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

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 back axle = 1.05 m
The various forces acting on the car are shown in the following figure.

Rf and Rb are the forces exerted by the level ground on the front and back wheels respectively. At translational equilibrium :

Rf + Rb = mg =

1800 × 9.8

= 17640 N ...(i)

For rotational equilibrium, on taking the torque about the C.G., we have :

Rf (1.05) = Rb (1.8 - 1.05)

Rf × 1.05 = Rb × 0.75

$\frac{R_f }{ R_b}$b =$\frac{ 0.75 }{ 1.05} =\frac{ 5 }{ 7}$

$\frac{R_b }{ R_f }$= $\frac{7 }{ 5}$

$R_b = 1.4 \,R_f...(ii)$

Solving equations (i) and (ii), we get : 1.4 Rf + Rf = 17640

Rf =$\frac{ 17640 }{ 2.4}$= 7350 N

∴ Rb = 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