Beginning Algebra With Applications, Chapter 3, 3.2, Section 3.2, Problem 196

An adult and a child are on a seesaw that is 14 ft long. The adult weighs 175 lb and the child weighs 70 lb. How many feet from the child must the fulcrum be places so that the see saw balances?

Using the Lever System Equation

$F_1 x = F_2 (d-x)$

Solving for $x$,


$
\begin{equation}
\begin{aligned}

F_1 x =& F_2 (d-x)
&& \text{Apply Distributive Property}
\\
\\
F_1 x + F_2 x =& F_2 d
&& \text{Add } F_2 x
\\
\\
x(F_1 + F_2) =& F_2 d
&& \text{Factor out } x
\\
\\
x =& \frac{F_2 d}{F_1 + F_2}
&& \text{Divide by } F_1 + F_2
\\
\\
x =& \frac{175(14)}{70+175}
&& \text{Substitute } F_1 = 70, F_2 = 175 \text{ and } d = 14
\\
\\
x =& \frac{2450}{245}
&& \text{Simplify}
\\
\\
x =& 10 \text{ ft}
&&

\end{aligned}
\end{equation}
$


The child must be 10 ft from the fulcrum so that the seesaw balances.