-\documentclass[a4paper]{article}
+\documentclass[a4paper, tikz, pstricks]{article}
\usepackage[a4paper,margin=2cm]{geometry}
\usepackage{array}
\usepackage{amsmath}
\usepackage{tcolorbox}
\usepackage{fancyhdr}
\usepackage{pgfplots}
+\usepackage{tikz}
+\usetikzlibrary{arrows,
+ calc,
+ decorations,
+ scopes,
+}
+\usepackage{pst-plot}
+\psset{dimen=monkey,fillstyle=solid,opacity=.5}
+\def\object{%
+ \psframe[linestyle=none,fillcolor=blue](-2,-1)(2,1)
+ \psaxes[linecolor=gray,labels=none,ticks=none]{->}(0,0)(-3,-3)(3,2)[$x$,0][$y$,90]
+ \rput{*0}{%
+ \psline{->}(0,-2)%
+ \uput[-90]{*0}(0,-2){$\vec{w}$}}
+}
+
\usepackage{tabularx}
+\usetikzlibrary{angles}
\usepackage{keystroke}
\usepackage{listings}
\usepackage{xcolor} % used only to show the phantomed stuff
\date{}
\maketitle
- \section{Forces}
+ \section{Resolution of forces}
+
+ \textbf{Resultant force} is sum of force vectors
+
+ \subsection{In angle-magnitude form}
+
+ \makebox[3cm]{Cosine rule:} \(c^2=a^2+b^2-2ab\cos\theta\)
+ \makebox[3cm]{Sine rule:} \(\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}\)
+
+ \subsection{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 \(\theta = \tan^{-1} \frac{b}{a}\)
+
+ \subsection{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 \(|\boldsymbol{r}|=\sqrt{\left(||\vec{OA}\right)^2 + \left(\perp\vec{OA}\right)^2},\quad \theta = \tan^{-1}\dfrac{\perp\vec{OA}}{||\vec{OA}}\)
+
+ \section{Newton's laws}
+
+ \begin{enumerate}
+ \item Velocity is constant without a net external velocity
+ \item \(\frac{d}{dt} \rho \propto \Sigma F \implies \boldsymbol{F}=m\boldsymbol{a}\)
+ \item Equal and opposite forces
+ \end{enumerate}
+
+ \subsection{Weight}
+ A mass of \(m\) kg has force of \(mg\) acting on it
+
+ \subsection{Momentum \(\rho\)}
+ \[ \rho = mv \tag{units kg m/s or Ns} \]
+
+ \subsection{Reaction force \(R\)}
+
+ \begin{itemize}
+ \item With no vertical velocity, \(R=mg\)
+ \item With upwards acceleration, \(R-mg=ma\)
+ \item With force \(F\) at angle \(\theta\), then \(R=mg-F\sin\theta\)
+ \end{itemize}
+
+ \subsection{Friction}
+
+ \[ F_R = \mu R \tag{friction coefficient} \]
+
+ \section{Inclined planes}
+
+ \[ \boldsymbol{F} = |\boldsymbol{F}| \cos \theta \boldsymbol{i} + |\boldsymbol{F}| \sin \theta \boldsymbol{j} \]
+ \def\iangle{30} % Angle of the inclined plane
+
+ \def\down{-90}
+ \def\arcr{0.5cm} % Radius of the arc used to indicate angles
+
+\begin{tikzpicture}[
+ >=latex',
+ scale=1,
+ 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},
+ ]
+ \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 ($(M.center)+({\Fnorme*sin(-\Fangle)}, 0)$);
+ % Draw axes and help lines
+
+ {[axis,->]
+ \draw (ymin) -- (ymax) node[right] {\(\boldsymbol{j}\)};
+ \draw (M) --(M-|xmax) node[right] {\(\boldsymbol{i}\)}; % mental note for me: change "right" to "above"
+ }
+
+ % Forces
+ {[force,->]
+ % Assuming that Mg = 1. The normal force will therefore be cos(alpha)
+ \draw (M.center) -- (N) node [right] {\(R\)};
+ \draw (M.center) -- (fr) node [left] {\(\mu R\)};
+ }
+% \draw [densely dotted, gray] (fr) |- (N) node [pos=.25, left] {\tiny$\lVert \vec F\rVert\cos\theta$} node [pos=.75, above] {\tiny$\lVert \vec F\rVert\sin\theta$};
+ \end{scope}
+ % Draw gravity force. The code is put outside the rotated
+ % scope for simplicity. No need to do any angle calculations.
+ \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] {\tiny\(\theta\)};
+ \end{tikzpicture}
+
+ \section{Connected particles}
+
+ \begin{itemize}
+ \item \textbf{Suspended pulley:} tension in both sections of rope are equal
+ \item \textbf{Linear connection:} find acceleration of system first
+ \item \textbf{Pulley on edge of incline:} find downwards force \(W_2\) and components of mass on plane
+ \end{itemize}
+\def\iangle{25} % Angle of the inclined plane
+
+\def\down{-90}
+\def\arcr{0.5cm} % Radius of the arc used to indicate angles
+
+{\begin{centering} {\begin{tikzpicture}[
+ force/.style={>=latex,draw=blue,fill=blue},
+ axis/.style={densely dashed,gray,font=\small},
+ M/.style={rectangle,draw,fill=lightgray,minimum size=0.6cm,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},
+ scale=1.5
+]
+
+\matrix[column sep=1cm] {
+ %% Sketch
+ \draw[plane] (0,-1) coordinate (base)
+ -- coordinate[pos=0.5] (mid) ++(\iangle:3) coordinate (top)
+ |- (base) -- cycle;
+ \path (mid) node[M,rotate=\iangle,yshift=0.3cm,font=\footnotesize] (M) {\(m_1\)};
+ \draw[pulley] (top) -- ++(\iangle:0.25) circle (0.25cm)
+ ++ (90-\iangle:0.5) coordinate (pulley);
+ \draw[string] (M.east) -- ++(\iangle:1.4cm) arc (90+\iangle:0:0.25)
+ -- ++(0,-1) node[m,font=\scriptsize] {\(m_2\)};
+
+ \draw[->] (base)++(\arcr,0) arc (0:\iangle:\arcr);
+ \path (base)++(\iangle*0.5:\arcr+5pt) node {\(\theta\)};
+ %%
+
+&
+ %% Free body diagram of m1
+ \begin{scope}[rotate=\iangle]
+ \node[M,transform shape] (M) {};
+ % Draw axes and help lines
+
+ {[axis,->]
+ \draw (0,-1) -- (0,2) node[right] {\(+\boldsymbol{i}\)};
+ \draw (M) -- ++(2,0) node[right] {\(+\boldsymbol{j}\)};
+ % Indicate angle. The code is a bit awkward.
+
+ \draw[solid,shorten >=0.5pt] (\down-\iangle:\arcr)
+ arc(\down-\iangle:\down:\arcr);
+ \node at (\down-0.5*\iangle:1.3*\arcr) {\(\theta\)};
+ }
+
+ % Forces
+ {[force,->]
+ % Assuming that Mg = 1. The normal force will therefore be cos(alpha)
+ \draw (M.center) -- ++(0,{cos(\iangle)}) node[above right] {$N$};
+ \draw (M.west) -- ++(-1,0) node[left] {\(F_R\)};
+ \draw (M.east) -- ++(1,0) node[above] {\(T_1\)};
+ }
+
+ \end{scope}
+ % Draw gravity force. The code is put outside the rotated
+ % scope for simplicity. No need to do any angle calculations.
+ \draw[force,->] (M.center) -- ++(0,-1) node[below] {\(m_1g\)};
+ %%
+
+&
+ %%%
+ % Free body diagram of m2
+ \node[m] (m) {};
+ \draw[axis,->] (m) -- ++(0,-2) node[left] {$+$};
+ {[force,->]
+ \draw (m.north) -- ++(0,1) node[above] {\(T_2\)};
+ \draw (m.south) -- ++(0,-1) node[right] {\(m_2g\)};
+ }
+
+\\
+};
+\end{tikzpicture}}\end{centering} }
+ \section{Equilibrium}
+
+ \[ \dfrac{A}{\sin a} = \dfrac{B}{\sin b} = \dfrac{C}{\sin c} \tag{Lami's theorem}\]
+
+ Three methods:
+ \begin{enumerate}
+ \item Lami's theorem (sine rule)
+ \item Triangle of forces or CAS (use to verify)
+ \item Resolution of forces (\(\Sigma F = 0\) - simultaneous)
+ \end{enumerate}
+
- \subsection{Resolution of forces}
+ \colorbox{cas}{On CAS:} use Geometry, lock known constants.
- The resolved part of a force \(P\) at angle \(\theta\) is equal to \(P \cos \Theta\)
\end{document}