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Three equal masses of m kg each are fixe...

Three equal masses of m kg each are fixed at the vertices of an equilateral triangle ABC.
(a) What is the force acting on a mass 2m placed at the centroid G of the triangle?
(b) What is the force if the mass at the vertex A is doubled?
Take AG=BG=CG=1m (see Fig.8.5)

Text Solution

Verified by Experts

The angle between GR and the positive x - axis is `30^(@)` . Similarly the angle between GQ and negative x - axis is `30^(@)`
Now , `F_(GP) = (G(2m)m)/1 hat(j) ............." "` (i.e ., along positive y - axis )
`F_(GQ) = (G(2m)m)/1 (-hat(i) cos 30^(@)- hat(j) sin 30^(@))`
` F_(GR) = (G(2m)m)/1 ( hat(i) cos 30^(@) - hat(j) sin 30^(@))`
By principle of superposition , we get
`vec(F)_(R) = vec(F)_(GP) = vec(F)_(GQ)+vec(F)_(GR)`
` :. " " vec(F)_(R) = 2Gm^(2) hat(j) +2Gm^(2)(-hat(i) cos 30^(@)- hat(j) sin 30^(@)) + 2Gm^(2) (hat(i) cos 30^(@) - hat(j) sin 30^(@)) = bar(0)`
(ii) ` :. vec(F)_(R) = 8Gm^(2)hat(j) + 2Gm^(2) hat(j) sin 30^(@) - 2Gm^(2) hat(j) sin 30^(@)`
` = 8Gm^(2)hat(j) - 4Gm^(2) sin 30^(@) `
` = 8Gm^(2)hat(j) - 2Gm^(2)hat(j)`
` = 6Gm^(2) hat(j)`
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