Question

1. Three bodies of masses m, 2m and 3m are placed at the corners of a triangle
having coordinates (1, 1.5), (2.5, 1.5) and (3, 3) respectively. Calculate the
coordinates of the centre of mass.
2. Three bodies of masses 5 kg, 4 kg and 2 kg have the position vectors as
2 + 3) -3k, i + j + k and 3î - 39 - 4. Find the coordinates of centre of
mass.
3. Find the coordinates of centre of mass of a square of side 1 m in which four
particles of masses 2m, 2m, 3m, 5m are placed at four corners.
4. The coordinates of centre of mass of three particles of masses 1 kg 2 kg,
3 kg are (2m, 2m, 2m), Where should a fourth
particle of mass 4 kg be placed
so that the coordinates of centre of mass are (Am, 4m, 4m)?

Answer 1

Answer:

The center of mass of an equilateral triangle lies at its geometrical center G.

The positions of the mass m1 , m2 and m3 are at positions A, B and C as shown in the Figure. From the given position of the masses, the coordinates of the masses m1 and m2 are easily marked as (0,0) and (1,0) respectively.

To find the position of m3 the Pythagoras theorem is applied. As the ∆DBC is a right angle triangle, 

BC2 = CD2 + DB2 

CD2 = BC2 – DB2

CD2 = 12 – (\(\frac{1}{2}\))2 = 1 – (\(\frac{1}{4}\)) = \(\frac{3}{2}\) 

CD =\(\frac{\sqrt 3}{2}\) 



The position of mass m is or (0.5, 0.5 \(\sqrt 3\))

X coordìnate of center of mass,



Explanation:

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