$A$ car weighs $1800 \; kg$. The distance between its front and back axles is $1.8 \; m$. Its centre of gravity is $1.05 \; m$ behind the front axle. Determine the force exerted by the level ground on each front wheel and each back wheel.

(N/A) Mass of the car,$m = 1800 \; kg$.
Distance between the front and back axles,$d = 1.8 \; m$.
Distance between the $C.G.$ (centre of gravity) and the front axle $= 1.05 \; m$.
Distance between the $C.G.$ and the back axle $= 1.8 - 1.05 = 0.75 \; m$.
Let $R_f$ be the total force on the front wheels and $R_b$ be the total force on the back wheels.
At translational equilibrium:
$R_f + R_b = mg = 1800 \times 9.8 = 17640 \; N \; \dots(i)$
For rotational equilibrium,taking torque about the $C.G.$:
$R_f \times 1.05 = R_b \times 0.75$
$R_f = R_b \times \frac{0.75}{1.05} = R_b \times \frac{5}{7} \; \dots(ii)$
Substituting $(ii)$ in $(i)$:
$R_b \times \frac{5}{7} + R_b = 17640$
$R_b \times \frac{12}{7} = 17640 \implies R_b = 10290 \; N$
$R_f = 17640 - 10290 = 7350 \; N$
Force on each front wheel $= \frac{7350}{2} = 3675 \; N$.
Force on each back wheel $= \frac{10290}{2} = 5145 \; N$.