Why do trucks have bigger brakes?

[Hurkyl]

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.

I can understand why one would wind resistance and the contact forces between tire and road to be special -- they are the only external forces in the picture. Brake pad on rotor, axle on bearings, those are all internal forces to the car. All important in ensuring braking happens correctly, of course, but they are the car acting on itself.

(and, of course, what is internal and what is external changes if we separate the car into parts)

 

[russ_watters]

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

It's an important enough distinction that it trips people up when learning about friction, as it has here. Teachers frame questions designed to flesh-out whether students grasp the difference. If you provide a student with a block sitting on an incline and a coefficient of friction, then ask to find the force of friction, you'll find out if they know the difference...

 

[phyzzle]

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

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

Yes, I did.

Okay, I finally get it. Sorry it seems pretty obvious to me now. I hadn't been thinking about the inequality, I was thinking about it conceptually and illogically so. I think I will go post this same question on the monster truck forum. I am sure I will get some equally astute responses.

 

[phyzzle]

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.

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.

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.

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

Yes I am aware of another formual which uses μ_s. In Tipler's college physics text, 3rd edition, chapter 5 he has an inequality for the static friction force and an equation for f_{s max}. At least I think that is what your asking, at this point I honestly am not sure of anything.

Anyhow thanks again for your help, I appreciate your patience while schooling me.

 

[phyzzle]

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.

In case you're interested here is the link for the formula. It will put you on page 73, just scroll up to page 71.

http://books.google.com/books?id=XF...QLG0ZXrBw&ved=0CEwQ6AEwAQ#v=onepage&q&f=false

 

[Dale]

In Tipler's college physics text, 3rd edition, chapter 5 he has an inequality for the static friction force and an equation for f_{s max}.

But that formula doesn't apply since you are explicitly considering forces less than f_{s max}.

 

[phyzzle]

But that formula doesn't apply since you are explicitly considering forces less than f_{s max}.

In theory why can't the forces be equal to f_{s max}? The forces would still be that of static friction as opposed to kinetic friction right?

 

[xxChrisxx]

The forces can be equal to fs_max. They don't have to be though.
EDIT: This is unless I've misunderstood what's going on. It's starting to become difficult to follow what is being discussed.
You seem to be having trouble with this, so just ignore the theory for now, and go with a graphic representation.

F=mu_s*N.
This gives that maximum force available to the tyre whilst rolling.
Draw a circle with that radius. If you are asking the tyre to apply a force that lies in the circle you are fine, if you are out of it you are knackered.

 

[russ_watters]

Try this: a 1kg block is placed on a 45 degree inclined plane. The static friction coefficient is 0.5. Does it slide?

 

[Dale]

In theory why can't the forces be equal to f_{s max}? The forces would still be that of static friction as opposed to kinetic friction right?

That doesn't make sense from what we have been discussing. If the car is at rest then the force is 0 as you have correctly pointed out multiple times. It cannot be both 0 and the maximum.

In order for the force to be at f max you would have to be braking as hard as possible using ideal antilock brakes.

 

[phyzzle]

That doesn't make sense from what we have been discussing. If the car is at rest then the force is 0 as you have correctly pointed out multiple times. It cannot be both 0 and the maximum.

In order for the force to be at f max you would have to be braking as hard as possible using ideal antilock brakes.

I am not sure what we are talking about anymore. You asked:

"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 fs≤μsfN which uses μs? If so, please post it with a reference."

and I responded with another formula, and it's reference. I do not think I ever implied that the static frictional force can be both 0 and the maximum, if I did I certainly did not intend to.
I came here with an honest question about something I did not understand, and thanks to your help and that of others I think I have clarity now. If your intention was to help my understanding you certainly did that, so...thank you.

 

[Dale]

You are welcome. Sorry if my last comment caused any confusion. I was just answering that small question I quoted. The force can be f max under certain conditions, but not those discussed here.