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When we take a spring,

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something that we are
so familiar with now,

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and the spring has length l
in a relaxed state,

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spring constant k,

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I can extend the spring
with some force that I apply.

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The spring, then, will counter
it with the spring force

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and it will be
in equilibrium there.

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I call this the zero position,
and let's call this now delta l

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instead of x,
which we have done before.

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If I double the force,
delta l will double.

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Hooke's Law says that the
force is linear with delta l;

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in other words, delta l
is proportional with F.

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Nothing new,
as long as Hooke's Law holds.

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If I make the spring
twice as long,

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I would get double
the extension,

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because when I have
two springs in series,

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each one, under
the influence of this force,

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will get longer by this amount.

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Since I have two springs
in series,

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I will get twice delta l.

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So delta l is also proportional
to the length of my spring.

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If I take two springs parallel,
one like so and one like so,

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relaxed length l,
both the same spring constant k,

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and if now I apply
a force on it,

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then each one of these spring
forces is only half this one.

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Together they counter
this force.

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In other words,
the extension delta l

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that I obtain
from this external force

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is now only half as much
as it would be with one spring,

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and if I had three springs
parallel, all identical,

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I would only get one-third of
the extension for a given force.

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In other words, delta l is
also inversely proportional

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to the number of springs
that I have,

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assuming that they
are identical springs.

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Now I'm going to use
a rod or wire

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which has cross-sectional area A
and length l,

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and I'm going to apply
a force here.

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As a result of that force

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it will get longer
by a certain amount delta l,

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exactly like the spring.

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And clearly, when I make
this force stronger,

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delta l will increase, and
as long as Hooke's Law holds

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that the spring force provided
by the rod balances this out--

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provided that the spring force
is linearly proportional

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with delta l--

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I have again that delta l will
be proportional with the force.

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Double the force,
I get twice delta l.

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If I put two
of these rods together--

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so I double the length
of the rod--

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then clearly I will get
twice delta l,

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because each rod
will experience the force,

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each rod will get longer
by delta l,

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and so two rods will get
longer by two delta l.

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So again, delta l is
proportional with l.

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Suppose, now, I have two of
these rods next to each other--

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notice the parallel with
the two parallel springs here--

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and I apply a force F.

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Then the spring force
on each one of them--

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if I call that
the spring force--

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will only have to be half
to counter this force.

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So they both have
a cross-sectional area A,

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and so now with double
the cross-sectional area,

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I get only half of delta l.

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And so now
we have a situation

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that if I made a rod whereby
this was 2A, just one rod,

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which is completely
equivalent to this situation,

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I'm only getting half delta l
for a given force.

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And so now we have
that delta l is proportional...

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inversely proportional to the
cross-sectional area of the rod.

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So now we can make up
the balance, and we can say,

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"Aha! Delta l is proportional
to the force, proportional to l,

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"and inversely proportional
to the area,

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the cross-sectional area."

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So F divided by A is then
proportional to delta l over l,

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and that proportionality
constant we give a name,

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and that is capital Y, and
that is called Young's modulus.

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So this is Young's modulus.

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F over A, which has
the dimension of pressure--

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force per unit area--
is what we call stress.

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And delta l over l,
which is dimensionless,

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we call that strain.

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If we compare two rods

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with different values
of Young's modulus,

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then the one with the smaller
value of Y for the same stress

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will give you a larger strain.

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In other words,
it's easier to make it longer.

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If Young's modulus is very high,
then the rod is extremely stiff.

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Then it is very difficult
to make the rod longer.

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I have here for you some numbers
which, of course, are on the Web

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so you don't have to copy them.

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And you have Young's modulus
there for various metals,

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and I also have it down there
for nylon,

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and today we will work
with that quite extensively.

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We could first do
a simple example

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just to get a feeling
for what is at stake here.

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I can take a rod with a radius
r, which is 0.5 centimeters.

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That would give me
a cross-sectional area

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of eight times ten
to the minus five square meters.

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So, yay thick, the rod.

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I make it very simple--
I make the length one meter,

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and I hang on the rod,
at the bottom,

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a mass M, which, let's say,
is 500 kilograms.

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Do I want 500 kilograms?

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Yes, I want 500 kilograms.

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In other words, the force
which I pull on the rod

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is about 5,000 newtons.

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So I can now ask myself

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how much longer
is this rod going to be?

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So we're going to have
that delta l is going to be

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F divided by A times
l divided by Young's modulus.

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And we do know what F is,

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we know what l is
and we know what A is.

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So in our case, this number is
6.3 times ten to the seventh

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for the numbers that we have.

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And so now we can look
at steel as an example.

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So we take steel.

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Y is 20 times ten to the tenth
newtons per square meter.

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And we substitute that in here,

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and we find that we get
an extension delta l

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which, I believe, is only
a third of a millimeter--

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indeed, 0.3 millimeters.

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Think about this:
steel rod, a centimeter thick,

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and it has a length
of one meter,

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and I hang 500 kilograms on it

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and it only gets longer
by three-tenths of a millimeter.

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You couldn't even see that.

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However, if you go
and make the rod,

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which would be really a rope...

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If you make it out of nylon,
which has a Young's modulus...

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Oh, well, it's 55 times lower.

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I don't have to use
that number.

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It's about 55 times lower,

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so the delta l will
then be 55 times larger.

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And so, instead
of a third of a millimeter,

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I will get something
like 17 millimeters.

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Andthat you can see
with your naked eye.

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I take a rope
yay thick of nylon.

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I hang 500 kilograms on it

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and I will see it gets longer
by 1.7 centimeters.

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You can see that right
in front of your eyes.

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If I start adding
more weight on these bars,

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then very interesting things
are going to happen

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as we will discuss today.

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One thing which is obvious
that will ultimately happen

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if you keep loading down,
keep adding mass, it will break.

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And the breaking point is given
there in the third column,

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and we call that
the ultimate tensile strength.

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So when F over A-- when this
value-- becomes too large,

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that's what it comes down to,
then it will simply break.

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If we take the case that
we have here, the steel rod,

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for which F over A is 6.3
times ten to the seventh--

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because l is one, remember--

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then you can look there under
steel that it wouldn't break.

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We would be very safe,
because it wouldn't break

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until F over A becomes
five times ten to the eighth.

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So we're a factor of ten away
from that-- there's no problem.

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Even nylon would be very safe,
because that wouldn't break

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until the stress is three
times ten to the eighth.

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However, if we chose aluminum,

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if we made the bar out of
aluminum, then you're very close

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to this number 6.3 times
ten to the seventh.

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And so if you add a little bit
more mass than 500 kilograms,

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your aluminum bar would break.

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So you see, the properties
of these metals

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are very, very different indeed.

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Now, before it breaks,
we reach a situation

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that the strain is no longer
proportional to the stress.

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In other words,
we abandon Hooke's Law,

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as I will show you also today
with a demonstration.

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The material begins to deform.

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It begins to flow, in a way.

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And even when you take
the weights off,

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it will no longer have
its original length.

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It will not return
to the original length

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but it will be much longer.

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And I will try
to sketch you here

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what a curve of stress versus
strain typically looks like.

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So here I'm going to plot delta
l over l, and here F over A.

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So this is the stress
and this is the strain.

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In the beginning, you will see
a portion that is linear.

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That's Hooke's Law.

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And then,
when you keep adding force--

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which we will do by gravity, we
will just hang weights on it--

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then it starts to bend over,
up to a point here

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which we call the elastic limit.

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And even though this portion
is no longer linear--

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so even though Hooke's Law
no longer holds--

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still, if you take the weight
off the wire, off the rod,

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it will still come back to zero.

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Once you are past this point,
that is no longer the case.

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You will now see also

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that a small increase of stress
will give a huge strain,

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so the rope will... the rod will
get very long all of a sudden

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with a little bit
of extra stress.

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And if now you were to take
the weight off the rod,

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it would not come back to zero

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but it would come back
somewhere to here

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so it's permanently deformed.

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And in general the rod gets hot,
and the work that you have done

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goes into heat and goes
into the deformation of the rod.

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And so it goes like this
and then it goes like this

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00:13:07 --> 00:13:13
and here it breaks.

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At this value for F over A,

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which is that third column
there, it will break.

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00:13:16 --> 00:13:22

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I'd like to discuss with you
this horizontal portion.

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With a little bit of luck

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we may actually be able to see
that with my demonstration,

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but it's hard to get
exactly at that point.

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If this part of the curve
is horizontal, it means

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that without any increase
of F over A,

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the wire will continue
to get longer and longer,

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and we call that plastic flow.

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So this whole portion
here is plastic flow.

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It almost becomes like a liquid.

222
00:13:57 --> 00:14:03

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00:14:00 --> 00:14:06
And then right here

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there is something very strange
that is happening.

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Then when it breaks,

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the stress here is
actually lower than there,

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00:14:12 --> 00:14:18
and this is something I can
never show you in class,

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but I can explain to you
why that happens.

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When you are past
this point here,

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the rod begins to pinch,
and you get this.

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It's unpredictable
where it will pinch,

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but somewhere
it will start to pinch

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so that the area here is A prime

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whereas the cross-sectional
area here is A.

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And so F divided by A prime

236
00:14:39 --> 00:14:45
will be larger
than F divided by A.

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If you could do this experiment
in a controlled fashion--

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and there are machines
designed to do that--

239
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then you could actually
lower F over A

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00:14:55 --> 00:15:01
so that the stress
actually goes down

241
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but that F over A prime
would still go up,

242
00:15:01 --> 00:15:07
and therefore
delta l will increase.

243
00:15:04 --> 00:15:10
And there are machines

244
00:15:06 --> 00:15:12
who are specially designed
to test these metals,

245
00:15:10 --> 00:15:16
and what they do is they go
up in very small steps of F--

246
00:15:14 --> 00:15:20
and so they trace
this whole curve--

247
00:15:17 --> 00:15:23
but by the time that they get
into the plastic flow area,

248
00:15:21 --> 00:15:27
they... before they increase the
force, they decrease it first.

249
00:15:27 --> 00:15:33
And if delta l gets larger
when they decrease it,

250
00:15:31 --> 00:15:37
they continue to decrease it,
and so that's the way

251
00:15:33 --> 00:15:39
that they can map out
also this portion.

252
00:15:35 --> 00:15:41
But we will not be able
to do that.

253
00:15:38 --> 00:15:44
There are metals
which are extremely brittle,

254
00:15:40 --> 00:15:46
and even though the curve
would look very similar--

255
00:15:43 --> 00:15:49
it has all
these characteristics--

256
00:15:45 --> 00:15:51
this point, then, would lie
all the way here.

257
00:15:50 --> 00:15:56
So this whole curve, then,
is squeezed

258
00:15:53 --> 00:15:59
into very small parameter space
of the strain.

259
00:16:01 --> 00:16:07
And so I want you to see most
of this, at least some of this,

260
00:16:05 --> 00:16:11
and we do that
with a demonstration

261
00:16:08 --> 00:16:14
which we have there,

262
00:16:10 --> 00:16:16
and I will make a drawing
of the basic idea.

263
00:16:13 --> 00:16:19
We have a copper wire--

264
00:16:15 --> 00:16:21
you will get the dimension
from me very shortly--

265
00:16:19 --> 00:16:25
and we attach to the copper wire
a rod, a solid rod.

266
00:16:26 --> 00:16:32
And at the end of the solid rod
is a mirror.

267
00:16:30 --> 00:16:36
This is a mirror.

268
00:16:32 --> 00:16:38

269
00:16:35 --> 00:16:41
And we're going
to hang weights on here.

270
00:16:39 --> 00:16:45
But this mirror is
on a little platform

271
00:16:42 --> 00:16:48
and can pivot at this point
but cannot lower itself.

272
00:16:47 --> 00:16:53
The platform is fixed.

273
00:16:49 --> 00:16:55
For those of you
who are very close,

274
00:16:51 --> 00:16:57
this is where that platform is.

275
00:16:53 --> 00:16:59
That is a fixed platform.

276
00:16:55 --> 00:17:01
And so the mirror can only tilt
but cannot go down.

277
00:16:58 --> 00:17:04
And so now I will show you

278
00:17:02 --> 00:17:08
what happens
when this wire gets longer.

279
00:17:06 --> 00:17:12
So here is the wire,
and this is where the...

280
00:17:12 --> 00:17:18

281
00:17:15 --> 00:17:21
This is this line here.

282
00:17:19 --> 00:17:25
But now the wire
has become longer.

283
00:17:22 --> 00:17:28
So the wire is now longer
by an amount delta l.

284
00:17:26 --> 00:17:32
So this is now the point

285
00:17:29 --> 00:17:35
where this bar, this rod
is attached.

286
00:17:33 --> 00:17:39
And so the mirror
will now be like this.

287
00:17:40 --> 00:17:46
We have tilted the mirror.

288
00:17:44 --> 00:17:50
And we have tilted the mirror
over an angle delta theta,

289
00:17:49 --> 00:17:55
and this length is l.

290
00:17:53 --> 00:17:59
And we're going to load it down
with a mass M,

291
00:17:57 --> 00:18:03
which we're going to increase.

292
00:17:59 --> 00:18:05
And then we have a laser beam
which we shine onto the mirror.

293
00:18:05 --> 00:18:11
The laser beam comes in like so.

294
00:18:07 --> 00:18:13

295
00:18:10 --> 00:18:16
And this is
the normal to the mirror,

296
00:18:15 --> 00:18:21
so this angle here
is delta theta.

297
00:18:20 --> 00:18:26
The laser beam bounces off
and returns... like so,

298
00:18:31 --> 00:18:37
and this angle, of course,
is also delta theta.

299
00:18:34 --> 00:18:40
That's the property of a mirror.

300
00:18:37 --> 00:18:43

301
00:18:40 --> 00:18:46
And this beam...

302
00:18:43 --> 00:18:49
This is the laser beam here
that we shine into the mirror.

303
00:18:46 --> 00:18:52
Here's the laser,
it goes into the mirror,

304
00:18:49 --> 00:18:55
and the spot there
of that return

305
00:18:51 --> 00:18:57
is all the way there
on the wall.

306
00:18:53 --> 00:18:59
So we are going to show you
where this spot is on the wall,

307
00:18:58 --> 00:19:04
and the wall is very far away.

308
00:19:00 --> 00:19:06
It's a distance capital L
from the wire.

309
00:19:06 --> 00:19:12
And you'll see very shortly
why we do it that way.

310
00:19:10 --> 00:19:16
Let me first give you the
dimensions of this instrument.

311
00:19:14 --> 00:19:20
L is 36 centimeters.

312
00:19:19 --> 00:19:25
It is copper.

313
00:19:21 --> 00:19:27
This bar, which has length d,
is 7½ centimeters.

314
00:19:27 --> 00:19:33

315
00:19:31 --> 00:19:37
The distance to the wall
is about 16 meters.

316
00:19:35 --> 00:19:41

317
00:19:40 --> 00:19:46
The radius of the wire...

318
00:19:44 --> 00:19:50
or actually, the diameter of
the wire, of the copper wire

319
00:19:48 --> 00:19:54
is 20/1,000 of an inch.

320
00:19:50 --> 00:19:56
I give it
in terms of the diameter

321
00:19:52 --> 00:19:58
because that's the way that
the manufacturer gives it to us.

322
00:19:55 --> 00:20:01
And that translates

323
00:19:57 --> 00:20:03
into a cross-sectional area
of that wire

324
00:20:01 --> 00:20:07
of 2.0 times ten to the
minus seven meters squared.

325
00:20:06 --> 00:20:12

326
00:20:11 --> 00:20:17
All right, notice
that delta theta, that angle...

327
00:20:19 --> 00:20:25
the angle here, delta theta,
is delta l divided by d.

328
00:20:26 --> 00:20:32
So delta theta equals
delta l divided by d.

329
00:20:32 --> 00:20:38

330
00:20:35 --> 00:20:41
The light that returns hits
the wall there at a distance l,

331
00:20:41 --> 00:20:47
so y at the wall,
or I can call it delta y--

332
00:20:43 --> 00:20:49
that is at the wall, I call that
displacement there delta y--

333
00:20:47 --> 00:20:53
divided by l
will be two delta theta.

334
00:20:50 --> 00:20:56
This is a small-angle
approximation;

335
00:20:53 --> 00:20:59
it's a very good approximation,
and theta is in radians.

336
00:20:58 --> 00:21:04
It is two delta theta,
because you see here

337
00:21:01 --> 00:21:07
that the change is
over an angle two delta.

338
00:21:07 --> 00:21:13
And so delta y equals
2L delta theta...

339
00:21:14 --> 00:21:20
2L delta theta times
delta l divided by d.

340
00:21:22 --> 00:21:28
And look what we have done now:

341
00:21:25 --> 00:21:31
We have convert something
that is immeasurable, delta l--

342
00:21:30 --> 00:21:36
which is fraction
of millimeters--

343
00:21:32 --> 00:21:38
we have convert that

344
00:21:34 --> 00:21:40
to something on the wall
that we can measure,

345
00:21:36 --> 00:21:42
because this ratio, 2L over d,

346
00:21:39 --> 00:21:45
in our case, for the dimensions
that I have chosen,

347
00:21:43 --> 00:21:49
is about 425; it's like
a magnification factor.

348
00:21:48 --> 00:21:54
So if we see a displacement
of that laser beam on the wall

349
00:21:52 --> 00:21:58
of 40 centimeters,
which is easy to see,

350
00:21:56 --> 00:22:02
it means that the wire got
longer by only one millimeter.

351
00:22:02 --> 00:22:08
So 40 centimeters
there translates

352
00:22:04 --> 00:22:10
to one millimeter there,

353
00:22:06 --> 00:22:12
and if we would see that
laser beam go up four meters,

354
00:22:10 --> 00:22:16
it would mean

355
00:22:11 --> 00:22:17
that the wire would only have
become one centimeter longer.

356
00:22:15 --> 00:22:21
So it is a wonderful way

357
00:22:16 --> 00:22:22
to magnify the effect
and to measure it.

358
00:22:20 --> 00:22:26

359
00:22:21 --> 00:22:27
So I will now make an attempt
to load down the copper wire.

360
00:22:31 --> 00:22:37

361
00:22:34 --> 00:22:40
Oh, we can actually leave this
here, so you can see that curve.

362
00:22:39 --> 00:22:45
So we start here with...
we load up with half kilograms.

363
00:22:43 --> 00:22:49
We will write down, then,

364
00:22:44 --> 00:22:50
how much that laser spot
goes up on the wall.

365
00:22:47 --> 00:22:53
And then, in between
increasing the weight,

366
00:22:50 --> 00:22:56
increasing the force

367
00:22:52 --> 00:22:58
we will take the masses off
to see whether they return...

368
00:22:56 --> 00:23:02
whether the length of the rod
returns to the original length.

369
00:23:00 --> 00:23:06
And you will see after a while

370
00:23:02 --> 00:23:08
that you get
permanent deformation,

371
00:23:04 --> 00:23:10
then it no longer comes back
to its original length.

372
00:23:07 --> 00:23:13
In other words

373
00:23:08 --> 00:23:14
the laser spot will not return
to zero on the wall

374
00:23:12 --> 00:23:18
but it will stay higher.

375
00:23:14 --> 00:23:20
So Ron, if you are there...

376
00:23:17 --> 00:23:23
Oh, boy,
you're hiding behind...

377
00:23:20 --> 00:23:26
Um, maybe we want to move
the view graph out of the way

378
00:23:25 --> 00:23:31
so that students can also see.

379
00:23:28 --> 00:23:34

380
00:23:34 --> 00:23:40
So, here is that copper wire

381
00:23:35 --> 00:23:41
which will be hard to see
for some of you--

382
00:23:38 --> 00:23:44
it's only 20/1
of an inch thick--

383
00:23:40 --> 00:23:46
and here is that mirror
which can pivot and can tilt.

384
00:23:43 --> 00:23:49
And Ron is going to put
weights on here,

385
00:23:46 --> 00:23:52
and then we will take
the weights off in between

386
00:23:49 --> 00:23:55
and we will try
to construct a curve

387
00:23:54 --> 00:24:00
of the stress versus strain,

388
00:23:58 --> 00:24:04
except that it is practical for
me to put here just the mass--

389
00:24:02 --> 00:24:08
how many kilograms
we have on there--

390
00:24:04 --> 00:24:10
because we know what A is,
so we can calculate F over A.

391
00:24:07 --> 00:24:13
That's not so important now.

392
00:24:08 --> 00:24:14
And here I simply have delta y.

393
00:24:10 --> 00:24:16
But keep in mind

394
00:24:11 --> 00:24:17
that delta y is always
425 times larger than delta l.

395
00:24:16 --> 00:24:22
So we're going to plot it,
and we're going to see

396
00:24:18 --> 00:24:24
whether we can come up
with a curve

397
00:24:20 --> 00:24:26
that is somewhat similar
to that one.

398
00:24:22 --> 00:24:28
So, Ron, if you put
on the first half kilogram...

399
00:24:26 --> 00:24:32

400
00:24:29 --> 00:24:35
The mirror always starts
to oscillate a little bit

401
00:24:33 --> 00:24:39
and so we have to be
a little patient,

402
00:24:36 --> 00:24:42
and in the beginning, you
may be bored because--tja!--

403
00:24:40 --> 00:24:46
we're going through that
linear part of the curve.

404
00:24:43 --> 00:24:49
So it goes up
very slowly, very gradually.

405
00:24:46 --> 00:24:52
We have five centimeters
for the first half kilogram.

406
00:24:50 --> 00:24:56
Could you remove
the half kilogram?

407
00:24:54 --> 00:25:00

408
00:25:02 --> 00:25:08
It returns practically to zero.

409
00:25:04 --> 00:25:10
Maybe it's a little higher,
but that's not very significant.

410
00:25:09 --> 00:25:15
Could you make it one kilogram?

411
00:25:11 --> 00:25:17

412
00:25:16 --> 00:25:22
Ah, ah, it's clearly higher.

413
00:25:21 --> 00:25:27
Ja, ja... oh, it's about
nine centimeters, nine, ten,

414
00:25:26 --> 00:25:32
so you see it's
in the linear part--

415
00:25:29 --> 00:25:35
nine to ten centimeters.

416
00:25:32 --> 00:25:38
Can you remove
the half kilogram, Ron?

417
00:25:36 --> 00:25:42
The one kilogram,
there was one kilogram on it.

418
00:25:40 --> 00:25:46
We have to just wait,
let it damp out a little.

419
00:25:42 --> 00:25:48
It's oscillating.

420
00:25:43 --> 00:25:49

421
00:25:48 --> 00:25:54
It's possible that you already
begin to see a small deformation

422
00:25:53 --> 00:25:59
which may be
one or 1½ centimeters.

423
00:25:56 --> 00:26:02
I put a question mark there--
it's possible.

424
00:26:00 --> 00:26:06

425
00:26:02 --> 00:26:08
Can you put 1½ kilograms on?

426
00:26:05 --> 00:26:11
Yeah, I think it is...

427
00:26:07 --> 00:26:13

428
00:26:08 --> 00:26:14
You can almost remove
the question mark.

429
00:26:12 --> 00:26:18

430
00:26:14 --> 00:26:20
So now we are at 1½.

431
00:26:17 --> 00:26:23

432
00:26:20 --> 00:26:26
So if it is strictly linear, you
would expect something like 15.

433
00:26:24 --> 00:26:30
Yeah, that's what it is, 15,
so it's still doing quite well.

434
00:26:28 --> 00:26:34
Can you take them off?

435
00:26:29 --> 00:26:35

436
00:26:31 --> 00:26:37
One and a half.

437
00:26:33 --> 00:26:39

438
00:26:39 --> 00:26:45
Ah, but you see
it no longer wants to return

439
00:26:44 --> 00:26:50
to its original length.

440
00:26:45 --> 00:26:51
It's clearly longer now.

441
00:26:47 --> 00:26:53
Permanent deformation
has already occurred,

442
00:26:50 --> 00:26:56
and so we're now something
like six centimeters.

443
00:26:53 --> 00:26:59
Can you make it two kilograms?

444
00:26:56 --> 00:27:02

445
00:27:10 --> 00:27:16
Two kilograms.

446
00:27:12 --> 00:27:18
If it's linear,
you would expect near 20.

447
00:27:15 --> 00:27:21
It's still amazingly linear.

448
00:27:18 --> 00:27:24
It's as close
as I can see it to 20,

449
00:27:20 --> 00:27:26
but all these readings are
no more accurate

450
00:27:23 --> 00:27:29
than half a centimeter or so.

451
00:27:24 --> 00:27:30
So can you remove
the two kilograms?

452
00:27:27 --> 00:27:33

453
00:27:31 --> 00:27:37
Oh, boy, look at that--
there's clear deformation now.

454
00:27:36 --> 00:27:42
It no longer returns
to zero, and it is...

455
00:27:39 --> 00:27:45
Oh, it's comfortably
ten centimeters long now.

456
00:27:44 --> 00:27:50
So can you make it 2½?

457
00:27:46 --> 00:27:52

458
00:27:51 --> 00:27:57
You're now slowly approaching

459
00:27:53 --> 00:27:59
the part that I hope
you are going to see,

460
00:27:56 --> 00:28:02
and that is that it is going
to take off like a rocket,

461
00:27:59 --> 00:28:05
that with a little bit
of extra weight,

462
00:28:01 --> 00:28:07
it will start to move
substantially.

463
00:28:04 --> 00:28:10
We haven't reached that point
yet, but we are close to it.

464
00:28:09 --> 00:28:15
We're now 26-- 25, 26...
it still looks quite linear.

465
00:28:14 --> 00:28:20
Can you take it off, Ron?

466
00:28:16 --> 00:28:22
Actually, there's no need
to take it off anymore

467
00:28:19 --> 00:28:25
because it's clear that we...

468
00:28:22 --> 00:28:28
that we have
permanent deformation,

469
00:28:24 --> 00:28:30
and there's no sense
in following that,

470
00:28:25 --> 00:28:31
so why don't you make it
three kilograms?

471
00:28:28 --> 00:28:34
So what was it?
What did I say it was?

472
00:28:31 --> 00:28:37
What was the number I said?

473
00:28:33 --> 00:28:39
20 what?

474
00:28:35 --> 00:28:41
25 or so?

475
00:28:39 --> 00:28:45
So we have three now?

476
00:28:41 --> 00:28:47

477
00:28:43 --> 00:28:49
Boy, this wire is hanging
in there, I must tell you.

478
00:28:47 --> 00:28:53
32, yeah.

479
00:28:48 --> 00:28:54
Can you make it four?

480
00:28:50 --> 00:28:56
Watch very closely
now on the wall,

481
00:28:53 --> 00:28:59
because the drama is
about to start now.

482
00:28:57 --> 00:29:03
What did I say, 30...?

483
00:28:59 --> 00:29:05

484
00:29:01 --> 00:29:07
I said 32.

485
00:29:03 --> 00:29:09
Four.

486
00:29:04 --> 00:29:10

487
00:29:09 --> 00:29:15
Ooh, still moving, still moving.

488
00:29:15 --> 00:29:21
52... settles at 53.

489
00:29:19 --> 00:29:25
Now, don't look at the board--
now, look at this spot now.

490
00:29:22 --> 00:29:28
Can you add... remember
the number, right? 53.

491
00:29:26 --> 00:29:32
Can you add one kilogram now?

492
00:29:28 --> 00:29:34

493
00:29:33 --> 00:29:39
And look at that.

494
00:29:34 --> 00:29:40
It became almost twice as long
and it is still moving.

495
00:29:39 --> 00:29:45
Still going... still going.

496
00:29:45 --> 00:29:51
I hope it will settle.

497
00:29:46 --> 00:29:52
I'm going to write down my 53.

498
00:29:49 --> 00:29:55

499
00:29:52 --> 00:29:58
Five kilograms.

500
00:29:54 --> 00:30:00

501
00:29:56 --> 00:30:02
We're at five, right?

502
00:29:58 --> 00:30:04

503
00:30:03 --> 00:30:09
97-- now put on six.

504
00:30:05 --> 00:30:11
97-- remind me, 97.

505
00:30:07 --> 00:30:13
Now, watch this point.

506
00:30:10 --> 00:30:16
Yay!

507
00:30:12 --> 00:30:18

508
00:30:14 --> 00:30:20
Now you're clearly
in that plastic flow portion.

509
00:30:17 --> 00:30:23
By adding one kilogram,

510
00:30:18 --> 00:30:24
look where that point is--
it's still moving.

511
00:30:20 --> 00:30:26
What was five? 97?

512
00:30:23 --> 00:30:29

513
00:30:25 --> 00:30:31
So we're now at six kilograms.

514
00:30:27 --> 00:30:33

515
00:30:30 --> 00:30:36
Oh, actually, that is
still easy for me to estimate.

516
00:30:33 --> 00:30:39
I would say it's
about double the length

517
00:30:36 --> 00:30:42
of that stick
that we have on the wall,

518
00:30:38 --> 00:30:44
and the stick is
two meters long.

519
00:30:40 --> 00:30:46
Is it moving? That's
still moving a little bit.

520
00:30:44 --> 00:30:50
It's a little more
than four meters.

521
00:30:46 --> 00:30:52
Close enough, four meters,
just for the idea.

522
00:30:49 --> 00:30:55
Four meters, so that's 400.

523
00:30:55 --> 00:31:01
Put on seven.

524
00:30:57 --> 00:31:03
It will go
through the ceiling now.

525
00:31:00 --> 00:31:06
So we'll lose it.

526
00:31:02 --> 00:31:08
But what I want to do now,

527
00:31:04 --> 00:31:10
I want to get
to the breaking point.

528
00:31:06 --> 00:31:12
We can no longer measure
the displacement,

529
00:31:10 --> 00:31:16
but we're very close
to the breaking point now.

530
00:31:12 --> 00:31:18
So we're going to load it up
to the point that it will break,

531
00:31:16 --> 00:31:22
and that allows us to measure
the ultimate tensile strength.

532
00:31:20 --> 00:31:26
We're at seven now--
can you put eight on?

533
00:31:22 --> 00:31:28

534
00:31:24 --> 00:31:30
Oh, we're running out of...
(chuckles )

535
00:31:29 --> 00:31:35
Oh, God!

536
00:31:30 --> 00:31:36
You could just see it sag
when the eight was put on.

537
00:31:33 --> 00:31:39
Did you actually look
at the wire?

538
00:31:35 --> 00:31:41
Okay, so at eight kilograms,
it breaks.

539
00:31:39 --> 00:31:45

540
00:31:43 --> 00:31:49
Okay, let's put
these numbers in here.

541
00:31:48 --> 00:31:54
So, we have 1½...

542
00:31:50 --> 00:31:56
or half a kilogram,
we have five centimeters.

543
00:31:53 --> 00:31:59
Let me do this in color.

544
00:31:56 --> 00:32:02
So that gives me a point here.

545
00:31:59 --> 00:32:05
And then at one,
we have about ten.

546
00:32:03 --> 00:32:09
And at 1½, we have about 15.

547
00:32:07 --> 00:32:13
And at two, we have about 20.

548
00:32:11 --> 00:32:17
And at 2½, we have 25.

549
00:32:15 --> 00:32:21

550
00:32:17 --> 00:32:23
Oh, I was a little bit
too high here, perhaps.

551
00:32:20 --> 00:32:26
I want to go
a little bit more carefully

552
00:32:22 --> 00:32:28
because this is
really terrific data.

553
00:32:27 --> 00:32:33
So this was at...

554
00:32:28 --> 00:32:34

555
00:32:33 --> 00:32:39
And then we have...
at 2½, we have 25.

556
00:32:36 --> 00:32:42

557
00:32:43 --> 00:32:49
Amazing how well...
how linear that is!

558
00:32:45 --> 00:32:51
And then at three, we have 32.

559
00:32:48 --> 00:32:54

560
00:32:55 --> 00:33:01
Ah!

561
00:32:57 --> 00:33:03
It looks like
it's beginning to bend over.

562
00:32:59 --> 00:33:05
At four, we have 53,
no question.

563
00:33:03 --> 00:33:09

564
00:33:10 --> 00:33:16
Oh, yeah.

565
00:33:13 --> 00:33:19
And then at five, we have 97.

566
00:33:18 --> 00:33:24
Here is five-- 97, that's here.

567
00:33:26 --> 00:33:32
And then our last point
that was...

568
00:33:28 --> 00:33:34
Oh, we even have a 400 here.

569
00:33:30 --> 00:33:36
Ooh, that's great.

570
00:33:32 --> 00:33:38

571
00:33:35 --> 00:33:41
Here is 400.

572
00:33:36 --> 00:33:42

573
00:33:39 --> 00:33:45
And what do we have there?

574
00:33:42 --> 00:33:48
A six.

575
00:33:43 --> 00:33:49

576
00:33:46 --> 00:33:52
One, two, three,
four, five, six.

577
00:33:49 --> 00:33:55

578
00:33:52 --> 00:33:58
That's our last point.

579
00:33:54 --> 00:34:00

580
00:33:56 --> 00:34:02
And then we don't have the rest.

581
00:34:00 --> 00:34:06
But look how wonderful
this is, isn't it?

582
00:34:02 --> 00:34:08
Isn't that a great curve?

583
00:34:04 --> 00:34:10
Very linear in the beginning.

584
00:34:06 --> 00:34:12

585
00:34:11 --> 00:34:17
Then it starts
to bend over, over.

586
00:34:15 --> 00:34:21
And you draw that line
anywhere you want to...

587
00:34:21 --> 00:34:27
and then it breaks.

588
00:34:23 --> 00:34:29
And so we can now actually
make an attempt

589
00:34:26 --> 00:34:32
to calculate Young's modulus
from the data.

590
00:34:32 --> 00:34:38
Of course, you'll have to select

591
00:34:35 --> 00:34:41
a portion where you think that
the data are reasonably linear.

592
00:34:39 --> 00:34:45
So Young's modulus itself...
equals F divided by A.

593
00:34:52 --> 00:34:58

594
00:34:54 --> 00:35:00
Here you have here
the equation.

595
00:34:56 --> 00:35:02
Young's modulus equals

596
00:34:59 --> 00:35:05
F divided by A
times l divided by delta l.

597
00:35:04 --> 00:35:10

598
00:35:08 --> 00:35:14
We know what A is,
we know what l is.

599
00:35:11 --> 00:35:17
It's still
on the blackboard there.

600
00:35:13 --> 00:35:19
And so now it's a matter

601
00:35:16 --> 00:35:22
of where do we think
that Hooke's Law still holds?

602
00:35:21 --> 00:35:27
I think this whole portion
would be fine,

603
00:35:24 --> 00:35:30
so we could take this point
as well as this point,

604
00:35:26 --> 00:35:32
because anywhere
on this straight line,

605
00:35:29 --> 00:35:35
you will get the same value
for Young's modulus.

606
00:35:31 --> 00:35:37
So I will use two kilograms
and 20 centimeters--

607
00:35:35 --> 00:35:41
I will use this point.

608
00:35:36 --> 00:35:42
So F equals 200 newtons,

609
00:35:41 --> 00:35:47
and then we have delta y
equals 20 centimeters.

610
00:35:48 --> 00:35:54
And so that means delta l equals
20 divided by 425 centimeters.

611
00:35:58 --> 00:36:04
So this is F divided by A,

612
00:36:02 --> 00:36:08
l, and then we have here
delta y times 425.

613
00:36:10 --> 00:36:16
And let's see what that is.

614
00:36:12 --> 00:36:18

615
00:36:16 --> 00:36:22
So F is 200-- that's right.

616
00:36:20 --> 00:36:26
No, no, F is 20--
ooh, ooh, ooh, ooh, ooh.

617
00:36:23 --> 00:36:29
I'm glad that we caught
that simultaneously.

618
00:36:27 --> 00:36:33
So F is 20-- 20 newtons,
that's right.

619
00:36:33 --> 00:36:39
The area is two times
ten to the minus seven,

620
00:36:38 --> 00:36:44
and I divide by the area.

621
00:36:39 --> 00:36:45
I multiply by the length,
which is .36.

622
00:36:44 --> 00:36:50
I multiply by 425
and I divide by delta y.

623
00:36:51 --> 00:36:57
But I need delta y
in centimeters, so...

624
00:36:54 --> 00:37:00
in meters, so that's .2.

625
00:36:57 --> 00:37:03
And I get 7.7 times
ten to the tenth.

626
00:37:01 --> 00:37:07

627
00:37:04 --> 00:37:10
And that is not bad at all

628
00:37:08 --> 00:37:14
because I think,
from what I remember,

629
00:37:11 --> 00:37:17
is that it is 11
times ten to the tenth.

630
00:37:14 --> 00:37:20
That's what it is.

631
00:37:15 --> 00:37:21
So that is quite amazing

632
00:37:17 --> 00:37:23
for such a crude measurement
that we do here.

633
00:37:21 --> 00:37:27
We can also measure
the ultimate tensile strength.

634
00:37:25 --> 00:37:31
That is, we can measure
the value for F over A

635
00:37:30 --> 00:37:36
when it broke.

636
00:37:32 --> 00:37:38
So that happened, I think...
wasn't that eight kilograms?

637
00:37:36 --> 00:37:42
So that will be 80 divided
by the area that we know,

638
00:37:42 --> 00:37:48
and so that gives me
80 divided by two times...

639
00:37:47 --> 00:37:53
divided by ten to the
minus seventh, and that is

640
00:37:50 --> 00:37:56
four times ten to the eighth
newtons per square meter.

641
00:37:56 --> 00:38:02
And this is also
newtons per square meter.

642
00:38:01 --> 00:38:07
And that's not bad.

643
00:38:03 --> 00:38:09
It's a little higher
than we have there,

644
00:38:05 --> 00:38:11
but it's
a very crude measurement.

645
00:38:06 --> 00:38:12
And don't forget, we were unable

646
00:38:09 --> 00:38:15
to follow the portion
when it was going down.

647
00:38:12 --> 00:38:18
We only went up
in mass, in weight,

648
00:38:15 --> 00:38:21
and so that value there
takes into account

649
00:38:18 --> 00:38:24
that the curve comes down,
something we could not do.

650
00:38:21 --> 00:38:27
We didn't have
the means of doing that.

651
00:38:24 --> 00:38:30

652
00:38:27 --> 00:38:33
The percentage strain is
actually extremely low

653
00:38:33 --> 00:38:39
in the portion
that the curve is linear.

654
00:38:37 --> 00:38:43
You can ask yourself
the question

655
00:38:39 --> 00:38:45
"What is delta l divided by l

656
00:38:43 --> 00:38:49
in terms of percentages
during this portion here?"

657
00:38:49 --> 00:38:55
Well, we know the length, 0.36,

658
00:38:57 --> 00:39:03
and when we take the case
where we sort of reach

659
00:39:01 --> 00:39:07
the end of the Hooke's Law
parameter space,

660
00:39:04 --> 00:39:10
we have a delta l which is
425 times lower than this,

661
00:39:09 --> 00:39:15
so we have 20 divided by 425,
and that is, um...

662
00:39:17 --> 00:39:23
four point seven
times ten to the minus two,

663
00:39:21 --> 00:39:27
but that is in terms
of centimeters

664
00:39:23 --> 00:39:29
and we want it in terms of
meters, so we have to multiply

665
00:39:28 --> 00:39:34
by another factor
ten to the minus two.

666
00:39:30 --> 00:39:36
So the exponent's changed--
minus two... divided by 0.36,

667
00:39:35 --> 00:39:41
and that gives me 1.3 times
ten to the minus three.

668
00:39:40 --> 00:39:46
In terms of percentages,
that would be 0.13%.

669
00:39:45 --> 00:39:51
So the wire, when it reaches

670
00:39:48 --> 00:39:54
the end of its Hooke's Law
parameter space,

671
00:39:52 --> 00:39:58
has only become longer by 0.13%.

672
00:39:56 --> 00:40:02
And when the wire breaks...

673
00:39:58 --> 00:40:04
In general, for metals

674
00:40:00 --> 00:40:06
it's maybe five or ten percent
longer than its original length.

675
00:40:05 --> 00:40:11
That's a typical value
for metals.

676
00:40:09 --> 00:40:15
Now, as long as I am in the
linear portion of the curve,

677
00:40:13 --> 00:40:19
I can generate
a simple harmonic motion.

678
00:40:18 --> 00:40:24

679
00:40:21 --> 00:40:27
I want to show you that.

680
00:40:23 --> 00:40:29
I need my wiper.

681
00:40:25 --> 00:40:31
Because in the linear portion
of the curve

682
00:40:29 --> 00:40:35
where Hooke's Law holds,

683
00:40:31 --> 00:40:37
I could hang a weight
on the rope... or on the rod

684
00:40:34 --> 00:40:40
and I could let it
oscillate vertically.

685
00:40:37 --> 00:40:43

686
00:40:40 --> 00:40:46
So the force that you have

687
00:40:44 --> 00:40:50
equals y times A times delta l
divided by l.

688
00:40:51 --> 00:40:57
That's the force that I apply.

689
00:40:54 --> 00:41:00
So the spring force
is opposing that,

690
00:40:56 --> 00:41:02
so the spring force
has a minus sign here

691
00:40:59 --> 00:41:05
to indicate the direction.

692
00:41:01 --> 00:41:07
And so this is similar
to F equals minus kX

693
00:41:06 --> 00:41:12
that you have seen
with the spring.

694
00:41:09 --> 00:41:15
This now is our k,
and delta l is our x.

695
00:41:14 --> 00:41:20
And so you can predict
that it will start to oscillate

696
00:41:18 --> 00:41:24
with an angular frequency

697
00:41:20 --> 00:41:26
of omega square root of k
divided by m

698
00:41:25 --> 00:41:31
and with a period
which is two pi...

699
00:41:30 --> 00:41:36
two pi times
the square root of m over k,

700
00:41:34 --> 00:41:40
and m is now the mass
that I'm hanging here.

701
00:41:37 --> 00:41:43

702
00:41:40 --> 00:41:46
If we take our copper wire,
we know what y is.

703
00:41:46 --> 00:41:52
If I take the value 11 instead
of the one that we found,

704
00:41:50 --> 00:41:56
but it's very close, anyhow...

705
00:41:52 --> 00:41:58
So if I take our copper wire,

706
00:41:56 --> 00:42:02
then I will find, depending upon
the mass that I hang on it...

707
00:42:00 --> 00:42:06
then I'll find in any case
for k, for this value k

708
00:42:05 --> 00:42:11
I find five times ten to the
fourth newtons per meter.

709
00:42:12 --> 00:42:18
So if now I hang on it
a mass of one kilogram,

710
00:42:18 --> 00:42:24
I can calculate the period
of one oscillation,

711
00:42:22 --> 00:42:28
and one over the period
is the frequency F,

712
00:42:27 --> 00:42:33
and I believe that is
something like 38 hertz.

713
00:42:30 --> 00:42:36

714
00:42:33 --> 00:42:39
So it would start to oscillate
like this, 38 times per second,

715
00:42:37 --> 00:42:43
and if I hang
two kilograms on it,

716
00:42:39 --> 00:42:45
then this frequency would become
something like 25 hertz,

717
00:42:44 --> 00:42:50
because a higher mass gives me
a longer period,

718
00:42:47 --> 00:42:53
gives me a lower frequency.

719
00:42:49 --> 00:42:55

720
00:42:53 --> 00:42:59
The speed of sound also depends
on Young's modulus.

721
00:43:00 --> 00:43:06
Without proof-- you will see
this if you ever take 803--

722
00:43:06 --> 00:43:12
I will tell you

723
00:43:07 --> 00:43:13
that the speed of sound is
the square root

724
00:43:12 --> 00:43:18
of Young's modulus divided by
the density of the material.

725
00:43:15 --> 00:43:21
And I have listed those there
on the view graph.

726
00:43:19 --> 00:43:25
Oh, this is by, yeah, the
square root-- I have that, yeah.

727
00:43:24 --> 00:43:30
And so the higher
Young's modulus is,

728
00:43:26 --> 00:43:32
the higher
the speed of sound is,

729
00:43:28 --> 00:43:34
and that is intuitively
sort of pleasing.

730
00:43:30 --> 00:43:36
I can sort of understand that

731
00:43:32 --> 00:43:38
although the square root
is hard to see.

732
00:43:36 --> 00:43:42
If I have here a rod, a bar

733
00:43:38 --> 00:43:44
and I give this bar here
a bang, I hit it,

734
00:43:41 --> 00:43:47
if this bar were
infinitely stiff--

735
00:43:44 --> 00:43:50
that means Young's modulus
were infinitely high--

736
00:43:48 --> 00:43:54
then the bar would
instantaneously move here

737
00:43:51 --> 00:43:57
when I hit it there.

738
00:43:53 --> 00:43:59
But if the bar is
not infinitely stiff,

739
00:43:56 --> 00:44:02
if it has a certain amount
of elasticity,

740
00:43:59 --> 00:44:05
then what happens... then I hit
it here and I produce here

741
00:44:04 --> 00:44:10
some kind
of local increased pressure

742
00:44:08 --> 00:44:14
which is going to propagate
to this end--

743
00:44:10 --> 00:44:16
it's like a pressure wave,
like sound is a pressure wave--

744
00:44:14 --> 00:44:20
and that takes time.

745
00:44:15 --> 00:44:21
And the larger
Young's modulus is,

746
00:44:18 --> 00:44:24
the stiffer the material is,
the faster that will go.

747
00:44:21 --> 00:44:27
If Young's modulus is very low,
the material is more elastic.

748
00:44:25 --> 00:44:31
It will take a longer time for
this pulse to reach this end.

749
00:44:29 --> 00:44:35
So it is sort of
intuitively pleasing for me

750
00:44:32 --> 00:44:38
that the stiffness
of the material

751
00:44:34 --> 00:44:40
is related
to the speed of sound.

752
00:44:38 --> 00:44:44

753
00:44:41 --> 00:44:47
I have here a magnesium bar,

754
00:44:46 --> 00:44:52
and the magnesium bar
has a length l.

755
00:44:50 --> 00:44:56

756
00:44:52 --> 00:44:58
And I can calculate
the speed of sound for magnesium

757
00:44:55 --> 00:45:01
by taking its Young's modulus,
which is there,

758
00:44:59 --> 00:45:05
divided by rho, which is there,

759
00:45:01 --> 00:45:07
and I come up
with the speed of sound,

760
00:45:04 --> 00:45:10
which is about
five kilometers per second.

761
00:45:08 --> 00:45:14
If you look at those numbers,

762
00:45:09 --> 00:45:15
they are all
substantially higher

763
00:45:11 --> 00:45:17
than the speed of sound in air,

764
00:45:13 --> 00:45:19
which is only some
340 meters per second.

765
00:45:16 --> 00:45:22
So the speed of sound
for magnesium

766
00:45:19 --> 00:45:25
is some 15 times larger
than the speed of sound in air.

767
00:45:25 --> 00:45:31
If this bar has a length l,

768
00:45:27 --> 00:45:33
this pressure disturbance
will start to move to the left,

769
00:45:33 --> 00:45:39
and then it'll come back here,

770
00:45:35 --> 00:45:41
so it will have made
a complete journey,

771
00:45:37 --> 00:45:43
which I call the period...

772
00:45:39 --> 00:45:45
It has made a complete journey
in so many seconds--

773
00:45:43 --> 00:45:49
2L divided
by the speed of sound.

774
00:45:46 --> 00:45:52
And so the frequency
of this bar--

775
00:45:49 --> 00:45:55
the frequency at which
it would like to oscillate

776
00:45:51 --> 00:45:57
when I give it a bang--

777
00:45:53 --> 00:45:59
is one over T, and that is the
speed of sound divided by 2L.

778
00:45:59 --> 00:46:05
Now, for those of you
who will take 803 in the future,

779
00:46:02 --> 00:46:08
you will see a much better
derivation of this frequency,

780
00:46:06 --> 00:46:12
so this is a poor man's version.

781
00:46:09 --> 00:46:15
The speed of sound is about
five kilometers per second.

782
00:46:13 --> 00:46:19
The length of the bar
is about 122 centimeters,

783
00:46:18 --> 00:46:24
and so that translates
into a frequency

784
00:46:21 --> 00:46:27
of roughly 2,200, 2,100 hertz.

785
00:46:25 --> 00:46:31
So the frequency
for this magnesium bar

786
00:46:29 --> 00:46:35
is about 2,100 hertz,
and you can hear that.

787
00:46:34 --> 00:46:40
It's a beautiful tone.

788
00:46:36 --> 00:46:42
I hold it here
and I will bang it here.

789
00:46:40 --> 00:46:46
(bar chimes, tone oscillates )

790
00:46:43 --> 00:46:49
Can you hear it? No?

791
00:46:48 --> 00:46:54
(bar chimes,
tone oscillates again )

792
00:46:52 --> 00:46:58
You hear it?
2,100 hertz-- beautiful tone.

793
00:46:56 --> 00:47:02

794
00:46:59 --> 00:47:05
Remember earlier in the course

795
00:47:01 --> 00:47:07
that we were wrestling
with this problem.

796
00:47:07 --> 00:47:13
We had a block and we had
two strings attached to it.

797
00:47:14 --> 00:47:20

798
00:47:18 --> 00:47:24
Here is one string,
here is the block,

799
00:47:23 --> 00:47:29
and here is another string.

800
00:47:26 --> 00:47:32
And we're going to pull
on this here with force F.

801
00:47:31 --> 00:47:37
Nothing was happening,
so the tension here, T, is F.

802
00:47:41 --> 00:47:47
Here, there is both the force F
plus the weight of the object,

803
00:47:47 --> 00:47:53
so here you have... call it
T prime equals F plus Mg.

804
00:47:54 --> 00:48:00
And so we argued

805
00:47:55 --> 00:48:01
if you increase F
and if the strings are identical

806
00:47:59 --> 00:48:05
that it should break
first here and then here,

807
00:48:02 --> 00:48:08
because the breaking point

808
00:48:03 --> 00:48:09
where the force is too large
for the string

809
00:48:06 --> 00:48:12
will occur first here,
then it occurs here,

810
00:48:08 --> 00:48:14
because you have
the extra amount Mg.

811
00:48:11 --> 00:48:17
But yet when we jerked on it
very fast, it would break here,

812
00:48:17 --> 00:48:23
and when we pulled very slowly--
I will do it again--

813
00:48:20 --> 00:48:26
it would break here.

814
00:48:21 --> 00:48:27
And now we can
fully understand that.

815
00:48:24 --> 00:48:30
Because what does it mean, that
this string is going to break?

816
00:48:29 --> 00:48:35
It means that string
has to get longer

817
00:48:31 --> 00:48:37
by a certain amount delta l

818
00:48:33 --> 00:48:39
before it says, "Sorry,
I got to go," and it breaks.

819
00:48:37 --> 00:48:43
For this string to get longer
by a certain amount delta l,

820
00:48:40 --> 00:48:46
this block has to come down.

821
00:48:43 --> 00:48:49
Now, if I pull here with a
certain force F-- F equals Ma--

822
00:48:47 --> 00:48:53
the block will come down
with an acceleration a,

823
00:48:52 --> 00:48:58
and in the time delta t,

824
00:48:54 --> 00:49:00
it will move over a distance
a delta t squared,

825
00:48:58 --> 00:49:04
which is the delta l
that this string will feel.

826
00:49:03 --> 00:49:09
But that takes time

827
00:49:05 --> 00:49:11
to reach the delta l
at which it wants to break,

828
00:49:08 --> 00:49:14
and so if I pull very fast,
I don't give it that time.

829
00:49:12 --> 00:49:18
And that's why, then, the one
at the bottom will break,

830
00:49:15 --> 00:49:21
and if I pull very slowly,
the one at the top will break.

831
00:49:19 --> 00:49:25
So here is a very thin wire--
you can't see it very well--

832
00:49:22 --> 00:49:28
and here is one, too.

833
00:49:23 --> 00:49:29
It's a string,
and they're identical,

834
00:49:26 --> 00:49:32
and this is my
emergency safety rope.

835
00:49:28 --> 00:49:34
And if I pull very fast,
you see the bottom one breaks.

836
00:49:32 --> 00:49:38
I have this in my hand,
and the top one is still safe.

837
00:49:37 --> 00:49:43
However, if I repeat this,
and I do it very gently...

838
00:49:44 --> 00:49:50
It's always hard to get this in.

839
00:49:46 --> 00:49:52

840
00:49:49 --> 00:49:55
There we go.

841
00:49:51 --> 00:49:57
Uh-oh...

842
00:49:52 --> 00:49:58

843
00:49:54 --> 00:50:00
Okay, so now I'm going
to slowly increase the force.

844
00:49:57 --> 00:50:03
I give the block plenty of time
to come down, plenty of time,

845
00:50:06 --> 00:50:12
and now the upper one goes.

846
00:50:08 --> 00:50:14
Have a good weekend.

847
00:50:09 --> 00:50:15
See you Monday.

848
00:50:10 --> 00:50:16.000