A car of $$800 \text{ kg}$$ is taking turn on a banked road of radius $$300 \text{ m}$$ and angle of banking $$30°$$. If coefficient of static friction is $$0.2$$ then the maximum speed with which car can negotiate the turn safely: $$(g = 10 \text{ m/s}^2, \sqrt{3} = 1.73)$$

Start from force resolution on a banked road at maximum speed. At this condition, friction acts down the plane (towards the center).

Forces:

• Normal reaction N (perpendicular to surface)
• Friction f = μN (down the slope)
• Weight mg (vertical)

Resolve along horizontal (towards center) and vertical.

Horizontal:
$$N\sinθ+μN\cosθ=mv^2/r$$

Vertical:
$$N\cosθ−μN\sinθ=mg$$

Now divide the horizontal equation by the vertical equation:

$$(N\sinθ+μN\cosθ)/(N\cosθ−μN\sinθ)=(mv^2/r)/(mg)$$

Cancel N and m:

$$(\sinθ+μ\cosθ)/(\cosθ−μ\sinθ)=v^2/(rg)$$

Now divide numerator and denominator by cosθ:

$$(\tanθ+μ)/(1−μ\tanθ)=v^2/(rg)$$

So,

$$\frac{v^2}{r}=g\cdot\frac{\tan\theta+\mu}{1-\mu\tan\theta}$$

Now substitute the given values

$$v^2=300\times\ 10\times\ \frac{\frac{1}{\sqrt{\ 3}}+0.2}{1-\frac{1}{\sqrt{\ 3}}\times\ 0.2}$$

$$v^2=300\times\ 10\times\ \frac{1.3464}{1.532}$$

$$v^2=3000\times0.8788$$

$$v^2=2636.4$$

$$v\approx\ 51.4\ \frac{m}{s}$$