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Three mass pulley system

  1. Sep 10, 2016 #1
    1. The problem statement, all variables and given/known data
    The system of masses M1, M2, and M3 in the sketch uses massless pulleys and ropes. The horizontal table is frictionless How are the accelerations related?

    2. Relevant equations

    3. The attempt at a solution

    I feel like the solution will come from the constraint equation involving the string of the pulley system. However, I am not sure how to do this. It seems that since M1 and M2 are different masses, they do not have to accelerate at the same rate, which means that M3 would have all kinds of funky motion, not just linear in the y-direction. How do I think about this problem?
  2. jcsd
  3. Sep 10, 2016 #2
    To think about these kinds of problems, I like to go to the extreme cases. One extreme case would be if M2 and M3 were equal and M1 was infinite. What would happen in that situation. What would the acceleration of M2 be relative to M3? Another case (which really isn't an extreme) would be if M1 and M2 were equal. What would the acceleration of M3 be relative to M1 and M2?

    Also, are there any forces available to move M3 in a direction other than vertical?
  4. Sep 10, 2016 #3
    So this is what I tried, where x1 is the length of the string from the first mass to the first pulley, and x2 is the length of the string from the first pulley to the middle pulley, and x3 is the length of the string from the third pulley to the third mass. Then

    ##l = x_1 + \pi r_1 + x_3 + \pi r_2 + x_3 + \pi r_3 + x_2##
    ##0 = \ddot{x}_1 + \ddot{x}_2 + 2 \ddot{x}_3##
    ##-2 \ddot{x}_3 = \ddot{x}_1 + \ddot{x}_2##

    Is this the correct relation between the accelerations of the masses?
  5. Sep 10, 2016 #4
    It looks right to me, but I haven't thought it all the way through. Here's how I see it in my own way of thinking. For a car driving down a road, the velocity of the point of the tire making contact with the road is 0. The midpoint of the tire/wheel is moving at the same velocity as the car. The top of the tire is moving twice as fast as the car. That is the situation we have here if M1 is infinite mass. The left side of the lower pulley is the road, the middle of the pulley is moving at some velocity and the right-most side of the pulley is moving twice the velocity of the center. So therefore your equation works in that case (because a1 = 0).

    For the case of M1 = M2, the velocity (and thus acceleration) of M1 and M2 has to be equal to M3 because there is no rotation of the lower pulley. It's as though the 2 sides of the rope going to the lower pulley are connected directly to M3. So once again, your equation works in that case.
  6. Sep 10, 2016 #5
    Well that's good. Also, can you explain why there is a negative sign? What does this mean physically? Is the negative sign supposed to be there?
  7. Sep 10, 2016 #6

    Charles Link

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    To answer your question, the position of the mass ## M_3 ## in the y-direction is ## y=x_3+b ## where ## y ## is measured from the plane of the table and ## b ## is some constant. Thereby ## \ddot{y}=\ddot{x_3} ##. If you want to call that a " ## -y ##" so that ## -y=x+b ## and ## \ddot{y}=-\ddot{x_3} ##, you can do that also. The sign of ## x_3 ## is determined by the length of the string: As ## x_1 ## and ## x_2 ## get longer, ## x_3 ## must get shorter, i.e. ## x_1+x_2+2 x_3=Constant ##.
  8. Sep 10, 2016 #7


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    You seem to have swapped the meanings of x2 and x3 at some point. According to your definitions, x1+2x2+x3 should be constant.
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