Physics Technology Update (4th Edition)

Published by Pearson
ISBN 10: 0-32190-308-0
ISBN 13: 978-0-32190-308-2

Chapter 9 - Linear Momentum and Collisions - Problems and Conceptual Exercises - Page 292: 48

Answer

a. tips to the left b. II is the best explanation

Work Step by Step

The location of the center of mass of a two-dimensional system of objects is defined as $\displaystyle \mathrm{X}_{\mathrm{c}\mathrm{m}}=\frac{m_{1}x_{1}+m_{2}x_{2}+\cdots}{m_{1}+m_{2}+\cdots}=\frac{\sum mx}{M}\qquad$9-14 and $\displaystyle \mathrm{Y}_{\mathrm{c}\mathrm{m}}=\frac{m_{1}\mathrm{y}_{1}+m_{2}y_{2}+\cdots}{m_{1}+m_{2}+\cdots}=\frac{\sum my}{M}\qquad$9-15 --- a. Let $m_{1}$ be the mass to the left of the fulcrum, and $m_{2}$ to the right. $x_{1}$ is farther to the left of the fulcrum than $x_{2}$ is, because it has almost double the length of $m_{2}.$ Since $m_{1}=m_{2},$ $\displaystyle \mathrm{X}_{\mathrm{c}\mathrm{m}}=\frac{m(x_{1}+x_{2})}{2m}=\frac{x_{1}+x_{2}}{2},$ which will be left of the fulcrum, so the object will tip to the left. b. I is wrong. The sheet will be level if the center of mass is directly above it. II is correct. III is wrong for the same reason as I.
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