Solving Balance Scale Math: X vs Y Weight Difference

[jackrabbit]

I was trying to figure out the following - assume you have an old fashioned balance scale, with two pans hanging from a lever with a fulcrum in the middle of the lever. On one side you have a weight of X, and on the other side you have a weight Y. If Y is big enough, that side of the scale will fall until it hits the desk holding the scale. But if Y is only slight more than X, the scale will only tip in Y's direction a small amount. That seems intuitive enough, but how does the math work? As long as there is any difference in weight between the two sides, which doesn't side Y fall the same amount in both cases?

I thought it had something to do with torque, but I can't get the calculations to work. For example, assume X is 5 lbs and Y is 5.5 lbs, and they are hanging from pans that are each 4 units from the fulcrum. The torque from X is (5*4* the sine of the angle between the force vector and the lever). The torque from Y is (5.5*4*the sine of the angle between the force vector and the lever). I thought that the differential between the two torques would equilibrate as the angles changed. However, as the angles are always supplementary to each other, the relevant sines are always the same, so the torque never ends up in equilibrium.

So, clearly my math and/or physics is wrong. Can someone help?

 

[BruceW]

There must be something else that forces the system back to equilibrium when it is displaced a small amount. I would guess it is because friction (at the connection between lever and fulcrum) increases slightly when the lever moves slightly from its initial position. Another possibility is that there is some kind of spring-like force (at the fulcrum connection) that returns the system to steady when there is only a small displacement. For example, maybe the connection gets slightly tighter when the lever rotates, so a small displacement is negated.