University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 11 - Equilibrium and Elasticity - Problems - Exercises - Page 357: 11.2

Answer

The 1.50-kg mass should be glued a distance of 9.53 cm to the left of the original center of gravity.

Work Step by Step

Let $m_1$ be the 5.00-kg object and let $m_2$ be the 1.50-kg mass. Let the original center of gravity be the origin. Let $r_2$ be the distance to the left of the original center of mass, where we will glue the 1.50-kg mass. $x_{cog} = \frac{m_1~r_1+m_2~r_2}{m_1+m_2}$ $m_2~r_2 = x_{cog}~(m_1+m_2)-m_1~r_1$ $r_2 = \frac{x_{cog}~(m_1+m_2)-m_1~r_1}{m_2}$ $r_2 = \frac{(2.20~cm)(5.00~kg+1.50~kg)-(5.00~kg)(0)}{1.50~kg}$ $r_2 = 9.53~cm$ The 1.50-kg mass should be glued a distance of 9.53 cm to the left of the original center of gravity.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.