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

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.

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

To find the centripetal force, you need to use the relationship between the centripetal force, the mass of the stone, and the radius of the tire.

The centripetal force acting on an object moving in a circular path is given by the equation:

Fc = (m * v^2) / r

Where:
- Fc is the centripetal force
- m is the mass of the stone
- v is the velocity of the stone
- r is the radius of the tire

In this case, you are given the mass of the stone as 3.0 × 10^-3 kg and the velocity of the tire surface as 10 m/s.

To find the centripetal force, substitute the known values into the equation:
Fc = (0.003 kg * (10 m/s)^2) / r

Now, you need to determine the value of the centripetal force when the stone flies out of the tread. Since the stone loses contact with the tread, the static friction no longer provides the centripetal force. Instead, the stone moves in a straight line tangent to the circular path of the tire.

At the point of release, the centripetal force is zero. Therefore, you can set Fc to zero in the equation and solve for r:

0 = (0.003 kg * (10 m/s)^2) / r

Now, rearrange the equation to solve for r:

r = (0.003 kg * (10 m/s)^2) / 0

However, dividing by zero is undefined. This means that there is no reliable information provided to determine the radius of the tire. The question or data might be incomplete or inaccurate, as the lack of information about the centripetal force prevents obtaining a meaningful solution.