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प्रश्न
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.
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उत्तर
Consider the diagram below, in which six point masses are placed at six verticles A, B, C, D, E and F.
AC = AG + GC = 2AG
= `2l cos 30^circ`
= `2l sqrt(3)/2`
= `sqrt(3)l`
= AE
AD = AH + HJ + JD
= `l sin 30^circ + l + l sin 30^circ`
= `2l`
Force on mass m at A due to mass m at B is, `f_1 = (Gmm)/l^2` = along AB.
Force on mass m at A due to mass m at C is, `f_2 = (Gm xx m)/(sqrt(3)l)^2 = (Gm^2)/(3l^2)` along AC. ......[∵ AC = `sqrt(2)`l]
Force on mass m at A due to mass m at D is, `f_3 = (Gm xx m)/(2l)^2 = (Gm^2)/(4l^2)` along AD ......[∵ AD = 2l]
Force on mass m at A due to mass m at E is, `f_4 = (Gm xx m)/(sqrt(3)l)^2 = (Gm^2)/(3l^2)` along AE.
Force on mass m at A due to mass m at F is, `f_5 = (Gm xx m)/l^2 = (Gm^2)/l^2`along AF.
Resultant force due to `f_1` and `f_5` is `F_1 = sqrt(f_1^2 + f_5^2 + 2f_1f_5 cos 120^circ) = (Gm^2)/l^2` along AD. ...........[∵ Angle between `f_1` and `f_2` = 120°]
Resultant force due to `f_2` and `f_4` is `F_2 = sqrt(f_2^2 + f_4^2 + 2f_2f_4 cos 60^circ) = (sqrt(3)Gm^2)/(3l^2) = (Gm^2)/(sqrt(3)l^2)`along AD.
So, net force along AD = `F_1 + F_2 + F_3`
= `(Gm^2)/l^2 + (Gm^2)/(sqrt(3)l^2) + (Gm^2)/(4l^2)`
= `(Gm^2)/l^2 (1 + 1/sqrt(3) + 1/4)`
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