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Beam supporting an object

Patrick Ford
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Hey, guys, let's check out this example of a horizontal beam that's being held against the wall with the rope. It also has a 500 kg weight or mass at its end here. So the beam has massive 400. We got the 500 there, so I'm gonna call 400 little M and the 500 will be big. The beam has a length out equals 8 m. I'm going to assume uniformed mass distribution on the beam, which means that the 400 um kilograms acts right here. So I have little mg will be 4000. I have big M G right here, which is going to be 5000. Um, this is a distance of half the length. 4 m. This is going to be 4 m as well. Um, notice how I have a tension here, Right? But because I'm gonna actually delete that t there because of equilibrium on the 500 this tension here equals 5000 equals DMG. It's just a connecting force. So you can just think that the bar is essentially being pulled right here with them, G. Okay, I'm gonna do that. So instead of having a mgmt. I just have the same one force. Okay, so imagine that. Treat this as if you had two mgs uh, little MGI and big m G right here. I'm also going to have a tension. I'm also gonna have attention here. So this tension, we're actually going to write it out, and this tension gets split into TX and see why. So this is going to be our T X right here. And this is going to be our t y right here. Okay, um, the angle between the cable on the horizontal is 53. So this angle right here is a 53 degrees and it says it connects 2 m from the right edge of the bean. So this distance here is 2 m, which means that this distance here will also be 2 m. Okay, so four to hands to. All right. A right of 500 kg. Object hangs from the right edge of the beam. We see that And when I calculate the magnitude of the net force, the hinge applies on the beam. So what ISS What is H nets? Remember, the hinge will apply an X force and a white force And so what we're gonna do is we're gonna have a church force this way. H X to cancel out the T X H y. We don't know if it's up or down, so we're going to assume h y is up. Let me write this down here. We're going to assume h y to be up and let's see what we get. Once I find a checks and h y, I'll be able to calculate H nets, which is this. Okay, so h net will be the square root of h X squared plus h y square. So what we're really looking for in this problem is we're looking for a checks and h y first so that we can find the magnitude of the net force here. Thanks. We'll be able to continue that. They're so how we're gonna find a checks in h Y. Well, like every torque, Um, every, um, equilibrium problem we're gonna right. The sum of all torques equals zero, and the sum of all forces equals zero. Let's start with forces because it's simpler. Some of all forces in the X equals zero, and some of all forces in the Y axes equals zero. There are only two forces in the X X is a checks and TX so they must cancel each other. H x equals T X And on the Y axis, I have two forces up and two forces down. I have t y plus h y equals little MGI plus big mg. Now I don't know t y r h white, but I know the twin GS they're going to be t y plus h y they r 9000. So it's 9000 total, and this is the best I could do. So far, I'm looking for a church y h X, but that requires knowing t X and I'm looking for a church y. But that requires knowing t why, Okay, so we're gonna have to write a third equation. All right, At this point, you can think of this question as not even you looking for a checks and h y, but you looking for t X and T Y. We're gonna right a third equation that has to be a torque equations. Some of all torques equals zero. We're going to pick the best possible points we're looking for T X and T y. So we wanna pick a point away from here so that we can actually find t X n t y. Okay, now we could also pick, um, so we could pick this one or this one. All right. And and here it really doesn't matter. Let's see. I'm gonna do this about I'm gonna do this about this point right here, okay? No, let's actually just do it. Let's do it over here. So the sum of all talks about this point here. So let's call this 0.0.1 to three and four. These are all the possible candidates. We're gonna go with this one because I wanna find t y and T X. So I wanna make sure my torque equation includes t y and T X. So I don't wanna pick point number three. That's a bad choice. If I wanna find t y and TX Andi, I wanna make sure I picked the point with the most where I will cancel the most forces. In this case. It's h one h, x, h y and HX. Though those two forces really Onley h x would be, uh, would give me a torque Onley h y would give me a torque. HX doesn't give you a torque. Okay, So really, all of these points, um, TX also doesn't give you a torque. So, really, all of these points on Lee have each one force that would give you a torque. So, you know, they're all sort of similarly um, Justus good or just bad. Harvey, one look at it. Um, so we're just gonna go with talk about 0.1? It would have worked with other points. Uh, don't sweat that too much. Okay, So talk about one means I have one here, and I have 4000 pulling this down. This is little MGI. I have 5000 big energy over here, and then I have t y to Why? We don't know. Both of these guys air providing are producing a torque in this direction. So this historic of little mg this historic of big mg. Both of these guys air clockwise, so they're both negative. And then this is gonna provide a counterclockwise torque positive. So I'm gonna be able to write that the torque of t y equals torque of little M G plus torque of big M G. And I'm gonna expand this equation here, it's gonna be t y our sign of theta notice that all the angles, all the angles are going to be 90. So this is the our vector for the little, um, Gene, this is the our vector for, um, big mg. And this is the our vector for T y. And all of these the angles are 90 degrees. So sign of 90. I'm m g are sign of big, um, gene are sign of 90 Sign of nineties one. So these guys go away the distance from the access which chose to t. Y. Is this whole thing here? If you look here, I got a four whips. I got a four with a two. So that's a six for little. MGI is just the four. And for big, um, gets the entire length. It's eight. Okay, so if you start plugging this in, you got 60 y equals little MGs 4000 times four big M G is times eight. Um, this is gonna be 56,000, and I want to divide that by six. So t y will be 56,005 by six, which is 93 33 Newton's. So that's t y now. Once I know T y I confined h y. Okay, so I'm gonna plug it into this equation right here. Okay? So, t y plus h y equals 9000. This means that h y will be 9000 minus t y or 9000 minus 93 Which means h y is negative. 33 three Newton's. So I got a negative. What the heck does that mean? Well, remember how we assumed h y to be up? We got a negative number. This means we assume this incorrectly. So all we gotta do here, it's cool. No problem. All we're gonna do here is we're gonna say h y is actually down. Boom. But nothing changes the magnitude. It's still 333 Okay, Nothing. Nothing to redo or undo. So we're just gonna keep going. I now have a Chuan I let me scroll up here. I now have a choice. Y so I just need a checks. Okay. I now have a choice. Y Andi, I have t y as well. I just need a checks to find a checks. I'm going to need t x So how do I get T X? Well, from T y. I could get t X. I can go from T y two t and then from t two t X and once I have t x t x is the same as a checks. Okay, so let's do that T y equals t sign of theta s o. I can figure out the tea I can find out the t equals t y divided by sine of data. In this case, the data we're talking about is the angle that he makes with the X axis, which is 53. So T y was 9333 divided by the sine of 53. The sign of 53 is, um, 0.8. And if you do this, you get 11666 Newton's alright. And now we're looking now we got t y we got t. We're gonna find t X t X equals t co sign of data. I now have tea so I can plug it in 11666 co sign of 53. And if you do this, you get 7000 so t X equals 7000, which means H X equals as well. Now that I have h y and I have a checks, I can plug it into that equation up top. Okay, so here, I'm gonna get the square root of a checks, which is 7000 squared. And then h y negative 333 Doesn't matter that it's negative because it's gonna get squared. Anyway, on this is gonna be 7000 and eight. So most of the vector came from here, Which makes sense. It's a much bigger number. Okay, So 7000 and eight Newtons is the Net force for the hinge. And this is the final answer. Okay, One point that I want to make here is in this problem the reason why we had a negative. The reason why we had a negative h y is because there's actually way more mass to the right of the cable, right? So if you were to calculate this the center of mass of the cable the center of mass of the system that's being held, which is beam plus the 500 kg is actually somewhere over here, Um, to the right of to the rights of the tension um this is the only time where you're gonna have a h y that is going down. But it's not worth trying to figure that out. Trying to guess, um, it's better to just always go with the h y going up because chances are h y will be up. I just wanted to show you an example where it doesn't go up basically to show you that nothing weird happens, right? Um and also we got to see how it works when we have sort of an object hanging from it, you have to objects in total. Alright, so that's it for this one. Let me know if you have any questions and let's keep going.

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