College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 5 - Problems - Page 186: 19


(a) We can see the free body diagram below. (b) $T_A = 3.64~N$ $T_B = 1.68~N$

Work Step by Step

(a) We can see the free body diagram below. (b) We can make an equation with the vertical components of the forces: $\sum F_y = 0$ $T_A~sin~30.0^{\circ}-T_B~sin~30.0^{\circ}-mg = 0$ $T_A~sin~30.0^{\circ} = T_B~sin~30.0^{\circ} + mg$ $T_A = T_B+\frac{mg}{sin~30.0^{\circ}}$ The sum of the horizontal components of the two tension forces provide the centripetal force to keep the ball moving in a circle: $T_A~cos~30.0^{\circ}+T_B~cos~30.0^{\circ} = m~\omega^2~r$ $T_A~cos~30.0^{\circ}+T_B~cos~30.0^{\circ} = m~\omega^2~(0.150~m)~cos~30.0^{\circ}$ $T_A+T_B = m~\omega^2~(0.150~m)$ $(T_B+\frac{mg}{sin~30.0^{\circ}})+T_B = m~\omega^2~(0.150~m)$ $2T_B = m~\omega^2~(0.150~m) - \frac{mg}{sin~30.0^{\circ}}$ $T_B = \frac{m~\omega^2~(0.150~m)}{2} - \frac{mg}{2sin~30.0^{\circ}}$ $T_B = \frac{(0.100~kg)(6.00\pi~rad/s)^2~(0.150~m)}{2} - \frac{(0.100~kg)(9.80~m/s^2)}{2sin~30.0^{\circ}}$ $T_B = 1.68~N$ We can find $T_A$: $T_A = T_B+\frac{mg}{sin~30.0^{\circ}}$ $T_A = (1.68~N)+\frac{(0.100~kg)(9.80~m/s^2)}{sin~30.0^{\circ}}$ $T_A = 3.64~N$
Small 1533395905
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.