A circular curve of highway is designed for traffic moving at 60 km/h. Assume the traffic consists of cars without negative lift. (a) If the radius of the curve is 150 m, what is the correct angle of banking of the road? (b) If the curve were not banked, what would be the minimum coefficient of friction between tires and road that would keep traffic from skidding out of the turn when traveling at 60 km/h?

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!!!

one person pulls 90n one way and the other has 100n pulling the other way.

so the net force = 90 - 100 = -10n,

or the ans can be 10n if the picture shows -90n and 100n instead.