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 6 - Dynamics I: Motion Along a Line - Exercises and Problems - Page 156: 39


$T_1 = 6396~N$ $T_2 = 4375~N$

Work Step by Step

The magnitude of the horizontal component of the tension in each rope must be equal. $T_1~sin(20^{\circ})=T_2~sin(30^{\circ})$ $T_1=\frac{T_2~sin(30^{\circ})}{sin(20^{\circ})}$ The sum of the vertical components of the tension in each rope must be equal in magnitude to the weight of the steel beam. $T_1~cos(20^{\circ})+T_2~cos(30^{\circ}) = mg$ $T_2~sin(30^{\circ})~cot(20^{\circ})+T_2~cos(30^{\circ}) = mg$ $T_2 = \frac{mg}{sin(30^{\circ})~cot(20^{\circ})~+~cos(30^{\circ})}$ $T_2 = \frac{(1000~kg)(9.80~m/s^2)}{sin(30^{\circ})~cot(20^{\circ})~+~cos(30^{\circ})}$ $T_2 = 4375~N$ We can use $T_2$ to find $T_1$. $T_1=\frac{T_2~sin(30^{\circ})}{sin(20^{\circ})}$ $T_1=\frac{(4375~N)~sin(30^{\circ})}{sin(20^{\circ})}$ $T_1 = 6396~N$
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