How is it possible that friction = centripetal force for turns?

by x86
Tags: centripetal, force, friction, turns
x86 is offline
Sep16-13, 09:22 PM
P: 86
So, I am confused. If a car is driving on a flat surface and it turns, it experiences centripetal force. Apparently, the friction of the bike is equal to this force. This doesn't make sense to me. I've drawn a few forces on an example picture here:

The car is going straight, and Fa is the force applied going straight (and friction in the opposite direction). The car then suddenly turns and its tires are perpendicular to its old direction.

I can only see four forces acting now:
1. Fa as depicted
2. Ff as depicted
3. Fi as depicted, which is in the new direction the tire is rolling towards.
4. Another frictional force acting in the same direction as Fi, which is going against the tire motion

The only problem is that this frictional force acting in the same direction as Fi is only affecting the tires, its slowing them down and shouldn't affect centripetal force. Its not actually accelerating toward the center of the circle, all it is doing is slowing down the tires.

So how come people say Ff = Fc for turns on flat roads? It doesnt make sense for me.

I can solve the problems easily, as I know the theory, but I just can't understand it for this specific scenario.

If we say that Ff = Fc, then how come we can't say that when a car is driving, the frictional force is actually in the direction that it is driving in (because of the tires).

Therefore, there would be two forces acting on the car.

1. The force making it go forward
2. Friction also making it go forward; but only affecting the tire speed.

Therefore, in freebody diagrams; if we are to follow the Ff=Fc logic, why do we draw it so frictional force is directed in the motion opposite of a car, when it really isnt? Is it just to make it easier for people to follow, since it is, after all, affecting the tires motion? (According to the Ff=Fc logic for this example)
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SteamKing is offline
Sep16-13, 10:15 PM
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What? You don't think tires won't develop friction if they start moving sideways to the direction of travel?
x86 is offline
Sep16-13, 10:29 PM
P: 86
Quote Quote by SteamKing View Post
What? You don't think tires won't develop friction if they start moving sideways to the direction of travel?
Yes there will be friction, but the friction won't be toward the center of the circle.

Nugatory is offline
Sep16-13, 11:05 PM
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How is it possible that friction = centripetal force for turns?

Quote Quote by x86 View Post
Yes there will be friction, but the friction won't be toward the center of the circle.
The physics of a car in a turn are surprisingly complex (which is why the people who build race cars are always redesigning and tweaking suspensions) but there's a simple model that works pretty well if the tires are holding so the car isn't skidding:

The tire contact patch, the part that's touching the ground, is not moving relative to the ground even though the tire is rolling; as the tire rolls it picks up the back of the patch and pushes more rubber down onto the road at the front of the patch. If you imagine a bug riding on the tire... The bug will be squashed flat but it won't be smeared sideways as the tire rotates.

OK, so at any given moment the tire contact patch is at rest relative to the surface of the road. Static friction stops it from sliding (if it did slide, we'd be skidding, not turning). We turn the steering wheel so that the contact patch moves to follow the curve. However, the car is still moving in a straight line under the influence of inertia so the tire sidewalls stretch as the contact patch moves away from a straight line while the wheel/axle tries to follow a straight line. Of course the tire resists this stretching, so it exerts a force on the wheel/axle... And this force is pretty much directly oriented towards the center of the turn.
voko is online now
Sep17-13, 12:06 AM
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What I think happens is in the attachment.
Attached Thumbnails
Delta˛ is offline
Sep17-13, 12:13 AM
P: 256
The vector sum of Ff and the force labeled as 4. in the original post has a direction towards center and its the centripetal force.

What confuses you is that Fa and Fi are fictional forces (not fRictional) and they dont really exist. Friction forces are the only ones that give translational acceleration to a car. The forces from the engine pistons all they do is to give rotational acceleration to the wheels.
rcgldr is offline
Sep17-13, 12:38 AM
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Just to be clear I understand the orginal post and that the original poster understands the issues:

The force that opposes tire rotation is called rolling resistance, not friction. It's mostly due to the inelastic deformation that occurs at the contact patch, where the force during recovery is less than the force during deformation, converting mechanical energy into heat.

If a car is not accelerating, then there's only enough static friction to overcome rolling resistance.

If a car is turning at constant speed, then the only acceleration is centripetal and related to the normally static (non-sliding) friction at the tires. There's a Newton third law pair of forces at the contact patches: the tires exert an outwards force onto the pavement, and the pavement exerts an inwards force (the centripetal force) onto the tires.
sophiecentaur is offline
Sep17-13, 04:46 AM
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I have to ask myself, what force other than friction is available to provide the centripetal motion? You surely have to fit your explanation to the facts and not the other way round.

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