Question

three equal point masses of 1.5 kg each fixed at the vertices of equilateral triangle of side 1m what is the force acting on a particle of mass 1kg placed at its centroid

Answer 1

For the solution to this problem, we have to refer to the sketch below.

We set up the diagram with each of the masses listed as $m_1,m_2, m_3$ respectively. We then impose an $x-y$ coordinate system. The mass $m_1$ applies a pulling force $F_1$ in the $y$ direction and no force in the $x$ direction . The mass $m_1$ applies a force equal to $-F_1cos(30)$. The angle is 30 degrees because the point masses are set up in an equilateral triangle. In the $y$-direction, the force at the center due to this point mass is $-F_1sin(30)$. The point mass $m_3$ applies a force   $F_3cos(30)$ in the positive $x$ direction and a force   $F_3sin(30)$ in the positive $x$ direction. Since the masses are all equal, we get that at the center, the magnitude of  $F_1$,  $F_2$,  $F_3$ are all equal.

In the $x$-direction,  the net forces are,

$F_{net}=-F_1cos(30)+F_3cos(30)=0.$.

In the x direction, the net forces are,

$F_{net}=-F_3sin(30)-F_2sin(30)+F_1=-0.5F_3-0.5F_2+F_1=0.$

The net force at the centroid is zero in both $x$ and  $y$directions, meaning that the net force is zero.

If one desires, they could have used the universal law of gravitation to work out the result. The argument and the result are the same is using the law.