# Friction of a Car in Motion

It is commonly known that friction opposes motion. For example, if a block is sliding down a wooden surface with no forces other than friction acting on it, friction acts in the direction opposite to the block's velocity.

Let's say a car is driving at constant speed along a perfectly circular loop road. Centripetal acceleration works toward the center of the circle, but the car is moving forward along the loop, perpendicularly to the car's acceleration. As we have already established, friction should be in the opposite direction of the car's motion. But no- apparently it acts as the centripetal force that is perpendicular to the velocity.

What is going on with this discrepancy?

Edit: new comment

Edit: Furthermore, the book I am using says that, when the loop is banked, friction ceases to act centripetally and the centripetal force is provided entirely by the horizontal component of the normal force (neither friction nor, to my surprise, parallel force.). Why is that? Thanks!

• *Friction opposes relative motion ;-) – Aaryan Dewan Jan 19 '17 at 14:07
• Friction opposes relative motion? The car is moving forward, and the ground is stationary on our frame of reference, so the friction would still act, based on that, along the tangent line passing through the car, in the reverse of the direction in which the car is moving. – George Bentley Jan 20 '17 at 12:51
• @GeorgeBentley The motion is not tangential to the car though. Just because the instantaneous velocity is only moving forward, it does not mean the motion of the object is forward. In fact, we know the motion is circular around the corner. This is because that velocity is still constantly changing direction due to the acceleration. The friction is acting in the reverse direction that the car is moving. You can't consider the car moving straight because there is more to it's motion than just the instantaneous velocity. – JMac Jan 20 '17 at 13:43
• @GeorgeBentley Thats a good question. I again stress the word relative. You have to see w.r.t the wheel. See, had there been oil on the road, then what would be the direction of friction force by the oil? w.r.t the ground the bottom most point will be moving backward ( it’s slipping ) so friction will be in the forward direction. Coming back to your question , the wheel of the car has a tendency to slip backwards so friction will be in the forward direction. Not the backward one – Aaryan Dewan Jan 20 '17 at 13:51
• @GeorgeBentley , actually the direction of friction is different in a bicycle/ car on the front and back wheels! – Aaryan Dewan Jan 20 '17 at 13:57

## 4 Answers

Friction is in the opposite direction of motion.

When you spin your wheels, you are trying to push the road backwards. The reaction to that friction is to push your wheel forward.

When you turn, you are changing the angle the friction is acting. Instead of just trying to push the road backwards, it also tries to push it to the side opposite that you want to turn. The reaction from that creates a component of the force acting in the direction you want to turn, leading to the circular motion.

It is commonly known that friction opposes motion.

If that were so you would not be able to move from rest.

Friction opposes or tries to oppose the relative movement between two surfaces.

If you have two blocks on top of one another and at rest and apply a force on the bottom block what will happen? The bottom block will start to increase its velocity and if the static frictional force is large enough the top block will also increase its velocity by the same amount.
Here static friction is producing the motion of the top block whilst trying to have no relative movement between the surfaces which are in contact.
If the static frictional force is not large enough and kinetic friction kicks in, the kinetic frictional force will still try and increase the velocity of the top block to reduce the relative movement between the two surfaces in contact.

With the car going around the corner the static friction force tries to have a situation that at the point of contact between the tyre and the road there is no slipping.

friction opposes motion

Yes!

But motion doesn't have to be there yet for friction to oppose it... Remember static friction. I personally better like saying "sliding" instead of "motion" to include the point about relative motion, which is discussed in the comments. So: Friction opposes sliding (or at least tries to).

As you said:

• If a block slides down an incline, friction tries to stop that sliding. (Called kinetic).

• But also, if a block rests on an incline with no motion, gravity still pulls in it and friction holds back to avoid sliding from starting. (Called static)

The backwards/forwards direction of the car might experience some kinetic friction (and also static because we are talking about wheels rolling, but that's not the point here.) The sideways direction experiences no kinetic friction. But when you turn that steering wheel, static friction appears in the sideways direction to avoid the tires to start sliding with a component along that direction. This static friction is a force that causes acceleration in this perpendicular direction in the next instance, because it isn't balanced - therefore the velocity turns in the next instance. Then this happens constantly continuously and that is how a circular path appears.

I believe that the whole point to answer your question is that you are dealing with two directions here - both might experience friction because there might be a tendency to sliding along both.

• The velocity doesn't get a sideways component. The velocity is always going directly forward at any instant. The friction force causes an acceleration which constantly changes the direction of the velocity to be in line with the curve. – JMac Jan 20 '17 at 16:17

There are two types of friction: static and kinetic.

Static friction tries to keep two things that aren't moving so that they aren't moving, while kinetic friction tries to take two things that are moving and slow them down so that they're not. This motion, of course, is relative.

The confusing thing about cars is that the car is moving relative to the ground, so shouldn't we be dealing with kinetic friction?

It's not actually the car that we're concerned with in terms of friction with the road, but its tires. Actually, a single point on each of its tires. This point is the point that is making contact with the road, which of course changes as the tire rotates.

When the tires are rolling, this is actually a static context, contrary to all intuition.

Consider a tank or a bulldozer: something with tracks. The end of each track has a semi-circle that is moving with respect to the ground, but but the long part of the track that is on the ground is stationary. The tank takes advantage of this static friction to propel itself forward. Now consider that we gradually shorten this long part that is stationary to the ground until it is only a single point. This is essentially what a tire is. (Actually, a small part of the tire will flatten out but for purposes of the model and all)

The static context's resistance to relative motion between the tire and the road is what allows the car to move in a circle. The tire will only rotate on one axis, and as it rotates on that axis, it moves in a different direction from the way the car wants to go. So the tire uses static friction to instead force the car to go the way it is pointed in exactly the same manner that it forces the car to accelerate when you push the gas pedal. And actually, this static context is what the car uses to decelerate when you push the brakes, too.

The kinetic context is what is better known as skidding. The coefficient of static friction will always* be higher than the coefficient of kinetic friction, which is why you accelerate and stop better when you're not skidding. Turning while skidding is even more unpredictable because rather than "enforcing" static contact, friction is now "coaxing" the car to go in the direction you want it to go, opposing the car's angular momentum.

How do the tires choose whether to roll or skid? It's a matter of the current state and the forces applied. Static friction and kinetic friction are governed by similar equations:

$$f_s = \mu_sF_n$$ $$f_k = \mu_kF_n$$

Where $\mu$ is coefficient of friction ($s$ and $k$ being static and kinetic, respectively). What the $f_s$ equation represents is the threshold for static friction. $f_s$ matches the lateral force applied but "maxes out" at $\mu_sF_n$. If you push something laterally with a greater force, it will swap to the kinetic context and begin to slide across the floor (or skid). At that point, the $f_k$ formula represents the force with which static friction opposes the sliding/skidding motion. The system will not return to a static context until the motion between the two bodies reaches $\vec 0$ (Note that this is different from the magnitudes of the motions being equal; they must also be in the same direction). This is why it's difficult to recover from a skid, and it's a good idea to let off the gas and steer against the skid.

*to my knowledge. If anyone can clarify or provide counterexamples, please do so.