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 8 - Dynamics II: Motion in a Plane - Exercises and Problems: 46

Answer

The tension in the bottom wire is 8.3 N The tension in the top wire is 14.2 N

Work Step by Step

We can find the angle $\theta$ that each wire makes with the vertical. $cos(\theta) = \frac{0.5~m}{1.0~m}$ $\theta = arccos(\frac{0.5~m}{1.0~m})$ $\theta = 60^{\circ}$ We can find the radius of the circle that the ball rotates. $\frac{r}{1.0~m} = sin(60^{\circ})$ $r = 0.866~m$ Let $T_t$ be the tension in the top wire. Let $T_b$ be the tension in the bottom wire. The vertical component of $T_t$ is equal to the sum of the vertical component of $T_b$ and the ball's weight. $T_t~cos(\theta) = T_b~cos(\theta) + mg$ $T_t = \frac{T_b~cos(\theta) + mg}{cos(\theta)}$ The sum of the horizontal component in each wire provides the centripetal force to keep the ball moving in a circle. $T_t~sin(\theta)+T_b~sin(\theta) = \frac{mv^2}{r}$ $(\frac{T_b~cos(\theta) + mg}{cos(\theta)})~sin(\theta)+T_b~sin(\theta) = \frac{mv^2}{r}$ $mg~tan(\theta)+2~T_b~sin(\theta) = \frac{mv^2}{r}$ $T_b = \frac{\frac{mv^2}{r}-mg~tan(\theta)}{2~sin(\theta)}$ $T_b = \frac{\frac{(0.300~kg)(7.5~m/s)^2}{0.866~m}-(0.300~kg)(9.80~m/s^2)~tan(60^{\circ})}{2~sin(60^{\circ})}$ $T_b = 8.3~N$ The tension in the bottom wire is 8.3 N We can use $T_b$ to find $T_t$. $T_t = \frac{T_b~cos(\theta) + mg}{cos(\theta)}$ $T_t = \frac{(8.3~N)~cos(60^{\circ}) + (0.300~kg)(9.80~m/s^2)}{cos(60^{\circ})}$ $T_t = 14.2~N$ The tension in the top wire is 14.2 N
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