Hello. In this video, we are going to see how to use the applet i-Cremona, to solve of a system of three forces. We will see how to introduce a force. I think that you remember this. We will see how to introduce supports. We will see how to activate and control the funicular polygon. And finally, we will see how we can obtain the internal forces which act within our structures. So, we have, here, in our applet, the person, whose known weight of 800 Newtons acts at the center of gravity. Let's say here. Then, we are going to look at a free-body, for example, here, which includes the person, which passes just under his feet, which cuts the rope, and which obviously includes the load which acts on the person. We do not have, naturally, supports, but where the structure has an intersection with the outside, where we are going to have an internal force, we can also introduce a support, either under the rope, either under the feet, I introduce first a support under the feet. I draw your attention on the fact that the first support that I introduce has a little one inside the triangle, this enables to know that it is the first support. You will see that it is important later. I introduce a second support, where the free-body cuts the rope. Now, I can start the solving, clicking on the button "funicular polygon". This button "funicular polygon", generates a funicular polygon, we are going to say, arbitrary. But, we are going to be able to control it by means of the following button, which is called initial tangent. And thanks to this initial tangent, if we come closer to the support one, there is a little line, a little black line which appears. We click on this black line, and we keep the button of the mouse pushed down. And it enables us to give the correct shape to the funicular polygon. That is to say that it follows the rope, here, and it passes, obviously, by the feet. And if we look at the part on the right, we can see that the construction of the funicular polygon has been done correctly. If we place the cursor on the middle of the segment, we can directly read the internal force. Thus, the internal force in the legs is equal to 827 Newtons. Then, in the part on the right, this force has enlightened in yellow, to indicate where it is located in the Cremona diagram. Likewise, here, the internal force in the horizontal part, in the rope, is equal to 225 red. That is to say in tension. And, they are going to appear. That is almost horizontal, 178 degrees instead of 180. If we place the cursor on the intersection of both forces, well, or of the three forces, then we can see that the equilibrium of these three forces is made by this polygon of forces. That is useful for certain occasions, especially when we will have more complicated structures. I am now going to quickly redo the same example, with the man who leans against a column. So, again, a load of 800 Newtons. Here, I do not draw the free-body, it is not necessary. I place a support under the feet, a support against the column. I activate the funicular polygon. I solve it. Sometimes, it is a little bit difficult. We can draw away as much as we want from the support. Well, we can imagine the internal force in the arms, it is something like that. So, we have, in the legs of the person, a compressive internal force of 860 Newtons. And in the arms, an internal force of roughly 250 Newtons. In this lecture, we have seen that to solve a system of three forces in the applet, it is possible to introduce only one of the forces, and two supports. These supports, when we use the resolution of the funicular polygon, will enable us, once we will have defined the proper geometry, to obtain the internal forces between each support and the force. We will be able to read their value and their inclination in the applet which will be used, for example, to later for a dimension the structural element.
