Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 8 - Further Applications of Integration - 8.3 Applications to Physics and Engineering - 8.3 Exercises - Page 607: 24

Answer

$M_y=-5$ $M_x=24$ So the center of mass is $\left(-0.28;1.33\right)$

Work Step by Step

$$M_y=m_1\cdot x_1+m_{2}\cdot x_2+m_{3}\cdot x_3+m_{4}\cdot x_4=5\cdot (-4)+4\cdot 0+3\cdot 3+6\cdot 1=-5$$ $$M_x=m_1\cdot y_1+m_{2}\cdot y_2+m_{3}\cdot y_3+m_{4}\cdot x_4=5\cdot 2+4\cdot 5+3\cdot 2+6\cdot (-2)=24$$ so the center of mass is: $$\left(\frac{-5}{5+4+3+6}\approx -0.28;\frac{24}{5+4+3+6}\approx 1.33\right)$$
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