Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 7 - Linear Momentum - Problems - Page 194: 52

Answer

$\frac{23}{6}(\mathcal{l_o})\approx 3.8 \mathcal{l_o}$.

Work Step by Step

By symmetry, the CM lies on the line connecting all of the centers, the x-axis. Let’s calculate the CM’s coordinate in the x-direction. The mass of each cube is volume times the density. $$m_1=\rho (\mathcal{l_o})^3$$ $$m_2=\rho (2\mathcal{l_o})^3$$ $$m_3=\rho (3\mathcal{l_o})^3$$ Locate the CM of each cube. $$x_1=0.5 \mathcal{l_o}$$ $$x_2=2 \mathcal{l_o}$$ $$x_3=4.5 \mathcal{l_o}$$ Finally, use the above numbers in equation 7–9a to calculate the CM as $\frac{23}{6}(\mathcal{l_o})\approx 3.8 \mathcal{l_o}$.
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