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>> Hi, this is Module 21 of An Introduction to Engineering Mechanics.

Â So let's take a look at the forest here, what I've called the forest before.

Â This is the overview of the course, as opposed to the trees as we go through each

Â lesson. We've gotten quite a ways along in the

Â course. We've, we've done the representation of

Â forces in 2 and 3D. We've went through the syst-, the part of

Â the course on particle equilibrium, we've learned about the concept of moments and

Â how to follow the moment of force about a point in a line or an act a line or an

Â axis, we've also discuss couples we're in the section about equilibrium equations

Â and equivalent systems, we're working on resultants and, where they act looking at

Â distributed forces finding centroids and then method we all use last time in what

Â I'm going to do again today is another example.

Â Who have using the method of composite parts.

Â Because it's a really important skill for engineers to know.

Â And then we'll go into the last block of instruction, starting with the next

Â module. So, the learning outcome for today is to

Â solve another problem using the method of composite parts to find a resultant force

Â and couple on a, on a, on a body. And so here, here's the problem we're

Â going to work with we want to, we have a, a, cantilever beam here, it's fixed at the

Â left hand end, it's like a diving board if you will, here's the beam itself and I've

Â got some distances on there and I've got some sort of a load, distributed load,

Â starting with a 100 pounds per foot up here, wrapping down to 60 pounds per foot,

Â leveling off, and then going down to zero at the end.

Â And so, we want to find an equivalent resultant force, and a couple at point o

Â over here on the left. And so, what I'd first like you to do for

Â practice is to break this load into. Using the techniques that we learned last

Â module, breaking it into standard shapes, and finding the resultant force for those

Â standard shapes, and the location of those resultant force for each of those standard

Â shapes. And so take, take some time to do that and

Â then come on back. Okay.

Â Just as a hint, in case you couldn't get started, what I want to do is I want to

Â show you the standard shapes that you should break this in to.

Â So, if we draw a line here, and a line here, and a line here.

Â Those are the standard shapes you should work with.

Â You should work with this triangle and this rectangle, this rectangle and this

Â triangle. And so, if you didn't finish up now is a

Â good time to go ahead and finish and find the result in forces and their locations

Â for each of these standard shapes. And then come on back.

Â Okay, these are the results you should have come up with.

Â For the upper triangle here, its magnitude is the area for a triangle one half the

Â base times the height or 100 pounds. Its location is 1 3rd from the left-hand

Â side, or 2 3rds from the right-hand side, so it's 1.67 feet from point o, 1 3rd of

Â five feet. This rectangle has an area of 300 pounds.

Â It's resultant force is acting in the middle of this rectangular section, which

Â is. 2.5 feet.

Â This rectangle is a, a base of 60 and a height of 4, for 240 pounds.

Â Its location is halfway along this four feet section so it's 5 plus 2, or 7 feet

Â from the left, left hand edge. And finally the rectangle here is one half

Â base times height, or 120 pounds, going from the left, you go 5 feet, 4 feet.

Â And then one-third of four feet. And so, if that's the result you got,

Â you're in pretty good shape for breaking the system into composite parts, and

Â finding those resultants, and their locations.

Â So now what I want to do is I want to come up with a single equivalent force and

Â couple at point o. So this will be system two.

Â So I'll draw my cantilever beam again. And we know from earlier lessons, when we

Â found equivalent systems where we work with equivalent systems that the sum of

Â the forces on system 1 has to be equal to the sum of the forces on system 2, and so

Â in this case. We sum the forces on system 1.

Â That's going to be, and I'll, I'll call down positive, so I'm going to have 100

Â pounds, plus 300 pounds, plus 240 pounds, plus 120 pounds equals the Total sum of

Â the forces on system, two. And so that ends up equaling, 760 pounds,

Â and so that's one of our answers. That's the result in force.

Â At point O, and so that would act down over here at point O on system 2 and will

Â be equal to 760 pounds. Okay, lets now keep on going and find or

Â satisfy the condition for sum of the moments being equal on both system 1 and

Â system 2 So let's start out by drawing system 21 here again.

Â What we've done so far, this is system two And this is point O.

Â We found that the result in force at point o is 760 pounds down.

Â And now we have to satisfy the other condition, which is the sum of the moments

Â About a common point on both systems, so this is point o in this case, have to be

Â equal. And so let's go ahead and sum moments

Â about o on system one. And so we've get, 100 is and the, and the

Â distance to that perpendicular distance to that line of action of that force is 1.67,

Â so I've got 100 times 1.67 and then I've got 300 times 2.5 and then I've got plus

Â 240. Perpendicular distance is 5 plus 2, or 7.

Â And plus 120 times the perpendicular distance to the line of action at that

Â force, which is 5, plus 4 is 9 plus 1.33, is 10.33.

Â And, all of that has to equal the moment about o on our second system.

Â So, the sum of the moments about O on system 2, if you do that math, is 3,836.

Â The units are. Foot, pounds.

Â And these individual resultants cause a clockwise rotation, so the moment of o or

Â the couple at o has to also be clockwise. And so that's my equivalent couple.

Â And I can draw it on the figure. Clockwise array, arrow and it's going to

Â be 3836 foot pounds. And so we've completed the problem, we've

Â replace the loading shown to the right with an equivalent resultant force and

Â couple at point O, and I'll see you next module.

Â