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.

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Solution

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.

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