Three particles of mass 1 kg, 2 kg and 3 k | vedclass

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Question

(A) The coordinates of the vertices are $A(0, 0)$,$B(1, 0)$,and $C(1/2, \sqrt{3}/2)$.The masses are $m_A = 1 \ kg$,$m_B = 2 \ kg$,and $m_C = 3 \ kg$.T

Vedclass · Accepted Answer

$\left(\frac{7}{12}, \frac{3 \sqrt{3}}{12}\right)$

Vedclass · Answer

(A) The coordinates of the vertices are $A(0, 0)$,$B(1, 0)$,and $C(1/2, \sqrt{3}/2)$.The masses are $m_A = 1 \ kg$,$m_B = 2 \ kg$,and $m_C = 3 \ kg$.The total mass $M = m_A + m_B + m_C = 1 + 2 + 3 = 6 \ kg$.The $x$-coordinate of the centre of mass is given by:$x_{cm} = \frac{m_A x_A + m_B x_B + m_C x_C}{M} = \frac{1(0) + 2(1) + 3(1/2)}{6} = \frac{0 + 2 + 1.5}{6} = \frac{3.5}{6} = \frac{7}{12}$.The $y$-coordinate of the centre of mass is given by:$y_{cm} = \frac{m_A y_A + m_B y_B + m_C y_C}{M} = \frac{1(0) + 2(0) + 3(\sqrt{3}/2)}{6} = \frac{3\sqrt{3}/2}{6} = \frac{3\sqrt{3}}{12}$.Thus,the centre of mass is $\left(\frac{7}{12}, \frac{3\sqrt{3}}{12}\right)$.

Three particles of mass $1 \ kg, 2 \ kg$ and $3 \ kg$ are placed at the vertices $A, B$ and $C$ respectively of an equilateral triangle $ABC$ of side $1 \ m$. The centre of mass of the system from vertex $A$ (located at origin) is

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