>> Hi, this is Module 25 of an Introduction to Engineering Mechanics.

Today we're going to, apply the 2D equilibrium equations that we've learned

use the tool of a free body diagram that we've learned about.

And we're going to solve for the force reactions and moment reactions acting on a

body that keep it in static equilibrium. So this is kind of the culmination of what

we try, been trying to get up to in, in the course.

So just as a review, static equilibrium equations, the forces and the moments must

be balanced in scale or form that leads to three equations of equilibrium, three

independent equations of equilibrium. The sum of the forces in the x and y equal

to zero, and the sum of the moments about any point that we choose must be equal to

zero. So here is the example that I had you do

last module, to, to draw the free body diagram and so this was the example and I

asked you to draw the free body diagram. This is the result that you should have

come up with, and so now that we have that result we're going to use the equilibrium

equations to find out. What are the force reactions, Ax, Ay and

By to hold this body in static equilibrium?

And you can see that I'm, I'm, I'm pleased with what I, I see so far, because I have

three unknowns and I have three independent equations, so hopefully we'll

be able to solve for all of those, those reactions.

And so let's start off with the sum of the forces in the x direction.

Fb equal to 0, I have to choose an arbitrary sign convention.

I'll say to the right is positive. And if I do that now on my free body

diagram I see Ax is to the right as I've drawn it, so that's positive.

We see that the 3 5ths component of the 600 Newton force is to the right, so

that's also positive, so we have plus 3 5ths Of 600 positive.

And that looks like all of the forces that we have in the x direction, so we're going

to set that equal to 0, that has to be balanced.

And if we do that we find out that ax equals minus 320.

And so, what that means is, I've drawn Ax on my free body diagram to the right.

I didn't know what direction it would be a priori.

But I found out that it's a negative value.

That means that ax is actually to the left.

And so I can write my. Answer for Ax, as a vector, as being 320

Newtons, to the left. Or if I want to put it in vectorial form

that would be minus 320, in the I direction.

If I is, is, to the right. And that would be Newtons.

And so There's my, one of my answers for the force reaction of Ax.

Okay so we've solve for the Ax force reaction using the sum of the forces in

the x direction equals to 0, now let's use the sum of the forces in the y direction

and see what that yields. And so.

Sum of the forces in the y direction equals zero.

I have to pick a, a, a positive direction for assembling my equation.

I'll say up is positive. And so in this case in my free body

diagram I have Ay. That's up.

I've got the 4 5ths component of the 600 Newton force is down.

So it's going to be negative in acc-, accordance with my sine convention.

So that's going to be minus 4 5ths times 600.

And then I've got the By force up. So that's plus By.

That's it for y direction forces. So that equals zero.

And we, we'll call that equation. Asterisk or star, because that's one

equation but I have two unknowns, so I can 't solve it.

So what we're going to have to do is to go to another equation.

And so in this case, I can sum moments about any point on the body.

I'll just pick a convenient point. Of.

A smart point to pick would be to pick point A and the reason I say that is, if I

pick point A, the line of action for Ax and the line of action for Ay causes no

moment about point A. And so the only unknown that's going to

cause a moment about point A is this. Force reaction By, so we'll be able to

solve for By directly by summing the moments about point A.

So I, I often refer to point A as what I call a smart point.

And so, you can do it about point B though in fact point B actually ends up being a

smart point as well as I, I noticed right now, because the line of action of Ax goes

through point B. And by goes through point B.

So if I sum moments about point B. I could solve, and you should try this on

your own. I could solve for ay directly.

All right. But we'll go ahead, and sum moments about

point a. Set it equal to zero.

I'll choose clockwise positive for. Assembling my equations.

As I just said, Ax and Ay, their line of actions go to, through point A, so there

is no perpendicular distance to the line of action of those forces.

And so there is no tendency to rotate about point A by those forces and so they

don't contribute to the moment equation. I don't know geometrically the

perpendicular distance from. Point A to this line of action to this 600

Newton Force, so I'm going to take advantage of Barry Neil's theorem and I'm

going to break this 600 Newton Force into its y component and its x component.

And by so doing that for the y component, it's going to tend to cause a clockwise

rotation about point A. So that's going to be positive in

accordance with the sign convention that I've written.

And so the force, the y-component of the 600 Newton force by similar triangle.

Is the 4 5ths component times 600. It's moment arm now is the perpendicular

distance between the line of action of the Fy force and point A, so that's going to

be five meters. As far as the Fx force is concerned or the

Fx component is concerned, if I neglect the, the thickness of the beam.

Then this Fx line of action also goes through point A, so it's not going to

cause a moment about point A. So the only other force reaction that's

going to cause a moment about point A is B sub I.

It's going to tend to cause a counter clockwise rotation, so that's going to be

a negative, in accordance with my sign retention.

By is the force. The moment on the perpendicular distance

is going to be 10, so it's going to be minus by times 10 and the only other thing

that's going to cause, tend to cause a rotation about point A is this 800 newton

meter couple. It's clockwise so it's positive.

So plus 800 equals zero. And if you solve that, you see that

there's only the b,y is the unknown. And we find that By equals 320.

Since it's positive that means it's in this, in the direction that I've chosen on

my free body diagram, so By vectorially is equal to, 320 newtons up, or 320 J

Newtons. And so I've got my second force reaction.

The last force reaction we need to find is a y.

We solve for Ax. We solve for By.

To solve for a Ay, all we have to do is substitute By equals 320 into equation

asterisk, sub By into asterisk and if I do that I find that Ay ends up being 160 its

possible so its in the direction that I've shown on a free body diagram up so Ay

vectorially is equal to a 160 Newtons up over 160.

J Newtons. And so I've solved for all three force

reactions and I know what they need to be to keep the body or the bar in static

equilibrium. Now, these are the equations of

equilibrium I used in solving this problem, they're independent.

As I mentioned in Module 13, and you may want to go back and review, you could also

use three other independent equations. For instance, you could use some of the

forces in one direction and two sums of the moments, or maybe three sums of, of

moments. And so, I'd like you to actually try that

and see that you get the same answer by summing forces in the x direction.

And then not only summing moments about the A point, but also summing moments

about the B point. And with those three equations you'll get

the same answer. I've put a PDF of the solution of that

problem in the module handouts. And you should check yourself once you

complete that problem. And the last thing I'd like you to do is

to solve this problem which is the cantilever beam that we did the free body

diagram for last module, I want you to determine in this case, not only the force

reaction but the moment reaction at A, due to the cantilever beam to support this

homogeneous bar, the bar weighs 100 pounds, you can neglect again the

thickness of the beam by the bar, and there's a 50 pound external force on the

right. And at, once again, I've put a PDF of the

solutions so you can check yourself out and make sure that you're solid on these

learning outcomes. And that's all for today's module.