A 600-kg car is going around a curve with a radius of 120 m that is banked at an angle of 20° with a speed of 24.5 m/s. What is the minimum coefficient of static friction required for the car not to skid?
A) 0.12
B) 0.24
C) 0.36
D) 0.48
E) 0.60

Question

Quizplus · Accepted Answer

The minimum coefficient of static friction required for the car not to skid is 0.12. This is because the force of friction is the only force that is preventing the car from sliding down the banked curve. The force of friction is equal to the coefficient of friction times the normal force. The normal force is the force that is perpendicular to the surface of the road. The force of friction is therefore equal to:$$f = \mu_s n$$where:* f is the force of friction* $\mu_s$ is the coefficient of static friction* n is the normal forceThe normal force is equal to the weight of the car minus the force of gravity that is pulling the car down the banked curve. The force of gravity is equal to:$$F_g = mg\sin	heta$$where:* $F_g$ is the force of gravity* m is the mass of the car* g is the acceleration due to gravity* $	heta$ is the angle of the banked curveThe normal force is therefore equal to:$$n = mg\cos	heta$$The force of friction is therefore equal to:$$f = \mu_s mg\cos	heta$$The car will not skid if the force of friction is greater than or equal to the force of gravity that is pulling the car down the banked curve. The force of gravity is equal to:$$F_g = mg\sin	heta$$The car will therefore not skid if:$$\mu_s mg\cos	heta \ge mg\sin	heta$$Simplifying this equation, we get:$$\mu_s \ge 	an	heta$$Substituting the given values into this equation, we get:$$\mu_s \ge 	an20^\circ = 0.12$$Therefore, the minimum coefficient of static friction required for the car not to skid is 0.12.

A 600-Kg Car Is Going Around a Curve with a Radius