A car weighs 1800 kg the …

Weight of car = 1800 Kg
Distance of COG from front axle = 1.05 m 
Distance of COG from back axle = 1.8 - 1.05 = 0.75 m 
Vertical forces are balanced , 
So,At translational equilibrium:
R₁ + R₂ = mg 
R₁ + R₂ = 1800 _{{tex}\times{/tex}} 9.8 = 17640

R₁ and R₂ are the forces exerted by the level ground on the front and back wheels respectively.
Angular momentum about centre of gravity is zero.
So, 
R1(1.05) = R2(1.8 - 1.05)
{tex}R _ { 1 } \times 1.05 = R _ { 2} \times 0.75{/tex}
{tex}\frac { R _ { 1 } } { R _ { 2 } } = \frac { 0.75 } { 1.05 } = \frac { 5 } { 7 }{/tex}
{tex}\frac { R _ { 1 } } { R _ { 2 } } = \frac { 7 } { 5 }{/tex}
R₁ = 1.4 R₂ …..(ii)
Solving equations (i) and (ii), we get:
1.4 R₂ + R₂ = 17640
{tex}R _ { 2 } = \frac { 17640 } { 2.4 } = 7350 \mathrm { N }{/tex}
{tex}\therefore{/tex} R₁ = 17640 - 7350 = 10290 N
Therefore, the force exerted on each front wheel {tex}=\frac { R_1 } { 2 }= \frac { 7350 } { 2 } = 3675 N{/tex}, and
The force exerted on each back wheel {tex}= \frac { R_2 } { 2 }=\frac { 10290 } { 2 } = 5145 N{/tex}