Beginning Algebra With Applications, Chapter 3, 3.2, Section 3.2, Problem 198

A 50-pond weight is applied to the left end of a seesaw that is 10 ft long. The fulcrum is 4 ft from the 50-pound weight. A weight of 30 lb is applied to the right end of the seesaw. Is the 30-pound weigh adequate to balance the seesaw?

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{50(10)}{30+50}
&&
\\
\\
x =& \frac{500}{80}
&&
\\
\\
x =& \frac{25}{4}
&&
\\
\\
x =& 6.25 \text{ ft}
&&

\end{aligned}
\end{equation}
$


The distance of a 30-pound weight to the fulcrum is $6.25$ ft or $\displaystyle 6 \frac{1}{4} $ ft. This means that the 30-pound weight is not adequate to balance the seesaw.