Question

Three masses each of mass $m$ are placed at the vertices of an equilateral triangle $ABC$ of side $I$ as shown in figure if the mass placed at vertex $A$ is doubled, then the force acting on the mass $2m$ placed at the centroid $O$ is

• A. zero
• B. $\frac{2Gm^{2}}{l^{2}}$
• C. $\frac{5Gm^{2}}{l^{2}}$
• D. $\frac{6Gm^{2}}{l^{2}}$

Answer 1

Answer: (D)

Force on mass $2m$ at $O$ due to mass $2m$ at $A$ is
$F'_{OA} = \frac{G\left(2m\right)\left(2m\right)}{\left(l /\sqrt{3}\right)^{2}} = \frac{12Gm^{2}}{l^{2}}$ along $OA$
$\therefore $ The resultant force on the mass $2m$ at $O$ due to masses at $A$, $B$ and $C$ is
$F'_{R} = F'_{OA} \left(F_{OB}\, sin30^{\circ} + F_{OC}\, sin30^{\circ}\right)$
$= \frac{12Gm^{2}}{l^{2}}-\left(\frac{6Gm^{2}}{l^{2}}\times\frac{1}{2}+\frac{6Gm^{2}}{l^{2}}\times \frac{1}{2}\right)$
$= \frac{12Gm^{2}}{l^{2}}-\frac{6Gm^{2}}{l^{2}}$
$= \frac{6Gm^{2}}{l^{2}}$ along $OA$