Question

Six point masses each of mass $m$ are placed at the vertices of a regular hexagon of side $l$. The force acting on any of the masses is

• A. $\frac{Gm^{2}}{l^{2}}\left[\frac{5}{4}+\frac{1}{\sqrt{3}}\right]$
• B. $\frac{Gm^{2}}{l^{2}}\left[\frac{3}{4}+\frac{1}{\sqrt{3}}\right]$
• C. $\frac{Gm^{2}}{l^{2}}\left[\frac{5}{4}-\frac{1}{\sqrt{3}}\right]$
• D. $\frac{Gm^{2}}{l^{2}}\left[\frac{3}{4}-\frac{1}{\sqrt{3}}\right]$

Answer 1

Answer: (A)

from figure,
$AC = AM + MC = 2 AM = 2lcos30^{\circ} = 2l \frac{\sqrt{3}}{2} = \sqrt{3}l$
Similarly, $AE = \sqrt{3}l$,
$AD = AO + ON + ND = l\, sin30^{\circ} + l + l\,sin30^{\circ}$
$= l \times \frac{1}{2}+l + \times \frac{1}{2} = 2l$
$AB = AF = l$
Force on mass $m$ at $A$ due to mass $m$ at $B$ is
$F_{AB} = \frac{Gmm}{\left(AB\right)^{2}} = \frac{Gmm}{l^{2}}$ along $AB$
Force on mass $m$ at $A$ due to mass $m$ at $C$ is
$F_{AC} = \frac{Gmm}{\left(AC\right)^{2}} = \frac{Gmm}{l^{2}}$ along $AC$
Force on mass $m$ at $A$ due to mass $m$ at $D$ is
$F_{AD} = \frac{Gmm}{\left(AD\right)^{2}} = \frac{Gmm}{l^{2}}$ along $AD$
Force on mass $m$ at $A$ due to mass $m$ at $E$ is
$F_{AE} = \frac{Gmm}{\left(AE\right)^{2}} = \frac{Gmm}{l^{2}}$ along $AE$
Force on mass $m$ at $A$ due to mass $m$ at $F$ is
$F_{AF} = \frac{Gmm}{\left(AF\right)^{2}} = \frac{Gmm}{l^{2}}$ along $AF$
Resultant force due to $F_{AB}$ and $F_{AF}$ is
$F_{R_1}= \sqrt{F^{2}_{AB}+F^{2}_{AF}+2F_{AB}F_{AF}\,cos120^{\circ}}$
$= \sqrt{\left(\frac{Gm^{2}}{l^{2}}\right)^{2}+\left(\frac{Gm^{2}}{l^{2}}\right)^{2}+2\left(\frac{Gm^{2}}{l^{2}}\right)^{2}\left(\frac{Gm^{2}}{l^{2}}\right)^{2}\left(-\frac{1}{2}\right)}$
$= \frac{Gm^{2}}{l^{2}}$ along $AD$
Resultant force due to $F_{AC}$ and $F_{AE}$ is
$F_{R_2}= \sqrt{F^{2}_{AC}+F^{2}_{AE}+2F_{AC}F_{AE}\,cos60^{\circ }}$
$= \sqrt{\left(\frac{Gm^{2}}{3l^{2}}\right)^{2}+\left(\frac{Gm^{2}}{3l^{2}}\right)^{2}+2\left(\frac{Gm^{2}}{3l^{2}}\right)^{2}\left(\frac{Gm^{2}}{3l^{2}}\right)^{2}\left(\frac{1}{2}\right)}$
$= \frac{\sqrt{3}Gm^{2}}{3l^{2}} = \frac{Gm^{2}}{\sqrt{3}l^{2}}$ along $AD$
Net force on mass $m$ along $AD$ is
$F_{R} = F_{R_1} +F_{R_2} +F_{AD} =\frac{Gm^{2}}{l^{2}}+\frac{Gm^{2}}{\sqrt{3}l^{2}}+\frac{Gm^{2}}{4l^{2}}$
$= \frac{Gm^{2}}{l^{2}}\left[1+\frac{3}{\sqrt{3}}+\frac{1}{4}\right]$
$= \frac{Gm^{2}}{l^{2}}\left[\frac{5}{4}+\frac{1}{\sqrt{3}}\right]$