91Ó°ÊÓ

The mass of the automobile, 1360 K g

The horizontal distance between the axels, L = 3.09 m

The horizontal distance from the front axle to the center of mass, $l\text{'}=1.78\text{m}$

The horizontal distance from the axle to the center of mass is,$l=\left(3.05-1.78\right)=1.27\text{m}$

Using the formula for torque, you can find the force F_1.Then by balancing forces for equilibrium, you can find F₂. The formula used is given below.$\mathrm{Ï„}=F×r$

The torque about the rear axle is,

$$\begin{gathered} \mathrm{τ}=F_1\left(2L\right) \\ =Mgl \\ {F}_1=\frac{Mgl}{2L} \\ =\frac{1360×9.8×1.27}{2×3.05} \\ =2.77×{10}^3\text{N} \\ \text{=}2.77\text{kN} \end{gathered}$$

Therefore, the magnitude of the force applied from the ground on each front wheel is 2.77 KN

At equilibrium,

$$\begin{gathered} 2F_1+2F_2=Mg \\ {F}_2=\frac{Mg-2F_1}{2} \\ =\frac{1360\left(9.8\right)-2\left(2.77×{10}^3\right)}{2} \\ =3.89×{10}^3\text{N} \\ \text{= 3.89}\text{kN} \end{gathered}$$

Therefore, the magnitude of the force applied from the ground on each rear wheel is 3.89 KN