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Scenario 1:

enter image description here

Here, $O$ is the center of mass of a uniform, rigid and solid block. $F_g$ is the force of gravity acting on the body vertically downward, and $N$ is the normal force acting on the body vertically upward. The coefficient of friction of the floor is $\mu$ (distinguishing between $\mu_s$ and $\mu_k$ is probably unnecessary here). So, if the body is pushed horizontally, then the frictional force will be $F_f=\mu N$.

Scenario 2:

enter image description here

Here, $O$ is the center of mass, and we are in space, so no force of gravity exists. The coefficient of friction of the floor is $\mu$. As no normal force exists in this scenario, the frictional force is zero.

My comments:

We can evidently see the two scenarios are different. One has friction, the other doesn't, but why is this the case. The net force on the block in both cases is zero, so why does friction exist in scenario 1 and doesn't exist in scenario 2? Aren't scenarios 1 and 2 essentially the same as the net force on the block is zero?

My question:

  1. Why does friction exist in scenario 1, but doesn't exist in scenario 2?
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  • $\begingroup$ Think of it this way... For scenario 2, the friction is there, but only if you can apply some downward force to the block. Friction is just resulting from the physical properties of objects, and frictional force is measurable when we can observe it in action. One could also argue that the friction force 'doesn't exist' in scenario 2 because gravity is absent. $\endgroup$
    – bleuofblue
    Apr 8, 2022 at 7:19
  • $\begingroup$ When you say in scenario 1 that the object "is pushed horizontally" do you mean the object is sliding on the surface? $\endgroup$
    – Bob D
    Apr 8, 2022 at 12:16
  • $\begingroup$ "Aren't scenarios 1 and 2 essentially the same as the net force on the block is zero?" Why do you say the net force in scenario 1 is zero? $\endgroup$
    – Bob D
    Apr 8, 2022 at 12:22
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    $\begingroup$ Does this answer your question? Why does frictional force depend on normal reaction and not the weight of a body? $\endgroup$
    – tryst with freedom
    Apr 12, 2022 at 13:48

4 Answers 4

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  1. Why does friction exist in scenario 1, but doesn't exist in scenario 2?

At the microscopic level, friction is usually modeled as due to irregularities (hills and valleys) of the contacting surfaces such that when you put two surfaces together they actually make contact at very few places.

In scenario 1, when trying to move the block horizontally to move it over the surface below, the high part parts on each surface get "stuck" on one another due to the continued downward force of gravity and upward normal reaction force that keeps the surfaces in contact with one another. The applied horizontal force keeps unsticking the high points of the surfaces to maintain relative horizontal motion.

In scenario 2, however, there is no force to maintain contact between the surfaces. This allows the block, initially in contact with the surface below, to ride up the high points and separate completely from the surface below losing contact.

In short, pressure needs to be maintained between the surfaces for there to be friction. That pressure does not exist in scenario 2.

Hope this helps.

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The intermolecular interactions between the top block and the bottom block exchange momentum. The direction of the momentum transferred to the top block from the bottom block in each interaction is random, but it can only point in an upward or lateral direction, never down. If there's no force pushing down on the block from above, then after the first few interactions, its net momentum is in the upward direction - that is, it is drifting very slowly upward, away from the interface.

Two objects without a normal force don't stay touching for any appreciable amount of time, and two objects that aren't touching can't have friction.

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The net force in each case is indeed zero. In the first case there are only vertical forces (provided you don't push the mass).

In the second case you have no forces in any direction and if you push the mass in the horizontal direction, there will still be no frictional force since there is no vertical force.

Put simply, without a force pressing an object onto a surface (like a normal force in scenario one), there cannot be a frictional force like in the second scenario. So the first scenario is different to the second since there exists a normal force.

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  • $\begingroup$ "So the first scenario is different to the second since there exists a normal force."- but isn't the normal force negated by the force of gravity? If a force is negated, then isn't it the same as the normal force not existing in the first place? So, how can scenarios 1 and 2 be different. By the way, thanks for your answer, sir! $\endgroup$
    – tryingtobeastoic
    Apr 8, 2022 at 8:09
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    $\begingroup$ The normal force is not actually negated. It is simply balanced by gravity. It is still there and exists. This can be demonstrated since if you push the mass, there will be a definite frictional force. Scenarios 1. and 2. are different in the sense that you have forces in equilibrium in 1. whereas in 2. there are no forces at all. And no worries at all. Cheers. $\endgroup$
    – joseph h
    Apr 8, 2022 at 8:13
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Friction happens due to asparities on the contacting surfaces interlocking and chemically adhering temporarily.

In your scenario 2, where you imagine a parallel motion of the block and then suddenly the normal force is removed, then the block will continue to experience friction-like forces for a brief moment. But it will then quickly distance itself slightly from the surface due to these surface actions causing an uneven resistance in the block as a whole. And the block will then (ideally) continue with no contact, just very close to the surface.

Also a note: in scenario 1, the formula $F_f=\mu n$ most likely implies kinetic friction (or the very limiting case of static friction), so it does matter that you distinguish between $\mu_k$ and $\mu_s$.

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