Center of Mass & Simple Balance - Online Tutor, Practice Problems & Exam Prep

Center of Mass & Simple Balance

Hey, guys. So 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 will stay balanced or tilt, at the edge of a surface. Let's check it out. So, first of all, remember that an object's weight, mg, always acts on the object's 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 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. 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 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 to 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. I'm sorry, like this will be uniform mass distribution. All right? So that's good news. If you have uniform mass distribution, the object's center of mass will be in its geometric center. What geometric center means is it's going to be in the middle, okay, middle. So 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. Okay. If you have an object sticking out of a surface like this, it will tilt if its center of mass is located beyond the support's 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 in the middle. And then here it is, 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 0. Right? So there's and this is at equilibrium, it won't tilt. Here, the object will tilt. There will be an acceleration that is not 0 and this is not equilibrium. So, if you want an object to tilt, if you want an object not to tilt, you want this situation here, and this is static equilibrium. So, some questions will ask, what's the farthest you can place this object so that it doesn't tilt? And we're gonna solve these problems using the center of mass equation, which I'll show you here, which is actually gonna be much simpler. These are 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, m₁ here and then m₂ 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. And 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. M are the masses, and x is the x position of that object. Alright. So, let's check out this example here. So here we have a 20-kilogram plank that is 10 meters long. So massive plank, 20, length of plank, 10. It's supported by two small blocks right here, 1, 2. 1 is at its left edge. So this is considered to be all the way at the left, even though it's a little, it's wide here. You can just think of it being right here at the very left. And 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, I'm gonna call it big M equals 60. And I wanna know, what is the farthest a person can get to the right of the rightmost support before the plank tips. So, I want to know, what is this distance here. Okay, what is this distance here? Alright, 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 the center of mass equation. And the idea is, if this person as this person changes position, the center of mass of the system will change. The system here is made up of Planck plus person. You can imagine if the guy is somewhere over here, don't draw this because I'm going to delete it. If the guy is somewhere here, the center of mass of the tube will be somewhere like here. Right? If this thing was really long, and the guy was, whoops, 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. Okay? So, what you want to find the right, the rightmost he can go, the farthest he can go, is you wanna know what position does he have to have so that the center, the center of mass of the system or the combination of the two, will end up here. This is the farthest that the center of mass can be before this thing tips. So basically, you wanna set the system's center of mass to be at this point, right, which is 7 meters from the left. Okay. So 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? Right? So that's what we're going to do. And what we're going to do to solve this is we're going to expand the x_cm equation. I have two objects. So it's going to be, m₁x₁ + m₂x₂ divided by m₁ + m₂. And this equals 7. And the tricky part here is going to be not the masses, but the xs. Alright? The distances. The first mass is 20. It's the mass of the plank. The x of the plank is where the plank is. Now the plank is an extended body. So where the plank is is really the plank's 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 the plane, because it has uniform mass distribution. I'm gonna assume this happens in the middle, mg, little mg. The guy has big mg over here. This happens at a distance of 5 meters, right down the middle. So I'm gonna put a 5 here. What about the guy? Well, the guy's position is over here, which is 7+x. I hope you see this is x and this whole thing here is 7. Right? This whole thing this whole thing here is 7. So this is going to be oops. Sorry. So this entire distance from the left is 7+x. So that's what we're going to do here. M₂ is the guy 67+x divided by the two masses which are 2-60, and this equals 7. This is a setup. If you got here, you're 99% done. We just got to get x out of here by using algebra. So, we're going to multiply these 2, 100. I'm going to distribute the 60. 60 times 7 is 420 plus 60 x. This is 80. If I multiply 7 times 80, I get 560. Okay. 7 let me put 7 times 80 here, and that's going to 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 460. I'm sorry, 560 minus these two, which is 520. And the answer here is or the result here is 40. So I have x equals 40 divided by 60 and 40 divided by 60 is four over six or two over three, 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, right? So, 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. So, 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. All right. So 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.

Non-Uniform Mass Distributions (Find Center of Mass)

Hey, everyone. So in the previous few videos, we've dealt with rigid bodies, not just tiny point masses, but rather extended objects. We assumed that they had what's called uniform mass distribution, which means that the mass is evenly spread out, and we can assume that the center of mass and the weight force acts in the middle. However, in some cases, you won't have that; you'll have what's called non-uniform mass distributions, like the problem we're going to work out here. So, I want to go ahead and show you how that works, and we're going to do this example together. Alright? So let's check this out here.

Unless your problems otherwise state, you can usually assume that a rigid body has uniform mass distribution. Remember, this uniformity just means that the mass is evenly spread out. For example, if I had a textbook, which I'm just going to imagine as a little rectangle, then if this mass is evenly distributed, one side of the textbook wouldn't be heavier than the other. This is good because we can usually just assume that the center of mass is in the middle, and that's where the weight force acts. We assume the weight force acts on its center of mass. Now, if you don't have uniform mass distribution, you cannot assume the location of its center of mass, and that's usually what these problems involve.

There are generally two broad types of problems that you're going to see. In some, you'll be given the center of mass, the location of it, and you'll be asked to calculate something else, like a force or a torque. In other problems, you might be given some information and asked to calculate where the center of mass is, and that's exactly what our problem will deal with. Alright? So let's just go ahead and get started with this.

In this problem, we have an 80 kilogram man who is 2 meters tall. So, I've got m=80 kg and l=2 m. But, we're told explicitly in this problem that human bodies generally do not have uniform mass distribution. This kind of makes sense if you think about it; our mass is a bit unevenly distributed, skewed or heavier towards our topside due to our arms and torso, and then we have our legs. As human beings, we do not usually have uniform mass distribution. We want to calculate how far from the man's head the center of mass is.

We've got two scales underneath this board, and remember scales measure the normal force. We're actually told what those normal forces are. The left one measures 320, so I'm going to call this normal left equals 320, and this one here measures 480. This right normal force is a little bigger, exerting more force, and this tells us that our mass is skewed more towards our head. There's also another force that's acting, which is the weight force of the man. The board has no mass, but the man does; he's 80 kilograms. We're going to draw that weight force, and if this were a uniform body, we would draw it in the center, but it's not uniform. If you didn't know that, you could sort of guess that because the normal forces are skewed toward the right, more of the weight force and more of the center of mass is skewed to that side.

Here's what we have: our mg force, and what we want to figure out is the distance between this center of mass and where the mg acts, and that's my x distance. So this is really what we want to find here, what x is. Usually, in these types of problems where you're given some kind of unknown distance, you want to write all of the other distances in terms of that variable. So if the distance here that you're looking for is x, then the other distance that I have to use will be this one here, which is really just the entire length minus the x. So, the whole length is 2, and this piece is x, this will be 2 minus x. These are our distances here. We want to calculate what this x is. This is a total equilibrium problem.

We know that the sum of forces and torques is going to be 0. Let's go ahead and get started with the easier one, sum of forces is 0. We've got two forces that point up, so you've got n_left and n_right, and then one that points down, which is mg, and that should equal 0. If you plug in these numbers quickly, 320 and 480, and then we're just going to use 10 for g. This is going to be 80 times 10. All you're going to get here is you're going to get 800 minus 800 equals 0. That's cute, but it doesn't really tell us anything. All it just says is that the forces cancel. This equation tells us nothing about the location of the center of mass; all it's just telling us is that the guy isn't lifting off into space or plummeting through the ground.

So, let's just go ahead and move on to the sum of all torques. The sum of all torques is equal to 0. Remember, when you are measuring torques or calculating torques, you have to pick a reference point, and you usually want to pick one of your reference points where one of the forces is acting. The reason for that is you want one of the torques to cancel out because its distance to the axis of rotation will be 0. That's a trick you can use to make one of your terms always cancel out. It doesn't matter which point you choose; you'll still get the right answer as long as you keep track of all the signs. But I think the easiest one to pick will be the first one here. So, we've got our object, and we've got a force that acts right here, that's n_left, and because this is the axis of rotation, this force will not produce a torque because there's no distance, their torque from n_left is equal to 0. What about the other forces? We've got an mg that's going to act somewhere over here, and this mg will produce a clockwise rotation. If you're pivoting it around at this point, like my hand over here, and you have a force this way, it's going to want to turn you clockwise. Remember, a clockwise rotation will be a negative torque, so this is going to be the torque of mg and that's going to be a negative. The last one here is going to be this torque over here from the normal from the normal right force, and that's going to produce a counterclockwise torque, and that's going to be a positive torque. So, we've got negative torque mg plus and we've got positive torque from n_r, and this equals 0. What I'm going to do is I'm just going to move this to the other side so that it becomes positive.

Remember, torques are f r sin theta. So, we've got normal force times some distance times the sin of theta equals the force here is mg, so we've got mg times r times the sin of theta. We just have to figure out what are the sin and r's that we plug in. If you look at these forces, remember we have to figure out where's the r vector. The r vector points from the axis of rotation out to that force. So this is going to be r_mg like this, and then the other one is going to be this one, and this is going to be r_nr. Notice how both of these r vectors make 90-degree angles with their forces, and so therefore the sine of the angle is just going to be 90 for both of them. Remember, the sine of 90 is just 1, so we can kind of just cancel that out, right? So we really just need to figure out what these lengths are, the r_mg, and the r for the normal force.

If you look at the first diagram that we drew, you'll notice that this distance right here from the feet to where the mg acts is going to be that distance, 2 minus x. So this is going to be 2 minus x here, that's what we plug into this equation. This is going to be 2 minus x, and then, for the mg, I'm sorry, and then for the right force, this is actually just going to be equal to 2 because it's the entire length of the man. So now, I'm just going to start plugging in some numbers. This is going to be 480 times 2, remember the sine goes away, equals, and this is going to be 80 times 10 for g, and this is going to be 2 minus x and then this is also going to be 1. This equation simplifies to 960 equals 1600 minus 800x.

Now all we have to do is solve for x. All the physics is done; we just have to do some algebra and cleaning up. We'll move the 800x to the right, and when you subtract 960 from 1600, you'll get 640. So your x distance is going to be 640 divided by 800, and if you cancel out a 0, you'll notice that these things are divisible by eights, this is going to be 0.8 meters. That is your answer. Going back to our diagram, what this tells us is that the center of mass is actually 0.8 meters away from the head, and that makes sense because we said that this distance here is going to be closer to the head than it is to the feet. If the human body is 2 meters long, then the middle point would be 1 meter, but the center of mass is closer towards the head, which means that this distance over here is actually going to be 1.2 meters.

So that should make some sense here. Alright, guys, that's it for this one. Let me know if you have any questions.

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/s².) 

• A. 260 N
• B. 472 N
• C. 519 N
• D. 943 N