Question

Sir particles of mass 1,2,3,4,5 and 6 kg are placed at the corners of a hexagon of side L. Find the distance between the center of this hexagon and centre of gravity.

Answer 1

Let the Centre of the hexagon be at coordinate (0,0).

Then the six individual points of the hexagon with the respective weight are at coordinates :  (-L/2, -L√3/2), (L/2, -L√3/2) ,(L,0) , (L/2, L√3/2), (-L/2, L√3/2), (-L,0),.

Hence X coordinate of the center of gravity =

[-(L/2 x 1) +(L/2 x 2) -( L x 3) + (L/2 x 4) -( L/2 x 5) - (L/2 x 6)]/21

= -L/7

Similarly Y coordinate of the center of gravity is =

[(-L√3/2 x 1) - (L√3/2 x 2) + 0x3 + (L√3/2 x 4) + (L√3/2 x 5) + 0x6 ]/21

= L√3/7

Hence distance between centre of hexagon (0,0) and centre of gravity (-L/7 , L√3/7)

= √(L²/49 + 3L²/49)

= 2L/7