We know that friction helps in driving a car, but does this mean that a car can move faster on rough surfaces? Since the coefficient of friction is higher on rough surfaces?
[Physics] Friction in driving car
forcesfrictionnewtonian-mechanics
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The problem with this question is that static friction and kinetic friction are not fundamental forces in any way-- they're purely phenomenological names used to explain observed behavior. "Static friction" is a term we use to describe the observed fact that it usually takes more force to set an object into motion than it takes to keep it moving once you've got it started.
So, with that in mind, ask yourself how you could measure the relative sizes of static and kinetic friction. If the coefficient of static friction is greater than the coefficient of kinetic friction, this is an easy thing to do: once you overcome the static friction, the frictional force drops. So, you pull on an object with a force sensor, and measure the maximum force required before it gets moving, then once it's in motion, the frictional force decreases, and you measure how much force you need to apply to maintain a constant velocity.
What would it mean to have kinetic friction be greater than static friction? Well, it would mean that the force required to keep an object in motion would be greater than the force required to start it in motion. Which would require the force to go up at the instant the object started moving. But that doesn't make any sense, experimentally-- what you would see in that case is just that the force would increase up to the level required to keep the object in motion, as if the coefficients of static and kinetic friction were exactly equal.
So, common sense tells us that the coefficient of static friction can never be less than the coefficient of kinetic friction. Having greater kinetic than static friction just doesn't make any sense in terms of the phenomena being described.
(As an aside, the static/kinetic coefficient model is actually pretty lousy. It works as a way to set up problems forcing students to deal with the vector nature of forces, and allows some simple qualitative explanations of observed phenomena, but if you have ever tried to devise a lab doing quantitative measurements of friction, it's a mess.)
I stumbled on a reference which summarizes what I think is the most correct answer to this question. You have several options, but I think we should only use terms that we have a physical reason to write. Here's the reference:
http://usna.edu/Users/physics/schneide/Buick.htm
They use a lot of unnecessary details like time between stopping that we're not interested in. I agree with their graph but not their equation. So here is my equation to explain their graph. I also limited it to flat roads (no hills).
$$F = \frac{P_0}{v} + \mu mg + c \rho A \frac{v^2}{2} $$
From the reference, known values about their car are:
- The weight, $m=1800 kg$
- The front area $A = 3 m^2$
I report these because there is no direct measurement available. I would then use their data to evaluate the three coefficients that determine the relative friction from each thing.
They put a constant term in the transmission factor (the 1/v term). I don't like this, because I want clean mathematics, so I'm bunching that long tail of transmission with the friction coefficient.
$$ \mu (1800 kg) (9.8 m/s^2) = 200 N + 250 N = 450 N \rightarrow \mu = 0.026 $$
$$ \frac{P_0}{15 m/s} = 500 N - 250 N = 250 N \rightarrow P_0 = 3750 W$$
$$ c (3 m^2) (1.3 kg/m^3) \frac{(31 m/s)^2}{2} = 500 N \rightarrow c = 0.27 $$
These are all consistent with what the link claimed, aside from the cases where I willfully used a different kind of definition. Another good thing that these all have physical interpretations, which those units are suggesting. I will avoid getting into the exact interpretation because I feel like there's space for quibbling.
I plotted this on Wolfram alpha. This is my altered version of that link.
This fits expectations fairly well. Take note, however, of the 1/v term. That represents fuel consumption due to constant loads (thus, units of power, of course). That might not be relevant if you're looking for a force, but it can be kind of interpreted as a force. It's a force the engine is exerting against itself (to some fraction of that number) due to idling. It is also the constant electronic loads on the battery... and the charging of the battery itself. It's not the friction of the wheels on the road of air on the car. If you're only interested in those then you might do well to just take the last two terms. If you do that, however, there is no concept of maximum gas mileage, nor should there be. What you're going to do with these terms depends on the application. I just believe this to be the best available option so I posted it.
Best Answer
The coefficient of friction $\mu_s$ might be higher for tires on rough surfaces, yes, but as you said yourself, it is friction $f_s$ that thrusts the car forward.
(And we are talking about static friction throughout, since we are talking about rolling wheels.)
And the coefficient of friction is not equal to friction. High coefficient of friction does not mean high friction.
It only means that there can be high friction. If necessary. It means that if the car starts gripping harder in the asphalt, then the asphalt can hold on. But only if. If the car grips the same amount (if you drive in the same manner) then you get the same friction, no matter what the coefficient of friction is (as long as it is not too small).
Mathematically that is shown in the formula for static friction:
$$f_s\lt \mu_s n$$
The friction $f_s$ does not have to rise, when $\mu_s$ is high. It can if it has too, but it doesn't if it doesn't have to. The coefficient of friction (times the normal force) just determines the maximum. That's all.