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. $\frac{2G{m}^2}{l^2}$
• C. $\frac{5G{m}^2}{l^2}$
• D. $\frac{6G{m}^2}{l^2}$

Answer 1

Answer: (D)

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