# The maximum force acting on a wheel that rolls without slipping as a function of the coefficient of friction

If we assume that our tire is a homogenous disk and that it is rolling without slipping, with a force $$F$$ exerted on its center of mass pulling it forward, how would I determine the maximum force as a function of $$\mu$$, the coefficient of (static) friction.

I know that the friction is $$\mu mg$$, where $$mg$$ is the normal force, and that for the wheel to roll without slipping $$v = R\omega$$, where $$v$$ is the velocity of the center of mass, $$R$$ is the radius of the tire and $$\omega$$ the angular speed. We also have that for a uniform disk, $$I = 1/2mR^2$$

How do I now find what the maximum force is such that the tire rolls without slipping?

## 1 Answer

I know that the friction is μmg

Not always true. $$\mu mg$$ corresponds to the maximum friction, $$f_{\rm max}$$. The static friction force $$f$$ can be anywhere between $$0$$ and $$f_{\rm max}$$:

$$0\leq f\leq f_{\rm max} = \mu N$$

To solve the problem, first, write down the net force and torque equations

$$ma=F-f \qquad \text{and}\qquad I\alpha = fR$$

Using $$v=\omega R\implies a=\alpha R$$, eliminate $$a$$ and $$\alpha$$ to get an equation relating $$f$$ and $$F$$. You can also rewrite the inertia as $$I=CmR^2$$ to cancel the $$R$$'s.

Then, suppose that $$f=f_\mathrm{max}=\mu mg$$. This implies that the force of friction is maxed out, such that if $$F$$ gets any larger, $$\mu$$ must also increase. This simply means that the friction force acting on the disk cannot get any larger without increasing $$\mu$$.

From there, solve for $$F$$ and you're done.

• Thank you! Just as a follow up question, how come we are interested in the static friction rather than kinetic friction? My thought is that since the tire is rolling without slipping the bottom of the tire is instantaneously at rest, which means that the friction at that resting point will be static? – StannisBa Jan 18 at 5:19
• That is precisely what happens. I have also illustrated that in my answer here: physics.stackexchange.com/questions/603126/… – user256872 Jan 18 at 5:27