Department of Energy Argonne National Laboratory Office of Science NEWTON's Homepage NEWTON's Homepage
NEWTON, Ask A Scientist!
NEWTON Home Page NEWTON Teachers Visit Our Archives Ask A Question How To Ask A Question Question of the Week Our Expert Scientists Volunteer at NEWTON! Frequently Asked Questions Referencing NEWTON About NEWTON About Ask A Scientist Education At Argonne Mass and Downhill Runs

Name: Eric
Status: other
Grade: other
Location: OR
Country: USA
Date: April 6, 2011

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 .

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

Click here to return to the Engineering Archives

NEWTON is an electronic community for Science, Math, and Computer Science K-12 Educators, sponsored and operated by Argonne National Laboratory's Educational Programs, Andrew Skipor, Ph.D., Head of Educational Programs.

For assistance with NEWTON contact a System Operator (, or at Argonne's Educational Programs

Educational Programs
Building 360
9700 S. Cass Ave.
Argonne, Illinois
60439-4845, USA
Update: June 2012
Weclome To Newton

Argonne National Laboratory