Basic Statics - A weird outcome. Contradictory?

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The discussion revolves around the static friction coefficient and its implications in mechanics. A scenario is presented where a beam's static coefficient is 0.3, leading to calculations that suggest the beam does not slide. The user questions the validity of applying the static friction formula Fs = N x μ without considering maximum static friction, leading to confusion about the normal force. It is clarified that the static friction force adjusts to maintain equilibrium and that the coefficients represent maximum values reached only at the verge of movement. Ultimately, the conclusion emphasizes that the static friction force is not equal to the product of the coefficient and normal force unless movement occurs.
Femme_physics
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This is not a HW question! THIS IS A DISCOVERY that I want to explore. I will explain it as follows:

Suppose I have this exercise, where I know the static coeffecient between the beam and the floor is 0.3, whereas in the wall there is no friction. I want to find out if the beam slides, or stays up. That's easy.

http://img703.imageshack.us/img703/4676/fsmaxtest.jpg

So by doing the calculations, I can see the beam doesn't slide. I solved the problem. Now, let's say I want to experiment further. What happens if I put the result of Fs I got when I didn't calculate Fs_max into the Fs = N x μ equation? I think it's a valid thing to do, since it's in the formula. But what happens then?

I'd get a different result for N. Which doesn't make sense, since N must be 30 [N] to resist the weight of the beam.

The friction coefficient sure can't change, since it is constant!

So, what did I just do? Did I find loop in mechanics theory?
 
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Femme_physics said:
What happens if I put the result of Fs I got when I didn't calculate Fs_max into the Fs = N x μ equation? I think it's a valid thing to do, since it's in the formula.

Not valid. If the surfaces were moving, you could use such a formula. In case of static objects, we can only get the maximum friction force.

When we are standing on the floor, there is a normal equal to our weight W . The surface has a μ too. Does it mean we are experiencing a horizontal force of μW on our shoes?!
 
When we are standing on the floor, there is a normal equal to our weight W . The surface has a μ too. Does it mean we are experiencing a horizontal force of μW on our shoes?!
Of course not. We don't feel friction without moving in the axis it exists.

Not valid. If the surfaces were moving, you could use such a formula. In case of static objects, we can only get the maximum friction force.

How come?

If you recall

http://img62.imageshack.us/img62/7107/frri.jpg

So clearly either the friction coeffecient or the normal force haven't reached their max value. One of them must change. Since it can't be the normal force, as that would defy mechanics, it must be the coefficient: This is my new assumption. The coefficients we see in charts therefor are maximum values. We reach that maximum value of that coefficient only on the verge of movement.

How about that for theory?
 
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Femme_physics said:
What happens if I put the result of Fs I got when I didn't calculate Fs_max into the Fs = N x μ equation?
You will get the minimum N needed to prevent sliding due to Fs.
Femme_physics said:
I'd get a different result for N.
No, it has nothing to do with the actual N.
Femme_physics said:
The coefficients we see in charts therefor are maximum values.
Yes, the static friction coefficient tells you the maximal horizontal force for a given normal force, for a static case.
 
Obviously.In every decent book I read,
they clearly state that Fstatic≤μstaticN
and NOT FstaticstaticN

That is,the friction force adjusts itself to keep the concerned body in equilibrium.And the maximum it can be in order to negate the applied forces is given by μstatic times N.

So the coefficient remains constant,but the maximum force of static friction,ie μstaticN is experienced only on the verge of the movement.
 
Got it sorted out. Thanks :)
 
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