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Question: Six point masses each of mass are placed at the vertices of a regular hexagon of side l. The force a...

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

A

B

C

D

Answer

Explanation

Solution

From figure,

Similarly, AE =

AD = AO + ON + ND = l sin 30∘30 ^ { \circ } + l + l sin 30∘30 ^ { \circ }

=1×12+1+1×12=21= 1 \times \frac { 1 } { 2 } + 1 + 1 \times \frac { 1 } { 2 } = 21

AB = AF = l

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

Force on mass m at A due to mass m at C is

FAC=Gmm(AC)2=Gmm(31)2=Gm2312\mathrm { F } _ { \mathrm { AC } } = \frac { \mathrm { Gmm } } { ( \mathrm { AC } ) ^ { 2 } } = \frac { \mathrm { Gmm } } { ( \sqrt { 3 } 1 ) ^ { 2 } } = \frac { \mathrm { Gm } ^ { 2 } } { 31 ^ { 2 } } along AC

Force on mass m at A due to mass m at D is

FAD=Gmm(AD)2=Gmm(2l)2=Gm24l2\mathrm { F } _ { \mathrm { AD } } = \frac { \mathrm { Gmm } } { ( \mathrm { AD } ) ^ { 2 } } = \frac { \mathrm { Gmm } } { ( 2 \mathrm { l } ) ^ { 2 } } = \frac { \mathrm { Gm } ^ { 2 } } { 4 \mathrm { l } ^ { 2 } } along AD

Force on mass m at A due to mass m at E is

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

FAF=Gmm(AF)2=Gm212\mathrm { F } _ { \mathrm { AF } } = \frac { \mathrm { Gmm } } { ( \mathrm { AF } ) ^ { 2 } } = \frac { \mathrm { Gm } ^ { 2 } } { 1 ^ { 2 } } along AF

Resultant force due to and FAF\mathrm { F } _ { \mathrm { AF } } is

FR1=FAB2+FAF2+2 FABFAFcos⁡120∘\mathrm { F } _ { \mathrm { R } 1 } = \sqrt { \mathrm { F } _ { \mathrm { AB } } ^ { 2 } + \mathrm { F } _ { \mathrm { AF } } ^ { 2 } + 2 \mathrm {~F} _ { \mathrm { AB } } \mathrm { F } _ { \mathrm { AF } } \cos 120 ^ { \circ } }

=Gm212= \frac { \mathrm { Gm } ^ { 2 } } { 1 ^ { 2 } } along AD

Resultant force due to FAC\mathrm { F } _ { \mathrm { AC } }and FAF\mathrm { F } _ { \mathrm { AF } }is

FR2=FAC2+FAE2+2 FACFAEcos⁡60∘\mathrm { FR } _ { 2 } = \sqrt { \mathrm { F } _ { \mathrm { AC } } ^ { 2 } + \mathrm { F } _ { \mathrm { AE } } ^ { 2 } + 2 \mathrm {~F} _ { \mathrm { AC } } \mathrm { F } _ { \mathrm { AE } } \cos 60 ^ { \circ } }

=(Gm2312)2+(Gm2312)2+2(Gm2312)(Gm2312)(12)= \sqrt { \left( \frac { \mathrm { Gm } ^ { 2 } } { 31 ^ { 2 } } \right) ^ { 2 } + \left( \frac { \mathrm { Gm } ^ { 2 } } { 31 ^ { 2 } } \right) ^ { 2 } + 2 \left( \frac { \mathrm { Gm } ^ { 2 } } { 31 ^ { 2 } } \right) \left( \frac { \mathrm { Gm } ^ { 2 } } { 31 ^ { 2 } } \right) \left( \frac { 1 } { 2 } \right) }

Net force on mass m along AD is