A car weighs 1800 kg. The distance between its front and back axles is 1.8 m. Its centre of gravity is 1 m behind the front axle. The force exerted by the level ground on each front wheel and each back wheel is (Take $\mathrm{g} = 10 {\mathrm{ms}}^{-1}$)

Here, mass of the car, M = 1800 kg
Distance between front and back axles = 1.8 m
Distance of gravity G behind the front axle = 1 m
Let ${\mathrm{R}}_{\mathrm{F}} \mathrm{and} {\mathrm{R}}_{\mathrm{B}}$, be the forces exerted by the level ground on each front wheel and each back wheel.

For translational equilibrium,
$2{\mathrm{R}}_{\mathrm{F}}+2{\mathrm{R}}_{\mathrm{B}} = \mathrm{Mg}$

or ${\mathrm{R}}_{\mathrm{F}}+{\mathrm{R}}_{\mathrm{B}} = \frac{\mathrm{Mg}}{2} = \frac{1800\times 10}{2} = 9000 \mathrm{N}----\left(\mathrm{i}\right)$

(As there are two front wheels and two back wheels)
For rotational equilibrium about G
$\left(2{\mathrm{R}}_{\mathrm{F}}\right)\left(1\right) = \left(2{\mathrm{R}}_{\mathrm{B}}\right)(0.8)$

$\frac{{\mathrm{R}}_{\mathrm{F}}}{{\mathrm{R}}_{\mathrm{B}}} = 0.8 = \frac{8}{10} = \frac{4}{5} \Rightarrow {\mathrm{R}}_{\mathrm{F}} = \frac{4}{5}{\mathrm{R}}_{\mathrm{B}}$

Substituting this in Eq. (i), we get
$\frac{4}{5}{\mathrm{R}}_{\mathbf{B}}+ {\mathrm{R}}_{\mathrm{B}} = 9000 \mathrm{or} \frac{9}{5}{\mathrm{R}}_{\mathrm{B}} = 9000$

${\mathrm{R}}_{\mathrm{B}} = \frac{9000\times 5}{9} = 5000 \mathrm{N}$

$\therefore {\mathrm{R}}_{\mathrm{F}} = \frac{4}{5}{\mathrm{R}}_{\mathrm{B}} = \frac{4}{5}\times 5000 \mathrm{N} = 4000 \mathrm{N}$