with the internal force of 8.3

Newtons in the

cable and the forces at the support V2 and

H2. We reuse

the vector

N2. Then, we introduce

H2, and we finish by the force

at the support V2. H2 is equal to

7 Newtons

and V2 is equal to 4.5

Newtons.

We can note that in this configuration, well,

the sum of the horizontal forces at the support

is not zero, since precisely, there is

a horizontal component in the load which is applied.

You maybe wondered, whether it be in this video, or in the

previous video, why, after having drawn the

load of 10 Newtons, I have systematically always

taken the internal force N2 which was on the right of this load, and afterwards the internal force N1.

And you maybe wondered, why

did not I do the reverse ?

Actually, I already applied, without officially saying it to you,

a convention which we are going to use for the rest of

this course, which is that, when we deal with the forces

which act on a free-body, we take them in a

systematic order which is the counterclockwise order.

So, we turn around the free-body, in the counterclockwise

order, starting by where the forces are known.

So, here, we know the load of 10 Newtons.

Afterwards, we will have the element two which is N2.

And finally, the element three which is N1.

And then, in the Cremona diagram,

this is the first element that we have traced.

Here, the second element.

Here, the third element.

We can, obviously, proceed with the opposite convention.

That is to say, to turn in the clockwise order.

Will the equilibrium be modified ?

No, since regarding the

treatment of each free-body, we insure the equilibrium,

and we can of course ensure the equilibrium by adding first N1, and afterwards N2.

Actually, that is what has been done in this figure, from the book The art of

structures, in which we have dealt with this free-body. And if we closely

look at this, indeed we have firstly dealt with the

force, then the internal force N1 and finally

the internal force N2. So, we have used the force which

is indicated by Q, then N1 and finally N2.

You can see that the diagram is similar

to the one we did before.

The results are obviously right, and are the same.

You can try to do this job by yourself.

To simplify the correction and the communication

within this course, we ask you,

for this course, for the exercises, for your

reasoning, also with the assistants or with your questions, to systematically

use the counterclockwise order, as I have

shown you in this lecture.

Within the framework of this video, we have generalized the approach

method to solve the equilibrium of a cable subjected to a transversal load.

We have first

dealt with the case of an eccentric load, then of an eccentric and inclined load.

We have seen that the order in which

we consider the internal forces and the loads which act

on a free-body, has an influence on the

shape and on the construction of the Cremona diagram.

For the rest of this course, we

will always use the counterclockwise

order for the construction of the Cremona diagrams.

We have seen how to

systematically obtain the internal forces in the segments

of cable, and how to copy them in

the structural sketch to communicate these internal forces.

Finally, we have seen how to obtain the forces

at the supports and their horizontal and vertical components.