Question

Three particles, each of mass m, are placed at the vertices of an equilateral triangle of side $a$. The net gravitational force on a body of mass $M$ at the centroid of the triangle is

• A. zero
• B. $\frac{GmM}{a^2}$
• C. $\frac{2GmM}{a^2}$
• D. $\frac{3GmM}{a^2}$

Answer 1

The correct option is A

zero

Given that AB = BC = CA = a. The perpendiculars from A, B and C on opposite sides meet at the centroid O and bisect the sides AB, BC and CA. Let AO = BO = CO = r. Centroid also divides the lines AD, BE and CF in the ratio 2 : 1, i.e. $AO=\frac{2}{3}AD,BO=\frac{2}{3}BE$ and $CO=\frac{2}{3}CF$.

In triangle ABD, $AD=asin{60}^{\circ }=\frac{\sqrt{3}a}{2}$
Similarly, BE = CF $=\frac{\sqrt{3}a}{2}$
$\therefore r=AO=BO=CO=\frac{2}{3}\times \frac{\sqrt{3}a}{2}=\frac{a}{\sqrt{3}}$
The net force exerted on a unit mass placed at O due to three equal masses m at vertices A, B and C.

Since the three masses are equal and their distances from O are also equal, they exert forces $F_A$, $F_B$ and $F_C$ of equal magnitude. Their directions are shown in the figure. It follows from the symmetry of forces that their resultant at point O is zero.

Hence, the correct choice is (a).