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

at the bottom of the ladder and we ask this person

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

between the rope and that rod to your advantage,

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

that center portion. Radius R,

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

friction to your advantage and get an enormous gain by

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

about that point P. The torque equals I about

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

careful than I was this morning, because the knife cut

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.

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.

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?

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

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?

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

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 …. ?

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) ?

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

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

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

EDIT: He later noticed it at 7:46.

Thank you Dr. Lewin. raphael santore

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….?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?

Professor, What are those rings you wear?

I also saw them in an interview of you.

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

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?

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.

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.

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

I have a newer version of the book that you have but I couldn't find what you told at 25:18………

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

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

Where can I download the book of this course ?

Sir what's in that cool red ring?

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

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

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?

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.

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

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

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

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

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

Professor So where should I refer for the derivation

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?

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?

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

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

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?

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

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.

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

Hello dear Lewin .

How i can find the page 361?

25:16

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

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 .

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 !😅

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

Thank you sir

PLEASE HELP

……At 1:58 , he drew two dotted lines but only marked one of them as b , why ? Why isn't the other one important? 4:04 , WHY is ther a normal force nq , WHY does it exist ?

3:17 What if there was friction at p ?

I don't know how to express a gratitude, this gyro was causing a head ache to me! Bloom! This makes it lot easier to understand and have imagination

I HAVE NO WORDS TO Thank you. im about to finish your lectures for second time they are so intersting i really enjoy.thaaanks