Essential University Physics: Volume 1 (4th Edition)

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

Chapter 12 - Exercises and Problems - Page 224: 52


$F_{right}=\fbox{22.89 N}$ $F_{left}= \fbox{16.34 N}$

Work Step by Step

We first integrate the function: $= \int_0^2 (1+x) dx = 4 $ We now integrate the function times x: $= \int_0^2 (x+x^2) dx = 4.67$ We now divide these two values to find the center of mass: $CM = 1.167$ Thus, we find the values of the scale readings to find what value will make the torque 0 to obtain: $1.167F_{left}=.833 F_{right}$ $F_{left}=.714 F_{right}$ We now use the fact that the scales add to the force of gravity to find: $F_{left}+F_{right}=4(9.81) $ $.714 F_{right}+F_{right}=4(9.81) $ $F_{right}=\fbox{22.89 N}$ $F_{left}=22.89 \times .714 = \fbox{16.34 N}$
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