00:01
So in this problem, we need to start with a free body diagram.
00:02
And our body here is this horizontal beam.
00:05
And so we can start with drawing that.
00:07
We're told to ignore its thickness, so we can just say that it's a line.
00:12
We have a pin at a, which means it doesn't support any moment, but it does support forces.
00:18
We can have ay and ax.
00:22
We're assuming needs to be in the positive direction.
00:25
And we'll also take this point to be our origin.
00:27
We have y positive upwards and x positive to the right here by convention.
00:36
We have our applied force, 26 kiloons, and we know that it's at an angle here of 12, 513.
00:48
So really convenient, we know that its x component is 1213s, or its y component is 1213s.
00:53
The force downwards here, and x component would be 513s by the similar triangles of the slope.
01:02
We have our applied force, f, which is also 40 kilonutons, and then a rope holding up this right end here.
01:12
Now, a rope's going to have a force acting on it or the tension acting through it in the same line as the rope.
01:18
And so we know that its slope is three, four, five like that.
01:25
So we want to find all of the forces holding up, all the reactions.
01:30
We have the answers, though.
01:30
We want to see, of course, how they work out.
01:33
So we have three unknowns here.
01:35
And when we do so, we can think of the equilibrium equations, sum of moments equals zero, and some of forces equals zero.
01:44
Now we know moment or torque, they can usually be interchanged, is equal to a force times a distance.
01:51
So the powerful thing about this equation is let's say the distance is zero, then that means the moment's equal to zero.
01:58
So let's look at taking moments about our origin here about point a.
02:01
Well, we know that ax and a both act through this point, they have zero distance to point a, which means they make no moment.
02:11
So if we take moves about point a, the only unknown we have is our tension t.
02:15
So let's start there because we have one equation, one unknown to solve for.
02:19
But we also need to put distances here.
02:20
We know this is two meters between x and 26 kitton and four meters for the rest of the beam.
02:31
So we said that we should take some of moments about point a.
02:35
This is an equilibrium.
02:36
So they add to zero because it's not rotating or anything.
02:40
And we take positive to be counterclockwise by the right -hand rule, but as long as you're consistent through the problem, it works either way.
02:49
So let's look at our 26 killion force here.
02:53
The x component acts through the beam, so it doesn't matter.
02:56
But the y component, though, let's imagine pin a is, a is pinned, like it actually is, but if it wasn't, we could imagine, and take our x component of our 26 killiontons, extended around a, keeping base basically the same radius.
03:10
We see this is a clockwise motion, which is the opposite way of our positive sign.
03:15
So this is a negative moment due to that.
03:17
The 40 kiloton force is the same thing, but the tension is rotating upwards here, and that's going to be a positive moment.
03:26
And like we said, our x components here acting through the beam, they have no moment.
03:29
We only have the y components, which we can find by 12, 13th, and 3 5ths.
03:35
So this is one of those convenient problem where we get the ratios.
03:39
So we said the 26th kilonuton force was a negative moment.
03:42
So we have the force, but we need its component just in the y, which was 1213s.
03:50
If we can get rid of that, there we go.
03:52
And then the distance, which was two meters...