Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson

Chapter 13 - Exercises and Problems - Page 240: 56

Answer

$1180\ N$

Work Step by Step

The maximum tension occurs at the bottom of the path. We know that the period is equal to 10 seconds from the equation in the book. From this, it follows that the velocity is: $v = \omega r = \frac{2\pi}{10}\times 25 = 15.7 \ m/s$ Thus, we find: $F_t = mg + \frac{mv^2}{r} = (9.81)(60)+\frac{(60)(15.7)^2}{25}=1180\ N$

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