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A motorcycle rides on the vertical walls around the perimeter of a large circular room. The friction coefficient between the motorcycle tires and the walls is μ. How does the minimum μ needed to prevent the motorcycle from slipping downwards change with the motorcycle’s speed, s?

I understand that:

$$N = \frac{mv^2}{r}$$

$$mg = \mu N$$

So I eventually get the correct answer which is that $μ ∝ \frac{1}{s^2}$ but I feel like I'm making a mistake. There's velocity in one direction which is perpendicular (?) to the gravitational force. How does this affect frictional force? Does the direction change or am I doing it right?

I'd really appreciate any help on this.

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Since the motorcycle is doing a circular trajectory, it is exerting a force $N=mv^2/r$ on the wall (and the wall is exerting the symmetric force on the motorcycle). You have a force because the velocity vector is changing, in this case the direction of the velocity is changing. And the change in the velocity, in this case, is not related with gravity but, instead, with the interaction between the wall and the motorcycle.

It is this force $N$ that is related with the frictional force. For example, if you press your hand against a wall and try to move it vertically, it is harder to move the stronger you push against the wall, i.e., the frictional force will be larger.

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  • $\begingroup$ kudos for saying "symmetric force" rather than reaction! $\endgroup$
    – Bill N
    Mar 20, 2019 at 15:21

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