Where does static friction come from in ABS?

[yosimba2000]

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?

 

[Cutter Ketch]

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.

 

[russ_watters]

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.

 

[Cutter Ketch]

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.

 

[russ_watters]

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?

 

[yosimba2000]

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.

Right, the wheel will spin, but it will not translate until it has grip. So the road provides the translation force, I think.

 

[russ_watters]

Right, the wheel will spin, but it will not translate until it has grip. So the road provides the translation force, I think.

It does indeed. So can you apply that knowledge to the subject of the thread?

 

[yosimba2000]

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?

I only typed what I did in post #6 in the context that no-slip MUST be observed. If the brakes apply CW torque, the wheel slows. If the whole car and wheel are translating at the same velocoty as it was before, then there will be slip due to the slowing of the wheels. If the no-slip has to happen, then the road has to counter the the brakes with a CCW torque of the same magnitude. But why does no-slip HAVE to happen? There's no law of nature saying this has to be so.

No, the car is just sliding now and experiencing opposing kinetic friction.

 

[russ_watters]

But why does no-slip HAVE to happen? There's no law of nature saying this has to be so.

You correctly pointed out the law of nature in post #1. You know why it slips or doesn't.

I don't see why you are arguing this though. Aren't you trying to understand why a car stops?

 

[yosimba2000]

It does indeed. So can you apply that knowledge to the subject of the thread?

In the context that the road provides translation force, then to stop the car going leftwards, the road has to provide a translation force to the right. But wouldn't this translation force also act as CCW torque on the wheels?

 

[russ_watters]

In the context that the road provides translation force, then to stop the car going leftwards, the road has to provide a translation force to the right. But wouldn't this translation force also act as CCW torque on the wheels?

Yes.

 

[yosimba2000]

Yes.

So the translation force works to keep the wheels spinning which keeps the car going left, but also slows down the car? How does that work? And if forces come in pairs, what's the other force that makes the road want to push rightwards?

 

[russ_watters]

So the translation force works to keep the wheels spinning...

Yes.

...which keeps the car going left but also slows down the car?

No! Is left clockwise or counterclockwise!? (at the contact patch)

And if forces come in pairs, what's the other force that makes the road want to push rightwards?

"Forces come in pairs" mean each object applies a force to the other through the same phenomena. In this case, that's friction at the contact patch.

 

[yosimba2000]

Yes.

No! Is left clockwise or counterclockwise!? (at the contact patch)

"Forces come in pairs" mean each object applies a force to the other through the same phenomena. In this case, that's friction at the contact patch.

Not sure what you're asking. If the bottom of the wheel is to go left, then the wheel turns CW. If the car is to go left, then the wheel is to go CCW, and CCW is the direction of the torque from the road on the wheel, keeping the wheel spinning.

Are you saying then the patch is wanting to move left, hence the rightward force from the road? If so, why would the patch want to move left if the wheel is going CCW?

 

[russ_watters]

Not sure what you're asking. If the bottom of the wheel is to go left, then the wheel turns CW. If the car is to go left, then the wheel is to go CCW, and CCW is the direction of the torque from the road on the wheel, keeping the wheel spinning longer.

Sorry, mixed two answers together: in post 40 you correctly said the road applies a force to the right. So: clockwise or counterclockwise?

Obviously the force cannot be both accelerating and decelerating *the car* at the same time.

 

[yosimba2000]

Sorry, mixed two answers together: in post 40 you correctly said the road applies a force to the right. So: clockwise or counterclockwise?

Obviously the force cannot be both accelerating and decelerating *the car* at the same time.

It will be CCW. But didn't you agree the CCW torque spins the wheel CCW? And also provides a rightward translation force in #41? So CCW wheel keeps the car moving left... but then I assume there's a split between how much force goes to rightward translation force and how much keeps the wheel going CCW.

 

[russ_watters]

It will be CCW. But didn't you agree the CCW torque spins the wheel CCW?

Of course! But that isn't the only torque being applied! Please start trying harder to carry forward what you already know as the discussion progresses. This has been one step forward another back, over and over.

And also provides a rightward translation force in #41?

Yes.

So CCW wheel keeps the car moving left...

How can a force to the right keep a car moving left? What are we trying to do here?

but then I assume there's a split between how much force goes to rightward translation and how much keeps the wheel going CCW.

It's one force and to be technical yes there is a split, but the net torque is negligible. Almost none of it goes toward keeping the wheel rotating (slightly less than zero, actually). Again, what are we trying to do?

 

[yosimba2000]

Of course! But that isn't the only torque being applied! Please start trying harder to carry forward what you already know as the discussion progresses. This has been one step forward another back, over and over.

Yes.
How can a force to the right keep a car moving left? What are we trying to do here?

It's one force and to be technical yes there is a split, but the net torque is negligible. Almost none of it goes toward keeping the wheel rotating (slightly less than zero, actually). Again, what are we trying to do?

I know the other torque is CW from the brakes, I just didn't mention it. But we also said before the brake torque and road torque to the right counteract each other, so no net torque is applied on the wheels.

The rightward force on the bottom of the wheel keeps it spinning CCW, and a CCW spinning wheel travels left. But the CCW wheel (no net torque on it now) that wants to make the car go left is overcome by part of the rightward translation force slowing the whole body down?

 

[russ_watters]

I know the other torque is CW from the brakes, I just didn't mention it. But we also said before the brake toeque and road torque to the right counteract each other.

They do... Causing a net torque of almost zero, clockwise.

A force on the bottom of the wheel keeps it spinning CCW longer, and a CCW spinning wheel travels left.

Yes, the car is traveling left. You specified that in post 1. Are you suggesting the force the ground applies to the wheel is what makes the car move left? It's not: the scenario you designed starts with no forces and a car moving to the left. Forces cause accelerations, not motion (directly).

 

[yosimba2000]

They do... Causing a net torque of almost zero, clockwise.

Yes, the car is traveling left. You specified that in post 1. Are you suggesting the force the ground applies to the wheel is what makes the car move left? It's not: the scenario you designed starts with no forces and a car moving to the left. Forces cause accelerations, not motion (directly).

I am. Isn't that what we said in #37?

 

[russ_watters]

I am. Isn't that what we said in #37?

This is what you said in post #36:

Right, the wheel will spin, but it will not translate until it has grip. So the road provides the translation force, I think.

Here you are describing a car initially at rest, which is not the scenario you described in post #1.

 

[Cutter Ketch]

Right, the wheel will spin, but it will not translate until it has grip. So the road provides the translation force, I think.

? What do you mean by won’t translate. I think you might mean that there is no tangential force which is correct. Translate means something else. The car and the attached wheel will be moving to the left at constant speed which is by definition translation

 

[yosimba2000]

Here you are describing a car initially at rest, which is not the scenario you described in post #1.

Can you show me when when I described a car at rest?

Let me look at it from a sliding standpoint to see if it's easier for me.

A car is sliding left with locked wheels. Kinetic friction by the road pushes to the right. This kinetic friction does two things: tries to spin the wheel CCW and pushes against the car rightwards. The wheel is relatively stable and doesn't turn due to the static friction offered by the brakes. We let go of the brakes and the wheels are now free to turn. But the wheels are still sliding, and the kinetic friction turns the wheels CCW. The wheels are sliding until the translation movement left is counteracted by the patch's rotational velocity to the right so that the contact patch has no velocity wrt the road. A moment later, the patch's velocity is no longer perfectly horizontal but now angled, so it has slightly less velocity to the right, so the translational velocity to the left is now larger, resulting in a net leftward velocity of the patch. The patch hits a small part of the road with a force not larger than the maximum static friction offered by the road, and the road provides a resulting force of the same magnitude rightward against the patch. This rightward force is mostly used to oppose the leftward movement of the car.

 

[yosimba2000]

? What do you mean by won’t translate. I think you might mean that there is no tangential force which is correct. Translate means something else. The car and the attached wheel will be moving to the left at constant speed which is by definition translation

Like a wheel spinning on ice just spins in place until it has grip to move forward (translation), which is offered by the tangential force from the road when the wheel pushes the ground.

 

[Cutter Ketch]

Like a wheel spinning on ice just spins in place until it has grip to move forward (translation), which is offered by the tangential force from the road when the wheel pushes the ground.

I think we may be talking at cross purposes. There are a few conversations and scenarios, and I must have attached your post to the wrong one. My apologies.

I was referring to tha case of a car coasting in neutral. Ignoring some small losses that would in reality slow the car down you can to first order say there are no tangential forces. The car (and the wheels) continue at constant speed from left to right. The wheels turn at constant speed staying in perfect contact with the road without slipping. Gravity pushes down, the road pushes up. Despite the perfect and continuous contact the frictional force at the contact patch is zero. There are no torques on the wheel and there is no accelerating/decelerating force on the car. Nothing happens until you step on the brake. Does that all make sense?

 

[yosimba2000]

I think we may be talking at cross purposes. There are a few conversations and scenarios, and I must have attached your post to the wrong one. My apologies.

I was referring to tha case of a car coasting in neutral. Ignoring some small losses that would in reality slow the car down you can to first order say there are no tangential forces. The car (and the wheels) continue at constant speed from left to right. The wheels turn at constant speed staying in perfect contact with the road without slipping. Gravity pushes down, the road pushes up. Despite the perfect and continuous contact the frictional force at the contact patch is zero. There are no torques on the wheel and there is no accelerating/decelerating force on the car. Nothing happens until you step on the brake. Does that all make sense?

It all makes sense except for how the car translates without tangential force.

 

[russ_watters]

Can you show me when when I described a car at rest?

The words "will not translate" mean it is not moving left or right.

Let me look at it from a sliding standpoint to see if it's easier for me.

A car is sliding left with locked wheels. Kinetic friction by the road pushes to the right. This kinetic friction does two things: tries to spin the wheel CCW and pushes against the car rightwards. The wheel is relatively stable and doesn't turn due to the static friction offered by the brakes.

All fine.

We let go of the brakes and the wheels are now free to turn. But the wheels are still sliding, and the kinetic friction turns the wheels CCW. The wheels are sliding until the translation movement left is counteracted by the patch's rotational velocity to the right so that the contact patch has no velocity wrt the road. A moment later, the patch's velocity is no longer perfectly horizontal but now angled, so it has slightly less velocity to the right, so the translational velocity to the left is now larger, resulting in a net leftward velocity of the patch. The patch hits a small part of the road with a force not larger than the maximum static friction offered by the road, and the road provides a resulting force of the same magnitude rightward against the patch. This rightward force is mostly used to oppose the leftward movement of the car.

In a non-ideal case where we include internal friction losses, this is correct. In a no-loss situation, when the scenario is completed there are no forces.

 

[russ_watters]

It all makes sense except for how the car translates without tangential force.

Newton's First Law tells us once in motion no force is needed to keep it in motion.

 

[yosimba2000]

The words "will not translate" mean it is not moving left or right.

In a non-ideal case where we include internal friction losses, this is correct. In a no-loss situation, when the scenario is completed there are no forces.

Oh, I was referring to a different scenario in that question, sorry.

How could the car ever stop in a no-loss situation?

 

[yosimba2000]

Newton's First Law tells us once in motion no force is needed to keep it in motion.

Ok, so a perfectly slippery road and the car was in motion beforehand. Makes sense so far then.