Center of Mass & Simple Balance

Concept: Center of Mass & Simple Balance

11m
Video Transcript

Hey guys! In this video I'm going to talk about the relationship between an object's center of mass and whether the object will balance itself on a surface, whether the object would stay balanced or tilt at the edge of the surface. Let's check it out. First of all, remember that an object's weight mg always acts on the objectÕs center of gravity. It's called center of gravity because that's where gravity acts. For most of you, most of the time center of gravity means the same thing as center of mass. If your professor has made a big deal about the difference between the two, then you need to know the difference between them. I'm not going to talk about it in this video. For a vast majority of you guys and for a vast majority of physics problems, all you need to know is that the two things are really the same. I'm going to call this center of gravity or center of mass. In fact, some of you will never really see a problem where they are different. Remember also that if an object has what's called uniform mass distribution, this means that mass is evenly distributed in an object. 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 uniform mass distribution, as opposed to if you have a bar that has way more mass here than in other parts. This is not a uniform mass distribution. Guess what? A vast majority of physics problems will be like this. It will be uniform mass distribution. That's good news. If you have uniform mass distribution, the objectÕs center of mass will be on its geometric center. What geometric center means is it's going to be in the middle. It's just going to be dead in the center right there and what that means is that mg 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 uniform mass distribution. If you have an object sticking out of a surface like this, it will tilt if its center of mass is located beyond the supportÕs edge. 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. In this case it's right down the middle and then here, it is beyond the table. What that means is that here, the object will not tilt. You can try this at home. But the acceleration will be zero and this is at equilibrium. It won't tilt. Here the object will tilt. There will be an acceleration. That is not zero and this is not equilibrium. If you want an object not to tilt, you want this situation here and this is static equilibrium. Some questions will 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 center of mass equation which I'll show you here, which is actually much simpler. These are not torque problems, though they show up in the middle of a bunch of torque equilibrium questions. The equation here is that, let's say if you have two objects, m1 is here and then m2 is here and you want to find the center of mass between them. The x position of the center of mass will be given by the sum of mx divided by the sum of m. What this means for two objects just to be very clear it's something like m1x1 + m2x2 divided by m1 + m2. If you had three objects, you keep going, m1x1, m2x2, m3x3. The masses in x is the x position of that object.

Let's check out this example here. Here we have a 20-kilogram plank that is 10 meters long, so mass of plank 20, length of plank 10. It's supported by two small blocks right here, one, two. One is at its left edge, so this is considered to be all the way at the left even though it's a little wide here, you can just think of it being right here at the very left. The other one is 3 meters from its right edge, so the right edge of the plank is here. This is 3 meters away. The entire thing is 10 meters, so if this is 3, this distance has to be 7. A 60-kilogram person walks on the plank so this guy right here IÕm going to call it big M = 60. I want to know what is the farthest the person can get to the right of the rightmost support before the plank tips. I want to know how far he can go to the right of this. I want to know what is this distance here. The idea is this is not really an equilibrium question weÕre gonna solve with torque. Instead, it's an equilibrium question we're going to solve with center of mass equation. The idea is as this person changes position, the center of mass of this system will change. The system here is made up of plank + person. You can imagine if the guy is somewhere over here, the center of mass of the two will be somewhere like here. 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 rightmost support point. It's past the edge. What you want to find the rightmost he can go, the farthest he can go, is you want to know what position does he have to have so that the center of mass of the system of the combination of the two will end up here. This is the farthest that the center of mass can be before this thing tips. Basically we want to set the systemÕs center of mass to be at this point right which is 7 meters from the left. The idea is if the center of mass can be as far as 7, what must x be? This distance here we're going to call this x. What must x be to achieve that? That's what we're going to do.

What we're going to do to solve this is weÕre going to expand the Xcm equation. I have two objects so it's going to be m1x1 + m2x2 / m1 + m2 and this equals 7. The tricky part here is going to be not the masses but the xÕs, the distances. The first mass is 20. It's the mass of the plank. The x of the plank is where the plank is. The plank is an extended body, so where the plank is really the plankÕs center of mass which because the plank has uniform mass distribution, it doesn't say this in the question but we can assume it. Because it has uniform mass distribution, I'm going to assume this happens in the middle, little mg. The guy has big Mg over here. This happens at a distance of 5 meters right down the middle. I'm going to put a 5 here. What about the guy? The guyÕs position is over here which is 7 + x. I hope you see this is x and this whole thing here is 7. This entire distance from the left is 7 + x, so that's what we're going to do here. m2 is the guy, 60 (7 + x) divided by the two masses which are 20 and 60 this equals to 7. This is a setup. If you got here, youÕre 99% done. We just got to get x out here by using algebra. IÕm gonna multiply these two, 100. IÕm going to distribute the 60. 60*7 is 420 + 60x. This is 80. If I multiply 7*80, I get 560. That' going to be 560. I forgot that this is 60x of course. I'm going to send these two guys to the other side so I'm going to get 60x = 560 - these two which is 520 and the answer here is or the result here is 40. I have x equals 40/60 and 40/60 is 4/6 or 2/3 which is 0.67 meters. This means that x is 0.67 meters. It's how much farther he can go beyond that point. That's not much. Even though this bar is 10 meters long and it's supported here, the guy can only walk a little bit more and that's because he's much heavier than the bar. This should make some sense if you can somehow picture a 10-meter long or a 30-foot long bar. You can only walk a few steps beyond its 7-meter point or 70% length of the bar before the bar starts tipping if you are much heavier than the bar. That's it. That's how you would find this and I hope it makes sense. Let me know if you have any questions and let's keep going.

Concept: Non-Uniform Mass Distributions (Find Center of Mass)

10m
Video Transcript

Hey guys! In most problems we've seen so far involving an extended object, so not a point mass a tiny point mass, but a long extended object or extended body, we've had uniform mass distribution which means that the center of mass of the object is in the middle. The object is even the distributed and the center of mass is assumed to be in the middle. That's not always going to be the case. In some problems you have what's called non-uniform mass distribution. I want to show you what this looks like in some examples. Let's check it out. Unless otherwise stated, you can safely assume that a rigid body, an extended body, will have uniform mass distribution. Uniform mass distribution, again, means that mass is evenly distributed. This is good news because then you can assume that the weight acts on the object's center. But if you don't have uniform mass distribution, it means that you cannot assume the location of the object's center of mass, therefore you can't assume where mg happens. There's two 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. I want to show you this example here so you get an idea of how this works.

I have an 80-kilogram man that is 2 meters tall and he lies horizontally as it's shown here on a 2-meter long board of negligible mass. The mass of the man is 80. Both him and the board are 2 meters long so we're going to assume that his head matches the right end and his feet are exactly touching the left end. He has a mass but the board doesn't have any mass. There are two scales that are placed under him so they're here. The readings of the scale at the ends, the reading of the scales are left 320 and right 480. The reading of the scale is a normal force so that means that there's a normal force up here. Normal one is 320, normal two is 480. Those are two forces. There's one more force. The board is massless so the board doesn't have an mg, but the guy has an mg. 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 feet, if you knew that you could just draw it there. If you didn't know that, you could tell by the numbers here by the by the normal forces. The fact that this is a greater force is telling you that the center of mass is closer to that right scale. I'm going to draw that a little closer to the head here. I want to know how far from his head is his center of mass. I wanna know this distance x here. In all of these problems where you're looking for a distance, these are tricky, you want to write every other distance as the variable that you're looking for. 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. If the whole thing is true and then this piece is x, then this piece is 2-x. To solve this problem this is a complete equilibrium question so I can use sum of all forces equals zero and I'm going to use sum of all torques equals zero. Let's start with the easy one, sum of all forces equals zero. I'm going to go through this one very quickly because all this equation does, itÕs gonna be pretty useless. It's just going to tell us that everything adds up. Check it out. I have N1 + N2 going up = mg going down. If you plug in these numbers, 320 + 480 = mg is 80 times 10, this just tells us that 800 = 800 which is cute but not really useful. I'm going to quickly jump into sum of all torques = 0. Remember, we can do this for any points but it might make sense to pick a point where a force happens, so either 1, 2 or 3. In this particular problem, it really doesn't matter which point you pick, you'll be able to find the answer either way. I'm just going to go ahead and pick one here sort of at random. Let's see what weÕre gonna get. Sum of all torques about point one = 0. This means that this is the axis of rotation and I have N1 happening here. N1 will not produce a torque because it happens on the axis of rotation. mg over here will produce a torque. It will produce a torque in this direction, torque of mg, that's because imagine the axis is here and you are pushing down so this thing is trying to spin this way. This is clockwise which is negative. Then Normal 2 right here is causing a torque in this direction. Again, you got the object here, held here, you push this way, you go this way which is in the direction of the unit circle which is counterclockwise so it is positive. Hopefully you caught that. Those are the only two torques I have. They have to cancel, so that means I can write that the torque of mg = the torque of N2. Torque is force, so mgr and then the sin_ and force N2 r sin_.

Let's draw our r vectors for these two forces. This is the axis of rotation. The r for mg is this one, r mg and the r for normal 2 is this one. Notice that both of them make 90 degrees with their respective forces, so this will be sine of 90 and sine of 90. They both become simply 1. Even if it wasn't one, they would cancel each other anyway. Then the distances, what's the distance to mg? It's this distance here which we called 2-x. This is the distance we're talking about right here. I'm going to put here 2-x and the distance to N2 is the entire length of this thing which is 2. See how there's an x here? That's my answer. We're going to get that in just a little bit. Now we're just going to plug it in and get the x out of there. Mass of the guy is 80, gravity 10 times 2-x = N2. I know N2, itÕs 480*2. This is 800, I'm going to distribute 800*2 is 1600 Ð 800x equals this which is 960. I got to move some stuff around. This guy is negative. I'm going to move it over here. But then I'm going to move this guy over here, so that the x is by itself. I'm gonna have 1600 Ð 960 = 800x. 1600 Ð 960 is 640 divided by 800. 640/800, that's my x. I didn't calculate this in advance but we're going to do this real quick. 64 and 80 both have 8 as a factor so this is 8*8 and this is 8*10. I do this, 8/10 is 0.8. That's my x. If the x is 0.8, the whole thing is 2, which means this will be 1.2. You can see how the middle would have been 1. You can see how this is sort of 80% of your height is where your center of mass is in this problem. This isn't necessarily fully accurate. It also changes from person to person, but that's it. That's where your center of mass is. This is how you can calculate a center of mass if you're given two measurements for supporting forces. A different version of this question by the way could have had tensions instead holding an object and then trying to figure out the center of mass of that object based on the difference between the two tensions. There's some shortcut you could have used here to solve this problem. For example, you could have used some ratios between 480 and 320 to figure out how far you have to be in terms of center of mass. If you do for example 480, that's this one here, divided by the whole thing, this is 0.6 which means youÕre 60% of the way. 0.6*2 meters would give you 1.2 and that would give you the distance from your feet up. You could have used that as a 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. Use the torque equation to solve it. That's it for this one. Hope it made sense. Let me know if you have any questions and let's keep going.

Problem: A 70 kg, 1.90 m man doing push-ups holds himself in place making 20° with the floor, as shown. His feet and arms are, respectively, 1.15 m below and 0.4 m above from 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? 

BONUS: How much force does the floor apply to each of his feet?

12m