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

Answer

(a) We can see the free body diagram below. (b) $T_A = 3.64~N$ $T_B = 1.68~N$
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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 $T_A$: $T_A = T_B+\frac{mg}{sin~30.0^{\circ}}$ $T_A = (1.68~N)+\frac{(0.100~kg)(9.80~m/s^2)}{sin~30.0^{\circ}}$ $T_A = 3.64~N$
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