How do you determine the distribution of frictional forces among the tires of a car?

[carrotstien]

When the car is moving in a circle at a constant velocity, how would you figure out the forces that come from each wheel (in the case that the max friction of each tire is much greater than anything encountered).

 

For a motorcycle i figured it out easily - it comes out to simple solving two equations:

 

sum of torques around center of mass = 0

sum of forces on body = centripetal acceleration = mv^2/R = mRw^2

 

This can even be solved for a two wheeled car, where instead of the other set of wheels, you have frictionless pegs.

 

Is there a way to solve it, or does it work in a way that the inside tires are put up to their extreme, and the outside tires try to make up anything left? Or vice versa? But either way, how would you figure out which tires starts to slip when if the car is turning with sand and debri lowering the friction of any chosen tire?

 

Roman

[swansont]

Do the inside and outside tires exert the same force on the ground? (think about the turning radius)

[carrotstien]

I understand that that the outside are accelerating faster than the inside.

But, why would anyone of these be invalid (when the mu's are high enough to sustain them)

 

~sum of forces from outside tires = mv^2/R and the other tires don't do anything.

where mass of car is M and radius of center of mass' motion is R

~vice versa

~each tire applies the same force as long as the sum of all the forces is mv^2/R and torques add to 0

 

I am not sure about the last one, but the first two are perfectly feasible since they are describing a 2 wheeled motorcycle with a center of mass off the line connecting the tires - which is algebraically solvable.

 

The reason i'm asking if one set of tires worked to its max first before the other in some way is because i was thinking of the following cases:

 

If you have block [A] and touching horizontally on a table like [A];

~you apply a force on block A horizontally to the right. What is the force of friction that excerts on the table? - well, that depends if the force of max friction of [A] has been brought to its limit. Only when [A] starts to try to move, then starts helping along.

~what about if in the same case, you have a rod connecting the two blocks together through their centers; then you apply a rightward force on the rod. You know that if its not slipping, then the sum of frictional forces have to be equal to your force, but that doesn't tell you the distribution. Would it be 50-50 until one can't add anymore? Or would it be something else entirely such as the increase of force of friction is a certain percent of the max of each simultaneously. So if block [A] has a max static friction of 100 newtons, and has a max of 10, then when [A] resists with 10 newtons, resists with 1, and when [A] resists with 20, resists with 2, and so on.