m1 and m2 are somewhat tricky. Suggest using two letters here: x1, x2 for motion relative to their pulley and X1 and X2 relative to the earth/laboratory. The mass m3 only needs one coordinate x3. The idea is basically to get 4 equations and 4 unknowns, etc. If you find you have fewer equations than unknowns, you may need to find one more equation. Perhaps the trickiest part (I needed to add this extra equation myself when I worked it) is to be sure and include the equation for the forces on pulley B.
     
    Last edited: Aug 19, 2016
  4. Aug 19, 2016 #3

    Simon Bridge

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    When in doubt: treat each mass separately, draw free body diagrams, apply Newton's Laws.
    Best practice is to use the letters provided to label forces etc and take care to define your positive direction.

    So the fbd for ##m_3## tells you that (1) ##T_C-m_3g = m_3a_3## and (2) ##T_C-(m_1+m_2)g = (m_1+m_2)a_B## taking "up" = positive.

    Notice that the notation is pretty clear? C is the light string about pulley D and so ##T_C## must be the tension in that string.
    Since the pulley is "light" the tension is the same on both sides. You don;t need to talk about "system 1" etc since the LHS of the pulley you only care about the acceleration of the pulley B. There's a slight anticipation, though, to realize that the effective mass pulling on the LHS is ##m_1+m_2## but you don't need to assume this: it is a consequence of the free body diagrams for ##m_1## and ##m_2## taken separately.

    Since the string C does not change length you get (3) ##a_B=-a_3## from the geometry: if mass 3 goes down, then pulley B goes up.
    Use (3) to eliminate ##a_B## from (2). Use (1) to eliminate ##T_C## from (2), then solve the resulting relation for ##a_3## to get your result vis:
    $$a_3 = \frac{(m_1+m_2-m_3)g}{m_1+m_2+m_3}$$ ... that's "just doing the algebra", and it does not take two pages, just a bit of discipline in the setup.

    Reality check: if ##m_1+m_2 > m_3## then ##a_3 > 0## which, by the adopted sign convention, means that mass 3 accelerates upwards. Is this consistent with what you understand about how this sort of machine should work?

    These "Atwood's machine" things are just "coupled mass" problems and are solved exactly the same way, when in doubt, go back to free body diagrams and Newton's Laws. However, you will probably have noticed that you can plug-n-chug the Atwood's machine relations in most cases.

    Aside: I wouldn't bother with coordinates since the problem only asks for accelerations.
    The problem is worded so it can be solved stepwise - but it is best practise to do it all in one go by finding the simultaneous equations describing the whole thing and solving them.
     
  5. Aug 19, 2016 #4
    Okay, but the acceleration of ##m_3## that you've given is incorrect, according to my textbook.

    acceleration of B = negative acceleration of m3
    As for part g) in the problem, I cannot solve anything until I have to correct approach.

    The textbook gives that the acceleration of m3 = [ (- 4*m1*m2 + m2*m3 + m1*m3) / (4*m1*m2 + m2*m3 + m1*m3) ] * g

    This is just chapter 5 in my textbook (Applying Newton's Laws). I'm mostly self taught and I wanted to go back 7 chapters or so to solve some problems (test my understanding) and this one in particular is just bothering me now.
     
  6. Aug 19, 2016 #5

    Charles Link

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    Since B is zero mass, what does that tell you about the tensions in the strings around B? i.e. Can you write an expression involving ## T_A ## and ## T_C ##? And meanwhile, I would recommend writing something like ## X1=x1-x3 ## for the coordinates, so that ## \ddot{X_1}=\ddot{x_1}-\ddot{x_3} ## etc. Notice also that ## x1 ## and ## x2 ## are related. I first solved it for ## \ddot{x_1} ## and worked my way from there (about 4 equations and 4 unknowns). And yes, I agree with the book's answer for ## \ddot{x_3} ##.
     
    Last edited: Aug 19, 2016
  7. Aug 19, 2016 #6

    Simon Bridge

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    That's what I wrote too.
    Are you saying that the magnitude is correct but the sign is wrong?
    In which case, your book has chosen or assume a convention where positive direction is "downwards". Check.

    You don't need the correct approach, just understand the physics. It's exactly the same approach as above.
    If you have trouble with massless pulleys then give the pulley a small mass ##m_B## and write out the equations ... then put ##m_B=0##.
     
  8. Aug 20, 2016 #7

    Charles Link

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    Because the masses on pulley B may be accelerating in some manner with the string of pulley B moving, this one is more complex than @Simon Bridge has for ## a_3 ## in post #3. I did get an answer that agrees with what the OP says his textbook gives. The equations around massless pulley B must be included and give the relation between the tension ## T_A ## and ## T_C ##. (It's a simple equation, but very necessary for the complete solution.) Masses ## m_1 ## and ## m_2 ## must be treated separately (and not as a single mass of ## m_1 ## + ## m_2 ##). There is the relation because they are connected that ## \ddot{x_1}=-\ddot{x_2} ##. ## \\ ## May I suggest the OP start with mass ## m_1 ## and write out the equations for the acceleration ## \ddot{X_1} ## where ## X_1=x_1-x_3 ##.## \\ ## (## x_1 ## is the motion of the string of mass ## m_1 ## relative to pulley B. Note: Sum of forces on mass 1 = ## m_1 \ddot{X_1} ## . ## \ ## ## X_1 ## is the motion of mass ## m_1 ## relative to the laboratory.) ## \\ ## And after that, do the same thing for mass ## m_2 ##.
     
    Last edited: Aug 20, 2016
  9. Aug 21, 2016 #8

    haruspex

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    Not so, for the reason Charles gives.
     
  10. Aug 21, 2016 #9

    Simon Bridge

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    Hmmm... missed a bit.
    Still dont see why positions are needed... but if it helps visualization...
     
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