A 1150 kg race car (assume no negative lift) is about to go around a curve with

Question

A 1150 kg race car (assume no negative lift) is about to go around a curve with radius of curvature 80 m and banked at an angle of 7 degrees. The coefficient of kinetic friction between the tires and the road is 0.48 and the coefficient of static friction between the tires and the road is 0.6.

(a) How fast can the car go around the curve without slipping?

(b) What is the normal force on the car in the curve at this max speed?

(c) What is the magnitude of the force on the car from the road in the curve at this max speed?

(d) While still at this same speed, the car goes over a hill in the road. What's the smallest radius of curvature that would allow the car to not leave the road?

Explanation / Answer

NcosT = mg + FfsinT Ncos(7)= (1150)(9.8) + (.6N)sin(7) N (.99)= 11270 + .073N .917N = 11270 N=12290.1 N (answer for part b) a) NsinT + .6NcosT = ma 12290.1 (.122) + (.6)(12198.5) = 1150a 1499.4 + 7319.1 = 1150a a = 7.67 m/s^2 = (v^2)/r (7.67)(80) = v^2 v = 24.77 m/s (answer) c) Ff = (mu)N = (0.6)(12290.1)^2 = 90627934.81 90627934.81 + (12290.1)^2 = 241674492.8 sqrt(241674492.8) = 15,545.88 N (answer)

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