A car rounds a circular curve of radius 80.0 m banked at 15 to the horizontal at 30.0 m/s. What is the coefficient of friction required for the car not to skid? Some hints. first, set the x-axis as horizontal, and the y-axis as vertical (this is different than how we normally handle inclines). Second, there will be 3 forces: the normal force (with x- and y- components), the force of gravity (y component only), and friction pointing down the slope (also with x- and y-components)

We must use Newton's second law, where we place a reference system along the horizontal x axis and the vertical Y axis, in this system the only force with components is normal and the force of friction, we will decompose it

sin15 = Nx / N

cos15 = Ny / N

Nx = N sin15

Ny = N cos 15

Note that the force of friction is perpendicular to the normal, therefore, the Angle of 15º is with respect to the axis

sin15 = fry / fr

cos15 = Frx / fr

frx = fr cos15

fry = fr sin 15

Let's write Newton's equations

When the skating car has to leave the curve, the friction force must go towards the center to oppose this movement.

X axis

Nx + frx = ma

Axis y

Ny -fry-W = 0

Let's replace

The acceleration is centripetal and the friction force has a formula

a = v2 / r

fr = μ N

N sin15 + μ N cos15 = m v² / r

N cos 15 - μ Nfr sin 15 = mg

N (sin 15 + μ cos 15) = m v² / r

N (cos 15- μ sin15) = mg

Let's solve the system of equations, let's divide the two

(Sin15 + μ cos 15) / (cos15 - μ sin 15) = v² / rg

(Sin15 + μ cos 15) = v² / rg (cos15 - μ sin 15)

μ cos15 + v² /rg μ sin 15 = v² /rg cos15 -sin 15

μ (v² / rg sin 15 + cos 15) = v² /rg cos15 - sin 15

Let's calculate the values

μ (30² /(80.0 9.8) sin 15 + cos 15) = 30² /(80.0 9.8) cos 15 - sin 15

μ (1,148 sin 15 + cos 15) = 1,148 cos 15 - sin 15

μ (0.2971 + 0.9659) = 1.109 - 0.2588

μ (1,263) = 0.8502

μ = 0.8502 / 1,263

μ = 0.673