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.

Given, weight of the car is 1800 kg, distance between its front and back axles is 1.8 m and its centre of gravity is 1.05 m behind the front axle.

The figure shows car of the given mass, the tires provide reactions R f and R b for the front and the back wheels.

Let m be the mass of the car.

The equilibrium equation for the force is,

R f + R b =mg

Substitute the values in the above expression.

R f + R b =1800×9.8 =17640 N (1)

The net moment about centre of gravity C.G. is,

R f ( 1.05 )= R b ( 1.8−1.05 ) R f ( 1.05 )= R b ( 0.75 ) R f =0.714 R b (2)

Substitute value of R f in the equation (1).

0.714 R b + R b =17640 1.714 R b =17640 R b =10291.715 N

Substitute the value of R b in the equation (1),

R f =7348.28 N

Hence, force exerted on each front wheel is,

R f 2 = 7348.28 2 ≈3675 N

Force exerted on each back wheel is,

R b 2 = 10291.715 2 ≈5145 N

Thus, the force on each front and back wheel is 3675 N and 5145 N respectively.