 # A car of weight 1800 kg

A car of weight 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 farce exerted by the level ground on each front wheel and each back wheel.

Total mass of the car = 1800 kg
Let m and (900 - m) kg be the masses of each front wheel and each back wheel respectively.
Distance of centre of gravity from the front axle = 1.05 m
.’. Distance of centre of gravity from the back axle = 1.80-1.05 = 0.75 m
Taking torque about centre of gravity,
m x 1.05 = (900 -m)x 0.75 or 1.05 m x 0.75 m = 900 x 0.75 or
1.80 m = 900x0.75 or
m = 900 x 0.75 / 1.80 = 375 kg
(900 -m) = 900 - 375 = 525 kg
.’. Weight of each front wheel =375 x 9.8
= 3675 N
Force exerted by the level ground on each front wheel. = force exerted by each front wheel on the level ground = 3675 N
Weight of each back wheel = 525 x 9.8 = 5145 N
Force exerted by the level ground on each back wheel.
= force exerted by each back wheel on level ground = 5145N.