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.

Options

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. `4500 N` on each front wheel, `4500 N` on each back wheel
D. `3000 N` on each front wheel, `6000 N` on each back wheel

Answer

A

Explanation

(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 `R_F` and `R_B` be the forces exerted by the level ground on each front wheel and each back wheel.
For translational equilibrium,
`2 R_F + 2 R_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`
`(2 R_F)(1) = (2 R_B) (0.8)`
`(R_F)/(R_B) =0.8 = (8)/(10) = (4)/(5) rArr R_F = (4)/(5) R_B` ...(ii)
Substituting this 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`
:. `R_F = (4)/(5) R_B = (4)/(5) xx 5000 N = 4000 N`.
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