The normal reaction $$N$$ for a vehicle of 800 kg mass, negotiating a turn on a 30$$^\circ$$ banked road at maximum possible speed without skidding is ___ $$\times 10^3$$ kg m s$$^{-2}$$.

Balancing forces vertically:  $$N \cos \theta - mg - f \sin \theta = 0$$

Using limiting friction:  $$f = \mu N$$, 

$$N \cos 30^\circ - \mu N \sin 30^\circ = mg$$

$$\implies N(\cos 30^\circ - \mu \sin 30^\circ) = mg$$

$$\implies N = \frac{mg}{\cos 30^\circ - \mu \sin 30^\circ}$$

$$\implies N = \frac{800 \times 10}{0.87 - 0.2(0.5)} = \frac{8000}{0.87 - 0.1} = \frac{8000}{0.77}$$

$$\implies N \approx 10389\ \text{N} \approx 10.2 \times 10^3\ \text{N}$$