Question

Consider again the problem of a car traveling along a banked turn. Sometimes roads have a "reversed" banking angle. That is, the road is tilted "away" from the center of curvature of the road. If the coefficient of static friction between the tires and the road is μ_s = 0.4, the radius of curvature is 25 m, and the banking angle is 14°, what is the maximum speed at which a car can safely navigate such a turn?

Answer 1

Solution) R = 25 m

Us = 0.4

Angle , theeta = 14°

V = ?

For safe turn

fs = (m(V^2))/(R)

fs = (Us)mg

(Us)mg = (m(V^2))/(R)

V^2 = (Us)(R)(g)

V^2 = (0.4)(25)(9.8)

V^2 = 98

V = (98)^(1/2)

V = 9.89 m/s