Static friction : Static friction opposes relative motion between two surfaces. The car is moving at some tangential velocity relative to the road. Thus, static friction should act in the opposite direction. Yet it acts down the incline rather than parallel to the motion of the car. If it acts down the incline, it's essentially helping the car maintain the same relative velocity w.r.t the incline rather than opposing this relative motion.

Kinetic friction : Consider a block placed on a long, rough plank which is moving to the right at some constant speed, and the block is moving to the right, but slower compared to the plank, so relative to the plank, the block is sliding towards the left. Will kinetic friction act to the left to reduce the kinetic energy of the block, or will it act towards the right to oppose the relative motion of the block and plank? If it acts to the right, doesn't that increase the kinetic energy of the block, which is something kinetic friction should not do?

  • $\begingroup$ When there is actual relative motion between two objects, the nature of friction can never be static. In your first case the friction should be kinetic. $\endgroup$ – Mitchell Oct 17 '17 at 19:04
  • $\begingroup$ @Mitchell Car tires are generally treated as static friction. There is always a single area of contact that doesn't really have relative motion with the road. Instead, the rotation of the tire creates a continuum of instant centres of rotation. $\endgroup$ – JMac Oct 17 '17 at 19:07
  • $\begingroup$ @Mitchell Are you sure? I'm quite certain my professor said static friction when he was explaining it.. $\endgroup$ – John Oct 17 '17 at 19:08
  • $\begingroup$ Well, if this question concerns rotational dynamics, thats a whole different story. $\endgroup$ – Mitchell Oct 17 '17 at 19:08
  • $\begingroup$ @Mitchell Noted. I've added that tag. $\endgroup$ – John Oct 17 '17 at 19:09

Just addressing the question in this comment about banked curves. Static friction is always going to oppose the motion that would happen if there were no friction. I will use the free-body diagram here as a reference for the case of no friction. The only two forces on the car are the normal force (N) and gravity (mg). The sum of these two forces is in the horizontal direction toward the center of the circle that the car is traveling around. This net force is what keeps the car traveling in a circle, and is equal to a component of the normal force. Now, if we consider the fourth equation on that page, which comes from considering $F_{net}=F_{centripetal}$:

$$mg\tan{\theta} = \frac{mv^2}{r}$$ And divide by $m$: $$g\tan{\theta} = \frac{v^2}{r}$$

This equation says for the car to stay in uniform circular motion (speed $v$ and radius $r$ don't change), there must be a balance between the four parameters in this equation. If, for example, speed $v$ is increased, radius $r$ must also increase given that $g$ and $\theta$ are constant. In the case that the car starts increasing its speed, it will start to slide up the incline. In this case, it will do so, and stop sliding sideways once the equation above is satisfied.

However, if we consider the case of an incline with friction, the situation changes. First, if the equation above is satisfied, then no friction will act sideways on the car tires (it isn't necessary, the car isn't trying to move sideways). However, there will be some friction on the tires in the forward direction, as in any case of "rolling without slipping." That phrase means the point of the tire that is in contact with the road at any instant in time is not moving w.r.t the road. It is "trying" to move backwards (think about a car at rest on ice. If you try to accelerate too quickly, the tires spin, and the point on the ice moves backwards. It is the same with a car in motion.), so the force from static friction must push forwards on the tire. This is what allows the car to accelerate forwards.

In the same scenario (rolling without slipping, $g\tan{\theta} = \frac{v^2}{r}$ initially satisfied) if the car's speed increases, then the equation will no longer be satisfied. But there is friction now, and as we found in the frictionless case, the car will "try" to move up the incline, and thus static friction will point down to oppose this motion. Vice versa, if the car decreases its speed, static friction will point up to oppose the car "trying" to slide down. (In this way, you can drive at a range of speeds for a given $g$, $\theta$, and $r$ if static friction is present.)

So, static friction always opposes attempted motion between two surfaces in contact.

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  • $\begingroup$ If the car's speed increases, the car will essentially move in spirals until it reaches a point where the radius is large enough for it to move in a circle. This means there is still some component of velocity in its original direction. Why does friction try to stop the sideways motion, but not this motion? i.e instead of acting up or down the incline, why doesn't it act forwards or backwards? Thanks. $\endgroup$ – John Oct 18 '17 at 0:30
  • $\begingroup$ The friction we are talking about here is sliding friction, the friction that occurs between two surfaces in contact that are sliding or trying to slide relative to each other. The forward motion of the car is not resisted by sliding friction because, in the forward direction, the car tire does not slide (this is an assumption we make, when we say "rolling without slipping"). It tries to rotate, but because of friction it applies a force to the ground. By Newton's 3rd law, the ground also applies a force to the tire in the direction the car is moving, this is how cars can accelerate. $\endgroup$ – PaulW Jan 5 '18 at 21:15

Why shouldn't kinetic friction increase the kinetic energy of the block?

It shouldn't increase the relative kinetic energy between the block and the plank; but because the friction isn't between the block and the static surroundings, there is no reason that it can't increase the kinetic energy of the block relative to the surroundings.

With the static friction on a care tire, it makes sense that the force points forward. The bottom of a tire is trying to slip backwards relative to the road. Static friction opposes this, and faces forward (also allowing the car to actually move). The way I think of it is that the car wants to pull the road backwards from underneath it.

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  • $\begingroup$ Thanks! I get the kinetic friction one. But in the banked curve case, static friction doesn't point forwards or backwards ; it points down the incline. $\endgroup$ – John Oct 17 '17 at 19:40
  • $\begingroup$ @John Oh, you're talking about banked inclines, sorry. Are you sure that's not just a component of friction they are considering in that case? $\endgroup$ – JMac Oct 17 '17 at 19:51
  • $\begingroup$ Pretty sure... I think he'd say something if it was only a component of friction. I googled it a bit, and it's the entire frictional force, not a component. Also, something I'm embarrassed to ask is how is it possible for a car to move at a constant speed? The only force exerted on the tires is friction, and by your logic, since the contact point of the wheels is 'pushing' backwards on the road, friction should point forwards, so it should never be 0 ( even at constant speed, a car's wheels rotate ) $\endgroup$ – John Oct 17 '17 at 20:09

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