# How is the centripetal force of a car when turning distributed over the wheels?

The centripetal force can easily be calculated as: $F = (M*v^2)/R = (M*v^2)*sin(\delta)/L$. But how is this force distributed over the (front and rear) wheels? My initial thought was to just divide it by 4 for each wheel, but when you turn your front wheels 90 degrees, there will be no force over the rear wheels. So when simply dividing by 4 is wrong, then how is the distribution in reality?

Is it also safe to assume the forces on the front wheels are equal to each other, and also the same for the rear wheels?

• Note that it's clear that the force on the inner wheels can drop to zero for real cars, since you can turn tightly enough that the inner wheels leave the road. – tfb Nov 18 at 13:22

You cannot simply divide by 4, no. If you set up that formula for each wheel, then you'll have to take into account that the speeds and distances are different.

The one thing all four wheels have in common is angular velocity $\omega$. Even at a 90 degree turn, the rear and front wheels spin equally fast (degrees per second) around the rotation point. Otherwise the car would be breaking apart.

The angular velocity relation $$v=\omega r$$ helps you calculate the linear speed $v$ for each individual wheel, since you know the distances $r$. With this $v$ and $r$ per wheel you can calculate the centripetal acceleration for each wheel.

I would then divide the mass by 4. That would be a necessary assumption, namely that each wheel "carries" equally much mass. Then the centripetal force on each wheel can be calculated.

I have not done the calculations, but I would expect all four forces to be different. And varying differently with the turning angle of the front wheels (some will be cosine and some sine to the angle). For a 90 degree turn you will have a rotation point located at the first rear wheel. That wheel spins around, but doesn't move linearly. It has zero speed $v$ and distance $r$. According to your formula (if you plug in the relation $r=v/\omega$ in place of $r$), zero speed indeed shows zero centripetal force.

Is it also safe to assume the forces on the front wheels are equal to eachother, and also the same for the rear wheels?

If the turn is big, then differences in the distance $r$ become negligible. The differences in speeds will as well become negligible. And then all forces are more or less equal, yes.

• But....wheels also have different camber, different static weight distributions (mostly F/B), then there are suspension (possibly active) dependent loading while turning at any radius, then you put a wing on the car and it goes way beyond distributing the mass of the car--the aero loads can be greater--you can pull 4+ lateral g's in a turn. Then put some electronic stability control and other assists that change power to each wheel. Bank the turn--assume there is a racing groove (position dependent static friction), finally add dirty air from the leader. this is a complex engineering question. – JEB Apr 17 '18 at 13:30

This problem is statically indeterminate : with the limited information provided it is not possible to determine how the static friction force is distributed between the 4 tyres. The total friction force on the vehicle must be $Mv^2/R$, but it can be distributed between the 4 tyres in an infinite number of ways all of which are consistent with the laws of friction and Newtonian Mechanics.

The friction force on each tyre can have any value up to $\mu N$, where $\mu$ is the coefficient of limiting static friction between the tyres and the road and $N$ is the normal reaction at each wheel. Even if you know the normal forces on each tyre, this only tells you the maximum friction force which can be provided at each tyre : it does not tell you what the actual friction force is, unless you know that the tyre in question is on the point of slipping.

Neither can it be assumed that the friction forces are the same on the two inner or two outer tyres, nor on the two front or two rear tyres. Even if the mass of the vehicle were distributed symmetrically, like a uniform cuboid, and the coefficients of friction were identical at each tyre, this does not prevent there being different friction forces on each tyre.

In practice there is of course a definite value of the friction force on each tyre, which depends on the preceding motion as well as such things as weight distribution and coefficients of static friction.

For example, suppose the inner tyres encounter a patch of ice. Temporarily they cannot provide any friction force : the outer tyres must provide the whole friction force. After the patch of ice has been passed the inner tyres could again provide friction, but if the outer tyres continue to grip well there is no need for the friction force to be re-distributed.

See the following similar questions about the distribution of friction across the surface of a block or between blocks connected by a string :