Does a ski racer with a larger mass have an advantage? Does a ski racer with a greater mass have an advantage over a racer with a lesser mass? 
If mass of one racer is 54 kg and the mass of a more slender racer is 44 kg I know the speed at which they descend should be equal if they were to fall in a vacuum. How does friction force, air resistance and momentum play a part in determining the advantage or disadvantage of larger mass in ski racing?
 A: Viscous drag will tend to favor the heavier person, since surface area scales slower than mass, i.e. $A \propto r^2$, $M \propto r^3$, $ F_D / F_G \propto M^{-1/3}$.  But I would generally expect this effect to be weak except at very high speeds.
Friction would cancel out if you were skiing on ice, however in soft snow the heavier person would tend to sink deeper in the snow and that extra compaction of the snow is going to take away more energy.  Unless the heavier skier has larger skis, i.e. it really would depend on the loading per area of the skis.
A: Stephan is on the right track but there is an additional part of the puzzle that many people ignore and was ignored above.  When you ski or skate you are not skiing on snow, but a very thin layer of water that is created from the pressure caused by your weight on the snow.  Remember, water is less dense than ice and pressure pushes the snow into it's most dense form.  Water has a much lower coefficient of friction than snow or ice so the faster it is created under the ski the lower the force of friction is.  
This explains several things.  First, this is why ski hills get icier as the day goes on, more skiers cause more melting and refreezing into thin layers of ice instead of fluffy snow.  Second, it explains why skiing on a very cold day is nearly impossible because it is too cold to melt the snow under your skis.  Thirdly, this is why you wax your ski, wax and water have one of the lowest coefficient of adhesion therefore less force required to unstick your ski from the thin layer of water that could otherwise act as a large suction cup.  Lastly it explains why there is an optimal length that downhill skiers can use for their weight because the more they spread their weight the harder it is to melt the snow.  The heavier you are the easier it is to melt the snow under ski and the larger the ski underside that is dominated by the lower coefficient of friction between water and the wax.  
As for the other portion of the problem it is best explained as arrested terminal velocity controlled by the ratio of mass to surface area.  Easily put, surface area is calculated as meters squared while mass is based on volume or meters cubed.  Increasing one dimension increases mass over surface area exponentially, therefore any slight increase in mass greatly outweighs the increase in surface area.  There's a famous quote from Haldane about this quandry "To the mouse and any smaller animal it presents practically no dangers. You can drop a mouse down a thousand-yard mine shaft; and, on arriving at the bottom, it gets a slight shock and walks away, provided that the ground is fairly soft. A rat is killed, a man is broken, a horse splashes."  SO... the larger you are the more you splash.
A: Not exactly. Friction makes no difference, as it is a force proportional to mass ($\mu_k mg\cos\theta$). Since $F=ma$, the $m$ cancels off and we get that the effect of friction on acceleration is constant for different masses. The same goes for gravity. Recall Galileo's famous drop-two-balls-from-the-leaning-tower experiment. 
There are only two forces left. Both depend upon the size and shape of the body and not the mass. Since we have to divide by mass to get acceleration, out of two people of the same shape, the heavier one gets the benefit. This can be said to be due to momentum/inertia. The two forces are buoyant force and viscous drag. Buoyant force is directly proportional to volume, and is pretty negligible. Viscous drag depends upon the size and shape and lots of tiny things. Generally, aheavier person will be broader as well, so this one can go either way.
All in all, it really doesn't matter what a skier's mass is, except for overweight people (too much drag I would think). The technique matters much more.
A: Experience and my rudimentary understanding of physics lead me to an answer different from the previous answers.
Any skier knows that heavier skiers tend to go faster downhill. If you ski you know that this is a big, obvious effect. Perhaps even clearer is how much one speeds if you crouch into the downhill tuck position. SO, drag thru the air is a big phenomenon. The answers  that 'doubt this is important' must not be skiers!
I believe that the reason is the drag (resistance) to falling through a gas or fluid (i.e., air) is proportional to the cross section area of the forward moving object, so you have the force of gravity, $$F= m*g-F_d$$ minus drag $F_d$. Assuming a spherical skier, gravitational force this is proportional to $r^3*g$. It is opposed by drag, which is not proportional to m but area, i.e. to $r^2$*(x = whatever else counts; air density etc) for the spherical skier. So the net force of producing acceleration downhill is proportional to $r^3*g - r^2*x$. Obviously as r increases, $r^3$ increases much faster than $r^2$, so the net force is more for larger $r$, i.e., the heavier skier. 
Al this changes somewhat if turning is considered, of course. 
That being said, I agree that ski racing depends on skill, strength, reflexes and courage more than on weight. These are the reasons Lindsey Vonn is the greatest downhill racer since Anne-Marie Moser-Pröll and Franz Klammer.
A: first identify the forces on a downhill skier:
$$m  a = m g \cos \left( \theta \right) − \mu m g \sin \left( \theta \right) − \frac{1}{2} \rho A v^{2} C_{d}$$
net force = force gravity - force friction - force air resistance
and if we solve for acceleration:
$$ a = \frac{m g \cos \left( \theta \right)}{m} − \frac{\mu m g \sin \left( \theta \right)}{m} − \frac{1}{2} \frac{\rho A v^{2} C_{d}}{m}$$
and we find that the mass cancels out on the first two forces so they become irrelvant when considering the mass of the skier.  The only variable that matters is the force of the drag.  Since all are constants, one only needs to consider the ratio of $\frac{A}{m}$ in the last equation.  $A$ is the silhouette or frontal area and $m$ is the mass of the skier.  Are they related?
The relationship of total body surface area to mass has been studied here!
Here is a graph of that relationship: (note the formula at bottom of graph)

A possible criticism will be this is a graph of total body surface area to weight (or mass) but we know that total body surface area is proportional to the silhouette and that reference can be found here!
With that concept, we can intuitively look at the graph and note that the slope of the graph exceeds $1/50$ on the left side of the graph and is less than $1/50$ on the right side of the graph.  Thus, there is greater drag per unit mass for lower mass because the differential of $\frac{A}{m}$ exceeds $1/50$ and there is less drag per unit mass when the differential of $\frac{A}{m}$ is less than $1/50$.  In summary, the smaller person has more drag per unit mass yielding less acceleration than a larger person.
You were given the equation at bottom of the graph: $A = 0.1173  m^{0.6466}$
and if we differentiate $A$ with respect to $M$,
$\frac{dA}{dm} = 0.1173 \cdot 0.6466  m^{-0.3534}$
This makes it obvious that as the change in mass increases, the change in area will not be as large.
I hope this helps to clarify,
dfdies
