Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson
ISBN 10: 0321993721
ISBN 13: 978-0-32199-372-4

Chapter 5 - Exercises and Problems - Page 88: 69


$T = \frac{u_k}{\sqrt{1+\mu_k^2}}$

Work Step by Step

In order to find the expression for the minimum tension needed, we take the derivative of the equation of the previous problem, which is: $\frac{T}{mg}=\frac{\mu_k}{cos\theta+\mu_ksin\theta}$ Taking the derivative gives the answer: $T = \frac{u_k}{\sqrt{1+\mu_k^2}}$
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