Why do trucks have bigger brakes?

[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.

 

[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.