# Why does friction cause a car to turn?

I've had a lot of difficulty conceptually understanding the physics of how a car turns on an unbanked curve, so I'm hoping you could help me out. When a car is moving in uniform circular motion, we know that $|\vec{a}| = \frac{v^2}{r}$, and the direction of acceleration is towards the radius of the circle about which the car is moving. Drawing a free body diagram for the car shows that there are only three forces acting on it: gravity $(\vec{F_g})$, the normal force $(\vec{F_n})$, and friction $(\vec{F_f})$. Since gravity and the normal force negate each other, the car isn't accelerating in the $y$ direction. Because it is in uniform circular motion we know it is accelerating in the $x$ direction, and summing up the forces in this direction yields $$\vec{F_{net x}}=m\vec{a}=\vec{F_f}$$ which implies that the centripetal acceleration is due to the frictional force.

What I am having difficulty understanding is why this intuitively makes sense. I've read some other people's answers on this question but I haven't found anything satisfactory. In particular, many people talk about how wheels "are pushing the pavement to the left or right", and this causes the pavement to exert a force on the car wheels by Newton's third law, but this hasn't made sense to me.

Another way of putting this might be that I don't understand why friction should be directed inwards towards the center of the circle about which one is turning. I would expect that, since the wheels have been turned, that friction would be directed in the opposite direction of where the car is moving to prevent the car from continuing to move forward and skidding on the road.

I hope this makes sense, thanks.

• How hard is it to push a car (in neutral, no brakes, level ground) in the direction the wheels are pointed, compared to moving it sideways under the same conditions? – DJohnM Oct 6 '13 at 20:39

## 3 Answers

I had fun trying to make this as intuitive as possible. I hope I've succeeded without doing the physics of the situation much injustice.

When a car is driving straight ahead, the plane in which the wheels are rotating is aligned with the direction of movement. Another way of saying this is that the rotation axis is perpendicular to the momentum vector $\vec{p}=m\vec{v}$ of the car. So the friction merely makes it harder for the car to move, which is part of the reason why you need to put your foot on the gas pedal to maintain a constant speed. At the same time, the friction is what allows you to maintain that constant speed because the rotating tires sort of grab onto the ground, which is the intuitive picture of friction. The tires grab the ground and pull/push it backwards beneath themselves, as you would do when dragging yourself over the floor (if it had handles to grab onto). Those grabbing and pulling/pushing forces are what keeps you going.

Things change when the wheels are turned. The plane in which they are rotating now is at an angle with the direction of motion. Alternatively but equivalently, we could say the rotation axis now makes an angle with the momentum vector of the car. To see how friction then makes the car turn, think again in terms of the wheels grabbing onto the ground. The fact that they now make an angle with the direction of motion, means the force the tires are exerting is also at an angle with the direction of motion - or equivalently, the momentum vector.

Now, a force is a change in momentum$^1$ and so (because the wheels are part of the rigid body that is a car) this force will change the direction of the car's momentum vector until it is aligned with the exerted force. Imagine dragging yourself forward on a straight line of handles on the floor and then suddenly grabbing hold of a handle slightly to one side instead of the one straight ahead. You'll steer yourself away from the original direction in which you were headed.

$^1$ Mathematically: $$\vec{F}=\frac{d\vec{p}}{dt}$$

The key IMO is that the "effective friction" of a wheel that is free to rotate is very different in different directions. Moving inline with the wheel the effective friction is low because the wheel can rotate. Moving perpendicular to the wheel the effective friction is simply the friction between wheel surface and ground which is high.

If the wheel is paralell to the direction of motion then it exerts a small force slowing the car down. If it is perpendicular to the car then it exerts a much larger force slowing the car down.

To see what happens when we turn the wheels digonally lets resolve the cars veolicty into components parlell and perpendicular the the wheel. Bot of these components will point diagonally forwards. That means that their coorespond frictional forces point digonally backwards.

due to the higher effective friction the component of the frictional force perpendicular to the wheel will be much stronger than the one paralell to the wheel. So lets only only consider that one from now on. if the front of the wheel is pointing right then the perpendicular frictional force will point backwards and right.

Resolving that force back to the car we see a force pointing backwards and a force pointing right.

If the car was 4 way symetrical and all wheels turned in the same direction then the car would just start moving diagonally but the wheels on a car don't do that. only the front wheels turn so the front of the car is pushed right causing the car to turn right.

If we want to keep a body in a circular orbit a central force is needed. If that is intuitive enough then,when you turn the wheel you are effectively changing the direction of the driving force ( if we are travelling at uniform speed this force is still there though it cancels off exactly with total frictional forces ) changing the force direction will not change the velocity direction instantly. if you make only one angle change of the driving force, it will take a finite amount of time to get the vehicle adjusted to the new angle. So it is the Inertia that gives this apparent slipping. Circular orbit is bit different, you keep on changing the direction of the driving force, now since inertia makes the vehicle slip all the time the Frictional force is active. But the maximum frictional force is limited by mu*R , R = mg , If the central force needed for our intended orbit is less than that my*R then vehicle sustains minimum slip, but if it is insufficient, slipping and turning occur successively until we reach a larger orbit with a higher Radius of curvature, that will make the central force lesser ( since it is mv^2/r ) so that it could be sustained by mu* R ( mu is the coefficient of friction ) Alternatively V could be made smaller so that the frictional force could sustain necessary central force. Main thing there are no two forces centrifugal and centripetal, It just makes this whole thing confusing. The thing is that central force that is needed for a particular orbit and a speed. It has to come from some external means.

P.S. Elaborate further on a sudden direction change to a vehicle which is traveling at a uniform speed. As soon as we turn it you can resolve the velocity in two directions. one along the new direction and perpendicular to that direction. Now the perpendicular component will push the vehicle outward, with a friction force in the direction opposite to slip, Because of that 'perpendicular velocity' will decrease. However if this remains the same, the constant speed of the vehicle in the new direction does not increase back to v because the driving force is barely sufficient to keep the velocity constant i.e. No acceleration. Because of the turn kinetic energy is lost. The key here is Inertia.

## protected by Qmechanic♦Mar 20 '14 at 17:33

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