Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 4 - Dynamics: Newton's Laws of Motion - Problems - Page 103: 35


The tension in cord C is 34 N

Work Step by Step

We know that the magnitude of the acceleration will be the same for both blocks. Let $F_T$ be the tension in the cord connected to both blocks. For the 1.2-kg block: $F_T - mg = ma$ $a = \frac{F_T-mg}{m}$ For the 3.2-kg block: $Mg - F_T = Ma$ $Mg - F_T = M(\frac{F_T-mg}{m})$ $Mmg - mF_T = MF_T - Mmg$ $F_T = \frac{2Mmg}{M+m}$ $F_T = \frac{(2)(3.2 ~kg)(1.2 ~kg)(9.8 ~m/s^2)}{3.2 ~kg+1.2 ~kg}$ $F_T = 17 ~N$ The tension in the cord connected to both blocks is 17 N. So, the tension in cord C is $2\times F_T$ which is equal to 34 N
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