15. Rotational Equilibrium

Center of Mass & Simple Balance

# Center of Mass & Simple Balance

Patrick Ford

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Hey, guys. So in this video, I'm going to talk about the relationship between objects center of mass and whether the object will balance itself on a surface, whether the object will stay balance or tilt at the edge of the surface. Let's check it out. So first of all, remember that an objects weight MGI always acts on the objects center of gravity. It's called center of gravity because that's where gravity acts. Okay, Now, for most of you, most of the time, center of gravity means the same thing. A center of mass. If you're Professor has made a big deal about the difference between the two, then you need to know the difference between them. I'm not gonna talk about it in this video for a vast majority of you guys and for a vast majority of physics problems, Um, all you need to know is that the two things are really the same. So I'm gonna call this center of gravity or center of mass. In fact, some of you will never really see a problem where they are different. Okay, So remember also that if an object has what's called uniforms mass distribution, this means that mass is evenly distributed in objects. Um, for example, if you have a bar, this means that you have the same amount of mass in every piece of the bar as opposed. So this is a uniformed mass distribution as opposed. If you have a bar that has way more mass here than in other parts, this is not a uniformed mass distribution. Guess what a vast majority physics problems will be like. This I'm sorry. Like this will be uniformed mass distribution. Alright, so that's good news. If you have uniformed mass distribution, um, your theology ECT center of mass will be on geometric center. What geometric center means that it's gonna be in the middle? Okay, middle eso It's just gonna be dead in the center right there. And what that means is that M G will act here. MG always acts on the center of gravity, and the center of gravity is almost always in the middle. It is in the middle. If you have uniformed mass distribution. Okay. If you have an object sticking out of a surface like this, it will tilt. If it's center of mass is located beyond the supports edge. So That's two situations here. I got the same bar on two desks, but this one is located here. The center of mass is within the table, Right. In this case, it's right down the middle. And then here it is. Um, it is beyond the table. What that means is that here, the object will not tilt. You can try this at home. Um, but the acceleration will be zero. Right? So there's and this is at equilibrium. It won't tilt here. The object will tilt. There will be no acceleration. That is not zero. And this is not equilibrium. So if you want an object that if you want a knob checked not to tilt you want this situation here and this is static equilibrium. So some questions we'll ask what's the farthest? You can place this object so that it doesn't tilt. And we're going to solve these problems using, um, center of mass equation, which I'll show you here, which is actually much simpler thes air, not torque problems, though they show up in the middle of a bunch of torque equilibrium questions. Okay, So the equation here is that let's say if you have two objects um, m ones here and then m choose here and you want to find the center of mass between them. The exposition of the center of mass will be given by the sum of X m x. Sorry, some of em X divided by the sum of them and what this means for two objects. Just to be very clear, it's something like M one x one plus m two x two divided by m one plus m two. If you have three objects, you keep going M one x 12 X, um three x three Um, em or the Masters and X is the exposition of that object. All right, so let's check out this example here. So here we have a 20 kg, um, plank that is 10 m long. So massive Plank. 20 length of plank. 10. Um, it's supported by two small blocks right here. 121 is that it's left edge. So this is considered to be all the way at the left. Even though it's a little, uh, Zweig here, you can just think of it being, um, right here at the very left. Um, and the other one is 3 m from its right edge. So the right edge of the plank is here. This is 3 m away. The entire thing is 10 m. So if this is three, this distance has to be seven. A 60 kg person walks on the plank. So this guy right here, I'm gonna call it Big M equals 60. And I want to know what is the farthest the person can get to the right of the right. Most support before the plank tips. So I wanna know how far he can go to the right of this. So I wanna know what is this distance here? Okay, what is this distance here? All right. And the idea is this is not really a torque. This is not really an equilibrium question we're gonna solve with torque. Instead, it's an equilibrium question we're gonna solve with center of mass equation in. The idea is, if this person, as this person changes position, the center of mass of this system will change. The system here is made up off plank plus person. You can imagine if the guy is somewhere over here. Don't draw this because I'm gonna delete it. If the guy somewhere here. The center of mass of the two will be somewhere like here. Right. Um, if this thing was really long and the guy was whips, if this thing was really long and the guy was here, you would imagine that the center of mass between the two would be somewhere here. Which means it would definitely tip because it's past the right most support point. It's past the edge. Okay, So what, you want to find the right thing? The right most. He can go. The farthest he can go is you want to know? What position does he have to have so that the center, the center of mass of the system of the combination of two we'll wind up here. This is the farthest at the center of mass can be before this thing tips. So basically want to set the system center of Mass to be at this point, right, which is 7 m from the left. Okay, so the idea is, if the center mass can be a sfar a seven. What? Must XB this distance here? We're gonna call this X. What must XB toe achieve that? Right. So that's what we're gonna do and what we're gonna do to solve this is we're gonna expand the X C m equation. I have to object. So it's gonna be m one x one plus m two x two divided by m one plus m two and this equals seven. And the tricky part here is going to be not the masses, but the X is all right. The distances the first mass is 20. It's the mass of the plank. Um, the X of the plank is where the plank is now the plank is an extended body. So where the plank is, really the planks center of mass. Which, because the plank has uniformed mass distribution. It doesn't say this in the question, but we can assume it because the plane, because it has uniform mass distribution, I'm gonna assume this happens in the middle mg little MGI. The guy has big mg over here. Um, this happens at a distance of 5 m right down the middle, so I'm gonna put a five year. What about the guy? Well, the guy's position is over here, which is seven plus x. I hope you see, this is X and this whole thing here is seven, right? This whole thing, this whole thing here is seven. So this is going to be whoops. Sorry. So this entire distance from the left is seven plus X. So that's what we're gonna do here. Um, em to is the guy 67 plus X divided by the two masses, which are 20 and 60 and this equals to seven. This is a set up. If you got here, you're 99% done. We just gotta get X out of here by using algebra. So I'm gonna multiply these 2. 100. I'm going to distribute the 60. 60 times seven is 4. 20 plus 60 x. This is 80. If I multiply seven times 80 I get 5. 60. Okay, Seven, Let me put seven times 80 here, and that's gonna be 5. 60. I forgot that. This is 60 X, of course. So I'm gonna send these two guys to the other side, so I'm gonna get 60. X equals 4. 60. I'm sorry. 5 60 minus thes two, which is five. 20. And the answer here is or the result here's 40. So I have X equals 40 divided by and 40 divided by 60 is 4/6 or 2/3, which is 0.67 meters. This means that X is 0.67 m. It's how much farther he can go beyond that point. That's not much right. So even though this bar, um, is 10 m long and it's supported here, the guy can Onley walk a little bit more, and that's because he's much heavier than the bar. So there should make some sense. If you can somehow picture a 10 m longer a 30 ft long bar, you can Onley walk a few steps beyond its 7 m point or 70% length of the bar before the bar starts tipping. If you are much heavier than the bar, all right, so that's it. That's how you would find this and hope. Hope it makes sense. Let me know if you have any questions and let's keep going

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