spec / dynamics.texon commit [methods] re-render methods notes (1dadb9e)
   1\documentclass[spec-collated.tex]{subfiles}
   2\begin{document}
   3
   4\section{Dynamics}
   5
   6\subsection*{Resolution of forces}
   7
   8\textbf{Resultant force} is sum of force vectors
   9
  10\subsubsection*{In angle-magnitude form}
  11
  12\makebox[3cm]{Cosine rule:} \(c^2=a^2+b^2-2ab\cos\theta\)
  13\makebox[3cm]{Sine rule:} \(\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}\)
  14
  15\subsubsection*{In \(\boldsymbol{i}\)---\(\boldsymbol{j}\) form}
  16
  17Vector 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.
  18
  19To find angle of an \(a\boldsymbol{i} + b\boldsymbol{j}\) vector, use \(\theta = \tan^{-1} \frac{b}{a}\)
  20
  21\subsubsection*{Resolving in a given direction}
  22
  23The resolved part of a force \(P\) at angle \(\theta\) is has magnitude \(P \cos \theta\)
  24
  25To convert force \(||\vec{OA}\) to angle-magnitude form, find component \(\perp\vec{OA}\) then:
  26\begin{align*}
  27  |\boldsymbol{r}| &= \sqrt{\left(||\vec{OA}\right)^2 + \left(\perp\vec{OA}\right)^2} \\
  28  \theta &= \tan^{-1}\dfrac{\perp\vec{OA}}{||\vec{OA}}
  29\end{align*}
  30
  31\subsection*{Newton's laws}
  32
  33\begin{theorembox}{}
  34  \begin{enumerate}[leftmargin=1mm]
  35    \item Velocity is constant without \(\Sigma F\)
  36    \item \(\frac{d}{dt} \rho \propto \Sigma F \implies \boldsymbol{F}=m\boldsymbol{a}\)
  37    \item Equal and opposite forces
  38  \end{enumerate}
  39\end{theorembox}
  40
  41\subsubsection*{Weight}
  42A mass of \(m\) kg has force of \(mg\) acting on it
  43
  44\subsubsection*{Momentum \(\rho\)}
  45\[ \rho = mv \tag{units kg m/s or Ns} \]
  46
  47\subsubsection*{Reaction force \(R\)}
  48
  49\begin{itemize}
  50  \item With no vertical velocity, \(R=mg\)
  51  \item With vertical acceleration, \(|R|=m|a|-mg\)
  52  \item With force \(F\) at angle \(\theta\), then \(R=mg-F\sin\theta\)
  53\end{itemize}
  54
  55\subsubsection*{Friction}
  56
  57\[ F_R = \mu R \tag{friction coefficient} \]
  58
  59\subsection*{Inclined planes}
  60
  61\[ \boldsymbol{F} = |\boldsymbol{F}| \cos \theta \boldsymbol{i} + |\boldsymbol{F}| \sin \theta \boldsymbol{j} \]
  62\begin{itemize}
  63  \item Normal force \(R\) is at right angles to plane
  64  \item Let direction up the plane be \(\boldsymbol{i}\) and perpendicular to plane \(\boldsymbol{j}\)
  65\end{itemize}
  66
  67\def\iangle{30} % Angle of the inclined plane
  68
  69\def\down{-90}
  70\def\arcr{0.5cm} % Radius of the arc used to indicate angles
  71
  72\tikzset{
  73  force/.style={->,draw=blue,fill=blue, thick},
  74  axis/.style={densely dashed,gray,font=\small},
  75  M/.style={rectangle,draw,fill=lightgray,minimum size=0.5cm,thin},
  76  m/.style={rectangle,draw=black,fill=lightgray,minimum size=0.3cm,thin},
  77  plane/.style={draw=black,fill=blue!10},
  78  string/.style={draw=red, thick},
  79  pulley/.style={thick}
  80}
  81\tikzset{
  82  % style to apply some styles to each segment of a path
  83  on each segment/.style={
  84    decorate,
  85    decoration={
  86      show path construction,
  87      moveto code={},
  88      lineto code={
  89        \path [#1]
  90        (\tikzinputsegmentfirst) -- (\tikzinputsegmentlast);
  91      },
  92      closepath code={
  93        \path [#1]
  94        (\tikzinputsegmentfirst) -- (\tikzinputsegmentlast);
  95      },
  96    },
  97  },
  98  % style to add an arrow in the middle of a path
  99  mid arrow/.style={postaction={decorate,decoration={
 100        markings,
 101        mark=at position .5 with {\arrow[#1]{stealth}}
 102      }}},
 103}
 104  \begin{center}\begin{tikzpicture}[scale=1.8]
 105
 106    \pgfmathsetmacro{\Fnorme}{2}
 107    \pgfmathsetmacro{\Fangle}{30}
 108
 109    \begin{scope}[rotate=\iangle]
 110      \node[M,transform shape] (M) {};
 111      \coordinate (xmin) at ($(M.south west)-({abs(1.1*\Fnorme*sin(-\Fangle))},0)$);
 112      \coordinate (xmax) at ($(M.south east)+({abs(1.1*\Fnorme*sin(-\Fangle))},0)$);
 113      \coordinate (ymax) at ($(M.center)+(0,{cos(\Fangle)})$);
 114      \coordinate (ymin) at ($(M.center)-(0,{cos(\Fangle)})$);
 115      \coordinate (axiscentre) at ($(M.south)+(0.5cm, 0.5cm)$);
 116      \draw[postaction={decorate, decoration={border, segment length=4pt, angle=-45},draw,red}] (xmin) -- (xmax);
 117      \coordinate (fr) at ($(M.center)+({\Fnorme*sin(-\Fangle)}, 0)$);
 118      {[axis,->]
 119      \draw ($(M)+(1,0)$) -- ($(M)+(1.5,0)$) node[above right] {\(\boldsymbol{i}\)};
 120      \draw ($(M)+(1,0)$) -- ($(M)+(1,0.5)$) node[above right] {\(\boldsymbol{j}\)};
 121      }
 122      {[force,->]
 123        \draw (M.center) -- (ymax) node [right] {\(R\)};
 124        \draw (M.center) -- (fr) node [left] {\(\mu R\)};
 125      }
 126    \end{scope}
 127    \draw[force,->] (M.center) -- ++(0,-1) node[below] {$mg$};
 128    \draw (xmin)+(0:\arcr) arc [start angle=0, end angle=\iangle, radius=\arcr] node [right, midway] {\footnotesize\(\theta\)};
 129      \coordinate (xbottom) at ($(1, {4*\Fnorme*-cos(\Fangle)})$);
 130      \draw [->] (xmin) -- ++($({1.35*\Fnorme*cos(\Fangle)}, 0)$);
 131      \begin{scope}[darkgray, rotate=\iangle] \path [draw=darkgray, postaction={on each segment={mid arrow}}] (M.center) -- (ymin) node [pos=0.5, right] {\(mg \cos \theta\)} -- ++(-0.5,0) node[pos=0.5, below right] {\(mg \sin \theta\)};
 132      \end{scope}
 133  \end{tikzpicture}\end{center}
 134
 135\subsection*{Connected particles}
 136
 137\def\boxwidth{0.5}
 138\tikzset{
 139  box/.style={rectangle,draw,fill=lightgray,minimum width=\boxwidth,thin},
 140  m/.style={rectangle,draw=black,fill=lightgray,minimum size=\boxwidth, thin}
 141}
 142
 143
 144\begin{center}
 145  \begin{tikzpicture}[scale=1.5]
 146
 147    \matrix {
 148      \begin{scope}[scale=1.5]
 149
 150        \coordinate (O) at (0,0);
 151        \coordinate (A) at ($({3*cos(\iangle)},{3*sin(\iangle)})$);
 152        \coordinate (B) at ($({3*cos(\iangle)},0)$);
 153        \coordinate (C) at ($({(1.5-0.5*\boxwidth)*cos(\iangle)},{(1.5-0.5*\boxwidth)*sin(\iangle)})$); % centre of box
 154        \coordinate (D) at ($(C)+(\iangle:\boxwidth)$);
 155        \coordinate (E) at ($(D)+(90+\iangle:0.5*\boxwidth)$);
 156        \coordinate (F) at ($(B)+(0,{1.5*sin(\iangle)})$);
 157        \coordinate (X) at ($(A)+(\iangle:0.5*\boxwidth)$); % centre of pulley
 158        \coordinate (Y) at ($(X)+(90+\iangle:0.5*\boxwidth)$); % chord of pulley
 159
 160        \draw[plane] (O) -- (A) -- (B) -- (O);
 161        \draw (O)+(\arcr,0) arc [start angle=0,end angle=\iangle,radius=\arcr] node [right, pos=.75] {\footnotesize\(\theta\)};
 162
 163        \draw [rotate=\iangle, m] (C) rectangle ++(\boxwidth,\boxwidth) node (z) [rotate=\iangle, midway] {\(m_1\)};
 164        \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] {};
 165        \draw [string] (E) -- (Y) arc (90+\iangle:0:0.25) -- ++($(0,{-1.5*sin(\iangle)-\boxwidth})$) node (p) {};
 166        \coordinate (Z) at ($(p.center)+({-0.5*\boxwidth},0)$);
 167        \draw [m] (Z) rectangle ++(\boxwidth, \boxwidth) node [midway] {\(m_2\)};
 168
 169      \end{scope}
 170\\
 171
 172      \begin{scope}[rotate=\iangle, scale=1.5]
 173
 174        \draw [m] ++(-0.5*\boxwidth,-0.5*\boxwidth) rectangle ++(\boxwidth,\boxwidth) node (n) [rotate=\iangle, midway] {\(m_1\)};
 175
 176        {[axis,-]
 177          \draw (0,-1) -- (0,0);
 178          \draw[solid,shorten >=0.5pt] (\down-\iangle:\arcr) arc(\down-\iangle:\down:\arcr);
 179          \node at (\down-0.5*\iangle:1.3*\arcr) {\(\theta\)};
 180        }
 181
 182        {[force,->]
 183          \draw (n.center) -- ++(0,{cos(\iangle)}) node[above right] {\(R_1\)};
 184          \draw (n.west) -- ++(-0.5,0) node[left] {\(\mu R_1\)};
 185          \draw (n.east) -- ++(1,0) node[above] {\(T_1\)};
 186        }
 187
 188        \draw[force,->, rotate=-\iangle] (M.center) -- ++(0,-1) node[below] {\(m_1 g\)};
 189
 190      \end{scope}
 191
 192      &
 193      \begin{scope}[scale=1.5]
 194
 195      \draw [m] ++(-0.5*\boxwidth,-0.5*\boxwidth) rectangle ++(\boxwidth,\boxwidth) node [midway] {\(m_2\)};
 196
 197      {[force,->]
 198        \draw (0,0.5*\boxwidth) -- ++(0,1) node[above] {\(T_2\)};
 199        \draw (0,-0.5*\boxwidth) -- ++(0,-1) node[right] {\(m_2 g\)};
 200      }
 201      \end{scope}
 202      \\
 203    };
 204  \end{tikzpicture}
 205  \end{center}
 206
 207\begin{itemize}
 208  \item \textbf{Suspended pulley:} \(T_1 = T_2\) \\
 209    \(|a| = g \dfrac{m_1 - m_2}{m_1 + m_2}\) where \(m_1\) accelerates down \\
 210    \[
 211    \left\{\begin{array}{lr}
 212      m_1g-T = m_1a\\
 213      T-m_2g = m_2a
 214    \end{array}\right\}
 215    \implies m_1 g - m_2 g = m_1 a + m_2 a
 216    \]
 217  \item \textbf{String pulling mass on inclined pane:} Resolve parallel to plane
 218    \[ T-mg \sin \theta = ma \]
 219  \item \textbf{Linear connection:} find acceleration of system first
 220  \item \textbf{Pulley on right angle:} \(a = \frac{m_2g}{m_1+m_2}\) where \(m_2\) is suspended (frictionless on both surfaces)
 221  \item \textbf{Pulley on edge of incline:} find downwards force \(W_2\) and components of mass on plane
 222\end{itemize}
 223
 224\begin{tabular}{rl}
 225  \parbox[t][][t]{8em}{In this example, note \(T_1 \ne T_2\):} &
 226  \parbox{12em}{
 227  \begin{tikzpicture}
 228
 229      \begin{scope}
 230
 231        \coordinate (O) at (0,0);
 232        \coordinate (A) at ($({3*cos(\iangle)},{3*sin(\iangle)})$);
 233        \coordinate (B) at ($({3*cos(\iangle)},0)$);
 234        \coordinate (C) at ($({(1-0.25*\boxwidth)*cos(\iangle)},{(1-0.25*\boxwidth)*sin(\iangle)})$); % centre of box
 235        \coordinate (D) at ($(C)+(\iangle:\boxwidth)$);
 236        \coordinate (E) at ($(D)+(90+\iangle:0.5*\boxwidth)$);
 237        \coordinate (F) at ($(B)+(0,{1.5*sin(\iangle)})$);
 238        \coordinate (G) at ($(A)+(\iangle:-2*\boxwidth)$);
 239        \coordinate (H) at ($(G)+(90+\iangle:0.5*\boxwidth)$);
 240        \coordinate (I) at ($(H)+(\iangle:-0.5*\boxwidth)$);
 241        \coordinate (J) at ($(H)+(\iangle:\boxwidth)$);
 242        \coordinate (X) at ($(A)+(\iangle:0.5*\boxwidth)$); % centre of pulley
 243        \coordinate (Y) at ($(X)+(90+\iangle:0.5*\boxwidth)$); % chord of pulley
 244
 245        \draw[plane] (O) -- (A) -- (B) -- (O);
 246        \draw (O)+(\arcr,0) arc [start angle=0,end angle=\iangle,radius=\arcr] node [right, pos=.75] {\footnotesize\(\theta\)};
 247
 248        \draw [rotate=\iangle, m] (C) rectangle ++(\boxwidth,\boxwidth) node (z) [rotate=\iangle, midway, font=\footnotesize] {\(m_1\)};
 249        \draw [rotate=\iangle, m] (G) rectangle ++(\boxwidth,\boxwidth) node (l) [rotate=\iangle, midway, font=\footnotesize] {\(m_2\)};
 250        \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] {};
 251        \draw [string] (E) -- (H) node [midway, above, font=\footnotesize, rotate=\iangle] {\(T_2\)};
 252        \draw [string] (J) -- (Y) node [midway, above, font=\footnotesize, rotate=\iangle] {\(T_1\)} arc (90+\iangle:0:0.25) -- ++($(0,{-1.5*sin(\iangle)})$) node [midway, above right, font=\footnotesize] {\(T_1\)} node[m] {\(m_3\)};
 253
 254      \end{scope}
 255
 256  \end{tikzpicture}}
 257\end{tabular}
 258
 259\subsection*{Equilibrium}
 260
 261\[ \dfrac{A}{\sin a} = \dfrac{B}{\sin b} = \dfrac{C}{\sin c} \tag{Lami's theorem}\]
 262\[ c^2 = a^2 + b^2 - 2ab \cos \theta \tag{cosine rule} \]
 263
 264Three methods:
 265\begin{enumerate}
 266  \item Lami's theorem (sine rule)
 267  \item Triangle of forces (cosine rule)
 268  \item Resolution of forces (\(\Sigma F = 0\) - simultaneous)
 269\end{enumerate}
 270
 271  \begin{cas}
 272    \textbf{To verify:} Geometry tab, then select points with normal cursor. Click right arrow at end of toolbar and input point, then lock known constants.
 273  \end{cas}
 274
 275\subsection*{Variable forces (DEs)}
 276
 277\[ a = \dfrac{d^2x}{dt^2} = \dfrac{dv}{dt} = v\dfrac{dv}{dx} = \dfrac{d}{dx} \left( \frac{1}{2} v^2 \right) \]
 278
 279\end{document}