question_answer 60)

                  Six point masses of mass \[m\] each are at the vertices of a regular hexagon of side 1. Calculate the force on any of the masses.

Answer:

                                  From figure, \[FM=FD=l\text{ }\cos 30{}^\circ =\frac{\sqrt{3}l}{2}\]                 \[\therefore \]\[FD=\,FB=\,\sqrt{3}\,l\]             \[FO\,=PC=l\,\cos \,{{60}^{o}}\,=\frac{l}{2}\]                               \[\therefore \]\[FC=\,\frac{l}{2}+\,\frac{l}{2}+\,l=2l\]                 Now, force on m at F due to m at D,                 \[{{f}_{4}}=\,\frac{G{{m}^{2}}}{3{{l}^{2}}}\]                 Force on m at F due to m at B, \[{{f}_{5}}=\frac{a{{m}^{2}}}{3{{l}^{2}}}\]                 \[\therefore \] Resultant force due to \[{{f}_{4}}\] and \[{{f}_{5}}\]                        \[=\,{{F}_{1}}=\,\sqrt{f_{4}^{2}+f_{5}^{2}+2{{f}_{4}}{{f}_{5}}\,\cos \,{{60}^{o}}}\]                 \[=\,\frac{\sqrt{3}\,G{{m}^{2}}}{3{{l}^{2}}}=\,\frac{G{{m}^{2}}}{\sqrt{3}{{l}^{2}}}\]                 Force on m at F due to at E, \[{{f}_{2}}=\frac{G{{m}^{2}}}{{{l}^{2}}}\]                 Force on m at F due to m at A, \[{{f}_{1}}\,=\,\frac{G{{m}^{2}}}{{{l}^{2}}}\]                 \ Resultant of \[{{f}_{1}}\] and \[{{f}_{2}}\].                 \[{{F}_{2}}=\,\sqrt{f_{1}^{2}+f_{2}^{2}+2{{f}_{1}}{{f}_{2}}\,\cos \,{{120}^{o}}}=\,\frac{G{{m}^{2}}}{{{l}^{2}}}\]                 Force on m at F due to m at C, \[{{F}_{3}}=\,\frac{G{{m}^{2}}}{4{{l}^{2}}}\]                 \[\therefore \] Net force, \[F=\,{{F}_{1}}\,+\,{{F}_{2}}+{{F}_{3}}\]                 \[=\,\frac{G{{m}^{2}}}{\sqrt{3}\,{{l}^{2}}}+\,\frac{G{{m}^{2}}}{{{l}^{2}}}+\,\frac{G{{m}^{2}}}{4{{l}^{2}}}\]                 \[=\frac{G{{m}^{2}}}{{{l}^{2}}}\,\,\left( \frac{1}{\sqrt{3}}+\,1+\frac{1}{4} \right)\]