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 7 - Newton's Third Law - Exercises and Problems - Page 188: 36


$T= \frac{M~m~g}{M+m}$

Work Step by Step

Let's consider the system of block $m$. We can set up a force equation for this block. $\sum F = ma$ $mg - T = ma$ $a = \frac{mg-T}{m}$ We can use this expression for $a$ in the force equation for the block of mass $M$. Note that both blocks have the same magnitude of acceleration. $T = Ma$ $T = (M)(\frac{mg-T}{m})$ $mT +MT= M~m~g$ $T= \frac{M~m~g}{M+m}$
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