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2 pulleys spinning 2 loads

  1. Sep 9, 2004 #1
    hie everybody, this is my first post on this interesting site... :biggrin:
    i am not sure how do i count the amount of force needed to turn such a pulley from the spindle attached to the dowel through the spindle. the pulleys are fixed to the dowel and are not movable. they just turn as the rope binds them together.
    the groove of the spindle is layed with a layer of elastic band so that the thin rope doesnt slip of the spindle when span. The front part of both the dowels are each lifting a different amount of weight.
    how do i count amount of force needed to spin from the spindle? what is the theory behind this? an illustration is attached to this thread. thanks a million... :smile:

    Attached Files:

  2. jcsd
  3. Sep 9, 2004 #2


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    One way to tackle this problem is energy conservation. I assume this is a homework problem?
  4. Sep 9, 2004 #3
    energy conversation? i guess is that hooke's law right?
    but i am only a high school student and i havent learn this yet. can u explain and show to me a little more please?
    this is not my homework. i asked this because i did an artefact which is similar to what i had drawn in the diagram and would like to know how much work done i have to apply to turn the spindles with 2 loads and find it interesting as well. :tongue2:
  5. Sep 10, 2004 #4
    I have thought of a solution. But i don"t know whether this is correct or not. But u can try and work out on this.

    If we assume both the pulleys have same radius (r) one roll of each pulley must lift the masses by a distance 2-pi-r. so the energy stored in the system is (m1+m2)g.2-pi-r for one rotation. If the distance of the handle of rotation of the spindle is R and the force required for rotation is F, the the torque required is F.R and the work done to rotate a full rotation is F.R. 2-pi. It follows that ,
    F.R. 2-pi = (m1+m2)g. 2-pi-r
    or, F.R = (m1+m2)gr
    So, F = (m1+m2)gr/R
    Now put the values and see the results if it tallies.
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