-\documentclass[a4paper]{article}
+\documentclass[a4paper, tikz]{article}
\usepackage[a4paper,margin=2cm]{geometry}
\usepackage{array}
\usepackage{amsmath}
\usepackage{tcolorbox}
\usepackage{fancyhdr}
\usepackage{pgfplots}
+\usepackage{tikz}
+\usetikzlibrary{arrows,
+ calc,
+ decorations,
+ scopes,
+}
\usepackage{tabularx}
\usepackage{keystroke}
\usepackage{listings}
\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}