A stone has a mass of 3.0 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 0.86. When the tire surface is rotating at 10 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.
well so far i THINK that since the rock is stuck between a wedge, the Fn would be 3.6, and since Fs = .86(3.6) = 3.096. And I see that since Fc = (mv^2)/ r, and this would mean that r = (mv^2) / Fc. The mass is .003 kg, and v = 10m/s.. Im having trouble finding Fc
3 answers
well apparently fc = fs, and i got fs as 3.096 as correct...but then that means that r = (.003 * 100)/ 3.096 which is .096889m..this doesn't make sense
One could approach it logically,
the centripetal force=force friction
m v^2 /r = 2*Fn*mu
solve for r
Now notice the 2 on the right. That is because you have friction working form Fn holding each side.
The solution for r indicates it is about 5 cm, a very small wheel for an automobile.
the centripetal force=force friction
m v^2 /r = 2*Fn*mu
solve for r
Now notice the 2 on the right. That is because you have friction working form Fn holding each side.
The solution for r indicates it is about 5 cm, a very small wheel for an automobile.
Actually the answer i put up was correct. I was just shocked because the wheel seemed to be so small..but i guess the numbers made it true