00:01
Once again, welcome to a new problem.
00:03
This time we have a sharp sign, and the sharp sign is pretty much hanging from a hinge, so it's horizontal.
00:17
You know, we have a uniform beam like that, and the weight of the beam, which we're going to call a w, we're going to call this wb.
00:34
That's the weight of the beam.
00:40
Is given us 155 newtons.
00:47
And the beam is uniform, meaning that the weight has equal distribution.
00:54
And then something else that happens is that we have a hinge right here.
00:59
And there is a force involved that's holding the beam.
01:03
At the hinge and the force happens to be directed in that direction.
01:12
Okay, the force is directed in that direction.
01:15
Of course, it's going to have its components, but we'll get to that.
01:19
There is also an additional weight which comes from the sign.
01:25
There is a sign on hanging at the edge of the beam itself and it's hanging right here.
01:35
And you know that sign happens to have a weight of 215 newtons so we'll call that w .s equals to 250 newtons the distance from the edge from the hinge itself up until the sign is given us so this whole distance all the way up until this edge is given as 1 .7 meters and then we also have another distance right here well this is halfway so it's you know it's not the entire the entire length of the beam so this is this is half with who so we're gonna call that you know one 1 .7 you want to divide that by two and maybe we could give them symbols just to make it easier for us when we're solving.
02:41
So we'll call this distance right here from the edge, from the hinge.
02:48
This is the hinge itself.
02:49
So that distance from the hinge up until the edge.
02:54
We're going to call that, or maybe we should have done it this way at the top.
02:58
Yeah.
02:59
So, you know, it's very clear.
03:01
We want to see that this is, let's call that x.
03:11
Xl, so that's the entire length of the beam.
03:15
So we're calling it xl, and that's 1 .7 meters.
03:19
And then we don't need this.
03:21
We only need half of it.
03:24
So, and that's where the uniform weight of the beam is hanging.
03:29
So we'll call that x mid.
03:35
It's kind of like the midpoint.
03:36
And if you take 1 .7 divided by 2, that ends up giving you 0 .85 meters.
03:45
So this is 0 .85 meters.
03:49
And also there's an additional, i don't want to call it talk, but there's a tension.
03:59
There's a tension.
04:00
Remember this is a wall right here.
04:03
So there's a wall right here.
04:06
And so there's a tension that's holding on.
04:16
So there's a string that's pulling that's pulling that way towards the wall.
04:23
And so, you know, we'll just make sure that we have a force right there.
04:28
It's pointing that way.
04:29
And there's an angle of 35 degrees that the string or the wire.
04:38
Mix with the with the wall so this is this happens to be 35 degrees there's a tension right there and so we're gonna call that the ft f subscript t and then also we're gonna call this fh that's the hinge itself pulling the hinge and so these two forces have components they have the horizontal components and they also have the vertical component this one also as a you know i have to clear this a little bit just to give it some space so i'm gonna give this some space right here so this one has a horizontal component this is a horizontal component that way and then there is also so this is a horizontal component that way and then there's another vertical component the force itself is ft.
05:52
Remember the angle is 35, so our theta equals to 35.
05:58
This vertical component will call it ft sign of theta.
06:04
And then the horizontal one, this one will call it ft cosine of theta.
06:10
This is the information we're given and this one will be fh cosine of theta and the vertical component would be ph, a sign of theta.
06:24
So, you know, just to make things a little bit clear, we do have a sign that's hanging from the beam, from the edge of the beam, and this is the sign itself, this one right here, and that's why we're calling it ws.
06:44
And then also the beam itself as a uniform weight, we're calling that wb.
06:50
And there's a, and there's a a hinge right here that's holding on to the beam and it's pulling that way, it's pushing that way, but it has components.
07:01
There are two components.
07:02
There's a horizontal component and there's a vertical component given.
07:06
And also there's a tension.
07:08
There's a tension due to the string that's pulling this way.
07:13
And then there's another one that's pulling that way.
07:18
And so that's the information were given and we want you know we want to find now as the question states we want to find the tension in the string ft okay and then we also want to find the horizontal and vertical forces exacted by the hinge so we want to find fh cosine of theta and we we also want to find fh sign of theta, you know, the horizontal and vertical components exerted by the hinge.
08:08
We don't have to call them, you know, fh sign of theta, fh, cosine of theta, because we don't know what the angle is.
08:16
So we want to use a different angle.
08:22
So instead of using theta, theta is for the tension, we can use, say, alpha, for example.
08:32
So we can call this angle alpha.
08:34
And that just means that some of the symbols will change...