# How does a vehicle's brake affect the friction between the vehicle and ground?

When brakes are activated in a vehicle it decelerates. Analysing its motion, there can be only one force that can be responsible for it: the friction between the car and the ground. So does braking affects the friction? Does it increase the coefficient of friction?

• It is complicated because the weight distribution on the wheels changes during braking. Commented Jun 24 at 14:08
• FYI: Coefficient of friction only depends on the material properties of the two surfaces (e.g., rubber and pavement) that are in contact with each other. Frictional force depends on the relative motion of the two surfaces, on their mutual coefficient of friction, and on the normal force that keeps them in contact with each other. Commented Jun 24 at 14:12
• Jon Custer said, "...weight distribution..." Specifically, when the brakes grab the turning wheels, that puts torque on the frame of the car, and the torque wants to rotate the whole vehicle forward, lifting weight off of the back wheels and putting more weight onto the front wheels. That means the front tires and brakes contribute significantly more to the "stopping power" than the rear tires and brakes contribute. Commented Jun 24 at 14:25
• To be technically correct, friction between Brake pads and wheel disk is responsible for stopping a car due to induced negative torque. Unless car is slipping on the ground, in this case stopping is due to wheel-ground kinetic friction, but usually modern day cars are equipped with ABS (Anti-lock braking system) which prevents that. Commented Jun 25 at 8:20
• @AgniusVasilauskas - well, ABS tries to prevent slipping. It isn't always successful at getting you to stop soon. Commented Jun 25 at 12:44

So does braking affects the friction? Does it increase the coefficient of friction?

Braking doesn't affect the coefficient of friction. But it can cause it to change from the coefficient of static friction to the coefficient of kinetic (sliding) friction, if the braking force exceeds the maximum possible static friction force responsible for traction between the wheel(s) and the road. This extends the stopping distance during braking.

Refer to the figure below.

During braking the torque applied to the wheel by the brakes results in a force acting forward on the road. As per Newton's 3rd law the road exerts an equal an opposite force on the wheel which is the force responsible for decelerating the vehicle.

If the braking force does not exceed the maximum possible static friction force between the tire and the road, braking will occur without skidding. The maximum static friction force is

$$f_{max}=\mu_{s}N$$

Where $$\mu_{s}$$ is the coefficient of static friction between the tire and the road and $$N$$ is the normal force acting on the road, which is the fraction of the weight of the vehicle supported by the wheel.

Although the braking force does not affect the coefficient of static friction, it can affect the maximum possible static friction force due the shifting of the vehicle weight among the wheels during braking, which can affect (reduce or increase) the normal force $$N$$ in the equation above.

Hope this helps.

• How does breaking without slipping work? Instant stop? Commented Jun 25 at 10:24
• No. Wheel keeps rotating while braking. Slipping means there is no relative motion between the point of contact of the tire with the road. Commented Jun 25 at 12:41
• Keep in mind, this is a Physics 101 model. 10-20% wheel slip results in maximum braking, so ABS tries to maintain that. It's complicated by the fact the optimal slip varies depending on surface, and state estimation is necessary when all wheels slip, which has caused issues with ABS on gravel, in particular. Commented Jun 25 at 15:44
• @user71659 As you said, this is a Physics 101 model. But that's applicable to this site. The application is in the realm of engineering. Commented Jun 25 at 15:55
• Some emergency brakes for rail vehicles (e.g. trams) do affect the coefficient of friction, by using something else than the wheels as the contact surface. Commented Jun 25 at 22:37

Friction coefficient is a function of the material and the condition (wear, wet,...) of wheel surface and ground. So, while braking without slip, friction coefficient is approximately constant. What changes is the maximum horizontal force exchanged between the wheels and the ground: this force provides pure rolling condition (if no slip occurs) and increases as a result of the internal torque applied by the brake pads on the brake disks, as you can prove solving dynamics of simple multi-body models of a vehicle.

While breaking, there is a weight transfer to the front wheels, increasing the normal reaction between the wheel and the ground, and thus increasing the maximum torque that can be applied to the front brakes without exceeding the maximum friction that provides pure rolling without slip (as a first approximation, while detailed dynamics of wheels is very complex: complex structures, non-linear dissipative medium, contact with ground, possible slip, and stick-slip regime,...)

As an example, bikes have larger brake disks at the front wheel (Ducati MotoGP uses from 320 to 350 $$mm$$ diameter 2-disk configuration at the front, 1-disk 220 $$mm$$ diameter configuration at the rear - front brake for braking, rear brake for stability and control); since drum brakes are cheaper and require less maintenance of disc brakes, while provides less braking force, some non-expensive car and electric vehicle (even not so cheap sometimes, energy recovery systems reduces the effort of the braking system) have drum brakes at the rear.

## Simple multi-body model

Anyway, I'm providing a simple mechanical model to study the 2-dimensional dynamics of a braking vehicle, built as a composition of three rigid bodies: two sets of wheels, front and read, and the chassis. These kind of models are usually presented in first courses about vehicle dynamics. Brake callipers are in red (go Brembo!:), and provide internal torques: breaking torques are shown in figure 2, in red, $$C_1$$ the front brake torque and $$C_2$$ the rear brake torque.

It's possible to study the system using a D'Alembert approach on each single component, after having introduced the constraint forces and torques (the one provided by the brakes), and writing the dynamical equations for translations (2 equations $$x$$, $$y$$ directions) and rotation for each body. I'm assuming no revolution friction and pure rolling here (to be check at the end, that $$F^{tang}_k < |F^{tang}_{k,max}| = \mu N_k$$) so that the system has only 1 degree of freedom.

I'm attaching a handwritten notes providing all the computations and presenting here only the qualitative results, assuming braking only at front wheels (limit condition).

• I can't remember the guy's name, but there is a current star of Moto-GP that seems to spend most of his braking time with the back wheel completely floating above the ground while he balances the entire bike on the front wheel. This demonstrates how little the back wheel is needed; i.e. zero. Commented Jun 25 at 7:16
• The size and style of brake have very little to do with stopping the car. As long as the brake has the ability stop the wheel from spinning and lock it up, or engage ABS, it doesn't matter if you have 320mm brakes or 32mm brakes, the braking force is the same. The brakes total brake force capacity might change, but the limit is set by the wheel's contact to the ground, not the brake, so the brake force the brake is capable of providing wont change stopping distance above a certain threshold. Commented Jun 25 at 17:04
• @DavidS "The size and style of brake have very little to do with stopping the car." except that torque is force times radius (the more distant the pads, the longer the leverage): same braking force, higher torque for larger discs. Otherwise it may be a coincidence that everyone has larger discs at the front, or all the people working on cars and motorbikes are stupid or you know things they don't. "the limit is set by the wheel's contact to the ground" false too. You can flip your motorbike if you brake too hard with front brake. The limit is weight transfer Commented Jun 25 at 17:31
• @basics When talking braking force, the force applied to the brakes, it has limits to how much it can take. Once the wheel stops spinning, there is no benefit, gain, or work done by any more brake force applied. This means that the size of the brakes is inconsequential as long as the brake has the ability to hit the threshold of locking the tires and beginning a slide. That is the limit to the braking force. Any more force beyond stopping the tire spin is wasted. Any added limitations, like flipping, only reduce the maximum brake force needed. Commented Jun 25 at 20:04
• @basics Larger discs aren't necessary for adding brake force. The larger discs add other performance characteristics that are very important, but not brake force. Things like heat dissipation, longevity, better performance under repeated high stress, etc. That lets you stop the car more reliably and more frequently. Once your brakes have the power to stop the wheel, the only thing left to stop the vehicle is the friction between the wheel and ground. Flipping the vehicle only changes the wheel for something less desirable in contact with the ground. Commented Jun 25 at 20:40

Initially when the vehicle is moving with a constant velocity, there is no relative velocity between the point of contact of tire and the ground. When there is no relative velocity between the tire and ground, the only friction in effect is rolling friction (close to zero in ideal cases). What braking does, is it induces a shear stress at the point of contact of the tire on the ground, introducing static friction to the equation. Coefficient of static friction is much greater in magnitude than that of rolling friction, and that helps us stop the vehicle faster.

If you think in terms of energy, braking converts the kinetic energy of the moving vehicle to thermal energy, heating up the brake pads and tires. Note that this dissipation is always there, but it happens at a faster rate when you apply the brakes.

It is true that both brake and wheel-ground systems work based on friction. However, these frictions are not the same. Let us look at them separately.

Brake system. In the brake system, friction is between the brake pad and the wheel. Brake pad is connected to the vehicle body and does not rotate, whereas the wheel rotates. Brake friction thus resists against rotation of the wheel. Note that the interaction between fixed brake pad and rotating wheel is kinetic (or dynamic) friction. In this system, friction causes the vehicle to stop.

Wheel-ground system. In the wheel-ground system, friction is between the wheel and ground. A torque $$T$$ is transferred by the drive train to the wheel. At the contact of wheel and ground, this torque creates a force $$F=\frac{T}{r}$$, where r is the radius of the wheel. With this force, the wheel pushes the ground away. In response to this force, the ground pushes the wheel back with friction force $$-F_f$$. I use negative sign to show that friction acts in the opposite direction of $$F$$. This friction force tends to stop the wheel from sliding on the ground. In other words, it momentarily keeps the ground contact point of the wheel fixed. Subsequently, a torque equal to $$-F_fr$$ is exerted from the ground to the wheel, which causes the center of the wheel to rotate about the ground contact point of the wheel momentarily. This momentary rotation becomes the forward translation of the vehicle with respect to the ground. Note that the interaction between fixed lower point of wheel and (fixed) ground is static friction. In this system, friction causes the vehicle to move.

Note that static friction $$F_f$$ follows Coulomb's law, that is: $$F_f \le \mu N$$, where $$\mu$$ is coefficient of friction and $$N$$ is the normal force between the two surfaces. In the wheel-ground system, $$\mu$$ depends on the material of the wheel and the ground; it does not change. $$N$$ is the weight of the vehicle; it also does not change. Therefore, the maximum friction force that the wheel-ground system can produce is a constant number than does not change with the speed of the vehicle.

When the vehicle is at rest on a rough horizontal surface, there is no torque on vehicle wheels, no horizontal force from wheels pushing the ground, and no friction force from the ground pushing the wheels back. At this time $$F_f=0$$. When the vehicle is moving on the same surface, $$F_f > 0$$. Indeed, as the vehicle speeds up, the wheel-ground friction increases; and as it slows down, the wheel-ground friction decreases. Note that these changes do not violate Coulomb's law, as $$\mu N$$ is the maximum possible value of the friction force of the wheel-ground system. $$F_f$$, which is the ground's reaction to the horizontal force exerted by the vehicle to the ground, varies between $$0$$ and this maximum value.

• Typo: it's BRAKE not BREAK Commented Jun 27 at 9:17
• @MikeB Oh! My bad, thank you Commented Jun 27 at 13:23

So does braking affects the friction? Does it increase the coefficient of friction?

The short answer to the first question is yes. The static friction increases. This is not the same as saying the coefficient of static friction changes. The coefficient of static friction is effectively the maximum limiting static friction, but the actual retarding friction force varies proportionally to the braking force. Some sources give the impression static friction force is constant, but that is only true for sliding friction. Static friction is variable and this is what changes when the brakes are applied as long as the limiting friction is not exceeded.

If you apply a gently increasing horizontal force to a block on a surface, initially it will not move and the static friction will adjust to match whatever force is applied up to a limiting friction and then it will start moving and now the friction becomes sliding friction which is generally less than the static friction. The important thing to note, is that static friction responds to the force trying to accelerate the object and is not constant like sliding friction.

For a rolling wheel, it is static friction that is relevant, because the contact patch of the tyre is at rest with the road at any instant. When the brakes are applied, the static friction increases up to the limiting friction. If even more braking force is applied to the brake disks, the wheel locks up and the contact patch starts sliding relative to the road. Since the sliding friction is less than the limiting friction, locking the wheels is not an efficient way to stop quickly and hence the development of anti lock brake systems.

The above diagram is from the Wikipedia entry on friction. It is the diagram for an increasing accelerating force applied to a block but it also applicable for an increasing decelerating braking force on a wheel and I have added the green line that represents rolling friction that is present before the brakes are applied. As the brakes are applied and gently increased, the static friction between the tyre and the road surface also increases. It increases up to the maximum limiting friction, at which point the tyre starts sliding relative to the road and now the friction becomes sliding/dynamic/kinetic friction which are just different names for the same thing. As can be seen from the graph, the sliding friction is less than the maximum static friction and unlike static friction, sliding friction is more or less constant. At a microscopic level, the road surface has lots of tiny bumps and the tyre can skip from location to location. This is why the curve for sliding friction wiggles up and down and is not an exact straight line. As an aside the above chart can be used to illustrate the opposite case of accelerating a vehicle. As the engine power is increased, the static friction between the tyre and the road increases to match up to the limiting friction. At that point the wheel spin occurs and the sliding friction is not as great as the maximum static friction. For this reason a drag racer tries to keep the power output just below the maximum static friction to maximise acceleration.

There is a transfer of weight towards the front of the vehicle during braking and this increase in the normal force increases the maximum limit of the static friction at the expense of reduced braking ability at the rear wheel/s. The disk brakes at the front of a vehicle are generally larger than at the rear, because a greater braking torque can be applied at the front before the wheels lock up and because greater surface area of the disks gives better dispersal of the heat created during braking.

When a vehicle is slowed down, it's kinetic energy is reduced and this energy has to go somewhere. When the brakes are initially applied the brake pads are sliding relative to the disk and a lot of heat is created at the pad/disk interface. The heat energy created is equal to the loss of kinetic energy of the vehicle. If the brakes lock up, heat ceases to be generated in the brake disk and now the heat is created at the tyre/road interface and the this result in rapid wear of the tyre, because now the tyre contact patch is sliding relative to the road.

What actually happens when you gently and progressively press on the brake pedal? The braking force of modern disc brake is a function of sliding friction given by $$F =\mu_s F_N$$ where $$\mu_s$$ is the coefficient of sliding friction and $$F_N$$ is the normal force between the brake pad and the disc. As the brake pedal is depressed, the hydraulic pressure in the brake system increases which increases the the normal force of the brake pad and in turn this increases the sliding friction of the disc pad in a proportional manner. The static friction between the tyre and the road increases correspondingly up to the static fraction limit.

Kinetic friction is when one object slides over another - in this case, it's either the tire skidding on the road or the brake calipers gently gripping but not stopping the rotor (braking surface). Kinetic friction coefficients multiply the normal force to describe the size of the friction force. Kinetic friction is a property of the two objects and while tires go bald and roads get slick, if you hold the objects constant, the kinetic friction remains constant. Static friction is when two objects do not slide - in the case of a wheel, that means either that the tire is rolling and maintaining contact with the road or that the brake calipers have completely clamped down on the rotor and stopped the wheel from turning. Static friction coefficients multiply the normal force to describe a maximum force before the objects begin to slide. In ordinary life, objects operating under static friction experience less force than this maximum amount most of the time. During kinetic friction, the objects sliding over each other will heat up, unlike during static friction.

If the caliper lightly grips the rotor (under kinetic friction), then the rotor gently slows the hub which slows the tire which slows the vehicle and the tires and the ground likely maintain static friction. The caliper and the rotors will heat up.

In the other case, the caliper tightly grips the rotor (under static friction) and the hub stops which stops the wheel which stops the tire while the vehicle slides over the road in kinetic friction. In this case, the meaningful heating will occur on the tire (while the larger area of the affected road will disperse the heating of the road).

Today, the more realistic case is that the anti-lock brakes use a fast-acting computer to push back on the brake pedal and release the brakes enough that they can keep rolling most of the time. Which means you have a combination of the two scenarios listed.

While answering another question, I found a neat way to think about this problem. What you are essentially asking, is how does the force applied by the brakes reach the ground. I think that root question is yet unanswered in this discussion so far.

Wheels bring the ground to you, for you to grip it better. - Me, probably

## Think of it as gears

Wheels act like gears between the road and the braking pad, transferring the stopping force of the braking pads to the road. Assuming no slip condition, the wheels act like a gear between one Big gear and a small gear.

The Big gear being the earth, and the small gear being the shaft/axle/ whatever connects the engine to the wheel. You don't really need wheels on a car, you could in theory just run it on the axles. Without wheels, for the same engine rev speed the rate at which you could rotate the earth is small. But with wheels, you could rotate the earth (circumference of wheel/circumference of axle) number of times faster.

So if you look at your braking process like a system of gears, it doesn't matter if you apply brakes on the Big gear or the small gear; its all a connected system. You could probably apply brake on the Big gear; (using some spikes and dragging it on the road) or you could apply brakes on the wheel (using braking pads). The point is, no matter where you apply the braking force, the torque gets transferred to the entire system, bringing everything to a halt.

The answer to your question is, the stopping force is provided by the braking pad. You could have pressed the braking pads directly to the ground as well, but that would have been so efficient. Wheels are anyways connected to the ground, so pressing against it is also good enough.

To say that the friction between wheel and road causes the vehicle to stop sounds wrong to me. Friction is important for the gears to function properly, but the reason why the gears stopped in the first place is the torque due to the braking pads.