Beam / Shelf Against a Wall

by Patrick Ford
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Hey, guys. So in this video, I'm going to start solving another type of static equilibrium or complete equilibrium questions in two dimensions. Now, in this type of problems, we're gonna have a beam or a shelf like object that's gonna be held against the wall, usually by a cable or by a rod. Let's check it out. So in some static equilibrium are completely or complete equal. Even problems will have shelf like or beam like objects being held against the wall. Attention against the wall like this. So first force I'm gonna draw here is the mass of the beam, which will happen in the middle. There is a tension here. T it's at an angle theta. So we decompose this tension into T X and T. Why all right now in these problems were going toe have the hinge apply a force against the beam. So notice how there's a t X to the left thes air equilibrium questions All forces have to cancel on the x and y all tor except cancel. So 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 gonna call this H X because the Hinch force on the X axis so the hands will always apply a horizontal force against the tension. Okay, So tensions going like this, which means TX is like this, which means they checks is like this, in fact, more specifically here, I can say against TX. Okay, The hinge will almost always apply a force h y on the beam so the hinge will also have a Y force on bits. Most of the time, it's gonna be helping hold it. 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, Um, h y 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. And what I mean by that's OK is that it doesn't mean your answer is correct. It doesn't mean that you have to start the problem again, okay, you just realize Oh, well, it was actually down, and then you fix that, but you don't have to restart the question with it pointing down. Okay, so you see how this works. When I start solving the question, there's an h y here. All right. So just before we do an example, I want to talk about why there has to be an h y force. Okay, so you don't need to really understand this to follow the question, but I just want to talk about this very briefly. Um, you have to have a t X so that you canceled today checks. That's easy. But why do you need a t y? If you already have a force canceling MGI, why do you need a T y? Well, why didn t h y rather well, let's say if you didn't have an h y, then t y would have to equal mg so that the forces canceled. The problem is T y is farther from the access at the hinge than MGs. And if they have, if they had the same forces this torque here, which would be much greater same forces, but a bigger our vector. So this talk would be greater. It's farther from the Axis. So this thing would spin like this. So that doesn't work. So in fact, what we need is we need T y to be less than M G. T. Y needs to be less than M. G so that they actually, because it's farther 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 of the Y axis. That's why. Wait. Why exists? Hopeful That makes sense. If it doesn't, don't worry about it too much. You don't really need that, um, in solving questions, we're just gonna assume that there's always an h y up. So let's solve this problem here. We have a beam of mass. 300 mass equals 300. Um, I'm gonna put that in the middle here, right there. So that's M G. I'm gonna use gravity is a 10. So 300 times 10 is 3000 m. G. Um, it's 4 m in length. That means that this distance here is 2 m, and in this distance here is 2 m and I have a tension. It's held horizontally against the vertical wall by a hinge Got a hinge here and a light cable show. So it's gonna attention here. Um, this is T It's gonna split into a T y and a T x t X right here. Thean angle here is 37 degrees. Okay. And I wanna know what is the A tension on the cable and be the net force that the hinge applies on the beam before we talk about be, uh, let me draw the hinge forces again. There's going to be always an h X to cancel out the T X and we're going to assume an h y up as always. Okay, now, net Force on the hinge is a combination of h Y HX. You have H y this way hx This way they're going to combine to form a H net, and that's what we want to know. Each Net will simply be we're looking for each net. The net force by the hinge will simply be the Pythagorean of its of its components. So hey, check square plus h y square. I like asking this question here because to find a checks, you basically have to know how to calculate everything else. So it's sort of a comprehensive question that guarantee that you know almost everything about this question. OK, so I got all my forces here. Remember, Once I decompose, um, once I decompose t into T x and T Y. I don't really have a team, same thing here. So we're gonna use the component forms t X t y h s and h y and not the total vectors. OK, So as with any, um, as with any complete equilibrium questions, I'm gonna start by writing some of our force equals zero sum of all torques equals zero. So some of our forces in the X axis equals zero sum of all torques in the Y axis equals zero, and they're only to force in the X axis. I have h x t x, they cancel second, right? That one equals the other. And on the Y X is I have to force is going up, and that's gonna cancel with the one force going down so t y and h y going up equals M G. First thing we're looking for is t um in this equation here and by the way, the way we're gonna find T is by finding t x or t y, and then we're gonna find T. Okay, Remember that t X equals t CO sign of China so I could technical replace this with t co sign of China. But I don't wanna do that. We're gonna leave everything in component for him because I think it's much easier this way. So we're gonna leave them as t x and t y. And then once you find TX, you're going to be able to get t because you know the angle. Or you could 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 confined in the easiest way, I cannot find TX because I don't know a checks. Um, I cannot find h y because even though I know M g, I don't know h y. So these two equations by themselves air insufficient. So I'm gonna have to write a third equation. So 123 The third equation will be some of all torques equals zero about a point, and we wanna make sure that we pick a clever points to solve this. If we're looking for t were first looking for either t X or T y, both of which act here. Okay, both of which act at this point here. So what we wanna do is we want to write a torque equation at a point away from that point where the tea is. So that tease will show up on my torque equation. Okay, so I have a few points that I could pick here, one at the hinge to at M G and three at the end where the tension happens. Remember, you always wanna pick torque. You always want to write the sum of all tor X equals zero equation. Um, at a point where there is a force. That's why it's either 12 or three. You want to write it at a point where the force you're looking for is not at. So I don't want to write this point. This is a bad point. I want to write it at one or two. And here one of the two is actually better than the other 0.1 is the best point because there are two forces h y and H X, And if I write it here, I will have fewer torques. Okay, so 20.1 is the best point too. Right now, hopefully, you're convinced that 0.1 is the best one. I want to sort of go off off to the side real quick and talk about a different point here, which is the fact that no matter which point you pick 12 or three, there will never be a torque by a checks. Okay? And there will never be a torque by T X. And that's because these guys air pointing in the direction of these points. Right? So both of these guys will never produce a torque. Okay, At least not if the hinge is vertical horizontal. Okay, so really, no matter what, you only had one torque to worry about here, which is h y and M G. And what that means is that you could have really picked either one of them. All right, But I had already convinced you about number one. So let's just keep going with that. It really doesn't matter at the end of the day. All right? So there are two talks about 20.1 Let's draw 0.1 right here. Remember, H y does not produce a talk because it's on the Axis. HX does not produce a talk because it's on the access and because they are vector would make a 90 degree angle anyway, if it had some length, I have m g halfway here and then I have t Y T x does not produce a torque because they are vector. I'm just gonna draw this real quick here, but I'm gonna lead it er vector between, um three angle between the our vector for T X and the TX itself makes a total of 1 80 design of 1 80 0, right? Think of a door if you push this way in the door, it doesn't spend So you really left with just these two guys, which is awesome. So m g is pushing down. So it does this and t y pushes up. So it does this. They go in opposite directions. The torque of M G is clockwise negative. The torque of T y is, um, close counterclockwise Positive. So I can write that torque MGI equals torque T y. And I'm gonna expand both sides of this equation. This is gonna be m g r sine of data, and this is going to be t y our sign of fada. Okay, data is not necessarily the 37. Be careful right there. Our vector for M G Looks like this. It has a length of half of the distance of the length, which is later the bar, which is two. And then this vector here is the entire thing. 4 m. Both of them. Notice make a 90 degree angle here. So sign is actually 90 not 37. Because we're not talking about tea. We're talking about T y. Okay, so that's one of the advantages of doing this in component form, leaving the TX and T y not using t. Okay, so both of them are 90. So this is gonna be one and one now the distances are m g two n T Y. Four. Okay. And I have that t y equals two mg, divided by four. In other words, mg over to. And this, by the way, should make sense. Some of you may have thought about this when I was talking about up here. That t y is double the distance so we can have the same magnitude as MG in fact, has to have half of the force of M. G so that the Tories balance and we just sort of proved that here. Okay, so in a question like this, where the tension is all the way at the end and you have a horizontal bar with no masses on it. So the standard, most most simplest form, um, you're gonna have the t y is mg over to I know mg. So t y is going to be 1500 Newton's Okay, so I got t y. Uh, that's good news, because remember, once we know T y will be able to find t by connecting by using the equation that connects the two t y is defined as t sign of data. So T is t y divided by sine of data. So it's 1500 divided by sine of 37. And if you do this, if you do this, you get a T of 2500. Newton's 25 1 Newton's. Okay, eso we got that t is 2500 newtons up here. Now we wanna know h nets to know h Net I need a checks and h wide. Okay, so let's look for H X. And let's look for h y in no particular order notice from this equation over here than h X s t x. I don't know t X, but I can get it now. I can get it. So HX is T X, which is t co sign of data, which is 2500. We just got that co sign of 37. And if you plug all of this in you get that this is Newton's. That's good. Now all I gotta do is find h y. And if you look around, you will see that there is. You'll see that there is an H Y equation right here trying to get an arrow there. That's gonna be hard. Um, there's an H Y equation right here. Okay. And now that I know t y and I know I'm g, I'm gonna be able to find a church y so no need to write a new equation already. Get that so t y plus h y equals m g. So h y is m g minus t y m g 3000 t y is 1500. So h y is 1500. There shouldn't be surprising. Um, that h y is 1500 same thing as t Y t y was holding half of the mg, so h y had to hold the other half of the mg. Okay, so now I can combine to find h h will be the square root of h X squared plus h y squared. And if I do this, I get h y is 1500 and then h this backwards H x is 2000 h wise 1500. And if you combine this, you get 2500 Newtons. Okay, um, this is the same as this, which is not a coincidence. It's not a coincidence. Okay, So this type of problem, I'm going to call this the base case because it's the simplest problem when you have the beam with attention at the edge, no masses on top. It's pretty interesting situation because this is very symmetric. There's a lot of symmetry in this problem, and the symmetry is that t y will equal h y. Obviously TX always equals hx because the wise and the exes are both the same. I'm gonna have that t will equal h. And the angle that team makes with the horizontal will be the same angle that h makes with the horizontal. So you see this angle right here? That angle here is going to be 37. Alright, So that angle the data of H in this problem, if you were calculated, you would get it 37. So there's a ton of symmetry here. Um, kind of interesting to notice. So that's it for this one? Hopefully made sense. Let me Do you have any questions?