A banked circular highway curve is designed for traffic moving at 63 km/h. The radius of the curve is 218 m. Traffic is moving along the highway at 42 km/h on a rainy day. What is the minimum coefficient of friction between tires and road that will allow cars to take the turn without sliding off the road? (Assume the cars do not have negative lift.)

This is a case of a curve with an angle of inclination, cant, to help the cars to take which, in a free body diagram, see that the normal is the force that we must decompose.

Let's start by reducing the speeds to the SI system

v1 = 63 km / h (1000m / 1km) (1h / 3600s) = 17.5 m / s

v2 = 42 km / h (1000m / 1km) (1h / 3600s) = 11.67 m / s

Let us write Newton's second law, for this type of case the located references the horizontal X axis and the vertical Y axis, so the only force that we must decompose is the Normal (N)

Axis y

Ny -W = 0

Ny = W

X axis

Nx = m a

a = v² / R

Let's use trigonometry to break down the force

sin θ = Nx / N

Nx = N sin θ

cos θ = Ny / N

Ny = N cos θ

We replace and calculate the angle of the cant

N cos θ = W = mg

N = mg / cos θ

N sin θ = m v² / R

(mg / cos θ) sin θ = m v² / R

tan θ = v² / gR (1)

θ = tan-1 (v² / gR)

θ = tan-1 (17.5² / 9.8 218)

θ = tan-1 (0.143)

θ = 8.16º

This is the angle of the curve that is constant

If the car goes this speed the friction force is zero, for different speeds this force begins to appear, in the second case the car goes slower, so the normal component would take it towards the center of the curve, whereby a force twelve appears that opposes this movement, consequently, the force of friction and directs towards the outside of the curve

We write Newton's law

Nx - fr = m a

The expression for force rubbing is

fr = μ N

N sin θ - μ N = mv² / R

N (sin θ - μ) = m v² / R

(mg / cos θ) (sin θ -μ) = (m v² / R)

sin θ - μ = v² cos θ / gR

μ = sin θ - v² cos θ / gR

Calculate

μ = sin 8.16 - 11.67²2 cos 8.16 / 9.8 218

μ = 0.1419 - 0.0631 .

μ = 0.0788