Essential University Physics: Volume 1 (3rd Edition)

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

Chapter 5 - Exercises and Problems - Page 86: 21


Please see the work below.

Work Step by Step

For left block $T=m_1gsin(60)$...........eq(1) For right block $T=m_2gsin(20)$...........eq(2) As the two blocks are in equilibrium, so we can equate the above two equations $m_1gsin(60)=m_2gsin(20)$ This can be rearranged as: $\frac{m_1}{m_2}=\frac{gsin(20)}{gsin(60)}$ $\implies \frac{m_1}{m_2}=0.394$ $m_1=0.394m_2$ Thus, the mass of left hand block is 0.394 times the right side block.
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