## Minggu, 01 Mei 2011

### Fisika untuk Universitas

Fisika untuk Universitas

Ditujukan untuk meningkatkan kualitas proses dan hasil perkuliahan Fisika di tingkat Universitas

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.

Ucapan Terima Kasih Kepada:

1. Para Dosen MIT di Departemen Fisika

a. Prof. Walter Lewin, Ph.D.

b. Prof. Bernd Surrow, Ph.D.

2. Para Dosen Pendidikan Fisika, FPMIPA, Universitas Pendidikan Indonesia.

Terima Kasih Semoga Bermanfaat dan mohon Maaf apabila ada kesalahan.