Why Physicists Don't Like To Talk About Friction

fm6 writes: "You would think that force required to overcome friction would be a function of the area of contact. But according to this Scientific American article, that's not true, and physicists don't have a really satisfying explanation." This is the sort of article that makes you want to go experiment with those teflon-coated disks made for moving furniture.

Comments

Wow.

[ktakki]

Three hours and not a single post.

Guess they don't like to talk about it.

k.

Bad science

[Alpha State]

I guess someone should post something serious.

This moving-crack theory is crap. I can't show it's not true, but a model of interlocking surfaces explains friction perfectly well. Consider two horizontal surfaces whose interface is a zig-zag. There is a force Fd holding these surfaces together and a horizontal force Fm on the top surface. The top surface will not move until it slides up to the peaks in the lower surface. It's quite trivial to show that the required force depends upon the degree of interlocking (the angle of the zig-zag) and the force Fd, which must be overcome to seperate the surfaces.

Hmm, is this harder than I am thinking

[labufadora]

It doesn't seem mysterious to me that it's related only to the force. The same force distributed over a wider area actually applies less force per square [your measure here]. So it's a wider area - big deal. It's compensated for by a proportionally smaller force per square area. Whatever atomic force is working at keeping the surfaces distinctly separated has to do less work at any single point when the force is acting in more places. The net effect? Surface area is irrelevant. Am I missing something? Is this explanation just way too simple? What's the catch?

Re:Hmm, is this harder than I am thinking

[kryzx]

The net effect? Surface area is irrelevant. Am I missing something? Is this explanation just way too simple? What's the catch?


The catch is that it's not true. The best example to disprove your hypothesis is car tires. If surface area were irrelevant it would not matter whether you had narrow or wide tires. A 1 inch wide tire and a 15 inch wide tire of the same material would acheive the same friction (and therefore acceleration/deceleration/turning power). This is obviously not true. There are definitely other factors involved, but I think it's clear that surface area has a significant effect.

Re:Hmm, is this harder than I am thinking

[jnik]

The best example to disprove your hypothesis is car tires. If surface area were irrelevant it would not matter whether you had narrow or wide tires. A 1 inch wide tire and a 15 inch wide tire of the same material would acheive the same friction (and therefore acceleration/deceleration/turning power). This is obviously not true.

You're forgetting something: size of the contact patch. If you have narrower tires, to support the weight of the car you need to either have a longer contact patch (i.e. the tires would basically look flat--the bottom of the tire would be straight for a long distance) or higher pressure on the tires. Pretty simple: size of contact patch times tire pressure times four equals weight of the car.

Re:Hmm, is this harder than I am thinking

[caffeinated_bunsen]

There's also the issue of the resistance of the tire material to separation at the surface. Assume that above some maximum tangent force Ft the surface of the rubber falls apart, leaving part of itself on the road surface. (note the black streaks left on the road from a car moving with locked wheels) Ft is obviously proportional to the contact area, since a larger area must tear off more rubber. If this mechanism for breaking static friction is assumed, then the friction is dependent on area.

For cases when both surfaces remain intact, friction per unit area is dependent on pressure, so total friction is (constant*force/area)*area = constant*force.

Re:Hmm, is this harder than I am thinking

[JohnsonJohnson]

There is a lot more going on with tires than simply the area of the contact patch. Put simply, wider tires help with acceleration and braking because of the stiffer sidewalls of low profile tires (which is actually what's important) and the increased surface area which reduces tire heating making it less likely that the compound will break down and liquefy. The better turn in performance of wider tires is primarily due to the shape of the contact patch (it's roughly elliptical) which on a wider tire provides a longer lever arm when turning. Actually on many high performance cars nowadays turn in is so abrupt that chassis engineers have to take other measures to reduce turn rate so normal drivers can control the cars.

Re:Hmm, is this harder than I am thinking

[labufadora]

The other thing I personally think has an effect with tires on pavement is that you are talking about an elastic material on a very rough surface - when the elastic rubber conforms to the road surface, you wind up with almost a gear like meshing of the two surfaces, which is capable of resisting shear force with more than just surface friction. I think this is borne out by the fact that, while a wider tire patch serves you well on pavement, it provides a much smaller benefit on ice. I think if you had two *perfectly* smooth and *perfectly* inelastic solids in contact, surface area wouldn't matter.

BTW, in light of this discussion, I find it amusing that so many early college physics problems include the following phrase somewhere in the setup: "assuming no friction, ..."

Is this a bad question?

[Kibo]

I think that certainly explains the static force of friction F sub s, but what of F sub k? Why should F sub K typically be so much higher than F sub s?

I suppose one might argue that a surface that experiences an F sub k might be assumed to have previously been at rest, and nestled firmly in the lowest state, or the trough if one likes, but why would this be SO much higher (typically)? Interaction between electrons at the surfaces? If there is a limited amount of interaction taking place, or the formation of weak bonds, why not view it as analogous to a crack (a very well understood phenomina)?

But back to F sub s, the static force of friction, wouldn't the surface fall foreward into the troughs of the supporting surface some of the time providing a slightly accelerating force of friction which would then turn decelerating as the atoms being supported tried to move up out of the trough against the force of gravity? Of course, that's not what we see, so it can't be the complete picture.

Why not move back to the formation of tenuious bonds between the surfaces (for a moment). If these bonds are being made occasionally, then stretched and broken, it would seem to my mind's eye that for a macro sized object F sub s would likely be a near constant (surface irregularities, pressure, whatnot would all play a part). Since the breaking of these bonds in a sence does change the surface properties, why not view it as a moving crack if it is convienent? Certainly we all except greater abstractions than this in our everyday life, if some scientists find it a helpful model is it worth belittling? Sometimes abstractions like this, reguardless of their accuracy, can be surprisingly useful. For my part, it is consistant with what I know to be true and seems to do a better job of explaining, at least for me, better than a classical speed bump theory. Your milage might vary; but so might theirs.

Re:Is this a bad question?

[caffeinated_bunsen]

If you assume that each surface is a series of circular arcs, instead of a zig-zag, then you get a similar result for static friction. If the surfaces are already moving, then they can't interlock as much as when static, and so the tangent force resulting from the normal force is reduced from that required to start from rest. But then this predicts that the sliding friction is a function of the extent of interlocking during movement, which depends on velocity. Oh well.

Re:Is this a bad question?

[superflex]

Well, it might be a bad question... if I'm correct in assuming that by F sub s you mean static force of friction, and F sub k means kinetic force of friction, then you're wrong about F sub k being higher that F sub s. The opposite is true.

The formulae for static and kinetic friction are:

F(s) (lessthan)= u(s)N *

F(k) = u(k)N

Where the u's are actually mu's (use your imagination), which are the coefficients of friction, and N is the weight of the object (mass*9.81 m/s^2)

If you look at a table of friction coefficients, you'll see that the coefficient of static friction is always less than the coefficient of kinetic friction.

As far as why static friction is always higher than kinetic friction, I always thought (IANAPhysicist) that it makes sense if you look at it on a microscopic scale. Surface roughness now looks like "mountains" sticking out of the surface of the two objects in question. When there is no relative motion between the objects, the mountains are fully interlocked, and it takes some extra force to get them unstuck. But once you get them moving, they bounce along off of each other, but they don't get fully interlocked of course, because on the scale of surface irregularities, 1 mm/sec is still pretty damn fast.

I always thought of friction coefficients as a statistical average of the "roughness" of two surfaces.

* sorry, I don't know the Unicode for a lessthan sign.

Teflon coated disks...

[Tower]

I find those work even better if you coat the entire floor with Mobil 1 first... add a pump and filter, and you never have to vacuum, either :)

Another link for this topic

[hubie]

Another place to read about this (complete with MPEGs of the self-healing crack) is at The PhysicsWeb.

car tire area and slipping

[raygundan]

Wider car tires work better because the rubber is less likely to tear due to the force. When stopping or starting, the force on the car tires is often enough to tear off rubber (hence the tracks you leave on the road). This means that the limiting factor in tire traction is not the actual coefficient of friction, but rather the strength of the tire. (Because we are sliding due to tearing rubber *before* we run exceed our force of friction) Since the strength of stuff *is* dependent on area (think 2x4s vs. a broomstick), wider tires will not tear as quickly, meaning more of the friction is available before we slide.

So it's not the friction that's changing due to area, but how quickly the tire tears.

For a reference (quick search on google) see:
http://www.cosm.sc.edu/~phys153/tirefriction.htm l

Why proportionate to area?

[MarkusQ]

I disagree with the very premise of this thread: I, in fact, would not think that friction should be proportionate to the area of contact.

Think about it. By simple everyday experence we know that friction is proportionate to the force of contact (typically, the weight of the load). So if you have to drag two identical crates, which are currently stacked one atop the other, and want to reduce the friction you can remove the top crate, cutting the friction in half. Note that you haven't affected the area of contact at all.

Now suppose you decice to move both crates at the same time, but not stacked. Each crate will have half the friction of the original load, and thus the whole will have the same friction, even though we have doubled the area of contact.

This is in the same category as the "you'd think lead bricks should fall faster than iron bricks of the same size and shape" or "which weighs more, a pound of lead or a pound of feathers"?

You might assume that friction was proportional to the area of contact, but you wouldn't think it.

-- MarkusQ

Re:Why proportionate to area?

[MarkusQ]

Um, no, actually each crate wouldn't have half as much friction. You're thinking of weight. Now weight (i.e. the force of gravity pulling the crate against the ground) is going to have some effect on the force required to move the crate along the ground (one crate would be easier to push than two), but it's not the same thing as friction.

No, I stand by my original statement, backed by any good physics book and everyday experence. Thing about it--putting a paperweight on a paper holds it in place because it increases the force (weight) pressing the paper to the desk; pulling a loaded sled is harder than pulling an empty sled; you can tie two stick together, end to end, but only if you tie them tightly. All of these things and many more reflect the fact that friction is proportional to the force normal to the plane of contact.

And given this, it can't be proportional to the area of contact, as my first post shows. This isn't rocket science, just simple basic physics.

-- MarkusQ