# MIT open course Physics 1 - Mechanics by Walter Lewin - Lecture 25 Static Equilibrium, Stability, Rope Walker

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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.

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.

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