# 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.

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.

• 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. Nov 26, 2013 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 ... Nov 26, 2013 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? Nov 27, 2013 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. Nov 27, 2013 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! Nov 5, 2015 at 5:09

Another answer from a duplicate post :

So you want to understand how the tire generates a force that is lateral to the vehicle direction.

You may want to take a look at cornering force. This force comes from the fact that :

1. The tire has a contact area with the ground. Forces between tire and ground can be exchanged over this area.
2. Tire is elastic and thus deformable. The lateral force comes from the deformation due to the difference between the vehicle direction and the tire orientation.

Let me explain this with a beautiful drawing. The friction is in opposite direction to the vehicle speed. If the tires are turned, they are deformed. They are elastic so they want to go back in their rest position. However, there is friction between the ground and the tire. This will cause a force on the ground, which I called here Elastic Force. This force is the source of the lateral force on the vehicle.

Read about cornering force and I guess you will understand it easily :)

• So you are saying that the vector sum of the friction and elastic force is what causes the centripetal acceleration. However, in the right turn, the vector sum of the friction force and the elastic force vector points southwest, which is away from the center of radius of a right turn. We want a vector that points towards the center, or a vector that is southeast. (also there is still static friction that prevents the tire from spinning which give the car the forward motion, but we are not dealing with that right now).
– john
Jul 12, 2022 at 14:18
• maybe the elastic force vector is in the wrong direction?
– john
Jul 12, 2022 at 14:22

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.

Start by looking at one wheel of a cornering car. It travels a curved path and experiences a lateral frictional force. If you take the wheel off of the car and roll it, it will only roll straight. To make it change direction and travel a curved path while it rolls, you must rotate it on a vertical axis by applying a vertical axis torque. Note that when you rotate the wheel on a vertical axis while it rolls, it will experience a lateral force. Here you have the wheel applying a force to the pavement and, with Newton’s third law, the pavement applying a force to the wheel.

So now we have a single wheel traveling a curved path and experiencing a lateral force. Here’s an unintuitive part but you will see it to be true in a bit. If all four wheels of the cornering car were experiencing continual vertical axis torques, they would all be traveling curved paths and experiencing lateral forces. There is a simple unnoticed mechanism that creates this vertical axis torque at each wheel when the front car wheels are pointed a different direction than the rear wheels.

In the single track bicycle model below, from a stop, the front and rear wheel will be biased to roll in one direction just as the single wheel described above was. This is because of the friction of the contact patch that resists vertical axis rotation. As you can see, with any amount of forward motion, the bicycle frame acts as a lever to apply a vertical axis torque at both the front and rear wheel. Both torques create a counter-clockwise vertical axis rotation turning both wheels to the left. The bike will continue along a curved path with both wheels experiencing lateral frictional force because the wheels individually change the direction they roll.

The video linked below shows this in more detail and with real life experiments.