Let the radius oh a circular segment be 110 m, the mass of the car 1508 kg, and the coefficient of the static friction between the road and tire 0.9. Find the magnitude of the normal force N which the road evers on the car at the optional speed (the spurred at which the frictional force is zero) of 85 km/h. Answer in units of N

To find the magnitude of the normal force, we need to consider the forces acting on the car at the optional speed when the frictional force is zero.

First, we need to convert the speed from km/h to m/s. Since 1 km/h is equal to 1000 m/3600 s, we can calculate:

85 km/h = (85 * 1000 m) / (3600 s) ≈ 23.61 m/s

When the frictional force is zero, the centripetal force Fc is equal to the product of the car's mass m and the centripetal acceleration ac:

Fc = m * ac

Given that the centripetal acceleration is given by ac = v^2 / r, where v is the speed and r is the radius of the circular segment:

ac = (23.61 m/s)^2 / 110 m ≈ 5.09 m/s^2

Now, we can calculate the centripetal force Fc:

Fc = (1508 kg) * (5.09 m/s^2) ≈ 7672.72 N

Since the centripetal force is provided by the normal force N, we have:

Fc = N

Therefore, the magnitude of the normal force N is approximately 7672.72 N.