Three point masses m_1 = 1.6 , kg,m_2 = 2.0 , | vedclass

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(C) The coordinates of the masses are as follows:$A(0, 0)$ with mass $m_1 = 1.6 \, kg$$B(1.2, 0)$ with mass $m_2 = 2.0 \, kg$$C(0, 1.0)$ with mass $m_

Vedclass · Accepted Answer

$(0.4, 0.4)$

Vedclass · Answer

(C) The coordinates of the masses are as follows:$A(0, 0)$ with mass $m_1 = 1.6 \, kg$$B(1.2, 0)$ with mass $m_2 = 2.0 \, kg$$C(0, 1.0)$ with mass $m_3 = 2.4 \, kg$The $x$-coordinate of the center of mass is given by:$x_{cm} = \frac{m_1 x_1 + m_2 x_2 + m_3 x_3}{m_1 + m_2 + m_3}$$x_{cm} = \frac{(1.6)(0) + (2.0)(1.2) + (2.4)(0)}{1.6 + 2.0 + 2.4} = \frac{2.4}{6.0} = 0.4 \, m$The $y$-coordinate of the center of mass is given by:$y_{cm} = \frac{m_1 y_1 + m_2 y_2 + m_3 y_3}{m_1 + m_2 + m_3}$$y_{cm} = \frac{(1.6)(0) + (2.0)(0) + (2.4)(1.0)}{1.6 + 2.0 + 2.4} = \frac{2.4}{6.0} = 0.4 \, m$Thus,the center of mass is located at $(0.4, 0.4) \, m$.

Three point masses $m_1 = 1.6 \, kg$,$m_2 = 2.0 \, kg$,and $m_3 = 2.4 \, kg$ are placed at the corners of a thin massless rectangular sheet $(1.2 \, m \times 1.0 \, m)$ as shown. The center of mass will be located at the point ........... $m$.

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