When we take a spring, something that we are
so familiar with now, and the spring has length l
in a relaxed state, spring constant k, I can extend the spring
with some force that I apply. The spring, then, will counter
it with the spring force and it will be
in equilibrium there. I call this the zero position,
and let's call this now delta l instead of x,
which we have done before. If I double the force,
delta l will double. Hooke's Law says that the
force is linear with delta l; in other words, delta l
is proportional with F. Nothing new,
as long as Hooke's Law holds. If I make the spring
twice as long, I would get double
the extension, because when I have
two springs in series, each one, under
the influence of this force, will get longer by this amount. Since I have two springs
in series, I will get twice delta l. So delta l is also proportional
to the length of my spring. If I take two springs parallel,
one like so and one like so, relaxed length l,
both the same spring constant k, and if now I apply
a force on it, then each one of these spring
forces is only half this one. Together they counter
this force. In other words,
the extension delta l that I obtain
from this external force is now only half as much
as it would be with one spring, and if I had three springs
parallel, all identical, I would only get one-third of
the extension for a given force. In other words, delta l is
also inversely proportional to the number of springs
that I have, assuming that they
are identical springs. Now I'm going to use
a rod or wire which has cross-sectional area A
and length l, and I'm going to apply
a force here. As a result of that force it will get longer
by a certain amount delta l, exactly like the spring. And clearly, when I make
this force stronger, delta l will increase, and
as long as Hooke's Law holds that the spring force provided
by the rod balances this out-- provided that the spring force
is linearly proportional with delta l-- I have again that delta l will
be proportional with the force. Double the force,
I get twice delta l. If I put two
of these rods together-- so I double the length
of the rod-- then clearly I will get
twice delta l, because each rod
will experience the force, each rod will get longer
by delta l, and so two rods will get
longer by two delta l. So again, delta l is
proportional with l. Suppose, now, I have two of
these rods next to each other-- notice the parallel with
the two parallel springs here-- and I apply a force F. Then the spring force
on each one of them-- if I call that
the spring force-- will only have to be half
to counter this force. So they both have
a cross-sectional area A, and so now with double
the cross-sectional area, I get only half of delta l. And so now
we have a situation that if I made a rod whereby
this was 2A, just one rod, which is completely
equivalent to this situation, I'm only getting half delta l
for a given force. And so now we have
that delta l is proportional... inversely proportional to the
cross-sectional area of the rod. So now we can make up
the balance, and we can say, "Aha! Delta l is proportional
to the force, proportional to l, "and inversely proportional
to the area, the cross-sectional area." So F divided by A is then
proportional to delta l over l, and that proportionality
constant we give a name, and that is capital Y, and
that is called Young's modulus. So this is Young's modulus. F over A, which has
the dimension of pressure-- force per unit area--
is what we call stress. And delta l over l,
which is dimensionless, we call that strain. If we compare two rods with different values
of Young's modulus, then the one with the smaller
value of Y for the same stress will give you a larger strain. In other words,
it's easier to make it longer. If Young's modulus is very high,
then the rod is extremely stiff. Then it is very difficult
to make the rod longer. I have here for you some numbers
which, of course, are on the Web so you don't have to copy them. And you have Young's modulus
there for various metals, and I also have it down there
for nylon, and today we will work
with that quite extensively. We could first do
a simple example just to get a feeling
for what is at stake here. I can take a rod with a radius
r, which is 0.5 centimeters. That would give me
a cross-sectional area of eight times ten
to the minus five square meters. So, yay thick, the rod. I make it very simple--
I make the length one meter, and I hang on the rod,
at the bottom, a mass M, which, let's say,
is 500 kilograms. Do I want 500 kilograms? Yes, I want 500 kilograms. In other words, the force
which I pull on the rod is about 5,000 newtons. So I can now ask myself how much longer
is this rod going to be? So we're going to have
that delta l is going to be F divided by A times
l divided by Young's modulus. And we do know what F is, we know what l is
and we know what A is. So in our case, this number is
6.3 times ten to the seventh for the numbers that we have. And so now we can look
at steel as an example. So we take steel. Y is 20 times ten to the tenth
newtons per square meter. And we substitute that in here, and we find that we get
an extension delta l which, I believe, is only
a third of a millimeter-- indeed, 0.3 millimeters. Think about this:
steel rod, a centimeter thick, and it has a length
of one meter, and I hang 500 kilograms on it and it only gets longer
by three-tenths of a millimeter. You couldn't even see that. However, if you go
and make the rod, which would be really a rope... If you make it out of nylon,
which has a Young's modulus... Oh, well, it's 55 times lower. I don't have to use
that number. It's about 55 times lower, so the delta l will
then be 55 times larger. And so, instead
of a third of a millimeter, I will get something
like 17 millimeters. And that you can see
with your naked eye. I take a rope
yay thick of nylon. I hang 500 kilograms on it and I will see it gets longer
by 1.7 centimeters. You can see that right
in front of your eyes. If I start adding
more weight on these bars, then very interesting things
are going to happen as we will discuss today. One thing which is obvious
that will ultimately happen if you keep loading down,
keep adding mass, it will break. And the breaking point is given
there in the third column, and we call that
the ultimate tensile strength. So when F over A-- when this
value-- becomes too large, that's what it comes down to,
then it will simply break. If we take the case that
we have here, the steel rod, for which F over A is 6.3
times ten to the seventh-- because l is one, remember-- then you can look there under
steel that it wouldn't break. We would be very safe,
because it wouldn't break until F over A becomes
five times ten to the eighth. So we're a factor of ten away
from that-- there's no problem. Even nylon would be very safe,
because that wouldn't break until the stress is three
times ten to the eighth. However, if we chose aluminum, if we made the bar out of
aluminum, then you're very close to this number 6.3 times
ten to the seventh. And so if you add a little bit
more mass than 500 kilograms, your aluminum bar would break. So you see, the properties
of these metals are very, very different indeed. Now, before it breaks,
we reach a situation that the strain is no longer
proportional to the stress. In other words,
we abandon Hooke's Law, as I will show you also today
with a demonstration. The material begins to deform. It begins to flow, in a way. And even when you take
the weights off, it will no longer have
its original length. It will not return
to the original length but it will be much longer. And I will try
to sketch you here what a curve of stress versus
strain typically looks like. So here I'm going to plot delta
l over l, and here F over A. So this is the stress
and this is the strain. In the beginning, you will see
a portion that is linear. That's Hooke's Law. And then,
when you keep adding force-- which we will do by gravity, we
will just hang weights on it-- then it starts to bend over,
up to a point here which we call the elastic limit. And even though this portion
is no longer linear-- so even though Hooke's Law
no longer holds-- still, if you take the weight
off the wire, off the rod, it will still come back to zero. Once you are past this point,
that is no longer the case. You will now see also that a small increase of stress
will give a huge strain, so the rope will... the rod will
get very long all of a sudden with a little bit
of extra stress. And if now you were to take
the weight off the rod, it would not come back to zero but it would come back
somewhere to here so it's permanently deformed. And in general the rod gets hot,
and the work that you have done goes into heat and goes
into the deformation of the rod. And so it goes like this
and then it goes like this and here it breaks. At this value for F over A, which is that third column
there, it will break. I'd like to discuss with you
this horizontal portion. With a little bit of luck we may actually be able to see
that with my demonstration, but it's hard to get
exactly at that point. If this part of the curve
is horizontal, it means that without any increase
of F over A, the wire will continue
to get longer and longer, and we call that plastic flow. So this whole portion
here is plastic flow. It almost becomes like a liquid. And then right here there is something very strange
that is happening. Then when it breaks, the stress here is
actually lower than there, and this is something I can
never show you in class, but I can explain to you
why that happens. When you are past
this point here, the rod begins to pinch,
and you get this. It's unpredictable
where it will pinch, but somewhere
it will start to pinch so that the area here is A prime whereas the cross-sectional
area here is A. And so F divided by A prime will be larger
than F divided by A. If you could do this experiment
in a controlled fashion-- and there are machines
designed to do that-- then you could actually
lower F over A so that the stress
actually goes down but that F over A prime
would still go up, and therefore
delta l will increase. And there are machines who are specially designed
to test these metals, and what they do is they go
up in very small steps of F-- and so they trace
this whole curve-- but by the time that they get
into the plastic flow area, they... before they increase the
force, they decrease it first. And if delta l gets larger
when they decrease it, they continue to decrease it,
and so that's the way that they can map out
also this portion. But we will not be able
to do that. There are metals
which are extremely brittle, and even though the curve
would look very similar-- it has all
these characteristics-- this point, then, would lie
all the way here. So this whole curve, then,
is squeezed into very small parameter space
of the strain. And so I want you to see most
of this, at least some of this, and we do that
with a demonstration which we have there, and I will make a drawing
of the basic idea. We have a copper wire-- you will get the dimension
from me very shortly-- and we attach to the copper wire
a rod, a solid rod. And at the end of the solid rod
is a mirror. This is a mirror. And we're going
to hang weights on here. But this mirror is
on a little platform and can pivot at this point
but cannot lower itself. The platform is fixed. For those of you
who are very close, this is where that platform is. That is a fixed platform. And so the mirror can only tilt
but cannot go down. And so now I will show you what happens
when this wire gets longer. So here is the wire,
and this is where the... This is this line here. But now the wire
has become longer. So the wire is now longer
by an amount delta l. So this is now the point where this bar, this rod
is attached. And so the mirror
will now be like this. We have tilted the mirror. And we have tilted the mirror
over an angle delta theta, and this length is l. And we're going to load it down
with a mass M, which we're going to increase. And then we have a laser beam
which we shine onto the mirror. The laser beam comes in like so. And this is
the normal to the mirror, so this angle here
is delta theta. The laser beam bounces off
and returns... like so, and this angle, of course,
is also delta theta. That's the property of a mirror. And this beam... This is the laser beam here
that we shine into the mirror. Here's the laser,
it goes into the mirror, and the spot there
of that return is all the way there
on the wall. So we are going to show you
where this spot is on the wall, and the wall is very far away. It's a distance capital L
from the wire. And you'll see very shortly
why we do it that way. Let me first give you the
dimensions of this instrument. L is 36 centimeters. It is copper. This bar, which has length d,
is 7½ centimeters. The distance to the wall
is about 16 meters. The radius of the wire... or actually, the diameter of
the wire, of the copper wire is 20/1,000 of an inch. I give it
in terms of the diameter because that's the way that
the manufacturer gives it to us. And that translates into a cross-sectional area
of that wire of 2.0 times ten to the
minus seven meters squared. All right, notice
that delta theta, that angle... the angle here, delta theta,
is delta l divided by d. So delta theta equals
delta l divided by d. The light that returns hits
the wall there at a distance l, so y at the wall,
or I can call it delta y-- that is at the wall, I call that
displacement there delta y-- divided by l
will be two delta theta. This is a small-angle
approximation; it's a very good approximation,
and theta is in radians. It is two delta theta,
because you see here that the change is
over an angle two delta. And so delta y equals
2L delta theta... 2L delta theta times
delta l divided by d. And look what we have done now: We have convert something
that is immeasurable, delta l-- which is fraction
of millimeters-- we have convert that to something on the wall
that we can measure, because this ratio, 2L over d, in our case, for the dimensions
that I have chosen, is about 425; it's like
a magnification factor. So if we see a displacement
of that laser beam on the wall of 40 centimeters,
which is easy to see, it means that the wire got
longer by only one millimeter. So 40 centimeters
there translates to one millimeter there, and if we would see that
laser beam go up four meters, it would mean that the wire would only have
become one centimeter longer. So it is a wonderful way to magnify the effect
and to measure it. So I will now make an attempt
to load down the copper wire. Oh, we can actually leave this
here, so you can see that curve. So we start here with...
we load up with half kilograms. We will write down, then, how much that laser spot
goes up on the wall. And then, in between
increasing the weight, increasing the force we will take the masses off
to see whether they return... whether the length of the rod
returns to the original length. And you will see after a while that you get
permanent deformation, then it no longer comes back
to its original length. In other words the laser spot will not return
to zero on the wall but it will stay higher. So Ron, if you are there... Oh, boy,
you're hiding behind... Um, maybe we want to move
the view graph out of the way so that students can also see. So, here is that copper wire which will be hard to see
for some of you-- it's only 20/1
of an inch thick-- and here is that mirror
which can pivot and can tilt. And Ron is going to put
weights on here, and then we will take
the weights off in between and we will try
to construct a curve of the stress versus strain, except that it is practical for
me to put here just the mass-- how many kilograms
we have on there-- because we know what A is,
so we can calculate F over A. That's not so important now. And here I simply have delta y. But keep in mind that delta y is always
425 times larger than delta l. So we're going to plot it,
and we're going to see whether we can come up
with a curve that is somewhat similar
to that one. So, Ron, if you put
on the first half kilogram... The mirror always starts
to oscillate a little bit and so we have to be
a little patient, and in the beginning, you
may be bored because--tja!-- we're going through that
linear part of the curve. So it goes up
very slowly, very gradually. We have five centimeters
for the first half kilogram. Could you remove
the half kilogram? It returns practically to zero. Maybe it's a little higher,
but that's not very significant. Could you make it one kilogram? Ah, ah, it's clearly higher. Ja, ja... oh, it's about
nine centimeters, nine, ten, so you see it's
in the linear part-- nine to ten centimeters. Can you remove
the half kilogram, Ron? The one kilogram,
there was one kilogram on it. We have to just wait,
let it damp out a little. It's oscillating. It's possible that you already
begin to see a small deformation which may be
one or 1½ centimeters. I put a question mark there--
it's possible. Can you put 1½ kilograms on? Yeah, I think it is... You can almost remove
the question mark. So now we are at 1½. So if it is strictly linear, you
would expect something like 15. Yeah, that's what it is, 15,
so it's still doing quite well. Can you take them off? One and a half. Ah, but you see
it no longer wants to return to its original length. It's clearly longer now. Permanent deformation
has already occurred, and so we're now something
like six centimeters. Can you make it two kilograms? Two kilograms. If it's linear,
you would expect near 20. It's still amazingly linear. It's as close
as I can see it to 20, but all these readings are
no more accurate than half a centimeter or so. So can you remove
the two kilograms? Oh, boy, look at that--
there's clear deformation now. It no longer returns
to zero, and it is... Oh, it's comfortably
ten centimeters long now. So can you make it 2½? You're now slowly approaching the part that I hope
you are going to see, and that is that it is going
to take off like a rocket, that with a little bit
of extra weight, it will start to move
substantially. We haven't reached that point
yet, but we are close to it. We're now 26-- 25, 26...
it still looks quite linear. Can you take it off, Ron? Actually, there's no need
to take it off anymore because it's clear that we... that we have
permanent deformation, and there's no sense
in following that, so why don't you make it
three kilograms? So what was it?
What did I say it was? What was the number I said? 20 what? 25 or so? So we have three now? Boy, this wire is hanging
in there, I must tell you. 32, yeah. Can you make it four? Watch very closely
now on the wall, because the drama is
about to start now. What did I say, 30...? I said 32. Four. Ooh, still moving, still moving. 52... settles at 53. Now, don't look at the board--
now, look at this spot now. Can you add... remember
the number, right? 53. Can you add one kilogram now? And look at that. It became almost twice as long
and it is still moving. Still going... still going. I hope it will settle. I'm going to write down my 53. Five kilograms. We're at five, right? 97-- now put on six. 97-- remind me, 97. Now, watch this point. Yay! Now you're clearly
in that plastic flow portion. By adding one kilogram, look where that point is--
it's still moving. What was five? 97? So we're now at six kilograms. Oh, actually, that is
still easy for me to estimate. I would say it's
about double the length of that stick
that we have on the wall, and the stick is
two meters long. Is it moving? That's
still moving a little bit. It's a little more
than four meters. Close enough, four meters,
just for the idea. Four meters, so that's 400. Put on seven. It will go
through the ceiling now. So we'll lose it. But what I want to do now, I want to get
to the breaking point. We can no longer measure
the displacement, but we're very close
to the breaking point now. So we're going to load it up
to the point that it will break, and that allows us to measure
the ultimate tensile strength. We're at seven now--
can you put eight on? Oh, we're running out of...
(chuckles) Oh, God! You could just see it sag
when the eight was put on. Did you actually look
at the wire? Okay, so at eight kilograms,
it breaks. Okay, let's put
these numbers in here. So, we have 1½... or half a kilogram,
we have five centimeters. Let me do this in color. So that gives me a point here. And then at one,
we have about ten. And at 1½, we have about 15. And at two, we have about 20. And at 2½, we have 25. Oh, I was a little bit
too high here, perhaps. I want to go
a little bit more carefully because this is
really terrific data. So this was at... And then we have...
at 2½, we have 25. Amazing how well...
how linear that is! And then at three, we have 32. Ah! It looks like
it's beginning to bend over. At four, we have 53,
no question. Oh, yeah. And then at five, we have 97. Here is five-- 97, that's here. And then our last point
that was... Oh, we even have a 400 here. Ooh, that's great. Here is 400. And what do we have there? A six. One, two, three,
four, five, six. That's our last point. And then we don't have the rest. But look how wonderful
this is, isn't it? Isn't that a great curve? Very linear in the beginning. Then it starts
to bend over, over. And you draw that line
anywhere you want to... and then it breaks. And so we can now actually
make an attempt to calculate Young's modulus
from the data. Of course, you'll have to select a portion where you think that
the data are reasonably linear. So Young's modulus itself...
equals F divided by A. Here you have here
the equation. Young's modulus equals F divided by A
times l divided by delta l. We know what A is,
we know what l is. It's still
on the blackboard there. And so now it's a matter of where do we think
that Hooke's Law still holds? I think this whole portion
would be fine, so we could take this point
as well as this point, because anywhere
on this straight line, you will get the same value
for Young's modulus. So I will use two kilograms
and 20 centimeters-- I will use this point. So F equals 200 newtons, and then we have delta y
equals 20 centimeters. And so that means delta l equals
20 divided by 425 centimeters. So this is F divided by A, l, and then we have here
delta y times 425. And let's see what that is. So F is 200-- that's right. No, no, F is 20--
ooh, ooh, ooh, ooh, ooh. I'm glad that we caught
that simultaneously. So F is 20-- 20 newtons,
that's right. The area is two times
ten to the minus seven, and I divide by the area. I multiply by the length,
which is .36. I multiply by 425
and I divide by delta y. But I need delta y
in meters, so that's .2. And I get 7.7 times
ten to the tenth. And that is not bad at all because I think,
from what I remember, is that it is 11
times ten to the tenth. That's what it is. So that is quite amazing for such a crude measurement
that we do here. We can also measure
the ultimate tensile strength. That is, we can measure
the value for F over A when it broke. So that happened, I think...
wasn't that eight kilograms? So that will be 80 divided
by the area that we know, and so that gives me
80 divided by two times... divided by ten to the
minus seventh, and that is four times ten to the eighth
newtons per square meter. And this is also
newtons per square meter. And that's not bad. It's a little higher
than we have there, but it's
a very crude measurement. And don't forget, we were unable to follow the portion
when it was going down. We only went up
in mass, in weight, and so that value there
takes into account that the curve comes down,
something we could not do. We didn't have
the means of doing that. The percentage strain is
actually extremely low in the portion
that the curve is linear. You can ask yourself
the question "What is delta l divided by l in terms of percentages
during this portion here?" Well, we know the length, 0.36, and when we take the case
where we sort of reach the end of the Hooke's Law
parameter space, we have a delta l which is
425 times lower than this, so we have 20 divided by 425,
and that is, um... four point seven
times ten to the minus two, but that is in terms
of centimeters and we want it in terms of
meters, so we have to multiply by another factor
ten to the minus two. So the exponent's changed--
minus two... divided by 0.36, and that gives me 1.3 times
ten to the minus three. In terms of percentages,
that would be 0.13%. So the wire, when it reaches the end of its Hooke's Law
parameter space, has only become longer by 0.13%. And when the wire breaks... In general, for metals it's maybe five or ten percent
longer than its original length. That's a typical value
for metals. Now, as long as I am in the
linear portion of the curve, I can generate
a simple harmonic motion. I want to show you that. I need my wiper. Because in the linear portion
of the curve where Hooke's Law holds, I could hang a weight
on the rope... or on the rod and I could let it
oscillate vertically. So the force that you have equals y times A times delta l
divided by l. That's the force that I apply. So the spring force
is opposing that, so the spring force
has a minus sign here to indicate the direction. And so this is similar
to F equals minus kX that you have seen
with the spring. And this now is our k,
and delta l is our x. And so you can predict
that it will start to oscillate with an angular frequency of omega square root of k
divided by m and with a period
which is two pi... two pi times
the square root of m over k, and m is now the mass
that I'm hanging here. If we take our copper wire,
we know what y is. If I take the value 11 instead
of the one that we found, but it's very close, anyhow... So if I take our copper wire, then I will find, depending upon
the mass that I hang on it... then I'll find in any case
for k, for this value k I find five times ten to the
fourth newtons per meter. So if now I hang on it
a mass of one kilogram, I can calculate the period
of one oscillation, and one over the period
is the frequency F, and I believe that is
something like 38 hertz. So it would start to oscillate
like this, 38 times per second, and if I hang
two kilograms on it, then this frequency would become
something like 25 hertz, because a higher mass gives me
a longer period, gives me a lower frequency. The speed of sound also depends
on Young's modulus. Without proof-- you will see
this if you ever take 803-- I will tell you that the speed of sound is
the square root of Young's modulus divided by
the density of the material. And I have listed those there
on the view graph. Oh, this is by, yeah, the
square root-- I have that, yeah. And so the higher
Young's modulus is, the higher
the speed of sound is, and that is intuitively
sort of pleasing. I can sort of understand that although the square root
is hard to see. If I have here a rod, a bar and I give this bar here
a bang, I hit it, if this bar were
infinitely stiff-- that means Young's modulus
were infinitely high-- then the bar would
instantaneously move here when I hit it there. But if the bar is
not infinitely stiff, if it has a certain amount
of elasticity, then what happens... then I hit
it here and I produce here some kind
of local increased pressure which is going to propagate
to this end-- it's like a pressure wave,
like sound is a pressure wave-- and that takes time. And the larger
Young's modulus is, the stiffer the material is,
the faster that will go. If Young's modulus is very low,
the material is more elastic. It will take a longer time for
this pulse to reach this end. So it is sort of
intuitively pleasing for me that the stiffness
of the material is related
to the speed of sound. I have here a magnesium bar, and the magnesium bar
has a length l. And I can calculate
the speed of sound for magnesium by taking its Young's modulus,
which is there, divided by rho, which is there, and I come up
with the speed of sound, which is about
five kilometers per second. If you look at those numbers, they are all
substantially higher than the speed of sound in air, which is only some
340 meters per second. So the speed of sound
for magnesium is some 15 times larger
than the speed of sound in air. If this bar has a length l, this pressure disturbance
will start to move to the left, and then it'll come back here, so it will have made
a complete journey, which I call the period... It has made a complete journey
in so many seconds-- 2L divided
by the speed of sound. And so the frequency
of this bar-- the frequency at which
it would like to oscillate when I give it a bang-- is one over T, and that is the
speed of sound divided by 2L. Now, for those of you
who will take 803 in the future, you will see a much better
derivation of this frequency, so this is a poor man's version. The speed of sound is about
five kilometers per second. The length of the bar
is about 122 centimeters, and so that translates
into a frequency of roughly 2,100 hertz. So the frequency
for this magnesium bar is about 2,100 hertz,
and you can hear that. It's a beautiful tone. I hold it here
and I will bang it here. (bar chimes, tone oscillates) Can you hear it? No? (bar chimes louder,
tone oscillates again) You can hear it?
2,100 hertz, beautiful tone. (Sound dies off) Remember earlier in the course that we were wrestling
with this problem. We had a block and we had
two strings attached to it. Here is one string,
here is the block, and here is another string. And we're going to pull
on this here with force F. Nothing was happening,
so the tension here, T, is F. Here, there is both the force F
plus the weight of the object, so here you have... call it
T prime equals F plus Mg. And so we argued if you increase F
and if the strings are identical that it should break
first here and then here, because the breaking point where the force is too large
for the string will occur first here,
then it occurs here, because you have
the extra amount Mg. But yet when we jerked on it
very fast, it would break here, and when we pulled very slowly--
I will do it again-- it would break here. And now we can
fully understand that. Because what does it mean, that
this string is going to break? It means that string
has to get longer by a certain amount delta l before it says, "Sorry,
I got to go," and it breaks. For this string to get longer
by a certain amount delta l, this block has to come down. Now, if I pull here with a
certain force F-- F equals Ma-- the block will come down
with an acceleration a, and in the time delta t, it will move over a distance
a delta t squared, which is the delta l
that this string will feel. But that takes time to reach the delta l
at which it wants to break, and so if I pull very fast,
I don't give it that time. And that's why, then, the one
at the bottom will break, and if I pull very slowly,
the one at the top will break. So here is a very thin wire--
you can't see it very well-- and here is one, too. It's a string,
and they're identical, and this is my
emergency safety rope. And if I pull very fast,
you see the bottom one breaks. I have this in my hand,
and the top one is still safe. However, if I repeat this,
and I do it very gently... It's always hard to get this in. There we go. Uh-oh... Okay, so now I'm going
to slowly increase the force. I give the block plenty of time
to come down, plenty of time, and now the upper one goes. Have a good weekend. See you Monday.
