Where does static friction come from in ABS?

  • #1
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Ok, so assume we have a wheel on the road travelling to the left. The wheel spins CCW and also has translation to the left.

If we let the wheel roll, assuming no slip, then the translation velocity to the left will counteract the rotational velocity to the right at the bottom of the wheel, so the section of the wheel touching the ground is not displaced in that moment.

So this means that the little part of the wheel is sitting on top of that piece of ground for a little moment... and then static friction? I'm confused here because isn't static friction a reaction to an opposing force? If that piece of wheel is just sitting there, like I'm sitting in a chair, what "pushing force" is that static friction trying to oppose?
 

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  • #2
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I think there are a few layers to your question, so let me answer in pieces.

First, static friction is a force of constraint. It is equal and opposite to whatever force is applied until the force is too great and the surface slips.

Second, yes, up until the force grows large enough that the contact patch slips you are dealing with static friction.

Third, yes as you describe it there could be moments when there is no force trying to make the contact patch slip, so the opposing force of static friction is also 0. However that is not the usual condition. There is always something trying to change the motion. If the car is coasting, air resistance is trying to slow the car, friction in the axles and bearings, too. There are forces in turning a corner or stepping on the accelerator and static friction keeps the tire “stuck” to the road. In turn that frictional reaction force is what accelerates, decelerated, and turns the car.

Since your post is titled with ABS I presume you are most interested in braking. The brakes try to stop the wheel from turning, but as long as the braking torque doesn’t exceed the maximum torque available from static friction the wheel keeps turning. Your post seems to suggest that as long as the tires don’t slip there is no friction force, but that is wrong. Static friction provides a rearward force that prevents the wheels from stopping under braking, and that rearward force is what stops the car. The force can be as large as the static friction allows but above that limit the tire starts slipping. (and the ABS detects the slipping and eases off the brakes)
 
  • #3
tnich
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Ok, so assume we have a wheel on the road travelling to the left. The wheel spins CCW and also has translation to the left.

If we let the wheel roll, assuming no slip, then the translation velocity to the left will counteract the rotational velocity to the right at the bottom of the wheel, so the section of the wheel touching the ground is not displaced in that moment.

So this means that the little part of the wheel is sitting on top of that piece of ground for a little moment... and then static friction? I'm confused here because isn't static friction a reaction to an opposing force? If that piece of wheel is just sitting there, like I'm sitting in a chair, what "pushing force" is that static friction trying to oppose?
The title of your post mentions ABS. Torque applied by the brakes results in a static frictional reaction force on the tires. The ABS reduces the braking force when necessary to prevent the tires from slipping.
 
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  • #4
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Static friction provides a rearward force that prevents the wheels from stopping under braking, and that rearward force is what stops the car.

Can you go step by step how the static friction "exists" if the contact patch is not accelerating to the left, and how the static friction prevents the wheels from stopping? If anything, it seems like the static friction helps to oppose leftwards translational acceleration, but doesn't do anything to impede a car otherwise.
 
  • #5
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At time 0 the car and the wheel are traveling to the left at v. The wheel is rotating counter-clockwise at w = v/(2 pi r). Through friction on the brake pads the brakes resist the motion of the wheel applying a clockwise torque T. To keep the wheel rolling without slipping the friction with the ground must apply an equal and opposite counter-clockwise torque -T = F r implying that through static friction the ground is pushing the contact patch to the right with a force F = -T/r. That force is what stops the car.
 
  • #6
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At time 0 the car and the wheel are traveling to the left at v. The wheel is rotating counter-clockwise at w = v/(2 pi r). Through friction on the brake pads the brakes resist the motion of the wheel applying a clockwise torque T. To keep the wheel rolling without slipping the friction with the ground must apply an equal and opposite counter-clockwise torque -T = F r implying that through static friction the ground is pushing the contact patch to the right with a force F = -T/r. That force is what stops the car.

It's becoming clearer, but now my question is what is enforcing the no-slip condition, and how does this stop the car? If the brake pads create CW torque, and the ground creates CCW torque of same magnitude, which is required to maintain no-slip, then essentially no torque is acting on the wheels so the wheels don't slow down. The ground torque works to turn the wheel, not stop the car, unless the wheels can no longer move independently of the car. But then the question still remains... what says the ground has to maintain no-slip?
 
  • #7
A.T.
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..isn't static friction a reaction to an opposing force?
No, in general it opposes (prevents) slippage. Only in the most trivial cases you can identify a single force that is balanced by the static friction. But in general it's the sum of all forces that determines how much static friction is required to prevent slippage.

If the brake pads create CW torque, and the ground creates CCW torque of same magnitude,
It's not the same magnitude, because the wheel has some rotational inertia too.
 
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  • #8
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It's becoming clearer, but now my question is what is enforcing the no-slip condition, and how does this stop the car? If the brake pads create CW torque, and the ground creates CCW torque of same magnitude, which is required to maintain no-slip, then essentially no torque is acting on the wheels so the wheels don't slow down. The ground torque works to turn the wheel, not stop the car, unless the wheels can no longer move independently of the car. But then the question still remains... what says the ground has to maintain no-slip?

Yes, that is the instantaneous condition I’ve described, no torque on the tire, so it doesn’t change speed, but that is only approximately correct. To achieve that condition I have a net force on the car. The car slows down, and the next instant everything is a little slower. Naturally the instantaneous condition shouldn’t be exactly no torque, but rather there will be a little residual torque so that the wheel slows down at the same rate the car decelerates. It’s only a little more difficult to solve correctly, but I didn’t want to confuse the issue.
 
  • #9
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... But then the question still remains... what says the ground has to maintain no-slip?

Static friction is a force of constraint. It is equal and opposite to whatever tangential force is applied. This is no different than rods or ropes or walls in statics problems. You lean on a wall. How hard does the wall push on you? Well, it CAN’T move, so it pushes back on you exactly however hard you push on it. The only difference with static friction is that there is a limit. It won’t allow movement and will match any tangential force until that force exceeds a limit and the contact slips.

Put your hand on a table with a decent normal force. Now apply some tangential force to try and make it slip. You can put a little force or a lot of force but in all cases your hand doesn’t slip. It isn’t moving so the sum of the forces must be zero. Just as the table is providing whatever equal and opposite normal force is necessary to keep your hand from going through the table, static friction must be providing whatever equal and opposite force is necessary to keep your hand from slipping. However, push hard enough and eventually your hand will slip.
 
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  • #10
A.T.
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You lean on a wall. How hard does the wall push on you? Well, it CAN’T move, so it pushes back on you exactly however hard you push on it. The only difference with static friction is that there is a limit.
The normal force on a wall has a limit too.
 
  • #11
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The normal force on a wall has a limit too.
True! And I have leaned on walls where the limit was, unfortunately, less than the applied force.
 
  • #12
russ_watters
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It's becoming clearer, but now my question is what is enforcing the no-slip condition, and how does this stop the car? If the brake pads create CW torque, and the ground creates CCW torque of same magnitude, which is required to maintain no-slip, then essentially no torque is acting on the wheels so the wheels don't slow down.
You aren't trying to stop the wheels, you are trying to stop the car. So to within an arbitrarily low limit, yes the net torque on the wheels can be zero.
The ground torque works to turn the wheel, not stop the car, unless the wheels can no longer move independently of the car. But then the question still remains...
It's probably helpful to think of the rolling case as being not fundamentally different from the sliding case.
... what says the ground has to maintain no-slip?
It doesn't have to, it just stops the car faster that way.
 
  • #13
Mister T
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I'm confused here because isn't static friction a reaction to an opposing force? If that piece of wheel is just sitting there, like I'm sitting in a chair, what "pushing force" is that static friction trying to oppose?

If you try to slide to the left, there will be a static friction force pushing you towards the right. You push to the left on the chair, the chair pushes to the right on you. That's an action-reaction pair of forces. Action-reaction pairs of forces don't oppose each other because they act on different objects.

Likewise, the tire pushes on the road and the road pushes on the tire.
 
  • #14
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If you try to slide to the left, there will be a static friction force pushing you towards the right. You push to the left on the chair, the chair pushes to the right on you. That's an action-reaction pair of forces. Action-reaction pairs of forces don't oppose each other because they act on different objects.

Likewise, the tire pushes on the road and the road pushes on the tire.

But the tire isn't pushing on the road because the that contact patch has no velocity from the road's perspective, right?

It's probably helpful to think of the rolling case as being not fundamentally different from the sliding case.

So let me try to compare the contact patch that is not moving relative to the ground, and a block sitting on the road. I push the block with a force lower than the maximum static friction, and the block doesn't move because it feels a force of the same magnitude pushing back on it. But ONLY until I push it does the block feel the counteracting static friction.

Is this not the same with the contact patch? The contact patch should feel no static friction force against it until a translational force is acting on the contact patch, right? So I don't see how static friction slows the car down when it should only exhibit itself when translational force is introduced.
 
  • #15
russ_watters
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But the tire isn't pushing on the road because the that contact patch has no velocity from the road's perspective, right?
Did you really mean to say that? When leaning against a wall, your velocity is zero, but you are applying forces, right?
So let me try to compare the contact patch that is not moving relative to the ground, and a block sitting on the road. I push the block with a force lower than the maximum static friction, and the block doesn't move because it feels a force of the same magnitude pushing back on it. But ONLY until I push it does the block feel the counteracting static friction.
Yes... but I don't see how this tells you anything useful about a car. Again: you are trying to stop the car, not the wheel. Whether the contact patch is stopped (rolling wheel) or moving (locked up wheel) doesn't tell you much about stopping the car (only different in magnitude).

You have stated repeatedly that the torques on the wheel are (roughly) balanced, so it doesn't slow down. That's correct! So can we talk about the car now? That's what you really want to know about!

Try this: a car with locked tires is like a sliding block, right? Why does it stop?
Is this not the same with the contact patch? The contact patch should feel no static friction force against it until a translational force is acting on the contact patch, right? So I don't see how static friction slows the car down when it should only exhibit itself when translational force is introduced.
The static friction force is translational, so this statement seems circular and empty. I'm not sure what you are trying to say.
 
  • #16
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Did you really mean to say that? When leaning against a wall, your velocity is zero, but you are applying forces, right?

Yes... but I don't see how this tells you anything useful about a car. Again: you are trying to stop the car, not the wheel. Whether the contact patch is stopped (rolling wheel) or moving (locked up wheel) doesn't tell you much about stopping the car (only different in magnitude).

You have stated repeatedly that the torques on the wheel are (roughly) balanced, so it doesn't slow down. That's correct! So can we talk about the car now? That's what you really want to know about!

Try this: a car with locked tires is like a sliding block, right? Why does it stop?

The static friction force is translational, so this statement seems circular and empty. I'm not sure what you are trying to say.

Sorry, I meant to say the contact patch doesn't seem to be applying any forces against the road. It has no velocity relative to the road so it's not sliding, so there's no kinetic friction against it... And there's no static friction against it since there's no force leftwards.

A car with locked tires slides, and the kinetic friction pushes in the opposite direction to slow it down until it stops.

I'm trying to understand HOW the static friction comes into play. ABS keeps the tires from slipping, so it doesn't go into the kinetic friction regime. But to manage it, it wants to keep the contact patch with no velocity wrt road. So now it's in the static friction regime, which would show if you were to apply a translation force on the contact patch to the left.
 
  • #17
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But the tire isn't pushing on the road because the that contact patch has no velocity from the road's perspective, right?

No relative motion does NOT mean no force! If you push on a wall the wall pushes back. Neither you nor the wall moves indicating the NET force on each of you is zero, but you are each exerting a force on each other. Somewhere there is a balancing force.

In the case of the tire the road exerts a torque on the tire, but the brake pad applied an opposing torque. The road applies a linear force on the tire. The momentum of the car applie an opposing linear force through the axle. The net effect is that the frictional force from the road is translated to the car and opposes the cars momentum so the car slows down by F=ma

Let me give another example. You are on ice skates facing a wall. You use a broom stick to push off of the wall. You push on the end of the broom stick with force F. The broomstick pushes on the wall with force F. The wall pushes back on the broom stick with force F. The broom stick pushes on you with force F. You accelerate away from the wall. Here there is no NET force on the broom stick because it gets pushed from both ends but there are definitely forces on the broom stick (and reaction forces from the broom stick). It does not move, but forces are being transmitted through the broom stick. You effectively push off the wall via the stationary broom stick.

The tire is like the broom stick. The car pushes off the road via the quasi stationary wheel. The wheel and tire have opposed and (nearly) cancelling forces on them, but those forces are real and they do stop the car.
 
  • #18
russ_watters
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Sorry, I meant to say the contact patch doesn't seem to be applying any forces against the road. It has no velocity relative to the road so it's not sliding, so there's no kinetic friction against it... And there's no static friction against it since there's no acceleration leftwards.
How does that follow from this:
I push the block with a force lower than the maximum static friction, and the block doesn't move because it feels a force of the same magnitude pushing back on it.
You seem to be confusing summing to zero with being zero. 1-1=0, but neither 1 nor 1 are zero.
A car with locked tires slides, and the kinetic friction pushes in the opposite direction to slow it down until it stops.
Great! So what makes you think anything has changed when not locked? Hint: how many torques are there and by what type of friction do they arise? What has really changed?
But to manage it, it wants to keep the contact patch with no velocity wrt road. So now it's in the static friction regime, which would show if you were to apply a translation force on the contact patch to the left.
Yes, with non locked up braking there is a static friction force applied to the wheel by the road, which stops the car. Or if preferred....
 
  • #19
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The wheel and tire have opposed and (nearly) cancelling forces on them, but those forces are real and they do stop the car.

Sorry, is the bolded part a typo?
 
  • #20
russ_watters
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Sorry, is the bolded part a typo?
It's not. Again: The forces on the wheel are equal and opposite. But if you want to know how the car stops, not the wheel, you need to start describing the forces acting on the car!
 
  • #21
russ_watters
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I have to go away from my phone for a bit, so I'll try to wrap it all together.

Case 1:
A skidding car tire with the brakes locked up has two forces applied to it by friction - one static and one dynamic - causing two torques, which roughly sum to zero. The car has one force applied to it and decelerates.

Case 2:
A rolling car tire with the brakes applied but not locked up has....

Can you finish case 2?

It really is as simple as it looks.
 
  • #22
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Great! So what makes you think anything has changed when not locked? Hint: how many torques are there and by what type of friction do they arise? What has really changed?

Yes, with non locked up braking there is a static friction force applied to the wheel by the road, which stops the car. Or if preferred....

With a non-locked tire, the tire is slowed down by the internal axles by kinetic friction, the applied brakes by kinetic friction, and the road pushing leftwards on the bottom of the CCW spining wheel.

The bolded part is my confusion. The contact patch is in the regime of static friction now, but it doesn't feel any static friction against it until something tries to translate the wheel.
 
  • #23
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I have to go away from my phone for a bit, so I'll try to wrap it all together.

Case 1:
A skidding car tire with the brakes locked up has two forces applied to it by friction - one static and one dynamic - causing two torques, which roughly sum to zero. The car has one force applied to it and decelerates.

Case 2:
A rolling car tire with the brakes applied but not locked up has....

Can you finish case 2?

It really is as simple as it looks.

I'll try to identify both cases just so we're onthe same page.

Case 1: The road applies kinetic friction rightwards on the bottom of the wheel, which acts as torque CCW on the wheel. The brakes applies a static friction against the wheel CW, and the wheel remains locked.

Case 2: The rolling tire experience two forces. The brakes provide a torque CW on the wheel, and the wheel slows. As the wheel keeps turning CCW, the bottom of the wheel pushes the road to the right, and the road pushes leftwards on the bottom of the wheel, creating a CW torque as well.
 
  • #24
russ_watters
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The contact patch is in the regime of static friction now, but it doesn't feel any static friction against it until something tries to translate the wheel.
I'm flumoxed: the brakes are being applied! The brakes apply one force and the road the other! I can't understand why you would think the road does not apply a force of static friction.
 
  • #25
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Sorry, is the bolded part a typo?
No. The thing we are discussing is a wheel and tire. I mention both because it is important to note the connection of the axle to the wheel, which might be confusing if I only say tire.
 
  • #26
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I'm flumoxed: the brakes are being applied! The brakes apply one force and the road the other! I can't understand why you would think the road does not apply a force of static friction.

Ok, I think I might have confused the instantenous V = 0m/s on the bottom of the tire to mean a non-rotating tire. So yes, the tire is rotating. Brakes apply CW torque, and the road does as well. The interaction between the road and tire is static friction?
 
  • #27
russ_watters
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Case 2: The rolling tire experience two forces. The brakes provide a torque CW on the wheel, and the wheel slows. As the wheel keeps turning CCW, the bottom of the wheel pushes the road to the right, and the road pushes leftwards on the bottom of the wheel, creating a CW torque as well.
If the wheel had to clockwise torques on it it would be accelerating. So which one is wrong?
 
  • #28
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No. The thing we are discussing is a wheel and tire. I mention both because it is important to note the connection of the axle to the wheel, which might be confusing if I only say tire.

Sorry, still confused. Is there a difference between a wheel and a tire?
*edit* Oh there is. Sorry, not really a car guy.
 
  • #29
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If the wheel had to clockwise torques on it it would be accelerating. So which one is wrong?

Wouldn't the CW torques oppose the CCW motion of the wheel? I don't see how it would accelerate.

If I have a rolling wheel with no brakes, then the road is what slows down the wheel, right?
If add brakes, then both the brakes and road will slow the wheel.
 
  • #30
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I'll try to identify both cases just so we're onthe same page.

Case 1: The road applies kinetic friction rightwards on the bottom of the wheel, which acts as torque CCW on the wheel. The brakes applies a static friction against the wheel CW, and the wheel remains locked.

Case 2: The rolling tire experience two forces. The brakes provide a torque CW on the wheel, and the wheel slows. As the wheel keeps turning CCW, the bottom of the wheel pushes the road to the right, and the road pushes leftwards on the bottom of the wheel, creating a CW torque as well.

Case 2 is a little messed up. Why would the bottom of the tire push the road right? Remember friction opposes the applied force. The brakes are trying to stop the wheel. If friction does nothing the tire will slow down or stop and be sliding from right to left. If friction is going to keep the tire rolling it has to oppose this by pushing from left to right, a CCW torque in opposition to the braking force.

Friction can only oppose. It never adds extra.
 
  • #31
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Case 2 is a little messed up. Why would the bottom of the tire push the road right? Remember friction opposes the applied force. The brakes are trying to stop the wheel. If friction does nothing the tire will slow down or stop and be sliding from right to left. If friction is going to keep the tire rolling it has to oppose this by pushing from left to right, a CCW torque in opposition to the braking force.

Friction can only oppose. It never adds extra.

Isn't that how a wheel rolls? A CCW spinning wheel travels to the left because the bottom of the wheel travels rightwards, and the wheel pushes the road with a rightward force. So the road pushes on the wheel with a leftwards force.

Also, is this correct:

A rolling wheel with no slip experiences static friction from the road against the wheel? This static friction is what allows the wheel to roll forward?
 
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  • #32
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Wouldn't the CW torques oppose the CCW motion of the wheel? I don't see how it would accelerate.

If I have a rolling wheel with no brakes, then the road is what slows down the wheel, right?
If add brakes, then both the brakes and road will slow the wheel.

Well, there are all sorts of things going on in reality, but those should only be added after you understand the basic idea. For our purposes let’s assume all the things which are almost negligible are exactly zero. Let’s say the road is flat. Let’s say that there is no wind resistance on the car. Let’s say that the bearings are very good and there is no friction at the axle. Finally let’s say the tire doesn’t deform and so there is no “rolling resistance”. That’s not friction at the contact. It’s something else. Anyhow all of these things are approximately true as evidenced by the fact that if I put my car in neutral on flat ground it will coast for a very long way slowing down very slowly compared to the stopping distance under braking. Why are they almost true? Because the engineers worked very hard to make them almost true.

In this only slightly idealized condition nothing slows down at all. Rolling on the road at a constant speed does not require any force or torque. The tire is already spinning and will keep spinning at the same rate due to its inertia. That keeps it in perfect time with the road so there is no slipping. The car continues moving under its inertia not accelerating or slowing down. The only forces are gravity downward countered by a restoring force upward.

Nothing happens until you step on the brake applying a CW torque on the wheel. At that time to keep the wheel from slipping static friction with the road applies a CCW torque on the wheel by applying a rightward tangential force at the contact patch. That force decelerates the car.
 
  • #33
russ_watters
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Wouldn't the CW torques oppose the CCW motion of the wheel? I don't see how it would accelerate.
You had this correct in post #6. I don't understand why your understanding is getting worse instead of better.

An object accelerates (decelerates) when there is a net force (or torque) on it. And to copy and paste again:

You aren't trying to stop the wheels, you are trying to stop the car.

This means you need to apply a net force to *the car*.
If I have a rolling wheel with no brakes, then the road is what slows down the wheel, right?
If you have a rolling wheel and no brakes, then there are no forces and no acceleration (deceleration).
If add brakes, then both the brakes and road will slow the wheel.
No! Reread the thread or draw yourself a picture if you've lost what you knew an hour ago. One of the torques is wrong: the net torque is roughly zero.

The directions of the forces are the same for a rolling wheel and skidding wheel, and the resulting net force on *the car* *the car* *the car* is the same. The only difference is which force is static friction and which is kinetic. *the car* stops because *the car* has a net force on it.
 
  • #34
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Isn't that how a wheel rolls? A CCW spinning wheel travels to the left because the bottom of the wheel travels rightwards, and the wheel pushes the road with a rightward force. So the road pushes on the wheel with a leftwards force.

Also, is this correct:

A rolling wheel with no slip experiences static friction from the road against the wheel? This static friction is what allows the wheel to roll forward?
See my simultaneous post, but the short answer is no. If it weren’t fore some small unavoidable losses in the system, to first order no force or torque is required to keep the wheel spinning. It is already spinning at the correct rate. Why would it not continue to do so? Of course you know in reality there are losses, but these can be minimized. The point is that there is no fundamental need to apply a force to keep a wheel turning.
 
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  • #35
russ_watters
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Despite my best efforts to focus you on *the car* you remain focused on stopping the wheels. So let's explore that:

You're traveling at 60mph and slam on the brakes in a car with no abs. A fraction of a second later the wheels stop spinning. Did you succeed in stopping the car?
 

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