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>> Hi, this is Module 27 of An Introduction to Engineering Mechanics.

Â For the learning outcomes, we're going to finish up that problem we started last

Â time with the crane applying 2D equilibrium equations to solve for the

Â force and moment reacting, reactions, and In the meantime we're also going to talk

Â about and define and recognize two-force members, which we actually started last

Â time. We came up with a second free body diagram

Â for this hydraulic arm, where we found out that the x components were equal to 0 at,

Â at point B and point D, and we only had y forces that remained.

Â And so, that's called a 2 force member, it's If this is a weightless member, where

Â there's only force application's at 2 friction-less pins, and we're assuming

Â that those are friction-less pins at point b and point d on the hydraulic alarm.

Â So, a 2 force member, is a weightless member again, 2 friction-less pins, it can

Â only be in-tangent or compression. And, so, In this case, it's going to be in

Â compression. It's acting down at the top, and up at the

Â bottom. Two force members could also be intention,

Â but that's all they can be. Those equilibrium require that the forces

Â had to be Equal to each other. They have to be in opposite directions,

Â and they have to be colinear. They have to be along the same line of

Â action. And so we found that By was equal to Dy.

Â They were equal, they were in opposite directions for equilibrium, and they're

Â colinear. And shape is not a factor and so, I've got

Â several examples. This could be a, an example of the

Â hydraulic arm, BD. And so, if I.

Â Have a compression at the top and a compression at the bottom, they can only

Â act opposite each other. And no matter what way I turn, the line of

Â action between those forces always stays the same.

Â Even if it was intention you can see that the force at the top and the force at the

Â bottom are equal opposite and along the same line of action.

Â And so that's what a 2 force member is all about.

Â Shapes not a factor, so even if I take a curved 2 force member like this It can

Â either be, in compression at the ends, or in tension, okay along the same line of

Â action. And just to show that shape doesn't need a

Â isn't a factor I've even took a, a weird s shape and again Either in tension or

Â compression. So if I, if I drew that s shape remember

Â it can be any shape at all. If I have 2 frictionless pins, one on this

Â side, and one on this side. The only thing that this can be is in.

Â Tension, like this along the line of action the line of action being the same

Â for both of those forces. Or, it would be in compression.

Â There's the 2 points. And the forces would, could be in

Â compression. Equal opposite and co-linear.

Â Okay? So, that should give you a good

Â understanding of 2 force members. When you can identify 2 force members, you

Â can get rid of all but, 2 reactions. Alright, using that information let's go

Â back and pick up from our original free body.

Â We've used this free body diagram of our hydraulic arm.

Â Now, let's go back to our original free body diagram and solve for the rest of the

Â unknowns. And so first thing we recognized is that

Â we found out that bx. Is equal to 0.

Â So let's sum forces in the X direction on this free body diagram.

Â I'll choose to the right positive, and I've got, I've got the 900 pound force,

Â it's X component will be the sign of 30 degrees.

Â And it's going to be to the left so it's going to be negative.

Â So minus 900 times sign of 30 degrees. The x is gone.

Â I've got an ax so I've got plus ax equals. Zero.

Â And so I find that Ax ends up equaling 450, since it's positive it's in the

Â direction that I've shown on my free body diagram.

Â So Ax is equal to 450 pounds to the right, where 400.

Â And 50 pounds in the i direction. And so there's another answer, that's

Â another one of my force reactions required to keep this crane in equilibrium.

Â Now let's keep going. Let's now do the forces in the y

Â direction. So we'll sum forces in the y direction.

Â I'll choose up as positive. To get the hang of this you should stop

Â the tape. Do it on your own and then check to see if

Â you get the same answer as I do. So we've got the y component at a 900

Â Pound forces is associated with the cosine of 30 degrees and it's down so it's going

Â to be negative in accordance with my sign convention.

Â So we have minus 900 times cosine of 30 degrees.

Â By is up so it's plus By. A y is up, plus A y, That's all the y

Â force, forces we have so that's equal to 0.

Â And so I get, from that equation, ay plus by equals 779.4.

Â And so, that's one equation. We have two unknowns.

Â So we're going to have to use some more equations.

Â But let's go ahead for, for now, and label this.

Â As equation asterisk, and we'll continue on, Ok so I've put it at the top here my

Â equation asterisk which I just solve for by summing force in the y direction lets

Â apply another equation of equilibrium in this case I have to sum moments about some

Â point, i'll go ahead and choose the sum of the moment about Point A equal to zero.

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And I'll choose clockwise as being positive.

Â And so, when I assemble that equation, I get.

Â Let's see here. About point A we've got this twenty

Â thousand pound, foot pound couple being applied.

Â It's clockwise so that's positive, in accordance with my sign convention.

Â So I have twenty thousand. By is going to cause a clockwise rotation.

Â It's. Its moment arm is going to be the line of

Â action. The, the perpendicular distance between

Â its line of action, and point A, so that's 10 feet.

Â So it's positive in accordance by sine coincidence.

Â It's going to be plus By times 10. Bx, remember, that was equal to 0 so we

Â don't have to worry about it. Then we have the 900 pound force.

Â And we'll work with each of the components separately.

Â So the cosign component is the y component, it's going to be down, it's

Â going to cause a counter clockwise rotation, it's going to be negative.

Â And so its magnitude is minus 900 times cosign of 30 degrees.

Â It's moment arm, this is the line of action okay?

Â Perpendicular distance to point A is going to be 25 feet.

Â And then we still have the x component of the 900 pound force which is going to be

Â to the left. So it's, and if it's to the left, it's

Â going to cause a counter-clockwise rotation.

Â Negative in accordance with my sign convention.

Â That will be 9, -900, and that's the sine component, so sine of 30 degrees, times

Â its moment arm. So now for the x component.

Â Here's the line of action, 'kay, horizontal.

Â The perpendicular distance between that line of action and the point about which

Â we're rotating is 30 feet, and so that is going to equal 0, because I have no other

Â forces or moments that's causing a rotation about point a.

Â And if I solve that for, for b I get By equals 1298.5.

Â It ends up being positive, so it's in the direction shown in my freebody diagram.

Â So By equals, if I round it off to three significant digits.

Â 1300 pounds up. Or 1300 j pounds.

Â That's another one of my force reactions. If By, let's look at the hydraulic arm

Â though. If By is down, then that means Dy is up.

Â And as we suspected from our diagram, that hydraulic arm is in compression.

Â And so I can say that the force. And BD that hydraulic arm is going to be,

Â 1300 pounds. And I'll call it in compression so I'm

Â going to put a c annotation behind it. So that was one of our questions.

Â Identify if the hydraulic arm is in compression or retention.

Â If By had come out to be neg-, excuse me. If By had come out to be negative.

Â Then that hydraulic arm would have been in tension.

Â And that would have happened, for instance.

Â It could have happened even in this case if this moment was large enough.

Â So really in advance we wouldn't know whether it is in tension or compression so

Â early id said oh yeah its going to be in compression but that is not always the

Â case, you have to be careful, be precise, find out what the reactions are, if they

Â come out to be positive then there in the direction that's shown on your diagram, if

Â they come out to be negative they are in the opposite direction but in this case

Â this hydrualic arm is indeed in compression, 'kay?

Â Last thing we need to do is solve for ay. That's the last reaction that we haven't

Â solved for yet, so we've got ay plus by equals 779.4 from an earlier equation.

Â We now have By, so we can substitute in Ay equals 779.4 minus By which is 1298.5 or

Â 519.1. So it's negative again.

Â It's going to be opposite the direction that I've drawn so it's going to be down.

Â So Ay. Factorially is 519 pounds down, or minus

Â 519 J. Pounds.

Â So that's all of the force reactions and we've solved the problem.

Â One thing I want you to note is that you could actually check your answer.

Â We sum moments about A we said that. If that was sum one of earlier for

Â something moments, that if the moments are 0 but any point on the body or off the

Â body has to be equal to 0 for all points and you could sum moments about B here

Â I've done that and you find out that 0 equals 0.

Â So that means that its in static equilibrium and.

Â It, you can check for any point. You can also sum moment about C or any

Â other point and it, it, it better equal 0 equals 0 so that there's no acceleration

Â in any directions. And so that's it.

Â Real neat problem that we've solved that's real world and, and something that

Â engineers need to know how to do.

Â