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During the past four lectures,

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we have dealt with
angular momentum, torques

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and rolling objects
and rotations.

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And many of you then think,
"Oh, my goodness,

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now we have to remember
a whole zoo of equations,"

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but that's not true.

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If you simply know

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how to make a conversion
from linear symbols to rotation,

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which is immediately trivial, of
course-- position becomes angle;

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velocity becomes
angular velocity;

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acceleration becomes angular
acceleration, and so on--

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then you can make the
conversions very easily.

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If you remember that the kinetic
energy is one-half m v squared,

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then the kinetic energy
of rotation then becomes

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one-half I omega squared.

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This is on the Web.

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Every view graph
that I show you in lectures

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is always on the Web,

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and you should look
under "lecture supplements,"

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and then you can make yourself
a hard copy.

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So I thought this might be
useful for you to remember.

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Today I want to discuss
in detail what it takes

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for an object to be
in complete static equilibrium.

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For an object to be
in static equilibrium,

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it is not enough that the
sum of all forces is zero.

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But what is also required,
that the sum of all torques

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relative to any point that
you choose is also zero.

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And that will be
the topic today.

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If this is an object
free in space

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and let's say
the center of mass is here,

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and I have a force on this
object in this direction

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and another force on this object
in opposite direction

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but equal in magnitude,

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then the sum of all forces
is zero.

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But you better believe it
that there is no equilibrium.

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There is a torque...

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and if this distance equals b,
then the torque on that object

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relative to any point
that you choose--

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it doesn't matter
which one you take--

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the magnitude of that
will be b times f.

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And if there is a torque,

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there's going to be
an angular acceleration.

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Torque is I alpha.

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And so it's going to rotate.

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In this case, it will rotate
about the center of mass,

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so it's not static equilibrium.

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The torque in this case would be
out of the blackboard.

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So it's going to rotate
like this.

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So we got to keep
a close eye on torques,

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as much as we have to
on the forces themselves.

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So, today I have chosen a ladder

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as my subject
of static equilibrium.

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I put a ladder against a wall.

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Here's the wall,
and this is the ladder.

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At point P, where the wall is,
there is almost...

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let's say there is no friction.

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Mu of P is zero.

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At point Q here, where
it's resting on the floor,

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there is friction.

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The static friction at point Q--
we'll simply call that mu.

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The ladder has a mass M,
and it has a length l.

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So here is the center of mass
of that ladder,

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right in the middle.

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And this angle equals alpha.

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We know from experience that
if this angle is too small

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that the ladder will slide,

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and so I want to make
the topic today

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"What should that angle be
so that it does not slide?"

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Well, we have here a force Mg; 
that's gravity.

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Then we have a normal force
here-- I call that NQ.

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We have friction
in this direction,

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because clearly the ladder wants
to slide like this,

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so the frictional force
will try to prevent that,

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will be in this direction.

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At point P there is no friction,

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so there can only be
a normal force.

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And I call that N of P.

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So these are the only forces
that act on this object.

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And now we can start
our exercise.

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We can say, all right,

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the sum of all forces in the
x direction have to be zero.

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So that means that N of P
must be in magnitude

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the same
as the frictional force.

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Now the sum of all forces in the
y direction have to be zero.

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So that means
that Mg must be N of Q.

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N of Q must be Mg.

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But now we need that the sum
of all torques have to be zero.

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It doesn't matter
which point you choose--

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you can pick any point.

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You can pick a point here,
or here, or there, or there.

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You can pick C, you can pick P.

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I choose Q.

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The reason why I choose Q is

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because then I lose both
this force and that force,

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and I only have to deal
with this one and that one.

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And so I'm going to take
the torque relative to point Q.

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So we have NP.

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We have the cross-product of the
position vector to this point

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times the force.

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Since the length is l
and the angle is alpha,

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this is l sine alpha,

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so I'm going to get
NP times l sine alpha.

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And I call the torque
that is in the blackboard,

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I call that
my positive direction,

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and the torque that is
out of the blackboard,

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I give that
a negative direction.

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So this one is
in the blackboard,

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so I call that positive.

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The next one, Mg, is going
to be negative torque,

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and so now I need

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the cross-product between
the position vector and Mg,

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so that is the length here,

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which is one-half
l cosine alpha times Mg--

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Mg one-half l cosine alpha,
and that now must be zero.

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So I find, then, that N of P--

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that is, the normal force
at point P--

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I lose my l,

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equals M divided by two times
the cotangent of alpha,

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and that must be
the frictional force.

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So I know what the frictional
force is given by this result.

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Now, I don't want
this ladder to slide,

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so now I have a requirement

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that the frictional force
must be less or equal

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to the maximum
frictional force possible.

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And the maximum frictional force
at this point Q

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is mu times the normal force.

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So my requirement now is

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that N over 2 times
the cotangent of alpha

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must be less or equal
to Mg times mu.

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Mu times Mg.

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Did I lose a g here?

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Yes, I did.

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I have here a g,
and I have here a g.

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Correct?

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Because if I have this force
here and this position vector,

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then I have Mg.

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So there is a g here.

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And so there is a g here,
and so I lose my Mg,

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and so you'll find that
the cotangent of alpha

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is smaller or equal than 2 mu,
or the tangent of alpha

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is larger or equal
than 1 over 2 mu.

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And so that is the condition
for the ladder to be stable.

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And when you look
at this result,

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it tells you
that the larger mu is...

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the larger mu is,
the smaller that angle,

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and that's very pleasing.

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That's exactly what you expect.

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And if mu is very low, then
the situation is very unstable.

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Then it will slide
almost at any angle.

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If we take some
numerical examples--

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for instance,
I take mu equals 0.5,

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then alpha would be 45 degrees,

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and if the angle is any less,
then it will slide.

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If you take mu is 0.25,

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then the critical angle where
it will start to slide, I think,

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is somewhere near 63 degrees,

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but you can check
that for yourself.

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So the angle has to be
larger than that number

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for the ladder to be stable.

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This result is very intuitive,

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namely that if the angle
is too small

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that the ladder starts to slide.

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I have a ladder here
against this wall.

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We have tried to make this here
as smooth as we possibly can.

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It's not perfect.

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So it's only a poor-man's
version of what I discuss there,

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but in any case, the friction
coefficient with the floor

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is substantially larger than
the friction coefficient here.

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And what is no surprise to you,

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that if the angle alpha is
too small,

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then there's no equilibrium.

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The angle of alpha has to be
larger than a certain value,

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as you see there,
and then it's stable.

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That's all I want you
to see now.

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But now we're going to make
the situation more interesting,

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and in a way I'd like
to test your intuition.

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Suppose I set
the angle of the ladder

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exactly at the critical point.

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In other words,
I'm going to set it

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so that the cotangent of alpha
is exactly 2 mu,

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so it is hanging in there
on its thumbs,

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just about to start sliding.

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And now I'm going to ask one
of you to walk up that ladder,

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to start here
and slowly walk up that ladder.

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Do you think that stepping
on that ladder, starting off,

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is super dangerous?

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That the ladder immediately
will start to slide?

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Or do you think

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that actually your being at the
bottom will make it more stable?

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Who thinks that it will
immediately start to slide?

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A few people.

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Who thinks that
it will not start to slide

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when you step on the lowest...

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All right.

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We'll see what happens later on.

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Now, this person is going
to climb the ladder,

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and then there comes a time

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that it passes point C
and reaches point P.

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Will it now be safe to do that,

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or do you think that now it's
going to be very dangerous?

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What do you think?

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Who thinks that you shouldn't
get too high up on the ladder?

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Who thinks it makes
no difference--

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you can go all the way
up to the end?

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There are always a few
very courageous people.

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Okay, so this is what I'm going
to analyze with you,

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and most of you have
the right intuition,

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but we're going to look at this
in a quantitative way

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as I know how to.

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So we're going to put a person
with mass little m

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on that ladder,
and we put the person here.

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So this force here is little mg,
and let's make this distance d.

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And now we're going to redo
all these calculations.

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We start completely
from scratch.

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The sum of all forces in the
x direction has to be zero.

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No change.

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N of P must be F of f.

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Now the sum of all forces in
the y direction has to be zero.

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Now there is a change.

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So now we have that
N of Q must become larger,

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must be equal to capital M
plus little m times g,

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so the maximum friction,
which is mu times NQ,

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now becomes mu times M
plus m times g.

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So the maximum friction goes up.

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Now we need that the torque,
and I pick my point Q--

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relative to point Q--
equals zero.

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Well, the first two terms
haven't changed,

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so I have N of P times
l sine alpha

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minus Mg times one-half l
times the cosine of alpha.

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But now we have a third term,
namely this position vector,
 

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and this force, and so now we're
going to get this distance--

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which is d cosine alpha--
times this force.

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So we have here
minus mg times d cosine alpha,

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and that now equals zero.

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So I'm going to take N of P
out of here.

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And I can take
g cosine alpha out.

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So if I take g cosine alpha out,

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I have Ml divided by two
left over

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plus little md, and then I have
to divide by l sine theta.

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Not theta, but alpha.

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And so now I can write

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for cosine alpha
divided by sine alpha,

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I can write cotangent alpha,

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and I'll bring
the l inside here, so I have

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g cotangent alpha times M over 2
plus little md divided by l.

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And that now must be
the frictional force,

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because NP is still
the frictional force.

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Let me make sure
that I have this right.

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Yes, I do.

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g cotangent alpha, M over 2,
little md over l--

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that is the frictional force.

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Notice that the frictional force
is going up,

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because this term is added, and
we didn't have that term before.

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Before we only had this term
that you see here.

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So you first thought may be

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that the situation
has become more dangerous,

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because if there is
more friction,

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well, the ladder was set
exactly at that critical point--

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it was hanging on there
by its thumbs.

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So if the friction goes up,
you may say,

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"My goodness, it will probably
start to slide."

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However, what
you overlook, then,

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is that the maximum friction
has also gone up,

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and so we have
to evaluate this now

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in comparison
with the maximum friction.

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And the best way to do this is

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to think of this first
as making d equal zero.

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So we said the person starts
at the bottom of the ladder

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and we ask this person
to gradually climb up.

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Now, notice when d equals zero

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that the frictional force
is exactly the same...

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what it was before--
there is no difference.

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That frictional force has not
changed when d equals zero.

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But what has changed is
the maximum friction.

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The maximum friction has
this little m in it,

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and that's independent of d.

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There is no d anywhere here.

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So if the maximum friction
goes up,

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and the friction itself
remains the same,

285
00:16:35 --> 00:16:41
clearly the ladder
has become more stable,

286
00:16:37 --> 00:16:43
and so you can step
on the lowest tread

287
00:16:40 --> 00:16:46
and nothing will happen.

288
00:16:41 --> 00:16:47
On the contrary, the situation
will become more stable.

289
00:16:45 --> 00:16:51
As the person starts to move up,

290
00:16:47 --> 00:16:53
the frictional force
gradually increases,

291
00:16:50 --> 00:16:56
because this term goes up,

292
00:16:52 --> 00:16:58
but the maximum friction
remains the same--

293
00:16:54 --> 00:17:00
it's independent of d,
and so now there comes a time

294
00:16:57 --> 00:17:03
that this force becomes larger
than the maximum friction

295
00:17:00 --> 00:17:06
and then the ladder
will start to slide,

296
00:17:04 --> 00:17:10
and that, of course, is what
we want to find out now.

297
00:17:06 --> 00:17:12
So now the situation is
only safe

298
00:17:09 --> 00:17:15
as long as the frictional force

299
00:17:12 --> 00:17:18
is smaller or equal to
the maximum value possible.

300
00:17:17 --> 00:17:23
And that's the case
when g cotangent alpha...

301
00:17:21 --> 00:17:27
but remember, we set it
at the critical angle,

302
00:17:25 --> 00:17:31
so that cotangent alpha is 2 mu.

303
00:17:28 --> 00:17:34
So I can replace this by 2 mu,

304
00:17:30 --> 00:17:36
because that's the way
I set up my experiment.

305
00:17:33 --> 00:17:39
I start that way.

306
00:17:34 --> 00:17:40
I don't start
with a random angle; 

307
00:17:35 --> 00:17:41
I start exactly at the angle

308
00:17:37 --> 00:17:43
so that the ladder is sort
of just hanging in there.

309
00:17:40 --> 00:17:46
So cotangent alpha is 2 mu.

310
00:17:44 --> 00:17:50
M divided by 2 plus m times d
divided by l,

311
00:17:51 --> 00:17:57
this now has to be
less or equal to this value--

312
00:17:56 --> 00:18:02
mu times M plus m times g.

313
00:18:02 --> 00:18:08
Notice I lose my g.

314
00:18:06 --> 00:18:12
I lose my mu.

315
00:18:08 --> 00:18:14
I have here 2 times M over 2,
which is M,

316
00:18:11 --> 00:18:17
and I have one M here,
so I lose my capital M,

317
00:18:14 --> 00:18:20
and so I find that 2md over l
has to be less or equal to m.

318
00:18:22 --> 00:18:28
I lose my m, and I find

319
00:18:25 --> 00:18:31
that d has to be smaller
or equal to l over 2.

320
00:18:29 --> 00:18:35
And that is not unlike
most of your intuition,

321
00:18:33 --> 00:18:39
namely as long as the person
who steps on the ladder

322
00:18:36 --> 00:18:42
stays on this side,

323
00:18:38 --> 00:18:44
the situation will become
more stable.

324
00:18:41 --> 00:18:47
Certainly when
the person starts here,

325
00:18:43 --> 00:18:49
the stability has
enormously increased.

326
00:18:46 --> 00:18:52
As you gradually approach
that point here--

327
00:18:49 --> 00:18:55
the center of mass, where
d is exactly l over 2--

328
00:18:52 --> 00:18:58
then, of course,

329
00:18:53 --> 00:18:59
the situation becomes
again extremely critical here,

330
00:18:57 --> 00:19:03
but when you're over this point,
it's no longer critical,

331
00:19:00 --> 00:19:06
and the ladder
will start to slide.

332
00:19:05 --> 00:19:11
So in a nutshell, then,

333
00:19:07 --> 00:19:13
we set the ladder
at the critical situation.

334
00:19:11 --> 00:19:17
It's about to start sliding.

335
00:19:14 --> 00:19:20
We put a person on here,
it becomes more stable.

336
00:19:17 --> 00:19:23
The person walks up slowly,

337
00:19:19 --> 00:19:25
the frictional force will
increase, because of this term,

338
00:19:23 --> 00:19:29
the maximum friction
will not change--

339
00:19:25 --> 00:19:31
it has already gone up because
the person is on the ladder--

340
00:19:29 --> 00:19:35
and as the person
approaches this point,

341
00:19:31 --> 00:19:37
the situation becomes less
and less stable all the time.

342
00:19:34 --> 00:19:40
Right at this point,
we are back to where we were,

343
00:19:37 --> 00:19:43
the situation is
about as critical

344
00:19:39 --> 00:19:45
as it was before
the person steps on it.

345
00:19:43 --> 00:19:49
And then the person proceeds;
then the ladder will slip.

346
00:19:47 --> 00:19:53
And I can show that to you--
at least I can make an attempt.

347
00:19:51 --> 00:19:57
I have here that same ladder,
and now what I will do is

348
00:19:55 --> 00:20:01
I will set the angle alpha not
exactly at the critical point

349
00:20:01 --> 00:20:07
but a little lower,

350
00:20:03 --> 00:20:09
so that when I let it go, we'll
all see that it will slide.

351
00:20:07 --> 00:20:13
So it's past the critical angle; 
the angle is smaller.

352
00:20:10 --> 00:20:16
But now I have
four kilograms here.

353
00:20:12 --> 00:20:18
I wasn't going to ask any one
of you to walk up this ladder,

354
00:20:17 --> 00:20:23
believe me, and I wasn't going
to do it myself either.

355
00:20:19 --> 00:20:25
So you see it is unstable.

356
00:20:23 --> 00:20:29
And now I put
the four kilograms on here.

357
00:20:26 --> 00:20:32
And I can let go, and
the ladder has become stable.

358
00:20:29 --> 00:20:35
Do I take it off--
there it goes.

359
00:20:31 --> 00:20:37
So the person standing very low
made it more stable,

360
00:20:35 --> 00:20:41
exactly consistent
with what we just saw.

361
00:20:38 --> 00:20:44
Now I make the angle alpha
a little larger than critical,

362
00:20:42 --> 00:20:48
so the ladder is happy.

363
00:20:44 --> 00:20:50
It's happy.

364
00:20:46 --> 00:20:52
But now the person is going
to do something dangerous.

365
00:20:49 --> 00:20:55
He's going to walk
and stand here--

366
00:20:52 --> 00:20:58
and he shouldn't do that,
as you see.

367
00:20:54 --> 00:21:00
That's exactly
what you have seen there.

368
00:20:58 --> 00:21:04
So the friction plays
an essential role

369
00:21:02 --> 00:21:08
for us to get
static equilibrium,

370
00:21:05 --> 00:21:11
and that is often the case.

371
00:21:07 --> 00:21:13
There are many examples
in our daily lives

372
00:21:10 --> 00:21:16
where friction can be used
to our advantage,

373
00:21:14 --> 00:21:20
and a very special case, which
I will discuss with you now,

374
00:21:19 --> 00:21:25
is often used by sailors--

375
00:21:22 --> 00:21:28
by holding, controlling
a very massive object,

376
00:21:27 --> 00:21:33
a very large force, controlling
it with a very small one.

377
00:21:32 --> 00:21:38
And you do that by making use
of friction.

378
00:21:35 --> 00:21:41
You wrap a rope around a pole,
or around a rod,

379
00:21:42 --> 00:21:48
and you use the friction
between the rope and that rod

380
00:21:46 --> 00:21:52
to your advantage,
and it works as follows.

381
00:21:50 --> 00:21:56
Here is this rod--

382
00:21:52 --> 00:21:58
it doesn't have to be horizontal
as this one--

383
00:21:57 --> 00:22:03
and I hang on this side
of the rod,

384
00:22:00 --> 00:22:06
I hang with a string
a very massive object.

385
00:22:06 --> 00:22:12
Capital M.

386
00:22:09 --> 00:22:15
So the tension here, T2,
would be Mg--

387
00:22:14 --> 00:22:20
assuming that
it is not accelerating,

388
00:22:16 --> 00:22:22
that it is at rest.

389
00:22:19 --> 00:22:25
I wrap this rope around here
several times,

390
00:22:25 --> 00:22:31
and here I hang an object which
is substantially less in mass,

391
00:22:31 --> 00:22:37
which is m, little m,
and the tension equals mg.

392
00:22:39 --> 00:22:45
If there is no friction at all
in this bar, then...

393
00:22:44 --> 00:22:50
and the rope is near massless,
then T1 will be the same as T2,

394
00:22:50 --> 00:22:56
so the situation will start
to accelerate.

395
00:22:53 --> 00:22:59
However, if we make use
of the friction,

396
00:22:56 --> 00:23:02
then we can have
a stable situation.

397
00:22:58 --> 00:23:04
We can have static equilibrium,

398
00:23:01 --> 00:23:07
so that this one is not moving
and this one is not moving,

399
00:23:04 --> 00:23:10
and then we can have T2 to be
way, way larger than T1,

400
00:23:09 --> 00:23:15
by using friction
to our advantage.

401
00:23:12 --> 00:23:18
Let's blow up
that center portion.

402
00:23:17 --> 00:23:23
Radius R,
and let us have this rope here--

403
00:23:24 --> 00:23:30
I'll give the rope a color.

404
00:23:27 --> 00:23:33
It doesn't have to come off
vertically, of course.

405
00:23:30 --> 00:23:36
It could be at any angle.

406
00:23:34 --> 00:23:40
There it is, and so here is
my T1 and here is my T2.

407
00:23:39 --> 00:23:45
And we take the situation

408
00:23:41 --> 00:23:47
that the pull on this side is
way larger than on that side

409
00:23:45 --> 00:23:51
so that this rope wants
to slip in this direction.

410
00:23:49 --> 00:23:55
That's what it would like to do.

411
00:23:52 --> 00:23:58

412
00:23:56 --> 00:24:02
If you look at these small
little sections of the rope,

413
00:24:02 --> 00:24:08
it is immediately obvious
if the rope wants

414
00:24:05 --> 00:24:11
to slide in this direction,
wants to start slipping,

415
00:24:08 --> 00:24:14
that the frictional forces
in these little pieces here

416
00:24:11 --> 00:24:17
are all in this direction...

417
00:24:16 --> 00:24:22
all the way around.

418
00:24:18 --> 00:24:24
And therefore,
it helps T1, so to speak,

419
00:24:23 --> 00:24:29
and so because of the friction,
which is opposing T2,

420
00:24:28 --> 00:24:34
T2 can now become
much larger than T1.

421
00:24:33 --> 00:24:39
What you have to do
to calculate this analytically,

422
00:24:36 --> 00:24:42
you have to evaluate these
individual frictional forces

423
00:24:40 --> 00:24:46
for these very small slices,
so that becomes an integral,

424
00:24:44 --> 00:24:50
and then you have to integrate
it over an angle,

425
00:24:48 --> 00:24:54
which I will call theta zero.

426
00:24:50 --> 00:24:56
And I remember when I last
lectured 801, that was in 1993,

427
00:24:55 --> 00:25:01
I derived that in class--
the relation between T2 and T1

428
00:25:00 --> 00:25:06
as a function
of this angle theta.

429
00:25:03 --> 00:25:09
And that took me
about five minutes,

430
00:25:05 --> 00:25:11
and after five minutes,
half the students were asleep.

431
00:25:08 --> 00:25:14
Now, I'm not sure whether you
want to sleep five minutes,

432
00:25:11 --> 00:25:17
but I don't think, frankly,
that you deserve it,

433
00:25:14 --> 00:25:20
so I decided
to not do the derivation

434
00:25:16 --> 00:25:22
but to refer you to the book,
which is page 361,

435
00:25:21 --> 00:25:27
and you will find, then,
that in the situation

436
00:25:23 --> 00:25:29
that the rope wants to start
slipping in this direction,

437
00:25:27 --> 00:25:33
that T2 divided by T1 is e to
the power mu times theta zero,

438
00:25:35 --> 00:25:41
if the friction coefficient here
is mu--

439
00:25:37 --> 00:25:43
it would be the static
friction coefficient.

440
00:25:40 --> 00:25:46
So that is the result.

441
00:25:41 --> 00:25:47
And notice

442
00:25:42 --> 00:25:48
that it is independent
of the radius of this bar,

443
00:25:45 --> 00:25:51
which is not so obvious,
strangely enough.

444
00:25:49 --> 00:25:55
Whether the bar is this small or
this small makes no difference.

445
00:25:52 --> 00:25:58
It only depends on this angle.

446
00:25:54 --> 00:26:00
This angle could be very large.

447
00:25:57 --> 00:26:03
You could wrap it
around ten times,

448
00:26:00 --> 00:26:06
as we will do very shortly.

449
00:26:02 --> 00:26:08
So there's no restriction
on theta.

450
00:26:05 --> 00:26:11
If there were
no friction at all,

451
00:26:07 --> 00:26:13
notice then at the moment
that it starts to slip,

452
00:26:10 --> 00:26:16
e to the power 0 is 1,
that's when T2 equals T1,

453
00:26:14 --> 00:26:20
that's obvious, so you can't
play this game if mu is zero.

454
00:26:18 --> 00:26:24
You need friction.

455
00:26:19 --> 00:26:25
That is at the heart
of this whole problem.

456
00:26:23 --> 00:26:29
And so now let's put in
some numbers

457
00:26:26 --> 00:26:32
so that we get an idea
of what we gain.

458
00:26:30 --> 00:26:36
So let us take the situation

459
00:26:32 --> 00:26:38
that we take
let's say three turns.
 

460
00:26:37 --> 00:26:43
We wrap the rope around
three times.
 

461
00:26:40 --> 00:26:46
So three turns.

462
00:26:42 --> 00:26:48
That means
theta zero equals six pi.

463
00:26:48 --> 00:26:54
And let the friction coefficient
mu be one-fifth, 0.2.

464
00:26:54 --> 00:27:00
So e to the power mu times
theta zero is now about 40.

465
00:27:02 --> 00:27:08
So what that means is that
with a force on this side

466
00:27:06 --> 00:27:12
which is 40 times smaller
than a force on that side,

467
00:27:10 --> 00:27:16
I have a balanced situation.

468
00:27:13 --> 00:27:19
I can hold this in my hand
and counter--

469
00:27:15 --> 00:27:21
if you want to think of it
that way--

470
00:27:18 --> 00:27:24
a force on this side
which is 40 times larger.

471
00:27:22 --> 00:27:28
But if I take six turns,

472
00:27:25 --> 00:27:31
then e to the power mu theta
zero will be about 2,000--

473
00:27:34 --> 00:27:40
2,000!

474
00:27:35 --> 00:27:41
So now I can really control
an elephant.

475
00:27:39 --> 00:27:45
Imagine now that I have here
an object, capital M,

476
00:27:45 --> 00:27:51
which would be
5,000 kilograms...

477
00:27:49 --> 00:27:55
make it 10,000 kilograms.

478
00:27:53 --> 00:27:59
That's what I'm hanging here.

479
00:27:55 --> 00:28:01
I can hang here now a mass which
is 2,000 times less massive.

480
00:28:01 --> 00:28:07
That means I could hang there
five kilograms.

481
00:28:08 --> 00:28:14
And the tension here would be
2,000 times less

482
00:28:11 --> 00:28:17
than the tension there.

483
00:28:13 --> 00:28:19
The system would be
about to start slipping,

484
00:28:16 --> 00:28:22
but it's not slipping.

485
00:28:18 --> 00:28:24
There is complete balance.

486
00:28:20 --> 00:28:26

487
00:28:22 --> 00:28:28
So imagine I hold this part
in my hand--

488
00:28:25 --> 00:28:31
here is this
10,000-kilogram weight

489
00:28:27 --> 00:28:33
and I hold this in my hand.

490
00:28:28 --> 00:28:34
All I need is a force
of about 50 newtons,

491
00:28:31 --> 00:28:37
and I'm standing there,

492
00:28:32 --> 00:28:38
and on the other side is
this 10,000-kilogram weight.

493
00:28:35 --> 00:28:41
And now I just make my force
a little less than 50 newtons,

494
00:28:40 --> 00:28:46
and what will happen now,
it will start to slip.

495
00:28:44 --> 00:28:50
Remember?

496
00:28:45 --> 00:28:51
Because we calculated here

497
00:28:47 --> 00:28:53
the requirement
for just not slipping.

498
00:28:49 --> 00:28:55
So now I let it go, and then

499
00:28:51 --> 00:28:57
the 10,000-kilogram on the other
side will slowly go down.

500
00:28:55 --> 00:29:01
I can just control it,
and I can stop it,

501
00:28:58 --> 00:29:04
and I can control it with
a very, very small force.

502
00:29:04 --> 00:29:10
Now comes a question for you.

503
00:29:06 --> 00:29:12
Suppose I wanted to raise
the 10,000-kilogram.

504
00:29:10 --> 00:29:16
Could I now pull a little bit
more than 50 newtons?

505
00:29:14 --> 00:29:20
Just a teeny weeny
little bit more?

506
00:29:17 --> 00:29:23
Would then that
10,000-kilogram come up?

507
00:29:20 --> 00:29:26
I see people shake their heads.

508
00:29:22 --> 00:29:28
Who thinks it would come up?

509
00:29:25 --> 00:29:31
Who thinks "no way"?

510
00:29:28 --> 00:29:34
Who doesn't think at all?

511
00:29:30 --> 00:29:36
Most of the people.

512
00:29:32 --> 00:29:38
All right-- sorry,
I didn't mean to be nasty.

513
00:29:36 --> 00:29:42
Um... there's no way
that you could pull it up,

514
00:29:40 --> 00:29:46
because if you want
to pull this up,

515
00:29:43 --> 00:29:49
the situation
completely reverses.

516
00:29:46 --> 00:29:52
Those frictional forces for this
rope to go in this direction

517
00:29:50 --> 00:29:56
will flip over.

518
00:29:52 --> 00:29:58
In other words,

519
00:29:53 --> 00:29:59
what is now T1 in our
calculations will become T2.

520
00:29:57 --> 00:30:03
So if you want this side
to go down,

521
00:30:00 --> 00:30:06
if you want to have it slip
in this direction,

522
00:30:02 --> 00:30:08
you're going to have
that T1 divided by T2 is now

523
00:30:06 --> 00:30:12
e to the power mu theta zero.

524
00:30:09 --> 00:30:15
And so now if you have
10,000 kilograms here

525
00:30:13 --> 00:30:19
and if you had six turns, then
the force that you need here

526
00:30:17 --> 00:30:23
is 2,000 times larger
than 10,000 kilograms,

527
00:30:22 --> 00:30:28
and so you would need
20 million kilograms.

528
00:30:25 --> 00:30:31
So it would be the dumbest thing
to do to lift it.

529
00:30:28 --> 00:30:34
You don't want to lift it.

530
00:30:30 --> 00:30:36
You use this device only
to balance a very strong force--

531
00:30:34 --> 00:30:40
that's the way it's used
by sailors--

532
00:30:37 --> 00:30:43
and even to control it, because
you can slowly release it,

533
00:30:41 --> 00:30:47
and then the force on the
other side will start to move.

534
00:30:46 --> 00:30:52
But you cannot use it
to lift something.

535
00:30:49 --> 00:30:55

536
00:30:50 --> 00:30:56
I have here a plastic bin,
and in this plastic bin

537
00:30:57 --> 00:31:03
are four of these
15-pound lead bricks.

538
00:31:01 --> 00:31:07
Would you do me a favor and come
up here and convince yourself

539
00:31:05 --> 00:31:11
that three are already in there?

540
00:31:06 --> 00:31:12
I don't want to put them
all four in,

541
00:31:09 --> 00:31:15
so I decided I'm only going
to put the last one in.

542
00:31:13 --> 00:31:19
You see those three?

543
00:31:14 --> 00:31:20
Thank you very much.

544
00:31:15 --> 00:31:21
And I'm going to put
the last one in.

545
00:31:16 --> 00:31:22
It's very heavy.

546
00:31:17 --> 00:31:23
Do you want to check this,
by the way?

547
00:31:19 --> 00:31:25
This is... Careful!

548
00:31:20 --> 00:31:26
It's very, very heavy.

549
00:31:21 --> 00:31:27
Yeah, okay.

550
00:31:24 --> 00:31:30
Okay, 15 pounds apiece--
60 pounds we have up there.

551
00:31:29 --> 00:31:35
And now I'm going to wrap this
around this bar,

552
00:31:34 --> 00:31:40
which is your bar.

553
00:31:35 --> 00:31:41
The radius doesn't matter.

554
00:31:36 --> 00:31:42
And why don't we start
with six rotations?

555
00:31:40 --> 00:31:46
One... two... three...

556
00:31:46 --> 00:31:52
(loud rustling
against microphone )

557
00:31:47 --> 00:31:53
Oh, my goodness.

558
00:31:50 --> 00:31:56
Four... five... six.

559
00:31:56 --> 00:32:02
Okay.

560
00:31:58 --> 00:32:04
Now I'm going to lower
this platform that we have

561
00:32:02 --> 00:32:08
that is holding it up.

562
00:32:04 --> 00:32:10
I can lower it.

563
00:32:07 --> 00:32:13
And it will shortly be hanging
now on my rope.

564
00:32:14 --> 00:32:20
There it is.

565
00:32:16 --> 00:32:22
I can remove this now.

566
00:32:18 --> 00:32:24
We don't need this anymore,
and we don't need this anymore.

567
00:32:23 --> 00:32:29
Effortless.

568
00:32:25 --> 00:32:31
30 kilograms--
(whistles softly )

569
00:32:29 --> 00:32:35
Hardly any force.

570
00:32:31 --> 00:32:37
Very gentle.

571
00:32:32 --> 00:32:38
Now let me lower it a little.

572
00:32:34 --> 00:32:40
There it goes.

573
00:32:36 --> 00:32:42
Just lower it.

574
00:32:37 --> 00:32:43

575
00:32:40 --> 00:32:46
Would be nice
if I had to do even less.

576
00:32:43 --> 00:32:49
Let's put a few more turns on.

577
00:32:45 --> 00:32:51
One... two... three.

578
00:32:49 --> 00:32:55
You know what we could do?

579
00:32:52 --> 00:32:58
We could put so many turns on

580
00:32:54 --> 00:33:00
that the weight of the rope
itself is enough to balance it.

581
00:32:59 --> 00:33:05
Let's try it.

582
00:33:02 --> 00:33:08
Not yet.

583
00:33:03 --> 00:33:09
Put a few more on.

584
00:33:05 --> 00:33:11
One... two... three... four.

585
00:33:11 --> 00:33:17
We have 12 windings now.

586
00:33:14 --> 00:33:20

587
00:33:18 --> 00:33:24
And the rope,
the weight of the rope

588
00:33:21 --> 00:33:27
is now sufficient
to balance the 30 kilograms.

589
00:33:24 --> 00:33:30
So you see this in action now.

590
00:33:27 --> 00:33:33
This is a marvel.

591
00:33:28 --> 00:33:34
Well, let me put a few more on,

592
00:33:30 --> 00:33:36
because I don't want it to come
down during the lecture.

593
00:33:33 --> 00:33:39
I want to be sure
that it stays there,

594
00:33:36 --> 00:33:42
and I'm going to secure it here.

595
00:33:38 --> 00:33:44

596
00:33:40 --> 00:33:46
So you see how you can use
friction to your advantage

597
00:33:44 --> 00:33:50
and get an enormous gain by
factors of thousands and more,

598
00:33:50 --> 00:33:56
and this is used
quite frequently.

599
00:33:54 --> 00:34:00

600
00:33:56 --> 00:34:02
So you saw
in a striking example

601
00:33:58 --> 00:34:04
of where friction helped us
to balance,

602
00:34:01 --> 00:34:07
and now I want to discuss
an object that is hanging

603
00:34:05 --> 00:34:11
and that can freely swing
due to gravity.

604
00:34:08 --> 00:34:14

605
00:34:11 --> 00:34:17
I want you to appreciate the
situation of static equilibrium

606
00:34:18 --> 00:34:24
that the sum of all forces
and the sum of all torques

607
00:34:23 --> 00:34:29
have to be zero.

608
00:34:25 --> 00:34:31
So here is this object
which I'm going to hang,

609
00:34:29 --> 00:34:35
maybe on the wall, and here is
a point P, and there is a...

610
00:34:34 --> 00:34:40
let's say a frictionless spin
that I put in the wall,

611
00:34:37 --> 00:34:43
so the object can
freely swing around.

612
00:34:40 --> 00:34:46
Let the center of mass be here.

613
00:34:45 --> 00:34:51
And so there is a force.

614
00:34:47 --> 00:34:53
If this object has a mass
little m, there is a force mg,

615
00:34:53 --> 00:34:59
and here is
the position vector r of P,

616
00:34:56 --> 00:35:02
and so it's clear that there is
a torque relative to point P.

617
00:35:02 --> 00:35:08
And if there is a torque
relative to point P,

618
00:35:06 --> 00:35:12
it would be r P cross mg.

619
00:35:12 --> 00:35:18
I put a vector sign over here,

620
00:35:13 --> 00:35:19
because you have to take the
cross-product between the two,

621
00:35:16 --> 00:35:22
so you take the...

622
00:35:18 --> 00:35:24
sine of this angle has
to be taken into account.

623
00:35:21 --> 00:35:27
This object is going to rotate.

624
00:35:23 --> 00:35:29
It's clearly going to rotate
about that point P.

625
00:35:28 --> 00:35:34
The torque equals I about
that point P times alpha,

626
00:35:33 --> 00:35:39
and alpha is the
angular acceleration.

627
00:35:38 --> 00:35:44
How can we ever get
a stable situation?

628
00:35:41 --> 00:35:47
Nature is now going to think
hard and is going to say,

629
00:35:46 --> 00:35:52
"Gee, I can only have
stable equilibrium

630
00:35:49 --> 00:35:55
"if the sum of all forces
equals zero

631
00:35:52 --> 00:35:58
"and if the sum of all torques
relative to any point

632
00:35:58 --> 00:36:04
equals zero."

633
00:35:59 --> 00:36:05
And nature knows how to do that.

634
00:36:03 --> 00:36:09
That always will happen
if P and the center of mass

635
00:36:09 --> 00:36:15
are along a vertical line.

636
00:36:12 --> 00:36:18
Because if that's the case, then
there will be here a force mg,

637
00:36:18 --> 00:36:24
and then here there will be
a force mg upwards.

638
00:36:22 --> 00:36:28
This object is hanging
on that pin, so the pin--

639
00:36:26 --> 00:36:32
action equals minus reaction--
will push upwards.

640
00:36:30 --> 00:36:36
So the sum of all forces is
zero, and there is no torque.

641
00:36:35 --> 00:36:41
Take any point you want to--
this point or that point

642
00:36:38 --> 00:36:44
or this point or that point
or that point--

643
00:36:40 --> 00:36:46
there is no torque anymore.

644
00:36:42 --> 00:36:48
And so now there is
complete equilibrium.

645
00:36:45 --> 00:36:51
So if you had an object,
for instance,

646
00:36:47 --> 00:36:53
that looked like this,
a thin rod--

647
00:36:50 --> 00:36:56
there's a very massive
object here,

648
00:36:53 --> 00:36:59
so that the center of mass
is almost there--

649
00:36:55 --> 00:37:01
then this situation
would be stable.

650
00:36:59 --> 00:37:05

651
00:37:01 --> 00:37:07
Or you might say this situation,

652
00:37:07 --> 00:37:13
because now we have the center
of mass, mP, vertical line.

653
00:37:13 --> 00:37:19
So also here
we have now mg down,

654
00:37:20 --> 00:37:26
so we have at the pin mg up.

655
00:37:22 --> 00:37:28
Sum of all forces is zero,
sum of all torques is zero.

656
00:37:26 --> 00:37:32
But there is a big difference

657
00:37:28 --> 00:37:34
between this situation
and this situation,

658
00:37:30 --> 00:37:36
which you immediately feel
in your stomach, of course,

659
00:37:33 --> 00:37:39
and that is that
this is highly unstable.

660
00:37:35 --> 00:37:41
If you just blow
on this a little,

661
00:37:37 --> 00:37:43
it will swing over,
it will start to rotate,

662
00:37:41 --> 00:37:47
whereas this is stable.

663
00:37:43 --> 00:37:49
If I move this to the side, it
will come back to that position.

664
00:37:48 --> 00:37:54
But the basic idea is, then,

665
00:37:50 --> 00:37:56
that the center of mass will
always in stable configuration

666
00:37:55 --> 00:38:01
line itself up below
the point of suspension.

667
00:37:59 --> 00:38:05

668
00:38:01 --> 00:38:07
I have here a triangle,

669
00:38:05 --> 00:38:11
and I have no clue
where the center of mass is.

670
00:38:08 --> 00:38:14
It may be somewhere here.

671
00:38:10 --> 00:38:16
It may even be in open space,
I don't know.

672
00:38:13 --> 00:38:19
What I can do now,
I can suspend it like this,

673
00:38:17 --> 00:38:23
and I know that the center of
mass must be below my finger.

674
00:38:22 --> 00:38:28
Did you notice?

675
00:38:23 --> 00:38:29
It actually rotated.

676
00:38:25 --> 00:38:31
So that the center of mass
lines itself up

677
00:38:28 --> 00:38:34
precisely below my finger,

678
00:38:30 --> 00:38:36
so I can now take a pencil
and draw a line here,

679
00:38:34 --> 00:38:40
or I can have a little string,

680
00:38:37 --> 00:38:43
and now I can change
my finger to here

681
00:38:40 --> 00:38:46
and I can again draw this line,

682
00:38:42 --> 00:38:48
and where the two intersect is
the center of mass.

683
00:38:45 --> 00:38:51
So the center of mass
will always find itself

684
00:38:47 --> 00:38:53
below the point of suspension.

685
00:38:49 --> 00:38:55
If now I take a piece of putty

686
00:38:52 --> 00:38:58
and I put the piece
of putty here,

687
00:38:54 --> 00:39:00
then clearly I change the
location of the center of mass

688
00:38:57 --> 00:39:03
quite substantially.

689
00:38:59 --> 00:39:05
It must have shifted enormously
in this direction.

690
00:39:03 --> 00:39:09
Where is the center of mass?

691
00:39:05 --> 00:39:11
I have no clue.

692
00:39:06 --> 00:39:12
But look.

693
00:39:08 --> 00:39:14
I balance it here.

694
00:39:10 --> 00:39:16
The center of mass must now be
somewhere here.

695
00:39:14 --> 00:39:20
I balance it here.

696
00:39:17 --> 00:39:23
The center of mass must be
somewhere here.

697
00:39:20 --> 00:39:26
And where
the two lines intersect--

698
00:39:22 --> 00:39:28
could be in this opening--

699
00:39:24 --> 00:39:30
that's where the center of mass
is located.

700
00:39:28 --> 00:39:34
So it's easy experimentally
to find the center of mass.

701
00:39:31 --> 00:39:37
Earlier in the course,

702
00:39:32 --> 00:39:38
we calculated where
the center of mass was.

703
00:39:35 --> 00:39:41
Perhaps you remember that,

704
00:39:36 --> 00:39:42
when we had some very special
geometrical configurations.

705
00:39:40 --> 00:39:46
I mentioned to you then already

706
00:39:41 --> 00:39:47
that there are ways that you can
experimentally determine it,

707
00:39:44 --> 00:39:50
and that can be done very easily
in the way I just showed you.

708
00:39:49 --> 00:39:55
Now... I want to show you some
examples of static equilibrium

709
00:39:55 --> 00:40:01
which are not so intuitive.

710
00:39:58 --> 00:40:04

711
00:40:00 --> 00:40:06
You may have seen some of them.

712
00:40:01 --> 00:40:07
You may actually have played
with some of them.

713
00:40:05 --> 00:40:11
If I take a pencil
and I take my pocket knife...

714
00:40:13 --> 00:40:19
So here is a pencil.

715
00:40:15 --> 00:40:21
I will show this to you shortly,
just a regular pencil.

716
00:40:21 --> 00:40:27
And I take my pocket knife

717
00:40:23 --> 00:40:29
and I put the knife,
I jam my knife in here.

718
00:40:27 --> 00:40:33
So this is the knife.

719
00:40:28 --> 00:40:34

720
00:40:31 --> 00:40:37
And this is now
my pocket knife, like so.

721
00:40:38 --> 00:40:44

722
00:40:42 --> 00:40:48
This arrangement
can easily be made

723
00:40:44 --> 00:40:50
such that the center of mass
is below this point,

724
00:40:48 --> 00:40:54
and so if now
I put my finger here,

725
00:40:51 --> 00:40:57
I can balance
this rather strange object

726
00:40:57 --> 00:41:03
and it will be
completely stable.

727
00:40:59 --> 00:41:05
It will arrange itself

728
00:41:00 --> 00:41:06
so that the center of mass
will fall below my finger,

729
00:41:03 --> 00:41:09
and it will be completely happy.

730
00:41:05 --> 00:41:11
And I want you to see that.

731
00:41:07 --> 00:41:13
I have here my pocket knife
and I have here a pencil.

732
00:41:12 --> 00:41:18
And you can see there--

733
00:41:14 --> 00:41:20
or the ones... those of you
who are close can see it here.

734
00:41:17 --> 00:41:23
I'll give you a better
light condition for that.

735
00:41:20 --> 00:41:26

736
00:41:22 --> 00:41:28
So there we are.

737
00:41:24 --> 00:41:30
So here is my pocket knife.

738
00:41:28 --> 00:41:34
I open it now.

739
00:41:30 --> 00:41:36
And here is the pencil.

740
00:41:32 --> 00:41:38
And I'm going to jam it in here.

741
00:41:35 --> 00:41:41
When you do that, be more
careful than I was this morning,

742
00:41:39 --> 00:41:45
because the knife cut
into my finger.

743
00:41:41 --> 00:41:47
That's why I have
this Band-Aid here.

744
00:41:44 --> 00:41:50
It can easily lead
to a nasty cut.

745
00:41:50 --> 00:41:56
So the knife is now
in this pencil,

746
00:41:53 --> 00:41:59
but now I have to get
that center of mass under...

747
00:41:58 --> 00:42:04

748
00:42:01 --> 00:42:07
under the pencil.

749
00:42:03 --> 00:42:09
And now I put my finger here,
and it's stable.

750
00:42:07 --> 00:42:13
So the center of mass
rearranges itself

751
00:42:10 --> 00:42:16
so that it goes
under the point of suspension

752
00:42:14 --> 00:42:20
and is completely stable.

753
00:42:16 --> 00:42:22
You can wiggle it
as much as you want to.

754
00:42:18 --> 00:42:24
That's not so intuitive.

755
00:42:20 --> 00:42:26

756
00:42:22 --> 00:42:28
I have here a... a very thin
bar, a wire, a very stiff wire.

757
00:42:33 --> 00:42:39
And I can put my hammer here
on the end of this wire.

758
00:42:38 --> 00:42:44

759
00:42:40 --> 00:42:46
And look at this.

760
00:42:42 --> 00:42:48
The center of mass is
way below my suspension point.

761
00:42:46 --> 00:42:52
This is stable like hell.

762
00:42:48 --> 00:42:54
It also hurts like hell,
by the way.

763
00:42:51 --> 00:42:57
You can even swing it.

764
00:42:53 --> 00:42:59
Center of mass is way here
below the suspension point.

765
00:42:58 --> 00:43:04
Sum of all torques is zero,

766
00:43:00 --> 00:43:06
if it's like this,
sum of all forces is zero.

767
00:43:04 --> 00:43:10
Center of mass below
the suspension point.

768
00:43:07 --> 00:43:13
They have to be exactly
vertically lined up.

769
00:43:10 --> 00:43:16
If they're not
vertically lined up, like now,

770
00:43:12 --> 00:43:18
then there is a net torque
relative to that point.

771
00:43:15 --> 00:43:21
If there is a net torque, there
will be an angular acceleration,

772
00:43:18 --> 00:43:24
and that's what you see.

773
00:43:20 --> 00:43:26
If they're exactly lined up,
there is no longer any torque.

774
00:43:25 --> 00:43:31
Aah.

775
00:43:27 --> 00:43:33

776
00:43:33 --> 00:43:39
Let's now turn to a rope walker.

777
00:43:36 --> 00:43:42
We have a rope walker.

778
00:43:38 --> 00:43:44

779
00:43:41 --> 00:43:47
Here is the rope walker.

780
00:43:44 --> 00:43:50
And the rope walker,
standing on the rope.

781
00:43:48 --> 00:43:54
I'll make the rope walker
a little smaller--

782
00:43:51 --> 00:43:57
you'll see shortly why I make it
a little smaller.

783
00:43:54 --> 00:44:00
This is the rope walker.

784
00:43:57 --> 00:44:03
And this is the rope.

785
00:43:59 --> 00:44:05
And let the center of mass of
the rope walker be right here--

786
00:44:05 --> 00:44:11
somewhere here.

787
00:44:07 --> 00:44:13
And the distance between the
center of mass and the rope--

788
00:44:12 --> 00:44:18
let that distance be one meter,

789
00:44:14 --> 00:44:20
and let the mass of the
rope walker be 70 kilograms.

790
00:44:23 --> 00:44:29
I build a light structure--
it could be built of wood--

791
00:44:26 --> 00:44:32
which the rope walker is going
to carry in her hands,

792
00:44:30 --> 00:44:36
or in his hands, just like this.

793
00:44:33 --> 00:44:39
And it goes down
quite a distance.

794
00:44:39 --> 00:44:45
And I put here
a mass of five kilograms,

795
00:44:44 --> 00:44:50
and I put here a mass
of five kilograms.

796
00:44:48 --> 00:44:54
And let's assume that this
structure is very lightweight,

797
00:44:51 --> 00:44:57
so that we can ignore that,

798
00:44:54 --> 00:45:00
and let this distance be
ten meters.

799
00:45:01 --> 00:45:07
Where is the center of mass
of this system?

800
00:45:05 --> 00:45:11
Seventy kilograms is
above the rope.

801
00:45:10 --> 00:45:16
One meter.

802
00:45:12 --> 00:45:18
Ten kilograms is ten meters
below the rope,

803
00:45:15 --> 00:45:21
so clearly the center of mass
of the system as a whole

804
00:45:19 --> 00:45:25
will be below the rope.

805
00:45:23 --> 00:45:29
And so that person is
as stable as anything--

806
00:45:26 --> 00:45:32
just as stable
as my hammer was--

807
00:45:28 --> 00:45:34
with one problem, that the
person could slide off the rope

808
00:45:33 --> 00:45:39
to the sides, and that would be
unfortunate, of course.

809
00:45:36 --> 00:45:42
And therefore
you would require, then,

810
00:45:38 --> 00:45:44
that if this is the rope,
the cross-section of the rope,

811
00:45:42 --> 00:45:48
then you would make yourself
soles on your shoes

812
00:45:46 --> 00:45:52
which are a little bit
like this.

813
00:45:49 --> 00:45:55
So that's only to prevent you

814
00:45:51 --> 00:45:57
from sliding off in this
direction and falling.

815
00:45:54 --> 00:46:00
But the center of mass is
below the rope,

816
00:45:56 --> 00:46:02
and so there is
really no danger.

817
00:45:58 --> 00:46:04
You can balance yourself
and walk on the rope

818
00:46:01 --> 00:46:07
and nothing would happen.

819
00:46:03 --> 00:46:09
You could even do this--
go back and forth on the rope.

820
00:46:07 --> 00:46:13
No problem.

821
00:46:08 --> 00:46:14
You always come back like this.

822
00:46:11 --> 00:46:17
Well, we have a very
special rope walker.

823
00:46:14 --> 00:46:20
This is our rope walker.

824
00:46:16 --> 00:46:22
And this rope walker has
indeed very special shoes.

825
00:46:20 --> 00:46:26
In fact, the shoes of the
rope walker is a little wheel,

826
00:46:25 --> 00:46:31
and this is what
the wheel looks like.

827
00:46:30 --> 00:46:36
And here is
where the rope will be,

828
00:46:32 --> 00:46:38
and this is the rope that you
see here in the lecture hall.

829
00:46:36 --> 00:46:42
And so the only reason why
we need this groove in the wheel

830
00:46:39 --> 00:46:45
is to prevent it
from sliding to the sides,

831
00:46:42 --> 00:46:48
as we need it for this person.

832
00:46:44 --> 00:46:50
But the center of mass--

833
00:46:46 --> 00:46:52
you see it when I balance it
on my finger--

834
00:46:49 --> 00:46:55
is way below
the suspension point.

835
00:46:51 --> 00:46:57
Because of these weights
on both sides,

836
00:46:53 --> 00:46:59
the center of mass
is somewhere here,

837
00:46:55 --> 00:47:01
so it is completely stable.

838
00:46:57 --> 00:47:03
You can even go like this--
no problem.

839
00:47:01 --> 00:47:07
This is what
the rope walker can do,

840
00:47:04 --> 00:47:10
provided it doesn't slide off.

841
00:47:06 --> 00:47:12
That's why we have this.

842
00:47:08 --> 00:47:14
Okay, let's ask this rope walker
to give us a demonstration.

843
00:47:12 --> 00:47:18
Let's make her walk.

844
00:47:15 --> 00:47:21
We don't need
that television anymore.

845
00:47:18 --> 00:47:24

846
00:47:20 --> 00:47:26
It would be nice
if one of the students...

847
00:47:22 --> 00:47:28
Would you do me a favor

848
00:47:24 --> 00:47:30
and welcome the rope walker
as she arrives here?

849
00:47:28 --> 00:47:34
Because we don't want her
to crash.

850
00:47:30 --> 00:47:36
Is that okay with you?

851
00:47:32 --> 00:47:38
So when she comes down,

852
00:47:33 --> 00:47:39
be gentle and give her
all the honors that she deserves

853
00:47:36 --> 00:47:42
and then you just make sure
that she doesn't fall,

854
00:47:39 --> 00:47:45
because that...
she will be damaged, right?

855
00:47:42 --> 00:47:48
We don't want that.

856
00:47:45 --> 00:47:51
Okay, so you have seen already
how stable the situation is.

857
00:47:51 --> 00:47:57
And there she goes.

858
00:47:52 --> 00:47:58
You ready for that?

859
00:47:54 --> 00:48:00

860
00:47:56 --> 00:48:02
Voilà! We can all become
rope walkers.

861
00:48:00 --> 00:48:06
Thank you very much.

862
00:48:02 --> 00:48:08
See you Friday.