Hi. This is an introduction to engineering
mechanics, and this is module 14. Today's learning outcomes are to, first of all,
describe the conditions for 2 systems to be con-, con-, considered, equivalent
systems, and then to solve a problem finding an unequivalent system. So first
of all, equivalent systems are defined as, as being such if the sum of the forces
acting on, on one system is equal to the sum of the forces acting on the other
system. And, if the sum of the moments about a common point p on the one, the
system one and the system two are the same. And, if both of those conditions are
met, we have an equivalent system. So let's go ahead and do an example. We're
going to find a force and a couple at point C here, which is together form a
system which is equivalent to this system, where we have a hundred foot pound couple
here and a two hundred pound force in the I direction and a three hundred and fifty
pound force in the J direction up here. So we'll call this system 1, and we want to
find an equivalent system too at point c. Let's go ahead and draw. I can't leave her
being there again, and this will be, be point C, and it's 3 feet from the left
edge. And so our condition is that the sum of the forces. First condition is that the
sum of the forces on system one equals the sum of the forces on system two. And so,
for the sum of the forces on system one, I only have two forces. I have a 200 pound
force in the i direction and a 350 pound force in the j direction. So we've got
200i plus 350j pounds equals the sum of the forces, on system 2 and so I can, go
ahead and draw that. On my system 2 now, so I have a 200 pound force here, and a
350 pound force here. So that's the first condition, now we have to apply the second
condition for the sum of the moments about the common point c between both systems.
And so I'll start off here. I've got my, my force condition on my system, s2. Now,
let's, moments about point c on system one. And make that equal to the sum of the
moments on sys tem 2 about point C. So, first of all I have on system 1, this 100
pound, foot pound couple acting, clockwise and so if it's clockwise by the right hand
rule. It's into the page, where that would be in the negative K direction. So I have
minus 100K, and then I've got two forces that are causing moments about point C.
And so I'll use the definition of the moment R cross F. So I've got plus R from
C to A for the 350 pound force. So that's crossed with 350 pounds in the j
direction. And then I also have my 200 pound force so plus r from C to A That's
the position vector from the point about which I'm rotating to a point on the line
of action of the force, crossed with 200i. By the way, just as a side note. Un, in
using R from C to A. For the 350 pound force I could have also used r from c to b
because point b also lies on the line of action for the 350 pound force, and I'd
get the same answer. And you should try that out on your own to make sure that you
do get the same answer. So r from c to a is just 7 units or 7 feet in the x
direction and 2 units or 2 feet in the y direction. So this is 7i plus 2j, R from C
to A over here is also 7i plus 2j, and if I do those cross products, I'll get minus
100. I forgot my vector symbol here so I'm going to put it in K so this is K. And
then if I cross this, I'm going to get plus, two hun-, 2,450 K. Because I crossed
j as k. 7 times 350 is 2,450. J cross j is 0. And then we're going to have, over
here, I cross I is 0. J cross I is minus k. So we're going to have a minus 400. K
equals, and up here I should have put, I also forgot equals the sum of the moment
about point c on system 2 right? So we had some of the moments about C on system 1 is
on the left hand side and then sum of the moments about system C on system 2 is on
the right hand side. So this is some of the moments about c on system 2. And so I
get some of the moments on system 2 about point c. Is equal to, 2450 minus 500 or
1950, in the k direction. And the units are force times dista nce, or foot,
pounds. Okay, so now we've satisfied the second condition. We have, not only, that
some of the force is equal to 0, or excuse me, that some of the force is equal to
same on both systems. I also have the, some of the Moments about point C the same
for both systems. And so, here it, it shows this, on, on my system 2, I've got
my 200 pound force, my 350 pound force, my 1950 foot-pound Moment at point C. we can
take one more step and, and, and do a resolution of, of the two forces into a
single direction force. So I'm going to say, note, I can take my. This is 200
pounds. This is 350 pounds, and I can make that a single vector. It's going to be on
a slope of 350 on, 2. 350 on 200, or 3.5 on 2. And its magnitude is going to be the
square root of the sum of the squares for these two orthogonal directions. So the
square root are the sum of the squares for 350 squared plus 200. Squared, which comes
out to be 403. So this magnitude is 403 pounds. And so I can draw a final picture
of my system 2, with a single force and couple is this. This is point C. This is
three feet. And we have our single force, which is up and to the right, 403 pounds.
It's on a 3.5 on 2 slope, and we also have our counterclockwise moment, 1950
foot-pounds. And that's our final answer. That's a force and couple at point C which
together form a system which is equivalent to the original system, system 1. And
that's it for today's module.
Dr. Wayne Whiteman, PESenior Academic Professional
