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: 27



Work Step by Step

As torque about pivot is zero $\implies mg\times x-F\times 22=0$ This can be rearranged as: $F=\frac{mgx}{22}$ We plug in the known values to obtain: $F=\frac{55\times 9.8\times 20.4}{22}$ $F=500N$
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