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Double Atwood Machine and acceleration

  1. Feb 12, 2008 #1
    1. The problem statement, all variables and given/known data
    The double Atwood machine shown in the figure has frictionless, massless pulleys and cords.


    Determine the acceleration of masses mA, mB, and mC.
    2. Relevant equations

    3. The attempt at a solution
    don't know how to start
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Feb 12, 2008 #2
    Start by drawing free-body diagrams for each box and pulley and write out their Newton's second law equations. You'll also need a few constraints based on the (constant) lengths of the cords that connect each pair of objects.
  4. Feb 13, 2008 #3
    so far i have FTA-mAg = mAa1 for mA, mBg-FTA=mBa1 for mB, FTC=2FTA for the pulley, and mCg-FTC=mCa2 for the mC block... am i on the right track? I don't quite understand what you mean by the constraints...
  5. Feb 13, 2008 #4
    Yeah, that's the right start. I can't exactly remember how to solve this problem though, so I won't be of too much help :p
  6. Oct 17, 2009 #5
  7. Oct 17, 2009 #6
    Start with the bottom section, forget the top. Determine the acceleration of the two masses, and determine the tension in the rope.

    Then the tension above the bottom section is 2T. This is the same as the tension pulling up mass "c". Use this to calculate the acceleration for mass "c".

    edit: good god, didn't notice the OP date!
  8. Oct 17, 2009 #7
    I think I asked this question 3 semesters ago :P was surprised to find an email in my inbox about the response! thanks for the help though :P
  9. Nov 24, 2009 #8
    Philnow, I'm glad you didn't notice the date, because I'm wrestling with this question, and I appreciate your help!

    I'm stuck on this point: why is the tension above the bottom section 2T? If the bottom pulley were in equilibrium, this would make sense, because we would have Ftc - 2Fta = 0 (where Ftc is the tension in the rope going around the upper pulley, and Fta is the tension in the rope going around the bottom pulley - I'm using the labels from the diagram given by the OP at http://session.masteringphysics.com/problemAsset/1057076/4/GIANCOLI.ch04.p56.jpg)

    However, the bottom pulley is accelerating, or might be at least, depending on the masses.

    I'm guessing that the answer is that because the pulleys are massless, we have Ftc - 2Fta = 0*a = 0. That is, Ftc - 2Fta = 0, even though the lower pulley isn't in equilibrium.

    Is that right? Intuitively, this is hard for me to accept, but it makes sense mathematically.
    Last edited: Nov 24, 2009
  10. Nov 24, 2009 #9
    Hi! Even with the lower pulley accelerating, by balancing the forces on the bottom pulley, we see that the tension in the upper string must be 2T. However... this tension will be dependent on the acceleration of the masses.

    The three F=ma equations:


    and the fact that (just by looking at the string set-up) the average position of mA and mB moves the same distance as the bottom pulley, which in turn moves the same distance (but in the opposite direction) as mC:

    aC = -(aB + aA)/2

    Should get you started!
  11. Nov 24, 2009 #10
    Thanks for the response. I'm still not sure that I understand why TUpperString = 2T, although I do understand how the rest of the solution you've outlined follows from that.

    My confusion has to do with what is meant here by "balancing the forces." When I do an FBD for the lower pulley itself, I get:

    TUpperString - 2T = mLowerPulley * aLowerPulley

    If the lower pulley is in equilibrium, then TUpperString balances 2T. But since aLowerPulley is nonzero, it would seem that TUpperString is not equal to 2T - that the forces on the lower pulley do not balance.

    However, since mLowerPulley = 0, I can see that TUpperString does indeed equal 2T. But this seems like a mathematical "trick" to me. I'm having trouble understanding the physical meaning of this.

    Is it correct to say that TUpperString = 2T is only true if mLowerPulley = 0? That if the lower pulley had mass, the forces on the pulley would not necessarily balance? I realize that if the lower pulley had mass, the problem would change in a number of ways (I'm just now learning about rotational dynamics in my physics class), but I just want to be sure I understand the reasoning behind your solution.

    Thanks again for your help.
  12. Nov 24, 2009 #11
    That's a great question. As an undergrad in physics myself, my hunch is that it's because the pulley has no mass, so ma=0. In fact, most Atwood machine problems that I have seen have been very careful to include that the pulley indeed has no mass... as for the physical meaning, I too would be interested in the answer.
  13. Nov 3, 2011 #12
    I would like to thank you guys. Even a significant time later this has helped me a lot. Especially considering I am only a high school AP student.
  14. Jul 8, 2012 #13
    Sorry to revive this post - could someone explain to me why we can't treat the free pulley as a system with mass mA+mB, solve for mC's acceleration as if it were a single Atwood machine? That would then give the acceleration of the free pulley-system, which we could use to compute the effective g of the free pulley-system due to its acceleration and solve that system as a single Atwood machine.
    Last edited: Jul 8, 2012
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