Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 12 - Rotation of a Rigid Body - Exercises and Problems: 31

Answer

The 4.0-kg cat should stand a distance of 1.5 meters to the left of the pivot.

Work Step by Step

Let $m_1$ be the mass of the bowl. Let $m_2$ be the mass of the 4.0-kg cat. Let $m_3$ be the mass of the 5.0-kg cat. To keep the seesaw balanced, the net torque must be zero. $\sum \tau = 0$ $(r_1\times m_1~g)+(d\times m_2~g)+(r_3\times m_3~g) = 0$ $d~m_2~g = r_3~m_3~g-r_1~m_1~g$ $d = \frac{r_3~m_3-r_1~m_1}{m_2}$ $d = \frac{(2.0~m)(5.0~kg)-(2.0~m)(2.0~kg)}{4.0~kg}$ $d = 1.5~m$ The 4.0-kg cat should stand a distance of 1.5 meters to the left of the pivot.
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