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 the car, m = 1800 kg
Distance between the front and back axles, d = 1.8 m
Distance between the C.G. (centre of gravity) and the front axle = 1.05 m
The various forces acting on the car are shown in the following figure.

$R_{f}$ and $R_{b}$ are the forces exerted by the level ground on the front and back wheels respectively.
At translational equilibrium:
$R_{f} + R_{b} = m g = 1800 \times 9.8 = 17640 N \dots (i)$
For rotational equilibrium, on taking the torque about the C.G., we have:
$R_{f} (1.05) = R_{b} (1.8 - 1.05) R_{f} \times 1.05 = R_{b} \times 0.75 \frac{R_{f}}{R_{b}} = \frac{5}{7} R_{b} = 1.4 R_{f} \dots$
Solving equations (i) and (ii), we get:
$1.4 R_{f} + R_{f} = 17640 R_{f} = \frac{17640}{2.4} = 7350 N ∴ = R_{b} = 17640 - 7350 = 10290 N$
Therefore, the force exerted on each front wheel $= \frac{7350}{2} = 3675 N$ . and
The force exerted on each back wheel $= \frac{10290}{2} = 5145 N$

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