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.