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Six point masses of mass m each are at t...

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.

A

`(Gm^(2))/(l^(2))[(5)/(4)+(1)/(sqrt(3))]`

B

`(Gm^(2))/(l^(2))[(3)/(4)+(1)/(sqrt(3))]`

C

`(Gm^(2))/(l^(2))[(5)/(4)-(1)/(sqrt(3))]`

D

`(Gm^(2))/(l^(2))[(3)/(4)-(1)/(sqrt(3))]`

Text Solution

Verified by Experts

The correct Answer is:
A
(a)
From Figure,
`AC=AM+MC=2AM=2lcos30^(@)=2l(sqrt(3))/(2)=sqrt(3)l`
Similarly, `AE=sqrt(3)l`
`AD=AO+ON+ND= l sin 30^(@)+l+ lsin 30^(@)`
`lxx(1)/(2)+l+lxx(1)/(2)=2l`
`AB=AF=l`
Force on mass m at A due to mass m at B is
`F_(AB)=(Gmm)/((AB)^(2))=(Gmm)/(l^(2))` along AB
Force on mass m at A due to mass m at C is
`F_(AC)=(Gmm)/((AC)^(2))=(Gmm)/(l^(2))` along AC
Force on mass m at A due to mass m at D is
`F_(AD)=(Gmm)/((AD)^(2))=(Gmm)/(l^(2))` along AD
Force on mass m at A due to mass m at E is
`F_(AE)=(Gmm)/((AE)^(2))=(Gmm)/(l^(2))` along AE
Force on mass m at A due to mass m at F is
`F_(AF)=(Gmm)/((AF)^(2))=(Gmm)/(l^(2))` along AF
Resultant force due to `F_(AB) " and " F_(AF)` is
`F_(R_(1))=sqrt(F_(AB)^(2)+F_(AF)^(2)+2F_(AB)F_(AF)cos 120^(@))`
`=sqrt(((Gm^(2))/(l^(2)))^(2)+((Gm^(2))/(l^(2)))^(2)+2((Gm^(2))/(l^(2)))((Gm^(2))/(l^(2)))(-(1)/(2)))`
`=(Gm^(2))/(l^(2))` along AD
Resultant force due to `F_(AC) " and " F_(AE)` is
`F_(R_(2))=sqrt(F_(AC)^(2)+F_(AE)^(2)+2F_(AC)F_(AE)cos 60^(@))`
`=sqrt(((Gm^(2))/(3l^(2)))^(2)+((Gm^(2))/(3l^(2)))^(2)+2((Gm^(2))/(3l^(2)))((Gm^(2))/(3l^(2)))((1)/(2)))`
`=(sqrt(3)Gm^(2))/(3l^(2))=(Gm^(2))/(sqrt(3)l^(2))` along AD ,brgt Net force on mass m along AD is
`F_(R)=F_(R_(1))+F_(R_(2))+F_(AD)=(Gm^(2))/(l^(2))+(Gm^(2))/(sqrt(3)l^(2))+(Gm^(2))/(4l^(2))`
`=(Gm^(2))/(l^(2))[1+(1)/(sqrt(3))+(1)/(4)]=(Gm^(2))/(l^(2))[(5)/(4)+(1)/(sqrt(3))]`
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