Center of Mass & Simple Balance - Video Tutorials & Practice Problems

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Center of Mass & Simple Balance

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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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Non-Uniform Mass Distributions (Find Center of Mass)

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Hey, guys. So in most problems we've seen so far involving a extended object, so not a point mass, a tiny point mass, but a long extended object or extended body. We've had uniformed mass distribution, which means that the center of mass of the object is in the middle. The object is evenly distributed in the center of Mass is assumed to be in the middle. Now that's not always going to be the case. And in some problems you have what's called non uniformed mass distribution. Andi, I want to show you what this looks like in some examples. Let's check it out. So unless otherwise stated, you can safely assume that a rigid body and extended body will have uniformed mass distribution. Okay, uniformed mass distribution again means that mass is evenly distributed. This is good news, because then you can assume that the weight acts on the objects center. Okay, but if you don't have uniformed mass distribution, it means that you cannot assume the location of the objects center of Mass. Therefore, you can't assume where MG happens. Okay, now there's 22 types of problems. I mean, there's more than two but two of the basic types of problems will be Either It will be giving you a center of mass and ask you to calculate something else or the other way around. It will give you some other information and ask you to calculate the center of mass. Alright, so I want to show you this example here. So you get a new idea of how this works. I have an 80 kg man that is 2 m tall and he lies horizontally as it's shown here on a 2 m long board off negligible mass. So the mass of the man is 80 both him in the border, 2 m long. So we're gonna assume that his head matches the right end and his feet are exactly touching the left end. Um, he has a mass. Uh, but the board doesn't have any mass there to scales that are placed under him. So they're here, and the readings of the scale at the ends. The reading the scales are left 3 20 right. Um, left 3. 20 and right for 80. The reading of the scale is the normal force. So that means that there's a normal force up here. Okay, so normal. One is 3. 20. Normal two is 4 80. Those are two forces. There's one more force the board is massless of. The board doesn't have an MG, but the guy has an MG. Now, if you didn't know where to draw his center of mass because that's what we're looking for, you could have just put in the middle. Or even better if you knew that the human body has its center of mass closer to the head than the feats. If you knew that, you could just draw out there. If you didn't know that, you could tell by the numbers here by the by the normal forces the fact that this, uh this is a greater force, um, is telling you that the center of mass is closer to that right scale. All right, so I'm gonna do it a little bit off to theme the little closer to the head here. So I wanna know how far from his head is the center of mass. So I wanna know this distance X here. Now, in all of these problems where you're looking for a distance these air tricky You wanna write every other distance as the variable that you're looking for. So here, if I'm looking for this piece and I call this piece X, I need to try to rewrite this other piece as a function of X. So in terms of X, So if the whole thing is to and then this piece is X, then this piece is two minus x. Okay. Now, to solve this problem, this is a a complete equilibrium. Questions so I can use I'm gonna use some of our forces equals zero, and I'm gonna use some of all topics equals zero. Let's start with the easy one. Some of all forces equals zero. I'm gonna go through this one very quickly because all this equation does it's gonna be pretty useless. Um, it's just gonna tell us that everything ends up, so check it out. I have n one plus into going up equals M g going down. And if you plug in these numbers 3 20 plus 4 um, M g is 80 times 10. This just tells us that 800 equals 800 which is cute, but not really useful. So I'm gonna quickly jump into some of all torques equals zero. Now, remember, we can do this for any points, but it might make sense to pick a point where it makes sense to pick a point with a force happen. So either one to or three. And in this particular problem, it really doesn't matter which point you pick, you'll be able to find the answer. Um, either way, Okay, So I'm just gonna go ahead and pick one here, um, sort of at random, and that's what we're gonna get. So some of all talks about 10.1 equals zero. This means that this is the axis of rotation and I have n one happening here, and the one will not produce a torque because it happens in the axis of rotation. Mg over here will produce a torque. It will produce a torque in this direction. Torque of M. G. Um, that's because imagine the access is here and you are pushing down. So this thing is trying to spend this way. This is clockwise, which is positive and then normal. Chew right here is, um causing it torque in this direction again. You got the project here held here. Then you push this way you go this way, which is in the direction of the unit circle, which is counter clockwise. Eso it is. I'm sorry. This is negative. Clockwise, and this is positive. Hopefully you caught. That's those are the only two talks I have. They have to cancel. So that means I can, Right? That's the torque of M G equals the torque of N two. Torque is force, so M g r. And then the sign of data and force and to our sign of data. So let's draw our our vectors. For these two forces, this is the axis of rotation. The R four mg is this one or M G. And they are for normal two. Is this one notice that both of them make 90 degrees with their respective forces. So this would be sign of 90 inside of 90. They both become simply one. Andi, Even if it wasn't one, they would cancel each other anyway. Okay? And then the distances. What's the distance to M. G? It's this distance here, which we called, um, to minus X right. This is the distance we're talking about right here. And so I'm gonna put here tu minus X in the distance toe and choose the entire length of this thing, which is to so see how there's an X here. That's my answer. Okay, so we're gonna get that in just a little bit. So now we're just gonna plug it in and get the X out of there. So massive, the guy is 80 gravity 10 times two minus X equals and to I know into its 4 80 for 80 times too. So we're gonna This is 800. I'm gonna distribute 800 times. Two is 1600 minus 800 X equals this, which is 9 60. Okay, so I gotta move some stuff around. This guy's negative. I'm gonna move it over here. But then I'm gonna move this guy over here so that the access by itself. So I'm gonna have 1600 minus 9. 60. Um, equals 800 x. So 1600 minus 9. 60 is 6 40. Divided by 800 6 40. Divided by 800. That's my ex. I didn't calculate this in advance, but we're gonna do this real quick. Um, 64 80 both have eight as a factor. So this is eight times eight on this is eight times 10. So I do this eight divided by 10.8. That's my X. Okay, So if the exes 0.8, the whole thing is, too. Which means this will be 1.2. You can see how the middle would've been one. Right? You can see how this is sort of 80% of your height is where your center of mass is in this problem, right? This isn't necessarily fully accurate, and it also changes from person to person. But that's it. That's where your center of masses and this is how you can calculate a center of mass. Um, if you're given two measurements for supporting forces, okay? A different version of this question, by the way, could have had tensions, instead holding an object and then trying to figure out center mass that object based on the difference between the two tensions. All right, um, there's some shortcut you could have used here to solve this problem. For example, you might have seen you could have used, like some ratios between 4. 80 and 3 20 to figure out your how far you have to be in terms of center of mass So if you do, for example, for 80 divided by, that's this one here divided by the whole thing, right, This is 0.6, which means it's 60% of your your 60% of the way now 600.6 times 2 m would give you 1.2, and that would give you the distance from your feet up. Right? So you couldn't use that a za quick shortcut. I'm showing that just because some of you might have noticed or wondered about that. But what I really recommend you do just to play it safe is just keep it standard to use the Tory equation to solve it. All right, so that's it for this one. Hope it makes sense. Let me know if you have any questions and let's keep going.

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Problem

Problem

A 70 kg, 1.90 m man doing push-ups holds himself in place making 20° with the floor. His feet and arms are, respectively, 1.15 m below and 0.4 m above his center of mass. You may model him as a thin, long board, and assume his arms and feet are perpendicular to the floor. How much force does the floor apply to each of his hands? (Use g=10 m/s^{2}.)

A

260 N

B

472 N

C

519 N

D

943 N

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