A 905 kg car moves in a circle of radius 13.1 meters in a flat parking lot. If the car can travel at 9.1 m/s without slipping, what is the force of friction between each tire and the parking lot? What is the coefficient of friction between each tire and the lot? (Remember that there are four tires on a car)

Question

A 905 kg car moves in a circle of radius 13.1 meters in a flat parking lot. If the car can travel at 9.1 m/s without slipping, what is the force of friction between each tire and the parking lot? What is the coefficient of friction between each tire and the lot? (Remember that there are four tires on a car)

Answer 1

do it for 1/4 of mass on one tire

F = m v^2/r = mu m g
where m = 905/4

Answer 2

To find the force of friction between each tire and the parking lot, we can use the centripetal force formula:

Fc = (m * v^2) / r

where Fc is the centripetal force, m is the mass of the car, v is the velocity of the car, and r is the radius of the circle.

Given:
m = 905 kg
v = 9.1 m/s
r = 13.1 m

Let's calculate Fc:

Fc = (905 kg * (9.1 m/s)^2) / 13.1 m

Fc = (905 kg * 82.81 m^2/s^2) / 13.1 m

Fc = 57770.05 kg m/s^2

Now, since there are four tires on a car, the total force of friction on the car is distributed equally between all the tires. Therefore, we need to divide the centripetal force by the number of tires:

Force of friction per tire = Fc / 4

Force of friction per tire = 57770.05 kg m/s^2 / 4

Force of friction per tire = 14442.51 kg m/s^2

So, the force of friction between each tire and the parking lot is approximately 14442.51 kg m/s^2.

To find the coefficient of friction, we can use the formula:

Coefficient of friction = Force of friction / Normal force

However, we need the normal force between each tire and the parking lot. In this case, the normal force is equal to the weight of the car since there is no vertical acceleration.

Weight = m * g

where g is the acceleration due to gravity (approximately 9.8 m/s^2).

Weight = 905 kg * 9.8 m/s^2

Weight = 8869 kg m/s^2

Now we can calculate the coefficient of friction:

Coefficient of friction = Force of friction per tire / Weight

Coefficient of friction = 14442.51 kg m/s^2 / 8869 kg m/s^2

Coefficient of friction = 1.63

Therefore, the force of friction between each tire and the parking lot is approximately 14442.51 kg m/s^2, and the coefficient of friction between each tire and the lot is approximately 1.63.