Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson
ISBN 10: 0321993721
ISBN 13: 978-0-32199-372-4

Chapter 12 - Exercises and Problems - Page 214: 19

Answer

a) .61 meters from the left end b) 1.42 meters from the left end

Work Step by Step

We find the torque necessary for these scale readings, and then we find where the boy should sit, using the same considerations as in problem 18. Doing this, we find: a) $\tau = rFsin\theta = (100)(1.6)=160 \ Nm$ Thus: $160 = (60)(9.81)(.4)+40(9.81)r$ $r=-.19\ m$ Thus, the boy should be .19 meters left of the pivot, which is the same as .61 meters right from the left end. $\tau = rFsin\theta = (300)(1.6)=480 \ Nm$ Thus: $480 = (60)(9.81)(.4)+40(9.81)r$ $r=.62\ m$ Thus, the boy should be .62 meters right of the pivot, which is the same as 1.42 meters right of the left end.
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