-\documentclass[a4paper, tikz, pstricks]{article}
-\usepackage[a4paper,margin=2cm]{geometry}
-\usepackage{array}
-\usepackage{amsmath}
-\usepackage{amssymb}
-\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
-\definecolor{cas}{HTML}{e6f0fe}
-
-\pagestyle{fancy}
-\fancyhead[LO,LE]{Year 12 Specialist - Dynamics}
-\fancyhead[CO,CE]{Andrew Lorimer}
-
-\setlength\parindent{0pt}
-
+\documentclass[spec-collated.tex]{subfiles}
\begin{document}
-\title{Dynamics}
-\author{}
-\date{}
-\maketitle
+\section{Dynamics}
-\section{Resolution of forces}
+\subsection*{Resolution of forces}
\textbf{Resultant force} is sum of force vectors
-\subsection*{In angle-magnitude form}
+\subsubsection*{In angle-magnitude form}
\makebox[3cm]{Cosine rule:} \(c^2=a^2+b^2-2ab\cos\theta\)
\makebox[3cm]{Sine rule:} \(\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}\)
-\subsection*{In \(\boldsymbol{i}\)---\(\boldsymbol{j}\) form}
+\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 \(\theta = \tan^{-1} \frac{b}{a}\)
-\subsection*{Resolving in a given direction}
+\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 \(|\boldsymbol{r}|=\sqrt{\left(||\vec{OA}\right)^2 + \left(\perp\vec{OA}\right)^2},\quad \theta = \tan^{-1}\dfrac{\perp\vec{OA}}{||\vec{OA}}\)
-\section{Newton's laws}
+\subsection*{Newton's laws}
\begin{tcolorbox}
- \begin{enumerate}
- \item Velocity is constant without a net external velocity
+ \begin{enumerate}[leftmargin=1mm]
+ \item Velocity is constant without a net external force
\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}
+\subsubsection*{Weight}
A mass of \(m\) kg has force of \(mg\) acting on it
-\subsection*{Momentum \(\rho\)}
+\subsubsection*{Momentum \(\rho\)}
\[ \rho = mv \tag{units kg m/s or Ns} \]
-\subsection*{Reaction force \(R\)}
+\subsubsection*{Reaction force \(R\)}
\begin{itemize}
\item With no vertical velocity, \(R=mg\)
- \item With upwards acceleration, \(R-mg=ma\)
+ \item With vertical acceleration, \(|R|=m|a|-mg\)
\item With force \(F\) at angle \(\theta\), then \(R=mg-F\sin\theta\)
\end{itemize}
-\subsection*{Friction}
+\subsubsection*{Friction}
\[ F_R = \mu R \tag{friction coefficient} \]
-\section{Inclined planes}
+\subsection*{Inclined planes}
\[ \boldsymbol{F} = |\boldsymbol{F}| \cos \theta \boldsymbol{i} + |\boldsymbol{F}| \sin \theta \boldsymbol{j} \]
\begin{itemize}
pulley/.style={thick}
}
-\begin{figure}[!htb]
- \centering
- \begin{tikzpicture}
+ \begin{center}\begin{tikzpicture}
\pgfmathsetmacro{\Fnorme}{2}
\pgfmathsetmacro{\Fangle}{30}
\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}
+ \end{tikzpicture}\end{center}
-\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}
+\subsection*{Connected particles}
\def\boxwidth{0.5}
\tikzset{
}
-\begin{figure}[!htb]
- \centering
+\begin{center}
\begin{tikzpicture}
\matrix[column sep=1cm] {
\\
};
\end{tikzpicture}
-\end{figure}
+ \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}
-\section{Equilibrium}
+ \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 or CAS (use to verify)
+ \item Triangle of forces (cosine rule)
\item Resolution of forces (\(\Sigma F = 0\) - simultaneous)
\end{enumerate}
-\colorbox{cas}{On CAS:} use Geometry, lock known constants.
+ \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}
-\section{Variable forces (DEs)}
+\subsection*{Variable forces (DEs)}
\[ a = \dfrac{d^2x}{dt^2} = \dfrac{dv}{dt} = v\dfrac{dv}{dx} = \dfrac{d}{dx} \left( \frac{1}{2} v^2 \right) \]