Essential University Physics: Volume 1 (4th Edition) Clone

Published by Pearson
ISBN 10: 0-134-98855-8
ISBN 13: 978-0-13498-855-9

Chapter 5 - Section 5.4 - Friction - Example - Page 87: 5.11

Answer

$ F_t = \frac{\mu_k mg}{cos\theta + \mu_k sin \theta}$

Work Step by Step

We first find the x and y components of each of the forces: $ F_n = mg - F_tsin\theta$ and $F_tcos\theta- \mu_k F_n= 0 $ Using substitution, we find: $F_tcos\theta- \mu_k F_n= 0 $ $F_tcos\theta- \mu_k (mg - F_tsin\theta)= 0 $ $ F_t = \frac{\mu_k mg}{cos\theta + \mu_k sin \theta}$
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