Do Amonton's laws apply to rubber?
Nothing I have written has caused more distress than the claim that friction is independent of apparent contact area. Many people claim that they can't understand how that could be true, which is true but irrelevant; friction is independent of apparent contact area even though they can't understand how that could be true. Some people who are slightly more informed about friction will remember from their high-school or college physics courses that friction is independent of apparent contact area but assert that that is true of bricks on boards, or more generally of solid objects, but not true of rubber. I've been so informed so many times, and often so obnoxiously, that I duplicated the high-school physics experiment to see if there could possibly be anything to it.The Setup
Find a board, find a nice smooth brick (I don't want raised or depressed lettering to affect the results), find a rubber inner tube. Cut the tube to get sheets of rubber that you can glue to the large and medium sides of the brick. My inner tube was smooth on the inside but textured on the outside so I glued it so that the smooth side was exposed. It also had some powder on the inside (to prevent it sticking to itself I presume) which I had to wash off with soap and hot water.The brick weighed 4 pounds 10 ounces (4.625 pounds). The large side was 28 square inches, the smaller side 17 square inches, about 61% of the large side.
Rather than pull the rubber-coated brick with a scale, I put the brick on the board and tipped the board, measuring the height of the free end. Obviously the greater the angle before the brick slides, the greater the friction. Using this technique also allowed me to compute the coefficient of friction, since the tangent of the angle where slippage starts is the coefficient of friction (as derived here).
The Result?
F=μN. The height to which I raised the free end did not change whether the brick was on its large side or its smaller side. Within the precision of the experiment, the coefficient of friction is independent of surface area (and is about .55 for that rubber on that board).Objections?
Several are possible. One is that the height to which I raised the end of the board varied by more than 5% over many trials. I did notice that any tremor in my hand would cause the brick to slide prematurely. This effect was similar over many trials on both the large and small sides. A more carefully-designed experiment would undertake to eliminate this factor and might be able to detect a smaller difference in slip angle.Objection: Inner-tube rubber??!! We don't ride on inner-tube rubber! Tire rubber is a lot harder and could have a much different result. Answer: You are welcome to find or make some tire rubber which is flat so that the experiment can be done; or change the experiment — perhaps with a scale pulling the weighted rubber on a horizontal surface — so that a curved piece of rubber can be used. Let me know your results. Meanwhile, perhaps you could think about this: You assert that rubber isn't a solid and isn't described by Amonton's laws as solids are. So why should a harder, more solid piece of tire rubber act less like a solid than softer rubber, which does act like a solid?
Objection: We don't ride on boards. You'd get a different result if you used actual pavement. Answer: Could be. You are welcome to do the experiment with a piece of roadway. Course, friction of other materials is described by the same laws whether the material is sliding on boards or pavement. But rubber could be different. Let me know your results.
Objection: Your 4.625-pound brick has a loading of .17 pounds per square inch on the large side, and .27 pounds per square inch on the smaller side. That isn't realistic for motorcycle tires on pavement. The results might be much different with a more realistic loading. Answer: True. You are welcome to do the experiment with a load more satisfactory to you. Let me know your results.
Objection: If I accept your word for the results of your experiment, then it is true that rubber friction is largely unaffected by apparent contact area under the conditions you describe. But "largely unaffected" just means that your experiment has insufficient precision to detect any dependence that might exist, less than say pure proportionality, where doubling the surface area would double the friction. Answer: True. If for instance friction were proportional to the square root of the apparent contact area, then doubling the friction would require quadrupling the contact area. If it's the cube root of the area then doubling friction would require multiplying the area by 8. And so on. My experiment could not possibly detect such dependence. But I doubt that it would be possible to detect such dependence with real tires on real motorcycles on real roads either. I changed the contact area by 40% and saw no change in friction. How much more "independent of area" do you need?
You're neglecting load sensitivity
Tire load sensitivity is a very interesting topic. I have assumed, as Amonton's Law does, that the coefficient of friction is a constant over a wide range of speeds and loads. More sensitive tests with rubber reveal that μ is not constant with increasing load; it declines slightly as the normal force increases. That is, if you double the load, you will not double the friction. Friction will increase, but not quite proportionally. I have seen estimates that F will increase as somewhere between N^.7 and N^.9. I do not know the provenance of those estimates, but if they're accurate, then doubling a load will result in friction increasing by a factor of 1.6 to 1.9, rather than a factor of 2.Aha, several people have told me! That's why wider tires provide better traction. If you add more contact area, the load on the tire will decrease, and due to tire load sensitivity, μ will increase, with resulting better traction!
Alas, the definition of load has subtly changed in that analysis. If my Bandit is designed so that the two tires share the weight equally, then as I ride the load on the two tires is about 325 pounds each. If I change to tires with twice the contact area, then the load on the two tires will be ... about 325 pounds each.
The pressure will have dropped, halved in fact. But pressure does not appear in the friction equation. The N in F=μN is a force, measured in Newtons, not a pressure, measured in Pascals (Newtons per area). Changing the tires doesn't change the weight of the bike (except for the difference in weight of the tires). Thus it does not change the traction either.