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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Â By to hold this body in static equilibrium?

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Â That means that ax is actually to the left.

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

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

Â that would be minus 320, in the I direction.

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

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

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

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

Â and see what that yields. And so.

Â Sum of the forces in the y direction equals zero.

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

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

Â diagram I have Ay. That's up.

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

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

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

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

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

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

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

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

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

Â I'll just pick a convenient point. Of.

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

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

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

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

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

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

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

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

Â through point B. And by goes through point B.

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

Â your own. I could solve for ay directly.

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

Â point a. Set it equal to zero.

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

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

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

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

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

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

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

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

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

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

Â accordance with the sign convention that I've written.

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

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

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

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

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

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

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

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

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

Â a negative, in accordance with my sign retention.

Â By is the force. The moment on the perpendicular distance

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

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

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

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

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

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

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

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

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

Â We solve for Ax. We solve for By.

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

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

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

Â vectorially is equal to a 160 Newtons up over 160.

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

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

Â equilibrium. Now, these are the equations of

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Â