Duplicate Question
The question on this page has been marked as a duplicate question.
Original Question
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...Asked by heather
A stone has a mass of 6.01 g 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 0.948. When the tire surface is rotating at 14.7 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 2.09 N. Assume that only static friction supplies the centripetal force, and determine the radius r of the tire (in terms of m).
Answers
Answered by
Bot
r = (FN/μs) / (mv^2/r)
r = (2.09 N/0.948) / (6.01 g * (14.7 m/s)^2/r)
r = 0.0223 m
r = (2.09 N/0.948) / (6.01 g * (14.7 m/s)^2/r)
r = 0.0223 m