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Possible to prove difference between Scissor Lift vs. Straight Lift?

  1. Feb 3, 2012 #1
    I've tried to come up with a simple model for scissor lifts

    and came to the conclusion that the force applied to the centre, by the screw must be the (payload + top arms)/tanθ

    I set payload + top arms = 981 N

    I then found the average value of the function by integrating from 15 to 90° and then dividing by pi/2-pi/12, and received an answer of 1013 N.

    A regular lift would have an average value of 981 N - top arms. It seems like using a scissor lift requires more effort, is the only advance that you can apply a load sideways?

    I'd appreciate any help.
    Last edited: Feb 3, 2012
  2. jcsd
  3. Feb 3, 2012 #2


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    Staff: Mentor

    The screw being represented by the horiz diagonal of your quadrilateral here?
  4. Feb 3, 2012 #3
    Yes sorry, the screw is the diagonal line that intersect the quadrilateral (scissor part).
  5. Feb 5, 2012 #4


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    Staff: Mentor

    There are two arms sharing the weight, so I think this will be load ÷ (2tanθ)

    Good question! :smile:

    If you use a basic screw, then with the scissor lift the ends of the screw can be allowed to extend out through the corners of the frame, allowing you to lift something from near floor level. Whereas, if you upend the screw and make it be the lift, then it will have to be able to extend through the middle of the load, or be able to retract into a hole in the ground, or comprise multiple telescoping sections, or something, to allow you to lift a load from near floor level. That's how it seems to me.
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