Question

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

Answer 1

For transnational equilibrium of car 

N_F + N_B = W = 1800 x 9.8 = 17640 N 

For rotational equilibrium of car 

1.05 N_F = 0.75 N_B 

1.05 N_F = 0.75(17640 – N_F) 

1.8 N_F = 13230 

N_F = 13230 / 1.8 = 7350 N 

N_B = 17640 – 7350 = 10290 N

Force on each front wheel = \(\frac{7350}{2}\) = 3675 N 

Force on each back wheel = \(\frac{10290}{2}\) = 5145 N