Hey guys. So in this video, I'm going to start solving another type of static equilibrium or complete equilibrium questions in 2 dimensions. Now, in these problems, we're going to have a beam or a shelf-like object that's going to be held against the wall, usually by a cable or by a rod. Let's check it out.

In some static equilibrium or complete equilibrium problems, we'll have shelf-like or beam-like objects being held against the wall, tensioned against the wall like this. The first force I'm going to draw here is the mass of the beam, which will happen in the middle. There is tension here, \( t \), at an angle \( \theta \). So we decompose this tension into \( t_x \) and \( t_y \). In these problems, we're going to have the hinge apply a force against the beam. Notice how there's a \( t_x \) to the left. These are equilibrium questions; all forces have to cancel on the X and the Y; all torques have to cancel. If there's a force to the left, there has to be a force to the right. And that's where the hinge comes in with a force on the right. So we're going to call this \( h_x \) because it's the hinge force on the x axis. So the hinge will always apply a horizontal force against the tension. The hinge will almost always apply a force \( h_y \) on the beam. The hinge will also have a y force. We're going to assume that \( h_y \) is up. The reason why I say assume is because sometimes it's going to be down. In fact, if you get a negative for \( h_y \), meaning you're calculating, and you end up with a negative number, that means that you guessed it incorrectly. You assumed the wrong direction. If you get a negative for \( h_y \), \( h_y \) was actually down, but that's okay.

Before we do an example, I want to talk about why there has to be an \( h_y \) force. You don't need to really understand this to solve the question, but I just want to talk about this very briefly. You have to have a \( t_x \) so that you cancel that \( h_x \). That's easy. But why do you need a \( t_y \) if you already have a force canceling \( mg \)? Let's say if you didn't have an \( h_y \), then \( t_y \) would have to equal \( mg \) so that the forces cancel. The problem is, \( t_y \) is farther from the axis at the hinge than \( mg \) is. If they had the same forces, this torque here would be much greater. It's farther from the axis. So this thing would spin like this. So in fact, what we need is we need \( t_y \) to be less than \( mg \). So that they balance in torque. But now it's not enough to hold it on the y axis. So there needs to be another force in the y axis. That's why \( h_y \) exists. Hopefully, that makes sense.

Now let's solve this problem here. We have a beam of mass 300, \( \text{mass} = 300 \). I'm going to put that in the middle here. Right there. So that's \( mg \). I'm going to use gravity as a 10. So \( 300 \times 10 = 3,000 \). \( MG \) is 4 meters in length. That means that this distance here is 2 meters. And then this distance here is 2 meters. And I have a tension. It's held horizontally against the vertical wall by a hinge. Got a hinge here and a light cable as shown. So it's going to be tension here. This is \( t \). It's going to split into a \( t_y \) and a \( t_x \). We'll put \( t_x \) right here. The angle here is 37 degrees. And I want to know what is the a, tension on the cable, and b, the net force that the hinge applies on the beam. </p>

Before we talk about b, let me draw the hinge forces. Again, there's always going to be an \( h_x \) to cancel out the \( t_x \) and we're going to assume an \( h_y \) up as always. Now the net force on the hinge is a combination of \( h_y \) and \( h_x \). You have \( h_y \) this way, \( h_x \) this way. They're going to combine to form an \( h_{\text{net}} \) and that's what we want to know. \( h_{\text{net}} \) will simply be the Pythagorean of its components. So \( h_x^2 + h_y^2 \). I like asking this question here because to find \( h_x \), you basically have to know how to calculate everything else. So it's sort of a comprehensive question that guarantees that you know almost everything about this question. Okay.

So I got all my forces here. Remember, once I decompose, once I decompose \( t \) into \( t_x \) and \( t_y \), I don't really have a \( t \). Same thing here. So we're going to use the component forms, \( t_x \), \( t_y \), \( h_x \) and \( h_y \), and not the total vectors. Okay? So as with any complete equilibrium questions, I'm going to start by writing the sum of all forces equals 0, the sum of all torques equals 0. So the sum of all forces in the x axis equals 0. So the sum of all forces in the y axis equals 0. And there are only 2 forces in the x axis. I have \( h_x \), \( t_x \). They cancel so I can write that one equals the other. And on the y axis, I have 2 forces going up and that's going to cancel with the one force going down. So \( t_y \) and \( h_y \) going up equals \( mg \). First thing we're looking for is \( t \), in this equation here. And by the way, the way we're going to find \( t \) is by finding \( t_x \) or \( t_y \) and then we're going to find \( t \). Okay? Remember that \( t_x \) equals \( t \cos(\theta) \). So I could technically replace this with \( t \cos(\theta) \), but I don't want to do that. We're going to leave everything in component form because I think it's much easier this way. So we're going to leave them as \( t_x \) and \( t_y \). And then once you find \( t_x \), you're going to be able to get \( t \) because you know the angle or you can get \( t_y \) first and then find \( t \). Okay? So really when we're looking for \( t \), we're really looking for \( t_x \) or \( t_y \) whichever one we can find in the easiest way. I cannot find \( t_x \) because I don't know \( h_x \). I cannot find \( h_y \) because even though I know \( mg \), I don't know \( h_y \). So these two equations by themselves are insufficient. So I'm going to have to write a third equation. The third equation will be the sum of all torques equals 0 about a point. And we want to make sure that we pick a clever point to solve this. If we're looking for \( t \), we're first looking for either \( t_x \) or \( t_y \), both of which act here. Both of which act at this point here. So what we want to do is we want to write a torque equation at a point away from that point where the \( t \) is so that \( t \)'s will show up on my torque equation. Okay. So I have a few points that I could pick here. 1 at the hinge, 2 at \( mg \) and 3 at the end where the tension happens. Remember, you always want to pick torque. You always want to write the sum of all torque equals 0 equation, at a point where there's a for