## Essential University Physics: Volume 1 (4th Edition)

$1180\ N$
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: $F_t = mg + \frac{mv^2}{r} = (9.81)(60)+\frac{(60)(15.7)^2}{25}=1180\ N$