# Torque on an open-ended rectangle

1. Jan 21, 2010

### Zula110100100

I was at work and we build a "fence" on a trailor to keep the load in place, and we got into a disagreement about how much stress would be on each corner if we hit a bump and the boards moved up unequally. Now, there's a lot more physics in that scenario than the one I am describing but I think this simplified version answers my question.

Which point would have more stress in this example:

I would imagine the only force we'd be calling stress on a corner would be the torque. Since lbs is a weight it already takes gravity into account so it would be 4 ft-lbs of torque on pt. A. On point be we would figure out the distance to weight with

a2+b2 = c2

4+4 = 8

c = 2.8284271247461900976033774484194

But the force of gravity is at a 45° angle to the radius so we use c sin 45 which is 2 ft.

So for pt. B, 2 lbs at 2 ft = 4 ft-lbs the same as A.

I tried with a couple of other rectangles and it was always the same...is that true?

Also, I am less sure of the physics but I think with a 4 sided rectangle if a force is applied parallel to one of the sides the torque on each corner would be equal?