University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 5 - Applying Newton's Laws - Problems - Exercises - Page 164: 5.51


The slanted cable has a tension of 1410 N. The horizontal cable has a tension of 6210 N.

Work Step by Step

We can find the speed $v$ when the rotation rate is 28.0 rpm. $v = (28.0~rpm)(\frac{1~min}{60~s})(\frac{(2\pi)(7.50~m)}{1~rev})$ $v = 22.0~m/s$ Let $T_1$ be the tension in the slanted cable. Note that the total weight of the passenger and the seat is 1080 N and the total mass $m$ is $\frac{1080~N}{g}$ $T_1~cos(40.0^{\circ}) = mg$ $T_1 = \frac{mg}{cos(40.0^{\circ})} = \frac{1080~N}{cos(40.0^{\circ})}$ $T_1 = 1410~N$ Let $T_2$ be the tension in the horizontal cable. $T_2 + T_1~sin(40.0^{\circ}) = \frac{mv^2}{r}$ $T_2 = \frac{mv^2}{r}- T_1~sin(40.0^{\circ})$ $T_2 = \frac{(1080~N)(22.0~m/s)^2}{(7.50~m)(9.80~m/s^2)}- (1410~N)~sin(40.0^{\circ})$ $T_2 = 6210~N$
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