Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 7 - Newton's Third Law - Exercises and Problems: 8

Answer

The maximum mass of block A is 12 kg

Work Step by Step

Since the angled rope is at an angle of $45^{\circ}$, the horizontal component of the tension is equal to the vertical component of the tension in this rope. At equilibrium, the vertical component of the tension will be equal in magnitude to the weight of block A. Then the tension in the rope attached to block B is equal to the weight of block A. To stay at equilibrium, the weight of block A can not exceed the maximum possible force of static friction acting on block B. $m_A~g = F_f$ $m_A~g = m_B~g~\mu_s$ $m_A = m_B~\mu_s$ $m_A = (20~kg)(0.60)$ $m_A = 12~kg$ The maximum mass of block A is 12 kg.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.