A car weighs $1800 k g$ . 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

Consider the free body diagram $(F B D)$ of the car with all the forces
The car is rotational equilibrium
$∴ Σ τ$ (around any stationery point)=0
Also $w = m g == (1800) \times (9.8) . . . . . . . . (1)$
Consider the torque around $F$ and $B$
$Σ ¯ τ (a b o u t F) = W (1.05) = F_{B} (1.8) = 0$
Where $F_{B}$ is the force on the back wheels.
From the above equation,
$F_{B} = \frac{m g (1.05)}{(1.8)} = 10290 N$
similarly $Σ F$ (about $B$ )=0
$\Rightarrow W (1.8 - 1.05) = F_{F} (1.8)$
$∴$ Force on front wheels,
$F_{F} = \frac{m g (0.75)}{(1.8)} = 7350 N$
Total force exerted by ground
$= F_{F} + F_{B} = 7350 + 10290$
= $17540 N$

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A car weighs 1800kg. The distance between its front and back axles is 1.8m. Its centre of gravity is 1.05m behind the front axle. Determine the force exerted by the level ground on each front wheel and each back wheel.

A car weighs $1800 k g$ . 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.