>> 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.