Friction depends on force, not surface area

[Bignate603]

I've discussed this with my father who's a mechanical engineer. The increased surface area does increase traction. The side effects of larger tires allow lower air pressure and softer rubber. This literally allows the tires to get more friction on a microscopic scale, because it can squish down farther into the tiny imperfections in the asphalt, so it increases the coefficient of friction. However, the best way to increase traction is to increase the normal force to the surface. But to make a car corner better you don't want to increase mass, so to get around this you need downforce created by spoilers and what not.

 

[tcsenter]

So no one knows why fatter tires grip better. Could it be that bumps in the surface of the rubber grip better with the bumpy surface of the road when there are more of both in contact?

Hey, why not?

Friction doesn't increase, but the capacity of the tire to resist the same amount of friction without surface failure does increase. Why? Essentially, because there's more tire! Same amount of friction, more tire (strength) to withstand that friction. See?

Its been a while since physics, so forgive any technical gaffes, especially terminology...

While friction is never directly dependent upon surface area, in this particular application, surface area matters because it is the most practical and advantageous of only a very limited number of ways to influence per unit of area pressure.

This question really boils down to the 'strength' of the tire composition and the capacity of any given portion of the tire's surface to withstand resistance before failing (wear). Loss of traction [assuming dry pavement, a layer of water or other intervening substrate changes everything] manifests in a thin layer of tire being stripped away (surface failure) by brute force.

The road's surface may also fail under stress, but since roads are stronger than rubber, we will assume that all failures are at the tire level, not the pavement level, which is generally the case in reality, anyway (though not always).

If you have a small contact area, pressure is distributed over a smaller area resulting in increased pressure per any given unit of contact area. This increases the burden that any given portion of the tire's contact area must support. The reverse occurs with a larger contact area, pressure is distributed over a greater contact area, resulting in reduced pressure per unit of contact area. This reduces the burden that any given portion of the tire's contact area must support.

If the pressure per unit of area exceeds that area's ability to withstand it (strength), the tire surface in that area fails, enabling the tire to spin.

There are also at least two other factors that come into play which dictate that larger tires result (not "give" but result) in better traction. One deals with the uniformity of the surfaces, or rather the non-uniformity of the surfaces, and how a larger contact area enhances uniformity.

The other deals with leverage, because this is an application that involves torque rather than a linear application of force, but I don't really feel comfortable getting into either because A. I'm not too sure I can get it substantially correct to avoid giving the technical nit-pickers a lot to do, and B. I'm not all that sure I even understand it.

 

[faZZter]

"Wrong. Pressure depends on surface area. Force does not. You have to exert the exact same amount of force on the glass whether you use your palms or your fingers. The PRESSURE on your fingers is greater because there is a certain force on a very small surface area. "- silverpig

Well since pressure is defined as force/area (like pounds/sq inch or PSI) I'd say they are all dependent on each other.

Force = area x pressure would follow of course.

As far as the window example goes, I think YOU are wrong. It would take more pounds of force to hold the window (without letting it slip) using just your fingers than it would when using the greater surface area of your palms. (assuming the coeffecient of friction was constant across your fingers/palms)

Why? Because the "grip" on the glass is related to the force times the surface contact area. More contact area requires less of the force to come out with the same amount of "grip" because of the greater surface area factor.

Do you think you could hold this window with two opposing pushpins? How much force on the pins would that take?? (assuming they were incredibly strong and wouldn't break) According to you it would be the same force. We are talking about how hard you would have to push on these pins (or whatever) to keep the glass from slipping.

This example would be the amount of force for the static coeffecient of friction. Once it slips it all changes.

(edited due to me being all messed up....lol)

 

[silverpig]

Originally posted by: faZZter
"Wrong. Pressure depends on surface area. Force does not. You have to exert the exact same amount of force on the glass whether you use your palms or your fingers. The PRESSURE on your fingers is greater because there is a certain force on a very small surface area. "- silverpig

Well since pressure is defined as force/area (like pounds/sq inch or PSI) I'd say they are all dependent on each other.

Force = area x pressure would follow of course.

As far as the window example goes, I think YOU are wrong. It would take more pounds of force to hold the window (without letting it slip) using just your fingers than it would when using the greater surface area of your palms. (assuming the coeffecient of friction was constant across your fingers/palms)

Why? Because the "grip" on the glass is related to the force times the surface contact area. More contact area requires less of the force to come out with the same amount of "grip" because of the greater surface area factor.

Do you think you could hold this window with two opposing pushpins? How much force on the pins would that take?? (assuming they were incredibly strong and wouldn't break) According to you it would be the same force. We are talking about how hard you would have to push on these pins (or whatever) to keep the glass from slipping.

This example would be the amount of force for the static coeffecient of friction. Once it slips it all changes.

(edited due to me being all messed up....lol)

Trust me dude. I'm right.

Force is a known effect. Pressure has been defined because of the existence of force. Pressure is a description we came up with to describe how much force a part of a surface feels. (pressure in Pa = F in newtons / area in m IIRC)

Pressure depends on force. You can easily find the force without knowing the pressure, but you can't easily find the pressure without first knowing the force.

Simply, the force between the tires and the road is the weight of the car + any aerodynamic lift/downforce. Easy to find.

How would you find the pressure between the tires and the road? The best way would be to find the above force, and then divide by the contact area. I can't even think of another simple way to do it right off the top of my head.

As far as the window example goes, I think YOU are wrong. It would take more pounds of force to hold the window (without letting it slip) using just your fingers than it would when using the greater surface area of your palms. (assuming the coeffecient of friction was constant across your fingers/palms)

Why don't you do an experiment? I've done it in high school.

Do you think you could hold this window with two opposing pushpins? How much force on the pins would that take?? (assuming they were incredibly strong and wouldn't break) According to you it would be the same force. We are talking about how hard you would have to push on these pins (or whatever) to keep the glass from slipping.

I sure could. It'd be difficult because you'd have to get them perfectly opposite each other, but it's not impossible.

And no, I never said it'd take the same force. The coefficient of friction between stainless steel and glass is likely to be less than that of skin and glass. However, if you were to use some kind of stainless steel "glove" and repeat the experiment, your results should match.

 

[Woodchuck2000]

He's right you know...
Pressure is a derived quantity defined in terms of force and surface area, not the other way around.

 

[SludgeFactory]

Originally posted by: tcsenter
If you have a small contact area, pressure is distributed over a smaller area resulting in increased pressure per any given unit of contact area. This increases the burden that any given portion of the tire's contact area must support. The reverse occurs with a larger contact area, pressure is distributed over a greater contact area, resulting in reduced pressure per unit of contact area. This reduces the burden that any given portion of the tire's contact area must support.

If the pressure per unit of area exceeds that area's ability to withstand it (strength), the tire surface in that area fails, enabling the tire to spin.

Seems like a good explanation to me. The unit load on an individual rubber "element" is lower when you have more of them (wider tire). The lower the load, the less the chance of failure, whether that be by rupture, fatigue, creep, etc.

Another thing is that the contact patch on a wide tire is wide and short, on a narrow tire it's thin and longer. With each wheel revolution, an arbitrary "element" of the wide tire would spend less time in contact with the ground than a particular one on the narrow tire. So the wide tire should experience comparatively less wear and better cooling, which would give longer tire life, better performance over that life, and/or allow for the use of softer, stickier tire compounds.

The coefficient of friction does change with the size and shape of the contact patch and we know wider tires are generally better, otherwise race cars would run around on skinny tires to enjoy the aero benefits and reduction in unsprung mass. I bet tire companies have reams of empirical data -- and a bunch of physics/engineering/tribology PhD's who are continually working on what's really going on at the tire/asphalt interface.

 

[ElFenix]

Originally posted by: GarlicBreath
Two reasons:

1) For offroad vehicles, a greater surface area equates to lower pounds per square inch of force, making the vehicle less likely to sink in sand or mud. Serious offroaders "air-down" their tires to further increase ground-contact area.

2) This one from Russ Steele, my high school physics teacher way back in 1982. As previous poster pointed out, if the force (weight of the vehicle) remains constant, the friction will remain constant regardless of surface area. Mr. Steel told us drag racers use fat tires to reduce wear. As rubber is abraded off the tire, the diameter decreases. Fat tires have a lot more rubber to wear off than skinny tires.

doesn't hurt grip that the drag tires weigh a whole lot more either

 

[gopunk]

Originally posted by: faZZter
"Wrong. Pressure depends on surface area. Force does not. You have to exert the exact same amount of force on the glass whether you use your palms or your fingers. The PRESSURE on your fingers is greater because there is a certain force on a very small surface area. "- silverpig

Well since pressure is defined as force/area (like pounds/sq inch or PSI) I'd say they are all dependent on each other.

Force = area x pressure would follow of course.

no, it doesn't work like that. for example, you could say that your grade in a course depends on how well you know the material. but it would be incorrect to say that how well you know the material depends on your grade in the course. you can get one from the other, but that doesn't mean it is dependant, you're just working backwards.

look it up in a freaking textbook okay. you'll see that we are right (friction does not depend on SA)

 

[Woody06]

I had a lot of trouble with this question back in high school, it pissed the hell out of me until i figured it out. Friction has nothing to do with surface area... unless there is a deformation of surfaces, The rubber conforms to the surface and the more surface in contact the more grip. Thats why soft tires grip better, they conform easier to the contours of the road. This is also why heating up the tires helps, it makes them sticky, once they get sticky the more surface area the better, but this is independant of the normal friction equations becuase there is a deformation of surfaces. I think someone might have already said this but im not sure.

 

[ChefJoe]

Hrm, I thought there was also an aspect to larger diameter tires where having the weight farther out on the tire helps it to keep its inertia.... I wonder if that improves its grip as well..

 

[Tominator]

This is a never ending argument. In 1969 I met a Goodyear Engineer who said, "Traction is independent of the tire footprint."

You figure it out....

 

[ElFenix]

Originally posted by: Tominator
This is a never ending argument. In 1969 I met a Goodyear Engineer who said, "Traction is independent of the tire footprint."

You figure it out....

traction, because you're less likely to have the whole contact patch leave the road surface or meet a part of the surface with too low a coefficient of friction, like paint or some oil.

 

[jteef]

doesn't hurt grip that the drag tires weigh a whole lot more either

i think if it were up to the drag teams, they would rather have the lightest wheels/tires they could. all that inertia impedes accelleration, which is the name of the game. It would be much better to place say a lead brick where you wanted it for mass/traction than to have it as part of the drivetrain.

jt

 

[Evadman]

Originally posted by: jteef

doesn't hurt grip that the drag tires weigh a whole lot more either

i think if it were up to the drag teams, they would rather have the lightest wheels/tires they could. all that inertia impedes accelleration, which is the name of the game. It would be much better to place say a lead brick where you wanted it for mass/traction than to have it as part of the drivetrain.

jt

unsprung mass = bad.