Answer:
a) 10.7° ≈ 11°
b) 0.19
Explanation:
If the road is banked at an angle, without seeking the help of friction, (i.e. frictionless road), the forces acting on the car are shown in the attached free body diagram to the question
In the y - direction
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 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 frictional force on the road has to balance the force keeping the car in circular motion
That is,
Fr = (mv²/r)
Fr = μN = μ mg
μ mg = mv²/r
μ = (v²/gr) = (16.667²/(9.8×150)) = 0.19
Hope this Helps!!!
