>> Hi and welcome to Module 22. For today we're going to look at
calculating the magnitude of a resultant force and its location for a force
distributed over a surface. So we're going to look at an example here.
And this is a special case, but we'll generalize it afterward.
Let's take a force that's perpendicular to the surface, so it's acting down.
Okay? And, the rectangular surface is shown here
as being flat and in the plane. So, this could be like a roof load, or a
snow load on a roof, or something like that, so I've.
All this, this pressure down acting a flat plane rectangular surface and the special
case that I'm going to look at is that the pressure is only going to vary in the x
direction. It's going to be constant in the y
direction. So what we're going to do is find the
resultant for a strip of infinitesimal width here dx and so my height here is p
of x. The, the force per area, the pressure, and
here's my width. And so if I multiply px by dx, I get a
distributed load here, and I can find the single force resulting for that in the
center. Which is the base, w, times the height px
dx. So that gives me a line down the center,
and so I've I can now look at that from the side direction and I've got a single
force per unit length over the distance l. So, that's pressure times The width, or
the w length here, so that's pressure is a force per area.
We've got a width of w here, so that ends up being a force per unit length f of x.
It acts over the entire length along the x-axis, which is l.
And so, now what I've done is I've reduced this problem to a, a forced along a, a,
the same as a force along a distributed er, force distributed along a line, and so
we can prosult, proceed with the final result using the techniques that we, we
used for finding the result in the longest straight line.
Now, if you didn't have pressure only varying in the X direction.
If you had pressure varying in the x and the y direction on any plane, flat
surface, you'd have to do a result in force location in the x direction by
integrating as shown here, and in the y direction as integrating shown here.
And so you're integrating over an area now, instead of over a line, so da is
equal to dx, dy where dx and dy can both vary.
If we then take and say that we have a constant pressure, these ps cancel, we can
pull them out from the integrals and they cancel.
And so what you get then is the, the, the resultant that you come up with is the
centroid for the area of that plane surface.
If we want to look at force distributed through a volume for example gravity or
electromagnetic force now our resultant force is the integral instead of over the
area it will be over the body so i've got integral over b or beta the body of the
force Which is, in this case I'm using g times little pieces of mass, dm, because
the, that acceleration due to gravity acts on each little piece of, of mass in the
body and so, g we can say to be constant, so we can pull it out of the interval, and
if we integrate over the entire body all these infinitesimal pieces of mass, that's
equal to the total mass. And so, the resultant force, as we would
expect, is mass times gravity, which is the weight.
Now, if we want to define the, the downward location of that resultant,
again, we would integrate. Now over the body of x dm divided by the
total mass, y dm divided by the total mass, and z dm tot-, divided by the total
mass. And that would give us our center of mass
for that resultant gravity force. Or, what's often referred to also as the
center of gravity. And, and one last note.
You'll, you'll find that dm Is equal to row dv where row is the mass density.
Okay? Or the mass per unit volume.
And if I look at a, a body then that has a constant mass density row, I can pull the
rows out. And now I get an integral over the volume
instead of over the entire body, and so, and, and, instead of using the integral
for, for mass and so I have the integral of x dV over the total volume, the
integral of y dV over the total volume, and the integral of z dV over the entire
volume, and that's the centroid of volume. And so you'll see that for a constant
density that the, the centroid of the volume and the centroid of the mass will,
will coincide. And so, that's just a quick lesson on how
to extend finding resultants and their locations from just along the distributed
line to. Having the ability to do it for surfaces,
which is, an application that's very relevant to real world engineering
problems. Thank you.
Dr. Wayne Whiteman, PESenior Academic Professional
