# Direction of friction when a car turns

When a ball moves to the right, friction acts to oppose the motion, in other words, to the left. However, when a car travels around a bend, the friction acts in the perpendicular direction to the car's velocity and provides the centripetal force. I just cannot understand why friction would act in that direction.

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similar question physics.stackexchange.com/questions/79852/… – nijankowski Nov 26 '13 at 15:31

The tires (or ball) want to travel in a straight line and friction (traction actually) make it deviate from the line.

Friction (acting on the direction of travel) is present on both cases. For the car there is the additional effect of traction which enforces the going around the bend part. The ball is going on a straight line and so no traction is needed.

Visualize the car on ice, making the tires slip. Relative to the bend (desired motion) the tires slip away and friction is opposing this motion. On normal roads you do not observe this slipping, but it happens. Look up "Tire Slip Angle" and you will learn a lot about it.

This slipping towards the outside of the circle is what generates the tire friction, and it is the same mechanism with you trying to slide a box in a straight line. Friction will oppose this motion.

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This is my attempt to illustrate what happens when the car wheel is turned:

Focus on the bit of the car tyre marked with a red spot, and the bit of the road marked with a green spot. If we could look at the contact patch between the tyre and the road we'd see something like the rectangle I've drawn on the left. When the wheel is straight the red spot on the tyre and the green spot on the road move together (I'm assuming the car is moving to the left so the ground is moving to the right).

but now suppose we turn the wheel to the left. The contact patch now looks like the rectangle on the right. Because the patch has been rotated relative to the road the red and green spots now don't move together, but instead the red spot on the tyre is scraped across the road surface. It's this lateral motion of the tyre surface across the road surface that causes the frictional force that turns the car.

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as the red spot on the tyre is scraped across the road surface, wouldn't the red spot experience a frictional force in the opposite direction (in the direction of red to green pot on second diagram)? In that case, the frictional force is not inwards and thus will not provide centripetal force. – Eliza Nov 26 '13 at 16:30
@eliza: in the diagrams above we are looking down (through the car) onto the contact patch between the tyre and the road. The car is move to the left, and the wheel has been turned to the left. The force the road is exerting on the tyre is towards the left (downwards in my diagram) so it is pulling the car to the left i.e. it's turning left as expected. Hmm, that made sense when I started ... – John Rennie Nov 26 '13 at 17:43
I can visualize it now. However, will the direction of the frictional force be opposite to the direction of the red arrow in the second diagram? – Eliza Nov 27 '13 at 3:40
@Eliza: the force acts along the line joining the red and green dots. so as you're looking at the diagram the force on the car is downwards and the force on the road surface is upwards. – John Rennie Nov 27 '13 at 7:54
@JohnRennie hey John, this is a great answer. I'm trying to understand it myself. What I don't understand is the tyre is turning left, and therefore left is upwards in your diagram. But you say "the force on the road is exerting on the tyre is towards the left (downwards)". Isn't it the opposite? The force the road is exerting on the tyre seems to be towards the right, because from what I see, it's opposing the wheel's turn. Thanks! – rb612 Nov 5 at 5:09