Four masses 1 kg, 1 kg, 2 kg and 2 kg are | vedclass

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(D) Let the square be in the $XY$-plane with $P(0, a), Q(a, a), R(a, 0)$ and $S(0, 0)$.The masses are $m_P = 1 \ kg, m_Q = 1 \ kg, m_R = 2 \ kg, m_S =

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$P$ and $Q$

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(D) Let the square be in the $XY$-plane with $P(0, a), Q(a, a), R(a, 0)$ and $S(0, 0)$.The masses are $m_P = 1 \ kg, m_Q = 1 \ kg, m_R = 2 \ kg, m_S = 2 \ kg$.The center of mass of $P$ and $S$ is at $O_1$ on the side $PS$ at a distance $y_1 = \frac{m_P \cdot a + m_S \cdot 0}{m_P + m_S} = \frac{1 \cdot a + 2 \cdot 0}{1 + 2} = \frac{a}{3}$ from $S$.The center of mass of $Q$ and $R$ is at $O_2$ on the side $QR$ at a distance $y_2 = \frac{m_Q \cdot a + m_R \cdot 0}{m_Q + m_R} = \frac{1 \cdot a + 2 \cdot 0}{1 + 2} = \frac{a}{3}$ from $R$.The total mass of the system is $M = 1 + 1 + 2 + 2 = 6 \ kg$.The center of mass of the whole system lies on the line joining $O_1$ and $O_2$ at a distance $x_{cm} = \frac{(m_P+m_S) \cdot 0 + (m_Q+m_R) \cdot a}{M} = \frac{3 \cdot 0 + 3 \cdot a}{6} = \frac{a}{2}$ from the $Y$-axis.Thus,the center of mass is at $(\frac{a}{2}, \frac{a}{3})$.Comparing distances from corners: The center of mass is closest to $R$ and $S$ (distance $\sqrt{(a/2)^2 + (a/3)^2}$) and farthest from $P$ and $Q$ (distance $\sqrt{(a/2)^2 + (2a/3)^2}$).

Four masses $1 \ kg, 1 \ kg, 2 \ kg$ and $2 \ kg$ are placed at the four corners $P, Q, R$ and $S$ of a square of side $a$ respectively. The center of mass of the system will be farthest from which of the following?

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