What is the magnitude of the total force between the bike tire and the road?

Question

A physicist rides his bike around a corner of radius 21 meters at a steady speed of 7.5 m/sec. The combined mass of the physicist and the bike is 98 kg. The coefficient of static friction between the bike and the road is μ_s = 0.37.

Answer 1

Given: r = 21 m

v = 7.5 m/s

m = 98 kg

μ_s = 0.37

centripetal force = F_c = mv² / r = (98 kg) (7.5 m/s)² / (21 m) = 262.5 N

normal force = F_N = mg = (98 kg) (9.8 m/s²) = 960.4 N

friction force = F_f = μ_s (F_{N) =} μ_s mg = .37 (98 kg) (9.8 m/s²) = 355.348 N

F_{x net} = F_f - F_c = 355.348 N - 262.5 N = 92.848 N

F_{y net} = F_N = 960.4 N

F_total = √[(F_{x net})² + (F_{y net})²] = √[(92.848 N)² + (960.4 N)²] = 964.878 N