Question: An automobile with a mass of 1360 kg has 3.05 m between the front and rear axles. Its center of gravity is located 1.78 m behind the front axle. With the automobile on level ground, determine the magnitude of the force from the ground on (a) each front wheel (assuming equal forces on the front wheels) and (b) each rear wheel (assuming equal forces on the rear wheels).

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.$\tau =F\times r$

The torque about the rear axle is,

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

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

At equilibrium,

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

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