# Friction between a banked road and tyres of a car

Let's say there is a car which is at rest on a banked track. The angle of banking is $$\theta$$ and coefficient of static friction between car's tyres and the track is $$tan\theta$$. Since coefficient of static friction is $$tan\theta$$, the car can just stay there on the track without sliding down the slope of the banked track.

I was trying to analyse this situation and from my understanding, the car can stay on the track alright, but it can't start moving. Because its tyres are already using the maximum available static friction to keep them from sliding down the slope of the banked track. If the car has to start moving, it will require more friction to accelerate itself from its state of rest. But since the tyres have already used up the maximum available static friction, I don't think the car should be able to accelerate without its tyres going into a skid.

Is my analysis of this situation correct? Thank you

• Your analysis is perfect! Commented May 18, 2021 at 15:52
• @Ritam_Dasgupta Thanks. Can I extend the same argument to banked tracks which are circular or curved? If a car is just resting on such a curved banked track and friction coefficient = $tan\theta$, then it can just stay on the track but can't move?
– 4d_
Commented May 18, 2021 at 18:04
• Yes, sure. At the initial instant it has some non-zero acceleration and zero velocity. This is not a feasible situation, because limiting friction won't allow it. Commented May 18, 2021 at 18:31
• @Ritam_Dasgupta "This is not a feasible situation because limiting friction won't allow it." Are you implying that limiting friction won't allow it because there is no more friction available for the car to use along the tangential direction to accelerate itself? Or did you mean something else?
– 4d_
Commented May 18, 2021 at 18:38
• Yes, by the same logic you used. Commented May 18, 2021 at 18:43

I would state you are almost there because the frictional force will equal the gravitational force pulling it down the slope. and the equation below states whether the car will slip.
$$F_{friction}\leq\mu F_{normal}$$ with $$\mu$$ being the coefficient of friction

so if the frictional force is equivalent to $$\mu F_{normal}$$ there is no more possible friction to use so the car will slip but if less then the car will move so it is $$F_{friction}\leq\mu F_{normal}$$ which defines whether there is any friction left to accelerate, you cannot just use the coefficient of friction to state weather it will be able to accelerate.

• I would state that he's way ahead of you! Because with the stated coefficient of static friction, the inequality becomes an equality. I think. Commented May 18, 2021 at 16:32
• You are correct Commented May 18, 2021 at 16:35
• Thank you. I thought if I chose the critical angle of repose as the angle of banking, it would be enough to explain what I was trying to ask. I should have included the inequality to make it clearer. Sorry
– 4d_
Commented May 18, 2021 at 18:05
• @Kristoffer Sjöö Can I extend the same argument to banked tracks which are circular or curved? If a car is just resting on such a curved banked track and static friction coefficient = tanθ, then it can just stay on the track but can't move?
– 4d_
Commented May 18, 2021 at 18:06
• Even in your original example, the car can in principle be in motion - it just can't start it stop moving. In a banked turn, the car would be able to move and accelerate slightly, because of centripetal acceleration. It still wouldn't be able to start from a standstill I believe. This is all very theoretical of course. Rubber wheels don't really work like this. Commented May 18, 2021 at 20:45