Consider six point masses with mass m placed at the vertices of a regular hexagon of side l. Now, find the force acting on any of the masses.

From figure,AE =AD=AM+MC=2AM=2lcos⁡30∘=2l32=3l Similary, AE=3lAD=AO+ON+ND=lsin30∘+l+lsin 300=2l AB = AF = lForce on mass m at A due to mass m at B isFAB=Gmm(AB)2=Gmml2 along AB FAF=Force on the mass m at a due to the mass m at F         =Gm2l2 along AFForce on mass m at A due to mass m at C isFAC=Gmm(AB)2=Gmm(3l)2=Gm23l2 along  ACFAD=Force on the mass m at A due to the mass m at D =Gm.m(2l)2 ⇒FAD=Gm24l2, along ADNet force on mass m along AD is given byFR=2FABcos 600+2FACcos 300+FAD        =2×Gm2l2×12+2×Gm2l2×32+Gm24l2       =Gm2l2(54+3)