Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 7 - Newton's Third Law - Exercises and Problems - Page 179: 36


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

Work Step by Step

We can find the acceleration of the system of both blocks as; $F = (M+m)~a$ $mg = (M+m)~a$ $a = \frac{mg}{M+m}$ Let's consider the system of only the block of mass $M$. The tension $T$ in the string provides the force to move this block with the acceleration we found above; $T = Ma$ $T = (M)(\frac{mg}{M+m})$ $T = \frac{M~m~g}{M+m}$
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