Where does static friction come from in ABS?

In summary, static friction is a force of constraint that is equal and opposite to the applied force until the force is too great and the surface slips. It is always present as long as the tires are not slipping, and it provides a rearward force that prevents the wheels from stopping under braking. This force is what ultimately stops the car, and it is maintained through the no-slip condition enforced by the interaction between the brake pads and the ground. The ground's torque works to turn the wheel, not stop the car, and static friction is a reaction to preventing slippage rather than an opposing force.
  • #36
Cutter Ketch said:
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.
 
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  • #37
yosimba2000 said:
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?
 
  • #38
russ_watters said:
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.
 
  • #39
yosimba2000 said:
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?
 
  • #40
russ_watters said:
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?
 
  • #41
yosimba2000 said:
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.
 
  • #42
russ_watters said:
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?
 
  • #43
yosimba2000 said:
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.
 
  • #44
russ_watters said:
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?
 
  • #45
yosimba2000 said:
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.
 
  • #46
russ_watters said:
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.
 
  • #47
yosimba2000 said:
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?
 
  • #48
russ_watters said:
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?
 
  • #49
yosimba2000 said:
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).
 
  • #50
russ_watters said:
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?
 
  • #51
yosimba2000 said:
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.
 
  • #52
yosimba2000 said:
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
 
  • #53
russ_watters said:
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.
 
  • #54
Cutter Ketch said:
? 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.
 
  • #55
yosimba2000 said:
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?
 
  • #56
Cutter Ketch said:
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.
 
  • #57
yosimba2000 said:
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.
 
  • #58
yosimba2000 said:
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.
 
  • #59
russ_watters said:
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?
 
  • #60
russ_watters said:
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.
 
  • #61
yosimba2000 said:
O
How could the car ever stop in a no-loss situation?
With no brakes and no other losses, the car will never stop, per Newton's First Law.
Ok, so a perfectly slippery road and the car was in motion beforehand. Makes sense so far then.
The road should not be thought of as perfectly slippery, otherwise we can't run your scenarios on it. The losses we're referring to include things like wind resistance, rolling resistance and drivetrain friction.
 
  • #62
russ_watters said:
The road should not be thought of as perfectly slippery, otherwise we can't run your scenarios on it. The losses we're referring to include things like wind resistance, rolling resistance and drivetrain friction.

Ok
 
  • #63
yosimba2000 said:
It all makes sense except for how the car translates without tangential force.

In this thread there have been several very similar exchanges that lead me to suspect you are confused about something very fundamental: Newton’s first law of motion. A body will continue in its present state of uniform motion until acted on by an external force. Once the car is moving at uniform speed you don’t need a force to keep it moving. It will keep moving all by itself. Once the wheel is spinning you don’t need a torque to keep it spinning. It will keep spinning all by itself.

This is counter intuitive to many people (including Aristotle) because we know in the real world nothing just keeps moving. However that is because we can’t easily eliminate all the outside forces. However we can do a pretty good job of minimizing them and in that case things will keep moving for a long time. My car will take 10 seconds to slow from 65 to 60 mph when coasting and that is mostly wind resistance. The bearing friction and rolling resistance are impressively small. The voyager spacecraft is at a point where even the gravity of the sun is weak, and it is going to keep on flying away forever without needing any additional force.

Force is not required for motion. Force is required to change motion.
 
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  • #64
yosimba2000 said:
Ok, so a perfectly slippery road and the car was in motion beforehand. Makes sense so far then.
No! The road does not need to be slippery! That’s what the wheels are for, the axle bearings need to be slippery.
 
  • #65
yosimba2000 said:
How could the car ever stop in a no-loss situation?
With regenerative braking you store the energy internally, instead of dissipating and loosing it.
 
  • #66
At the risk of getting things even more confused, try this.
The following is a bit simplified but hopefully easy enough to follow.
  • Car weighs 4000 pounds, so the load on each tire is 1000 pounds. Static Coefficient Of Friction (COF) between tire and road is 0.6.
  • Car coasting left, tires turning CCW matching car speed. Result is no sliding between tires and road.
  • Brakes are partially applied, meaning the brake pads/shoes are sliding on the brake rotor/drum, just for discussion let's say the brakes are applied sufficient to supply 500 pounds of retarding force at the outer diameter of the tire.
  • The car is still moving left taking the tires along with it. Due to the retarding force of the brakes, there is now 500 a pound sliding force on each tire due to the inertia of the moving car. Since the tire patch is not sliding on the road, there can be up to 1000 x 0.6 = 600 pounds of sliding force before the tire starts to slide. Result is the car slows.
The energy taken from the moving car to slow it is converted to heat at the brake pads/shoes.

Now let's investigate what happens if the brakes are applied more forcefully.
  • Same vehicle as above but we take into account the Sliding Coefficient Of Friction (COF) between tire and road as 0.4. Static COF remains 0.6
  • This time the brakes are applied sufficient to supply 700 pounds of retarding force at the outer diameter of the tire.
  • The car is still moving left taking the tires along with it. Due to the retarding force of the brakes, there is 700 a pound sliding force on each tire due to the inertia of the moving car. The 700 pound sliding force experienced by the tire patch now exceeds the 600 pounds of static friction between tire and road, so the tire stops rotating and starts sliding.
  • Note that due to the Sliding COF, being only 0.4, the tires will continue sliding until the retarding force of the brakes is decreased to less than 400 pounds at the outer diameter of the tire.

The above also explains why it is better to avoid locking the brakes in a panic stop. Once the tires start sliding, they supply less retarding force than if they kept rolling. That's something that is hard to remember in a panic situation!

Hope this helps.

Cheers,
Tom

p.s. ABS works by sensing the rotational speed of each tire during braking. If a tire slows appreciatively in relation to the others, the ABS reduces the hydraulic pressure to the brake on that tire until it stops sliding.
 
  • #67
Is it not about time one of you posted a diagram showing the relevant forces?
 
  • #68
AZFIREBALL said:
Is it not about time one of you posted a diagram showing the relevant forces?
Yes! The OP should.
 

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