-\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,
+ angles
+}
+\usetikzlibrary{calc}
+\usetikzlibrary{angles}
+\usetikzlibrary{datavisualization.formats.functions}
+\usetikzlibrary{decorations.markings}
+\usepgflibrary{arrows.meta}
+\usetikzlibrary{decorations.markings}
+\usepgflibrary{arrows.meta}
+\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
\begin{document}
- \title{Dynamics}
- \author{}
- \date{}
- \maketitle
+\title{Dynamics}
+\author{}
+\date{}
+\maketitle
+
+\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{tcolorbox}
+ \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}
+\end{tcolorbox}
+
+\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}
+
+\[ \boldsymbol{F} = |\boldsymbol{F}| \cos \theta \boldsymbol{i} + |\boldsymbol{F}| \sin \theta \boldsymbol{j} \]
+\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{figure}[!htb]
+ \centering
+ \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{figure}
+
+\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 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}
+
+\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{figure}[!htb]
+ \centering
+ \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{figure}
+
+\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}
- \section{Forces}
+\colorbox{cas}{On CAS:} use Geometry, lock known constants.
- \subsection{Resolution of forces}
+\section{Variable forces (DEs)}
- The resolved part of a force \(P\) at angle \(\theta\) is equal to \(P \cos \Theta\)
+\[ a = \dfrac{d^2x}{dt^2} = \dfrac{dv}{dt} = v\dfrac{dv}{dx} = \dfrac{d}{dx} \left( \frac{1}{2} v^2 \right) \]
\end{document}