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\begin{center}
{\LARGE \underline{Coefficient of Friction of an Air Track}}
\bigskip \\
{\Large \it{8s63 Experiment Report}}
\bigskip \\
{\large Juliet Bourne, December 1988}
\end{center}

\bigskip \bigskip 
{\large \noindent \underline{Theory (Basis for Calculations)}}
\medskip \\

A cart is placed on an inclined air track. The forces acting on this
cart are shown in the diagram below.
\vspace*{1.5in}

The acceleration, ${}a$, of the cart can be found by setting the
resultant force on the cart to ${}ma$.

\begin{eqnarray}
F &=& m g \sin{} \theta \\
f &=& \mu m g cos \theta \\
F_{net} &=& F - f \\
ma &=& m g \sin{} \theta - \mu m g cos \theta \\
a &=& g \sin{} \theta - \mu g cos \theta \\
a &=& g \sin{} \theta (1 - \mu \cot{} \theta)
\end{eqnarray}

This value for ${}a$ can be substituted into the general position
function of x in terms of time.

\begin{eqnarray}
x &=& x_0 + v_0t + \frac{1}{2}at^2 \\
x &=& x_0 + v_0t + \frac{g \sin{} \theta}{2}(1 - \mu \cot{} \theta)t^2
\end{eqnarray}

The coefficients given by the polychi program map into this equation
like so:
\begin{eqnarray} 
x &=& a_0 + a_1t + a_2t^2
\end{eqnarray}

Therefore, the value of ${}a_2$ given by the polychi program is equivalent
to 
\begin{eqnarray} 
\frac{g \sin{} \theta}{2}(1 - \mu \cot{} \theta) 
\end{eqnarray}

Solving for ${}u$:

\begin{eqnarray}
a_2 &=& \frac{g \sin{} \theta}{2}(1 - \mu \cot{} \theta) \\
1 - \mu \cot{} \theta &=& \frac{2 a_2}{g \sin{} \theta} \\
\mu \cot{} \theta &=& 1 - \frac{2 a_2}{g \sin{} \theta} \\
\mu &=& \frac{1}{\cot{} \theta} - \frac{2 a_2}{g \sin{} \theta \cot{} \theta}
\\
\mu &=& \tan{} \theta - \frac{2 a_2 \tan{} \theta}{g \sin{} \theta}
\end{eqnarray}

The value for ${}\tan{} \theta$ is determined through trigonometry:
\vspace*{1.5in}

\begin{eqnarray}
\tan{} \theta = \frac{b - a}{c}
\end{eqnarray}

\newpage
The error for this value is:

\begin{eqnarray}
\frac{\Delta \tan{} \theta}{\tan{} \theta} &=& \sqrt{\left(
\frac{0.1}{b - a} \right)^2 + \left( \frac{0.1}{c} \right)^2} \\
\Delta \tan{} \theta &=& \sqrt{\left( \frac{0.1}{b - a} \right)^2 +
\left( \frac{0.1}{c} \right)^2} \left( \tan{} \theta \right) .
\end{eqnarray}

In equation 15, since the angle used in this experiment is so small,
${}\tan{} \theta \approx \sin{} \theta$. Therefore,

\begin{eqnarray}
\mu &=& \tan{} \theta - \frac{2 a_2}{g} \pm \Delta \tan{} \theta .
\end{eqnarray}

\newpage

{\Large \noindent \underline{Data and Results}}
\medskip \\

\noindent
{\underline {Average ${}a_2$} :}  \\
2nd order fit: 2.06 x 10${}^{-1} \pm$ 5.54 x 10${}^{-2}$ \\
3rd order fit: 3.76 x 10${}^{-1} \pm$ 1.33 x 10${}^{-1}$ \\

\medskip \noindent 
{\underline {Average ${}\chi^2$} :} \\
2nd order fit: 1.389 x 10${}^{-3} \pm$ 7.66 x 10${}^{-4}$ \\
3rd order fit: 3.132 x 10${}^{-4} \pm$ 2.56 x 10${}^{-4}$ \\

\medskip \noindent
{\underline {Average ${}a_3$} : -9.30 x 10${}^{-2} \pm$ 7.27 x 10${}^{-2}$} \\

\medskip \noindent
{\underline{Run 1} :} \\
$\tan{} \theta$ = 5.18E-02 \\
$\Delta \tan{} \theta$ = $\pm$ 1.725E-04 \\
$\mu$ : \\
(2nd order) = 9.75E-03 $\pm$ 5.54E-02 \\
(3rd order) = -2.50E-02 $\pm$ 1.33E-01 \\

\medskip \noindent
{\underline{Runs 2-5} :} \\
$\tan{} \theta$ = 6.17E-02 \\
$\Delta \tan{} \theta$ = $\pm$ 6.97E-04 \\
$\mu$ : \\
(2nd order) = 1.96E-02 $\pm$ 5.54E-02 \\
(3rd order) = -1.33E-02 $\pm$ 1.33E-01 \\

\medskip
{\large \noindent \underline{Comments}} 

The coefficients of friction noted above are very interesting in that
they have negative values.  However, the margin of error is so large
that these negative values can be justified to a degree.  Many things
may have happened during this experiment, such as rough spots on the
air track, bad data analysis, bad set-up, to cause these results.  The
cubic terms are consisently negative and roughly of the same order of
magnitude, so it would seem that the ``drag'' part of experiment is
behaving the way it ought to, within experimental error.  

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