A 1125 kg race car (assume no negative lift) is about to go around a curve with

Question

A 1125 kg race car (assume no negative lift) is about to go around a curve with radius of curvature 70 m and banked at an angle of 8 degrees. The coefficient of kinetic friction between the tires and the road is 0.43 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?

(a) max speed allowed on curve: m/s
(b) normal force on curve: N
(c) magnitude of force from road on curve: N
(d) least possible radius of curvature of hill: m

Explanation / Answer

Greetings...

For part (a):
The maximum allowed speed on curve = sqrt[rg(Sin+sCos)/(Cos-sSin)

Where: s = static friction = 0.43

Speed (max)=sqrt[70*9.8(Sin8o+0.43*Cos8o)/(Cos8o-0.43*Sin8o)

Speed(max)= sqrt[ 686* (0.565 )/( 0.9304)] =20.4 m/s

For part(b):

Normal Force= m*v2/r(Sin+sCos)= 1125*(20.4)2/70*(Sin8o+0.43*Cos8o)=11837.7 N

I'll try to solve other parts, if I figure out how to solve it.

Best Regards!

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