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Normal force for a lever model

  1. May 23, 2013 #1
    My model is a lever on a table top. One arm is horizontal on the table, while the other arm is raised at an angle alpha. I'm assuming the weight of the horizontal lever (F) acts at its center of gravity at a distance f from the pivot, while the weight of the raised lever (W) acts at its center of gravity at a distance s from the pivot. Finally, the normal force is N. At static equilibrium the torque equation would be:

    f x F - N = s x W x Cos α

    To simplify the equation, I need to eliminate the normal force.
    Can I assume that the torque of the raised lever (s x W x Cos α) is such that it ever so slightly exceeds the term (f x F), so that N is approximately zero yet the system still remains at static equilibrium? I realize this is a contradiction of terms, but I can't think of another way to assume N=0.

    Any suggestions will be appreciated.

    Stan
     
  2. jcsd
  3. May 23, 2013 #2

    tech99

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    Gold Member

    Is N the reaction of the table on the lever arm?
     
  4. May 23, 2013 #3
    Yes, N is the force of the table opposing that of the weight of the resting lever arm when the latter is not balanced by the raised arm.
     
  5. May 23, 2013 #4

    tech99

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    Gold Member

    When the system is just balancing on its "corner", N acts thru the fulcrum so exerts no torque and can be ignored.
    When the horizontal arm is resting firmly on the table, N acts thru the CG of the complete system (including both arms). So calculate the location of this point and then take moments about it.
    You have here a machine whose geometry changes depending on loading.
     
  6. May 23, 2013 #5
    I suppose I can assume that N acts thru the fulcrum even when the horizontal arm is on the table & therefore, ignore it. Otherwise the equation becomes useless for my purpose.
     
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