Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson
ISBN 10: 0321993721
ISBN 13: 978-0-32199-372-4

Chapter 12 - Exercises and Problems - Page 216: 35


86.67 kilograms

Work Step by Step

We know that the net torque must be 0. Thus, calling the bottom of the log the axis of rotation, we find: $2.1Mgsin63^{\circ}+6.3m_cgsin63^{\circ}=(6.3)(.92)(M+m_c)(g)sin27^{\circ}$ $2.1(340)(9.81)sin63^{\circ}+6.3m_c(9.81)sin63^{\circ}=(6.3)(.92)(340+m_c)(9.81)sin27^{\circ}$ $m_c=\fbox{86.67 kilograms }$
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