# I Can someone explain how a car turns on a curve?

1. Mar 5, 2016

### edgarpokemon

I am having a hard time understand why the frictional force of the tires of the car point to the center of the circle (centripetral force). http://physics.stackexchange.com/qu...centripetal-force-during-the-turning-of-a-car . Here is the picture of a car making a turn on a curve, if the car is driving straight, like the blue arrow points, then shouldn't the friction be backwards, or opposite in direction of the blue arrow? why is the friction pointing to the center as the wheel turns? I already spend like a day thinking about this help!!

2. Mar 5, 2016

### edgarpokemon

sorry the blue arrow is not pointing straight, but why isn't the friction opposite in direction? why is it sideways

3. Mar 6, 2016

### Drakkith

Staff Emeritus
The little patch of tire in contact with the ground is not moving relative to the ground, so when driving straight and with no acceleration, there is no force from friction. But when a car turns, friction acts inward and supplies the centripetal force necessary to make the car move in a circle.

4. Mar 6, 2016

### CrazyNinja

To get a proper idea about this, you must understand that when the car wheels roll there exists no friction. Friction opposes relative motion between two surfaces in contact. In the case of a wheel (assume almost uniform cross-section), the point in contact with the road does not move relative to the road. Hence, friction does not act. Now you know why the wheel is said to be an awesome invention.

As for why friction is necessary for the turning of a car, Drakkith has given a good explanation. The car requires a centripetal force to undergo circular motion. Which of the forces present can do this task but friction? If you want to improve your understanding of this, look up the turning of a car on a sloped surface. Have you wondered why curves on highways and race tracks are steeper on one side than the other?

5. Mar 6, 2016

### edgarpokemon

I still not get the idea since I don't know alot about tires, is my first year in a physics class in college and I haven't taken any physics courses before!! So I was thinking that when the drivers turn the wheels of the car, the wheels will be at an angle, having an y component and a x component. The car needs to apply a force to turn the wheels right? so the wheels will be at an angle on a curve. The y component of the force will be pointing straight, so friction is backwards. and on the x component the force will be pointing left or right depending on the curve, so the friction will be pointed opposite to that force, and friction will then point to the circle. is that a good way to understand it??

6. Mar 6, 2016

### Drakkith

Staff Emeritus
Yes, but this is irrelevant to the original question. Once the car's tires are turned as far as they need to go, no more force is exerted on them and the car continues to move in a circle.

No. How the car originally started to turn doesn't matter. We are only talking about what's happening after the car is in a turn.

7. Mar 6, 2016

### rcgldr

In NASCAR, most (if not all) of the banked tracks have a constant slope angle on banked turns for the main driven part of the track, with a curved transition on the inner, normally non-driven part of the track.

That diagram doesn't include the fact that the rear tires also generate an inwards centripetal force.

8. Mar 6, 2016

### Drakkith

Staff Emeritus
That's a nice catch. I didn't even realize that.

9. Mar 6, 2016

### edgarpokemon

I still dont understand why friction is pointing in the same direction as the turning direction of the car. maybe I need to know about car engineering? I understand centripetal force in a string and stuff, but not in a car.

10. Mar 6, 2016

### edgarpokemon

In a right curve for example, right at the moment when the wheel makes the turn, if there was not friction, then the car will move to the outward part of the curve right?? So friction will act inward, to the center of the circle correct?? But if the wheels of the car make a right turn, then somehow there must be a force on the car that is also pointing to the right. So in the road, it doesn't matter what the engine of the car does? only the tires and road??

11. Mar 6, 2016

### Staff: Mentor

Are you sure about that "no force from friction?" You're neglecting air resistance, right?

chet

12. Mar 6, 2016

### Drakkith

Staff Emeritus

13. Mar 6, 2016

### Drakkith

Staff Emeritus
Without friction the car would simply move in a straight line, no matter how you turn the wheels.

I'm not sure what you mean here. Are you thinking of what happens as the car transitions from a straight path into a turn?

Indeed. We're just looking at what happens in a turn. You could consider a car moving in neutral gear if you'd like.

14. Mar 6, 2016

### rcgldr

The engine doesn't matter, the car could be in neutral, coasting, and still make turns. At the contact patches, there is a Newton third law pair of forces, the tires exert an outwards force onto the pavement, and the pavement exerts an inwards force on the tires. The reaction of the surface of the earth is to accelerate outwards by a very tiny amount, since the earth is so massive, while the car accelerates inwards.

When turning, the axis of the front and rear tires is not parallel, and an imaginary extension of the front and rear axis cross at some point to the inside of the car (ignoring Ackerman type issues here), and that point would be the "geometric" center of the circle that the car follows based on the non-parallel axis. The actual radius is a bit larger due to the contact patches deforming a bit outwards in response to the cornering forces.

In the case of a bike turning while leaning, the point on the pavement above the point where the leaned and non-parallel axis cross would be the "geometric" center of the circle that the bike would follow (again contact patch deformation affects this).

15. Mar 6, 2016

### Staff: Mentor

I just wanted to be sure the OP understood that, if there were air resistance, there would have to be an equal and opposite forward static frictional force exerted by the road on the drive tires.

Chet

16. Mar 6, 2016

### edgarpokemon

thank you!! your second paragraph make things more clear for me!! So only the front wheels turn, since they are not powered by the engine right? But the backward wheels are powered by the engine, so when making a right turn on a curve, the front wheels will follow the path of the curve, but the back wheels will still want to go straight, creating a centrifugal force outwards, so friction will be inwards. Right?

17. Mar 6, 2016

### rcgldr

Both the front and back wheels turn the car. A front wheel or all wheel drive car still corners using the same principle. Note as a car turns it "yaws" (rotates about a vertical axis), so that the rear tires turn as the car turns. This Ackerman diagram may help to understand this.

http://en.wikipedia.org/wiki/Ackermann_steering_geometry

Note that all (both front and rear tire) axis are perpendicular to the path of the car, all of them point towards the "geometric" center of the circular path the car follows. There's a Newton third law pair of forces at both the front and rear tires, tires outwards on pavement, pavement inwards on tires.

18. Mar 6, 2016

### edgarpokemon

right, but the in the ackermann diagram, only the front wheels have that thing in the middle (yellow part) and the back wheels don't have anything. so the back wheels will follow the front wheels all the time when going straight, but when the car make a turn, the front wheels will move, and the back wheels will also move, but will also have a centrifugal force since they still want to follow the same straight path, causing a centrifugal force at the instant that the front wheels turn. correct?