\section{M-10   Power and Friction}

\noindent APPARATUS: Prony brake, scale, timer, stopwatch, 15 meter tape.
\begin{center} {\bf Part A - POWER } \end{center}
\begin{enumerate}
\item \latex{\vspace*{-.05in}}Measure your horsepower (1 hp = 746 watts)
by use of the Prony brake which is mounted on 
the wall.  Try a slow rate that you could keep up all day and another as fast 
as you can turn the wheel.  Figure out what data are required and how to use 
them.  Check with your instructor to be sure you have analyzed the problem 
correctly.  Many calculate the frictional force incorrectly.

\item Measure the horsepower you develop running up a flight of 
stairs.
\end{enumerate}

\begin{center} {\bf Part B - FRICTION} \end{center}

\noindent INTRODUCTION:
\begin{enumerate}
\item[] \latex{\vspace*{-.05in}}We measure the  
coefficient of kinetic friction, $\mu$, between rope and wheel of the Prony 
brake and test whether $\mu$ is independent of velocity and of normal force.
(By the way, the Prony brake is named after Gaspard Clair Francois Marie Riche de Prony, who invented it in Paris in 
1821 to measure the power of engines.  He was the director of the Ecole des Ponts et Chausses, a position he chose over
joining Napoleon's army that invaded Egypt.)

\noindent Let {\bf f = frictional force/unit length} and {\bf n = normal 
force/unit length} 
along the rope.  For an element $\Delta$s = $r\Delta \theta$ along the rope, 
let $T'$
and $T$ be the tensions at opposite ends of $\Delta$s.  For small $\Delta 
\theta$, tensions
$\vec{T}'$ and $\vec{T}$
will be almost antiparallel and differ in magnitude, $T' - T = \Delta T$, only by 
the frictional force $f\Delta s$.  Thus  


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 \noindent
\begin{equation} \Delta T = f \Delta s \end{equation}
The reaction of the wheel must supply a normal force $n \Delta s$ 
sufficient to 
balance the vector sum $\vec{T}' + \vec{T}$, or 
$n \Delta s = |\vec{T}' + \vec{T}|$.  But from the figure we see that 
$|\vec{T}' + \vec{T}| \simeq T \Delta \theta$ so
 \begin{equation} T \Delta \theta = n \Delta s \end{equation} 
Eliminating $\Delta$s between (1) and (2) gives $T \Delta \theta = 
n \Delta T/f =  \Delta T/\mu$
since $f = \mu n$.  Hence
$$ \mu \Delta \theta = \frac{\Delta T}{T}  \mbox{~~~~~~and~~~~~~}
 \mu d \theta = \frac{dT}{T}~~. $$
For the Prony brake, we integrate between $0$ and $\pi$:
$$\mu \int_0^{\pi} d \theta = \int_{T_1}^{T_2} dT/T \hspace{.2in} \mbox{and so}
\hspace*{.2in} \mu = \frac{1}{\pi} \ell n \left( \frac{T_2}{T_1} \right) $$. }
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{\noindent
{\em Note \#1}: For the Prony brake, $\theta = \pi$.  More generally, the
string could go around any number of turns.  Suppose it goes around $x$ times,
where $x$ is of course not necessarily an integer.  The string then wraps
through a total angle $\theta = 2\pi x$. Then $dT/T = \mu d \theta$ integrates to 
$$\ell n (T/T_0) = \mu \theta \latex{\hspace*{.2in}}
\mbox{and hence} \latex{\hspace*{.2in}} T = T_0 e^{\mu \theta}$$
which holds also for a slipping string if $\mu = \mu_k$.
}
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\end{enumerate}

\noindent {SUGGESTIONS FOR PART B}:
\begin{enumerate} 
\item  \latex{\vspace*{-.05in}}Calculate your value of $\mu$.
\item  Test whether $\mu$ is independent of velocity by turning 
at different 
speeds and noting whether $T_2$ or $T_1$ changes appreciably.
\item Test whether $\mu$ is independent of the normal force by using a 
different tension setting for the brake. (Use different link setting in the 
chain to vary the tension).
\end{enumerate}

\noindent {QUESTIONS}:
\begin{enumerate} 
\item For a stationary wheel the two spring balances will not read the same 
if 
\begin{enumerate}
\item[(a)]  \latex{\vspace*{-.05in}}there are zero errors in the balances, 
\item[(b)] there is friction in the wheel bearing.  
\item[(c)] the handle for turning the brake is not directly above or below the 
wheel's axis.
\end{enumerate}
{\em Discuss the errors resulting from these effects}.
\item How might one understand a change in $\mu$ as the normal 
force increases 
in \#3 of Part B? 
\end{enumerate}

{\em Note \#2:} 
\begin{enumerate}
\item[] \latex{\vspace*{-.05in}}Friction arises from intermolecular forces between the atoms (or 
molecules) of the two surfaces, and so frictional forces are very
complex.  It is therefore surprising that the simple empirical relations 
for friction (e.g. that $\mu$ is a constant) hold as well as they do.
\end{enumerate}
