Question

Three point masses m, = 2 kg, m, = 4 kg and m = 6 kg are kept at the three corners of an
equilateral triangle of side 1 m. Find the location of their centre of mass.
​

Answer 1

SoluTion:

Let -

• $\sf{m_1}$ be at the origin.

• x - axis along the line joining $\sf{m_1}$ to $\sf{m_2}$ [ Refer to the attachment ].

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†It is clear that, x-y coordinates of $\sf{m_1}$ is $\sf{\bigg( x_{1} , y_{1} \bigg) = \bigg( 0,0 \bigg)}$,

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†Same of $\sf{m_2}$ is $\sf{\bigg( x_{2} , y_{2} \bigg) = \bigg( 1,0 \bigg)}$,

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†Same of $\sf{m_3}$ is $\sf{\bigg( x_{3} , y_{3} \bigg) = \bigg( \dfrac{1}{2} , \dfrac{\sqrt{3}}{2} \bigg)}$

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We know that,

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$\sf{x_{CM} = \dfrac{m_{1} x_{1} + m_{2} x_{2} + m_{3} x_{3}}{m_{1} + m_{2} + m_{3}}}$

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$\longrightarrow$ $\sf{x_{CM} = \dfrac{2 \times 0 + 4 \times 1 + 6 \times \dfrac{1}{2}}{2+4+6}}$

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$\longrightarrow$ $\sf{x_{CM} = \dfrac{7}{12}\:m}$

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$\longrightarrow$ $\sf{y_{CM} = \dfrac{2 \times 0 + 4 \times 0 + 6 \times \dfrac{\sqrt{3}}{2}}{2+4+6}}$

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$\longrightarrow$ $\sf{y_{CM} = \dfrac{3 \sqrt{3}}{12}}$

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$\longrightarrow$ $\sf{y_{CM} = \dfrac{\sqrt{3}}{4}\:m}$

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$\therefore$ $\large{\boxed{\sf{CM\:is\:at\:\bigg( \dfrac{7}{12} , \dfrac{\sqrt{3}}{4} \bigg)}}}$