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Rotate pulleys using gravity.

  1. Dec 3, 2006 #1
    Cross section of pulley1, pulley2 , pulley3 and pulley4 are as shown in figure. Centres of Pulley 1 and pulley 2 are placed on same horizontal line. Centres of pulley 1 and pulley 2 are fixed. Pulley 1 and pulley 2 can rotate about their centre. Centres of pulley3 and pulley4 are also placed on one horizontal line. Pulley 3 and pulley 4 can also rotate about their centre. Centre of pulley 3 and pulley4 are joined with one stationary plate.Pulley 3 and pulley4 are simply supported on pulley1 and pulley2.Now we apply some load or weight on stationary plate. Finaly this force is applied on pulley 1 and pulley 2 in vertical downward direction because pulley 3 and pulley4 are simply supported on pulley 1 and pulley2. Tengential componants of these forces helps to rotate pulley1 , pulley2 , pulley3 and pulley4.Pl visit [crackpot link deleted]
     

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    Last edited by a moderator: Dec 3, 2006
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  3. Dec 3, 2006 #2

    russ_watters

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    Staff: Mentor

    If you are here to learn, great, but be advised that we will not support crackpottery here.

    The flaw in your reasoning is simple: force on a single contact point of a pulley can only be along the line connecting their centers. To figure out what the forces are everywhere, draw a free body diagram at each pulley center.
     
  4. Dec 4, 2006 #3

    FredGarvin

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    I can't even tell what he is trying to do here.
     
  5. Dec 4, 2006 #4

    russ_watters

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    Staff: Mentor

    It appears to me that he thinks that since weight acts straight down while the line of force between the touching pulleys is at an angle that there is a force component at the contact point of the pulleys, tangential to them. The error is drawing a fbd (if you can call it that) that shows the weight acting from the edge of the pulley instead of its center.

    Regardless, any 2nd-grader playing with Legos could tell you this won't do anything.
     
    Last edited: Dec 4, 2006
  6. Dec 4, 2006 #5
    Hi Guys,
    And that ain't the only error! He also labels those forces as each mg/2 (so he is obviously not accounting for the weight of pulleys 3 & 4 and the stationary plate that conjoins them).

    Sure enough... but I too would like to know just exactly what problem he thinks he is solving with this Rube Goldberg special?

    Rainman
     
  7. Dec 4, 2006 #6

    Danger

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    It looks like the basis for a fairly decent printing press... :uhh:
     
  8. Dec 17, 2006 #7
    Just another version of the classic counterweighted perpetual motion machine. Works as well as any other.
    See Gardner D. Hiscox's "970 Mechanical Appliances and Novelties of Construction" Copyright 1904.
     
  9. Dec 18, 2006 #8

    Mech_Engineer

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