Question

Three point masses 'm' each are placed at the three vertices of an equilateral traingle of side 'a'. Find net gravitational force on any point mass.

Answer 1

Given ;

Mass of three objects = m kg

Placed at vertices of equilateral triangle.

Side = a meter

To find ;

Net gravitational force on any point mass

Solution;

$\textbf{\Large Gravitational force F =}\frac{\textbf{\Large Gm}_1\textbf{\Large m}_2}{\textbf{\Large r}^2}$                    (equation 1)

In this case both masses = m kg

and r = a meter

So,

$\textbf{\Large F =}\frac{\textbf{\Large Gm}^2}{\textbf{\Large a}^2}$                                                                          (equation 2)

The net force by both points are same = F

so the net resultant force on third point

where angle between resultant force and points = 30°

$F_N_E_T = \sqrt{F^2 + F^2 + 2F\cos30}$

$\textbf{\Large F}_N_E_T = \sqrt{3}\ \textbf{\Large F}$

Putting the value of F from (equation 2)

So the net force on any point = $\textbf{\Large F}_N_E_T = \sqrt{3}\ \frac{\textbf{\Large Gm}^2}{\textbf{\Large a}^2}$