Here goes.....
1) Racing tires (tyres?) are not only mechanically stuck on the track, but
chemically, as well. Those little clumps that Indycars leave all
over Lakeshore Blvd. are REALLY STICKY, like chewing gum. Friction is
never independent of mass (or more properly, weight). Slide a book across
a table. Now have Murray Walker sit on it. Which is harder to move??
2) Friction's force is a product of a friction coefficient, cleverly called
'the coefficient of friction', and has two values, static and kinetic.
static is higher, always, and is effective whenever an object is at rest
and a force is trying to move it. Since the velocity of a tire is zero
(at the bottom of the tire, where it touches the ground- else it would
spin, like a burnout), static applies. Since kinetic is less, a spinning
tire is easier to keep spinning than it is to start the thing spinning.
That out of the way, friction is governed by the equation F=uN.
F is frictional force, u is the coefficient (called 'mu') and N is the
normal force (force perpendicular to the plane of motion). Note that
both F and N are vectors, and F is parallel to the plane of motion.
In this case, N is the WEIGHT (not mass) of the object, let's
call it 1/4 of the car's weight for simplicity. Now, 1/4 of the weight will
give F=u*(mass of car/4)*g, where g is gravity. If mass is increased, say
due to Prost having a nosejob OR running with a full fuel load, than the
normal force will be larger by 1/4 of the larger amount, and thus higher
friction will result. Thus (whew!) friction is NEVER independent of weight
(assuming fixed 'g', it's mass dependent also) SO.... the heavier car has
better grip through corners, all other things being equal. It might have a
greater tendency to slide off, even with more frictional adhesion. This
is covered in Chapter 2, Inertia and the Chicane.
3) Using your a=F/m arguement as it applies to friction...here goes...
a=F/m F=uN
a=uN/m (replace F with uN, as the friction eqn. allows)
a=umg/m (replace normal force (N) with m*g, the weight)
a=g (acceleration in this case is gravity)
g=umg/m
1=u (you have now proven that all friction coeff's are =1)!
Obviously incorrect, as I once found out on a dynamics exam.
4) The friction caused by the tires is the maximum limit of force that they
can exert on the track. This is (hopefully) never exceeded (are you
listening, Gerhard?!??), since the tires spin if it is. As far as a heavier
car accelerating slower- the force the car can impart to the track is more or
less constant during a race- so F=ma here, but in a different direction
(accelerating forward) and with the force constant, and a higher
mass (fuel, or maybe Nigel's donut) the acceleration will drop. There
are more issues involving kinetic energy and inertia, but I'll leave that
for another physics student....
Sigh. Time for coffee.
Jeff