A car drives around a curve with radius \(410 \mathrm{m}\) at a speed of $32 \mathrm{m} / \mathrm{s} .$ The road is not banked. The mass of the car is \(1400 \mathrm{kg} .\) (a) What is the frictional force on the car? (b) Does the frictional force necessarily have magnitude $\mu_{\mathrm{s}} N ?$ Explain.

Given: Mass of the car, \(m = 1400 \mathrm{kg}\) Radius of the curve, \(r = 410 \mathrm{m}\) Speed of the car, \(v = 32 \mathrm{m/s}\)

Centripetal force is the force needed to keep an object moving in a circular path. The formula for centripetal force is: \(F_c = m\frac{v^2}{r}\) Plug in the given values: \(F_c = 1400 \cdot \frac{32^2}{410}\) Calculate the centripetal force: \(F_c \approx 3626.83 \mathrm{N}\) The centripetal force required to keep the car moving in a circular path is \(3626.83\) N.

Since there is no other force acting on the car in the horizontal direction, the frictional force provides the required centripetal force to keep the car on the circular path. Therefore, the frictional force is equal to the centripetal force: \(F_f = F_c = 3626.83 \mathrm{N}\) The frictional force on the car is \(3626.83\) N.

The maximum static frictional force is equal to \(\mu_{\mathrm{s}} N\), but the frictional force acting on the car doesn't have to be the same as the maximum static frictional force. The frictional force is just enough to supply the necessary centripetal force required to keep the car moving in its circular path. If the car were to drive at a higher speed, the frictional force would have to increase to maintain the motion, but there would be a limit at which the static frictional force reaches its maximum and can no longer supply the necessary force, resulting in the car skidding. So, the frictional force does not necessarily have magnitude \(\mu_{\mathrm{s}} N\). It will be equal to or less than the maximum static frictional force, depending on the speed and the required centripetal force for the car to stay on the curved path.