Do the brakes provide friction to tyres of bicycle/car?

[LogU16]

Hello, members.

I know that the cars and bicycles move on the road simply due to friction. How do the brakes of bicycle and car work to stop them?
Do the brakes provide friction to tyres of bicycle/car to stop them from moving?

I'm confused, could you please explain to me why do the brakes work and what's the physics behind it?

Your help would really be appreciated.

Thanks in advance.

 

[Snapp'tappin27]

Brakes work on the inside of the wheel of a car. The brake pads are pushed against the wheel to slow it down. You can look up diagrams online or maybe even a youtube video.
Brakes on a bike are rubber-like pads that push against the wheel from either side onto the wheel.

So yes, friction is a big part of how brakes work.

 

[DrClaude]

The question is surprising. Haven't you ever ridden a bicycle, and thought about what happens when you brake?

A tire needs friction to grip the road, but also needs to roll to keep going, that's the basic principle of a wheel. Brakes rub against the wheel to hinder its rotation, therefore slowing down a vehicle. When fully applied, a brake should completely stop a wheel turning, which then essentially doesn't act as a wheel anymore.

 

[LogU16]

Brakes work on the inside of the wheel of a car. The brake pads are pushed against the wheel to slow it down. You can look up diagrams online or maybe even a youtube video.
Brakes on a bike are rubber-like pads that push against the wheel from either side onto the wheel.

So yes, friction is a big part of how brakes work.

Thank you snap. Does the road (on which bicycle is moving) provide friction to the wheels of bicycle to cause it to stop or bicycle pads do this? It is the friction by which we can move a bicycle or any other vehicle on the road, why do we need another friction to stop it from moving and why can we not stop a bicycle by the help of the friction we used to move it on the road? Kindly help me clarify this, it sounds a bit confusing to me. Thanks.

 

[jack action]

To stop a car/bicycle, you use the friction between the tire and the road.

The necessary wheel torque to create that friction can be done anyway you can think of. The usual way is to create that torque by friction between pads and a disc or drum. But, if you wish, you could use an electric or hydraulic motor to create that torque (for example, in a regenerative brake system).

 

[Snapp'tappin27]

Thank you snap. Does the road (on which bicycle is moving) provide friction to the wheels of bicycle to cause it to stop or bicycle pads do this? It is the friction by which we can move a bicycle or any other vehicle on the road, why do we need another friction to stop it from moving and why can we not stop a bicycle by the help of the friction we used to move it on the road? Kindly help me clarify this, it sounds a bit confusing to me. Thanks.

Obviously, the friction between the tire and the road contribute to the stop. The bike would just slide into infinity because no other force is acting on it. But the brakes don't add friction to the tires themselves. They make the wheel slow down, (and the car come to a stop) while friction between the tires and the pavement keep the car from sliding.

You can use the concept of driving on ice to help you picture this if it helps. The brakes use friction to stop the wheels, but if a car is on ice, the car won't stop because there is no friction between the tires and the ground.

As simply as I can think to put it, the brake pads remove the cars velocity, but only because friction is an opposite force on the vehicle, making it stop. Without both friction (to actually stop the car) and the brakes (to stop the wheels which resist friction otherwise), vehicles couldn't stop.

I don't know if this is actually helping or if I'm just "spinning my wheels", but I hope that this at least helps give you some concept.

 

[davenn]

But the brakes don't add friction to the tires themselves. They make the wheel slow down,

true not directly to the tyre ... for a motor vehicle/motorbike they supply friction to the brake drum which is connected to the axle which is connected to the tyre
that friction directly slows/stops the wheel moving

for a pushbike, the brake pad applies friction to the metal wheel rim that the tyre is mounted on
again the direct application of pressure to the wheel rim slows/stops the movement

The pressure of the brakes isn't supplied directly to the tyre as it would result in too much wear and tear on the rubber tyreDave

 

[Ranger Mike]

Do the brakes provide friction to tyres of bicycle/car?
Yes. This is a good point to discuss how and why.
All of the above posts are very good by the way!

Friction is the Force that resists movement of a sliding object. Once the object ( in this case a 3000 pound automobile) has been accelerated to a constant velocity, it has kinetic energy.
The object can be stopped by removing the energy source that maintains the constant velocity ( turn off the engine). Once you turn off the power the vehicle will immediately begin to decelerate from that constant velocity to zero velocity. The vehicle will eventually stop because of the large amount friction caused by of moving through a curtain of air that weighs 14.7 pounds per square inch. at sea level. There is also friction in the form of rolling resistance of the tires to asphalt contact. There are some other nit-noid things causing the vehicle to stop like rotating friction of suspension parts, parasitic drag and the like but you get the idea.

If you wanted to stop the vehicle faster ( in a shorter amount of time) you could direct it to contact a non moveable object (like a large tree). The kinetic energy would be converted to potential energy quite rapidly but you could only do this one time and this is not the recommended solution.

We could install a brake system that would reduce kinetic energy when the driver actuated it. We could construct a closed brake system that immediately locked all 4 wheels and prevent them from rotating. This would cause the tires to slide on the asphalt and scrub off the kinetic energy. See attached photos. This has several outcomes. The good one is that the car will stop in the shortest amount of time. The not good outcomes are the tires have 4 flat spots on them. You will have out of balance tires and non pleasant ride. There is limited control of the vehicle while sliding and you may reach that non moveable object as consequence. Finally , anyone following behind you in another 3000 pound vehicle will not be happy with your decision to stop.

The recommended solution - We could install a brake system that would reduce kinetic energy when the driver actuated it IN A CONTROLLED MANNER. This is the current automobile brake system. Driver actuated hydraulic brake system that applied friction to the wheels rotation in a linear manner. The farther the brake pedal is depressed the more pressure is applied to the brake pad/ shoe and thus more friction. You are converting kinetic energy to heat energy and velocity is reduced. Apply too much pressure and lock the brakes.

Follow the force vectors. Automobile is at speed, Drivers foot applies pressure to hydraulic master cylinder that is connected to wheel cylinders at each wheel. Each wheel cylinder applies brake shoe/pad to rotating brake drum/rotor that is mounted to the rotating wheel. Wheel with attached tire contacts asphalt and we have controlled friction.

More automobile races are won by out braking the other driver versus driving around the other guy!

 

[jack action]

We could construct a closed brake system that immediately locked all 4 wheels and prevent them from rotating. This would cause the tires to slide on the asphalt and scrub off the kinetic energy. See attached photos. This has several outcomes. The good one is that the car will stop in the shortest amount of time.

Not true. To stop in the shortest amount of time, you need to put just enough force such that the tire begins sliding, but still rolling. It's by maintaining that very critical point, each time they brake, that good drivers win races.

The following shows that you need about 20% slip to get the highest friction coefficient:

 

[Ranger Mike]

Jack..as always...you are correct...i was a little over zealous on that point. I get that when i ramble...I was still thinking in the old school mind set before ABS. You point is well stated and thank you..

 

[rcgldr]

The following shows that you need about 20% slip to get the highest friction coefficient ...

Much of the slip percentage is related to longitudinal tire deformation (which coexists with longitudinal tire force), which is related to tire stiffness, so this will vary depending on the tire.

 

[Randy Beikmann]

To clarify why several friction forces are involved in a "normal stop" (tires still rolling), note that the whole idea of braking is to dissipate kinetic energy, converting it to heat. The sliding (kinetic) friction in the brakes does this, but the friction between the tire and road (mainly static friction) does not. That's why you could roll for a long time with the brakes off - no kinetic energy dissipation! I hope that helps.

 

[OmCheeto]

Not true. To stop in the shortest amount of time, you need to put just enough force such that the tire begins sliding, but still rolling. It's by maintaining that very critical point, each time they brake, that good drivers win races.

The following shows that you need about 20% slip to get the highest friction coefficient:

I must be getting old.
This graph makes absolutely no sense to me.
Although it makes sense as an algorithm for an ABS system, it doesn't make sense from a first term physics problem point of view, as to how to most quickly stop a vehicle.

To clarify why several friction forces are involved in a "normal stop" (tires still rolling), note that the whole idea of braking is to dissipate kinetic energy, converting it to heat. The sliding (kinetic) friction in the brakes does this, but the friction between the tire and road (mainly static friction) does not. That's why you could roll for a long time with the brakes off - no kinetic energy dissipation! I hope that helps.

You seem to know what you're talking about. Isn't static friction always greater than kinetic friction? That's what I remember from physics class. (It's been 30 years, btw.)

I have to admit, that I've never heard of this "% Slip" concept before, and it strikes me as more a "skip" function.
At 0% slip, the tires will be in 100% static friction mode.
At 100% slip, the tires will be in 100% kinetic friction mode.
In between, the tires, as far as I can comprehend, will be in some hybrid, hopscotchey mode.

To minimize stopping distance, one would have to be at 0% slip, IMHO.

 

[cjl]

You're correct about what the percent slip means (though it doesn't mean the tire is skipping - the rubber is deforming, allowing the wheel speed to be slower than the road speed. As for minimizing stopping distance, you do want some slip, since the simplified model of static and kinetic friction (with static higher than kinetic) is not entirely correct when applied to real tires. As that graph shows, in reality, you want about 20% slip on high-grip surfaces to maximize deceleration, and you want a bit less on slippery surfaces (you can see you want very little slip on ice, and maybe 15% or so on a wet surface). Having a bit of deformation in the tire actually aids grip.

 

[OmCheeto]

You're correct about what the percent slip means (though it doesn't mean the tire is skipping - the rubber is deforming, allowing the wheel speed to be slower than the road speed. As for minimizing stopping distance, you do want some slip, since the simplified model of static and kinetic friction (with static higher than kinetic) is not entirely correct when applied to real tires. As that graph shows, in reality, you want about 20% slip on high-grip surfaces to maximize deceleration, and you want a bit less on slippery surfaces (you can see you want very little slip on ice, and maybe 15% or so on a wet surface). Having a bit of deformation in the tire actually aids grip.

I'm still not getting it.

"allowing the wheel speed to be slower than the road speed"​

In my tiny little brain, this could happen for maybe a fraction of a second.

 

[Randy Beikmann]

OmCheeto, the explanation cjl gives is correct in relating the tire's traction to the slip. Note that "slip" doesn't necessarily mean "tread sliding." When you accelerate away from a stoplight, the carcass of the tire twists, stretching behind the tread, and "bunching up" in front of the tread. The tread blocks bend too. To make a long story short, the wheel center doesn't go as far as you'd calculate from its rotation and the tire radius: x < theta*radius, because of this. Overall, the tire must deform during acceleration, so at first the more acceleration and traction force, the more the carcass twists, and the more slip - still with largely static friction between tire and road. But when you "overdo it" and the slip gets too high, the tread slides a lot, and the friction (traction) drops, being mostly kinetic at that point.
When you brake, the same thing happens, except the slip is so that the wheel center moves faster than the rotation would indicate.
There are several good references for this phenomenon: "Theory of Ground Vehicles" by Wong, "Fundamentals of Vehicle Dynamics" by Gillespie, "Race Car Vehicle Dynamics" by Milliken, and "Physics for Gearheads" by myself. Gillespie's and mine are more basic, while I would rate Wong's and Milliken's as more advanced.
In any case, pictures of this are worth a thousand words!

 

[OmCheeto]

OmCheeto, the explanation cjl gives is correct in relating the tire's traction to the slip. Note that "slip" doesn't necessarily mean "tread sliding." When you accelerate away from a stoplight, the carcass of the tire twists, stretching behind the tread, and "bunching up" in front of the tread. The tread blocks bend too. To make a long story short, the wheel center doesn't go as far as you'd calculate from its rotation and the tire radius: x < theta*radius, because of this. Overall, the tire must deform during acceleration, so at first the more acceleration and traction force, the more the carcass twists, and the more slip - still with largely static friction between tire and road. But when you "overdo it" and the slip gets too high, the tread slides a lot, and the friction (traction) drops, being mostly kinetic at that point.
When you brake, the same thing happens, except the slip is so that the wheel center moves faster than the rotation would indicate.
There are several good references for this phenomenon: "Theory of Ground Vehicles" by Wong, "Fundamentals of Vehicle Dynamics" by Gillespie, "Race Car Vehicle Dynamics" by Milliken, and "Physics for Gearheads" by myself. Gillespie's and mine are more basic, while I would rate Wong's and Milliken's as more advanced.
In any case, pictures of this are worth a thousand words!

Thanks! I'll have to study this further.
I think my confusion stems from the wording.
(study, study study)
Ah ha!
Plastic deformation?

Slip (materials science)
In materials science, a slip system describes the set of symmetrically identical slip planes and associated family of slip directions for which dislocation motion can easily occur and lead to plastic deformation.

So, am I interpreting this correctly with the following image:

The slip here being defined as the angular difference between the contact surface and axis of the relaxed tire, versus the tire under acceleration? (green and or blue sections)

 

[jack action]

It is more of a static friction vs kinetic friction kind of thing. Friction forces are at maximum when there is no relative movement between two parts. When there is a relative motion, kinetic friction comes into play and the friction force is lower. The rubber-road case do not obey pure Coulomb friction theory because of the rubber deformation, hence why the peak friction coefficient happens when there is a small relative motion.

Suggested reading:

• Rubber Friction;
• http://www.airporttech.tc.faa.gov/safety/downloads/Braking_Performance_of_Aircraft_Tires.pdf .

 

[Randy Beikmann]

OmCheeto, unfortunately the "slip" in the reference you found is a difference type. Dislocations like it describes are a plastic deformation, but plastic deformations are permanent. While the mechanics of rubber deformation isn't elastic, it is viscoelastic, not plastic.
But I'll reiterate that carcass deflection (the fabric) of the tire is just as important as the rubber deflection.

 

[Ranger Mike]

not wanting to continue beating this dead horse, i recommend you look at the following link, explains tire action under braking better than this old racer canhttp://www.insideracingtechnology.com/tirebkexerpt2.htm

 

[OmCheeto]

OmCheeto, unfortunately the "slip" in the reference you found is a difference type. Dislocations like it describes are a plastic deformation, but plastic deformations are permanent. While the mechanics of rubber deformation isn't elastic, it is viscoelastic, not plastic.
But I'll reiterate that carcass deflection (the fabric) of the tire is just as important as the rubber deflection.

You know too many big words.
And now I'm feeling like a mansplainer*.

not wanting to continue beating this dead horse, i recommend you look at the following link, explains tire action under braking better than this old racer can

http://www.insideracingtechnology.com/tirebkexerpt2.htm

Good grief!
Who would have guessed that mere friction could be so complicated.-----------
*Mansplainer: Someone who talks like he knows what he's talking about, but really has no clue.

I'm trained as an electrician, and one day, I was helping my friend try and get his solar powered water pump working.
A mansplainer showed up to help.

Mansplainer; "Om, it's all about the amps".
Om; "What? Um, I think it's more complicated than that".
Mansplainer; "You see, blah blah blah... ad absurdum...".
Om; "Thanks! That makes total sense now".