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Two pulley system

  1. Oct 3, 2003 #1
    Mass m1 is attached to a rope that goes over an ideal pulley. At the other end of the rope is another pulley. A rope goes over this pulley connecting two masses m2 and m3. Find the acceletation of the masses and the tensions in the ropes.

    When considering the first mass, I pretend that masses m2 and m3 are one mass and I call it M, so
    I substract the two equations and get,
    At this point I stop, because the book already says I'm wrong. The book get a=g(4m2m3-m1m2-m1m3)/(4m2m3+m1m2+m1m3).
    What did I do wrong?
  2. jcsd
  3. Oct 3, 2003 #2

    Chi Meson

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    According to the information given, your answer is correct. Double-check to see if your information is complete. As I see it...

    there are three distinct masses

    m1 is hanging vertically

    string goes up, over a pulley, across to another pulley then down

    string is supporting m2 and m3. If this is how it goes, you have the correct answer: in "Atwood machine" problems, acceleration is the difference in weights divided by total mass.

    evidently something's missing.
  4. Oct 3, 2003 #3
    I attached the figure from the book.
    I'll retype the question word for word.
    A mass m1 hangs from one end of a string passing over a frictionless, massless pulley. A second frictionless, massless pulley hangs from the other end of the string (Figure 6.38). Masses m2 and m3 hang from a second string passing over this second pulley. Find the acceleration of the three masses, and find the tensions in the two strings.

    (The question has three stars next to it - it supposed to be really difficult)

    Ah, I see you have misunderstood me. This should clear things up.
    Perhaps I am wrong to assume that the force pulling down on the rope on the right hand side is Mg. I think it should be 2T2 instead.

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  5. Oct 4, 2003 #4
    Remember that the pulley & masses on the right are not an inertial system. That system is accelerating at the same rate as m1, so I think you should let
    a = acceleration of m1
    a2 = acceleration of right pulley
    a + a2 = acceleration of m2
    a - a2 = acceleration of m3
    and see where that gets you. You'll have to figure out how to set up the weights & tension on the right side to be consistent with the accelerations.

    (I'm not sure, but it seems to make sense.)
  6. Oct 5, 2003 #5
    Well, using that approach I came up with
    a1 = g(4m2m3 - m1m2 - m1m3)/(-m1m2 - m1m3)

    or, to put it slightly differently:
    a1 = -g(4m2m3 - m1m2 - m1m3)/(m1m2 + m1m3)

    Did you get that result?
    It's a little different from the answer in your book, but I've looked over my work thoroughly & can't find a mistake. Maybe the grad student who solved it for your textbook made the mistake.
    Anyway, here's my work. Maybe you can find something wrong in it. As you can see, the difference is in what happens to the "2m2m3" terms in the denominator. In mine there's a +2m2m3 and a -2m2m3 so they cancel. In his, apparently they were both +. But they're so similar that I can't imagine that we're working with different equations to begin with. One of us must have made a mistake.

    Here are the original equations:
    m1g - 2t = m1a1
    m2g - t = m2(a1+a2)
    t - m3g = m3(a1+a2)

    And here's the rest:
    1> -2t - m1a1 + m1g = 0
    2> -t - m2a1 - m2a2 + m2g = 0
    3> t - m3a1 - m3a2 - m3g = 0

    4> -(m1+2m3)a1 - 2m3a2 + (m1-2m3)g = 0 ... [eq 1 + 2*eq3]
    5> -(m2+m3)a1 - (m2+m3)a2 + (m2-m3)g = 0 ... [eq2 + eq3]

    6> -(m1+2m3)(m2+m3)a1 - 2m3(m2+m3)a2 + (m1-2m3)(m2+m3)g = 0 ... [eq4*(m2+m3)]
    7> 2m3(m2+m3)a1 + 2m3(m2+m3)a2 -2m3(m2-m3)g = 0 ... [eq5*-2m3]

    8> a1(2m3(m2+m3)-(m1+2m3)(m2+m3)) - g(2m3(m2-m3)-(m1-2m3)(m2+m3)) = 0 ... [eq6+eq7]

    a1 = g(2m2m3-2m3m3-m1m2+2m2m3-m1m3+2m3m3)/(2m2m3+2m3m3-m1m2-2m2m3-m1m3-2m3m3)

    a1 = g(4m2m3 - m1m2 - m1m3)/(-m1m2 - m1m3)
  7. Oct 5, 2003 #6
    Here are the original equations:
    m1g - 2t = m1a1
    m2g - t = m2(a1+a2)
    t - m3g = m3(a1+a2)
    Can you explain to me how you got these three equations? Somehow you have written it with three unknowns. The best I ever did was to write it with 5 unknowns. I had,
    T1- m1g=m1a1
    T2 - m2g=m2a2
    T2 - m3g=m3a3
    There are two ropes, so there are two tensions. There are three masses so there are three accelerations.
  8. Oct 5, 2003 #7
    m2 and m3 are at the two ends of the same rope, so t2 and t3 must be equal, and also (relative to their pulley) a2 and a3 must be equal.

    Therefore, I dropped a3 and called both of them a2.

    As for the tensions, if you look at the lower pulley, there are two ropes pulling down and one pulling up. Well, the two pulling down are really two ends of the same rope, so those two tensions must be equal. Next, I decided that the tension in the rope that pulls up must be equal to the sum of the tensions pulling down. I don't know how to prove that to you. It just seems right to me. The best analogy I can come up with is, if you have 3 ropes knotted together at one point, and one rope-end is attached to the ceiling, and weights are hung from the other two rope-ends (and the ropes are weightless), obviously the tension in the first rope will equal the sum of the tensions in the other two. I know a pulley isn't the same as a knot, but it seems that, at least for an ideal pulley, this should be true.

    (If anybody else in the forum can explain this aspect more rigorously, please do.)

    So I called the tension in the "lower" rope (the one attached to m2 and m3) t, and therefore the tension in the rope attached to m1 is 2t.

    So now I'm dealing with just 3 unknowns: t, a1 and a2.

    All I have to worry about is ... did I set up the equations correctly?

    I didn't. While thinking about how to explain my reasoning to you, I found my mistake.

    More on that in a little while.
  9. Oct 5, 2003 #8
    This is the part I can't agree with. You assume that the pulley is motionless (ie, net force is zero). Even so, if the net force isn't zero, then won't the pulley have an infinite acceleration since it has no mass?
  10. Oct 5, 2003 #9
    OK. I finally figured out the correct signs for the acceleratons. Here are my new equations.

    m1g - 2t = m1a1 [this one hasn't changed]
    m2g - t = m2(a2 - a1)
    t - m3g = m3(a2 +a1)

    Now, taking positive g to be downward, if m1g > 2t, a1 will be positive, meaning m1 accelerates downward, and the movable pulley on the right accelerates upward.

    If m2g > t, a2 is positive, so m2 accelerates downward relative to the movable pulley. But if positive a2 is down for m2, and positive a1 is up for m2, when I combine the accelerations, they must have opposite signs. So the combined acceleration for m2 must be (a2 - a1).

    Similarly, in the equation for m3, if t > m3g, a2 is positive, so, for m3, positive a2 means up relative to the movable pulley, and positive a1 means the pulley itself accelerates up. So combining the accelerations for m3 requires the same sign for both. Therefore, m3's combined acceleration is (a2 + a1).

    After these corrections, my result for a1 is:

    a1 = -g(4m2m3 - m1m2 - m1m3)/(4m2m3 + m1m2 + m1m3)

    The same as the answer in your book, except for the - sign. I guess that's because I defined positive a1 to be down for m1, and they want down to be represented by a negative value.
  11. Oct 5, 2003 #10
    I think I'm only assuming that the pulley is motionless with respect to the two ropes. In fact, I think you just came up with the best argument to support my claim. The net force on the pulley must be zero, and therefore, the sum of the two tensions pulling down, plus the one tension pulling up, must be zero. And therefore, the magnitude of t1 = 2t2.
  12. Oct 5, 2003 #11
    But if the pulley on the right accelerates upward, we know that t1-2(t2)>0. t1-2(t2)=0 only in the reference frame that is at rest wrt the movable pulley. Can you use this result to determine the accelerations of the bodies in the reference frame that is stationary wrt the first pulley? I agree with all other parts of your solution, I just can't see how this part is valid.
  13. Oct 6, 2003 #12
    How about ... tstationary - tmovable = m3*a1 ?

    So I take care of the difference between t in the two frames by including the acceleration of one frame wrt the other.

    Sorry, that sounds so lame it doesn't even convince myself, but it's the best I can come up with. (Of course, there's also the fact that it gives the correct result, according to your book.) :smile:

    If you come up with anything better, please tell me about it,
  14. Oct 8, 2003 #13
    Here's a better argument to prove that t1 = 2*t2. Newton's 2nd Law requires it.

    F = ma and m=0, so &sum;F must equal 0. Mathematically, it can't be otherwise.

    Do you accept that?
  15. Oct 8, 2003 #14
    Yes, I accept that. I mean, I certainly believe you since you have the right answer. But what about the acceleration of the pulley! If the sum of the forces and the mass are zero, what can we say about its acceleration?

    Bottom line is pulleys should have mass!
  16. Oct 9, 2003 #15
    But this pulley has no mass (it's an "ideal" pulley), so F=0 and m=0, and therefore a can be any number. It's acceleration is determined by the acceleration of m1.

    You want the pulleys to have mass? If the pulleys have mass, the problem becomes much more complicated. You could not say that t1=2t2, and you would have to deal with not only the mass of the movable pulley, but also the inertia of both pulleys, the torques exerted by the ropes about the axes of the pulleys, the linear acceleration of the movable pulley and the angular accelerations of both pulleys.

    Have fun!
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