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Tough Pulley System Problem

  1. Sep 26, 2014 #1
    1. The problem statement, all variables and given/known data

    Note this problem is from Morin 3.28.
    The atwood machine has two masses, m. The axle of the bottom pulley has two string ends attached to it, as shown. Find the accelerations of the masses. A small diagram is attached below. Note, at the two loose ends of string, there is a mass m. And the blue dot on the bottom pulley represents the axle.

    2. Relevant equations



    3. The attempt at a solution

    T_1 - mg = ma_1
    T_2 - mg = ma_2

    So I figure since you have two separate strings attached, the respective tensions should be different. Although, trying to find an equation that relates a_1 to a_2 seems very challenging though. Can anyone possible steer me in the right direction?
     

    Attached Files:

  2. jcsd
  3. Sep 26, 2014 #2

    gneill

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    Staff: Mentor

    The axle of the bottom pulley is offset from its center? That will complicate things. You'll need to find moments about the center of rotation.... you'll need some specifics about the dimensions of the lower pulley and the position of the axle.
    Fig1.gif
     
  4. Sep 26, 2014 #3
    I apologize for the bad drawing. Basically, everything should be centred. The blue axle should be at the centre of the large pulley. The strings should roll over the two smaller pulleys. Hope that clarifies things.
     
  5. Sep 26, 2014 #4

    gneill

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    Staff: Mentor

    Okay, maybe I can help with your diagram. How does this look:

    Fig1.gif
     
  6. Sep 26, 2014 #5
    Everything is correct, except the axle at the bottom pulley has two strings attached to it, instead of the one long string you included. One string extends from one side of the small pulley (on the top) to the centre axle, and the other from the centre axle to the mass m.
     
    Last edited: Sep 26, 2014
  7. Sep 26, 2014 #6
    Hi NATURE.M.
    I suggest you to draw the free body diagrams of two masses and pulley separately.
    Then set up equations using 2nd law.
    As there is only one string so there should be only one tension.
     
  8. Sep 26, 2014 #7
    Sorry I did't see your #post5 before replying.Please ignore #post6
     
  9. Sep 26, 2014 #8

    gneill

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    So,... something like this?

    Fig2.gif
     
  10. Sep 26, 2014 #9
    Exactly. The axle separating the two strings is what confuses me. I assume the tension in the two strings would be different, hence my initial equations. But theres something I'm missing.
     
  11. Sep 27, 2014 #10

    gneill

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    Staff: Mentor

    Think of the string coming down to the axle as terminating there. Instead of a hanging weight on a string below the axle, associate the weight with the axle itself, or transform the string holding the weight to the axle into a rigid support member. Either way you want to shed the idea that the two strings joining the axle are in some way related other than by happening to be connected to the same location.
     
  12. Sep 27, 2014 #11
    Ok so if I assume the bottom mass to be at the axle, the its weight is just mg. But then, should I just forgot about the tension on the 2nd string ? If so, I end up with basically the same equation, and when I solve for a_1 and a_2, there equal, which I know is not right.
     
  13. Sep 27, 2014 #12

    gneill

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    You can ignore the tension in the "second string" (supporting the mass attached to the axle), since you've incorporated that mass into the axle itself. Or you can leave the mass dangling on that string but realize that it has fixed length and moves as a unit with the big pulley. The string looping over the pulleys supports the big pulley at three places, and since the pulleys are assumed to have zero rotational inertia, the tension in all those string segments should be the same.

    What you'll need to do is draw Free Body Diagrams for the masses (one of which is now the big pulley) and determine how the movement (acceleration) of one relates to the other. The latter is done by imagining how far one mass would move for a given movement of the other --- remember that the string is unstretchable.
     
  14. Sep 28, 2014 #13
    Ok so doing as you said I obtain:

    3T-mg = ma_1
    T-mg = ma_2
    and the acceleration relation is: x_2 = -3x_1 => a_2 = -3a_1.

    From this the accelerations are easy to solve and are only in terms g. But I'd generally just like to verify that my above equations are correct ?
     
  15. Sep 28, 2014 #14

    gneill

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    Staff: Mentor

    Your equations look okay to me.
     
  16. Sep 28, 2014 #15
    Thanks a lot gneill.
     
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