Essential University Physics: Volume 1 (3rd Edition)

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

Chapter 12 - Exercises and Problems - Page 215: 25


a) 40 Nm b) 1.3 kN

Work Step by Step

a) The torques are due to the arm and due to the weight. Thus, we find: $\tau = rFsin\theta \\ \tau = (.21)(4.2)(9.81)sin(105)+(.56)(6)(9.81)sin105 \approx 40 \ Nm$ b) $\tau = rF sin \theta \\ F = \frac{\tau}{rsin\theta}=\frac{40.19}{(.18)sin(170)}\approx 1.3 \ kN$
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