Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 8 - Section 8.3 - Applications to Physics and Engineering. - 8.3 Exercises - Page 567: 24

Answer

The moment about the $x$-axis is $$ \begin{aligned} M_{x}&=\sum_{i=1}^{4} m_{i} y_{i} \\ &=24 \end{aligned} $$ The moment about the $y$-axis is $$ \begin{aligned} M_{y}&=\sum_{i=1}^{4} m_{i} x_{i} \\ &=-5 \end{aligned} $$ The center of mass is $$ \begin{aligned} (\overline{x}, \overline{y})& =\left(\frac{M_{y}}{m}, \frac{M_{x}}{m}\right) \\ &=\left(\frac{M_{y}}{\sum_{i=1}^{4} m_{i}}, \frac{M_{x}}{\sum_{i=1}^{4} m_{i}}\right)\\ &=\left(\frac{-5}{18}, \frac{24}{18}\right) \\ &=\left(\frac{-5}{18}, \frac{4}{3}\right) . \end{aligned} $$

Work Step by Step

the system: $$ \begin{array}{l}{m_{1}=5, m_{2}=4, m_{3}=3, m_{4}=6} \\ {P_{1}(-4,2), P_{2}(0,5), P_{3}(3,2), P_{4}(1,-2)}\end{array} $$ The moment about the $x$-axis is $$ \begin{aligned} M_{x}&=\sum_{i=1}^{4} m_{i} y_{i} \\ &=m_{1} y_{1}+m_{2} y_{2}+m_{3} y_{3}+m_{4} y_{4} \\ &=5\cdot (2)+4\cdot (5)+3\cdot (2) +6\cdot (-2) \\ &=24 \end{aligned} $$ The moment about the $y$-axis is $$ \begin{aligned} M_{y}&=\sum_{i=1}^{4} m_{i} x_{i} \\ &=m_{1} x_{1}+m_{2} x_{2}+m_{3} x_{3} +m_{4} x_{4}\\ &=5\cdot (-4)+4\cdot (0)+3\cdot (3)+6\cdot (1) \\ &=-5 \end{aligned} $$ The center of mass is $$ \begin{aligned} (\overline{x}, \overline{y})& =\left(\frac{M_{y}}{m}, \frac{M_{x}}{m}\right) \\ &=\left(\frac{M_{y}}{\sum_{i=1}^{4} m_{i}}, \frac{M_{x}}{\sum_{i=1}^{4} m_{i}}\right)\\ &=\left(\frac{M_{y}}{m_{1} +m_{2} +m_{3}+m_{4}}, \frac{M_{x}}{m_{1} +m_{2} +m_{3}+m_{4} }\right)\\ &=\left(\frac{M_{y}}{5+4 +3+6}, \frac{M_{x}}{5+4 +3+6}\right)\\ &=\left(\frac{-5}{18}, \frac{24}{18}\right) \\ &=\left(\frac{-5}{18}, \frac{4}{3}\right) . \end{aligned} $$
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