How do you find the moments $M_{x}$ $M_x$ and $M_{y}$ $M_y$ and the center of mass of the system?

Question

$m_{1} = 4, m_{2} = 3, m_{3} = 2, m_{4} = 5;$ $m_1 = 4, m_2 = 3, m_3 = 2, m_4 = 5;$

$P_{1} (- 4, 2), P_{2} (0, 5), P_{3} (3, 2), P_{4} (1, - 2)$ $P_1(−4, 2), P_2(0, 5), P_3(3, 2), P_4(1, −2)$

$M_{x} =$ $M_x =$ ???

$M_{y} =$ $M_y =$ ???

(x̄, ȳ) = ???

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Answer 1 · 2017-06-24T06:47:28

$¯ y = \frac{17}{14}$ $bary=17/14$

$¯ x = - \frac{5}{14}$ $barx=-5/14$

$M_{y} = 13$ $M_y=13$
$M_{x} = - 5$ $M_x=-5$

to find centre of mass

$¯ x \sum m = \sum m x$ $barxsum m=summx$

$¯ y \sum m = \sum m y$ $barysum m=summy$

$\sum m = 4 + 3 + 2 + 5 = 14$ $summ=4+3+2+5=14$

$14 \times ¯ y = (4 \times 2) + (3 \times 5) + (2 \times 2) + (5 \times - 2)$ $14xx bary= (4xx2)+(3xx5)+(2xx2)+(5xx -2)$

$14 \times ¯ y = 8 + 15 + 4 + - 10$ $14xxbary=8+15+4+ -10$

$14 ¯ y = 13$ $14bary=13$

$M_{y} = 13$ $M_y=13$

$¯ y = \frac{17}{14}$ $bary=17/14$

$14 \times ¯ x = (4 \times - 4) + (3 \times 0) + (2 \times 3) + (5 \times 1)$ $14xxbarx=(4xx-4)+(3xx0)+(2xx3)+(5xx1)$

$14 ¯ x = - 16 + 0 + 6 + 5$ $14barx=-16+0+6+5$

$¯ x = - \frac{5}{14}$ $barx=-5/14$

$M_{x} = - 5$ $M_x=-5$