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.

Explanation

`implies` Here each vertex has equal mass (m)

`AC = AG + CG =2AG`
` =2l cos30^@ = 2lsqrt3 //2`
`=3sqrt3`
Similarly `AE = AC = sqrt3l`
`AD +AH +HJ +JD`
`lsin30^@ +l +lsin30^@ = 2l`
Force on mass m at A due to mass m at B is
`F_1 = (Gmm)/(l^2) ` (along `vec(AB)` )
Force on mass m at A due to mass m at C is
`F_(2) = (Gmxxm)/(sqrt3l^2)=(Gm^2)/(3l^2) ` (along `vecAC` )
Force on mass m at A due to mass m at D is
`F_(3) = (Gmxxm)/(sqrt2l^2)=(Gm^2)/(4l^2) ` (along `vecAD` )
Force on mass m at A due to mass m at E is
`F_(4) = (Gmxxm)/(sqrt3l^2)=(Gm^2)/(3l^2) ` (along `vecAE` )
Force on mass m at A due to mass m at F is
`F_(5) = (Gmxxm)/(l^2)=(Gm^2)/(l^2) ` (along `vecAF` )
Resultant force due to `F_1 and F_5` is
`F._1=sqrt(F_1^2+F_5^2+2F_1F_5cos120^@)=(Gm^2)/(l^2)` (along `vec(AD))`
[ `:.` Angle between `F_1 and F_5 = 120^@` ]
Resultant force due to `F_2 and F_4` is
`F._2=sqrt(F_2^2+F_4^2+2F_2F_4cos60^(@))`
`=(sqrt3Gm^2)/(3l^2)=(Gm^2)/sqrt(3l^2) = ` (along `vecAD`)
[ `:.` Angle between `F_2 and F_4 =60^@` ]
Net force,
`=F._1 +F._2+F._3=(Gm^2)/l^2+(Gm^2)/sqrt(3l^2)+(Gm^2)/(4l^2)=(Gm^2)/l^2`
`(1+1/sqrt3+1/4)`