\documentclass[spec-collated.tex]{subfiles}
\begin{document}

\section{Dynamics}

\subsection*{Resolution of forces}

\textbf{Resultant force} is sum of force vectors

\subsubsection*{In angle-magnitude form}

\makebox[3cm]{Cosine rule:} \(c^2=a^2+b^2-2ab\cos\theta\)
\makebox[3cm]{Sine rule:} \(\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}\)

\subsubsection*{In \(\boldsymbol{i}\)---\(\boldsymbol{j}\) form}

Vector of \(a\) N at \(\theta\) to \(x\) axis is equal to \(a \cos \theta \boldsymbol{i} + a \sin \theta \boldsymbol{j}\). Convert all force vectors then add.

To find angle of an \(a\boldsymbol{i} + b\boldsymbol{j}\) vector, use \(\theta = \tan^{-1} \frac{b}{a}\)

\subsubsection*{Resolving in a given direction}

The resolved part of a force \(P\) at angle \(\theta\) is has magnitude \(P \cos \theta\)

To convert force \(||\vec{OA}\) to angle-magnitude form, find component \(\perp\vec{OA}\) then:
\begin{align*}
  |\boldsymbol{r}| &= \sqrt{\left(||\vec{OA}\right)^2 + \left(\perp\vec{OA}\right)^2} \\
  \theta &= \tan^{-1}\dfrac{\perp\vec{OA}}{||\vec{OA}}
\end{align*}

\subsection*{Newton's laws}

\begin{tcolorbox}
  \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}

\subsubsection*{Weight}
A mass of \(m\) kg has force of \(mg\) acting on it

\subsubsection*{Momentum \(\rho\)}
\[ \rho = mv \tag{units kg m/s or Ns} \]

\subsubsection*{Reaction force \(R\)}

\begin{itemize}
  \item With no vertical velocity, \(R=mg\)
  \item With vertical acceleration, \(|R|=m|a|-mg\)
  \item With force \(F\) at angle \(\theta\), then \(R=mg-F\sin\theta\)
\end{itemize}

\subsubsection*{Friction}

\[ F_R = \mu R \tag{friction coefficient} \]

\subsection*{Inclined planes}

\[ \boldsymbol{F} = |\boldsymbol{F}| \cos \theta \boldsymbol{i} + |\boldsymbol{F}| \sin \theta \boldsymbol{j} \]
\begin{itemize}
  \item Normal force \(R\) is at right angles to plane
  \item Let direction up the plane be \(\boldsymbol{i}\) and perpendicular to plane \(\boldsymbol{j}\)
\end{itemize}

\def\iangle{30} % Angle of the inclined plane

\def\down{-90}
\def\arcr{0.5cm} % Radius of the arc used to indicate angles

\tikzset{
  force/.style={->,draw=blue,fill=blue},
  axis/.style={densely dashed,gray,font=\small},
  M/.style={rectangle,draw,fill=lightgray,minimum size=0.5cm,thin},
  m/.style={rectangle,draw=black,fill=lightgray,minimum size=0.3cm,thin},
  plane/.style={draw=black,fill=blue!10},
  string/.style={draw=red, thick},
  pulley/.style={thick}
}

  \begin{center}\begin{tikzpicture}

    \pgfmathsetmacro{\Fnorme}{2}
    \pgfmathsetmacro{\Fangle}{30}

    \begin{scope}[rotate=\iangle]
      \node[M,transform shape] (M) {};
      \coordinate (xmin) at ($(M.south west)-({abs(1.1*\Fnorme*sin(-\Fangle))},0)$);
      \coordinate (xmax) at ($(M.south east)+({abs(1.1*\Fnorme*sin(-\Fangle))},0)$);
      \coordinate (ymax) at ($(M.north)+(0, {abs(1.1*\Fnorme*cos(-\Fangle))})$);
      \coordinate (ymin) at ($(M.south)-(0, 1cm)$);
      \coordinate (axiscentre) at ($(M.south)+(0.5cm, 0.5cm)$);
      \draw[postaction={decorate, decoration={border, segment length=2pt, angle=-45},draw,red}] (xmin) -- (xmax);
      \coordinate (N) at ($(M.center)+(0,{\Fnorme*cos(-\Fangle)})$);
      \coordinate (fr) at ($(M.center)+({\Fnorme*sin(-\Fangle)}, 0)$);
      {[axis,-]
      \draw (ymin) -- (M.center);
      }
      {[axis,->]
      \draw ($(M)+(1,0)$) -- ($(M)+(2,0)$) node[above right] {\(\boldsymbol{i}\)};
      \draw ($(M)+(1,0)$) -- ($(M)+(1,1)$) node[above right] {\(\boldsymbol{j}\)};
      }
      {[force,->]
        \draw (M.center) -- (N) node [right] {\(R\)};
        \draw (M.center) -- (fr) node [left] {\(\mu R\)};
      }
    \end{scope}
    \draw[force,->] (M.center) -- ++(0,-1) node[below] {$mg$};
    \draw (M.center)+(-90:\arcr) arc [start angle=-90,end angle=\iangle-90,radius=\arcr] node [below, pos=.5] {\footnotesize\(\theta\)};
  \end{tikzpicture}\end{center}

\subsection*{Connected particles}

\def\boxwidth{0.5}
\tikzset{
  box/.style={rectangle,draw,fill=lightgray,minimum width=\boxwidth,thin},
  m/.style={rectangle,draw=black,fill=lightgray,minimum size=\boxwidth, font=\footnotesize, thin}
}


\begin{center}
  \begin{tikzpicture}

    \matrix[column sep=1cm] {
      \begin{scope}

        \coordinate (O) at (0,0);
        \coordinate (A) at ($({3*cos(\iangle)},{3*sin(\iangle)})$);
        \coordinate (B) at ($({3*cos(\iangle)},0)$);
        \coordinate (C) at ($({(1.5-0.5*\boxwidth)*cos(\iangle)},{(1.5-0.5*\boxwidth)*sin(\iangle)})$); % centre of box
        \coordinate (D) at ($(C)+(\iangle:\boxwidth)$);
        \coordinate (E) at ($(D)+(90+\iangle:0.5*\boxwidth)$);
        \coordinate (F) at ($(B)+(0,{1.5*sin(\iangle)})$);
        \coordinate (X) at ($(A)+(\iangle:0.5*\boxwidth)$); % centre of pulley
        \coordinate (Y) at ($(X)+(90+\iangle:0.5*\boxwidth)$); % chord of pulley

        \draw[plane] (O) -- (A) -- (B) -- (O);
        \draw (O)+(\arcr,0) arc [start angle=0,end angle=\iangle,radius=\arcr] node [right, pos=.75] {\footnotesize\(\theta\)};

        \draw [rotate=\iangle, m] (C) rectangle ++(\boxwidth,\boxwidth) node (z) [rotate=\iangle, midway, font=\footnotesize] {\(m_1\)};
        \draw [pulley] (A) -- (X) ++(0.5*\boxwidth, 0) arc[rotate=\iangle, start angle=0, delta angle=360, x radius=0.25, y radius=0.25] node(r) [midway, rotate=\iangle] {};
        \draw [string] (E) -- (Y) arc (90+\iangle:0:0.25) -- ++($(0,{-1.5*sin(\iangle)})$) node[m] {\(m_2\)};

      \end{scope}

      &

      \begin{scope}[rotate=\iangle]

        \draw [m] ++(-0.5*\boxwidth,-0.5*\boxwidth) rectangle ++(\boxwidth,\boxwidth) node (m1) [rotate=\iangle, midway, font=\footnotesize] {\(m_1\)};

        {[axis,-]
          \draw (0,-1) -- (0,0);
          \draw[solid,shorten >=0.5pt] (\down-\iangle:\arcr) arc(\down-\iangle:\down:\arcr);
          \node at (\down-0.5*\iangle:1.3*\arcr) {\(\theta\)};
        }

        {[force,->]
          \draw (M.center) -- ++(0,{cos(\iangle)}) node[above right] {\(R_1\)};
          \draw (M.west) -- ++(-0.5,0) node[left] {\(\mu R_1\)};
          \draw (M.east) -- ++(1,0) node[above] {\(T_1\)};
        }

        \draw[force,->, rotate=-\iangle] (M.center) -- ++(0,-1) node[below] {\(m_1 g\)};

      \end{scope}

      &

      \draw [m] ++(-0.5*\boxwidth,-0.5*\boxwidth) rectangle ++(\boxwidth,\boxwidth) node [midway, font=\footnotesize] {\(m_2\)};

      {[force,->]
        \draw (0,0.5*\boxwidth) -- ++(0,1) node[above] {\(T_2\)};
        \draw (0,-0.5*\boxwidth) -- ++(0,-1) node[right] {\(m_2 g\)};
      }
      \\
    };
  \end{tikzpicture}
  \end{center}

\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: \\
    \[ \begin{cases}m_1 g - T = m_1 a\\ T - m_2 g = m_2 a\end{cases} \\ \implies m_1 g - m_2 g = m_1 a + m_2 a \]
  \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
  \item \textbf{Pulley on right angle:} \(a = \frac{m_2g}{m_1+m_2}\) where \(m_2\) is suspended (frictionless on both surfaces)
  \item \textbf{Pulley on edge of incline:} find downwards force \(W_2\) and components of mass on plane
\end{itemize}

\hspace{2em}\parbox{8em}{In this example, note \(T_1 \ne T_2\):}
  \begin{tikzpicture}

      \begin{scope}

        \coordinate (O) at (0,0);
        \coordinate (A) at ($({3*cos(\iangle)},{3*sin(\iangle)})$);
        \coordinate (B) at ($({3*cos(\iangle)},0)$);
        \coordinate (C) at ($({(1-0.25*\boxwidth)*cos(\iangle)},{(1-0.25*\boxwidth)*sin(\iangle)})$); % centre of box
        \coordinate (D) at ($(C)+(\iangle:\boxwidth)$);
        \coordinate (E) at ($(D)+(90+\iangle:0.5*\boxwidth)$);
        \coordinate (F) at ($(B)+(0,{1.5*sin(\iangle)})$);
        \coordinate (G) at ($(A)+(\iangle:-2*\boxwidth)$);
        \coordinate (H) at ($(G)+(90+\iangle:0.5*\boxwidth)$);
        \coordinate (I) at ($(H)+(\iangle:-0.5*\boxwidth)$);
        \coordinate (J) at ($(H)+(\iangle:\boxwidth)$);
        \coordinate (X) at ($(A)+(\iangle:0.5*\boxwidth)$); % centre of pulley
        \coordinate (Y) at ($(X)+(90+\iangle:0.5*\boxwidth)$); % chord of pulley

        \draw[plane] (O) -- (A) -- (B) -- (O);
        \draw (O)+(\arcr,0) arc [start angle=0,end angle=\iangle,radius=\arcr] node [right, pos=.75] {\footnotesize\(\theta\)};

        \draw [rotate=\iangle, m] (C) rectangle ++(\boxwidth,\boxwidth) node (z) [rotate=\iangle, midway, font=\footnotesize] {\(m_1\)};
        \draw [rotate=\iangle, m] (G) rectangle ++(\boxwidth,\boxwidth) node (l) [rotate=\iangle, midway, font=\footnotesize] {\(m_2\)};
        \draw [pulley] (A) -- (X) ++(0.5*\boxwidth, 0) arc[rotate=\iangle, start angle=0, delta angle=360, x radius=0.25, y radius=0.25] node(r) [midway, rotate=\iangle] {};
        \draw [string] (E) -- (H) node [midway, above, font=\footnotesize, rotate=\iangle] {\(T_2\)};
        \draw [string] (J) -- (Y) node [midway, above, font=\footnotesize, rotate=\iangle] {\(T_1\)} arc (90+\iangle:0:0.25) -- ++($(0,{-1.5*sin(\iangle)})$) node [midway, above right, font=\footnotesize] {\(T_1\)} node[m] {\(m_3\)};

      \end{scope}

  \end{tikzpicture}
\subsection*{Equilibrium}

\[ \dfrac{A}{\sin a} = \dfrac{B}{\sin b} = \dfrac{C}{\sin c} \tag{Lami's theorem}\]
\[ c^2 = a^2 + b^2 - 2ab \cos \theta \tag{cosine rule} \]

Three methods:
\begin{enumerate}
  \item Lami's theorem (sine rule)
  \item Triangle of forces (cosine rule)
  \item Resolution of forces (\(\Sigma F = 0\) - simultaneous)
\end{enumerate}

  \begin{cas}
    \textbf{To verify:} Geometry tab, then select points with normal cursor. Click right arrow at end of toolbar and input point, then lock known constants.
  \end{cas}

\subsection*{Variable forces (DEs)}

\[ a = \dfrac{d^2x}{dt^2} = \dfrac{dv}{dt} = v\dfrac{dv}{dx} = \dfrac{d}{dx} \left( \frac{1}{2} v^2 \right) \]

\end{document}