## Solving Levers with the Teeter-Totter Principle.

9sep10

\begin{document} \textbf{Question:} Where should the fulcrum be if a weight of $+2$ is placed at $A$ and a weight of $-3$, a balloon for example, is at $B$? What if it's $+3, -2$ instead? \textbf{Solution:} \subsection{Analyse the known version of this problem} In the lesson on barycentric coordinates, the weight $a$ and $A$ and $b$ at $B$ become the coefficients in $C = \frac{2A+1B}{2+1} = \frac{2}{3}A + \frac{1}{3} B$. The placement of the fulcrum $C$ is obvious. Substitute a variable for a constant to conjecture the general rule. Let's use a $\frac{1}{3} = t$ and write $ C =_1 A + t(B-A) =_2 A(1-t) + tB . $ Equation 1 places the fulcrum a third of the way from $A$ to $B$. Equation 2 identifies this point's barycentric coordinates. \subsection{Generalize and check the validity} Writing $X = \frac{2A + (-3)B}{2-3}$ leads to rewriting rewriting $X = A + 2(B-A)$. This places the fulcrum twice as far from $B$ making $A$ three times as far from the fulcrum. The lever equation you learned in high school physics becomes $ 3 \times 2 = 2 \times 3 $. \subsection{Do it again to make sure.} Exchanging the weights leads to a surprising result. Following the same recipe leads to $Y = 3A + (-2)B = A + 2(A-B) $, which places the fulcrum on the other side. The lever equation remains the same. \end{document}