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Leveraging with opposite forces

  1. Aug 14, 2010 #1
    Suppose you have 3" tube acting as a post. A piece of metal is inserted into the pipe. The part inserted is 2 1/2" and the outside part 18". Your need to lift a 70 lb. platform resting on the external part of the piece of metal. If a spring is placed under the inside end of the piece it would pull down on the short end that is only 2 1/2" and needs to lift 70 lbs. You may add an additional spring to help pulling down on the short end or place a spring-out device (sping that pushes out or shock absorber pushig out) pushing up on the outside of the metal piece. As a result you would have a spring pulling down on the short end and a spring pushing up just a few inches from the pivot point. What would be better, using 2 springs on the inside end or a spring on the inside end pulling down and another on the immediate outside pushing up? Is there a formula to determine this?
  2. jcsd
  3. Aug 15, 2010 #2
    It sounds like a fairly simple exercise in 'moment of force'. But your description leaves too many questions to be answered simply.

    One thing that's obvious is that the 'inside' spring would need to be very strong to take the force on it. (about 500lb). A weaker spring placed directly under the platform would only need to take 70lb.
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