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Mass and Downhill Runs
Name: Eric
Status: other
Grade: other
Location: OR
Country: USA
Date: April 6, 2011
Question:
My friend and I are coasting down an incline on nice road bikes. He weighs 190 and I'm AT 150. Why does he out coast me even switching bikes? I thought objects accelerated the same under gravitational pull regardless of weight .
Replies:
The reason is that while the gravitational force is proportional to your mass, the air resistance is not. Air resistance is a very significant force for bicyclists traveling with an airspeed over about 7 mi/h. Basically, when you are coasting downhill at a constant speed, the air resistance pushing back on you is exactly as strong as the downhill component of your weight pushing you forward. Since your friend has more weight, he can go faster (giving a greater drag force) before his coasting speed maxes out.
Richard E. Barrans Jr., Ph.D., M.Ed.
Department of Physics and Astronomy
University of Wyoming
The drag and friction your friend experiences are smaller
fractions of the net force pulling him downhill. The drag force
on your friend is relatively smaller than yours because it is
proportional to area, while mass is proportional to volume.
Before we start, let us simplify a little: your mass is 150/g, where
g is the acceleration due to gravity. I am going to drop the g in what
follows, because we only need ratios of forces to solve the problem.
Also, the force pulling you downhill is 150*sin(angle of incline).
I am going to choose an incline of around 6 degrees, where
sin(angle)=0.1. At this angle, the force pulling you downhill is
150*0.1=15 pounds; the force pulling your friend is 19 pounds.
Friction is simpler, and it illustrates one aspect of the problem, so
let's start with that. Your friend squashes the tires only a little
more than you do, so the frictional force working against him is almost
the same as yours. For simplicity, let us say it is 1 pound of force for
you both. His net acceleration (force/mass) is proportional to
(19-1)/190=.0947; yours is proportional to (15-1)/150=.0933. The ratio
is 1.015; you lose by around 1.5%.
Drag is a larger effect, and the difference increases with speed.
Your friend is wider, taller, and thicker than you, and all three
factors increase the force pulling him downhill. But drag depends
on the area he presents to the wind, so only his width and height
figure into the force opposing his downhill motion.
If y'all were spheres, he would have a radius proportional to 190^(1/3)
(mass is proportional to radius cubed), while your radius would be
proportional to 150^(1/3). His drag force would be proportional to
his radius squared: 190^(2/3)=33; yours would be proportional to
150^(2/3)=28. His drag acceleration would then be proportional to
33/190=.174, yours to 28/150=.187. To compare net forces, we have to
choose a speed. At some speed, your drag force will equal the 15 pound
force pulling you downhill, and that will be your top speed, because the
net force acting on you will then be zero. At the same speed, his drag
force would be 15*(33/28)=17.7, smaller than the force pulling him
downhill. He will continue to accelerate until his drag force is 19 pounds,
while yo wi'll continue moving at constant speed.
--
Tim Mooney
I think you are confused about the force of gravity (F):
F = G m x M/R^2 where m and M are the masses of the object and the
Earth respectively.
So the force DOES depend on the mass of the body (your friend or you).
Having said that I suspect the issues are more complicated. The weight
(mass) of the rider changes the profile of the tire against the ground, once
the motion has started, the pressure on the wheel bearings will be
different, and probably other factors as well. So we probably cannot fault
Newton without other experimental controls.
Vince Calder
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