And at their intersection, acts

the load of 10 Newtons.

On the left and on the right, we have elements which are

fixed supports. They are supports which does not

enable any movements.

Here, likewise, in the

structural sketch, we have on the left and on the right a fixed support.

Sometimes the fixed support is also symbolized

in this way, like a small triangle, the support being located at the top of the triangle.

We draw again the structural sketch

here. That is quite easy because within

the framework of this construction, the distance

between the supports, which we

call the span, which is designated by the letter

l, is equal to 1.2 meters. And the

vertical distance between the supports

and the lower point of the cable, which we

call the rise,

f, is equal to 0.6 meters.

Thus, the angle is 45 degrees, which facilitated the drawing.

We have here, acting, a load of 10

Newtons. We first take an interest in the

free-body of the load and we draw it again

below. So, we have a segment

of cable parallel to the piece of cable on

the left, a segment parallel to the one of the cable on the right,

a load of 10 Newtons, and then both

internal forces on the left and on the right which are in

the extension of the cut segments of cable

and as I indicated before, we are going to have N1 and N2.

Let's now continue with the Cremona diagram in which we want to draw

the force of 10 Newtons which acts downwards.

I draw it with a length of 10 centimeters to make a drawing to scale.

So, I have 10 Newtons, equal to, we are going to say 100 millimeters.

For the free-body of the load to be in

equilibrium, it is necessary that the three forces which act

on this free-body then, the load of 10 Newtons, the internal force

N1 and the internal force N2, should be converging on only one point.

Well, that is good, we can see well that indeed,

they are converging on the point of hanging of the load.

Then, on the other hand, that their vectorial

sum should be zero, so we are going to construct the

vectorial sum of the load of 10 Newtons and of both internal forces N1 and N2.

So, I trace straight lines at 45 degrees

in my graph which correspond to the inclination

of the internal forces N1 and N2, thus

this segment here is parallel to this and also to this.

This segment here is parallel to the segment on

the left in the free-body or in the overall system.

I notice,

indeed, that the internal force N2 acts

in the direction I had supposed, that is to say that it pulls

on the free-body. An internal force which pulls on the free-body is

a tensile internal force. And the tension, for us, is red,

then I am going to draw this arrow again,

here, in red. And, since I have drawn to scale,

I can see that the internal force, here, is equal to 7.1 Newtons.

So, as I have drawn to scale, I directly obtain the numerical

value of the internal forces. Likewise,

for the internal force

N1 that I also measure at

7.1 Newtons. Seventy-one

millimeters give 7.1 Newtons. We can

now copy these internal forces in the real system.

That is to say that in the real system, we are going to replace

the basic color of each bar by

the conventional color for the tension, red.

And we will write, we will sketch the internal force drawing

a kind of small piece of cable with an element of tension and then,

here, we will write 7.1 Newtons.

Likewise,

for the segment on the left, also

7.1 Newtons. We now want to take an interest in

what happens near the supports. We are going to

study a free-body which cuts the cable and which

stops just before the support on the left. I draw again this

free-body. Once again, there will be a segment

of cable which is parallel to the segment on the right.

We are going to draw the beginning of the support. What is the internal force in

this cable ?

Well, we know it, it is the internal force N1 which is equal to 7.1 Newtons.