Calculating Weight from Lever Position?

June 4, 2009 1:51 PM

Given a first class lever the fulcrum in the middle of the crossbar. If you know the weight of one of the loads (C) on the lever and the angle (A) the lever is resting at and the length of the cross bar (L), how do you calculate the weight of the second item (W)?

I can get the difference in height pretty simply with a little trig, and I can get the difference in torque even, but frankly, I'm lost where I convert those into the relative weights of the item. I dabbled with some potential energy calculations there, but I'm sorta missing where you tie this together.

And I so wish this was homework, last time I had a physics class was over a decade ago.

posted by gte910h to Science & Nature (8 answers total)

When you say "difference in torque", do you mean the torque measured about the fulcrum? Torque is just a measure of force x radius, where presumably the force would be the weight of each load.

I'd need to know a little bit more about the situation, such as where the angle is measured from and whether the weights could be assumed to be at the very end of the lever, but if the weight "C" is the heavier weight then I think the formula would be:

TC/TW = C * (L/2) * sin(A) / W * (L/2) * cos(A)

so the formula for W would be

 W = (TW/TC) * C * sin(A) / cos(A)

Of course, if W is the heavier weight then flip the sine and cosine.
posted by muddgirl at 2:04 PM on June 4, 2009

Additionally, make sure your units are correct! If your torque is in lbs-ft then make sure you use L in feet and you'll get a weight in lbs. If your torque is in N-m then use L in meters and you'll get the force-weight in newtons. Divide by standard gravitational acceleration (9.8 m/s^2) to get the mass of the items.
posted by muddgirl at 2:06 PM on June 4, 2009

I like this question, because I can't answer it. Why is it not the case that the heavier side continues to fall until the bar is vertical? (Excepting the effects of friction at the pivot)
posted by mjg123 at 2:11 PM on June 4, 2009

Best answer: This is an indeterminate problem. In order for the bar to rest at a nonzero angle, there must be a frictional moment on the joint between the fulcrum and the crossbar in order to support the difference in moments. We've only got one equation (the sum of the moments about the fulcrum equals zero) and two unknowns (the weight of item W and the moment about the fulcrum).

Or, if you'd like, two equations (Sum of moments=0 and sum of forces=0) and three unknowns (weight of W, force on fulcrum, moment about fulcrum).
posted by mbd1mbd1 at 2:28 PM on June 4, 2009

Oh yeah, I'm totally wrong...
posted by muddgirl at 2:45 PM on June 4, 2009

Force * distance = force * distance. You have to know 3 of these to calculate the 4th.
posted by neuron at 3:07 PM on June 4, 2009

The question does not really make sense. A first class lever sitting on a plane, in a gravitational field (in ideal physics land with no friction) can be in 3 states. Left side on the ground, right side on the ground, and level. To figure out (relative) weights, you move the fulcrum until the beam is level. Then the distances from the fulcrum will tell you the relative weights (force x distance = force x distance), you can convert force to weight since it's being applied by a mass in a gravitational field.
http://en.wikipedia.org/wiki/Lever

Is this question based on something you see, or something you thought up? We need more details!
posted by defcom1 at 3:55 PM on June 4, 2009

Response by poster: Thanks guys, we went with something else.

Concept was flawed, prototype had just approximated a reasonably close answer by using a high friction pivot point.
posted by gte910h at 5:19 PM on June 4, 2009