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Three identical particles each of mass m...

Three identical particles each of mass m are placed at the vertices of an equilateral triangle of side a. Fing the force exerted by this system on a particle P of mass m placed at the

(a) the mid point of a side
(b) centre of the triangle

Text Solution

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Using the superposition principle, the net gravitational force on P is `vec(F)=vec(F)_(A)+vec(F)_(B)+vec(F)_(C )`
(a) As shown in the figure, when P is at the mid point of a side, `vec(F)_(A)` and `vec(F)_(B)` will be equal in magnitude but opposite in direction. Hence they will cancel each other. So the net force on the particle P will be the force due to the particle placed at C only.

`rArr F = F_(C ) = G(m.m)/((CP)^(2))=G(m^(2))/((asin60)^(2)) = (4Gm^(2))/(3a^(2))` along PC.
(b) At the centre of triangle O, the forces `vec(F)_(A), vec(F)_(B)` and `vec(F)_(C )` will be equal in magnitude and will subtend `120^(@)` with each other. Hence the resultant on P at O is `vec(F) = vec(F)_(A)+vec(F)_(B)+vec(F)_(C ) = 0`.
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