Friction as a function of area

[rbayer]

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.

[Gadget470]

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

[Adam Y.]

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.

[Suneet]

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

[SlamminSammy]

Originally posted by Gadget470
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=u*N*A*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 .

[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.


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.

[Adam Y.]

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

[Madison]

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.

[Suneet]

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.

[Gadget470]

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

In Black: The wheel's treads
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.

[rbayer]

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?

[Madison]

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.

[Adam Y.]

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?

[Madison]

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.

[Adam Y.]

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.