Kinda OT - a simple "mechanics" question

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.

A real beam will bend between F1 and F2, and pull away from F3.

jsw

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It's a moment and force balance, yes, but torque isn't moment of inertia.

To help visualize, redraw simply replacing F2/F3 w/ ropes tied to the ground, say, and their resulting tensions...

| | 1 2 3 V_______________________________________ /\ | | / \ | | ----------------------V----------V- XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX

Besides the above, there's the balance of forces that

F1 = F2+F3

Substitute for F1 above gives

(F2+F3)*L1 = F2*L2 + F3*L3 F2L1 + F3L1 = F2L2 + F3L3

collecting terms in F1, F3

F2(L1-L2) = F3(L3-L1)

F2/F3 = (L3-L1)/(L1-L2)

So their ratio is the ratio of the overall spacing...

The answer is that there is nothing that fundamentally sets the ratio of tensions at the two tie-downs. By lengthing one tie-down a smidgen, you reduce the tension at that point and increase it at the other.

So then you will ask, "what if the lever is infinitely stiff as are both tie-downs, and both tie downs have exactly equal length?" This is the same class of question as "what happens when an infinite force meets an infinite mass?" When you talk about a non-real world, you have to make up your own rules. Depending on the rules of physics you create for infinite stiffness, you might say there is no tension at 3 since it has all been relieved at point 2.

OK. Or, make the tie downs less absolute. See my reply to myself. Bob

Oh yeah - not what you'd first expect. Thanks. Bob

On 7/5/2013 6:13 PM, dpb wrote: [another drawing that didn't quote right]

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I don't think so ... consider if L2 & L3 were much greater than L1. Then F2 & F3 would be much less than F1 & still balance.

Thanks anyhow, Bob

Thanks - that makes sense. Instead of making up rules, I need to re-state the problem less abstractly. More at 11:00. Thanks, Bob

Having been enlightened by knowledgeable replies, let me re-state the problem.

Instead of points 2 & 3 being perfectly fixed points, let them be springs, with equal spring constants. So the forces will be proportional to the deflections at those points. If I still assume a perfectly rigid lever, then the deflections will be proportional to the distances and it's solvable (statically determinable, if that's the way it's put).

If A (alpha) is the angular deflection and k is the spring constant then:

F2 = k * L2 * sinA (for really small A) F3 = k * L3 * sinA and F2 = F3*L2/L3 etc

Which is exactly opposite of my original intuition that the forces would be _inversely_ proportional to the distance.

As k gets really large (and A small, but k * sinA finite, it starts to look like my original problem of points 2 & 3 fixed. But I suppose that the limit of k*sinA is not determinate and it's not _exactly_ the same case.

I await your destruction of my "thesis", Bob

Make a physical model with fishing scales and experiment. I think you will find that the total of F2 and F3 force* moment is a constant but either force can be any value between zero and the total. When you pull one scale harder the other one will back off proportionally. As I pointed out F3 can even become negative.

If you hang weights from the lever you have an infinite combination of L2 and L3 that combine to balance F1. Weights don't change their pull for small lever deflections, making them theoretically simple, the limiting case of infinite compliance. Limiting cases can be extremely useful to simplify a problem.

The practical problem with rigid supports is that the force distribution changes very rapidly and unpredictably as they and the lever yield. That can be handled to some extent by assuming every connection has been stressed until it yields enough to pick up its share, which the designers of the I-35 bridge failed to do correctly.

Here is a theoretical basis:

In your last post F2*L2 and F3*L3 equal F1*L1, not each other. jsw

You already had your answer the first time you posted this silly problem.

"Then F2 & F3 could be any combination of forces, as long as it balances."

F2 and F3 are any combination that balances the stupid thing. One can even be weight applied above, while the other is tension from below.

Why don't we make it interesting by turning it all upside down? Then you might see that there are 4 forces acting on the lever, not 3.

F1, F2, and F3 are floats. The lever is a boat, and you're sitting in it at the fulcrum.

That *may* make the solution easier, but doesn't make the structure statically determinate. A bit easier if, for example, you make the supports springs with a known rate and assume the lever is sufficiently stiff in comparison with the springs that it can be treated as a rigid body. Which is the converse of the usual structural situation where the supports of an elastic beam are assumed to be perfectly rigid. If neither of those assumptions applies, in other words, you have to account for the elasticity of both the lever and supports, the problem is somewhat more complicated than your original.

A valid simplification to demonstrate this is to make L2 = L3 and the springs fishing scales, the ones you keep in the vehicles to check weight at the scrapyard. You can determine the spring constant with the built-in tape measure. Mine extend 1" at 45 lbs.

jsw