The angle of banking of the road is 11° and The minimum coefficient of friction between tires and road is 0.19
The attached free-body diagram to the query illustrates the forces operating on the car if the road is banked at an angle without utilizing friction (i.e., frictionless road).
mg = N cos θ (Eqn 1)
mg = weight of the car.
N = normal reaction of the plane on the car
And in the direction parallel to the inclined plane,
(mv²/r) = N sin θ (Eqn 2)
(mv²/r) = force keeping the car in a circular motion
Divide (Eqn 2) by (Eqn 1)
(v²/gr) = Tan θ
v = velocity of car = 60 km/h = 16.667 m/s
g = acceleration due to gravity
r = 150 m
(16.667²/(9.8×150)) = Tan θ
θ = Tan⁻¹ (0.18896)
θ = 10.7° ≈ 11°
b) In the absence of banking, the force maintaining the car in a circular motion must be balanced by the frictional force of the road.
That is,
Fr = (mv²/r)
Fr = μN = μ mg
μ mg = mv²/r
μ = (v²/gr) = (16.667²/(9.8×150)) = 0.19
Know more about the coefficient of friction at:
https://brainly.com/question/19291861
#SPJJ4
