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, \[F=\,{{F}_{1}}\,+\,{{F}_{2}}+{{F}_{3}}\]
\[=\,\frac{G{{m}^{2}}}{\sqrt{3}\,{{l}^{2}}}+\,\frac{G{{m}^{2}}}{{{l}^{2}}}+\,\frac{G{{m}^{2}}}{4{{l}^{2}}}\]
\[=\frac{G{{m}^{2}}}{{{l}^{2}}}\,\,\left(
\frac{1}{\sqrt{3}}+\,1+\frac{1}{4} \right)\]
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