Direction of friction: deceleration and turning

[ikjadoon]

Hi! :) I'm confused about the direction of the friction vector when we decelerate and when we turn. I've memorized the directions (if I decelerate going east, friction is aimed west; if I am in a flat circle and going around it, friction points to the middle), but they don't exactly make sense to me. I do know that friction opposes relative motion.

Issue 1:

So, there is a car accelerating to the east. The tires are spinning clockwise. Basically, the tire is pushing LEFT against the road. Friction opposes this and thus the friction vector points RIGHT (towards the east). I understand that pretty well. But when we decelerate (say slowing down while still moving east), the friction vector now points to the left?

So the same car is decelerating to the east. The tires are still spinning clockwise. The tires are now pushing RIGHT (somehow? ). Friction opposes this and thus the friction vector now points LEFT (towards the west...somehow? ).

I don't understand what motion friction is opposing when the car is decelerating? In the acceleration one, it is opposing the force of the tires pushing left. But in the deceleration one, isn't that the same motion? And thus friction seems to point forward?

Issue 2

So, friction allows cars to turn and its vector points towards the center of the circle (the circle they would "draw" if they went all the way around). But, if I apply the "wheels push against the road-and-friction-opposes-that", then it seems as if the friction force should point the same direction the wheels are pointing. Like here:

http://img855.imageshack.us/img855/2621/57fba08c2966484b9cf8885.png

But that isn't right at all. Why does this "wheels push against the road-and-friction-opposes-that" rule only work for the car accelerating forward? If that isn't a good way to "visualize" this or wrap my head around it (ostensibly so), any suggestions?

I have heard that if the net force is friction, it will be in the same direction as acceleration (through Newton's 2nd Law). But, I don't know how to say this, but that sounds like the result and not the steps in understanding it.

 

[Doc Al]

Hi! :) I'm confused about the direction of the friction vector when we decelerate and when we turn. I've memorized the directions (if I decelerate going east, friction is aimed west; if I am in a flat circle and going around it, friction points to the middle), but they don't exactly make sense to me. I do know that friction opposes relative motion.

Better to think of friction opposing slipping between surfaces.

Issue 1:

So, there is a car accelerating to the east. The tires are spinning clockwise. Basically, the tire is pushing LEFT against the road. Friction opposes this and thus the friction vector points RIGHT (towards the east). I understand that pretty well. But when we decelerate (say slowing down while still moving east), the friction vector now points to the left?

So the same car is decelerating to the east. The tires are still spinning clockwise. The tires are now pushing RIGHT (somehow? ). Friction opposes this and thus the friction vector now points LEFT (towards the west...somehow? ).

I don't understand what motion friction is opposing when the car is decelerating? In the acceleration one, it is opposing the force of the tires pushing left. But in the deceleration one, isn't that the same motion? And thus friction seems to point forward?

Realize that unless the tires are slipping (sliding) there is no movement between the surfaces. We are talking static friction here. What matters is the direction of the acceleration. To accelerate, the tires must push against the road (thanks to friction) and the road pushes back. Same thing when the tires decelerate. The tires slow down and end up pushing the road in the forward direction, causing the road to push back opposite to the direction of motion of the car.

Issue 2

So, friction allows cars to turn and its vector points towards the center of the circle (the circle they would "draw" if they went all the way around). But, if I apply the "wheels push against the road-and-friction-opposes-that", then it seems as if the friction force should point the same direction the wheels are pointing. Like here:

<snip>

But that isn't right at all. Why does this "wheels push against the road-and-friction-opposes-that" rule only work for the car accelerating forward? If that isn't a good way to "visualize" this or wrap my head around it (ostensibly so), any suggestions?

I have heard that if the net force is friction, it will be in the same direction as acceleration (through Newton's 2nd Law). But, I don't know how to say this, but that sounds like the result and not the steps in understanding it.

Again, unless the tires are slipping there is no relative motion between the surfaces. No tangential force is needed to keep the tires moving at constant speed in circle, but unless friction acts to push the car towards the center, the car will just keep going at constant velocity. Again, it is static friction that opposes the tendency to fly outward due to inertia.

 

[ikjadoon]

Thank you for your taking the time to write a reply, Doc Al! :) I should warn you: I'm not good at understanding physics, but if I don't understand it, then unless a problem looks identical to the one before, I'm at a bit of a loss, so here's hoping that I can develop this "physical intuition." :D

Friction opposes slipping between surfaces; that makes sense. But, OK, we have just a rolling tire on a surface with friction (no forces besides friction). If it is static, there is no slipping. What is friction opposing here, then?

If I have a box on surface with friction that is pushed by a force less than the force of static friction and let go of that force, after letting go of that force, the force of static friction is zero because nothing is pushing it, right?

--a little sidebar below--

So, then, are tires are fundamentally different than boxes when we talk about friction?

On a frictionless surface: if I push a box to the right and then remove my force, it will slide at a constant velocity to the right. If, on this same frictionless surface, I push a tire to the right at its center of mass, it will slide at a constant velocity on the right. If I push the tire on the top of the tire, though, it'll spin clockwise in place.

On a friction surface: say a box and a tire have the same mass and the same coefficient of static friction. My pushing force is greater than this calculated force of static friction. If I push the box and let go, the box will slide right (but then stop because of the force of kinetic friction, opposing the slipping and pointing left). If I push the tire (say on the top) and let go, it will again spin clockwise (but have no translational motion because that requires force less than the force of static friction). Then, the tire will slow its clockwise spinning because of the force of kinetic friction (which is opposing the slipping and thus is pointing right). And then the tire will have translation motion to the right and thereafter will soon stop because of the force of static friction. (Not sure if it will actually have translational motion).

 

[Doc Al]

Friction opposes slipping between surfaces; that makes sense. But, OK, we have just a rolling tire on a surface with friction (no forces besides friction). If it is static, there is no slipping. What is friction opposing here, then?

Friction acts to oppose or prevent slipping. If a tire is just rolling without slipping on a horizontal surface (ignoring any dissipative forces) then the friction force required to maintain the motion is zero. If the tire accelerates, as when rolling down a hill, then non-zero friction is required to keep it rotating at the proper rate to prevent slipping as it picks up speed. The friction is static friction, even though the tire as a whole is obviously moving.

If I have a box on surface with friction that is pushed by a force less than the force of static friction and let go of that force, after letting go of that force, the force of static friction is zero because nothing is pushing it, right?

Say the max static friction is given by μW, where W is the weight of the box. If you push the side of the box with a force F (less than μW) to the right, then the static friction will push the box to the left with an equal force F to prevent slipping. Remove that applied force and the friction goes back to zero.

--a little sidebar below--

So, then, are tires are fundamentally different than boxes when we talk about friction?

Tires can rotate, so things get more complicated.

On a frictionless surface: if I push a box to the right and then remove my force, it will slide at a constant velocity to the right.

OK.

If, on this same frictionless surface, I push a tire to the right at its center of mass, it will slide at a constant velocity on the right.

OK. (Once you stop pushing.)

If I push the tire on the top of the tire, though, it'll spin clockwise in place.

No. That push at the top will make it both slide and spin.

On a friction surface: say a box and a tire have the same mass and the same coefficient of static friction. My pushing force is greater than this calculated force of static friction. If I push the box and let go, the box will slide right (but then stop because of the force of kinetic friction, opposing the slipping and pointing left).

OK.

If I push the tire (say on the top) and let go, it will again spin clockwise (but have no translational motion because that requires force less than the force of static friction). Then, the tire will slow its clockwise spinning because of the force of kinetic friction (which is opposing the slipping and thus is pointing right). And then the tire will have translation motion to the right and thereafter will soon stop because of the force of static friction. (Not sure if it will actually have translational motion).

To figure out the motion of the tire you must consider Newton's 2nd law for both translation and rotation. That will allow you to figure out the friction force required to prevent slipping and whether that required friction force is less than the maximum available, which is μW.