There is an easy way to visualize that this situation is not uniquely determined: say that the two men at the ends are holding digital scales upside down up against the ends of the bars. The scales should read the solutions of your original problem. Now have that third kid come along with his digital scale, assign him a place to stand and have him start pushing gently. As he applies a force, the other forces change. But there is a range of forces he can apply. You can specify his force and solve for the others. You have him increase his force until one of the original scales reads zero (or even both if the third kid is in the right place).
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From: phys-l-bounces@carnot.physics.buffalo.edu [phys-l-bounces@carnot.physics.buffalo.edu] On Behalf Of Bob Sciamanda [treborsci@verizon.net]
Sent: Sunday, March 14, 2010 5:02 PM
To: phys-l@carnot.physics.buffalo.edu
Subject: [Phys-l] Statics conundrum
Here is a common elementary statics problem:
A perfectly rigid and uniform beam of given weight and length (W and L) is suported by two men, exerting upward vertical forces F1 and F2, one at each end of the bar. Determine F1 and F2 in terms of W and L.
This is easily solved by imposing translational and rotational equilibrium:
F1 + F2 = W
W*L/2 = F2*L => F2 = W/2 and F1 = W/2 (independent of L)
One can even add other given loads at given positions on the bar, and the problem is still easily solved.
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But a curious student might uncover the following conundrum:
If one adds a third man exerting a third upward force F3 at a given location (say L/4 from one end), The two equilibrium equations are insufficient to solve for the values of the three unknons, F1 F2 and F3.
The three man experiment can be performed and the forces measured (even with added loads on the beam). They ARE physically determined.
How does one analytically predict this result?
Please discuss. Is it the perfect rigidity which must be relaxed? Why?
How explain this to the curious student? (and to me)