Question

Three masses each of mass $m$ are placed at the vertices of an equilateral triangle $ABC$ of side $l$ as shown in figure. The force acting on a mass $2m$ placed at the centroid $O$ of the triangle is

• A. zero
• B. $\frac{3Gm^{2}}{l^{2}}$
• C. $\frac{5Gm^{2}}{l^{2}}$
• D. $\frac{7Gm^{2}}{l^{2}}$

Answer 1

Answer: (A)

Draw a perpendicular $AD$ to the side $BC$.
$\therefore AD = ABsin60^{\circ} = \frac{\sqrt{3}}{2}l$
Distance $AO$ of the centroid $O$ from $A$ is $\frac{2}{3} AD$.
$\therefore AO = \frac{2}{3}\left(\frac{\sqrt{3}}{2}l\right) = \frac{l}{\sqrt{3}}$
By symmetry, $AO = BO = CO = \frac{l}{\sqrt{3}}$
Force on mass $2m$ at $O$ due to mass $m$ at $A$ is
$F_{OA} = \frac{Gm\left(2m\right)}{\left(l/ \,\sqrt{3}\right)^{2}} = \frac{6Gm^{2}}{l^{2}}$ along $OA$.
Force on mass $2m$ at $O$ due to mass $m$ at $B$ is
$ F_{OB}= \frac{Gm\left(2m\right)}{\left(l/ \,\sqrt{3}\right)^{2}} = \frac{6Gm^{2}}{l^{2}}$ along $OB$
Force on mass $2m$ at $O$ due to mass $m$ at $C$ is
$F_{OC} = \frac{Gm\left(2m\right)}{\left(l/ \,\sqrt{3}\right)^{2}} = \frac{6Gm^{2}}{l^{2}}$ along $OC$
Draw a line $PQ$ parallel to $BC$ passing through $O$. Then
$∠BOP = 30^{\circ} = ∠COQ $
Resolving $\vec{F}_{OB}$ and $\vec{F}_{OC}$ into two components.
Components acting along $OP$ and $OQ$ are equal in magnitude and opposite in direction. So, they will cancel out while the components acting along $OD$ will add up.
$\therefore $ The resultant force on the mass $2m$ at $O$ is
$F_{R} = F_{OA} - \left(F_{OB}\, sin30^{\circ} + F_{OC}\, sin30^{\circ}\right)$
$= \frac{6Gm^{2}}{l^{2}}-\left(\frac{6Gm^{2}}{l^{2}}\times\frac{1}{2}+\frac{6Gm^{2}}{l^{2}}\times\frac{1}{2}\right) = 0$