When you take your 1200 kg car out for a spin, you go around a corner of radius 54.5 m with a speed of 16.2 m/s. The coefficient of static friction between the car and the road is 0.95. Assuming your car doesn't skid, what is the force exerted on it by static friction?

Answer 1

F(fr) = k•N =k•m•g =0.95•1200•9.8 =11172 N,
friction force = centripetal force to keep the car on road
mv^2/R =1200•(16.2)^2/54.5 =5778.5 N
Net force = 11172-5778.5 =5393.5N