I have heard that walking on a frictionless surface is not possible because you can't grip, but does friction itself affect the force that you apply? if i walk on a rough surface vs a smooth surface, which surface would i be able to walk on faster? if i apply 1000N on a smooth surface, will the force be 2000N on a surface that has 2x friction as the smooth surface?

  • $\begingroup$ I feel like your first statement answers your question. If you can move, then you're already moving faster than if you were not moving. $\endgroup$ – Aaron Stevens Oct 25 at 12:42
  • $\begingroup$ I think we can't walk on smooth surface $\endgroup$ – VK_fan Oct 25 at 12:42
  • $\begingroup$ @Aaron Stevens not on a frictionless surface. $\endgroup$ – Zheer Oct 25 at 12:52
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    $\begingroup$ Right.... So it seems like you already know your answer $\endgroup$ – Aaron Stevens Oct 25 at 12:54
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    $\begingroup$ Re, "which surface would I be able to walk on faster?" Have you ever tried walking on ice? It's not simply a question of how fast you can walk, it's more like, a question of how fast you can walk without falling down. The reason you walk slowly on ice is, you have to walk carefully. People tend to fall if their feet slip when they weren't expecting their feet to slip. $\endgroup$ – Solomon Slow Oct 25 at 13:01

The maximum force that friction can apply is given by the equation $F = μ N$. Here, $N$ is the normal force, which on a flat rigid surface is equal in magnitude to the weight of an object. The coefficient of friction is $μ$, which is related to how much "grip" the surface has - how rough or smooth it is. Ice has a low coefficient of friction, while sandpaper has a higher one. You can see that as the coefficient of friction increases, the maximum frictional force also increases - this essentially allows you to push off of rough surfaces harder. If you try to push off a surface with a force that exceeds $μN$, the reactive friction force will still only be $μN$, so your foot will slip backwards, since the friction force is unable to push on you as hard as you are pushing on the surface.

The frictional force will be equal and opposite to the force you apply to the surface, up to the point where you exceed the maximum possible frictional force, at which point you start slipping. On a smooth surface, this happens with lower force than with a rough surface. If you try to take accelerate to a sprint while on ice, your feet will slide backwards. If you do the same on rough ground, the larger frictional force will allow you to accelerate faster (exert a larger force without slipping).

So long as the force you apply is below $μN$, the coefficient of friction does not change how hard a surface pushes on you. It will always be equal and opposite to how hard you are pushing on the surface. The only thing that changes is the threshold for when slipping starts to occur - at which point the frictional force cannot match the force you're applying. When stepping forward with a small amount of force, ice and sandpaper will both apply the exact same reactive force - if you're not slipping, the surface must be pushing on you as hard as you're pushing on it, regardless of what it's made of. The important difference is that ice fails to give that same reactive force when attempting to push off with greater force.

  • $\begingroup$ What about two surfaces, one of them is much rougher than the other one,with coefficent of friction high enough that my feet doesn't slip on both of the surfaces, will the force that i apply on both of the surfaces be the same?will F₁=F₂? $\endgroup$ – Zheer Oct 25 at 13:13
  • $\begingroup$ The coefficient of static friction determines the maximum force you can apply. What force you actually push is up to you. You can walk at the same pace and apply the same force on two surfaces with different coefficient of friction. Or you can push harder on the surface with lower coefficient, as long as the pushing force does not exceeds the maximum static friction. $\endgroup$ – nasu Oct 25 at 13:18
  • $\begingroup$ @nasu that was the answer that i looking for, thank you very much. $\endgroup$ – Zheer Oct 25 at 13:21

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