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Component of Force downward on Loppers (the kind for trimming bushes)

  1. Jan 20, 2016 #1
    Hi people,

    I have a lopper (the kind for trimming bushes or tree branches). I have it rigged up like the diagram below. The blades are on the bottom left. The bottom arm is straight while the top arm is at an angle theta. I have a load cell which is applying a constant force downward and compressing the arm, thus changing the theta.

    I noticed that as I apply the load cell force downward, it starts slipping down the upper handle as it compresses. In any case, if F is the force that I apply downward, does it make sense to componentize it into F_x and F_y components? If so, what would that be? I figured it may be some sort of arctan equation but haven't quite wrapped my head around how to go about doing it.

    diagram.jpg

    Thanks for your help.
     

    Attached Files:

  2. jcsd
  3. Jan 20, 2016 #2

    mfb

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    Staff: Mentor

    F, as you draw it, does not have a (non-zero) x component. You can split the force in a component orthogonal and a component parallel to the diagonal bar, however. Sine and cosine give the corresponding force components.
     
  4. Jan 21, 2016 #3
    Okay, so if I were the break up F into the orthogonal and parallel components to the diagonal bar (upper handle) would it be:

    Parallel: F*csc(theta)
    Orthogonal:F*csc(theta+90)
    ?

    Not sure if that is right...

    Edit: No, it seems the orthogonal is Fcos(theta)..
     
    Last edited: Jan 21, 2016
  5. Jan 21, 2016 #4

    mfb

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    And the other one is F sin(theta).
     
  6. Jan 21, 2016 #5
    Perfect. Thank you!
     
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