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Question

Three particles of masses 1 kg,2 kg and 3 kg are situated at the corners of an equilateral triangle of side b. The (x,y) coordinates respectively for the centre of mass of the system of particles will be:

A
[7b12,3b12]
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B
[33b12,7b12]
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C
[7b12,33b12]
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D
[7b12, 33b4]
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Solution

The correct option is C [7b12,33b12]
The coordinates of points A, B and C are
A(0,0)
B(b,0)
C(b2, bsin60)=(b2, 3b2)


Applying the formula for coordinates of COM of system:
xCM=m1x1+m2x2+m3x3m1+m2+m3
xCM=(1×0)+(2×b)+(3×b2)1+2+3=7b12
Simillarly,
yCM=m1y1+m2y2+m3y3m1+m2+m3
yCM=(1×0)+(2×0)+(3×3 b2)1+2+3=33 b12
So, the coordinates of the centre of mass are [7b12,33b12]

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