I was thinking about the maximal acceleration of a bicycle, I found a similar question.

What remains unclear is that I found two coefficients of friction, one for rolling and one for static friction.

So far I believed that I calculate the friction force like

$$F = \nu m g$$

where $\nu$ is the rolling resistance, but if I use the formula from the linked question I get the

$$ a = \nu g = 0.004 \cdot 10\,\text{m/s}^2 = 0.04\,\text{m/s}^2 $$

where $\nu$ is the rolling resistance between a tire and asphalt road.

Now that number seems to be rather low. This video states I have to use static resistance (0.4-0.6 can't be rolling resistance for tires). If I use the static resistance, the numbers don't match with the tire power consumption tests I found. What do I miss, how should I calculate the maximal acceleration? How do I calculate the power loss from friction? ($P =F_\text{friction} \cdot v$, but force of rolling or static friction?)


1 Answer 1


You are talking about two different things here.

When a car is moving at a constant velocity, it constantly needs to overcome the rolling friction (along with air resistance) in order to maintain its speed. This is the power that is used to maintain velocity. Therefore, the power loss is related to rolling friction.

If you want to calculate the maximum acceleration, you need to use the static friction coefficient. If you apply a force greater than $F=mg\nu_\text{static}$, the wheels will start to slip. So this is the largest force the tyres can transmit to the ground.

These two types of friction have very little to do with each other. One is trying to slow the car down when it is already moving, the other is how hard the car can "push forward against the ground".

  • $\begingroup$ Okay, it's clear now, thank you very much! $\endgroup$
    – H. Csaba
    Mar 11, 2019 at 11:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.