Indoor, Outdoor & Kids' Trampolines

# 8.01x – Lect 25 – Static Equilibrium, Stability, Rope Warker

We…during the past four lectures, we have dealt with
angular momentum, torques and rolling objects
and rotations. And many of you then think,
“Oh, my goodness, now we have to remember
a whole zoo of equations,” but that’s not true. If you simply know how to make a conversion
from linear symbols to rotation, which is immediately trivial, of
course– position becomes angle; velocity becomes
angular velocity; acceleration becomes angular
acceleration, and so on– then you can make the
conversions very easily. If you remember that the kinetic
energy is one-half m v squared, then the kinetic energy
of rotation then becomes one-half I omega squared. This is on the Web. Every view graph
that I show you in lectures is always on the Web, and you should look
under “lecture supplements,” and then you can make yourself
a hard copy. So I thought this might be
useful for you to remember. Today I want to discuss
in detail what it takes for an object to be
in complete static equilibrium. For an object to be
in static equilibrium, it is not enough that the
sum of all forces is zero. But what is also required,
that the sum of all torques relative to any point that
you choose is also zero. And that will be
the topic today. If this is an object
free in space and let’s say
the center of mass is here, and I have a force on this
object in this direction and another force on this object
in opposite direction but equal in magnitude, then the sum of all forces
is zero. But you better believe it
that there is no equilibrium. There is a torque… and if this distance equals b,
then the torque on that object relative to any point
that you choose– it doesn’t matter
which one you take– the magnitude of that
will be b times f. And if there is a torque, there’s going to be
an angular acceleration. Torque is I alpha. And so it’s going to rotate. In this case, it will rotate
about the center of mass, so it’s not static equilibrium. The torque in this case would be
out of the blackboard. So it’s going to rotate
like this. So we got to keep
a close eye on torques, as much as we have to
on the forces themselves. So, today I have chosen a ladder as my subject
of static equilibrium. I put a ladder against a wall. Here’s the wall,
and this is the ladder. At point P, where the wall is,
there is almost… let’s say there is no friction. Mu of P is zero. At point Q here, where
it’s resting on the floor, there is friction. The static friction at point Q–
we’ll simply call that mu. The ladder has a mass M,
and it has a length l. So here is the center of mass
of that ladder, right in the middle. And this angle equals alpha. We know from experience that
if this angle is too small that the ladder will slide, and so I want to make
the topic today “What should that angle be
so that it does not slide?” Well, we have here a force Mg;
that’s gravity. Then we have a normal force
here– I call that NQ. We have friction
in this direction, because clearly the ladder wants
to slide like this, so the frictional force
will try to prevent that, will be in this direction. At point P there is no friction, so there can only be
a normal force. And I call that N of P. So these are the only forces
that act on this object. And now we can start
our exercise. We can say, all right, the sum of all forces in the
x direction have to be zero. So that means that N of P
must be in magnitude the same
as the frictional force. Now the sum of all forces in the
y direction have to be zero. So that means
that Mg must be N of Q. N of Q must be Mg. But now we need that the sum
of all torques have to be zero. It doesn’t matter
which point you choose– you can pick any point. You can pick a point here,
or here, or there, or there. You can pick C, you can pick P. I choose Q. The reason why I choose Q is because then I lose both
this force and that force, and I only have to deal
with this one and that one. And so I’m going to take
the torque relative to point Q. So we have NP. We have the cross-product of the
position vector to this point times the force. Since the length is l
and the angle is alpha, this is l sine alpha, so I’m going to get
NP times l sine alpha. And I call the torque
that is in the blackboard, I call that
my positive direction, and the torque that is
out of the blackboard, I give that
a negative direction. So this one is
in the blackboard, so I call that positive. The next one, Mg, is going
to be negative torque, and so now I need the cross-product between
the position vector and Mg, so that is the length here, which is one-half
l cosine alpha times Mg– Mg one-half l cosine alpha,
and that now must be zero. So I find, then, that N of P– that is, the normal force
at point P– I lose my l, equals M divided by two times
the cotangent of alpha, and that must be
the frictional force. So I know what the frictional
force is given by this result. Now, I don’t want
this ladder to slide, so now I have a requirement that the frictional force
must be less or equal to the maximum
frictional force possible. And the maximum frictional force
at this point Q is mu times the normal force. So my requirement now is that N over 2 times
the cotangent of alpha must be less or equal
to Mg times mu. Mu times Mg. Did I lose a g here? Yes, I did. I have here a g,
and I have here a g. Correct? Because if I have this force
here and this position vector, then I have Mg. So there is a g here. And so there is a g here,
and so I lose my Mg, and so you’ll find that
the cotangent of alpha is smaller or equal than 2 mu,
or the tangent of alpha is larger or equal
than 1 over 2 mu. And so that is the condition
for the ladder to be stable. And when you look
at this result, it tells you
that the larger mu is… the larger mu is,
the smaller that angle, and that’s very pleasing. That’s exactly what you expect. And if mu is very low, then
the situation is very unstable. Then it will slide
almost at any angle. If we take some
numerical examples– for instance,
I take mu equals 0.5, then alpha would be 45 degrees, and if the angle is any less,
then it will slide. If you take mu is 0.25, then the critical angle where
it will start to slide, I think, is somewhere near 63 degrees, but you can check
that for yourself. So the angle has to be
larger than that number for the ladder to be stable. This result is very intuitive, namely that if the angle
is too small that the ladder starts to slide. I have a ladder here
against this wall. We have tried to make this here
as smooth as we possibly can. It’s not perfect. So it’s only a poor-man’s
version of what I discuss there, but in any case, the friction
coefficient with the floor is substantially larger than
the friction coefficient here. And what is no surprise to you, that if the angle alpha is
too small, then there’s no equilibrium. The angle of alpha has to be
larger than a certain value, as you see there,
and then it’s stable. That’s all I want you
to see now. But now we’re going to make
the situation more interesting, and in a way I’d like
to test your intuition. Suppose I set
the angle of the ladder exactly at the critical point. In other words,
I’m going to set it so that the cotangent of alpha
is exactly 2 mu, so it is hanging in there
on its thumbs, just about to start sliding. And now I’m going to ask one
of you to walk up that ladder, to start here
and slowly walk up that ladder. Do you think that stepping
on that ladder, starting off, is super dangerous? That the ladder immediately
will start to slide? Or do you think that actually your being at the
bottom will make it more stable? Who thinks that it will
immediately start to slide? A few people. Who thinks that
it will not start to slide when you step on the lowest… All right. We’ll see what happens later on. Now, this person is going
to climb the ladder, and then there comes a time that it passes point C
and reaches point P. Will it now be safe to do that, or do you think that now it’s
going to be very dangerous? What do you think? Who thinks that you shouldn’t
get too high up on the ladder? Who thinks it makes
no difference– you can go all the way
up to the end? There are always a few
very courageous people. Okay, so this is what I’m going
to analyze with you, and most of you have
the right intuition, but we’re going to look at this
in a quantitative way as I know how to. So we’re going to put a person
with mass little m on that ladder,
and we put the person here. So this force here is little mg,
and let’s make this distance d. And now we’re going to redo
all these calculations. We start completely
from scratch. The sum of all forces in the
x direction has to be zero. No change. N of P must be F of f. Now the sum of all forces in
the y direction has to be zero. Now there is a change. So now we have that
N of Q must become larger, must be equal to capital M
plus little m times g, so the maximum friction,
which is mu times NQ, now becomes mu times M
plus m times g. So the maximum friction goes up. Now we need that the torque,
and I pick my point Q– relative to point Q–
equals zero. Well, the first two terms
haven’t changed, so I have N of P times
l sine alpha minus Mg times one-half l
times the cosine of alpha. But now we have a third term,
namely this position vector, and this force, and so now we’re
going to get this distance– which is d cosine alpha–
times this force. So we have here
minus mg times d cosine alpha, and that now equals zero. So I’m going to take N of P
out of here. And I can take
g cosine alpha out. So if I take g cosine alpha out, I have Ml divided by two
left over plus little md, and then I have
to divide by l sine theta. Not theta, but alpha. And so now I can write for cosine alpha
divided by sine alpha, I can write cotangent alpha, and I’ll bring
the l inside here, so I have g cotangent alpha times M over 2
plus little md divided by l. And that now must be
the frictional force, because NP is still
the frictional force. Let me make sure
that I have this right. Yes, I do. g cotangent alpha, M over 2,
little md over l– that is the frictional force. Notice that the frictional force
is going up, because this term is added, and
we didn’t have that term before. Before we only had this term
that you see here. So you first thought may be that the situation
has become more dangerous, because if there is
more friction, well, the ladder was set
exactly at that critical point– it was hanging on there
by its thumbs. So if the friction goes up,
you may say, “My goodness, it will probably
start to slide.” However, what
you overlook, then, is that the maximum friction
has also gone up, and so we have
to evaluate this now in comparison
with the maximum friction. And the best way to do this is to think of this first
as making d equal zero. So we said the person starts
to gradually climb up. Now, notice when d equals zero that the frictional force
is exactly the same… what it was before–
there is no difference. That frictional force has not
changed when d equals zero. But what has changed is
the maximum friction. The maximum friction has
this little m in it, and that’s independent of d. There is no d anywhere here. So if the maximum friction
goes up, and the friction itself
remains the same, clearly the ladder
has become more stable, and so you can step
on the lowest tread and nothing will happen. On the contrary, the situation
will become more stable. As the person starts to move up, the frictional force
gradually increases, because this term goes up, but the maximum friction
remains the same– it’s independent of d,
and so now there comes a time that this force becomes larger
than the maximum friction and then the ladder
will start to slide, and that, of course, is what
we want to find out now. So now the situation is
only safe as long as the frictional force is smaller or equal to
the maximum value possible. And that’s the case
when g cotangent alpha… but remember, we set it
at the critical angle, so that cotangent alpha is 2 mu. So I can replace this by 2 mu, because that’s the way
I set up my experiment. I start that way. I don’t start
with a random angle; I start exactly at the angle so that the ladder is sort
of just hanging in there. So cotangent alpha is 2 mu. M divided by 2 plus m times d
divided by l, this now has to be
less or equal to this value– mu times M plus m times g. Notice I lose my g. I lose my mu. I have here 2 times M over 2,
which is M, and I have one M here,
so I lose my capital M, and so I find that 2md over l
has to be less or equal to m. I lose my m, and I find that d has to be smaller
or equal to l over 2. And that is not unlike
most of your intuition, namely as long as the person
who steps on the ladder stays on this side, the situation will become
more stable. Certainly when
the person starts here, the stability has
enormously increased. As you gradually approach
that point here– the center of mass, where
d is exactly l over 2– then, of course, the situation becomes
again extremely critical here, but when you’re over this point,
it’s no longer critical, and the ladder
will start to slide. So in a nutshell, then, we set the ladder
at the critical situation. It’s about to start sliding. We put a person on here,
it becomes more stable. The person walks up slowly, the frictional force will
increase, because of this term, the maximum friction
will not change– it has already gone up because
the person is on the ladder– and as the person
approaches this point, the situation becomes less
and less stable all the time. Right at this point,
we are back to where we were, the situation is
about as critical as it was before
the person steps on it. And then the person proceeds;
then the ladder will slip. And I can show that to you–
at least I can make an attempt. I have here that same ladder,
and now what I will do is I will set the angle alpha not
exactly at the critical point but a little lower, so that when I let it go, we’ll
all see that it will slide. So it’s past the critical angle;
the angle is smaller. But now I have
four kilograms here. I wasn’t going to ask any one
of you to walk up this ladder, believe me, and I wasn’t going
to do it myself either. So you see it is unstable. And now I put
the four kilograms on here. And I can let go, and
the ladder has become stable. Do I take it off–
there it goes. So the person standing very low
made it more stable, exactly consistent
with what we just saw. Now I make the angle alpha
a little larger than critical, so the ladder is happy. It’s happy. But now the person is going
to do something dangerous. He’s going to walk
and stand here– and he shouldn’t do that,
as you see. That’s exactly
what you have seen there. So the friction plays
an essential role for us to get
static equilibrium, and that is often the case. There are many examples
in our daily lives where friction can be used
to our advantage, and a very special case, which
I will discuss with you now, is often used by sailors– by holding, controlling
a very massive object, a very large force, controlling
it with a very small one. And you do that by making use
of friction. You wrap a rope around a pole,
or around a rod, and you use the friction
and it works as follows. Here is this rod– it doesn’t have to be horizontal
as this one– and I hang on this side
of the rod, I hang with a string
a very massive object. Capital M. So the tension here, T2,
would be Mg– assuming that
it is not accelerating, that it is at rest. I wrap this rope around here
several times, and here I hang an object which
is substantially less in mass, which is m, little m,
and the tension equals mg. If there is no friction at all
in this bar, then… and the rope is near massless,
then T1 will be the same as T2, so the situation will start
to accelerate. However, if we make use
of the friction, then we can have
a stable situation. We can have static equilibrium, so that this one is not moving
and this one is not moving, and then we can have T2 to be
way, way larger than T1, by using friction
to our advantage. Let’s blow up
and let us have this rope here– I’ll give the rope a color. It doesn’t have to come off
vertically, of course. It could be at any angle. There it is, and so here is
my T1 and here is my T2. And we take the situation that… the pull on this side is
way larger than on that side so that this rope wants
to slip in this direction. That’s what it would like to do. If you look at these small
little sections of the rope, it is immediately obvious
if the rope wants to slide in this direction,
wants to start slipping, that the frictional forces
in these little pieces here are all in this direction… all the way around. And therefore,
it helps T1, so to speak, and so because of the friction,
which is opposing T2, T2 can now become
much larger than T1. What you have to do
to calculate this analytically, you have to evaluate these
individual frictional forces for these very small slices,
so that becomes an integral, and then you have to integrate
it over an angle, which I will call theta zero. And I remember when I last
lectured 801, that was in 1993, I derived that in class–
the relation between T2 and T1 as a function
of this angle theta. And that took me
about five minutes, and after five minutes,
half the students were asleep. Now, I’m not sure whether you
want to sleep five minutes, but I don’t think, frankly,
that you deserve it, so I decided
to not do the derivation but to refer you to the book,
which is page 361, and you will find, then,
that in the situation that the rope wants to start
slipping in this direction, that T2 divided by T1 is e to
the power mu times theta zero, if the friction coefficient here
is mu– it would be the static
friction coefficient. So that is the result. And notice that it is independent
of the radius of this bar, which is not so obvious,
strangely enough. Whether the bar is this small or
this small makes no difference. It only depends on this angle. This angle could be very large. You could wrap it
around ten times, as we will do very shortly. So there’s no restriction
on theta. If there were
no friction at all, notice then at the moment
that it starts to slip, e to the power 0 is 1,
that’s when T2 equals T1, that’s obvious, so you can’t
play this game if mu is zero. You need friction. That is at the heart
of this whole problem. And so now let’s put in
some numbers so that we get an idea
of what we gain. So let us take the situation that we take
let’s say three turns. We wrap the rope around
three times. So three turns. That means
theta zero equals six pi. And let the friction coefficient
mu be one-fifth, 0.2. So e to the power mu times
theta zero is now about 40. So what that means is that
with a force on this side which is 40 times smaller
than a force on that side, I have a balanced situation. I can hold this in my hand
and counter– if you want to think of it
that way– a force on this side
which is 40 times larger. But if I take six turns, then e to the power mu theta
zero will be about 2,000– 2,000! So now I can really control
an elephant. Imagine now that I have here
an object, capital M, which would be
5,000 kilograms… make it 10,000 kilograms. That’s what I’m hanging here. I can hang here now a mass which
is 2,000 times less massive. That means I could hang there
five kilograms. And the tension here would be
2,000 times less than the tension there. The system would be
about to start slipping, but it’s not slipping. There is complete balance. So imagine I hold this part
in my hand– here is this
10,000-kilogram weight and I hold this in my hand. All I need is a force
of about 50 newtons, and I’m standing there, and on the other side is
this 10,000-kilogram weight. And now I just make my force
a little less than 50 newtons, and what will happen now,
it will start to slip. Remember? Because we calculated here the requirement
for just not slipping. So now I let it go, and then the 10,000-kilogram on the other
side will slowly go down. I can just control it,
and I can stop it, and I can control it with
a very, very small force. Now comes a question for you. Suppose I wanted to raise
the 10,000-kilogram. Could I now pull a little bit
more than 50 newtons? Just a teeny weeny
little bit more? Would then that
10,000-kilogram come up? I see people shake their heads. Who thinks it would come up? Who thinks “no way”? Who doesn’t think at all? Most of the people. All right– sorry,
I didn’t mean to be nasty. Um… there’s no way
that you could pull it up, because if you want
to pull this up, the situation
completely reverses. Those frictional forces for this
rope to go in this direction will flip over. In other words, what is now T1 in our
calculations will become T2. So if you want this side
to go down, if you want to have it slip
in this direction, you’re going to have
that T1 divided by T2 is now e to the power mu theta zero. And so now if you have
10,000 kilograms here and if you had six turns, then
the force that you need here is 2,000 times larger
than 10,000 kilograms, and so you would need
20 million kilograms. So it would be the dumbest thing
to do to lift it. You don’t want to lift it. You use this device only
to balance a very strong force– that’s the way it’s used
by sailors– and even to control it, because
you can slowly release it, and then the force on the
other side will start to move. But you cannot use it
to lift something. I have here a plastic bin,
and in this plastic bin are four of these
15-pound lead bricks. Would you do me a favor and come
up here and convince yourself that three are already in there? I don’t want to put them
all four in, so I decided I’m only going
to put the last one in. You see those three? Thank you very much. And I’m going to put
the last one in. It’s very heavy. Do you want to check this,
by the way? This is… Careful!
It’s very, very heavy. Yeah, okay. Okay, 15 pounds a piece–
60 pounds we have up there. And now I’m going to wrap this
around this bar, which is your bar. The radius doesn’t matter. And why don’t we start
with six rotations? One… two… three… (loud rustling
against microphone) Oh, my goodness. Four… five… six. Okay. Now I’m going to lower
this platform that we have that is holding it up. I can lower it. And it will shortly be hanging
now on my rope. There it is. I can remove this now. We don’t need this anymore,
and we don’t need this anymore. Effortless. 30 kilograms–
(whistles softly) Hardly any force. Very gentle. Now let me lower it a little. There it goes. Just lower it. Would be nice
if I had to do even less. Let’s put a few more turns on. One… two… three. You know what we could do? We could put so many turns on that the weight of the rope
itself is enough to balance it. Let’s try it. Not yet. Put a few more on. One… two… three… four. We have 12 windings now. And the rope,
the weight of the rope is now sufficient
to balance the 30 kilograms. So you see this in action now. This is a marvel. I’m going to…
Well, let me put a few more on, because I don’t want it to come
down during the lecture. I want to be sure
that it stays there, and I’m going to secure it here. So you see how you can use
factors of thousands and more, and this is used
quite frequently. So you saw
in a striking example of where friction helped us
to balance, and now I want to discuss
an object that is hanging and that can freely swing
due to gravity. I want you to appreciate the
situation of static equilibrium that the sum of all forces
and the sum of all torques have to be zero. So here is this object
which I’m going to hang, maybe on the wall, and here is
a point P, and there is a… let’s say a frictionless spin
that I put in the wall, so the object can
freely swing around. Let the center of mass be here. And so there is a force. If this object has a mass
little m, there is a force mg, and here is
the position vector r of P, and so it’s clear that there is
a torque relative to point P. And if there is a torque
relative to point P, it would be r P cross mg. I put a vector sign over here, because you have to take the
cross-product between the two, so you take the… sine of this angle has
to be taken into account. This object is going to rotate. It’s clearly going to rotate
that point P times alpha, and alpha is the
angular acceleration. How can we ever get
a stable situation? Nature is now going to think
hard and is going to say, “Gee, I can only have
stable equilibrium “if the sum of all forces
equals zero “and if the sum of all torques
relative to any point equals zero.” And nature knows how to do that. That always will happen
if P and the center of mass are along a vertical line. Because if that’s the case, then
there will be here a force mg, and then here there will be
a force mg upwards. This object is hanging
on that pin, so the pin– action equals minus reaction–
will push upwards. So the sum of all forces is
zero, and there is no torque. Take any point you want to–
this point or that point or this point or that point
or that point– there is no torque anymore. And so now there is
complete equilibrium. So if you had an object,
for instance, that looked like this,
a thin rod– there’s a very massive
object here, so that the center of mass
is almost there– then this situation
would be stable. Or you might say this situation, because now we have the center
of mass, mP, vertical line. So also here
we have now mg down, so we have at the pin mg up. Sum of all forces is zero,
sum of all torques is zero. But there is a big difference between this situation
and this situation, which you immediately feel
in your stomach, of course, and that is that
this is highly unstable. If you just blow
on this a little, it will swing over,
it will start to rotate, whereas this is stable. If I move this to the side, it
will come back to that position. But the basic idea is, then, that the center of mass will
always in stable configuration line itself up below
the point of suspension. I have here a triangle, and I have no clue
where the center of mass is. It may be somewhere here. It may even be in open space,
I don’t know. What I can do now,
I can suspend it like this, and I know that the center of
mass must be below my finger. Did you notice? It actually rotated. So that the center of mass
lines itself up precisely below my finger, so I can now take a pencil
and draw a line here, or I can have a little string, and now I can change
my finger to here and I can again draw this line, and where the two intersect is
the center of mass. So the center of mass
will always find itself below the point of suspension. If now I take a piece of putty and I put the piece
of putty here, then clearly I change the
location of the center of mass quite substantially. It must have shifted enormously
in this direction. Where is the center of mass? I have no clue. But look. I balance it here. The center of mass must now be
somewhere here. I balance it here. The center of mass must be
somewhere here. And where
the two lines intersect– could be in this opening– that’s where the center of mass
is located. So it’s easy experimentally
to find the center of mass. Earlier in the course, we calculated where
the center of mass was. Perhaps you remember that, when we had some very special
geometrical configurations. I mentioned to you then already that there are ways that you can
experimentally determine it, and that can be done very easily
in the way I just showed you. Now… I want to show you some
examples of static equilibrium which are not so intuitive. You may have seen some of them. You may actually have played
with some of them. If I take a pencil
and I take my pocket knife… So here is a pencil. I will show this to you shortly,
just a regular pencil. And I take my pocket knife and I put the knife,
I jam my knife in here. So this is the knife. And this is now
my pocket knife, like so. This arrangement
can easily be made such that the center of mass
is below this point, and so if now
I put my finger here, I can balance
this rather strange object and it will be
completely stable. It will arrange itself so that the center of mass
will fall below my finger, and it will be completely happy. And I want you to see that. I have here my pocket knife
and I have here a pencil. And you can see there– or the ones… those of you
who are close can see it here. I’ll give you a better
light condition for that. So there we are. So here is my pocket knife. I open it now. And here is the pencil. And I’m going to jam it in here. When you do that, be more
into my finger. That’s why I have
this Band-Aid here. It can easily lead
to a nasty cut. So the knife is now
in this pencil, but now I have to get
that center of mass under… under the pencil. And now I put my finger here,
and it’s stable. So the center of mass
rearranges itself so that it goes
under the point of suspension and is completely stable. You can wiggle it
as much as you want to. That’s not so intuitive. I have here a… a very thin
bar, a wire, a very stiff wire. And I can put my hammer here
on the end of this wire. And look at this. The center of mass is
way below my suspension point. This is stable like hell. It also hurts like hell,
by the way. You can even swing it. Center of mass is way here
below the suspension point. Sum of all torques is zero, if it’s like this,
sum of all forces is zero. Center of mass below
the suspension point. They have to be exactly
vertically lined up. If they’re not
vertically lined up, like now, then there is a net torque
relative to that point. If there is a net torque, there
will be an angular acceleration, and that’s what you see. If they’re exactly lined up,
there is no longer any torque. Aah. Let’s now turn to a rope walker. We have a rope walker. Here is the rope walker. And the rope walker,
standing on the rope. I’ll make the rope walker
a little smaller– you’ll see shortly why I make it
a little smaller. This is the rope walker. And this is the rope. And let the center of mass of
the rope walker be right here– somewhere here. And the distance between the
center of mass and the rope– let that distance be one meter, and let the mass of the
rope walker be 70 kilograms. I build a light structure–
it could be built of wood– which the rope walker is going
to carry in her hands, or in his hands, just like this. And it goes down
quite a distance. And I put here
a mass of five kilograms, and I put here a mass
of five kilograms. And let’s assume that this
structure is very lightweight, so that we can ignore that, and let this distance be
ten meters. Where is the center of mass
of this system? Seventy kilograms is
above the rope. One meter. Ten kilograms is ten meters
below the rope, so clearly the center of mass
of the system as a whole will be below the rope. And so that person is
as stable as anything– just as stable
as my hammer was– with one problem, that the
person could slide off the rope to the sides, and that would be
unfortunate, of course. And therefore
you would require, then, that if this is the rope,
the cross-section of the rope, then you would make yourself
soles on your shoes which are a little bit
like this. So that’s only to prevent you from sliding off in this
direction and falling. But the center of mass is
below the rope, and so there is
really no danger. You can balance yourself
and walk on the rope and nothing would happen. You could even do this–
go back and forth on the rope. No problem. You always come back like this. Well, we have a very
special rope walker. This is our rope walker. And this rope walker has
indeed very special shoes. In fact, the shoes of the
rope walker is a little wheel, and this is what
the wheel looks like. And here is
where the rope will be, and this is the rope that you
see here in the lecture hall. And so the only reason why
we need this groove in the wheel is to prevent it
from sliding to the sides, as we need it for this person. But the center of mass– you see it when I balance it
on my finger– is way below
the suspension point. Because of these weights
on both sides, the center of mass
is somewhere here, so it is completely stable. You can even go like this–
no problem. This is what
the rope walker can do, provided it doesn’t slide off. That’s why we have this. Okay, let’s ask this rope walker
to give us a demonstration. Let’s make her walk. We don’t need
that television anymore. It would be nice
if one of the students… Would you do me a favor and welcome the rope walker
as she arrives here? Because we don’t want her
to crash. Is that okay with you? So when she comes down, be gentle and give her
all the honors that she deserves and then you just make sure
that she doesn’t fall, because that…
she will be damaged, right? We don’t want that. Okay, so you have seen already
how stable the situation is. And there she goes. You ready for that? Voilà! We can all become
rope walkers. Thank you very much. See you Friday.

1. Antonio Romero says:

Professor Walter Lewin, I am a mechanical engineer from Brasil and I teach phisycs as a voluteer, i would like to thank you for all that you have done for education. You have no idea how your lectures are inspire me. You are really making difference in this crazy word. Thanks a lot for help me to help others.

2. Dr. Science Sc.D says:

Hello Dr. Lewin, in the static equilibrium example with the person walking up the ladder, how come we do not consider the normal force from the contact between the person and the ladder or is this normal force combined with the normal force the ladder makes with the ground?

3. Intom D says:

Hi professor. How can I see the derivation at minute 25:00?

4. vijay R says:

Hello Professor Lewin. I was unable to find the derivation at 25:00 on page 361, a page which seems to be concerned with the conservation of linear momentum.The PDF link I am using is http://dl1.ponato.com/eb1/719__y4H4nGR.pdf. Is it possible that it is the wrong book?

5. Ustat Khullar says:

at 24:14 why we divided rope in little parts to calculate friction as friction is independent of area of contact?

6. SKGaming Box says:

Hello Professor Lewin Can you explain me in one or two sentences what is difference between maximum friction force and just friction force? I understand the equations but in this problem I dont very see that, mainly in 16:10 when you said that friction force and maximum friction force are higher… well if friction force is higher doesnt that mean that the ladder need to overcome more friction force so it is more stable …. ?

7. Sharath Chandar says:

What would have happened if the bicycle wheel (after giving a good spin with the motor) had been suspended onto the rope horizontally ( plane of wheel parallel to floor) ?

8. Commentor 643 says:

can't afford your book sir, where can I start for the tension 1 and 2 derivations

9. Nikola Ivanovic says:

Hello, professor. Why isn't there a vertical component in the point P (the ladder example)?

10. Rainer Wahnsinn says:

At 6:42 he lost a g in the process?!

EDIT: He later noticed it at 7:46.

11. Raphael Santore says:

Thank you Dr. Lewin. raphael santore

12. Giant Miller says:

Ik hoop dat u nog een beetje Nederlands kent 😉 > Waarom is er bij Ff max al geen rekening gehouden met de afstand waar de persoon zich op de trap bevind (d)? Immers aan het begin van de trap komt de massa van de persoon geheel op punt Q, dus dan kan de gehele massa wel worden meegenomen (M+m) maar waarom niet halverwege de trap een half (of zoiets afhankelijk van de hoek) maal m nemen? Zal misschien uiteindelijk niets uitmaken, maar toch….?

13. Sword Seraph says:

What does it mean for a torque to have a direction? I understand its mathematical significance in figuring out a torque's sign in the torque balance, but what does it mean physically for the torque to go out of the board? Does the torque generate an angular acceleration in that direction?

14. Chetan Kochar says:

Professor, What are those rings you wear?
I also saw them in an interview of you.

15. manas srivastava says:

Sir in 6:32 you made a derivation of critical angle at which ladder can stand without standing. I do not understand why did you take l/2 in mg.cos a.l/2. Please explain me

16. Captain Calculus says:

At the 48:00…wouldn't the angular momentum of the wheel also provide stability, in exactly the same way a bicycle is stable only when it is moving?

17. Neven Jakopovic says:

I'm sure it's something very obvious, but I just can't see the geometry at (A) 5:45 and (similar!) at time (B) 6:15.

Torque (vectors): τ = r x F; so magnitude is τ = r * F * sin(angle r->F)

I drew the situation and looked at the angles (going completely "by the book", i.e. evaluating angles from the direction vector towards the force vector, I used sin(180-α) = sin(α) for QC and sin(90-α)=cos(α) for QP). I got the same result, but with a lot more work.

Prof. Lewin just takes the orthogonal projection of r and I can't see: why does it work?

I've re-watched the lecture on vectors again but it did not help me here.

18. William Ibarra says:

Professor Lewin, at 16:56, is it true that friction can never exceed the maximum friction? You say that there comes a time when friction is larger than maximum friction.

19. Surendar Kumar says:

Why a tightrope walker spreads her hands while walking on rope ? How it helps to increase her stability ?

20. YD says:

I have a newer version of the book that you have but I couldn't find what you told at 25:18………

21. Steve Marks says:

Why don't the students ever ask questions, is this just for video? Surely some students don't understand everything first time?

22. Ahmed Hosny says:

what does (e) refer to in the equa T1/T2=e^mu theta?

23. The Greatest says:

24. Abhishek cherath says:

Sir what's in that cool red ring?

25. anugu arun says:

sir i had a doubt regarding statics A frictionless surface is in the shape of a function which has its endpoints at
the same height but is otherwise arbitrary. A chain of uniform mass per unit
Figure 1.10 length rests on this surface (from end to end; see Fig. 1.10). Show that the
chain will not move. this is the question taken from the book david morin introduction to classical mechanics proble 3 chapter 1 could u pls solve it for me

this is not my homework

26. Cocoa says:

Which book are they using in this class? I want to see the derivation at ~25:00

27. md65000 says:

I would have liked to see ladder example go one step further: If you put a bunch of weight on the bottom step of a ladder, then can a person climb past the center of mass to the top without the ladder falling?

28. Malkavian Khaoz says:

For those who want to see the derivation mentioned at 25:00, it's here https://drive.google.com/file/d/1ZfZBgvbJBeMQvAQEU2d2zyh01FHBdSl9/view?usp=sharing
Please tell me if you find any error.

29. Pascal Kayosde says:

Goof job. Highly motivating. Creative,pragamatic and creative…..,.learning made easy! Another Richard Feyman

30. rakshit sohlot says:

Hello professor which book did you refer in the video which edition

31. rakshit sohlot says:

Professor to which chapter did you refer to because I have 3rd edition not 2nd

32. rakshit sohlot says:

Professor why did not you take the force exerted by the man on the ladder to walk up the ladder in the problem on 10:24

33. rakshit sohlot says:

I referred to the index and chapter 'statics and elasticity' but didn't find the derivation

34. rakshit sohlot says:

Professor So where should I refer for the derivation

35. Prakhar Bhalla says:

sir if large number of forces are being applied on the body. then is there any way to predict the point about which body will rotate assuming that there is resultant force not equal to zero. this type of problem will arise when man on ladder will climb just more than half of the length. then about which point ladder will rotate and will its center of mass also move in direction away perpendicular to wall?

36. Prakhar Bhalla says:

for a rectangular object in vertical plane if a small force is applied on top of it then a equal but opposite friction will act on the bottom surface. now to cancel the torque of these two forces the normal force from ground should shift parallel in the direction of force. Is this the way phenomenon occur or not?

37. チュimoc says:

I was afraid he'd say "Now, I want to demonstrate that to you." at 19:05 XD

38. HABIB UL HOQUE says:

This is very similar to Irodov's PROBLEMS IN GENERAL PHYSICS 1.93..

39. Prakhar Bhalla says:

sir, if net force on body is zero but there is net torque not equal to zero then this net torque at 2:15 will be equal to (moment of inertia about center of mass *angular acceleration) which is same about any point we choose. is it true?

40. Abdullah Alsakka says:

24:00 where can I find the derivation sir. Cause I don’t have the book 😪😪

41. Trishit Chandra says:

Sir, I have a question about a toy of mine which is a good example of static equilibrium. But still I am not able to find the mystery of the physics of that toy. And I am trying through various social communication platforms to show you the video of that toy. But nothing happened with those contact information. So please help me to reach you and find the mystery and I'm pretty sure that you would like it too.

42. nilesh rathod says:

what will happen if Fs is greater than Ff max? why the condition can't be Fs >= Ff max?

43. Mansour Gendezli says:

Hello dear Lewin .
How i can find the page 361?
25:16

44. Curious George says:

Hi Prof- If an object has net Forces = 0, net Torques = 0, but has a constant velocity, is that object in static equilibrium? Or does the object need to have v = 0 to be in static equilibrium? Thank you

45. Michel M. says:

I love these lectures ! But I have a question about the rotational equilibrium condition : it simply seems very artificial to me that the rotational equilibrium of an object is dictated by the fact that the sum of all the forces times their distances( taken perpendicularly to the forces) to a specific point has to be zero for every point in space . I understand it is an experimental fact , but is there any way of proving it only based on theory ? I understand that torque is proportional to angular acceleration and so if we make sure for an object , that the total Torque that is acting upon him is zero relative to any point in space , it means that the angular accelerations relative to all these points must also be zero , and so rotational equilibrium is reached .

46. SOUVIK MAHATO says:

12:36 ,
i thought we have to consider the normal force applied by the man on the ladder, instead of its full weight !

thought it would become complex !😅

47. AKRAM A. A. AL-KHAZZAR says:

the guy @30:00 like a robot if mr lewin asked me something i ll jump of my chair running to do it

48. Rayeed Hasan says:

Thank you sir

49. Rayeed Hasan says:

50. Rayeed Hasan says:
51. Yashu Sphinx says:
52. morteza khoshbin says: