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The Complete Equilibrium

Six point mas...

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:

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

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

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\frac{G m^{2}}{l^{2}} [\frac{5}{4} - \frac{1}{\sqrt{3}}]

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\frac{G m^{2}}{l^{2}} [\frac{3}{4} - \frac{1}{\sqrt{3}}]

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Solution

The correct option is C $\frac{G m^{2}}{l^{2}} [\frac{5}{4} + \frac{1}{\sqrt{3}}]$
From figure, $A C = A M + M C = 2 A M = 2 l cos 30 = 2 l \frac{\sqrt{3}}{2} = \sqrt{3} l$ Similarly, $A E = \sqrt{3} l$ ,
$A D = A O + O N + N D = l sin 30 + l + l sin 30 = l \times \frac{1}{2} + l + l \times \frac{1}{2} = 2 l$
$A B = A F = l$
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 AB
Force on mass m at A due to mass m at C is $F_{A C} = \frac{G m m}{{(A C)}^{2}} = \frac{G m m}{{(\sqrt{3} l)}^{2}} = \frac{G m^{2}}{{3 l}^{2}}$ along AC
Force on mass m at A due to mass m at D is $F_{A D} = \frac{G m m}{{(A D)}^{2}} = \frac{G m m}{{(2 l)}^{2}} = \frac{G m^{2}}{{4 l}^{2}}$ along AD
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 AE
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 AF
Resultant force due to $F_{A B}$ and $F_{A F}$ is $F_{R 1} = \sqrt{F_{A B}^{2} + F_{A F}^{2} + 2 F_{A B} F_{A F} cos 120}$
$= \sqrt{{(\frac{G m^{2}}{l^{2}})}^{2} + {(\frac{G m^{2}}{l^{2}})}^{2} + 2 (\frac{G m^{2}}{l^{2}}) (\frac{G m^{2}}{l^{2}}) (- \frac{1}{2})}$
$= \frac{G m^{2}}{l^{2}}$ along AD
Resultant force due to $F_{A C}$ and $F_{A E}$ is $F_{R 2} = \sqrt{F_{A C}^{2} + F_{A E}^{2} + 2 F_{A C} F_{A E} cos 60}$
$= \sqrt{{(\frac{G m^{2}}{3 l^{2}})}^{2} + {(\frac{G m^{2}}{3 l^{2}})}^{2} + 2 (\frac{G m^{2}}{3 l^{2}}) (\frac{G m^{2}}{3 l^{2}}) (\frac{1}{2})}$
$= \frac{\sqrt{3} G m^{2}}{3 l^{2}} = \frac{G m^{2}}{\sqrt{3} l^{2}}$ along AD
Net force on mass M along AD is $F_{R} = F_{R 1} + F_{R 2} + F_{A D} = \frac{G m^{2}}{l^{2}} + \frac{G m^{2}}{\sqrt{3} l^{2}} + \frac{G m^{2}}{4 l^{2}}$
$= \frac{G m^{2}}{l^{2}} [1 + \frac{1}{\sqrt{3}} + \frac{1}{4}] = \frac{G m^{2}}{l^{2}} [\frac{5}{4} + \frac{1}{\sqrt{3}}]$

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Similar questions

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