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: 2

Answer

$T_1 = 86.6~N$ $T_2 = 50.0~N$

Work Step by Step

The vertical component of $T_3$ is equal in magnitude to $T_2$. We can find the magnitude of $T_2$ as; $T_2 = T_3~sin(\theta)$ $T_2 = (100~N)~sin(30^{\circ})$ $T_2 = 50.0~N$ The horizontal component of $T_3$ is equal in magnitude to $T_1$. We can find the magnitude of $T_1$ as; $T_1 = T_3~cos(\theta)$ $T_1 = (100~N)~cos(30^{\circ})$ $T_1 = 86.6~N$
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