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I Leverage and Forces on a Pivot System

  1. Feb 21, 2017 #1
    The diagram drawn below shows a lever system fixed to a point on the left with a pivot point (circle) on the line which attaches to vertical line. I have dimensions for Y and Z and the forces pulling the lever down on the right, and the force acting upwards from below. I am trying to work out the dimension for X to determine how long the lever needs to be to compress the upwards force, but I am unsure due to the varying angle as the lever is pulled down. Any help is greatly appreciated.
    Forces Sketch.png
    Before lever is pulled down.
    upload_2017-2-21_16-13-3.png
    After lever is pulled down.

    Thanks.
     
  2. jcsd
  3. Feb 21, 2017 #2

    BvU

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    Hi Saints,

    I sense some contradiction between the Y (FIxed Dim) and the Z piece seemingly staying vertical. Y is clearly shorter in the lower picture. Is the point where XY touches Z fixed on XY (and X also a 'Fixed Dim') or not ?
     
  4. Feb 22, 2017 #3
    Y is a fixed dim (may not be to scale on the image). The point on the left of line Y is fixed, and the point where XY touches Z is fixed. However, X is a variable dim as I am trying to work out the length of leverage required to compress the force acting up.

    I hope this clears it up a bit better.
    Thanks.
     
  5. Feb 22, 2017 #4

    BvU

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    Point on the left of Y is fixed. Y is a fixed dim ##\Rightarrow## point where XY touches describes a circle when the 'varying angle' varies. Correct ?
     
  6. Feb 22, 2017 #5
    Yes. The angle varies when the lever is pulled down, as Z always stays vertical. The circle represents a pivot point.
     
  7. Feb 22, 2017 #6

    BvU

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    You miss my point. The x coordinate of the pivot point moves in a circle, so Z can not stay vertical

    upload_2017-2-22_15-24-44.png
     
  8. Feb 22, 2017 #7
    I understand the point moves in a circle, however, where Z touches XY it is a pivot point. The pivot allows Z to stay vertical as it moves around the red circle marked on.
     
  9. Feb 22, 2017 #8

    BvU

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    Let me try again. In the left picture the arm is at 45 degrees, in the right it is horizontal. Do you see the distance between the left fixed point and the vertical Z has changed ? So either the leeft fixed point has moved to the left or Z has moved to the right and down.

    upload_2017-2-22_15-40-10.png
     
  10. Feb 22, 2017 #9
    I can see that the length of the arrows have changed as the lever moves down. However, the Y dimension has not changed- the length of the yellow line has stayed the same. From my understanding the pivot allows Z to stay at the same length and vertical, and also allows Y to stay at a fixed length?

    upload_2017-2-22_14-49-0.png
     
  11. Feb 22, 2017 #10
    The point on the left is always fixed.
     
  12. Feb 23, 2017 #11
    Has anyone got any other suggestions as to how this would be solved? Thanks.
     
  13. Feb 23, 2017 #12

    Nidum

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    The problem as stated can only be solved for the case where the lever almost doesn't move . Applying down pressure on the lever end can then apply down force on the column without the point of contact moving significantly .
     
  14. Feb 23, 2017 #13
    I'm struggling to work out what the equation would be. I want to find out the minimum length X would have to be to move the lever down, with a specified force pulling the lever down.
     
  15. Feb 23, 2017 #14

    Nidum

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    For what you describe - and assuming the small movement idea is ok - and if the 'variable angle' shown in your diagram is set to be somewhere near 90 degrees - the essential mechanism is just that of a simple lever .

    Solution can be obtained by taking moments about the fixed pivot point at left end of lever .

    Force down on column multiplied by distance Y = Force on end of lever multiplied by distance ( X + Y )

    Example :

    X = 2 metre , Y = 1 metre and force on end of lever = 100 N

    Force on column * 1 = 100 * ( 2 +1 ) . Therefore force on column = 300 N
     
  16. Feb 23, 2017 #15
    Thanks for your help.

    However, the varying angle can be between 45 and 90 degrees. Does that have an affect on the force on the Z column? And also does the length of Z change the force created?
     
  17. Feb 23, 2017 #16

    Nidum

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    I am struggling now . Perhaps we have to define the mechanism more clearly .

    When you say variable angle do you mean :

    (a) that the lever has to swing through that angle when the force is applied ?
    or
    (b) that the lever could be preset to different angles but doesn't then swing significantly when force is applied ?
     
  18. Feb 23, 2017 #17
    The column Z has to stay horizontal when the lever is pulled down. Therefore when the lever is up the angle will be aprrox. 45 degrees, and once its been pulled down it will end up at approx. 90 degrees.

    The mechanism is waste compactor, similar to the one below.
    41wF6gK1mQL._SX300_.jpg
     
  19. Feb 23, 2017 #18

    Nidum

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    So the vertical link is not actually fixed in position after all . Mystery solved .

    For the lever being horizontal the calculation in #14 is still ok .

    You could do a calculation for the lever at other angles but since the reaction force from squashing the waste is least when the lever is 45 deg up and greatest when it is somewhere near horizontal I think you could just use that one answer for horizontal for design purposes .

    Any calculations are going to be approximate anyway due to the varying properties of waste material .
     
  20. Feb 23, 2017 #19
    That's great. So as the force will be greatest at 90 degrees I can assume that it would be worst case?

    Yes, they will be approximate. I have done some testing to get an approximate force required to compact household waste.
     
  21. Feb 23, 2017 #20

    Nidum

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    Yes - I think you'd be quite safe to use lever being horizontal as worst case .
     
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