# Why do trucks have bigger brakes?

1. May 26, 2012

### phyzzle

I am a little confused about why a passenger truck like a Chevy 2500 or a Ford F250 requires brake rotors with larger diameters than a car like a Toyota. I have a degree in Physics, and although I have not used it in 10 years I specifically remember learning that mass does not effect braking distance. The reason is as follows. A car stops due to frictional force between road and tire. When comparing stopping distance between two cars any difference in mass can be ignored because a greater momentum in a car with greater mass will also have a greater frictional force by the same magnitude, since friction equals normal force times co-efficiant of friction (kinetic or otherwise). For example if you have 2 Toyota Camrys and one is made out of aluminum and the other iron and they are allowed to coast to a stop from an initial velocity of 60mph they will come to rest at the same time. I understand that mass distribution plays a role in 18 wheelers, but assuming mass is distributed uniformally why does a truck need bigger brake rotors???

Paul Tippler's College Physics Text shows mathematically that mass cancels when calculating stopping distance.

Thanks!

2. May 26, 2012

### fzero

The coefficient of friction is proportional to the common area of contact, in this case between the brake rotor and pad. To increase the frictional force generated by the braking system, we have some options. We can increase the normal force applied to the brake rotor, by using a more powerful hydraulic system. We can use special materials in the brake pad that have a higher amount of friction with the rotor. Or we can increase the area by using a larger rotor, which then allows you to use a larger brake pad. The solution used is the most cost-effective combination of all three options.

Note that the normal force applied to the braking surface is generated by the hydraulics, not by the mass of the vehicle. We're not just coasting to a stop, we are applying an additional force with the brakes in order to stop more quickly. This additional force that must be added to the one that you're talking about. When both frictional forces are included, the mass will not drop out.

3. May 26, 2012

### Staff: Mentor

If you go back in your textbook to the calculation of stopping distance it is probably considering sliding friction. Does that apply to a car braking?

4. May 26, 2012

### phyzzle

You are mistaken, coefficient of friction has nothing to do with surface area. Further more suface area of contact points does not increase or decrease friction. Greater surface area simply distributes force but does not increase it. A larger rotor can be used to increase torque by allowing the brake pad to be placed further from the axis of rotation, but it does not increase friction by merit of its size.

The frictional force between the rotor and the pad does not stop the vehicle. If you think it does imagine stopping on ice, or oil, or a fictional surface with no friction. As long as the wheel is rotating on the road the stopping force is static friction between road and tire. Nothing else. Friction between rotor and pad serves to increase friction between road and tire. Anti lock brakes ensure that the friction stays static. My example of coasting to a stop was to illustrate that the greater momentum of the heavier car is cancelled by its greater friction of the same magnitude. So if they are going to coast to a stop in the same distance and time, then they should require the same amount of braking force to produce equally shorter stopping distances. In the calculations I had to perform in college the mass always cancelled.

Last edited: May 26, 2012
5. May 26, 2012

### phyzzle

I did a quick google search and found the example from my text book. It actually talks about both static and kinetic (sliding) friction. He shows mathematicaly that mass cancels.

Since I am a nooby it will not allow me to post a link. But if you google search:
'paul tipler stopping distance' it is the second result.

6. May 26, 2012

### Staff: Mentor

Also, when a car stops, that kinetic energy has to go somewhere. Where do you think it goes?

7. May 26, 2012

### Pkruse

There are a lot of things happening here that complicate the design engineers have to consider that are beyond the scope of the simplified analysis you saw in the text book.

Actual forces required to stop the truck are higher. Wheels have a larger radius. Both of these mean that you need more braking torque from the rotor. The rotor will have to be larger to be strong enough to strong enough to take the loads.

Additionally, more heat will be generated, so you need a bigger heat sink to dump it into.

Since the coefficient of friction is the same on the pads, as you say. So the only way to get more braking torque is to put the pads at a larger radius. Again, the obvious answer is to make the rotor bigger.

For more force, you need bigger hydraulic cylinders. Making the pads bigger is a simple solution to making them fit the bigger cylinder. Again, this also results in a larger rotor.

Keeping the brake pad temperatures into acceptable limits for the material works better when they and the rotor are bigger.

In the end, the design engineer has dozens of concerns to trade off, the preferred solution to each is to make the whole system physically larger for a heavier vehicle.

8. May 26, 2012

### xxChrisxx

9. May 26, 2012

### 256bits

That would be more true in a vacuum environment.
You cannot neglect the air fricton.
The lighter vehicle coasts to a stop more quickly in that case.
Ergo a heavier vehicle needs a more robust breaking system to absorb the kinetic energy of a moving vehicle.

10. May 26, 2012

### D H

Staff Emeritus
Of course it does. If not, why do brakes exist?

The coefficient of static friction between road and tire is a limiting factor of the effectiveness of the brakes. See below.

Tire inflation, tire quality, tire makeup, the nature the road surface, and a slew of other factors change the coefficient of static friction. Applying the brakes does not change a single one of these factors. There is no change in the coefficient of static friction that results from applying the brakes.

It is the brakes rather than tire and road that do the work of slowing the vehicle. The coefficient of static friction between tire and road limits the amount of work that can be done by the brakes. Brake too hard and the tires will lock. When this happens it is the tires rather than the brakes that are doing the work, but this is a very undesirable situation. Braking distance is considerable longer when the tires are locked.

Assuming the vehicle doesn't have a regenerative braking system, the brakes perform work by converting kinetic energy to heat. More work is needed to stop a larger vehicle moving at the same speed as a smaller one. More work means more heating. Larger vehicles need larger brakes to better dissipate the generated heat.

11. May 26, 2012

### Hurkyl

Staff Emeritus
On this particular point, he's technically right. The frictional force between the road and tires is the force that decelerates the car. The frictional force between the rotor and pad just ensures that there will be friction between the road and tires.

12. May 26, 2012

### mrspeedybob

A larger vehicle has more kinetic energy. If that energy were converted to heat in a small rotor the temperature of the rotor would rise to a level that would damage the metallurgy.

Getting sufficient braking torque from a small rotor to stop a large vehicle would mean higher clamping forces. Higher clamping force would require either larger caliper pistons or higher hydraulic pressure. Larger pistons would be inconvenient to design around a small rotor and higher hydraulic pressure would require more robust lines, fittings, master cylinders, and brake boosters.

Attempting to stop a large vehicle with small brakes would cause the pads and rotors to wear out quickly. Do you want to pay for a brake job with every oil change?

I'm sure there are more reasons.

13. May 26, 2012

### Staff: Mentor

Oh, you are right. This is a very misleading question IMO, you are justified in your confusion.

The formula for static friction is an inequality: f≤μsfN. So you cannot use it to calculate a single stopping distance, but rather a range of stopping distances. There is not one single stopping distance implied by that formula, and that fact is critical for being able to bring a car to a controlled stop.

Since the static friction formula gives a range of forces, then what is it that determines a car's stopping distance in a controlled stop? Well, it should be obvious that it is the brakes. The brakes work by applying friction between two pieces of material, so the friction formula should apply to the brakes as well. If we think about it we realize that in a controlled stop the different parts of the brake are sliding past each other, so the correct formula would be the coefficient of dynamic friction: f=μdfN

So, the question is, what is fN in that equation? If it were the weight of the car then it would be constant and so we would always be braking at a constant force, i.e. in an uncontrolled manner. Since that is not the case, we know immediately that the normal force of interest is not the weight of the car.

Instead, the fN which is of interest is the force exerted by the hydraulic mechanism of the brakes. This force can be varied by changing the pressure in the brake fluid, thus allowing for a varying braking force, and permitting controlled braking.

I hope that helps.

Last edited: May 26, 2012
14. May 26, 2012

### Staff: Mentor

By this same line of reasoning the frictional force between the tire and the road does not stop a vehicle. If you think it does then imagine stopping without any friction between the rotor and pads. (I am sure you have seen many TV shows or movies that dealt with exactly that situation )

Both frictional forces are required for controlled braking.

15. May 26, 2012

### Q_Goest

Hi phyzzle, welcome to the board.

You're asking 2 different things here. On one hand you're talking about the coefficient of friction between tires and road and stopping distance, and then you're talking about larger brakes on larger vehicles. The two considerations are mutually exclusive.
So the questions, why should a truck have larger brake rotors than the small car, is not a question about stopping distance. mrspeedybob has it essentially correct.
During braking, the kinetic energy of the vehicle is converted to heat. The amount of kinetic energy that has to be converted to heat is proportional to the mass at any given velocity. Double the mass, you double the kinetic energy that has to be converted. To do that, you need more force on the rotor which is generally done by making the pistons in the calipers larger. With more heat, you also need to cool the brakes better and have more thermal mass in the brake to absorb that heat energy. Larger vehicles need larger brakes to 1) create the larger forces 2) absorb the energy as heat 3) reject that heat to the atmosphere.

16. May 26, 2012

### phyzzle

Thank you everybody for the responses! I will take time to read through them all and see if it makes sense.

17. May 26, 2012

### phyzzle

Thanks that makes sense. I hadn't considered the radius of the tires. The increased rotaional inertia of the truck tires would require a greater torque. I guess I did not state it in my OP but I am aware that the rotors are bigger to dissipate heat. I got that. My confusion lies in the fact that if the force required to stop two vehicles of different mass were the same then the heat created would be the same as well. I am under the impression (mistakenly obviously) that the force required would be the same so I did not mention heat.

To be clear I understand that if a vehicle of greater mass requires more force to stop, then said vehicle will need larger rotors. What I do not understand is why does the heavier vehicle require more force to stop?! - I guess that should have been my original question.

Thanks again!

18. May 26, 2012

### phyzzle

Very true, are you implying that the reason for larger rotors comes down to air drag?

19. May 27, 2012

### truesearch

The brakes do not stop the car!!
They stop the wheels rotating
Friction between tyres and road (+ air resistance) bring the car to a halt.
Someone mentioned being on ice in an earlier post to bring this point home.

20. May 27, 2012

### D H

Staff Emeritus
That is akin to saying that the engine doesn't accelerate the car. That start the wheels rotating. It is friction between tires and road bring the car up to speed.

Which is nonsense. It's putting the cart before the horse.

That friction between tire and road is of course a necessary ingredient in making the car speed up or slow down, but it is the engine and the brakes that are ultimately responsible for making the car change speed.

21. May 27, 2012

### phyzzle

Again, the frictional force between the rotor and the pad does not stop the vehicle. If I drove my car off of a really high cliff, and hit the brakes it would do nothing to slow my horizontal velocity. Friction between the tire and road is what stops a car, whether it is static or kinetic.

“The coefficient of static friction between road and tire is a limiting factor of the effectiveness of the brakes. See below.”

Just to be clear we all understand that static friction occurs when tires are rotating right? Kinetic friction occurs when the wheels lock. Static friction is not a limiting factor of the effectiveness of the brakes. In fact it is not a factor of anything. It is a product of the normal force and the coefficient of static friction. Furthermore static friction is not always constant. It can not be as forces act in pairs, action/reaction, equal/opposite. It varies with the applied force until it reaches its threshold at which point it becomes kinetic friction. There IS a change in static friction when applying the brakes. Normal force times the coefficient of static friction is equal to the maximum static friction, however the actual magnitude of the frictional force can be between 0 and that maximum. A gentler brake application will result in less static friction.

I understand the relationship between mass and kinetic energy, however based on Tipler’s example I am still under the impression that the greater kinetic energy is canceled by the greater friction of the same magnitude.
Thanks but I am still confused.

22. May 27, 2012

### phyzzle

Thanks for the welcome! I like this place already. I do not think I am asking two questions. I understand that if you double the mass you double the kinetic energy, I also understand that if you double the mass you double the friction between the road and tires, and this is why Tipler in his diagram pointed out that mass cancels and has nothing to do with stopping distance. Stopping distance is a measure of how quickly and effectively kinetic energy is being converted to heat. So I think the question why should a truck have larger brake rotors than the small car, is a question about stopping distance.

Thanks again.

23. May 27, 2012

### phyzzle

Dalespam, thanks for seeing the validity in my question. I agree with most of what you say. If I may take it a step further though, any varying friction in the brake system will result in the same varience of friction between road and tire. The variable frictional force is not a result of the normal force exerted by car on road but rather a result of the coefficient of static fricton which is variable and is directly proportional and opposite to the applied force by the tire on the road. If the car is at rest the friction between the tire and road is 0 - since we know the normal force is constant it must be the coefficient of friction that is changing. The frictional force is in the horizontal direction, therefore if the tire is not applying a horizontal force to the road there will be no frictional force. The frictional force develops the moment the tire starts to move. I hope that makes sense, and that I understood what you were trying to get across.

24. May 27, 2012

### D H

Staff Emeritus
Yes, it is a constraint. Static friction is a constraint force.

Imagine a block of wood of mass M on a ramp. If the ramp is horizontal the frictional force exerted by the ramp on the block of wood is zero. Raise one end of the ramp a small angle θ so the block remains stationary and the frictional force on the block is M*g*sin(θ). Note well: It is not M*g*μ*cos(θ), where μ is the coefficient of static friction. Not yet, at least. That sin(θ) factor increases and cos(θ) factor decreases as you continue raising that end of the ramp. The block starts sliding down the ramp at the critical angle θcrit which is given by tan(θcrit)=μ. This is the point at which the force required to keep the block stationary just exceeds what static friction can achieve.

Now think of a car. Touch the brakes lightly and the car slowly decelerates. The force exerted by the road on the tires is less than M*g*μ. That M*g*μ is the maximum frictional force that the road can exert on the tires before the tires start slipping.

25. May 27, 2012

### D H

Staff Emeritus
What I'm getting at is that coefficient of static friction limits how hard you can apply the brakes before the wheels lock. The wheels lock at the point at which the brakes attempt to slow the rotation wheels at a faster rate than static friction can keep pace and keep the wheels rotating. Drive on a slippery surface and you need to apply the brakes gingerly lest the wheels lock. When you are on a NASCAR course you can hit the brakes much, much harder and still not have the wheels lock.

Sorry. I meant that the coefficient of static friction is constant. Sloppy typing.