Question

As we saw in the chapter , wings on race cars push them into the track. The increased normal force makes large friction forces possible. At one formula one racetracks, cars turn around a half circle with diameter 190 m at 68 m/s. For a 610 kg vehicle, the approx. minimum static friction force is complete this turn is A 6000 N

B 15,000 N
C 18,000 N
D 24,000 N
E 30,000 N

Answer 1

The minimum static friction force required for a car to complete a turn without slipping can be calculated by determining the centripetal force exerted on the car and finding the frictional force that provides the necessary centripetal force.

Step-by-step explanation:

In order to determine the minimum static friction force required for a car to complete a turn without slipping, we need to calculate the centripetal force exerted on the car and then find the frictional force that provides the necessary centripetal force. The centripetal force is given by Fc = m × ω² × r, where m is the mass of the car, ω is the angular velocity, and r is the radius of the turn. In this case, the mass of the car is 610 kg, the speed is 68 m/s, and the radius is 190 m.

First, calculate the angular velocity ω = v / r = 68 m/s / 190 m = 0.358 rad/s. Then, calculate the centripetal force Fc = 610 kg × (0.358 rad/s)² × 190 m = 22,824 N. Since the frictional force provides the centripetal force, the minimum static friction force required to complete the turn is approximately 22,824 N.