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 =


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