Solving Levers with the Teeter-Totter Principle.
9sep10

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\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.

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