Chapter 11.6, Problem 14P

In Exercise 13, what if the 75 lb fifth grader and a friend of equal weight sit 4 ft from the center of the seesaw to balance the 90 lb fifth grader? How far from the center should the 90 lb fifth grader be?

To determine

To calculate : how far from the centre should the $90lb$ fifth grader be

Answer to Problem 14P

  $90lb$ Fifth grader be sit $6.67ft$ approx from the centre

Explanation of Solution

Given information : If the $75lb$ fifth grader and a friend of equal weight sit $4ft$ from the centre of the seesaw to balance the $90lb$ fifth grader

Calculation :

One fifth grader and a friend of equal

  $\begin{array}{l}w\alpha \frac{1}{d}\\ w=\frac{k}{d}\\ w_1=75lb\\ w_1=150ft\\ d_1=4ft\\ w_2=90lb\\ d_2=?\\ w_1=\frac{k}{d_1}\\ k=w_1{d}_1\mathrm{......}\left(1\right)\\ w_2=\frac{k}{d_2}\\ k=w_2{d}_2\mathrm{......}\left(2\right)\\ \text{from equation }\left(\text{1}\right)\text{ \& }\left(\text{2}\right)\\ w_1{d}_1=w_2{d}_2\\ d_2=\frac{w_1{d}_1}{w_2}\\ d_2=\frac{\left(150\right)\left(4\right)}{90}\\ d_2=6.67ft\end{array}$