How to Determine if Friction Will Act?

[Acala]

Let's consider a system of three boxes stacked vertically, resting on the ground. There is friction acting between all surfaces involved.

I push on the middle box, but without enough force to move the system at all. If we are to analyze the middle box, we'll see a static friction force coming from the boxes both above and below it. The friction from the top box must equal zero, though, because if it were nonzero, the Newton's third law paired friction force from the middle box would cause the top box to accelerate.

My question is this: if we were not given the information that nothing in the system is accelerating, and rather simply given the information about the applied force and friction forces, how would we know that no friction were coming from the top box? Also, why is there no friction from this box in the first place?

 

[tiny-tim]

Hi Acala!

The friction from the top box must equal zero, though, because if it were nonzero, the Newton's third law paired friction force from the middle box would cause the top box to accelerate.

That's completely correct.

if we were not given the information that nothing in the system is accelerating, and rather simply given the information about the applied force and friction forces, how would we know that no friction were coming from the top box? Also, why is there no friction from this box in the first place?

I don't understand why you're asking. You obviously understand the F = ma principle applied to the top box. Everything in mechanics ultimately boils down to F = ma.

Why should there be friction if nobody's pushing the top box?

 

[Acala]

I mean that if we were given no information about the acceleration of the system, then we would have to assume that there would be a friction force from the top box, whatever value it may be. I say this because I calculated the friction force from the top box under the assumption that it was not accelerating.

I am wondering why all of the friction is coming from the bottom box in the first place; why is the top box not contributing friction and then just sliding off? Does it have something to do with the normal force?

 

[tiny-tim]

… why is the top box not contributing friction and then just sliding off? Does it have something to do with the normal force?

there was friction from the top box when you started pushing …

but as you approached cruising speed, the acceleration reduced, and so did the friction,

and when you reached cruising speed, the acceleration was zero, and so was the friction

 

[cepheid]

Hi Acala! That's completely correct. I don't understand why you're asking. You obviously understand the F = ma principle applied to the top box. Everything in mechanics ultimately boils down to F = ma.

Why should there be friction if nobody's pushing the top box?

I think what the OP is asking is, if you push on the middle box, then you're trying to make two pairs of surfaces slide relative to each other. One pair happens to be the top of the middle box, and the bottom of the top box. So why isn't there static friction between the top and middle boxes in the case when you push, but not hard enough to make anything move?

I tried this with a stack of three books. I pushed on the middle one. If I didn't push hard enough, nothing happened.

If I pushed hard enough, static friction between the middle book and the bottom book was overcome, and the middle book started moving. The top book moved along with the middle book in this situation. So, clearly, in the situation where the middle book moves, there IS static friction between the middle book and the top book, because the middle one carries the top one along with it. Yet, in the case where nothing moved, there was clearly no static friction between the middle and top book, because this would lead to an unbalanced force on the top book. How do contact surfaces between the middle and top book "know" that no friction is necessary in the latter case?

 

[cepheid]

Thinking about it some more, I think the answer is this:

In the case where you push on the middle box, but too weakly for anything to happen, it will always be true that static friction between the middle and bottom boxes is stronger than static friction between the top and middle boxes. The reason is that for the former case, the normal force is equal to the combined weight of the middle and top boxes. For the latter case, the normal force is only equal to the weight of the top box.

The middle box can't slide if the bottom one is holding it in place. And if the bottom one is holding it in place, the top one has no need to, since, from its "point of view" the middle box is not trying to slide relative to it.

 

[Acala]

Yes, cepheid's interpretation of my question was correct.

The middle box can't slide if the bottom one is holding it in place. And if the bottom one is holding it in place, the top one has no need to, since, from its "point of view" the middle box is not trying to slide relative to it.

This does make sense, and it's quite insightful, but why, in that case, was the bottom box the first box that was called upon to provide the static friction? It seems that even if its maximum potential static friction force were lower, it could simply be overcome and the top box would be called upon to provide the friction.

Is there something more fundamental causing this? Perhaps, since gravity seems to be the force running this situation, it's because gravity points downward? And if that sounds reasonable, could it be (or has it been) generalized?