Three point particles of masses 1.0 ; kg,1.5 | vedclass

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(B) Let the $1.0 \; kg$ mass be at the origin $(0, 0)$.The coordinates of the three masses are:$m_1 = 1.0 \; kg$ at $(0, 0) \; cm$$m_2 = 1.5 \; kg$ at

Vedclass · Accepted Answer

$0.9 \; cm$ right and $2.0 \; cm$ above $1 \; kg$ mass

Vedclass · Answer

(B) Let the $1.0 \; kg$ mass be at the origin $(0, 0)$.The coordinates of the three masses are:$m_1 = 1.0 \; kg$ at $(0, 0) \; cm$$m_2 = 1.5 \; kg$ at $(3, 0) \; cm$$m_3 = 2.5 \; kg$ at $(0, 4) \; cm$The $x$-coordinate of the center of mass is:$x_{cm} = \frac{m_1x_1 + m_2x_2 + m_3x_3}{m_1 + m_2 + m_3} = \frac{1.0(0) + 1.5(3) + 2.5(0)}{1.0 + 1.5 + 2.5} = \frac{4.5}{5.0} = 0.9 \; cm$The $y$-coordinate of the center of mass is:$y_{cm} = \frac{m_1y_1 + m_2y_2 + m_3y_3}{m_1 + m_2 + m_3} = \frac{1.0(0) + 1.5(0) + 2.5(4)}{1.0 + 1.5 + 2.5} = \frac{10.0}{5.0} = 2.0 \; cm$Thus,the center of mass is at $0.9 \; cm$ to the right and $2.0 \; cm$ above the $1.0 \; kg$ mass.

Three point particles of masses $1.0 \; kg$,$1.5 \; kg$,and $2.5 \; kg$ are placed at three corners of a right-angled triangle of sides $4.0 \; cm$,$3.0 \; cm$,and $5.0 \; cm$ as shown in the figure. The center of mass of the system is at a point:

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