Question

Three particles of masses 1.0 kg, 2.0 kg and 3.0 kg are placed at the corners A, B and C respectively of an equilateral triangle ABC of edge 1 m. Locate the centre of mass of the system.

Answer 1

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Explanation:

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Answer 2

The center of mass of the system located at the points $\frac{7}{12} , \frac{3\sqrt{3} }{12}$

Explanation:

Step 1:

Given data in the question  

3 particles of mass 1.0 kg, 2.0 kg and 3.0 kg are placed at the corners A, B and C respectively of an edge 1 m equilateral ABC triangle

That means  

m₁ = 1 kg  

m₂ = 2 kg  

m₃ = 3 kg

x₁ = 0          

x₂ = 1        

x₃ = 1/2

y₁ = 0        

y₂ = 0          

y₃ = √3/2

Step 2:

So, The center position of the mass is =

$\frac{(m_1x_1+m_2x_2+m_3x_3)}{(m_1+m_2+m_3)} , \frac{m_1y_1+m_2y_2+m_3y_3}{m_1+m_2+m_3}$

Substitute the respective values in above equation

$\frac{(1\times 0+2\times 1+3\times \frac{1}{2} )}{(1+2+3)} , \frac{(1\times 0+2\times 0+3\times \frac{\sqrt{3} }{2}}{(1+2+3)}$

$\frac{( 0+2 + \frac{3}{2} )}{(6)} , \frac{( 0+ 0+3\times \frac{\sqrt{3} }{2}}{(6)}$

$\frac{7}{12} , \frac{3\sqrt{3} }{12}$

(From point B)

Three particles of masses  1.0 kg, 2.0 kg and 3.0 kg are placed at the corners A, B and C respectively of an edge 1 m equilateral ABC triangle .For this condition the center of mass of the system located at the point $\frac{7}{12} , \frac{3\sqrt{3} }{12}$