A stone has a mass of 8.0 10-3 kg and is wedged into the tread of an automobile tire, as the drawing shows. The coefficient of static friction between the stone and each side of the tread channel is 0.83. When the tire surface is rotating at 17 m/s, the stone flies out of the tread. The magnitude FN of the normal force that each side of the tread channel exerts on the stone is 1.8 N. Assume that only static friction supplies the centripetal force, and determine the radius r of the tire in meters.
I got 1.55 m but that is wrong. I first found static friction (I got 1.49) then pluggd that in as the centripetal force and got 1.55 for r.
What did I do wrong?
2006-10-16 13:00:19 · 2 answers · asked by Confused 1 in Science & Mathematics ➔ Physics
First, thanx for actually trying to do the problem.
FYI, I get r=0.774 m, so maybe you just missed by a factor of two. I believe the static friction should be twice what you suggest, since each side of the stone feels the force. If you reached in and tried to pull the stone out, it would be harder than if you first lubed up one side of the stone with WD-40, which would result in a single side of friction (assuming you prevented it from rotating while pulling it out).
Hope this helps.
2006-10-16 15:39:45 · answer #1 · answered by SAN 5 · 0⤊ 0⤋
http://en.wikipedia.org/wiki/Coefficient_of_friction
2006-10-17 00:43:07 · answer #2 · answered by Surya M. 3 · 0⤊ 0⤋