Dependency of friction on contact area 3

[Tunalover]

Folks-
I've always wondered how first-order statics and dynamics calculations allow that the friction force F=uN where N is the normal force and u is the coefficient of friction. It's obvious that the contact area plays a role, for example, a wide tire vs. a narrow tire provides better grip. Am I overlooking something?


Tunalover

 

[Heckler]

Are you trying to analyze crack initiation & propagation with regards to fretting fatigue?

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[Twoballcane]

hmmm...u=F/N...depends on the normal force pushing down on a block and how much pulling (F) force to move the block. This ratio will give you the coef of friction. Now depending on the interface btw the block and floor, you will have diff pulling forces to move the bock. So if the interface is smooth F will be small, if rough F will be bigger. Now there are two pulling forces. One is to get the block to "break" from the interface and the other is to continuasly slide it across the floor. The "breaking" force is bigger than the sliding force.

Now this is off the top of my head...

Tobalcane
"If you avoid failure, you also avoid success."

 

[Theophilus]

Methinks the better grip is due to non-ideal conditions (in the tire example) such as intermediary material(s) like water or particles. Thus in this case, the greater the surface area, the greater degree to which u=F/N as an ideal (because of the greater degree to which the material truly makes contact with the surface per normal force applied).

It seems intuitive to see surface area as playing a proportional role, but I don't believe this is the case (apart from the exceptions above, which aren't truly exceptions).



Jeff Mowry

http://www.industrialdesignhaus.com

Reason trumps all. And awe transcends reason.

 

[reidh]

That is all true about friction, but I don't think it answers Tunalovers question.


Check out this discussion that has already taken place on this forum:

http://www.eng-tips.com/viewthread.cfm?qid=102250&page=1

-Reidh

 

[SnTMan]

It looks to me like tunalover is asking "Why should the friction force, uN, be indpendent of area of contact?"

Perhaps it is not, really, if N is considered as the product of pressure and area, N = P*A. Then, for the normal force the contact area is implicit, or for given N, P*A = constant.

This does not address something like the contact patch of the tire mentioned in tunalover's original post, whose behavior is complex.

My best guess, I probably need to dig out the old physics book.

Regards,

Mike

 

[GregLocock]

There are at least 4 mechanisms by which friction is developed at the contact patch of a tire, it is a very bad example to use for discussing simple friction.

In no particular order there is chemical adhesion (micro welds), cogging, the work required to overcome damping in the rubber as it drags over the surface, and one which I have forgotten or never knew.

It is by no means necessarily true that a wider tire develops more grip. Usually it is, since a wider tire can be made of softer rubber and still have a reasonable life. Soft rubber tends to have more chemical adhesion, whether that is because it squooges into the declivities of the road surface, and so has more bonding sites in proximity, I am not sure. Typically it has less damping, but if the slip velocities are higher (due to the squooging) then more work may be performed, even though the damping per unit volume is less.

Static friction is even more complicated. I do not know much about it.






Cheers

Greg Locock

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[JStephen]

I've never seen a discussion of it in detail, but always got the idea that it was a pretty gross approximation. You might consider F=Kx for spring force for a tire- due to geometry, this ought to be a considerable departure from the linear approximation as well.

 

[MintJulep]

I have participated in some brake dynamometer testing that demonstrated that friction is reasonably (but not entirely) independant of contact pressure.

That is, for the same friction material, with the same normal force, friction changed a little, but not a lot, with a change in area.

However, if you are disapating the same energy, through a smaller contact patch, the temperature will increase more than a large contact area.

Friction is highly dependant on temperature.

 

[corus]

I've already asked this question with regard to friction and area in a previous post, but in the case of a tyre (note correct spelling* for non-brits) doesn't the deformation of the tyre whilst in motion have an effect on the rolling resistance, and hence the apparent effect on the force required to overcome friction with area?

*The question of tyres is something I'll never tire of.

corus

 

[JakeRAD]

Assuming this is NOT related to tires, intuitively the area of friction contact is related to the strength of the friction force. However, this is incorrect.
Assume you have a force of 100N acting on 4 feet of a chair. These feet have a small surface area, of 100mm^2. The total pressure is 1N/mm^2. If you increase the area, you DECREASE the pressure, since the force is constant. There is a tradeoff between contact area and the pressure which is generating the friction.
Obviusly, thick heavy tires have higher friction force that thin and light tires

 

[GregLocock]

A fourth grip mechanism is work done on the mating surface by the tire. This is not generally significant, except on ice and snow, where it provides surprisingly high grip levels in the right circumstances. mu on cold snow can be as high as 0.6 Rather contradicting that, ice racers have a grip of about 0.3, using studded tires.

Incidentally, what sort of friction is involved when you 'ring' two measuring blocks together?





Cheers

Greg Locock

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[Compositepro]

A basic principle of Physics is that friction does not depend on contact area. It is usually best not to rewrite the Physics books to solve a problem. In the real world there are other effects besides pure friction involved in almost any situation. A tire is a good example. Tire grip is due to many factors besides friction.
Oil lubrication is another example. People often refer to the friction in a bearing. With a complete oil film there is viscous drag which will increase with velocity. This is not friction as defined in Physics which does not depend on velocity.
The real problem is just that the word friction is very commonly misused and this leads to a lot of confusion. Similar to the distinction between force and pressure which have very precise and different definitions in Physics but are often interchanged speech (as occured above).

 

[btrueblood]

That's the Casimir (sp?) force, Greg. Results when surfaces are so smooth that the contact zone is only a few hundred nanometers or less; the Pauli exclusion principle from quantum physics comes into play (can't have more than x electrons in any h^3 region of space), the number/frequency/enery of particles between the two blocks is reduced, and a net attraction force develops between the two blocks. Source of the force is very similar to Van Der Waal's forces in chemistry/physics of gas molecules.

 

[JStephen]

Composite- to be honest, I don't recall physics classes or textbooks ever making any kind of distinction as to these different forces. One of the old handbooks I've got here defines "Friction (F) is that component of the total reaction (R) which is tangent to the surface" which is how it's being used above.

 

[GregLocock]

One more important mechanism is the direct shearing of material off the tire, and indeed, shearing material off the road surface..

Cheers

Greg Locock

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[KENAT]

I've heard numerous explantions of why surface area/pressure doesn't have (a significant) effect on frictional force.

I do know this. In A level physics (last 2 years of high school) we did some experiments that showed it to be true that F=uR and changes in surface area had no measurable effect.

 

[eromlignod]

I haven't seen it mentioned yet that the contact area of a tire with the road is a function of the weight of the car and the pressure in the tire only. It is independent of the width of the tire. In other words if a tire has 1000 lbs of weight on it and is pressurized at 35 psi, then the rubber will compress until an area of approximately 1000/35 = 28.6 sq. in. is in contact with the pavement. A wider tire will mean that the contact area will be wider, but it will also not be as high because the area should be the same.

This is why your mechanic can pressurize your tires when your car is up on a pneumatic lift (with no weight on the tires) to 35 psi each and when you let it down onto the ground, the tire pressures will still be 35 psi, even with the weight of the car on them.

Don
Kansas City

 

[jdubu]

My two cents; The friction force and the normal force are both somewhat proportional to the contact area however the cof of friction is not. That is what I beleive KENAT's high school experiments proved. Think in terms of the units

 

[Tunalover]

Thanks for the inputs folks!


Tunalover