\subsection*{Newton's laws}
-\begin{tcolorbox}
+\begin{theorembox}{}
\begin{enumerate}[leftmargin=1mm]
\item Velocity is constant without \(\Sigma F\)
\item \(\frac{d}{dt} \rho \propto \Sigma F \implies \boldsymbol{F}=m\boldsymbol{a}\)
\item Equal and opposite forces
\end{enumerate}
-\end{tcolorbox}
+\end{theorembox}
\subsubsection*{Weight}
A mass of \(m\) kg has force of \(mg\) acting on it
\begin{itemize}
\item \textbf{Suspended pulley:} tension in both sections of rope are equal \\
- \(|a| = g \frac{m_1 - m_2}{m_1 + m_2}\) where \(m_1\) accelerates down \\
- With tension:
- \[ \begin{cases}m_1 g - T = m_1 a\\ T - m_2 g = m_2 a\end{cases} \\ \implies m_1 g - m_2 g = m_1 a + m_2 a \]
+ \(|a| = g \dfrac{m_1 - m_2}{m_1 + m_2}\) where \(m_1\) accelerates down \\
+ \[
+ \left\{\begin{array}{lr}
+ m_1g-T = m_1a\\
+ T-m_2g = m_2a
+ \end{array}\right\}
+ \implies m_1 g - m_2 g = m_1 a + m_2 a
+ \]
\item \textbf{String pulling mass on inclined pane:} Resolve parallel to plane
\[ T-mg \sin \theta = ma \]
\item \textbf{Linear connection:} find acceleration of system first