### Q.9:- 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:-**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.

*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} = mg

= 1800 × 9.8

= 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} / *R*_{f} = 7 / 5

*R*_{b} = 1.4 *R*_{f} ……**(ii)**

Solving equations (*i*) and (*ii*), we get:

1.4*R*_{f} + *R*_{f} = 17640

*R*_{f} = 7350 N

∴ *R*_{b} = 17640 – 7350 = 10290 N

Therefore, the force exerted on each front wheel = 7350 / 2 = 3675 N, and

The force exerted on each back wheel = 10290 / 2 = 5145 N