Where does static friction come from in ABS?

[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.

 

[russ_watters]

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.

 

[yosimba2000]

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

 

[Cutter Ketch]

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.

 

[Cutter Ketch]

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.

 

[A.T.]

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.

 

[Tom.G]

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.

 

[AZFIREBALL]

Is it not about time one of you posted a diagram showing the relevant forces?

 

[russ_watters]

Is it not about time one of you posted a diagram showing the relevant forces?

Yes! The OP should.