A turn of radius 20 m is banked for vehicles travelling at a speed...

Given:

Radius of the turn, $R = 20 \text{ m}$

Optimum speed, $v_0 = 36 \text{ km/h} = 36 \times \dfrac{5}{18} \text{ m/s} = 10 \text{ m/s}$

Coefficient of static friction, $\mu_s = 0.4$

Acceleration due to gravity, $g = 10 \text{ m/s}^2$

The angle of banking $\theta$ is given by:

$\tan \theta = \dfrac{v_0^2}{Rg} = \dfrac{10^2}{20 \times 10} = 0.5$

The minimum speed to avoid slipping down is:

v_\min = \sqrt{Rg \left( \dfrac{\tan \theta - \mu_s}{1 + \mu_s \tan \theta} \right)}

v_\min = \sqrt{20 \times 10 \left( \dfrac{0.5 - 0.4}{1 + 0.4 \times 0.5} \right)} = \sqrt{200 \left( \dfrac{0.1}{1.2} \right)} = \sqrt{\dfrac{50}{3}} \text{ m/s}

Converting to $\text{km/h}$:

v_\min = \sqrt{\dfrac{50}{3}} \times \dfrac{18}{5} \approx 14.7 \text{ km/h}

The maximum speed to avoid skidding up is:

v_\max = \sqrt{Rg \left( \dfrac{\tan \theta + \mu_s}{1 - \mu_s \tan \theta} \right)}

v_\max = \sqrt{20 \times 10 \left( \dfrac{0.5 + 0.4}{1 - 0.4 \times 0.5} \right)} = \sqrt{200 \left( \dfrac{0.9}{0.8} \right)} = \sqrt{225} = 15 \text{ m/s}

Converting to $\text{km/h}$:

v_\max = 15 \times \dfrac{18}{5} = 54 \text{ km/h}

The possible speeds are between $14.7 \text{ km/h}$ and $54 \text{ km/h}$.