From: Andrew Lorimer Date: Sun, 25 Aug 2019 10:02:40 +0000 (+1000) Subject: [spec] diagrams for connected particles X-Git-Tag: yr12~55 X-Git-Url: https://git.lorimer.id.au/notes.git/diff_plain/b4a3bef863d66c9b749e7c42de7b51085d826fef?ds=sidebyside [spec] diagrams for connected particles --- diff --git a/spec/dynamics.pdf b/spec/dynamics.pdf index 4703147..031b21c 100644 Binary files a/spec/dynamics.pdf and b/spec/dynamics.pdf differ diff --git a/spec/dynamics.tex b/spec/dynamics.tex index c219f62..83eb5a2 100644 --- a/spec/dynamics.tex +++ b/spec/dynamics.tex @@ -11,7 +11,15 @@ 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{% @@ -37,204 +45,204 @@ \begin{document} - \title{Dynamics} - \author{} - \date{} - \maketitle +\title{Dynamics} +\author{} +\date{} +\maketitle + +\section{Resolution of forces} + +\textbf{Resultant force} is sum of force vectors - \section{Resolution of forces} +\subsection*{In angle-magnitude form} - \textbf{Resultant force} is sum of force vectors +\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 angle-magnitude form} +\subsection*{In \(\boldsymbol{i}\)---\(\boldsymbol{j}\) 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. - 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 \(\theta = \tan^{-1} \frac{b}{a}\) - 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} - \subsection{Resolving in a given direction} +The resolved part of a force \(P\) at angle \(\theta\) is has magnitude \(P \cos \theta\) - 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 \(|\boldsymbol{r}|=\sqrt{\left(||\vec{OA}\right)^2 + \left(\perp\vec{OA}\right)^2},\quad \theta = \tan^{-1}\dfrac{\perp\vec{OA}}{||\vec{OA}}\) - 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} - \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} + +\[ F_R = \mu R \tag{friction coefficient} \] - \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 +\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 -{\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 +\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) {}; - % Draw axes and help lines + \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 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} +} - {[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) {\(\theta\)}; +\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) {\(\theta\)}; } - % 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] {\(T_1\)}; + \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] {\(T_1\)}; } - \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} - - - \colorbox{cas}{On CAS:} use Geometry, lock known constants. + \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} +\colorbox{cas}{On CAS:} use Geometry, lock known constants. \end{document}