Essential University Physics: Volume 1 (4th Edition) Clone

Published by Pearson
ISBN 10: 0-134-98855-8
ISBN 13: 978-0-13498-855-9

Chapter 12 - Exercises and Problems - Page 222: 30

Answer

$12.2KN$

Work Step by Step

We know that the net torque is supposed to be into page and it is zero $\implies \sum\tau=m_1g+0.38+m_2g\times 3.98+m_3g\times 8.6-F_2\times 0.76$ This simplifies to: $F_2=\frac{321\times 9.8\times 0.38+175\times 9.8\times 3.98+64.7\times 8.6}{0.76}$ $F_2=17.7KN$ Now $F_1=17.7KN-(m_1+m_2+m_3)g$ We plug in the known values to obtain: $F_1=17.7KN-(321+175+64.7)$ $F_1=12.2KN$
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