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A car weighs 1800 kg. The distance betwe...

A car weighs 1800 kg. The distance between its front and back axles is 1.8 m. Its centre of gravity is 1 m behind the front axle. The force exerted by the level ground on each front wheel and each back wheel is (Take `g : 10 ms^(-1)`)

A

4000 N on each front wheel, 5000 N on each back wheel

B

5000 N on each front wheel, 4000 N an each back wheel

C

4500N on each front wheel, 4500 N on each back wheel

D

3000 N on each front wheel, 6000 N on each back wheel

Text Solution

Verified by Experts

The correct Answer is:
A
Here, mass of the car, M= 1800 kg
Distance between front and back axles = 1.8 m Distance of gravity G behind the front axle = 1 m
Let RF and RB be the forces exerted by the level ground on each front wheel and each back wheel.

For translational equilibrium,
`2R_(F) + 2R_(B) = Mg`
or `R_(F) + R_(B) = (Mg)/2 =(1800 xx 10)/2 = 9000 N`.....(i)
(As there are two front wheels and two back wheels) For rotational equilibrium about G
`(2R_(F))(1) =(2R_(B)) (0.8)`
`R_(F)/R_(B) = 0.8 = 8/10 = 4/5 rArr R_(F) = 4/5 R_(B)`........(ii)
Substituting this in Eq. (i), we get
`4/5 R_(B) + R_(B) = 9000` or `9/5 R_(B)= 9000`
`R_(B) = (900 xx 5)/9 = 5000 N`
`therefore R_(F) = 4/5 R_(B) = 4/5 xx 5000 N = 4000 N`
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