physics problem

Originally posted by PBE6
Here's my solution, angie:

First draw a free body diagram for the cyclists on a slope. There are two forces present, the force of gravity and the normal force. We know that the force of gravity acts straight down, but the normal force acts perpendicular to the slope surface. These two forces sum to give a resultant force on the cyclist parallel to the slo ...[text shortened]... frictionless conditions is a slope of 12.0 degrees.

Hope that's right, good luck angie!

Very excellent.

I reduced the equations a little different but got the same answer.


F(g) = mg = N cos(theta)
F(r) = mv(squared)/r = N sin(theta)

N sin(theta) / N cos(theta) = tan (theta) = v(square)/rg

tan (theta) = (36 km/hr)squared / (50 m)(9.8m/sec2)

*convert seconds to hours to cancel, convert km to meters to cancel.

Theta = 11.5 degrees

Originally posted by Doctor Rat
It's still possible, even in the case of a frictionless track. Imagine the case of a tilted track and a frictionless surface. Now you magically place a cyclist and his bike onto the track with giant invisible genie fingers (or Green Lanter's ring, etc), you place the cyclist aligned perpendicular to the surface of the track (neither leaning left nor right ...[text shortened]... onless, angled track, you will find a point of equilibrium wh

I have said too much already.

No. to find the optimal angle you need the lowest coefficient of static friction that will occur which is why rubberjaw30 asked how slippery the track might be