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

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

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

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

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

time. We came up with a second free body diagram

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

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

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

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

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

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

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

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

bottom. Two force members could also be intention,

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

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

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

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

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

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

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

hydraulic arm, BD. And so, if I.

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

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

action between those forces always stays the same.

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

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

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

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

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

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

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

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

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

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

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

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

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

compression. Equal opposite and co-linear.

Okay? So, that should give you a good

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

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

back and pick up from our original free body.

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

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

unknowns. And so first thing we recognized is that

we found out that bx. Is equal to 0.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

to be negative in accordance with my sign convention.

So we have minus 900 times cosine of 30 degrees.

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

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

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

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

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

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

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

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

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

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.

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

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

thousand pound, foot pound couple being applied.

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

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

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

action. The, the perpendicular distance between

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

So it's positive in accordance by sine coincidence.

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

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

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

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

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

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

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

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

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

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

going to cause a counter-clockwise rotation.

Negative in accordance with my sign convention.

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

its moment arm. So now for the x component.

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

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

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

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

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

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

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

1300 pounds up. Or 1300 j pounds.

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

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

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

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

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

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

Identify if the hydraulic arm is in compression or retention.

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

Then that hydraulic arm would have been in tension.

And that would have happened, for instance.

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

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

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

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

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

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

this hydrualic arm is indeed in compression, 'kay?

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

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

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

519.1. So it's negative again.

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

So Ay. Factorially is 519 pounds down, or minus

519 J. Pounds.

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

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

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

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

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

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

So that means that its in static equilibrium and.

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

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

in any directions. And so that's it.

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

engineers need to know how to do.