But the moment equations are not independent. One can solve for F1 and F2
in terms of F3, but you really can't get F3. There are other
considerations. The physical example of 3 scales shows quite clearly that
F1 & F2 depend on F3, but doing the actual algebra to get the solution
should also be a convincing example. If you propose a flexible bar with
known constant it should also be possible to readily get a solution if you
also propose the 3 points of contact are in a straight line.
I think the problem is soluble if you assume 3 springs holding up the bar
with the relaxed length of all 3 the same. You should be able to solve it.
Presumably the springs do not even have to have the same constant, as long
as you assume something like they all have the same relaxed length, and a
rigid bar. Indeed one should be able to solve for n springs holding up the
bar.
An interesting problem that I would propose to students is to ask them what
happens if the bar is tilted say 30 degrees. How many students would chug
through and get the solution, and how many would just use a little
proportional reasoning and have the solution immediately?
John M. Clement
Houston, TX
-----Original Message-----
From: phys-l-bounces@carnot.physics.buffalo.edu
[mailto:phys-l-bounces@carnot.physics.buffalo.edu] On Behalf Of LaMontagne,
Bob
Sent: Sunday, March 14, 2010 9:33 PM
To: Bob Sciamanda; Forum for Physics Educators
Subject: Re: [Phys-l] Statics conundrum
OK - What am I missing here?
F1 and F2 are applied at the ends.
F3 is applied, say, L/4 from the left end.
F1+F2+F3=mg where m is the mass of the rigid bar.
Moments from right end
F1 + F3*3L/4 - mgL/2 = 0
Moments from left end
F3*L/4+F2*L-mgL/2 = 0
Solve to get F1=mg/4, F3=mg/3, and F2=5mg/12
Why are you restricting the solution to only one moment equation?
Bob at PC
________________________________________
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.
*****************
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)