Why do trucks have bigger brakes?

[Lsos]

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.

 

[truesearch]

'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

 

[K^2]

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.

 

[Dale]

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: f_s≤μ_sf_N

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?

 

[Dale]

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 f_N between the tire and road.

The confusion is that the larger f_N between the tire and road is not the determining factor for controlled braking. The f_N that determines the braking force during controlled braking is the f_N between the pad and disk.

 

[phyzzle]

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: f_s≤μ_sf_N

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?

I was not saying that the force on the left was proportional to the force on the right of the inequality. It is an inequality because the force on the right is the maximum static friction. The actual friction will be an equation and the applied force will be equal to the static friction and directly proportional to the coeffeciant of static friction.

This paragraph from GSU physics department mentions it:

'frictional force is also presumed to be proportional to the coefficient of friction. However, the amount of force required to move an object starting from rest is usually greater than the force required to keep it moving at constant velocity once it is started. Therefore two coefficients of friction are sometimes quoted for a given pair of surfaces - a coefficient of static friction and a coefficent of kinetic friction. The force expression above can be called the standard model of surface friction and is dependent upon several assumptions about friction.'

I will maintain that the coefficient of static friction is not constant. Think if the car is not moving, is there any friction in the horizontal direction?

 

[D H]

I will maintain that the coefficient of static friction is not constant. Think if the car is not moving, is there any friction in the horizontal direction?

You need to gain a better understanding of constraint forces. Static friction is zero in this case. The coefficient of static friction is not. The coefficient of static friction for a pair of objects yields the maximum of static friction for a given load before the objects start slipping.

 

[Dale]

This paragraph from GSU physics department mentions it:

'frictional force is also presumed to be proportional to the coefficient of friction. However, the amount of force required to move an object starting from rest is usually greater than the force required to keep it moving at constant velocity once it is started. Therefore two coefficients of friction are sometimes quoted for a given pair of surfaces - a coefficient of static friction and a coefficent of kinetic friction. The force expression above can be called the standard model of surface friction and is dependent upon several assumptions about friction.'

That paragraph mentions only two coefficients of friction for a given pair of surfaces, a static and a kinetic coefficient. I agree completely. That is the "standard model" of friction.

Under normal braking conditions, as I have been discussing, only the static coefficient applies.

I will maintain that the coefficient of static friction is not constant. Think if the car is not moving, is there any friction in the horizontal direction?

This claim is not correct and is not supported by the reference above nor by your example of a stationary car.

Recall the inequality governing static friction: f_s≤μ_sf_N

Typical coefficients of friction on a dry road are 0.7 (ref). A typical car might weigh about 4000 lb. You correctly point out that when the car is not moving the frictional force is 0, so let's plug in these numbers into the inequality.

f_s≤μ_sf_N
0 lb ≤ 0.7 4000 lb
0 lb ≤ 2800 lb
true

So, we see that the inequality holds for a stationary car without varying the coefficient of friction. Your example therefore does not demonstrate a varying coefficient of friction.

 

[K^2]

phyzzle, static friction coefficient is a constant. The force of friction is NOT proportional to it, however. Kinetic friction is proportional to kinetic friction coefficient. But for kinetic friction to take over, the wheel must slip, which is not normal braking condition.

Inequality Dale quoted is correct. Friction force is less than or equal to a quantity that is proportional to static friction coefficient. The actual static friction force is a constraint force, and it's magnitude is governed by the constraint itself, and not by any coefficient or magnitude of normal force.

 

[phyzzle]

Sorry about the above post, I was a little rushed. After looking at the mentioned reference a little closer it was not what I thought it was.

Newton's 3rd law clearly shows that static friction is variable and proportional to the applied force. Because the normal force of the car is constant it must be the coefficient of static friction which is variable. If you push a car in the positive direction with half of the amount of force required to move it, then it will not move.
What is the frictional force at this point? It is not Normal force times the (max)coefficient of static friction. That would be the maximum friction and the actual friction is half of that. The coefficient of static friction does not jump from 0 to maximum. If it did, then when you push the car with half of the force required to move it in the positive direction it would actually move in the negative direction. Of course this never happens. Does anyone see what I am saying?

Check this link: http://www.astro.sunysb.edu/iyers/home/Quiz2_Sol.pdf

 

[xxChrisxx]

I will maintain that the coefficient of static friction is not constant.

Doesn't really make sense to call it a coefficient then does it.
As a primary assumption, you can assume that the coefficient of friction is constant, it's why values for it are empirically determined.
As you seem to be having trouble with the basics, load sensitivity will really just screw things up.


The bottom line is:
F=mu_s*N
Defines the operating envelope.

 

[russ_watters]

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.

I share DH's way of viewing it: I prefer to follow the energy. All of the energy to stop the car gets dissipated by the brakes. The tires/wheels are basically just a way of attaching the brakes (and the engine!) to the road.

$.02

 

[phyzzle]

Thanks for sticking with me on this. I will get it eventually.

 

[xxChrisxx]

It's got a bit out of hand though.
I've just skim read the thread since I last posted and you are all going round in circles.

Deal with one thing at once.And thank christ engineers design cars and not physicists.

 

[phyzzle]

Doesn't really make sense to call it a coefficient then does it.
As a primary assumption, you can assume that the coefficient of friction is constant, it's why values for it are empirically determined.
As you seem to be having trouble with the basics, load sensitivity will really just screw things up.


The bottom line is:
F=mu_s*N
Defines the operating envelope.

Values are not empirically determined, maximum values are determined. And those maximums are very useful.

 

[phyzzle]

It's got a bit out of hand though.
I've just skim read the thread since I last posted and you are all going round in circles.

Deal with one thing at once.


And thank christ engineers design cars and not physicists.

Yes indeed, this thread needed a little humor.

 

[xxChrisxx]

Values are not empirically determined, maximum values are determined. And those maximums are very useful.

I can't tell, are you being serious?

 

[xxChrisxx]

Starting from step 1:

The maximum force a tyre can apply longitudinally before slipping (wheel spin) is determined by it's static coefficient of friction. We shall ignore cornering (lateral forces) for this thread.

Ft = Fl * mu_s
Tractive force = load * static coefficient of friction.

Load in this case is defined as gross vehicle weight.
Do you agree with this?

 

[K^2]

Newton's 3rd law clearly shows that static friction is variable and proportional to the applied force. Because the normal force of the car is constant it must be the coefficient of static friction which is variable.

You are trying to define F_s=μ_sN. This is NOT TRUE. There is no such equality. Static friction coefficient is NOT the proportionality constant between force of static friction and normal force. That is absolutely NOT how it is defined.

 

[truesearch]

how is coefficient of static friction defined? Text book definition would be good so that it can be checked.
My book says (limiting frictional force)/(normal reaction) = coeff of static friction...is my book correct?

 

[xxChrisxx]

You are trying to define F_s=μ_sN. This is NOT TRUE. There is no such equality. Static friction coefficient is NOT the proportionality constant between force of static friction and normal force. That is absolutely NOT how it is defined.

This is overly nit-picky to the point of being confusing, as it made me go 'huh'?

The equality is how I'd describe it (as I have above), just with the crucial point that it's a maximum value. The force value can be less, which is why it's usually shown as an inequality (you aren't always braking or accelerating at the maximum).

More than getting to, and arguing about definitions. I feel the crucial thing to take away is that the the equation above is the maximum 'potential' force that can be applied by a Tyre-surface system. It is always likely to be less.
Operating envelope for Tyre, or GG diagram. As force is proportional to acceleration, the graph would look the same for force.
http://hpwizard.com/gg-diagram.html

 

[K^2]

how is coefficient of static friction defined? Text book definition would be good so that it can be checked.
My book says (limiting frictional force)/(normal reaction) = coeff of static friction...is my book correct?

Yes, your book is correct. Key word here is "limiting". It's the maximum amount of friction force you can have. The actual force of static friction is whatever it needs to be up to that limiting force.

This is overly nit-picky to the point of being confusing, as it made me go 'huh'?

The equality is how I'd describe it (as I have above), just with the crucial point that it's a maximum value.

Trouble is, phyzzle isn't talking about the maximum force. He's talking about whatever the force of friction is, and he insists that static friction coefficient changes to match it, which is simply not how it is defined. It's not nitpicking. It's pointing out the actual flaw in logic that sends phyzzle along a completely wrong route.

 

[phyzzle]

This is overly nit-picky to the point of being confusing, as it made me go 'huh'?

The equality is how I'd describe it (as I have above), just with the crucial point that it's a maximum value. The force value can be less, which is why it's usually shown as an inequality (you aren't always braking or accelerating at the maximum).

More than getting to, and arguing about definitions. I feel the crucial thing to take away is that the the equation above is the maximum 'potential' force that can be applied by a Tyre-surface system. It is always likely to be less.



Operating envelope for Tyre, or GG diagram. As force is proportional to acceleration, the graph would look the same for force.
http://hpwizard.com/gg-diagram.html

Awesome, I agree with your post completely. And I think we are seeing eye to eye.

In your statement;
'I feel the crucial thing to take away is that the the equation above is the maximum 'potential' force that can be applied by a Tyre-surface system. It is always likely to be less.'

Can I rephrase it as saying that the force that can be applied by a Tyre-surface system is variable? Since you said it has a maximum but is always likely to be less? Variable?

And if so, since there are 2 factors that determine the friction of a tyre-surface system one being normal force or weight of vehicle which we know is constant, then the other factor has to be variable. Yes?

The other factor is of course the coefficient.

 

[K^2]

Friction force divided by normal force is a nonsensical quantity. It has no physical meaning. It's not a coefficient of anything.

 

[Dale]

Newton's 3rd law clearly shows that static friction is variable and proportional to the applied force.

Newton's 3rd shows that the friction of the tire on the road is equal and opposite to the friction of the road on the tire. I don't know if you are calling one of those the "applied force" or if you are referring to some other force. If you mean one of the friction forces then it is not just proportional, it is equal and opposite, which is a much stronger statement. If you mean some other force then that is not true in general.

Because the normal force of the car is constant it must be the coefficient of static friction which is variable.

Why? There is only one equation that I know of which involves the coefficient of static friction and it allows for a range of forces for a constant coefficient. Are you aware of any formula other than f_s≤μ_sf_N which uses μ_s? If so, please post it with a reference.

It is not Normal force times the (max)coefficient of static friction. That would be the maximum friction and the actual friction is half of that.

Agreed, but this fact does not support your argument. See my above rebuttal, and simply substitute 1400 lb on the left hand side where I had 0 lb, the same process refutes this argument as your previous one.

The coefficient of static friction does not jump from 0 to maximum. If it did, then when you push the car with half of the force required to move it in the positive direction it would actually move in the negative direction. Of course this never happens. Does anyone see what I am saying?

Yes, I see exactly what you are saying, but you are simply wrong.

 

[phyzzle]

Okay I understand now. Thanks for putting up with me everyone.


For anyone that is interested:

http://books.google.com/books?id=BM...AKUh-XGDw&ved=0CFMQ6AEwAw#v=onepage&q&f=false

 

[Dale]

how is coefficient of static friction defined? Text book definition would be good so that it can be checked.
My book says (limiting frictional force)/(normal reaction) = coeff of static friction...is my book correct?

In symbols, your book says
f_{s max}/f_N = μ_s
which can be rearranged to
f_{s max} = μ_s f_N
and since
f_s ≤ f_{s max}
we have
f_s ≤ μ_s f_N
in which, as both K^2 and I have pointed out, μ_s is not a coefficient of proportionality.

 

[phyzzle]

In symbols, your book says
f_{s max}/f_N = μ_s
which can be rearranged to
f_{s max} = μ_s f_N
and since
f_s ≤ f_{s max}
we have
f_s ≤ μ_s f_N
in which, as both K^2 and I have pointed out, μ_s is not a coefficient of proportionality.

Dalespam, I have resolved my initial question thanks.

If an object is being pushed without moving, the opposing frictional force is equal and opposite to that force pushing on the object. Increase the force a little bit without moving the object and the opposing frictional force will increase to match it. The normal force has not changed, so what has changed to cause an increase in the opposing frictional force?

Did you check out this link? Please take a look.

http://www.astro.sunysb.edu/iyers/home/Quiz2_Sol.pdf

 

[Dale]

Dalespam, I have resolved my initial question thanks.

Excellent, although it seems like you still have a misunderstanding about the meaning of the inequality in the static friction force equation.

If an object is being pushed without moving, the opposing frictional force is equal and opposite to that force pushing on the object. Increase the force a little bit without moving the object and the opposing frictional force will increase to match it. The normal force has not changed, so what has changed to cause an increase in the opposing frictional force?

Nothing. Remember that the equation is an inequality:
f_s ≤ μ_s f_N

So nothing has to change. Specifically, a change in f_s does not in any way imply that there must have been a change in either μ_s or f_N.

Do you understand what ≤ means? Do you see the difference in form between that and the equation for kinetic friction: f_k = μ_k f_N

Did you check out this link? Please take a look.

http://www.astro.sunysb.edu/iyers/home/Quiz2_Sol.pdf

Yes, I did.

 

[K^2]

so what has changed to cause an increase in the opposing frictional force?

Push against a wall. It pushes back with the same force. Push harder. It pushes back harder. What changed? Certainly not a friction coefficient of any kind, since no friction is involved.