Question

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.

Answer 1

Here, `m = 1800 kg`
Distance between front and black axles `= 1.8 m`
Distance of centre of gravity `(C )` behind the axle `= 1.05 m`.
Let `R_(1),R_(2)` be the forces exerted by the level ground on each front wheel and back wheel. As is clear from Fig.
`R_(1) + R_(2) = mg = 1800 xx 9.8` ...(i)
For rotational equilibrium about `C, R_(1) xx 1.05 = R_(2) (1.8 - 1.05) = R_(2) xx 0.75`
`(R_(1))/(R_(2)) = (0.75)/(1.05) = (5)/(7)`
Put in (i), `(5)/(7) R_(2) + R_(2) = 1800 xx 9.8`
`R_(2) = (7xx1800 xx 9.8)/(12) = 10290 N`
`R_(1) = (5)/(7)R_(2) = (5)/(7) xx 10290 = 7350 N`