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in Physics by (64.7k points)

A car weighs 1800 kg. The distance between its front and back axle 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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Here m = 1800 kg

Distance between the front and rear axles = 1.8 m

Distance of the centre of gravity from the front axle = 1.05 m

Let R1 and R2 be the reactions of ground on the front and rear wheels, respectively.

Now, R1 + R2 = mg = 1800 x 9.8    ...(i)

For rotational equilibrium about G

R1 x 1.05 = R2 (1.8 - 1.05)

= R2 x 0.75

or, R1/R2 = 0.75/1.05 = 5/7

or, R1 = 5/7 R2

Putting R1 = 5/7 R2 in (i) we get

5/7 R2 + R2 = 1800 x 9.8

or, R2 = {7 x 1800 x 9.8}/{12} = 10290 N

and R1 = 5/7 x 10290 = 7350 N

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