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\begin{flushright}
	Alex Bliskovsky\\
	Jacobs 4\\
\end{flushright}

\section{Experimental Design}

The purpose of this lab was calculate the weight of an object using two spring scales which support the object at two angles.

\noindent
We used the following equipment:

	\begin{enumerate}
		\item
		An Unknown Weight
		\item
		Two or Three Spring Scales
		\item
		String
		\item
		Meter Stick
	\end{enumerate}

\begin{figure}[here]
\includegraphics[scale=0.55]{Coke_Bottle_Weights}
\caption{$\triangle ABC$ is formed with $\overline{AC}$ and $\overline{AB}$ as the strings with spring scales and $\overline{CB}$ as the ceiling.}
\label{Setup}
\end{figure}



\section{Observations}

\begin{tabular}{|l|l|}
\hline
\multicolumn{2}{|c|}{Measurements}\\
\hline
Component & Measurement\\
\hline
$h$ & 1.15 m \\
$\overline{CB}$ & Unmeasured \\
$\overline{CA}$ & 1.3 m \\
$\overline{AB}$ & 1.86 m \\
$T_1$ & 1.7 N\\
$T_2$ & 1.1 N\\
\hline
\end{tabular}
\section{Analysis}

Because the object was at rest, we knew that the sum of the forces on the object was zero. This means that $T_1 + T_2 + mg = 0$, and by algebra $T_1 + T_2 = -mg$. If our measurements are accurate, $T_{1_{horizontal}} + T_{2_{horizontal}} = 0$, because the tensions are the only things acting on the object in the horizontal direction. Calculating the components of $T$ is simple trigonometry. To calculate the components, we need a few angles.

\noindent
First, let point $X$ be defined as the point where $\overline{CB}$ and $\overline{T_1+T_2}$ intersect.

\[\angle CAX = \arccos{(\frac{h}{\overline{CA}})} = \theta_1\]
\[\angle XAB = \arccos{(\frac{h}{\overline{AB}})} = \theta_2\]

Next, we can calculate the vertical components of $T$:

\[T_{1_v} = T_1 \cos{\theta_1}\]
\[T_{2_v} = T_2 \cos{\theta_2}\]

Because we know that $T_{1_h} + T_{2_h} = 0$, we don't need to calculate their components. As proven:
\[T_{1_v} + T_{2_v} = T_1 + T_2 = -mg\]



\section{Error Analysis}
\subsection{Sources of Error}


\subsection{Significant Digits}

\section{Results}

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