## Essential University Physics: Volume 1 (3rd Edition)

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.