Question

Four particles of masses m, m, 2m and 2m are placed at the four corners of square of side a. Find the centre of mass of the system.

Answer 1

Explanation:

see attached figure, here Four particles of masses m, 2m, 3m , 4m are kept in square ABCD at the corners of a square of side a. a particle of mass m is placed at centre O of square.

now, force exerted by A on O = Gm²/(a/√2)² = 2Gm²/a²

force exerted by B on O = 2Gm²/(a/√2)²

= 4Gm²/a²

force exerted by C on O = 3Gm²/(a/√2)² = 6Gm²/a²

force exerted by D on O = 4Gm²/(a/√2)²

= 8Gm²/a²

Let us consider F = 2Gm²/a²

then, force exerted by A on O = F

force exerted by B on O = 2F

force exerted by C on O = 3F

force exerted by D on O = 4F

see figure , resultant force of B and D= 4F - 2F = 2F along CO

and resultant force of A and C = 3F - F = 2F along DO

now, net force acting on O = √{(2F)² + (2F)²} = 2√2F

putting F = 2Gm²/a²

so, net force acting on O = 4√2Gm²/a²

Answer 2

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