Why do trucks have bigger brakes?

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In summary, the brake rotors on a passenger truck like a Chevy 2500 or a Ford F250 are larger than those on a car like a Toyota because the truck needs greater braking force to stop quickly.
  • #1
phyzzle
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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!
 
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  • #2
phyzzle said:
friction equals normal force times co-efficiant of friction (kinetic or otherwise)

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
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
fzero said:
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.

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 canceled 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.

Thanks for your reply!
 
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  • #5
DaleSpam said:
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?

I did a quick google search and found the example from my textbook. 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
Also, when a car stops, that kinetic energy has to go somewhere. Where do you think it goes?
 
  • #7
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 textbook.

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
Thread needs only two words.

Brake Fade.
 
  • #9
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.

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
phyzzle said:
The frictional force between the rotor and the pad does not stop the vehicle.
Of course it does. If not, why do brakes exist?

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

Friction between rotor and pad serves to increase friction between road and tire.
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
D H said:
Of course it does. If not, why do brakes exist?
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
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
phyzzle said:
I did a quick google search and found the example from my textbook. 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.
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.
 
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  • #14
phyzzle said:
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.
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 :smile:)

Both frictional forces are required for controlled braking.
 
  • #15
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.
phyzzle said:
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 understand that mass distribution plays a role in 18 wheelers, but assuming mass is distributed uniformally why does a truck need bigger brake rotors?
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.
mrspeedybob said:
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.
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
Thank you everybody for the responses! I will take time to read through them all and see if it makes sense.
 
  • #17
Pkruse said:
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 textbook.

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.

Thanks that makes sense. I hadn't considered the radius of the tires. The increased rotational 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
256bits said:
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.

Very true, are you implying that the reason for larger rotors comes down to air drag?
 
  • #19
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
truesearch said:
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.
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
D H said:
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.

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
Q_Goest said:
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.

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
DaleSpam said:
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.

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
phyzzle said:
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.
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
phyzzle said:
Sorry I have never heard of a constraint force. I am not really sure what you are getting at.
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.

Originally you said static friction is constant, and your example above shows that it varies with the applied force.
Sorry. I meant that the coefficient of static friction is constant. Sloppy typing.
 
  • #26
phyzzle said:
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.

it does stop the vehicle. else as said, why bother having brakes ?
what purpose do you think the brakes have ?
The friction of the tyres to the road only apply IF the tyres are locked (not rotating) and they are what is trying to stop the forward(reverse) motion of the vehicle. Whilst the tyres are still rotating, its the applied frictional contact surface between the brake pad and the brake disc (drum) that is slowing the rotation of the tyres and thereby slowing the motion of the vehicle. The more massive the vehicle the more kinetic energy the more friction and heat generation between the brake pad and the disc

driving off a cliff and applying the brakes is irrelevant to the discussion. I'm really strugging to see why you are not understanding the answers give to you by DH and others

Dave
 
  • #27
The engine does not accelerate the car, it rotates the wheels. It is friction between the wheels and road that causes the acceleration. No friction...no acceleration I would refer to what happens on ice.
The brakes stop the wheels rotating. it is friction between the tyres and the road that causes the car to stop. No friction...no stopping
Again I would refer to what happens on ice.
 
  • #28
truesearch said:
The engine does not accelerate the car, it rotates the wheels. It is friction between the wheels and road that causes the acceleration. No friction...no acceleration I would refer to what happens on ice.
The brakes stop the wheels rotating. it is friction between the tyres and the road that causes the car to stop. No friction...no stopping
Again I would refer to what happens on ice.

OK so we will stop producing vehicles with brakes, since in your mind they don't do anything

D
 
  • #29
phyzzle said:
why does the heavier vehicle require more force to stop?!
F=ma.
 
  • #30
I think, truesearch and phyzzle trying to get at the fact that a truck and a car, having identical tire surface, should have identical stopping distances regardless of their mass, since that's the limiting factor. In that sense, "brakes don't stop the car" makes perfect sense.

However, that still produces torque on the wheels that needs to be compensated. That torque has to be compensated by the brakes. And the amount of torque that has to be supplied by brakes of the heavy car is much greater than that of a light car. Hence, the much larger brakes.
 
  • #31
Hurkyl said:
F=ma.

Lots of valid points in this thread, but the above it the single most important reason why trucks have bigger brakes. They simply have more mass, and therefore need more force to stop them. The simplest ways to do that are by either increasing the rotor diameter or by squeezing it harder.

Another important reason is that a truck has more kinetic energy, which a larger rotor can absorb before overheating.

All other factors are secondary. They might explain how the truck is able to apply more force due to greater friction between the tire and road etc., but the ultimate point is that a truck simply needs more F=ma.
 
  • #32
'OK so we will stop producing vehicles with brakes, since in your mind they don't do anything'
Please don't suggest that, if anyone takes it up there will be no way to stop the wheels rotating.
I did not say that brakes 'do nothing'...it is clear what I said and clear what I meant...they stop the wheels from turning.
@k^2...sorry you think wrong, I am not trying to get at anything other than what I stated about brakes...wheels...friction.
Any ideas that are posted here are supposed to reflect conventional physics which can be backed up by recognised textbooks (one of the rules I gather).
I would like to see some independent references to the ideas that have appeared here so that I can check my understanding of car braking etc.
I am a teacher...I have been teaching this stuff for years, it is basic physics (I know where F=ma comes into the story) if I have been getting it wrong I should know about it... and so should the exam boards and question makers and solution producers for those exams.

I will make no further posts to this thread, it is in the wrong category
 
  • #33
truesearch, you need to brush up on constraint forces. If you insist that brakes aren't stopping the car, which is technically correct, then you need to realize that the force applied to wheels doesn't stop the car, either. It stops the wheels. You say, no friction, no stopping. I say, wheels not attached to car, no stopping. It's the same idea, really.

Friction provides constraint force for rolling constraint. Normal force between axle and bearings provides constraint force for wheels moving along with the car. It's the force on the bearings that actually brings the car to a stop, not friction on tires. Now, you might think the distinction is academic, but just recently, I applied brakes on my car, which reduced the wheel rolling speed, which applied stopping force to the wheel via friction. Bur rather than stopping my car, the wheel came off due to a faulty ball joint. The car still came to a stop, but in a manner that's rather different than intended.

Point is, there are two ways we can talk about constraint forces. We can take your route, and then we have to examine all constraints, and see whether they actually hold. Tire might slip. Wheel might come off. Many things can go wrong. Or we can assume that constraints do hold. If we do that, we can completely disregard the actual constraint forces. You don't need to worry about the force applied to bearings and joints, because I grant you the constraint that wheel stays attached, and I'm going to overlook friction, because you grant me a rolling constraint. After all, normal braking is supposed to be performed under rolling conditions. If your wheels slipped, you are doing something wrong. Yes, even when driving on ice.

With rolling constraint in place, the friction force is rather irrelevant. It's not really what stops the car. Rather, because the car stops, the friction must be acting on the wheel to enforce constraint. And the car stops, of course, because we applied additional torque via brakes.
 
  • #34
phyzzle said:
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.
The first part is correct, but the second part is is not correct. The coefficient of static friction does not change under normal conditions (i.e. good road and tire conditions). Things that cause the coefficient of friction to vary are changes in the surfaces, e.g. ice or water. Applying the brakes does not make ice or water appear to change the coefficient of friction.

Also, it is incorrect to say that the coefficient of static friction is proportional to the force. Remember, the governing equation for static friction is an inequality: fs≤μsfN

Since it is an inequality, it is not correct to say that the force on the left hand side is proportional to the force on the right hand side. Does that make sense?
 
  • #35
phyzzle said:
What I do not understand is why does the heavier vehicle require more force to stop?!
As Hurkyl mentioned, this is a direct consequence of Newton's 2nd law: f=ma

We want the heavier vehicle to stop as quickly as the smaller vehicle, so a is the same for both vehicles. But m is larger for the heavier vehicle, and therefore f must also be larger. The homework problem that spawned your confusion was not claiming that f was the same for both vehicles, it was just pointing out that the larger vehicle automatically has a larger fN between the tire and road.

The confusion is that the larger fN between the tire and road is not the determining factor for controlled braking. The fN that determines the braking force during controlled braking is the fN between the pad and disk.
 
<h2>1. Why do trucks have bigger brakes?</h2><p>Trucks have bigger brakes because they are designed to carry heavier loads and travel at higher speeds compared to smaller vehicles. This means that they require more stopping power to safely come to a stop.</p><h2>2. What are the benefits of bigger brakes on trucks?</h2><p>The main benefit of bigger brakes on trucks is improved stopping power. This is especially important when carrying heavy loads or traveling at high speeds, as it allows the driver to safely slow down and stop the vehicle. Bigger brakes also tend to last longer and require less maintenance.</p><h2>3. How do bigger brakes affect the handling of trucks?</h2><p>Bigger brakes can affect the handling of trucks in a few ways. They can increase the weight of the vehicle, which can impact acceleration and turning. However, they also provide better control and stability when braking, which can improve overall handling and safety.</p><h2>4. Do all trucks have bigger brakes?</h2><p>No, not all trucks have bigger brakes. Smaller trucks, such as pickup trucks, may have similar brake sizes to cars. However, larger trucks, such as semi-trucks and heavy-duty trucks, typically have bigger brakes to accommodate their heavier loads and higher speeds.</p><h2>5. Are there any downsides to having bigger brakes on trucks?</h2><p>One potential downside of bigger brakes on trucks is the added weight, which can decrease fuel efficiency. Additionally, bigger brakes may be more expensive to replace and require specialized maintenance. However, the benefits of improved safety and handling often outweigh these potential downsides.</p>

1. Why do trucks have bigger brakes?

Trucks have bigger brakes because they are designed to carry heavier loads and travel at higher speeds compared to smaller vehicles. This means that they require more stopping power to safely come to a stop.

2. What are the benefits of bigger brakes on trucks?

The main benefit of bigger brakes on trucks is improved stopping power. This is especially important when carrying heavy loads or traveling at high speeds, as it allows the driver to safely slow down and stop the vehicle. Bigger brakes also tend to last longer and require less maintenance.

3. How do bigger brakes affect the handling of trucks?

Bigger brakes can affect the handling of trucks in a few ways. They can increase the weight of the vehicle, which can impact acceleration and turning. However, they also provide better control and stability when braking, which can improve overall handling and safety.

4. Do all trucks have bigger brakes?

No, not all trucks have bigger brakes. Smaller trucks, such as pickup trucks, may have similar brake sizes to cars. However, larger trucks, such as semi-trucks and heavy-duty trucks, typically have bigger brakes to accommodate their heavier loads and higher speeds.

5. Are there any downsides to having bigger brakes on trucks?

One potential downside of bigger brakes on trucks is the added weight, which can decrease fuel efficiency. Additionally, bigger brakes may be more expensive to replace and require specialized maintenance. However, the benefits of improved safety and handling often outweigh these potential downsides.

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