Question

An equilateral triangle $ABC$ of side $l$ having three point masses $m$ at the three vertices is shown in figure. Calculate the force acting on a mass $2m$ , if it is placed at the centroid $O$ of the triangle.

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

Answer 1

Answer: (A)

Draw a perpendicular $AD$ to the side $BC$ .
$AD=ABsin60^o=\frac{\sqrt{2}}{3}l$
Distance $AO$ of the centroid $O$ from $A$ is $\frac{2}{3}$ $AD$ .
$AO=\frac{2}{3}\left(\frac{\sqrt{3}}{2}l\right)=\frac{l}{\sqrt{3}}$
By symmetry, $AO=BO=CO=\frac{1}{\sqrt{3}}$
Force on mass $2m$ at $O$ due to mass m at A is
$F_{OA}=\frac{Gm(2m)}{(l/\sqrt{3}{)}^2}=\frac{6G{m}^2}{l^2}alongOA$
Force on mass $2m$ at $O$ due to mass m at $B$ is
$F_{OB}=\frac{Gm(2m)}{(l/\sqrt{3}{)}^2}=\frac{6G{m}^2}{l^2}alongOA$
Force on mass $2m$ at $O$ due to mass $m$ at $C$ is
$F_{OC}=\frac{Gm(2m)}{(l/\sqrt{3}{)}^2}=\frac{6G{m}^2}{l^2}alongOC$
Draw a line PQ parallel to BC passing through O. Then $\angle BOP=30º=\angle COQ$
Resolving ${\overrightarrow{F}}_{OB}$ and ${\overrightarrow{F}}_{OC}$ 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.
$\therefore$ he resultant force on the mass $2m$ at $O$ is
$F_R=F_{OA}-(F_{OB}sin30º+F_{OC}sin30º)$ .
$=\frac{6G{m}^2}{l^2}-(\frac{6G{m}^2}{l^2}\times \frac{1}{2}+\frac{6G{m}^2}{l^2}\times \frac{1}{2})=0$