Oh, Waverly, I would avoid the concept of adding accelerations as it often tends to confuse people and lead to just these kinds of problems. In fact, I think that was why some of your posts were misunderstood. It is better to just calculate the force due to gravity where g is a conversion factor and only convert back to acceleration when you have a resultant force. My 2 cents from common mistakes I see.
Trunk, I think I can help. I see a point you may have overlooked:
Yes, in a vaccuum, the heavier sphere is pulled downward more strongly than the lighter sphere. However, this doesn’t mean it accelerates more. It has a greater resistance to acceleration due to its inertia. The net result is that the heavy sphere and the light sphere fall at the same rate in a vaccuum, even though the heavier sphere is being pulled more strongly.
heavy sphere in vaccuum:
a) large downward gravitational force
b) large inertia opposing acceleration
light sphere in vacuum:
a) small downward gravitational force
b) small inertia opposing acceleration
In both of these, a+b works out to the same acceleration.
In the atmosphere, those same forces exist: the light sphere is pulled gently by gravity, and the heavy sphere is pulled hard. All other things being equal, they would accelerate the same. However, they also both have an equal drag force pushing upward on them:
heavy sphere in air:
a) large downward gravitational force
b) large inertia opposing acceleration
c) drag force (upward)
light sphere in air:
a) small downward gravitational force
b) small inertia opposing acceleration
c) drag force (upward)
We already know that adding up a+b in both cases results in the same acceleration, so now what happens if we add in the additional force c to each? Well, a=f/m, so the same force produces a different acceleration on bodies of different masses. The more massive body is upwardly accelerated less, so it falls faster.
Man, I need to preview to catch posts that have showed up while I am typing. Hoodoo Ulove, traction is indeed better for heavier riders. The tractin can be simply represented as the coefficient of friction (unchanged for different riders) times the normal force. This is simply the weight of the rider and bike (with some reductions when the rider is accelerating downhill), so the heavier rider gets better traction. Remember, however, that he has more inertia to resist as well, so though the traction is better, the actual resistance to sliding will be similar. It gets compllex when you take all these little things into account.
No fancy degree here but I can tell you that when I was a kid my heavier matchbox cars went down ramps faster than my lighter ones. 
Quoth Padeye:
During the Tour de France, some of my office-mates and I looked up some of Lance Armstrong’s stats. During the uphill mountain parts of the race, he was putting out nearly half a horsepower just in lifting his mass, without regard even for power wasted on friction or drag.
Quoth Princhester:
A constant frustration, in teaching. The first lesson of science, which so many fail to grasp, is that you should observe what actually happens, and what happens is the truth. A balloon falls more slowly than a ball of lead the same size and shape, as anyone who observes knows. If equations and theory tell you otherwise, then either you are misinterpreting the theory, or the theory is wrong.
That was what I meant. Actually, with rubber tires on pavement, the coefficient of friction is a bit better with lower unit load. That’s why dragsters run big tires.
I can’t agree with your point 1 there, and this may be the source of our confusion: this statement either ignores air resistance, or treats acceleration as a vector that can be added with other accelerations to find the final acceleration. You can add forces as vectors because an object can have multiple forces acting on it at the same time, but you can’t do it with acceleration because an object can’t have several different accelerations at the same time.
scr4,
You may want to double check that. An object may have more than one acceleration. In linear motion it’s possible to simplify as per flight’s suggestion above. In non linear motion, you have to consider them.
Yeah ** scr4 ** you are wrong on this. After all an acceleration vector is nothing more than a force vector divided by a scalar (mass).
You can do this if you want to fail your physics exam. The normal force times the coeffecient of friction is the MAXIMUM force due to friction not necessarily the friction force.
Of course treis, but I did not say friction force, I said traction. Though I suppose I could be misunderstood to meant that, I was quantifying the ability of the tire to stick to the road without slippage, which is neatly expressed with the max friction force. This would, course, be using the coefficient of static friction as sliding motion is assumed not to have started yet.
I was differing with Vunderbob’s answers, specifically no. 1. A couple of othe posts appeared while I was writing mine - I really should get into the habit of quoting what I’m replying to!
Doesn’t matter. Even if your big guy and little guy are in identical aerodynamic crouches, the bigger guy will have a larger weight:surface-area ratio, so he will go faster freewheeling down a hill.
Actually, I was just trying to clarify what scr4 was saying.
He made it sound like gravity was accelerating a heavier object faster just because it was heavier.
I think he had done an intermediate step in his head that he was assuming some of use had picked up on.
He might have been applying the mass to what treis called Far, and basically, what you listed as ‘c’, but he was only mentioning what you called ‘a’ and ‘b’.
Or at least that’s how I read it.
treis and zut really clarified it with some equations.
Thanks all.
Why, thanks. Would it be too late to point out that I’ve got a PhD in mechanical engineering? Not in fluids, but still. I also happen to teach a kinematics class, and I’ve assigned a project to computationally determine the motion of falling objects, taking into account the air resistance.
It’s due tonight. 
Oh yeah? Well I… got nothing really to compare to that. Actually I did write a class paper on the aerodynamics of falling water droplets, but that was way back as an undergrad. Man, you think it is hard to calculate drag on a rigid body? Fluid is a bitch!
Concerning the glass ball and cannon ball rolling down a plank:
This reasoning is correct, as far as it goes, but I’d thought I’d illuminate the situation for the TM with a clarifying bit of math:
A uniform disc is simpler than a ball, so we’ll use that; all the conclusions will apply to a sphere as well with some slightly different factors.
Consider a uniform disk at the top of an incline. It has mass M and radius R. At the top, before it moves, it has gravitational potential energy, given by PE = Mgh, where h is the height, and g is the acceleration due to gravity (9.8 m/s^2).
The disk rolls down the plank, and its gravitational potential energy is converted to kinetic energy, both linear and rotational. So at the bottom, it has.
KE = 1/2Mv^2 + 1/2Iw^2
Where I is the moment of inertia ( 1/2MR^2 for a uniform disk) and w is the angular velocity.
For a disk that is spinning with angular velocity w, a point on the edge has tangential (i.e., linear) velocity given by v = wr (you’ll find this in any freshman physics book). If we consider our disk to roll without slipping, this tangential velocity of a point on the edge is equal to the linear velocity of the center of mass. Thus our KE term is:
KE = 1/2Mv^2 + (1/2)(1/2)MR^2w^2
KE = 1/2M(Rw)^2 + (1/2)(1/2)MR^2w^2 using v= wr
KE = 3/4 M(v)^2.
Now we began with PE, and wound up with KE. So, in classic physics style,
KE = PE
3/4Mv^2 = Mgh
v^2 = 4/3gh
v = sqrt(4/3gh)
Note that the mass of the disk (ball) does not appear in the expression for the velocity.
This treatment, obviously, ignores effects due to air resistance, which, for reasons explained above, will cause the heavier ball to roll faster.
Don, you are a scholar!
I on the other hand am a mathematical lightweight and a lazy, lazy man…
If this is a hollow glass sphere, the concentration of the mass at the surface gives it a higher polar moment of inertia compared to its mass, and thus it will roll more slowly down the plank. This is neglecting aerodynamic drag, which of course would slow it further compared to the cannonball.
I disagree with Quercus, Treis, and perhaps a few others on the matter of rolling resistance. Treis assumes frictionless bearings and ignores the deformation of the tires. From there on the comparison is interesting but theoretical, because real bearings have friction, and real tires behave differently under heavier load.
Quercus seems to say that the lighter rider would have an advantage in rolling resistance. That cannot be true. Bearings first. Consider that bicycle wheel bearing are quite efficient for their tiny size, but even if they were much bigger, a bearing’s friction increases under load. Consider, too, that the heavier rider nears the safe capacity of a bicycle wheel bearing. Surely, when the load approaches the bearing’s capacity, the friction (not to mention the possibility of failure) increases.
I’m assuming the lighter rider weighs 160 pounds, and the heavier rider, 260.
Now, tires. We cannot assume that a wheel, with tire, is a disc. The wheel is pretty stable, but the tire deforms at the contact patch (the “footprint” actually in contact with the pavement.) If you roll the empty bike along the street, the contact patch is about 1/4 inch wide. With a rider aboard, it’s radically different. At the forward edge of that patch, the sidewalls squish down, and at the trailing edge, they rebound to original shape. That flexing takes energy, and you subtract that from forward-motion energy. (Most of it is expended as heat.)
Your 160 lb. rider has a contact patch that looks, at rest, like a stretched-out football (US). The 260 lb. rider has a longer, fatter contact patch, and the deformation necessary to make it takes more energy than for the lighter rider. The distance between wheel and ground gets perilously close to zero, and the sidewalls get hot from this abuse.
In the overall scheme of things, rolling resistance might not amount to much, but it is different for the two riders, and more for the heavier rider.
But is it proportionately higher? If not, then the benefit is to the big guy.
Agreed that the inclusion of rolling resistance makes this a more complicated problem. My gut feel, although I’m willing to be wrong, is that the rolling resistance would scale with (rider + cycle) weight, close enough.
However, from what I’m able to find, aerodynamic drag at reasonable speeds is much greater than rolling resistance. For example, “In fact, at 8 mph (3.5 meters/second) the aerodynamic drag of a bicycle and rider is greater than the rolling resistance (wheels on the ground). At 20 mph (11 m/s), the aerodynamic drag is more than 80% of the total drag.”
So if you were to, say, graph the speed of two bicycles coasting down a hill from a dead stop, you’d expect the lighter one to pull ahead first (due to rolling resistance) and the heavier one to catch up later (due to aerodynamic drag). Makes for a complicated problem, and depends a lot on your assumptions.