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.
Given,
Weight of the car, W = (1800×g) N = 1800×9.8 = 17640 N
Distance between front and back wheels, d = 1.8 m
Distance between centre of gravity from front axle = 1.05 m
The free body diagram of car can be drawn as,
From the figure,
For transitional equilibrium,
Rf + Rb = W
Rf + Rb = 17640 N ……..(1)
For rotational equilibrium, on taking torque about centre of gravity as,
Rb = 1.4Rf ……..(2)
By solving equations 1 and 2 we get,
1.4Rf + Rf = 17640 N
∴
Rb = 17640 – 7350 = 10290 N
Hence, The force exerted on each of the front wheel =
The force exerted on each of the rear wheel =