Three equal masses of $m \; kg$ each are fixed at the vertices of an equilateral triangle $ABC.$
$(a)$ What is the force acting on a mass $2 \; m$ placed at the centroid $G$ of the triangle?
$(b)$ What is the force if the mass at the vertex $A$ is doubled?
Take $AG = BG = CG = 1 \; m$.

(N/A) The angle between $GC$ and the positive $x$-axis is $30^{\circ}$,and so is the angle between $GB$ and the negative $x$-axis. The individual forces in vector notation are:
$F_{GA} = \frac{G m (2m)}{1^2} \hat{j} = 2Gm^2 \hat{j}$
$F_{GB} = \frac{G m (2m)}{1^2} (-\cos 30^{\circ} \hat{i} - \sin 30^{\circ} \hat{j}) = 2Gm^2 (-\frac{\sqrt{3}}{2} \hat{i} - \frac{1}{2} \hat{j}) = -Gm^2 \sqrt{3} \hat{i} - Gm^2 \hat{j}$
$F_{GC} = \frac{G m (2m)}{1^2} (+\cos 30^{\circ} \hat{i} - \sin 30^{\circ} \hat{j}) = 2Gm^2 (\frac{\sqrt{3}}{2} \hat{i} - \frac{1}{2} \hat{j}) = Gm^2 \sqrt{3} \hat{i} - Gm^2 \hat{j}$
From the principle of superposition,the resultant gravitational force $F_R$ on $(2m)$ is:
$F_R = F_{GA} + F_{GB} + F_{GC} = 2Gm^2 \hat{j} + (-Gm^2 \sqrt{3} \hat{i} - Gm^2 \hat{j}) + (Gm^2 \sqrt{3} \hat{i} - Gm^2 \hat{j}) = 0$
Alternatively,by symmetry,the resultant force is zero.
$(b)$ If the mass at vertex $A$ is doubled,the new force $F'_{GA} = \frac{G (2m) (2m)}{1^2} \hat{j} = 4Gm^2 \hat{j}$.
The forces $F_{GB}$ and $F_{GC}$ remain unchanged.
$F'_R = F'_{GA} + F_{GB} + F_{GC} = 4Gm^2 \hat{j} - Gm^2 \hat{j} - Gm^2 \hat{j} = 2Gm^2 \hat{j}$ (directed towards $A$).