# Friction force dilema. A new physics breakthrough?

1. Apr 13, 2007

Friction force dilema. A new physics breakthrough? :)

Sometimes during lunch time, strange talks appear :). And then great doubts about life and physics...

Imagine two cars: one with wide tires and the other not that wide. It is obvious that the car with wider tires will curve better and brake faster, so the friction force will be greater, correct?

The friction force (Ff) is (N -> normal force; uf --> friction coeficient):
Ff = N . uf

The uf is only dependent on the material of the two surfaces, if we have a dynamic or static friction (not on the area of contact).

The N (normal force) is dependent on the gravitacional force multiplied by a coeficient that relates to the position it is being applied. For braking in a horizontal surface, you may assume that the Normal force is equal to the gravitacional force. And the gravitacional force is the mass multiplied by the gravitacional acceleration :).

My point is that today's formulas to calculate the friction force are independent on the area of the contact surface and thus wrong. They say that if we have a 1 ton car with big tires and a 1 ton car with very small tires braking or curving at 100km/h, they will have the same friction force on the tires and consequently brake within the same distance and time, and be able to curve the same way.

I believe this is wrong. I think that the friction force should be dependent on the normal force*area of contact and not just on the normal force.

So, what do you think? Am I running for the nobel prize :) or did I miss something?

Cheers,
Rui

2. Apr 13, 2007

### ZapperZ

Staff Emeritus
You missed something. The coefficient of friction is typically phenomenological value, and the contact surface is taken into account there. That is why you seldom see the values of coefficient of friction being tabulated. It not only depends on the material and contact area, but also surface roughness which can vary tremendously.

Going off topic, I never complain about people's nickname on here, but I'd like to say that in light of what is going on around certain parts of the world, that your nickname is highly inappropriate and in bad taste.

Zz.

Last edited: Apr 13, 2007
3. Apr 13, 2007

### lpfr

No, it is far from obvious. It is true that a larger surface of contact on tires gives better grip on the road, but just a little more, not two times.

Friction forces depend, among others things, on surface ruggedness. Road surfaces are far from flat and the formula about the normal force does not apply perfectly.
Imagine the "friction force" between two teflon gears. This is an extreme case, but the "friction" force is independent of the normal force.

The formulas in the physics books are good for flat surfaces, and you will not find a book which says that the formulas are good in all cases (well, if you find one, you can throw it in the garbage can).

Do not think that you are the first one to remark such kind of evident facts. There have been millions of physics students before you, and some of them where almost as clever as you are.

4. Apr 13, 2007

### Staff: Mentor

Also, since tires deform, the friction force varies with pressure. A tire with a larger surface area won't "sink in" to the bumps in the road as much.

5. Apr 13, 2007

ZapperZ, I've had this nickname for 7 years, and began when fighting for my life and the death of others in the on-line Unreal Tournament matches. It has nothing to do with current events. I always use it so that I don't forget it.
Also, ZapperZ, I think you missed my point, I was not talking about the type of the materials, simply the area of the surface, which is not taken in account in the tables of coeficient of friction (see http://www.roymech.co.uk/Useful_Tables/Tribology/co_of_frict.htm)

lpfr, you got my point :). However, I think you are not being open enough and you are trusting too much on what your books told you (and you labeled that when things don't match that's because the real conditions are different from the theoretical). I don't think that flatness is the explanation or the way out of this messy question.
Imagine a cork in a bottle: if you have a long cork, the instantaneous force applied to remove it from the bottle is greater than if you have a small cork (explanation: the contact surface area).
Still not convinced??

lpfr, it has been many years since I was a physics student :). And I'm not that clever. This discussion came up when talking with my work colleagues and we all think there's a gap here, the solution of "normal force*area of contact" was a conclusion.

6. Apr 13, 2007

### ZapperZ

Staff Emeritus
No, I think it is you who missed both my point, the the specific nature of the reference you gave. Note that the values given are working values for a specific geometry and configuration. In other words, you do not find such a thing given for, let's say, material M on top of material Y. You cannot do that table because you can get a whole range of values depending on not just surface area of contact, but surface roughness as well. So in essence, you question itself is incomplete, because you could also add that not only is friction different with different surface area, but also when you have the same surface area, the roughness could be different, which results in different friction!

Again, the coefficient of friction is phenomenology. Someone has measured this already for a specific configuration, be it a nut around a bolt, or a block sliding on a plane. I cannot take that same value and use it elsewhere under a different set of conditions even if the same materials are moving against each other.

Zz.

7. Apr 13, 2007

### lpfr

Sorry, the way you posted your question was typical of a young student.

For me, one of the differences between mathematics and physics is that physicist know that they can just approach the reality. They simplify, in order to obtain "easily" just approximate results. But they know that the results have not an absolute precision. But, even so, they can send probes to Saturn and land on Titan.

There are a lot of subjects that are over simplified in text books. The results are not perfect, but describe the reality approximately.

As for the cork, it is a very bad example. I have uncorked many a bottle of wine in my life and I can assert that you never know how much force you will need next time.
In first approximation, and all things identical, the force to uncork is proportional to the length of the cork. Of course, as the length shortens, vertical deformation is facilitated and the force lessens more and more. For lengths far bigger than the cork diameter the proportionality should be respected.

8. Apr 13, 2007

### lpfr

I'm not sure if I have been clear about the corks.
The force to push two bricks is twice the force to push one brick.
What you pretend is that the force is lesser if the two bricks lie on the table than if a brick is over the other.
In your cork example you pretended that to push two bricks is harder just because the surface is bigger. Then let's put one over the other, and you would have proved that the force diminishes with the surface.

9. Apr 13, 2007

### Stingray

The friction model taught in introductory physics (called Coulomb friction) is not very accurate for many systems. It's extremely poor for rubber on rough surfaces, so you shouldn't try to apply it to tires.

If a car is designed for it, wider tires do tend to corner better up to a point. This is a complicated effect that is not summarized in any simple equation. Also note that a wider tire doesn't necessarily have a larger contact patch. The area of that patch is mostly determined by the pressure and normal force, so you're really changing its shape more than its size.

The study of tire properties is a very active branch of automotive engineering. Most data is kept secret, unfortunately, but there are still entire books you can buy on the subject. Here's a very nice chapter out of a textbook (Wong: Theory of Ground Vehicles) available from the publisher: http://media.wiley.com/product_data/excerpt/19/04713546/0471354619.pdf

10. Apr 13, 2007

### mr200backstrok

I'm not sure if this is what he meant in the beginning, but this is a question i have after reading this:

Say I have a brick that weighs 10kg (arbitrary), with a ground contact area of 1 square meter. This is placed on a table. I have another brick, also 10kg, but with a ground contact area of 2 square meters. This is placed on the same table. For the sake of simplicity, lets assume that the coefficient of friction (µ) is the same for both (and equals 1). the equation provided shows that force of friction = 9.8(m/s^2)*10kg*1 = 98 newtons for both. However, logic says that the brick with a larger contact area would have a higher force of friction, right?

11. Apr 14, 2007

### lpfr

Maybe that, at first, intuition (not logic) tells that bigger surface means more friction force. I can't remember what I thought the first time. But intuition is deceived by the fact that bigger surface also means lower pressure. This is the reason why, coarsely, the friction force doesn't depends on the surface.

Edit: I have never heard or read about the dependence of the friction coefficient with pressure. You find everywhere that $$\mu$$ is constant. Obviously this is not the case, at least for tires.

Last edited: Apr 14, 2007
12. May 9, 2008

### sableFerret

Friction

As some people pointed out friction is quite a complex subject (the area of study is called Tribology) In general for materials F = μW, μ being the coeffecient of friction. As stated the frictional force is only dependant on the normal force applied. I realise this is counter-intuitive for alot of people and to really understand it you have to look at a microscopic level at what is going on.

In reality surfaces that look smooth are infact tiny mountain ranges with millions of troughs and peaks that are nanometres-micrometers tall and when two surfaces to into contact it is only a few of the highest of these points (known as asperities) that are actually touching so when you think of say 1 square meter of contact its actually more like a few hundredths of a square millemeter that are touching.

The friction comes when the asperities try and move past each other but cant and cold weld to each other, meaning that in order to carry on sliding the asperities from the weaker material have to actually break off (which is why surfaces in contact wear down).

However this doesn't really apply to elastomers (of which tyre rubber is one). Elastomers are a special case where the material is actually a series of very long chains of molecules that are all intertwined. When you try and slide an elastomer over another surface tiny ripples are formed at the surface known as Schallamach waves which travel through the material and set off the elastomers visco-elastic effects. The long and short of it is that the rubber in a (very simplified) way melts to the road surface causing it to bind closely with the cracks and creates lots of contacting asperities (again simplifying).

This means that unlike normal materials where no matter what you do the contact is only on a few points, with elastomers the more rubber you put down the more contact you get so in this very special case tyre size is roughly proportional to frictional force.

Friction is a cruel mistress and no matter how good us engineers get we always must remember the dictum

There is never enough when you want it [grip] and always too much when you dont [wear]

13. May 9, 2008

### rcgldr

Friction in real life in complicated. For example, take a look at the second half of video # 2 from this web site:

http://www.gyroscopes.org/1974lecture.asp

As mentioned already, the coefficient of friction for tires decreases with an increase in normal force. Double the downforce on a tire, and grip increases, but not by twice as much. A wider tire reduces the pressure at the contact patch (less force per unit area), which helps improve grip. A wider tire also provides more surface for cooling, and handles irregularities and contamination between tire and pavement. The cons for a wider tire are more weight and aerodynamic drag.

14. May 9, 2008

### DeepSeeded

This is not true. Two vehicles, 1 with small tires and 1 with large tires (both with emergency brake set) will slide down a hill at the same angle.

15. May 9, 2008

### sableFerret

Agreed they will both start sliding at the same point but the one with smaller tyres will hit the ground first

ie They both have identical static friction coefficients but kinetic coefficient change with visco-elastic effects

16. May 9, 2008

### rcgldr

Based on the previous posts and from other web sites regarding tire friction versus normal force, the vehicle with the wider tire will require a larger angle before it starts to slide, assuming that the pressure in the wider tire is lowered so it has the same relative deformation at the contact patch that the narrower tires does, which gives the wider tire a wider contact patch than the narrower tire.

As mentioned before, the coefficient of friction for tires decreases as the normal force increases. Reducing the force per unit area of the contact patch increases coefficient of friction.

17. May 9, 2008

### atom888

woooo, i'm here to save the day. Me think.

We must establish our boundary first. Assuming the surface contact is homogenius. That means I don't care how big your tire is, the area that it touches have same roughness as the other little tire case. It is true that you'll get a different coefficient if you smooth the surface down. Therefore, it's not a law that road,wood,metal have exact coefficient.

Anyway, Ff=uxN . The big tire have bigger surface contact, BUT the pressure or distributed normal force is LESS if you wanna go down and dirty. The little tire have less surface contact, BUT the pressure or distributed normal force is MORE. Now if you wanna do some $$\int$$$$^{N_{0}}$$u dn . Be my guess, you'll get same F both cases.

18. May 10, 2008

### reilly

Consider two boxes, each with 100 pounds of, say, sand. But one has a large contact area, while the other has half the contact area of the first. Push both along a concrete floor; what differences do you expect? Try it out, 15 or 20 pounds should do quite nicely.

Don't forget the flywheel; lots of inertia there.

And, note that the basics of motion with friction go back probably 200 years or more. Given that these basics are important for autos, airplanes,..... one might well think that there are no flies in the ointment, because these basics have served us well over time.

Questioning is good; informed questioning is better.
Regards,
Reilly Atkinson

19. May 10, 2008

### rcgldr

This isn't about boxes, it's about tires, which have a load sensitivity. At small load factors, the coefficient of friction is constant, but at higher load factors the coefficient of friction decreases. Link to Wiki article:

Race cars take advantage of load senstivity by using springs and anti roll bars to make one end of the car relatively stiffer than the other end, which changes the relative grip, lowering it at the relatively stiffer end, which is used to adjust understeer / oversteer. For example, if the load distribution differnential while turning is increased by stiffening the springs and/or anti froll bars from 350/150 lbs (outside/inside tire) to 375/125lbs, then the cornering grip is decreased at that end of the race car.

20. May 12, 2008

### DeepSeeded

Try this experiment. Hold a somewhat heavy cylinder verticaly with all 5 fingers using the same pressure and remove one finger at a time until you only have your thumb and index finger holding it. Then try to say that frictional force is independent of surface area.

I think the only reason we find that it is; is because our experiments are done with an object that is sitting on the ground being pulled by gravity in an orthagonal (?? I think I spelled that right) direction. So as the surface area is reduced the pressure is increased on the remaining surface area. This would not be the case in space or with a cylinder in your hand, etc.

So I am starting to think that with any surface, the surface area is proportional to the friction as well as the pressure (not the mass).

Last edited: May 12, 2008