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. 

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_f × 1.05 = R_b × 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

R_f =\(\frac{ 17640 }{ 2.4}\)= 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