I did some calculations and discovered that a person’s weight is irrelevant- and that the only two factors needed are the angle of the hill and the coefficient of kinetic friction (which, for simplicity’s sake, I assume is generally constant between snow and most sleds). I didn’t use any numbers- rather, I just wrote a formula to calculate acceleration as a function of their weight, and the weight ended up cancelling out. Any help?
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Weight is NOT a factor in the attraction between the lower point and a higher point (gravity); however weight, even when sledding, does affect the terminal velocity (discovered by the ancient Greeks).
However, well done! We all have to confirm what we know to make it our own idea!
Yes, that makes sense. A heavier person would be more affected by gravity than a lighter person, but they also have more inertia (resistance to a change in speed). The two end up cancelling out, and the lighter and heavier person will go down the hill at the same rate.
For more information about this, watch the short video at this link: http://www.youtube.com/watch?v=_mCC-68Ly…
One possible factor that I can think of to throw off the balance is that the heavier persons sled would push more forcefully into the ground. This could increase the friction and slow them down, letting the lighter person go ahead.
You are correct.
I can tell you from years of making pinewood derby cars that the ones that are heaviest start down the ramp slower but develop more kinetic energy which helps maintain the speed longer on the level part of the track (so they win). However, that involves a ramp and a level part afterwards and it is rolling friction not sliding friction so might not be exactly the same situation. I suppose, with sledding, if it was a constant slope (with no level parts) a lighter person might go faster if the sled runners didn’t dig so far into the snow (less resistance). In as much as it is winter time, maybe try to prove your theory (you’ll need the help of your skinniest & fattest friends).
No it does not. You are correct. So the 4 man luge will come down just a fast as the single man sled, all other things equal.
In fact the acceleration down a slope with incline theta degrees and coefficient of friction k can be shown to be a = g sin(theta) – kg cos(theta) = g(sin(theta) – k cos(theta)). As you can see, there is no weight W = Mg anywhere in this relationship. And here’s why.
Multiply both sides by M, the sledder, and we have the net force along the slope as f = Ma = Mg (sin(theta) – k cos(theta)) = W sin(theta) – k W cos(theta) = w – kN; where w is the weight component down the slope and kN is retarding friction force where N is the normal weight. So the a = equation comes from the balance of forces and the fact that the mass, M, cancels out of every term. QED.