On Fri, 05 Jul 2013 17:47:20 -0400, Bob Engelhardt
wrote:

Let's say there's a lever with 3 forces acting on it:

1 2 3
V_________________________V____________V
/\

let F1, F2, & F3 be the forces and L1, L2, & L3 be the distances of
those forces from the fulcrum.

Then
F1*L1 = F2*L2 + F3*L3

Simple enough. Now, let's say that F1 is a fixed force (e.g., a weight
sitting on the lever). Then F2 & F3 could be any combination of forces,
as long as it balances.

Now say that points 2 & 3 are fixed. I.e., the lever is "tied down" at
those points. What are the forces acting on those tie downs? Is the
torque (moment of inertia, if you will) attributable to each equal, so
that the forces are inversely proportional to the distances? I.e.:
F2*L2 = F3*L3 (= 1/2 F1*L1)

It seems intuitively so, but if it is, how do you show that it is?

This has been bugging me for days, so any resolution that you can offer
will be greatly appreciated.

Bob

If 2 and 3 are fixed the problem is "statically indeterminate." In
other words it can't be solved with simple algebra; the elastic
properties of the lever must be considered. In a structural context
the lever would be considered a continuous beam.

--
Ned Simmons