%
%  2004/Lecture notes/Lecture12/Lecture12.tex
%
\documentclass[pdf,mmoll,slideColor%,colorBG
]{prosper}
\RequirePackage{color}

\usepackage[T1]{fontenc}
\usepackage{amsmath,amssymb,natbib}
%\usepackage[T1,noTS1]{mathtime}
\SetMathAlphabet{\mathbf}{normal}{\encodingdefault}{ptm}{b}{n}
\usepackage{subfigure}
\usepackage{rotating}
\usepackage{graphicx}
%\usepackage[dvips]{color}
\usepackage{epsfig}
\graphicspath{{../figs/}}

\newcommand{\ee}{\end{equation}} 
\newcommand{\be}{\begin{equation}} 
\newcommand{\ec}{\end{center}} 
\newcommand{\bc}{\begin{center}} 
\newcommand{\eea}{\end{eqnarray}} 
\newcommand{\bea}{\begin{eqnarray}} 
%\newcommand{\sumprime}{\mathop{{\sum}'}} 
\newcommand{\bd}{\begin{description}} 
\newcommand{\ed}{\end{description}} 
\newcommand{\bi}{\begin{itemize}} 
\newcommand{\ei}{\end{itemize}} 
\newcommand{\pa}{\partial} 
\newcommand{\hor}{\left(\frac{h}{R_0}\right)} 

\newcommand{\code}[0]{\ttfamily}
\newcommand{\co}[0]{\hbox{c}}
\newcommand{\si}[0]{\hbox{s}}
\newcommand{\pprime}[0]{\prime\prime}
\newcommand{\ppprime}[0]{\prime\prime\prime}
\newcommand{\conv}{\operatorname{conv}}
\newcommand{\pos}{\operatorname{pos}}
\newcommand{\lin}{\operatorname{lin}}
\newcommand{\vertop}{\operatorname{vert}}
\newcommand{\spanop}{\operatorname{span}}
\newcommand{\supp}{\operatorname{supp}}
\newcommand{\intoper}{\operatorname{int}}
\newcommand{\half}{\operatorname{half}}
\newcommand{\rad}{\operatorname{rad}}
\newcommand{\diam}{\operatorname{diam}}
\newcommand{\push}{\operatorname{push}}
\newcommand{\sqz}{\operatorname{sqz}}
\newcommand{\sgn}{\operatorname{sgn}}
%\renewcommand{\Re}{\Bbb{R}}
\renewcommand{\Re}{\mathbf{R}}
\newcommand{\vv}[1]{\mathbf{#1}}
\newcommand{\uu}[1]{\hat {\vv #1}}
\newcommand{\defterm}[1]{\textbf{#1}}
\newcommand{\rot}{\operatorname{rot}}

\newcommand{\btimes}{\times}
\newcommand{\bomega}{\omega}

 
\newcommand{\bis}{\begin{itemstep}} 
\newcommand{\eis}{\end{itemstep}} 

\makeatletter
\define@key{PDF}{Movie}{\pdf@addtoks{#1}{Movie}}
\define@key{PDF}{Activation}{\pdf@addtoks{#1}{Activation}}
\newcommand{\moviewithpreview}[3]{% args: width, preview, movie
   \pdfmark[{\includegraphics[width=#1]{#2}}]{%
   pdfmark=/ANN,Subtype=/Movie,Movie=<< /F (#3) >>,%
   Activation=<< /ShowControls true /Mode /Repeat >>}}
\newcommand{\movie}[3]{% args:width, height, movie
   \pdfmark[{\hbox to #1 {\vbox to #2 { }}}]{%
   pdfmark=/ANN,Subtype=/Movie,Movie=<< /F (#3) /Poster true >>,%
   Activation=<< /ShowControls true /Mode /Repeat >>}}
\makeatother

\title{12.  Foundations of Statics}
\subtitle{ Mechanics of Manipulation}
\author{Matt Mason}
\institution{Carnegie Mellon}
\email{matt.mason@cs.cmu.edu \\
http://www.cs.cmu.edu/\symbol{"7E}mason}
\slideCaption{Lecture 12. \hfill Mechanics of Manipulation }

\begin{document}

\maketitle

%=======================================================================
%
% SLIDE 2
%

\begin{slide}[R]{Lecture 12.  Foundations of statics.}
\vspace{-0.5in}
\includegraphics[height=3.0in]{TOC12}
\end{slide}

%=======================================================================
%
% SLIDE 3
%
%

\begin{slide}[R]{Outline.}
              
Summary:
\begin{itemize}
\item              
According to Newton, particles interact through forces.

              \item
Rigid bodies interact through \emph{wrenches}.

              \item
Wrenches and twists are dual.
\end{itemize}

Outline:              
\begin{enumerate}
\item Foundations.
\item Equivalence theorems. 
\item Line of action.
\item Poinsot's theorem.
\item Wrenches.
\end{enumerate}

\end{slide}

%=======================================================================
%
% SLIDE 4
%
%
\begin{slide}[R]{What is force?}
\begin{trivlist}
\item[]You cannot measure force, only its effects:  deformation of structures, acceleration.
\item[]We could start from Newton's laws, but instead we hypothesize:
\begin{itemize}
\item[]A force applied to a particle is a vector.
\item[]Motion is determined by vector sum.  
\item[]A particle remains at rest only if total force acting on it is zero.
\end{itemize}
\end{trivlist}
\end{slide}

%=======================================================================
%
% SLIDE 5
%
%
\overlays{2}{
\begin{slide}[R]{Moment of force}
\begin{trivlist}
\item Moment of force about a line:
\begin{itemize}
\item[]Let $l$ be line through origin with direction $\uu l$, 
\item[]Let $\vv f$ act at $\vv x$.  
\item[]\defterm{moment of force} or \defterm{torque} of $f$ about $l$:

\begin{equation*}
        n_l = \uu l \cdot (\vv x \btimes \vv f)
\end{equation*}
\end{itemize}
\fromSlide{2}{\item Moment of force about a point:
\begin{itemize}
\item[]\defterm{moment of force} or \defterm{torque} of $f$ about $O$:
\begin{equation*}
        \vv n_O = (\vv x - \vv O) \btimes \vv f 
\end{equation*}
\item[]

If the origin is $\vv O$ this reduces to $\vv n = \vv x \btimes \vv f$.

\item[]

If $\vv n$ is moment about the origin, and $n_l$ is moment 
about $l$ through the origin, 
\begin{equation*}
        n_l = \uu l \cdot \vv n
\end{equation*}
\end{itemize}}
\end{trivlist}
\end{slide}}

%=======================================================================
%
% SLIDE 6
%
%
\begin{slide}[R]{Rigid body force}
\begin{trivlist}
\item[]Consider a rigid body, and a set of forces $\{\vv f_i\}$ acting at $\{\vv x_i\}$ resp.
\item[]\defterm{Total force $\vv F$}:  sum of all external forces.
\item[]\defterm{Total moment $\vv N$}:  sum of all corresponding moments.
\begin{align*}
        \vv F  = & \sum \vv f_i  \\      
        \vv N  = & \sum \vv x_i \btimes \vv f_i   
\end{align*} 

\item[]Two systems of forces are \defterm{equivalent} if they have equal $\vv F$ and $\vv N$.
\item[](Equivalent, specifically, because they would have the same effect on a rigid body, according to Newton.
\end{trivlist}

\end{slide}

%=======================================================================
%
% SLIDE 7
%
%
\begin{slide}[R]{Resultant}
\begin{trivlist}
\item[]\defterm{Resultant} of a system of forces:  a system comprising a single force, equivalent to  the given system.
\end{trivlist}
\end{slide}

%=======================================================================
%
% SLIDE 8
%
%
\begin{slide}[R]{Line of action}
\begin{minipage}[t]{6cm}
  \begin{trivlist}
  \item[]Consider a force $\vv f$ applied at some point $\vv x_1$.
  \item[]Total force: $\vv F = \vv f$ 
  \item[]Total moment: $\vv N = \vv x_1 \btimes \vv f$.
  \item[]Consider line parallel to $\vv f$ through $\vv x_1$, and a
    second point $\vv x_2$ on the line.
  \item[]Force $\vv f$ through $\vv x_2$ is \emph{equivalent} to
    force $\vv f$ through $\vv x_1$. 
  \item[]So \emph{point} of application is more than you need to know $\ldots$
  \item[]\defterm{Line of action} of a force:  line through point of
    application parallel to force. 
  \item[]\defterm{Bound vector}, \defterm{free vector}, \defterm{line
      vector}, \defterm{point vector}. 
  \end{trivlist}
\end{minipage}
\hfill
\begin{minipage}[t]{4cm}
\begin{figure}[h!]
\includegraphics[width=4cm]{actline}
%\includegraphics[height=3cm]{*file*}
\end{figure}\end{minipage}
\end{slide}


%=======================================================================
%
% SLIDE 9
%
%
\begin{slide}[R]{Resultant of two forces}
\begin{minipage}[t]{6cm}
\begin{trivlist}
\item[]Let $\vv f_1$ and $\vv f_2$ act along $L_1$ and $L_2$ 
respectively.
\item[] Slide $\vv 
f_1$ and $\vv f_2$ along their respective lines of action to the 
intersection (if any)
\item[] Resultant:  the vector sum $\vv f_1 + \vv f_2$, acting at the 
intersection. 
\end{trivlist}
\end{minipage}
\hfill
\begin{minipage}[t]{4cm}
\begin{figure}[h!]
\includegraphics[width=4cm]{resultant2}
%\includegraphics[height=3cm]{*file*}
\end{figure}\end{minipage}
\end{slide}

%=======================================================================
%
% SLIDE 10
%
%
\begin{slide}[R]{Change of reference}
\begin{trivlist}
\item[]Using reference $Q$ or $R$, a system is described by 
\begin{align*}
        \vv F_Q  =  \sum f_i \quad \quad &
        \vv N_Q  =  \sum (\vv x_i - \vv Q) \btimes \vv f_i \\
        \vv F_R  =  \sum f_i \quad \quad &
        \vv N_R  =  \sum (\vv x_i - \vv R) \btimes \vv f_i
\end{align*}

\item[]From which it follows
\begin{align*}
        \vv F_R  =&  \vv F_Q  \\
        \vv N_R - \vv N_Q  =&  \sum (\vv Q - \vv R) \btimes \vv f_i
\end{align*}
which gives
\begin{align*}
        \vv N_R  = & \vv N_Q + (\vv Q - \vv R) \btimes \vv F 
\end{align*}

\end{trivlist}

\end{slide}

%=======================================================================
%
% SLIDE 11
%
%
\overlays{2}{
\begin{slide}[R]{Couple}
\begin{minipage}[t]{6cm}
  \begin{trivlist}
  \item[]Is a moment like a force?  Can you apply a moment?  Does it
    have a line of action? 
  \item[]A \defterm{couple} is a system of forces whose total force
    $\vv F = \sum \vv f_i$ is zero.
  \item[]I.e. a pure moment.
  \item[]Notice that the moment $\vv N$ of a couple is independent of
    reference point.
  \item[]For an arbitrary couple, can you construct an equivalent
    system of just two forces?
\end{trivlist}
\end{minipage}
\hfill
\begin{minipage}[t]{4cm}
\begin{figure}[h!]
\fromSlide{2}{\includegraphics[width=4cm]{forcencouple}}
%\includegraphics[height=3cm]{*file*}
\end{figure}
\end{minipage}
\end{slide}}

%=======================================================================
%
% SLIDE 12
%
%

\begin{slide}[R]{Equivalence theorems}
  \begin{trivlist}
  \item[]Our goal: defining \emph{wrench}, and showing every system of
    forces is equivalent to some wrench.
  \item[]Analogous to the program for kinematics, resulting in
    definition of \emph{screw}.
  \item[]Theorem: For any reference point $Q$, any system of forces is
    equivalent to a single force through $Q$, plus a couple.

    Proof:
    \begin{itemize}
      \item[] Let $\vv F$ be the total force;
        \item[] let $\vv N_Q$ be the total moment about $Q$.
          \item[] Apply $\vv F$ at $Q$;
            \item[] construct a couple with moment $\vv N_Q$.
              \end{itemize}

\end{trivlist}
\end{slide}

%=======================================================================
%
% SLIDE 13
%
%
\begin{slide}[R]{Two forces are sufficient}
  
  Theorem: Every system of forces is equivalent to a system of just
  two forces.
  \\[6pt]
  Proof:
  \begin{itemize}
  \item[]Given arbitrary $\vv F$ and $\vv N$, construct equivalent
    force and couple, comprising three forces in total.
  \item[]Move couple so that one of its forces acts at same point as
    $\vv F$.
  \item[]Replace those two forces with their resultant.
\end{itemize}

\end{slide}


%=======================================================================
%
% SLIDE 14
%
%

\begin{slide}[R]{Planar system with nonzero $\vv F$ has a resultant}
\begin{minipage}[t]{6cm}
  Theorem: A system consisting of a single non-zero force plus a
  couple in the same plane, i.e.\ a torque vector perpendicular to the
  force, has a resultant.
  \\[6pt]
  Proof:
  \begin{itemize}
  \item[]Let $\vv F$ be the force, acting at $P$\@.
  \item[]Let $\vv N$ be the moment of the couple.
  \item[]Construct an equivalent couple as in the figure.
  \item[]Translate the couple so $-\vv F$ is applied at $P$\@.
\end{itemize}
\end{minipage}
\hfill
\begin{minipage}[t]{4cm}
\begin{figure}[h!]
\includegraphics[width=4cm]{oneforce}
%\includegraphics[height=3cm]{*file*}
\end{figure}\end{minipage}
\end{slide}

%=======================================================================
%
% SLIDE x15
%
%
\begin{slide}[R]{Poinsot's theorem}
  Every system of forces is equivalent to a single force, plus a
  couple with moment parallel to the force. \\[6pt]
  Proof:
  \begin{itemize}
  \item[]Let $\vv F$ and $\vv N$ be the given force and moment.
  \item[]Decompose the moment: $\vv N_\parallel$ parallel to $\vv F$,
    and $\vv N_\perp$ perpendicular to $\vv F$\@.
  \item[]Since planar system with nonzero force has a resultant,
    replace $\vv F$ and $\vv N_\perp$ by a single force $\vv F^\prime$
    parallel to $\vv F$\@.
  \item[]Now we construct a couple with moment $\vv N_\parallel$ to
    obtain the desired result: a force and a couple with moment
    parallel to the force.
\end{itemize}

\end{slide}

%=======================================================================
%
% SLIDE 16
%
%
\begin{slide}[R]{Wrench}
  \begin{trivlist}
  \item[]\defterm{Wrench}: a screw plus a scalar magnitude, giving a
    force along the screw axis plus a moment about the screw axis.
  \item[]The force magnitude is the wrench magnitude, and the moment
    is the twist magnitude times the pitch.
  \item[]Thus the pitch is the ratio of moment to force.
    
  \item[]Poinsot's theorem is succinctly stated: every system of rigid
    body forces reduces to a wrench along some screw.
\end{trivlist}

\end{slide}

%=======================================================================
%
% SLIDE 17
%
%
\begin{slide}[R]{Screw coordinates for wrenches}
  \begin{trivlist}
  \item[]Let $f$ be the magnitude of the force acting along a line
    $l$,
  \item[]Let $n$ be the magnitude of the moment about $l$.
  \item[]The magnitude of the wrench is $f$.
  \item[]Recall definition in terms of Pl\"ucker coordinates:
    \begin{align*}
      \vv w &= f \vv q \\
      \vv w_0 &= f \vv q_0 + f p \vv q
    \end{align*}
    where $(\vv q, \vv q_0)$ are the normalized Pl\"{u}cker
    coordinates of the wrench axis $l$, and $p$ is the pitch,
    which is defined to be
    \begin{equation*}
      p =  n / f        
    \end{equation*}
  \end{trivlist}
\end{slide}

%=======================================================================
%
% SLIDE 18
%
%
\begin{slide}[R]{Screw coordinates for wrenches demystified}
  \begin{trivlist}
  \item[]Let $\vv r$ be some point on the wrench axis
    \begin{equation*}
      \vv q_0 = \vv r \btimes \vv q
    \end{equation*}
  \item[]With some substitutions $\ldots$
    \begin{align*}
      \vv w &= \vv f \\
      \vv w_0 &= \vv r \btimes \vv f + \vv n
    \end{align*}
  \item[]which can be written:
    \begin{align*}
      \vv w &= \vv f \\
      \vv w_0 &= \vv n_0
    \end{align*}
    where $\vv n_0$ is just the moment of force at the origin.  
  \item[]Screw coordinates of a wrench are actually a familiar
    representation $(\vv f, \vv n_0)$.
  \item[]Wrenches for a vector space.  You can scale and add them,
    just as with differential twists.
  \end{trivlist}

\end{slide}


%=======================================================================
%
% SLIDE 19
%
%
\begin{slide}[R]{Reciprocal product of twist and wrench}
  \begin{trivlist}
  \item[] Reciprocal product: 
    \begin{equation*}
      (\vv \bomega, \vv v_0) * (\vv f, \vv n_0) 
      = \vv f \cdot \vv v_0 + \vv n_0 \cdot \vv \bomega
    \end{equation*}
  \item[]The power produced by the wrench $(\vv f, \vv n_0)$ and
    differential twist $(\vv \bomega, \vv v_0)$.
  \item[]A differential twist is reciprocal to a wrench if and only if
    no power would be produced.\item[]Repelling if and only if
    positive power.
  \item[]Contrary if and only if negative power.
  \end{trivlist}

\end{slide}

%=======================================================================
%
% SLIDE 20
%
%
\begin{slide}[R]{Force versus motion}
  \begin{trivlist}
  \item[]Wrench coordinates and twist coordinates seem to use
    different conventions.
  \item[]Rotation or translation first?  Pitch is translation over
    rotation, or its inverse?
  \item[]Not just a peculiar convention.  Roles of translation and
    rotation are reversed!
  \item[]Example:  rotation axis versus line of action.
\end{trivlist}

\end{slide}

%=======================================================================
%
% SLIDE 21
%
%
\begin{slide}[R]{Comparing motion and force}
%\begin{unnumtable}
  \begin{tabular}{@{}p{.4\linewidth}@{\hspace{0.25in}}p{.4\linewidth}@{}}
%  \toprule
    {\bf Motion} & {\bf Force} \\
%  \colrule
    A zero-pitch twist is a pure rotation.  & A zero-pitch wrench is a
        pure force.  \\[6pt] 
        For a pure translation, the direction of the axis is
        determined, but the location is not.  
        & For a pure moment, the direction of the axis is 
        determined, but the location is not.  \\[6pt] 
        A differential translation is equivalent to a rotation about
        an axis at infinity.  & 
        A couple is equivalent to a force along a line at 
        infinity.  \\[12pt] 
        In the plane, any motion can be described as a rotation about some 
        point, possibly at infinity. 
        & In the plane, any system of forces reduces to a 
        single force, possibly at infinity.
%        \botrule
\end{tabular}
%\end{unnumtable}

\end{slide}


%=======================================================================
%
% SLIDE last
%

\begin{slide}[R]{Next:  Cones.}
\vspace{-0.5in}
\includegraphics[height=3.0in]{TOC13}
\end{slide}
\end{document}



\end{document}
