Hey, guys. Let's check out this example of a horizontal beam that's been held against the wall with a rope. It also has a 500-kilogram weight or mass at its end here. So the beam has a mass of 400 kg. We got the 500 there. So I'm going to call it 400 little m and the 500 will be big M. The beam has a length l equal to 8 meters. I'm going to assume uniform mass distribution on the beam, which means that the 400 kg acts right here. So I have little mg will be 4,000. I have big Mg right here, which is going to be 5,000. This is a distance of half the length, 4 meters. This is going to be 4 meters as well. Notice how I have a tension here. Right? But because I'm going to actually delegate that to you there. Because of equilibrium on the 500, this tension here equals 5,000 equals the Mg. It's just a connecting force. So you can just think that the bar is essentially being pulled right here with Mg. Okay? I'm going to do that so instead of having Mg and t, I just have the same one force. Okay? So imagine that, treat this as if you had 2 mgs, little mg and big mg right here. I'm also going to have a tension. I'm also going to have a tension here. So this tension, we're actually going to write it out and this tension gets split into tx and ty. So this is going to be our tx right here and this is going to be our ty right here. Okay. The angle between the cable and the horizontal is 53 degrees. So this angle right here is 53 degrees. And it says connect 2 meters from the right edge of the beam. So this distance here is 2 meters which means that this distance here will also be 2 meters. Okay? So 4, 2 and 2. Alright? A 500-kilogram object hangs from the right edge of the beam. We see that. And we're going to calculate the magnitude of the net force, the hinge applies on the beam. So what is hnet? Remember, the hinge will apply an x force and a y force. And so what we're going to do is we're going to have an h force this way, hx to cancel out the tx. hy, we don't know if it's up or down so we're going to assume hy is up. Let me write this down here. We're going to assume hy to be up and let's see what we get. Once I find hx and hy, I'll be able to calculate hnet, which is this. Okay? So hnet will be the square root of hx+hy squared. So what we're really looking for in this problem is we're looking for hx and hy first so that we can find the magnitude of the net force here. Okay. So we'll be able to continue that there. So how are we going to find hx and hy? Well, like every torque, every equilibrium problem, we're going to write the sum of all torques equals 0 and the sum of all forces equals 0. Let's start with forces because it's simpler. Sum of all forces in the x equals 0 and sum of all forces in the y axis equals 0. There are only 2 forces in the x-axis, hx and tx, so they must cancel each other. hx equals tx. And on the y-axis, I have 2 forces up and 2 forces down. I have ty plus hy equals littlemg plus bigmg. Now I don't know ty or hy, but I know the 2 mgs. They're going to be, ty plus hy. They are 4,000 and 5,000. So it's 9,000 total. And this is the best I can do so far. I'm looking for hx, but that requires knowing tx. And I'm looking for hy but that requires knowing ty. Okay. So we're going to have to write a third equation. Alright. At this point, you can think of this question as not even you looking for hx and hy, but you looking for tx and ty. We're going to write a third equation that has to be a torque equation. Sum of all torques equals 0. We're going to pick the best possible points. We're looking for tx and ty. So we want to pick a point away from here so that we can actually find tx and ty. Okay. Now we could also pick, so we could pick this one or this one. Alright. And in here, it doesn't really matter. Let's see. I'm going to do this about I'm going to 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 torques about this point here. So let's call this point 1, 2, 3, and 4. These are all the possible candidates. We're going to go with this one because I want to find ty and tx. So I want to make sure my torque equation includes ty and tx. So I don't want to pick point number 3. That's a bad choice if I wanna find ty and tx. And I want to make sure, I pick the point with the most where I will cancel the most forces. In this case, it's hone and hx, hy and hx. Though those two forces really only hx would be, would give me a torque. Only hy would give me a torque. hx doesn't give you a torque. Okay? So really, all of these points, tx also doesn't give you a torque. So really, all of these points only have each one force that would give you a torque. So, you know, they're all sort of similarly, just as good or just as bad however you want to look at it. So we're just going to go with the torque of about point 1. It would have worked with other points. Don't sweat that too much. Okay? So torque about point 1 means I have 1 here and I have 4,000 pulling this down. This is little mg. I have 5,000 big Mg over here, and then I have ty. ty, we don't know. Both of these guys are providing, are producing a torque in this direction. So this is the torque of little mg. This is the torque of big Mg. Both of these guys are clockwise so they're both negative. And then this is going to provide a counterclockwise torque positive. So I'm going to be able to write that the torque of ty equals the torque of little mg plus the torque of big Mg. And I'm going to expand this equation here. It's going to be tyrsin(θ). Notice that all the angles are going to be 90 degrees. So this is the r vector for the little mg. This is the r vector for big Mg, and this is the r vector for ty. In all of these, the angles are 90 degrees.
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Beam / Shelf Against a Wall
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Beam / Shelf Against a Wall practice set
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