Lec 06: Newton’s First, Second, and Third Laws | 8.01 Classical Mechanics, Fall 1999 (Walter Lewin)

Lec 06: Newton’s First, Second, and Third Laws | 8.01 Classical Mechanics, Fall 1999 (Walter Lewin)

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Last time we discussed
that an acceleration is caused by a push or by a pull. Today we will express this
more qualitatively in three laws which
are called Newton’s Laws. The first law really goes back to the first part
of the 17th century. It was Galileo who expressed what he called
the law of inertia and I will read you his law. “A body at rest remains at rest “and a body in motion
continues to move “at constant velocity
along a straight line unless acted upon
by an external force.” And now I will read to you Newton’s own words
in his famous book,Principia. “Every body perseveres
in its state of rest “or of uniform motion
in a right line “unless it is compelled
to change that state by forces impressed upon it.” Now, Newton’s First Law is clearly against
our daily experiences. Things that move don’t move
along a straight line and don’t continue to move, and
the reason is, there’s gravity. And there is another reason. Even if you remove gravity then there is friction,
there’s air drag. And so things will
always come to a halt. But we believe, though, that
in the absence of any forces indeed an object, if it had
a certain velocity would continue along a straight
line forever and ever and ever. Now, this law,
this very fundamental law does not hold
in all reference frames. For instance, it doesn’t hold
in a reference frame which itself is
being accelerated. Imagine that I accelerate
myself right here. Either I jump on my horse,
or I take my bicycle or my motorcycle or my car and you see me being accelerated
in this direction. And you sit there and you say,
“Aha, his velocity is changing. “Therefore, according
to the First Law, there must be a force on him.” And you say, “Hey, there,
do you feel that force?” And I said, “Yeah, I do! “I really feel that,
I feel someone’s pushing me.” Consistent with the first law. Perfect, the First Law works
for you. Now I’m here. I’m being accelerated
in this direction and you all come towards me being accelerated
in this direction. I say, “Aha, the First Law
should work so these people
should feel a push.” I say, “Hey, there! Do you feel the push?” And you say, “I feel nothing. There is no push,
there is no pull.” Therefore, the First Law doesn’t
work from my frame of reference if I’m being accelerated
towards you. So now comes the question,
when does the First Law work? Well, the First Law works
when the frame of reference is what we call an “inertial”
frame of reference. And an inertial frame
of reference would then be a frame in which there are
no accelerations of any kind. Is that possible? Is 26.100…
is this lecture hall an inertial reference frame? For one, the earth rotates
about its own axis and 26.100 goes with it. That gives you
a centripetal acceleration. Number two, the earth
goes around the sun. That gives it
a centripetal acceleration including the earth, including
you, including 26.100. The sun goes around the Milky
Way, and you can go on and on. So clearly 26.100 is not
an inertial reference frame. We can try to make an estimate on how large
these accelerations are that we experience
here in 26.100 and let’s start with the one that is due
to the earth’s rotation. So here’s the earth… rotating
with angular velocity omega and here is the equator, and
the earth has a certain radius. The radius of the earth…
this is the symbol for earth. Now, I know that 26.100 is here but let’s just take the worst
case that you’re on the equator. You’re… (no audio ) You go around like this
and in order to do that you need a centripetal
acceleration, a c which, as we have seen last
time, equals omega squared R. How large is that one? Well, the period of rotation
for the earth is 24 hours times 3,600 seconds so omega equals two pi
divided by 24 times 3,600 and that would then be
in radians per second. And so you can calculate now
what omega squared R earth is if you know that the radius
of the earth is about 6,400 kilometers. Make sure you convert this
to meters, of course. And you will find, then that the centripetal
acceleration at the equator which is the worst case–
it’s less here– is 0.034 meters
per second squared. And this is way, way less–
this is 300 times smaller than the gravitational
acceleration that you experience
here on Earth. And if we take the motion
of the earth around the sun then it is an additional factor
of five times lower. In other words,
these accelerations even though they’re real
and they can be measured easily with today’s
high-tech instrumentation– they are much, much lower
than what we are used to which is
the gravitational acceleration. And therefore,
in spite of these accelerations we will accept this hall as a reasonably good
inertial frame of reference in which the First Law
then should hold. Can Newton’s Law be proven? The answer is no, because
it’s impossible to be sure that your reference frame
is without any accelerations. Do we believe in this? Yes, we do. We believe in it
since it is consistent within the uncertainty
of the measurements with all experiments
that have been done. Now we come to the Second Law,
Newton’s Second Law. I have a spring… Forget gravity for now– you can do this somewhere
in outer space. This is the relaxed length
of the spring and I extend the spring. I extend it over a certain
amount, a certain distance– unimportant how much. And I know that I when I do that
that there will be a pull– non-negotiable. I put a mass, m1, here,
and I measure the acceleration that this pull causes
on this mass immediately after I release it. I can measure that. So I measure
an acceleration, a1. Now I replace this object
by mass m2 but the extension is the same,
so the pull must be same. The spring doesn’t know what the
mass is at the other end, right? So the pull is the same. I put m2 there, different mass and I measure
the new acceleration, a2. It is now an experimental fact
that m1 a1 equals m2 a2. And this product, ma,
we call the force. That is our definition
of force. So the same pull
on a ten times larger mass would give a ten times
lower acceleration. The Second Law
I will read to you: “A force action on a body
gives it an acceleration which is in the direction
of the force…” That’s also important– the acceleration is in
the direction of the force. “And has a magnitude
given by ma.” ma is the magnitude and the direction is
the direction of the force. And so now we will write
this in all glorious detail. This is the Second Law by Newton perhaps the most important law
in all of physics but certainly in all of 801: F equals ma. The units of this force are kilograms times meters
per second squared. In honor of the great man,
we call that “one newton.” Like the First Law,
the Second Law only holds in inertial reference frames. Can the Second Law be proven? No. Do we believe in it? Yes. Why do we believe in it? Because all experiments
and all measurements within the uncertainty
of the measurements are in agreement
with the Second Law. Now you may object
and you may say “This is strange,
what you’ve been doing. “How can you
ever determine a mass “if there is no force somewhere? “Because if you want
to determine the mass “maybe you put it on a scale, “and when you put it on a scale
to determine the mass “you made use
of gravitational force “so isn’t that some kind
of a circular argument that you’re using?” And your answer is “No.” I can be somewhere
in outer space where there is no gravity. I have two pieces of cheese;
they are identical in size. This is cheese without holes,
by the way. They are identical in size. The sum of the two has
double the mass of one. Mass is determined
by how many molecules– how many atoms I have. I don’t need gravity to have
a relative scale of masses so I can determine the relative
scale of these masses without ever using the force. So this is a very legitimate way of checking up
on the Second Law. Since all objects in
this lecture hall and the earth fall with the constant
acceleration, which is g we can write down
that the gravitational force would be m times
this acceleration, g. Normally I write an “a” for it,
but I make an exception now because gravity, I call it
“gravitational force.” And so you see
that the gravitational force due to the earth
on a particular mass is linearly proportional
with the mass. If the mass becomes
ten times larger then the force due to gravity
goes up by a factor of ten. Suppose I have here
this softball in my hands. In the reference frame… 26.100 we will accept to be
an inertial reference frame. It’s not being accelerated
in our reference frame. That means the force on it
must be zero. So here is that ball. And we know if it has mass, m– which in this case
is about half a kilogram– that there must be
a force here, mg which is about five newtons,
or half a kilogram. But the net force is zero. Therefore it is very clear that I, Walter Lewin,
must push up with a force from my hand onto the ball,
which is about the same… which is exactly the same,
five newtons. Only now is
there no acceleration so I can write down
that force of Walter Lewin plus the force of gravity
equals zero. Because it’s
a one-dimensional problem you could say that the force
of Walter Lewin equals minus mg. F equals ma. Notice that there is
no statement made on velocity or speed. As long as you know f
and as long as you know m a is uniquely specified. No information is needed
on the speed. So that would mean,
if we take gravity and an object was falling down
with five meters per second that the law would hold. If it would fall down
with 5,000 meters per second it would also hold. Will it always hold? No. Once your speed approaches
the speed of light then Newtonian mechanics
no longer works. Then you have to use Einstein’s
theory of special relativity. So this is only valid
as long as we have speeds that are substantially smaller,
say, than the speed of light. Now we come
to Newton’s Third Law: “If one object exerts
a force on another “the other exerts the same force in opposite direction
on the one.” I’ll read it again. “If one object exerts
a force on another “the other exerts the same force in opposite direction
on the one.” And I normally summarize that
as follows, the Third Law as “Action equals
minus reaction.” And the minus sign indicates,
then, that it opposes so you sit on your seats and you are pulled down on
your seats because of gravity and the seats will push back
on you with the same force. Action equals minus reaction. I held the baseball in my hand. The baseball pushes on my hand
with a certain force. I push on the baseball
with the same force. I push against the wall
with a certain force. The wall pushes back
in the opposite direction with exactly the same force. The Third Law always holds. Whether the objects
are moving or accelerated makes no difference. All moments in time, the force– we call it actually the “contact
force” between two objects– one on the other is always
the same as the other on one but in the opposite direction. Let us work out
a very simple example. We have an object
which has a mass, m1. We have object number one
and m1 is five kilograms. And here, attached to it,
is an object two and m2 equals 15 kilograms. There is a force and the force is coming in
from this direction. This is the force– and the magnitude
of the force is 20 newtons. What is the acceleration
of this system? F equals ma. Clearly the mass is
the sum of the two– this force acts on both– so we get m1 plus m2 times a. This is 20, this is 20 so a equals
one meters per second squared in the same direction as f. So the whole system
is being accelerated with one meters
per second squared. Now watch me closely. Now I single out this object– here it is… object number two. Object number one, while
this acceleration takes place must be pushing
on object number two. Otherwise object number two
could never be accelerated. I call that force f12 the force
that one exerts on two. I know that number two
has an acceleration of one. That’s a given already. So here comes f equals ma. f12 equals m2 times a. We know a is one,
we know m2 is 15 so we see that the magnitude
of the force 12 is 15 newtons. This force is 15. Now I’m going to isolate
number one out. Here is number one. Number one experiences
this force, f, which was the 20 and it must experience
a contact force from number two. Somehow, number two must
be pushing on number one if one is pushing on number two. And I call that force “f21.” I know that number one
is being accelerated and I know the magnitude is
one meter per second squared. That’s non-negotiable, and so we have that f,
this one, plus f21 must be m1 times a. This is one, this is five,
this is 20 and so this one, you can
already see, is minus 15. F21 is in this direction and the magnitude is exactly
the same as f12. So you see? One is pushing on two with
15 newtons in this direction. Two is pushing back on one
with 15 newtons and the whole system
is being accelerated with one meter
per second squared. Now, in these two examples– the one whereby I had
the baseball on my hand– you saw that it was consistent
with the Third Law. In this example, you also see that it’s consistent
with the Third Law. The contact force
from one on the other is the same as
from the other on one but in opposite signs. Is this a proof? No. Can the Third Law be proven? No. Do we believe in it? Yes. Why do we believe in it? Because all measurements,
all experiments within the uncertainties are
consistent with the Third Law. Action equals minus reaction. It is something
that you experience every day. I remember I had
a garden hose on the lawn and I would open the faucet and the garden hose would start
to snake backwards. Why? Water squirts out. The garden hose pushes onto
the water in this direction. The water pushes back onto the
garden hose and it snakes back. Action equals minus reaction. You take a balloon. You take a balloon
and you blow up the balloon and you let the air out. The balloon pushes onto the air. The air must push
onto the balloon. And therefore,
when you let it go the balloon will go
in this direction which is the basic idea
behind the rocket. (huffing and puffing ) I love to play with balloons,
don’t you? So, if I do it like this,
and I let it go the air will come out
in this direction and so then it means the balloon is pushing on the air
in this direction. the air must be pushing on
the balloon in this direction. There it goes. (whistles ) It didn’t make it to the moon but you saw the idea
of a rocket. Action equals minus reaction. If you fire a gun, the gun
exerts a force on the bullet the bullet exerts
an equal force on the gun which is called the recoil. You feel that in your hands
and your shoulder. I have here a marvelous device which is a beautiful example of
“action equals minus reaction.” I show you from above
what it looks like. You’ll see more details later. This rotates about this axis
rather freely– the axis is vertical– and we have here a reservoir
of water, which we will heat up. It turns into steam and these are hollow tubes
and the steam will squirt out. And so when the steam squirts
out in this direction the tube exerts a force
on the steam in this direction so the steam exerts an equal
force in the opposite direction and so the thing will start
to rotate like this. And I would like
to demonstrate that. You can see it now there. With a little bit of luck,
there you see it. So we’re going to heat it. (torch hissing ) Walking. When you walk,
you push against the floor. The floor pushes back at you and if the floor
wouldn’t push back at you you couldn’t even walk,
you couldn’t go forwards. If you walk on ice,
very slippery– you can’t go anywhere, because
you can’t push on the ice so the ice won’t push back
on you. That’s another example
where you see action equals minus reaction. This engine is called
“Hero’s engine.” Hero, according
to the Greek legend was a priestess of Aphrodite. Let’s first look at it. She was a priestess of Aphrodite
and her lover, Leander would swim across the Hellespont
every night to be with her. And then one night
the poor guy drowned and Hero threw herself
into the sea. Very romantic thing to do but, of course, also
not a very smart thing to do. On the other hand, it must
have been a smart lady if she invented,
really, this engine. Yesterday, I looked
at the Web, “ask.com.” It’s wonderful–
you can ask any question. You can say, “How old am I?” Now, you may not get
the right answer but you can ask any question. And I typed in, “Hero’s engine.” And out popped a very nice high-
tech version of Hero’s engine. A soda can– you pop four holes
in the soda can at the bottom. So here’s your soda can. You pop four holes in here,
but when you put a nail in there you bend every time
the nail to the same side so the holes are slanted. You put it in water you lift it out of water
and you have a Hero’s engine. And I made it for you–
it took me only five minutes. I went to one of MIT’s machines,
got myself a soda put the holes in it,
and here it is. It’s in the water there. When I lift it out,
you will see the water squirts. There it goes. High-tech version
of Hero’s engine. Also makes a bit of a mess,
but okay. All right. Try to make one– it’s fun
and it’s very quick. It doesn’t take
much time at all. There are some bizarre
consequences of these laws. Imagine that an object
is falling towards the earth. An apple is falling
towards the earth from a height, say, of,
hmm, I’d say 100 meters. And let’s calculate
how long it takes for this apple to hit the earth which should for you
be trivial, of course. So here’s the earth… and the mass of the earth is about 6 times 10
to the 24 kilograms. And here at a distance, h– for which we will take
100 meters– is this apple, m, which, say,
has a mass of half a kilogram. There’s a force
from the earth onto the apple and this is that force. And the magnitude of that force
is mg and that is 5 newton. I make g ten and
just round it off a little. Now, how long does it take
this object to hit the earth? So, we know that
1/2 gt squared equals h. It doesn’t start with any
initial speed, so that is 100. G is 10, this is 5,
so t squared is 20. So t is about 4½ seconds. So after 4½ seconds, it hits
the earth– so far, so good. But now, according
to the Third Law the earth must experience exactly the same force
as the apple does but in opposite direction. So therefore the earth will
experience this same force, f– 5 newton, in this direction. What is the earth going to do? Well, the earth is going to fall
towards the apple– f equals ma. So the force on the earth
is the mass of the earth times the acceleration
of the earth. The force, we know, is 5. We know the mass,
6 times 10 to the 24 so the acceleration will be 5
divided by 6 times 10 to the 24 which is about 8 times 10 to the minus 25 meters
per second squared. How long will the earth fall? Well, the earth will fall
roughly 4½ seconds before they collide. How far does the earth move
in the 4½ seconds? Well, it moves
one-half a earth t squared. That’s the distance
that it moves. We know a and we know t squared,
which is 20. One-half times 20 is 10 so that means this distance
becomes that number times 10. It’s about 8 times 10
to the minus 24 meters. The earth moves 8 times 10
to the minus 24 meters. That, of course, is
impossible to measure. But just imagine what
a wonderful concept this is! When this ball falls back to me the earth and you and I and MIT
are falling towards the ball. Every time that the ball
comes down we’re falling towards the ball. Imagine the power I have
over you and over the earth! But you may want
to think about this– if I throw the ball up,
going to be away from the earth I’ll bet you anything that the earth will also
go away from the ball. So as I do this,
casually playing– believe me, man,
what a glorious feeling it is– earth is going down, earth
is coming towards the ball. The earth is going down
and I’m part of the earth and I’m shaking this earth
up and down by simply playing
with this ball. That is the consequence
of Newton’s Third Law even though the amount
by which the earth moves is, of course,
too small to be measured. I now want to work out with you
a rather detailed example of something in which we combine
what we have learned today– a down-to-earth problem– the kind of a problem
that you might see on an exam
or on an assignment. We hang an object on two strings and one string makes an angle
of 60 degrees with the vertical and the other makes an angle
of 45 degrees with the vertical. So this is the one
that makes an angle… oh, 60 degrees with the horizon,
30 degrees with the vertical and this one, 45 degrees. Let’s assume that the strings
have negligible mass. So they are attached here
to the ceiling and I hang here an object, m. Well, if there’s an object m for sure there will be
a force mg, gravitational force. This object is hanging there,
it’s not being accelerated so the net acceleration
must be zero. And so one string must
be pulling in this direction and the other string must
be pulling in this direction so that the net force
on the system is zero. Let’s call this pull,
for now, “T1.” We’ll call that the tension
in the string and we call the tension
in this string “T2.” And the question now is how
large is T1 and how large is T2? There are various ways
you can do this. One way that always works–
pretty safe– you call this the x direction. You may choose which direction
you call “plus.” I call this plus,
I call this negative. And you could call this
the y direction and you may call this plus
and this negative. I know, from Newton’s
Second Law– F equals ma– that there is no acceleration,
so this must be zero so the sum of all forces
on that mass must be zero. These three forces must eat
each other up, so to speak.
Well, if that’s the case,
then the sum of all forces in the x direction
must also be zero because there’s no acceleration
in the x direction and the sum of all forces
in the y direction must be zero. And so I am going
to decompose them– something we have done before. I am going to decompose
the forces into an x and
into a y direction. So here comes
the x component of T1 and its magnitude is T1
times the cosine of 60 degrees. Now I want to know
what this one is. This one is T1 times
the sine of 60 degrees. This projection, T2,
cosign 45 degrees and the y component,
T2 times the sine of 45 degrees. So we go into the x direction. In the x direction
I have T1 cosign 60 degrees minus T2 cosign 45 degrees
equals zero– that’s one equation. The cosine of 60 degrees
is one-half and the cosine of 45 degrees
is one-half square root two. Now I go to the y direction. This is plus, this is minus,
so we get one component here which is T1 times
the sine of 60 degrees plus T2 times the sine
of 45 degreesminus mg. It’s in the opposite direction–
must be zero. That’s my second equation. The sine of 60 degrees equals
one-half the square root three and the sine of 45 degrees is the same as the cosine
one-half square root two. Notice I have two equations
with two unknowns. If you tell me what m is I should be able to solve
for T1 and for T2. In fact, if we add them up it’s going to be very easy
because we lose this because we have both
one-half square root two. And so you see immediately here
that one-half times T1 plus one-half square root three
times T1 equals mg and so you find that
the tension 1 equals two mg divided by one
plus the square root of three. I can go back now
to this equation– T1 times one-half equals T2 times one-half
square root of two. I lose my half and so T2 equals T1 divided
by the square root of two. So the bottom line is,
you tell me what m is I’ll tell you what T1 is
and I’ll tell you what T2 is. Suppose we take a mass
of four kilograms– m equals four kilograms,
so mg is about 40 if we make g ten for simplicity. Then T1, if you put in
the numbers, is about 29.3 and T2… 29.3 newtons and T2 is about 20.7 newtons,
I believe. It’s very difficult to rig
this up as an experiment but I’ve tried that. I’ll show you in a minute. I want you to know
that there is another method which is perhaps
even more elegant and which you may consider in which there is
no decomposition in the two directions. Here is mg– that’s a given. And we know that the other
directions are also given– this angle of 30 degrees here
and this angle of 45 degrees. If these two forces
must cancel out this one why don’t I flip this one over? Here it comes. I flip it over. There it is. T1 and T2 now, together,
must add up to this one. Then the problem is solved,
then the net force is zero. Well, that’s easy– I do this. And now I have constructed a complete fair construction
of T1 and of T2. No physics anymore now,
it’s all over. You know this angle here, 45
degrees, so this is 45 degrees. This is 30, this is 30. You know all the angles and
you know this magnitude is mg so it’s a high school problem. You have a triangle with
all the angles and one side; you can calculate
the other sides and you should find exactly
the same answer, of course. We made an attempt to rig it up. How do we measure tension? Well, we put in these lines,
scales, tension meters and that is problematic,
believe me. We put in here a tension meter,
we put in here a tension meter and the bottom one, we hang on a
string with a tension meter and then here we put
four kilograms. These scales are not masses. That’s already problematic. The scales are
not very accurate so we may not even come close
to these numbers. For sure,
if I put four kilograms here then I would like this one
to read 40 newtons or somewhere
in that neighborhood depending on
how accurate my meters are. These are springs,
and the springs extend and when the springs extend,
you see a handle… a hand go. You can clearly see
how that works because if there is a force
on that bottom scale in this direction, which is mg,
and it’s not being accelerated then the string
must pull upwards and so… in order to make
the net force zero. And if you have a pull down here
and you have a pull up here and you have in here a spring then you see you have
a way of measuring that force. We often do that– we measure with springs
the tension in strings. For whatever it’s worth, I will
show you what we rigged up. Now a measurement without
knowledge of uncertainties is meaningless–
I told you that. So maybe this is meaningless,
what I am going to do now. Let me do something meaningless
for once. And remember, when I show it,
you can always close your eyes so that you haven’t seen it. So we have here something
that approaches this 60 degrees and this approaches
the 45 degrees and we’re going to hang
four kilograms at the bottom. There it is, and here it is. All right, this one–
it’s not too far from 40. It’s not an embarrassment. This one is not too far
from 20.7. This one is a bit
on the low side. Maybe I can push it up a little. I think that’s close to 30;
it’s not bad. So you see, it’s very difficult
to get these angles right but it’s not too far off. So let’s remove this again because this will block
your view. These scales were calibrated
in newtons, as you could see. Now we come
to something very delicate. Now I need your alertness
and I need your help. I have a block–
you see it there– and that block weighs
two kilograms. A red block. So here it is. It’s red. And I have two strings. It’s hanging from a black string
here and a black string there. Ignore that red string,
that is just a safety. But it’s avery thin thread
here and here. And they are as close
as we can make them the same. They come from the same batch. This one has a mass
of two kilograms and this string has no mass. This is two kilograms. So what will be the tension
in the upper string which is string number one? This is string number two. Well, this string must be able
to carry this two kilograms so the tension has to be
20 newtons. So you will find here
the tension– call it T1– which is about 20 newtons. So it’s pulling
up on this object. It’s also pulling down
from the ceiling, by the way. Think about it, it’s pulling
from the ceiling. The tension is here, 20 newtons. We could put in here
one of these scales and you would see
approximately 20 newtons. What is the tension here? Well, the tension here
is very close to zero. There’s nothing hanging on it
and the string has no weight so there’s no tension there–
you can see that. Now I am going to pull on here and I’m going to increase
the tension on the bottom one until one of the two breaks. So this tension goes up and up and therefore, since this object
is not being accelerated– we’re going to get a force
down now on this object– this tension must increase,
right? You see that? If I have a force on this one… so there’s a force here,
and there is mg then, of course, this string
must now be mg plus this force.
So the tension will go up here
and the tension will go up here. The strings are as identical
as they can be. Which of the strings
will break first? What do you think? LEWIN:
Excuse me? (student answers
unintelligibly ) I can’t hear you. STUDENT:
The one on top. LEWIN:
The one on top. Who is in favor
of the one on top? Who says no, the bottom one? (Student answers
unintelligibly ) LEWIN:
Who says they won’t break
at all? Okay, let’s take a look at it. The one on top–
that’s the most likely, right? Three, two, one, zero. The bottom one broke. My goodness. Newton’s Second Law is at stake. Newton’s Third Law is at stake. The whole world is at stake! Something is not working. I increased tension here,
this one didn’t break. This one’s stronger, perhaps. No, I don’t cheat on you;
I’m not a magician. I want to teach you physics. Did we overlook something? You know,
I’ll give you a second chance. We’ll do it again. Let’s have another vote. So I’ll give you a chance
to change your minds. It’s nothing wrong in life,
changing your mind. It’s one of the greatest things
that you can do. What do you think
will happen now? Who is in favor still
of the top one? Seeing is believing. You still insist on the top one? Who is now in favor
of the bottom one? Ah, many of you
got converted, right? Okay, there we go. Three, two, one, zero. The top one broke. So some of you were right. Now I’m getting so confused. I can’t believe it anymore. First we argued
that the top one should break but it didn’t–
the bottom one broke. Then we had another vote
and then the top one broke. Is someone pulling our leg? I suggest we do it
one more time. I suggest we do it one more time and whatever’s going to happen,
that’s the winner. If the top one breaks,
that’s the winner. If the bottom one breaks, well,
then, we have to accept that. But I want you to vote again. I want you to vote again
on this decisive measurement whether the top one will break
first or the bottom one? Who is in favor of the top one? Many of you are scared, right? You’re notvoting anymore! (class laughs ) LEWIN:
I can tell, you’re not voting. Who is in favor
of the bottom one? Only ten people are voting. (class laughs ) LEWIN:
Let’s do this
in an undemocratic way. You may decide–
what’s your name? Alicia? Georgia, close enough. (laughter ) You may decide
whether the top one or the bottom one will break. Isn’t that great?
Doesn’t it give you
a fantastic amount of power? The bottom one. The bottom one. You ready? Three, two, one, zero. The bottom one broke. You were right. You will pass this course. Thank you,
and see you Wednesday. By the way, think about this,
think about this.

36 thoughts on “Lec 06: Newton’s First, Second, and Third Laws | 8.01 Classical Mechanics, Fall 1999 (Walter Lewin)

  • Abdul AB Post author

    the director is the worst, why the fuck he is filming the students!! I cannot concentrate. 

  • Iamdynamite Post author

    can someone explain the last experiment? it was hard to see with the bad quality of the video. The 2kg block is hanging by the top string, how does the bottom string get involved?

  • Louie McConnell Post author

    are there any problem sets/exams?

  • qualquan Post author

    Professor Lewin isn't the 3rd Law of Newton (action = reaction) just conservation of momentum and not a new law at all? I see no difference.

  • follow the light Post author

    I've never understood this. If a particle A exerts a force on particle B and it's equal and opposite. Say, both these particles had significantly different masses. Then the force exerted by a on b pushes the particle b in the direction of the force right? But then particle A remains stationary and if particle B is exerting the same force as A just in the opposite direction then why does particle A not move. I can understand that there is difference of mass and stuff. But mathematically it doesn't hold bc particle a remains stationary and has an accelatrion of 0 which makes the force acting on it 0. where as the equal and opposite force is not 0?!?!?!?! Someone put me out of my misery, ive not understood this since 6th grade!

  • Seineele Moagi Post author

    lmao,akward ending

  • Ilic Sorrentino Post author

    he is the best. the physics teacher I've always wanted having at school. a pity for the MIT case. I discovered him a little bit late in my life but he really improved my knowledge.

  • Nikhil Banka Post author

    Physics.Yes I'm in love with it <3

  • Ksuy A Post author

    i was this was HD Walter Lewin's <3

  • Alex De La Torre Post author

    Wouldn't F1on2 be 5N? because the mass is 5kg and the acceleration is 1m/s^2?

  • MrDJRObert1234 Post author

    wait, can someone explain to me the balloon example. When i release the balloon the air inside pushes outward so the reaction is in the opposite way which makes it go forward. But why is it going forward? isnt it supposed to stay still? since action equals minus reaction?

  • Jayden.theKid Post author

    Question: I'm not from america but I'm kinda wondering: is this what you'd learn in college? or is this part of high school syllabus as well?

  • Arthur Santana Post author

    Does anybody know what is 26100 he keeps mentioning?

  • mipatcheu charles Post author

    i just love this lectures

  • Alp Bartu Post author

    I think the physics at the end of the lecture takes us back to ''Some thing breaks because the magnitude of acceleration becomes too high'', even tho he exerts the same force, if the impact time differs, the results are pretty plausible.

  • Allye Baker Post author

    How the hell did he make those dots so fast, for the diameter of the sun?! (4:12ish)

  • Rohit Sai Post author

    why did bottom string broke?

  • colton ellis Post author

    These are arguably 3 of the most powerful laws of physics ever, especially the 2nd law.

  • Eric Thies Post author

    Love this guy…

    12:50, yup that's what my class looks like.

    I'm glad to see that even the best lecturers get this response form students.

  • Prashik Naik Post author

    WoW superb lecture

  • H.Bannai Post author

    can someone please explain why the strings holding the red cube at the end behaved that way ?

  • girish kulkarni Post author

    That's not the correct statement of Newton's second law the second law is The rate of change of momentum of a body is directly proportional to the applied force. you only arrive at F=MA if and only if the mass dosent varies with time which in cases like rocket propulsion does vary with time!

  • Medhawini Kapoor Post author


  • FUMBANANI SOKO Post author


  • Paul Edward Post author

    Brazil is here.

  • a j Post author

    Very strange, first a force is defined as mass times acceleration and then one step later it is stated as a law. So when you say F:= m * a ofcourse holds F = m * a. Unbelievable what a mess those physicist make. And it is written more or less the same in all those "advanced" textbooks. Humanity has still a long way to go.

  • Kamlesh Kumar Kamal Post author

    Sir when will the tension remain same? What are the conditions for it to be same?

  • John Gormley Post author

    "Is someone pulling our leg"

  • Shallu Goyal Post author

    At 30:28 he said that earth would go away from the ball it means gravitational force would become repulsive
    Isn't the earth should go towards ball because force is acting still in the same direction

  • Miguel Monroy Post author

    can act the first and third law at the same time?

  • Naresh Kumar Post author

    Sir there is question in third law we say action and reaction acts on two differrent bodies hence dont cancel each other just in example of ball thrown on wall returns back but how then book resting on table has zero net force i.e how action reaction are cancelled in this case

  • alistair lee Post author

    I have this physics conundrum which I cant answer, please help! https://medium.com/@alee250485/newtons-third-law-of-motion-is-it-always-applicable-11a4ee0e6b7c

  • Dalton Lucas Post author

    Good tech all ways teach in order student understand nyc for your lecture.

  • Nadja Resch Post author

    What does 'twenty six one hundered' mean?

  • Priyam Chauhan Post author

    Please someone tell me the introduction of this man… lots of gratitude to him…. he made my concepts clear… please post such lectures daily… we really need it…

  • krishnaa sharma Post author

    Tanks a lot.. Sr I m a big 💪 fan of yours.. My I have ur 👉👤. Mail 📧I'd sr please

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