Car - brake locking, stopping distance

[ChessEnthusiast]

I have been reading quite a lot about the Physics behind car dynamics and I have gotten to a point which keeps confusing me, namely - brake locking.

I understand brake locking as the situation when the wheels of the car do not rotate anymore but the car is still moving - this causes the tires to rub against the asphalt.
This implies that the force used to decelerate the car will be the force of kinetic friction exerted by asphalt on the tires (I am not taking air resistance into account)
If the wheels weren't locked, the stopping force would be the force of rolling resistance on the wheels plus the friction between the braking pads and the rotor. (Please, correct me if I'm wrong, as I am not entirely sure)
How can we prove that the stopping force is larger when the wheels are not locked?

Apart from that, I read that the wheel-locking appears when the braking force is larger than the asphalt can provide - I am not entirely sure how to interpret this.

 

[russ_watters]

If the wheels weren't locked, the stopping force would be the force of rolling resistance on the wheels plus the friction between the braking pads and the rotor.

Not rolling resistance, static friction. And it isn't "plus the friction between the brake pads and rotor": the rotor isn't touching the ground!

How can we prove that the stopping force is larger when the wheels are not locked?

If kinetic friction were larger than static friction, there would be a step-change up in friction when you start sliding an object along the ground. What would that look like? Would the object repeatedly start and then stop again?

Apart from that, I read that the wheel-locking appears when the braking force is larger than the asphalt can provide - I am not entirely sure how to interpret this.

Can you elaborate on your question? Static friction has a current value and a maximum value. If the current value "tries" to exceed the maximum value, it breaks and the object starts sliding.

 

[ChessEnthusiast]

Not rolling resistance, static friction. And it isn't "plus the friction between the brake pads and rotor": the rotor isn't touching the ground!

Can we say that in this case the friction between the wheels and the ground is equal to the friction between the pads and the rotor?
Besides, if the wheels are still spinning, why doesn't rolling resistance play any role?

 

[russ_watters]

Can we say that in this case the friction between the wheels and the ground is equal to the friction between the pads and the rotor?

No. The rotor is much smaller than the wheels, so the force is much higher. They are related, though, by torque.

Besides, if the wheels are still spinning, why doesn't rolling resistance play any role?

It doesn't play any more or less of a role than when the car isn't braking. It can be a significant fraction of the force needed to keep the car moving (perhaps 10-20%) but a far smaller fraction of the force applied during rapid braking (as little as 1%).

 

[ChessEnthusiast]

@russ_watters
Thank you for your explanations, now I'm really starting to see it clearly.

The last thing that is puzzling me - you mentioned that the force which stops the car is actually the static friction.
Is it the static friction between this part of the tire which is in contact with the ground and the ground itself? It doesn't seem to be the case because the wheels are still rotating.

 

[sophiecentaur]

Besides, if the wheels are still spinning, why doesn't rolling resistance play any role?

It will be 'noticeable' at the highest speeds but, once the vehicle has slowed down, it will be a smaller and smaller proportion of the stopping force - not at all, when the car has actually stopped. But you can get an impression of the effect of rolling resistance by just taking your foot off the accelerator pedal. The car does slow down a bit but it's insignificant in an emergency.

 

[sophiecentaur]

It doesn't seem to be the case because the wheels are still rotating.

The speed of the tyre surface relative to the ground is zero if you are not skidding whilst braking.

 

[ChessEnthusiast]

The speed of the tyre surface relative to the ground is zero if you are not skidding whilst braking.

What about the relative speed of the tyre surface to the ground when you are not applying brakes at all? Does it change?

 

[sophiecentaur]

What about the relative speed of the tyre surface to the ground when you are not applying brakes at all? Does it change?

Very much so. If you are driving, the force is forward acting on the wheels (third law pair with the force that the wheels are pushing backwards). When you are coasting, the force will be small, acting against the motion of the car (rolling resistance) and so will the force from air resistance. When you are braking, the force will be large and acting against the motion of the car.
The rolling resistance is hard to visualise when you are accelerating and it represents work done by the engine that is not transferred to work done on the air resistance and increasing Kinetic Energy. It has to be a force times a velocity and I look upon it as the tyres driving 'uphill' constantly, against the inelastic compression of the tyre, which doesn't return all the energy due to its distortion.

 

[russ_watters]

The last thing that is puzzling me - you mentioned that the force which stops the car is actually the static friction.
Is it the static friction between this part of the tire which is in contact with the ground and the ground itself? It doesn't seem to be the case because the wheels are still rotating.

The only thing touching the car that can apply force to it is the ground touching the tire. And as said, the part of the tire touching the ground is always changing, but is stationary while touching the ground.

 

[sophiecentaur]

Is it the static friction between this part of the tire which is in contact with the ground and the ground itself?

Using the word "static" can get in the way of understanding in a 'real' situation like a motor tyre. The tyre is not rigid and parts of the surface are slipping along the ground whilst most of the footprint is not. Is that static or dynamic? Does it matter what you call it?

 

[A.T.]

Is it the static friction between this part of the tire which is in contact with the ground and the ground itself? It doesn't seem to be the case because the wheels are still rotating.

Static friction means that the contact patches have no relative motion, not that the entire body is static.

 

[ChessEnthusiast]

Okay, so now let's come up with some numbers.
I infer that the maximal thowing force a car is able to experience is
$$F= f_{static}mg$$.
Does it imply that the maximal braking force is $$-F$$?

 

[jbriggs444]

Does it imply that the maximal braking force is $$-F$$?

Just F. There is no need for a minus sign since you have not established a sign convention.

 

[ChessEnthusiast]

Thanks!
My last question is - what is responsible for the change in the braking force once the brakes are applied if the wheels are still spinning? In what way does the rotor rubbing against brake pads increase the force of static friction between the wheels and asphalt?

 

[sophiecentaur]

In what way does the rotor rubbing against brake pads increase the force of static friction between the wheels and asphalt?

Why should it? The (so called) static friction between tyres and road is the maximum braking force available if the brake servo has been designed correctly and the driver has strong enough legs for that design. That is not necessarily the case for older or low performance cars. If the brakes are applied lightly then the braking force on the road is much less.
I think you are ignoring the fact that the part of the tyre that's instantaneously in contact with the road is more or less stationary under nearly all conditions - apart from the leading and trailing 'squish' areas where the tyre is either just approaching the road surface or just leaving it. Those parts of the tyre are flexing so they are not necessarily stationary.

 

[A.T.]

The (so called) static friction between tyres and road is the maximum braking force available...

This is not the naming convention I am familiar with. To me "static friction" means the actual force of static friction, which is limited by normal force time friction coefficient.

 

[ChessEnthusiast]

But when the car is accelerating, the force of static friction is the thowing force and when the vehicle is braking, it's the stopping force - what causes the force to 'change' direction if the wheels are still spinning in the same direction?

 

[A.T.]

...what causes the force to 'change' direction...

What do you think "causes" the change in direction of the normal force on your butt, when you try to get up from a sticky chair?

Constraint forces take whatever value is needed to prevent relative motion and/or penetration at contact (within some bounds). On the microscopic level it's the deformation that changes direction, and is resisted by the material.

 

[sophiecentaur]

This is not the naming convention I am familiar with. To me "static friction" means the actual force of static friction, which is limited by normal force time friction coefficient.

Yes. You are right and I didn't put it the right way. The static part of the friction force against the road can be low and it will increase as the brake pad friction increases to a maximum limit.
The problem - and the reason for so many conversations about this topic - is the appreciation of what 'rolling resistance' is caused by and that is not a static friction phenomenon, despite the observation that the wheels are not slipping. It is due to slippage at the edges and hysteresis in the tyres as they distort etc.. Rolling resistance always acts against the motion of the car, whether the brakes or the accelerator are being applied.

But when the car is accelerating, the force of static friction is the thowing force and when the vehicle is braking, it's the stopping force - what causes the force to 'change' direction if the wheels are still spinning in the same direction?

Friction is a Reactive Force, acting against the applied force. It is misleading to say that friction just slows things down. There is no confusion if that's borne in mind.

 

[A.T.]

It is misleading to say that friction just slows things down.

It opposes or prevents the local relative motion of the contact patches. Whether that slows down or speeds up an object globally, depends on the situation and the reference frame.

 

[sophiecentaur]

It opposes or prevents the local relative motion of the contact patches. Whether that slows down or speeds up an object globally depends on the reference frame.

Exactly. (A translation of that for the uninitiated could be useful, though because it is the uninitiated who have problems with it)

 

[ChessEnthusiast]

I guess I've finally grasped it.

1. A very small area of the tire is in contact with the ground. As the wheel is rotating, the area "tries to move" the asphalt backwards, as a result a N3 force of static friction is generated - the force which is towing the car.
As the car is speeding up, the axle provides more and more torque, but it cannot exceed the torque generated by the force of friction on the wheel, otherwise the tires will lose grip.

2. If the car is coasting, with every revolution of the wheels some energy is lost due to deformations of rubber and asphalt (rolling resistance) which combined with air resistance slowly yet steadily are slowing the car down.

3. When one applies brakes, the rotor experiences a large force of friction due to the brake pads, generating "negative torque" which eventually brings the vehicle to a halt.

Could you, please, point to any flaws in my explanation?

 

[rcgldr]

The speed of the tyre surface relative to the ground is zero if you are not skidding whilst braking.

The rate of rotation of the tire will be a bit less when braking than when coasting at any speed. Under braking, the tire surface speeds up during the transition into the contact patch, then slows slows back down during transition out of the contact patch.

https://en.wikipedia.org/wiki/Slip_ratio

 

[A.T.]

Could you, please, point to any flaws in my explanation?

I would rather suggest you draw a free body diagram with all forces for each case.