# Friction as a function of area

Here’s a question I ran accross recently while trying to teach the build team everything they need to know…

In physics class, we are taught that the frictional force f is given by f=uN and we even verified this experimentally.

Now, this equation obviously doesn’t involve area and our results from lab also showed that area didn’t matter. However, as soon as two surfaces “stick” together, area does come into play.

My question is what is the equation relating area, u, and N? Is it just f=uNA? The units on this aren’t correct, though. Also, how does one go about determining whether A is a factor? For example, hot rubber (such as Indy car tires) seem to care, but cool rubber doesn’t.

Since pressure P = f/A, the area contacting the ground is inversely proportional to the pressure applied at each point of contact <edit> (f = P*A) </edit>

f in f = P * A is the normal force.
Ex:
100 oz/in^2 = P oz * 4 in^2; P = 25
100 oz/in^2 = P oz * 5 in^2; P = 20

When you increase the area, you also relieve pressure

f = uN is equivalent to f = u (P * A) because Normal Force = Pressure * Area

Woops we answered the question at the same time. Heheh. Anyway traction is what he was thinking of which is dependent on surface area.

The traction in tires is different than the standard force of friction that most physics textbooks talk about. In automobile tires, the edges of the treads are a major factor in preventing sliding. They account for much more traction or friction than the coefficient of friction of the rubber on the pavement. Of course, a wider tire would usually have more treads than a thin tire.

The reason why they make racing tires hot is that they need to get sticky or else there is only the traction from the edges of the tire.

For example, hot rubber (such as Indy car tires) seem to care, but cool rubber doesn’t.

I think this is so because conduction of heat depends directly on area of contact, and the tires are so wide so they stay cooler (don’t compeltely melt).

I really don’t think you need to worry about area when dealing with friction… If the field is made of sand this season (for example) we’ll need to worry about pressure so the wheels don’t sink into the sand, but I think you only worry about pressue for stuff like that.

Hehe, the Q&A works by not allowing extra discussion until a Mod has answered. Then you get the extra link. That’s so there’s a qualified answer first

**f = uN is equivalent to f = u (P * A) because Normal Force = Pressure * Area **

Well, I’ve never liked pressure so I’ll use my own formula. Static friction is the force due to partial cold-welding of the two surfaces, so I think the formula should be ‘f=uNA*c’ where c is an experimentally derived constant equal the amount of cold-welding per unit area. With c being measured in 1/m^2 and A measured in m^2, the units cancel out. My formula seems logical, but I am not responsible for any injuries that are a result of an incorrect formula :p.

To clarify:
f=uN for simple surfaces. This does not involve area. However, for FIRST, area DOES matter. For example, try dragging a 1mm wide tire accross carpet and then try a 1m wide strip. Assuming both have the same mass, or at least the same net force acting into the carpet, the 1m wide will be harder to pull. My question is why.

To wysywig: the traction vs friction thing doesn’t really make much sense to me. If the traction is not from friction, where does the force come from? Forces can’t appear out of nowhere. As to the edge thing, are you saying that each edge provides a small part of the net force so more edges equals more net force? In that case, area does matter as the edges are nothing more than area.

Im not sure but why not use the original equation because it’s surface area and friction are independent. I get it the treads give the tire it’s traction including the force of friction of rubber. Wider tires= more treads thus more traction. http://www.physlink.com/Education/AskExperts/ae200.cfm Lol a search for traction and surface area brought this up. http://www.wheelchairjunkie.com/traction.html

Mu, as a coefficient of friction, is applicable to two smooth surfaces interacting with one another. For our purposes, when considering the force wheel chair wheels exert unto the carpet, F=uN is fine.

However, as Matt Leese states in this thread, traction can increase when two surfaces mesh with one another. In such instances, this occurs because there is contact occurring between the surfaces that is perpendicular to the direction of movement.

Think about a set of spur gears, if you will. Their ability to transmit torque isn’t limited by the friction that occurs between those two surfaces because they mesh.

On our robots, belting or file cards or other irregular surfaces often serve a similar purpose. Last season, our robot featured treads with over 1 ft. sq. of surface area contacting the carpet. Because the carpet is made of small loops of fiber and our belting material was made of similar PVC hooks, the materials meshed, giving us superior traction.

Unfortunately, I don’t know if there’s a single universal formula that applies to the force two meshed surfaces can incur before slipping. I would imagine that, like mu, it is dependent on some experimentally derived constant that varies with each surface.

Sorry I couldn’t help more.

*Originally posted by rbayer *
**To clarify:
f=uN for simple surfaces. This does not involve area. However, for FIRST, area DOES matter. For example, try dragging a 1mm wide tire accross carpet and then try a 1m wide strip. Assuming both have the same mass, or at least the same net force acting into the carpet, the 1m wide will be harder to pull. My question is why. **

Humn. If my understanding is correct, the two strips will have the same resistance to sliding motion caused by friction, as long as the pair of surfaces involves the same materials, and the normal force between the strip and the ground is the same. The 1mm tire and 1m strip do not have similar normal forces with the carpet. If you did this by placing different width strips under the same weights and dragging the weight with only the strip in contact with the ground, the force of friction would be the same in the two cases.

At the bottom is a quick mspaint explaining what I mean the difference between traction and friction.

In Red: A surface such as carpet that can be molded by pressure
In Green: Points of frictional force application
As you will notice, the left picture has 2 posts, and the right picture has 8. When pressure force is applied, the carpet will mold into the treads.

The top pictures represent a “0-weight” situation, where the wheels are just barely touching the carpet, and where frictional force is applied. The center picture represents when weight is applied how the wheels will go “in to” the carpet. The points where the carpet is compressed is now in the treads. The bottom picture shows how there are more points of friction even after the pressure is applied.

This is how “Traction” is different from “Friction” Traction the friction across different planes.

*Originally posted by Suneet *
**Humn. If my understanding is correct, the two strips will have the same resistance to sliding motion caused by friction, as long as the pair of surfaces involves the same materials, and the normal force between the strip and the ground is the same. The 1mm tire and 1m strip do not have similar normal forces with the carpet. If you did this by placing different width strips under the same weights and dragging the weight with only the strip in contact with the ground, the force of friction would be the same in the two cases. **

Yes, they do. If you look at my post, I specified that they have the same mass, thus the same normal force. (Please don’t say this isn’t true. N=mg).

Anyway, there’s a reason I posted in Q&A: none of the info here is anything new to me. What I’m asking is how will increasing our surface area increase our traction/friction/whatever you want to call it. If we double the width of our wheel, for example, does our traction double? Does it go up by root 2? The cube root of 2? Or is it not possible to quantify?

*Originally posted by rbayer *
**Anyway, there’s a reason I posted in Q&A: none of the info here is anything new to me. What I’m asking is how will increasing our surface area increase our traction/friction/whatever you want to call it. If we double the width of our wheel, for example, does our traction double? Does it go up by root 2? The cube root of 2? Or is it not possible to quantify? **

If your wheel is a smooth surface (i.e., aluminum, two wheel chair wheels), double the area will do nothing to aid your traction at all. It will increase by a factor of 1.

If the material on your wheel is not a smooth surface, in that it interacts with the carpet fibers, your traction will increase. There is no set factor that determines how effective this is.

Whereas the smooth surface of a wheelchair wheel propels the robot due to friction, an irregular surface that meshes with the carpet fibers propels the robot due to both friction (the sliding force between the smooth portions of the wheel and the carpet) and torque (the contact between two surfaces perpendicular to the direction of movement)

Again, think of two spur gears meshed together. Assume that the one on top is your wheel and the one below is the carpet. If each ‘gear’ had no teeth, power transmission occurs because of friction between the two surfaces. When you add teeth to each gear, friction along the surface of the gear is not a factor. Instead, the force is transferred from the face of one tooth on one gear to the face of a tooth on the second gear via torque.

When the motors output a torque that is greater than the force of friction between two smooth surfaces, the surfaces slip. When those same surfaces are meshed, they cannot slip - so power is transferred more efficiently (or, the teeth break ).

Gears are simplified example of the phenomenon that is taking place when irregularly shaped objects interact with the carpet. Because the carpet and belting materials are flexible and irregular, the relationship is similar, but not quite the same.

EDIT: So, to really answer your question; No, the equation you provided isn’t correct. There isn’t really any simple equation to characterize this behavior because the contact surfaces vary at any given movement in myriad ways, including contact surface area and angle, among others. If determining a mathematical relationship between meshed services were an absolute necessity, I would conduct experiments and determine “mu,” realizing that it isn’t friction at all that we’re dealing with. Instead, you’re encapsulating the interaction of the many faces into one system and treating that system like friction. It’s not friction, however. I hope that helps.

This is utterly confusing. Hear is what I understand. Friction is not dependent upon surface area almost every single website says that. But other websites I’ve searched describe surface area as a way to improve traction. Well aparently traction and friction aren’t the same thing so what is traction?

*Originally posted by wysiswyg *
**This is utterly confusing. Hear is what I understand. Friction is not dependent upon surface area almost every single website says that. But other websites I’ve searched describe surface area as a way to improve traction. Well aparently traction and friction aren’t the same thing so what is traction? **

Traction is a word. There are no physical units associated with traction in the same way that there are for friction, or mass, or speed (i.e. friction is measured in Newtons, but traction isn’t <i>measured</i> in anything.)

Friction, on the other hand, is a force measured in Newtons. It occurs between two surfaces sliding together.

Good ‘traction’ comes from best harnessing the output torque of our motors and gearboxes. This means, esssentially, that we’re trying to maximize the force that is used to propel the robot along the floor while avoiding two situations - stalling the motors or slipping the wheels.

Never mind you answered the question with the gear scenario. The more surface area you have the more room you have to mesh with the tiny little bumps in the ground. Heheh btw I found this little tidbit about rubber tires which is interesting.

Tires are actually their grippiest when there is about 5% slippage involved.

There are three basic types of friction: kinetic static and rolling. When you slid the mass blocks around in physics, you were learning about kinetic and static friction, and all that F=µN stuff is true.

What rob is talking about is rolling friction. All forms of friction ARE dependent of surface area, but in kinetic and static, only very slightly. The force dependent on area is several orders of magnitude weaker than the µN force, so the formula is Ff = µN + really really small number. Rolling friction is almost entirely dependent on the weak forces. This is because the tire is technically not moving against the road. If it was, it would be skidding. Watch a wheel go for a bit – the point in contact with the road stays on the same bit of road until it get lifted off of it. So when your tires start spinning, you lose the effect of having the extra area.

I will go look in my new physics C text tonight. Maybe i can clear it up. And if i am dead wrong, which i probably am, i apologize. But i do hold that you need to consider the difference between kinetic and rolling friction.

Take a look at these sites.

http://www.school-for-champions.com/science/frictionrolling2.htm

http://hyperphysics.phy-astr.gsu.edu/hbase/frict2.html

*Originally posted by M. Krass *
**
If the material on your wheel is not a smooth surface, in that it interacts with the carpet fibers, your traction will increase. There is no set factor that determines how effective this is.

Whereas the smooth surface of a wheelchair wheel propels the robot due to friction, an irregular surface that meshes with the carpet fibers propels the robot due to both friction (the sliding force between the smooth portions of the wheel and the carpet) and torque (the contact between two surfaces perpendicular to the direction of movement)

**

OK - Tomorrow night when my daughter (IBApril180 on this board) is over we will post the results of her science experiment here, which I must admit surprised me. Using Brecoflex belts and FIRST spec carpet, the traction increased by DECREASING the amount of contact area! This appeared to be because the impact of additional deformation of the carpet due to higher local pressure was a higher order effect than the additional quantity of “nubs” interacting with the carpet. It threw our F = muN + nuA hypothesis out the window. Stay tuned.

OK - Tomorrow night when my daughter (IBApril180 on this board) is over we will post the results of her science experiment here, which I must admit surprised me. Using Brecoflex belts and FIRST spec carpet, the traction increased by DECREASING the amount of contact area! This appeared to be because the impact of additional deformation of the carpet due to higher local pressure was a higher order effect than the additional quantity of “nubs” interacting with the carpet. It threw our F = muN + nuA hypothesis out the window. Stay tuned.

That can’t be right. The formula isn’t right but there some correlation between surface area and traction. Ahhh I see if you have two objects with the same pressures the one with more surface area will have greater traction than the one that doesn’t. Which is why tank treads work and you need the bogies in the treads so that there is pressure on the track all the way through.

Originally posted by Gary Dillard *
**OK - Tomorrow night when my daughter (IBApril180 on this board) is over we will post the results of her science experiment here, which I must admit surprised me. Using Brecoflex belts and FIRST spec carpet, the traction increased by DECREASING the amount of contact area! This appeared to be because the impact of additional deformation of the carpet due to higher local pressure was a higher order effect than the additional quantity of “nubs” interacting with the carpet. It threw our F = mu
N + nu*A hypothesis out the window. Stay tuned. **

Oooh. Bring it on

Your daughter can have a white paper on her hands. I’m looking forward to reading that, and it’s just in time, too.

Are you sure that there aren’t characteristics of the belting that may be contributing to that discovery in ways other materials might not? Do the results hold true for the wheelchair wheels, for example?

Last year we did a similar test with neoprene pad material (available from spi). Our test was slightly different in that we had two wheels surfaced with this material and also a large plate surfaced with the same stuff. We were using about 70 lb. of dead weight and some FIRST carpet. Our test indicated that the double wheel (still at 70 lb. of normal force) slightly outperformed the single wheel and the sheet underperformed both. We also tested all three of these at different weights, but I won’t bore you with the details.

My conclusion was that there is an optimum amount of surface area for a corresponding weight for these two materials. Too much surface area and the tread floats on top of the carpet fiber. Too little and you don’t have as much fiber engagement. It also seemed to me that the coefficient of friction verses contact area (at a fixed normal force) curve has a pretty flat top. Here’s my ASCII drawing of what I think it looks like based on my somewhat limited experiments where mu is the coefficient of friction and SA is surface area (again for a constand normal force):

``````
mu
|   ________
|  /        \_____
| /
|/_____________ SA
``````