College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 8 - Problems - Page 312: 30

Answer

A force of $98.0~N$ must be applied to lift the load.

Work Step by Step

To lift the load, the torque exerted by the applied force about the fulcrum must be equal in magnitude to the torque exerted by the load's weight about the fulcrum. We can find the applied force $F_a$: $r_a~F_a = r_l~F_l$ $F_a = \frac{r_l~F_l}{r_a}$ $F_a = \frac{r_l~mg}{r_a}$ $F_a = \frac{(1.2~m)(20.0~kg)(9.80~m/s^2)}{2.4~m}$ $F_a = 98.0~N$ A force of $98.0~N$ must be applied to lift the load.
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