\input{labpreamble}
\begin{document}
	\parskip=.18in
\subsubsection*{PHSX 101L\hfill Experiment 3\hfill Name:\underline{\hspace*{2in}}}


\underline{Title:} Forces as Vectors

\noindent
\underline{Object:} To study how simultaneous forces 
combine together to produce a resultant force.

\underline{Introduction:} A basic concept in the study of physics is 
that forces can be represented theoretically by vectors.  This means 
that when several forces act simultaneously on the same object, they 
combine together into one resultant force.  This resultant force can 
be calculated theoretically using the rules of vector addition.  The 
following are two theoretical methods for combining vectors:

\vspace*{-.2in}
\begin{enumerate}
\item \underline{By diagram (graphically):} Several vectors 
representing concurrent forces can be added by diagramming the vectors 
head-to-tail using rulers and protractors to measure the magnitude and 
direction of each vector.  The resultant vector is found by drawing 
an arrow from the tail of the first vector in the series to the head 
of the last.

\item \underline{By component (numerically):} The resultant of 
several concurrent vectors can also be found by breaking each vector 
(of magnitude $r$ and direction $\theta$)
up into its $x$- and $y$-components.  
\begin{equation}\label{eqn:xy}
x=r\cos\theta\ \ \ \ \ \ \ y=r\sin\theta
\end{equation}
\noindent
Corresponding components of the 
several forces are then added together separately to obtain the $x$- and 
$y$-components of the resultant.  The magnitude and direction of this 
resultant vector are given by:
\begin{equation}\label{eqn:rth}
r = \sqrt{x^{2}+y^{2}}\ \ \ \ \ \ \theta = \tan^{-1}\left({\frac{y}{x}}\right)
\end{equation}
\end{enumerate}

 
\underline{Procedure:} Using a force table, set up the
forces as described in the following list.  Determine an additional
force so as to balance the resultant of the given forces.  This
balancing force, which you will find by trial and error, is the
\emph{reverse} of the resultant force you are looking for.  There are
only two forces to be added in trial 1; three in trial 2.  Use 
equation \ref{eqn:xy} to get the $x$- and $y$-components for each 
vector, and after summing them use equation \ref{eqn:rth} to find the 
magnitude and direction.

\hspace{.5in}\underline{\ \ Trial 1\ \ }\ \ \ \ 
\hfill\underline{($x$-comp.,\ \ \ $y$-comp.)}\hfill\underline{Apparatus:} 
Draw the force table.\hspace{.2in}
\vspace{-.15in}

$\boldsymbol{A}$ = 135 g at 35\degree\hspace{.4in}$\rightarrow$\hspace{.37in} ( \hspace{.5in} , \hspace{.5in} )
\vspace{-.15in}

$\boldsymbol{B}$ = 150 g at 110\degree\hspace{.32in}$\rightarrow$\hspace{.37in} ( \hspace{.5in} , \hspace{.5in} )

\vspace{-.3in}\hspace{2.2in}\underline{\hspace{1.5in}}
\vspace{-.15in}

sum = \hspace{1.25in}$\leftarrow$\hspace{.26in} ( \hspace{.5in} , \hspace{.5in} )

\vspace{.25in}
\hspace{.5in}\underline{\ \ Trial 2\ \ }
\vspace{-.15in}

$\boldsymbol{A}$ = 110 g at 20\degree\hspace{.4in}$\rightarrow$\hspace{.37in} ( \hspace{.5in} , \hspace{.5in} )
\vspace{-.15in}

$\boldsymbol{B}$ = 100 g at -75\degree\hspace{.33in}$\rightarrow$\hspace{.37in} ( \hspace{.5in} , \hspace{.5in} )
\vspace{-.15in}

$\boldsymbol{C}$ = 150 g at 120\degree\hspace{.32in}$\rightarrow$\hspace{.37in} ( \hspace{.5in} , \hspace{.5in} )

\vspace{-.3in}\hspace{2.2in}\underline{\hspace{1.5in}}
\vspace{-.15in}

sum = \hspace{1.25in}$\leftarrow$\hspace{.26in} ( \hspace{.5in} , \hspace{.5in} )

\newpage
\noindent
\underline{Theoretical Results:} As in most scientific work, the 
results of the experiment (in this case the resultant force obtained 
from the force table) are to be compared with theoretical 
predictions.  The theory of vectors described above is 
to be applied to the given forces, and a resultant is predicted.
\vspace{-.2in}
\begin{enumerate}
\item Record the magnitude and direction of each resultant force 
predicted by the \emph{diagram} method.

\hfill Trial 1: \hspace{.4in}$r_{1}$ = \underline{\hspace*{1.2in}}\hspace{.4in} $\theta_{1}$ = 
\underline{\hspace*{1.2in}}\hfill\ 

\vspace{.1in}
\hfill Trial 2: \hspace{.4in}$r_{2}$ = \underline{\hspace*{1.2in}}\hspace{.4in} $\theta_{2}$ = 
\underline{\hspace*{1.2in}}\hfill\ 

\item Record the magnitude and direction of each resultant force 
predicted by the \emph{component} method.

\hfill Trial 1: \hspace{.4in}$r_{1}$ = \underline{\hspace*{1.2in}}\hspace{.4in} $\theta_{1}$ = 
\underline{\hspace*{1.2in}}\hfill\ 

\vspace{.1in}
\hfill Trial 2: \hspace{.4in}$r_{2}$ = \underline{\hspace*{1.2in}}\hspace{.4in} $\theta_{2}$ = 
\underline{\hspace*{1.2in}}\hfill\ 

\end{enumerate}

\noindent
\underline{Experimental Results:} 
What was the result (vector sum) of each trial as measured experimentally?  
(Remember, the resultant force is the opposite or negative of the 
balancing force.)

\hfill Trial 1: \hspace{.4in}$r_{1}$ = \underline{\hspace*{1.2in}}\hspace{.4in} $\theta_{1}$ = 
\underline{\hspace*{1.2in}}\hfill\ 

\vspace{.1in}
\hfill Trial 2: \hspace{.4in}$r_{2}$ = \underline{\hspace*{1.2in}}\hspace{.4in} $\theta_{2}$ = 
\underline{\hspace*{1.2in}}\hfill\ 

\underline{Analysis:}  Find the percent difference between the experimental vector \emph{magnitudes} 
obtained on the force table and those obtained by each theoretical method 
for each trial.  
\vspace{-.2in}

\begin{tabbing}
\hspace{1.2in} \=  \hspace{0.8in} \=       \hspace{1.5in}        \=                            \kill
               \>                 \> \hspace{.2in} Diagram       \> \hspace{.15in}  Component  \\
                                                                                               \\
               \>  Trial 1:       \>\underline{\hspace*{1.2in}}  \>\underline{\hspace*{1.2in}} \\
                                                                                               \\
               \>  Trial 2:       \>\underline{\hspace*{1.2in}}  \>\underline{\hspace*{1.2in}}  
\end{tabbing}


\noindent
\underline{Question:} Find the net force (both magnitude and direction) for the two 
forces shown below:

\vspace{.1in}
\noindent
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\put(0,50){\line(1,0){150}} % horizontal axis
\put(40,0){\line(0,1){100}} % vertical axis
\put(40,50){\vector(3,2){40}}
\put(75,57){\makebox(0,0){33.\degree}}
\put(90,85){\makebox(0,0){48 N}}
\put(40,50){\vector(2,-1){100}}
\put(80,43){\makebox(0,0){-26.6\degree}}
\put(100,5){\makebox(0,0){112 N}}
\end{picture}

\vspace*{-1.3in}
\ \ \ \ Resultant: 

\ \ \ \ magnitude (r) = \underline{\hspace*{1.2in}}

\ \ \ \ direction ($\theta$) = \underline{\hspace*{1.2in}}


\vspace{.2in}
\underline{Conclusions:} 



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