College Physics (4th Edition)

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

Review & Synthesis: Chapters 6-8 - MCAT Review - Page 326: 3


The correct answer is: B. $6.5\times 10^{-7}~N$

Work Step by Step

Since the system is in equilibrium, the torques about the fulcrum must be equal in magnitude. We can find the downward force $F$: $r_L~F = r_R~mg$ $(0.60~m)~F = (0.40~m)~(1.0\times 10^{-7}~kg)(9.80~m/s^2)$ $F = \frac{(0.40~m)~(1.0\times 10^{-7}~kg)(9.80~m/s^2)}{0.60~m}$ $F = 6.5\times 10^{-7}~N$ The correct answer is: B. $6.5\times 10^{-7}~N$
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