take .png files generated from .svg out of version control

[light-and-matter.git] / sn / ch03 / ch03.rbtex

<%
  require "../eruby_util.rb"
%>
<%
  chapter(
    '03',
    %q{Conservation of Momentum},
    'ch:3',
    %q{Forces transfer momentum to the girl.},
    {'opener'=>'sledding','sidecaption'=>true,'anonymous'=>true}
  )
%>

\epigraphlong{I think, therefore I am.\par\noindent{}I hope that posterity will judge me kindly, not only as to the things which I have explained, but also to those which I have intentionally omitted so as to leave to others the pleasure of discovery. }{Ren\'e Descartes}\index{Descartes, Ren\'e}
<% begin_sec("Momentum in One Dimension",0) %>\index{momentum}
<% begin_sec("Mechanical momentum") %>
        In the martial arts movie \emph{Crouching Tiger, Hidden Dragon},
        those who had received mystical enlightenment are able to violate the
        laws of physics. Some of the violations are obvious, such as their ability
        to fly, but others are a little more subtle. The rebellious young
        heroine/antiheroine Jen Yu gets into an argument while sitting
        at a table in a restaurant. A young tough, Iron Arm Lu, comes running toward her at full
        speed, and she puts up one arm and effortlessly makes him bounce
        back, without even getting out of her seat or bracing herself against
        anything. She does all this between bites.
        
<% marg(50) %>
<%
  fig(
    'mechanicalsystems',
    %q{%
      Systems consisting of material
              particles that interact through an energy $U(r)$. \emph{Top:\/} The galaxy M100. Here
              the ``particles'' are stars. \emph{Middle:\/} The pool balls don't interact until they
              come together and become compressed; the energy $U(r)$ has a sharp upturn when
              the center-to-center distance $r$ gets small enough for the balls to be in contact.
              \emph{Bottom:\/} A uranium nucleus undergoing fission. The energy $U(r)$ 
              has a repulsive contribution from the electrical interactions of the protons, plus
              an attractive one due to the strong nuclear interaction.
              \photocredit{M100: Hubble Space Telescope image.}
    }
  )
%>
<% end_marg %>
        \label{pconsproof1d}
        Although kinetic energy doesn't depend on the direction of motion,
        we've already seen on page \pageref{subsec:predictingdirection} how
        conservation of energy combined with Galilean relativity allows us to
        make some predictions about the direction of motion. One of the examples
        was a demonstration that it isn't possible for a hockey puck to
        spontaneously reverse its direction of motion. In the scene from the
        movie, however, the woman's assailant isn't just gliding through
        space. He's interacting with her, so the previous argument doesn't 
        apply here, and we need to generalize it to more than
        one object.
        We consider the case of a physical system composed of pointlike material
        particles, in which every particle interacts with every other particle
        through an energy $U(r)$ that depends only on the distance $r$ between them.
        This still allows for a fairly general \emph{mechanical system},\index{mechanical system}
        by which I mean roughly a system made of matter, not light.
        The characters
        in the movie are made of protons, neutrons, and electrons, so they
        would constitute such a system if the interactions among all these
        particles were of the form $U(r)$.\footnote{Electrical and magnetic interactions
        \emph{don't} quite behave like this, which is a point we'll take up
        later in the book.} We might even be able to get away with thinking of each person
        as one big particle, if it's a good approximation to say that
        every part of each person's whole body moves in the same
        direction at the same speed.

        The basic insight can be extracted from the special case where there
        are only two particles interacting, and they only move in one dimension,
        as in the example shown in figure \figref{poolballs}.
        Conservation of energy says
        \begin{equation*}
                K_{1i}+K_{2i}+U_i = K_{1f}+K_{2f}+U_f \qquad .
        \end{equation*}
        For simplicity, let's assume that the interactions start after the time we're calling
        initial, and end before the instant we choose as final. This is true 
        in figure \figref{poolballs}, for example. Then $U_i=U_f$, and we can subtract the
        interaction energies from both sides, giving.
        \begin{align*}
                K_{1i}+K_{2i} &= K_{1f}+K_{2f}  \\
                \frac{1}{2}m_1v_{1i}^2+\frac{1}{2}m_2v_{2i}^2 
                                &= \frac{1}{2}m_1v_{1f}^2+\frac{1}{2}m_2v_{2f}^2 \qquad .
        \end{align*}
        As in the one-particle argument on page  \pageref{subsec:predictingdirection}, the
        trick is to require conservation of energy not just in one particular frame of reference,
        but in every frame of reference. In a frame of reference moving at velocity $u$
        relative to the first one, the velocities all have $u$ added onto them:\footnote{We 
        can now see that the derivation would
        have been equally valid for $U_i\ne U_f$. The two observers agree
        on the distance between the particles, so they also agree on the interaction energies,
        even though they disagree on the kinetic energies.}
        \begin{multline*}
                        \frac{1}{2}m_1(v_{1i}+u)^2+\frac{1}{2}m_2(v_{2i}+u)^2 
                                = \frac{1}{2}m_1(v_{1f}+u)^2+\frac{1}{2}m_2(v_{2f}+u)^2 
        \end{multline*}
        When we square a quantity like $(v_{1i}+u)^2$, we get the same $v_{1i}^2$ that
        occurred in the original frame of reference, plus two $u$-dependent terms,
        $2v_{1i}u+u^2$. Subtracting the original conservation of energy equation
        from the  version in the new frame of reference, we have
        \begin{align*}
                m_1v_{1i}u+m_2v_{2i}u = m_1v_{1f}u+m_2v_{2f}u \qquad , \\
                \intertext{or, dividing by $u$,}
                m_1v_{1i}+m_2v_{2i} = m_1v_{1f}+m_2v_{2f} \qquad .
        \end{align*}
        This is a statement that when you add up $mv$ for the whole
        system, that total remains constant over time. In other words, this is a conservation law.
        The quantity $mv$ is called \emph{momentum}\index{momentum}, notated
        $p$ for obscure historical reasons. Its units are $\kgunit\cdot\munit/\sunit$.
<% marg(120) %>
<%
  fig(
    'poolballs',
    %q{%
      A collision between two pool balls is seen in two different
              frames of reference. The solid ball catches up with the striped ball.
              Velocities are shown with arrows. The second observer is moving to the left at velocity
              $u$ compared to the first observer, so all the velocities in the second frame
              have $u$ added onto them. The two observers must agree on conservation of energy.
    }
  )
%>
<% end_marg %>

        Unlike kinetic energy, momentum depends on the direction of motion, since the
        velocity is not squared. In one dimension, motion in the same direction as the
        positive $x$ axis is represented with positive values of $v$ and $p$. 
        Motion in the opposite direction has negative $v$ and $p$.

        \enlargethispage{-\baselineskip}

        \begin{eg}{Jen Yu meets Iron Arm Lu}
                \egquestion
                Initially, Jen Yu is at rest, and Iron Arm Lu is charging to the left, toward her,
                at 5 m/s. Jen Yu's mass is 50 kg, and Lu's is 100 kg. After the collision,
                the movie shows Jen Yu still at rest, and Lu rebounding at 5 m/s to the right.
                Is this consistent with the laws of physics, or would it be impossible in real
                life?

                \eganswer
                This is perfectly consistent with conservation of mass (50 kg+100 kg=50 kg+100 kg),
                and also with conservation of energy, since neither person's kinetic energy changes,
                and there is therefore no change in the total energy. (We don't have to worry about
                interaction energies, because the two points in time we're considering are ones at
                which the two people aren't interacting.)
                To analyze whether the scene violates conservation of momentum, we have to pick a
                coordinate system. Let's define positive as being to the right.
                The initial momentum is
                (50 kg)(0 m/s)+(100 kg)($-$5 m/s)=$-$500 $\kgunit\cdot\munit/\sunit$, and the final
                momentum is (50 kg)(0 m/s)+(100 kg)(5 m/s)=500 $\kgunit\cdot\munit/\sunit$.
                This is a change of 1000 $\kgunit\cdot\munit/\sunit$, which is impossible if
                the two people constitute a closed system.

                One could argue that they're not
                a closed system, since Lu might be exchanging momentum with the floor, and
                Jen Yu might be exchanging momentum with the seat of her chair. This is
                a reasonable objection, but in the following section we'll see that there are
                physical reasons why, in this situation, the force of
                friction would be relatively weak, and would not be able
                to transfer that much momentum in a fraction of a second.
        \end{eg}

        This example points to an intuitive interpretation of conservation of momentum,
        which is that interactions are always mutual. That is, Jen Yu can't change Lu's
        momentum without having her own momentum changed as well.
                                
        \begin{eg}{A cannon}
                \egquestion
                A cannon of mass 1000 kg fires a 10-kg shell at a
                velocity of 200 m/s. At what speed does the cannon recoil?

                \eganswer
                The law of conservation of momentum tells us that
                \begin{equation*}
                         p_{cannon,i} +  p_{shell,i} 
                                =  p_{cannon,f} +  p_{shell,f}   \qquad .
                \end{equation*}
                Choosing a coordinate system in which the cannon points in
                the positive direction, the given information is
                \begin{align*}
                         p_{cannon,i}  &= 0 \\
                         p_shell,i   &= 0 \\
                         p_{shell,f}  &= 2000\ \zu{kg}\cdot\zu{m/s}   \qquad .
                \end{align*}
                We must have $p_{cannon,f}=-\zu{2000 kg}\cdot\zu{m/s}$, so the recoil velocity
                of the cannon is 2 m/s.

        \end{eg}

<% marg(170) %>
<%
  fig(
    'ion-drive',
    %q{%
      The ion drive engine of the NASA Deep Space 1 probe, shown
                      under construction (top) and being tested in a vacuum
                      chamber (bottom) prior to its October 1998 launch. Intended
                      mainly as a test vehicle for new technologies, the craft
                      nevertheless also carried out a scientific program
                      that included a        rendezvous with a comet in 2004.\photocredit{NASA}
    }
  )
%>
<% end_marg %>

        \begin{eg}{Ion drive}\index{ion drive}\index{Deep Space 1}
                \egquestion
                 The experimental solar-powered ion drive of the
                Deep Space 1 space probe expels its xenon gas exhaust at a
                speed of 30,000 m/s, ten times faster than the exhaust
                velocity for a typical chemical-fuel rocket engine. Roughly
                how many times greater is the maximum speed this spacecraft
                can reach, compared with a chemical-fueled probe with the
                same mass of fuel (``reaction mass'') available for pushing
                out the back as exhaust?

                \eganswer
                Momentum equals mass multiplied by velocity. Both
                spacecraft are assumed to have the same amount of reaction
                mass, and the ion drive's exhaust has a velocity ten times
                greater, so the momentum of its exhaust is ten times
                greater. Before the engine starts firing, neither the probe
                nor the exhaust has any momentum, so the total momentum of
                the system is zero. By conservation of momentum, the total
                momentum must also be zero after all the exhaust has been
                expelled. If we define the positive direction as the
                direction the spacecraft is going, then the negative
                momentum of the exhaust is canceled by the positive momentum
                of the spacecraft. The ion drive allows a final speed that
                is ten times greater. (This simplified analysis ignores the
                fact that the reaction mass expelled later in the burn is
                not moving backward as fast, because of the forward speed of
                the already-moving spacecraft.)
        \end{eg}
        
<% end_sec() %>
<% begin_sec("Nonmechanical momentum") %>\index{momentum!nonmechanical}
        So far, it sounds as though conservation of momentum can be proved mathematically,
        unlike conservation of mass and energy, which are entirely based on observations.
        The proof, however, was only for a mechanical system, with
        interactions of the form $U(r)$. Conservation of momentum can be extended to
        other systems as well, but this generalization is based on experiments, not
        mathematical proof. Light is the most important example of momentum that doesn't
        equal $mv$ --- light doesn't have mass at all, but it does have momentum. For
        example, a flashlight left on for an hour would absorb about
        $10^{-5}\ \kgunit\cdot\munit/\sunit$ of momentum as it recoiled.\index{light!momentum of}
        \index{momentum!of light}

        \begin{eg}{Halley's comet}\index{Halley's comet}\index{comet}
                Momentum is not always equal to $mv$. Halley's comet, shown
                in figure \figref{halley}, has a very elongated elliptical orbit, like those of
                many other comets. About once per century, its orbit brings
                it close to the sun. The comet's head, or nucleus, is
                composed of dirty ice, so the energy deposited by the
                intense sunlight gradually removes ice from the surface and turns it into water vapor. The bottom photo
                shows a view of the water coming off of the nucleus from
                the European Giotto space probe, which passed within 596 km
                of the comet's head on March 13, 1986.

                The sunlight does not
                just carry energy, however. It also carries momentum. Once
                the steam comes off, the momentum of the sunlight impacting
                on it pushes it away from the sun, forming a tail as shown
                in in the top image.
                The tail always points away from the sun, so when the comet is
                receding from the sun, the tail is in front.
                By analogy with matter, for which momentum equals $mv$, you
                would expect that massless light would have zero momentum,
                but the equation $p= mv$ is not the correct one for light, and
                light does have momentum. (Some comets also have a second
                tail, which is propelled by electrical forces rather than by
                the momentum of sunlight.)
        \end{eg}

<% marg(145) %>
<%
  fig(
    'halley',
    %q{%
      Halley's comet. Top: A photograph made from earth.
              Bottom: A view of the nucleus from the Giotto space probe.
              \photocredit{W. Liller and European Space Agency}
    }
  )
%>
<% end_marg %>%
        The reason for bringing this up is not so that you can
        plug numbers into formulas in these exotic situations.
        The point is that the conservation laws have proven so
        sturdy exactly because they can easily be amended to fit
        new circumstances. The momentum of light will be a natural consequence
        of the discussion of the theory of relativity in chapter \ref{ch:rel}.

<% end_sec() %>
<% begin_sec("Momentum compared to kinetic energy") %>
        \index{momentum!compared to kinetic energy}\index{kinetic energy!compared to momentum}
        Momentum and kinetic energy are both measures of the
        quantity of motion, and a sideshow in the Newton-Leibniz
        controversy over who invented calculus was an argument over
        whether $mv$ (i.e., momentum) or $mv^2$ (i.e., kinetic energy
        without the 1/2 in front) was the ``true'' measure of motion.
        The modern student can certainly be excused for wondering
        why we need both quantities, when their complementary nature
        was not evident to the greatest minds of the 1700s. The
        following table highlights their differences.
        
        \begin{tabular}{|p{45mm}|p{45mm}|}
                \hline                

                        Kinetic energy...        & Momentum... \\
                \hline\hline
                doesn't depend on direction.                & depends on direction.\\
                \hline                

                is always positive, and cannot cancel out. &
                cancels with momentum in the opposite direction.\\
                \hline                

                can be traded for forms of energy that do not involve motion. 
                Kinetic energy is not a conserved quantity by itself.         &
                is always conserved in a closed system. \\
                \hline                

                is quadrupled if the velocity is doubled. &
                        is doubled if the velocity is doubled.\\
                \hline                

        \end{tabular}

        \enlargethispage{-2\baselineskip}
        
        Here are some examples that show the different behaviors of
        the two quantities.

                \begin{eg}{A spinning top}\label{eg:spinning-top}
                A spinning top has zero total momentum, because for every
                moving point, there is another point on the opposite side
                that cancels its momentum. It does, however, have kinetic
                energy.
                \end{eg}

<% marg(5) %>

<%
  fig(
    'eg-top-and-tuning-fork',
    %q{%
      Examples \ref{eg:spinning-top} and \ref{eg:tuning-fork}. The momenta cancel, but the energies don't.
    }
  )
%>
<% end_marg %>

\begin{eg}{Why a tuning fork has two prongs}\label{eg:tuning-fork}
A tuning fork is made with two prongs so that they can vibrate in opposite directions,
canceling their momenta. In a hypothetical version with only one prong, the momentum
would have to oscillate, and this momentum would have to come from somewhere, such as
the hand holding the fork. The result would be that vibrations would be transmitted
to the hand and rapidly die out. In a two-prong fork, the two momenta cancel, but the
energies don't.
\end{eg}
                
                \begin{eg}{Momentum and kinetic energy in firing a rifle}
                The rifle and bullet have zero momentum and zero kinetic
                energy to start with. When the trigger is pulled, the bullet
                gains some momentum in the forward direction, but this is
                canceled by the rifle's backward momentum, so the total
                momentum is still zero. The kinetic energies of the gun and
                bullet are both positive numbers, however, and do not
                cancel. The total kinetic energy is allowed to increase,
                because kinetic energy is being traded for other forms of
                energy. Initially there is chemical energy in the gunpowder.
                This chemical energy is converted into heat, sound, and
                kinetic energy.
                The gun's ``backward'' kinetic energy
                does not refrigerate the shooter's shoulder!
                \end{eg}

        \enlargethispage{-2\baselineskip}

                \begin{eg}{The wobbly earth}
                As the moon completes half a circle around the earth, its
                motion reverses direction. This does not involve any change
                in kinetic energy. The reversed velocity 
                does, however, imply a reversed momentum, so
                conservation of momentum in the closed earth-moon system
                tells us that the earth must also reverse its momentum. In
                fact, the earth wobbles in a little ``orbit'' about a point
                below its surface on the line connecting it and the moon.
                The two bodies' momenta always point in opposite
                directions and cancel each other out.
                \end{eg}
        
                
        <% marg(0) %>
<%
  fig(
    'earthmoondivorce',
    %q{Example \ref{eg:earthmoondivorce}.}
  )
%>
<% end_marg %>
                \begin{eg}{The earth and moon get a divorce}\label{eg:earthmoondivorce}
                Why can't the moon suddenly decide to fly off one way and
                the earth the other way? It is not forbidden by conservation
                of momentum, because the moon's newly acquired momentum in
                one direction could be canceled out by the change in the
                momentum of the earth, supposing the earth headed off in the
                opposite direction at the appropriate, slower speed. The
                catastrophe is forbidden by conservation of energy, because
                their energies would have to increase greatly.
                \end{eg}
                
                \begin{eg}{Momentum and kinetic energy of a glacier}
                A cubic-kilometer glacier would have a mass of about $10^{12}$
                kg. If it moves at a speed of $10^{-5}$ m/s, then its momentum is
                $10^7\ \kgunit\cdot\munit/\sunit$.
                This is the kind of heroic-scale result we
                expect, perhaps the equivalent of the space shuttle taking
                off, or all the cars in LA driving in the same direction at
                freeway speed. Its kinetic energy, however, is only 50 J,
                the equivalent of the calories contained in a poppy seed or
                the energy in a drop of gasoline too small to be seen
                without a microscope. The surprisingly small kinetic energy
                is because kinetic energy is proportional to the square of
                the velocity, and the square of a small number is an even
                smaller number.
                \end{eg}

\startdqs

\begin{dq}
If all the air molecules in the room settled down in a
thin film on the floor, would that violate conservation of
momentum as well as conservation of energy?
\end{dq}

\begin{dq}
A  refrigerator has coils in back that get hot, and heat
is molecular motion. These moving molecules have both energy
and momentum. Why doesn't the refrigerator need to be tied
to the wall to keep it from recoiling from the momentum
it loses out the back?
\end{dq}

<% end_sec() %>
<% begin_sec("Collisions in one dimension",4) %>
<% marg(20) %>
<%
  fig(
    'colliding-galaxies',
    %q{%
      This Hubble Space 
              Telescope photo shows a small galaxy
              (yellow blob in the lower right) that has collided with a
              larger galaxy (spiral near the center), producing a wave of
              star formation (blue track) due to the shock waves passing
              through the galaxies' clouds of gas. This is considered a
              collision in the physics sense, even though it is
              statistically certain that no star in either galaxy ever
              struck a star in the other --- the stars are
              very small compared to the distances between 
              them.\photocredit{NASA}
    }
  )
%>
<% end_marg %>%
        Physicists employ the term ``collision'' in a broader sense
        than in ordinary usage, applying it to any situation where
        objects interact for a certain period of time. A bat hitting
        a baseball, a cosmic ray damaging DNA,
        and a gun and a bullet going their separate ways are all
        examples of collisions in this sense. Physical contact is
        not even required. A comet swinging past the sun on a
        hyperbolic orbit is considered to undergo a collision, even
        though it never touches the sun. All that matters is that
        the comet and the sun interacted gravitationally with each
        other.
        
        The reason for broadening the term ``collision'' in this way
        is that all of these situations can be attacked
        mathematically using the same conservation laws in similar
        ways. In our first example, conservation of momentum is all
        that is required.
        
        \begin{eg}{Getting rear-ended}
        \egquestion
        Ms.\ Chang is rear-ended at a stop light by Mr.\ Nelson,
        and sues to make him pay her medical bills. He
        testifies that he was only going 55 km per hour when he
        hit Ms.\ Chang. She thinks he was going much faster than
        that. The cars skidded together after the impact, and
        measurements of the length of the skid marks show that their joint velocity
        immediately after the impact was 30 km per hour. Mr.\
        Nelson's Nissan has a mass of 1400 kg, and Ms.\ Chang 's
        Cadillac is 2400 kg. Is Mr.\ Nelson telling the
        truth?
        
        \eganswer
        Since the cars skidded together, we can write down
        the equation for conservation of momentum using only two
        velocities, $v$ for Mr.\ Nelson's velocity before the crash,
        and $v'$ for their joint velocity afterward:
        \begin{equation*}
                 m_{N} v =  m_{N} v' + m_{C} v' \qquad   .
        \end{equation*}
        Solving for the unknown, $v$, we find
        \begin{align*}
                 v &=  \left(1+\frac{ m_C}{ m_{N}}\right) v' \\
                        &= \zu{80\ km/hr} \qquad  .
        \end{align*}
        He is lying.
        \end{eg}
        
        The above example was simple because both cars had the same
        velocity afterward. In many one-dimensional collisions,
        however, the two objects do not stick. If we wish to predict
        the result of such a collision, conservation of momentum
        does not suffice, because both velocities after the
        collision are unknown, so we have one equation in two
        unknowns.
        
        Conservation of energy can provide a second equation, but
        its application is not as straightforward, because kinetic
        energy is only the particular form of energy that has to do
        with motion. In many collisions, part of the kinetic energy
        that was present before the collision is used to create heat
        or sound, or to break the objects or permanently bend
        them. Cars, in fact, are carefully designed to crumple in a
        collision. Crumpling the car uses up energy, and that's good
        because the goal is to get rid of all that kinetic energy in
        a relatively safe and controlled way. At the opposite
        extreme, a superball is ``super'' because it emerges from a
        collision with almost all its original kinetic energy,
        having only stored it briefly as interatomic electrical energy while it
        was being squashed by the impact.
        
        Collisions of the superball type, in which almost no kinetic
        energy is converted to other forms of energy, can thus be
        analyzed more thoroughly, because they have $K_f=K_i$, as
        opposed to the less useful inequality $K_f<K_i$ for a case
        like a tennis ball bouncing on grass.
        
\begin{eg}{Pool balls colliding head-on}\label{eg:pool-balls}
\egquestion Two pool balls collide head-on, so that the
collision is restricted to one dimension. Pool balls are
constructed so as to lose as little kinetic energy as
possible in a collision, so under the assumption that no
kinetic energy is converted to any other form of energy,
what can we predict about the results of such a collision?
<% marg(40) %>
\formatlikecaption{%
\begin{flushleft}\textit{Gory Details of the Proof in Example \ref{eg:pool-balls}}\end{flushleft}
The equation $A+B = C+D$ says that
the change in one ball's velocity is
equal and opposite to the change
in the other's. We invent a symbol
$x=C-A$ for the change in ball 1's velocity. The second equation can
then be rewritten as
$A^2+B^2 = (A+x)^2+(B-x)^2$.
Squaring out the quantities in parentheses and then simplifying, we get
$0 = Ax-Bx+x^2$.
The equation has the trivial solution $x=0$, i.e., neither ball's velocity
is changed, but this is physically
impossible because the balls can't travel through each other like
ghosts. Assuming $x\ne 0$, we can divide by $x$ and solve for $x=B-A$. This
means that ball 1 has gained an
amount of velocity exactly sufficient to mat\-ch ball 2's
initial velocity, and vice-versa. The
balls must have swap\-ped velocities.}%
<% end_marg %>

\eganswer Pool balls have identical masses, so we use the
same symbol $m$ for both. Conservation of energy and no loss
of kinetic energy give us the two equations
\begin{align*}
  mv_{1i}+mv_{2i} &=  mv_{1f}+mv_{2f} \\
  \frac{1}{2}mv_{1i}^2+\frac{1}{2}mv_{2i}^2 &=  \frac{1}{2}mv_{1f}^2+\frac{1}{2}mv_{2f}^2
\end{align*}
The masses and the factors of 1/2 can be divided out, and we
eliminate the cumbersome subscripts by replacing the symbols
$v_{1i}$,... with the symbols $A$, $B$, $C$, and $D$:
\begin{align*}
 A+B &= C+D \\
 A^2+B^2 &= C^2+D^2 \qquad .
\end{align*}
A little experimentation with numbers shows that given
values of $A$ and $B$, it is impossible to find $C$ and $D$
that satisfy these equations unless $C$ and $D$ equal $A$
and $B$, or $C$ and $D$ are the same as $A$ and $B$ but
swapped around. A formal proof of this fact is given in the
sidebar.
In the special case where ball 2 is initially at
rest, this tells us that ball 1 is stopped dead by the
collision, and ball 2 heads off at the velocity originally
possessed by ball 1. This behavior will be familiar to players of pool.
\end{eg}
        
        Often, as in example \ref{eg:pool-balls}, the details of the algebra
        are the least interesting part of the problem, and
        considerable physical insight can be gained simply by
        counting the number of unknowns and comparing to the number
        of equations. Suppose a beginner at pool notices a case
        where her cue ball hits an initially stationary ball and
        stops dead. ``Wow, what a good trick,'' she thinks. ``I bet I
        could never do that again in a million years.'' But she tries
        again, and finds that she can't help doing it even if she
        doesn't want to. Luckily she has just learned about
        collisions in her physics course. Once she has written down
        the equations for conservation of momentum and no loss of
        kinetic energy, she really doesn't have to complete the
        algebra. She knows that she has two equations in two
        unknowns, so there must be a well-defined solution. Once she
        has seen the result of one such collision, she knows that
        the same thing must happen every time. The same thing would
        happen with colliding marbles or croquet balls. It doesn't
        matter if the masses or velocities are different, because
        that just multiplies both equations by some constant factor.
        
        <% begin_sec("The discovery of the neutron") %>\label{chadwick-as-example-of-collision}
        This was the type of reasoning employed by James Chadwick\index{Chadwick, James}\index{neutron!discovery of}
        in his 1932 discovery of the neutron. At the time, the atom was
        imagined to be made out of two types of fundamental
        particles, protons and electrons. The protons were far more
        massive, and clustered together in the atom's core, or
        nucleus. Electrical attraction caused the electrons
        to orbit the nucleus in circles, in much the same way that
        gravity kept the planets from cruising out of
        the solar system. Experiments showed, for example,
        that twice as much energy was required to strip the last
        electron off of a helium atom as was needed to remove the
        single electron from a hydrogen atom,
        and this was explained by saying that
        helium had two protons to hydrogen's one. The trouble was
        that according to this model, helium would have two
        electrons and two protons, giving it precisely twice the
        mass of a hydrogen atom with one of each. In fact, helium
        has about four times the mass of hydrogen.
        
        Chadwick suspected that the helium nucleus possessed two
        additional particles of a new type, which did not
        participate in electrical interactions at all, i.e., were
        electrically neutral. If these particles had very nearly the
        same mass as protons, then the four-to-one mass ratio of
        helium and hydrogen could be explained.  In 1930, a new type
        of radiation was discovered that seemed to fit this
        description. It was electrically neutral, and seemed to be
        coming from the nuclei of light elements that had been
        exposed to other types of radiation. At this time, however,
        reports of new types of particles were a dime a dozen, and
        most of them turned out to be either clusters made of
        previously known particles or else previously known
        particles with higher energies. Many physicists believed
        that the ``new'' particle that had attracted Chadwick's
        interest was really a previously known particle called a
        gamma ray, which was electrically neutral. Since gamma rays
        have no mass, Chadwick decided to try to determine the new
        particle's mass and see if it was nonzero and approximately
        equal to the mass of a proton.
<%
  fig(
    'chadwick',
    %q{%
      Chadwick's subatomic pool table. A disk of the naturally
              occurring metal polonium provides a source of radiation
              capable of kicking neutrons out of the beryllium nuclei. The
              type of radiation emitted by the polonium is easily absorbed
              by a few mm of air, so the air has to be pumped out of the
              left-hand chamber. The neutrons, Chadwick's mystery
              particles, penetrate matter far more readily, and fly out
              through the wall and into the chamber on the right, which is
              filled with nitrogen or hydrogen gas. When a neutron
              collides with a nitrogen or hydrogen nucleus, it kicks it
              out of its atom at high speed, and this recoiling nucleus
              then rips apart thousands of other atoms of the gas. The
              result is an electrical pulse that can be detected in the
              wire on the right. Physicists had already calibrated this
              type of apparatus so that they could translate the strength
              of the electrical pulse into the velocity of the recoiling
              nucleus. The whole apparatus shown in the figure would fit
              in the palm of your hand, in dramatic contrast to today's
              giant particle accelerators.
    },
    {'width'=>'wide'}
  )
%>
        Unfortunately a subatomic particle is not something you can
        just put on a scale and weigh. Chadwick came up with an
        ingenious solution. The masses of the nuclei of the various
        chemical elements were already known, and techniques had
        already been developed for measuring the speed of a rapidly
        moving nucleus. He therefore set out to bombard samples of
        selected elements with the mysterious new particles. When a
        direct, head-on collision occurred between a mystery
        particle and the nucleus of one of the target atoms, the
        nucleus would be knocked out of the atom, and he would
        measure its velocity.
        
        Suppose, for instance, that we bombard a sample of hydrogen
        atoms with the mystery particles. Since the participants in
        the collision are fundamental particles, there is no way for
        kinetic energy to be converted into heat or any other form
        of energy, and Chadwick thus had two equations in three
        unknowns:
        
        equation \#1: conservation of momentum
        
        equation \#2: no loss of kinetic energy
        
        unknown \#1: mass of the mystery particle
        
        unknown \#2: initial velocity of the mystery particle
        
        unknown \#3:  final velocity of the mystery particle
        
        The number of unknowns is greater than the number of
        equations, so there is no unique solution. But by creating
        collisions with nuclei of another element, nitrogen, he
        gained two more equations at the expense of only one more
        unknown:
        
        equation \#3: conservation of momentum in the new collision
        
        equation \#4: no loss of kinetic energy in the new collision
        
        unknown \#4: final velocity of the mystery particle in the
        new collision

        \enlargethispage{\baselineskip}
        
        He was thus able to solve for all the unknowns, including
        the mass of the mystery particle, which was indeed within 1\%
        of the mass of a proton. He named the new particle the
        neutron, since it is electrically neutral.

\startdqs

\begin{dq}
Good pool players learn to make the cue ball spin, which can
cause it not to stop dead in a head-on collision with a
stationary ball. If this does not violate the laws of
physics, what hidden assumption was there in the example in the
text where it was proved that the cue ball must stop?
\end{dq}

<% end_sec() %>
<% end_sec() %>
<% begin_sec("The center of mass") %>
        Figures \figref{highjumper} and \figref{wrench} show two examples where
        a motion that appears complicated actually has a very simple feature. In both
        cases, there is a particular point, called the center of mass\index{center of mass},
        whose motion is surprisingly simple. The highjumper flexes his body as he passes
        over the bar, so his motion is intrinsically very complicated, and yet his center of
        mass's motion is a simple parabola, just like the parabolic arc of a pointlike
        particle. The wrench's center of mass travels in a straight line as seen from above,
        which is what we'd expect for a pointlike particle flying through the air.

<% marg(20) %>
<%
  fig(
    'highjumper',
    %q{%
      The highjumper's body passes over the bar,
              but his center of mass passes under it.\photocredit{Dunia Young}
    }
  )
%>
\spacebetweenfigs
<%
  fig(
    'cmballs',
    %q{Two pool balls collide.}
  )
%>
<% end_marg %>

<%
  fig(
    'wrench',
    %q{%
      In this multiple-flash photograph, we see the wrench from
              above as it flies through the air, rotating as it goes. Its
              center of mass, marked with the black cross, travels along a
              straight line, unlike the other points on the wrench, which
              execute loops. \photocredit{PSSC Physics}
    },
    {
      'width'=>'wide'
    }
  )
%>%
        The highjumper and the wrench are both complicated systems,
        each consisting of
        zillions of subatomic particles. To understand what's going on, let's instead look at a nice
        simple system, two pool balls colliding.
        We assume the balls are a closed system (i.e., their interaction with the felt
        surface is not important) and that their rotation is unimportant, so that
        we'll be able to treat each one as a single particle.
        By symmetry, the only place their center of mass can be is half-way in between,
        at an $x$ coordinate equal to the average of the two balls' positions,
        $x_{cm}=(x_1+x_2)/2$.

        Figure \figref{cmballs} makes it appear that the center
        of mass, marked with an $\times$, moves with constant velocity to the right,
        regardless of the collision, and we can easily prove this using conservation
        of momentum:
        \begin{align*}
                v_{cm}         &= \der{}x_{cm}/\der{}t \\
                                &= \frac{1}{2}(v_1+v_2) \\
                                & = \frac{1}{2m}(mv_1+mv_2) \\
                                & = \frac{p_{total}}{m_{total}}
        \end{align*}
        Since momentum is conserved, the last expression is constant,        which proves that $v_{cm}$ is constant.

        Rearranging this a little, we have $p_{total}=m_{total}v_{cm}$.
        In other words, the total momentum of the system is the same
        as if all its mass was concentrated at the center of mass
        point.  
<% marg(40) %>
<%
  fig(
    'pears',
    %q{%
      No matter what point you hang the
      pear from, the string lines up with the
      pear's center of mass. The center of
      mass can therefore be defined as the
      intersection of all the lines made by
      hanging the pear in this way.  Note that
      the X in the figure should not be
      interpreted as implying that the center
      of mass is on the surface --- it is
      actually inside the pear.
    }
  )
%>
\spacebetweenfigs
<%
  fig(
    'trapeze',
    %q{%
      The circus performers hang with the ropes passing
      through their centers of mass.
    }
  )
%>
<% end_marg %>

        <% begin_sec("Sigma notation") %>\index{sigma notation}
        When there is a large, potentially unknown number of particles,
        we can write sums like the ones occurring above using symbols like ``$+\ldots$,''
        but that gets awkward. It's more convenient to use the Greek
        uppercase sigma, $\Sigma$, to indicate addition. For example,
        the sum $1^2+2^2+3^2+4^2=30$ could be written as
        \begin{equation*}
                \sum_{j=1}^n{j^2} = 30 \qquad ,
        \end{equation*}
        read ``the sum from $j=1$ to $n$ of $j^2$.'' The variable $j$ is
        a dummy variable, just like the $\der{}x$ in an integral that tells you you're
        integrating with respect to $x$, but has no significance outside the integral.
        The $j$ below the sigma
        tells you what variable is changing from one term of the sum to the
        next, but $j$ has no significance outside the sum.  

        As an example,
        let's generalize the proof of $p_{total}=m_{total}v_{cm}$ to
        the case of an arbitrary number $n$ of identical particles moving in one
        dimension, rather than just two particles. The center of mass is at
        \begin{align*}
                x_{cm}        &= \frac{1}{n}\sum_{j=1}^{n}{x_j} \qquad ,
        \end{align*}
        where $x_1$ is the mass of the first particle, and so on. The velocity of the center of mass is
        \begin{align*}
                v_{cm}         &= \der{}x_{cm}/\der{}t \\
                                &= \frac{1}{n}\sum_{j=1}^{n}{v_j} \\
                                &= \frac{1}{nm}\sum_{j=1}^{n}{mv_j} \\
                                & = \frac{p_{total}}{m_{total}}
        \end{align*}

        What about a system containing objects with unequal masses,
        or containing more than two objects? The reasoning above can
        be generalized to a weighted average:
        \begin{align*}
                x_{cm}        &= \frac{\sum_{j=1}^{n}{m_jx_j}}{\sum_{j=1}^{n}{m_j}}
        \end{align*}
<% marg(70) %>
<%
  fig(
    'unbalanced-wheel',
    %q{%
      An improperly balanced wheel has a
      center of mass that is not at its
      geometric center. When you get a new
      tire, the mechanic clamps little weights
      to the rim to balance the wheel.
    }
  )
%>
\spacebetweenfigs
<%
  fig(
    'jumping-toy',
    %q{%
      This toy was intentionally designed so that the mushroom-shaped
      piece of metal on top would throw off the center of mass. When you wind it
      up, the mushroom spins, but the center of mass doesn't want to move, so the
      rest of the toy tends to counter the mushroom's motion, causing the whole
      thing to jump around.
    }
  )
%>
<% end_marg %>

        \begin{eg}{The solar system's center of mass}
                In the discussion of the sun's gravitational field on
                page \pageref{subsec:sungrav}, I mentioned in a footnote that the sun
                doesn't really stay in one place while the planets orbit around it.
                Actually, motion is relative, so it's meaningless to ask whether the
                sun is absolutely at rest, but it is meaningful to ask whether it
                moves in a straight line at constant velocity.
                We can now see that since the solar system is a closed system,
                its total momentum must be constant, and
                $p_{total}= m_{total}v_{cm}$ then tells us
                that it's the solar system's center of mass that has constant velocity,
                not the sun. The sun wobbles around this point irregularly due to
                its interactions with the planets, Jupiter in particular.
        \end{eg}                

        \begin{eg}{The earth-moon system}\label{eg:earthmooncg}
                The earth-moon system is much simpler than the solar system because it contains
                only two objects. Where is the center of mass of this system? Let $x$=0 be
                the earth's center, so that the moon lies at 
                $x=3.8\times10^5\  \zu{km}$. Then
                \begin{align*}
                         x_{cm}        &= \frac{\sum_{ j=1}^{2}{ m_{j} x_{j}}}
                                {\sum_{ j=1}^{2}{ m_j}} \\
                                &= \frac{ m_{1} x_{1}+ m_2 x_{2}}
                                        { m_{1}+ m_2} \qquad , \\
                \intertext{and letting 1 be the earth and 2 the moon, we have}
                         x_{cm}        
                                &= \frac{ m_{earth}\times0+ m_{moon} x_{moon}}
                                        { m_{earth}+ m_{moon}} \\
                                &= 4600\ \zu{km} \qquad ,
                \end{align*}
                or about three quarters of the way from the earth's center to its surface.
        \end{eg}                

\pagebreak[4]

\begin{eg}{The center of mass as an average}\label{eg:cm-examples}
\egquestion Explain how we know that the center of mass of each
object is at the location shown in figure \figref{cm-examples}.
<%
  fig(
    'cm-examples',
    %q{Example \ref{eg:cm-examples}.},
    {
      'width'=>'wide',
      'sidecaption'=>true
    }
  )
%>

\eganswer The center of mass is a sort of average, so the height
of the centers of mass in 1 and 2 has to be midway between the
two squares, because that height is the average of the heights
of the two squares. Example 3 is a combination of examples 1
and 2, so we can find its center of mass by averaging the horizontal
positions of their centers of mass. In example 4, each 
square has been skewed a little, but just as much mass has been
moved up as down, so the average vertical position of the mass
hasn't changed. Example 5 is clearly not all that different from
example 4, the main difference being a slight clockwise rotation,
so just as in example 4, the center of mass must be hanging in
empty space, where there isn't actually any mass. Horizontally,
the center of mass must be between the heels and toes, or else
it wouldn't be possible to stand without tipping over.
\end{eg}

\begin{eg}{Momentum and Galilean relativity}
        The principle of Galilean relativity states that the
        laws of physics are supposed to be equally valid in all
        inertial frames of reference. If we first calculate some
        momenta in one frame of reference and find that momentum is
        conserved, and then rework the whole problem in some other
        frame of reference that is moving with respect to the first,
        the numerical values of the momenta will all be different.
        Even so, momentum will still be conserved. All that matters
        is that we work a single problem in one consistent frame of
        reference.
        
        One way of proving this is to apply the equation
        $p_{total}=m_{total}v_{cm}$. If the velocity of one frame relative to
        the other is $u$, then the only effect of changing frames of
        reference is to change $v_{cm}$ from its original value to
        $v_{cm}+u$. This adds a constant onto the momentum,
        which has no effect on conservation of momentum.
\end{eg}
        
<% self_check('gymnastics-wheel',<<-'SELF_CHECK'
The figure shows a gymnast holding onto the inside of a big wheel.
From inside the wheel, how could he make it roll one way or the
other?
  SELF_CHECK
  ) %>
<% marg(50) %>
<%
  fig(
    'gymnastics-wheel',
    %q{Self-check \ref{sc:gymnastics-wheel}.}
  )
%>

<% end_marg %>
        

<% end_sec() %>
<% end_sec() %>
<% begin_sec("The center of mass frame of reference",4) %>\index{center of mass frame}
<% marg(10) %>
<%
  fig(
    'cmballs2',
    %q{%
      The same collision of two pools balls,
              but now seen in the center of mass frame of reference.
    }
  )
%>%
        \spacebetweenfigs%
        %
<%
  fig(
    'slingshot',
    %q{The sun's frame of reference.}
  )
%>%
                \label{eg:slingshot1}%
        \spacebetweenfigs%
        %
<%
  fig(
    'slingshotcm',
    %q{The c.m. frame.}
  )
%>%
                \label{eg:slingshot2}
<% end_marg %>%
        A particularly useful frame of reference in many cases is
        the frame that moves along with the center of mass, called
        the center of mass (c.m.) frame. In this frame, the total
        momentum is zero. The following examples show how the center
        of mass frame can be a powerful tool for simplifying our
        understanding of collisions.
        
        \begin{eg}{A collision of pool balls viewed in the c.m. frame}\label{eg:cmballs2}
        If you move your head so that your eye is always above the
        point halfway in between the two pool balls, as in figure \figref{cmballs2}, you are viewing
        things in the center of mass frame. In this frame, the balls
        come toward the center of mass at equal speeds. By symmetry,
        they must therefore recoil at equal speeds along the lines
        on which they entered. Since the balls have essentially
        swapped paths in the center of mass frame, the same must
        also be true in any other frame. This is the same result
        that required laborious algebra to prove previously without
        the concept of the center of mass frame.
        \end{eg}
        
        \begin{eg}{The slingshot effect}\label{eg:slingshot}
        It is a counterintuitive fact that a spacecraft can pick up
        speed by swinging around a planet, if it arrives in the
        opposite direction compared to the planet's motion. Although
        there is no physical contact, we treat the encounter as a
        one-dimensional collision, and analyze it in the center of
        mass frame. Since Jupiter is so much more massive than the
        spacecraft, the center of mass is essentially fixed at
        Jupiter's center, and Jupiter has zero velocity in the
        center of mass frame, as shown in figure \ref{eg:slingshot2}. The c.m. frame
        is moving to the left compared to the sun-fixed frame used
        in figure \ref{eg:slingshot1}, so the spacecraft's initial velocity is greater in
        this frame than in the sun's frame.
        
        Things are simpler in the center of mass frame, because it
        is more symmetric. In the sun-fixed frame, the incoming leg
        of the encounter is rapid, because the two bodies are
        rushing toward each other, while their separation on the
        outbound leg is more gradual, because Jupiter is trying to
        catch up. In the c.m. frame, Jupiter is sitting still, and
        there is perfect symmetry between the incoming and outgoing
        legs, so by symmetry we have $v_{1f}=- v_{1i}$. Going back to the
        sun-fixed frame, the spacecraft's final velocity is
        increased by the frames' motion relative to each other. In
        the sun-fixed frame, the spacecraft's velocity has increased
        greatly.
\end{eg}
        
\begin{eg}{Einstein's motorcycle}
        We've assumed we were dealing with a system of material objects,
        for which the equation $ p=mv$ was true. What if our system contains only light
        rays, or a mixture of light and matter? 
         As a college
        student, Einstein kept worrying about was what a beam of light
        would look like if you could ride alongside it on a motorcycle. In other
        words, he imagined putting himself in the light beam's center of mass frame.
        Chapter \ref{ch:rel} discusses Einstein's resolution of this problem, but the basic point is
        that you \emph{can't} ride the motorcycle alongside the light beam, because material
        objects can't go as fast as the speed of light. A beam of light has no center of
        mass frame of reference.
\end{eg}

\startdqs

\begin{dq}
Make up a numerical example of two unequal masses moving
in one dimension at constant velocity, and verify the
equation $p_{total}=m_{total}v_{cm}$ over a time
interval of one second.
\end{dq}

\begin{dq}
A  more massive tennis racquet or baseball bat makes the
ball fly off faster. Explain why this is true, using the
center of mass frame. For simplicity, assume that the
racquet or bat is simply sitting still before the collision,
and that the hitter's hands do not make any force large
enough to have a significant effect over the short
duration of the impact.
 \end{dq}

<% end_sec() %>
<% end_sec() %>
<% begin_sec("Force in One Dimension") %>
<% begin_sec("Momentum transfer") %>
        For every conserved quantity, we can define an associated rate of flow.
        An open system can have mass transferred in or out of it, and we can measure the
        rate of mass flow, $\der{}m/\der{}t$ in units of kg/s. Energy can flow in or out, and
        the rate of energy transfer is the power, $P=\der{}E/\der{}t$, measured in
        watts.\footnote{Recall that uppercase $P$ is power, while lowercase $p$ is momentum.}
         The rate of
        \emph{momentum} transfer is called force,\index{force!defined}
        \begin{equation*}
                F = \frac{\der{}p}{\der{}t} \qquad \text{[definition of force]} \qquad .
        \end{equation*}
        The units of force are $\kgunit\unitdot\munit/\sunit^2$, which can be abbreviated
        as newtons, $1\ \nunit=\kgunit\unitdot\munit/\sunit^2$.\index{newton (unit)}
        Newtons are unfortunately not as familiar as watts. A newton
        is about how much force you'd use to pet a dog. The most powerful rocket engine
        ever built, the first stage of the Saturn V that sent
        astronauts to the moon, had a thrust of about 30 million newtons.
        In one dimension, positive and negative signs indicate the direction of the force ---
        a positive force is one that pushes or pulls in the direction
        of the positive $x$ axis.  
<% marg(70) %>
<%
  fig(
    'e-and-p-transfer',
    %q{%
      Power and force are the rates at which
      energy and momentum are transferred.
    }
  )
%>
<% end_marg %>

        \begin{eg}{Walking into a lamppost}
        \egquestion
        Starting from rest, you begin walking, bringing
        your momentum up to 100 $\momunit$. You walk straight into a
        lamppost. Why is the momentum change of $-100\ \momunit$ so much
        more painful than the change of $+100\ \momunit$ when you started
        walking?
        
        \eganswer
        The forces are not really constant, but for this type of qualitative
        discussion we can pretend they are, and approximate $\der{} p/\der{} t$
        as $\Delta{} p/\Delta{} t$.
        It probably takes you
        about 1 s to speed up initially, so the ground's force on
        you is $F=\Delta{} p/\Delta{} t\approx100\ \zu{N}$.
        Your impact with the lamppost,
        however, is over in the blink of an eye, say 1/10 s or less.
        Dividing by this much smaller $\Delta{} t$ gives a much larger force,
        perhaps thousands of newtons (with a negative sign).
        \end{eg}
        
        This is also the principle of airbags in cars. The time
        required for the airbag to decelerate your head is fairly
        long: the time it takes your face to travel 20 or 30 cm.
        Without an airbag, your face would have hit the
        dashboard, and the time interval would have been the much
        shorter time taken by your skull to move a couple of
        centimeters while your face compressed. Note that either
        way, the same amount of momentum is transferred:
        the entire momentum of your head.
<% marg(100) %>
<%
  fig(
    'airbag',
    %q{%
      The airbag increases $\Delta t$ so as to reduce $F=\Delta p/\Delta t$.
    }
  )
%>
<% end_marg %>

        Force is defined as a derivative, and the derivative of a sum is the sum
        of the derivatives. Therefore force is additive: when more than one force
        acts on an object, you add the forces to find out what happens. An important special case
        is that forces can cancel. Consider your body sitting in a chair as you read
        this book. Let the positive $x$ axis be upward. The chair's upward force
        on you is represented with a positive number, which cancels out with
        the earth's downward gravitational force, which is negative. The total rate
        of momentum transfer into your body is zero, and your body doesn't
        change its momentum.

\begin{eg}{Finding momentum from force}
\egquestion An object of mass $m$ starts at rest at $t=t_\zu{o}$. 
A force varying as $F=bt^{-2}$,
where $b$ is a constant, begins acting on it. Find the greatest speed it will
ever have.

\eganswer
\begin{align*}
F &= \frac{\der p}{\der t} \\
\der p &= F \der t \\
p &= \int F \der t + p_\zu{o}\\
  &= -\frac{b}{t}+p_\zu{o} \qquad ,
\end{align*}
where $p_\zu{o}$ is a constant of integration.
The given initial condition is that $p=0$ at $t=t_\zu{o}$, so we find that $p_\zu{o}=b/t_\zu{o}$. The negative term gets closer
to zero with increasing time, so the maximum momentum is achieved by letting $t$ approach infinity. That is, the
object will never stop speeding up, but it will also never surpass a certain speed. In the limit $t\rightarrow\infty$,
we identify $p_\zu{o}$ as the momentum that the object will approach asymptotically. The maximum
velocity is $v=p_\zu{o}/m=b/mt_\zu{o}$.
\end{eg}

\startdq

\begin{dq}
 Many collisions, like the collision of a bat with a
baseball, appear to be instantaneous. Most people also would
not imagine the bat and ball as bending or being compressed
during the collision. Consider the following possibilities:

(1) The collision is instantaneous.\\
(2) The collision takes a finite amount of time, during
which the ball and bat retain their shapes and remain in contact.\\
(3) The collision takes a finite amount of time, during
which the ball and bat are bending or being compressed.

How can two of these be ruled out based on energy or
momentum considerations?
\end{dq}

<% end_sec() %>
<% begin_sec("Newton's laws") %>
        Although momentum is the third conserved quantity we've encountered,
        historically it was the first to be discovered. Isaac Newton formulated a
        complete treatment of mechanical systems in terms of force and momentum.
        Newton's theory was based on three laws of motion, which we now
        think of as consequences of conservation of mass, energy, and momentum.
<% marg(20) %>
<%
  fig(
    'newton',
    %q{%
      Isaac Newton (1643-1727).
    }
  )
%>
<% end_marg %>


        \begin{tabular}{|p{90mm}|}
        \hline
        Newton's laws in one dimension:\\
        \hline
        \hline
        \noindent\textsl{Newton's first law:\/} If there is no force acting on
        an object, it stays in the same state of motion.\\
        \hline
        \noindent\textsl{Newton's second law:\/} The sum of all the forces acting on an
        object determines the rate at which its momentum changes, $F_{total}=\der{}p/\der{}t$.\\
        \hline
        \noindent\textsl{Newton's third law:\/} Forces occur in opposite pairs.
        If object A interacts with object B, then A's force on B and B's force on
        A are related by $F_{AB}=-F_{BA}$.\\
        \hline
        \end{tabular}

        \noindent{}The second law is the definition of force, which we've already
        encountered.\footnote{This is with the benefit of hindsight. At the time,
        the word ``force'' already had certain connotations, and people thought
        they understood what it meant and how to measure it, e.g., by using a spring
        scale. From their point of view, $F=\der{}p/\der{}t$ was not a definition but a
        testable --- and controversial! --- statement.} The first law is a special case 
        of the second
        law ---
        if $\der{}p/\der{}t$ is
        zero, then $p=mv$ is a constant, and since mass is conserved, constant $p$
        implies constant $v$.
         The third law is a restatement of conservation of momentum: for two objects
        interacting, we have constant total momentum, so $0=\frac{\der}{\der{}t}(p_A+p_B)
        =F_{BA}+F_{AB}$.

        \begin{eg}{a=F/m}
        Many modern textbooks restate Newton's second law as $a= F/ m$, i.e.,
        as an equation that predicts an object's acceleration based on the force exerted on it.
        This is easily derived from Newton's original form as follows:
        $a=\zu{d} v/\zu{d} t=(\zu{d} p/\zu{d} t)/ m= F/ m$.
        \end{eg}
        \begin{eg}{Gravitational force related to g}
        As a special case of the previous example, consider an object in free fall, and
        let the $x$ axis point down. Then $a=+g$, and
        $F= ma= mg$. For example, the gravitational force on a 1 kg mass
        at the earth's surface is about 9.8 N. Even if other forces act on the object, and it
        isn't in free fall, the gravitational force on it is still the same, and can still be
        calculated as $mg$.
        \end{eg}

        \begin{eg}{Changing frames of reference}
        Suppose we change from one frame reference into another, which is moving relative
        to the first one at a constant velocity $u$. If an object of mass $m$ is moving
        at velocity $v$ (which need not be constant), then the effect is to change its
        momentum from $mv$ in one frame to $mv+mu$ in the other. Force is defined
        as the derivative of momentum with respect to time, and the derivative of a constant
        is zero, so adding the constant $mu$ has no effect on the result. We therefore conclude
        that observers in different inertial frames of reference agree on forces.
        \end{eg}

<% begin_sec("Using the third law correctly") %>
        If you've already accepted Galilean relativity in your heart, then there is nothing
        really difficult about the first and second laws. The third law, however, is more
        of a conceptual challenge. The first hurdle is that it is counterintuitive. Is it really
        true that if a fighter jet collides with a mosquito,
        the mosquito's force on the jet is just as strong as the jet's force on the mosquito?
        Yes, it is true, but it is hard to believe at first. That amount of force simply has more of an effect
        on the mosquito, because it has less mass.
<% marg(90) %>
<%
  fig(
    'third-law-magnets',
    %q{%
      Two magnets 
              exert forces on each other.
    }
  )
%>%
        \spacebetweenfigs%
        %
<%
  fig(
    'bartab',
    %q{%
      It doesn't make sense for the man to talk about the woman's
              money canceling out his bar tab, because there is no good
              reason to combine his debts and her assets.
    }
  )
%>%
        \spacebetweenfigs%
        %
<%
  fig(
    'skaters',
    %q{%
      Newton's third law does not mean that forces always cancel
              out so that nothing can ever move. If these two ice skaters, initially at rest,
              push against each other, they will both move.
    }
  )
%>
<% end_marg %>%
        A more humane
        and practical experiment is shown in figure \figref{third-law-magnets}.
        A large magnet and a small magnet
        are weighed separately, and then one magnet is hung from the
        pan of the top balance so that it is directly above the
        other magnet. There is an attraction between the two
        magnets, causing the reading on the top scale to increase
        and the reading on the bottom scale to decrease. The large
        magnet is more ``powerful'' in the sense that it can pick up a
        heavier paperclip from the same distance, so many people
        have a strong expectation that one scale's reading will
        change by a far different amount than the other. Instead, we
        find that the two changes are equal in magnitude but
        opposite in direction, so the upward force of the top magnet
        on the bottom magnet is of the same magnitude as the
        downward force of the bottom magnet on the top magnet.

        \enlargethispage{-\baselineskip}

        To students, it often sounds as though Newton's third law
        implies nothing could ever change its motion, since the two
        equal and opposite forces would always cancel. 
        As illustrated in figure \figref{bartab}, the fallacy arises
        from assuming that we can add things that it doesn't make sense to add.
        It only makes sense to
        add up forces that are acting on the same object, whereas
        two forces related to each other by Newton's third law are
        always acting on two different objects.
         If two objects are interacting via a force and no other forces are
        involved, then \emph{both\/} objects will accelerate --- in opposite
        directions, as shown in figure \figref{skaters}!    

Here are some suggestions for avoiding misapplication of Newton's third law:
\begin{enumerate}
\item It always relates exactly two forces, not more.
\item The two forces involve exactly two objects, in the pattern A on B, B on A.
\item The two forces are always of the same type, e.g., friction and friction, or gravity and gravity.
\end{enumerate}
<% marg(60) %>
<%
  fig(
    'swimming',
    %q{%
      A swimmer doing the breast stroke pushes backward
      against the water. By Newton's third law, the water pushes
      forward on him. 
    }
  )
%>
<% end_marg %>

<% end_sec() %>
<% begin_sec("Directions of forces") %>
We've already seen that momentum, unlike energy, has a direction in space. Since force is defined in terms
of momentum, force also has a direction in space. For motion in one dimension, we have to pick a coordinate
system, and given that choice, forces and momenta will be positive or negative. We've already used signs
to represent directions of forces in Newton's third law, $F_{AB}=-F_{BA}$.

        \enlargethispage{-\baselineskip}

There is, however, a complication with force that we were able to avoid with momentum. If an object is moving
on a line, we're guaranteed that its momentum is in one of two directions: the two directions along the
line. But even an object that stays on a line may still be subject to forces that act perpendicularly to the
line. For example, suppose a coin is sliding to the right across a table, \figref{unavoidable-crosswise-forces}, and let's choose a positive
$x$ axis that points to the right. The coin's motion is along a horizontal line, and its momentum is
positive and decreasing. Because the momentum is decreasing, its time derivative $\der p/\der t$ is negative. This
derivative equals the horizontal force of friction $F_1$, and its negative sign tells us that this force on the coin is to the left.
<% marg(30) %>
<%
  fig(
    'unavoidable-crosswise-forces',
    %q{%
      A coin slides across a table. Even for motion in one dimension, some of the forces may not lie along the line of the motion.
    }
  )
%>
<% end_marg %>

But there are also vertical forces on the coin. The Earth exerts a downward gravitational force $F_2$ on it, and
the table makes an upward force $F_3$ that prevents the coin from sinking into the wood. In fact, without these vertical
forces the horizontal frictional force wouldn't exist: surfaces don't exert friction against one another unless
they are being pressed together.

To avoid mathematical complication, we want to postpone the full three-dimensional treatment of force and
momentum until section \ref{sec:motion-in-three-d}. For now, we'll limit ourselves to examples like the coin,
in which the motion is confined to a line, and any forces perpendicular to the line cancel each other out.
<% end_sec() %>

\startdqs

\begin{dq}
Criticize the following incorrect statement:\\
``If an object is at rest and the total force on it is zero,
it stays at rest. There can also be cases where an object is
moving and keeps on moving without having any total force on
it, but that can only happen when there's no friction,
like in outer space.''
\end{dq}

\pagebreak

\begin{dq}\label{dq:johnson}
The table gives laser timing data for Ben
Johnson's 100 m dash at the 1987 World Championship in
Rome. (His world record was later revoked because he tested
positive for steroids.) How does the total force on him
change over the duration of the race?
\end{dq}

<% marg(40) %>
\formatlikecaption{%
\quad\begin{tabular}{ll}
$x$ (m)        & $t$ (s) \\
10                & 1.84 \\
20                & 2.86 \\
30                & 3.80 \\
40                & 4.67 \\
50                & 5.53 \\
60                & 6.38 \\
70                & 7.23 \\
80                & 8.10 \\
90                & 8.96 \\
100                & 9.83 
\end{tabular}\\
Discussion question \ref{dq:johnson}.
}%
<% end_marg %>

\begin{dq}
You hit a tennis ball against a wall. Explain any and all
incorrect ideas in the following description of the physics
involved: ``According to Newton's third law, there has to be
a force opposite to your force on the ball. The opposite
force is the ball's mass, which resists acceleration, and
also air resistance.''
\end{dq}

\begin{dq}
Tam Anh grabs Sarah by the hand and tries to pull her.
She tries to remain standing without moving. A student
analyzes the situation as follows. ``If Tam Anh's force on
Sarah is greater than her force on him, he can get her to
move. Otherwise, she'll be able to stay where she is.''
What's wrong with this analysis?
\end{dq}

<% end_sec() %>
<% begin_sec("What force is not") %>

Violin teachers have to endure their beginning students'
screeching. A frown appears on the woodwind teacher's face
as she watches her student take a breath with an expansion
of his ribcage but none in his belly. What makes physics
teachers cringe is their students' verbal statements about
forces. Below I have listed several dicta about what force is not.

<% begin_sec("Force is not a property of one object.") %>

A great many of students' incorrect descriptions of forces
could be cured by keeping in mind that a force is an
interaction of two objects, not a property of one object.

\begin{egnoheader}
\emph{Incorrect statement:\/} ``That magnet has a lot of force.''

\noindent<% x_mark %> If the magnet is one millimeter away from a steel ball
bearing, they may exert a very strong attraction on each
other, but if they were a meter apart, the force would be
virtually undetectable. The magnet's strength can be rated
using certain electrical units $(\zu{ampere}-\zu{meters}^2)$, but
not in units of force.
\end{egnoheader}

<% end_sec() %>
<% begin_sec("Force is not a measure of an object's motion.") %>

If force is not a property of a single object, then it
cannot be used as a measure of the object's motion.

\begin{egnoheader}
\emph{Incorrect statement:\/} ``The freight train rumbled down the
tracks with awesome force.''

\noindent<% x_mark %> Force is not a measure of motion. If the freight train
collides with a stalled cement truck, then some awesome
forces will occur, but if it hits a fly the force will be small.
\end{egnoheader}

\index{force!distinguished from energy}
\index{energy!distinguished from force}
<% end_sec() %>
<% begin_sec("Force is not energy.") %>

\begin{egnoheader}
\emph{Incorrect statement:\/} ``How can my chair be making an upward
force on my rear end? It has no power!''

\noindent<% x_mark %> Power is a concept related to energy, e.g., a 100-watt
lightbulb uses up 100 joules per second of energy. When you
sit in a chair, no energy is used up, so forces can exist
between you and the chair without any need for a source of power.
\end{egnoheader}

<% end_sec() %>
<% begin_sec("Force is not stored or used up.") %>

Because energy can be stored and used up, people think force
also can be stored or used up.

\begin{egnoheader}
\emph{Incorrect statement:\/} ``If you don't fill up your tank with
gas, you'll run out of force.''

\noindent<% x_mark %> Energy is what you'll run out of, not force.
\end{egnoheader}

<% end_sec() %>
<% begin_sec("Forces need not be exerted by living things or machines.") %>

Transforming energy from one form into another usually
requires some kind of living or mechanical mechanism. The
concept is not applicable to forces, which are an interaction
between objects, not a thing to be transferred or transformed.

\begin{egnoheader}
\emph{Incorrect statement:\/} ``How can a wooden bench be making an
upward force on my rear end? It doesn't have any springs or
anything inside it.''

\noindent<% x_mark %> No springs or other internal mechanisms are required. If
the bench didn't make any force on you, you would obey
Newton's second law and fall through it. Evidently it does
make a force on you!
\end{egnoheader}

<% end_sec() %>
<% begin_sec("A force is the direct cause of a change in motion.") %>

I can click a remote control to make my garage door change
from being at rest to being in motion. My finger's force on
the button, however, was not the force that acted on the
door. When we speak of a force on an object in physics, we
are talking about a force that acts directly. Similarly,
when you pull a reluctant dog along by its leash, the leash
and the dog are making forces on each other, not your hand
and the dog. The dog is not even touching your hand.

<% self_check('force-or-not',<<-'SELF_CHECK'
Which of the following things can be correctly described in terms of force?

(1) A nuclear submarine is charging ahead at full steam.

(2) A nuclear submarine's propellers spin in the water.

(3) A nuclear submarine needs to refuel its reactor periodically.
  SELF_CHECK
  ) %>

\startdqs

\begin{dq}
Criticize the following incorrect statement: ``If you
shove a book across a table, friction takes away more and
more of its force, until finally it stops.''
\end{dq}

\begin{dq}
You hit a tennis ball against a wall. Explain any and all
incorrect ideas in the following description of the physics
involved: ``The ball gets some force from you when you hit
it, and when it hits the wall, it loses part of that force,
so it doesn't bounce back as fast. The muscles in your arm
are the only things that a force can come from.''
\end{dq}

<% end_sec() %>
<% end_sec() %>
<% begin_sec("Forces between solids",nil,'forcesbetweensolids') %>
        Conservation laws are more fundamental than Newton's laws,
        and they apply where Newton's laws don't, e.g., to light and to
        the internal structure of atoms. However, there are certain problems
        that are much easier to solve using Newton's laws. As a trivial example,
        if you drop a rock, it could conserve momentum and energy by levitating,
        or by falling in the usual manner.\footnote{This pathological solution was
        first noted on page \pageref{hoveringrefback}, and discussed in more detail
        on page \pageref{hovering}.} With Newton's laws, however, we can
        reason that $a=F/m$, so the rock must respond to the gravitational force
        by accelerating.

        Less trivially, suppose a person is hanging onto a rope, and we want to
        know if she will slip. Unlike the case of the levitating rock, here the
        no-motion solution could be perfectly reasonable if her grip is strong
        enough. We know that her hand's interaction with the rope is fundamentally
        an electrical interaction between the atoms in the surface of her palm and
        the nearby atoms in the surface of the rope. For practical problem-solving,
        however, this is a case where we're better off forgetting the fundamental
        classification of interactions at the atomic level and working with a more
        practical, everyday classification of forces. In this practical scheme, we
        have three types of forces that can occur between solid objects in contact:
        \index{friction!kinetic}\index{kinetic friction}\index{friction!static}\index{static friction}
        \index{normal force}\index{force!normal}

        \noindent\begin{tabular}{|lp{65mm}|}
                \hline
                \emph{A normal force, $F_n$,} 
                        & is perpendicular to the surface of contact, and prevents objects
                                from passing through each other by becoming as strong
                                as necessary (up to the point where the objects break).
                                ``Normal'' means perpendicular. \\
                \hline
                \emph{Static friction, $F_s$,} 
                        & is parallel to the surface of contact, and prevents the surfaces from
                                starting to slip by becoming as strong as necessary, up
                                to a maximum value of $F_{s,max}$. 
                                ``Static'' means not moving, i.e., not slipping.\\
                \hline
                \emph{Kinetic friction, $F_k$,} 
                        & is parallel to the surface of contact, and tends to slow down
                                any slippage once it starts.
                                ``Kinetic'' means moving, i.e., slipping. \\
                \hline
        \end{tabular}

        <% self_check('frictionnormal',<<-'SELF_CHECK'
Can a frictionless surface exert a normal force? Can
        a frictional force exist without a normal force?
  SELF_CHECK
  ) %>

        If you put a coin on this page, which is horizontal, gravity pulls down on the coin,
        but the atoms in the paper and the coin repel each other electrically, and the atoms
        are compressed until the repulsion becomes strong enough to stop the downward
        motion of the coin. We describe this complicated and invisible atomic process by
        saying that the paper makes an upward normal force on the coin, and the coin
        makes a downward normal force on the paper. (The two normal forces are related by
        Newton's third law. In fact, Newton's third law only relates forces that are of the same
        type.)

        If you now tilt the book a little, static friction keeps the coin from slipping. The
        picture at the microscopic level is even more complicated than the previous
        description of the normal force. One model is to think of the tiny bumps and
        depressions in the coin as settling into the similar irregularities in the paper.
        This model predicts that rougher surfaces should have more friction, which is
        sometimes true but not always. Two very smooth, clean glass surfaces or very well
        finished machined metal surfaces may actually stick \emph{better} than
        rougher surfaces would, the probable explanation being that there is some kind of
        chemical bonding going on, and the smoother surfaces allow more atoms to be
        in contact.

        Finally, as you tilt the book more and more, there comes a point where static
        friction reaches its maximum value. The surfaces become unstuck, and the
        coin begins to slide over the paper. Kinetic friction slows down this slipping motion
        significantly. In terms of energy, kinetic friction is converting mechanical energy
        into heat, just like when you rub your hands together to keep warm. One model
        of kinetic friction is that the tiny irregularities in the two surfaces bump against
        each other, causing vibrations whose energy rapidly converts to heat and sound ---
        you can hear this sound if you rub your fingers together near your ear.
<% marg(120) %>
<%
  fig(
    'waiter',
    %q{%
      Static friction: the tray doesn't slip on
      the waiter's fingers.
    }
  )
%>
\spacebetweenfigs
<%
  fig(
    'skid',
    %q{Kinetic friction: the car skids.}
  )
%>

<% end_marg %>%

        For \emph{dry} surfaces, experiments show that the following equations 
        usually work fairly well:
        \begin{equation*}
                F_{s,max} \approx \mu_sF_n \qquad ,
        \end{equation*}
        and
        \begin{equation*}
                F_k \approx \mu_kF_n \qquad ,
        \end{equation*}
        where $\mu_s$, the coefficient of
        static friction,\index{static friction!coefficient of}\index{coefficient of static friction}
         and $\mu_k$, the coefficient of kinetic
        friction,\index{kinetic friction!coefficient of}\index{coefficient of kinetic friction}
         are constants that depend on the properties of the
        two surfaces, such as what they're made of and how rough they are.

<% self_check('static-or-kinetic',<<-'SELF_CHECK'
1. When a baseball player slides in to a base, is the
friction static, or kinetic?

\noindent 2. A mattress stays on the roof of a slowly accelerating
car. Is the friction static, or kinetic?

\noindent 3. Does static friction create heat? Kinetic friction?
  SELF_CHECK
  ) %>

\pagebreak

        \begin{eg}{Maximum acceleration of a car}
        \egquestion
        Rubber on asphalt gives $\mu_k\approx0.4$ and $\mu_s\approx 0.6$.
        What is the upper limit on a car's acceleration on a flat road, assuming that the engine
        has plenty of power and that air friction is negligible?

        \eganswer
        This isn't a flying car, so we don't expect it to accelerate vertically. The vertical forces acting on the car should cancel out.
        The earth makes a downward gravitational force on the car whose absolute value is $mg$, so the road apparently makes an upward normal force
        of the same magnitude, $F_n= mg$.

        Now what about the horizontal motion? As is always true, the coefficient
        of static friction is greater than the coefficient of kinetic friction, so the maximum acceleration
        is obtained with static friction, i.e., the driver should try not to burn rubber.
        The maximum force of static friction is $F_{s,max}=\mu_{s} F_{n} =\mu_s mg$. The maximum acceleration is 
        $a= F_s/ m=\mu_{s} g\approx6\ \munit/\sunit^2$. This
        is true regardless of how big the tires are, since the experimentally determined
        relationship $F_{s,max}=\mu_{s} F_n$ is independent of
        surface area.
        \end{eg}

        
<% self_check('find-directions-of-forces',<<-'SELF_CHECK'
Find the direction of each of the forces in 
figure \figref{sc-find-directions-of-forces}.
  SELF_CHECK
  ) %>

<%
  fig(
    'sc-find-directions-of-forces',
    %q{%
      1. The
      cliff's normal force on the climber's feet.
      2. The track's static frictional force on the wheel of
      the accelerating dragster.
      3. The ball's normal force on the bat.
    },
    {
      'width'=>'wide',
      'sidecaption'=>true
    }
  )
%>

\begin{eg}{Locomotives}\label{eg:locomotives}
Looking at a picture of a locomotive, \figref{locomotive}, we notice
two obvious things that are different from an automobile. Where a car
typically has two drive wheels, a locomotive normally has many --- ten
in this example. (Some also have smaller, unpowered wheels in front of and
behind the drive wheels, but this example doesn't.) Also, cars these days are generally built to be as light as possible
for their size, whereas locomotives are very massive, and no effort seems to be made
to keep their weight low. (The steam locomotive in the photo is from about 1900, but this
is true even for modern diesel and electric trains.)
<% fig('locomotive','Example \ref{eg:locomotives}.',
    {
      'width'=>'wide',
      'sidecaption'=>true
    }
) %>

The reason locomotives are built to be so heavy is for traction. The upward normal force of
the rails on the wheels, $F_N$, cancels the downward force of gravity, $F_W$, so ignoring plus and minus
signs, these two forces are equal in absolute value, $F_N=F_W$. Given this amount of normal
force, the maximum force of static friction is $F_s=\mu_s F_N=\mu_s F_W$. This static frictional
force, of the rails pushing forward on the wheels, is the only force that can accelerate the
train, pull it uphill, or cancel out the force of air resistance while cruising at constant speed.
The coefficient of static friction for steel on steel is about 1/4, so no locomotive can pull
with a force greater than about 1/4 of its own weight. If the engine is capable of supplying
more than that amount of force, the result will be simply to break static friction and spin the wheels.

The reason this is all so different from the situation with a car is that a car isn't pulling
something else. If you put extra weight in a car, you improve the traction, but you also
increase the inertia of the car, and make it just as hard to accelerate. In a train, the inertia
is almost all in the cars being pulled, not in the locomotive.

The other fact we have to explain is the large number of driving wheels. First, we have to
realize that increasing the number of driving wheels neither increases nor decreases the
total amount of static friction, because static friction is independent of the amount of
surface area in contact. (The reason four-wheel-drive is good in a car is that if one or
more of the wheels is slipping on ice or in mud, the other wheels may still have traction.
This isn't typically an issue for a train, since all the wheels experience the same conditions.)
The advantage of having more driving wheels on a train is that it allows us to increase the weight of the locomotive
without crushing the rails, or damaging bridges.
\end{eg}
        

<% end_sec() %>
<% begin_sec("Fluid friction") %>\index{friction!fluid}

Try to drive a nail into a waterfall and you will be
confronted with the main difference between solid friction
and fluid friction. Fluid friction is purely kinetic; there
is no static fluid friction. The nail in the waterfall may
tend to get dragged along by the water flowing past it, but
it does not stick in the water. The same is true for gases
such as air: recall that we are using the word ``fluid'' to
include both gases and liquids.

Unlike kinetic friction between solids, fluid friction
increases rapidly with velocity. It also depends
on the shape of the object, which is why a fighter jet is
more streamlined than a Model T.
For objects of the same shape but different sizes, fluid friction
typically scales up with the cross-sectional area of the
object, which is one of the main reasons that an SUV gets
worse mileage on the freeway than a compact car.
<% marg(80) %>
<%
  fig(
    'hummer-vs-prius',
    %q{%
      The wheelbases of the Hummer H3 and the Toyota Prius are surprisingly similar,
      differing by only 10\%. The main difference in shape is that the Hummer is much taller and wider.
      It presents a much greater cross-sectional area to the wind, and this is the main
      reason that it uses about 2.5 times more gas on the freeway.
    }
  )
%>
<% end_marg %>

\startdq

\begin{dq}
Criticize the following analysis: ``A book is sitting on a table. I shove it, overcoming
static friction. Then it slows down until it has less force than static friction, and it stops.''
\end{dq}

<% end_sec %>
<% begin_sec("Analysis of forces",0,'analysis-of-forces') %>\index{force!analysis of forces}

Newton's first and second laws deal with the total of all
the forces exerted on a specific object, so it is very
important to be able to figure out what forces there are.
Once you have focused your attention on one object and
listed the forces on it, it is also helpful to describe all
the corresponding forces that must exist according to
Newton's third law. We refer to this as ``analyzing the
forces'' in which the object participates.

\begin{egwide}{A barge}
A barge is being pulled along a canal by teams of horses on
the shores. Analyze all the forces in which the barge participates.

\vspace{2mm}

\begin{tabular}{|p{70mm}|p{70mm}|}
\hline
\emph{force acting on barge}  &   \emph{force related to it by Newton's third law} \\
\hline
ropes' forward normal forces on barge  &  barge's backward normal force on ropes\\
\hline
water's backward fluid friction force on barge  &  barge's forward fluid friction force on water\\
\hline
planet earth's downward gravitational force on barge  &  barge's upward gravitational force on earth\\
\hline
water's upward ``floating'' force on barge  &  barge's downward ``floating'' force on water\\
\hline
\end{tabular}

\vspace{2mm}

Here I've used the word ``floating'' force as an example of
a sensible invented term for a type of force not classified
on the tree in the previous section. A more formal technical
term would be ``hydrostatic force.''

Note how the pairs of forces are all structured as ``A's
force on B, B's force on A'': ropes on barge and barge on
ropes; water on barge and barge on water. Because all the
forces in the left column are forces acting on the barge,
all the forces in the right column are forces being exerted
by the barge, which is why each entry in the column
begins with ``barge.''
\end{egwide}

Often you may be unsure whether you have missed one of
the forces. Here are three strategies for checking your list:

\begin{indentedblock}
\noindent 1. See what physical result would come from the forces
you've found so far. Suppose, for instance, that you'd
forgotten the ``floating'' force on the barge in the example
above. Looking at the forces you'd found, you would have
found that there was a downward gravitational force on the
barge which was not canceled by any upward force. The barge
isn't supposed to sink, so you know you need to find a
fourth, upward force.

\noindent 2. Whenever one solid object touches another, there will
be a normal force, and possibly also a frictional force; check for
both.

\noindent 3. Make a drawing of the object, and draw a dashed boundary
line around it that separates it from its environment. Look
for points on the boundary where other objects come in
contact with your object. This strategy guarantees that
you'll find every contact force that acts on the object,
although it won't help you to find non-contact forces.

\end{indentedblock}

The following is another example in which we can profit by
checking against our physical intuition for what should be happening.

\begin{egwide}{Rappelling}
As shown in the figure below, Cindy is rappelling down a
cliff. Her downward motion is at constant speed, and she
takes little hops off of the cliff, as shown by the dashed
line. Analyze the forces in which she participates at a
moment when her feet are on the cliff and she is pushing off.

\begin{tabular}{|p{70mm}|p{70mm}|}
\hline
\emph{force acting on Cindy}  &   \emph{force related to it by Newton's third law} \\
\hline
planet earth's downward gravitational force on Cindy  &  Cindy's upward gravitational force on earth\\
\hline
ropes upward frictional force on Cindy (her hand)  &  Cindy's downward frictional force on the rope\\
\hline
cliff's rightward normal force on Cindy  &  Cindy's leftward normal force on the cliff\\
\hline
\end{tabular}

The two vertical forces cancel, which is what they should be
doing if she is to go down at a constant rate. The only
horizontal force on her is the cliff's force, which is not
canceled by any other force, and which therefore will
produce an acceleration of Cindy to the right. This makes
sense, since she is hopping off. (This solution is a little
oversimplified, because the rope is slanting, so it also
applies a small leftward force to Cindy. As she flies out to
the right, the slant of the rope will increase, pulling her
back in more strongly.)
\end{egwide}

\anonymousinlinefig{eg-rappelling}

I believe that constructing the type of table described in
this section is the best method for beginning students. Most
textbooks, however, prescribe a pictorial way of showing all
the forces acting on an object. Such a picture is called a
free-body diagram. It should not be a big problem if a
future physics professor expects you to be able to draw such
diagrams, because the conceptual reasoning is the same. You
simply draw a picture of the object, with arrows representing
the forces that are acting on it. Arrows representing
contact forces are drawn from the point of contact,
noncontact forces from the center of mass. Free-body
diagrams do not show the equal and opposite forces exerted
by the object itself.

\startdqs

\begin{dq}
When you fire a gun, the exploding gases push outward in
all directions, causing the bullet to accelerate down the
barrel. What Newton's-third-law pairs are involved? [Hint: Remember
that the gases themselves are an object.] 
\end{dq}

\begin{dq}
In the example of the barge going down the canal, I
referred to a ``floating'' or ``hydrostatic'' force that
keeps the boat from sinking. If you were adding a new branch
on the force-classification tree to represent this
force, where would it go?
\end{dq}

\pagebreak

\begin{dq}
A pool ball is rebounding from the side of the pool
table. Analyze the forces in which the ball participates
during the short time when it is in contact with the side of the table.
\end{dq}

\begin{dq}\label{dq:shovel}
The earth's gravitational force on you, i.e., your weight,
is always equal to $mg$, where $m$ is your mass. So why
can you get a shovel to go deeper into the ground by jumping
onto it? Just because you're jumping, that doesn't mean your
mass or weight is any greater, does it?
\end{dq}

<% end_sec() %>
<% begin_sec("Transmission of forces by low-mass objects",nil,'transmitforces') %>%
\index{force!transmission}\index{transmission of forces}

You're walking your dog. The dog wants to go faster than you
do, and the leash is taut. Does Newton's third law guarantee
that your force on your end of the leash is equal and
opposite to the dog's force on its end? If they're not
exactly equal, is there any reason why they should be
approximately equal?

If there was no leash between you, and you were in direct
contact with the dog, then Newton's third law would apply,
but Newton's third law cannot relate your force on the leash
to the dog's force on the leash, because that would involve
three separate objects. Newton's third law only says that
your force on the leash is equal and opposite to the
leash's force on you,
\begin{equation*}
        F_{yL}  =  - F_{Ly}  ,
\end{equation*}

and that the dog's force on the leash is equal and opposite
to its force on the dog
\begin{equation*}
        F_{dL}  =  - F_{Ld}  .
\end{equation*}
Still, we have a strong intuitive expectation that whatever
force we make on our end of the leash is transmitted to the
dog, and vice-versa. We can analyze the situation by
concentrating on the forces that act on the leash, $F_{dL}$
and $F_{yL}$. According to Newton's second law, these relate
to the leash's mass and acceleration:
\begin{equation*}
        F_{dL} + F_{yL}  =  m_La_L  .
\end{equation*}
The leash is far less massive then any of the other objects
involved, and if $m_L$ is very small, then apparently the
total force on the leash is also very small, $F_{dL}$ +
$F_{yL}\approx 0$, and therefore
\begin{equation*}
 F_{dL}\approx - F_{yL} \qquad .
\end{equation*}
Thus even though Newton's third law does not apply directly
to these two forces, we can approximate the low-mass leash
as if it was not intervening between you and the dog. It's
at least approximately as if you and the dog were acting
directly on each other, in which case Newton's third
law would have applied.

In general, low-mass objects can be treated approximately as
if they simply transmitted forces from one object to
another. This can be true for strings, ropes, and cords, and
also for rigid objects such as rods and sticks.
<%
  fig(
    'tension',
    %q{%
      If we imagine dividing a taut
      rope up into small segments, then any segment has forces
      pulling outward on it at each end. If the rope is of
      negligible mass, then all the forces equal
      $+T$ or $-T$, where $T$, the tension, is a single number.
    },
    {
      'width'=>'wide'
    }
  )
%>

If you look at a piece of string under a magnifying glass as
you pull on the ends more and more strongly, you will see
the fibers straightening and becoming taut. Different parts
of the string are apparently exerting forces on each other.
For instance, if we think of the two halves of the string as
two objects, then each half is exerting a force on the other
half. If we imagine the string as consisting of many small
parts, then each segment is transmitting a force to the next
segment, and if the string has very little mass, then all
the forces are equal in magnitude. We refer to the magnitude
of the forces as the tension in the string, $T$.\index{tension}

The term ``tension'' refers only to internal forces within the string.
If the string makes
forces on objects at its ends, then those forces are typically
normal or frictional forces (example \ref{eg:types-of-forces-made-by-ropes}).

\begin{eg}{Types of force made by ropes}\label{eg:types-of-forces-made-by-ropes}
\egquestion Analyze the forces in figures \subfigref{tension-is-not-a-type-of-force}{1}
and \subfigref{tension-is-not-a-type-of-force}{2}.

\eganswer
In all cases, a rope can only make ``pulling'' forces, i.e., forces that are parallel to its own length
and that are toward itself, not away from itself. You can't push with a rope!

In \subfigref{tension-is-not-a-type-of-force}{1}, the rope passes through a type of hook, called a
carabiner, used in rock climbing and mountaineering. Since the rope can only pull along its own length,
the direction of its force on the carabiner must be down and to the right. This is perpendicular to
the surface of contact, so the force is a normal force.
<% marg(80) %>
<%
  fig(
    'tension-is-not-a-type-of-force',
    %q{%
      Example \ref{eg:types-of-forces-made-by-ropes}. The forces between the rope and other objects are normal and frictional forces.
    }
  )
%>
<% end_marg %>

\begin{tabular}{|p{52mm}|p{52mm}|}
\hline
\emph{force acting on carabiner}  &   \emph{force related to it by Newton's third law} \\
\hline
rope's normal force on carabiner \hfill \anonymousinlinefig{../../../share/misc/arrows/5-oclock} &
          carabiner's normal force on rope \hfill \anonymousinlinefig{../../../share/misc/arrows/11-oclock} \\
\hline
\end{tabular}

(There are presumably other forces acting on the carabiner from other hardware above it.)

In figure \subfigref{tension-is-not-a-type-of-force}{2}, the rope can only exert a net force at its end that is
parallel to itself and in the pulling direction, so its force on the hand is down and to the left.
This is parallel
to the surface of contact, so it must be a frictional force. If the rope isn't slipping through the hand,
we have static friction. Friction can't exist without normal forces. These forces are perpendicular to the
surface of contact.
For simplicity, we show only two pairs of these normal forces, as if the hand were a pair of pliers.

\begin{tabular}{|p{52mm}|p{52mm}|}
\hline
\emph{force acting on person}  &   \emph{force related to it by Newton's third law} \\
\hline
rope's static frictional force on person \hfill \anonymousinlinefig{../../../share/misc/arrows/8-oclock} &
          person's static frictional force on rope \hfill \anonymousinlinefig{../../../share/misc/arrows/2-oclock} \\
\hline
rope's normal force on \linebreak[4] person \hfill \anonymousinlinefig{../../../share/misc/arrows/11-oclock} &
          person's normal force on \linebreak[4]  rope \hfill \anonymousinlinefig{../../../share/misc/arrows/5-oclock} \\
\hline
rope's normal force on \linebreak[4] person \hfill \anonymousinlinefig{../../../share/misc/arrows/5-oclock} &
          person's normal force on \linebreak[4] rope \hfill \anonymousinlinefig{../../../share/misc/arrows/11-oclock} \\
\hline
\end{tabular}

(There are presumably other forces acting on the person as well, such as gravity.)

\end{eg}


If a rope goes over a pulley or around some other object,
then the tension throughout the rope is approximately equal
so long as the pulley has negligible mass and there is not too much friction. A rod or stick
can be treated in much the same way as a string, but it is
possible to have either compression or tension.

\startdq

\begin{dq}
When you step on the gas pedal, is your foot's force being
transmitted in the sense of the word used in this section?
\end{dq}

<% end_sec() %>

\vspace{20mm}

<% begin_sec("Work") %>
        <% begin_sec("Energy transferred to a particle") %>
        To change the kinetic energy, $K=(1/2)mv^2$, of a particle moving in one dimension, 
        we must change its velocity. That will entail a change in its momentum, $p=mv$, as
        well, and since force is the rate of transfer of momentum, we conclude that the only
        way to change a particle's kinetic energy is to apply a force.\footnote{The converse
        isn't true, because kinetic energy doesn't depend on the direction of motion, but
        momentum does. We can change a particle's momentum without changing its
        energy, as when a pool ball bounces off a bumper, reversing the sign of $p$.}
        A force in the same direction as the motion speeds it up, increasing the kinetic
        energy, while a force in the opposite direction slows it down.

        Consider an infinitesimal time interval during which the particle moves
        an infinitesimal distance $\der{}x$, and its kinetic energy changes by $\der{}K$.
        In one dimension,
        we represent the direction of the force and the direction of the motion with
        positive and negative signs for $F$ and $\der{}x$, so the relationship among the
        signs can be summarized as follows:
        \begin{center}
                \begin{tabular}{|l|l|l|}
                \hline
                $F>0$        & $\der{}x>0$        & $\der{}K>0$ \\
                \hline
                $F<0$        & $\der{}x<0$        & $\der{}K>0$ \\
                \hline
                $F>0$        & $\der{}x<0$        & $\der{}K<0$ \\
                \hline
                $F<0$        & $\der{}x>0$        & $\der{}K<0$ \\
                \hline
                \end{tabular}
        \end{center}
        This looks exactly like the rule for determining the sign of a product, and
        we can easily show using the chain rule that this is indeed a multiplicative
        relationship:\label{originalworkderiv}
        \begin{align*}
                \der{}K        &= \frac{\der{}K}{\der{}v}\frac{\der{}v}{\der{}t}\frac{\der{}t}{\der{}x}\der{}x
                                                \qquad \text{[chain rule]} \\
                                &= (mv)(a)(1/v)\der{}x \\
                                & = m\,a\,\der{}x \\
                                & = F\,\der{}x \qquad \text{[Newton's second law]}
        \intertext{We can verify that force multiplied by distance has units of energy:}
                \nunit\unitdot\munit &= \frac{\kgunit\unitdot\munit/\sunit}{\sunit}\times\munit \\
                                &= \kgunit\unitdot\munit^2/\sunit^2 \\
                                &= \junit
        \end{align*}

        \begin{eg}{A TV picture tube}
        \egquestion
        At the back of a typical TV's picture tube, electrical forces accelerate each electron
        to an energy of $5\times10^{-16}\ \junit$ over a distance of about 1 cm.
        How much force is applied to a single electron? (Assume the force is constant.)
        What is the corresponding acceleration?

        \eganswer
        Integrating
        \begin{align*}
                \der{} K        &=  F\der{} x \qquad , 
        \intertext{we find}
                 K_{f}- K_{i}        &=  F( x_{f}- x_{i}) 
        \intertext{or}
                \Delta K &=  F\Delta x \qquad .
        \intertext{The force is}
                 F &= \Delta K/\Delta x \\
                        &= \frac{5\times10^{-16}\ \junit}{ 0.01\ \zu{m}} \\
                        & = 5\times10^{-14}\ \nunit \qquad .
        \intertext{This may not sound like an impressive force, but it's enough to supply
                an electron with a spectacular acceleration. Looking up the mass of an
                electron on p. \pageref{subatomicparticlesdata}, we find}
                 a        &=  F/ m \\
                                &= 5\times10^{16}\ \munit/\sunit^2 \qquad .
        \end{align*}
        \end{eg}

        \begin{eg}{An air gun}
                \egquestion
                An airgun, figure \figref{airgun}, uses compressed air to accelerate
                a pellet. As the pellet moves from $x_1$ to $x_2$,
                the air decompresses, so the force is not constant. Using methods
                from chapter \ref{ch:thermo}, one can show that the air's force
                on the pellet is given by $F= bx^\zu{-7/5}$. A typical high-end
                airgun used for competitive target shooting has
                \begin{align*}
                         x_1        &=  0.046\ \zu{m} \qquad , \\
                         x_2        &=  0.41\ \zu{m} \qquad , \\
                \intertext{and}
                         b        &=  4.4\ \zu{N}\unitdot\munit^\zu{7/5} \qquad . 
                \end{align*}
                What is the kinetic energy of the pellet when it leaves the muzzle?
                (Assume friction is negligible.)

<% marg(90) %>
<%
  fig(
    'airgun',
    %q{A simplified drawing of an airgun.}
  )
%>
<% end_marg %>

                \eganswer
                Since the force isn't constant, it would be incorrect to 
                do $F = \Delta K/\Delta x$.
                Integrating both sides of the equation $\der{} K= F\der{} x$, we have
                \begin{align*}
                        \Delta K        &= \int_{ x_{1}}^{ x_{2}} F\der{} x \\
                                        &= -\frac{5 b}{2}\left( x_2^\zu{-2/5}
                                                                - x_1^\zu{-2/5}\right) \\
                                        &= 22\ \zu{J}
                \end{align*}
        \end{eg}

        In general, when energy is transferred by a force,\footnote{The part of the
        definition about ``by a force'' is meant to exclude the transfer of energy
        by heat conduction, as when a stove heats soup.} we use the term
        \emph{work}\index{work!defined} to refer to the amount of energy transferred.
        This is different from the way the word is used in ordinary speech. If you
        stand for a long time holding a bag of cement, you get tired, and
        everyone will agree that you've worked hard, but you haven't changed the
        energy of the cement, so according to the definition of the physics term,
        you haven't done any work on the bag. There has been an energy transformation
        inside your body, of chemical energy into heat, but this just means that
        one part of your body did positive work (lost energy) while another part
        did a corresponding amount of negative work (gained energy).
<% marg(-30) %>
<%
  fig(
    'blackbox',
    %q{The black box does work by reeling in its cable.}
  )
%>
<% end_marg %>

        <% end_sec() %>
<% begin_sec("Work in general") %>
        I derived the expression $F\der{}x$ for one particular type of kinetic-energy
        transfer, the work done in accelerating a particle,
        and then defined work as a more general term. Is the equation
        correct for other types of work as well? For example,
        if a force lifts a mass $m$ against the resistance of gravity at constant velocity,
        the increase in the mass's gravitational energy is $\der{}(mgy)=mg\der{}y=F\der{}y$,
        so again the equation works, but this still doesn't prove that the equation is \emph{always}
        correct as a way of calculating energy transfers.

        Imagine a black box\footnote{``Black box'' is a traditional engineering term
        for a device whose inner workings we don't care about.}, containing a gasoline-powered
        engine, which is designed to reel in a steel cable, exerting a certain force $F$.
        For simplicity, we imagine that this force is always constant, so we can talk about
        $\Delta{}x$ rather than an infinitesimal $\der{}x$. If this black box is used to
        accelerate a particle (or any mass without internal structure), and no other forces
        act on the particle, then the original derivation applies, and the work done by the
        box is $W=F\Delta{}x$. Since $F$ is constant, the box will run out of gas after
        reeling in a certain amount of cable $\Delta{}x$. The chemical energy inside the box
        has decreased by $-W$, while the mass being accelerated has gained $W$ worth
        of kinetic energy.\footnote{For conceptual simplicity, we ignore the transfer of
        heat energy to the outside world via the exhaust and radiator. In reality, the
        sum of these energies plus the useful kinetic energy transferred would equal $W$.}

        Now what if we use the black box to pull a plow? The energy increase in the outside
        world is of a different type than before; it
        takes the forms of (1) the gravitational energy of the dirt that has been lifted out to the
        sides of the furrow, (2) frictional heating of the dirt and the plowshare,
        and (3) the energy needed to break up the dirt clods (a form of
        electrical energy involving the attractions among the atoms in the clod). The box,
        however, only communicates with the outside world via the hole through which its
        cable passes. The amount of chemical energy lost by the gasoline can therefore
        only depend on $F$ and $\Delta{}x$, so it is the same $-W$ as when the
        box was being used to accelerate a mass, and thus by conservation of energy, the
        work done on the outside world is again $W$.

<% marg(100) %>
<%
  fig(
    'kcm',
    %q{%
      The wheel spinning in the air has $K_{cm}=0$. The space
              shuttle has all its kinetic energy in the form of center of mass motion, $K=K_{cm}$.
              The rolling ball has some, but not all, of its energy in the form of center
              of mass motion, $K_{cm}<K$.\photocredit{Space Shuttle photo by NASA}
    }
  )
%>
<% end_marg %>
        This is starting to sound like a proof that the force-times-distance method
        is always correct, but there was one subtle assumption involved, which was that
        the force was exerted at one point (the end of the cable, in the black box example). 
        Real life often isn't like that. For example, a cyclist exerts forces on both pedals
        at once. Serious cyclists use toe-clips, and the conventional wisdom is that one
        should use equal amounts of force on the upstroke and downstroke, to make full
        use of both sets of muscles. This is  a two-dimensional example, since the
        pedals go in circles. We're
         only discussing one-dimensional motion right now, so let's just
        pretend that the upstroke and downstroke are both executed in straight lines.
        Since the forces are in opposite directions, one is positive and one is negative.
        The cyclist's \emph{total} force on the crank set is zero, but the work done isn't
        zero. We have to add the work done by each stroke, $W=F_1\Delta{}x_1+F_2\Delta{}x_2$.
        (I'm pretending that both forces are constant, so we don't have to do integrals.)
        Both terms are positive; one is a positive number multiplied by a positive number,
        while the other is a negative times a negative.

        This might not seem like a big deal --- just remember not to use the total force ---
        but there are many situations where the total force is all we can measure. The ultimate
        example is heat conduction. Heat conduction is not supposed to be counted as a form
        of work, since it occurs without a force. But at the atomic level, there are forces, and
        work is done by one atom on another. When you hold a hot potato in your hand, the
        transfer of heat energy through your skin takes place with a total force that's extremely
        close to zero. At the atomic level, atoms in your skin are interacting electrically with
        atoms in the potato, but the attractions and repulsions add up to zero total force.
        It's just like the cyclist's feet acting on the pedals, but with zillions of forces involved
        instead of two. There is no practical way to measure all the individual forces, and
        therefore we can't calculate the total energy transferred.

        To summarize, $\sum{F_j\der{}x_j}$ is a correct way of calculating work, where
        $F_j$ is the individual force acting on particle $j$, which moves a distance
        $\der{}x_j$. However, this is only useful if
        you can identify all the individual forces and determine the distance moved at
        each point of contact. For convenience, I'll refer to this as the
        \emph{work theorem}\index{work theorem}. (It doesn't have a standard name.)

        There is, however, something useful we can do with the total force. We can use it
        to calculate the part of the work done on an object that consists of a change in the
        kinetic energy it has due to the motion of its center of mass. The proof is essentially
        the same as the proof on p.~\pageref{originalworkderiv}, except that now we don't
        assume the force is acting on a single particle, so we have to
        be a little more delicate. Let the object consist of $n$ particles. Its total kinetic
        energy is $K=\sum_{j=1}^n{(1/2)m_jv_j^2}$, but this is what we've already
        realized \emph{can't} be calculated using the total force. The kinetic energy it
        has due to motion of its center of mass is
        \begin{align*}
                K_{cm}        &= \frac{1}{2}m_{total}v_{cm}^2 \qquad .
        \end{align*}
        Figure \figref{kcm} shows some examples of the distinction between $K_{cm}$ and $K$.
        Differentiating $K_{cm}$, we have
        \begin{align*}
                \der{}K_{cm}        &= m_{total}v_{cm}\der{}v_{cm} \\
                                &= m_{total}v_{cm}
                                                \frac{\der{}v_{cm}}{\der{}t}\frac{\der{}t}{\der{}x_{cm}}\der{}x_{cm}
                                                \qquad \text{[chain rule]} \\
                                &= m_{total}\frac{\der{}v_{cm}}{\der{}t}\der{}x_{cm} 
                                                \qquad \text{[$\der{}t/\der{}x_{cm}=1/v_{cm}$]} \\
                                &= \frac{\der{}p_{total}}{\der{}t}\der{}x_{cm} 
                                                \qquad \text{[$p_{total}=m_{total}v_{cm}$]} \\
                                &= F_{total}\der{}x_{cm}  
        \end{align*}
        I'll call this the \emph{kinetic energy theorem}\index{kinetic energy theorem} ---
        like the work theorem, it has no standard name. 
        
        \enlargethispage{-\baselineskip}

        \begin{eg}{An ice skater pushing off from a wall}
        The kinetic energy theorem tells us how to calculate
        the skater's kinetic energy if we know the amount of force
        and the distance her center of mass travels while she is
        pushing off.
        
        The work theorem tells us that the wall does no work on
        the skater, since the point of contact isn't moving.
        This makes sense, because the wall does not  have
        any source of energy.
        \end{eg}
        
        \begin{eg}{Absorbing an impact without recoiling?}
        \egquestion
        Is it possible to absorb an impact without
        recoiling? For instance, if a ping-pong ball hits
        a brick wall, does the wall ``give'' at all?
        
        \eganswer
        There will always be a recoil. In the example
        proposed, the wall will surely have some energy transferred
        to it in the form of heat and vibration. The work theorem
        tells us that we can only have an energy transfer if the distance
        traveled by the point of contact is nonzero.
        \end{eg}
        
        
        
        \begin{eg}{Dragging a refrigerator at constant velocity}
        The fridge's momentum is constant, so there is no net
        momentum transfer, and the total force on it
        must be zero: your force is canceling the
        floor's kinetic frictional force. The kinetic energy
        theorem is therefore true but useless. It tells us that
        there is zero total force on the refrigerator, and that the
        refrigerator's kinetic energy doesn't change.
        
        The work theorem tells us that the work you do equals
        your hand's force on the refrigerator multiplied by the
        distance traveled. Since we know the floor has no source of
        energy, the only way for the floor and refrigerator to gain
        energy is from the work you do. We can thus calculate the
        total heat dissipated by friction in the refrigerator and
        the floor.
        
        Note that there is no way to find how much of the heat is
        dissipated in the floor and how much in the refrigerator.
        \end{eg}

        \begin{eg}{Accelerating a cart}
        If you push on a cart and accelerate it, there are two
        forces acting on the cart: your hand's force, and the static
        frictional force of the ground pushing on the wheels in the
        opposite direction.
        
        Applying the work theorem to your force tells us how to
        calculate the work you do.
        
        Applying the work theorem to the floor's force tells us
        that the floor does no work on the cart. There is no motion
        at the point of contact, because the atoms in the floor are
        not moving. (The atoms in the surface of the wheel are also
        momentarily at rest when they touch the floor.) This makes
        sense, because the floor does not have any source of energy.
        
        The kinetic energy theorem refers to the total force,
        and because the floor's backward force cancels part of your
        force, the total force is less than your force. This tells
        us that only part of your work goes into the kinetic energy
        associated with the forward motion of the cart's center of
        mass. The rest goes into rotation of the wheels.
        \end{eg}

\startdqs

\begin{dq}
Criticize the following incorrect statement: ``A force
doesn't do any work unless it's causing the object to move.''
\end{dq}
\begin{dq}\label{dq:brakingdistance}
To stop your car, you must first have time to react, and
then it takes some time for the car to slow down. Both of
these times contribute to the distance you will travel
before you can stop. The figure shows how the average
stopping distance increases with speed. Because the stopping
distance increases more and more rapidly as you go faster,
the rule of one car length per 10 m.p.h. of speed is not
conservative enough at high speeds. In terms of work and
kinetic energy, what is the reason for the more rapid
increase at high speeds?
\end{dq}
<%
  fig(
    'brakingdistance',
    %q{Discussion question \ref{dq:brakingdistance}.},
    {
      'width'=>'wide'
    }
  )
%>

<% end_sec() %>
<% end_sec() %>

\pagebreak

<% begin_sec("Simple machines") %>

<% marg(-55) %>
<%
  fig(
    'pulley1',
    %q{The force is transmitted to the block.}
  )
%>%
                \spacebetweenfigs%
                %
<%
  fig(
    'pulley2',
    %q{A mechanical advantage of 2.}
  )
%>%
                \spacebetweenfigs%
                %
<%
  fig(
    'pulley3',
    %q{An inclined plane.}
  )
%>%
<% end_marg %>

        Conservation of energy provided the necessary tools for analyzing some
        mechanical systems, such as the seesaw on page \pageref{eg:seesaw} and the
        pulley arrangements of the homework problems on page \pageref{hw:atwood-energy-sn}, but we could
        only analyze those machines by computing the total energy of the system.
        That approach wouldn't work for systems like the biceps/forearm
        machine on page \pageref{fig:biceps}, or the one in figure \figref{pulley1},
        where the energy content of the person's body is impossible to compute
        directly. Even though the seesaw and the biceps/forearm system were clearly
        just two different forms of the lever, we had no way to treat them both on the
        same footing.
        We can now successfully attack such problems using the work and
        kinetic energy theorems.

        \begin{eg}{Constant tension around a pulley}
        \egquestion
        In figure \figref{pulley1}, what is the relationship between the
        force applied by the person's hand and the force exerted on the block?

        \eganswer
        If we assume the rope and the pulley are ideal, i.e., frictionless and
        massless, then there is no way for them to absorb or release energy,
        so the work done by the hand must be the same as the work done
        on the block. Since the hand and the block move the same distance,
        the work theorem tells us the two forces are the same.

        Similar arguments provide an alternative justification for the statement
        made in section \ref{subsec:transmitforces} that
        show that an idealized rope
        exerts the same force, the tension,\index{tension} anywhere it's attached to something, and the same
        amount of force is also exerted by each segment of the rope on the neighboring
        segments. Going around an ideal pulley also has no effect on the tension.

        This is an example of a simple machine,\index{simple machine}
         which is any mechanical system
        that manipulates forces to do work. This particular machine reverses the
        direction of the motion, but doesn't change the force or the speed of motion.
        \end{eg}

        \begin{eg}{A mechanical advantage}
        The idealized pulley in figure \figref{pulley2} has negligible
        mass, so its kinetic energy is zero, and the kinetic energy theorem
        tells us that the total force on it is zero. We know, as in the preceding
        example, that the two forces
        pulling it to the right are equal to each other, so the force on the left
        must be twice as strong. This simple machine doubles the applied force,
        and we refer to this ratio as a \emph{mechanical advantage} (M.A.) of 2.
        There's no such thing as a free lunch, however; the distance traveled by
        the load is cut in half, and there is no increase in the amount of work done.
        \end{eg}
        
        \begin{eg}{Inclined plane and wedge}\label{eg:inclinedplanework}
        In figure \figref{pulley3}, the force applied by the hand is equal
        to the one applied to the load, but there is a mechanical advantage compared
        to the force that would have been required to lift the load straight up.
        The distance traveled up the inclined plane is greater by a factor of 1/sin $\theta$,
        so by the work theorem, the force is smaller by a factor of sin $\theta$,
        and we have M.A.=1/sin $\theta$. The wedge, \figref{wedge}, is similar.
        \end{eg}
        
        \begin{eg}{Archimedes' screw}
        In one revolution, the crank travels a distance $2\pi{} b$, and the water
        rises by a height $h$. The mechanical advantage is $2\pi{} b/ h$.
        \end{eg}
<% marg(40) %>
<%
  fig(
    'wedge',
    %q{A wedge.}
  )
%>%
\spacebetweenfigs%
<%
  fig(
    'archimedesscrew',
    %q{Archimedes' screw}
  )
%>
<% end_marg %>%

<% end_sec() %>
<% begin_sec("Force related to interaction energy") %>
        In section \ref{gravphenomenasection}, we saw that there were two equivalent
        ways of looking at gravity, the gravitational field and the gravitational energy.
        They were related by the equation $\der{}U=mg\der{}r$, so if we knew the
        field, we could find the energy by integration, $U=\int{mg\der{}r}$, and if we
        knew the energy, we could find the field by differentiation, $g=(1/m)\der{}U/\der{}r$.

        The same approach can be applied to other interactions, for example
        a mass on a spring. The main difference is that only in gravitational interactions
        does the strength of the interaction depend on the mass of the object, so
        in general, it doesn't make sense to separate out the factor of $m$ as in the
        equation $\der{}U=mg\der{}r$. Since $F=mg$ is the gravitational force,
        we can rewrite the equation in the more suggestive
        form $\der{}U=F\der{}r$. This form no longer refers to gravity specifically, and
        can be applied much more generally. The only remaining detail is that I've been
        fairly cavalier about positive and negative signs up until now. That wasn't such
        a big problem for gravitational interactions, since gravity is always attractive, but
        it requires more careful treatment for nongravitational forces, where we don't
        necessarily know the direction of the force in advance, and we need to use
        positive and negative signs carefully for the direction of the force.

        In general, suppose that forces are acting on a particle --- we can think of them
        as coming from other objects that are ``off stage'' --- and that the interaction between
        the particle and the off-stage objects can be characterized by an
        interaction energy, $U$, which depends only on the particle's position, $x$.
        Using the kinetic energy theorem, we have $\der{}K=F\der{}x$. (It's not necessary
        to write $K_{cm}$, since a particle can't have any other kind of kinetic energy.)
        Conservation of energy tells us $\der{}K+\der{}U=0$, so the relationship between
        force and interaction energy is $\der{}U=-F\der{}x$, or
        \begin{equation*}
                F = -\frac{\der{}U}{\der{}x} 
                        \qquad \text{[relationship between force and interaction energy]} \qquad .
        \end{equation*}
        
        \begin{eg}{Force exerted by a spring}\label{eg:springforce}
                \egquestion
                A mass is attached to the end of a spring, and the energy of the spring
                is $U=(1/2) kx^2$, where $x$ is the position of the mass,
                and $x=0$ is defined to be the equilibrium position. What is the force
                the spring exerts on the mass? Interpret the sign of the result.

                \eganswer
                Differentiating, we find
                \begin{align*}
                         F        &= -\frac{\der{} U}{\der{} x} \\
                                        &= - kx \qquad .
                \end{align*}
                If $x$ is positive, then the force is negative, i.e., it acts so as to bring the
                mass back to equilibrium, and similarly for $x<0$ we have $F>0$.

                Most books do the $F=- kx$ form before the $U=(1/2) kx^2$
                form, and call it Hooke's law.\index{Hooke's law} Neither form is really more
                fundamental than the other --- we can always get from one to the other by integrating
                or differentiating.
        \end{eg}
        
        \begin{eg}{Newton's law of gravity}
                \egquestion
                Given the equation $U=- Gm_1 m_{2}/ r$ for the energy
                of gravitational interactions, find the corresponding equation for the gravitational force
                on mass $m_2$. Interpret the positive and negative signs.

                \eganswer
                We have to be a little careful here, because we've been taking $r$ to be
                positive by definition, whereas the position, $x$, of mass $m_2$
                could be positive or negative, depending on which side of
                $m_1$ it's on.

                For positive $x$, we have $r= x$, and differentiation gives
                \begin{align*}
                         F        &= -\frac{\der{} U}{\der{} x} \\
                                        &= - Gm_1 m_{2}/ x^2 \qquad .
                \end{align*}
                As in the preceding example, we have $F<0$ when $x$
                is positive, because the object is being attracted back toward $x=0$.

                When $x$ is negative, the relationship between $r$ and $x$ becomes
                $r=- x$, and the result for the force is the same as before, but with
                a minus sign. We can combine the two equations by writing
                \begin{equation*}
                        | F|        =  Gm_1 m_{2}/ r^2 \qquad ,
                \end{equation*}
                and this is the form traditionally known as Newton's law of gravity.
                As in the preceding example, the $U$ and $F$ equations contain
                equivalent information, and neither is more fundamental than the other.
        \end{eg}

        \begin{eg}{Equilibrium}
                I previously described the condition for equilibrium as a local maximum
                or minimum of $U$. A differentiable function has a zero derivative
                at its extrema, and we can now relate this directly to force: zero force
                acts on an object when it is at equilibrium.
        \end{eg}

<% end_sec() %>
<% end_sec() %>

\vfill

<% begin_sec("Resonance",4,'resonance') %>
<% marg(0) %>
<%
  fig(
    'swingimpulseatres',
    %q{%
      An $x$-versus-$t$  graph for a swing pushed at resonance.
    }
  )
%>
\spacebetweenfigs
<%
  fig(
    'swingimpulsedblres',
    %q{A swing pushed at twice its resonant frequency.}
  )
%>
\spacebetweenfigs
<%
  fig(
    'drivingimpulsive',
    %q{The $F$-versus-$t$ graph for an impulsive driving force.}
  )
%>
\spacebetweenfigs
<%
  fig(
    'drivingsine',
    %q{A sinusoidal driving force.}
  )
%>
<% end_marg %>
        Resonance is a phenomenon in which an oscillator responds most strongly\label{swing-resonance}
        to a driving force that matches its own natural frequency of vibration.
        For example, suppose a child is on a playground swing with a natural
        frequency of 1 Hz. That is, if you pull the child away from equilibrium, release
        her, and then stop doing anything for a while, she'll oscillate at 1 Hz.
        If there was no friction, as we assumed in section \ref{sec:oscillations},
        then the sum of her gravitational and kinetic energy would remain constant,
        and the amplitude would be exactly the same from one oscillation to the next.
        However, friction is going to convert these forms of energy into heat, so her
        oscillations would gradually die out. To keep this from happening, you might
        give her a push once per cycle, i.e., the frequency of your pushes would be
        1 Hz, which is the same as the swing's natural frequency. As long as you stay
        in rhythm, the swing responds quite well. If you start the swing from rest,
        and then give pushes at 1 Hz, the swing's amplitude
        rapidly builds up, as in figure \figref{swingimpulseatres}, until after a while it reaches a steady state
        in which friction removes just as much energy as you put in over the course
        of one cycle.

<% self_check('swing-energy',<<-'SELF_CHECK'
In figure \figref{swingimpulseatres}, compare the amplitude of the cycle immediately following the
first push to the amplitude after the second. Compare the energies as well.
  SELF_CHECK
  ) %>


        What will happen if you try pushing at 2 Hz? Your first push puts in
        some momentum, $p$, but your second push happens after only half a cycle,
        when the swing is coming right back at you, with momentum $-p$!
        The momentum transfer from the second push is exactly enough to stop
        the swing. The result is a very weak, and not
        very sinusoidal, motion, \figref{swingimpulsedblres}.

\formatlikesubsubsection{Making the math easy}\\
        This is a simple and physically transparent example of resonance: the swing
        responds most strongly if you match its natural rhythm. However, it has some
        characteristics that are mathematically ugly and possibly unrealistic.
        The quick, hard pushes are known as \emph{impulse} forces, \figref{drivingimpulsive},
        and they lead to an $x$-$t$ graph that has nondifferentiable kinks. Impulsive
        forces like this are not only badly behaved mathematically, they are usually undesirable
        in practical terms. In a car engine, for example, the engineers
        work very hard to make the force on the pistons change smoothly, to avoid excessive
        vibration. Throughout the rest of this section, we'll assume a driving force
        that is sinusoidal, \figref{drivingsine}, i.e., one whose $F$-$t$ graph is either a sine
        function or a function that differs from a sine wave in phase, such as a cosine.
        The force is positive for half of each cycle and negative for the other half, i.e.,
        there is both pushing and pulling. Sinusoidal
        functions have many nice mathematical characteristics (we can differentiate and
        integrate them, and the sum of sinusoidal functions that have the same frequency
        is a sinusoidal function), and they are also used in many practical situations.
        For instance, my garage door zapper sends out a sinusoidal radio wave, and the
        receiver is tuned to resonance with it.
        
        A second mathematical issue that I glossed over in the swing example was how
        friction behaves. In section \ref{subsec:forcesbetweensolids}, about forces between
        solids, the empirical equation for kinetic friction was independent of velocity.
        Fluid friction, on the other hand, is velocity-dependent. For a child on a swing,
        fluid friction is the most important form of friction, and is approximately proportional
        to $v^2$. In still other situations, e.g., with a low-density gas or friction between solid
        surfaces that have been lubricated with a fluid such as oil, we may find that the frictional
        force has some other dependence on velocity, perhaps being proportional to $v$, or having
        some other complicated velocity dependence that can't even be expressed with a simple
        equation.
        It would be extremely complicated to have to treat all of these different possibilities
        in complete generality, so for the rest of this section, we'll assume friction proportional
        to velocity
        \begin{equation*}
                F        = -bv        \qquad ,
        \end{equation*}
        simply because the resulting equations happen to be the easiest to solve.
        Even when the friction doesn't behave in exactly this way, many of our results
        may still be at least qualitatively correct.

__incl(text/damped_oscillator_sn)
<% begin_sec("The quality factor") %>

        It's usually impractical to measure $b$ directly and determine $c$ from the equation
        $c=b/2m$. For a child on a swing, measuring $b$ would require putting the child
        in a wind tunnel! It's usually much easier to characterize the amount of damping by
        observing the actual damped oscillations and seeing how many cycles it takes for
        the mechanical energy to decrease by a certain factor. The unitless
        \emph{quality factor}\index{quality factor}, $Q$, is defined as 
        $Q=\omega_\zu{o}/2c$, and in the limit of weak damping, where
        $\omega\approx\omega_\zu{o}$, this can be interpreted as
        the number of
        cycles required for the mechanical energy to fall off by a factor of $e^{2\pi}=535.49\ldots$
        Using this new quantity, we can rewrite the equation for the frequency of damped
        oscillations in the slightly more elegant form
        $\omega_f        = \omega_\zu{o}\sqrt{1-1/4Q^2}$.

        <% self_check('q-is-when-dead',<<-'SELF_CHECK'
        What if we wanted to make a simpler definition of $Q$, as the number of oscillations required for
        the vibrations to die out completely, rather than the number required for the energy to
        fall off by this obscure factor?
          SELF_CHECK
  ) %>

        \begin{eg}{A graph}
        The damped motion in figure \figref{damping-effect-on-frequency} has $Q\approx 4.5$,
        giving $\sqrt{1-1/4Q^2}\approx 0.99$, as claimed at the end of the preceding subsection.
        \end{eg}

        \begin{eg}{Exponential decay in a trumpet}
        \egquestion
         The vibrations of the air column inside a trumpet
        have a $Q$ of about 10. This means that even after the trumpet
        player stops blowing, the note will keep sounding for a
        short time. If the player suddenly stops blowing, how will
        the sound intensity 20 cycles later compare with the sound
        intensity while she was still blowing?
        
        \eganswer
         The trumpet's $Q$ is 10, so after 10 cycles the
        energy will have fallen off by a factor of 535. After
        another 10 cycles we lose another factor of 535, so the
        sound intensity is reduced by a factor of 
        $535\times535= 2.9\times10^5$.
        \end{eg}

        The decay of a musical sound is part of what gives it its
        character, and a good musical instrument should have the
        right $Q$, but the $Q$ that is considered desirable is different
        for different instruments. A guitar is meant to keep on
        sounding for a long time after a string has been plucked,
        and might have a $Q$ of 1000 or 10000. One of the reasons why
        a cheap synthesizer sounds so bad is that the sound suddenly
        cuts off after a key is released.

<% end_sec() %>
<% begin_sec("Driven motion") %>
<% marg(60) %>
\formatlikecaption{%
\noindent{\textbf{Summary of Notation}}\\
\begin{tabular}{lp{41mm}}
$k$ & spring constant \\
$m$ & mass of the oscillator\\
$b$ & sets the amount of damping, $F=-bv$ \\
$T$ & period \\
$f$ & frequency, $1/T$ \\
$\omega$ & (Greek letter omega), angular frequency, $2\pi f$, often referred to simply as ``frequency'' \\
$\omega_\zu{o}$ & frequency the oscillator would have without damping, $\sqrt{k/m}$ \\
$\omega_f$ & frequency of the free vibrations \\
$c$ & sets the time scale for the exponential decay envelope $e^{-ct}$ of the free vibrations \\
$F_m$ & strength of the driving force, which is assumed to vary sinusoidally with frequency $\omega$ \\
$A$ & amplitude of the steady-state response \\
$\delta$ & phase angle of the steady-state response \\
\end{tabular}
}
<% end_marg %>
        The driven case is extremely important in science, technology, and engineering.
        We have an external driving force $F=F_m \sin \omega t$, where the constant
        $F_m$ indicates the maximum strength of the force in either direction. The equation
        of motion is now
        \begin{multline}\label{eqn:resonancemotion}
                ma+bv+kx = F_m \sin \omega t \\
                                \qquad \text{[equation of motion for a driven oscillator]} \qquad .
        \end{multline}
        After the driving force has been applied for a while, we expect that the amplitude
        of the oscillations will approach some constant value. This motion is known as the
        \emph{steady state}\index{steady state}\index{oscillations!steady state}, and it's the most
        interesting thing to find out; as we'll see later, the most general type of motion is
        only a minor variation on the steady-state motion. For the steady-state motion,
        we're going to look for a solution of the form
        \begin{equation*}
                x = A \sin (\omega{}t+\delta) \qquad .
        \end{equation*}
        In contrast to the undriven case, here it's not possible to sweep $A$ and $\delta$ 
        under the rug. The amplitude of the steady-state motion, $A$, is actually the
        most interesting thing to know about the steady-state motion, and it's not true that we
        still have a solution no matter how we fiddle with $A$; if we have a solution for
        a certain value of $A$, then multiplying $A$ by some constant would break the
        equality between the two sides of the equation of motion. It's also no longer true
        that we can get rid of $\delta$ simply be redefining when we start the clock; here
        $\delta$ represents a \emph{difference} in time between the start of one cycle of the driving
        force and the start of the corresponding cycle of the motion.

        The velocity and
        acceleration are $v=\omega{}A\cos(\omega  t+\delta)$ and
        $a=-\omega^2A\sin(\omega t+\delta)$, and if we plug these into the equation
        of motion, \eqref{eqn:resonancemotion}, and simplify a little, we find
        \begin{equation}\label{eqn:steadystate}
                (k-m\omega^2)\sin (\omega t+\delta)
                         +\omega b \cos (\omega t+\delta) 
                        = \frac{F_m}{A} \sin \omega t \qquad .
        \end{equation}
        The sum of any two sinusoidal functions with the same frequency is also
        a sinusoidal, so the whole left side adds up to a sinusoidal. By fiddling with
        $A$ and $\delta$ we can make the amplitudes and phases of the two sides
        of the equation match up.
        <% begin_sec("Steady state, no damping") %>
        $A$ and $\delta$ are easy to find in the case where there is no damping at all.
        There are now no cosines in equation \eqref{eqn:steadystate} above, only sines,
        so if we wish we can set $\delta$ to zero, and we find
        $A=F_m/(k-m\omega^2)=F_m/m(\omega_\zu{o}^2-\omega^2)$.
        This, however, makes $A$ negative for $\omega>\omega_\zu{o}$. 
        The variable $\delta$ was designed to represent this kind of phase relationship,
        so we prefer to keep $A$ positive and set $\delta=\pi$ for $\omega>\omega_\zu{o}$.
        Our results are then
        \begin{align*}
                A                &= \frac{F_m}{m\left|\omega^2-\omega_\zu{o}^2\right|} \\
        \intertext{and}
                \delta        &= \left\{\begin{array}{ll} 0, & \omega<\omega_\zu{o}\\
                                                \pi, & \omega>\omega_\zu{o}\end{array}\right. \qquad .
        \end{align*}
<% marg(50) %>
<%
  fig(
    'resonance-undamped',
    %q{%
      Dependence of the amplitude and phase angle
              on the driving frequency, for an undamped oscillator. The amplitudes
              were calculated with $F_m$, $m$, and $\omega_\zu{o}$,  all set to 1.
    }
  )
%>
<% end_marg %>

        The most important feature of the result is that there is a resonance: the
        amplitude becomes greater
        and greater, and approaches infinity,
        as $\omega$ approaches the resonant frequency $\omega_\zu{o}$.
        This is the physical behavior we anticipated on page \pageref{swing-resonance} in the
        example of pushing a child on a swing. If the driving frequency matches the frequency of
        the free vibrations, then the driving force will always be in the right direction to
        add energy to the swing. At a driving frequency very different from the resonant frequency,
        we might get lucky and push at the right time during one cycle, but our next push would
        come at some random point in the next cycle, possibly having the effect of slowing the swing
        down rather than speeding it up.

        The interpretation of the infinite amplitude at $\omega=\omega_\zu{o}$ is that there really isn't any steady
        state if we drive the system exactly at resonance --- the amplitude will just keep
        on increasing indefinitely. In real life, the amplitude can't be infinite both because there is always some damping and because
        there will always be some difference, however small, between
        $\omega$ and $\omega_\zu{o}$. Even though the infinity is unphysical, it has entered into
        the popular consciousness, starting with the eccentric Serbian-American inventor and physicist Nikola Tesla.\index{Tesla!Nikola}
        Around 1912, the tabloid newspaper \emph{The World Today} credulously reported a story which Tesla probably fabricated --- or wildly exaggerated ---
        for the sake of publicity. Supposedly he created a steam-powered device ``no larger than an alarm clock,'' containing a piston
        that could be made to vibrate at a tunable and precisely controlled frequency.
        ``He put his little vibrator in his coat-pocket and went out to hunt a half-erected steel building. Down in the Wall
        Street district, he found one --- ten stories of steel framework without a brick or a stone laid around it. He clamped
        the vibrator to one of the beams, and fussed with the adjustment [presumably hunting for the building's resonant frequency] until he got it.
        Tesla said finally the structure began to creak and weave and the steel-workers came to the ground panic-stricken, believing
        that there had been an earthquake. Police were called out. Tesla put the vibrator in his pocket and went away. Ten
        minutes more and he could have laid the building in the street. And, with the same vibrator he could have dropped the Brooklyn Bridge into
        the East River in less than an hour.''

        The phase angle  $\delta$ also exhibits surprising behavior. As the frequency is tuned upward past resonance,
        the phase abruptly shifts so that the phase of the response is opposite to that of the driving force.
        There is a simple interpretation for this.
        The system's mechanical energy can only
        change due to work done by the driving force, since there is no damping
        to convert mechanical energy to heat. In the steady state, then, the power
        transmitted by the driving force over a full cycle of motion must average out to zero.
        In general, the work theorem $\der{}E=F\der{}x$ can always be divided by $\der{}t$
        on both sides to give the useful relation $P=Fv$. If $Fv$ is to average out to zero,
        then $F$ and $v$ must be out of phase by $\pm\pi/2$, and since $v$ is ahead of
        $x$ by a phase angle of $\pi/2$, the phase angle between $x$ and $F$ must be
        zero or $\pi$.

        Given that  these are the two possible phases, why is there a difference in behavior between $\omega<\omega_\zu{o}$
        and $\omega>\omega_\zu{o}$? At the low-frequency limit, consider $\omega=0$, i.e., a constant force.
        A constant force will simply displace the oscillator to one side, reaching an equilibrium that is offset from
        the usual one. The force and the response are in phase, e.g., if the force is to the right, the equilibrium
        will be offset to the right. This is the situation depicted in the amplitude graph of figure \figref{resonance-undamped}
        at $\omega=0$. The response, which is not zero, is simply this static displacement of the oscillator to one side.

        At high frequencies, on the other hand, imagine shaking the poor child on the swing back and forth with a force
        that oscillates at 10 Hz. This is so fast that there is essentially no time for the force $F=-kx$ from gravity and
        the chain to act from one cycle to the next. The problem becomes equivalent to the oscillation of a \emph{free}
        object. If the driving force varies like $\sin(\omega t)$, with $\delta=0$, then the acceleration is also
        proportional to the sine. Integrating, we find that the velocity goes like minus a cosine, and a second
        integration gives a position that varies as minus the sine --- opposite in phase to the driving force.
        Intuitively, this mathematical result corresponds to the fact that at the moment when the object has reached
        its maximum displacement to the \emph{right}, that is the time when the greatest force is being applied to the
        \emph{left}, in order to turn it around and bring it back toward the center.

\vfill

       \begin{eg}{A practice mute for a violin}\label{eg:violin-mute}
       The amplitude of the driven vibrations, $A=F_m/(m|\omega^2-\omega_\zu{o}^2|)$, contains
       an inverse proportionality to the mass of the vibrating object. This is simply because
       a given force will produce less acceleration when applied to a more massive object.
       An application is shown in figure \ref{eg:violin-mute}.  

       In a stringed instrument, the strings themselves don't have
       enough surface area to excite sound waves very efficiently.
       In instruments of the violin family, as the strings vibrate from left to right, they cause the
       bridge (the piece of wood they pass over)
       to wiggle clockwise and counterclockwise, and this motion is transmitted to the top panel
       of the instrument, which vibrates and creates sound waves in the air.

       A string player who wants to practice
       at night without bothering the neighbors can add some mass to the bridge.
       Adding mass to the bridge causes the
       amplitude of the vibrations to be smaller, and the sound to be much softer.
       A similar effect is seen when an electric guitar is used without an amp. The body
       of an electric guitar is so much more massive than the body of an acoustic guitar that
       the amplitude of its vibrations is very small.
       \end{eg}
<%
  fig(
    'viola-mute',
    %q{%
      Example \ref{eg:violin-mute}: a viola without a mute (left), and
              with a mute (right). The mute doesn't touch the strings themselves.
    },
    {
      'width'=>'wide',
      'sidecaption'=>true
    }
  )
%>
<% end_sec() %>
\pagebreak % For some reason, this doesn't work if I just do it as a ,4 in the following begin_sec -- page break appears after the title
<% begin_sec("Steady state, with damping") %>
        The extension of the analysis to the damped case involves some lengthy
        algebra, which I've outlined on page \pageref{misc:steadystate} in
        appendix \ref{miscappendix}. The results are shown in figure 
        \figref{resonance}. It's not surprising that the steady state response is
        weaker when there is more damping, since the steady state is reached when
        the power extracted by damping matches the power input by the driving force.
        The maximum amplitude, at the peak of the resonance curve, is approximately proportional to $Q$.
<%
  fig(
    'resonance',
    %q{%
      Dependence of the amplitude and phase angle
              on the driving frequency. The undamped case is $Q=\infty$, 
              and the other curves represent $Q$=1, 3, and 10. $F_m$, $m$, and $\omega_\zu{o}$ are all set to 1.
    },
    {'width'=>'wide','sidecaption'=>true}
  )
%>

<% self_check('a-propto-q',<<-'SELF_CHECK'
From the final result of the analysis on page \pageref{misc:steadystate}, substitute $\omega=\omega_\zu{o}$,
and satisfy yourself that the result is proportional to $Q$. Why is $A_{res}\propto Q$ only an approximation?
  SELF_CHECK
  ) %>
<% marg(0) %>
<%
  fig(
    'fwhm-omega',
    %q{%
      The definition of $\Delta\omega$,
                              the full width at half maximum.
    }
  )
%>
<% end_marg %>%

        What is surprising is that the amplitude is strongly affected by damping close
        to resonance, but only weakly affected far from it. In other words, the shape of
        the resonance curve is broader with more damping, and even if we were to scale up
        a high-damping curve so that its maximum was the same as that of a low-damping
        curve, it would still have a different shape. The standard way of describing the shape
        numerically is to give the quantity $\Delta\omega$, called the
         \emph{full width at half-maximum}\index{full width at half-maximum},
        or FWHM\index{FWHM}, which is defined in figure \figref{fwhm-omega}. Note that
        the $y$ axis is energy, which is proportional to the square of the amplitude.
        Our previous observations amount to a statement that $\Delta\omega$ is greater
        when the damping is stronger, i.e., when the $Q$ is lower. It's not hard to
        show from the equations on page  \pageref{misc:steadystate} that for
        large $Q$, the FWHM is given approximately by
        \begin{equation*}
                \Delta\omega \approx \omega_\zu{o}/Q \qquad .
        \end{equation*}

        Another thing we notice in figure \figref{resonance} is that for small values of $Q$
        the frequency $\omega_{res}$ of the maximum $A$ is
        less than $\omega_\zu{o}$.\footnote{The relationship is $\omega_{max\ A}/\omega_\zu{o}=\sqrt{1-1/2Q^2}$,
        which is similar in form to the equation for the frequency of the free vibration, $\omega_{f}/\omega_\zu{o}=\sqrt{1-1/4Q^2}$. A subtle point here is that
        although the maximum of $A$ and the maximum of $A^2$ must occur at the same frequency, the maximum energy does not occur, as we might expect, at the same
        frequency as the maximum of $A^2$. This is because the interaction energy is proportional to $A^2$ regardless of frequency, but the kinetic energy is
        proportional to $A^2\omega^2$. The maximum energy actually occurs are precisely $\omega_\zu{o}$.}
        At even lower values of $Q$, like $Q=1$, the $A-\omega$ curve doesn't even have a maximum
        near $\omega>0$.

        \begin{eg}{An opera singer breaking a wineglass}
        In order to break a wineglass by singing, an opera singer
        must first tap the glass to find its natural frequency of
        vibration, and then sing the same note back, so that her driving force will produce
        a response with the greatest possible amplitude. If she's shopping for the right
        glass to use for this display of her prowess, she should look for one that
        has the greatest possible $Q$, since the resonance curve has a higher
        maximum for higher values of $Q$. 
        \end{eg}

\pagebreak

        \begin{eg}{Collapse of the Nimitz Freeway}
        Figure \figref{nimitz} shows a section
        of the Nimitz Freeway in Oakland, CA, that collapsed during an
        earthquake in 1989.
        An earthquake consists of many low-frequency vibrations that
        occur simultaneously, which is why it sounds like a rumble
        of indeterminate pitch rather than a low hum. The
        frequencies that we can hear are not even the strongest
        ones; most of the energy is in the form of vibrations in the
        range of frequencies from about 1 Hz to 10 Hz.
<% marg(0) %>
<%
  fig(
    'nimitz',
    %q{The collapsed section of the Nimitz Freeway}
  )
%>
<% end_marg %>
        
        All the structures we build are resting on geological
        layers of dirt, mud, sand, or rock. When an earthquake wave
        comes along, the topmost layer acts like a system with a
        certain natural frequency of vibration, sort of like a cube
        of jello on a plate being shaken from side to side. The
        resonant frequency of the layer depends on how stiff it is
        and also on how deep it is. The ill-fated section of the
        Nimitz freeway was built on a layer of mud, and analysis by
        geologist Susan E. Hough of the U.S. Geological Survey shows
        that the mud layer's resonance was centered on about 2.5 Hz,
        and had a width covering a range from about 1 Hz to 4 Hz.
        
        When the earthquake wave came along with its mixture of
        frequencies, the mud responded strongly to those that were
        close to its own natural 2.5 Hz frequency. Unfortunately, an
        engineering analysis after the quake showed that the
        overpass itself had a resonant frequency of 2.5 Hz as well!
        The mud responded strongly to the earthquake waves with
        frequencies close to 2.5 Hz, and the bridge responded
        strongly to the 2.5 Hz vibrations of the mud, causing
        sections of it to collapse.
        \end{eg}
        
        <% end_sec() %>
<% begin_sec("Physical reason for the relationship between Q and the FWHM") %>
        What is the reason for this surprising relationship between the damping and
        the width of the resonance? Fundamentally, it has to do with the fact that
        friction causes a system to lose its ``memory'' of its previous state. If the Pioneer
        10 space probe, coasting through the frictionless vacuum of interplanetary space,
        is detected by aliens a million years from now, they will be able to trace its
        trajectory backwards and infer that it came from our solar system. On the other hand,
        imagine that I shove a book along a tabletop, it comes to rest, and then someone
        else walks into the room. There will be no clue as to which direction the book was moving
        before it stopped --- friction has erased its memory of its motion.
        Now consider the playground swing driven at twice its natural frequency,
        figure \figref{swingimpulsedamp}, where
        the undamped case is repeated from figure \figref{swingimpulsedblres} on page
        \pageref{fig:swingimpulsedblres}. 
        In the undamped case, the first push starts the swing moving with momentum $p$,
        but when the
        second push comes,  if there
        is no friction at all, it now has a momentum of exactly $-p$, and the momentum transfer
        from the second push is exactly enough to stop it dead. With moderate damping, however,
        the momentum on the rebound is not quite $-p$, and the second push's effect isn't
        quite as disastrous. With very strong damping, the swing comes essentially to rest long before
        the second push. It has lost all its memory, and the second push puts energy into the
        system rather than taking it out. Although the detailed mathematical results with this
        kind of impulsive driving force are different,\footnote{For example, the graphs calculated
        for sinusoidal driving have resonances that are somewhat below the natural frequency, 
        getting lower with increasing damping, until for $Q\le1$ the maximum response occurs
        at $\omega=0$. In figure \figref{swingimpulsedamp}, however, we can see that impulsive
        driving at $\omega=2\omega_\zu{o}$ produces a steady state with more energy
        than at $\omega=\omega_\zu{o}$.}
        the general results are the same
        as for sinusoidal driving: the less damping there is, the greater the penalty you pay
        for driving the system off of resonance.
<% marg(80) %>
<%
  fig(
    'swingimpulsedamp',
    %q{%
      An $x$-versus-$t$ graph of the steady-state
                      motion of a swing being
                      pushed at twice its resonant frequency by an impulsive force.
    }
  )
%>
<% end_marg %>

        \begin{eg}{High-Q speakers}\label{eg:highqspeakers}
                Most good audio speakers have $Q\approx1$, but
                the resonance curve for a higher-$Q$ oscillator always lies above the
                corresponding curve for one with a lower $Q$, so people who want their
                car stereos to be able to rattle the windows of the neighboring cars
                will often choose speakers that have a high $Q$. Of course they could
                just use speakers with stronger driving magnets to increase $F_m$, but
                the speakers might be more expensive, and a high-$Q$ speaker also
                has less friction, so it wastes less energy as heat.

                One problem with this is that whereas the resonance
                curve of a low-$Q$ speaker (its ``response curve'' 
                or ``frequency response'' in audiophile lingo) is
                fairly flat, a higher-$Q$ speaker tends to emphasize the frequencies
                that are close to its natural resonance. In audio, a flat response curve
                gives more realistic reproduction of sound, so a higher quality factor, $Q$,
                really corresponds to a \emph{lower\/}-quality speaker.

                Another problem with high-$Q$ speakers is discussed in example
                \ref{eg:boomyspeakers} on page \pageref{eg:boomyspeakers} .
        \end{eg}

        \begin{eg}{Changing the pitch of a wind instrument}
        \egquestion
         A saxophone player normally selects which note to
        play by choosing a certain fingering, which gives the
        saxophone a certain resonant frequency. The musician can
        also, however, change the pitch significantly by altering
        the tightness of her lips. This corresponds to driving the
        horn slightly off of resonance. If the pitch can be altered
        by about 5\% up or down (about one musical half-step) without
        too much effort, roughly what is the $Q$ of a saxophone?
        
        \eganswer
         Five percent is the width on one side of the
        resonance, so the full width is about 10\%, $\Delta f/f_\zu{o}\approx 0.1$.
        The equation $\Delta\omega=\omega_\zu{o}/ Q$ is defined in terms of
        angular frequency, $\omega=2\pi\ f$, and we've been given our
        data in terms of ordinary frequency, $f$. The factors of $2\pi$ end up canceling out,
        however:
        \begin{align*}
                 Q        &= \frac{\omega_\zu{o}}{\Delta\omega} \\
                                & = \frac{2\pi f_\zu{o}}{2\pi\Delta f} \\
                                & = \frac{f_\zu{o}}{f} \\
                                & \approx 10
        \end{align*}
        In other words, once the musician stops
        blowing, the horn will continue sounding for about 10 cycles
        before its energy falls off by a factor of 535. (Blues and
        jazz saxophone players will typically choose a mouthpiece
        that gives a low $Q$, so that they can produce the bluesy
        pitch-slides typical of their style. ``Legit,'' i.e.,
        classically oriented players, use a higher-$Q$ setup because
        their style only calls for enough pitch variation to produce
        a vibrato, and the higher $Q$ makes it easier to play in tune.)
        \end{eg}                

        \begin{eg}{Q of a radio receiver}
        \egquestion
         A radio receiver used in the FM band needs to be
        tuned in to within about 0.1 MHz for signals at about 100
        MHz. What is its $Q$?
        
        \eganswer
        As in the last example, we're given data in terms of $f$s, not
        $\omega$s, but the factors of $2\pi$ cancel. The resulting
        $Q$ is about 1000, which is extremely high compared to the
        $Q$ values of most mechanical systems.
        \end{eg}

        <% end_sec() %>
<% begin_sec("Transients") %>
        What about the motion before the steady state is achieved? When we computed
        the undriven motion numerically on page 
        \pageref{freedampednumerical}, the program had to initialize the position and
        velocity. By changing these two variables, we could have gotten any of an
        infinite number of simulations.\footnote{If you've learned about differential equations,
        you'll know that any second-order differential equation requires the specification
        of two boundary conditions in order to specify solution uniquely.}
        The same is true when we have an equation of motion with a driving term,
        $ma+bv+kx = F_m \sin \omega{}t$ (p. \pageref{eqn:resonancemotion}, equation
        \eqref{eqn:resonancemotion}). The steady-state solutions, however, have no
        adjustable parameters at all --- $A$ and $\delta$ are uniquely determined
        by the parameters of the driving force and the oscillator itself. If the oscillator
        isn't initially in the steady state, then it will not have the steady-state motion
        at first. What kind of motion will it have?

        The answer comes from realizing that if we start with the solution to the driven
        equation of motion, and then add to it any solution to the free equation of motion,
        the result,
        \begin{equation*}
                        x = A \sin (\omega t+\delta) + A' e^{-ct}\sin (\omega_f t+\delta') \qquad ,
        \end{equation*}
        is also a solution of the driven equation. Here, as before,
        $\omega_f$
        is the frequency of the free oscillations ($\omega_f\approx\omega_\zu{o}$
        for small $Q$), $\omega$ is the frequency of the driving force,
        $A$ and $\delta$ are related as usual to the parameters of the driving
        force, and $A'$ and $\delta'$ can have any values at all. Given the initial
        position and velocity, we can always choose $A'$ and $\delta'$ to reproduce
        them, but this is not something one often has to do in real life. What's more
        important is to realize that the second term dies out exponentially over time,
        decaying at the same rate at which a free vibration would. For this reason,
        the $A'$ term is called a transient. A high-$Q$ oscillator's transients take a
        long time to die out, while a low-$Q$ oscillator always settles down to its
        steady state very quickly.  

        \begin{eg}{Boomy bass}\label{eg:boomyspeakers}
                In example \ref{eg:highqspeakers} on page \pageref{eg:highqspeakers},
                I've already discussed one of the drawbacks of a high-$Q$ speaker,
                which is an uneven response curve.
                Another problem is that in a high-$Q$ speaker, transients take a long
                time to die out. The bleeding-eardrums crowd tend to focus mostly
                on making their bass loud, so it's usually their woofers that have
                high $Q$s. The result is that bass notes, 
                ``ring'' after the onset of the note, a phenomenon referred to as ``boomy bass.''
        \end{eg}

        <% end_sec() %>
<% begin_sec("Overdamped motion") %>
        The treatment of free, damped motion on page \pageref{freedampedanalytic}
        skipped over a subtle point: in the equation
        $\omega_f = \sqrt{k/m-b^2/4m^2} = 
                \omega_\zu{o}\sqrt{1-1/4Q^2}$, $Q<1/2$ results in an answer that
        is the square root of a negative number. For example, suppose we had
        $k=0$, which corresponds to a neutral equilibrium. A physical example
        would be a mass sitting in a tub of syrup. If we set it in motion,
        it won't oscillate --- it will simply slow to a stop. This system
        has $Q=0$. The equation of motion
        in this case is $ma+bv=0$, or, more suggestively,
        \begin{equation*}
                m\frac{\der{}v}{\der{}t}+bv=0 \qquad .
        \end{equation*}
        One can easily verify that this has the solution
        $v=\text{(constant)}e^{-bt/m}$, and integrating, we find 
        $x=\text{(constant)}e^{-bt/m}+\text{(constant)}$. In other words,
        the reason $\omega_f$ comes out to be mathematical nonsense\footnote{Actually,
        if you know about complex numbers and Euler's theorem, it's not quite so
        nonsensical.} is that we were incorrect in assuming a solution that oscillated at a frequency
        $\omega_f$. The actual motion is not oscillatory at all.

        In general, systems with $Q<1/2$, called
        overdamped\index{oscillations!overdamped!mechanical}%
        \index{overdamped oscillations!mechanical}\index{damped oscillations!overdamped!mechanical}
         systems, do not
        display oscillatory motion. Most cars' shock absorbers are designed with
        $Q\approx1/2$, since it's undesirable for the car to undulate up and down for
        a while after you go over a bump. (Shocks with extremely low values of $Q$ are not good
        either, because such a system takes a very long time to come back to
        equilibrium.) It's not particularly important for our purposes, but for completeness
        I'll note, as you can easily verify, that the general solution to the equation of motion
        for $0<Q<1/2$ is of the form $x=Ae^{-ct}+Be^{-dt}$, while
        $Q=1/2$, called the critically damped\index{critically damped}\index{damping!critical}%
        \index{damped oscillations!critically damped} case, gives $x=(A+Bt)e^{-ct}$.\label{overdamped}

        
<% end_sec() %>
<% end_sec() %>
<% end_sec %>
<% begin_sec("Motion in Three Dimensions",4,'motion-in-three-d') %>
<% begin_sec("The Cartesian perspective") %>
        When my friends and I were bored in high school, we used to play a paper-and-pencil
        game which, although we never knew it, was Very Educational --- in fact, it
        pretty much embodies the entire world-view of classical physics. To play the
        game, you draw a racetrack
        on graph paper, and try to get your car around the track before anyone else.
        The default is for your car to continue at constant speed in a straight
        line, so if it moved three squares to the right and one square up on your last turn, it will do the
        same this turn. You can also control the car's motion by changing its $\Delta x$
        and $\Delta y$ by up to one unit. If it moved three squares to the right last turn,
        you can have it move anywhere from two to four squares to the right this turn.

<% marg(50) %>
<%
  fig(
    'cargame',
    %q{%
      The car can
                      change its $x$ and $y$ motions by one
                      square every turn.
    }
  )
%>
<% end_marg %>

<%
  fig(
    'descartesstamp',
    %q{%
      French mathematician Ren\'e Descartes invented analytic geometry;
                              Cartesian ($xyz$) coordinates
                              are named after him. He did work in philosophy, and was particularly
                              interested in the mind-body problem.
                              He was a skeptic and an antiaristotelian,
                              and, probably for fear of religious persecution,
                              spent his adult life in the Netherlands, where
                              he fathered a daughter with a Protestant peasant whom he could not marry.
                              He kept his daughter's existence secret from his enemies in France to avoid giving them
                              ammunition, but he was crushed when she died of
                              scarlatina at age 5.
                              A pious Catholic, he was
                              widely expected to be sainted. 
                              His body was buried in Sweden but then reburied several
                              times in France, and along the way everything but a few fingerbones was stolen
                              by peasants who
                              expected the body parts to become holy relics. 
    },
    {
      'width'=>'wide',
      'narrowfigwidecaption'=>true,
      'float'=>false
    }
  )
%>

\enlargethispage{-2\baselineskip}

        \index{Descartes, Ren\'e}
        The fundamental way of dealing with the direction of an object's motion in physics
        is to use conservation of momentum, since momentum depends on direction.
        Up until now, we've only done momentum in one dimension. How does this
        relate to the racetrack game? In the game, the motion of a car from one turn to the
        next is represented by its $\Delta x$ and $\Delta y$. In one dimension, we would only need
        $\Delta x$, which could be related to the velocity, $\Delta x/\Delta t$, and the momentum,
        $m\Delta x/\Delta t$. In two dimensions, the rules of the game amount to a statement
        that if there is no momentum transfer, then both $m\Delta x/\Delta t$
        and $m\Delta y/\Delta t$ stay the same. In other words, there are two flavors of
        momentum, and they are \emph{separately} conserved. 
        All of this so far has been done with an artificial division of time into ``turns,'' but
        we can fix that by redefining everything in terms of derivatives, and for motion in
        three dimensions rather than two, we augment $x$ and $y$ with $z$:
        \begin{align*}
                v_x &= \der x/\der t        & v_y &= \der y/\der t        & v_z &= \der z/\der t \\
        \intertext{and} 
                p_x &= mv_x        &        p_y &= mv_y        &p_z &= mv_z        
        \end{align*}
        We call these the $x$, $y$, and $z$ components\index{component}
        of the velocity and the momentum.

        
        There is both experimental and theoretical evidence that the  $x$, $y$, and $z$
        momentum components are separately conserved, and that a momentum transfer
        (force) along one axis has no effect on the momentum components along the other
        two axes. On page \pageref{subsec:predictingdirection}, for example, I argued that
        it was impossible for an air hockey puck to make a 180-degree turn spontaneously,
        because then in the frame moving along with the puck, it would have begun moving
        after starting from rest. Now that we're working in two dimensions, we might wonder
        whether the puck could spontaneously make a 90-degree turn, but exactly the same
        line of reasoning shows that this would be impossible as well, which proves that
        the puck can't trade $x$-momentum for $y$-momentum. A more general proof
        of separate conservation will be given on page \pageref{separatepconsproof},
        after some of the appropriate 
        mathematical techniques have been introduced.
<%
  fig(
    'bullets',
    %q{Bullets are dropped and shot at the same time.},
    {
      'width'=>'wide',
      'floatpos'=>'t'
    }
  )
%>

        As an example of the experimental evidence for separate conservation of the momentum
        components, figure \figref{bullets} shows correct and incorrect predictions of what happens
        if you shoot a rifle and arrange for a second
        bullet to be dropped from the same height at exactly the same
        moment when the first one left the barrel. 
        Nearly everyone expects that the dropped
        bullet will reach the dirt first, and Aristotle would have
        agreed, since he believed that the bullet had to lose its horizontal motion before
        it could start moving vertically. In reality, we find that the vertical momentum transfer
        between the earth and the bullet is completely unrelated to the horizontal momentum.
        The bullet ends up with $p_y<0$, while the planet picks up an upward momentum
        $p_y>0$, and the total momentum in the $y$ direction remains zero. 
        Both bullets hit the ground at the same time. This is much simpler
        than the Aristotelian version!\index{Aristotle}

        \begin{eg}{The Pelton waterwheel}
        \egquestion
        There is a general class of machines that either 
        do work on  a gas or
        liquid, like a boat's propeller, or have work done
        on them by a gas or liquid, like the turbine in a hydroelectric
        power plant. Figure \figref{peltonwheel} shows two types of
        surfaces that could be attached to the circumference of an
        old-fashioned waterwheel. Compare the force exerted by the water
        in the two cases.
        
        \eganswer
        Let the $x$ axis point to the right, and the $y$ axis up.
        In both cases, the stream of water rushes down onto the surface
        with momentum $p_{y,i}=- p_\zu{o}$,
        where the subscript $i$ stands for ``initial,'' i.e., before the
        collision.

        In the case of surface 1, the streams of water leaving the
        surface have no momentum in the $y$ direction, and their
        momenta in the $x$ direction cancel. The final momentum
        of the water is zero along both axes, so its entire momentum,
        $- p_\zu{o}$, has been transferred to the waterwheel.

        When the water leaves surface 2, however, its momentum
        isn't zero. If we assume there is no friction,
        it's $p_{y,f}= +p_\zu{o}$, with the positive
        sign indicating upward momentum. The change in the
        water's momentum is $p_{y,f}- p_{y,i}=2 p_\zu{o}$,
        and the momentum transferred to the waterwheel is $-2 p_\zu{o}$.

        Force is defined as the rate of transfer of momentum, so
        surface 2 experiences double the force. A waterwheel constructed
        in this way is known as a Pelton waterwheel.\index{Pelton waterwheel}
        \end{eg}

<% marg(110) %>
<%
  fig(
    'peltonwheel',
    %q{%
      Two surfaces that could be used
              to extract energy from a stream of water.
    }
  )
%>
\spacebetweenfigs
<%
  fig(
    'yarkovsky',
    %q{%
      An asteroid absorbs visible light from the sun, and gets rid
              of the energy by radiating infrared light.
    }
  )
%>
<% end_marg %>

        
        \begin{eg}{The Yarkovsky effect}\label{eg:yarkovsky}\index{Yarkovsky effect}\index{asteroid}
        We think of the planets and asteroids as inhabiting their orbits permanently, but
        it is possible for an orbit to change over periods of millions or billions of years,
        due to a variety of effects. For asteroids with diameters
        of a few meters or less, an important mechanism is the Yarkovsky effect,
        which is easiest to understand if we consider an asteroid
        spinning about an axis that is exactly perpendicular to its orbital plane.  

        The illuminated side of the asteroid is relatively hot, and radiates more infrared
        light than the dark (night) side. Light has momentum, and a total force away from
        the sun is produced by combined effect
        of the sunlight hitting the asteroid and the imbalance between the momentum
        radiated away on the two sides. This force, however, doesn't cause the asteroid's
        orbit to change over time, since it simply cancels a tiny fraction of the sun's gravitational
        attraction. The result is merely a tiny, undetectable violation of Kepler's law of
        periods.

        Consider the sideways momentum transfers, however. In figure
        \figref{yarkovsky}, the part of the asteroid on the right has been illuminated
        for half a spin-period (half a ``day'') by the sun, and is hot. It radiates
        more light than the morning side on the left. This imbalance produces a total
        force in the $x$ direction which points to the left. If the asteroid's orbital
        motion is to the left, then this is a force in the same direction as the motion,
        which will do positive work, increasing the asteroid's energy and boosting it
        into an orbit with a greater radius. On the other hand, if the asteroid's spin and
        orbital motion are in opposite directions, the Yarkovsky push brings the asteroid
        spiraling in closer to the sun.

        Calculations show that it takes on the order of $10^7$ to $10^8$ years
        for the Yarkovsky effect to move an asteroid out of the asteroid belt and into the
        vicinity of earth's orbit, and this is about the same as the typical age of a meteorite
        as estimated by its exposure to cosmic rays. The Yarkovsky effect doesn't remove
        all the asteroids from the asteroid belt, because many of them have orbits that are
        stabilized by gravitational interactions with Jupiter. However, when collisions occur,
        the fragments can end up in orbits which are not stabilized in this way, and they
        may then end up reaching the earth due to the Yarkovsky effect. The cosmic-ray
        technique is really telling us how long it has been since the fragment was broken
        out of its parent.
        
        \end{eg}

\startdqs

\begin{dq}\label{dq:target-shooting}
The following is an incorrect explanation of a fact
about target shooting:

``Shooting a high-powered rifle with a high muzzle velocity
is different from shooting a less powerful gun. With a less
powerful gun, you have to aim quite a bit above your target,
but with a more powerful one you don't have to aim so high
because the bullet doesn't drop as fast.''

What is the correct explanation?
\end{dq}
%
<%
  fig(
    'dq-target-shooting',
    %q{Discussion question \ref{dq:target-shooting}.},
    {
      'width'=>'wide'
    }
  )
%>

\begin{dq}
You have thrown a rock, and it is flying through the air
in an arc.  If the earth's gravitational force on it is
always straight down, why doesn't it just go straight down
once it leaves your hand?
\end{dq}

\begin{dq}
Consider the example of the bullet that is dropped at the
same moment another bullet is fired from a gun. What would
the motion of the two bullets look like to a jet pilot
flying alongside in the same direction as the shot bullet
and at the same horizontal speed?
\end{dq}

<% end_sec() %>
<% begin_sec("Rotational invariance",nil,'rotationalinvariance') %>
        \index{rotational invariance}\index{invariance!rotational}
        The Cartesian approach requires that we choose $x$, $y$, and $z$ axes.
        How do we choose them correctly? The answer is that it had better not matter
        which directions the axes point (provided they're perpendicular to each other),
        or where we put the origin, because  if it did matter,
        it would mean that space was asymmetric. If there was a certain point in the universe
        that was the right place to put the origin, where would it be? The top of
        Mount Olympus? The United Nations headquarters? We find that
        experiments come out the same no matter where we do them, and regardless of
        which way the laboratory is oriented, which indicates that no location in space or
        direction in space is special in any way.\footnote{Of course, 
        you could tell in a sealed laboratory
        which way was down, but that's because there happens to be a big planet  nearby,
        and the planet's gravitational field reaches into the lab, not because space itself
        has a special down direction. Similarly, if your experiment was sensitive to
        magnetic fields, it might matter which way the building was oriented, but that's
        because the earth makes a magnetic field, not because space itself comes
        equipped with a north direction.}

        This is closely related to the idea of Galilean
        relativity stated on page \pageref{sec:galileanrelativity}, from which we already
        know that the absolute \emph{motion} of a frame of reference is irrelevant and undetectable.
        Observers using frames of reference that are in motion relative to each other will
        not even agree on the permanent identity of a particular point in space, so it's not
        possible for the laws of physics to depend on where you are in space. For instance,
        if gravitational energies were proportional to $m_1m_2$ in one location but to
        $(m_1m_2)^{1.00001}$ in another, then it would be possible to determine when you
        were in a state of absolute motion, because the behavior of gravitational interactions
        would change as you moved from one region to the other.

        \index{Galileo!Galilean relativity}\index{relativity!Galilean}
        Because of this close relationship, we restate the principle of Galilean
        relativity in a more general form.
         This extended principle of Galilean relativity states that the laws of physics
        are no different in one
        time and place than in another, and that they also don't depend on your
        orientation or your motion, provided that your motion is in a straight line and at
        constant speed.

        The irrelevance of time and place could have been stated in chapter \ref{ch:1}, but since this
        section is the first one in which we're dealing with three-dimensional physics
        in full generality, the irrelevance of orientation is what we really care about right
        now. This property of the laws of physics is called rotational invariance. The word
        ``invariance'' means a lack of change, i.e., the laws of physics don't change when
        we reorient our frame of reference.

        \begin{eg}{Rotational invariance of gravitational interactions}
        Gravitational energies depend on the quantity 1/$r$, which by the
        Py\-thag\-or\-e\-an theorem equals
        \begin{equation*}
                 \frac{1}{\sqrt{\Delta x^2+\Delta y^2+\Delta z^2}}
                        \qquad .
        \end{equation*}
        Rotating a line segment doesn't change its length, so this expression
        comes out the same regardless of which way we orient our coordinate axes.
        Even though $\Delta x$, $\Delta y$, and $\Delta z$ are different
        in differently oriented coordinate systems, $r$ is the same.
        \end{eg}

<% marg(0) %>
<%
  fig(
    'rolldowncone',
    %q{Two balls roll down a cone and onto a plane.}
  )
%>
<% end_marg %>
        \begin{eg}{Kinetic energy}\label{eg:ke3d}
                Kinetic energy equals $\zu{(1/2)} mv^2$, but what does that mean
                in three dimensions, where we have $v_x$, $v_y$, and
                $v_z$? If you were tempted to add the components and calculate
                $K=(1/2) m( v_x+ v_y+ v_z)^2$,
                figure \figref{rolldowncone} should convince you otherwise. Using that
                method, we'd have to assign a kinetic energy of zero to ball number 1,
                since its negative $v_y$ would exactly cancel its positive $v_x$,
                whereas ball number 2's kinetic energy wouldn't be zero. This would violate
                rotational invariance, since the balls would behave differently.

                The only possible way to generalize kinetic energy to three dimensions,
                without violating rotational invariance, is to  use an expression that
                resembles the Pythagorean theorem,
                \begin{equation*}
                         v=\sqrt{ v_{x}^2+ v_{y}^2+ v_z^2} \qquad ,
                \end{equation*}
                which results in
                \begin{equation*}
                         K=\frac{1}{2} m
                        \left( v_{x}^2+ v_{y}^2+ v_z^2\right)
                        \qquad .
                \end{equation*}
                Since the velocity components are squared, the positive and negative signs
                don't matter, and the two balls in the example behave the same way.
        \end{eg}

<% end_sec() %>
<% begin_sec("Vectors",4) %>
        Remember the title of this book?
        It would have been possible to obtain the result of example \ref{eg:ke3d}
        by applying the Pythagorean theorem to
        $\der x$, $\der y$, and $\der z$, and then dividing by $\der t$,
        but the rotational invariance approach is \emph{simpler}, and is 
        useful in a much broader context. Even with a quantity you presently
        know nothing about, say the magnetic field, you can infer that if the
        components of the magnetic field are $B_x$, $B_y$, and $B_z$, then the 
        physically useful way to talk about the strength of the magnetic field is
        to define it as $\sqrt{B_x^2+B_y^2+B_z^2}$. Nature knows your
        brain cells are precious, and doesn't want you to have to waste them
        by memorizing mathematical rules that are different for magnetic fields
        than for velocities.

        When mathematicians see that the same set of techniques is useful in
        many different contexts, that's when they start making definitions that
        allow them to stop reinventing the wheel. The ancient Greeks, for example,
        had no general concept of fractions. They couldn't say that a circle's
        radius divided by its diameter was equal to the number 1/2. They had to
        say that the radius and the diameter were in the ratio of one to two. With this
        limited number concept, they couldn't have said that water was dripping out
        of a tank at a rate of 3/4 of a barrel per day; instead, they would have had to say
        that over four days, three barrels worth of water would be lost. Once enough of
        these situations came up, some clever mathematician finally realized that
        it would make sense to define something called a fraction, and that one could
        think of these fraction thingies as numbers that lay in the gaps between the traditionally
        recognized numbers like zero, one, and two. Later generations of mathematicians
        introduced further subversive generalizations of the number concepts, inventing
        mathematical creatures like negative numbers, and the square root of two, which
        can't be expressed as a fraction.

        In this spirit, we define a \emph{vector}\index{vector!defined} as any quantity that has
        both an amount and a direction in space. In contradistinction, a
        \emph{scalar}\index{scalar!defined} has an amount, but no direction. Time and
        temperature are scalars. Velocity, acceleration, momentum, and force are vectors.
        In one dimension,
        there are only two possible directions, and we can use positive and negative numbers
        to indicate the two directions. In more than one dimension, there are infinitely many
        possible directions, so we can't use the two symbols $+$ and $-$ to
        indicate the direction of a vector. Instead, we can specify the three components
        of the vector, each of which can be either negative or positive.
        We represent vector quantities in handwriting by writing an
         arrow above them, so for example the
        momentum vector looks like this, $\vec{p}$,
        but the arrow looks ugly in print, so in books vectors are usually
         shown in bold-face type: \vc{p}.
        A straightforward way of thinking about vectors is that a vector equation really
        represents three different equations. For instance, conservation of momentum
        could be written in terms of the three components,
        \begin{align*}
                \Delta p_x        &= 0 \\
                \Delta p_y        &= 0 \\
                \Delta p_z        &= 0 \qquad ,
        \end{align*}
        or as a single vector equation,\footnote{The zero here is really a zero
        \emph{vector}, i.e., a vector whose components are all zero, so we should
        really represent it with a boldface \vc{0}. There's usually not much danger of
        confusion, however, so most books, including this one, don't use boldface for the
        zero vector.}
        \begin{equation*}
                \Delta \mathbf{p} = 0 \qquad .
        \end{equation*}
        The following table summarizes some vector operations.

        \begin{tabular}{|l|p{65mm}|}
        \hline
        \textsl{operation}        & \textsl{definition} \\
        \hline
        $|\mathbf{vector}|$        & $\sqrt{vector_x^2+vector_y^2+vector_z^2}$ \\
        $\mathbf{vector}+\mathbf{vector}$        & Add component by component. \\
        $\mathbf{vector}-\mathbf{vector}$        & Subtract component by component. \\
        $\mathbf{vector}\ \cdot\ scalar        $        & Multiply each component by the scalar. \\
        $\mathbf{vector}\ /\ scalar        $        & Divide each component by the scalar. \\
        \hline
        \end{tabular}

        \noindent{}The first of these is called the \emph{magnitude} of the vector; in one dimension,
        where a vector only has one component, it amounts to taking the absolute value,
        hence the similar notation.
        \index{vector!magnitude of}\index{vector!addition}\index{vector!subtraction}
        \index{vector!multiplication by a scalar}\index{vector!division by a scalar}

        <% self_check('translatefvector',<<-'SELF_CHECK'
Translate the equations $F_{x}\zu{=} ma_x$, 
        $F_{y}\zu{=} ma_y$, and $F_{z}\zu{=} ma_z$
         into a single equation in vector notation.
  SELF_CHECK
  ) %>

<% marg(0) %>
<%
  fig(
    'kaboom',
    %q{Example \ref{eg:kaboom}.}
  )
%>
<% end_marg %>
        \begin{eg}{An explosion}\label{eg:kaboom}
        \egquestion 
        Astronomers observe the planet Mars as the
        Martians fight a nuclear war. The Martian bombs are so
        powerful that they rip the planet into three separate pieces
        of liquefied rock, all having the same mass. If one fragment
        flies off with velocity components $v_{1 x}=0$,
        $v_{1 y}\zu{=1.0x10}^4$ km/hr,
        and the second with $v_{2 x}\zu{=1.0x10}^4$ km/hr, 
        $v_{2 y}=0$, what is the
        magnitude of the third one's velocity?
        
        \eganswer 
        We work the problem in the center of mass frame,
        in which the planet initially had zero momentum. After the
        explosion, the vector sum of the momenta must still be zero.
        Vector addition can be done by adding components, so
        \begin{align*}
                 mv_{1 x} +  mv_{2 x} +  mv_{3 x}  &=  0   \\
        \intertext{and}
                 mv_{1 y} +  mv_{2 y} +  mv_{3 y} &=  0   \qquad ,
        \end{align*}
        where we have used the same symbol $m$ for all the terms,
        because the fragments all have the same mass. The masses can
        be eliminated by dividing each equation by $m$, and we find
        \begin{align*}
                 v_{3 x}  &=  -\zu{1.0x10}^4\ \zu{km/hr} \\
                 v_{3 y}  &=  -\zu{1.0x10}^4\ \zu{km/hr} \qquad ,
        \end{align*}
        which gives a magnitude of
        \begin{align*}
                |\vc{v}_3|        &= \sqrt{ v_{3 x}^2+ v_{3 y}^2} \\
                &=  \zu{1.4x10}^4\ \zu{km/hr} \qquad .
        \end{align*}
        \end{eg}

<% marg(30) %>
<%
  fig(
    'toppling-box',
    %q{Example \ref{eg:toppling-box}.}
  )
%>
\spacebetweenfigs
<%
  fig(
    'flashlight',
    %q{The geometric interpretation of a vector's components.}
  )
%>
<% end_marg %>
\begin{eg}{A toppling box}\label{eg:toppling-box}
If you place a box on a frictionless surface, it will fall
over with a very complicated motion that is hard to predict
in detail. We know, however, that its center of mass's motion is
related to its momentum, and the rate at which momentum is transferred
is the force. Moreover, we know that these relationships apply separately
to each component. Let $x$ and $y$ be horizontal, and $z$ vertical. There
are two forces on the box, an upward force from the table and a downward gravitational force.
Since both of these are along the $z$ axis, $p_z$ is the only component of the
box's momentum that can change. We conclude that the center
of mass travels vertically. This is true even if the box
bounces and tumbles. [Based on an example by Kleppner and Kolenkow.]
\end{eg}

        <% begin_sec("Geometric representation of vectors") %>
        A vector in two dimensions can be easily visualized by
        drawing an arrow whose length represents its magnitude and
        whose direction represents its direction. The $x$ component of
        a vector can then be visualized, \figref{flashlight}, as the length of the shadow
        it would cast in a beam of light projected onto the $x$ axis,
        and similarly for the $y$ component. Shadows with arrowheads
        pointing back against the direction of the positive axis
        correspond to negative components.
        
        In this type of diagram, the negative of a vector is the
        vector with the same magnitude but in the opposite
        direction. Multiplying a vector by a scalar is represented
        by lengthening the arrow by that factor, and similarly for
        division.

        <% self_check('graphicalscalarmult',<<-'SELF_CHECK'
Given vector \vc{Q} represented by an arrow
        below, draw arrows representing the vectors 1.5\vc{Q} and $-\vc{Q}$.\\
        \begin{center}\anonymousinlinefig{graphicalscalarmult}\end{center}
  SELF_CHECK
  ) %>

        \begin{eg}[4]{A useless vector operation}
        The way I've defined the various vector operations above aren't as arbitrary as they seem.  
        There are many different
        vector operations that we could define, but only some of the possible definitions
        are mathematically useful. Consider the operation of
        multiplying two vectors component by component to produce a
        third vector:
        \begin{align*}
                 R_{x} & =   P_{x}  Q_x        \\
                 R_{y} & =   P_{y}  Q_y        \\
                 R_{z} & =   P_{z}  Q_z
        \end{align*}
<% marg(100) %>
<%
  fig(
    'useless',
    %q{%
      Two vectors, 1, to which we apply the same operation
                      in two different frames of reference, 2 and 3.
    }
  )
%>
<% end_marg %>

        As a simple example, we choose vectors \vc{P} and \vc{Q} to have
        length 1, and make them perpendicular to each other, as
        shown in figure \figref{useless}/1. If we compute the result of our new
        vector operation using the coordinate system shown in \figref{useless}/2,
        we find:
        \begin{align*}
                 R_{x}  &=  0 \\
                 R_{y}  &=  0 \\
                 R_z  &=  0 
        \end{align*}
        The $x$ component is zero because $P_x =0$, the $y$ component is
        zero because $Q_y=0$, and the $z$ component is of course zero
        because both vectors are in the $x$-$y$ plane. However, if we
        carry out the same operations in coordinate system \figref{useless}/3,
        rotated 45 degrees with respect to the previous one, we find
        \begin{align*}
                 R_{x}  &=  -\zu{1/2} \\
                 R_{y}  &=  \zu{1/2} \\
                 R_z  &=  0 
        \end{align*}
        The operation's result depends on what coordinate system we
        use, and since the two versions of \vc{R} have different lengths
        (one being zero and the other nonzero), they don't just
        represent the same answer expressed in two different
        coordinate systems. Such an operation will never be useful
        in physics, because experiments show physics works the same
        regardless of which way we orient the laboratory building!
        The useful vector operations, such as addition and scalar
        multiplication, are rotationally invariant, i.e., come out
        the same regardless of the orientation of the coordinate
        system.
        \end{eg}

        All the vector techniques can be applied to any kind of vector, but
        the graphical representation of vectors as arrows is particularly
        natural for vectors that represent lengths and distances.
        We define a vector called $\vc{r}$ whose components are the
        coordinates of a particular point in space, $x$, $y$, and $z$.
        The $\Delta\vc{r}$ vector,
        whose components are $\Delta x$, $\Delta y$,
        and $\Delta z$, can then be used to represent motion that starts
        at one point and ends at another. Adding two $\Delta \vc{r}$ vectors
        is interpreted as a trip with two legs: by computing the $\Delta \vc{r}$
        vector going from point A to point B plus the vector from B to C,
        we find the vector that would have taken us directly from A to C.

        <% end_sec() %>
<% begin_sec("Calculations with magnitude and direction") %>
        If you ask someone where Las Vegas is compared to Los
        Angeles, she is unlikely to say that the $\Delta x$ is 290 km and
        the $\Delta y$ is 230 km, in a coordinate system where the positive
        $x$ axis is east and the $y$ axis points north. She will
        probably say instead that it's 370 km to the northeast. If
        she was being precise, she might specify the direction as
        $38\degunit$ counterclockwise from east. In two dimensions, we can
        always specify a vector's direction like this, using a
        single angle. A magnitude plus an angle suffice to specify
        everything about the vector. The following two examples show
        how we use trigonometry and the Pythagorean theorem to go
        back and forth between the $x$-$y$ and magnitude-angle
        descriptions of vectors.

<% marg(0) %>
<%
  fig(
    'eg-la-vegas',
    %q{Example \ref{eg:comptopolar}.}
  )
%>
<% end_marg %>
        \begin{eg}{Finding magnitude and angle from components}\label{eg:comptopolar}
        \egquestion
         Given that the $\Delta\vc{r}$ vector from LA to Las Vegas has $\Delta x$=290 km 
        and $\Delta y$=230 km,
        how would we find the magnitude and direction of $\Delta\vc{r}$?
        
        \eganswer
         We find the magnitude of $\Delta\vc{r}$ from the Pythagorean theorem:
        \begin{align*}
                |\Delta\vc{r}|        &= \sqrt{\Delta x^2+\Delta y^2} \\
                        &= 370\ \zu{km}
        \end{align*}
        We know all three sides of the triangle, so the angle $\theta$ can be found using 
        any of the inverse trig functions. For example, we know the opposite and adjacent sides, so
        \begin{align*}
                \theta         &= \zu{tan}^{-1} \frac{\Delta y}{\Delta x}\\
                        &= 38\degunit   \qquad .
        \end{align*}
        
        
        \end{eg}
        \begin{eg}{Finding the components from the magnitude and angle}\label{eg:la-vegas-components}
        \egquestion
         Given that the straight-line distance from Los Angeles to Las Vegas is 370 \zu{km}, and
         that the angle $\theta$ in the figure is 38\degunit, how can the $x$ and $y$ components
         of the $\Delta\vc{r}$ vector be found?
        
        \eganswer
         The sine and cosine of $\theta$ relate the given information to the information we wish to find:
        \begin{align*}
                \zu{cos}\ \theta        &= \frac{\Delta x}{|\Delta\vc{r}|}\\
                \zu{sin}\ \theta        &= \frac{\Delta y}{|\Delta\vc{r}|}
        \end{align*}
        Solving for the unknowns gives
        \begin{align*}
                \Delta x        &= |\Delta\vc{r}|\:\zu{cos}\ \theta\\
                        &= 290\ \zu{km}\\
                \Delta y        &= |\Delta\vc{r}|\:\zu{sin}\ \theta\\
                        &= 230\ \zu{km}
        \end{align*}
        \end{eg}

        The following example shows the correct handling of the plus and minus signs, 
        which is usually the main cause of mistakes by students.
        
        \begin{eg}{Negative components}\label{eg:sdla}
        \egquestion
         San Diego is 120 km east and 150 km south of Los Angeles. An airplane pilot 
        is setting course from San Diego to Los Angeles. At what angle should she set her course,
         measured counterclockwise from east, as shown in the figure?
        
<% marg(0) %>
<%
  fig(
    'sdla',
    %q{Example \ref{eg:sdla}.}
  )
%>
<% end_marg %>
        \eganswer
         If we make the traditional choice of coordinate axes, with $x$ pointing to the right and $y$
         pointing up on the map, then her $\Delta x$ is negative, because her final $x$ value 
        is less than her initial $x$ value. Her $\Delta y$ is positive, so we have
        \begin{align*}
                \Delta x         &= -120\ \zu{km}\\
                \Delta y        &= 150\ \zu{km}  \qquad  .
        \end{align*}
        
        If we work by analogy with the example \ref{eg:comptopolar}, we get
        \begin{align*}
                \theta        &= \zu{tan}^{-1} \frac{\Delta y}{\Delta x}\\
                        &= \zu{tan}^{-1}\left(- 1.25\right) \\
                        &= -51\degunit   \qquad .
        \end{align*}
        According to the usual way of defining angles in trigonometry, a negative result means 
        an angle that lies clockwise from the $x$ axis, which would have her heading for the Baja
         California. What went wrong? The answer is that when you ask your calculator to take
         the arctangent of a number, there are always two valid possibilities differing by 
        180\degunit. That is, there are two possible angles whose tangents equal -1.25:
        \begin{align*}
                \zu{tan}\ 129\degunit &= - 1.25 \\
                \zu{tan}\left(-51\degunit\right) &= - 1.25
        \end{align*}
        You calculator doesn't know which is the correct one, so it just picks one. In this case,
         the one it picked was the wrong one, and it was up to you to add $180\degunit$ to it to
         find the right answer.
        \end{eg}

\begin{eg}{A shortcut}\label{eg:component-shortcut}
\egquestion A split second after nine o'clock, the hour hand on a clock dial
has moved clockwise past the nine-o'clock position by some imperceptibly small
angle $\phi$. Let positive $x$ be to the right and positive $y$ up.
If the hand, with length $\ell$, is represented by a $\Delta\vc{r}$ vector
going from the dial's center to the tip of the hand,
find this vector's $\Delta x$.

\eganswer The following shortcut is the easiest way to work out examples like
these, in which a vector's direction is known relative to one of the axes.
We can tell that $\Delta\vc{r}$ will have a large, negative $x$ component
and a small, positive $y$.
Since $\Delta x<0$,  there are really only
two logical possibilities: either $\Delta x = -\ell \cos\phi$, or
$\Delta x = -\ell \sin\phi$. Because $\phi$ is small, $\cos\phi$ is large
and $\sin\phi$ is small. We conclude that $\Delta x = -\ell \cos\phi$.

A typical application of this technique to force vectors is given in
example \ref{eg:layback} on p.~\pageref{eg:layback}.
\end{eg}

        <% end_sec() %>
<% begin_sec("Addition of vectors given their components") %>\label{subsec:vector-addition}\index{analytic addition of vectors}\index{vector addition!analytic}
        The easiest type of vector addition is when you are in possession of the components, 
        and want to find the components of their sum.
        
        \begin{eg}{San Diego to Las Vegas}\label{eg:sdvegas}
        \egquestion
         Given the $\Delta x$ and $\Delta y$ values from the previous examples, 
        find the $\Delta x$ and $\Delta y$ from San Diego to Las Vegas.

        \eganswer
        \begin{align*}
                \Delta x_{total}        &= \Delta x_1 + \Delta x_2 \\
                        &= -120\ \zu{km} + 290\ \zu{km} \\
                        &= 170\ \zu{km} \\
                \Delta y_{total}        &= \Delta y_1 + \Delta y_2 \\
                        &= 150\ \zu{km} + 230\ \zu{km} \\
                        &= 380 
        \end{align*}
        \end{eg}
<% marg(90) %>
<%
  fig(
    'eg-sd-vegas',
    %q{Example \ref{eg:sdvegas}.}
  )
%>
<% end_marg %>
        
        \subsubsection{Addition of vectors given their 
                        magnitudes and directions}
        In this case, you must first translate the magnitudes and directions into components,
        and the add the components.
        
        <% end_sec() %>
<% begin_sec("Graphical addition of vectors") %>\label{subsec:vector-addition-graphical}\index{graphical addition of vectors}\index{vector addition!graphical}
        Often the easiest way to add vectors is by making a scale drawing on a piece of paper.
         This is known as graphical addition, as opposed to the analytic techniques discussed 
        previously.
<%
  fig(
    'eg-sd-vegas-graphical',
    %q{Example \ref{eg:sd-vegas-graphical}.},
    {
      'width'=>'wide',
      'sidecaption'=>true
    }
  )
%>

        \begin{eg}{From San Diego to Las Vegas, graphically}\label{eg:sd-vegas-graphical}
        \egquestion
         Given the magnitudes and angles of the $\Delta\vc{r}$ vectors from 
        San Diego to Los Angeles and
         from Los Angeles to Las Vegas, find the magnitude and angle of the $\Delta\vc{r}$ vector 
        from San Diego to Las Vegas.

        \eganswer
         Using a protractor and a ruler, we make a careful scale drawing, as shown in figure \figref{eg-sd-vegas-graphical} on page \pageref{fig:eg-sd-vegas-graphical}.
         A scale of 1 cm$\leftrightarrow$10 \zu{km} was chosen for this solution.
        With a ruler, we measure the
         distance from San Diego to Las Vegas to be 3.8 cm, which corresponds to 380 \zu{km}.
         With a protractor, we measure the angle $\theta$ to be 71\degunit.
        \end{eg}
        
        Even when we don't intend to do an actual graphical calculation with a ruler and protractor, 
        it can be convenient to diagram the addition of vectors in this way, as shown in figure \figref{tip-to-tail}. With $\Delta\vc{r}$
         vectors, it intuitively makes sense to lay the vectors tip-to-tail and draw the sum vector
         from the tail of the first vector to the tip of the second vector. We can do the same
         when adding other vectors such as force vectors.
<% marg(28) %>
<%
  fig(
    'tip-to-tail',
    %q{%
      Adding vectors graphically by placing them tip-to-tail, like a train.
    }
  )
%>
<% end_marg %>

        <% end_sec() %>
<% begin_sec("Unit vector notation") %>
        When we want to specify a vector by its components, it can be cumbersome to have to
        write the algebra symbol for each component:
        \begin{equation*}
                \Delta x = 290\ \zu{km},\quad \Delta y = 230\ \zu{km}
        \end{equation*}
        A more compact notation is to write
        \begin{equation*}
                \Delta\vc{r} = (290\ \zu{km})\hat{\vc{x}} + (230\ \zu{km})\hat{\vc{y}}   ,
        \end{equation*}
        where the vectors $\hat{\vc{x}}$, $\hat{\vc{y}}$, and $\hat{\vc{z}}$, called the unit vectors,
         are defined as the vectors that have magnitude equal to 1 and directions lying along 
        the $x$, $y$, and $z$ axes. In speech, they are referred to as ``x-hat,''
        ``y-hat,'' and ``z-hat.''
        
        A slightly different, and harder to remember, version of this notation is unfortunately 
        more prevalent. In this version, the unit vectors are called
         $\hat{\vc{i}}$, $\hat{\vc{j}}$, and $\hat{\vc{k}}$:
        
        \begin{equation*}
                \Delta\vc{r} = (290\ \zu{km})\hat{\vc{i}} + (230\ \zu{km})\hat{\vc{j}} \qquad   .
        \end{equation*}

        
        <% end_sec() %>
<% begin_sec("Applications to relative motion, momentum, and force") %>
        Vector addition is the correct way to generalize the
        one-dimensional concept of adding velocities in relative
        motion, as shown in the following example:\index{velocity!addition of!vector}
        
\begin{eg}{Velocity vectors in relative motion}\label{eg:boat}
\egquestion You wish to cross a river and arrive at a dock
that is directly across from you, but the river's current
will tend to carry you downstream. To compensate, you must
steer the boat at an angle. Find the angle $\theta $, given
the magnitude, $|\vc{v}_{WL}|$, of the water's velocity relative
to the land, and the maximum speed, $|\vc{v}_{BW}|$, of which the
boat is capable relative to the water.

\eganswer The boat's velocity relative to the land equals
the vector sum of its velocity with respect to the water and
the water's velocity with respect to the land,
\begin{equation*}
            \vc{v}_{BL} =  \vc{v}_{BW}+  \vc{v}_{WL} \qquad     .
\end{equation*}
If the boat is to travel straight across the river, i.e.,
along the $y$ axis, then we need to have $\vc{v}_{BL,x}=0$. This $x$
component equals the sum of the $x$ components of the other two vectors,
\begin{equation*}
            \vc{v}_BL,x  =  \vc{v}_BW,x +  \vc{v}_WL,x   \qquad  ,
\end{equation*}
or
\begin{equation*}
            0  =  -|\vc{v}_{BW}| \sin  \theta  + |\vc{v}_{WL}|   \qquad   .
\end{equation*}
Solving for $\theta $, we find
\begin{align*}
            \sin  \theta  &=  |\vc{v}_{WL}|/|\vc{v}_{BW}|   \qquad ,\\
\intertext{so}
            \theta &= \sin^{-1}\frac{|\vc{v}_{WL}|}{\vc{v}_{BW}}\qquad   .
\end{align*}
\end{eg}
<% marg(130) %>
<%
  fig(
    'greyhound',
    %q{The racing greyhound's velocity vector is in the direction of its motion, i.e., tangent to
       its curved path.}
  )
%>
\spacebetweenfigs
<%
  fig(
    'boat',
    %q{Example \ref{eg:boat}}
  )
%>
<% end_marg %>
        
        \begin{eg}{How to generalize one-dimensional equations}
        \egquestion
        How can the one-dimensional relationships
        \begin{gather*}
                 p_{total} =  m_{total}  v_{cm}
        \intertext{and}
                 x_{cm} = \frac{\sum_{j}{ m_{j} x_j}}{\sum_{j}{ m_j}}
        \end{gather*}
        be generalized to three dimensions?

        \eganswer
        Momentum and velocity are vectors, since they have
        directions in space. Mass is a scalar. If we rewrite the first 
        equation to show the appropriate quantities notated as
        vectors,
        \begin{equation*}
                \vc{p}_{total} =  m_{total} \vc{v}_{cm} \qquad ,
        \end{equation*}
        we get a valid mathematical operation, the multiplication of
        a vector by a scalar. Similarly, the second equation becomes
        \begin{equation*}
                \vc{r}_{cm} = \frac{\sum_j{ m_{j}\vc{r}_j}}{\sum_{j}{ m_j}}
                                \qquad ,
        \end{equation*}
        which is also valid. Each term in the sum on top contains a vector multiplied by a scalar,
        which gives a vector. Adding up all these vectors gives a vector, and dividing by the
        scalar sum on the bottom gives another vector.

        This kind of wave-the-magic-wand-and-write-it-all-in-bold-face technique
        will always give the right generalization from one dimension to three,
        provided that the result makes sense mathematically --- if you find yourself
        doing something nonsensical, such as adding a scalar to a vector, then
        you haven't found the generalization correctly.\label{magicwand}
        \end{eg}

\begin{eg}{Colliding coins}
\egquestion Take two identical coins, put one down on a piece of paper, and slide
the other across the paper, shooting it fairly rapidly so that it hits the
target coin off-center. If you trace the initial and final positions of the
coins, you can determine the directions of their momentum vectors after the
collision. The angle between these vectors is always fairly close to, but a
little less than, 90 degrees. Why is this?

\eganswer Let the velocity vector of the incoming coin be $\vc{a}$, and let
the two outgoing velocity vectors be $\vc{b}$ and $\vc{c}$. Since the masses
are the same, conservation of momentum amounts to $\vc{a}=\vc{b}+\vc{c}$,
which means that it has to be possible to assemble the three vectors into
a triangle. If we assume that no energy is converted into heat and sound, then
conservation of energy gives (discarding the common factor of $m/2$)
$a^2=b^2+c^2$ for the magnitudes of the three vectors. This is the
Pythagorean theorem, which will hold only if the three vectors form a
right triangle.

The fact that we observe the angle to be somewhat less than 90 degrees shows
that the assumption used in the proof is only approximately valid: a little energy \emph{is}
converted into heat and sound. The opposite case would be a collision between
two blobs of putty, where the maximum possible amount of energy is converted
into heat and sound, the two blobs fly off together, giving an angle of
zero between their momentum vectors. The real-life experiment interpolates
between the ideal extremes of 0 and 90 degrees, but comes much closer to 90.
\end{eg}

        Force is a vector, and we add force vectors when more than one force
        acts on the same object.

<% marg(70) %>
                %
<%
  fig(
    'rampforces',
    %q{Example \ref{eg:rampforces}.}
  )
%>%
        
<% end_marg %>
        \begin{eg}{Pushing a block up a ramp}\label{eg:rampforces}
        \egquestion
        Figure \figref{rampforces}/1 shows a block being pushed up a
        frictionless ramp at constant speed by an applied force $F_a$.
        How much force is required, in terms of the block's mass, $m$,
        and the angle of the ramp, $\theta$?
        
        \eganswer
        We analyzed this simple machine in example \ref{eg:inclinedplanework}
        on page \pageref{eg:inclinedplanework} using the concept of work.
        Here we'll do it using vector addition of forces.
        Figure \figref{rampforces}/2 shows the other two forces acting on
        the block: a normal force, $F_n$,  created by the ramp, and the
        gravitational force, $F_g$. Because
        the block is being pushed up at constant speed, it has zero
        acceleration, and the total force on it must be zero. In
        figure \figref{rampforces}/3, we position all the force vectors tip-to-tail
        for addition. Since they have to add up to zero, they must join up without
        leaving a gap, so they form a triangle. Using trigonometry we find
        \begin{align*}
                 F_{a}         &=  F_g\ \zu{sin}\:\theta \\
                        &=  mg\ \zu{sin}\:\theta \qquad   .
        \end{align*}
        \end{eg}

        \begin{eg}{Buoyancy, again}\label{eg:buoyancy}
        \index{buoyancy}\index{Archimedes' principle}
        In example \ref{eg:buoyancycube} on page \pageref{eg:buoyancycube}, we found that
        the energy required to raise a cube immersed in a fluid is as if the
        cube's mass had been reduced by an amount equal to the mass of the fluid
        that otherwise would have been in the volume it occupies (Archimedes' principle).
        From the energy perspective, this effect occurs because raising the cube allows
        a certain amount of fluid to move downward, and the decreased gravitational
        energy of the fluid tends to offset the increased gravitational energy of the cube. The proof
        given there, however, could not easily be extended to other shapes.

<% marg(50) %>
<%
  fig(
    'buoyancy2',
    %q{%
      Archimedes' principle works regardless of whether the
                      object is a cube. The fluid makes a force on every square millimeter of the object's
                      surface.
    }
  )
%>
<% end_marg %>
        Thinking in terms of force rather than energy, it becomes easier to give a proof
        that works for any shape. A certain upward force is needed to support the object
        in figure \figref{buoyancy2}. If this force was applied, then the object would be
        in equilibrium: the vector sum of all the forces acting on it would be zero. These
        forces are $\vc{F}_a$, the upward force just mentioned,
        $\vc{F}_g$, the downward force of gravity,
        and $\vc{F}_f$, the total force from the fluid:
        \begin{equation*}
                \vc{F}_a+\vc{F}_{g}+\vc{F}_f = 0
        \end{equation*}
        Since the fluid is under more pressure at a greater
        depth, the part of the fluid underneath the object tends to make more force than the
        part above, so the fluid tends to help support the object.

        Now suppose the object was removed, and instantly replaced with an equal volume
        of fluid. The new fluid would be in equilibrium without any force applied to hold it up, so
        \begin{equation*}
                \vc{F}_{gf}+\vc{F}_f = 0 \qquad ,
        \end{equation*}
        where $\vc{F}_{gf}$, the weight of the fluid, is not the same as $\vc{F}_g$, the
        weight of the object, but $\vc{F}_f$ is the same as before, since the pressure of
        the surrounding fluid is the same as before at any particular depth. We therefore have
        \begin{equation*}
                \vc{F}_a=-\left(\vc{F}_g-\vc{F}_{gf}\right) \qquad ,
        \end{equation*}
        which is Archimedes' principle in terms of force: the force required to support the
        object is lessened by an amount equal to the weight of the fluid that would have occupied
        its volume.
        \end{eg}

        By the way, the word ``pressure'' that I threw around casually in the preceding
        example has a precise technical definition: force per unit area. The SI units of
        pressure are $\nunit/\munit^2$, which can be abbreviated as pascals,
        1 Pa = 1 $\nunit/\munit^2$. Atmospheric pressure is about 100 kPa.
        By applying the equation $\vc{F}_{g}+\vc{F}_f = 0$ to the top and bottom
        surfaces of a cubical volume of fluid, one can easily prove that the difference
        in pressure between two different depths is $\Delta P=\rho g\Delta y$. (In physics,
        ``fluid'' can refer to either a gas or a liquid.)\index{fluid!defined}
        \index{pressure!defined}\index{pressure!as a function of depth}\index{pascal (unit)}
        Pressure is discussed in more detail in chapter \ref{ch:thermo}.

        \begin{eg}{A solar sail}\label{eg:solarsail}
                A solar sail\index{solar sail}, figure \figref{solarsail}/1, allows a spacecraft
                to get its thrust without using  internal stores of energy or having to carry along
                mass that it can shove out the back like a rocket. Sunlight strikes the sail
                and bounces off, transferring momentum to the sail.
                A working 30-meter-diameter solar sail, Cosmos 1, was built by
                an American company, and was supposed to be launched into
                orbit aboard
                a Russian booster launched from a submarine, but launch attempts in 2001 and
                2005 both failed.
<% marg(0) %>
<%
  fig(
    'solarsail',
    %q{Example \ref{eg:solarsail}.}
  )
%>
          \spacebetweenfigs
<%
  fig(
    'solar-sail-art',
    %q{An artist's rendering of what Cosmos 1 would have looked like in orbit.}
  )
%>
<% end_marg %>

                In this example, we will calculate the optimal orientation of the sail, 
                assuming that ``optimal'' means changing the vehicle's energy as rapidly as possible.
                For simplicity, we model the complicated shape of the sail's surface as a disk,
                seen edge-on in figure \figref{solarsail}/2, and we assume that the craft is
                in a nearly circular orbit around the sun, hence the 90-degree angle between the
                direction of motion and the incoming sunlight. We assume that the sail is
                100\% reflective. The orientation
                of the sail is specified using the angle $\theta$ between the
                incoming rays of sunlight and the perpendicular to the sail. In other words,
                $\theta\zu{=0}$ if the sail is catching the sunlight full-on, while $\theta\zu{=90}\degunit$
                means that the sail is edge-on to the sun.

                Conservation of momentum gives
                \begin{align*}
                        \vc{p}_{light,i} &= \vc{p}_{light,f}+\Delta\vc{p}_{sail} \qquad , \\
                \intertext{where $\Delta\vc{p}_{sail}$ is the change in momentum picked up
                        by the sail. Breaking this down into components, we have}
                        0 &= p_{light,f,x}+\Delta p_{sail,x} \qquad \text{and} \\
                        p_{light,i,y} &= p_{light,f,y}+\Delta p_{sail,y} \qquad .
                \end{align*}
                As in example \ref{eg:yarkovsky} on page \pageref{eg:yarkovsky}, the component
                of the force that is directly away from the sun (up in figure \figref{solarsail}/2)
                doesn't change the energy of the craft, so we only care about
                $\Delta p_{sail,x}$, which equals $- p_{light,f,x}$. The outgoing
                light ray forms an angle of 2$\theta$ with the negative $y$ axis, or
                270$\degunit-2\theta$ measured counterclockwise from the $x$
                axis, so the useful thrust depends on $-\zu{cos}(270\degunit-2\theta)
                =\zu{sin}\:2\theta$.

                However, this is all assuming a given amount of light strikes the sail.
                During a certain time period, the amount of
                sunlight striking the sail depends on the cross-sectional area the sail presents
                to the sun, which is proportional to cos $\theta$. For $\theta\zu{=90}\degunit$,
                cos $\theta$ equals zero, since the sail is edge-on to the sun.  

                Putting together these two factors, the useful thrust is proportional to
                $\zu{sin}\ 2\theta\ \zu{cos}\:\theta$, and this quantity is maximized
                for $\theta\approx35\degunit$. A counterintuitive fact about this
                maneuver is that as the spacecraft spirals outward, its total energy
                (kinetic plus gravitational) increases, but its kinetic energy actually
                decreases!
        \end{eg}
 % vectors

\begin{eg}{A layback}\label{eg:layback}
The figure shows a rock climber 
using a technique called a layback. He can make the normal forces
$\vc{F}_{N1}$ and $\vc{F}_{N2}$ large, which has the side-effect of increasing
the frictional forces $\vc{F}_{F1}$ and $\vc{F}_{F2}$, so that he doesn't slip
down due to the gravitational (weight) force $\vc{F}_W$. The purpose of the
problem is not to analyze all of this in detail, but simply to practice
finding the components of the forces based on their magnitudes.
To keep the notation simple, let's write $F_{N1}$ for 
$|\vc{F}_{N1}|$, etc. The crack overhangs by a small, positive angle $\theta\approx9\degunit$.

In this example, we 
 determine the
$x$ component of $\vc{F}_{N1}$. The other nine components are left as an exercise to
the reader (problem \ref{hw:layback}, p.~\pageref{hw:layback}).

The easiest method is the one demonstrated in example \ref{eg:component-shortcut}
on p.~\pageref{eg:component-shortcut}. Casting vector $\vc{F}_{N1}$'s shadow on the ground,
we can tell that it would point to the left, so its $x$ component is
negative. The only two possibilities for its $x$ component are therefore
$-F_{N1}\cos\theta$ or $-F_{N1}\sin\theta$. We expect this force to have
a large $x$ component and a much smaller $y$. Since $\theta$ is small,
$\cos\theta\approx 1$, while $\sin\theta$ is small. Therefore
the $x$ component must be $-F_{N1}\cos\theta$.
\end{eg}

<%
  fig(
    'hw-layback',
    %q{Example \ref{eg:layback} and problem \ref{hw:layback} on p.~\pageref{hw:layback}.},
    {
      'width'=>'wide',
      'sidecaption'=>false
    }
  )
%>

\startdqs

\begin{dq}
An object goes from one point in space to another. After
it arrives at its destination, how does the magnitude of its
$\Delta\zb{r}$ vector compare with the distance it traveled?
\end{dq}

\begin{dq}
In several examples, I've dealt with vectors having negative components.
Does it make sense as well to talk about negative and
positive vectors?
\end{dq}

\begin{dq}
If you're doing \emph{graphical} addition of vectors,
does it matter which vector you start with and which vector
you start from the other vector's tip?
\end{dq}

\begin{dq}
If you add a vector with magnitude 1 to a vector of
magnitude 2, what magnitudes are possible for the vector sum?
\end{dq}

<% marg(0) %>
<%
  fig(
    'dq-check-tip-to-tail',
    %q{Discussion question \ref{dq:tiptotail}.}
  )
%>
<% end_marg %>
\begin{dq}\label{dq:tiptotail}
Which of these examples of vector addition are correct,
and which are incorrect?
\end{dq}

\begin{dq}
Is it possible for an airplane to maintain a constant
velocity vector but not a constant $|\zb{v}|$?  How about the
opposite -- a constant $|\zb{v}|$ but not a constant velocity vector? Explain.
\end{dq}

\pagebreak

\begin{dq}
New York and Rome are at about the same latitude, so the
earth's rotation carries them both around nearly the same
circle. Do the two cities have the same velocity vector
(relative to the center of the earth)? If not, is there any
way for two cities to have the same velocity vector?
\end{dq}

<% marg(0) %>
<%
  fig(
    'rollercoaster',
    %q{Discussion question \ref{dq:rollercoaster}.}
  )
%>
<% end_marg %>
\begin{dq}\label{dq:rollercoaster}
The figure shows a roller coaster car rolling down and
then up under the influence of gravity. Sketch the car's
velocity vectors and acceleration vectors. Pick an
interesting point in the motion and sketch a set of force
vectors acting on the car whose vector sum could have
resulted in the right acceleration vector.
\end{dq}

\begin{dq}
The following is a question commonly asked by students:

``Why does the force vector always have to point in the same
direction as the acceleration vector? What if you suddenly
decide to change your force on an object, so that your force
is no longer pointing in the same direction that the object is accelerating?''

What misunderstanding is demonstrated by this question?
Suppose, for example, a spacecraft is blasting its rear main
engines while moving forward, then suddenly begins firing
its sideways maneuvering rocket as well. What does the
student think Newton's laws are predicting?
\end{dq}

\begin{dq}
Debug the following  \emph{incorrect} solutions to this vector addition
problem.

\emph{Problem}: Freddi Fish$^\zu{TM}$ swims 5.0 km northeast, and then
12.0 km in the direction 55 degrees west of south. How far does she
end up from her starting point, and in what direction is she from her
starting point?

\emph{Incorrect solution \#1:}\\
5.0 km+12.0 km=17.0 km

\emph{Incorrect solution \#2:}\\
$\sqrt{(5.0\ \zu{km})^2+(12.0\ \zu{km})^2}$=13.0 km

\emph{Incorrect solution \#3:}\\
Let \zb{A} and \zb{B} be her two $\Delta\zb{r}$ vectors, and
let $\zb{C}=\zb{A}+\zb{B}$. Then 
\begin{align*}
        A_x        &= (5.0 \ \zu{km}) \cos 45\degunit = 3.5 \ \zu{km}\\
        B_x        &= (12.0 \ \zu{km}) \cos 55\degunit = 6.9 \ \zu{km} \\
        A_y        &= (5.0 \ \zu{km}) \sin 45\degunit = 3.5 \ \zu{km} \\
        B_y        &= (12.0 \ \zu{km}) \sin 55\degunit  = 9.8 \ \zu{km} \\
        C_x        &= A_x+B_x \\
                &= 10.4 \ \zu{km} \\
        C_y        &= A_y+B_y \\
                &= 13.3 \ \zu{km} \\
        |\zb{C}| &= \sqrt{C_x^2+C_y^2} \\
                &= 16.9 \ \zu{km} \\
        \zu{direction} &= \tan^{-1} (13.3/10.4) \\
                        &= 52 \degunit \ \text{north of east}
\end{align*}

\pagebreak

\emph{Incorrect solution \#4:}\\
(same notation as above) 
\begin{align*}
        A_x        &= (5.0 \ \zu{km}) \cos 45\degunit = 3.5 \ \zu{km}\\
        B_x        &= -(12.0 \ \zu{km}) \cos 55\degunit = -6.9 \ \zu{km} \\
        A_y        &= (5.0 \ \zu{km}) \sin 45\degunit = 3.5 \ \zu{km} \\
        B_y        &= -(12.0 \ \zu{km}) \sin 55\degunit  = -9.8 \ \zu{km} \\
        C_x        &= A_x+B_x \\
                &= -3.4 \ \zu{km} \\
        C_y        &= A_y+B_y \\
                &= -6.3 \ \zu{km} \\
        |\zb{C}| &= \sqrt{C_x^2+C_y^2} \\
                &= 7.2 \ \zu{km} \\
        \zu{direction} &= \tan^{-1} (-6.3/-3.4) \\
                        &= 62 \degunit\ \text{north of east}
\end{align*}

\emph{Incorrect solution \#5:}\\
(same notation as above) 
\begin{align*}
        A_x        &= (5.0 \ \zu{km}) \cos 45\degunit = 3.5 \ \zu{km}\\
        B_x        &= -(12.0 \ \zu{km}) \sin 55\degunit = -9.8 \ \zu{km} \\
        A_y        &= (5.0 \ \zu{km}) \sin 45\degunit = 3.5 \ \zu{km} \\
        B_y        &= -(12.0 \ \zu{km}) \cos 55\degunit  = -6.9 \ \zu{km} \\
        C_x        &= A_x+B_x \\
                &= -6.3 \ \zu{km} \\
        C_y        &= A_y+B_y \\
                &= -3.4 \ \zu{km} \\
        |\zb{C}| &= \sqrt{C_x^2+C_y^2} \\
                &= 7.2 \ \zu{km} \\
        \zu{direction} &= \tan^{-1} (-3.4/-6.3) \\
                        &= 28 \degunit\ \text{north of east}
\end{align*}

\end{dq}

<% end_sec() %>
<% end_sec() %>
<% begin_sec("Calculus with vectors") %>
<% marg(-60) %>
<%
  fig(
    'deltav',
    %q{Visualizing the acceleration vector.}
  )
%>
<% end_marg %>
        <% begin_sec("Differentiation") %>
        In one dimension, we define the velocity as the derivative of the position
        with respect to time, and we can think of the derivative as what we get
        when we calculate $\Delta x/\Delta t$ for very short time intervals.
        The quantity $\Delta x=x_f-x_i$ is calculated by subtraction.
        In three dimensions, $x$ becomes \vc{r}, and the $\Delta\vc{r}$ vector
        is calculated by \emph{vector} subtraction, $\Delta\vc{r}=\vc{r}_f-\vc{r}_i$.
        Vector subtraction is defined component by component, so when we take
        the derivative of a vector, this means we end up taking the derivative
        component by component,
        \begin{equation*}
                v_x =  \frac{\der x}{\der t} , \quad v_y =  \frac{\der y}{\der t}, \quad
                v_z =  \frac{\der z}{\der t} 
        \end{equation*}
        or
        \begin{equation*}
                \frac{\der\vc{r}}{\der t} 
                        = \frac{\der x}{\der t}\hat{\vc{x}}
                        +\frac{\der y}{\der t}\hat{\vc{y}}+\frac{\der z}{\der t}\hat{\vc{z}} \qquad .
        \end{equation*}
        All of this reasoning applies equally well to any derivative of a vector, so for
        instance we can take the second derivative,
        \begin{equation*}
                a_x =  \frac{\der v_x}{\der t} , \quad 
                a_y =  \frac{\der v_y}{\der t} , \quad 
                a_z =  \frac{\der v_z}{\der t} 
        \end{equation*}
        or
        \begin{equation*}
                \frac{\der\vc{v}}{\der t} 
                        = \frac{\der v_x}{\der t}\hat{\vc{x}}+\frac{\der v_y}{\der t}\hat{\vc{y}}
                        +\frac{\der v_z}{\der t}\hat{\vc{z}} \qquad .
        \end{equation*}

        A counterintuitive consequence of this is that the acceleration vector does not
        need to be in the same direction as the motion. The velocity vector 
        points in the direction of motion, but by Newton's second law,
        $\vc{a}=\vc{F}/m$, the acceleration vector points in the same direction as the force,
        not the motion. This is easiest to understand if we take velocity vectors
        from two different moments in the motion, and visualize subtracting them
        graphically to make a $\Delta\vc{v}$ vector.
        The direction of the $\Delta\vc{v}$
        vector tells us the direction of the acceleration vector as well, since
        the derivative $\der\vc{v}/\der t$ can be approximated as $\Delta\vc{v}/\Delta t$.
        As shown in figure \figref{deltav}/1, a change in the magnitude of the velocity vector implies
        an acceleration that is in the direction of motion. A change in the direction
        of the velocity vector produces an acceleration perpendicular to the motion,  \figref{deltav}/2.

        \index{circular motion}
        \begin{eg}{Circular motion}\label{eg:circularaccel}
        \egquestion
        An object moving in a circle of radius $r$ in the $x$-$y$ plane has
        \begin{align*}
                 x        &=  r\ \zu{cos}\ \omega t \qquad \text{and}\\
                 y        &=  r\ \zu{sin}\ \omega t \qquad ,
        \end{align*} 
        where $\omega$ is the number of radians traveled per second, and the positive or negative
        sign indicates whether the motion is clockwise or counterclockwise.
        What is its acceleration?

        \eganswer
        The components of the velocity are
        \begin{align*}
                 v_{x}        &= -\omega r\ \zu{sin}\ \omega t \qquad \text{and}\\
                 v_{y}        &= \ \ \ \omega r\ \zu{cos}\ \omega t \qquad ,
        \end{align*} 
        and for the acceleration we have
        \begin{align*}
                 a_{x}        &= -\omega^2 r\ \zu{cos}\ \omega t \qquad \text{and}\\
                 a_{y}        &= -\omega^2 r\ \zu{sin}\ \omega t \qquad .
        \end{align*} 
        The acceleration vector has cosines and sines in the same places as the
        \vc{r} vector, but with minus signs in front, so it points in the opposite direction,
        i.e., toward the center of the circle. By Newton's second law, \vc{a}=\vc{F}/$m$,
        this shows that the force must be inward as well; without this force, the object
        would fly off straight.
<% marg(30) %>
<%
  fig(
    'hammer',
    %q{%
      This figure shows an intuitive justification for the fact proved mathematically in the example,
      that the direction of the force and acceleration in circular motion is inward.
      The heptagon, 2, is a better approximation to a circle than the
              triangle, 1. To make an infinitely good approximation to circular motion,
              we would need to use an infinitely large number of infinitesimal taps, which
              would amount to a steady inward force.
    }
  )
%>
<% end_marg %>

        The magnitude of the acceleration is
        \begin{align*}
                |\vc{a}|        &= \sqrt{ a_x^2+ a_{y}^2}\\
                                &= \omega^2 r \qquad .
        \end{align*}
        It makes sense that $\omega$ is squared, since reversing the sign of $\omega$
        corresponds to reversing the direction of motion, but the acceleration is toward the
        center of the circle, regardless of whether the motion is clockwise or counterclockwise.
        This result can also be rewritten in the form
        \begin{equation*}
                |\vc{a}|        = \frac{|\vc{v}|^2}{r} \qquad .
        \end{equation*}
        \end{eg}

        Although I've relegated the results $a=\omega^2 r=|\vc{v}|^2/r$ to an example because they are a straightforward corollary
        of more general principles already developed, they are important and useful enough to record for later use.
        \index{circular motion!inward force}
        \index{circular motion!no forward force}
        These results are
        counterintuitive as well. Until Newton, physicists and laypeople alike had assumed
        that the planets would need a force to push them \emph{forward} in their orbits.
        Figure \figref{hammer} may help to make it more plausible that only an inward force
        is required. A forward force might be needed in order to cancel out a backward
        force such as friction, \figref{carcircleforces}, but the total force in the forward-backward
        direction needs to be exactly zero for constant-speed motion.
        \index{circular motion!no outward force}
        When you are in a car undergoing circular motion, there is also a strong illusion of an
        \emph{outward} force. But what object could be making such a force? The car's seat
        makes an inward force on you, not an outward one. There is no object that could be
        exerting an outward force on your body. In reality, this force is an illusion that comes
        from our brain's intuitive efforts to interpret the situation within a noninertial frame of
        reference. As shown in figure \figref{truckcircular}, we can describe everything perfectly
        well in an inertial frame of reference, such as the frame attached to the sidewalk.
        In such a frame, the bowling ball goes straight because there is \emph{no} force
        on it. The wall of the truck's bed hits the ball, not the other way around.
<% marg(100) %>
<%
  fig(
    'carcircleforces',
    %q{%
      The
                       total force in the forward-backward direction is zero in both
              cases.
    }
  )
%>
\spacebetweenfigs
<%
  fig(
    'truckcircular',
    %q{%
      There is no outward force on the bowling ball, but in the noninertial
                      frame it seems like one exists.
    }
  )
%>
<% end_marg %>

        <% end_sec() %>
<% begin_sec("Integration") %>
        An integral is really just a sum of many infinitesimally small terms. Since vector
        addition is defined in terms of addition of the components, an integral of a vector
        quantity is  found by doing integrals component by component.

        \begin{eg}{Projectile motion}
        \egquestion
        Find the motion of an object whose acceleration vector is constant, for instance
        a projectile moving under the influence of gravity.

        \eganswer
        We integrate the acceleration to get the velocity, and then integrate the velocity to
        get the position as a function of time. Doing this to the $x$ component of the
        acceleration, we find
        \begin{align*}
                 x         &= \int{\left(\int{a_x\ \zu{d}t}\right)\ \zu{d}t} \\
                                &= \int{\left(a_xt+v_{x\zu{o}}\right)\zu{d}t} \qquad ,\\
        \intertext{where $v_{ x\zu{o}}$ is a constant of integration, and}
                x        &= \frac{1}{2}a_xt^2
                                                + v_{x\zu{o}}t + x_\zu{o} \qquad .
        \end{align*}
        Similarly, $y\zu{=(1/2)} a_{y} t^2+  v_{ y\zu{o}} t +  y_\zu{o}$
        and $z=\zu{(1/2)} a_{z} t^2+  v_{ z\zu{o}} t +  z_\zu{o}$.
        Once one has gained a little confidence, it becomes natural to do the whole thing
        as a single vector integral,
        \begin{align*}
                \vc{r}        &= \int{\left(\int{\vc{a}\ \zu{d} t}\right)\ \zu{d} t} \\
                                &= \int{\left(\vc{a} t+\vc{v}_\zu{o}\right)\zu{d} t} \\
                        &= \frac{1}{2}\vc{a} t^2+\vc{v}_\zu{o} t+\vc{r}_\zu{o} \qquad ,
        \end{align*}
        where now the constants of integration are vectors.
        \end{eg}

\startdqs

<% marg(0) %>
<%
  fig(
    'crack-the-whip',
    %q{Discussion question \ref{dq:crack-the-whip}.}
  )
%>
<% end_marg %>
\begin{dq}\label{dq:crack-the-whip}
 In the game of crack the whip, a line of people stand
holding hands, and then they start sweeping out a circle. 
One person is at the center, and rotates without changing
location.  At the opposite end is the person who is running
the fastest, in a wide circle.  In this game, someone always
ends up losing their grip and flying off.  Suppose the
person on the end loses her grip.  What path does she follow
as she goes flying off?  (Assume she is going so fast that
she is really just trying to put one foot in front of the
other fast enough to keep from falling; she is not able to
get any significant horizontal force between her feet and the ground.)
\end{dq}

\begin{dq}
Suppose the person on the outside is still holding on,
but feels that she may loose her grip at any moment.  What
force or forces are acting on her, and in what directions
are they?  (We are not interested in the vertical forces,
which are the earth's gravitational force pulling down, and
the ground's normal force pushing up.)
Make a table in the format shown in __subsection_or_section(analysis-of-forces).
\end{dq}

\begin{dq}
Suppose the person on the outside is still holding on,
but feels that she may loose her grip at any moment.  What
is wrong with the following analysis of the situation? 
``The person whose hand she's holding exerts an inward force
on her, and because of Newton's third law, there's an equal
and opposite force acting outward.  That outward force is
the one she feels throwing her outward, and the outward
force is what might make her go flying off, if it's strong enough.''
\end{dq}

\begin{dq}
If the only force felt by the person on the outside is an
inward force, why doesn't she go straight in?
\end{dq}

<% marg(0) %>
<%
  fig(
    'dq-tilt-a-whirl',
    %q{Discussion question \ref{dq:tilt-a-whirl}.}
  )
%>
<% end_marg %>
\begin{dq}\label{dq:tilt-a-whirl}
In the amusement park ride shown in the figure, the
cylinder spins faster and faster until the customer can pick
her feet up off the floor without falling. In the old Coney
Island version of the ride, the floor actually dropped out
like a trap door, showing the ocean below. (There is also a
version in which the whole thing tilts up diagonally, but
we're discussing the version that stays flat.) If there is
no outward force acting on her, why does she stick to the
wall? Analyze all the forces on her.
\end{dq}

\begin{dq}
What is an example of circular motion where the inward
force is a normal force?  What is an example of circular
motion where the inward force is friction?  What is an
example of circular motion where the inward force is the sum
of more than one force? 
\end{dq}

\begin{dq}
Does the acceleration vector always change continuously
in circular motion? The velocity vector?
\end{dq}

\begin{dq}
A certain amount of force is needed to provide the
acceleration of circular motion. What if we are exerting a
force perpendicular to the direction of motion in an attempt
to make an object trace a circle of radius $r$, but the
force isn't as big as $m|\zb{v}|^2/r$?
\end{dq}

\begin{dq}
Suppose a rotating space station is built that gives its
occupants the illusion of ordinary gravity. What happens
when a person in the station lets go of a ball? What happens
when she throws a ball straight ``up'' in the air (i.e.,
towards the center)?
\end{dq}

<% end_sec() %>
<% end_sec() %>
<% begin_sec("The dot product",nil,'dotproduct') %>
        How would we generalize the mechanical work
        equation $\der E=F \der x$ to three dimensions?
        Energy is a scalar, but force and distance are vectors, so it might
        seem at first that the kind of ``magic-wand'' generalization discussed
        on page \pageref{magicwand} failed here, since we don't know of any way
        to multiply two vectors together to get a scalar. Actually, this is Nature giving
        us a hint that there is such a multiplication operation waiting for us to invent it,
        and since Nature is simple, we can be assured that this
        operation will work just fine in any situation where a similar generalization is
        required.  

        How should this operation be defined? Let's consider what we would get
        by performing this operation on various combinations of the unit vectors
        $\hat{\vc{x}}$, $\hat{\vc{y}}$, and $\hat{\vc{z}}$. The conventional notation
        for the operation is to put a dot, $\cdot$, between the two vectors, and the
        operation is therefore called
        the \emph{dot product}.\index{dot product}\index{vector!dot product}
        Rotational invariance requires that we handle the three coordinate axes
        in the same way, without giving special treatment to any of them, so we must
        have $\hat{\vc{x}}\cdot\hat{\vc{x}}=\hat{\vc{y}}\cdot\hat{\vc{y}}=\hat{\vc{z}}\cdot\hat{\vc{z}}$
        and $\hat{\vc{x}}\cdot\hat{\vc{y}}=\hat{\vc{y}}\cdot\hat{\vc{z}}=\hat{\vc{z}}\cdot\hat{\vc{x}}$.
        This is supposed to be a way of generalizing ordinary multiplication, so for consistency
        with the property $1\times1=1$ of ordinary numbers, the result
        of multiplying a magnitude-one vector by itself had better be the scalar 1, so
        $\hat{\vc{x}}\cdot\hat{\vc{x}}=\hat{\vc{y}}\cdot\hat{\vc{y}}=\hat{\vc{z}}\cdot\hat{\vc{z}}=1$.
        Furthermore, there is no way to satisfy rotational invariance unless we define the mixed
        products to be zero,
        $\hat{\vc{x}}\cdot\hat{\vc{y}}=\hat{\vc{y}}\cdot\hat{\vc{z}}=\hat{\vc{z}}\cdot\hat{\vc{x}}=0$;
        for example, a 90-degree rotation of our frame of reference about the $z$ axis
        reverses the sign of $\hat{\vc{x}}\cdot\hat{\vc{y}}$, but rotational invariance requires
        that $\hat{\vc{x}}\cdot\hat{\vc{y}}$ produce the same result either way, and zero is the
        only number that stays the same when we reverse its sign. Establishing these six
        products of unit vectors suffices to define the operation in general, since any
        two vectors that we want to multiply can be broken down into components, e.g.,
        $(2\hat{\vc{x}}+3\hat{\vc{z}})\cdot\hat{\vc{z}}
        =2\hat{\vc{x}}\cdot\hat{\vc{z}}+3\hat{\vc{z}}\cdot\hat{\vc{z}}=0+3=3$. Thus by
        requiring rotational invariance and consistency with multiplication of ordinary
        numbers, we find that there is only one possible way to define a multiplication
        operation on two vectors that gives a scalar as the result.\footnote{There is, however,
        a different operation, discussed in the next chapter, which multiplies two vectors
        to give a vector.} The dot product has all of the properties we normally associate with
        multiplication, except that there is no ``dot division.''

        \begin{eg}{Dot product in terms of components}
        If we know the components of any two vectors \vc{b} and \vc{c}, we can find
        their dot product:
        \begin{align*}        
                \vc{b}\cdot\vc{c}        &= 
                \left( b_{x}\hat{\vc{x}}+ b_{y}\hat{\vc{y}}+ b_z\hat{\vc{z}}\right)
                \cdot
                \left( c_{x}\hat{\vc{x}}+ c_{y}\hat{\vc{y}}+ c_z\hat{\vc{z}}\right) \\
                        &=  b_x c_x+ b_y c_y+ b_z c_z \qquad .
        \end{align*}
        \end{eg}

        \begin{eg}{Magnitude expressed with a dot product}\label{eg:magnitudedot}
        If we take the dot product of any vector \vc{b} with itself, we find
        \begin{align*}        
                \vc{b}\cdot\vc{b}        &= 
                \left( b_{x}\hat{\vc{x}}+ b_{y}\hat{\vc{y}}+ b_z\hat{\vc{z}}\right)
                \cdot
                \left( b_{x}\hat{\vc{x}}+ b_{y}\hat{\vc{y}}+ b_z\hat{\vc{z}}\right) \\
                        &=  b_{x}^2+ b_{y}^2+ b_z^2 \qquad ,\\
        \intertext{so its magnitude can be expressed as}
                |\vc{b}| &= \sqrt{\vc{b}\cdot\vc{b}} \qquad .
        \end{align*}
        We will often write $b^2$ to mean $\vc{b}\cdot\vc{b}$, when the context makes it
        clear what is intended. For example, we could express kinetic energy as
        $\zu{(1/2)} m|\vc{v}|^2$, $\zu{(1/2)} m\vc{v}\cdot\vc{v}$,
        or $\zu{(1/2)} m v^2$. In the third version, nothing but context tells
        us that $v$ really stands for the magnitude of some vector $\vc{v}$.
        \end{eg}

<% marg(0) %>
<%
  fig(
    'dotgeom',
    %q{The geometric interpretation of the dot product.}
  )
%>
<% end_marg %>
        \begin{eg}{Geometric interpretation}\label{eg:dot-product-interp}
        In figure \figref{dotgeom}, vectors \vc{a}, \vc{b}, and \vc{c} represent the sides of a triangle,
        and $\vc{a}=\vc{b}+\vc{c}$. The law of cosines gives
        \begin{equation*}
                |\vc{c}|^2 = |\vc{a}|^2+|\vc{b}|^2-2|\vc{a}||\vc{b}|\ \zu{cos}\ \theta \qquad .
        \end{equation*}
        Using the result of example \ref{eg:magnitudedot}, we can also write this as
        \begin{align*}
                |\vc{c}|^2        &= \vc{c}\cdot\vc{c} \\
                                        &= (\vc{a}-\vc{b})\cdot(\vc{a}-\vc{b}) \\
                                        &= \vc{a}\cdot\vc{a}+\vc{b}\cdot\vc{b}-2\vc{a}\cdot\vc{b} \qquad .
        \end{align*}
        Matching up terms in these two expressions, we find
        \begin{equation*}
                \vc{a}\cdot\vc{b} = |\vc{a}||\vc{b}|\ \zu{cos}\ \theta \qquad ,
        \end{equation*}
        which is a geometric interpretation for the dot product.         \end{eg}

        The result of example \ref{eg:dot-product-interp} is very useful.
        It gives us a way to find the angle between two
        vectors if we know their components. It can be used to show that the dot product
        of any two perpendicular vectors is zero.
        It also leads to a nifty
        proof that the dot product is rotationally invariant --- up until now I've only
        proved that if a rotationally invariant product exists, the dot product is it ---
        because angles and lengths aren't affected by a rotation, so the right side of the
        equation is rotationally invariant, and therefore so is the left side.

<% marg(30) %>
<%
  fig(
    'breakingtrail',
    %q{%
      Breaking trail, by Walter E. Bohl.
              The pack horse is not doing any work on the pack, because
              the pack is moving in a horizontal line at constant speed, and therefore there is
              no kinetic or gravitational energy being transferred into or out of it.
    }
  )
%>
<% end_marg %>
        I introduced the whole discussion of the dot product by way of generalizing
        the equation $\der E=F\der x$ to three dimensions. In terms
        of a dot product, we have
        \begin{align*}
                \der E        &= \vc{F}\cdot\der\vc{r} \qquad .\\
        \intertext{If \vc{F} is a constant, integrating both sides gives}
                \Delta E        &= \vc{F}\cdot\Delta\vc{r} \qquad .
        \end{align*}
        (If that step seemed like black magic, try writing it out in terms
        of components.) If the force is perpendicular to the motion, as in figure
        \figref{breakingtrail}, then the work done is zero. The pack horse is doing
        work within its own body, but is not doing work on the pack.

        \begin{eg}{Pushing a lawnmower}
        \egquestion
        I push a lawnmower with a force $\vc{F}\zu{=(110 N)}\hat{\vc{x}}-\zu{(40 N)}\hat{\vc{y}}$,
        and the total distance I travel is $\zu{(100 m)}\hat{\vc{x}}$. How much work do I do?

        \eganswer
        The dot product is 11000 $\nunit\unitdot\munit$ = 11000 J.
        \end{eg}

        \label{separatepconsproof}
        A good application of the dot product is to allow us to write a simple, streamlined
        proof of separate conservation of the momentum components. (You can skip the
        proof without losing the continuity of the text.) The argument is
        a generalization of the one-dimensional proof on page \pageref{pconsproof1d},
        and makes the same assumption about the type of system of particles we're
        dealing with. The kinetic energy of one of the particles
        is $(1/2)m\vc{v}\cdot\vc{v}$, and when we
        transform into a different frame of reference moving with velocity \vc{u} relative
        to the original frame, the one-dimensional rule $v\rightarrow v+u$ turns into
        vector addition, $\vc{v}\rightarrow \vc{v}+\vc{u}$. In the new frame of reference,
        the kinetic energy is $(1/2)m(\vc{v}+\vc{u})\cdot(\vc{v}+\vc{u})$. For a system
        of $n$ particles, we have
        \begin{align*}
                K        &= \sum_{j=1}^{n}{\frac{1}{2}m_j(\vc{v}_j+\vc{u})\cdot(\vc{v}_j+\vc{u})} \\
                        &= \frac{1}{2}\left[\sum_{j=1}^{n}{m_j\vc{v}_j\cdot\vc{v}_j}
                                        +2\sum_{j=1}^{n}{m_j\vc{v}_j\cdot\vc{u}}
                                        +\sum_{j=1}^{n}{m_j\vc{u}\cdot\vc{u}}\right]  \qquad .
        \end{align*}
        As in the proof on page \pageref{pconsproof1d}, the first sum is simply the
        total kinetic energy in the original frame of reference, and the last sum is
        a constant, which has no effect
        on the validity of the conservation law. The middle sum can be rewritten as
        \begin{align*}
                2\sum_{j=1}^{n}{m_j\vc{v}_j\cdot\vc{u}}
                        &= 2\ \vc{u}\cdot\sum_{j=1}^{n}{m_j\vc{v}_j}\\
                        &= 2\ \vc{u}\cdot\sum_{j=1}^{n}{\vc{p}_j} \qquad ,
        \end{align*}
        so the only way energy can be conserved for all values of \vc{u} is
        if the vector sum of the momenta is conserved as well.

<% end_sec() %>
<% begin_sec("Gradients and line integrals (optional)") %>\label{gradandlineintegral}
        This subsection introduces a little bit of vector calculus. It can be omitted without
        loss of continuity, but the techniques will be needed in our study of electricity and
        magnetism, and it may be helpful to be exposed to them in easy-to-visualize
        mechanical contexts before applying them to invisible electrical and magnetic
        phenomena.

<% marg(0) %>
<%
  fig(
    'sea-of-arrows-wind',
    %q{An object moves through a field of force.}
  )
%>
<% end_marg %>
        In physics we often deal with fields of force, meaning situations where the force
        on an object depends on its position. For instance, figure \figref{sea-of-arrows-wind}
        could represent a map of the trade winds affecting a sailing ship, or a chart of
        the gravitational forces experienced by a space probe entering a double-star
        system. An object moving under the influence of this force will not necessarily
        be moving in the same direction as the force at every moment. The sailing
        ship can tack against the wind, due to the force from the water on the keel.
        The space probe, if it entered from the top of the diagram at high speed,
        would start to curve around to the right, but its inertia would carry it forward, and
        it wouldn't instantly swerve to match the direction of the gravitational force.
        For convenience, we've defined the gravitational field, \vc{g}, as the force
        \emph{per unit mass}, but that trick only leads to a simplification because the
        gravitational force on an object is proportional to its mass. Since this subsection
        is meant to apply to any kind of force, we'll discuss everything in terms of the
        actual force vector, \vc{F}, in units of newtons.

        If an object moves through the field of force along some curved path from point $\vc{r}_1$
        to point $\vc{r}_2$,
        the force will do a certain amount of work on
        it. To calculate this work, we can 
        break the path up into infinitesimally short segments,
        find the work done along each segment, and add them all up.
        For an object traveling along a nice straight $x$ axis, we use the symbol
        $\der x$ to indicate the length of any infinitesimally short segment.
        In three dimensions, moving along a curve, each segment is
        a tiny vector $\der\vc{r}=\hat{\vc{x}}\der x+\hat{\vc{y}}\der y+\hat{\vc{z}}\der z$.
        The work theorem can be expressed as a dot product, so the work done
        along a segment is $\vc{F}\cdot\der\vc{r}$. We want to integrate this, but
        we don't know how to integrate with respect to a variable that's a vector,
        so let's define a variable $s$ that indicates the distance traveled so far along
        the curve, and integrate with respect to it instead. The expression
        $\vc{F}\cdot\der\vc{r}$ can be rewritten as $|\vc{F}|\:|\der\vc{r}|\:\cos\theta$,
        where $\theta$ is the angle between \vc{F} and $\der\vc{r}$. But
        $|\der\vc{r}|$ is simply $\der s$, so the amount of work done becomes
        \begin{equation*}
                \Delta E = \int_{\vc{r}_1}^{\vc{r}_2}{|\vc{F}| \cos\theta}\ \der s \qquad .
        \end{equation*}
        Both \vc{F} and $\theta$ are functions of $s$.
        As a matter of notation, it's cumbersome to have to write the integral like this.
        Vector notation was designed to eliminate this kind of drudgery. We therefore
        define the line integral\index{line integral}
        \begin{align*}
                \int_C{\vc{F}\cdot\der\vc{r}}
        \end{align*}
        as a way of notating this type of integral. The `C' refers to the curve along which
        the object travels. If we don't know this curve then we typically can't evaluate
        the line integral just by knowing the initial and final positions $\vc{r}_1$ and $\vc{r}_2$.

        The basic idea of calculus is that integration undoes differentiation, and
        vice-versa. In one dimension, we could describe an interaction either in terms of
        a force or in terms of an interaction energy. 
        We could integrate force with respect to position
         to find minus the energy, or
        we could find the force by taking minus the derivative of the energy. In the line
        integral, position is represented by a vector. What would it mean to take a derivative
        with respect to a vector? The correct way to generalize the derivative
        $\der U/\der x$ to three dimensions is to replace it with the following vector,
        \begin{equation*}
                \frac{\der U}{\der x}\hat{\vc{x}}
                +\frac{\der U}{\der y}\hat{\vc{y}}
                +\frac{\der U}{\der z}\hat{\vc{z}} \qquad ,
        \end{equation*}
        called the \emph{gradient}\index{gradient} of $U$, and written with an upside-down
        delta\footnote{The symbol $\nabla$ is called a ``nabla.'' Cool word!}
         like this, $\nabla U$.
        Each of these three derivatives is really what's known as a
        partial derivative.\index{derivative!partial}\index{partial derivative}\label{partial-der} What that means
        is that when you're differentiating $U$ with respect to $x$, you're supposed to treat
        $y$ and $z$ and constants, and similarly when you do the other two derivatives.
        To emphasize that a derivative is a partial derivative, it's customary to write it using
        the symbol $\partial$ in place of the differential d's. Putting all this notation together, we have
        \begin{equation*}
                \nabla U = \frac{\partial U}{\partial x}\hat{\vc{x}}
                +\frac{\partial U}{\partial y}\hat{\vc{y}}
                +\frac{\partial U}{\partial z}\hat{\vc{z}} \qquad \text{[definition of the gradient]} \qquad .
        \end{equation*}
        The gradient looks scary, but it has a very simple physical interpretation. It's a vector
        that points in the direction in which $U$ is increasing most rapidly, and it tells you
        how rapidly $U$ is increasing in that direction. For instance, sperm cells in plants
        and animals find the egg cells by traveling in the direction of the gradient of the
        concentration of certain hormones. When they reach the location 
        of the strongest hormone concentration, they find their destiny. In terms of the
        gradient, the force corresponding to a given interaction energy is
        $\vc{F}=-\nabla U$.

        \begin{eg}{Force exerted by a spring}
        In one dimension, Hooke's law is $U=\zu{(1/2)} kx^2$.
        Suppose we tether one end of a spring to a post, but it's free to stretch and
        swing around in a plane. Let's say its equilibrium length is zero, and let's
        choose the origin of our coordinate system to be at the post.
        Rotational invariance requires that its energy only depend on the magnitude
        of the \vc{r} vector, not its direction, so in two dimensions we have
        $U=\zu{(1/2)} k|\vc{r}|^2
                =\zu{(1/2)} k\left( x^2+ y^2\right)$.
        The force exerted by the spring is then
        \begin{align*}
                \vc{F}        &= -\nabla  U \\
                                &= -\frac{\partial U}{\partial x}\hat{\vc{x}}
                                        -\frac{\partial U}{\partial y}\hat{\vc{y}}\\
                                &= - kx\hat{\vc{x}}- ky\hat{\vc{y}} \qquad .
        \end{align*}
        The magnitude of this force vector is $k|\vc{r}|$, and its direction is toward
        the origin.
        \end{eg}

\backofchapterboilerplate{3}

 % ============================= homework ==============================  

<% end_sec() %>
<% end_sec() %><% begin_hw_sec %>

<% begin_hw('keintermsofp',0) %>__incl(hw/keintermsofp)<% end_hw() %>

<% begin_hw('rowboat') %>__incl(hw/rowboat)<% end_hw() %>

<% begin_hw('gun') %>__incl(hw/gun)<% end_hw() %>

<% begin_hw('funkosity') %>__incl(hw/funkosity)<% end_hw() %>

<% begin_hw('twotoonecollision') %>__incl(hw/twotoonecollision)<% end_hw() %>

<% begin_hw('recoil-double-speed') %>__incl(hw/recoil-double-speed)<% end_hw() %>

<% begin_hw('max-heat-released-in-collision') %>__incl(hw/max-heat-released-in-collision)<% end_hw() %>

<% begin_hw('exhaust',2,{'calc'=>true}) %>__incl(hw/exhaust)<% end_hw() %>

<% begin_hw('no-force') %>__incl(hw/no-force)<% end_hw() %>

<% begin_hw('trailer',0) %>__incl(hw/trailer)<% end_hw() %>

<% begin_hw('elevatortension') %>__incl(hw/elevatortension)<% end_hw() %>

<% begin_hw('fridge-recoil') %>__incl(hw/fridge-recoil)<% end_hw() %>


<% begin_hw('copter') %>__incl(hw/copter)<% end_hw() %>

<% begin_hw('blimp') %>__incl(hw/blimp)<% end_hw() %>
        
<% begin_hw('time-to-brake') %>__incl(hw/time-to-brake)<% end_hw() %>
        
<% begin_hw('oldlady') %>__incl(hw/oldlady)<% end_hw() %>
<% marg(50) %>
<%
  fig(
    'hw-oldlady',
    %q{Problem \ref{hw:oldlady}.}
  )
%>
<% end_marg %>

<% begin_hw('earthaccelup') %>__incl(hw/earthaccelup)<% end_hw() %>

<% begin_hw('gravityvsnormal') %>__incl(hw/gravityvsnormal)<% end_hw() %>

<% marg(0) %>
<%
  fig(
    'elevator',
    %q{Problem \ref{hw:elevator}.}
  )
%>
<% end_marg %>

<% begin_hw('elevator') %>__incl(hw/elevator)<% end_hw() %>

\pagebreak
        
<% begin_hw('tugboat') %>
        A tugboat of mass $m$ pulls a ship of mass $M$, accelerating
        it. Ignore fluid friction acting on their hulls, although
        there will of course need to be fluid friction acting on the
        tug's propellers.\hwendpart
        (a) If the force acting on the tug's propeller is
        $F$, what is the tension, $T$, in the cable connecting the two
        ships? \hwhint{hwhint:tugboat}\answercheck\hwendpart
        (b) Interpret your answer in the special cases of $M=0$ and
        $M=\infty$.
<% end_hw() %>
        
<% begin_hw('kineticstrongerthanstatic') %>__incl(hw/kineticstrongerthanstatic)<% end_hw() %>



<% begin_hw('alphastopping') %>__incl(hw/alphastopping)<% end_hw() %>

<% begin_hw('mass-or-weight') %>__incl(hw/mass-or-weight)<% end_hw() %>
<% marg(30) %>
<%
  fig(
    'hw-mass-or-weight',
    %q{Problem \ref{hw:mass-or-weight}, part c.}
  )
%>
<% end_marg %>

<% begin_hw('sally-spacehound') %>__incl(hw/sally-spacehound)<% end_hw() %>

<% begin_hw('rice-sticks') %>__incl(hw/rice-sticks)<% end_hw() %>

<% begin_hw('rope-over-edge',2,{'calc'=>true}) %>__incl(hw/rope-over-edge)<% end_hw() %>

\noindent \emph{In problems \ref{hw:magnetundercar}-\ref{hw:airplaneforces}, analyze the forces using a table in the
format shown in section \ref{subsec:analysis-of-forces}. Analyze the forces in which the
italicized object participates.}

<% begin_hw('magnetundercar') %>__incl(hw/magnetundercar)<% end_hw() %>

<% begin_hw('nonormal') %>__incl(hw/nonormal)<% end_hw() %>

<% begin_hw('row') %>__incl(hw/row)<% end_hw() %>
<% marg(20) %>
<%
  fig(
    'hw-row',
    %q{Problem \ref{hw:row}.}
  )
%>
\spacebetweenfigs
<%
  fig(
    'hw-airplaneforces',
    %q{Problem \ref{hw:airplaneforces}.}
  )
%>
\spacebetweenfigs
<%
  fig(
    'hw-stacked-blocks',
    %q{Problem \ref{hw:stacked-blocks}}
  )
%>
\spacebetweenfigs
<%
  fig(
    'hw-dogsled',
    %q{Problem \ref{hw:dogsled}.}
  )
%>
<% end_marg %>

<% begin_hw('cowcrush') %>__incl(hw/cowcrush)<% end_hw() %>

<% begin_hw('airplaneforces') %>__incl(hw/airplaneforces)<% end_hw() %>



<% begin_hw('stacked-blocks') %>__incl(hw/stacked-blocks)<% end_hw() %>

<% begin_hw('dogsled') %>__incl(hw/dogsled)<% end_hw() %>

<% begin_hw('third-law-partners') %>__incl(hw/third-law-partners)<% end_hw() %>

<% begin_hw('skidmarks') %>__incl(hw/skidmarks)<% end_hw() %>

<% begin_hw('coasting-skater') %>__incl(hw/coasting-skater)<% end_hw() %>

<% begin_hw('youngmodulus') %>__incl(hw/youngmodulus)<% end_hw() %>

<% marg(0) %>
<%
  fig(
    'hw-bondstretching',
    %q{Problem \ref{hw:cubiclattice}}
  )
%>
<% end_marg %>

<% begin_hw('cubiclattice') %>__incl(hw/cubiclattice)<% end_hw() %>

<% begin_hw('three-pulleys') %>This problem has been deleted.<% end_hw() %>

<% begin_hw('swimbladder',0) %>__incl(hw/swimbladder)<% end_hw() %>



<% begin_hw('maxampatdc') %>__incl(hw/maxampatdc)<% end_hw() %>

\vspace{15mm}

<% begin_hw('qsix') %>__incl(hw/qsix)<% end_hw() %>

\vspace{15mm}

<% begin_hw('braginskii',2) %>__incl(hw/braginskii)<% end_hw() %>



<% begin_hw('firework') %>__incl(hw/firework)<% end_hw() %>

<% begin_hw('hockey-pucks') %>__incl(hw/hockey-pucks)<% end_hw() %>
<% marg(50) %>
<%
  fig(
    'hw-hockey-pucks',
    %q{Problem \ref{hw:hockey-pucks}}
  )
%>
<% end_marg %>

<% begin_hw('planes') %>__incl(hw/planes)<% end_hw() %>

<% begin_hw('tossup') %>__incl(hw/tossup)<% end_hw() %>
        
<% begin_hw('misslettuce') %>__incl(hw/misslettuce)<% end_hw() %>

<% begin_hw('niagara') %>__incl(hw/niagara)<% end_hw() %>



<% begin_hw('baseballpitch') %>__incl(hw/baseballpitch)<% end_hw() %>
<%
  fig(
    'hw-baseballpitch',
    %q{Problem \ref{hw:baseballpitch}.},
    {
      'width'=>'wide',
      'sidecaption'=>true
    }
  )
%>        

<% begin_hw('baseballrange') %>__incl(hw/baseballrange)<% end_hw() %>

<% begin_hw('baseballrangeair',2) %>__incl(hw/baseballrangeair)<% end_hw() %>
        
\pagebreak        

<% begin_hw('walking') %>
        If you walk 35 km at an angle 25\degunit counterclockwise from
        east, and then 22 km at 230\degunit counterclockwise from east,
        find the distance and direction from your starting point to
        your destination. \answercheck
<% end_hw() %>

<% marg(2) %>
<%
  fig(
    'hw-tiptotail',
    %q{Problem \ref{hw:tiptotail}.}
  )
%>
<% end_marg %>
<% begin_hw('tiptotail',0) %>__incl(hw/tiptotail)<% end_hw() %>

<% begin_hw('bangkok',0) %>__incl(hw/bangkok)<% end_hw() %>

<% begin_hw('oppositevanda') %>__incl(hw/oppositevanda)<% end_hw() %>

<%
  fig(
    'hw-fossil',
    %q{Problem \ref{hw:fossil}.},
    {
      'width'=>'wide'
    }
  )
%>
<% begin_hw('fossil') %>__incl(hw/fossil)<% end_hw() %>

<% begin_hw('bird') %>__incl(hw/bird)<% end_hw() %>

\pagebreak

<% marg(0) %>
<%
  fig(
    'hw-pressblock',
    %q{Problem \ref{hw:pressblock}}
  )
%>
<% end_marg %>
<% begin_hw('pressblock') %>__incl(hw/pressblock)<% end_hw() %>

<% begin_hw('skier') %>__incl(hw/skier)<% end_hw() %>

<% begin_hw('cartesianbullet') %>__incl(hw/cartesianbullet)<% end_hw() %>

<% begin_hw('annieoakley') %>__incl(hw/annieoakley)<% end_hw() %>
<% marg(-30) %>
<%
  fig(
    'cargoplane',
    %q{Problem \ref{hw:cargoplane}}
  )
%>
\spacebetweenfigs
<%
  fig(
    'wagon',
    %q{Problem \ref{hw:wagon}}
  )
%>
\spacebetweenfigs
<%
  fig(
    'hw-mixer',
    %q{Problem \ref{hw:mixer}.}
  )
%>
<% end_marg %>

<% begin_hw('cargoplane') %>__incl(hw/cargoplane)<% end_hw() %>

\enlargethispage{\baselineskip}

<% begin_hw('wagon') %>__incl(hw/wagon)<% end_hw() %>

<% begin_hw('reposeasteroid') %>__incl(hw/reposeasteroid)<% end_hw() %>

\enlargethispage{\baselineskip}

\vfill

<% begin_hw('mixer') %>__incl(hw/mixer)<% end_hw() %>

\vfill

<% begin_hw('circularaccelunits',0) %>__incl(hw/circularaccelunits)<% end_hw() %>

\vfill

<% begin_hw('loop') %>__incl(hw/loop)<% end_hw() %>

\vfill

<% begin_hw('anglebetween',0) %>__incl(hw/anglebetween)<% end_hw() %>


<% begin_hw('ropeslopes') %>__incl(hw/ropeslopes)<% end_hw() %>

<% marg(50) %>
<%
  fig(
    'ropeslopes',
    %q{Problem \ref{hw:ropeslopes}.}
  )
%>
<% end_marg %>

\pagebreak

\enlargethispage{\baselineskip}

<% begin_hw('spider-oscillations') %>__incl(hw/spider-oscillations)<% end_hw() %>
<%
  fig(
    'hw-spider-oscillations',
    %q{Problem \ref{hw:spider-oscillations}.},
    {'width'=>'wide'}
  )
%>

\pagebreak

\enlargethispage{\baselineskip}

<% begin_hw('skier-hits-dirt',2) %>__incl(hw/skier-hits-dirt)<% end_hw() %>

<% begin_hw('microwaveice') %>__incl(hw/microwaveice)<% end_hw() %>

<% begin_hw('atwood-redux') %>
Problem \ref{ch:2}-\ref{hw:atwood-energy-sn} on page \pageref{hw:atwood-energy-sn} was intended to be solved using conservation
of energy. Solve the same problem using Newton's laws.
<% end_hw() %>

<% begin_hw('spiral-terminal-velocity',2) %>__incl(hw/spiral-terminal-velocity)<% end_hw() %>

<% begin_hw('bikeloop') %>__incl(hw/bikeloop)<% end_hw() %>

\pagebreak

<% begin_hw('maserati') %>__incl(hw/maserati)<% end_hw() %>

<% begin_hw('board-on-counterrotating-wheels',2) %>__incl(hw/board-on-counterrotating-wheels)<% end_hw() %>
<% marg(20) %>
<%
  fig(
    'hw-board-on-counterrotating-wheels',
    %q{%
      Problem \ref{hw:board-on-counterrotating-wheels}.
    }
  )
%>
<% end_marg %>

\pagebreak

<% begin_hw('climbing-anchors') %>__incl(hw/climbing-anchors)<% end_hw() %>
<% marg(80) %>
<%
  fig(
    'hw-climbing-anchors',
    %q{Problem \ref{hw:climbing-anchors}.}
  )
%>
\spacebetweenfigs
<%
  fig(
    'hw-crevasse',
    %q{Problem \ref{hw:crevasse}.}
  )
%>
<% end_marg %>

<% begin_hw('crevasse') %>__incl(hw/crevasse)<% end_hw() %>

\pagebreak

<%
  fig(
    'hw-layback',
    %q{Problem \ref{hw:layback}.},
    {
      'width'=>'wide',
      'sidecaption'=>true
    }
  )
%>

<% begin_hw('layback') %>__incl(hw/layback)<% end_hw() %>

<% begin_hw('wall-of-death') %>__incl(hw/wall-of-death)<% end_hw() %>
<% marg(73) %>
<%
  fig(
    'hw-wall-of-death',
    %q{Problem \ref{hw:wall-of-death}.}
  )
%>
<% end_marg %>

\pagebreak


<% begin_hw('death-triangle') %>__incl(hw/death-triangle)<% end_hw() %>
<% marg(50) %>
<%
  fig(
    'hw-death-triangle',
    %q{Problem \ref{hw:death-triangle}.}
  )
%>
<% end_marg %>

<% begin_hw('beer-stability') %>__incl(hw/beer-stability)<% end_hw() %>

<% begin_hw('haul-bag') %>__incl(hw/haul-bag)<% end_hw() %>

<% end_hw_sec %>

 % ==================================================================== 
 % ==================================================================== 
 % ==================================================================== 
\begin{exsection}

\extitle{A}{Force and Motion}

Equipment:

\begin{indentedblock}
    2-meter pieces of butcher paper

    wood blocks with hooks

    string

    masses to put on top of the blocks to increase friction

    spring scales (preferably calibrated in Newtons)
\end{indentedblock}

Suppose a person pushes a crate, sliding it across the floor
at a certain speed, and then repeats the same thing but at a
higher speed. This is essentially the situation you will act
out in this exercise. What do you think is different about
her force on the crate in the two situations? Discuss this
with your group and write down your hypothesis:

\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_

1. First you will measure the amount of friction between the
wood block and the butcher paper when the wood and paper
surfaces are slipping over each other. The idea is to attach
a spring scale to the block and then slide the butcher paper
under the block while using the scale to keep the block from
moving with it. Depending on the amount of force your spring
scale was designed to measure, you may need to put an extra
mass on top of the block in order to increase the amount of
friction. It is a good idea to use long piece of string to
attach the block to the spring scale, since otherwise one
tends to pull at an angle instead of directly horizontally.

First measure the amount of friction force when sliding the
butcher paper as slowly as possible:\_\_\_\_\_\_\_\_\_

Now measure the amount of friction force at a significantly
higher speed, say 1 meter per second. (If you try to go too
fast, the motion is jerky, and it is impossible to get an
accurate reading.) \_\_\_\_\_\_\_\_\_

Discuss your results. Why are we justified in assuming that
the string's force on the block (i.e., the scale reading) is
the same amount as the paper's frictional force on the block?

2. Now try the same thing but with the block moving and the
paper standing still. Try two different speeds.

Do your results agree with your original hypothesis? If not,
discuss what's going on. How does the block ``know'' how fast to go?

 % ==================================================================== 
 % ==================================================================== 
 % ==================================================================== 
\vfill\pagebreak[4]

\extitle{B}{Vibrations}

\noindent Equipment:
\begin{itemize}
\item air track and carts of two different masses
\item springs
\item spring scales
\end{itemize}

<%
  fig(
    'ex-vibrations',
    '',
    {
      'width'=>'wide',
      'anonymous'=>true,
      'float'=>false
    }
  )
%>

\noindent Place the cart on the air track and attach springs so that it can vibrate.

1. Test whether the period of vibration depends on amplitude. Try at
least two moderate amplitudes, for which the springs do not go slack,
and at least one amplitude that is large enough so that they do go
slack.

2. Try a cart with a different mass. Does the period change by the
expected factor, based on the equation $T=2\pi\sqrt{m/k}$?

3. Use a spring scale to pull the cart away from equilibrium, and
make a graph of force versus position. Is it linear? If so, what is
its slope?

4. Test the equation $T=2\pi\sqrt{m/k}$ numerically.

\vfill\pagebreak[4]

 % ==================================================================== 
 % ==================================================================== 
 % ====================================================================  

\extitle{C}{Worksheet on Resonance}

1. Compare the oscillator's energies at A, B, C, and D.

\widefignocaptionnofloat{ex-resonance-1}

2. Compare the Q values of the two oscillators.

\widefignocaptionnofloat{ex-resonance-2}

3. Match the x-t graphs in \#2 with the amplitude-frequency graphs below.

\widefignocaptionnofloat{ex-resonance-3}

\vfill\pagebreak[4]

 % ==================================================================== 
 % ==================================================================== 
 % ====================================================================  

\vspace{100mm}

Exercise D is on the following two pages.

\pagebreak

\extitle{D}{Vectors and Motion}

Each diagram on page \pageref{fig:ex-knightish} shows the motion of an object in
an $x-y$ plane. Each dot is one location of the object at
one moment in time. The time interval from one dot to the
next is always the same, so you can think of the vector that
connects one dot to the next as a $\vc{v}$ vector, and subtract
to find $\Delta\vc{v}$ vectors.

1. Suppose the object in diagram 1 is moving from the top
left to the bottom right. Deduce whatever you can about the
force acting on it. Does the force always have the same
magnitude? The same direction?

Invent a physical situation that this diagram could represent.

What if you reinterpret the diagram, and reverse the
object's direction of motion?

2. What can you deduce about the force that is acting in diagram 2?

Invent a physical situation that diagram 2 could represent.

3. What can you deduce about the force that is acting in diagram 3?

Invent a physical situation.

\vfill\pagebreak[4]

\fullpagewidthfignocaption{ex-knightish}\label{fig:ex-knightish}

\vfill\pagebreak[4]

\end{exsection}
<% end_chapter() %>