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 218: 57



Work Step by Step

As the bracket is in equilibrium hence $\sum \tau_{pivot}=0$ $\implies F_{sx}\times 7.2+Mg\times 9+mg\times 28=0$ This simplifies to: $F_{sx}=-\frac{9Mg+28mg}{7.2}$ We plug in the known values to obtain: $F_{sx}=\frac{(9\times 0.85+28\times 4.2)\times 9.8}{7.2}=-170N$ The negative sign shows that the force is in the opposite direction.
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