WEBVTT
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We've seen previously that
if a rope is under tension
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and we approximate
the rope is massless
00:00:08.590 --> 00:00:11.620
that the tension is then uniform
everywhere along the rope,
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even if the rope is accelerated.
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
00:00:47.070 --> 00:00:50.720
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
00:01:14.780 --> 00:01:19.720
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
00:01:24.611 --> 00:01:26.860
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
00:01:40.229 --> 00:01:41.880
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
00:01:45.700 --> 00:01:46.520
at the bottom.
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And if we wanted to work
out at some distance
00:01:49.509 --> 00:01:53.370
from the ceiling--
let's call it x--
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at some position x-- say
here, what the tension is
00:01:56.660 --> 00:01:59.610
at that point-- one way of
figuring that out would just
00:01:59.610 --> 00:02:02.540
be imagine we cut the
rope at this point
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and then asked
what tension force
00:02:05.140 --> 00:02:09.253
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
00:03:13.210 --> 00:03:15.780
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
00:08:09.030 --> 00:08:21.830
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,
00:09:08.440 --> 00:09:10.980
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
00:09:45.210 --> 00:09:47.685
can be solved very simply
by a technique called
00:09:47.685 --> 00:09:50.030
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
00:09:55.440 --> 00:09:57.610
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.
00:10:21.033 --> 00:10:23.210
And I'll integrate this side.
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I'm going to do a
definite integral
00:10:25.180 --> 00:10:28.670
over the continuous
distribution that I'm studying.
00:10:28.670 --> 00:10:34.370
So on the right-hand side, I'm
going to start at x equals 0
00:10:34.370 --> 00:10:37.466
and go to my position
of interest, which is x.
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.
00:10:40.680 --> 00:10:42.280
This is a dummy variable.
00:10:42.280 --> 00:10:48.900
So x prime here represents
all the values of position,
00:10:48.900 --> 00:10:54.030
ranging from my first endpoint
0 up to my other endpoint x.
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So x here represents
a particular position,
00:10:56.410 --> 00:10:59.290
whereas dx prime
is a placeholder
00:10:59.290 --> 00:11:01.330
for all the positions
between the two
00:11:01.330 --> 00:11:04.820
endpoints in my infinite
sum, which is an integral.
00:11:04.820 --> 00:11:06.440
So that's on the
right-hand side.
00:11:06.440 --> 00:11:09.590
On the left-hand side, I'm
integrating the tension t,
00:11:09.590 --> 00:11:11.130
with respect to the tension t.
00:11:11.130 --> 00:11:14.080
My limits need to
correspond to the limits
00:11:14.080 --> 00:11:16.160
on the right-hand side integral.
00:11:16.160 --> 00:11:19.750
So on the right-hand side, my
lower limit is at x equals 0.
00:11:19.750 --> 00:11:23.010
So on the left-hand
side, I want to have
00:11:23.010 --> 00:11:26.400
my lower limit be the tension
at that position x equals 0.
00:11:26.400 --> 00:11:29.990
So that's t of 0.
00:11:29.990 --> 00:11:32.880
And then the upper
limit of the integral
00:11:32.880 --> 00:11:35.930
is the tension at the upper
limit of the position integral,
00:11:35.930 --> 00:11:38.754
so that's t of x.
00:11:38.754 --> 00:11:40.170
And again, to avoid
confusion, I'm
00:11:40.170 --> 00:11:42.909
actually going to call the
integration variable dt prime.
00:11:42.909 --> 00:11:43.950
This is a dummy variable.
00:11:43.950 --> 00:11:46.360
It's a placeholder
for all the values
00:11:46.360 --> 00:11:49.560
of tension from the
tension at x equals 0
00:11:49.560 --> 00:11:53.860
to the tension at my
position of interest x.
00:11:53.860 --> 00:11:56.370
And so this integral
now tells me
00:11:56.370 --> 00:11:59.790
how the tension is varying
in a continuous fashion
00:11:59.790 --> 00:12:02.190
along this continuous
mass distribution.
00:12:02.190 --> 00:12:05.340
So now I can evaluate
both integrals.
00:12:05.340 --> 00:12:06.750
On this side, I
have the integral
00:12:06.750 --> 00:12:09.410
of a constant with
respect to the tension.
00:12:09.410 --> 00:12:17.850
And so that's just going to
give me t of x minus t of 0,
00:12:17.850 --> 00:12:22.700
and that's equal
to minus mg over l.
00:12:22.700 --> 00:12:28.930
The interval of dx prime
from 0 to x is just x.
00:12:28.930 --> 00:12:34.410
So this tells me how the tension
changes from the endpoint
00:12:34.410 --> 00:12:37.810
to some arbitrary position x.
00:12:37.810 --> 00:12:41.240
If I want to actually
solve for t of x,
00:12:41.240 --> 00:12:44.241
I need to specify what the
tension is at the endpoint.
00:12:44.241 --> 00:12:46.240
But we know what the
tension is at the endpoint.
00:12:46.240 --> 00:12:47.400
We found that earlier.
00:12:47.400 --> 00:12:51.960
We solved from the simple
argument that at the endpoint,
00:12:51.960 --> 00:12:56.410
the tension here is
just equal to the weight
00:12:56.410 --> 00:12:59.700
of the entire length of
rope below that point.
00:13:03.000 --> 00:13:08.550
So we know that t at x
equals 0 is just equal to mg.
00:13:08.550 --> 00:13:11.570
And therefore, the
tension at position
00:13:11.570 --> 00:13:15.800
x is just-- if I bring
[INAUDIBLE] of 0 to this side
00:13:15.800 --> 00:13:26.050
is just mg minus mg x over l.
00:13:26.050 --> 00:13:35.640
And so I can just write that
as mg times 1 minus x over l.
00:13:35.640 --> 00:13:38.520
So that tells me
what the tension
00:13:38.520 --> 00:13:40.460
is as a function of position.
00:13:40.460 --> 00:13:44.810
Note again that if I put in
x equals 0, I just get mg.
00:13:44.810 --> 00:13:49.370
If I put in x equals l,
I get the tension is 0.
00:13:49.370 --> 00:13:53.500
And for any point in between,
we see that the tension varies
00:13:53.500 --> 00:13:56.890
smoothly between mg and 0.
00:13:56.890 --> 00:13:58.930
This is exactly the same
result we found earlier
00:13:58.930 --> 00:14:02.500
by just cutting the
rope at x and asking
00:14:02.500 --> 00:14:05.040
how do we balance the
weight of the remaining
00:14:05.040 --> 00:14:07.030
rope below that point.
00:14:07.030 --> 00:14:08.690
But the advantage
of this technique
00:14:08.690 --> 00:14:10.580
which is a little
bit more complicated,
00:14:10.580 --> 00:14:13.630
is that it's a very
powerful technique
00:14:13.630 --> 00:14:16.540
applicable to any continuous
distribution of mass.
00:14:16.540 --> 00:14:21.690
So in this specific problem, if
instead of the uniform density
00:14:21.690 --> 00:14:25.570
rope that we had here, imagine
we had a clumpy rope where
00:14:25.570 --> 00:14:29.670
the mass of the rope wasn't
distributed smoothly,
00:14:29.670 --> 00:14:33.730
but there were clumps,
little parts of the rope that
00:14:33.730 --> 00:14:37.100
were heavier than others,
and so the density varied
00:14:37.100 --> 00:14:38.841
with position along the rope.
00:14:38.841 --> 00:14:40.590
In that case, we would
represent that when
00:14:40.590 --> 00:14:43.420
we were writing what our
mass element delta m is.
00:14:43.420 --> 00:14:47.230
Delta m, instead of just
being delta x over l times m,
00:14:47.230 --> 00:14:50.330
where here, delta x over
l was the uniform density
00:14:50.330 --> 00:14:51.800
of the rope, we
would have to put
00:14:51.800 --> 00:14:54.070
in some position-dependent
density.
00:14:54.070 --> 00:14:57.550
So delta n would
depend upon position.
00:14:57.550 --> 00:15:00.110
And then when I did this
integral, instead of
00:15:00.110 --> 00:15:03.010
a constant out front with
my integral of dx prime,
00:15:03.010 --> 00:15:05.812
I would have some thing
that was a function of x,
00:15:05.812 --> 00:15:07.770
and I'd get a different
value for the integral.
00:15:07.770 --> 00:15:09.950
But the technique
would still work.
00:15:09.950 --> 00:15:12.720
So this differential
analysis technique
00:15:12.720 --> 00:15:17.140
is applicable to any system
where we have a continuous mass
00:15:17.140 --> 00:15:17.880
distribution.
00:15:17.880 --> 00:15:20.960
And we're going to actually use
this technique over and over
00:15:20.960 --> 00:15:21.850
again in this course.
00:15:21.850 --> 00:15:23.300
We'll see it a number of times.
00:15:23.300 --> 00:15:25.600
This is just the first
instance we're using it.
00:15:25.600 --> 00:15:29.420
We wanted to introduce you
carefully to the approach.