From: Andrew Lorimer <andrew@lorimer.id.au>
Date: Thu, 15 Aug 2019 06:27:38 +0000 (+1000)
Subject: [spec] pulley-mass system diagram
X-Git-Tag: yr12~62
X-Git-Url: https://git.lorimer.id.au/notes.git/diff_plain/f9515eeaedea8ea9d610a3e42bf1c885f611f410

[spec] pulley-mass system diagram
---

diff --git a/spec/dynamics.pdf b/spec/dynamics.pdf
index a2e7446..4703147 100644
Binary files a/spec/dynamics.pdf and b/spec/dynamics.pdf differ
diff --git a/spec/dynamics.tex b/spec/dynamics.tex
index e22575b..c219f62 100644
--- a/spec/dynamics.tex
+++ b/spec/dynamics.tex
@@ -1,4 +1,4 @@
-\documentclass[a4paper, tikz]{article}
+\documentclass[a4paper, tikz, pstricks]{article}
 \usepackage[a4paper,margin=2cm]{geometry}
 \usepackage{array}
 \usepackage{amsmath}
@@ -12,7 +12,18 @@
     decorations,
     scopes,
 }
+\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
@@ -50,6 +61,8 @@
 
   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{enumerate}
@@ -128,12 +141,100 @@
         \draw (M.center)+(-90:\arcr) arc [start angle=-90,end angle=\iangle-90,radius=\arcr] node [below, pos=.5] {\tiny\(\theta\)};
     \end{tikzpicture}
 
-  \subsection{Connected particles}
+  \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
+
+\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
+    \begin{scope}[rotate=\iangle]
+        \node[M,transform shape] (M) {};
+        % Draw axes and help lines
+
+        {[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) {\(\theta\)};
+        }
+
+        % 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] {\(T_1\)};
+        }
+
+    \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.
+
 
 \end{document}