Each side of a rhombus has length a. Four par | vedclass

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(B) Let the masses at the vertices be $m_1 = m, m_2 = 2m, m_3 = 3m, m_4 = 4m$.Based on the geometry of the rhombus in the $x-y$ plane:$r_1 = (0, 0)$$r

Vedclass · Accepted Answer

$\left( 0.95a, \frac{\sqrt{3}}{4}a \right)$

Vedclass · Answer

(B) Let the masses at the vertices be $m_1 = m, m_2 = 2m, m_3 = 3m, m_4 = 4m$.Based on the geometry of the rhombus in the $x-y$ plane:$r_1 = (0, 0)$$r_2 = (a \cos 60^o, a \sin 60^o) = (a/2, a\sqrt{3}/2)$$r_3 = (a + a \cos 60^o, a \sin 60^o) = (3a/2, a\sqrt{3}/2)$$r_4 = (a, 0)$The center of mass $(X_{cm}, Y_{cm})$ is given by:$X_{cm} = \frac{m_1x_1 + m_2x_2 + m_3x_3 + m_4x_4}{m_1 + m_2 + m_3 + m_4} = \frac{m(0) + 2m(a/2) + 3m(3a/2) + 4m(a)}{10m} = \frac{0 + am + 4.5am + 4am}{10m} = \frac{9.5am}{10m} = 0.95a$$Y_{cm} = \frac{m_1y_1 + m_2y_2 + m_3y_3 + m_4y_4}{m_1 + m_2 + m_3 + m_4} = \frac{m(0) + 2m(a\sqrt{3}/2) + 3m(a\sqrt{3}/2) + 4m(0)}{10m} = \frac{am\sqrt{3} + 1.5am\sqrt{3}}{10m} = \frac{2.5a\sqrt{3}}{10} = \frac{\sqrt{3}}{4}a$Thus,the center of mass is $(0.95a, \frac{\sqrt{3}}{4}a)$.

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