# Question on circular motion — car turning on road

Lets say a car is turning on a road that has friction. I made this small diagram

where the two parallel lines indicate the tires of the car. In order for the car to keep turning in a circle, there must be an force directed to the center of curvature — the centripetal force. I have read that this force comes as a result of friction.

However, in order for there to be a frictional force towards the center, doesn't there have to be a force directly opposite pulling away from the center, as frictional forces cannot exist by themselves — they have to act in order to counter an opposite force.

Thus, how can there be a frictional force towards the center without a corresponding force pulling away from the center? Does this have something to do with the centrifugal pseudo-force? Further, where exactly does this frictional force come from? Also, why is the frictional force directed towards the center? Shouldn't it be opposite the direction the car is traveling at the moment -- thus in a direction opposite the tangential?

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Friction forces act as a response, and opposite, to velocity, not force (that would be normal forces).

The car has a liner velocity in the forward direction, and it keeps moving indefinitely, ignoring any residual friction. Then, if the steering wheel is turned left, the front tires are rotated to the left, thus there appears a frictional force perpendicular to those tires. This force is caused by the tires resisting the movement, just as any other friction. The interesting thing is that tires can rotate freely only in one direction, but not in the perpendicular. Thus the friction appears only in that non-rotating direction.

This force will point not perpendicular to the car, but perpendicular to the tires. The difference is small, since the actual angle that the tires are rotated is quite small, particularly at high speeds. Also, once the car is rotated from the straight line, a small lateral friction will appear also in the back tires, because the velocity will no longer be aligned with the axis of the tires.

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But in your diagram, the velocity of the cars (and the tires) is still in the forward direction. Thus shouldn't the frictional force go behind, not perpendicular? –  user2612743 Jul 7 at 18:44
@user2612743: I added a bit more of explanation. Let me do another drawing... –  rodrigo Jul 7 at 18:46
I see that there will indeed be a lateral force, but it seems that the amount would be too small to create any meaningful frictional force. Also, there would still be friction in the direction opposite the non-lateral (forward) velocity, right? One last question -- if Friction forces act as a response, and opposite, to velocity, not force, for an object on an incline plane there is still friction acting on it but no velocity (there is however a force pulling it down the ramp as a result of gravity). So in this case isn't it a response to force? –  user2612743 Jul 7 at 18:57
@user2612743: The friction in the forward direction is very small because the wheels turn and their axis is lubricated. The other force is relatively small but enough to force the tires move in their pointing direction, more or less, and make the car go round. –  rodrigo Jul 7 at 19:09
Ok got it -- What about the last question in my above comment: if Frictional forces act as a response, and opposite, to velocity, not force, for an object on an incline plane there is still friction acting on it but no velocity (there is however a force pulling it down the ramp as a result of gravity). So in this case isn't it a response to force? –  user2612743 Jul 7 at 19:13

Something moving in a circle at steady speed is experiencing constant inward accelleration. From F = mA, we know that this requires a force on a object to accellerate it, and that force is proportional to the mass of the object and the accelleration.

In the case of a car going around at circle at steady speed, that force comes from the ground pushing on the tires in a radially inward direction. This force is made possible by the friction of the tires against the ground. There is no corresponding force pushing outwards on the car, although the fictitional "centifugal force" is sometimes invoked to rationalize it. The only forces on the car are the ground pushing up to counteract gravity, and the ground pushing inwards to provide the force for its radially inward accelleration.

If a portion of the circle the car was traveling over were replaced by a icy patch (ideal 0 friction), then the car would stop going in a circle and continue on whatever tanget it was on when it hit the icy patch.

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