I'm trying to understand the mechanics which determine if a car making a turn will skid.

Are the following correct or incorrect:

  1. A vehicle making a turn will skid unless the centripetal force is adequate to produce the centripetal acceleration. In a simple turn, where $r$ is constant, the force needed is $mr\omega^2$.

  2. If the road is flat, this force can only come from friction created by turning the wheel. The maximum force is $\text{weight}\times\mu_\text{static}$. All of this force will be directed centripetally, and will therefore be available to prevent the skid.

  3. If the road is banked by $\phi$, the friction force will be lessened by $\cos\phi$, but there will also be a centripetal component of the normal force, equal to $\sin\phi\cos\phi$.

Are these all correct? I believe they are but my results using them don't seem to work.

UPDATE: I believe my mistake was in the direction friction will be in. Determining the direction of friction in this case is tricky, because it has to both counteract the sliding due to normal force of the banked road, and also accelerate in the centripetal direction. It requires 3 dimensions. How do I determine the direction of friction, with respect to $r$ and $\theta$ vectors?

  • $\begingroup$ Hi, and welcome to Physics Stack Exchange! This is not a bad question at all, but we usually prefer our questions to invite more than just a yes-or-no answer. Can you elaborate on, say, what you think might be wrong with these statements? $\endgroup$
    – David Z
    Jul 10, 2012 at 3:22
  • $\begingroup$ @David - With kbeta's help, I now know the specific area that I don't understand: determining the direction of friction with respect to $r$ and $\theta$. Question updated. $\endgroup$ Jul 12, 2012 at 14:15

2 Answers 2


(1) is correct, but (2) and (3) are not. Consider a car turning counterclockwise around a track that is banked inwards. There are three forces acting on the vehicle: the weight of the car $\vec{W}$, the normal force $\vec{N}$, and the force of friction $\vec{f}$. Breaking these up into Cartesian coordinates gives:

$$\vec{W} = -mg\hat{y} \\ \vec{N} = N \left(\cos\phi \hat{y} - \sin\phi \hat{x}\right) \\ \vec{f} = f \left(-\sin\phi \hat{y} - \cos\phi \hat{x}\right)$$

Getting the direction correct for the friction is a bit subtle. We know from the $\phi \rightarrow 0$ limit that the friction has to point roughly in the direction of the turn, and also be perpendicular to the normal force, so for these coordinates, it must point down and to the left.

Since the car is moving in a circle in the horizontal plane, the forces in the $\hat{y}$ direction must cancel. Setting these equal to each other and considering the limiting case $f = \mu_s N$ gives an equation that can be solved for $N$. Then you can sum up the horizontal forces an insert this value to get the centripitel force.

  • $\begingroup$ Can you please explain the f equation? $\endgroup$ Jul 10, 2012 at 19:03
  • $\begingroup$ Thanks for adding the vector symbols. What is the basis of the f equation? What does f represent? $\endgroup$ Jul 10, 2012 at 20:48
  • $\begingroup$ Hopefully this edit is more clear. $\endgroup$
    – kbeta
    Jul 10, 2012 at 22:15
  • $\begingroup$ Yes, the edit helps: now I understand where my mistake was: the direction of friction. I'm still confused, however - can you look at the original question, which I've updated to ask a more pointed question? $\endgroup$ Jul 12, 2012 at 14:13

As a side note skidding is not a yes or no state with tires. See this answer for more details.

The sum of the normal force and friction force that act on the car is the reaction force to the sum of the weight and centrifugal force of the car on the road.

We can equate them in the coordinate frame parallel and perpendicular to the road.

enter image description here

$$N=m\,g\, \cos(\theta) + m\,r\,\omega^2\,\sin(\theta)$$

In order for the car to not skid the maximum available frictional force must exceed the virtual centripetal force in the direction parallel to the road.

$$\mu_s\, N \gt |m\,r\,\omega^2\,\cos(\theta)-m\,g\, \sin(\theta)|$$

Combining yields:

$$\mu_s (m\,g\, \cos(\theta) + m\,r\,\omega^2\,\sin(\theta)) \gt |m\,r\,\omega^2\,\cos(\theta)-m\,g\, \sin(\theta)|$$

$$\mu_s \gt \frac{ |r\,\omega^2\,\cos(\theta)-g\, \sin(\theta)|}{g\, \cos(\theta) + r\,\omega^2\,\sin(\theta)}$$


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