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00:00:03,250 --> 00:00:06,260
We've seen previously that
if a rope is under tension

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and we approximate
the rope is massless

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that the tension is then uniform
everywhere along the rope,

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even if the rope is accelerated.

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00:00:13,680 --> 00:00:17,370
However, if the rope
has nonzero mass,

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then its tension will
vary along its length.

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We can see that by
looking at this example.

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Imagine we have a massive
rope of length l suspended

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from a ceiling.

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Now, it's easy to see that,
at the top of the rope,

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a little element of
rope right at the top

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has to support the entire
weight of the entire rope.

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So if I were to examine just a
little piece at the top here,

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the weight of the entire rope
would be acting downwards.

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And for the rope to
remain stationary,

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there has to be
a tension upward.

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I'll call this t-top, because
this is at the top of the rope.

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And so we see
immediately from that

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that at the top of
the rope, the tension

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is just equal to mg, where m
is the mass of the entire rope.

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Similarly, if we asked
what the tension is

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at the bottom of the
rope, an element right

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at the bottom of the rope
isn't supporting any weight,

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because there is
no weight below it.

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And so at the
bottom, the tension

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is 0, because there
is no weight pulling

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on the bottom of the rope.

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So the tension is going to vary
from mg at the top down to 0

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at the bottom.

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And if we wanted to work
out at some distance

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from the ceiling--
let's call it x--

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at some position x-- say
here, what the tension is

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at that point-- one way of
figuring that out would just

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be imagine we cut the
rope at this point

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and then asked
what tension force

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would be necessary to support
this bottom part of the rope.

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This length is l
minus x, so the mass

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of that fraction of the
rope is l minus x over l.

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And the mass of that
fraction of the rope

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is that fraction times m.

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And so the tension at
this point, t at x,

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is equal to the weight
of the length of rope

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below that point.

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So that's this mass times g.

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And I can rewrite that as
1 minus x over l times mg.

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And this gives me mg
if I put in x equals 0,

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and it gives me 0 if
I put in x equals l.

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So that's one way of
figuring this out.

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I'd like to use this
same example, though,

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to introduce another, more
elegant way of analyzing

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what the tension as a function
of a position on this rope.

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The advantage-- so this is
going to be a little bit more

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complicated, but it's a
much more powerful method,

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and it's one that
we can generally

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use for any continuous
distribution of mass as opposed

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to a point mass.

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So let's consider the same
example again of a hanging mass

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of rope of length l and mass m.

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Here's the length
l and the mass m.

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And the approach
we're going to use

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is called differential analysis.

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It's a technique from
calculus, and it's

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applicable to any continuous
distribution of mass.

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What we're going
to do is imagine

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our continuous
distribution of mass

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is made up of a whole bunch of
little pieces, little elements,

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examine f equals ma acting
on a single element,

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and then generalize
to the entire mass.

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So let's do that here.

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What we'll do is we'll examine
a piece at some position x.

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So I'm measuring x from the top.

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And let's say I define
a little element here,

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which is that of position x and
which has an extent delta x.

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So this thickness is
delta x, and we'll

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call the mass of that little
piece, that little element,

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delta m.

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Now, one thing I want to
point out to begin with

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is that we want to pick an
element that's somewhere

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in the middle of
the distribution,

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not at one end or the other.

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The endpoints are
special cases, so we

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want to pick a general case.

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So choose some x somewhere in
the middle of our distribution

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which has a finite extent.

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That extent is delta x.

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In this case, we'll assume
that that piece delta x

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has a mass delta m.

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So let's blow that up
here and analyze it.

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So here is my element.

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And it's going to have--
let's analyze what

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the forces acting on it are.

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So there's gravity
acting downwards,

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which will be delta m times g.

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There is a tension
acting upwards,

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exerted by the rope above it.

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I'll write that as t of x,
because it's at the location x.

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There's also a tension exerted
in the downward direction

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by the remainder of the
rope below the mass element.

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And I'll call that
t of x plus delta x.

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We expect the tension
to vary along the rope.

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And because this element
does have a finite extent,

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the tension at the
bottom is going

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to be slightly different
than the tension at the top-

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so t of x upwards, t of
x plus delta x downwards,

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and then the weight downwards.

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So let's now-- that's
our free body diagram.

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Let's write down
Newton's second law,

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f equals ma for
that mass element.

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So in the positive
direction, which is downward,

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we have t of x plus
delta x plus delta mg.

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And then in the upward
or minus direction,

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we have minus t of x.

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So those are the
combined forces.

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Now, this rope is suspended.

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It's not moving.

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And so mass object
acceleration is just 0.

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I'm going to rearrange that.

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I can write that as t of x
plus delta x minus t of x

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is equal to minus
delta m times g.

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And by the way,
let me remind you--

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if this were a massless rope,
then delta n would be 0.

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And so the right-hand
side would be 0,

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and the tension
would be uniform.

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We'd have the same
tension above and below.

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But because the
rope does have mass,

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and in particular, this element
has a nonzero mass delta m,

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there is a difference
in the tensions.

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OK.

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Now, our delta m can
be represented in terms

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of what this length is.

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Notice that that mass, delta
m-- so note that delta m

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is just a fraction
of the total mass

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that's in that
particular mass element.

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Well, the fraction
of the total rope

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is just the length of
this element, delta x,

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divided by the length of
the entire rope, which is l.

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So that's the fraction,
and I multiply that

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by the total mass.

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So that is my mass delta
m in terms of the length.

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So now I can rewrite
this equation

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as t of x plus
delta x minus t of x

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is equal to minus delta
x m over l times g.

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Now, I'm going to
rearrange this by dividing

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both sides by delta x.

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I'll do that over here.

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So we have t of x
plus delta x minus t

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of x divided by delta x is
equal to minus mg over l.

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This tells us how the
tension is varying

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over this small but finite-sized
mass element, delta x.

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Now I'd like to
examine what happens

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if I go to the limit of
a small-massed element--

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the limit as delta x goes
to 0, or in other words,

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the limit of an infinitesimally
small mass element.

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So I'm going to take the
limit of this equation

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as delta x approaches 0.

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Now, the left-hand side of this
equation should look familiar.

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It's just an expression for
the derivative of the tension

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t as a function of position.

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So I can write that
as dt dx, and that's

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equal to minus mg over l.

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This is an example of a
differential equation,

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or an equation that
involves a derivative.

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This particular
differential equation

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can be solved very simply
by a technique called

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the separation of
variables, where I just

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take each part of the
integral-- the dt and the dx--

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and put it on different
sides of the equation

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and then integrate both
sides of the equation

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to find the solution.

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So in this case, I'll multiply
both sides of the equation

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by dx.

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So I get dt on
the left-hand side

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is equal to minus mg over l dx.

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And now I want to
integrate both sides.

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So I'll integrate this side.

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And remember, mg
over l is a constant,

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so I can keep it
outside the integral.

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And I'll integrate this side.

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I'm going to do a
definite integral

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over the continuous
distribution that I'm studying.

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So on the right-hand side, I'm
going to start at x equals 0

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and go to my position
of interest, which is x.

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00:10:37,466 --> 00:10:38,840
And to avoid
confusion, I'm going

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to call the integration
viable dx prime.

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00:10:40,680 --> 00:10:42,280
This is a dummy variable.

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So x prime here represents
all the values of position,

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ranging from my first endpoint
0 up to my other endpoint x.

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So x here represents
a particular position,

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whereas dx prime
is a placeholder

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for all the positions
between the two

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00:11:01,330 --> 00:11:04,820
endpoints in my infinite
sum, which is an integral.

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00:11:04,820 --> 00:11:06,440
So that's on the
right-hand side.

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00:11:06,440 --> 00:11:09,590
On the left-hand side, I'm
integrating the tension t,

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with respect to the tension t.

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00:11:11,130 --> 00:11:14,080
My limits need to
correspond to the limits

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00:11:14,080 --> 00:11:16,160
on the right-hand side integral.

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So on the right-hand side, my
lower limit is at x equals 0.

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00:11:19,750 --> 00:11:23,010
So on the left-hand
side, I want to have

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00:11:23,010 --> 00:11:26,400
my lower limit be the tension
at that position x equals 0.

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00:11:26,400 --> 00:11:29,990
So that's t of 0.

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00:11:29,990 --> 00:11:32,880
And then the upper
limit of the integral

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00:11:32,880 --> 00:11:35,930
is the tension at the upper
limit of the position integral,

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00:11:35,930 --> 00:11:38,754
so that's t of x.

204
00:11:38,754 --> 00:11:40,170
And again, to avoid
confusion, I'm

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00:11:40,170 --> 00:11:42,909
actually going to call the
integration variable dt prime.

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00:11:42,909 --> 00:11:43,950
This is a dummy variable.

207
00:11:43,950 --> 00:11:46,360
It's a placeholder
for all the values

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00:11:46,360 --> 00:11:49,560
of tension from the
tension at x equals 0

209
00:11:49,560 --> 00:11:53,860
to the tension at my
position of interest x.

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00:11:53,860 --> 00:11:56,370
And so this integral
now tells me

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00:11:56,370 --> 00:11:59,790
how the tension is varying
in a continuous fashion

212
00:11:59,790 --> 00:12:02,190
along this continuous
mass distribution.

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00:12:02,190 --> 00:12:05,340
So now I can evaluate
both integrals.

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00:12:05,340 --> 00:12:06,750
On this side, I
have the integral

215
00:12:06,750 --> 00:12:09,410
of a constant with
respect to the tension.

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00:12:09,410 --> 00:12:17,850
And so that's just going to
give me t of x minus t of 0,

217
00:12:17,850 --> 00:12:22,700
and that's equal
to minus mg over l.

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00:12:22,700 --> 00:12:28,930
The interval of dx prime
from 0 to x is just x.

219
00:12:28,930 --> 00:12:34,410
So this tells me how the tension
changes from the endpoint

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00:12:34,410 --> 00:12:37,810
to some arbitrary position x.

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00:12:37,810 --> 00:12:41,240
If I want to actually
solve for t of x,

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00:12:41,240 --> 00:12:44,241
I need to specify what the
tension is at the endpoint.

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00:12:44,241 --> 00:12:46,240
But we know what the
tension is at the endpoint.

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00:12:46,240 --> 00:12:47,400
We found that earlier.

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00:12:47,400 --> 00:12:51,960
We solved from the simple
argument that at the endpoint,

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the tension here is
just equal to the weight

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00:12:56,410 --> 00:12:59,700
of the entire length of
rope below that point.

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So we know that t at x
equals 0 is just equal to mg.

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00:13:08,550 --> 00:13:11,570
And therefore, the
tension at position

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00:13:11,570 --> 00:13:15,800
x is just-- if I bring
[INAUDIBLE] of 0 to this side

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00:13:15,800 --> 00:13:26,050
is just mg minus mg x over l.

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00:13:26,050 --> 00:13:35,640
And so I can just write that
as mg times 1 minus x over l.

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00:13:35,640 --> 00:13:38,520
So that tells me
what the tension

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00:13:38,520 --> 00:13:40,460
is as a function of position.

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00:13:40,460 --> 00:13:44,810
Note again that if I put in
x equals 0, I just get mg.

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00:13:44,810 --> 00:13:49,370
If I put in x equals l,
I get the tension is 0.

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00:13:49,370 --> 00:13:53,500
And for any point in between,
we see that the tension varies

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00:13:53,500 --> 00:13:56,890
smoothly between mg and 0.

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00:13:56,890 --> 00:13:58,930
This is exactly the same
result we found earlier

240
00:13:58,930 --> 00:14:02,500
by just cutting the
rope at x and asking

241
00:14:02,500 --> 00:14:05,040
how do we balance the
weight of the remaining

242
00:14:05,040 --> 00:14:07,030
rope below that point.

243
00:14:07,030 --> 00:14:08,690
But the advantage
of this technique

244
00:14:08,690 --> 00:14:10,580
which is a little
bit more complicated,

245
00:14:10,580 --> 00:14:13,630
is that it's a very
powerful technique

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applicable to any continuous
distribution of mass.

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So in this specific problem, if
instead of the uniform density

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rope that we had here, imagine
we had a clumpy rope where

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the mass of the rope wasn't
distributed smoothly,

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but there were clumps,
little parts of the rope that

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were heavier than others,
and so the density varied

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with position along the rope.

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In that case, we would
represent that when

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00:14:40,590 --> 00:14:43,420
we were writing what our
mass element delta m is.

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00:14:43,420 --> 00:14:47,230
Delta m, instead of just
being delta x over l times m,

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where here, delta x over
l was the uniform density

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of the rope, we
would have to put

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in some position-dependent
density.

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So delta n would
depend upon position.

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And then when I did this
integral, instead of

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a constant out front with
my integral of dx prime,

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I would have some thing
that was a function of x,

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and I'd get a different
value for the integral.

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But the technique
would still work.

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So this differential
analysis technique

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is applicable to any system
where we have a continuous mass

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distribution.

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And we're going to actually use
this technique over and over

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again in this course.

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00:15:21,850 --> 00:15:23,300
We'll see it a number of times.

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This is just the first
instance we're using it.

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We wanted to introduce you
carefully to the approach.