Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 12 - Rotation of a Rigid Body - Exercises and Problems - Page 331: 30


$F_1 = 120~N$ $F_2 = 80~N$

Work Step by Step

Let's consider the torque about an axis located at the position of the force $F_2$. $\sum \tau = 0$ $-(3.0~m)(40~N)+(1.0~m)~F_1 = 0$ $F_1 = \frac{(3.0~m)(40~N)}{1.0~m}$ $F_1 = 120~N$ We can use the vertical forces to find $F_2$. $\sum F_y = 0$ $40~N-F_1+F_2 = 0$ $F_2 = F_1-40~N$ $F_2 = 120~N-40~N$ $F_2 = 80~N$
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