# Car turning on an unbanked (level) surface

1. Dec 21, 2012

### Bipolarity

I've been trying to understand this:
http://www.batesville.k12.in.us/physics/phynet/mechanics/circular motion/an_unbanked_turn.htm

What I can't understand is why the frictional force equals the centripetal force. As the car turns, the frictional force should be opposite (antiparallel) to its velocity, and thus normal to the centripetal force, since the centripetal force is normal to the velocity.

What am I missing here? All help is appreciated thanks!

BiP

2. Dec 22, 2012

### Staff: Mentor

The frictional force is acting between the road surface and the tire patch that is in contact with the road. That tire contact patch is (as long as the car is not sliding) at rest relative to the road, and no matter which direction you try to push it, the frictional force will resist. The car wants to go in a straight line, pulling the tire patch towards the outside of curve, and the frictional force between tire patch and road is resisting.

You might also want to read #21 in this thread. https://www.physicsforums.com/showthread.php?p=3984064&highlight=Tire#post3984064

Last edited: Dec 22, 2012
3. Dec 22, 2012

### Bipolarity

If the frictional force is opposite the (tangential velocity), it should be normal to the centripetal acceleration. I don't understand why this is not so.

BiP

4. Dec 22, 2012

### Staff: Mentor

It's not opposite to the tangential velocity, because there is no tangential (or radial) velocity to be opposite to - the bit of tire that is touching the road is momentarily stationary.

I edited my first response above to add a pointer to an older thread on this topic:
https://www.physicsforums.com/showthread.php?p=3984064&highlight=Tire#post3984064

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