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.

Explanation

`m=1800 kg`
Distance between front and back axles = 1.8 m Distance of centre of gravity (C) from axle = 1.05 m
From the figure above
`R_(1)+R_(2)=mg " "....(i)`
For rotational equilibrium about C.
`R_(1)xx(1.8-1.05)=R_(2)xx(1.05)`
`R_(1)xx0.75=R_(2)xx1.05`
`(R_(1))/(R_(2))=(0.75)/(1.05)=(75)/(105)=(15)/(21)=(5)/(7)`
`R_(1)=(5)/(7)R_(2)`
Put in (i) `(5)/(7)R_(2)+R_(2)=mg`
`(12R_(2))/(7)=1800xx9.8`
`R_(2)=(1800xx9.8xx7)/(12)=10290 N`
`:.` Reaction is 5145N on each front wheal
`R_(1)=(5)/(7)R_(2)=(5)/(7)xx10290=7350N`
`:.` Reaction is 3675 N on each back wheel.