91Ó°ÊÓ

• Length of each side of the square is L = 2.0 m
• Gravitational constant$G=6.67×{10}^{-11}\mathrm{N}.{\mathrm{m}}^2/{\mathrm{kg}}^2$

When the force is attracted by different object within a mass is termed asGravitational force.

The gravitational force between two different objects is given by,

$\mathrm{F}=\frac{{\mathrm{Gm}}_1{\mathrm{m}}_2}{{\mathrm{r}}^2}$

Where $m_1$and $m_2$are the masses of the two objects, G is the gravitational constant and ris the separation between them.

Let the masses be A, B, C, and D, and they are placed as shown

Let us calculate the force on the mass A due to the other three objects at B, C, and D.

Force on the mass A due to force on B,

Substituting the given values in equation (1)

$$\begin{gathered} {\mathrm{F}}_{\mathrm{AB}}=\frac{\mathrm{G}×800\mathrm{kg}×800\mathrm{kg}}{{\left(2.0\right)}^2{\mathrm{m}}^2} \\ {\mathrm{F}}_{\mathrm{AB}}=1.067×{10}^{-5}\mathrm{N} \end{gathered}$$

Similarly, the force on the mass A due to the force on D,

${\mathrm{F}}_{\mathrm{AD}}=1.067×{10}^{-5}\mathrm{N}$

Now, the resultant of the two forces will be along with the diagonal AC of the square, with which each side of the square makes an angle of ${45}^{°}$.

Therefore, the resultant force along with AC due to each a$F_{AB}$and $F_{AD}$is

$F_{res}=\left(\cos \mathrm{θ}×F_{AB}\right)+\left(\cos \mathrm{θ}×F_{AD}\right)$

Now substituting the values, we get,

$$\begin{gathered} {\mathrm{F}}_{\mathrm{res}}=\left(\cos {45}^{°}×1.067×{10}^{-5}\mathrm{N}\right)+\left(\cos {45}^{°}×1.067×{10}^{-5}\mathrm{N}\right) \\ {\mathrm{F}}_{\mathrm{res}}=1.509×{10}^{-5}\mathrm{N} \end{gathered}$$

Now, the force on mass A due to the force on mass C is,

$$\begin{gathered} {\mathrm{F}}_{\mathrm{AC}}=\frac{\mathrm{G}×800\mathrm{kg}×800\mathrm{kg}}{\left(2.0+2.0\right){\mathrm{m}}^2} \\ {\mathrm{F}}_{\mathrm{AC}}=1.067×{10}^{-5}\mathrm{N} \end{gathered}$$

Adding the values to $F_{res}$and $F_{AC}$we get the net force on the mass A.

So, the net force is,

$$\begin{gathered} {\mathrm{F}}_{\mathrm{net}}={\mathrm{F}}_{\mathrm{res}}+{\mathrm{F}}_{\mathrm{AC}} \\ {\mathrm{F}}_{\mathrm{net}}=\left(1.509×{10}^{-5}\mathrm{N}\right)+\left(1.067×{10}^{-5}\mathrm{N}\right) \\ {\mathrm{F}}_{\mathrm{net}}=2.576×{10}^{-5}\mathrm{N} \end{gathered}$$

The net gravitational force on mass A is$2.576×{10}^{-5}\mathrm{N}$.

The direction of this force will be along the diagonal of the square.