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

1. Jan 20, 2016

### noblerare

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.

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2. Jan 20, 2016

### 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.

3. Jan 21, 2016

### noblerare

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
4. Jan 21, 2016

### Staff: Mentor

And the other one is F sin(theta).

5. Jan 21, 2016

### noblerare

Perfect. Thank you!