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Hello, this

video is dedicated to the method of the auxiliary cable which will enable us,

on the one hand, to directly and freely calculate the shape of a cable,

and on the other hand, to obtain, without needing

a model, the position and the magnitude of the resultant of a

system of arbitrary forces, and particularly of

forces which are parallel one to each other.

We will see that the method of the auxiliary

cable enables us to determine the shape of a cable.

We will see how to construct the auxiliary cable,

what are the arbitrary choices that we have to make to

be able to make this construction, and finally,

we will see how to use this construction

to obtain the position of the resultant of a system of forces.

In this video, you recognize the system with two loads.

We now add a red chain on

which we hang up two weights in its middle.

We can note that the red chain and the white chain are perfectly

superimposed, on the left and right

parts, between the support and the first force,

respectively between the last force and the support.

However, in the middle, the shape is different.

We can see how to solve these systems.

I am not going to take lots of time to

do that, because we already did it several times before.

For the system on the top, we have

two loads, a load of 10 Newtons

which is in equilibrium with a horizontal

internal force and an internal force which goes up towards the

left. So, we have these internal forces.

We could read their value, but we have

already determined them another time before, thus, it will not be necessary.

Thereafter, on the right part, we also have 10

Newtons, and

we have

obtained the internal forces in the segments, on the left and on

the right, as well as in the intermediate segment of cable.

If we now look at the red chain,

on the bottom. It is subjected to a load of

20 Newtons

with, in its right part, an inclination which

goes up towards the right, and on its left

part, and inclination which goes up towards the left.

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We can note big similarities between these two

constructions. Actually, 10 plus 10 equals 20.

So, the effect of this force is equal to the effect of these two forces.

We can note that this segment has the same length than this other

segment, as well as this segment, here, has the same length than this

other segment.

In other words, the internal forces, in the inclined parts of the white

cable, are identical to the internal forces in the inclined parts of the red cable.

The only difference is that the red cable

does not have any horizontal segment in its middle.

What does it mean ?

It means that for the red chain and the white

chain, in the part between the support and the first load, for example,

the internal force is identical.

The link, both links, if we

consider them individually, they have exactly the same internal force.

What does it mean ?

Well, it means that the force of 20 Newtons

has exactly the same effect than both forces of 10 Newtons.

What does it mean ?

Remember the definition I gave

you for the resultant, when I introduced

this concept?

The resultant it is a force whose the effect on the system

is the same than the one of all the considered forces.

Well, clearly here, we are dealing with a resultant.

What is also interesting is to see

where is located the line of action of the resultant.

I think that

we already agreed to say that the resultant

was going to have a magnitude of 20 Newtons.

But now, what we can also see,

it is that this resultant has its line of action

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which is on the intersection

of the extension of the first and of the last segment of the

cable, the ones which were indicated by three and by two before.

The line of action of the resultant passes where they cross each other.

Let's look at this second case, it is the same cable,

but now, with a load of 10 Newtons on the left, and two loads of

10 Newtons on the right. I make quickly the construction.

So, the load of 10 Newtons, and I know that I will have the load

of 20 Newtons afterwards.

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and we can note once again that these two elements,

here, and here, respectively here,

and here, are identical. So, once again,

the force, here, of 30 Newtons and the resultant

of two forces of 10 and of 20 Newtons.

Its line of action

is located, one more time, on the intersection of the extension

of the first segment, and of the last segment of the

cable. We are going to use this property

which is general to define a general method,

to determine the position of the line of action of the resultant.

Let's start with a simple example.

Now, we deal with the resultant of two forces.

We do not have any cables.

We can, in the Cremona diagram,

because we know that we will

need it, we can already draw these two forces,

the force of 10 Newtons, and the force

of 20 Newtons.

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We are going to work with what we call an auxiliary cable.

We can also call it a fictional cable, because it is going to be useful,

and afterwards, once it will have

used it, we will be able to forget it,

respectively, to erase it if we wish to.

This cable will start from a support, here, and will have a certain initial inclination.

I choose,

because it is easier for me, to

give it an initial inclination of 45 degrees.

But this, that is totally arbitrary, we shall return to this point later.

It corresponds, in the Cremona diagram, to this part, and that is also

very easy with an angle of 45 degrees. We know that the internal force in this cable

is somewhere, here.

Afterwards, we have a second choice

to make, either we choose a second angle.

For example, we could say that the cable is horizontal

after having touched the force of 10 Newtons.

Alternatively, we can choose a point in the Cremona

diagram to represent the internal force in the intermediate part of the cable.

So, now, I prefer

to choose that the cable is horizontal between both forces.

You remember that it was not the

case in the cable which we have constructed previously.

It was horizontal when there was only one force.

But actually, we can find a cable whose the part, here, is horizontal.

What does it mean ?

Well, it means that in the Cremona

diagram, this part, here, must be horizontal, and then,

the internal force in the first

segment is now determined. Then, we

also have automatically determined, in the Cremona

diagram, the inclination of the last segment of cable, which is

like this. And, we can finish the diagram

of forces.

I copy this inclination in the structural sketch,

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well, we

obtain, on the intersection, a point

which is on the line

of action of the resultant.

We come back to the Cremona diagram.

What is the value of the resultant ?

Well, it is equal to the sum of both forces.

I draw it, a little bit staggered for the drawing to stay clear.

But it is clear that the resultant, here, is equal to 30 Newtons.

But where does it pass in the real system ?

Well, this resultant will be vertical.

It will pass by the point which we have determined before.

So, here, I draw the line of action of the resultant,

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and then, the resultant itself, I can represent it where I want, for example,

here. I indicate R is equal to 30 Newtons.

For this construction, we are going to construct the resultant of the

forces, starting by the force which will touch the cable first.

For the moment, it is a little bit obscure, but you will see it, you will understand it a

little bit later.

The first thing consists in choosing a support for the

auxiliary cable, then in choosing an initial inclination for the

auxiliary cable, which is going to give us the first segment, like

the last time, but this, it is simply because it suits me well.

Sometimes, it could not be a good solution.

I choose 45 degrees, and I indicate this segment by 1.

We now have to work in the Cremona diagram.

So, I am going to copy the

force of

20 Newtons in the Cremona diagram, and I

also copy the first

segment of cable, which I call 1.

I do not know exactly where it is going

to stop yet.

I now choose, either on the right, the point of intersection

in the Cremona diagram, either the inclination of the second segment.

To change, I am going to choose the

point of intersection in the Cremona diagram.

I am going to choose this point, here. On the basis of this point, here, it

gives me a second segment, which is not horizontal anymore as before.

Then, I copy this segment

into the real system. I indicate it

by 2. I can now add the second

force which is equal to 10 Newtons, and determine the

the rest of the Cremona diagram, with this

inclination here, which I indicate by 3. And I

copy this inclination into the

real system, and I can put

a support. I directly extend

this first line of action.

Here, I put the support about which I talked.

Likewise, I extend the first line

of action.

I have obtained a point on the line of action of the resultant.

The resultant, I get its magnitude, and its orientation, R

equals 30 Newtons, in the Cremona diagram, and afterwards,

I copy this orientation, and I obtain, here,

the line of action of the resultant. And the

resultant itself, I can place it here, on

this line of action, R equals 30 Newtons.

One more time, this method is very general.

It will also apply to several forces, we are going to see it

later, and also to forces which would not be

parallel.

Let's precisely look at the case of the resultant of three parallel forces.

I first draw these three forces in the Cremona diagram.

The first one, of 10 Newtons, the second one, of 20 Newtons,

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I choose to place a support.

Note that in this system, the lines of

action of the forces have already been placed, which is

very useful because usually, we are not going to directly intersect

the force, but we will always intersect its line of action.

To change a little bit, we are going to assume that the first segment of the cable is

horizontal. We also draw it

in the Cremona diagram. Actually, I can directly

draw it like this, and I choose this point.

So, it directly gives me the inclination of the second

segment of the cable, which I am going to copy in the real system.

So, this is the segment 1. And this is the segment 2.

The third segment has this

inclination

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So, this is the segment number 4. Also here, I also

indicate the support which ends the cable. The resultant itself,

in the Cremona diagram, it is vertical, and it has a

magnitude of 60 Newtons. And we can copy this

vertical line of action in the real

system representing the

resultant, here, which is equal to 60 Newtons.

So, we have seen that this system of construction, which

I had shown for two forces, can be applied generally for three

forces, and actually, for any number of forces,

I mean at least as many as you will have the patience for.

In this lecture on the auxiliary cable, we have seen

that this method enables us to determine the shape of a

cable in an arbitrary way, deciding, for example, on the orientation

of the first segment of a cable, even of the first two segments.

We have seen

how to systematically proceed for this construction.

We have seen that this construction, by the intersection

of the first and of the last segment of cable,

gives us a point on the line of action of

the resultant, and it thus enables us to determine

the resultant for any number of forces.

We have seen that this method is very general, and that it can be applied

for parallel forces, or for non-parallel forces.