Question

Three equal masses m are placed at the three corners of an equilateral triangle of side a. Find the force exerted by this system on another particle of mass m placed at (a) the mid-point of a side, (b) at the centre of the triangle.

Answer 1

(a)

If 'm' is placed at mid point of a side
then, ${\overrightarrow{F}}_{OA}=\frac{4G{m}^2}{a^2}$ in OA direction
${\overrightarrow{F}}_{OB}=\frac{4G{m}^2}{a^2}$ in OB direction
Since equal and opposite forces cancel each other
${\overrightarrow{F}}_{OC}=\frac{4G{m}^2}{\left\{\right(3\left)a^2\right\}}$
$=\frac{4G{m}^2}{3a^2}$ in OC direction
(b) If 'm' is placed at O (centroid)

Resultant ${\overrightarrow{F}}_{OA}+{\overrightarrow{F}}_{OB}+{\overrightarrow{F}}_{OC}=0$
(Because forces are equal and opposite.) Net gravitation force = 0