Masses of 8 kg, 2 kg, 4 kg, and 2 kg are | vedclass

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(B) Let the side length of the square be $a$. The diagonal is $d = a\sqrt{2} = 80 \ cm$,so $a = 40\sqrt{2} \ cm$.Place $A$ at $(0, 0)$,$B$ at $(a, 0)$

Vedclass · Accepted Answer

$30$

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(B) Let the side length of the square be $a$. The diagonal is $d = a\sqrt{2} = 80 \ cm$,so $a = 40\sqrt{2} \ cm$.Place $A$ at $(0, 0)$,$B$ at $(a, 0)$,$C$ at $(a, a)$,and $D$ at $(0, a)$.The masses are $m_A = 8 \ kg, m_B = 2 \ kg, m_C = 4 \ kg, m_D = 2 \ kg$.The coordinates of the center of mass $(X, Y)$ are:$X = \frac{m_A x_A + m_B x_B + m_C x_C + m_D x_D}{m_A + m_B + m_C + m_D} = \frac{8(0) + 2(a) + 4(a) + 2(0)}{8 + 2 + 4 + 2} = \frac{6a}{16} = \frac{3a}{8}$.$Y = \frac{m_A y_A + m_B y_B + m_C y_C + m_D y_D}{m_A + m_B + m_C + m_D} = \frac{8(0) + 2(0) + 4(a) + 2(a)}{16} = \frac{6a}{16} = \frac{3a}{8}$.Substituting $a = 40\sqrt{2} \ cm$:$X = \frac{3(40\sqrt{2})}{8} = 15\sqrt{2} \ cm$ and $Y = 15\sqrt{2} \ cm$.The distance from $A(0, 0)$ is $r = \sqrt{X^2 + Y^2} = \sqrt{(15\sqrt{2})^2 + (15\sqrt{2})^2} = \sqrt{450 + 450} = \sqrt{900} = 30 \ cm$.

Masses of $8 \ kg, 2 \ kg, 4 \ kg,$ and $2 \ kg$ are placed at the corners $A, B, C,$ and $D$ respectively of a square $ABCD$ with a diagonal of $80 \ cm$. The distance of the center of mass from $A$ is ....... $cm$.

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