Quote:
Originally Posted by dfox
in saying mass times "normal force", i will admit i was incorrect, but only in semantics, it is the normal acceleration. f is always equal to ma. i initially wrote f=coefficient time mass time gravity, however that's not always true, especially if you're on a banked turn, the acceleration actually has two parts, and will be higher then the acceleration due to gravity.
in nothing else stated do I agree with, and chances are, I'll never change your mind.
F(friction)=Mu(coefficient of friction) x m (mass of object) a (acceleration of object normal (perpendicular) to the force of friction). You show me in that formula, where area is. The mass of the object cannot change. The acceleration of the object is typically the force of gravity, however as I mentioned above it can be slightly larger if you're on a banked road. I see no area in that calculation.
As was pointed out, it is very true that F(force) = p (pressure) times a (area). The problem comes when you assume that the force changes due to area. When the area goes down, the pressure goes up. What are the units of pressure? pounds per square inch is one unit. If you have a given force, a reduction in contact area proportionally increases the pressure. If you as a person, lay on a scale, is your weight any different than if you're standing? no. does the pressure you exert on each square inch of the scale change? yes.
does contact patch help deal with inconsistencies in the road better? yes. does it increase the frictional force and allow you to turn harder? NO!
my credentials? a professional engineer who has taken a significant amount of static and dynamic physics classes at the undergrad and graduate level. now, you want to start talking about the coefficient of kinetic friction, THAT gets complicated, because now you're starting to talk about contact area making a difference, but at the microscopic level.
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Hey guys, just wanted to jump in here... dfox is quite CORRECT about the physics of this situation. The static friction between the motorcycle's tires and the road will vary ONLY with: a) the coefficient of static friction between the tires and the road, and b) the normal force acting on the motorcycle.
The contact patch does nothing DIRECTLY for your friction (grip) between the road and the bike. In fact, the contact patch is simply a direct visual representation of the combination of tire material, tire pressure, bike mass, and temperature.
The reason that higher performance bikes and cars have wider tires is to spread this contact area of the tires over a larger area of pavement, to increase the probability that overall grip will be maintained.
For example: If you have tires that are exactly 3" wide, and you were to run over a patch of ice that was exactly 3" wide... you're a lot more likely to lose grip than if your tires were 5" wide, because those extra 2" are able to provide traction on non-iced ground. It's a similar concept.
Wider tires are also generally made of a softer tire compound (for a better coefficient of friction), and they consequently require a wider tire footprint to distribute the bike's weight without collapsing.
Also... wider tires look cooler.
Credentials!: I am also an engineer with lots of boring physics experience.