From: Andrew Lorimer <andrew@lorimer.id.au>
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=inline

[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}