-\documentclass[a4paper]{article}
-\usepackage[a4paper,margin=2cm]{geometry}
-\usepackage{array}
-\usepackage{amsmath}
-\usepackage{amssymb}
-\usepackage{tcolorbox}
-\usepackage{fancyhdr}
-\usepackage{pgfplots}
-\usepackage{tabularx}
-\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}
+
+\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{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{theorembox}
+
+\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, thick},
+ 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}
+}
+\tikzset{
+ % style to apply some styles to each segment of a path
+ on each segment/.style={
+ decorate,
+ decoration={
+ show path construction,
+ moveto code={},
+ lineto code={
+ \path [#1]
+ (\tikzinputsegmentfirst) -- (\tikzinputsegmentlast);
+ },
+ closepath code={
+ \path [#1]
+ (\tikzinputsegmentfirst) -- (\tikzinputsegmentlast);
+ },
+ },
+ },
+ % style to add an arrow in the middle of a path
+ mid arrow/.style={postaction={decorate,decoration={
+ markings,
+ mark=at position .5 with {\arrow[#1]{stealth}}
+ }}},
+}
+ \begin{center}\begin{tikzpicture}[scale=1.8]
+
+ \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 ($(M.center)+(0,{cos(\Fangle)})$);
+ \coordinate (ymin) at ($(M.center)-(0,{cos(\Fangle)})$);
+ \coordinate (axiscentre) at ($(M.south)+(0.5cm, 0.5cm)$);
+ \draw[postaction={decorate, decoration={border, segment length=4pt, angle=-45},draw,red}] (xmin) -- (xmax);
+ \coordinate (fr) at ($(M.center)+({\Fnorme*sin(-\Fangle)}, 0)$);
+ {[axis,->]
+ \draw ($(M)+(1,0)$) -- ($(M)+(1.5,0)$) node[above right] {\(\boldsymbol{i}\)};
+ \draw ($(M)+(1,0)$) -- ($(M)+(1,0.5)$) node[above right] {\(\boldsymbol{j}\)};
+ }
+ {[force,->]
+ \draw (M.center) -- (ymax) node [right] {\(R\)};
+ \draw (M.center) -- (fr) node [left] {\(\mu R\)};
+ }
+ \end{scope}
+ \draw[force,->] (M.center) -- ++(0,-1) node[below] {$mg$};
+ \draw (xmin)+(0:\arcr) arc [start angle=0, end angle=\iangle, radius=\arcr] node [right, midway] {\footnotesize\(\theta\)};
+ \coordinate (xbottom) at ($(1, {4*\Fnorme*-cos(\Fangle)})$);
+ \draw [->] (xmin) -- ++($({1.35*\Fnorme*cos(\Fangle)}, 0)$);
+ \begin{scope}[darkgray, rotate=\iangle] \path [draw=darkgray, postaction={on each segment={mid arrow}}] (M.center) -- (ymin) node [pos=0.5, right] {\(mg \cos \theta\)} -- ++(-0.5,0) node[pos=0.5, below right] {\(mg \sin \theta\)};
+ \end{scope}
+ \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, thin}
+}
+
+
+\begin{center}
+ \begin{tikzpicture}[scale=1.5]
+
+ \matrix {
+ \begin{scope}[scale=1.5]
+
+ \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] {\(m_1\)};
+ \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) -- ++($(0,{-1.5*sin(\iangle)-\boxwidth})$) node (p) {};
+ \coordinate (Z) at ($(p.center)+({-0.5*\boxwidth},0)$);
+ \draw [m] (Z) rectangle ++(\boxwidth, \boxwidth) node [midway] {\(m_2\)};
+
+ \end{scope}
+\\
+
+ \begin{scope}[rotate=\iangle, scale=1.5]
+
+ \draw [m] ++(-0.5*\boxwidth,-0.5*\boxwidth) rectangle ++(\boxwidth,\boxwidth) node (n) [rotate=\iangle, midway] {\(m_1\)};
+
+ {[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 (n.center) -- ++(0,{cos(\iangle)}) node[above right] {\(R_1\)};
+ \draw (n.west) -- ++(-0.5,0) node[left] {\(\mu R_1\)};
+ \draw (n.east) -- ++(1,0) node[above] {\(T_1\)};
+ }
+
+ \draw[force,->, rotate=-\iangle] (M.center) -- ++(0,-1) node[below] {\(m_1 g\)};
+
+ \end{scope}
+
+ &
+ \begin{scope}[scale=1.5]
+
+ \draw [m] ++(-0.5*\boxwidth,-0.5*\boxwidth) rectangle ++(\boxwidth,\boxwidth) node [midway] {\(m_2\)};
+
+ {[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{scope}
+ \\
+ };
+ \end{tikzpicture}
+ \end{center}
+
+\begin{itemize}
+ \item \textbf{Suspended pulley:} \(T_1 = T_2\) \\
+ \(|a| = g \dfrac{m_1 - m_2}{m_1 + m_2}\) 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
+ \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}
+
+\begin{tabular}{rl}
+ \parbox[t][][t]{8em}{In this example, note \(T_1 \ne T_2\):} &
+ \parbox{12em}{
+ \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}}
+\end{tabular}
+
+\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}
- \section{Forces}
+ \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{Resolution of forces}
+\subsection*{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}