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

Mass of car $m = 1800 k g$
Distance between front and back axles $d = 1.8 m$
Distance between center of gravity and front axle $= 1.05 m$
Let $R_{b}$ and $R_{f}$ be the forces exerted by the level ground on the back and front wheels respectively.

At translational equilibrium:
$R_{f} + R_{b} = m g = 17640 N . . . . . (i)$
For rotational equilibrium, on taking the torque about the C.G., we have:
$R_{f} (1.05) = R_{b} (1.8 - 1.05)$
$⟹ R_{b} = 1.4 R_{f} . . . . . . . (i i)$
Solving $(i)$ and $(i i)$ gives
$R_{f} = 7350 N$
$R_{b} = 10290 N$

The force exerted on each front wheel = $7350 / 2 = 3675 N$
The force exerted on each back wheel = $10290 / 2 = 5145 N$

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