Consider the free body diagram (FBD) of the car with all the forces
The car is in rotational equilibrium
∴ Σ τ (around any stationery point) = 0
Also w = mg = (1800) × (9.8) …………. (1)
consider the torque around F and B
Σ \(\bar τ \)(about F) = W (1.05) – FB (1.8) = 0
where FB is the force on the back wheels.
From the above equation,
FB = \(\frac {mg (1.05)}{(1.8)}\)= 10290 N
similarly ΣF (about B) = 0
⇒ W (1.8 – 1.05) = FF (1.8)
∴ Force on front wheels,
FF = \(\frac {mg (0.75)}{(1.8)}\) = 7350 N
Total force exerted by ground
= FF + FB = 7350 + 10290
=17640 N.
