Six-point masses of mass $m$ each are at the vertices of a regular hexagon of side $l$ . Calculate the force on any of the masses.

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Solution

Step 1: Draw the diagram as per the given information.

Step 2: Calculate the different forces.

From figure, $A C = A M + M C = 2 A M$

$A C = 2 l cos 30^{\circ} = 2 l \frac{\sqrt{3}}{2} = \sqrt{3 l}$

As we know,

$A E = A C$

$∴ A E = \sqrt{3 l}$

$A D = A O + O N + N D$

$= 1 sin 30^{\circ} + 1 + 1 sin 30^{\circ}$

$= 1 \times \frac{1}{2} + 1 + 1 \times \frac{1}{2} = 2 l$

$∴ A D = 2 l, A B = A F = l$

Force on mass $m$ at $A$ due to mass $m$ at $B$ is

$F_{A B} = \frac{G m m}{(A B)^{2}} = \frac{G m m}{l^{2}}$ along $A B$

Force on mass $m$ at $A$ due to mass $m$ at $D$ is
$F_{A D} = \frac{G m m}{(A D)^{2}} = \frac{G m m}{(2 I)^{2}} = \frac{G m^{2}}{4 l^{2}}$ along $A D$

Force on mass $m$ at $A$ due to mass $m$ at $E$ is
$F_{A E} = \frac{G m m}{(A E)^{2}} = \frac{G m m}{(\sqrt{3 l})^{2}} = \frac{G m^{2}}{3 l^{2}}$ along $A E$

Force on mass $m$ at $A$ due to mass $m$ at $F$ is

$F_{A F} = \frac{G m m}{(A F)^{2}} = \frac{G m^{2}}{l^{2}}$ along $A F$

Step 3: Calculate the resultant forces.

Therefore, resultant force due to $F_{A B}$ and $F_{A F}$ is

$F_{R_{1}} = \sqrt{F_{A B}^{2} + F_{A F}^{2} + 2 F_{A B} F_{A F} cos 120^{\circ}}$

By putting the values of different forces obtained we get,

$F_{R_{1}} = \sqrt{{(\frac{G m^{2}}{l^{2}})}^{2} + {(\frac{G m^{2}}{l^{2}})}^{2} + 2 (\frac{G m^{2}}{l^{2}}) (\frac{G m^{2}}{l^{2}}) (- \frac{1}{2})}$
$∴ F_{R_{1}} = \frac{G m^{2}}{l^{2}}$ along $A D$

Resultant force due to $F_{A C}$ and $F_{A E}$ is

$F_{R_{2}} = \sqrt{F_{A C}^{2} + F_{A E}^{2} + 2 F_{A C} F_{A E} cos 60^{\circ}}$

$F_{R_{2}} = \sqrt{{(\frac{G m^{2}}{3 l^{2}})}^{2} + {(\frac{G m^{2}}{3 l^{2}})}^{2} + 2 (\frac{G m^{2}}{3 l^{2}}) (\frac{G m^{2}}{3 l^{2}}) (\frac{1}{2})}$

$F_{R_{2}} = \frac{\sqrt{3} G m^{2}}{3 l^{2}} = \frac{G m^{2}}{\sqrt{3} l^{2}}$ along $A D$

Step 4: Calculate the net force.

Net force on mass $M$ along $A D$ is

$F_{R} = F_{R_{1}} + F_{R_{2}} + F_{A D}$

$F_{R} = \frac{G m^{2}}{l^{2}} + \frac{G m^{2}}{\sqrt{3} l^{2}} + \frac{G m^{2}}{4 l^{2}}$

$F_{R} = \frac{G m^{2}}{l^{2}} [1 + \frac{1}{\sqrt{3}} + \frac{1}{4}]$

Therefore, net force is given by

$F_{R} = \frac{G m^{2}}{l^{2}} [\frac{5}{4} + \frac{1}{\sqrt{3}}]$

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