From: Andrew Lorimer <andrew@lorimer.id.au>
Date: Fri, 9 Aug 2019 04:39:55 +0000 (+1000)
Subject: [spec] dynamics notes
X-Git-Tag: yr12~70
X-Git-Url: https://git.lorimer.id.au/notes.git/diff_plain/3731517d3dc1ac7a694452137aff3c2836cb6590

[spec] dynamics notes
---

diff --git a/spec/dynamics.pdf b/spec/dynamics.pdf
new file mode 100644
index 0000000..a2e7446
Binary files /dev/null and b/spec/dynamics.pdf differ
diff --git a/spec/dynamics.tex b/spec/dynamics.tex
index 7708043..e22575b 100644
--- a/spec/dynamics.tex
+++ b/spec/dynamics.tex
@@ -1,4 +1,4 @@
-\documentclass[a4paper]{article}
+\documentclass[a4paper, tikz]{article}
 \usepackage[a4paper,margin=2cm]{geometry}
 \usepackage{array}
 \usepackage{amsmath}
@@ -6,6 +6,12 @@
 \usepackage{tcolorbox}
 \usepackage{fancyhdr}
 \usepackage{pgfplots}
+\usepackage{tikz}
+\usetikzlibrary{arrows,
+    calc,
+    decorations,
+    scopes,
+}
 \usepackage{tabularx}
 \usepackage{keystroke}
 \usepackage{listings}
@@ -25,10 +31,109 @@
   \date{}
   \maketitle
 
-  \section{Forces}
+  \section{Resolution of forces}
 
-  \subsection{Resolution of forces}
+  \textbf{Resultant force} is sum of force vectors
 
-  The resolved part of a force \(P\) at angle \(\theta\) is equal to \(P \cos \Theta\)
+  \subsection{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}
+
+  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}
+
+  The resolved part of a force \(P\) at angle \(\theta\) is has magnitude \(P \cos \theta\)
+
+  \section{Newton's laws}
+  
+  \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}
+
+  \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}
+
+  \subsection{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}
 
 \end{document}